Philpot MoM 2nd Ch01-06 ISM

ISM for

Chapter 1 Stress 1.1 Introduction 1.2 Normal Stress Under Axial Loading 1.3 Direct Shear Stress 1.4 Bearing Stress 1.5 Stresses on Inclined Sections 1.6 Equality of Shear Stresses on Perpendicular Planes Chapter 2 Strain 2.1 Displacement, Deformation, and the Concept of Strain 2.2 Normal Strain 2.3 Shear Strain 2.4 Thermal Strain Chapter 3 Mechanical Properties of Materials 3.1 The Tension Test 3.2 The Stress‐Strain Diagram 3.3 Hooke’s Law 3.4 Poisson’s Ratio Chapter 4 Design Concepts 4.1 Introduction 4.2 Types of Loads 4.3 Safety 4.4 Allowable Stress Design 4.5 Load and Resistance Factor Design Chapter 5 Axial Deformation 5.1 Introduction 5.2 Saint‐Venant’s Principle 5.3 Deformations in Axially Loaded Bars 5.4 Deformations in a System of Axially Loaded Bars 5.5 Statically Indeterminate Axially Loaded Members 5.6 Thermal Effects on Axial Deformation 5.7 Stress Concentrations Chapter 6 Torsion 6.1 Introduction 6.2 Torsional Shear Strain 6.3 Torsional Shear Stress 6.4 Stresses on Oblique Planes 6.5 Torsional Deformations 6.6 Torsion Sign Conventions 6.7 Gears in Torsion Assemblies 6.8 Power Transmission 6.9 Statically Indeterminate Torsion Members 6.10 Stress Concentrations in Circular Shafts Under Torsional Loadings 6.11 Torsion of Noncircular Sections 6.12 Torsion of Thin‐Walled Tubes: Shear Flow

Chapter 7 Equilibrium of Beams 7.1 Introduction 7.2 Shear and Moment in Beams 7.3 Graphical Method for Constructing Shear and Moment Diagrams 7.4 Discontinuity Functions to Represent Load, Shear, and Moment Chapter 8 Bending 8.1 Introduction 8.2 Flexural Strains 8.3 Normal Strains in Beams 8.4 Analysis of Bending Stresses in Beams 8.5 Introductory Beam Design for Strength 8.6 Flexural Stresses in Beams of Two Materials 8.7 Bending Due to Eccentric Axial Load 8.8 Unsymmetric Bending 8.9 Stress Concentrations Under Flexural Loadings Chapter 9 Shear Stress in Beams 9.1 Introduction 9.2 Resultant Forces Produced by Bending Stresses 9.3 The Shear Stress Formula 9 .4 The First Moment of Area Q 9.5 Shear Stresses in Beams of Rectangular Cross Section 9.6 Shear Stresses in Beams of Circular Cross Section 9.7 Shear Stresses in Webs of Flanged Beams 9.8 Shear Flow in Built‐Up Members Chapter 10 Beam Deflections 10.1 Introduction 10.2 Moment‐Curvature Relationship 10.3 The Differential Equation of the Elastic Curve 10.4 Deflections by Integration of a Moment Equation 10.5 Deflections by Integration of Shear‐Force or Load Equations 10.6 Deflections Using Discontinuity Functions 10.7 Method of Superposition Chapter 11 Statically Indeterminate Beams 11.1 Introduction 11.2 Types of Statically Indeterminate Beams 11.3 The Integration Method 11.4 Use of Discontinuity Functions for Statically Indeterminate Beams 11.5 The Superposition Method

Chapter 12 Stress Transformations 12.1 Introduction 12.2 Stress at a General Point in an Arbitrarily Loaded Body 12.3 Equilibrium of the Stress Element 12.4 Two‐Dimensional or Plane Stress 12.5 Generating the Stress Element 12.6 Equilibrium Method for Plane Stress Transformation 12.7General Equations of Plane Stress Transformation 12.8 Principal Stresses and Maximum Shear Stress 12.9 Presentation of Stress Transformation Results 12.10 Mohr’s Circle for Plane Stress 12.11 General State of Stress at a Point Chapter 13 Strain Transformations 13.1 Introduction 13.2 Two‐Dimensional or Plane Strain 13.3 Transformation Equations for Plane Strain 13.4 Principal Strains and Maximum Shearing Strain 13.5 Presentation of Strain Transformation Results 13.6 Mohr’s Circle for Plane Strain 13.7 Strain Measurement and Strain Rosettes 13.8 Generalized Hooke’s Law for Isotropic Materials Chapter 14 Thin‐Walled Pressure Vessels 14.1 Introduction 14.2 Spherical Pressure Vessels 14.3 Cylindrical Pressure Vessels 14.4 Strains in Pressure Vessels Chapter 15 Combined Loads 15.1 Introduction 15.2 Combined Axial and Torsional Loads 15.3 Principal Stresses in a Flexural Member 15.4 General Combined Loadings 15.5 Theories of Failure Chapter 16 Columns 16.1 Introduction 16.2 Buckling of Pin‐Ended Columns 16.3 The Effect of End Conditions on Column Buckling 16.4 The Secant Formula 16.5 Empirical Column Formulas & Centric Loading 16.6 Eccentrically Loaded Columns

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1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a

compression member. If the normal stress in the member must be limited to 200 MPa, determine the

maximum load P that the member can support.

Solution

The cross-sectional area of the stainless steel tube is

2 2 2 2 2( ) [(60 mm) (50 mm) ] 863.938 mm4 4

A D d

The normal stress in the tube can be expressed as

P

A

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable

normal stress, rearrange this expression to solve for the maximum load P

2 2

max allow (200 N/mm )(863.938 mm ) 172,788 172.8 kN NP A Ans.




1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip

load. If the normal stress in the member must be limited to 18 ksi, determine the wall thickness

required for the tube.

Solution

From the definition of normal stress, solve for the minimum area required to support a 27-kip load

without exceeding a stress of 18 ksi

2

min

27 kips1.500 in.

18 ksi

P PA

A

The cross-sectional area of the aluminum tube is given by

2 2( )4

A D d

Set this expression equal to the minimum area and solve for the maximum inside diameter d

2 2 2

2 2 2

2 2 2

max

[(2.50 in.) ] 1.500 in.4

4(2.50 in.) (1.500 in. )

4(2.50 in.) (1.500 in. )

2.08330 in.

d

d

d

d

The outside diameter D, the inside diameter d, and the wall thickness t are related by

2D d t

Therefore, the minimum wall thickness required for the aluminum tube is

min

2.50 in. 2.08330 in.0.20835 in. 0.208 in.

2 2

D dt

Ans.




1.3 Two solid cylindrical rods (1) and (2) are

joined together at flange B and loaded, as shown in

Fig. P1.3. The diameter of rod (1) is d1 = 24 mm

and the diameter of rod (2) is d2 = 42 mm.

Determine the normal stresses in rods (1) and (2).

Fig. P1.3

Solution

Cut a FBD through rod (1) that includes the free end of the rod at A.

Assume that the internal force in rod (1) is tension. From equilibrium,

1 180 kN 0 80 kN (T)xF F F

Next, cut a FBD through rod (2) that includes the free

end of the rod A. Assume that the internal force in rod

(2) is tension. Equilibrium of this FBD reveals the

internal force in rod (2):

2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F F

From the given diameter of rod (1), the cross-sectional area of rod (1) is

2 2

1 (24 mm) 452.3893 mm4

A

and thus, the normal stress in rod (1) is

11 2

1

(80 kN)(1,000 N/kN)176.8388 MPa

452.389176

3 m.8 M )

mPa (T

F

A Ans.


2 2

2 (42 mm) 1,385.4424 mm4

A

Accordingly, the normal stress in rod (2) is

22 2

2

( 200 kN)(1,000 N/kN)144.3582 MPa

1,385.4144.4 MPa (

424C)

mm

F

A

Ans.





joined together at flange B and loaded, as shown in

Fig. P1.4. If the normal stress in each rod must be

limited to 120 MPa, determine the minimum

diameter required for each rod.

Fig. P1.4

Solution

Cut a FBD through rod (1) that includes the free end of the rod at A.

Assume that the internal force in rod (1) is tension. From equilibrium,

1 180 kN 0 80 kN (T)xF F F

Next, cut a FBD through rod (2) that includes the free

end of the rod A. Assume that the internal force in rod

(2) is tension. Equilibrium of this FBD reveals the

internal force in rod (2):

2 2140 kN 140 kN 80 kN 0 200 kN 200 kN (C)xF F F

If the normal stress in rod (1) must be limited to 120 MPa, then the minimum cross-sectional area that

can be used for rod (1) is

211,min 2

(80 kN)(1,000 N/kN)666.6667 mm

120 N/mm

FA

The minimum rod diameter is therefore

2 2

1,min 1 1666.6667 mm 29.1346 mm 29. mm4

1 A d d

Ans.

Similarly, the normal stress in rod (2) must be limited to 120 MPa. Notice that rod (2) is in

compression. In this situation, we are concerned only with the magnitude of the stress; therefore, we

will use the magnitude of F2 in the calculations for the minimum required cross-sectional area.

222,min 2

(200 kN)(1,000 N/kN)1,666.6667 mm

120 N/mm

FA

The minimum diameter for rod (2) is therefore

2 2

2,min 2 21,666.6667 mm 46.0659 mm 46. mm4

1 A d d

Ans.





joined together at flange B and loaded, as

shown in Fig. P1.5. If the normal stress in

each rod must be limited to 40 ksi,

determine the minimum diameter required

for each rod.

Fig. P1.5

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A. As a

matter of course, we will assume that the internal force in rod (1) is tension (even

though it obviously will be in compression). From equilibrium,

1

1

15 kips 0

15 kips 15 kips (C)

yF F

F

Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we

will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals

the internal force in rod (2):

2

2

30 kips 30 kips 15 kips 0

75 kips 75 kips (C)

yF F

F

Notice that rods (1) and (2) are in compression. In this situation, we are

concerned only with the stress magnitude; therefore, we will use the force

magnitudes to determine the minimum required cross-sectional areas. If the

normal stress in rod (1) must be limited to 40 ksi, then the minimum cross-

sectional area that can be used for rod (1) is

211,min

15 kips0.375 in.

40 ksi

FA


2 2

1,min 1 10.375 in. 0.6909 0.691 9 i4

inn. .A d d

Ans.

Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of

222,min

75 kips1.875 in.

40 ksi

FA

The minimum diameter for rod (2) is therefore

2 2

2,min 2 21.875 in. 1.54509 1.545 in.7 in.4

A d d

Ans.




1.6 Two solid cylindrical rods (1) and (2) are joined

together at flange B and loaded, as shown in Fig.

P1.6. The diameter of rod (1) is 1.75 in. and the

diameter of rod (2) is 2.50 in. Determine the normal

stresses in rods (1) and (2).

Fig. P1.6

Solution

Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We

will assume that the internal force in rod (1) is tension (even though it obviously will

be in compression). From equilibrium,

1

1

15 kips 0

15 kips 15 kips (C)

yF F

F

Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we

will assume that the internal force in rod (2) is tension. Equilibrium of this FBD

reveals the internal force in rod (2):

2

2

30 kips 30 kips 15 kips 0

75 kips 75 kips (C)

yF F

F


2 2

1 (1.75 in.) 2.4053 in.4

A

and thus, the normal stress in rod (1) is

11 2

1

15 kips6.23627 ksi

2.4053 in6.24 ksi )

.(C

F

A

Ans.


2 2

2 (2.50 in.) 4.9087 in.4

A

Accordingly, the normal stress in rod (2) is

22 2

2

75 kips15.2789 ksi

2.4053 in.15.28 ksi (C)

F

A

Ans.




1.7 Axial loads are applied with rigid bearing plates to the

solid cylindrical rods shown in Fig. P1.7. The diameter of

aluminum rod (1) is 2.00 in., the diameter of brass rod (2) is

1.50 in., and the diameter of steel rod (3) is 3.00 in.

Determine the normal stress in each of the three rods.

Fig. P1.7

Solution

Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal

force in rod (1) is tension (even though it obviously will be in compression). From equilibrium,

1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F

FBD through rod (1)

FBD through rod (2)

FBD through rod (3)

Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal

force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F

Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in

rod (3) is:

3

3

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

yF F

F





2 2

1 (2.00 in.) 3.1416 in.4

A

and thus, the normal stress in aluminum rod (1) is

11 2

1

16 kips5.0930 ksi

3.1416 in5.09 ksi (C)

.

F

A

Ans.


2 2

2 (1.50 in.) 1.7671 in.4

A

Accordingly, the normal stress in brass rod (2) is

22 2

2

14 kips7.9224 ksi

1.7671 in.7.92 ksi (T)

F

A Ans.

Finally, the cross-sectional area of rod (3) is

2 2

3 (3.00 in.) 7.0686 in.4

A

and the normal stress in the steel rod is

33 2

3

26 kips3.6782 ksi

7.0686 in3.68 ksi (C)

.

F

A

Ans.




1.8 Axial loads are applied with rigid bearing plates to the solid

cylindrical rods shown in Fig. P1.8. The normal stress in

aluminum rod (1) must be limited to 18 ksi, the normal stress in

brass rod (2) must be limited to 25 ksi, and the normal stress in

steel rod (3) must be limited to 15 ksi. Determine the minimum

diameter required for each of the three rods.

Fig. P1.8

Solution

The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that

includes the free end A. We will assume that the internal force in rod (1) is tension (even though it

obviously will be in compression). From equilibrium,

1 18 kips 4 kips 4 kips 0 16 kips 16 kips (C)yF F F

FBD through rod (1)

FBD through rod (2)

FBD through rod (3)

Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal

force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

2 28 kips 4 kips 4 kips 15 kips 15 kips 0 14 kips 14 kips (T)yF F F

Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in

rod (3) is:




3

3

8 kips 4 kips 4 kips 15 kips 15 kips 20 kips 20 kips 0

26 kips 26 kips (C)

yF F

F

Notice that two of the three rods are in compression. In these situations, we are concerned only with the

stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-

sectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be

limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is

211,min

1

16 kips0.8889 in.

18 ksi

FA


2 2

1,min 1 10.8889 in. 1.0638 in 1.064 in..4

A d d

Ans.

The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of

22

2,min

2

14 kips0.5600 in.

25 ksi

FA

which requires a minimum diameter for rod (2) of

2 2

2,min 2 20.5600 in. 0.8444 in 0.844 in..4

A d d

Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required

for this rod is:

23

3,min

3

26 kips1.7333 in.

15 ksi

FA


2 2

3,min 3 31.7333 in. 1.4856 in 1.486 in..4

A d d

Ans.




1.9 Two solid cylindrical rods support

a load of P = 50 kN, as shown in Fig.

P1.9. If the normal stress in each rod

must be limited to 130 MPa, determine

the minimum diameter required for

each rod.

Fig. P1.10

Solution

Consider a FBD of joint B. Determine the angle between

rod (1) and the horizontal axis:

4.0 m

tan 1.600 57.99462.5 m

and the angle between rod (2) and the horizontal axis:

2.3 m

tan 0.7188 35.70673.2 m

Write equilibrium equations for the sum of forces in the

horizontal and vertical directions. Note: Rods (1) and (2)

are two-force members.

2 1cos(35.7067 ) cos(57.9946 ) 0xF F F (a)

2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the

substitution method, Eq. (b) can be solved for F2 in terms of F1:

2 1

cos(57.9946 )

cos(35.7067 )F F

(c)

Substituting Eq. (c) into Eq. (b) gives

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P PF

For the given load of P = 50 kN, the internal force in rod (1) is therefore:

1

50 kN40.6856 kN

1.2289F




Backsubstituting this result into Eq. (c) gives force F2:

2 1

cos(57.9946 ) cos(57.9946 )(40.6856 kN) 26.5553 kN

cos(35.7067 ) cos(35.7067 )F F

The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area

required for rod (1) is

21

1,min 2

1

(40.6856 kN)(1,000 N/kN)312.9664 mm

130 N/mm

FA


2 2

1,min 1 1312.9664 mm 19.9620 19.4

96 mmmmA d d

Ans.

The minimum area required for rod (2) is

22

2,min 2

2

(26.5553 kN)(1,000 N/kN)204.2718 mm

130 N/mm

FA


2 2

2,min 2 2204.2718 mm 16.1272 16.4

13 mmmmA d d

Ans.




1.10 Two solid cylindrical rods

support a load of P = 27 kN, as

shown in Fig. P1.10. Rod (1) has a

diameter of 16 mm and the diameter

of rod (2) is 12 mm. Determine the

normal stress in each rod.

Fig. P1.10

Solution

Consider a FBD of joint B. Determine the angle between

rod (1) and the horizontal axis:

4.0 m

tan 1.600 57.99462.5 m

and the angle between rod (2) and the horizontal axis:

2.3 m

tan 0.7188 35.70673.2 m

Write equilibrium equations for the sum of forces in the

horizontal and vertical directions. Note: Rods (1) and (2)

are two-force members.

2 1cos(35.7067 ) cos(57.9946 ) 0xF F F (a)

2 1sin(35.7067 ) sin(57.9946 ) 0yF F F P (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the

substitution method, Eq. (b) can be solved for F2 in terms of F1:

2 1

cos(57.9946 )

cos(35.7067 )F F

(c)

Substituting Eq. (c) into Eq. (b) gives

1 1

1

1

cos(57.9946 )sin(35.7067 ) sin(57.9946 )

cos(35.6553 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 )

cos(57.9946 ) tan(35.7067 ) sin(57.9946 ) 1.2289

F F P

F P

P PF

For the given load of P = 27 kN, the internal force in rod (1) is therefore:

1

27 kN21.9702 kN

1.2289F




Backsubstituting this result into Eq. (c) gives force F2:

2 1

cos(57.9946 ) cos(57.9946 )(21.9702 kN) 14.3399 kN

cos(35.7067 ) cos(35.7067 )F F

The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is:

2 2

1 (16 mm) 201.0619 mm4

A

and the normal stress in rod (1) is:

211 2

1

(21.9702 kN)(1,000 N/kN)109.2710 N/mm

201.0109.3 MPa (T)

619 mm

F

A Ans.

The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is:

2 2

2 (12 mm) 113.0973 mm4

A

and the normal stress in rod (2) is:

22

2 2

2

(14.3399 kN)(1,000 N/kN)126.7924 N/mm

113.0126.8 MPa (T)

973 mm

F

A Ans.




1.11 A simple pin-connected truss is loaded

and supported as shown in Fig. P1.11. All

members of the truss are aluminum pipes that

have an outside diameter of 4.00 in. and a wall

thickness of 0.226 in. Determine the normal

stress in each truss member.

Fig. P1.11

Solution

Overall equilibrium:

Begin the solution by determining the

external reaction forces acting on the

truss at supports A and B. Write

equilibrium equations that include all

external forces. Note that only the

external forces (i.e., loads and

reaction forces) are considered at this

time. The internal forces acting in the

truss members will be considered

after the external reactions have been

computed. The free-body diagram

(FBD) of the entire truss is shown.

The following equilibrium equations

can be written for this structure:

2 kips

2 ki

0

ps

x

x

xF A

A

(6 ft) (5 kips)(14 ft) (2 kips)(7 ft)

14 kips

0

y

A yB

B

M

5 kips 0

9 kips

y y y

y

F A B

A

Method of joints:

Before beginning the process of determining the internal forces in the axial members, the geometry of

the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use

the definition of the tangent function to determine AC and BC:

7 fttan 0.50 26.565

14 ft

7 fttan 0.875 41.186

8 ft

AC AC

BC BC




Joint A:

Begin the solution process by considering a FBD of joint A. Consider

only those forces acting directly on joint A. In this instance, two axial

members, AB and AC, are connected at joint A. Additionally, two

reaction forces, Ax and Ay, act at joint A. Tension forces will be

assumed in each truss member.

cos(26.565 ) 0x AC AB xF F F A (a)

sin(26.565 ) 0y AC yF F A (b)

Solve Eq. (b) for FAC:

9 kips

sin(26.565 ) sin(26.520.125 kip

65 )s

y

AC

AF

and then compute FAB using Eq. (a):

cos(26.565 )

(20.125 kips)cos(26.5 16.000 kips65 ) ( 2 kips)

AB AC xF F A

Joint B:

Next, consider a FBD of joint B. In this instance, the equilibrium

equations associated with joint B seem easier to solve than those that

would pertain to joint C. As before, tension forces will be assumed in

each truss member.

cos(41.186 ) 0x AB BCF F F (c)

sin(41.186 ) 0y BC yF F B (d)

Solve Eq. (d) for FBC:

14 kips

sin(41.186 ) sin(41.1821.260 kip

6s

)

y

BC

BF

Eq. (c) can be used as a check on our calculations:

cos(41.186 )

( 16.000 kips) ( 21.260 kips)cos(41.186 ) 0

x AB BCF F F

Checks!

Section properties:

For each of the three truss members:

2 2 24.00 in. 2(0.226 in.) 3.548 in. (4.00 in.) (3.548 in.) 2.67954 in.4

d A

Normal stress in each truss member:

2

16.000 kips5.971 ksi

2.679545.97 ksi (C)

in.

ABAB

AB

F

A

Ans.

2

20.125 kips7.510 ksi

2.679547.51 ksi (T)

in.

ACAC

AC

F

A Ans.

2

21.260 kips7.934 ksi

2.679547.93 ksi (C)

in.

BCBC

BC

F

A

Ans.







have an outside diameter of 60 mm and a wall

thickness of 4 mm. Determine the normal


Fig. P1.12

Solution



external reaction forces acting on the truss at

supports A and B. Write equilibrium

equations that include all external forces.

Note that only the external forces (i.e., loads

and reaction forces) are considered at this

time. The internal forces acting in the truss

members will be considered after the external

reactions have been computed. The free-

body diagram (FBD) of the entire truss is

shown. The following equilibrium equations

can be written for this structure:

12 k

12

N 0

kNx

x xF A

A

(1 m) (15 kN)(4.3 m) 0

64.5 kNy

A yB

B

M

15 kN

49.5 kN

0

y

y y yF

A

A B

Method of joints:


the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use

the definition of the tangent function to determine AB and BC:

1.5 mtan 1.50 56.310

1.0 m

1.5 mtan 0.454545 24.444

3.3 m

AB AB

BC BC




Joint A:




reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed

in each truss member.

cos(56.310 ) 0x AC AB xF F F A (a)

sin(56.310 ) 0y y ABF A F (b)

Solve Eq. (b) for FAB:

49.5 kN

sin(56.310 ) sin(56.310 )59.492 kN

y

AB

AF

and then compute FAC using Eq. (a):

cos(56.310 )

( 59.492 kN)cos(56.3 45.000 10 ) ( 12 kN) kN

AC AB xF F A

Joint C:

Next, consider a FBD of joint C. In this instance, the equilibrium

equations associated with joint C seem easier to solve than those that

would pertain to joint B. As before, tension forces will be assumed in

each truss member.

cos(24.444 ) 12 kN 0x AC BCF F F (c)

sin(24.444 ) 15 kN 0y BCF F (d)

Solve Eq. (d) for FBC:

15 kN

sin(24.444 )36.249 kNBCF

Eq. (c) can be used as a check on our calculations:

cos(24.444 ) 12 kN 0

(45.000 kN) ( 36.249 kN)cos(24.444 ) 12 kN 0

x AC BCF F F

Checks!

Section properties:


2 2 260 mm 2(4 mm) 52 mm (60 mm) (52 mm) 703.7168 mm4

d A


2

( 59.492 kN)(1,000 N/kN)84.539 MPa

7084.5 MPa (C)

3.7168 mm

ABAB

AB

F

A

Ans.

2

(45.000 kN)(1,000 N/kN)63.946 MPa

7063.9 MPa

3.7168)

mm(TAC

AC

AC

F

A Ans.

2

( 36.249 kN)(1,000 N/kN)51.511 MPa

7051.5 MPa (C)

3.7168 mm

BCBC

BC

F

A

Ans.







have an outside diameter of 42 mm and a wall

thickness of 3.5 mm. Determine the normal


Fig. P1.13

Solution


Begin the solution by determining the external

reaction forces acting on the truss at supports A

and B. Write equilibrium equations that include all

external forces. Note that only the external forces

(i.e., loads and reaction forces) are considered at

this time. The internal forces acting in the truss

members will be considered after the external

reactions have been computed. The free-body

diagram (FBD) of the entire truss is shown. The

following equilibrium equations can be written for

this structure:

30 kN

30 kN

0y

y

yF A

A

(30 kN)(4.5 m) (15 kN)(1.6 m) (5.6 m)

19.821 kN

0

x

A x

B

M B

15 kN 0

15 kN 15 kN ( 19.821 kN 34.821 ) kN

x x

x x

x

x

F A B

A AB

Method of joints:


the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use

the definition of the tangent function to determine AC and BC:

1.6 mtan 0.355556 19.573

4.5 m

4 mtan 0.888889 41.634

4.5 m

AC AC

BC BC




Joint A:






cos(19.573 ) 0x x ACF A F (a)

sin(19.573 ) 0y y AC ABF A F F (b)

Solve Eq. (a) for FAC:

34.821 kN

cos(19.573 ) cos(19.573 )36.957 kNx

AC

AF

and then compute FAB using Eq. (b):

sin(19.573 )

(30.000 kN) (36.957 kN)sin(19. 17.619 573 ) kN

AB y ACF A F

Joint B:




each truss member.

cos(41.634 ) 0x x BCF B F (c)

sin(41.634 ) 0y BC ABF F F (d)

Solve Eq. (c) for FBC:

( 19.821 kN)

cos(41.634 ) cos(41.634 )26.520 kNx

BC

BF

Eq. (d) can be used as a check on our calculations:

sin(41.634 )

( 26.520 kN)sin(41.634 ) (17.619 kN) 0

y BC ABF F F

Checks!

Section properties:


2 2 242 mm 2(3.5 mm) 35 mm (42 mm) (35 mm) 423.3296 mm4

d A


2

(17.619 kN)(1,000 N/kN)41.620 MPa

4241.6 MPa

3.3296)

mm(TAB

AB

AB

F

A Ans.

2

(36.957 kN)(1,000 N/kN)87.301 MPa

4287.3 MPa

3.3296)

mm(TAC

AC

AC

F

A Ans.

2

( 26.520 kN)(1,000 N/kN)62.647 MPa

4262.6 MPa (C)

3.3296 mm

BCBC

BC

F

A

Ans.




1.14 The members of the truss shown in Fig.

P1.14 are aluminum pipes that have an outside

diameter of 4.50 in. and a wall thickness of

0.237 in. Determine the normal stress in each

truss member.

Fig. P1.14

Solution



external reaction forces acting on the truss at

supports A and B. Write equilibrium

equations that include all external forces.

Note that only the external forces (i.e., loads

and reaction forces) are considered at this

time. The internal forces acting in the truss

members will be considered after the

external reactions have been computed. The

free-body diagram (FBD) of the entire truss

is shown. The following equilibrium

equations can be written for this structure:

(15 kips)cos5

9.642 k

0

ip

0

sx

x xF A

A

(4 ft) (15 kips)(4 ft)cos50 (15 kips)(18 ft)sin50 0

61.350 kips

(15 kips)sin50

49.859 kips

0

A y

y y y

y

y

M B

F A B

B

A

Method of joints:


the truss will be used to determine the magnitude of the inclination angles of members AB, AC, and BC.

Use the definition of the tangent function to determine AB, AC, and BC:

6 fttan 1.5 56.3099

4 ft

4 fttan 0.222222 12.5288

18 ft

10 fttan 0.714286 35.5377

14 ft

AB AB

AC AC

BC BC




Joint A:






cos(12.5288 ) cos(56.3099 ) 0x AC AB xF F F A (a)

sin(12.5288 ) sin(56.3099 ) 0y AC AB yF F F A (b)

Solve Eqs. (a) and (b) simultaneously to obtain:

49.948 kips

38.259 kips

AB

AC

F

F

Joint B:




each truss member.

cos(35.5377 ) cos(56.3099 ) 0x BC ABF F F (c)

sin(35.5377 ) sin(56.3099 ) 0y BC AB yF F F B (d)

Solve Eq. (c) for FBC:

cos(56.3099 ) cos(56.3099 )

( 49.9484)cos(35.5377 ) cos(35.5377

34.048 k s)

ipBC ABF F

Eq. (d) can be used as a check on our calculations:

sin(35.5377 ) sin(56.3099 )

( 34.0485 kips)sin(35.5377 ) ( 49.9484 kips)sin(56.3099 ) 61.350 kips 0

y BC AB yF F F B

Checks!

Section properties:


2 2 24.50 in. 2(0.237 in.) 4.026 in. (4.50 in.) (4.026 in.) 3.17405 in.4

d A


2

49.948 kips15.736 ksi

3.17405 15.74 ksi (C

in)

.

ABAB

AB

F

A

Ans.

2

38.259 kips12.054 ksi

3.17405 12.05 ksi (T

n)

i .

ACAC

AC

F

A Ans.

2

34.048 kips10.727 ksi

3.17405 10.73 ksi (C

in)

.

BCBC

BC

F

A

Ans.




1.15 Bar (1) in Fig. P1.15 has a cross-

sectional area of 0.75 in.2. If the stress in bar

(1) must be limited to 30 ksi, determine the

maximum load P that may be supported by

the structure.

Fig. P1.15

Solution

Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi,

the maximum force that may be carried by bar (1) is

2

1,max 1 1 (30 ksi)(0.75 in. ) 22.5 kipsF A

Consider a FBD of ABC. From the moment equilibrium

equation about joint A, the relationship between the force in

bar (1) and the load P is:

1

1

(6 ft) (10 ft) 0

6 ft

10 ft

AM F P

P F

Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that

may be applied to the structure:

1

6 ft 6 ft(22.5 kips)

10 ft 10 ft13.50 kipsP F Ans.




1.16 Two 6 in. wide wooden boards are to

be joined by splice plates that will be fully

glued on the contact surfaces. The glue to

be used can safely provide a shear strength

of 120 psi. Determine the smallest

allowable length L that can be used for the

splice plates for an applied load of P =

10,000 lb. Note that a gap of 0.5 in. is

required between boards (1) and (2).

Fig. P1.16

Solution

Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied

load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as

V, equilibrium in the horizontal direction requires

0

10,000 lb5,000 lb

2

xF P V V

V

In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be

provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on

board (2) must be at least

2

min

5,000 lb41.6667 in.

120 psiA

The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least

2

glue joint

41.6667 in.6.9444 in.

6 in.L

Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1)

and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the

boards in order to resist the 10,000 lb applied load.

The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore,

the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a

0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer.

Altogether, the length of the splice plates must be at least

min 6.9444 in. 6.9444 in. 0.5 in 14.39 in..L Ans.




1.17 For the clevis connection shown in Fig. P1.17,

determine the shear stress in the 22-mm diameter bolt

for an applied load of P = 90 kN.

Fig. P1.17

Solution

Consider a FBD of the bar that is connected by the clevis,

including a portion of the bolt. If the shear force acting on each

exposed surface of the bolt is denoted by V, then the shear force

on each bolt surface is

90 kN

0 45 kN2

xF P V V V

The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:

2 2 2

bolt bolt (22 mm) 380.1327 mm4 4

A d

Therefore, the shear stress in the bolt is

2

2

bolt

(45 kN)(1,000 N/kN)118.3797 N/mm

380.1118.4 MPa

327 mm

V

A Ans.




1.18 For the clevis connection shown in Fig.

P1.18, the shear stress in the 3/8 in. diameter

bolt must be limited to 36 ksi. Determine the

maximum load P that may be applied to the

connection.

Fig. P1.18

Solution

Consider a FBD of the bar that is connected by the clevis,

including a portion of the bolt. If the shear force acting on each

exposed surface of the bolt is denoted by V, then the shear force

on each bolt surface is related to the load P by:

0 2xF P V V P V

The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt:

2 2 2 2

bolt bolt (3 / 8 in.) (0.3750 in.) 0.1104466 in.4 4 4

A d

If the shear stress in the bolt must be limited to 36 ksi, the maximum shear force V on a single cross-

sectional surface must be limited to

2

bolt (36 ksi)(0.1104466 in. ) 3.976078 kipsV A

Therefore, the maximum load P that may be applied to the connection is

2 2(3.976078 kips) 7.952156 k 7ips .95 kipsP V Ans.




1.19 For the connection shown in Fig. P1.19,

determine the average shear stress in the 7/8-in.

diameter bolts if the load is P = 45 kips.

Fig. P1.19

Solution

The bolts in this connection act in single shear. The cross-sectional area of a single bolt is

2 2 2 2

bolt bolt (7 / 8 in.) (0.875 in.) 0.6013205 in.4 4 4

A d

Since there are five bolts, the total area that carries shear stress is

2 2

bolt5 5(0.6013205 in. ) 3.006602 in.VA A

Therefore, the shear stress in each bolt is

2

45 kips14.96706 ksi

3.006602 in.14.97 ksi

V

P

A Ans.




1.20 The five-bolt connection shown in Fig.

P1.20 must support an applied load of P = 300

kN. If the average shear stress in the bolts must

be limited to 225 MPa, determine the minimum

bolt diameter that may be used in the connection.

Fig P1.20

Solution

To support a load of 300 kN while not exceeding an average shear stress of 225 MPa, the total shear

area provided by the bolts must be at least

2

2

(300 kN)(1,000 N/kN)1,333.3333 mm

225 N/mmV

PA

Since there are five single-shear bolts in this connection, five cross-sectional surfaces carry shear stress.

Consequently, each bolt must provide a minimum area of

2

2

bolt

1,333.3333 mm266.6667 mm

5 5

VAA

The minimum bolt diameter is therefore

2 2

bolt bolt bolt 18.43 266.6667 mm 18.42 mm64 mm4

A d d

Ans.




1.21 The three-bolt connection shown in Fig. P1.21

must support an applied load of P = 40 kips. If the

average shear stress in the bolts must be limited to

24 ksi, determine the minimum bolt diameter that

may be used in the connection.

Fig. P1.21

Solution

The shear force V that must be provided by the bolts equals the applied load of P = 40 kips. The total

shear area required is thus

240 kips1.66667 in.

24 ksiV

VA

The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are

available to transmit shear stress.

2

2

bolt

1.66667 in.0.27778 in. per surface

(2 surfaces per bolt)(3 bolts) 6 surfaces

VAA

The minimum bolt diameter must be

2 2

bolt bolt0.27778 in. 0.59471 in. 0.595 i .4

nd d

Ans.




1.22 For the connection shown in Fig. P1.22,

the average shear stress in the 12-mm-diameter

bolts must be limited to 160 MPa. Determine

the maximum load P that may be applied to the

connection.

Fig. P1.22

Solution

The cross-sectional area of a 12-mm-diameter bolt is

2 2 2

bolt bolt (12 mm) 113.097355 mm4 4

A d

This is a double-shear connection. Therefore, the three bolts provide a total shear area of

2 2

bolt2(3 bolts) 2(3 bolts)(113.097355 mm ) 678.58401 mmVA A

Since the shear stress must be limited to 160 MPa, the total shear force that can be resisted by the three

bolts is

2 2

max (160 N/mm )(678.58401 mm ) 108,573.442 NVV A

In this connection, the shear force in the bolts is equal to the applied load P; therefore,

max 108.6 kNP Ans.




1.23 A hydraulic punch press is used to

punch a slot in a 0.50-in. thick plate, as

illustrated in Fig. P1.23. If the plate shears

at a stress of 30 ksi, determine the

minimum force P required to punch the

slot.

Fig. P1.23

Solution

The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is

given by

perimeter 2(3.00 in.) + (0.75 in.) 8.35619 in.

Thus, the area subjected to shear stress is

2perimeter plate thickness (8.35619 in.)(0.50 in.) 4.17810 in.VA

Given that the plate shears at = 30 ksi, the force required to remove the slug is therefore

2

min (30 ksi)(4.17810 in. ) 125.343 kips 125.3 kipsVP A Ans.




1.24 A coupling is used to connect a 2 in. diameter

plastic pipe (1) to a 1.5 in. diameter pipe (2), as

shown in Fig. P1.24. If the average shear stress in

the adhesive must be limited to 400 psi, determine

the minimum lengths L1 and L2 required for the joint

if the applied load P is 5,000 lb.

Fig. P1.24

Solution

To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is

2

adhesive

5,000 lb12.5 in.

400 psiV

VA

Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the

circumference C1 of pipe (1) is

1 1 (2.0 in.) 6.2832 in.C D

The minimum length L1 is therefore

2

1

1

12.5 in.1.9894 in.

6.2832 i1.989 i

n n.

.

VAL

C Ans.

Consider the coupling on pipe (2). The circumference C2 of pipe (2) is

2 2 (1.5 in.) 4.7124 in.C D

The minimum length L2 is therefore

2

2

2

12.5 in.2.6526 in.

4.7124 2.65 in.

in.

VAL

C Ans.




1.25 A lever is attached to a shaft with a

square shear key, as shown in Fig. P1.25.

The force applied to the lever is P = 400 N.

If the shear stress in the key must not exceed

90 MPa, determine the minimum dimension

“a” that must be used if the key is 15 mm

long.

Fig. P1.25

Solution

To determine the shear force V that must be resisted by the shear key, sum moments about the center of

the shaft (which will be denoted O):

50 mm

(400 N)(750 mm) 0 12,000 N2

OM V V

Since the shear stress in the key must not exceed 90 MPa, the shear area required is

2

2

12,000 N133.3333 mm

90 N/mmV

VA

The shear area in the key is given by the product of its length L (i.e., 15 mm) and its width a. Therefore,

the minimum key width a is

2133.3333 mm

8.8889 mm15

8.89 mmmm

VAa

L Ans.




1.26 A common trailer hitch connection is shown in

Fig. P1.26. The shear stress in the pin must be limited

to 30,000 psi. If the applied load is P = 4,000 lb,

determine the minimum diameter that must be used

for the pin.

Fig. P1.26

Solution

The shear force V acting in the hitch pin is equal to the applied load; therefore, V = P = 4,000 lb. The

shear area required to support a 4,000 lb shear force is

24,000 lb

0.1333 in.30,000 psi

V

VA

The hitch pin is used in a double-shear connection; therefore, two cross-sectional areas of the pin are

subjected to shear stress. Thus, the cross-sectional area of the pin is given by

2

2

pin pin

0.1333 in.2 0.0667 in.

2 2

V

V

AA A A

and the minimum pin diameter is

2 2

pin pin0.0667 in. 0.2913 in. 0.291 n.4

id d

Ans.




1.27 An axial load P is supported by a short steel

column, which has a cross-sectional area of

11,400 mm2. If the average normal stress in the steel

column must not exceed 110 MPa, determine the

minimum required dimension “a” so that the bearing

stress between the base plate and the concrete slab does

not exceed 8 MPa.

Fig. P1.27

Solution

Since the normal stress in the steel column must not exceed 110 MPa, the maximum column load is

2 2

max (110 N/mm )(11,400 mm ) 1,254,000 NP A

The maximum column load must be distributed over a large enough area so that the bearing stress

between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate area

is

2

min 2

1,254,000 N156,750 mm

8 N/mmb

PA

Since the plate is square, the minimum plate dimension a must be

2

min

396

156,

mm

750 mm

395.9167 mm

A a a

a

Ans.




1.28 The steel pipe column shown in Fig. P1.28 has an

outside diameter of 8.625 in. and a wall thickness of 0.25

in. The timber beam is 10.75 in wide, and the upper plate

has the same width. The load imposed on the column by

the timber beam is 80 kips. Determine

(a) The average bearing stress at the surfaces between the

pipe column and the upper and lower steel bearing

plates.

(b) The length L of the rectangular upper bearing plate if

its width is 10.75 in. and the average bearing stress

between the steel plate and the wood beam is not to

exceed 500 psi.

(c) The dimension “a” of the square lower bearing plate if

the average bearing stress between the lower bearing

plate and the concrete slab is not to exceed 900 psi.

Fig. P1.28

Solution

(a) The area of contact between the pipe column and one of the bearing plates is simply the cross-

sectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d:

2 2 8.625 in. 2(0.25 in.) 8.125 in.D d t d D t

The pipe cross-sectional area is

2 2 2 2 2

pipe (8.625 in.) (8.125 in.) 6.5777 in.4 4

A D d

Therefore, the bearing stress between the pipe and one of the bearing plates is

2

80 kips12.1623 ksi

6.5777 in.12.16 ksib

b

P

A Ans.

(b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi

(i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least

280 kips

160 in.0.5 ksi

b

b

PA

If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be

2160 in.

14.8837 in.beam width 10.75

14.88 in

.in.

bAL Ans.

(c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi

(i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least

280 kips

88.8889 in.0.9 ksi

b

b

PA

Since the lower bearing plate is square, its dimension a must be

288.8889 in. 9.4 9.43 in281 n. .ibA a a a Ans.




1.29 A vertical shaft is supported by a thrust

collar and bearing plate, as shown in Fig.

P1.29. The average shear stress in the collar

must be limited to 18 ksi. The average bearing

stress between the collar and the plate must be

limited to 24 ksi. Based on these limits,

determine the maximum axial load P that can

be applied to the shaft.

Fig. P1.29

Solution

Consider collar shear stress: The area subjected to shear stress in the collar is equal to the product of

the shaft circumference and the collar thickness; therefore,

2shaft circumference collar thickness (1.0 in.)(0.5 in.) 1.5708 in.VA

If the shear stress must not exceed 18 ksi, the maximum load that can be supported by the vertical shaft

is:

2(18 ksi)(1.5708 in. ) 28.2743 kipsVP A

Consider collar bearing stress: We must determine the area of contact between the collar and the

plate. The overall cross-sectional area of the collar is

2 2

collar (1.5 in.) 1.7671 in.4

A

is reduced by the area taken up by the shaft

2 2

shaft (1.0 in.) 0.7854 in.4

A

Therefore, the area of the collar that actually contacts the plate is

2 2 2

collar shaft 1.7671 in. 0.7854 in. 0.9817 in.bA A A

If the bearing stress must not exceed 24 ksi, the maximum load that can be supported by the vertical

shaft is:

2(24 ksi)(0.9817 in. ) 23.5619 kipsb bP A

Controlling P: Considering both shear stress in the collar and bearing stress between the collar and the

plate, the maximum load that can be supported by the shaft is

max 23.6 kipsP Ans.




1.30 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial

load of 150 kN. Determine the maximum normal and shear stresses in the bar.

Solution

The maximum normal stress in the steel bar is

max

(150 kN)(1,000 N/kN)

(25 mm)(75 mm)80 MPa

F

A Ans.

The maximum shear stress is one-half of the maximum normal stress

maxmax

240 MPa

Ans.




1.31 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The maximum

stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the required

diameter for the rod.

Solution

Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is

2

min

max

92 kips3.0667 in.

30 ksi

FA

For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be

2

min

max

92 kips3.8333 in.

2 2(12 ksi)

FA

Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the

normal and shear stress limits.

The minimum rod diameter D is therefore

2 2

min min3.8333 in. 2.2092 in. 2.21 in.4

d d

Ans.




1.32 An axial load P is applied to the

rectangular bar shown in Fig. P1.32. The

cross-sectional area of the bar is 400 mm2.

Determine the normal stress perpendicular to

plane AB and the shear stress parallel to

plane AB if the bar is subjected to an axial

load of P = 70 kN.

Fig. P1.32

Solution

The angle for the inclined plane is 35°. The

normal force N perpendicular to plane AB is

found from

cos (40 kN)cos35 57.3406 kNN P

and the shear force V parallel to plane AB is

sin (70 kN)sin35 40.1504 kNV P

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is

2

2400 mm488.3098 mm

cos cos35n

AA

The normal stress n perpendicular to plane AB is

2

(57.3406 kN)(1,000 N/kN)117.4268 MPa

488117.4 MP

.3098 mman

n

N

A Ans.

The shear stress nt parallel to plane AB is

2

(40.1504 kN)(1,000 N/kN)82.2231 MPa

482.2 MP

88.309 a

8 mmnt

n

V

A Ans.




1.33 An axial load P is applied to the 1.75 in.

by 0.75 in. rectangular bar shown in Fig.

P1.33. Determine the normal stress

perpendicular to plane AB and the shear stress

parallel to plane AB if the bar is subjected to

an axial load of P = 18 kips.

Fig. P1.33

Solution

The angle for the inclined plane is 60°. The

normal force N perpendicular to plane AB is

found from

cos (18 kips)cos60 9.0 kipsN P


sin (18 kips)sin60 15.5885 kipsV P

The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane

AB is

2

21.3125 in./ cos 2.6250 in.

cos60nA A


2

9.0 kips3.4286 ksi

2.6250 in3.43 ks

.in

n

N

A Ans.


2

15.5885 kips5.9385 ksi

2.62505.94 ks

ii

n.nt

n

V

A Ans.




1.34 A compression load of P = 80 kips is applied to a 4 in.

by 4 in. square post, as shown in Fig. P1.34. Determine the

normal stress perpendicular to plane AB and the shear stress

parallel to plane AB.

Fig. P1.34

Solution

The angle for the inclined plane is 55°. The normal force N

perpendicular to plane AB is found from

cos (80 kips)cos55 45.8861 kipsN P


sin (80 kips)sin55 65.5322 kipsV P

The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area

along inclined plane AB is

2

216 in./ cos 27.8951 in.

cos55nA A


2

45.8861 kips1.6449 ksi

27.8951 1.645 ksi

in.n

n

N

A Ans.


2

65.5322 kips2.3492 ksi

27.89512.35 ks

ii

n.nt

n

V

A Ans.




1.35 Specifications for the 50 mm × 50 mm square bar

shown in Fig. P1.35 require that the normal and shear

stresses on plane AB not exceed 120 MPa and 90 MPa,

respectively. Determine the maximum load P that can be

applied without exceeding the specifications.

Fig. P1.35

Solution

The general equations for normal and shear stresses on an inclined plane in terms of the angle are

(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm

2, and the angle for plane AB is

55°.

The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be

supported by the square bar is found from Eq. (a):

2 22 2(2,500 mm )(120 N/mm )

911,882 N1 cos2 1 cos2(55 )

nAP

The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear

stress limit is

2 22 2(2,500 mm )(90 N/mm )

478,880 Nsin 2 sin 2(55 )

ntAP

Thus, the maximum load that can be supported by the bar is

max 479 kNP Ans.




1.36 Specifications for the 6 in. × 6 in. square post shown in

Fig. P1.36 require that the normal and shear stresses on plane

AB not exceed 800 psi and 400 psi, respectively. Determine

the maximum load P that can be applied without exceeding

the specifications.

Fig. P1.36

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The cross-sectional area of the square post is A = (6 in.)2 = 36 in.

2, and the angle for plane AB is 40°.

The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be

supported by the square post is found from Eq. (a):

22 2(36 in. )(800 psi)

49,078 lb1 cos2 1 cos2(40 )

nAP

The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear

stress limit is

22 2(36 in. )(400 psi)

29,244 lbsin 2 sin 2(40 )

ntAP

Thus, the maximum load that can be supported by the post is

max 29,200 l 29.2 ipb k sP Ans.




1.37 A 90 mm wide bar will be used to carry an axial

tension load of 280 kN. The normal and shear stresses

on plane AB must be limited to 150 MPa and 100 MPa,

respectively. Determine the minimum thickness t

required for the bar.

Fig. P1.37

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The angle for plane AB is 50°.

The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A

required to support P = 280 kN can be found from Eq. (a):

2

2

(280 kN)(1,000 N/kN)(1 cos 2 ) (1 cos 2(50 )) 771.2617 mm

2 2(150 N/mm )n

PA

The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A

required to support P = 280 kN can be found from Eq. (b):

2

2

(280 kN)(1,000 N/kN)sin 2 sin 2(50 ) 1,378.7309 mm

2 2(100 N/mm )nt

PA

To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin =

1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be

2

min

1,378.7309 mm15.3192 mm

90 m15.3

m2 mmt Ans.




1.38 A rectangular bar having width w = 6.00

in. and thickness t = 1.50 in. is subjected to a

tension load P. The normal and shear stresses

on plane AB must not exceed 16 ksi and 8 ksi,

respectively. Determine the maximum load P

that can be applied without exceeding either

stress limit. Fig. P1.38

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The angle for inclined plane AB is calculated from

3

tan 3 71.56511

The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2.

The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from

Eq. (a):

22 2(9.0 in. )(16 ksi)

1,440 ksi1 cos2 1 cos2(71.5651 )

nAP

The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear

stress limit is

22 2(9.0 in. )(8 ksi)

240 kipssin 2 sin 2(71.5651 )

ntAP

Thus, the maximum load that can be supported by the bar is

max 240 kipsP Ans.




1.39 In Fig. P1.39, a rectangular bar having width

w = 1.25 in. and thickness t is subjected to a

tension load of P = 30 kips. The normal and shear

stresses on plane AB must not exceed 12 ksi and 8

ksi, respectively. Determine the minimum bar

thickness t required for the bar.

Fig. P1.39

Solution


(1 cos 2 )2

n

P

A (a)

and

sin 22

nt

P

A (b)

The angle for inclined plane AB is calculated from

3

tan 3 71.56511

The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A

required to support P = 30 kips can be found from Eq. (a):

230 kips

(1 cos2 ) (1 cos2(71.5651 )) 0.2500 in.2 2(12 ksi)n

PA

The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required

to support P = 30 kips can be found from Eq. (b):

230 kips

sin 2 sin 2(71.5651 ) 1.1250 in.2 2(8 ksi)nt

PA

To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin =

1.1250 in.2. Since the bar width is 1.25 in., the minimum bar thickness t must be

2

min

1.1250 in.0.900 in.

1.25 in0.900 .

. int Ans.




1.40 The rectangular bar has a width of w = 3.00

in. and a thickness of t = 2.00 in. The normal

stress on plane AB of the rectangular block

shown in Fig. P1.40 is 6 ksi (C) when the load P

is applied. Determine:

(a) the magnitude of load P.

(b) the shear stress on plane AB.

(c) the maximum normal and shear stresses in

the block at any possible orientation.

Fig. P1.40

Solution

The general equation for normal stress on an inclined plane in terms of the angle is

(1 cos 2 )2

n

P

A (a)

and the angle for inclined plane AB is

3

tan 0.75 36.86994

The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2.

(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from

Eq. (a):

22 2(6.0 in. )(6 ksi)

56.25 kips1 cos2 1 co

56.3 kips2(36.8699 )

snAP

Ans.

(b) The general equation for shear stress on an inclined plane in terms of the angle is

sin 22

nt

P

A

therefore, the shear stress on plane AB is

2

56.25 kipssin 2(36.8699 )

2(6.00 in. )4.50 ksint Ans.

(c) The maximum normal stress at any possible orientation is

max 2

56.25 kips9.3750 ksi

6.00 i9.38 k

n si

.

P

A Ans.

and the maximum shear stress at any possible orientation in the block is

max 2

56.25 kips4.6875 ksi

2 2(6.00 4.69 ksi

in. )

P

A Ans.




1.41 The rectangular bar has a width of w = 100

mm and a thickness of t = 75 mm. The shear stress

on plane AB of the rectangular block shown in

Fig. P1.41 is 12 MPa when the load P is applied.

Determine:


(b) the normal stress on plane AB.

(c) the maximum normal and shear stresses in the

block at any possible orientation.

Fig. P1.41

Solution

The general equation for shear stress on an inclined plane in terms of the angle is

sin 22

nt

P

A (a)

and the angle for inclined plane AB is

3

tan 0.75 36.86994

The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2.

(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated

from Eq. (a):

2 22 2(7,500 mm )(12 N/mm )

187,500 Nsin 2 sin 2(36.869

187.5 k9

N)

ntAP

Ans.

(b) The general equation for normal stress on an inclined plane in terms of the angle is

(1 cos 2 )2

n

P

A

therefore, the normal stress on plane AB is

2

187,500 N(1 cos2(36.8699 ))

2(7,500 mm16.00 MPa

)n Ans.

(c) The maximum normal stress at any possible orientation is

max 2

187,500 N

7,500 mm25.0 MPa

P

A Ans.

and the maximum shear stress at any possible orientation in the block is

max 2

187,500 N

2 2(7,500 mm12.50 MPa

)

P

A Ans.




2.1 When an axial load is applied to the

ends of the bar shown in Fig. P2.1, the total

elongation of the bar between joints A and C

is 0.15 in. In segment (2), the normal strain

is measured as 1,300 in./in. Determine:

(a) the elongation of segment (2).

(b) the normal strain in segment (1) of

the bar.

Fig. P2.1

Solution

(a) From the definition of normal strain, the elongation in segment (2) can be computed as

6

2 2 2 (1,300 10 )(90 in. 0.1170 i) n.L Ans.

(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that

occurs in segment (1) must be

1 total 2 0.15 in. 0.1170 in. 0.0330 in.

The strain in segment (1) can now be computed:

11

1

0.0330 in.0.000825 in./in.

40 in.825 μin./in.

L Ans.




2.2 A rigid steel bar is supported by three rods, as

shown in Fig. P2.2. There is no strain in the rods before

the load P is applied. After load P is applied, the

normal strain in rod (2) is 1,080 in./in. Assume initial

rod lengths of L1 = 130 in. and L2 = 75 in. Determine:

(a) the normal strain in rods (1).

(b) the normal strain in rods (1) if there is a 1/32-in. gap

in the connections between the rigid bar and rods

(1) at joints A and C before the load is applied.

(c) the normal strain in rods (1) if there is a 1/32-in. gap

in the connection between the rigid bar and rod (2)

at joint B before the load is applied.

Fig. P2.2

Solution

(a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated:

6

2 2 2 (1,080 10 )(75 in.) 0.0810 in.L

Since rod (2) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must

move downward by an amount equal to the deformation of rod (2); therefore,

2 0.0810 in. (downward)Bv

By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods

(1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The

deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal

strain in rods (1) can now be calculated as:

11

1

0.0810 in.0.0006231 in./in.

130 in.623 μin./in.

L Ans.

(b) We can assume that the bolted connection at B is

perfect; therefore, vB = 0.0810 in. (downward).

Further, the rigid bar must remain horizontal as it

deflects downward by virtue of symmetry.

Therefore, the deflection downward of joints A and C

is still equal to 0.0810 in.

What effect is caused by the gap at A and C? When

joint A (or C) moves downward by 0.0810 in., the

first 1/32-in. of this downward movement does not

stretch rod (1)—it just closes the gap. Therefore, rod

(1) only gets elongated by the amount

1 0.03125 in.

0.0810 in. 0.03125 in.

0.04975 in.

Av




This deformation creates a strain in rod (1) of:

11

1

0.04975 in.0.0003827 in./in.

130 383 μin./in.

in.L Ans.

(c) The gap is now at joint B. We know the strain

in rod (2); hence, we know its deformation must

be 0.0810 in. However, the first 1/32-in. of

downward movement by the rigid bar does not

elongate the rod—it simply closes the gap. To

elongate rod (2) by 0.0810 in., joint B must move

down:

2 0.03125 in.

0.0810 in. 0.03125 in.

0.11225 in.

Bv

Again, since the rigid bar remains horizontal, vB =

vA = vC. The joints at A and C are assumed to be

perfect; thus,

1 0.11225 in.Av

and the normal strain in rod (1) is:

11

1

0.11225 in.0.00086346 in./in.

130 i863 μin./i

.n.

nL Ans.




2.3 A rigid steel bar is supported by three rods, as

shown in Fig. P2.3. There is no strain in the rods

before the load P is applied. After load P is

applied, the normal strain in rods (1) is 860 m/m.

Assume initial rod lengths of L1 = 2,400 mm and L2

= 1,800 mm. Determine:

(a) the normal strain in rod (2).

(b) the normal strain in rod (2) if there is a 2-

mm gap in the connections between the rigid

bar and rods (1) at joints A and C before the

load is applied.

(c) the normal strain in rod (2) if there is a 2-

mm gap in the connection between the rigid bar

and rod (2) at joint B before the load is applied. Fig. P2.3

Solution

(a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated:

6

1 1 1 (860 10 )(2,400 mm) 2.064 mmL

Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must

move downward by an amount equal to the deformation of rod (1); therefore,

1 2.064 mm (downward)A Cv v

By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod

(2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of

rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can

now be calculated as:

22

2

2.064 mm0.0011467 mm/mm 1,147 μmm/m

1,800 mm

mL Ans.

(b) We know the strain in rod (1); hence, we know its

deformation must be 2.064 mm. However, the joints

at A and C are not perfect connections. The first 2

mm of downward movement by the rigid bar does not

elongate rod (1)—it simply closes the gap. To

elongate rod (1) by 2.064 mm, joint A must move

down:

1 2 mm 2.064 mm 2 mm 4.064 mmAv

By symmetry, the rigid bar remains horizontal;

therefore, vB = vA = vC. The joint at B is assumed to be

perfect; thus, any downward movement of the rigid

bar also elongates rod (2) by the same amount:

2 4.064 mmBv

and the normal strain in rod (2) is:




22

2

4.064 mm0.0022578 mm/mm 2,260 μmm/m

1,800 mm

mL Ans.

(c) We now assume that the bolted connection at A

(or C) is perfect; therefore, the rigid bar deflection

as calculated in part (a) must be vA = 2.064 mm

(downward). Further, the rigid bar must remain

horizontal as it deflects downward by virtue of

symmetry; thus, vB = vA = vC. Therefore, the

deflection downward of joint B is vB = 2.064 mm

What effect is caused by the gap at B? When joint

B moves downward by 2.064 mm, the first 2 mm

of this downward movement does not stretch rod

(2)—it just closes the gap. Therefore, rod (2) only

gets elongated by the amount

2 2 mm

2.064 mm 2 mm

0.064 mm

Bv

This deformation creates a strain in rod (2) of:

22

2

0.064 mm0.0000356 mm/mm

1,800 m35.6 μmm/mm

mL Ans.




2.4 A rigid bar ABCD is supported by two bars as

shown in Fig. P2.4. There is no strain in the

vertical bars before load P is applied. After load P

is applied, the normal strain in rod (1) is −570

m/m. Determine:

(a) the normal strain in rod (2).

(b) the normal strain in rod (2) if there is a 1-mm

gap in the connection at pin C before the load is

applied.

(c) the normal strain in rod (2) if there is a 1-mm

gap in the connection at pin B before the load is

applied.

Fig. P2.4

Solution

(a) From the strain given for rod (1)

1 1 1

6( 570 10 )(900 mm)

0.5130 mm

L

Therefore, vB = 0.5130 mm (downward).

From the deformation diagram of rigid bar ABCD

240 mm (240 mm 360 mm)

600 mm(0.5130 mm) 1.2825 mm

240 mm

B C

C

v v

v

Therefore, 2 = 1.2825 mm (elongation), and thus, from the definition of strain:

62

2

2

1.2825 mm855 10 mm/mm

1,500 m855 με

mL Ans.

(b) The 1-mm gap at C doesn’t affect rod (1);

therefore, 1 = −0.5130 mm. The rigid bar

deformation diagram is unaffected; thus, vB =

0.5130 mm (downward) and vC = 1.2825 mm

(downward).

The rigid bar must move downward 1 mm at C

before it begins to elongate member (2).

Therefore, the elongation of member (2) is

2 1 mm

1.2825 mm 1 mm

0.2825 mm

Cv




and so the strain in member (2) is

62

2

2

0.2825 mm188.3 10 mm/mm

1,500 mm188.3 με

L Ans.

(c) From the strain given for rod (1), 1 =

−0.5130 mm. In order to contract rod (1) by this

amount, the rigid bar must move downward at B

by

0.5130 mm 1 mm 1.5130 mmBv

From deformation diagram of rigid bar ABCD

240 mm (240 mm 360 mm)

600 mm(1.5130 mm) 3.7825 mm

240 mm

B C

C

v v

v

and so

62

2

2

3.7825 mm2,521.7 10 mm/mm

1,5002,520 ε

μ

mmL Ans.




2.5 In Fig. P2.5, rigid bar ABC is supported by a

pin connection at B and two axial members. A slot

in member (1) allows the pin at A to slide 0.25-in.

before it contacts the axial member. If the load P

produces a compression normal strain in member

(1) of −1,300 in./in., determine the normal strain

in member (2).

Fig. P2.5

Solution

From the strain given for member (1)

1 1 1

6( 1,300 10 )(32 in.)

0.0416 in.

L

Pin A has to move 0.25 in. before it

contacts member (1); therefore, vA =

1.080 mm (downward).

0.0416 in. 0.25 in.

0.2916 in.

0.2916 in. (to the left)

Av

From the deformation diagram of

rigid bar ABC

12 in. 20 in.

20 in.(0.2916 in.) 0.4860 in. (downward)

12 in.

A C

C

v v

v

Therefore, 2 = 0.4860 in. (elongation). From the definition of strain,

62

2

2

0.4860 in.3,037.5 10 in./in.

160 in3,040 με

.L Ans.




2.6 The sanding-drum mandrel shown in Fig.

P2.6 is made for use with a hand drill. The

mandrel is made from a rubber-like material that

expands when the nut is tightened to secure the

sanding sleeve placed over the outside surface. If

the diameter D of the mandrel increases from 2.00

in. to 2.15 in. as the nut is tightened, determine

(a) the average normal strain along a diameter of

the mandrel.

(b) the circumferential strain at the outside surface

of the mandrel.

Fig. P2.6

Solution

(a) The change in strain along a diameter is found from

2.15 in. 2.00 in.

2.00 in.0.075 in./in.D

D

D Ans.

(b) Note that the circumference of a circle is given by D. The change in strain around the

circumference of the mandrel is found from

(2.15 in.) (2.00 in.)

(2.00 in.)0.075 in./in.C

C

C Ans.




2.7 The normal strain in a suspended bar of material of varying cross section due to its own weight is

given by the expression y/3E where is the specific weight of the material, y is the distance from the

free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E,

(a) the change in length of the bar due to its own weight.

(b) the average normal strain over the length L of the bar.

(c) the maximum normal strain in the bar.

Solution

(a) The strain of the suspended bar due to its own weight is given as

3

y

E

Consider a slice of the bar having length dy. In general, = L.

Applying this definition to the bar slice, the deformation of slice dy is

given by

3

yd dy dy

E

Since this strain expression varies with y, the total deformation of the bar must be found by integrating

d over the bar length:

2 2

0 063 3 2

LLL

E

y ydy

E E Ans.

(b) The average normal strain is found by dividing the expression above for by the bar length L

2

avg

/ 6

6

L E

L

L

E Ans.

(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum

value of y, that is, at y = L:

max33y L

y L

y

E

L

E Ans.




2.8 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A

uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point,

the weight of the cable produces an additional normal strain that is proportional to the length of the cable

below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700

in./in., determine

(a) the strain in the cable at a depth of 500 ft.

(b) the total elongation of the cable.

Solution

Call the vertical coordinate y and establish the origin of the y

axis at the lower end of the steel cable. The strain in the cable

has a constant term (i.e., = 250 ) and a term (we will call it

k) that varies with the vertical coordinate y.

6250 10 k y

The problem states that the normal strain at the cable drum (i.e.,

y = 2,000 ft) is 700 in./in. Knowing this value, the constant k

can be determined

6 6

6 66

700 10 250 10 (2,000 ft)

700 10 250 100.225 10 /ft

2,000 ft

k

k

Substituting this value for k in the strain expression gives

6 6250 10 (0.225 10 /ft) y

(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft

is

6 6 6250 10 (0.225 10 /ft)(1,500 ft) 587.5 1 588 0 με Ans.

(b) The total elongation is found by integrating the strain expression over the cable length; thus,

2000

6 6

0

20006

6 2

0

250 10 (0.225 10 /ft)

0.225 10(250 1

0.950 ft 11.40 in.

0 )2

dy y dy

y y

Ans.




2.9 The 16 × 22 × 25-mm rubber blocks shown

in Fig. P2.9 are used in a double U shear mount

to isolate the vibration of a machine from its

supports. An applied load of P = 690 N causes

the upper frame to be deflected downward by 7

mm. Determine the average shear strain and the

shear stress in the rubber blocks.

Fig. P2.9

Solution

Consider the deformation of one block. After a downward

deflection of 7 mm:

7 mm

tan 0.4375 0.412410 rad16 mm

and thus, the shear strain in the block is

0.412 rad 412,000 μrad Ans.

Note that the small angle approximation most definitely does not

apply here!

The applied load of 690 N is

carried by two blocks; therefore,

the shear force applied to one

block is V = 345 N. The area

subjected to shear stress is the area

that is parallel to the direction of

the shear force; that is, the 22 mm

by 25 mm surface of the block.

The shear stress is

345 N

(22 mm)(25 mm)

0.627 M

62

Pa

7 kPa Ans.




2.10 A thin polymer plate PQR is

deformed such that corner Q is displaced

downward 1/16-in. to new position Q’ as

shown in Fig. P2.10. Determine the shear

strain at Q’ associated with the two edges

(PQ and QR).

Fig. P2.10

Solution

Before deformation, the angle of the plate at

Q was 90° or /2 radians. We must now

determine the plate angle at Q′ after

deformation. The difference between these

angles is the shear strain.

After point Q displaces downward by 1/16-

in., the angle ′ is

25 in. 25 in.tan

(10 in. 1 / 16-in.) 10.0625 in.

68.075215

and the angle ′ is

4 in. 4 in.tan

(10 in. 1 /16-in.) 10.0625 in.

21.678589

After deformation, the angle of the plate at Q′ is

68.075215 21.678589 89.753804 1.566499 rad

The difference in the plate angle at Q before and after deformation is the shear strain:

1.566499 rad 0.004297 ra 4,300 μrad d2

Ans.




2.11 A thin polymer plate PQR is

deformed so that corner Q is displaced

downward 1.0 mm to new position Q’ as

shown in Fig. P2.11. Determine the

shear strain at Q’ associated with the two

edges (PQ and QR).

Fig. P2.11

Solution

Before deformation, the angle of the plate

at Q was 90° or /2 radians. We must

now determine the plate angle at Q′ after

deformation. The difference between

these angles is the shear strain.

After deformation, the angle ′ is

120 mm 120 mmtan

(300 mm 1 mm) 301 mm

21.735741


750 mm 750 mmtan

(300 mm 1 mm) 301 mm

68.132764

Therefore, after deformation, the angle of the plate at Q′ is

21.735741 68.132764 89.868505 1.568501 rad

The difference in the angle at Q before and after deformation is the shear strain:

1.568501 rad 0.002295 ra 2,300 μrad d2

Ans.




2.12 A thin rectangular plate is uniformly

deformed as shown in Fig. P2.12.

Determine the shear strain xy at P.

Fig. P2.12

Solution

Before deformation, the angle of the plate at P

was 90° or /2 radians. We must now

determine the plate angle at P after deformation.

The difference between these angles is the shear

strain.

After deformation, the angle that side PQ makes

with the horizontal axis is

1.4 mmtan

1,100 mm

0.072922

and the angle that side PR makes with the

vertical axis is

0.7 mmtan

600 mm

0.066845

Therefore, the angle of the plate at P that was initially 90° has been reduced by the sum of and :

0.072922 0.066845 0.139767 0.002439 rad

Since shear strain is defined as the change in angle between two initially perpendicular lines, the shear

strain in the plate at P is

0.002439 ra 2,440 radd μP Ans.

Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved

the extra steps involved in using the inverse tangent function. Thus,

1.4 mm

tan 0.001273 rad1,100 mm

0.7 mm

tan 0.001167 rad600 mm

0.001273 rad 0.001167 rad 0.002439 ra 2,440 μrad dP Ans.




2.13 A thin rectangular plate is uniformly

deformed as shown in Fig. P2.13.

Determine the shear strain xy at P.

Fig. P2.13

Solution

Before deformation, the angle of the plate at P was 90° or /2

radians. We must now determine the plate angle at P after

deformation. The difference between these angles is the shear strain.

After deformation, the angle that side PQ makes with the horizontal

axis is

0.125 in.

tan 0.39788118 in.

and the angle that side PR makes with the vertical axis is

0.125 in.

tan 0.28647725 in.

By inspection, the angle of the plate at P has been increased by

and decreased by ; thus, the angle at P after deformation is:

90 90 0.397881 0.286477 90.111404

The angle in the plate at P that was initially 90° has been increased to 90.111404° = 1.572741 rad. The

shear strain in the plate at P is thus

1.572741 rad 0.001944 ra 1,944 μrd d2

aP Ans.

Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved

the extra steps involved in using the inverse tangent functions. Thus,

0.125 in.

tan 0.006944 rad18 in.

0.125 in.

tan 0.005000 rad25 in.

By inspection, the angle of the plate at P has been increased by and decreased by ; thus, the angle at

P after deformation is:

0.006944 rad 0.005000 rad 0.001944 rad2 2 2

Therefore, the shear strain in the plate at P is P = −0.001944 rad = −1,944 rad. Ans.




2.14 A thin polymer plate PQR is deformed

so that corner Q is displaced downward 1.0

mm to new position Q’ as shown in Fig.

P2.14. Determine the shear strain at Q’

associated with the two edges (PQ and QR).

Fig. P2.14

Solution

Before deformation, the angle of the plate at Q was 90° or

/2 radians. We must now determine the plate angle at Q′

after deformation. The difference between these angles is

the shear strain.

After deformation, the angle ′ is

180 mm 1.0 mm 181 mmtan

300 mm 300 mm

31.103981


500 mm 1.0 mm 499 mmtan

300 mm 300 mm

58.985614

Therefore, after deformation, the angle of the plate at Q′ is

31.103981 58.985614 90.089595 1.572360 rad

The difference in plate angle before and after deformation is the shear strain:

1.572360 rad 0.001564 ra 1d ,562

4 μrad Ans.




2.15 An airplane has a half-wingspan of 33 m. Determine the change in length of the aluminum

alloy [ A = 22.5×10-6

/°C] wing spar if the plane leaves the ground at a temperature of 15°C and

climbs to an altitude where the temperature is –55°C.

Solution

The change in temperature between the ground and the altitude in flight is

final initial 55°C 15°C 70°CT T T

The thermal strain is given by

6(22.5 10 /°C)( 70°C) 0.001575 mm/mmT T

and thus the deformation in the 33-m wing is

( 0.001575 mm/mm)(33 m) 0.0520 m 52.0 mmT L Ans.




2.16 A large cement kiln has a length of 400 ft and a diameter of 20 ft. Determine the change in

length and diameter of the structural steel [ S = 6.5×10-6

/°F] shell caused by an increase in

temperature of 350°F.

Solution

The thermal strain is given by

6(6.5 10 /°F)( 350°F) 0.002275 in./in.T T

The change in length of the 225-ft kiln length is

( 0.002275 in./in.)(400 ft) 0.9100 ft 10.92 in.TL L Ans.

The change in diameter of the 20-ft diameter is

( 0.002275 in./in.)(20 ft) 0.004550 ft 0.546 in.TD L Ans.




2.17 A cast iron pipe has an inside diameter of d = 208 mm and an outside diameter of D = 236 mm. The

length of the pipe is L = 3.0 m. The coefficient of thermal expansion for cast iron is I = 12.1×10-6

/°C.

Determine the dimension changes caused by an increase in temperature of 70°C.

Solution

The thermal strain caused by a temperature increase of 70°C is given by

6(12.1 10 /°C)(70°C) 0.000847 mm/mmT T

The dimension changes caused by a temperature increase of 70°C are

(0.000847 mm/mm)(236 mm) 0.1999 mmTD D Ans.

(0.000847 mm/mm)(208 mm) 0.1762 mmTd d Ans.

(0.000847 mm/mm)(3.0 m) 0.002541 m 2.54 mmTL L Ans.




2.18 At a temperature of 40°F, a 0.08-in.

gap exists between the ends of the two

bars shown in Fig. P2.18. Bar (1) is an

aluminum alloy [ = 12.5 × 10−6

/°F] and

bar (2) is stainless steel [ = 9.6 ×

10−6

/°F]. The supports at A and C are

rigid. Determine the lowest temperature

at which the two bars contact each other.

Fig. P2.18

Solution

Write expressions for the temperature-induced deformations and set this equal to the 0.08-in. gap:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

0.08 in.

0.08 in.

0.08 in. 0.08 in.77.821°F

(12.5 10 / °F)(40 in.) (9.6 10 / °F)(55 in.)

T L T L

T L L

TL L

Since the initial temperature is 40°F, the temperature at which the gap is closed is 117.8°F. Ans.




2.19 At a temperature of 5°C, a 3-mm gap

exists between two polymer bars and a rigid

support, as shown in Fig. P2.19. Bars (1) and

(2) have coefficients of thermal expansion of

= 140 × 10−6

/°C and = 67 × 10−6

/°C,

respectively. The supports at A and C are rigid.

Determine the lowest temperature at which the

3-mm gap is closed.

Fig. P2.19

Solution

Write expressions for the temperature-induced deformations and set this equal to the 3-mm gap:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

3 mm

3 mm

3 mm 3 mm30.084°C

(140 10 / °C)(540 mm) (67 10 / °C)(360 mm)

T L T L

T L L

TL L

Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.084°C = 35.1°C. Ans.




2.20 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at

the same temperature is 5 mm longer. At what temperature will the aluminum pipe be 15 mm

longer than the steel pipe? Assume that the coefficient of thermal expansion for the aluminum is

22.5×10-6

/°C and that the coefficient of thermal expansion for the steel is 12.5×10-6

/°C.

Solution

The length of the aluminum pipe after a change in temperature can be expressed as

final initial

initial initial

A A A

AA A

L L L

L T L (a)

Similarly, the length of the steel pipe after the same change in temperature is given by

final initial

initial initial

S S S

SS S

L L L

L T L (b)

From the problem statement, we are trying to determine the temperature change that will cause the final

length of the aluminum pipe to be 15 mm longer than the steel pipe. This requirement can be expressed

as

final final 15 mm

A SL L (c)

Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship

initial initial initial initial 15 mmA SA A S S

L T L L T L

Collect the terms with T on the left-hand side of the equation

initial initial initial initial 15 mmA SA S S A

T L T L L L

Factor out T

initial initial initial initial 15 mmA SA S S A

T L L L L

and thus, T can be expressed as

initial initial

initial initial

15 mmS A

A SA S

L LT

L L

Convert all length dimensions to units of millimeters and solve

6 6

60,005 mm 60,000 mm 15 mm

(22.5 10 /°C)(60,000 mm) (12.5 10 /°C)(60,005 mm)

20 mm 20 mm

1.35 mm/°C 0.75 mm/°C 0.60 mm/°C

33.3°C

T

Initially, the pipes were at a temperature of 10°C. With the temperature change determined above, the

temperature at which the aluminum pipe is 15 mm longer than the steel pipe is

final initial 4310°C 3 .3°C3.3°CT T T Ans.




2.21 Determine the movement of the pointer

of Fig. P2.21 with respect to the scale zero in

response to a temperature increase of 60°F.

The coefficients of thermal expansion are

6.6×10-6

/°F for the steel and 12.5×10-6

/°F for

the aluminum.

Fig. P2.21

Solution

In response to the 60°F temperature increase, the steel pieces elongate by the amount

6

Steel Steel Steel (6.6 10 /°F)(60°F)(12 in.) 0.004752 in.T L

and the aluminum piece elongates by

6

Alum Alum Alum (12.5 10 /°F)(60°F)(12 in.) 0.009000 in.T L

From the deformation diagram of the pointer, the scale reading can be determined from similar triangles

pointerAlum Steel

pointer Alum Steel

1.5 in. 8.5 in.

8.5 in.5.6667 0.009000 in. 0.004752 in.

1.50.0241 in. (upward)

in.

v

v Ans.




2.22 Determine the horizontal movement of

point A of Fig. P2.22 due to a temperature

increase of 75°C. Assume that member AE

has a negligible coefficient of thermal

expansion. The coefficients of thermal

expansion are 11.9×10-6

/°C for the steel and

22.5×10-6

/°C for the aluminum alloy.

Fig. P2.22

Solution

In response to the 75°C temperature increase, the steel piece elongates by the amount

6

Steel Steel Steel (11.9 10 /°C)(75°C)(300 mm) 0.267750 mmT L

Thus, joint C moves to the right by 0.267750 mm.

Next, calculate the deformation of the aluminum piece:

6

Alum Alum Alum (22.5 10 /°C)(75°C)(300 mm) 0.50625 mmT L

Joint E moves to the right by 0.50625 mm.

Take the initial position of E as the origin. The coordinates of E in the deflected position are (0.50625

mm, 0) and the coordinates of C are (0.267750 mm, 25 mm). Use the deflected-position coordinates of

E and C to determine the slope of the pointer.

25 mm 0 25 mm

104.8218030.267750 mm 0.50625 mm 0.2385 mm

C E

C E

y yslope

x x

A general equation for the deflected pointer can be expressed as

104.821803y mx b x b




Use the known coordinates of point E to determine b:

0 104.821803(0.50625 mm)

53.066038 mm

b

b

Thus, the deflected pointer can be described by the line

104.821803 53.066038 mmy x

or rearranging to solve for x

(53.066038 mm)

104.821803

yx

At pointer tip A, y = 275 mm; therefore, the x coordinate of the pointer tip is

(53.066038 mm 275 mm) 221.933962 mm

2.11725 mm104.821803 104.82

2.118

m0

m3

2 x Ans.

The x coordinate is the same as the horizontal movement since we took the initial position of joint E as

the origin.




2.23 At a temperature of 25°C, a cold-rolled red brass [ B = 17.6×10-6

/°C] sleeve has an inside

diameter of dB = 299.75 mm and an outside diameter of DB = 310 mm. The sleeve is to be placed

on a steel [ S = 11.9×10-6

/°C] shaft with an outside diameter of DS = 300 mm. If the temperatures

of the sleeve and the shaft remain the same, determine the temperature at which the sleeve will slip

over the shaft with a gap of 0.05 mm.

Solution

The inside diameter of the brass sleeve after a change in temperature can be expressed as

final initial

initial initial

B B B

BB B

d d d

d T d (a)

Similarly, the outside diameter of the steel shaft after the same change in temperature is given by

final initial

initial initial

S S S

SS S

D D D

D T D (b)

From the problem statement, we are trying to determine the temperature change that will cause the

inside diameter of the brass sleeve to be 0.05 mm greater than the outside diameter of the steel shaft.

This requirement can be expressed as

final final 0.05 mm

B Sd D (c)

Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship

initial initial initial initial 0.05 mmB SB B S S

d T d D T D

Collect the terms with T on the left-hand side of the equation

initial initial initial initial 0.05 mmB SB S S B

T d T D D d

Factor out T

initial initial initial initial 0.05 mmB SB S S B

T d D D d

and thus, T can be expressed as

initial initial

initial initial

0.05 mmS B

B SB S

D dT

d D

Solve for the temperature change

6 6

300 mm 299.75 mm 0.05 mm

(17.6 10 /°C)(299.75 mm) (11.9 10 /°C)(300 mm)

0.30 mm 0.30 mm

0.005276 mm/°C 0.003570 mm/°C 0.001706 mm/°C

175.9°C

T

Initially, the shaft and sleeve were at a temperature of 25°C. With the temperature change determined

above, the temperature at which the sleeve fits over the shaft with a gap of 0.05 mm is

final initial 225° 01°CC 175.9°CT T T Ans.




3.1 At the proportional limit, a 2-inch gage length of a 0.375-in.-diameter alloy bar has elongated 0.0083

in. and the diameter has been reduced 0.0005 in. The total tension force on the bar was 4.75 kips.

Determine the following properties of the material:

(a) the modulus of elasticity.

(b) Poisson’s ratio.

(c) the proportional limit.

Solution

(a) The bar cross-sectional area is

2 2 2(0.375 in.) 0.110447 in.4 4

A D

and thus, the normal stress corresponding to the 4.75-kip force is

2

4.75 kips43.007204 ksi

0.110447 in.

The longitudinal strain in the bar is

long

0.0083 in.0.004150 in./in.

2 in.L

The modulus of elasticity is therefore

long

43.007204 ksi10,363.2 ksi

0.004150 i10,360 k

n. .s

ni

/iE

Ans.

(b) The longitudinal strain in the bar was calculated previously as

long 0.004150 in./in.

The lateral strain can be determined from the reduction of the diameter:

lat

0.0005 in.0.001333 in./in.

0.375 in.

D

D

Poisson’s ratio for this specimen is therefore

lat

long

0.001333 in./in.0.321285

0.004150 in./in.0.321

Ans.

(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,

43.0 ksiPL Ans.




3.2 At the proportional limit, a 30 mm wide × 12 mm thick bar elongates 2.0 mm under an axial load of

41.5 kN. The bar is 1.5-m long. If Poisson’s ratio is 0.33 for the material, determine:


(b) the proportional limit.

(c) the change in each lateral dimension.

Solution


2(30 mm)(12 mm) 360 mmA

and thus, the normal stress corresponding to the 41.5-kN axial load is

2

(41.5 kN)(1,000 N/kN)115.2778 MPa

360 mm


long

2.0 mm0.001333 mm/mm

(1.5 m)(1,000 mm/m)L


long

115.2778 MPa86,458 MPa

0.00133386.5 GP

mm/mmaE

Ans.

(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore,

115.3 MPaPL Ans.

(c) Poisson’s ratio is given as = 0.33. The longitudinal strain in the bar was calculated previously as

long 0.001333 mm/mm

The corresponding lateral strain can be determined from Poisson’s ratio:

lat long (0.33)(0.001333 mm/mm) 0.000440 mm/mm

Using this lateral strain, the change in bar width is

latwidth (width) ( 0.000440 mm/mm)(30 0mm) .01320 mm Ans.

and the change in bar thickness is

latthickness (thickness) ( 0.000440 mm/mm)(12 mm 0.00528 ) mm Ans.




3.3 At an axial load of 22 kN, a 45-mm-wide × 15-mm-thick polyimide polymer bar elongates 3.0 mm

while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the

polymer bar is less than its proportional limit. Determine:


(b) Poisson’s ratio.

(c) the change in the bar thickness.

Solution


2(15 mm)(45 mm) 675 mmA

and thus, the normal stress corresponding to the 22-kN axial load is

2

(22 kN)(1,000 N/kN)32.592593 MPa

675 mm


long

3.0 mm0.0150 mm/mm

200 mmL


long

32.592593 MPa2,173 MPa

0.01502.17 GP

m ma

m / mE

Ans.

(b) The longitudinal strain in the bar was calculated previously as

long 0.0150 mm/mm

The lateral strain can be determined from the reduction of the bar width:

lat

width 0.25 mm0.005556 mm/mm

width 45 mm

Poisson’s ratio for this specimen is therefore

lat

long

0.005556 mm/mm0.370370

0.0150 mm/mm0.370

Ans.

(c) The change in bar thickness can be found from the lateral strain:

latthickness (thickness) ( 0.005556 mm/mm)(15 mm) 0.0833 mm Ans.




3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width

of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the

longitudinal (x) and transverse (y) directions: x = 840 and y = −250 .

(a) Determine Poisson’s ratio for this

specimen.

(b) If the measured strains were produced

by an axial load of P = 32 kips, what is the

modulus of elasticity for this specimen?

Fig. P3.4

Solution

(a) Poisson’s ratio for this specimen is

lat

long

0.29250 με

840 με8

y

x

Ans.

(b) The bar cross-sectional area is

2(3.0 in.)(0.75 in.) 2.25 in.A

and so the normal stress for an axial load of P = 32 kips is

2

32 kips14.222222 ksi

2.25 in.

The modulus of elasticity is thus

long

14.222222 ksi16,931.2 ksi

1 in./in.(840 με)

1,000,000

16,930 ks

με

iE

Ans.




3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width

of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in the

longitudinal (x) and transverse (y) directions: x = 900 and y = −275 .

(a) Determine Poisson’s ratio for this

specimen.

(b) If the measured strains were produced

by an axial load of P = 19 kN, what is the

modulus of elasticity for this specimen?

Fig. P3.5

Solution

(a) Poisson’s ratio for this specimen is

lat

long

0.30275 με

900 με6

y

x

Ans.

(b) The bar cross-sectional area is

2(30 mm)(6 mm) 180 mmA

and so the normal stress for an axial load of P = 19 kN is

2

(19 kN)(1,000 N/kN)105.556 MPa

180 mm

The modulus of elasticity is thus

long

105.556 MPa117,284.0 MPa

1 mm/mm(900 με)

1,000,000 μ

117.3 GPa

ε

E

Ans.




3.6 A copper rod [E = 110 GPa] originally 350-mm long is pulled in tension with a normal stress of 180

MPa. If the deformation is entirely elastic, what is the resulting elongation?

Solution

Since the deformation is elastic, the strain in the rod can be determined from Hooke’s Law,

180 MPa

0.0016364 mm/mm110,000 MPaE

The elongation in the rod is thus

(0.0016364 mm/mm)(350 m 0m) .573 mmL Ans.




3.7 A 6061-T6 aluminum tube [E = 10,000 ksi; = 0.33] has an outside diameter of 4.000 in. and a wall

thickness of 0.065 in.

(a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract

by 0.005 in.

(b) If the tube is 84-in. long, what is the overall elongation?

Solution

(a) The lateral strain associated with the given diameter contraction is

lat

0.005 in.0.001250 in./in.

4.000 in.

From the given Poisson’s ratio, the longitudinal strain in the tube must be

lat

long

0.001250 in./in.0.003788 in./in.

0.33

and from Hooke’s Law, the normal stress can be calculated as

long (10,000 ksi)(0.003788 in./in.) 37.878788 ksiE

The cross-sectional area of the tube is needed to determine the tension force. Given that the outside

diameter of the tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is

3.870 in. The tube cross-sectional area is thus

2 2 2(4.000 in.) (3.870 in.) 0.803541 in.4

A

and the force applied to the tube is

2(37.878788 ksi)(0.803541 in. ) 30.437154 kips 30.4 kipsF A Ans.

(b) The longitudinal strain was calculated previously. Use the longitudinal strain to determine the

overall elongation:

long (0.003788 in./in.)(84 i 0n.) .318 in.L Ans.




3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in

tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the

fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent

reduction in area.

Solution

Percent elongation is simply the longitudinal strain at fracture:

(3.08 in. 2.000 in.) 1.08 in.0.54 in./in.

2.000 in. 2.000 in.

percent elonga 54%tion

L

Ans.

The initial cross-sectional area of the specimen is

2 2

0 (0.500 in.) 0.196350 in.4

A

The final cross-sectional area at the fracture location is

2 2(0.260 in.) 0.053093 in.4

fA

The percent reduction in area is

0

0

2 2

2

percent reduction of area (100%)

(0.196350 in. 0.053093 in. )(100%)

0.196350 in73.0%

.

fA A

A

Ans.




3.9 A portion of the stress-strain curve for a

stainless steel alloy is shown in Fig. P3.9. A 350-

mm-long bar is loaded in tension until it elongates

2.0 mm and then the load is removed.

(a) What is the permanent set in the bar?

(b) What is the length of the unloaded bar?

(c) If the bar is reloaded, what will be the

proportional limit?

Fig. P3.9

Solution

(a) The normal strain in the specimen is

2.0 mm

0.005714 mm/mm350 mmL

Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of =

0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the

permanent set:

permanent set 0.0035 mm/mm Ans.

(b) The length of the unloaded bar is therefore:

(0.0035 mm/mm)(350 mm) 1.225 mm

350 mm 1.225 mm 351.225 mmf

L

L

Ans.

(c) From the stress-strain curve, the reload proportional limit is 444 MPa . Ans.




3.10 The 16 by 22 by 25-mm rubber blocks

shown in Fig. P3.10 are used in a double U

shear mount to isolate the vibration of a

machine from its supports. An applied load of

P = 285 N causes the upper frame to be

deflected downward by 5 mm. Determine the

shear modulus G of the rubber blocks.

Fig. P3.10

Solution

Consider a FBD of the upper U frame. The

downward force P is resisted by two upward shear

forces V; therefore, V = 285 N / 2 = 142.5 N.

Next, consider a FBD of one of the rubber blocks.

The shear force acting on one rubber block is V =

142.5 N. The area of the rubber block that is

parallel to the direction of V is

2(22 mm)(25 mm) 550 mmVA

Consequently, the shear stress in one rubber block is

2

142.5 N0.259091 MPa

550 mmV

V

A

The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by:

5 mm

tan 0.312500 0.302885 rad16 mm

From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed:

0.259091 MPa

0.855411 MPa0.30288

0.855 M5 a

Par d

G G

Ans.




3.11 Two hard rubber blocks are used in an anti-vibration

mount to support a small machine, as shown in Fig. P3.11.

An applied load of P = 150 lb causes a downward deflection

of 0.25 in. Determine the shear modulus of the rubber blocks.

Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in.

Fig. P3.11

Solution

Determine the shear strain from the angle formed by the

downward deflection and the block thickness a:

0.250 in.

tan 0.500 0.463648 rad0.50 in.

Note: The small angle approximation tan is not applicable in this instance.

Determine the shear stress from the applied load P and the block area.

Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block

is half of the total load: V = P/2 = 75 lb.

To determine the area needed here, consider the surface that is bonded to the plate. This area has

dimensions of bc. The shear stress acting on a single block is therefore:

75 lb

30 psi(1.0 in.)(2.5 in.)

V

A

The shear modulus G can be calculated from Hooke’s law for shear stress and strain:

30 psi

64.704 psi0.463648

64.7 psirad

G G

Ans.




3.12 Two hard rubber blocks [G = 350 kPa] are used in an anti-

vibration mount to support a small machine, as shown in Fig.

P3.12. Determine the downward deflection that will occur for an

applied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and

c = 80 mm.

Fig. P3.12

Solution

Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block

is half of the total load: V = P/2 = 450 N.

Determine the shear stress from the shear force V and the block area. To determine the area needed

here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress

acting on a single block is therefore:

450 N

0.112500 MPa(50 mm)(80 mm)

V

A

Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress

and shear strain:

0.112500 MPa

0.321429 rad0.350 MPa

GG

From the angle and the block thickness a, the downward deflection of the block can be determined

from:

tan tan (20 mm) tan(0.321429 rad) 6.65951 6.662 mmm maa

Ans.




3.13 A load test on a 6 mm diameter by 225 mm long aluminum alloy rod found that a tension load of

4,800 N caused an elastic elongation of 0.52 mm in the rod. Using this result, determine the elastic

elongation that would be expected for a 24-mm-diameter rod of the same material if the rod were 1.2 m

long and subjected to a tension force of 37 kN.

Solution

The cross-sectional area of the 6-mm-diameter rod is

2 2(6 mm) 28.274334 mm4

A

Thus, the normal stress in the rod due to a 4,800-N load is

2

4,800 N169.76527 MPa

28.274334 mm

F

A

The strain in the 225-mm long rod associated with a 0.52-mm elongation is

0.52 mm

0.0023111 mm/mm225 mmL

Therefore, the elastic modulus of the aluminum alloy is

169.76527 MPa

73,456.128 MPa0.0023111 mm/mm

E

The cross-sectional area of the 24-mm-diameter rod is

2 2(24 mm) 252.389342 mm4

A

Thus, the normal stress in the 24-mm-diameter rod due to a 37-kN load is

2

(37 kN)(1,000 N/kN)81.787957 MPa

452.389342 mm

F

A

From Hooke’s Law, the strain that would be expected is

81.787957 MPa

0.0011134 mm/mm73,456.128 MPaE

Since the 42-mm-diameter rod is 1.2-m long, the expected elongation is

(0.0011134 mm/mm)(1.2 m)(1,000 mm/m) 1.336111 mm 1.336 mmL Ans.




3.14 The stress-strain diagram for a

particular stainless steel alloy is shown in

Fig. P3.14. A rod made from this material is

initially 800 mm long at a temperature of

20°C. After a tension force is applied to the

rod and the temperature is increased by

200°C, the length of the rod is 804 mm.

Determine the stress in the rod and state

whether the elongation in the rod is elastic

or inelastic. Assume the coefficient of

thermal expansion for this material is 18 ×

10−6

/°C.

Fig. P3.14

Solution

The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The

200°C temperature increase causes a normal strain of:

6(18 10 / )(200 ) 0.003600 mm/mmT T C C

which means that the rod elongates

(0.003600 mm/mm)(800 mm) 2.8800 mmT T L

The portion of the 4-mm total elongation due to load is therefore

4 mm 2.8800 mm 1.1200 mmT

The strain corresponding to this elongation is

1.1200 mm

0.001400 mm/mm800 mmL

By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is

elastic in this instance.

For the linear region, the elastic modulus can be determined from the lower scale plot:

(400 MPa 0)

200,000 MPa(0.002 mm/mm 0)

E

Using Hooke’s Law (or directly from the - diagram), the stress corresponding to the 0.001400

mm/mm strain is

(200,000 MPa)(0.001400 mm/mm 280 MP) aE Ans.




3.15 In Fig. P3.15, rigid bar ABC is supported by

axial member (1), which has a cross-sectional area

of 400 mm2, an elastic modulus of E = 70 GPa, and

a coefficient of thermal expansion of = 22.5 ×

10−6

/°C. After load P is applied to the rigid bar

and the temperature rises 40°C, a strain gage

affixed to member (1) measures a strain increase of

2,150 . Determine:

(a) the normal stress in member (1).

(b) the magnitude of applied load P.

(c) the deflection of the rigid bar at C.

Fig. P3.15

Solution

(a) The strain measured in member (1) is due to both the internal force in the member and the

temperature change. The strain caused by the temperature change is

6(22.5 10 / °C)(40°C) 0.000900 mm/mmT T

Since the total strain is = 2,150 = 0.002150 mm/mm, the strain caused by the internal force in

member (1) must be

0.002150 mm/mm 0.000900 mm/mm 0.001250 mm/mmT

The elastic modulus of member (1) is E = 70 GPa; thus, from Hooke’s Law, the stress in the member is:

1 (70,000 MPa)(0.001250 mm/mm) 87.5 MPaE Ans.

(b) If the normal stress in member (1) is 87.5 MPa, the axial force in the member is

2 2

1 1 1 (87.5 N/mm )(400 mm ) 35,000 NF A

Consider moment equilibrium of rigid bar ABC about joint

A to determine the magnitude of P:

(1.75 m)(35,000 N) (3.0 m) 0

20,417 N 20.4 kN

AM P

P

Ans.

(c) The strain in member (1) was measured as = 2,150 = 0.002150 mm/mm; therefore, the total

elongation of member (1) is

1 1 1 (0.002150 mm/mm)(3,750 mm) 8.0625 mmL

The deflection of the rigid bar at B is equal to this elongation; therefore, vB = 1 = 8.0625 mm

(downward). By similar triangles, the deflection of the rigid bar at C is given by:

1.75 m 3.0 m

3.0 m 3.0(8.0625 mm) 13.821429 mm

1.75 m 1.13.82

7m

5m

B C

C B

v v

v v

Ans.




3.16 A tensile test specimen of 1045 hot-rolled

steel having a diameter of 0.505 in. and a gage

length of 2.00 in. was tested to fracture. Stress

and strain data obtained during the test are

shown in Fig. P3.16. Determine



(c) the ultimate strength.

(d) the yield strength (0.20% offset).

(e) the fracture stress.

(f) the true fracture stress if the final diameter of

the specimen at the location of the fracture was

0.392 in.

Fig. P3.16

Solution

From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.0019.

(Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.)

(a) The modulus of elasticity is

60 ksi

0.0019 in./in.31,600 ksiE

Ans.

(b) From the diagram, the proportional limit is taken as

60 ksiPL Ans.

(c) The ultimate strength is

105 ksiU Ans.

(d) The yield strength is

68 ksiY Ans.

(e) The fracture stress is

fracture 98 ksi Ans.

(f) The original cross-sectional area of the specimen is

2 2

0 (0.505 in.) 0.200296 in.4

A

The cross-sectional area of the specimen at the fracture location is

2 2(0.392 in.) 0.120687 in.4

fA

The true fracture stress is therefore

2

fracture 2

0.200296 in.true (98 ksi)

0.120687 in.162.6 ksi Ans.




3.17 A tensile test specimen of stainless

steel alloy having a diameter of 0.495 in.

and a gage length of 2.00 in. was tested to

fracture. Stress and strain data obtained

during the test are shown in Fig. P3.17.

Determine:






(f) the true fracture stress if the final

diameter of the specimen at the location of

the fracture was 0.350 in.

Fig. P3.17

Solution

From the stress-strain curve, the proportional limit will be taken as = 60 ksi at a strain of = 0.002.



60 ksi

0.002 in./in.30,000 ksiE

Ans.


60 ksiPL Ans.


159 ksiU Ans.


80 ksiY Ans.


fracture 135 ksi Ans.


2 2

0 (0.495 in.) 0.192442 in.4

A


2 2(0.350 in.) 0.096211 in.4

fA


2

fracture 2

0.192442 in.true (135 ksi)

0.096211 in270 i

. ks Ans.




3.18 A bronze alloy specimen having a

diameter of 12.8 mm and a gage length

of 50 mm was tested to fracture. Stress

and strain data obtained during the test

are shown in Fig. P3.18. Determine:






(f) the true fracture stress if the final

diameter of the specimen at the location

of the fracture was 10.5 mm.

Fig. P3.18

Solution

From the stress-strain curve, the proportional limit will be taken as = 210 MPa at a strain of = 0.002.



210 MPa

0.002 in./in.105,000 MPaE

Ans.


210 MPaPL Ans.


380 MPaU Ans.


290 MPaY Ans.


fracture 320 MPa Ans.


2 2

0 (12.8 mm) 128.679635 mm4

A


2 2(10.5 mm) 86.590148 mm4

fA


2

fracture 2

128.679635 mmtrue (320 MPa)

86.590148 mm476 MPa Ans.




3.19 An alloy specimen having a diameter of 12.8

mm and a gage length of 50 mm was tested to

fracture. Load and deformation data obtained

during the test are given. Determine:





(e) the yield strength (0.20% offset).

(f) the fracture stress.

(g) the true fracture stress if the final diameter of


11.3 mm.

Load Change in

Length Load

Change in

Length

(kN) (mm) (kN) (mm)

0 0

7.6 0.02 43.8 1.50

14.9 0.04 45.8 2.00

22.2 0.06 48.3 3.00

28.5 0.08 49.7 4.00

29.9 0.10 50.4 5.00

30.6 0.12 50.7 6.00

32.0 0.16 50.4 7.00

33.0 0.20 50.0 8.00

33.3 0.24 49.7 9.00

36.8 0.50 47.9 10.00

41.0 1.00 45.1 fracture

Solution

The plot of the stress-strain data is shown below.




(a) Using the data point for 28.5 kN load and 0.08 mm elongation, the modulus of elasticity can be

calculated as

221.48 MPa

0.0016 mm/mm138,400 MPaE

Ans.


234 MPaPL Ans.


2

(50.7 kN)(1,000 N/kN)

(12.8 mm)4

394 MPaU Ans.

(d) The yield strength by the 0.05% offset method is

239 MPaY Ans.

(e) The yield strength by the 0.2% offset method is

259 MPaY Ans.

(f) The fracture stress is

fracture

2

(45.1 kN)(1,000 N/kN350 MP

)350.48 aMPa

(12.8 mm)4

Ans.


2 2

0 (12.8 mm) 128.679635 mm4

A


2 2(11.3 mm) 100.287492 mm4

fA


2

fracture 2

128.679635 mmtrue (356 MPa)

100.287492 m457 a

m MP Ans.




3.20 A 1035 hot-rolled steel specimen with a

diameter of 0.500 in. and a 2.0-in. gage length was

tested to fracture. Load and deformation data

obtained during the test are given. Determine:









0.387 in.

Load Change in

Length Load

Change

in Length

(lb) (in.) (lb) (in.)

0 0 12,540 0.0209

2,690 0.0009 12,540 0.0255

5,670 0.0018 14,930 0.0487

8,360 0.0028 17,020 0.0835

11,050 0.0037 18,220 0.1252

12,540 0.0042 18,820 0.1809

13,150 0.0046 19,110 0.2551

13,140 0.0060 19,110 0.2968

12,530 0.0079 18,520 0.3107

12,540 0.0098 17,620 0.3246

12,840 0.0121 16,730 0.3339

12,840 0.0139 16,130 0.3385

15,900 fracture

Solution





(a) Using the data point for 12,540 lb load and 0.0042 in. elongation, the modulus of elasticity can be

calculated as

63.866 ksi

0.0021 in./in.30,400 ksiE

Ans.


63.8 ksiPL Ans.


97.3 ksiU Ans.

(d) The yield strength using the 0.05% offset method is

65.4 ksiY Ans.

(e) The yield strength using the 0.2% offset method is

63.8 ksiY Ans.


fracture

2

15,900 lb80,978 psi

(0.500 in.)4

81.0 ksi

Ans.

(g) The original cross-sectional area of the specimen is

2 2

0 (0.500 in.) 0.196350 in.4

A


2 2(0.387 in.) 0.117628 in.4

fA


2

fracture 2

0.196350 in.true (80,978 psi) 135,172 psi

0.117628 135

in.2 ksi

. Ans.




3.21 A 2024-T4 aluminum test specimen with a

diameter of 0.505 in. and a 2.0-in. gage length was

tested to fracture. Load and deformation data

obtained during the test are given. Determine:









0.452 in.

Load Change in

Length Load

Change in

Length

(lb) (in.) (lb) (in.)

0 0.0000 11,060 0.0139

1,300 0.0014 11,500 0.0162

2,390 0.0023 12,360 0.0278

3,470 0.0032 12,580 0.0394

4,560 0.0042 12,800 0.0603

5,640 0.0051 13,020 0.0788

6,720 0.0060 13,230 0.0974

7,380 0.0070 13,450 0.1159

8,240 0.0079 13,670 0.1391

8,890 0.0088 13,880 0.1623

9,330 0.0097 14,100 0.1994

9,980 0.0107 14,100 0.2551

10,200 0.0116 14,100 0.3200

10,630 0.0125 14,100 0.3246

14,100 fracture

Solution





(a) Using the data point for 6,720 lb load and 0.0060 in. elongation, the modulus of elasticity can be

calculated as

33.55 ksi

0.003 in./in.11,180 ksiE

Ans.


33.6 ksiPL Ans.


70.4 ksiU Ans.

(d) The yield strength using the 0.05% offset method is

44.4 ksiY Ans.

(e) The yield strength using the 0.2% offset method is

54.5 ksiY Ans.


fracture 70.4 ksi Ans.

(g) The original cross-sectional area of the specimen is

2 2

0 (0.505 in.) 0.200296 in.4

A


2 2(0.452 in.) 0.160460 in.4

fA


2

fracture 2

0.200296 in.true (70.4 ksi)

0.160460 in.87.9 ksi Ans.




3.22 A 1045 hot-rolled steel tension test specimen

has a diameter of 6.00 mm and a gage length of 25

mm. In a test to fracture, the stress and strain data

below were obtained. Determine:









4.65 mm.

Load Change in

Length Load

Change in

Length

(kN) (mm) (kN) (mm)

0.00 0.00 13.22 0.29

2.94 0.01 16.15 0.61

5.58 0.02 18.50 1.04

8.52 0.03 20.27 1.80

11.16 0.05 20.56 2.26

12.63 0.05 20.67 2.78

13.02 0.06 20.72 3.36

13.16 0.08 20.61 3.83

13.22 0.08 20.27 3.94

13.22 0.10 19.97 4.00

13.25 0.14 19.68 4.06

13.22 0.17 19.09 4.12

18.72 fracture

Solution





(a) Using the data point for 8.52 kN load and 0.03 mm elongation, the modulus of elasticity can be

calculated as

301.3 MPa

0.0012 mm/mm251,000 MPaE

Ans.


400 MPaPL Ans.


2

(20.72 kN)(1,000 N/kN)732.82MPa

(6

733

.00 mm4

P

)

M aU Ans.

(d) The yield strength by the 0.05% offset method is

465 MPaY Ans.

(e) The yield strength by the 0.2% offset method is

465 MPaY Ans.


fracture

2

(18.72 kN)(1,000 N/kN)662.08 MPa

(6.00

662 MP

m )

a

m4

Ans.


2 2

0 (6 mm) 28.274334 mm4

A


2 2(4.65 mm) 16.982272 mm4

fA


2

fracture 2

28.274334 mmtrue (662.08 MPa)

16.981,102 MP

ma

2272 m Ans.




3.23 Rigid bar BCD in Fig. P3.23 is

supported by a pin at C and by aluminum

rod (1). A concentrated load P is applied to

the lower end of aluminum rod (2), which is

attached to the rigid bar at D. The cross-

sectional area of each rod is A = 0.20 in.2

and the elastic modulus of the aluminum

material is E = 10,000 ksi. After the load P

is applied at E, the strain in rod (1) is

measured as 1,350 (tension).

(a) Determine the magnitude of load P.

(b) Determine the total deflection of point E

relative to its initial position.

Fig. P3.23

Solution

(a) From the measured strain, the stress in rod (1) is

6

1 1 1 (10,000 ksi)(1,350 10 in./in.) 13.5 ksiE

and thus, the force in rod (1) is

2

1 1 1 (13.5 ksi)(0.20 in. ) 2.70 kips (T)F A

Consider the equilibrium of the rigid bar, and write a

moment equilibrium equation about C to determine the

magnitude of load P:

(20 in.)(2.70 kips) (30 in.) 0

1.8 kips

CM P

P

Ans.

(b) From the measured strain, the elongation of rod (1) is

6

1 1 1 (1,350 10 in./in.)(50 in.) 0.0675 in.L

From similar triangles, the deflection of the rigid bar at D can

be expressed in terms of the deflection at B:

20 in. 30 in.

30 in. 30 in.(0.0675 in.) 0.10125 in.

20 in. 20 in.

B D

D B

v v

v v




The elongation of rod (2) due to the 1.8-kip load must be determined, also. The stress in rod (2) is

22 2

2

1.8 kips9 ksi

0.2 in.

F

A

and consequently, the strain in rod (2) is

22

2

9 ksi0.000900 in./in.

10,000 ksiE

From the strain, the elongation in rod (2) can be computed:

2 2 2 (0.000900 in./in.)(100 in.) 0.0900 in.L

The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2):

2 0.10125 in. 0.09 in. 0.19125 in. 0.1913 in.E Dv v Ans.




3.24 Rigid bar ABC is supported by pin-connected axial member (1) and by a 0.75-in.-diameter double

shear pin connection at C as shown in Fig. P3.24. Member (1) is a 2.75 in. wide by 1.25 in. thick

rectangular bar made of aluminum with an elastic modulus of 10,000 ksi. When a concentrated load P is

applied to the rigid bar at A, the normal strain in member (1) is measured as –880 .

(a) Determine the magnitude of

applied load P.

(b) Determine the average shear

stress in the pin at C.

Fig. P3.24

Solution

Member (1) is a two-force member that is oriented at

with respect to the horizontal axis:

30 in.

tan 0.75 36.87040 in.

From a FBD of rigid structure ABC, the following

equilibrium equations can be written:

1 cos36.870 0x xF F C (a)

1 sin36.870 0y yF P F C (b)

1(80 in.) ( sin36.870 )(56 in.) 0CM P F (c)

(a) The force in member (1) can be determined from the measured strain of –880 . From Hooke’s

law, the stress in member (1) is:

1 1 6

1 in./in.(10,000 ksi)( 880 με) 8.800 ksi

1 10 μεE

and thus, the force in member (1) is:

1 1 1 ( 8.800 ksi)(2.75 in.)(1.25 in.) 30.250 kipsF A

Substitute this value into Eq. (c) to compute P:

1( sin36.870 )(56 in.) ( 30.250 kips)(sin36.870 )(56 in.)

80 in. 80 in.

12.705 kips 12.71 kips

FP

Ans.




(b) Substitute the values for P and F1 into Eqs. (a) and (b) to obtain Cx and Cy:

1 cos36.870 ( 30.250 kips)cos36.870 24.200 kipsxC F

1 sin36.870 12.705 kips ( 30.250 kips)sin36.870 5.445 kipsyC P F

The resultant pin force at C is found from Cx and Cy:

2 2 2 2( 24.200 kips) ( 5.445 kips) 24.805 kipsx yC C C

The pin at C is supported in a double-shear connection. The cross-sectional area of the 0.75-in.-

diameter pin is:

2 2

pin (0.75 in.) 0.4421786 in.4

A

The average shear stress in the pin is thus:

2

24.805 kips28.074 ksi

(2 surfaces)(0.441786 in. /surfa28.1 ks

)i

ceV

C

A Ans.




3.25 A concentrated load P is supported by two

bars as shown in Fig. P3.25. Bar (1) is made of

cold-rolled red brass [E = 16,700 ksi; = 10.4 ×

10−6

/°F] and has a cross-sectional area of 0.225

in.2. Bar (2) is made of 6061-T6 aluminum [E =

10,000 ksi; = 13.1 × 10−6

/°F] and has a cross-

sectional area of 0.375 in.2. After load P has been

applied and the temperature of the entire assembly

has increased by 50°F, the total strain in bar (1) is

measured as 1,400 (elongation). Determine:


(b) the total strain in bar (2).

Fig. P3.25

Solution

Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore,

the equilibrium equations can be written as:

2 1cos55 cos 40 0xF F F (a)

2 1sin55 sin 40 0yF F F P (b)

From Eq. (a):

2 1

cos 40

cos55F F

(c)

Substitute this expression into Eq. (b) to obtain:

1 1

1

1

cos40sin55 sin 40

cos55

cos40 tan55 sin 40

1.736812

F F P

F P

P F

(d)

(a) The total strain in bar (1) is measured as 1,400 (elongation). Part of this strain is due to the stress

acting in the bar and part of this strain is due to the temperature increase. The strain caused by the

temperature change is

6 6(10.4 10 / °F)(50°F) 520 10 in./in.T T

Since the total strain is = 1,400 = 0.001400 in./in., the strain caused by the stress in bar (1) must be:

0.001400 in./in. 0.000520 in./in. 0.000880 in./in.T

From Hooke’s law, the stress in bar (1) is therefore:

1 1 (16,700 ksi)(0.000880 in./in.) 14.696 ksiE

and thus, the force in bar (1) is:

2

1 1 1 (14.696 ksi)(0.225 in. ) 3.3066 kipsF A




From Eq. (d), the applied load P is:

1 1.736812 (3.3066 kips)(1.736812) 5.7429 kips 5.74 kipsP F Ans.

(b) From Eq. (c), the force in bar (2) is:

2 1

cos40(3.3066 kips)(1.335558) 4.4162 kips

cos55F F

The normal stress in bar (2) is therefore:

22 2

2

4.4162 kips11.7764 ksi

0.375 in.

F

A

The normal strain due to this stress is:

62

2

2

11.7764 ksi1,177.64 10 in./in.

10,000 ksiE

The strain caused in bar (2) by the temperature change is:

6 6(13.1 10 / °F)(50°F) 655 10 in./in.T T

and thus, the total strain in bar (2) is:

6 6

6

1,177.64 10 in./in. 655 10 in./in.

1,832.64 10 in./in 1,833 με.

T

Ans.




3.26 The rigid bar AC in Fig. P3.26 is

supported by two axial bars (1) and (2).

Both axial bars are made of bronze [E =

100 GPa; = 18 × 10−6

/°C]. The cross-

sectional area of bar (1) is A1 = 240 mm2

and the cross-sectional area of bar (2) is A2

= 360 mm2. After load P has been applied

and the temperature of the entire assembly

has increased by 30°C, the total strain in

bar (2) is measured as 1,220

(elongation). Determine:


(b) the vertical displacement of pin A.

Fig. P3.26

Solution

(a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in

temperature. The strain caused by the 30°C temperature increase is:

6(18 10 / C)(30 C) 0.000540 mm/mmT T

The strain caused by the axial force in the bar is thus:

2, 2 2, 0.001220 mm/mm 0.000540 mm/mm 0.000680 mm/mmT

The stress in bar (2) is

2 2 2, (100,000 MPa)(0.000680 mm/mm) 68 MPaE

and the force in bar (2) is

2 2

2 2 2 (68 N/mm )(360 mm ) 24,480 NF A

Next, consider a FBD of the rigid bar AC. Equilibrium

equations for this FBD are:

1 2

2

0

(1,400 mm) (500 mm) 0

y

A

F F F P

M F P

which can be solve simultaneously to give:

2

500 mm0.357143

1,400 mmF P P

and

1 0.642857F P

The applied load P can be expressed in terms of F2 as

2 2

12.8

0.357143P F F




and so the magnitude of load P is

22.8 2.8(24,480 N) 68,544 N 68.5 kNP F Ans.

(b) The force in bar (1) is

1 0.642857 (0.642857)(68,544 N) 44,064 NF P

Thus, the stress in bar (1) is

11 2

1

44,064 N183.60 MPa

240 mm

F

A

The normal strain due to the axial force in bar (1) is

11,

1

183.60 MPa0.001836 mm/mm

100,000 MPaE

The normal strain caused by the 30°C temperature increase is:

6

1, (18 10 / C)(30 C) 0.000540 mm/mmT T

Therefore, the total strain in bar (1) is

1 1, 1, 0.001836 mm/mm 0.000540 mm/mm 0.002376 mm/mmT

and the elongation in bar (1) is

1 1, 1 (0.002376 mm/mm)(1,300 mm) 3.0888 mmL

Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by

an amount equal to the elongation of bar (1); therefore,

1 3.09 mmAv Ans.




3.27 The rigid bar in Fig. P3.27 is supported

by axial bar (1) and by a pin connection at

C. Axial bar (1) has a cross-sectional area of

A1 = 275 mm2, an elastic modulus of E =

200 GPa, and a coefficient of thermal

expansion of = 11.9 × 10−6

/°C. The pin at

C has a diameter of 25 mm. After load P has

been applied and the temperature of the

entire assembly has been increased by 20°C,

the total strain in bar (1) is measured as 925

(elongation). Determine:


(b) the average shear stress in pin C.

Fig. P3.27

Solution

The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:

T

The normal strain due to the increase in temperature is:

6(11.9 10 / C)(20 C) 0.000238 mm/mmT T

Therefore, the normal stress in bar (1) causes a normal strain of:

0.000925 mm/mm 0.000238 mm/mm 0.000687 mm/mmT

From Hooke’s law, the normal stress in bar (1) can be calculated as:

1 (200,000 MPa)(0.000687 mm/mm) 137.4 MPaE

and thus the axial force in bar (1) must be:

2 2

1 1 1 (137.4 N/mm )(275 mm ) 37,785 NF A

Next, consider a free-body diagram of the rigid bar. Write

a moment equilibrium equation about pin C:

1(260 mm) (540 mm)

(260 mm)(37,785 N) (540 mm) 0

18,192 18. .19 k8 N N

CM F P

P

P

Ans.

Now that P is known, the horizontal and vertical reactions

at C can be calculated:

1 10 37,785 N

0 19,192.8 N

x x x

y y y

F C F C F

F C P C P

The resultant force acting on pin C is:

2 2 2 2(37,785 N) (18,192.8 N) 41,936.659 Nx yC C C

Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the

resultant force: V = 20,968.330 N. The area of one shear plane of the 25-mm-diameter pin at C (in

other words, the cross-sectional area of the pin) is:




2 2

pin (25 mm) 490.874 mm4

A

and thus the average shear stress in pin C is:

2

2

20,968.330 N42.716 N/mm

490.874 42.7 MPa

mmC

V

V

A Ans.




3.28 The rigid bar in Fig. P3.28 is

supported by axial bar (1) and by a pin

connection at C. Axial bar (1) has a cross-

sectional area of A1 = 275 mm2, an elastic

modulus of E = 200 GPa, and a coefficient

of thermal expansion of = 11.9 × 10−6

/°C. The pin at C has a diameter of 25 mm.

After load P has been applied and the

temperature of the entire assembly has

been decreased by 30°C, the total strain in

bar (1) is measured as 925 (elongation).

Determine:


(b) the average shear stress in pin C.

Fig. P3.28

Solution

The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:

T

The normal strain due to the decrease in temperature is:

6(11.9 10 / C)( 30 C) 0.000357 mm/mmT T

Therefore, the normal stress in bar (1) causes a normal strain of:

0.000925 mm/mm ( 0.000357 mm/mm) 0.001282 mm/mmT

From Hooke’s law, the normal stress in bar (1) can be calculated as:

1 (200,000 MPa)(0.001282 mm/mm) 256.4 MPaE

and thus the axial force in bar (1) must be:

2 2

1 1 1 (256.4 N/mm )(275 mm ) 70,510 NF A

Next, consider a free-body diagram of the rigid bar. Write

a moment equilibrium equation about pin C:

1(260 mm) (540 mm)

(260 mm)(70,510 N) (540 mm)

33.9 k

0

33,949.3 N N

CM F P

P

P

Ans.

Now that P is known, the horizontal and vertical reactions

at C can be calculated:

1 10 70,510 N

0 33,949.3 N

x x x

y y y

F C F C F

F C P C P

The resultant force acting on pin C is:

2 2 2 2(70,510 N) (33,949.3 N) 78,257.3 Nx yC C C




Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the

resultant force: V = 39,128.7 N. The area of one shear plane of the 25-mm-diameter pin at C (in other

words, the cross-sectional area of the pin) is:

2 2

pin (25 mm) 490.874 mm4

A

and thus the average shear stress in pin C is:

2

2

39,128.7 N79.712 N/mm

490.874 m79.7 MP

maC

V

V

A Ans.




4.1 A stainless steel alloy bar 25 mm wide by

16 mm thick is subjected to an axial load of

P = 145 kN. Using the stress-strain diagram

given in Fig. P4.1, determine:

(a) the factor of safety with respect to the

yield strength defined by the 0.20% offset

method.

(b) the factor of safety with respect to the

ultimate strength.

Fig. P4.1

Solution

(a) From the stress-strain curve, the yield strength defined by the 0.20% offset method is

550 MPaY

The normal stress in the bar is

(145 kN)(1,000 N/kN)

362.5 MPa(25 mm)(16 mm)

F

A

Therefore, the factor of safety with respect to yield is

actual

550 MPaFS

362.5 MPa1.517Y

Ans.

(b) From the stress-strain curve, the ultimate strength is

1,100 MPaU

Therefore, the factor of safety with respect to the ultimate strength is

actual

1,100 MPaFS

362.5 MPa3.03U

Ans.




4.2 Six bolts are used in the connection between the

axial member and the support, as shown in Fig.

P4.2. The ultimate shear strength of the bolts is 300

MPa, and a factor of safety of 4.0 is required with

respect to fracture. Determine the minimum

allowable bolt diameter required to support an

applied load of P = 475 kN.

Fig. P4.2

Solution

The allowable shear stress for the bolts is

allow

300 MPa75 MPa

FS 4

U

To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed:

2

2

allow

475,000 N6,333.333 mm

75 N/mmV

PA

There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress

is

2

bolt(6 bolts)(2 surfaces per bolt)4

VA d

The minimum bolt diameter can be found be equating the two expressions for AV:

2 2

bolt

bolt

(12 surfaces) 6,333.333 mm4

25.923 m 25.9 mm m

d

d

Ans.




4.3 A 14-kip load is supported by two bars as

shown in Fig. P4.3. Bar (1) is made of cold-rolled

red brass (Y = 60 ksi) and has a cross-sectional

area of 0.225 in.2. Bar (2) is made of 6061-T6

aluminum (Y = 40 ksi) and has a cross-sectional

area of 0.375 in.2. Determine the factor of safety

with respect to yielding for each of the bars.

Fig. P4.3

Solution

Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore,

the equilibrium equations can be written as:

2 1cos50 cos35 0xF F F (a)

2 1sin50 sin35 14 kips 0yF F F (b)

From Eq. (a):

2 1

cos35

cos50F F

Substitute this expression into Eq. (b) to obtain:

1 1

1

1

1

cos35sin50 sin35 14 kips

cos50

cos35 tan50 sin35 14 kips

1.549804 14 kips

9.033401 kips

F F

F

F

F

Backsubstitute to obtain F2:

2 1

cos35(9.033401 kips)(1.274374) 11.511935 kips

cos50F F

The normal stress in bar (1) is

11 2

1

9.033401 kips40.148452 ksi

0.225 in.

F

A

Therefore, the factor of safety in bar (1) is

1

1

60 ksiFS 1.494454

40.148452 ks1

i.494Y

Ans.


22 2

2

11.511935 kips30.698495 ksi

0.375 in.

F

A

and thus, the factor of safety in bar (2) is

2

2

40 ksiFS 1.302995

30.698495 ks1

i.303Y

Ans.




4.4 A steel plate is to be attached to a support with three bolts as

shown in Fig. P4.4. The cross-sectional area of the plate is 560

mm2 and the yield strength of the steel is 250 MPa. The ultimate

shear strength of the bolts is 425 MPa. A factor of safety of 1.67

with respect to yield is required for the plate. A factor of safety of

4.0 with respect to the ultimate shear strength is required for the

bolts. Determine the minimum bolt diameter required to develop

the full strength of the plate. (Note: consider only the gross cross-

sectional area of the plate—not the net area.)

Fig. P4.4

Solution

The allowable normal stress is

allow

250 MPa149.7006 MPa

FS 1.67

Y

The full strength of the plate (based on the gross cross-sectional area) is therefore:

2

max allow (149.7006 MPa)(560 mm ) 83,832.3353 NP A

The allowable shear stress of the bolts is

allow

425 MPa106.25 MPa

FS 4.0

U

The minimum shear area required to support the maximum load P is

2max

2

allow

83,832.3353 N789.0102 mm

106.25 N/mmV

PA

There are three bolts, each acting in single shear, which provide a shear area of

2 2

bolt bolt(3 bolts)(1 surface per bolt) (3 surfaces)4 4

VA d d

Equating the two expressions for the shear area, the minimum bolt diameter can be computed:

2 2

bolt

bolt

(3 surfaces) 789.0102 mm4

18.2994 mm 18.30 mm

d

d

Ans.




4.5 In Fig. P4.5, member (1) is a steel bar with a cross-

sectional area of 1.35 in.2 and a yield strength of 50 ksi.

Member (2) is a pair of 6061-T6 aluminum bars having a

combined cross-sectional area of 3.50 in.2 and a yield

strength of 40 ksi. A factor of safety of 1.6 with respect to

yield is required for both members. Determine the

maximum allowable load P that may be applied to the

structure. Report the factors of safety for both members at

the allowable load.

Fig. P4.5

Solution

Consider a FBD of joint B and write the following equilibrium equations:

2 1cos65 0xF F F (a)

2 sin 65 0yF F P (b)

From Eq. (b):

2sin 65

PF

(c)

Substituting Eq. (c) into Eq. (a) gives an expression for P in terms of F1:

2 1

1

1

cos65 0

cos65 0sin 65

tan 65

F F

PF

P F

(d)

and rearranging Eq. (c) gives an expression for P in terms of F2:

2 sin 65P F (e)

The allowable normal stress in member (1) is

,1

allow,1

1

50 ksi31.25 ksi

FS 1.6

Y

and the allowable force in member (1) is

2

allow,1 allow,1 1 (31.25 ksi)(1.35 in. ) 42.1875 kipsF A (f)

The allowable normal stress in member (2) is

,2

allow,2

2

40 ksi25.00 ksi

FS 1.6

Y

and the allowable force in member (2) is

2

allow,2 allow,2 2 (25.00 ksi)(3.50 in. ) 87.50 kipsF A (g)

Substitute the allowable force in member (1) from Eq. (f) into Eq. (d) to obtain the maximum load P

based on the capacity of member (1):

allow,1 tan 65 (42.1875 kips) tan 65 90.471386 kipsP F (h)




Repeat with the allowable force in member (2) from Eq. (g) substituted into Eq. (e) to obtain the

maximum load P based on the capacity of member (2):

allow,2 sin 65 (87.50 kips)sin 65 79.301931 kipsP F (i)

Compare the results in Eqs. (h) and (i) to find that the maximum load P that may be applied is controlled

by the capacity of member (2):

max 79.3 kipsP Ans.

The factor of safety in member (2) is thus:

2FS 1.6 Ans.

For an applied load of P = 79.3 kips, the force in member (1) can be computed from Eq. (d):

1

79.301931 kips36.979096 kips

tan 65 tan 65

PF

which results in a normal stress of

11 2

1

36.979096 kips27.391924 ksi

1.350 in.

F

A

and thus, its factor of safety is:

,1

1

1

50 ksiFS

27.391924 ksi1.825

Y

Ans.




4.6 The rigid structure ABD in Fig. P4.6 is supported at B by a 35-mm-diameter tie rod (1) and at A by a

30-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mm-

diameter pins used in double shear connections. Tie rod (1) has a yield strength of 250 MPa, and each of

the pins has an ultimate shear strength of 330 MPa. A concentrated load of P = 50 kN acts as shown at

D. Determine:

(a) the normal stress in rod (1).

(b) the shearing stress in the pins at A and B.

(c) the factor of safety with respect to the yield

strength for tie rod (1).

(d) the factor of safety with respect to the

ultimate strength for the pins at A and B.

Fig. P4.6

Solution

Member (1) is a two-force member that is

oriented at with respect to the horizontal axis:

8 mtan 1.509434

5.3 m

56.4755

From a FBD of rigid structure ABD, the following equilibrium equations can be written:

1cos60 cos56.4755 0x xF P F A (a)

1sin 60 sin56.4755 0y yF P F A (b)

1( cos56.4755 )(8 m) ( cos60 )(3.4 m) ( sin 60 )(7.5 m) 0AM F P P (c)

From Eq. (c):

1

(3.4 m)cos60 (7.5 m)sin 60 (3.4 m)cos60 (7.5 m)sin 60(50 kN) 92.7405 kN

(8 m)cos56.4755 (8 m)cos56.4755F P

Substitute F1 into Eq. (a) to obtain Ax:

1 cos56.4755 cos60 (92.7405 kN)cos56.4755 (50 kN)cos60 26.2199 kNxA F P

and substitute F1 into Eq. (b) to obtain Ay:

1 sin56.4755 sin 60 (92.7405 kN)sin56.4755 (50 kN)sin 60 120.6144 kNyA F P




The resultant pin force at A is found from Ax and Ay:

2 2 2 2(26.2199 kN) (120.6144 kN) 123.4314 kNx yA A A

(a) The cross-sectional area of the 35-mm-diameter tie rod is:

2 2

1 (35 mm) 962.112750 mm4

A

and thus, the normal stress in rod (1) is:

11 2

1

92,740.5 N96.3925 MPa

962.1127596.4 MPa

0 mm

F

A Ans.

(b) The 30-mm-diameter single shear pin at A has a shear area of

2 2

, (30 mm) 706.858347 mm4

V AA

Consequently, the shear stress in pin A is

2

123,431.4 N174.6197 MPa

706.858347 m174. M a

m6 PA Ans.

The 24-mm-diameter double shear pins at B and C have a shear area of

2 2

, (2 surfaces) (24 mm) 904.778685 mm4

V BA

The shear stress in pins B and C is

2

92,740.5 N102.5008 MPa

904.778685 mm102.5 MPaB Ans.

(c) The factor of safety for tie rod (1) is

1

1

250 MPaFS

96.3925 MPa2.59Y

Ans.

(d) The factor of safety with respect to the ultimate strength for the pin at A is

330 MPa

FS174.6197 MPa

1.890UA

A

Ans.

and the factor of safety for pin B is

330 MPa

FS102.5008 MP

3.2a

2UB

B

Ans.




4.7 Rigid bar ABD in Fig. P4.7 is supported by a pin

connection at A and a tension link BC. The 8-mm-

diameter pin at A is supported in a double shear

connection, and the 12-mm-diameter pins at B and C are

both used in single shear connections. Link BC is 30-mm

wide and 6-mm thick. The ultimate shear strength of the

pins is 330 MPa and the yield strength of link BC is 250

MPa.

(a) Determine the factor of safety in pins A and B with

respect to the ultimate shear strength.

(b) Determine the factor of safety in link BC with respect

to the yield strength.

Fig. P4.7

Solution

Consider a free-body diagram of the rigid bar. Link BC is a

two-force member that is oriented at an angle of with

respect to the horizontal axis:

0.6 m

tan 53.13010.45 m

The equilibrium equations for the rigid bar can be written as:

cos53.1301 (8.2 kN)cos70 0

sin53.1301 (8.2 kN)sin70 0

(0.75 m) cos53.1301 (0.45 m) sin53.1301

(1.85 m)(8.2 kN)sin70 (0.75 m)(8.2 kN)cos70 0

x BC x

y BC y

A BC BC

F F A

F F A

M F F

Solving these three equations simultaneously gives:

20.1958 kN 9.3129 kN 8.4511 kNBC x yF A A

The resultant force at pin A is:

2 2 2 2(9.3129 kN) ( 8.4511 kN) 12.5758 kNx yA A A

Before the factors of safety can be determined, we must compute the shear stresses in pins A and B as

well as the normal stress in link BC.

Pin shear stresses:

The 8-mm-diameter pin at A is supported in a double shear connection; therefore, the shear force acting

on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant

force at pin A: VA = 6.2879 kN. The cross-sectional area of the pin at A is:

2 2

(8 mm) 50.265 mm4

pin AA

and therefore, the shear stress in pin A is:

2

2

(6.2879 kN)(1000 N/kN)125.0940 N/mm 125.0940 MPa

50.265 mm

AA

V

V

A




The 12-mm-diameter pin at B and C is supported in a single shear connection and so the shear force

acting on one shear plane is the entire force in link BC: VB = 20.1958 kN. The cross-sectional area of

the pin at B is:

2 2

(12 mm) 113.097 mm4

pin BA

and therefore, the shear stress in pin B is:

2

2

(20.1958 kN)(1000 N/kN)178.5697 N/mm 178.5697 MPa

113.097 mm

BB

V

V

A

Normal stress in link:

The normal stress in link BC is:

2(20.1958 kN)(1000 N/kN)

112.1986 N/mm 112.1986 MPa(30 mm)(6 mm)

BCBC

BC

F

A

Factors of safety:

For pin A, the factor of safety is:

330 MPa

FS125.0940 MP

2.6a

4UA

A

Ans.

For pin B, the factor of safety is:

330 MPa

FS178.5697 MPa

1.848UB

B

Ans.

The factor of safety in link BC is:

250 MPa

FS112.1986 M

2.23Pa

YBC

BC

Ans.




4.8 In Fig. P4.8, davit ABD is supported at A by a single

shear pin connection and at B by a tie rod (1). The pin at

A has a diameter of 1.25 in. and the pins at B and C are

each 0.75-in.-diameter pins. Tie rod (1) has an area of

1.50 in.2. The ultimate shear strength in each pin is 80 ksi,

and the yield strength of the tie rod is 36 ksi. A

concentrated load of 25 kips is applied as shown to the

davit structure at D. Determine:

(a) the normal stress in rod (1).

(b) the shearing stress in the pins at A and B.

(c) the factor of safety with respect to the yield strength

for tie rod (1).

(d) the factor of safety with respect to the ultimate

strength for the pins at A and B.

Fig. P4.8

Solution

Member (1) is a two-force member that is oriented at with


9 ft

tan 0.75 36.87012 ft

From a FBD of rigid structure ABD, the following equilibrium

equations can be written:

1(25 kips)cos60 cos36.870 0x xF F A (a)

1(25 kips)sin 60 sin 36.870 0y yF F A (b)

1( cos36.870 )(9 ft) (25 kips)cos 60 (11 ft)

(25 kips)sin 60 (7 ft) 0

AM F

(c)

From Eq. (c):

1

(11 ft)cos60 (7 ft)sin 60(25 kips) 40.1465 kips

(9 ft)cos36.870F


1 cos36.870 (25 kips)cos60 (40.1465 kips)cos36.870 (25 kips)cos60 19.6172 kipsxA F


1 sin 36.870 (25 kips)sin 60 (40.1465 kips)sin 36.870 (25 kips)sin 60 45.7386 kipsyA F


2 2 2 2(19.6172 kips) (45.7386 kips) 49.7680 kipsx yA A A

(a) The normal stress in tie rod (1) is:

11 2

1

40.1465 kips26.7643 ksi

1.50 i26.8 k

n si

.

F

A Ans.




(b) The 1.25-in.-diameter single shear pin at A has a shear area of

2 2

, (1.25 in.) 1.2272 in.4

V AA

Consequently, the shear stress in pin A is

2

49.7680 kips40.5541 ksi

1.2272 in40.6 i

.ksA Ans.

The 0.75-in.-diameter double shear pins at B and C have a shear area of

2 2

, (2 surfaces) (0.75 in.) 0.8836 in.4

V BA

The shear stress in pins B and C is

2

40.1465 kips45.4352 ksi

0.8836 in45.4 i

.ksB Ans.

(c) The factor of safety for tie rod (1) is

1

1

36 ksiFS

26.7643 ksi1.345Y

Ans.

(d) The factor of safety with respect to the ultimate strength for the pin at A is

80 ksi

FS40.5541 ksi

1.973UA

A

Ans.

and the factor of safety for pin B is

80 ksi

FS45.4352 ksi

1.761UB

B

Ans.




4.9 The pin-connected structure is subjected to a load P

as shown in Fig. P4.9. Inclined member (1) has a cross-

sectional area of 250 mm2 and has a yield strength of

255 MPa. It is connected to rigid member ABC with a

16-mm-diameter pin in a double shear connection at B.

The ultimate shear strength of the pin material is 300

MPa. For inclined member (1), the minimum factor of

safety with respect to the yield strength is FSmin = 1.5.

For the pin connections, the minimum factor of safety

with respect to the ultimate strength is FSmin = 3.0.

(a) Based on the capacity of member (1) and pin B,

determine the maximum allowable load P that may be

applied to the structure.

(b) Rigid member ABC is supported by a double shear

pin connection at A. Using FSmin = 3.0, determine the

minimum pin diameter that may be used at support A.

Fig. P4.9

Solution

The allowable normal stress for inclined member (1) is

allow

255 MPa170.00 MPa

FS 1.50

Y

and the allowable shear stress for the pins is

allow

300 MPa100.0 MPa

FS 3.00

U

(a) The allowable axial force in member (1) based on normal stress is

2

allow allow 1 (170.00 MPa)(250 mm ) 42,500 NF A (a)

Member (1) is connected with a 16-mm-diameter double shear pin. The shear area of this pin is:

2 2(2 surfaces) (16 mm) 402.1239 mm4

VA

Consequently, the shear force V (which is applied by the inclined member) that can be applied to the pin

is limited to

2

allow allow (100.0 MPa)(402.1239 mm ) 40,212.4 NVV A (b)

Comparing the two values given in Eq. (a) and Eq. (b), the maximum allowable force in member (1) is

1 40,212.4 NF (c)






1.4 m

tan 0.6087 31.32872.3 m



1 cos31.3287 0x xF A F P (d)

1 sin 31.3287 0y yF A F (e)

1(3.2 m) ( cos31.3287 )(1.4 m) 0AM P F (f)

Substitute the value of F1 from Eq. (c) into Eq. (f) to obtain

the maximum load P:

1

(1.4 m)cos31.3287

3.2 m

(1.4 m)cos31.3287(40,212.4 N)

3.2 m

15,027.8 N 15.03 kN

P F

Ans.

Substitute F1 and P into Eq. (d) to obtain Ax:

1 cos31.3287 (15,027.8 N) (40,212.4 N)cos31.3287 19,321.5 NxA P F

and substitute F1 into Eq. (e) to obtain Ay:

1 sin31.3287 (40,212.4 N)sin31.3287 20,908.3 NyA F


2 2 2 2( 19,321.5 N) (20,908.3 N) 28,468.9 Nx yA A A

The total shear area required to support the resultant pin force is

2

2

allow

28,468.9 N284.689 mm

100 N/mmV

AA

Since the pin at A is in a double shear connection, the shear area provided by the pin is

2

pin(2 surfaces)4

VA d

Equate these two expressions and solve for the required minimum pin diameter:

2 2

pin

pin

(2 surfaces) 284.689 mm4

13.46 mm

d

d

Ans.




4.10 After load P is applied to the pin-connected structure shown in Fig. P4.10, a normal strain of

= +550 is measured in the longitudinal direction of member (1). The cross-sectional area of

member (1) is A1 = 0.60 in.2, its elastic modulus is E1 = 29,000 ksi, and its yield strength is 36 ksi.

(a) Determine the axial force in

member (1), the applied load P, and the

resultant force at pin B.

(b) The ultimate shear strength of the

steel pins is 54 ksi. Determine the

minimum diameter for the pin at B if a

factor of safety of 2.5 with respect to

the ultimate shear strength is required.

(c) Compute the factor of safety for

member (1) with respect to its yield

strength.

Fig. P4.10

Solution

(a) Given the strain in member (1), its stress can be computed from Hooke’s law:

1 1 1 (29,000 ksi)(0.000550 in./in.) 15.9500 ksiE

The force in member (1) is the product of the normal stress and the cross-sectional area:

2

1 1 1 (15.9500 ksi)(0.60 in 9.57 kip. ) sF A Ans.

Next, consider a free-body diagram of

horizontal member ABC. Member (1) makes

an angle of with respect to the horizontal

axis:

3.50 ft

tan 36.38444.75 ft

Note that member (1) is a two-force member.

Equilibrium equations for horizontal member

ABC can be written as:

1

1

1

cos36.3844 0

sin36.3844 0

(6.50 ft) sin36.3844 (9.50 ft) 0

x x

y y

B

F B F

F B F P

M F P

Substituting the value F1 = 14.79 kips into these equations gives:

7.7044 kips 9.5611 kips 3.8842 kips 3.88 kipsx yB B P Ans.

The resultant force at pin B is:

2 2 2 2( 7.7044 kips) (9.5611 kips) 12.2789 kips 12.28 kipsx yB B B Ans.

(b) The pin at B is supported in a double shear connection. Therefore, the shear force acting on one

shear plane of the pin is half of the resultant force: VB = 6.1395 kips. The allowable shear stress for the

pin is computed from the ultimate shear strength and the factor of safety:

54 ksi

21.6 ksi2.5

Ua

FS




Next, the cross-sectional area required for a single shear plane can be determined:

26.1395 kips0.2842 in.

21.6 ksi

B Ba B

B a

V VA

A

To provide AB, the pin at B must have a diameter of at least:

2 20.2842 in. 0.602 in.4

B Bd d

Ans.

(c) The factor of safety for member (1) with respect to its yield strength is:

1

1

36 ksiFS

15.9500 ksi2.26Y

Ans.




4.11 The simple pin-connected structure carries a

concentrated load P as shown in Fig. P4.11. The

rigid bar is supported by strut AB and by a pin

support at C. The steel strut AB has a cross-

sectional area of 0.25 in.2

and a yield strength of

60 ksi. The diameter of the steel pin at C is 0.375

in., and the ultimate shear strength is 54 ksi. If a

factor of safety of 2.0 is required in both the strut

and the pin at C, determine the maximum load P

that can be supported by the structure.

Fig. P4.11

Solution

From a FBD of the rigid bar, the following equilibrium


1 0x xF F C (a)

0y yF C P (b)

1(10 in.) (22 in.) 0CM F P (c)

From Eq. (c), express F1 in terms of the unknown load P:

1

22 in.2.2

10 in.F P P (d)

Substitute this result into Eq. (a) to express Cx in terms of P:

1 2.2xC F P

The resultant reaction force at pin C can now be expressed as a function of P:

2 2 2 2(2.2 ) ( ) 2.41661x yC C C P P P (e)

Strut AB:

The allowable normal stress for strut AB is

allow

60 ksi30 ksi

FS 2.0

Y

Therefore, the allowable axial force for strut AB is

2

allow,1 allow 1 (30 ksi)(0.25 in. ) 7.50 kipsF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the strut

normal stress is

allow,1

max

7.50 kips3.4091 kips

2.2 2.2

FP (f)




Pin C:

The allowable shear stress for the pin at C is

allow

54 ksi27 ksi

FS 2.0

U

The 0.375-in.-diameter double shear pin at C has a shear area of

2 2(2 surfaces) (0.375 in.) 0.2209 in.4

VA

thus, the allowable reaction force at C is

2

allowallow(27 ksi)(0.2209 in. ) 5.9641 kipsVC A

From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin

shear stress is

allowmax

5.9641 kips2.4680 kips

2.41661 2.41661

CP (g)

Maximum load P:

Comparing the results in Eqs. (f) and (g), the maximum load P that may be applied to the rigid bar is

max 2.4680 ki 2.47 ipsps kP Ans.




4.12 In Fig. P4.12, rigid tee-beam ABC is supported at

A by a single shear pin connection and at B by a strut,

which consists of two 30 mm wide by 8 mm thick steel

bars. The pins at A, B, and D are each 12 mm in

diameter. The yield strength of the steel bars in strut (1)

is 250 MPa, and the ultimate shear strength of each pin

is 500 MPa. Determine the allowable load P that may

be applied to the rigid bar at C if an overall factor of

safety of 3.0 is required. Use L1 = 1.1 m and L2 = 1.3 m.

Fig. P4.12

Solution

From a FBD of the rigid beam ABC, the following


0x xF A (a)

1 0y yF A F P (b)

1 1 1 2( ) 0AM L F L L P (c)


1 21

1

1.1 m 1.3 m2.181818

1.1 m

L LF P P P

L

(d)

Substitute this result into Eq. (b) to express Ay in terms of P:

1 ( 2.181818 ) 1.181818yA P F P P P

The resultant reaction force at pin A can now be expressed as a function of P:

2 2 2 2(0) ( 1.181818 ) 1.181818x yA A A P P (e)

The allowable normal stress for strut (1) is

allow

250 MPa83.333333 MPa

FS 3.0

Y

Therefore, the allowable axial force for strut (1) is

allow,1 allow 1 (83.333333 MPa)(2 30 mm 8 mm) 40,000 NF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress

limitation is

allow,1

max

40,000 N18,333 N

2.181818 2.181818

FP (f)

The allowable shear stress for the pins at A, B, and D is

allow

500 MPa166.666667 MPa

FS 3.0

U

The cross-sectional area of a 12-mm-diameter pin is

2 2

pin (12 mm) 113.097336 mm4

A




The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is

2

allow ,allow(166.666667 MPa)(113.097336 mm ) 18,849.5560 NV AA A

From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is

allowmax

18,849.5560 N15,949.627 N

1.181818 1.181818

AP (g)

The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut

(1) based on the capacity of the pins at B and D is

2

allow,1 allow , (166.666667 MPa)(2)(113.097336 mm ) 37,699.1121 NV BF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins

at B and D is

allow,1

max

37,699.1121 N17,278.7611 N

2.181818 2.181818

FP (h)

Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar

is

max 15,949.62 15.7 9 N 5 kNP Ans.




4.13 In Fig. P4.13, rigid bar ABC is supported at A by a

single shear pin connection and at B by a strut, which

consists of two 2 in. wide by 0.25 in. thick steel bars.

The pins at A, B, and D each have a diameter of 0.5 in.

The yield strength of the steel bars in strut (1) is 36 ksi,

and the ultimate shear strength of each pin is 72 ksi.

Determine the allowable load P that may be applied to

the rigid bar at C if an overall factor of safety of 3.0 is

required. Use L1 = 36 in. and L2 = 24 in.

Fig. P4.13

Solution

From a FBD of the rigid beam ABC, the following


0x xF A (a)

1 0y yF A F P (b)

1 1 1 2( ) 0AM L F L L P (c)


1 21

1

36 in. 24 in.1.666667

36 in.

L LF P P P

L

(d)

Substitute this result into Eq. (b) to express Ay in terms of P:

1 ( 1.666667 ) 0.666667yA P F P P P

The resultant reaction force at pin A can now be expressed as a function of P:

2 2 2 2(0) ( 0.666667 ) 0.666667x yA A A P P (e)

Strut (1):

The allowable normal stress for strut (1) is

allow

36 ksi12.0 ksi

FS 3.0

Y

Therefore, the allowable axial force for strut (1) is

allow,1 allow 1 (12.0 ksi)(2 2 in. 0.25 in.) 12.0 kipsF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress

limitation is

allow,1

max

12.0 kips7.20 kips

1.666667 1.666667

FP (f)




Pins:

The allowable shear stress for the pins at A, B, and D is

allow

72 ksi24.0 ksi

FS 3.0

U

The cross-sectional area of a 0.5-in.-diameter pin is

2 2

pin (0.5 in.) 0.196350 in.4

A

Pin A:

The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is

2

allow ,allow(24.0 ksi)(0.196350 in. ) 4.7124 kipsV AA A

From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is

allowmax

4.7124 kips7.0686 kips

0.666667 0.666667

AP (g)

Pin B and D:

The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut

(1) based on the capacity of the pins at B and D is

2

allow,1 allow , (24.0 ksi)(2)(0.196350 in. ) 9.4248 kipsV BF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins

at B and D is

allow,1

max

9.4248 kips5.6549 kips

1.666667 1.666667

FP (h)

Maximum load P:


is

max 5.6549 ki 5.65 ipsps kP Ans.




4.14 A concentrated load of P = 70 kips is applied to

beam AB as shown in Fig. P4.14. Rod (1) has a

diameter of 1.50 in., and its yield strength is 60 ksi. Pin

A is supported in a double shear connection, and the

ultimate shear strength of pin A is 80 ksi.

(a) Determine the normal stress in rod (1).

(b) Determine the factor of safety with respect to the

yield strength for rod (1).

(c) If a factor of safety of 3.0 with respect to the

ultimate strength is specified for pin A, determine the

minimum required pin diameter.

Fig. P4.14

Solution

Tie rod (1) is a two-force member that is oriented at


8.5 ft

tan 0.7083 35.311212 ft

From a FBD of beam AB, the following equilibrium


1 cos35.3112 0x xF A F (a)

1 sin35.3112 0y yF A F P (b)

1

1

10 in.( cos35.3112 )

12 in./ft

( sin35.3112 )(12 ft) (8 ft) 0

AM F

F P

(c)

From Eq. (c):

1

(70 kips)(8 ft)73.5272 kips

(cos35.3112 ) 0.833333 ft (sin35.3112 )(12 ft)F


1 cos35.3112 (73.5272 kips)cos35.3112 60.0 kipsxA F


1 sin35.3112 70 kips (73.5272 kips)sin35.3112 27.50 kipsyA P F



(a) The cross-sectional area of the 1.5-in.-diameter tie rod (1) is

2 2

1 (1.5 in.) 1.7671 in.4

A




The normal stress in tie rod (1) is:

11 2

1

73.5272 kips41.6079 ksi

1.7671 41.6 ks

ii

n.

F

A Ans.

(b) The factor of safety for tie rod (1) is

1

1

60 ksiFS

41.6079 ksi1.442Y

Ans.

(c) The allowable shear stress is:

allow

80 ksi26.6667 ksi

FS 3.0

U

The shear area required for pin A is thus

2

ult

66.0019 kips2.4751 in.

26.6667 ksiV

AA

Since pin A is in a double shear connection, the minimum required pin area is

2

2

pin

2.4751 in.1.2375 in.

2 shear surfaces 2

VAA

Thus, the diameter of pin A must be

2 21.2375 in.

1.255 in.

4A

A

d

d

Ans.




4.15 Beam AB is supported as shown in Fig. P4.15. Tie

rod (1) is attached at B and C with double shear pin

connections while the pin at A is attached with a single

shear connection. The pins at A, B, and C each have an

ultimate shear strength of 54 ksi, and tie rod (1) has a

yield strength of 36 ksi. A concentrated load of P = 16

kips is applied to the beam as shown. A factor of safety

of 3.0 is required for all components. Determine:

(a) the minimum required diameter for tie rod (1).

(b) the minimum required diameter for the double shear

pins at B and C.

(c) the minimum required diameter for the single shear

pin at A.

Fig. P4.15

Solution

Tie rod (1) is a two-force member that is oriented at


8.5 ft

tan 0.7083 35.311212 ft

From a FBD of beam AB, the following equilibrium


1 cos35.3112 0x xF A F (a)

1 sin35.3112 0y yF A F P (b)

1

1

10 in.( cos35.3112 )

12 in./ft

( sin35.3112 )(12 ft) (8 ft) 0

AM F

F P

(c)

From Eq. (c):

1

(16 kips)(8 ft)16.8062 kips

(cos35.3112 ) 0.833333 ft (sin35.3112 )(12 ft)F


1 cos35.3112 (16.8062 kips)cos35.3112 13.7143 kipsxA F


1 sin35.3112 16 kips (16.8062 kips)sin35.3112 6.2857 kipsyA P F



(a) The allowable normal stress for tie rod (1) is

allow

36 ksi12.0 ksi

FS 3.0

Y




The minimum cross-sectional area required for the tie rod is

21

min

allow

16.8062 kips1.400518 in.

12 ksi

FA

Therefore, the minimum tie rod diameter is

2 2

1 11.400518 in. 1.335363 in 1.335 in..4

d d

Ans.

(b) The allowable shear stress for the pins at A, B, and C is

allow

54 ksi18 ksi

FS 3.0

U

The double-shear pin connection at B and C must support a load of F1 =16.8062 kips . The shear area

AV required for these pins is

21

allow

16.8062 kips0.933679 in.

18 ksiV

FA

The pin diameter can be computed from

2 2

pin pin(2 surfaces) 0.933679 in. 0.7710 in. 0.771 in.4

d d

Ans.

(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin

is

2

allow

15.0861 kips0.838119 in.

18 ksiV

AA


2 2

pin pin(1 surface) 0.838119 in. 1.0330 in. 1.033 .4

ind d

Ans.




4.16 In Fig. P4.16, the rigid member ABDE is supported at A by a

single shear pin connection and at B by a tie rod (1). The tie rod is

attached at B and C with double shear pin connections. The pins at

A, B, and C each have an ultimate shear strength of 80 ksi, and tie

rod (1) has a yield strength of 60 ksi. A concentrated load of

P = 24 kips is applied perpendicular to DE, as shown. A factor of

safety of 2.0 is required for all components. Determine:

(a) the minimum required diameter for the tie rod.

(b) the minimum required diameter for the pin at B.

(c) the minimum required diameter for the pin at A.

Fig. P4.16

Solution

Member (1) is a two-force member that is oriented at with respect to

the horizontal axis:

9 ft

tan 1.5 56.3106 ft

From a FBD of rigid structure ABDE, the following equilibrium


1cos70 cos56.310 0x xF P F A (a)

1sin 70 sin 56.310 0y yF P F A (b)

1( cos56.310 )(9 ft) ( cos 70 )[13 ft (8 ft)sin 20 ]

( sin 70 )[(8 ft) cos 20 ] 0

AM F P

P

(c)

From Eq. (c):

1

(cos70 )[13 ft (8 ft)sin 20 ] (sin70 )[(8 ft)cos20 ]

(cos56.310 )(9 ft)

5.382084 ft 7.064178 ft(24 kips) 59.834286 kips

4.992293 ft

F P

(d)


1 cos56.310 cos70

(59.834286 kips)cos56.310 (24 kips)cos70

24.981548 kips

xA F P


1 sin 56.310 sin 70

(59.834286 kips)sin 56.310 (24 kips)sin 70

72.337797 kips

yA F P





2 2 2 2(24.981548 kips) (72.337797 kips) 76.529959 kipsx yA A A (e)

(a) The allowable normal stress for tie rod (1) is

allow

60 ksi30.0 ksi

FS 2.0

Y

The minimum cross-sectional area required for the tie rod is

21

min

allow

59.834286 kips1.994476 in.

30.0 ksi

FA

Therefore, the minimum tie rod diameter is

2 2

1 11.994476 in. 1.593564 in 1.594 in..4

d d

Ans.

(b) The allowable shear stress for the pins at A, B, and C is

allow

80 ksi40.0 ksi

FS 2.0

U

The double-shear pin connection at B and C must support a load of F1 = 59.834286 kips . The shear

area AV required for these pins is

21

allow

59.834286 kips1.495857 in.

40.0 ksiV

FA


2 2


d d

Ans.

(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin

is

2

allow

76.529959 kips1.913249 in.

40.0 ksiV

AA


2 2

pin pin(1 surface) 1.913249 in. 1.560777 in. 1.561 in.4

d d

Ans.




4.17 Rigid bar ABC is subjected to a concentrated load P as

shown in Fig. P4.17. Inclined member (1) has a cross

sectional area of A1 = 2.250 in.2 and is connected at ends B

and D by 1.00-in.-diameter pins in double shear

connections. The rigid bar is supported at C by a 1.00-in.-

diameter pin in a single shear connection. The yield

strength of inclined member (1) is 36 ksi, and the ultimate

strength of each pin is 60 ksi. For inclined member (1), the

minimum factor of safety with respect to the yield strength

is FSmin = 1.5. For the pin connections, the minimum factor

of safety with respect to the ultimate strength is FSmin = 2.0.

Determine the maximum load P that can be supported by

the structure. Fig. P4.17

Solution

Member (1) is a two-force member that is oriented at with


40 in.

tan 0.8 38.659850 in.

From a FBD of rigid bar ABC, the following equilibrium


1 cos38.6598 0x xF F C (a)

1 sin38.6598 0y yF P F C (b)

1(80 in.) ( sin38.6598 )(50 in.) 0CM P F (c)


1

(80 in.)2.561250

(50 in.)sin38.6598

PF P

(d)

Substitute this result into Eq. (a) to express Cx in terms of P:

1 cos38.6598 ( 2.561250 )cos38.6598 2.0xC F P P

And similarly, express Cy in terms of P from Eq. (b):

1 sin38.6598 ( 2.561250 )sin38.6598 0.60yC P F P P P

The resultant reaction force at pin C can now be expressed as a function of P:

2 2 2 2(2.0 ) (0.60 ) 2.088061x yC C C P P P (e)

Inclined member (1):

The allowable normal stress for inclined member (1) is

allow

36 ksi24 ksi

FS 1.5

Y

Therefore, the allowable axial force for inclined member (1) is

2

allow,1 allow 1 (24 ksi)(2.25 in. ) 54 kipsF A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the

inclined member normal stress is

allow,1

max

54 kips21.083455 kips

2.561250 2.561250

FP (f)




Pin B:

The allowable shear stress for the pins is

allow

60 ksi30 ksi

FS 2.0

U

A 1.0-in.-diameter pin has a cross-sectional area of

2 2

pin (1.0 in.) 0.785398 in.4

A

For a double shear connection, the corresponding shear area is

2 2

pin2 2(0.785398 in. ) 1.570796 in.VA A

Thus, the maximum allowable force in member (1) based on the pin at B is:

2

allow,1 allow (30 ksi)(1.570796 in. ) 47.123880 kipsV

F A

From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the pin

at B is

allow,1

max

47.123880 kips18.398782 kips

2.561250 2.561250

FP (g)

Pin C:

The pin at C is in a single shear connection; therefore, the shear area of pin C is equal to the cross-

sectional area of the 1.0-in.-diameter pin. The maximum allowable shear force in this pin is thus

2

allow,C allow (30 ksi)(0.785398 in. ) 23.561940 kipsV

F A

From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin

shear stress is

allowmax

23.561940 kips11.284125 kips

2.088061 2.088061

CP (h)

Maximum load P:


is

max 11.284125 k 11.28ips kipsP Ans.




4.18 Rigid bar ABC is supported by pin-connected axial member (1) and by a pin connection at C as

shown in Fig. P4.18. A 6,300-lb concentrated load is applied to the rigid bar at A. Member (1) is a 2.75

in. wide by 1.25 in. thick rectangular bar made of steel with a yield strength of Y = 36,000 psi. The pin

at C has an ultimate shear strength of U = 60,000 psi.

(a) Determine the axial force in

member (1).

(b) Determine the factor of safety in

member (1) with respect to its yield

strength.

(c) Determine the magnitude of the

resultant reaction force acting at pin C.

(d) If a minimum factor of safety of FS

= 3.0 with respect to the ultimate shear

strength is required, determine the

minimum diameter that may be used

for the pin at C.

Fig. P4.18

Solution



30 in.

tan 0.75 36.87040 in.



1 cos36.870 0x xF F C (a)

16,300 lb sin 36.870 0y yF F C (b)

1

(6,300 lb)(80 in.)

( sin36.870 )(56 in.) 0

CM

F

(c)

(a) From Eq. (c), the axial force in member (1) is:

1

(6,300 lb)(80 in.)15,000 lb

(56 in.)sin15,000 lb

36. (C)

870F

Ans.

(b) The normal stress magnitude in member (1) is

1

1

1

15,000 lb4,363.6364 psi

(2.75 in.)(1.25 in.)

F

A

Therefore, its factor of safety with respect to the 36,000 psi yield stress is

1

1

36,000 psiFS

4,363.6364 psi8.25Y

Ans.

(c) Substitute F1 into Eq. (a) to obtain Cx:

1 cos36.870 ( 15,000 lb)cos36.870 12,000 lbxC F

and substitute F1 into Eq. (b) to obtain an expression for Ay in terms of the unknown load P:

16,300 lb sin 36.870 6,300 lb ( 15,000 lb)sin 36.870 2,700 lbyC F

The resultant pin force at C is found from Cx and Cy:




2 2 2 2( 12,000 lb) ( 2,700 lb 12,300 ) lbx yC C C Ans.

(d) The allowable shear stress for the pins at C is

allow

60,000 psi20,000 psi

FS 3.0

U

The double-shear pin connection at C must support a load of 12,300 lb. The minimum shear area AV

required to support this load is

2

allow

12,300 lb0.6150 in.

20,000 psiV

CA


2 2


d d

Ans.




4.19 A rectangular steel plate is used as an axial member to support a dead load of 70 kips and a live

load of 110 kips. The yield strength of the steel is 50 ksi.

(a) Use the ASD method to determine the minimum cross-sectional area required for the axial member if

a factor of safety of 1.67 with respect to yielding is required.

(b) Use the LRFD method to determine the minimum cross-sectional area required for the axial member

based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and

load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution

(a) The service load on the axial member is

70 kips 110 kips 180 kipsP D L


allow

50 ksi29.940 ksi

FS 1.67

Y

The minimum cross-sectional area required to support the service load is

min

allow

2180 kips

29.940 ks6.01 in

i .

PA

Ans.

(b) The ultimate load for LRFD is

1.2 1.6 1.2(70 kips) 1.6(110 kips) 260 kipsU D L

The design equation for an axial member subjected to tension can be written in LRFD as

t Y A U

Consequently, the minimum cross-sectional area required for the tension member is

2

min

260 kips

0.9(50 ks5.78 i .

i)n

t Y

UA

Ans.




4.20 A 20-mm-thick steel plate will be used as an axial member to support a dead load of 150 kN and a

live load of 220 kN. The yield strength of the steel is 250 MPa.

(a) Use the ASD method to determine the minimum plate width b required for the axial member if a

factor of safety of 1.67 with respect to yielding is required.

(b) Use the LRFD method to determine the minimum plate width b required for the axial member based

on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load

factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution


150 kN 220 kN 370 kNP D L


allow

250 MPa149.70 MPa

FS 1.67

Y


2

min 2

allow

(370 kN)(1,000 N/kN)2,471.61 mm

149.70 N/mm

PA

Since the plate is 20-mm-thick, the minimum plate width is therefore:

2

minmin

2,471.61 mm

20 m12 . mm

m3 6

Ab

t Ans.


1.2 1.6 1.2(150 kN) 1.6(220 kN) 532 kNU D L


t Y A U


2

min 2

(532 kN)(1,000 N/kN)2,364.44 mm

0.9(250 N/mm )t Y

UA

Since the plate is 20-mm-thick, the minimum plate width is therefore:

2

minmin

2,364.44 mm

20 m11 . mm

m8 2

Ab

t Ans.




4.21 A round steel tie rod is used as a tension member to support a dead load of 30 kips and a live load

of 15 kips. The yield strength of the steel is 46 ksi.

(a) Use the ASD method to determine the minimum diameter required for the tie rod if a factor of safety

of 2.0 with respect to yielding is required.

(b) Use the LRFD method to determine the minimum diameter required for the tie rod based on yielding

of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2

and 1.6 for the dead and live loads, respectively.

Solution


30 kips 15 kips 45 kipsP D L


allow

46 ksi23.0 ksi

FS 2.0

Y


2

min

allow

45 kips1.956522 in.

23.0 ksi

PA

and thus, the minimum tie rod diameter is

2 2

min min1.956522 in 1.578 in.4

.d d

Ans.


1.2 1.6 1.2(30 kips) 1.6(15 kips) 60 kipsU D L


t Y A U


2

min

60 kips1.449275 in.

0.9(46 ksi)t Y

UA


2 2

min min1.449275 in 1.358 in.4

.d d

Ans.




4.22 A round steel tie rod is used as a tension member to support a dead load of 190 kN and a live load

of 220 kN. The yield strength of the steel is 320 MPa.

(a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of

safety of 2.0 with respect to yielding is required.

(b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on

yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors

of 1.2 and 1.6 for the dead and live loads, respectively.

Solution


190 kN 220 kN 410 kNP D L


allow

320 MPa160 MPa

FS 2.0

Y


2

min 2

allow

(410 kN)(1,000 N/kN)2,562.500 mm

160 N/mm

PA


2 2

min min2,562.500 mm4

57.1 mmd d

Ans.


1.2 1.6 1.2(190 kN) 1.6(220 kN) 580 kNU D L


t Y A U


2

min 2

(580 kN)(1,000 N/kN)2,013.889 mm

0.9(320 N/mm )t Y

UA


2 2

min min2,013.889 mm4

50.6 mmd d

Ans.




5.1 A steel [E = 200 GPa] rod with a circular cross section is 7.5-m long. Determine the minimum

diameter required if the rod must transmit a tensile force of 50 kN without exceeding an allowable stress

of 180 MPa or stretching more than 5 mm.

Solution

If the normal stress in the rod cannot exceed 180 MPa, the cross-sectional area must equal or exceed

2

2

(50 kN)(1,000 N/kN)277.7778 mm

180 N/mm

PA

If the elongation must not exceed 5 mm, the cross-sectional area must equal or exceed

2

2

(50 kN)(1,000 N/kN)(7,500 mm)375.0000 mm

(200,000 N/mm )(5 mm)

PLA

E

Therefore, the minimum cross-sectional area that may be used for the rod is Amin = 375 mm2. The

corresponding rod diameter is

2 2

rod rod375 mm 21.8510 mm 2 m4

1.9 md d

Ans.




5.2 An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than

0.25 in. when the tension in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 12

ksi, determine:

(a) the smallest diameter that can be used for the rod.

(b) the corresponding maximum length of the rod.

Solution

(a) If the normal stress in the rod cannot exceed 20 ksi, the cross-sectional area must equal or exceed

22.20 kips0.1833 in.

12 ksi

PA

The corresponding rod diameter is

2 2

rod rod0.1833 in. 0.4834

in.d d

Ans.

(b) If the elongation must not exceed 0.25 in., the aluminum control rod having a cross-sectional area of

A = 0.1833 in.2 can have a length no greater than

2(0.1833 in. )(10,000 ksi)(0.25 in.)

208.3333 in.2.20 kips

208 in.AE

LP

Ans.




5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2)

is connected to a 30-mm-wide by 8-mm-thick

rectangular aluminum [E = 70 GPa] bar (1), as

shown in Fig. P5.3. Determine the force P required

to stretch the assembly 10.0 mm.

Fig. P5.3

Solution

The elongations in the two axial members are expressed by

1 1 2 21 2

1 1 2 2

andF L F L

A E A E

The total elongation of the assembly is thus

1 1 2 21 2

1 1 2 2

C

F L F Lu

A E A E

Since the internal forces F1 and F2 are equal to external load P, this expression can be simplified to

1 2

1 1 2 2

C

L Lu P

A E A E

For rectangular aluminum bar (1), the cross-sectional area is

2

1 (30 mm)(8 mm) 240 mmA

and the cross-sectional area of steel rod (2) is

2 2

2 (12 mm) 113.0973 mm4

A

The force P required to stretch the assembly 10.0 mm is thus

1 2

1 1 2 2

2 2 2 2

10.0 mm

450 mm 1,300 mm

(240 mm )(70,000 N/mm ) (113.09739 mm )(200,000 N/mm )

118,682.6 N

= 118.7 kN

CuP

L L

A E A E

Ans.




5.4 Two polymer bars are connected to

a rigid plate at B, as shown in Fig. P5.4.

Bar (1) has a cross-sectional area of

1.65 in.2 and an elastic modulus of

2,400 ksi. Bar (2) has a cross-sectional

area of 0.975 in.2 and an elastic modulus

of 4,000 ksi. Determine the total

deformation of the bar.

Fig. P5.4

Solution

Draw a FBD that cuts through member (1) to find

that the internal axial force in member (1) is

1 8 kips (T)F

Similarly, draw a FBD that cuts through member (2)

to find that the internal axial force in member (2) is

2 14 kips (T)F

The elongation in bar (1) can be computed as

1 11 2

1 1

(8 kips)(20 in.)0.040404 in.

(1.65 in. )(2,400 ksi)

F L

A E

and the elongation in bar (2) can be computed as

2 22 2

2 2

(14 kips)(30 in.)0.107692 in.

(0.975 in. )(4,000 ksi)

F L

A E

The total elongation of the bar is thus

1 2 0.040404 in. 0.107692 in 0.14 1. 8 in. Ans.




5.5 An axial member consisting of two polymer

bars is supported at C as shown in Fig. P5.5. Bar

(1) has a cross-sectional area of 540 mm2 and an

elastic modulus of 28 GPa. Bar (2) has a cross-

sectional area of 880 mm2 and an elastic

modulus of 16.5 GPa. Determine the deflection

of point A relative to support C.

Fig. P5.5

Solution

Draw a FBD that cuts through member (1) to find that the internal

axial force in member (1) is

1 35 kN (T)F

Similarly, draw a FBD that cuts through member

(2) and includes the free end of the axial member.

From this FBD, the equilibrium equation is

235 kN 50 kN 50 kN 0xF F

Therefore, the internal axial force in member (2) is

2 65 kN 65 kN (C)F

The deformation in bar (1) can be computed as

1 11 2 2

1 1

(35 kN)(1,000 N/kN)(850 mm)1.9676 mm

(540 mm )(28,000 N/mm )

F L

A E

and the deformation in bar (2) can be computed as

2 22 2 2

2 2

( 65 kN)(1,000 N/kN)(1,150 mm)5.1481 mm

(880 mm )(16,500 N/mm )

F L

A E

The deflection of point A relative to the support at C is the sum of these two deformations:

1 2 1.9676 mm ( 5.1481 mm) 3.18 mm 3.18 mmAu Ans.




5.6 The roof and second floor of a building are supported

by the column shown in Fig. P5.6. The column is a

structural steel W10 × 60 wide-flange section [E = 29,000

ksi; A = 17.6 in2]. The roof and floor subject the column to

the axial forces shown. Determine:

(a) the amount that the first floor will settle.

(b) the amount that the roof will settle.

Fig. P5.6

Solution

(a) Draw a FBD that cuts through member (1) and includes the free

end of the column. From this FBD, the sum of forces in the

vertical direction can be written as

1110 kips 155 kips 0yF F


1 270 kips 270 kips (C)F

The settlement of the first floor is determined by the deformation

(i.e., contraction in this case) that occurs in member (1).

1 11

1 1

2

( 270 kips)(16 ft)(12 in./ft)

(17.6 in. )(29,000 ksi)

0.101567 in. 0.1016 in.

B

F Lu

A E

Ans.

(b) Draw a FBD that cuts through member (2) and includes the free

end of the column. From this FBD, the equilibrium equation is

2115 kips 0xF F


2 115 kips 115 kips (C)F

The deformation in member (2) (which will be contraction in this

instance) can be computed as

2 22 2

2 2

( 115 kips)(14 ft)(12 in./ft)0.037853 in.

(17.6 in. )(29,000 ksi)

F L

A E

The settlement of the roof is found from the sum of the contractions in the two members:

1 2 0.101567 in. ( 0.037853 0.139420 in. 0.1394 in.in .)Cu Ans.




5.7 Aluminum [E = 70 GPa] member ABC supports a

load of 28 kN, as shown in Fig. P5.7. Determine:

(a) the value of load P such that the deflection of joint C

is zero.

(b) the corresponding deflection of joint B.

Fig. P5.7

Solution

Cut a FBD that exposes the internal axial force in member (2):

2 228 kN 0 28 kNyF F F

Similarly, cut a FBD that exposes the internal axial force in

member (1):

1 128 kN 0 28 kNyF P F F P

The deflection at joint C, which must ultimately equal zero, can be

expressed in terms of the member deformations 1 and 2:

1 1 2 21 2

1 1 2 2

0C

F L F Lu

A E A E

or

1 1 2 2

1 1 2 2

2 2 2 2

(28,000 N )(1,000 mm) (28,000 N)(1,300 mm)0

(50 mm) (70,000 N/mm ) (32 mm) (70,000 N/mm )4 4

C

F L F Lu

A E A E

P

The only unknown in this equation is P, which can be computed as

80.5841 k 80. kNN 6P Ans.

The corresponding deflection at joint B can be found from the deformation in member (1):

1 11

2 21 1

(28,000 N 80,584 N)(1,300 mm)

(50 mm) (70,000 N/m

0.497

m

mm 0.497 mm

)4

B

F Lu

A E

Ans.




5.8 A solid brass [E = 100 GPa] axial member is loaded and

supported as shown in Fig. P5.8. Segments (1) and (2) each have a

diameter of 25 mm and segment (3) has a diameter of 14 mm.

Determine:

(a) the deformation of segment (2).

(b) the deflection of joint D with respect to the fixed support at A.

(c) the maximum normal stress in the entire axial member.

Fig. P5.8

Solution

(a) Draw a FBD that cuts through segment (2) and includes the free

end of the axial member. From this FBD, the sum of forces in the

vertical direction reveals the internal force in the segment:

2 214 kN 28 kN 0 42 kN (T)yF F F

The cross-sectional area of the 25-mm-diameter segment is

2 2

2 (25 mm) 490.8739 mm4

A

The deformation in segment (2) is thus

2 22 2 2

2 2

(42,000 N)(1,200 mm)

(490.8739 mm )(100,000 N/mm )

1.026740 1.027 mmmm

F L

A E

Ans.

(b) The internal forces in segments (1) and (3) must be determined at

the outset. From a FBD that cuts through segment (1) and includes

the free end of the axial member:

1 140 kN 14 kN 28 kN 0 82 kN (T)yF F F

The deformation in segment (1) can be computed as

1 11 2 2

1 1

(82,000 N)(1,800 mm)3.006883 mm

(490.8739 mm )(100,000 N/mm )

F L

A E




Similarly, consider a FBD that cuts through segment (3) and includes the free end of

the axial member:

3 328 kN 0 28 kN (T)yF F F

The cross-sectional area of the 14-mm-diameter segment is

2 2

3 (164 mm) 153.9380 mm4

A


3 33 2 2

3 3

(28,000 N)(1,600 mm)2.810262 mm

(153.9380 mm )(100,000 N/mm )

F L

A E

The deflection of joint D with respect to the fixed support at A is found from the sum of the three

segment deformations:

1 2 3

3.006883 mm 1.026740 mm 2.810262 mm

6.943885 6.94 mm mm

Du

Ans.

(c) Since segments (1) and (2) have the same cross-sectional area, the maximum normal stress in these

two segments occurs where the axial force is greater; that is, in segment (1):

11 2

1

82,000 N167.0490 MPa (T)

490.8739 mm

F

A

The normal stress in segment (3) is

33 2

3

28,000 N181.8914 MPa (T)

153.9380 mm

F

A

Therefore, the maximum normal stress in the axial member occurs in segment (3):

max 181.9 MPa (T) Ans.




5.9 A hollow steel [E = 30,000 ksi] tube (1) with

an outside diameter of 2.75 in. and a wall

thickness of 0.25 in. is fastened to a solid

aluminum [E = 10,000 ksi] rod (2) that has a 2-

in.-diameter and a solid 1.375-in.-diameter

aluminum rod (3). The bar is loaded as shown in

Fig. P5.9. Determine:

(a) the change in length of steel tube (1).

(b) the deflection of joint D with respect to the

fixed support at A.

(c) the maximum normal stress in the entire axial

assembly.

Fig. P5.9

Solution

Before proceeding, it is convenient to determine the internal forces in each of the three axial segments.

Segment (1): Draw a FBD that cuts through

segment (1) and includes the free end of the axial

member. From this FBD, the sum of forces in the

horizontal direction gives the force in segment (1):

1

1

2(34 kips) 2(18 kips) 25 kips 0

57 kips

xF F

F





2

2

2(18 kips) 25 kips 0

11 kips

xF F

F





3

3

25 kips 0

25 kips

xF F

F

(a) The cross-sectional area of the hollow steel tube (inside diameter = 2.25 in.) is

2 2 2

1 (2.75 in.) (2.25 in.) 1.963495 in.4

A

The deformation in segment (1) is thus

1 11 2

1 1

( 57 kips)(60 in.)0.058060 in.

(1.963495 in. )0.0581

(30 in

,000 ks.

i)

F L

A E

Ans.

(b) For segment (2), the cross-sectional area of the 2-in.-diameter aluminum rod is

2 2

2 (2 in.) 3.141593 in.4

A


2 22 2

2 2

(11 kips)(40 in.)0.014006 in.

(3.141593 in. )(10,000 ksi)

F L

A E




For segment (3), the cross-sectional area of the 1.375-in.-diameter aluminum rod is

2 2

3 (1.375 in.) 1.484893 in.4

A


3 33 2

3 3

( 25 kips)(30 in.)0.050509 in.

(1.484893 in. )(10,000 ksi)

F L

A E

The deflection of joint D with respect to the fixed support at A is found from the sum of the three

segment deformations:

1 2 3 0.058060 in. 0.014006 in. 0.050509 in.

0.094563 0.0946 iin. n. 0.0946 in.

Du

Ans.

(c) Compute the normal stress in each of the three segments:

11 2

1

22 2

2

33 2

3

57 kips29.0299 ksi (C)

1.963495 in.

11 kips3.5014 ksi (T)

3.141593 in.

25 kips16.8362 ksi (C)

1.484893 in.

F

A

F

A

F

A

Therefore, the maximum normal stress in the axial member occurs in segment (1):

max 29.0 ksi (C) Ans.




5.10 A solid 5/8-in. steel [E = 29,000 ksi] rod

(1) supports beam AB, as shown in Fig. P5.10.

If the stress in the rod must not exceed 30 ksi

and the maximum deformation in the rod must

not exceed 0.25 in., determine the maximum

load P that may be supported.

Fig. P5.10

Solution

The area of the 5/8-in.-diameter rod is

2 2

1 (0.625 in.) 0.306796 in.4

A

If the stress in the rod must not exceed 30 ksi, then the maximum force that can be applied to rod (1) is

2

1 allow,1 1 (30 ksi)(0.306796 in. ) 9.203880 kipsF A

The length of rod (1) is

2 2

1 (20 ft) (16 ft) 25.612497 ft 307.35 in.L

If the maximum deformation in the rod must not exceed 0.25 in., the maximum internal force in the rod

must be limited to

2

1 1 11

1

(0.25 in.)(0.306796 in. )(29,000 ksi)7.236932 kips

307.35 in.

A EF

L

Therefore, the maximum internal force in rod (1) must not exceed 7.237 kips.

Consider a FBD of the rigid beam. Rod (1) is a two-

force member at an orientation defined by:

16 ft

tan 0.8 38.66020 ft

Write the equilibrium equation for the sum of moments

about A to find the relationship between F1 and P:

1(20 ft)( sin38.660 ) (8 ft) 0

(20 ft)(7.236932 kips)sin38.66011. 11302236 kips

8 ft.30 kips

AM F P

P

Ans.




5.11 A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and = 0.320 lb/in3] hangs

vertically while suspended from one end. Determine the change in length of the bar due to its own

weight.

Solution

An incremental length dy of the bar has an incremental deformation given by

( )F y

d dyAE

The force in the bar can be expressed as the product of the unit density of the bronze (bronze) and the

volume of the bar below the incremental slice dy:

bronzeF y A y

Therefore, the incremental deformation can be expressed as

bronze bronzeA y yd dy dy

AE E

Integrate this expression over the entire length L of the bar:

2

2bronze bronze bronze bronze

00 0

1

2 2

L L Ly Ldy y dy y

E E E E

The change in length of the bar due to its own weight is therefore:

2 3 2

bronze (0.320 lb/in. )(16 ft 12 in./ft)0.000393216 in.

2 2(10.000393 in

5,000,000 ps.

i)

L

E

Ans.




5.12 A homogenous rod of length L and elastic modulus E is a truncated cone with diameter that varies

linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of

the rod, as shown in Fig. P5.12. Assume that the taper of the cone is slight enough for the assumption of

a uniform axial stress distribution over a cross section to be valid.

(a) Determine an expression for the stress distribution on an arbitrary cross section at x.

(b) Determine an expression for the elongation of the rod.

P5.12

Solution

(a) Determine an expression for the diameter of the

truncated cone through the use of the similar triangles

concept:

0 0 0

0

2 ( )

( ) 1

d d d x d

L x

xd x d

L

(a)

Next, consider equilibrium of the truncated cone:

( ) 0 ( )xF F x P F x P

The internal force F(x) can be expressed in terms of stress and area:

( ) ( ) ( )A

F x dA x A x

Equating this expression to the equilibrium equation gives:

( ) ( ) ( )

( )( )

F x x A x P

Px

A x

From Eq. (a), the area of the cone at any location x is expressed as

2

2

2 0 1( )

( )4 4

xd

d x LA x

Therefore, the stress can be written as

2

2

0

4

1

( )( )

P

xd

L

Px

A x

Ans.




(b) The elongation in the axial member is found from Eq. (5.5):

0 0

( )

( )

L L F xd dx

A x E

For the truncated cone axial member:

2 220 00

2

0

2

0

2

0

0

4 4

1 1

4 2

1

L L

L

P P dxdx

Edx xEd

L L

P L

x

L

d EE

L

P

d

Ans.




5.13 Determine the extension, due to its own weight, of

the conical bar shown in Fig. P5.13. The bar is made of

aluminum alloy [E = 10,600 ksi and = 0.100 lb/in.3].

The bar has a 2-in. radius at its upper end and a length of

L = 20 ft. Assume the taper of the bar is slight enough for

the assumption of a uniform axial stress distribution over

a cross section to be valid.

Fig. P5.13

Solution

An incremental length dy of the bar has an incremental deformation given by

( )F y

d dyAE

The force in the bar can be expressed as the product of the unit density of the aluminum (alum) and the

volume of the bar below the incremental slice dy. The volume below the slice dy is a cone. At y, the

cross-sectional area of the base of the cone is:

2

y

yA r

L

and the volume of a cone is given by:

1

(area of base)(altitude)3

V

The internal force at y can be expressed as

alum

1( )

3yF y A y

The incremental deformation can be expressed as

alum alum

3 3

y

y

A y yd dy dy

A E E

Integrate this expression over the entire length L of the bar:

2

2alum alum alum alum

00 0

1

3 3 3 2 6

L L Ly Ldy y dx y

E E E E

The change in length of the bar due to its own weight is therefore:

2 3

62

5alum (0.100 lb/in. )(20 ft 12 in./ft)9.0566 10 in.

6 6(90.6 10 in

10,600,000 psi).

L

E

Ans.




5.14 Rigid bar ABCD is loaded and supported as

shown in Fig. P5.14. Steel [E = 30,000 ksi] bars

(1) and (2) are unstressed before the load P is

applied. Bar (1) has a cross-sectional area of

0.75 in.2 and bar (2) has a cross-sectional area of

0.425 in.2. After load P is applied, the strain in

bar (1) is found to be 780 . Determine:

(a) the stresses in bars (1) and (2).

(b) the vertical deflection of point D.

(c) the load P.

Fig. P5.14

Solution

From the strain in bar (1), the deformation in bar (1) is

6

1 1 1 (780 10 in./in.)(60 in.) 0.0468 in.L

Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of

bar (1):

1 0.0468 in.Bv

From a deformation diagram of the rigid bar,

the vertical deflection of joint C is related to B

by similar triangles:

40 in. 80 in.

80 in.2

40 in.

2(0.0468 in.) 0.0936 in.

B C

C B B

v v

v v v

The joint at C is also a perfect connection; therefore, the downward displacement of C also causes an

equal deformation in bar (2):

2 0.0936 in.Cv

(a) Now that the deformations in both bars are known, the normal stresses in each can be computed.


1 1 1 1 1 11 1

1 1 1 1

(0.0468 in.)(30,000 ksi)

60 23.4 ks T

i .i ( )

n

F L L E

A E E L Ans.

and the normal stress in bar (2) is

2 2 2 2 2 22 2

2 2 2 2

(0.0936 in.)(30,000 ksi)

90 31.2 ks T

i .i ( )

n

F L L E

A E E L Ans.

Alternatively, the stress in (1) could be calculated from Hooke’s Law:

6

1 1 1 (30,000 ksi)(780 10 in./in.) 23.4 ksi (T)E




(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C


40 in. 80 in. 110 in.

110 in.2.75(0.0 0.1287 in.468 in.)

40 in.

B C D

D B

v v v

v v Ans.

(c) The force in each bar can be determined from the stresses computed previously:

2

1 1 1

2

2 2 2

(23.4 ksi)(0.75 in. ) 17.55 kips

(31.2 ksi)(0.425 in. ) 13.26 kips

F A

F A

Consider a FBD of the rigid bar. The

equilibrium equation for the sum of moments

about A can be used to determine the load P:

1 2(40 in.) (80 in.) (110 in.) 0

(40 in.)(17.55 kips) (80 in.)(13.26 kips)16.0255 kips

110 in.16.03 kips

AM F F P

P Ans.





shown in Fig. P5.15. Bars (1) and (2) are

unstressed before the load P is applied. Bar (1) is

made of bronze [E = 100 GPa] and has a cross-

sectional area of 520 mm2. Bar (2) is made of

aluminum [E = 70 GPa] and has a cross-sectional

area of 960 mm2. After the load P is applied, the

force in bar (2) is found to be 25 kN (in tension).

Determine:


(b) the vertical deflection of point A.

(c) the load P.

Fig. P5.15

Solution

Given that the axial force in bar (2) is 25 kN (in tension), the deformation can be computed as:

2 22 2 2

2 2

(25,000 N)(800 mm)0.297619 mm

(960 mm )(70,000 N/mm )

F L

A E

Since the pin at C is a perfect connection, the deflection of the rigid bar at C is equal to the deformation

of bar (2):

2 0.297619 mmCv

From a deformation diagram of the rigid bar, the

vertical deflection of joint C is related to B by

similar triangles:

1.6 m 0.5 m

1.6 m3.2

0.5 m

3.2(0.297619 mm) 1.116071 mm

B C

B C C

v v

v v v

The joint at B is also a perfect connection; therefore, the downward displacement of B also causes an

equal contraction in bar (1):

1 1.116071 mmBv

Note that a downward displacement at B causes contraction (and hence, compression) in bar (1).

(a) The normal stress in aluminum bar (2) can be computed from the known force in the bar:

22 2

2

25,000 N26.041667 MPa

960 mm26.0 MPa (T)

F

A Ans.

The normal stress in bronze bar (1) can be computed from its contraction:

1 1 1 1

1

1 1 1

2

1 1

1

1

( 1.116071 mm)(100,000 N/mm )186.011905 MPa

600 m186.0 M

mPa (C)

F L L

A E E

E

L Ans.

(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A is related to joint B by

similar triangles:




2.0 m 0.5 m

2.0 m4 4(0.297619 mm) 1.488 m1.488095 mm

0.5 mm

A C

A C C

v v

v v v Ans.

(c) The force in bar (2) is given as F2 = 25 kN. The force in bar (1) can be determined from the stress

computed previously:

2 2

1 1 1 ( 186.011905 N/mm )(520 mm ) 96,726.1905 N 96.7262 kNF A



about D can be used to determine the load P:

1 2(1.6 m) (0.5 m) (2.0 m) 0

(0.5 m)(25 kN) (1.6 m)( 96.7262 kN)

2.0 m

77. 77.55446 kN kN

DM F F P

P

Ans.




5.16 In Fig. P5.16, aluminum [E = 70

GPa] links (1) and (2) support rigid beam

ABC. Link (1) has a cross-sectional area

of 300 mm2 and link (2) has a cross-

sectional area of 450 mm2. For an

applied load of P = 55 kN, determine the

rigid beam deflection at point B.

Fig. P5.16

Solution

From a FBD of the rigid beam, write two equilibrium

equations:

1 2 55 kN 0yF F F (a)

2(2,200 mm) (1,400 mm)(55 kN) 0AM F (b)

Solve Eq. (b) for F2:

2

(1,400 mm)(55 kN)35 kN

2,200 mmF

and backsubstitute into Eq. (a) to obtain F1:

1 255 kN 55 kN 35 kN 20 kNF F

Next, determine the deformations in links (1) and (2):

1 11 2 2

1 1

(20,000 N)(2,500 mm)2.380952 mm

(300 mm )(70,000 N/mm )

F L

A E

2 22 2 2

2 2

(35,000 N)(4,000 mm)4.444444 mm

(450 mm )(70,000 N/mm )

F L

A E

Since the connections at A and C are perfect, the rigid beam deflections at these joints are equal to the

deformations of links (1) and (2), respectively:

1 22.380952 mm and 4.444444 mmA Cv v

The deflection at B can be determined

from similar triangles:

2,200 mm 1,400 mm

C A B Av v v v

Solve this expression for vB:




1,400 mm( )

2,200 mm

1,400 mm(4.444444 mm 2.380952 mm) 2.380952 mm

2,200 mm

(0.636364)(2.053492 mm) 2.3

3.6

80952 mm

= 9 mm

B C A Av v v v

Ans.




5.17 Rigid bar ABC is supported by bronze rod (1) and aluminum

rod (2), as shown in Fig P5-17. A concentrated load P is applied

to the free end of aluminum rod (3). Bronze rod (1) has an elastic

modulus of E1 = 15,000 ksi and a diameter of d1 = 0.50 in.

Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and

a diameter of d2 = 0.75 in. Aluminum rod (3) has a diameter of d3

= 1.0 in. The yield strength of the bronze is 48 ksi and the yield

strength of the aluminum is 40 ksi.

(a) Determine the magnitude of load P that can safely be applied

to the structure if a minimum factor of safety of 1.67 is required.

(b) Determine the deflection of point D for the load determined

in part (a).

(c) The pin used at B has an ultimate shear strength of 54 ksi. If a

factor of safety of 3.0 is required for this double shear pin

connection, determine the minimum pin diameter that can be

used at B.

Fig. P5.17

Solution

Before beginning, the cross-sectional areas of the three rods can be calculated from the specified

diameters:

2 2 2

1 2 30.196350 in. 0.441786 in. 0.785398 in.A A A

The allowable stress of the bronze is

,bronze

allow,1

48 ksi28.742515 ksi

FS 1.67

Y

and the allowable stress of the aluminum is

,alum

allow,2 allow,3

40 ksi23.952096 ksi

FS 1.67

Y

(a) From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P.

From a FBD of the rigid bar, write two equilibrium equations:

1 2 3 1 2 0yF F F F F F P (a)

2 3 2(4 ft) (2.5 ft) (4 ft) (2.5 ft) 0AM F F F P (b)

Solve Eq. (b) for F2 in terms of the unknown load P:

2

2.5 ft0.6250

4 ftF P P (c)

Backsubstitute this expression into Eq. (a) to obtain F1 in terms of

the unknown load P:

1 2 0.6250 0.3750F P F P P P (d)

Rearrange Eqs. (c) and (d) to express P in terms of F2 and F1, respectively:

21.6000P F (e)

12.6667P F (f)




Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod

(1) is:

2

1 allow,1 1 (28.742515 ksi)(0.196350 in. ) 5.643579 kipsF A

Thus, from Eq. (f), the corresponding maximum load P is:

12.6667 2.6667(5.643579 kips) 15.049545 kipsP F (g)

Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for

the aluminum is:

2


which leads to a corresponding maximum load P from Eq. (e):

21.6000 1.6000(10.581711 kips) 16.930738 kipsP F (h)

Finally, we should also check the capacity of rod (3):

2


which means that

18.811931 kipsP (i)

From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is

max 15.049545 k 15.05ips kipsP Ans.

Based on this maximum load P, the internal forces in the three rods are:

1

2

3

0.3750 0.3750(15.049545 kips) 5.643579 kips

0.6250 0.6250(15.049545 kips) 9.405966 kips

15.049545 kips

F P

F P

F P

(b) Next, determine the deformations in rods (1), (2), and (3):

1 11 2

1 1

(5.643579 kips)(6 ft)(12 in./ft)0.137964 in.

(0.196350 in. )(15,000 ksi)

F L

A E

2 22 2

2 2

(9.405966 kips)(8 ft)(12 in./ft)0.204391 in.

(0.441786 in. )(10,000 ksi)

F L

A E

3 33 2

3 3

(15.049545 kips)(3 ft)(12 in./ft)0.068982 in.

(0.785398 in. )(10,000 ksi)

F L

A E

Since the connections at A and C are perfect, the rigid bar deflections at these joints are equal to the

deformations of rods (1) and (2), respectively:

1 20.137964 in. and 0.204391 in.A Cv v

The rigid bar deflection at B can be

determined from similar triangles:

4 ft 2.5 ft

C A B Av v v v




Solve this expression for vB:

2.5 ft( )

4 ft

2.5 ft(0.204391 in. 0.137964 in.) 0.137964 in.

4 ft

(0.6250)(0.066427 in.) 0.137964 in.

0.179481 in.

B C A Av v v v

The deflection of joint D is equal to the deflection of the rigid bar at B plus the deformation in rod (3):

3 0.179481 in. 0.068982 in. 0.248463 in. 0.248 in.D Bv v Ans.

(c) The pin at B has an allowable shear stress of

allow

54 ksi18 ksi

FS 3.0

U

The force tending to shear this pin is equal to the load P = 15.049545 kips. The shear area AV required

for this pin is thus

2

allow

15.049545 kips0.836086 in.

18 ksiV

VA

Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional

area of the pin:

2

pin pin2 24

VA A d

and so the minimum pin diameter is

2 2

pin pin2 0.836086 in. 0.729568 in.4

0.730 in.d d Ans.




5.18 Solve problem 5.14 when there is a

clearance of 0.05 in. in the pin connection at C.

5.14 (Repeated here for convenience). Rigid

bar ABCD is loaded and supported as shown in

Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2)

are unstressed before the load P is applied. Bar

(1) has a cross-sectional area of 0.75 in.2 and bar

(2) has a cross-sectional area of 0.425 in.2. After

load P is applied, the strain in bar (1) is found to

be 780 . Determine:


(b) the vertical deflection of point D.

(c) the load P.

Fig. P5.14 (repeated)

Solution

From the strain in bar (1), the deformation in bar (1) is

6

1 1 1 (780 10 in./in.)(60 in.) 0.0468 in.L

Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of

bar (1):

1 0.0468 in.Bv

From a deformation diagram of the rigid bar,

the vertical deflection of joint C is related to B


40 in. 80 in.

80 in.2

40 in.

2(0.0468 in.) 0.0936 in.

B C

C B B

v v

v v v

There is a 0.05-in. clearance in the connection at C; therefore,

2

2

0.05 in.

0.05 in. 0.0936 in. 0.05 in. 0.0436 in.

C

C

v

v

(a) Now that the deformations in both bars are known, the normal stresses in each can be computed.


1 1 1 1 1 11 1

1 1 1 1

(0.0468 in.)(30,000 ksi)

60 23.4 ks T

i .i ( )

n

F L L E

A E E L Ans.

and the normal stress in bar (2) is

2 2 2 2 2 22 2

2 2 2 2

(0.0436 in.)(30,000 ksi)14.5333 ksi

90 14.53 ksi

in.(T)

F L L E

A E E L Ans.




(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C


40 in. 80 in. 110 in.

110 in.2.75(0.0 0.1287 in.468 in.)

40 in.

B C D

D B

v v v

v v Ans.

(c) The force in each bar can be determined from the stresses computed previously:

2

1 1 1

2

2 2 2

(23.4 ksi)(0.75 in. ) 17.55 kips

(14.5333 ksi)(0.425 in. ) 6.1767 kips

F A

F A



about A can be used to determine the load P:

1 2(40 in.) (80 in.) (110 in.) 0

(40 in.)(17.55 kips) (80 in.)(6.1767 kips)10.8739 kips

110 in.10.87 kips

AM F F P

P Ans.




5.19 The rigid beam in Fig. P5.19 is supported by

links (1) and (2), which are made from a polymer

material [E = 16 GPa]. Link (1) has a cross-

sectional area of 400 mm2 and link (2) has a cross-

sectional area of 800 mm2. Determine the

maximum load P that may by applied if the

deflection of the rigid beam is not to exceed 20

mm at point C.

Fig. P5.19

Solution

Equilibrium: Consider a FBD of the rigid beam

and assume tension in each link.

1 2 0yF F F P (a)

2(600 mm) (900 mm) 0AM F P (b)

From Eq. (b):

2

900 mm1.5

600 mmF P P (c)

Backsubstituting into Eq. (a):

1 2 1.5 0.5F F P P P P (d)

Force-deformation relationships: The relationship between internal force and member deformation

for links (1) and (2) can be expressed as:

1 11

1 1

F L

A E and 2 2

2

2 2

F L

A E (e)

Geometry of deformations: Consider a deformation diagram of

the rigid beam. Using similar

triangles, one way to express the

relationships between vA, vB, and vC

is:

900 mm 300 mm

A C C Bv v v v (f)

Note: Here, vA, vB, and vC are

treated as unsigned magnitudes in

the directions shown on the

deformation diagram.

The rigid beam deflection at A will equal the deformation that occurs in link (1):

1Av

and similarly at B:




2Bv

Equation (f) can now be rewritten in terms of e1 and e2 as:

1 2

900 mm 300 mm

C Cv v

or

1 2 2

900 mm( ) 3 3

300 mmC C Cv v v

Solving for vC gives:

1 2 1 22 3 0.5 1.5C Cv v

Substitute the force-deformation relationships from Eq. (e) to obtain:

1 1 2 21 2

1 1 2 2

0.5 1.50.5 1.5C

F L F Lv

A E A E

and then substitute Eqs. (c) and (d) to derive an expression for vC in terms of the unknown load P:

1 2 1 2

1 1 2 2 1 1 2 2

0.5(0.5 ) 1.5(1.5 ) 0.25 2.25C

P L P L L Lv P

A E A E A E A E

The unknown load P is thus related to the rigid beam deflection at C by:

1 2

1 1 2 2

0.25 2.25Cv

PL L

A E A E

Substituting the appropriate values into this relationship gives the maximum load P that may be applied

to the rigid beam at C without causing more than 20 mm deflection:

2 2 2 2

20 mm

(0.25)(1,000 mm) (2.25)(1,250 mm)

(400 mm )(16,000 N/mm ) (800 mm )(16,000 N/mm )

77,283 N 77.3 kN

P

Ans.




5.20 Three aluminum [E = 10,000 ksi] bars are

used to support the loads shown in Fig. P5.20. The

elongation in each bar must be limited to 0.25 in.

Determine the minimum cross-sectional area

required for each bar.

Fig. P5.20

Solution

Determine the inclination angles for bars (1), (2), and (3):

1 1

2 2

3 1

7 fttan 34.992

10 ft

3 fttan 18.435

9 ft

8 fttan 57.995

5 ft

The bar lengths are:

2 2

1

2 2

2

2 2

3

(7 ft) (10 ft) 12.206556 ft 146.4787 in.

(3 ft) (9 ft) 9.486833 ft 113.8420 in.

(8 ft) (5 ft) 9.433981 ft 113.2078 in.

L

L

L

Equilibrium: Consider a FBD of joint B and write two equilibrium

equations:

1 2

1 2

cos34.992 cos18.435 0

sin34.992 sin18.435 31 kips 0

x

y

F F F

F F F

Solve these two equations simultaneously to obtain F1 = 36.61967 kips

and F2 = 31.62278 kips.

Next, consider a FBD of joint C and write two additional equilibrium

equations:

2 3

2 3

cos18.435 cos57.995 0

sin18.435 sin57.995 38 kips 0

x

y

F F F

F F F

From this, the internal force in bar (3) is F3 = 56.60389 kips.

The minimum cross-sectional area required for each bar is:

11

21

1 1

(36.61967 kips)(146.4787 in.)

(0.25 in.)(10,000 ks2.15 in.

i)

F LA

e E Ans.

22

22

2 2

(31.62278 kips)(113.8420 in.)

(0.25 in.)(10,000 ksi1.440 in.

)

F LA

e E Ans.




33

23

3 3

(56.60389 kips)(113.2078 in.)

(0.25 in.)(10,000 ks2.56 in.

i)

F LA

e E Ans.




5.21 A tie rod (1) and a pipe strut (2) are used to

support an 80-kip load as shown in Fig. P5.21. Pipe

strut (2) has an outside diameter of 6.625 in. and a

wall thickness of 0.280 in. Both the tie rod and the

pipe strut are made of structural steel with a modulus

of elasticity of E = 29,000 ksi and a yield strength of

Y = 36 ksi. For the tie rod, the minimum factor of

safety with respect to yield is 1.5 and the maximum

allowable axial elongation is 0.20 in.

(a) Determine the minimum diameter required to

satisfy both constraints for tie rod (1).

(b) Draw a deformation diagram showing the final

position of joint B. Fig. P5.21

Solution

The angles of inclination for members (1) and (2) are:

1 1

2 2

12 fttan 26.565

24 ft

30 fttan 51.340

24 ft

The member lengths are:

2 2

1

2 2

2

(12 ft) (24 ft) 26.8328 ft 321.9938 in.

(30 ft) (24 ft) 38.4187 ft 461.0249 in.

L

L

(a) Equilibrium: Consider a FBD of joint B and write two equilibrium

equations:

1 2

1 2

cos26.565 cos51.340 0

sin 26.565 sin51.340 80 kips 0

x

y

F F F

F F F

Solve these two equations simultaneously to obtain F1 = 51.1103 kips and

F2 = −73.1786 kips.

The allowable normal stress for tie rod (1) is

allow

36 ksi24 ksi

FS 1.5

Y

The minimum cross-sectional area required to satisfy the normal stress requirement is

21

allow

51.1103 kips2.1296 in.

24 ksi

FA

The maximum allowable axial deformation for the tie rod is 0.20 in. The minimum cross-sectional area

required to satisfy the deformation requirement is

21 1

1 1

(51.1103 kips)(321.9938 in.)2.8374 in.

(0.20 in.)(29,000 ksi)

F LA

E

To satisfy both the stress and deformation requirements, the tie rod must have a minimum cross-

sectional area of A = 2.8374 in.2. The corresponding rod diameter is




2 2

rod rod2.8374 in 1.901 in. .4

d d Ans.

(b) The pipe has a cross-sectional area of A2 = 5.5814 in.2. The pipe deformation is

2 22 2

2 2

( 73.1786 kips)(461.0249 in.)0.2084 in.

(5.5814 in. )(29,000 ksi)

F L

A E

Rod (1) elongates 0.20 in. and pipe (2) contracts

0.2084 in. Therefore, the deformation diagram

showing the final position of joint B is shown.




5.22 Two axial members are used to support a load of P = 72

kips as shown in Fig. P5.22. Member (1) is 12-ft long, it has a

cross-sectional area of A1 = 1.75 in.2, and it is made of structural

steel [E = 29,000 ksi]. Member (2) is 16-ft long, it has a cross-

sectional area of A2 = 4.50 in.2, and it is made of an aluminum

alloy [E = 10,000 ksi].

(a) Compute the normal stress in each axial member.

(b) Compute the deformation of each axial member.

(c) Draw a deformation diagram showing the final position of

joint B.

(d) Compute the horizontal and vertical displacements of joint B.

Fig. P5.22

Solution

(a) Consider a FBD of joint B and write two equilibrium equations:

1 2

2

cos55 0

sin55 72 kips 0

x

y

F F F

F F

Solve these two equations simultaneously to obtain F1 = 50.4149 kips

and F2 = 87.8958 kips.

The normal stress in member (1) is:

11 2

1

50.4149 kips

1.75 in.28.8 ksi

F

A Ans.

and the normal stress in member (2) is:

22 2

2

87.8958 kips

4.50 in.19.53 ksi

F

A Ans.

(b) The member deformations are

1 11 2

1 1

(50.4149 kips)(12 ft)(12 in./ft)

(1.75 in. )(29,000 ks0.1430 in.

i)

F L

A E Ans.

2 22 2

2 2

(87.8958 kips)(16 ft)(12 in./ft)

(4.50 in. )(10,000 ksi)0.375 in.

F L

A E Ans.




(c) Member (1) elongates 0.1430 in.

and member (2) elongates 0.375 in.

Therefore, the deformation diagram

showing the final position of joint B is

shown.

(d) The horizontal displacement of B is

0.1430 in.x Ans.

The vertical displacement is found from

0.375 in. 0.1430cos55sin55

0.375 in. 0.1430cos55 0.375 in. 0.0820 in.

sin55 sin0.558 i .

5n

5

y

y Ans.




5.23 The 200 × 200 × 1,200-mm oak [E = 12 GPa] block (2)

shown in Fig. P5.23 was reinforced by bolting two 6 × 200 × 1,200

mm steel [E = 200 GPa] plates (1) to opposite sides of the block. A

concentrated load of 360 kN is applied to a rigid cap. Determine:

(a) the normal stresses in the steel plates (1) and the oak block (2).

(b) the shortening of the block when the load is applied.

Fig. P5.23

Solution

Equilibrium

Consider a FBD of the rigid cap. Sum forces in the vertical direction to obtain:

1 22 360 kN 0yF F F (a)

Geometry of Deformations Relationship

For this configuration, the deformations of both members will be equal;

therefore,

1 2 (b)

Force-Deformation Relationships

The relationship between the internal force and the deformation of an axial member can be stated for

members (1) and (2):

1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)

Compatibility Equation

Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to

derive the compatibility equation:

1 1 2 2

1 1 2 2

F L F L

A E A E (d)

Solve the Equations

Solve Eq. (d) for F2, recognizing that steel plates (1) and oak block (2) have the same length:

1 2 2 1 2 2 2 22 1 1 1

1 1 2 2 1 1 1 1

L A E L A E A EF F F F

A E L L A E A E (e)

Substitute this expression into equilibrium equation (a) and solve for F1:

2 2 2 21 2 1 1 1

1 1 1 1

2 2 2 360 kNy

A E A EF F F F F F

A E A E (f)




For this structure, the areas and elastic moduli are given below:

2 2 2

1 2

1 2

(6 mm)(200 mm) 1,200 mm (200 mm) 40,000 mm

200 GPa 12 GPa

A A

E E

Substitute these values into Eq. (f) and calculate F1 = −90 kN. Backsubstitute into Eq. (e) to calculate

F2 = −180 kN.

Normal Stresses

The normal stresses in each axial member can now be calculated:

11 2

1

( 90 kN)(1,000 N/kN)75 MPa

175 MPa (C)

,200 mm

F

A Ans.

22 2

2

( 180 kN)(1,000 N/kN)4.50 MPa

40,4.50 M

000 a (

mP C)

m

F

A Ans.

(b) The shortening of the block is equal to the deformation (i.e., contraction in this instance) of member

(2):

2 22 2 2

2 2

( 180,000 N)(1,200 mm)0.450 mm

(40,000 mm )(12,000 N/mm )0.450 mmB

F Lu

A E Ans.




5.24 Two identical steel [E = 200 GPa] pipes, each with a

cross-sectional area of 1,475 mm2, are attached to

unyielding supports at the top and bottom, as shown in

Fig. P5.24. At flange B, a concentrated downward load of

120 kN is applied. Determine:

(a) the normal stresses in the upper and lower pipes.

(b) the deflection of flange B.

Fig. P5.24

Solution

(a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical

direction to obtain:

1 2 120 kN 0yF F F (a)

Geometry of Deformations:

1 2 0 (b)

Force-Deformation Relationships:

1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation:

1 1 2 2

1 1 2 2

0F L F L

A E A E (d)

Solve the Equations: Solve Eq. (d) for F2:

1 2 2 1 2 22 1 1

1 1 2 2 1 1

L A E L A EF F F

A E L L A E (e)

and substitute this expression into Eq. (a) to determine F1:

1 2 2 1 2 21 2 1 1 1

2 1 1 2 1 1

1 120 kNy

L A E L A EF F F F F F

L A E L A E (f)

Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as:

11 2 2

2 1 1

120 kN 120 kN 120 kN66.268657 kN

3.0 m 1.8108111 1+

3.7 m

FL A E

L A E

and from Eq. (e), F2 has a value of

1 2 22 1

2 1 1

3.0 m(66.268657 kN) 53.731343 kN

3.7 m

L A EF F

L A E




Normal Stresses: The normal stresses in each axial member can now be calculated:

11 2

1

66.268657 N44.927903 MPa

1,475 mm44.9 MPa (T)

F

A Ans.

22 2

2

53,731.343 N36.428029 MPa

1,475 m36.4 MPa (C)

m

F

A Ans.

(b) The deflection of flange B is equal to the deformation (i.e., contraction in this instance) of member

(2):

2 22 2 2

2 2

( 53.731343 N)(3,700 mm)0.673919 mm

(1,475 mm )(200,000 N/m0.674

mmm

)B

F Lu

A E Ans.




5.25 Solve Problem 5.24 if the lower support in Fig. P5.24 yields

and displaces downward 1.0 mm as the load P is applied.

5.24 (Repeated here for convenience) Two identical steel [E =

200 GPa] pipes, each with a cross-sectional area of 1,475 mm2,

are attached to unyielding supports at the top and bottom, as

shown in Fig. P5.24. At flange B, a concentrated downward load

of 120 kN is applied. Determine:

(a) the normal stresses in the upper and lower pipes.


Fig. P5.24 (repeated)

Solution

(a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical


1 2 120 kN 0yF F F (a)


1 2 1.0 mm (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

1.0 mmF L F L

A E A E (d)


1 1 2 2 2 2 1 2 22 1

1 1 2 2 2 1 1

(1.0 mm)1.0 mm

F L A E A E L A EF F

A E L L L A E (e)


2 2 1 2 21 2 1 1

2 2 1 1

(1.0 mm)120 kNy

A E L A EF F F F F

L L A E (f)




Simplify this expression, gathering terms with F1 on the left-hand side of the equation:

2 2 1 2 2

1 1

2 2 1 1

1 2 2 2 2

1

2 1 1 2

(1.0 mm)120 kN

(1.0 mm)1 120 kN

A E L A EF F

L L A E

L A E A EF

L A E L

Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as:

2 2

1

(1.0 mm)(1,475 mm )(200,000 N/mm )120,000 N

199,729.7297 N3,700 mm110,298.5075 N

3,000 mm 1.8108111

3,700 mm

F

and from Eq. (a), F2 has a value of

2 1 120 kN 110,298.507 N 120,000 N 9,701.493 NF F


11 2

1

110,298.507 N74.778649 MPa

1,475 mm74.8 MPa (T)

F

A Ans.

22 2

2

9,701.493 N6.577283 MPa

1,475 m6.58 MPa (C)

m

F

A Ans.

(b) The downward deflection of flange B is equal to the deformation of member (1):

1 11 2 2

1 1

(110,298.507 N)(3,000 mm)1.121680 mm

(1,475 mm )(200,000 N/mm1.122 mm

) B

F Lu

A E Ans.




5.26 A load P is supported by a structure consisting of

rigid bar ABC, two identical solid bronze [E = 15,000

ksi] rods, and a solid steel [E = 30,000 ksi] rod, as

shown in Fig. P5.26. The bronze rods (1) each have a

diameter of 0.75 in. and they are symmetrically

positioned relative to the center rod (2) and the

applied load P. Steel rod (2) has a diameter of 0.50 in.

If all bars are unstressed before the load P is applied,

determine the normal stresses in the bronze and steel

rods after a load of P = 20 kips is applied.

Fig. P5.26

Solution

Equilibrium

By virtue of symmetry, the forces in the two bronze rods (1) are

identical. Consider a FBD of the rigid bar. Sum forces in the

vertical direction to obtain:

1 22 0yF F F P (a)


For this configuration, the deformations of all rods will be equal;

therefore,

1 2 (b)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)




1 1 2 2

1 1 2 2

F L F L

A E A E (d)

Solve the Equations

Solve Eq. (d) for F2:

1 2 2 1 2 22 1 1

1 1 2 2 1 1

L A E L A EF F F

A E L L A E (e)


1 2 2 1 2 21 2 1 1 1

2 1 1 2 1 1

2 2 2y

L A E L A EF F F F F F P

L A E L A E (f)




For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below:

1 2

2 2 2 2

1 2

1 2

50 in. 90 in.

(0.75 in.) 0.441786 in. (0.50 in.) 0.196350 in.4 4

15,000 ksi 30,000 ksi

L L

A A

E E

Substitute these values into Eq. (f) and calculate F1 = 8.019802 kips. Backsubstitute into Eq. (e) to

calculate F2 = 3.960396 kips.

Normal Stresses


11 2

1

8.019802 kips

0.441786 in18. 5

.1 ksi

F

A Ans.

22 2

2

3.960396 kips

0.196350 i20.2 ks

.i

n

F

A Ans.




5.27 An aluminum alloy [E = 10,000 ksi] pipe

with a cross-sectional area of A1 = 4.50 in.2 is

connected at flange B to a steel [E = 30,000 ksi]

pipe with a cross-sectional area of A2 = 3.20 in.2.

The assembly (shown in Fig. P5.27) is connected

to rigid supports at A and C. For the loading

shown, determine:

(a) the normal stresses in aluminum pipe (1) and

steel pipe (2).


Fig. P5.27

Solution

(a) Equilibrium: Consider a FBD of flange B. Sum forces

in the horizontal direction to obtain:

1 2 90 kips 0xF F F (a)


1 2 0 (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

0F L F L

A E A E (d)


1 2 2 1 2 22 1 1

1 1 2 2 1 1

L A E L A EF F F

A E L L A E (e)


1 2 2 1 2 21 2 1 1 1

2 1 1 2 1 1

1 90 kipsx


L A E L A E (f)

F1 can be computed as:

1 21 2 2

22 1 1

90 kips 90 kips 90 kips35.273159 kips

2.551515160 in. 3.20 in. 30,000 ksi1 1+220 in. 4.50 in. 10,000 ksi

FL A E

L A E


2 1 90 kips 35.273159 kips 90 kips 54.726841 kipsF F


11 2

1

35.273159 kips7.838480 ksi

4.50 in7.84

.ksi (T)

F

A Ans.




22 2

2

54.726841 kips17.102138 ksi

3.20 in.17.10 ksi (C)

F

A Ans.

(b) The deflection of flange B is equal to the deformation of member (1):

1 11 2

1 1

(35.273159 kips)(160 in.)0.125416 in.

(4.50 in. )(100.1254 in

,000 ).

ksiB

F Lu

A E Ans.




5.28 The concrete [E = 29 GPa] pier shown in

Fig. P5.28 is reinforced using four steel [E =

200 GPa] reinforcing rods, each having a

diameter of 19 mm. If the pier is subjected to an

axial load of 670 kN, determine:

(a) the normal stress in the concrete and in the

steel reinforcing rods.

(b) the shortening of the pier.

Fig. P5.28

Solution

(a) Equilibrium: Consider a FBD of the pier, cut around the upper end.

The concrete will be designated member (1) and the reinforcing steel bars

will be designated member (2). Sum forces in the vertical direction to

obtain:

1 2 670 kN 0yF F F (a)


1 2 (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

F L F L

A E A E (d)


1 2 2 1 2 22 1 1

1 1 2 2 1 1

L A E L A EF F F

A E L L A E (e)


1 2 2 1 2 21 2 1 1 1

2 1 1 2 1 1

1 670 kNy


L A E L A E (f)

Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to

2 21

1 1

1 670 kNA E

FA E

(g)

The gross cross-sectional area of the pier is (250 mm)2 = 62,500 mm

2; however, the reinforcing bars

take up a portion of this area. The cross-sectional area of four 19-mm-diameter reinforcing bars is




2 2

2 (4 bars) (19 mm) 1,134.115 mm4

A

and thus the area of the concrete is

2 2 2

1 62,500 mm 1,134.115 mm 61,365.885 mmA

F1 can now be computed as:

2

1 2

1

1,134.115 mm 200 GPa1 670 kN

61,365.885 mm 29 GPa

670 kN594.258 kN

1.127457

F

F


2 1 670 kN ( 594.248 kN) 670 kN 75.742 kNF F

Normal Stresses: The normal stresses in each axial member can now be calculated. In the concrete, the

normal stress is:

11 2

1

594,258 N9.68 MPa

61,365.8859

mm.68 MPa (C)

F

A Ans.

and the normal stress in the reinforcing bars is

22 2

2

75,742 N66.8 MPa

1,134.115 66.8 MPa (C)

mm

F

A Ans.

(b) The shortening of the 1.5-m-long pier is equal to the deformation (i.e., contraction in this instance) of

member (1) or member (2):

1 11 2 2

1 1

( 594,258 N)(1,500 mm)0.501 mm

(61,365.885 mm )(29,000 N/mm )0.501 mm

F Lu

A E Ans.




5.29 The concrete [E = 29 GPa] pier

shown in Fig. P5.29 is reinforced using

four steel [E = 200 GPa] reinforcing rods.

If the pier is subjected to an axial force of

670 kN, determine the required diameter D

of each rod so that 20% of the total load is

carried by the steel.

Fig. P5.29

Solution

(a) Equilibrium: Consider a FBD of the pier, cut around the upper end. The

concrete will be designated member (1) and the reinforcing steel bars will be

designated member (2). Sum forces in the vertical direction to obtain:

1 2 670 kN 0yF F F (a)


1 2 (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

F L F L

A E A E (d)


2 1 1 2 1 11 2 2

2 2 1 1 2 2

L A E L A EF F F

A E L L A E (e)


2 1 1 2 1 21 2 2 2 2

1 2 2 1 2 2

1 670 kNy


L A E L A E (f)

Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to

1 12

2 2

1 670 kNA E

FA E

(g)

From the problem statement, 20% of the total load must be carried by the reinforcing steel; therefore,

2 0.20( 670 kN) 134 kNF




The gross cross-sectional area of the pier is Agross = (250 mm)2 = 62,500 mm

2; however, the reinforcing

bars take up a portion of this area. Therefore, the areas of the concrete (1) and steel (2) are related by:

gross 1 2A A A

and thus, the area of the concrete can be expressed as A1 = Agross − A2. Equation (g) can now be restated

and solved for the area of the reinforcing steel bars:

2

2

2

2

2

2

2

2

62,500 mm 29 GPa( 134 kN) 1 670 kN

200 GPa

62,500 mm27.586207

2,186.369 mm

A

A

A

A

A

Knowing the area required from four bars, the diameter of each bar can be computed:

2 2

bar bar(4 bars) 2,186.369 mm4

26.4 mmd d Ans.




5.30 A load of P = 100 kN is supported by a structure

consisting of rigid bar ABC, two identical solid bronze

[E = 100 GPa] rods, and a solid steel [E = 200 GPa]

rod, as shown in Fig. P5.30. The bronze rods (1) each

have a diameter of 20 mm and they are symmetrically

positioned relative to the center rod (2) and the applied

load P. Steel rod (2) has a diameter of 24 mm. All bars

are unstressed before the load P is applied; however,

there is a 3-mm clearance in the bolted connection at

B. Determine:

(a) the normal stresses in the bronze and steel rods.

(b) the downward deflection of rigid bar ABC.

Fig. P5.30

Solution

(a) Equilibrium: By virtue of symmetry, the forces in the

two bronze rods (1) are identical. Consider a FBD of the rigid

bar. Sum forces in the vertical direction to obtain:

1 22 0yF F F P (a)

Geometry of Deformations: For this configuration, the

deflections of joints A, B, and C are equal:

A B Cv v v (b)

The pin connections at A and C are ideal; therefore, the deflection of joints A and C will cause an

identical deformation of rods (1):

1Av (c)

The pin at B has a 3-mm clearance; thus, the deformation of rod (2) is related to rigid bar deflection vB

by:

2 3 mmBv (d)

Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation:

1 2 3 mm (e)

Force-Deformation Relationships: The relationship between the internal force and the deformation of

an axial member can be stated for members (1) and (2):

1 1 2 21 2

1 1 2 2

F L F L

A E A E (f)

Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of

deformation relationship (e) to derive the compatibility equation:

1 1 2 2

1 1 2 2

3 mmF L F L

A E A E (g)




Solve the Equations: Solve Eq. (g) for F1:

2 2 1 1 2 1 1 1 11 2

2 2 1 1 2 2 1

(3 mm)3 mm

F L A E L A E A EF F

A E L L A E L

Substitute this expression into Eq. (a)

2 1 1 1 11 2 2 2

1 2 2 1

(3 mm)2 2y

L A E A EF F F F F P

L A E L

and derive an expression for F2:

2 1 1 1 12

1 2 2 1

1 1

12

2 1 1

1 2 2

(3 mm)2 1 2

(3 mm)2

2 1

L A E A EF P

L A E L

A EP

LF

L A E

L A E

(h)

For this structure, P = 100 kN = 100,000 N, and the lengths, areas, and elastic moduli are given below:

1 2

2 2 2 2

1 2

1 2

3,000 mm 1,500 mm

(20 mm) 314.159265 mm (24 mm) 452.389342 mm4 4

100,000 MPa 200,000 MPa

L L

A A

E E

Substitute these values into Eq. (h) and calculate F2 = 27,588.728 N. Backsubstitute into Eq. (a) to

calculate F1 = 36,205.636 N.

Normal Stresses: The normal stresses in each rod can now be calculated:

11 2

1

36,205.636 N

314.1593 mm115.2 MPa

F

A Ans.

22 2

2

27,588.728 N

452.3893 m61. a

m0 MP

F

A Ans.

(b) The downward deflection of the rigid bar can be determined from the deformation of rods (1):

1 11 2

1 1

(36,205.636 N)(3,000 mm)

(314.1593 mm )(100,003.46 mm

0 MPa)A B C

F Lv v v

A E Ans.




5.31 Two steel [E = 30,000 ksi] pipes (1) and (2) are

connected at flange B, as shown in Fig. P5.31. Pipe (1)

has an outside diameter of 6.625 in. and a wall

thickness of 0.28 in. Pipe (2) has an outside diameter of

4.00 in. and a wall thickness of 0.226 in. If the normal

stress in each steel pipe must be limited to 18 ksi,

determine:

(a) the maximum downward load P that may be applied

at flange B.

(b) the deflection of flange B at the load determined in

part (a).

Fig. P5.31

Solution

(a) Pipe section properties: The pipe cross-sectional areas are:

1 1

2 2 2

1

2 2

2 2 2

2

6.625 in. 6.625 in. 2(0.28 in.) 6.0650 in.

(6.625 in.) (6.0650 in.) 5.5814 in.4

4.00 in. 4.00 in. 2(0.226 in.) 3.5480 in.

(4.00 in.) (3.5480 in.) 2.6795 in.4

D d

A

D d

A

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical


2 1 0yF F F P (a)


1 2 0 (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

0F L F L

A E A E (d)




For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1

and 2:

1 1 2 2

1 2

0L L

E E (e)

Solve the Equations: Solve Eq. (e) for 1:

2 11 2

1 2

L E

L E (f)

The elastic moduli of the two pipes are equal; therefore, E1/E2 = 1. The normal stresses in pipes (1) and

(2) are limited to 18 ksi. Assume that the normal stress in pipe (2) controls, meaning that 2 = 18 ksi.

Calculate the corresponding stress in pipe (1):

2 11 2

1 2

16 ft(18 ksi) 28.80 ksi

10 ft

L E

L E

Since this stress magnitude is greater than 18 ksi, our assumption is proven incorrect. We now know

that the normal stress in pipe (1) actually controls; thus, 1 = −18 ksi (negative by inspection). The

normal stress in pipe (1) can be found from Eq. (f)

1 22 1

2 1

10 ft( 18 ksi) 11.250 ksi

16 ft

L E

L E (f)

Now that the stresses are known, the allowable forces F1 and F2 can be computed:

2

1 1 1

2

2 2 2

( 18 ksi)(5.5814 in. ) 100.4652 kips

(11.250 ksi)(2.6795 in. ) 30.1444 kips

F A

F A

Substitute these values into Eq. (a) to obtain the allowable load P:

max 2 1 30.1444 kips ( 100.4652 kips) 130.6 kipsP F F Ans.

(b) The deflection of flange B is equal to the deformation of pipe (2):

2 22

2

(11.250 ksi)(16 ft)(12 in./ft)

30,000 ksi0.0720 in.B

Lu

E Ans.




5.32 A solid aluminum [E = 70 GPa] rod (1) is connected to a solid

bronze [E = 100 GPa] rod at flange B, as shown in Fig P5.32. Aluminum

rod (1) has an outside diameter of 35 mm and bronze rod (2) has an

outside diameter of 20 mm. The normal stress in the aluminum rod must

be limited to 160 MPa, and the normal stress in the bronze rod must be

limited to 110 MPa. Determine:

(a) the maximum downward load P that may be applied at flange B.

(b) the deflection of flange B at the load determined in part (a).

Fig. P5.32

Solution

(a) Rod section properties: The rod cross-sectional areas are:

2 2 2 2

1 2(35 mm) 962.1128 mm (20 mm) 314.1593 mm4 4

A A

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to

obtain:

2 1 0yF F F P (a)


1 2 0 (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

0F L F L

A E A E (d)

For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1

and 2:

1 1 2 2

1 2

0L L

E E (e)

Solve the Equations: Solve Eq. (e) for 1:

2 11 2

1 2

L E

L E (f)

The normal stress in aluminum rod (1) is limited to 160 MPa. We will assume that the aluminum rod

controls, and then, we will calculate the corresponding stress in bronze rod (2):




1 22 1

2 1

175 mm 100 GPa(160 MPa) 117.647 MPa

340 mm 70 GPa

L E

L E

This stress magnitude is greater than the 110 MPa allowable stress for the bronze rod; therefore, this

calculation shows that the stress in the bronze rod actually controls. If the stress in the bronze rod is

limited to 110 MPa, the normal stress in aluminum rod (1) can be found from Eq. (f)

2 11 2

1 2

340 mm 70 GPa(110 MPa) 149.600 MPa

175 mm 100 GPa

L E

L E (f)


2

1 1 1

2

2 2 2

( 149.600 MPa)(962.1128 mm ) 143,932.1 N 143.9321 kN

(110 MPa)(314.1593 mm ) 34,557.5 N 34.5575 kN

F A

F A


max 2 1 34.5575 kN ( 143.9321 kN) 178.5 kNP F F Ans.

(b) The deflection of flange B is equal to the deformation of rod (2):

2 22

2

(110 MPa)(340 mm)

100,000 MPa0.374 mmB

Lu

E Ans.




5.33 A pin-connected structure is supported and

loaded as shown in Fig. P5.33. Member ABCD

is rigid and is horizontal before the load P is

applied. Bars (1) and (2) are both made from

steel [E = 30,000 ksi] and both have a cross-

sectional area of 1.25 in.2. A concentrated load

of P = 25 kips acts on the structure at D.

Determine:

(a) the normal stresses in both bars (1) and (2).

(b) the downward deflection of point D on the

rigid bar.

Fig. P5.33

Solution

Equilibrium

Consider a FBD of the rigid bar. Assume tension forces in

members (1) and (2). A moment equation about pin A gives

the best information for this situation:

1 2(66 in.) (168 in.) (212 in.) 0AM F F P (a)


Draw a deformation diagram of the rigid bar. The deflections

of the rigid bar are related by similar triangles:

66 in. 168 in.

B Cv v (b)

Since there are no gaps, clearances, or other misfits at pins B

and C, the deformation of member (1) will equal the

deflection of the rigid bar at B and the deformation of member

(2) will equal the deflection of the rigid bar at C. Therefore,

Eq. (b) can be rewritten in terms of the member deformations

as:

1 2

66 in. 168 in. (c)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (d)


Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to





1 1 2 2

1 1 2 2

1 1

66 in. 168 in.

F L F L

A E A E (e)

Solve the Equations

Solve Eq. (e) for F1:

2 1 1 2 1 11 2 2

2 2 1 1 2 2

66 in. 66 in.

168 in. 168 in.

L A E L A EF F F

A E L L A E (f)


1 2

2 1 1

2 2

1 2 2

2 1 1

2

1 2 2

(66 in.) (168 in.) (212 in.) 0

66 in.(66 in.) (168 in.) (212 in.)

168 in.

66 in.(66 in.) (168 in.) (212 in.)

168 in.

AM F F P

L A EF F P

L A E

L A EF P

L A E (g)


1 2

2 2

1 2

1 2

80 in. 120 in.

1.25 in. 1.25 in.

30,000 ksi 30,000 ksi

L L

A A

E E

Substitute these values into Eq. (g) and calculate F2 = 25.617124 kips. Backsubstitute into Eq. (f) to

calculate F1 = 15.095805 kips.

Normal Stresses


11 2

1

15.095805 kips

1.25 in.12.08 ksi

F

A Ans.

22 2

2

25.617124 kips

1.25 in20. i

.5 ks

F

A Ans.

Deflections of the rigid bar

Calculate the deformation of one of the axial members, say member (1):

1 11 2

1 1

(15.095805 kips)(80 in.)0.032204 in.

(1.25 in. )(30,000 ksi)

F L

A E (h)

Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1);

therefore, vB = 1 = 0.032204 in. (downward). From similar triangles, the deflection of the rigid bar at D

is related to vB by:

66 in. 212 in.

B Dv v (i)

From Eq. (i), the deflection of the rigid bar at D is:

212 in. 212 in.

(0.032204 in.)66 in. 66 i

0.1034 i .n.

nD Bv v Ans.




5.34 A pin-connected structure is supported and

loaded as shown in Fig. P5.34. Member ABCD

is rigid and is horizontal before the load P is

applied. Bars (1) and (2) are both made from

steel [E = 30,000 ksi] and both have a cross-

sectional area of 1.25 in.2. If the normal stress in

each steel bar must be limited to 18 ksi,

determine the maximum load P that may be

applied to the rigid bar.

Fig. P5.34

Solution

Equilibrium


members (1) and (2). A moment equation about pin A

gives the best information for this situation:

1 2(66 in.) (168 in.) (212 in.) 0AM F F P (a)


Draw a deformation diagram of the rigid bar. The

deflections of the rigid bar are related by similar triangles:

66 in. 168 in.

B Cv v (b)

Since there are no gaps, clearances, or other misfits at pins

B and C, the deformation of member (1) will equal the

deflection of the rigid bar at B and the deformation of

member (2) will equal the deflection of the rigid bar at C.

Therefore, Eq. (b) can be rewritten in terms of the member

deformations as:

1 2

66 in. 168 in. (c)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (d)







1 1 2 2

1 1 2 2

1 1

66 in. 168 in.

F L F L

A E A E (e)

Solve the Equations

Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress:

1 1 2 2

1 2

66 in.

168 in.

L L

E E (f)

and solve for 2:

1 22 1

2 1

168 in.

66 in.

L E

L E

Since both bars have the same elastic modulus, E1/E2 = 1. Substitute 1 = 18 ksi and solve for 2:

1 22 1

2 1

168 in. 168 in. 80 in.(18 ksi) 30.54546 ksi 18 ksi N.G.

66 in. 66 in. 120 in.

L E

L E

It is now evident that the stress in bar (2) controls. Substitute 2 = 18 ksi and solve for 1:

2 11 2

1 2

66 in. 66 in. 120 in.(18 ksi) 10.60714 ksi 18 ksi OK

168 in. 168 in. 80 in.

L E

L E


2

1 1 1

2

2 2 2

(10.60714 ksi)(1.25 in. ) 13.25893 kips

(18 ksi)(1.25 in. ) 22.5 kips

F A

F A


max

(66 in.)(16.875 kips) (168 in.)(22.5 kips) (212 in.) 0

21.95797 kips 22.0 kips

AM P

P Ans.




5.35 The pin-connected structure shown in Fig.

P5.35 consists of a rigid beam ABCD and two

supporting bars. Bar (1) is an aluminum alloy [E =

70 GPa] with a cross-sectional area of A1 = 800

mm2. Bar (2) is a bronze alloy [E = 100 GPa] with

a cross-sectional area of A2 = 500 mm2. The

normal stress in the aluminum bar must be limited

to 70 MPa, and the normal stress in the bronze rod

must be limited to 90 MPa. Determine:

(a) the maximum downward load P that may be

applied at B.

(b) the deflection of the rigid beam at B.

Fig. P5.35

Solution

Equilibrium

Consider a FBD of the rigid bar. Assume tension forces

in members (1) and (2). A moment equation about pin D

gives the best information for this situation:

1 2(3.5 m) (1.2 m) (2.7 m) 0DM F F P (a)



deflections of the rigid bar are related by similar

triangles:

3.5 m 1.2 m

CAvv

(b)

Since there are no gaps, clearances, or other misfits at

pins A and C, the deformation of member (1) will equal

the deflection of the rigid bar at A and the deformation of


Therefore, Eq. (b) can be rewritten in terms of the

member deformations as:

1 2

3.5 m 1.2 m (c)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (d)







1 1 2 2

1 1 2 2

1 1

3.5 m 1.2 m

F L F L

A E A E (e)

Solve the Equations

Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress:

1 1 2 2 2 2

1 2 2

3.5 m2.9167

1.2 m

L L L

E E E (f)

and solve for 1:

2 11 2

1 2

2.9167L E

L E

Substitute 2 = 90 MPa and solve for 1:

2 11 2

1 2

3.3 m 70 GPa2.9167 2.9167(90 MPa) 209.1 MPa 70 MPa N.G.

2.9 m 100 GPa

L E

L E

It is now evident that the stress in bar (1) controls. Substitute 1 = 70 MPa and solve for 2:

1 22 1

2 1

1 1 2.9 m 100 GPa(70 MPa) 30.13 MPa 90 MPa O.K.

2.9167 2.9167 3.3 m 70 GPa

L E

L E


2

1 1 1

2

2 2 2

(70 MPa)(800 mm ) 56,000 N 56.00 kN

(30.13 MPa)(500 mm ) 15,065 N 15.065 kN

F A

F A


max

(3.5 m)(56.00 kN) (1.2 m)(15.065 kN) (2.7 m)

7

0

79.288 kN 9.3 kN

DM P

P Ans.

(b) Deflections of the rigid bar


1 11 2

1 1

(56,000 N)(2,900 mm)2.9000 mm

(800 mm )(70,000 MPa)

F L

A E (g)

Since there are no gaps at pin A, the rigid bar deflection at

A is equal to the deformation of member (1); therefore, vA

= 1 = 2.9000 mm (downward). From similar triangles, the

deflection of the rigid bar at B is related to vA by:

3.5 m 2.7 m

A Bv v (h)

From Eq. (h), the deflection of the rigid bar at B is:

2.7 m 2.7 m

(2.9000 mm)3.5 m 3.5

2.24 m

mm

B Av v Ans.





P5.36 consists of a rigid beam ABCD and two

supporting bars. Bar (1) is an aluminum alloy [E =

70 GPa] with a cross-sectional area of A1 = 800

mm2. Bar (2) is a bronze alloy [E = 100 GPa] with

a cross-sectional area of A2 = 500 mm2. All bars

are unstressed before the load P is applied;

however, there is a 3-mm clearance in the pin

connection at A. If a load of P = 54 kN is applied

at B, determine:


(b) the normal strains in bars (1) and (2).

(c) determine the downward deflection of point A

on the rigid bar.

Fig. P5.36

Solution

Equilibrium


members (1) and (2). A moment equation about pin D gives


1 2(3.5 m) (1.2 m) (2.7 m) 0DM F F P (a)




3.5 m 1.2 m

CAvv

(b)

Since there is no clearance at pin C, the deformation of


However, there is a 3-mm clearance at A. Consequently, not

all of the rigid beam deflection at A will go toward elongating

bar (1). The relationship between rigid beam deflection and

axial member deformation can be expressed:

1 3 mmAv (c)

Equation (c) can be rewritten in terms of the member deformations as:

1 23 mm

3.5 m 1.2 m (d)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (e)





Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to


1 1 2 2 2 2

1 1 2 2 2 2

3.5 m3 mm 2.9167

1.2 m

F L F L F L

A E A E A E (f)

Solve the Equations

Solve Eq. (f) for F2:

2 2 1 1 1 2 2 2 22 1

2 1 1 2 1 1 2

(3 mm)3 mm

2.9167 2.9167 2.9167

A E F L L A E A EF F

L A E L A E L

and substitute into Eq. (a) to solve for F1:

1 2

1 2 2 2 21 1

2 1 1 2

1 2 2 2 21 1

2 1 1 2

1

(3.5 m) (1.2 m) (2.7 m)(54 kN)

(3 mm)(3.5 m) (1.2 m) (2.7 m)(54 kN)

2.9167 2.9167

(3 mm)(3.5 m) (1.2 m) (2.7 m)(54 kN) (1.2 m)

2.9167 2.9167

(2.7 m)(

F F

L A E A EF F

L A E L

L A E A EF F

L A E L

F

2 2

2

1 2 2

2 1 1

(3 mm)54 kN) (1.2 m)

2.9167

(3.5 m) (1.2 m)2.9167

A E

L

L A E

L A E

The value of F1 is thus calculated as:

2 2

1 2

2

(3 mm)(500 mm )(100,000 N/mm )(2.7 m)(54,000 N) (1.2 m)

2.9167(3,300 mm)

2.9 m 500 mm 100 GPa3.5 m (1.2 m)

2.9167(3.3 m) 800 mm 70 GPa

33,247.4 N 33.2474 kN

F

and backsubstituting into Eq. (a) gives F2:

12

(2.7 m)(54 kN) (3.5 m)

1.2 m

(2.7 m)(54 kN) (3.5 m)(33.2474 kN)

1.2 m

24.5285 kN

FF

(a) Normal stresses:

11 2

1

33,247.4 N

8041.6 MPa (T)

0 mm

F

A Ans.

22 2

2

24,528.5 N

5049.1 MPa (T)

0 mm

F

A Ans.




(b) Normal strains:

61

1

1

41.5592 MPa593.7 10 mm/mm

70,000 MPμε

a594

E Ans.

62

2

2

49.0570 MPa490.6 10 mm/mm

100,000 MPμε

a491

E Ans.

(c) Deflections of the rigid bar

Calculate the deformation of member (1):

1 11 2

1 1

(33,247.4 N)(2,900 mm)1.721739 mm

(800 mm )(70,000 MPa)

F L

A E

The relationship between the rigid bar deflection at A and

the deformation of member (1) was expressed in Eq. (c).

Therefore, the rigid bar deflection at A is:

1 3 mm

1.721739 mm 3 mm

4.72 mm

Av

Ans.





P5.37 consists of a rigid bar ABCD and two axial

members. Bar (1) is bronze [E = 100 GPa] with a

cross-sectional area of A1 = 620 mm2. Bar (2) is

an aluminum alloy [E = 70 GPa] with a cross-

sectional area of A2 = 340 mm2. A concentrated

load of P = 75 kN acts on the structure at D.

Determine:


(b) the downward deflection of point D on the

rigid bar.

Fig. P5.37

Solution

Equilibrium

Consider a FBD of the rigid bar. Assume tension

forces in members (1) and (2). A moment equation

about pin C gives the best information for this situation:

1 2(825 mm) (325 mm)

(1,100 mm) 0

CM F F

P (a)




triangles:

825 mm 325 mm 1,100 mm

A B Dv v v (b)

Since there are no gaps or clearances at either pin A or

pin B, the deformations of members (1) and (2) will

equal the deflections of the rigid bar at A and B,

respectively.

1 2

825 mm 325 mm (c)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (d)







1 1 2 2 2 2

1 1 2 2 2 2

825 mm2.538462

325 mm

F L F L F L

A E A E A E (e)

Solve the Equations


2 1 11 2

1 2 2

2.538462L A E

F FL A E

(f)


1 2

2 1 1

2 2

1 2 2

2

2 2

2

(825 mm) (325 mm) (1,100 mm)

(825 mm)(2.538462) (325 mm) (1,100 mm)

540 mm 620 mm 100 GPa(825 mm)(2.538462) 325 mm (1,100 mm)(75 kN)

900 mm 340 mm 70 GPa

22.9273 kN

F F P

L A EF F P

L A E

F

F

Backsubstitute this result into Eq. (f) to compute F1:

2

1 2

540 mm 620 mm 100 GPa(2.538462) (22.9273 kN)

900 mm 340 mm 70 GPa

90.9680 kN

F


11 2

1

90,968.0 N

62146.7 MPa (

0 mT)

m

F

A Ans.

22 2

2

22,927.3 N

3467.4 MPa (T)

0 mm

F

A Ans.



1 11 2

1 1

(90,968.0 N)(900 mm)1.3205 mm

(620 mm )(100,000 MPa)

F L

A E

Since the pin at A is assumed to have a perfect connection, vA = 1 = 1.3205 mm. From Eq. (b),

1,100 mm

1.333333(1.3205 mm)82

1.761mm

m5

m

D Av v Ans.





P5.38 consists of a rigid bar ABCD and two axial

members. Bar (1) is bronze [E = 100 GPa] with a

cross-sectional area of A1 = 620 mm2. Bar (2) is

an aluminum alloy [E = 70 GPa] with a cross-

sectional area of A2 = 340 mm2. All bars are

unstressed before the load P is applied; however,

there is a 3-mm clearance in the pin connection at

A. If a concentrated load of P = 90 kN acts on the

structure at D, determine:



(c) the downward deflection of point D on the

rigid bar.

Fig. P5.38

Solution

Equilibrium




1 2(825 mm) (325 mm)

(1,100 mm) 0

CM F F

P (a)




triangles:

825 mm 325 mm 1,100 mm

A B Dv v v (b)

Since there is no clearance at pin B, the deformation of

member (2) will equal the deflection of the rigid bar at

B. However, there is a 3-mm clearance in the

connection at A. Consequently, not all of the rigid bar

deflection at A will go toward elongating bar (1). The

relationship between rigid bar deflection and axial

member deformation can be expressed:

1 3 mmAv (c)

Equation (b) can be rewritten in terms of the member deformations as:

1 23 mm

825 mm 325 mm (d)







1 1 2 21 2

1 1 2 2

F L F L

A E A E (e)


Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to


1 1 2 2 2 2

1 1 2 2 2 2

825 mm3 mm 2.538462

325 mm

F L F L F L

A E A E A E (f)

Solve the Equations

Solve Eq. (f) for F2:

2 2 1 1 1 2 2 2 22 1

2 1 1 2 1 1 2

(3 mm)3 mm

2.538462 2.538462 2.538462

A E F L L A E A EF F

L A E L A E L


1 2

1 2 2 2 21 1

2 1 1 2

1 2 2 2 21 1

2 1 1 2

1

(825 mm) (325 mm) (1,100 mm)

(3 mm)(825 mm) (325 mm) (1,100 mm)

2.538462 2.538462

(3 mm)(825 mm) (325 mm) (1,100 mm) (325 mm)

2.538462 2.538462

(1

F F P

L A E A EF F P

L A E L

L A E A EF F P

L A E L

F

2 2

2

1 2 2

2 1 1

(3 mm),100 mm) (325 mm)

2.538462

825 mm (325 mm)2.538462

A EP

L

L A E

L A E

The value of F1 is thus calculated as:

2 2

1 2 2

2 2

(3 mm)(340 mm )(70,000 N/mm )(1,100 mm)(90,000 N) (325 mm)

2.538462(540 mm)

900 mm 340 mm 70,000 N/mm825 mm (325 mm)

2.538462(540 mm) 620 mm 100,000 N/mm

90,495.6 N 90.4956 kN

F

and backsubstituting into Eq. (a) gives F2:

12

(1,100 mm)(90 kN) (825 mm)

325 mm

(1,100 mm)(90 kN) (825 mm)(90.4956 kN)

325 mm

74.8957 kN

FF





11 2

1

90,495.6 N

62146.0 MPa (

0 mT)

m

F

A Ans.

22 2

2

74,895.7 N

3220 MPa (T)

40 mm

F

A Ans.

(b) Normal strains:

61

1 2 2

1 1

90,495.6 N1,459.6 10 mm/mm

(620 mm )(100,0001,460 μ

N/ε

mm )

F

A E Ans.

62

2 2 2

2 2

74,895.7 N3,146.9 10 mm/mm

(340 mm )(70,0003,150 μ

N/ε

mm )

F

A E Ans.

(c) Deflections of the rigid bar


2 22 2

2 2

(74,895.7 N)(540 mm)1.6993 mm

(340 mm )(70,000 MPa)

F L

A E

Since the pin at B is assumed to have a perfect connection, vB = 2 = 1.6993 mm. From Eq. (b),

1,100 mm

3.384615(1.6993 m 5.75 mm)3 m

m25 m

D Bv v Ans.




5.39 The assembly shown in Fig. P5.39 consists of

a solid aluminum alloy [E = 70 GPa] post (2)

surrounded by a bronze [E = 100 GPa] tube (1).

Before the load P is applied, there is a clearance of

2 mm between the post and the tube. The yield

stress for the aluminum post is 260 MPa and the

yield stress for the bronze tube is 340 MPa.

Determine:

(a) the maximum load P that may be applied to the

assembly without causing yielding of either the

post or the tube.

(b) the downward displacement of rigid cap B.

(c) the normal strain in the bronze tube.

Fig. P5.39

Solution

Section properties:

2 2 2

1 1 1

2 2

2 2

35 mm 29 mm (35 mm) (29 mm) 1,206.372 mm

15 mm (15 mm) 706.858 mm

R r A

R A

Equilibrium: Consider a FBD around rigid cap B after the gap has been closed.

Sum forces in the vertical direction to obtain:

1 2 0yF F F P (a)


2 1 2 mm (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)




2 2 1 1

2 2 1 1

2 mmF L F L

A E A E (d)

Solve the Equations

Since allowable stresses are specified, it is convenient to express Eq. (d) in terms of stress:

2 2 1 1

2 1

2 mmL L

E E (e)

and solve for 2:




1 2 22 1

2 1 2

1

(2 mm)

600 mm 70,000 MPa 70,000 MPa(2 mm)

602 mm 100,000 MPa 602 mm

L E E

L E L

Substitute 1 = 340 MPa and solve for 2:

2

70,000 MPa0.697674(340 MPa) (2 mm) 469.8 MPa 260 MPa N.G.

602 mm

It is now evident that the stress in post (2) controls. Substitute 2 = 260 MPa and solve for 1:

2 11 2

1 2

70,000 MPa(2 mm)

602 mm

70,000 MPa 602 mm 100,000 MPa260 MPa (2 mm)

602 mm 600 mm 70,000 MPa

39.333 MPa 340 MPa OK

L E

L E

(a) Maximum load: Now that the stresses are known, the allowable forces F1 and F2 can be computed:

2

1 1 1

2

2 2 2

(39.333 MPa)(1,206.372 mm ) 47,450 N 47.450 kN

(260 MPa)(706.858 mm ) 183,783 N 183.783 kN

F A

F A

Substitute these values into Eq. (a) to obtain the allowable load P. By inspection, the forces in the post

and the tube must be compression; therefore:

1 2

max

0

( 47.450 kN 231 ) ( 183.783 kN kN)

yF F F P

P Ans.

(b) Displacement of cap B: The contraction of tube (1) is:

1 11 2 2

1 1

( 47,451 N)(600 mm)0.2360 mm

(1,206.372 mm )(100,000 N/mm )

F L

A E

The displacement of cap B is 2 mm greater:

0.2360 mm 2 mm 2.2360 mm 2.24 mmBu Ans.

(c) Normal strain in tube (1): The strain in bronze tube (1) is:

11

1

39.333 MPa

100,000 MPa393 με

E Ans.




5.40 A 4.5-m-long aluminum tube (1) is to be connected to a 2.4-m-

long bronze pipe (2) at B. When put in place, however, a gap of 8

mm exists between the two members, as shown in Fig P5.40.

Aluminum tube (1) has an elastic modulus of 70 GPa and a cross-

sectional area of 2,000 mm2. Bronze pipe (2) has an elastic modulus

of 100 GPa and a cross-sectional area of 3,600 mm2. If bolts are

inserted in the flanges and tightened so that the gap at B is closed,

determine:

(a) the normal stresses produced in each of the members.

(b) the final position of flange B with respect to support A.

Fig. P5.40

Solution

Equilibrium: Consider a FBD of flange B after the bolts have been tightened

and the gap at B has been closed. Sum forces in the vertical direction to obtain:

1 2 1 20yF F F F F (a)


1 2 8 mm (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

8 mmF L F L

A E A E (d)

Solve the Equations: Substitute Eq. (a) into Eq. (d) and solve for F1:

1 2

1

1 1 2 2

1

2 2 2 2

8 mm

8 mm

4,500 mm 2,400 mm

(2,000 mm )(70,000 N/mm ) (3,600 mm )(100,000 N/mm )

206,135.0 N

L LF

A E A E

F

(f)


11 2

1

206,135.0 N103.0675 MPa

2,000 m103.1

m MPa (T)

F

A Ans.

22 2

2

206,135.0 N57.2597 MPa

3,600 mm57.3 MPa (T)

F

A Ans.




(b) The deflection of flange B is equal to the deformation of bronze pipe (2):

2 22 2 2

2 2

(206,135.0 N)(2,400 mm)

(3,600 mm )(100,000 N/mm1.374 mm

)

F L

A E Ans.

In its final position, flange B is located 2,401.374 mm away from support A.




5.41 The assembly shown in Fig. P5.41 consists of a steel [E1 =

30,000 ksi; A1 = 1.25 in.2] rod (1), a rigid bearing plate B that is

securely fastened to rod (1), and a bronze [E2 = 15,000 ksi; A2 =

3.75 in.2] post (2). The yield strengths of the steel and bronze are

62 ksi and 75 ksi, respectively. A clearance of 0.125 in. exists

between the bearing plate B and bronze post (2) before the

assembly is loaded. After a load of P = 65 kips is applied to the

bearing plate, determine:

(a) the normal stresses in bars (1) and (2).

(b) the factors of safety with respect to yield for each of the

members.

(c) the vertical displacement of bearing plate B.

Fig. P5.41

Solution

Deformation in rod (1) alone: Check to see if the bearing plate attached to rod (1) will contact the

bronze post for the 65-kip load.

1 11 2

1 1

(65 kips)(14 ft)(12 in./ft)0.2912 in. 0.125 in. gap contact will occur

(1.25 in. )(30,000 ksi)

F L

A E

This calculation proves that the bearing plate attached to rod (1) will contact the bronze post when the

65-kip load is applied; therefore, this structure must be analyzed as a statically indeterminate structure.

Equilibrium: Consider a FBD of flange B after the gap at B has been closed.

Sum forces in the vertical direction to obtain:

1 2 0yF F F P (a)


1 2 0.125 in. (b)


1 1 2 21 2

1 1 2 2

F L F L

A E A E (c)


1 1 2 2

1 1 2 2

0.125 in.F L F L

A E A E (d)


2 2 1 1 1 1 2 1 11 2

2 2 1 1 1 2 2

0.125 in. (0.125 in.)F L A E A E L A E

F FA E L L L A E

and substitute the resulting expression into Eq. (a) to determine an expression for F2:




1 1 2 1 12 2

1 1 2 2

2 1 1 1 12

1 2 2 1

1 1

12

2 1 1

1 2 2

(0.125 in.) 65 kips 0

1 (0.125 in.) 65 kips

(0.125 in.) 65 kips

1

A E L A EF F

L L A E

L A E A EF

L A E L

A E

LF

L A E

L A E

The axial force in post (2) is thus:

2

2 2

2

(1.25 in. )(30,000 ksi)(0.125 in.) 65 kips

(168 in.)32.4609 kips

36 in. 1.25 in. 30,000 ksi1

168 in. 3.75 in. 15,000 ksi

F

and the axial force in rod (1) is:

1 2 32.4609 kips 65 kips 32.5391 kipsF F P


11 2

1

32.5391 kips26.0313 ksi

1.25 in.26.0 ksi (T)

F

A Ans.

22 2

2

32.4609 kips8.6563 ksi

3.75 in8.6

.6 ksi (C)

F

A Ans.

(b) Factors of safety:

1 22.3862 ksi 75 ksi

FS FS26.0313 ksi 8.6563 ksi

8.66 Ans.

(c) Displacement of plate B: The displacement of plate B is equal to the deformation of rod (1).

1 1

1 2

1 1

(32.5391 kips)(168 in.)0.145775 in.

(1.25 in. )(30,00

0.1458 in

0 )

.

ksi

B

F L

A E

u

Ans.




5.42 A hollow steel [E = 30,000 ksi] tube (1) with


thickness of 0.216 in. is fastened to a solid 2-in.-

diameter aluminum [E = 10,000 ksi] rod. The

assembly is attached to unyielding supports at the

left and right ends and is loaded as shown in Fig.

P5.42. Determine:

(a) the stresses in all parts of the axial structure.

(b) the deflections of joints B and C.

Fig. P5.42

Solution

Section properties: The steel tube cross-sectional area is:

1 1

2 2 2

1

3.50 in. 3.50 in. 2(0.216 in.) 3.068 in.

(3.50 in.) (3.068 in.) 2.2285 in.4

D d

A

and the aluminum rod has a cross-sectional area of:

2 3

2 2

2 3

2.00 in.

(2.00 in.) 3.1416 in.4

D D

A A

Equilibrium: Consider a FBD of flange B. Sum forces in

the horizontal direction to obtain:

1 2 34 kips 0xF F F (a)

Consider a FBD of flange C. Sum forces in the horizontal


2 3 26 kips 0xF F F (b)


1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

F L F L F L

A E A E A E (d)

Compatibility Equation: Substitute Eqs. (d) into Eq. (c) to derive the compatibility equation:

3 31 1 2 2

1 1 2 2 3 3

0F LF L F L

A E A E A E (e)

Solve the Equations: Solve Eq. (a) for F1:

1 2 34 kipsF F (f)

Solve Eq. (b) for F3:

3 2 26 kipsF F (g)




Substitute Eqs. (f) and (g) into compatibility equation (e):

2 32 1 2 2

1 1 2 2 3 3

( 26 kips)( 34 kips)0

F LF L F L

A E A E A E

and expand terms:

2 3 32 1 1 2 2

1 1 1 1 2 2 3 3 3 3

(26 kips)(34 kips)0

F L LF L L F L

A E A E A E A E A E

Regroup terms:

3 31 2 12

1 1 2 2 3 3 1 1 3 3

(26 kips)(34 kips)L LL L LF

A E A E A E A E A E

and solve for F2:

31

1 1 3 32

31 2

1 1 2 2 3 3

2 2

2 2

(26 kips)(34 kips)

(34 kips)(48 in.) (26 kips)(60 in.)

(2.2285 in. )(30,000 ksi) (3.1416 in. )(10,000 ksi)

48 in. 60 in.

(2.2285 in. )(30,000 ksi) (3.1416 in. )(10,000 k

LL

A E A EF

LL L

A E A E A E

2

60 in.

si) (3.1416 in. )(10,000 ksi)

16.3227 kips

Backsubstitute into Eqs. (f) and (g) to obtain F1 and F3:

1 2

3 2

34 kips 16.3227 kips 34 kips 17.6773 kips

26 kips 16.3227 kips 26 kips 9.6773 kips

F F

F F

(a) Normal stresses: The normal stresses in each axial member can now be calculated:

11 2

1

17.6773 kips7.9325 ksi

2.2285 i7.93 ksi (C)

n.

F

A Ans.

22 2

2

16.3227 kips5.1957 ksi

3.1416 i5.20 ksi (

n.T)

F

A Ans.

33 2

3

9.6773 kips3.0834 ksi

3.1416 in3.0

.8 ksi (C)

F

A Ans.

(b) Joint deflections: The deflection of flange B is equal to the deformation (i.e., contraction in this

instance) of member (1):

1 11 2

1 1

( 17.6773 kips)(48 in.)0.012692 in.

(2.2285 0.01269 in

in. )(30,000 ksi).B

F Lu

A E Ans.

The deflection of flange C is equal to the sum of the deformations in members (1) and (2). The

deformation of member (2) is:




2 22 2

2 2

(16.3227 kips)(60 in.)0.031174 in.

(3.1416 in. )(10,000 ksi)

F L

A E

And thus, the deflection of flange C is:

1 2 0.012692 in. 0.031174 in. 0.018482 in 0.0184 in.. 8Cu Ans.




5.43 Rigid bar ABCD in Fig. P5.43 is supported by a

pin connection at A and by two axial bars (1) and (2).

Bar (1) is a 30-in.-long bronze [E = 15,000 ksi] bar

with a cross-sectional area of 1.25 in.2. Bar (2) is a

40-in.-long aluminum alloy [E = 10,000 ksi] bar with

a cross-sectional area of 2.00 in.2. Both bars are

unstressed before the load P is applied. If a

concentrated load of P = 27 kips is applied to the rigid

bar at D, determine:


(b) the deflection of the rigid bar at point D.

Fig. P5.43

Solution

Equilibrium


in members (1) and (2). A moment equation about pin

A gives the best information for this situation:

1 2(36 in.) (84 in.) (98 in.) 0AM F F P (a)




triangles:

36 in. 84 in. 98 in.

CB Dvv v

(b)

There are no gaps, clearances, or other misfits at the

pins in this structure. The vertical deflection of the

rigid bar at C will produce deformation in member (2);

however, and the vertical deflection of the rigid bar at B

will create contraction in member (1) (see the

discussion under the heading Structures with a

rotating rigid bar in the text and examine Fig. 5.11).

Therefore, Eq. (b) can be rewritten in terms of the


1 2

36 in. 84 in. (c)




1 1 2 21 2

1 1 2 2

F L F L

A E A E (d)







1 1 2 2

1 1 2 2

1 1

36 in. 84 in.

F L F L

A E A E (e)

Solve the Equations


2 1 1 2 1 11 2 2

2 2 1 1 2 2

36 in. 36 in.

84 in. 84 in.

L A E L A EF F F

A E L L A E (f)


1 2

2 1 12 2

1 2 2

2 1 12

1 2 2

(36 in.) (84 in.) (98 in.) 0

36 in.(36 in.) (84 in.) (98 in.)

84 in.

3(36 in.) (84 in.) (98 in.)

7

AM F F P

L A EF F P

L A E

L A EF P

L A E (g)


1 2

2 2

1 2

1 2

30 in. 40 in.

1.25 in. 2.00 in.

15,000 ksi 10,000 ksi

L L

A A

E E

Substitute these values into Eq. (g) and calculate F2 = 25.6183 kips. Backsubstitute into Eq. (f) to

calculate F1 = −13.7241 kips.

(a) Normal Stresses


11 2

1

13.7241 kips

1.10.98 ksi (C)

25 in.

F

A Ans.

22 2

2

25.6183 kips

2.12.81 ksi (

00 iT

n)

.

F

A Ans.



2 22 2

2 2

(25.6183 kips)(40 in.)0.051237 in.

(2.00 in. )(10,000 ksi)

F L

A E (h)

Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2);

therefore, vC = 2 = 0.051237 in. (downward). From similar triangles [Eq. (b)], the deflection of the

rigid bar at D is therefore:

98 in. 98 in.

(0.051237 in.)84 in. 84 in

0.0598 in..

D Cv v Ans.




5.44 A 25-mm-diameter by 3.5-m-long steel rod (1)

is stress free after being attached to rigid supports,

as shown in Fig. P5.44. At A, a 16-mm-diameter

bolt is used to connect the rod to the support.

Determine the normal stress in steel rod (1) and the

shear stress in bolt A after the temperature drops

60°C. Use E = 200 GPa and = 11.9 × 10−6

/°C.

Fig. P5.44

Solution

Section properties:

For the 25-mm-diameter rod, the cross-sectional area is:

2 2

1 (25 mm) 490.873852 mm4

A

Force-Temperature-Deformation Relationship

The relationship between internal force, temperature change, and deformation of an axial member is:

1 11 1 1

1 1

F LT L

A E

Since the rod is attached to rigid supports, 1 = 0.

1 11 1

1 1

0F L

T LA E

Thus, the force produced in the rod by the temperature drop is:

1 1

1 1 1 1 1 1

1

6 2 2(11.9 10 / C)( 60 C)(490.873852 mm )(200,000 N/mm )

70,096.786 N

A EF T L T A E

L

The normal stress in the steel rod is:

1 2

70,096.786 N

490.873142.8

852 MPa

mm (T) Ans.

The 16-mm-diameter bolt has a cross-sectional area of:

2 2

bolt (16 mm) 201.061930 mm4

A

Since the bolt is loaded in double shear, the shear stress in the bolt is

bolt 2

70,096.786 N

2(201.061930 mm174 MPa

).3 Ans.




5.45 A 0.875-in.-diameter by 15-ft-long steel rod

(1) is stress free after being attached to rigid

supports. A clevis-and-bolt connection, as shown in

Fig. P5.45, connects the rod with the support at A.

The normal stress in the steel rod must be limited to

18 ksi, and the shear stress in the bolt must be

limited to 42 ksi. Assume E = 29,000 ksi and

= 6.6 × 10−6

/°F and determine:

(a) the temperature decrease that can be safely

accommodated by rod (1) based on the allowable

normal stress.

(b) the minimum required diameter for the bolt at A

using the temperature decrease found in part (a).

Fig. P5.45

Solution

Section properties:

For the 0.875-in.-diameter rod, the cross-sectional area is:

2 2

1 (0.875 in.) 0.601320 in.4

A


The relationship between internal force, temperature change, and deformation of an axial member is:

1 11 1 1

1 1

F LT L

A E

Since the rod is attached to rigid supports, 1 = 0.

1 11 1

1 1

0F L

T LA E

which can also be expressed in terms of the rod normal stress:

11 1 1

1

0L

T LE

Solve for T corresponding to a 24 ksi normal stress in the steel rod:

1 1

1

1 1 1 1 1

6

1

18 ksi

(6.6 10 / F)(29,000 ksi)94.0 F

LT

E L E

Ans.

The normal force in the steel rod is:

2

1 (18 ksi)(0.601320 in. ) 10.823768 kipsF

If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear

bolt is

2 2

bolt

bolt

10.823768 kips2 0.257709 in.

4 42 k

0.405 in.

sid

d

Ans.




5.46 A steel [E = 29,000 ksi and = 6.6×10-6

/°F] rod containing a turnbuckle has its ends attached

to rigid walls. During the summer when the temperature is 82°F, the turnbuckle is tightened to

produce a stress in the rod of 5 ksi. Determine the stress in the rod in the winter when the

temperature is 10°F.

Solution

The normal strain in the rod can be expressed as:

TE

Since the rod is attached to rigid walls, the rod strain after the temperature change is = 0.

0TE

The change in temperature between the summer and winter is

winter summer 10 F 82 F 72 FT T T

Solve for the stress increase created by the 72°F drop in temperature.

6(6.6 10 / F)( 72 F)(29,000 ksi) 13.78 ksi (T)

T E

In the summer, the rod had a tension normal stress of 5 ksi; therefore, the rod stress in the winter is:

winter 5 ksi 13.78 18.78 ks ksi i (T) Ans.




5.47 A high-density polyethylene [E = 120 ksi and = 78 ×

10−6

/°F] block (1) is positioned in a fixture as shown in Fig.

P5.47. The block is 2-in. by 2-in. by 32-in.-long. At room

temperature, a gap of 0.10 in. exists between the block and

the rigid support at B. Determine:

(a) the normal stress in the block caused by a temperature

increase of 100°F.

(b) the normal strain in block (1) at the increased

temperature.

Fig. P5.47

Solution

If the polyethylene block were completely free to elongate, a temperature change of 100°F would cause

an elongation of

6

1 1 1 (78 10 / F)(100 F)(32 in.) 0.2496 in.T L

Since this elongation is greater than the 0.10-in. gap, the temperature change will cause the polyethylene

block to contact the support at B, which will create normal stress in the block.


The relationship between the internal force, temperature change, and the deformation of an axial

member is:

1 11 1 1 1

1 1

F LT L

A E

In this situation, the deformation of member (1) equals the 0.10-in. gap:

1 11 1 1

1 1

0.10 in.F L

T LA E

This relationship can be stated in terms of normal stress as

1 11 1 1

1

0.10 in.L

T LE

(a) Normal stress:

The normal stress in the block due to the 100°F temperature increase is:

1

1 1 1 1

1

6

0.10 in.

120 ksi0.10 in. (78 10 / F)(100 F)(32 in.) 0.561 ksi

32561 psi (C)

in.

ET L

L

Ans.

(b) Normal strain:

The normal strain in the polyethylene block is:

11

1

0.10 in.0.003125 in./i 3,125 μn.

32 in.ε

L

Ans.




5.48 The assembly shown in Fig. P5.48 consists of a brass shell

(1) fully bonded to a solid ceramic core (2). The brass shell [E

= 115 GPa, = 18.7 × 10−6

/°C] has an outside diameter of 50

mm and an inside diameter of 35 mm. The ceramic core [E =

290 GPa, = 3.1 × 10−6

/°C] has a diameter of 35 mm. At a

temperature of 15°C, the assembly is unstressed. Determine the

largest temperature increase that is acceptable for the assembly

if the normal stress in the longitudinal direction of the brass

shell must not exceed 80 MPa.

Fig. P5.48

Solution

Section properties: The cross-sectional areas of brass shell (1) and ceramic core (2) are:

2 2 2 2 2

1 2(50 mm) (35 mm) 1,001.3827 mm (35 mm) 962.1128 mm4 4

A A

Equilibrium

Consider a FBD cut through the assembly. Sum forces in the

horizontal direction to obtain:

1 2 2 10xF F F F F (a)


For this configuration, the deformations of both members will be equal; therefore,

1 2 (b)

Force-Temperature-Deformation Relationships

The relationship between internal force, temperature change, and deformation of an axial member can be

stated for members (1) and (2):

1 1 2 21 1 1 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)




1 1 2 21 1 2 2

1 1 2 2

F L F LT L T L

A E A E (d)

Solve the Equations

Since a limiting stress is specified for brass shell (1), express Eq. (d) in terms of normal stress:

1 21 1 1 2 2 2

1 2

L LT L T L

E E (e)

Based on Eq. (a), the normal stress 2 can be expressed in terms of 1 as:

2 1 1 1 12 1

2 2 2 1 2

F F F A A

A A A A A

Substitute this expression into Eq. (e) to obtain

1 1 21 1 1 1 2 2

1 2 2

L A LT L T L

E A E




Rearrange terms

1 1 21 2 2 1 1 2 2 1 1

1 2 2

( )L A L

T L T L L L TE A E

and solve for T, recognizing that both the shell and the core have the same length:

1 1 2 11 1

1 2 2 1 2 2

2 2 1 1 2 1

1 1L A L A

E A E E A ET

L L

(f)

Substitute the problem data into Eq. (f) to compute T that will produce a normal stress of 80 MPa in

the brass shell:

2

2

6 6

1 1,001.3827 mm 1( 80 MPa)

115,000 MPa 962.1128 mm 290,000 MPa

3.1 10 / C 18.7 10 / C

62.9983 C

T

Since the problem asks for the largest temperature increase,

max 63.0 CT Ans.




5.49 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown in Fig.

P5.49. Bar (1) is an aluminum alloy [E = 10,000 ksi, = 0.32, = 12.5 × 10−6

/°F] bar with a width of 3

in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi, = 0.12, = 9.6 × 10−6

/°F] bar

with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine:

(a) the lowest temperature at which the two bars

contact each other.

(b) the normal stress in the two bars at a


(c) the normal strain in the two bars at 250°F.

(d) the change in width of the aluminum bar at a


Fig. P5.49

Solution

(a) Lowest Contact Temperature

Before the gap is closed, only thermal strains and the associated axial elongations exist. Write

expressions for the temperature-induced elongations and set this equal to the 0.04-in. gap:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

0.04 in.

0.04 in.

0.04 in. 0.04 in.48.6381°F

(12.5 10 / °F)(32 in.) (9.6 10 / °F)(44 in.)

T L T L

T L L

TL L

Since the initial temperature is 60°F, the temperature at which the gap is closed is 108.6°F. Ans.

(b) Equilibrium

From the results obtained for part (a), we know that the gap will be closed at 250°F, making this a

statically indeterminate axial configuration. Knowing this, consider a FBD at joint B.

Assume that both internal axial forces will be tension, even

though we know intuitively that both F1 and F2 will turn out

to be compression.

1 2 0xF F F (a)


For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap:

1 2 0.04 in. (b)



member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)


Substitute the force-temperature-deformation relationships (c) into the geometry of deformation

relationship (b) to derive the compatibility equation:




1 1 2 21 1 1 2 2 2

1 1 2 2

0.04 in.F L F L

T L T LA E A E

(d)

Solve the Equations

This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:

1 1 1 21 1 1 2 2 2

1 1 2 2

1 1 1 21 1 1 2 2 2

1 1 2 2

1 21 1 1 1 2 2 2

1 1 2 2

1 1 1 2 2 21

1 2

1 1 2 2

0.04 in.

0.04 in.

0.04 in.

0.04 in.

F L F LT L T L

A E A E

F L F LT L T L

A E A E

L LF T L T L

A E A E

T L T LF

L L

A E A E

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the

same for both members; therefore, T1 = T2 = T = 190°F:

1 2

2 2

1 2

1 2

6 6

1 2

32 in. 44 in.

(3 in.)(0.75 in.) 2.25 in. (2 in.)(0.75 in.) 1.50 in.

10,000 ksi 28,000 ksi

12.5 10 / °F 9.6 10 / °F

L L

A A

E E

Substitute these values into Eq. (e) and calculate F1 = −47.0702 kips. Backsubstitute into Eq. (a) to

calculate F2 = −47.0702 kips. Note that the internal forces are compression, as expected.

Normal Stresses


1

1 2

1

2

2 2

2

20.9 k47.070

si (C)

31.4 k

2 kips20.9201 ksi

2.25 in.

47.0702 kips31.3801 ksi

1.50 isi

n. (C)

F

A

F

A

Ans.

(c) Normal Strains

The force-temperature-deformation relationships were expressed in Eq. (b). By definition, = /L.

Therefore, the normal strain for each axial member can be determined by dividing the relationships in

Eq. (c) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT T

A E A E (f)

Substitute the appropriate values to calculate the normal strains in each member:




11 1 1

1 1

6

2

47.0702 kips(12.5 10 / °F)(190 F)

(2.25 in. )(10,000 ksi)

0.000283 283 με in./in.

FT

A E

Ans.

22 2 2

2 2

6

2

47.0702 kips(9.6 10 / °F)(190 F)

(1.50 in. )(28,000 ksi)

0.000703 703 με in./in.

FT

A E

Ans.

(d) The change in width of the aluminum bar (1) is caused partly by the Poisson effect and partly by the

temperature change. The longitudinal strain in the aluminum bar caused by the internal force F1 is:

61

long, 2

1 1

47.0702 kips2,092.01 10 in./in.

(2.25 in. )(10,000 ksi)

F

A E

The accompanying lateral strain due to the Poisson effect is thus

6 6

lat, long, (0.32)( 2,092.01 10 in./in.) 669.44 10 in./in.

The lateral strain caused by the temperature change is

6 6

lat, 1 1 (12.5 10 /°F)(190 F) 2,375 10 in./in.T T

Therefore, the total lateral strain in aluminum bar (1) is

lat lat, lat,

6 6 6669.44 10 in./in. 2,375 10 in./in. 3,044.44 10 in./in.

T

The change is width of the aluminum bar is thus

6

latwidth (width) (3,044.44 10 in./in.)(3 0in.) .00913 in. Ans.




5.50 At a temperature of 5°C, a 3-mm gap exists

between two polymer bars and a rigid support, as

shown in Fig. P5.50. Bar (1) is 50-mm wide and

20-mm thick [E = 800 MPa, = 140 × 10−6

/°C].

Bar (2) is 75-mm wide and 25-mm thick [E = 2.7

GPa, = 67 × 10−6

/°C] bar. The supports at A and

C are rigid. Determine:

(a) the lowest temperature at which the 3-mm gap

is closed.

(b) the normal stress in the two bars at a

temperature of 60°C.

(c) the normal strain in the two bars at 60°C.

Fig. P5.50

Solution

(a) Before the gap is closed, only thermal strains and the associated axial elongations exist. Write

expressions for the temperature-induced elongations and set this equal to the 3-mm gap:

1 1 2 2

1 1 2 2

-6 -6

1 1 2 2

3 mm

3 mm

3 mm 3 mm24.038°C

(140 10 / °C)(700 mm) (67 10 / °C)(400 mm)

T L T L

T L L

TL L

Since the initial temperature is 5°C, the temperature at which the gap is closed is 29.038°C = 29.0°C. Ans.

Equilibrium

(b) From the results obtained for part (a), we know that the gap will be closed at 60°C, making this a

statically indeterminate axial configuration. Knowing this, consider a FBD at joint B.

Assume that both internal axial forces will be tension, even

though we know intuitively that both F1 and F2 will turn out

to be compression.

1 2 0xF F F (a)


For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap:

1 2 3 mm (b)



member can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)


Substitute the force-temperature-deformation relationships (c) into the geometry of deformation

relationship (b) to derive the compatibility equation:

1 1 2 21 1 1 2 2 2

1 1 2 2

3 mmF L F L

T L T LA E A E

(d)

Solve the Equations

This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives:




1 1 1 21 1 1 2 2 2

1 1 2 2

1 1 1 21 1 1 2 2 2

1 1 2 2

1 21 1 1 1 2 2 2

1 1 2 2

1 1 1 2 2 21

1 2

1 1 2 2

3 mm

3 mm

3 mm

3 mm

F L F LT L T L

A E A E

F L F LT L T L

A E A E

L LF T L T L

A E A E

T L T LF

L L

A E A E

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the

same for both members; therefore, T1 = T2 = T = 55°C:

1 2

2 2

1 2

1 2

6 6

1 2

700 mm 400 mm

(50 mm)(20 mm) 1,000 mm (75 mm)(25 mm) 1,875 mm

800 MPa 2,700 MPa

140 10 / °C 67 10 / °C

L L

A A

E E

Substitute these values into Eq. (e) and calculate F1 = −4.050262 kN. Backsubstitute into Eq. (a) to

calculate F2 = −4.050262 kN. Note that the internal forces are compression, as expected.

Normal Stresses


1

1 2

1

2

2 2

2

4.05 M( 4.050

Pa (C)

2.16 M

262 kN)(1,000 N/kN)4.050262 MPa

1,000 mm

( 4.050262 kN)(1,000 N/kN)2.160140 MPa

1,875Pa (C)

mm

F

A

F

A

Ans.

Normal Strains

The force-temperature-deformation relationships were expressed in Eq. (c). By definition, = /L.


Eq. (c) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT T

A E A E (f)


11 1 1

1 1

6

2 2

( 4.050262 kN)(1,000 N/kN)(140 10 / °C)(55 C)

(1,000 mm )(800 N/mm )

0.00263717 2,6 mm/m 0 μεm 4

FT

A E

Ans.




22 2 2

2 2

6

2 2

( 4.050262 kN)(1,000 N/kN)(67 10 / °C)(55 C)

(1,875 mm )(2,700 N/mm )

0.00288495 2,8 mm/m 80 μεm

FT

A E

Ans.




5.51 The axial assembly shown in Fig. P5.51

consists of a solid 1-in.-diameter aluminum alloy

[E = 10,000 ksi, = 0.32, = 12.5 × 10−6

/°F] rod

(1) and a solid 1.5-in.-diameter bronze [E =

15,000 ksi, = 0.15, = 9.4 × 10−-6

/°F] rod (2). If

the supports at A and C are rigid and the assembly

is stress free at 0°F, determine:

(a) the normal stress in both rods at 160°F.

(b) the displacement of flange B.

(c) the change in diameter of the aluminum rod.

Fig. P5.51

Solution

Equilibrium

Consider a FBD at joint B. Assume that both internal

axial forces will be tension.

1 2 1 20xF F F F F (a)


1 2 0 (b)


1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)


1 1 2 21 1 1 2 2 2

1 1 2 2

0F L F L

T L T LA E A E

(d)

Solve the Equations

From Eq. (a), F1 = F2. The temperature change is the same for both members; therefore, T1 = T2 =

T. Eq. (d) then can be written as:

1 1 1 21 1 2 2

1 1 2 2

0F L F L

T L T LA E A E

Solving for F1:

1 1 1 21 1 2 2

1 1 2 2

1 21 1 1 2 2

1 1 2 2

1 1 2 2

1

1 2

1 1 2 2

F L F LT L T L

A E A E

L LF T L L

A E A E

T L LF

L L

A E A E

(e)




For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given

below.

1 2

2 2 2 2

1 2

1 2

6 6

1 2

15 in. 22 in.

(1 in.) 0.7854 in. (1.5 in.) 1.7671 in.4 4

10,000 ksi 15,000 ksi

12.5 10 / °F 9.4 10 / °F

L L

A A

E E

Substitute these values along with T = +160°F into Eq. (e) and calculate F1 = −23.0263 kips. From Eq.

(a), F2 = F1 = −23.0263 kips.

(a) Normal Stresses

The normal stresses in each rod can now be calculated:

11 2

1

23.0263 kips29.318 ksi

0.7854 i29.3 ksi (C)

n.

F

A

Ans.

22 2

2

23.0263 kips13.030 ksi

1.7671 in13.03 ksi (C

.)

F

A

Ans.

(b) Displacement of Flange B

The displacement of flange B is equal to the deformation (i.e., contraction in this instance) of rod (1).

The deformation of rod (1) is given by:

1 1

1 1 1 1

1 1

6

2

( 23.0263 kips)(15 in.)(12.5 10 / °F)(160°F)(15 in.) 0.013977 in.

(0.7854 in. )(10,000 ksi)

F LT L

A E

The displacement of flange B is thus:

1 0.01398 in.Bu Ans.

(c) Change in diameter of the aluminum rod

The change in diameter of aluminum rod (1) is caused partly by the Poisson effect and partly by the

temperature change. The longitudinal strain in the aluminum rod caused by the internal force F1 is:

61

long, 2

1 1

23.0263 kips2,931.78 10 in./in.

(0.7854 in. )(10,000 ksi)

F

A E

The accompanying lateral strain due to the Poisson effect is thus

6 6

lat, long, (0.32)( 2,931.78 10 in./in.) 938.17 10 in./in.

The lateral strain caused by the temperature change is

6 6

lat, 1 1 (12.5 10 /°F)(160 F) 2,000 10 in./in.T T

Therefore, the total lateral strain in aluminum rod (1) is

lat lat, lat,

6 6 6938.17 10 in./in. 2,000 10 in./in. 2,938.17 10 in./in.

T

The change in diameter of the aluminum rod is thus

6

1 lat 1 (2,938.17 10 in./in.)(1 in.) 0.00294 in.d d Ans.





P5.52 consists of a rigid bar ABC, a solid bronze

[E = 100 GPa, = 16.9 × 10−6

/°C] rod (1), and a

solid aluminum alloy [E = 70 GPa, =

22.5×10−6

/°C] rod (2). Bronze rod (1) has a

diameter of 24 and aluminum rod (2) has a

diameter of 16 mm. The bars are unstressed

when the structure is assembled at 25°C. After

assembly, the temperature of rod (2) is

decreased by 40°C while the temperature of rod

(1) remains constant at 25°C. Determine the

normal stresses in both rods for this condition. Fig. P5.52

Solution

Equilibrium


members (1) and (2). A moment equation about pin B gives


1 2(200 mm) (350 mm) 0BM F F (a)




200 mm 350 mm

A Cv v (b)

There are no gaps, clearances, or other misfits at pins A and

C; therefore, Eq. (c) can be rewritten in terms of the member

deformations as:

1 2

200 mm 350 mm

(c)


1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)


1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

200 mm 350 mm

F L F LT L T L

A E A E

(e)




Solve the Equations

Solve Eq. (a) for F1:

1 2 2

350 mm1.75

200 mmF F F (f)

Substitute this result into Eq. (e):

2 1 2 21 1 1 2 2 2

1 1 2 2

1 ( 1.75 ) 1

200 mm 350 mm

F L F LT L T L

A E A E

Note that T1 = 0°C; therefore, the equation simplifies to:

2 1 2 22 2 2

1 1 2 2

1.75 200 mm

350 mm

F L F LT L

A E A E

Solve this equation for F2:

2 1 2 2

2 2 2

1 1 2 2

1 2

2 2 2 2

1 1 2 2

2 2 2

21 2

1 1 2 2

1.75 200 mm

350 mm

1.75 200 200

350 350

200

3501.75 200

350

F L F LT L

A E A E

L LF T L

A E A E

T L

FL L

A E A E

(g)

For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given

below.

1 2

2 2 2 2

1 2

1 2

6 6

1 2

250 mm 500 mm

(24 mm) 452.3893 mm (16 mm) 201.0619 mm4 4

100,000 MPa 70,000 MPa

16.9 10 / °C 22.5 10 / °C

L L

A A

E E

Substitute these values along with T2 = −40°C into Eq. (g) and calculate F2 = −8,579.65 N. From Eq.

(f), calculate F1 = −15,014.39 N.

Normal Stresses


11 2

1

15,014.39 N

452.3893 mm33.2 MPa (C)

F

A

Ans.

22 2

2

8,579.65 N

201.42.7 MPa (T

0619 mm)

F

A Ans.




5.53 Rigid bar ABC is supported by two identical solid bronze

[E = 100 GPa, = 16.9 × 10−6

/°C] rods, and a solid steel [E =

200 GPa, = 11.9×10−6

/°C] rod, as shown in Fig. P5.53. The

bronze rods (1) each have a diameter of 16 mm and they are

symmetrically positioned relative to the center rod (2) and the

applied load P. Steel rod (2) has a diameter of 20 mm. The

bars are unstressed when the structure is assembled at 30°C.

When the temperature decreases to –20°C, determine:


(b) the normal strains in the bronze and steel rods.

Fig. P5.53

Solution

Equilibrium: By virtue of symmetry, the forces in the two

bronze rods (1) are identical. Consider a FBD of the rigid


1 22 0yF F F (a)

Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal:

A B Cv v v (b)

All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical

deformation of rods (1):

1Av (c)

and the rigid bar deflection vB will cause an identical deformation of rod (2):

2Bv (d)


1 2 (e)

Force-Temperature-Deformation Relationships: The relationship between internal force,

temperature change, and deformation can be stated for members (1) and (2):

1 1 2 21 1 1 2 2 2

1 1 2 2

F L F LT L T L

A E A E (f)



1 1 2 21 1 1 2 2 2

1 1 2 2

F L F LT L T L

A E A E (g)

Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2:

1 1 2 22 1 1 1 2 2 2

1 1 2

1 2 2 2 21 1 1 2 2

2 1 1 2

F L A EF T L T L

A E L

L A E A EF T L L

L A E L





1 2 2 2 21 1 1 1 2 2

2 1 1 2

2 0L A E A E

F F T L LL A E L

and derive an expression for F1, noting that L1 = L2:

2 2

1 1 2 2 2

1 1

1 2 2 2

12 2

1 1

2

2

A EF T A E

A E

T A EF

A E

A E

(h)

For this structure, the areas, coefficients of thermal expansion, and elastic moduli are given below:

2 2 2 2

1 2

1 2

6 6

1 2

(16 mm) 201.0619 mm (20 mm) 314.1593 mm4 4

100,000 MPa 200,000 MPa

16.9 10 / °C 11.9 10 / °C

A A

E E

Substitute these values along with T = −50°C into Eq. (h) and calculate F1:

1 2 2 2

12 2

1 1

6 6 2 2

2

2

2

( 50 C) 16.9 10 11.9 10 (452.3893 mm )(200,000 N/mm )

314.1593 mm 200,000 MPa2

201.0619 mm 100,000 MPa

3,064.97 N

T A EF

A E

A E

Backsubstitute this result into Eq. (a) to calculate F2 = –6,129.94 N.

(a) Normal Stresses: The normal stresses in each rod can now be calculated:

11 2

1

3,064.97 N

201.015.24 MPa (

619 T

m)

m

F

A Ans.

22 2

2

6,129.94 N

314.1593 mm19.51 MPa (C)

F

A

Ans.

(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By

definition, = /L. Therefore, the normal strain for each axial member can be determined by dividing

the relationships in Eq. (f) by the respective member lengths:

1 1 2 21 1 1 1 1 2 2 2 2 2

1 1 1 2 2 2

F FT T T T

A E E A E E

(f)


1

1 1 1

1

615.2439 MPa(16.9 10 / °C)( 50 C)

100,000

693

MPa

0.00069256 mm/mm με

TE

Ans.




2

2 2 2

2 2

619.5122 MPa(11.9 10 / °C)( 5

69

0 C)200,000 MPa

0.00 3 με069256 mm/mm

FT

A E

Ans.




5.54 A load of P = 85 kN is supported by a structure

consisting of rigid bar ABC, two identical solid bronze

[E = 100 GPa, = 16.9 × 10−6

/°C] rods, and a solid

steel [E = 200 GPa, = 11.9×10−6

/°C] rod, as shown

in Fig. P5.54. The bronze rods (1) each have a

diameter of 20 mm and they are symmetrically

positioned relative to the center rod (2) and the applied

load P. Steel rod (2) has a diameter of 14 mm. The

bars are unstressed when the structure is assembled.

After the temperature of the structure has been

increased by 45°C, determine:


(b) the normal strains in the bronze and steel rods.

(c) the downward deflection of rigid bar ABC. Fig. P5.54

Solution

Equilibrium: By virtue of symmetry, the forces in the two

bronze rods (1) are identical. Consider a FBD of the rigid


1 22 0yF F F P (a)

Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal:

A B Cv v v (b)

All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical


1Av (c)

and the rigid bar deflection vB will cause an identical deformation of rod (2):

2Bv (d)


1 2 (e)

Force-Temperature-Deformation Relationships: The relationship between internal force,

temperature change, and deformation can be stated for members (1) and (2):

1 1 2 21 1 1 2 2 2

1 1 2 2

F L F LT L T L

A E A E (f)



1 1 2 21 1 1 2 2 2

1 1 2 2

F L F LT L T L

A E A E (g)




Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2:

1 1 2 22 1 1 1 2 2 2

1 1 2

1 2 2 2 21 1 1 2 2

2 1 1 2

F L A EF T L T L

A E L

L A E A EF T L L

L A E L


1 2 2 2 21 1 1 1 2 2

2 1 1 2

2L A E A E

F F T L L PL A E L

and derive an expression for F1:

1 2 2 2 21 1 1 2 2

2 1 1 2

2 21 1 2 2

21

1 2 2

2 1 1

2

2

L A E A EF P T L L

L A E L

A EP T L L

LF

L A E

L A E

(h)

For this structure, P = 85 kN = 85,000 N, and the lengths, areas, and elastic moduli are given below:

1 2

2 2 2 2

1 2

1 2

6 6

1 2

1,200 mm 2,200 mm

(20 mm) 314.1593 mm (14 mm) 153.9380 mm4 4

100,000 MPa 200,000 MPa

16.9 10 / °C 11.9 10 / °C

L L

A A

E E

Substitute these values along with T = +45°C into Eq. (h) and calculate F1:

2 21 1 2 2

21

1 2 2

2 1 1

2 26 6

2

2

2

(153.9380 mm )(200,000 N/mm )(85,000 N) (45 C) (16.9 10 )(1,200) (11.9 10 )(2,200)

2,200 mm

1,200 mm 153.9380 mm 200,000 MPa2

2,200 mm 314.1593 mm 100

A EP T L L

LF

L A E

L A E

,000 MPa

35,002.53 N

Backsubstitute this result into Eq. (a) to calculate F2 = 14,994.94 N.


11 2

1

35,002.53 N

314.1593 mm111.4 MPa

F

A Ans.

22 2

2

14,994.94 N

153.9380 m97. a

m4 MP

F

A Ans.




(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By

definition, = /L. Therefore, the normal strain for each axial member can be determined by dividing

the relationships in Eq. (f) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT T

A E A E (f)


11 1 1

1 1

6

2 2

35,002.53 N(16.9 10 / °C)(45 C)

(314.1593 mm )(100,000 N/mm )

0.00187467 1,875 μεmm/mm

FT

A E

Ans.

22 2 2

2 2

6

2 2

14,994.94 N(11.9 10 / °C)(45 C)

(153.9380 mm )(200,000 N/mm )

0.0010225 1,023 μεmm/mm

FT

A E

Ans.

(c) Rigid bar deflection: The downward deflection of the rigid bar can be determined from the


1 1

1 1 1 1

1 1

6

2

(35,002.53N)(1,200 mm)(16.9 10 / °C)(45°C)(1,200 mm)

(314.1593 mm )(100,000 MPa

2.2

)

5 mm

B

F Lv T L

A E

Ans.




5.55 A solid aluminum [E = 70 GPa, = 22.5 ×

10−6

/°C] rod (1) is connected to a solid bronze

[E = 100 GPa, = 16.9 × 10−6

/°C] rod at flange

B, as shown in Fig. P5.55. Aluminum rod (1)

has a diameter of 40 mm and bronze rod (2) has

a diameter of 120 mm. The bars are unstressed

when the structure is assembled at 30°C. After

the 300-kN load is applied to flange B, the

temperature increases to 45°C. Determine:

(a) the normal stresses in rods (1) and (2).


Fig. P5.55

Solution



1 2 300 kN 0xF F F (a)


1 2 0 (b)


1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)


1 1 2 21 1 1 2 2 2

1 1 2 2

0F L F L

T L T LA E A E

(d)

Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (d) for F1:

2 1 1 1 11 2 1 1 2 2

1 2 2 1

L A E A EF F T L L

L A E L

Substitute this expression into Eq. (a):

2 1 1 1 12 1 1 2 2 2

1 2 2 1

300 kNL A E A E

F T L L FL A E L

and solve for F2:

2 1 1 1 12 1 1 2 2

1 2 2 1

1 11 1 2 2

12

2 1 1

1 2 2

1 300 kN

300 kN

1

L A E A EF T L L

L A E L

A ET L L

LF

L A E

L A E

(f)




Member lengths, areas, elastic moduli, and coefficients of thermal expansion are given below:

1 2

2 2 2 2

1 2

1 2

6 6

1 2

2,600 mm 1,000 mm

(40 mm) 1,256.637 mm (120 mm) 11,309.734 mm4 4

70,000 MPa 100,000 MPa

22.5 10 / °C 16.9 10 / °C

L L

A A

E E

Substitute these values along with T = +15°C into Eq. (f) and compute F2 = −328.4395 kN.

Backsubstitute this result into Eq. (a) to find F1 = −28.4395 kN.


11 2

1

28,439.5 N

1,256.637 mm22.6 MPa (C)

F

A

Ans.

22 2

2

328,439.5 N

11,309.734 m29.0 MPa (C

m)

F

A

Ans.

(b) Deflection of Flange B:

The deflection of flange B can be determined from the deformation of rod (1):

1 11 1 1 1

1 1

6

2

( 28,439.5 N)(2,600 mm)(22.5 10 / °C)(15°C)(2,600 mm)

(1,256.637 m

0.0369

m )(70,000 MPa

m

)

m

B

F Lu T L

A E

Ans.




5.56 A steel [E = 30,000 ksi, = 6.6 ×

10−6

/°F] pipe column (1) with a cross-

sectional area of A1 = 5.60 in.2 is connected at

flange B to an aluminum alloy [E = 10,000

ksi, = 12.5 × 10−6

/°F] pipe (2) with a cross-

sectional area of A2 = 4.40 in.2. The assembly

(shown in Fig. P5.56) is connected to rigid

supports at A and C. It is initially unstressed at

a temperature of 90°F.

(a) At what temperature will the normal stress

in steel pipe (1) be reduced to zero?

(b) Determine the normal stresses in steel pipe

(1) and aluminum pipe (2) when the

temperature reaches –10°F.

Fig. P5.56

Solution



1 2 60 kips 0xF F F (a)


1 2 0 (b)

Force-Temperature-Deformation Relationships:

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (c)


1 1 2 21 1 1 2 2 2

1 1 2 2

0F L F L

T L T LA E A E

(d)

Solve the Equations: Set F1 = 0 and solve Eq. (a) to find F2 = 60 kips. Substitute these values for F1

and F2 into Eq. (d) along with the observation that the temperature change for both axial members is the

same (i.e., T1 = T2 = T) and solve for T:

1 1 2 2

2

1 1 2 2

6 6

1 1 2 2

(60 kips)(144 in.)0

(4.40 in. )(10,000 ksi)75.758 F

(6.6 10 / F)(120 in.) (12.5 10 / F)(144 in.)

F L F L

A E A ET

L L

Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel

pipe (1) is reduced to zero is

90 F 75.748 F 14.24 FT Ans.

(b) Solve Eq. (a) for F2 to obtain

2 1 60 kipsF F (e)

When the temperature reaches −10°F, the total change in temperature is T = −100°F. Substitute this

value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for F1:




1 1 1 21 1 2 2

1 1 2 2

1 2 21 1 1 2 2

1 1 2 2 2 2

1 2 21 1 1 2 2

1 1 2 2 2 2

21 1 2 2

2 21

1 2

1 1 2 2

( 60 kips)

(60 kips)

(60 kips)

(60 kips)

F L F LT L T L

A E A E

L L LF T L L

A E A E A E

L L LF T L L

A E A E A E

LT L L

A EF

L L

A E A E

and compute F1:

6 6

2

1

2 2

6

(60 kips)(144 in.)( 100 F) (6.6 10 )(120 in.) (12.5 10 )(144 in.)

(4.40 in. )(10,000 ksi)

120 in. 144 in.

(5.60 in. )(30,000 ksi) (4.40 in. )(10,000 ksi)

0.259200 in. 0.196364 in.

714.2857 10 i

F

6 6

0.062836 in.

n./kip 3,272.7273 10 in./kip 3,987.0130 10 in./kip

15.7602 kips

From Eq. (a), F2 has a value of

2 1 60 kips 15.7602 kips 60 kips 75.7602 kipsF F

(a) Normal Stresses: The normal stresses in each axial member can now be calculated:

11 2

1

15.7602 kips2.8143 ksi

5.60 in2.81

.ksi (T)

F

A Ans.

22 2

2

75.7602 kips17.2182 ksi

4.40 in.17.22 ksi (T)

F

A Ans.




5.57 A load P will be supported by a structure

consisting of a rigid bar ABCD, a polymer [E =

2,300 ksi, = 2.9 × 10−6

/°F] bar (1) and an

aluminum alloy [E = 10,000 ksi, = 12.5 ×

10−6

/°F] bar (2), as shown in Fig. P5.57. Each bar

has a cross-sectional area of 2.00 in.2. The bars are

unstressed when the structure is assembled at

30°F. After a concentrated load of P = 26 kips is

applied and the temperature is increased to 100°F,

determine:


(b) the vertical deflection of joint D. Fig. P5.57

Solution

Equilibrium



about pin A gives the best information for this situation:

1 2(30 in.) (84 in.) (66 in.) 0AM F F P (a)




triangles:

30 in. 84 in.

B Dv v (b)

Since there are no gaps, clearances, or other misfits

at pins B and D, the deformation of member (1) will

equal the deflection of the rigid bar at B and the

deformation of member (2) will equal the deflection

of the rigid bar at D. Therefore, Eq. (b) can be

rewritten in terms of the member deformations as:

1 2

30 in. 84 in.

(c)


The relationship between the internal force, temperature change, and deformation of an axial member

can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)







1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

30 in. 84 in.

F L F LT L T L

A E A E

(e)

Solve the Equations

Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq.

(e) for F1:

1 1 2 22 2 1 1

1 1 2 2

2 1 1 1 11 2 2 2 1 1 1

1 2 2 1

30 in.

84 in.

30 in. 30 in.

84 in. 84 in.

F L F LT L T L

A E A E

L A E A EF F T L T A E

L A E L

(f)

Substitute this expression into equilibrium equation (a):

2 1 1 1 12 2 2 1 1 1 2

1 2 2 1

30 in. 30 in.(30 in.) (84 in.) (66 in.) 0

84 in. 84 in.

L A E A EF T L T A E F P

L A E L

and solve for F2:

2

1 12 2 1 1 1

12 2

2 1 1

1 2 2

(30 in.)(66 in.) (30 in.)

84 in.

(30 in.)84 in.

84 in.

A EP T L T A E

LF

L A E

L A E

(g)

For this structure, P = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal

expansion are listed below:

1 2

2 2

1 2

1 2

6 6

1 2

72 in. 96 in.

2.00 in. 2.00 in.

2,300 ksi 10,000 ksi

2.9 10 / °F 12.5 10 / °F

L L

A A

E E

Substitute these values along with T = 70°F into Eq. (g) and calculate F2 = 19.3218 kips.

Backsubstitute into Eq. (f) to calculate F1 = 3.0991 kips.

Normal Stresses


11 2

1

3.0991 kips

2.01.550 ksi (T)

0 in.

F

A Ans.

22 2

2

19.3218 kips

2.9.66 ksi (T)

00 in.

F

A Ans.



62 2

2 2 2 2 2

2 2

(19.3218 kips)(96 in.)(12.5 10 / °F)(70°F)(96 in.) 0.1767 in.

(2.00 in. )(10,000 ksi)

F LT L

A E (h)

Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2);

therefore:

2 0.1767 in.Dv Ans.




5.58 A load P will be supported by a structure

consisting of a rigid bar ABCD, an aluminum alloy

[E = 10,000 ksi, = 12.5 × 10−6

/°F] bar (1) and a

steel [E = 29,000 ksi, = 6.5 × 10−6

/°F] bar (2), as

shown in Fig. P5.58. Each bar has a cross-

sectional area of 2.00 in.2. The bars are unstressed

when the structure is assembled at 80°F. After a

concentrated load of P = 42 kips is applied and the

temperature is decreased to –10°F, determine:


(b) the vertical deflection of joint D.

Fig. P5.58

Solution

Equilibrium



about pin A gives the best information for this situation:

1 2(30 in.) (84 in.) (66 in.) 0AM F F P (a)




triangles:

30 in. 84 in.

B Dv v (b)

Since there are no gaps, clearances, or other misfits

at pins B and D, the deformation of member (1) will

equal the deflection of the rigid bar at B and the

deformation of member (2) will equal the deflection

of the rigid bar at D. Therefore, Eq. (b) can be

rewritten in terms of the member deformations as:

1 2

30 in. 84 in.

(c)



can be stated for members (1) and (2):

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)




1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

30 in. 84 in.

F L F LT L T L

A E A E

(e)




Solve the Equations

Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq.

(e) for F1:

1 1 2 22 2 1 1

1 1 2 2

2 1 1 1 11 2 2 2 1 1 1

1 2 2 1

30 in.

84 in.

30 in. 30 in.

84 in. 84 in.

F L F LT L T L

A E A E

L A E A EF F T L T A E

L A E L

(f)

Substitute this expression into equilibrium equation (a):

2 1 1 1 12 2 2 1 1 1 2

1 2 2 1

30 in. 30 in.(30 in.) (84 in.) (66 in.) 0

84 in. 84 in.

L A E A EF T L T A E F P

L A E L

and solve for F2:

2

1 12 2 1 1 1

12 2

2 1 1

1 2 2

(30 in.)(66 in.) (30 in.)

84 in.

(30 in.)84 in.

84 in.

A EP T L T A E

LF

L A E

L A E

(g)

For this structure, P = 42 kips, and the lengths, areas, elastic moduli, and coefficients of thermal

expansion are listed below:

1 2

2 2

1 2

1 2

6 6

1 2

72 in. 96 in.

2.00 in. 2.00 in.

10,000 ksi 29,000 ksi

12.5 10 / °F 6.5 10 / °F

L L

A A

E E

Substitute these values along with T = –90°F into Eq. (g) and calculate F2 = 25.4609 kips.

Backsubstitute into Eq. (f) to calculate F1 = 21.1094 kips.

Normal Stresses


11 2

1

21.1094 kips

2.10.55 ksi (

00 iT

n)

.

F

A Ans.

22 2

2

25.4609 kips

2.12.73 ksi (

00 iT

n)

.

F

A Ans.



2 2

2 2 2 2

2 2

6

2

(25.4609 kips)(96 in.)(6.5 10 / °F)( 90°F)(96 in.) 0.01402 in.

(2.00 in. )(29,000 ksi)

F LT L

A E

(h)

Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2);

therefore:

2 0.01402 in.Dv Ans.





P5.59 consists of a rigid bar ABCD and two

axial members. Bar (1) is steel [E = 200 GPa,

= 11.7 × 10−6

/°C ] with a cross-sectional area of

A1 = 400 mm2. Bar (2) is an aluminum alloy [E

= 70 GPa, = 22.5 × 10−6

/°C] with a cross-

sectional area of A2 = 400 mm2. The bars are

unstressed when the structure is assembled.

After a concentrated load of P = 36 kN is

applied and the temperature is increased by

25°C, determine:


(b) the deflection of point D on the rigid bar.

Fig. P5.59

Solution

Equilibrium




1 2(950 mm) (600 mm)

(720 mm)(36 kN) 0

CM F F

(a)




triangles:

950 mm 600 mm 720 mm

A B Dv v v (b)




respectively.

1 2

950 mm 600 mm

(c)



can be stated for members (1) and (2)

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)







1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

950 mm 600 mm

F L F LT L T L

A E A E

(e)

Solve the Equations


2 2 1 1

1 2 2 2 1 1 1

2 2 1

950

600

F L A EF T L T L

A E L

(f)

and substitute this expression into Eq. (a):

2 2 1 1

2 2 2 1 1 1 2

2 2 1

950(950 mm) (600 mm) (720 mm)(36 kN)

600

F L A ET L T L F

A E L

(g)

For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed

below:

1 2

2 2

1 2

1 2

6 6

1 2

900 mm 900 mm

400 mm 400 mm

200,000 MPa 70,000 MPa

11.7 10 / °C 22.5 10 / °C

L L

A A

E E

Substitute these values along with T1 = T2 = +25°C into Eq. (g) and calculate F2 = –3,989.18 N.

Backsubstitute into Eq. (f) to calculate F1 = 29,803.69 N.


11 2

1

29,803.69 N

474.5 MPa (T

m)

00 m

F

A Ans.

22 2

2

3,989.18 N

49.97 MPa (C)

00 mm

F

A

Ans.



1 1

1 1 1 1

1 1

6

2

(29,803.69 N)(900 mm)(11.7 10 / C)(25 C)(900 mm)

(400 mm )(200,000 MPa)

0.5985 mm

F LT L

A E

Since the pin at A is assumed to have a perfect connection, vA = 1 = 0.5985 mm. From Eq. (b),

720 mm

0.757895(0.5985 mm)950

0.454 m mm

mD Av v Ans.





P5.60 consists of a rigid bar ABCD and two

axial members. Bar (1) is steel [E = 200 GPa,

= 11.7 × 10−6

/°C ] with a cross-sectional area of

A1 = 400 mm2. Bar (2) is an aluminum alloy [E

= 70 GPa, = 22.5 × 10−6

/°C] with a cross-

sectional area of A2 = 400 mm2. The bars are

unstressed when the structure is assembled.

After a concentrated load of P = 36 kN is

applied and the temperature is decreased by

50°C, determine:


(b) the deflection of point D on the rigid bar.

Fig. P5.60

Solution

Equilibrium




1 2(950 mm) (600 mm)

(720 mm)(36 kN) 0

CM F F

(a)




triangles:

950 mm 600 mm 720 mm

A B Dv v v (b)




respectively.

1 2

950 mm 600 mm

(c)



can be stated for members (1) and (2)

1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)







1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

950 mm 600 mm

F L F LT L T L

A E A E

(e)

Solve the Equations


2 2 1 1

1 2 2 2 1 1 1

2 2 1

950

600

F L A EF T L T L

A E L

(f)

and substitute this expression into Eq. (a):

2 2 1 1

2 2 2 1 1 1 2

2 2 1

950(950 mm) (600 mm) (720 mm)(36 kN)

600

F L A ET L T L F

A E L

(g)

For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed

below:

1 2

2 2

1 2

1 2

6 6

1 2

900 mm 900 mm

400 mm 400 mm

200,000 MPa 70,000 MPa

11.7 10 / °C 22.5 10 / °C

L L

A A

E E

Substitute these values along with T1 = T2 = –50°C into Eq. (g) and calculate F2 = 23,855.47 N.

Backsubstitute into Eq. (f) to calculate F1 = 12,217.60 N.


11 2

1

12,217.60 N

430.5 MPa (T

m)

00 m

F

A Ans.

22 2

2

23,855.47 N

459.6 MPa (T

m)

00 m

F

A Ans.



1 1

1 1 1 1

1 1

6

2

(12,217.60 N)(900 mm)(11.7 10 / C)( 50 C)(900 mm)

(400 mm )(200,000 MPa)

0.3891 mm

F LT L

A E

Since the pin at A is assumed to have a perfect connection, vA = 1 = –0.3891 mm (i.e., joint A moves to

the left). From Eq. (b),

720 mm

0.757895( 0.3891 mm)950

0.295 mmmm

D Av v Ans.





shown in Fig. P5.61. Bar (1) is made of bronze [E

= 100 GPa, = 16.9 × 10−6

/°C] and has a cross-

sectional area of 400 mm2. Bar (2) is made of

aluminum [E = 70 GPa, = 22.5 × 10−6

/°C] and

has a cross-sectional area of 600 mm2. Bars (1)

and (2) are initially unstressed. After the

temperature has increased by 40°C, determine:


(b) the vertical deflection of point A.

Fig. P5.61

Solution

Equilibrium


in members (1) and (2). A moment equation about pin

D gives the best information for this situation:

1 2 2 1(3 m) (1 m) 0 3DM F F F F (a)




triangles:

4 m 3 m 1 m

CA Bvv v

(b)

There are no gaps, clearances, or other misfits at pins B

and C; therefore, Eq. (b) can be rewritten in terms of the


1 21 23

3 m 1 m

(c)

Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically

indeterminate rigid bar configurations with opposing members.


1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)


1 1 2 21 1 1 2 2 2

1 1 2 2

3F L F L

T L T LA E A E

(e)

Solve the Equations

For this situation, T1 = T2 = T = 40°C. Substitute Eq. (a) into Eq. (e):




1 21 1

1 1 2 2

1 1 2 2

33

F LF LT L T L

A E A E

and solve for F1:

2 2 1 1

11 2

1 1 2 2

6 6

2 2 2 2

3

9

(40 C) 3(22.5 10 / C)(920 mm) (16.9 10 / C)(840 mm)

840 mm 9(920 mm)

(400 mm )(100,000 N/mm ) (600 mm )(70,000 N/mm )

13,990 N 13.990 kN

T L LF

L L

A E A E

Backsubstitute into Eq. (a) to find F2 = −41.970 kN.

(a) Normal Stresses


11 2

1

13,990 N

4035.0 MPa (C)

0 mm

F

A

Ans.

22 2

2

41,970 N

6070.0 MPa (C)

0 mm

F

A

Ans.

(b) Deflection of the rigid bar at A


1 1

1 1 1 1

1 1

6

2 2

( 13,990 N)(840 mm)(16.9 10 / °C)(40 C)(840 mm)

(400 mm )(100,000 N/mm )

0.27405 mm

F LT L

A E

Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1);

therefore, vB = 1 = 0.27405 mm (upward). From similar triangles, the deflection of the rigid bar at A is

related to vB by:

4 m 3 m

A Bv v

The deflection of the rigid bar at A is thus:

4 m 4 m

(0.27405 mm)3 m 3

0.365 mmm

A Bv v Ans.




5.62 Rigid bar ABCD in Fig. P5.62 is supported by a

pin connection at A and by two axial bars (1) and (2).

Bar (1) is a 30-in.-long bronze [E = 15,000 ksi, =

9.4 × 10−6

/°F] bar with a cross-sectional area of 1.25

in.2. Bar (2) is a 40-in.-long aluminum alloy [E =

10,000 ksi, = 12.5 × 10−6

/°F] bar with a cross-

sectional area of 2.00 in.2. Both bars are unstressed

before the load P is applied. If a concentrated load of

P = 27 kips is applied to the rigid bar at D and the

temperature is decreased by 100°F, determine:



(c) the deflection of the rigid bar at point D.

Fig. P5.62

Solution

Equilibrium


members (1) and (2). A moment equation about pin A gives


1 2(36 in.) (84 in.) (98 in.)(27 kips) 0AM F F (a)




36 in. 84 in.

CBvv

(b)

There are no gaps, clearances, or other misfits at pins B and

C; therefore, Eq. (b) can be rewritten in terms of the member

deformations as:

1 2

36 in. 84 in.

(c)

Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically



1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)


1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

36 in. 84 in.

F L F LT L T L

A E A E

(e)

Solve the Equations





2 2 1 11 2 2 2 1 1 1

2 2 1

2 2 1 12 2 2 1 1 1

2 2 1

36 in.

84 in.

3 3

7 7

F L A EF T L T L

A E L

F L A ET L T L

A E L

Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −100°F:

1 2

2 2 1 12 2 2 1 1 1 2

2 2 1

(36 in.) (84 in.) (98 in.)(27 kips)

3 3(36 in.) (84 in.) (98 in.)(27 kips)

7 7

F F

F L A ET L T L F

A E L

2 1 12

1 2 2

1 1 1 12 2 2 1 1 1

1 1

2

22

3(36 in.) 84 in. (98 in.)(27 kips)

7

3(36 in.) (36 in.)

7

3 40 in. 1.25 in. 15,000 ksi(36 in.) 84 in. (98 in.)(27 kips)

7 30 in. 2.00 in. 10,000 ksi

(36 i

L A EF

L A E

A E A ET L T L

L L

F

2

6

26

3 (1.25 in. )(15,000 ksi)n.) (12.5 10 )( 100 F)(40 in.)

7 30 in.

(1.25 in. )(15,000 ksi)(36 in.)(9.4 10 )( 100 F)(30 in.)

30 in.

2

2,646 kip-in. 482.1429 kip-in. 634.5000 kip-in. 3,762.6429 kip-in.

19.2857 in. 84 in. 103.2857 in.

36.4295 kips

F

Backsubstitute into Eq. (a) to find F1:

1

(84 in.)(36.4295 kips) (98 in.)(27 kips)11.5022 kips

36 in.F

Normal Stresses


11 2

1

11.5022 kips

1.9.20 ksi (T)

25 in.

F

A Ans.

22 2

2

36.4295 kips

2.18.21 ksi (

00 iT

n)

.

F

A Ans.

(b) Normal Strains

The force-temperature-deformation relationships were expressed in Eq. (d). By definition, = /L.


Eq. (d) by their respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT T

A E A E





11 1 1

1 1

6

2

6

11.5022 kips(9.4 10 / °F)( 100 F)

(1.25 in. )(15,000

32

ksi)

326.55 10 in./in . 7 με

FT

A E

Ans.

22 2 2

2 2

6

2

6

36.4295 kips(12.5 10 / °F)( 100 F)

(2.00 in. )(10,000 ksi)

571.48 10 571 με in./in.

FT

A E

Ans.

(c) Deflection of the rigid bar at D


62 22 2 2 2 2

2 2

(36.4295 kips)(40 in.)(12.5 10 / °F)( 100 F)(40 in.)

(2.00 in. )(10,000 ksi)

0.022859 in.

F LT L

A E

This deformation can also be determined from the strain in member (2):

6

2 2 2 (571.48 10 in./in.)(40 in.) 0.022859 in.L


therefore, vC = 2 = 0.022859 in. (downward). From similar triangles, the deflection of the rigid bar at D

is related to vC by:

84 in. 98 in.

C Dv v

The deflection of the rigid bar at D is thus:

98 in. 98 in.

(0.022859 in.)84 in. 84 in

0.0267 in..

D Cv v Ans.





P5.63 consists of a rigid bar ABC, a steel bar

(1), and a steel rod (2). The cross-sectional area

of bar (1) is 1.5 in.2 and the diameter of rod (2)

is 0.75 in. Assume E = 30,000 ksi and = 6.6

× 10−6

/°F for both axial members. The bars are

unstressed when the structure is assembled at

70°F. After application of a concentrated force

of P = 20 kips, the temperature is decreased to

30°F. Determine:

(a) the normal stresses in bar (1) and rod (2).

(b) the normal strains in bar (1) and rod (2).

(c) the deflection of pin C from its original

position.

Fig. P5.63

Solution

Equilibrium


forces in members (1) and (2). A moment

equation about pin B gives the best information

for this situation:

1 2(12 in.) (20 in.)

(15 in.)(20 kips) 0

BM F F

(a)




triangles:

12 in. 20 in.

CAvv

(b)

There are no gaps, clearances, or other misfits at

pins A and C; therefore, Eq. (b) can be rewritten

in terms of the member deformations as:

1 2

12 in. 20 in.

(c)

Note: To understand the negative sign associated with 1, see Section 5.5 for a discussion of statically



1 1 2 21 1 1 1 2 2 2 2

1 1 2 2

F L F LT L T L

A E A E (d)





1 1 2 21 1 1 2 2 2

1 1 2 2

1 1

12 in. 20 in.

F L F LT L T L

A E A E

(e)

Solve the Equations


2 2 1 11 2 2 2 1 1 1

2 2 1

2 2 1 12 2 2 1 1 1

2 2 1

12 in.

20 in.

12 12

20 20

F L A EF T L T L

A E L

F L A ET L T L

A E L

Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −40°F:

1 2

2 2 1 12 2 2 1 1 1 2

2 2 1

(12 in.) (20 in.) (15 in.)(20 kips)

12 12(12 in.) (20 in.) (15 in.)(20 kips)

20 20

F F

F L A ET L T L F

A E L

Note that the area of rod (2) is A2 = /4(0.75 in.)2 = 0.4418 in.

2.

2 1 12

1 2 2

1 1 1 12 2 2 1 1 1

1 1

2

22

12(12 in.) 20 in. (15 in.)(20 kips)

20

12(12 in.) (12 in.)

20

12 80 in. 1.5 in. 30,000 ksi(12 in.) 20 in. (15 in.)(20 kips

20 32 in. 0.4418 in. 30,000 ksi

L A EF

L A E

A E A ET L T L

L L

F

2

6

26

)

12 (1.5 in. )(30,000 ksi)(12 in.) (6.6 10 )( 40 F)(80 in.)

20 32 in.

(1.5 in. )(30,000 ksi)(12 in.)(6.6 10 )( 40 F)(32 in.)

32 in.

2

300 kip-in. 213.84 kip-in. 142.56 kip-in. 656.40 kip-in.

61.113626 in. 20 in. 81.1136267 in.

8.092 kips

F

Backsubstitute into Eq. (a) to find F1:

1

(20 in.)(8.092 kips) (15 in.)(20 kips)11.513 kips

12 in.F




(a) Normal Stresses


11 2

1

11.513 kips

1.5 in.7.68 ksi (C)

F

A

Ans.

22 2

2

8.092 kips

0.4418.32 ksi (

18 iT

n)

.

F

A Ans.

(b) Normal Strains

The force-temperature-deformation relationships were expressed in Eq. (d). By definition, = /L.


Eq. (d) by the respective member lengths:

1 21 1 1 2 2 2

1 1 2 2

F FT T

A E A E


11 1 1

1 1

6

2

6

11.513 kips(6.6 10 / °F)( 40 F)

(1.5 in. )(30,000 ksi)

519.8 10 52 i 0 μn./in. ε

FT

A E

Ans.

22 2 2

2 2

6

2

6

8.092 kips(6.6 10 / °F)( 40 F)

(0.4418 in. )(30,000 ksi)

346.5 10 in./in. 347 με

FT

A E

Ans.

(c) Deflection of the rigid bar at C


2 2

2 2 2 2

2 2

6

2

(8.092 kips)(80 in.)(6.6 10 / °F)( 40 F)(80 in.)

(0.4418 in. )(30,000 ksi)

0.027723 in.

F LT L

A E

This deformation can also be determined from the strain in member (2):

6

2 2 2 (346.5 10 in./in.)(80 in.) 0.027723 in.L


therefore:

2 0.0277 in.Cv Ans.




5.64 Three rods of different materials are connected

and placed between rigid supports at A and D, as

shown in Fig. P5.64. Properties for each of the three

rods are given below. The bars are initially

unstressed when the structure is assembled at 70°F.

After the temperature has been increased to 250°F,

determine:

(a) the normal stresses in the three rods.

(b) the force exerted on the rigid supports.

(c) the deflections of joints B and C relative to rigid

support A.

Fig. P5.64

Aluminum (1)

L1 = 10 in.

A1 = 0.8 in.2

E1 = 10,000 ksi

= 12.5×10-6

/°F

Cast Iron (2)

L2 = 5 in.

A2 = 1.8 in.2

E2 = 22,500 ksi

= 7.5×10-6

/°F

Bronze (3)

L3 = 7 in.

A3 = 0.6 in.2

E3 = 15,000 ksi

= 9.4×10-6

/°F

Solution

Equilibrium

Consider a FBD at joint B. Assume that both internal axial

forces will be tension.

1 2 1 20xF F F F F (a)

Similarly, consider a FBD at joint C. Assume that both internal


2 3 3 20xF F F F F (b)


1 2 3 0 (c)


1 1 2 2 3 31 1 1 1 2 2 2 2 3 3 3 3

1 1 2 2 3 3

F L F L F LT L T L T L

A E A E A E (d)


3 31 1 2 21 1 1 2 2 2 3 3 3

1 1 2 2 3 3

0F LF L F L

T L T L T LA E A E A E

(e)

Solve the Equations

From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members;

therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:

2 1 2 32 2

1 1 2 2 3 3

1 1 2 2 3 3

0F L F LF L





Solving for F2:

3 32 1 2 21 1 2 2 3 3

1 1 2 2 3 3

31 22 1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

231 2

1 1 2 2 3 3

F LF L F LT L T L T L

A E A E A E

LL LF T L L L

A E A E A E

T L L LF

LL L

A E A E A E

(f)

Substitute the problem data along with T = +180°F into Eq. (f) and calculate F1 = −19.1025 kips.

From Eq. (a), F1 = −19.1025 kips and from Eq. (b), F3 = −19.1025 kips.

(a) Normal Stresses


11 2

1

19.1025 kips23.878 ksi

0.8 in.23.9 ksi (C)

F

A

Ans.

22 2

2

19.1025 kips10.613 ksi

1.8 in.10.61 ksi (C)

F

A

Ans.

33 2

3

19.1025 kips31.838 ksi

0.6 in.31.8 ksi (C)

F

A

Ans.

(b) Force on Rigid Supports

The force exerted on the rigid supports is equal to the internal axial force:

19.10 kipsA DR R Ans.

(c) Deflection of Joints B and C

The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The

deformation of rod (1) is given by:

1 1

1 1 1 1

1 1

6

2

( 19.1025 kips)(10 in.)(12.5 10 / °F)(180°F)(10 in.) 0.001378 in.

(0.8 in. )(10,000 ksi)

F LT L

A E

The deflection of joint B is thus:

1 0.001378 in.Bu Ans.


2 2

2 2 2 2

2 2

6

2

( 19.1025 kips)(5 in.)(7.5 10 / °F)(180°F)(5 in.) 0.004392 in.

(1.8 in. )(22,500 ksi)

F LT L

A E

The deflection of joint C is:

2 0.001378 in. 0.004392 in. 0.00301 in.C Bu u Ans.




5.65 Three rods of different materials are connected

and placed between rigid supports at A and D, as

shown in Fig. P5.65. Properties for each of the three

rods are given below. The bars are initially

unstressed when the structure is assembled at 20°C.

After the temperature has been increased to 100°C,

determine:

(a) the normal stresses in the three rods.

(b) the force exerted on the rigid supports.

(c) the deflections of joints B and C relative to rigid

support A.

Fig. P5.65

Aluminum (1)

L1 = 440 mm

A1 = 1,200 mm2

E1 = 70 GPa

= 22.5×10-6

/°C

Cast Iron (2)

L2 = 200 mm

A2 = 2,800 mm2

E2 = 155 GPa

= 13.5×10-6

/°C

Bronze (3)

L3 = 320 mm

A3 = 800 mm2

E3 = 100 GPa

= 17.0×10-6

/°C

Solution

Equilibrium

Consider a FBD at joint B. Assume that both internal axial

forces will be tension.

1 2 1 20xF F F F F (a)

Similarly, consider a FBD at joint C. Assume that both internal


2 3 3 20xF F F F F (b)


1 2 3 0 (c)


1 1 2 2 3 31 1 1 1 2 2 2 2 3 3 3 3

1 1 2 2 3 3

F L F L F LT L T L T L

A E A E A E (d)


3 31 1 2 21 1 1 2 2 2 3 3 3

1 1 2 2 3 3

0F LF L F L


(e)

Solve the Equations

From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members;

therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:

2 1 2 32 2

1 1 2 2 3 3

1 1 2 2 3 3

0F L F LF L





Solving for F2:

3 32 1 2 21 1 2 2 3 3

1 1 2 2 3 3

31 22 1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

231 2

1 1 2 2 3 3

F LF L F LT L T L T L

A E A E A E

LL LF T L L L

A E A E A E

T L L LF

LL L

A E A E A E

(f)

Substitute the problem data along with T = +80°C into Eq. (f) and calculate F1 = −148.80 kN. From

Eq. (a), F1 = −148.80 kN and from Eq. (b), F3 = −148.80 kN.

(a) Normal Stresses


11 2

1

148,800 N124.00 MPa

1,200 m124.0 MPa (C)

m

F

A

Ans.

22 2

2

148,800 N53.143 MPa

2,800 m53.1 MPa (C)

m

F

A

Ans.

33 2

3

148,800 N186.0 MPa

800 mm186.0 MPa (C)

F

A

Ans.

(b) Force on Rigid Supports

The force exerted on the rigid supports is equal to the internal axial force:

148.8 kNA DR R Ans.

(c) Deflection of Joints B and C

The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The

deformation of rod (1) is given by:

1 1

1 1 1 1

1 1

6

2 2

( 148,800 N)(440 mm)(22.5 10 / °C)(80°C)(440 mm) 0.01257 mm

(1,200 mm )(70,000 N/mm )

F LT L

A E

The deflection of joint B is thus:

1 0.01257 mmBu Ans.


2 2

2 2 2 2

2 2

6

2 2

( 148,800 N)(200 mm)(13.5 10 / °C)(80°C)(200 mm) 0.14743 mm

(2,800 mm )(155,000 N/mm )

F LT L

A E

The deflection of joint C is:

2 0.01257 mm 0.14743 mm 0.1600 mmC Bu u Ans.




5.66 The machine part shown in Fig. P5.66

is 3/8-in. thick and is made of cold-rolled

18-8 stainless steel (see Appendix D for

properties). Determine the maximum safe

load P if a factor of safety of 2.5 with

respect to failure by yield is specified.

Fig. P5.66

Solution

allow

165 ksi66 ksi

FS 2.5

Y

Fillet:

4.0 in. 0.5 in.

2.0 0.25 1.822.0 in. 2.0 in.

D rK

d d

allow minallow

(66 ksi)(2.0 in.)(0.375 in.)27.2 kips

1.82

AP

K

(a)

Hole:

1.25 in.

0.3125 2.344.0 in.

dK

D

allow netallow

(66 ksi)(4.0 in. 1.25 in.)(0.375 in.)29.1 kips

2.34

AP

K

(b)

Controlling Load:

The fillet controls in this case; therefore,

allow 27.2 kipsP Ans.





is 12 mm thick and is made of SAE 4340

heat-treated steel (see Appendix D for

properties). The holes are centered in the

bar. Determine the maximum safe load P

if a factor of safety of 3.0 with respect to

failure by yield is specified.

Fig. P5.67

Solution

allow

910 MPa303.33 MPa

FS 3.0

Y

Small Hole:

15 mm

0.1 2.73100 mm

dK

D

2

allow netallow

(303.33 N/mm )(100 mm 10 mm)(12 mm)120,000 N 120.0 kN

2.73

AP

K

(a)

Large Hole:

35 mm

0.35 2.29100 mm

dK

D

2

allow netallow

(303.33 N/mm )(100 mm 35 mm)(12 mm)103,319 N 103.3 kN

2.29

AP

K

(b)

Controlling Load:

The large hole controls in this case; therefore,

allow 103.3 kNP Ans.




5.68 A 100-mm-wide by 8-mm-thick steel

bar is transmitting an axial tensile load of

3,000 N. After the load is applied, a 4-

mm-diameter hole is drilled through the

bar, as shown in Fig. P5.68. The hole is

centered in the bar.

(a) Determine the stress at point A (on the

edge of the hole) in the bar before and

after the hole is drilled.

(b) Does the axial stress at point B on the

edge of the bar increase or decrease as the

hole is drilled? Explain.

Fig. P5.68

Solution

(a) Stress at point A:

Before hole is drilled:

3,000 N

(100 mm)(8 mm)3.75 MPaA

P

A Ans.

After hole is drilled:

4 mm

0.04 2.89100 mm

dK

D

net

(3,000 N)(2.89)

(100 mm 4 mm)(8 mm)11.29 MPaA

PK

A

Ans.

(b) Stress at point B:

The axial stress at point B decreases. Since the average stress changes very little with the introduction

of the small hole and since the stress at A is larger than the average stress, the axial stress far from the

hole must be less than the average stress.





is 90-mm-wide by 12-mm-thick and is

made of 2014-T4 aluminum (see Appendix

D for properties). The hole is centered in

the bar. Determine the maximum safe load

P if a factor of safety of 1.50 with respect

to failure by yield is specified.

Fig. P5.69

Solution

allow

290 MPa193.33 MPa

FS 1.50

Y

Hole:

30 mm

0.33 2.3190 mm

dK

D

2

allow netallow

(193.33 N/mm )(90 mm 30 mm)(12 mm)60,260 N 60.3 kN

2.31

AP

K

(a)

Notches:

10 mm 90 mm

0.20 1.8 2.2750 mm 50 mm

r DK

d d

2

allow minallow

(193.33 N/mm )(50 mm)(12 mm)51,101 N 51.1 kN

2.27

AP

K

(b)

Controlling Load:

The notches control in this case; therefore,

allow 51.1 kNP Ans.




5.70 The machine part shown in Fig.

P5.70 is 8-mm thick and is made of

AISI 1020 cold-rolled steel (see

Appendix D for properties). Determine

the maximum safe load P if a factor of

safety of 3 with respect to failure by

yield is specified.

Fig. P5.70

Solution

allow

427 MPa142.33 MPa

FS 3

Y

Fillets:

120 mm 16 mm

1.5 0.2 1.8380 mm 80 mm

D rK

d d

allow minallow

(142.33 MPa)(80 mm)(8 mm)49.8 kN

1.83

AP

K

(a)

Hole:

15 mm

0.125 2.67120 mm

dK

D

allow netallow

(142.33 MPa)(120 mm 15 mm)(8 mm)44.8 kN

2.67

AP

K

(b)

Notches:

10 mm 800 mm

0.167 1.333 2.2860 mm 60 mm

r DK

d d

allow minallow

(142.33 MPa)(60 mm)(8 mm)30.0 kN

2.28

AP

K

(c)

Controlling Load:

The notches control in this case; therefore,

allow 30.0 kNP Ans.





is 10 mm thick, is made of AISI 1020 cold-

rolled steel (see Appendix D for

properties), and is subjected to a tensile

load of P = 45 kN. Determine the minimum

radius r that can be used between the two

sections if a factor of safety of 2 with

respect to failure by yield is specified.

Round the minimum fillet radius up to the

nearest 1-mm multiple.

Fig. P5.71

Solution

allow

427 MPa213.5 MPa

FS 2

Y

2

allow min

allow

(213.5 N/mm )(40 mm)(10 mm)1.90

45,000 N

AK

P

80 mm

240 mm

D

d

Then, from Fig. 5.15

min min

0.214

(0.214) (0.214)(40 mm) 8.56 mm say 9 mm

r

d

r d r

Ans.




5.72 The 0.25-in.-thick bar shown in Fig.

P5.72 is made of 2014-T4 aluminum (see

Appendix D for properties) and will be

subjected to an axial tensile load of P =

1,500 lbs. A 0.5625-in.-diameter hole is

located on the centerline of the bar.

Determine the minimum safe width D for

the bar if a factor of safety of 2.5 with

respect to failure by yield must be

maintained.

Fig. P5.72

Solution

allow

42 ksi16.8 ksi 16,800 psi

FS 2.5

Y

By definition:

max

nom

K

In this instance, the maximum stress equals the allowable stress:

maxnom

nom nom

16,800 psi16,800 psiK K

If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w must

satisfy:

nom

net

16,800 psi(16,800 psi)

P P K PK K K w

A wt t (a)

And the bar width D is the sum of the net width w and the hole diameter d:

D w d (b)

For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined

until D is known, and since the calculation of D from Eqs. (a) and (b) depends on K, a trial-and-error

process can be established. One possible sequence is summarized in the table below.

Trial value of d

D

K

[from Fig. 5.14]

w (in.)

[from Eq. (a)]

D (in.)

[from Eq. (b)] Resulting value of

d

D

0.25 2.43 0.868 1.430 0.393

0.39 2.25 0.804 1.366 0.412

0.41 2.23 0.796 1.359 0.414

(agrees with trial value)

Therefore, the minimum bar width is:

min 1.359 in.D Ans.




5.73 The stepped bar with a circular hole,

shown in Fig. P5.73, is made of annealed

18-8 stainless steel. The bar is 12 mm thick

and will be subjected to an axial tensile

load of P = 70 kN. The normal stress in the

bar is not to exceed 150 MPa. To the

nearest millimeter, determine:

(a) the maximum allowable hole diameter

d.

(b) the minimum allowable fillet radius r.

Fig. P5.73

Solution

(a) Maximum hole diameter d:

By definition:

max

nom

K

In this instance, the maximum stress equals the 150-MPa allowable stress:

maxnom

nom nom

150 MPa150 MPaK K

If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w can be

expressed as w = D – d. The nominal stress at the hole location depends on Anet, which in turn can be

expressed in terms of w:

2

nom 2

net

150 N/mm ( )( ) (150 N/mm )

P P P K PK K K K D d

A wt D d t t

From this relationship, the hole diameter d is:

2(150 N/mm )

K Pd D

t (a)

For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined

until d is known, and since the calculation of d from Eq. (a) depends on K, a trial-and-error process can

be established. One possible sequence is summarized in the table below.

Trial value of d

D

K

[from Fig. 5.14]

d (mm)

[from Eq. (a)] Resulting value of

d

D

0.10 2.73 23.83 0.183

0.18 2.56 30.44 0.234

0.23 2.47 33.94 0.261

0.26 2.42 35.89 0.276

0.28 2.39 37.06 0.285

0.29 2.37 37.83 0.291

(agrees with trial value)

Therefore, the maximum hole diameter is:

max 37 mmd Ans.




(b) Minimum fillet radius r:

130 mm

1.3100 mm

D

d

Therefore, the curve for D/d = 1.30 on Fig. 5.15 will be used to determine the stress concentration

factor.

2

max allow

nom min

150 N/mm2.571

/ (70,000 N) / (100 mm)(12 mm)K

P A

From Fig. 5.15,

min min

0.047

0.047(100 mm) 4.7 mm say 5 mm

r

d

r r

Ans.




6.1 A solid circular steel shaft having an outside diameter of d = 0.75 in. is subjected to a pure torque of

T = 650 lb-in. Determine the maximum shear stress in the shaft.

Solution

The polar moment of inertia for the shaft is

4 4(0.75 in.) 0.031063 in.32

J

The maximum shear stress in the steel shaft is found from the elastic torsion formula:

max 4

(650 lb-in.)(0.75 in. / 2)7,846.93 psi

0.07,850 ps

31063 in.i

Tc

J Ans.




6.2 A hollow aluminum shaft with an outside diameter of 80 mm and a wall thickness of 5 mm has an

allowable shear stress of 75 MPa. Determine the maximum torque T that may be applied to the shaft.

Solution


4 4 4 4 4(80 mm) (70 mm) 1,664,062 mm32 32

J D d

Rearrange the elastic torsion formula to determine the maximum torque T:

2 4

allow (75 N/mm )(1,664,062 mm )3,120,117 N-mm

80 mm3,120 N-

2m

/

JT

c

Ans.




6.3 A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm is subjected

to a pure torque of T = 5,500 N-m.

(a) Determine the maximum shear stress in the hollow shaft.

(b) Determine the minimum diameter of a solid steel shaft for which the maximum shear stress is the

same as in part (a) for the same torque T.

Solution


4 4 4 4 4(100 mm) (80 mm) 5,796,238 mm32 32

J D d

(a) The maximum shear stress in the hollow steel shaft is found from the elastic torsion formula:

max 4

(5,500 N-m)(100 mm / 2)(1,000 mm/m)47.445 MPa

5,796,238 m47.4

mMPa

Tc

J Ans.

(b) The polar moment of inertia for a solid shaft can be expressed as

4

32J d

Rearrange the elastic torsion formula to group terms with d on the left-hand side:

4

32 ( / 2)

d T

d

and simplify to

3

16

d T

From this equation, the unknown diameter of the solid shaft can be expressed as

316T

d

To support a torque of T = 5,500 N-m without exceeding the maximum shear stress determined in part

(a), a solid shaft must have a diameter of

3 32

16 16(5,500 N-m)(1,000 mm/m)83.891 mm

(47.445 N/m83.9 m

mm

)

Td

Ans.




6.4 A compound shaft consists of two pipe

segments. Segment (1) has an outside diameter of

200 mm and a wall thickness of 10 mm. Segment

(2) has an outside diameter of 150 mm and a wall

thickness of 10 mm. The shaft is subjected to

torques TB = 42 kN-m and TC = 18 kN-m, which act

in the directions shown in Fig. P6.4. Determine the

maximum shear stress magnitude in each shaft

segment.

Fig. P6.4

Solution

Equilibrium:

1 142 kN-m 18 kN-m 0 24 kN-mxM T T

2 218 kN-m 0 18 kN-mxM T T

Section properties:

4 4 4 4 4

1 1 1 (200 mm) (180 mm) 54,019,686 mm32 32

J D d

4 4 4 4 4

2 2 2 (150 mm) (130 mm) 21,551,281 mm32 32

J D d

Shear stress magnitudes:

1 4

(24 kN-m)(200 mm / 2)(1,000 N/kN)(1,000 mm/m)

54,01944.4 MPa

,686 mm Ans.

2 4

(18 kN-m)(150 mm / 2)(1,000 N/kN)(1,000 mm/m)

21,66162.3 MPa

,281 mm Ans.




6.5 A compound shaft consists of two pipe

segments. Segment (1) has an outside diameter of

10.75 in. and a wall thickness of 0.365 in. Segment

(2) has an outside diameter of 6.625 in. and a wall

thickness of 0.280 in. The shaft is subjected to

torques TB = 60 kip-ft and TC = 24 kip-ft, which act

in the directions shown in Fig. P6.5. Determine the

maximum shear stress magnitude in each shaft

segment.

Fig. P6.5

Solution

Equilibrium:

1 160 kip-ft 24 kip-ft 0 36 kip-ftxM T T

2 224 kip-ft 0 24 kip-ftxM T T

Section properties:

4 4 4 4 4

1 1 1 (10.75 in.) (10.02 in.) 321.4685 in.32 32

J D d

4 4 4 4 4

2 2 2 (6.625 in.) (6.0650 in.) 56.2844 in.32 32

J D d


1 4

(36 kip-ft)(10.75 in. / 2)(12 in./ft)

321.46857.

2

in2 ks

. i Ans.

2 4

(24 kip-ft)(6.625 in. / 2)(12 in./ft)

56.2844 16.9

in.5 ksi Ans.




6.6 A compound shaft (Fig. P6.6) consists of brass segment (1) and

aluminum segment (2). Segment (1) is a solid brass shaft with an outside

diameter of 0.625 in. and an allowable shear stress of 6,000 psi. Segment

(2) is a solid aluminum shaft with an outside diameter of 0.50 in. and an

allowable shear stress of 9,000 psi. Determine the magnitude of the largest

torque TC that may be applied at C.

Fig. P6.6

Solution

Section properties:

4 4

1 (0.625 in.) 0.014980 in.32

J

4 4

2 (0.50 in.) 0.006136 in.32

J

Allowable internal torques:

4

1 11

1

(6,000 psi)(0.014980 in. )287.621 lb-in.

0.625 in./2

JT

c

4

2 22

2

(9,000 psi)(0.006136 in. )220.893 lb-in. controls

0.50 in./2

JT

c

Equilibrium:

The internal torque magnitude in each segment equals the external torque; therefore, T1 = T2 = TC. The

controlling internal torque is T2 = 220.893 lb-in.; therefore, the maximum external torque TC that may be

applied to the compound shaft is

220.893 lb-i 221 ln. b-in.CT Ans.





aluminum segment (2). Segment (1) is a solid brass shaft with an

allowable shear stress of 60 MPa. Segment (2) is a solid aluminum shaft

with an allowable shear stress of 90 MPa. If a torque of TC = 23,000 N-m

is applied at C, determine the minimum required diameter of (a) the brass

shaft and (b) the aluminum shaft.

Fig. P6.7

Solution

The polar moment of inertia for a solid shaft can be expressed as

4

32J d


4

32 ( / 2)

d T

d

and simplify to

3

16

d T


316T

d

Equilibrium:

For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T1 =

T2 = TC = 23,000 N-m.

Minimum shaft diameters:

(a) Brass shaft (1)

133

1 2

allow,1

16 16(23,000 N-m)(1,000 mm/m)

(60 N/mm )125.0 mm

Td

Ans.

(b) Aluminum shaft (2)

233

2 2

allow,2

16 16(23,000 N-m)(1,000 mm/m)

(90 N/mm )109.2 mm

Td

Ans.




6.8 A solid 0.75-in.-diameter shaft is subjected to

the torques shown in Fig. P6.8. The bearings shown

allow the shaft to turn freely.

(a) Plot a torque diagram showing the internal

torque in segments (1), (2), and (3) of the shaft. Use

the sign convention presented in Section 6-6.

(b) Determine the maximum shear stress magnitude

in the shaft.

Fig. P6.8

Solution

Equilibrium:

1 110 lb-ft 0 10 lb-ftxM T T

2 210 lb-ft 50 lb-ft 0 40 lb-ftxM T T


Section properties:

4 4

1 2 3 (0.75 in.) 0.031063 in.32

J J J


1 11 4

1

(10 lb-ft)(0.75 in. / 2)(12 in./ft)1,448.7 psi

0.031063 in.

T c

J

2 22 4

2

(40 lb-ft)(0.75 in. / 2)(12 in./ft)5,794.7 psi

0.031063 in.

T c

J

3 33 4

3

(30 lb-ft)(0.75 in. / 2)(12 in./ft)4,346.0 psi

0.031063 in.

T c

J

The maximum shear stress in the shaft occurs in segment (2):

max 5,790 psi Ans.




6.9 A solid constant-diameter shaft is subjected to

the torques shown in Fig. P6.9. The bearings shown


(a) Plot a torque diagram showing the internal

torque in segments (1), (2), and (3) of the shaft. Use


(b) If the allowable shear stress in the shaft is 80

MPa, determine the minimum acceptable diameter

for the shaft.

Fig. P6.9

Solution

3 3160 N-m 0 160 N-mxM T T

2 2160 N-m 380 N-m 0 220 N-mxM T T

1 1160 N-m 380 N-m 330 N-m 0 110 N-mxM T T

The maximum torque magnitude in the shaft occurs in segment (2): Tmax = 220 N-m.

(b) The elastic torsion formula gives the relationship between shear stress and torque in a shaft.

Tc

J

In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic

torsion formula, putting the known terms on the right-hand side of the equation:

J T

c

Express the left-hand side of this equation in terms of the shaft diameter D:

4

332

/ 2 16

dT

dd

and solve for the minimum acceptable diameter:

3 3

2

16 16(220 N-m)(1,000 mm/m)14,005.635 mm

(80

24.1 mm

N/mm )

Td

d

Ans.




6.10 A solid circular steel shaft having an outside diameter of 1.25 in. is subjected to a pure torque of T

= 2,200 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine:

(a) the maximum shear stress in the shaft.

(b) the magnitude of the angle of twist in a 6-ft length of shaft.

Solution


4 4(1.25 in.) 0.239684 in.32

J

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula:

max 4

(2,200 lb-in.)(1.25 in. / 2)5, 5.74 ks736.7 psi

0.239684 in.i

Tc

J Ans.

(b) The magnitude of the angle of twist in a 6-ft length of shaft is

4

(2,200 lb-in.)(6 ft)(12 in./ft)0.055072 rad

(0.239684 in. )(12,000,0000.0551 rad 3.16

psi)

TL

JG Ans.




6.11 A solid circular steel shaft having an outside diameter of 35 mm is subjected to a pure torque of T =

640 N-m. The shear modulus of the steel is G = 80 GPa. Determine:


(b) the magnitude of the angle of twist in a 1.5-m length of shaft.

Solution


4 4(35 mm) 147,323.515 mm32

J


max 4

(640 N-m)(35 mm / 2)(1,000 mm/m)76.023 MPa

147,323.515 m76.0

mMPa

Tc

J Ans.

(b) The magnitude of the angle of twist in a 1.5-m length of shaft is

2

4 2

(640 N-m)(1.5 m)(1,000 mm/m)0.081453 rad

(147,323.515 m0.0815 r

m )(80,ad

000 N/mm.67

)4

TL

JG Ans.




6.12 A hollow steel shaft with an outside diameter of 85 mm and a wall thickness of 10 mm is subjected

to a pure torque of T = 7,000 N-m. The shear modulus of the steel is G = 80 GPa. Determine:


(b) the magnitude of the angle of twist in a 2.5-m length of shaft.

Solution


4 4 4 4 4(85 mm) (65 mm) 3,372,303 mm32 32

J D d


max 4

(7,000 N-m)(85 mm / 2)(1,000 mm/m)88.219 MPa

3,372,303 m88.2

mMPa

Tc

J Ans.

(b) The magnitude of the angle of twist in a 2.5-m length of shaft is

2

4 2

(7,000 N-m)(2.5 m)(1,000 mm/m)0.064867 rad

(3,372,303 m0.0649 r

m )(80,ad

000 N/mm.72

)3

TL

JG Ans.




6.13 A solid stainless steel [G = 12,500 ksi] shaft that is 72-in. long will be subjected to a pure torque of

T = 900 lb-in. Determine the minimum diameter required if the shear stress must not exceed 8,000 psi

and the angle of twist must not exceed 5°. Report both the maximum shear stress and the angle of twist

at this minimum diameter.

Solution

Consider shear stress:


4

32J d

The elastic torsion formula can be rearranged to gather terms with d:

4 3

32 ( / 2) 16

d d T

d


316T

d

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:

316(900 lb-in.)

0.831 in.(8,000 psi)

d

Consider angle of twist:

Rearrange the angle of twist equation:

4

32

TL TLJ d

JG G

and solve for the minimum diameter that will satisfy the angle of twist limitation:

4 432 32(900 lb-in.)(72 in.)

0.882 in.(5 )( rad/180 )(12,500,000 psi)

TLd

G

Therefore, the minimum diameter that could be used for the shaft is

min 0.882 in.d Ans.

The angle of twist for this shaft is = 5° = 0.087266 rad. To compute the shear stress in a 0.882-in.-

diameter shaft, first compute the polar moment of inertia:

4 4(0.882 in.) 0.059404 in.32

J

The shear stress in the shaft is thus:

max 4

(900 lb-in.)(0.882 in. / 2)6,681.12 psi

0.0596,680 p

40si

4 in. Ans.




6.14 A solid stainless steel [G = 86 GPa] shaft that is 2.0 m long will be subjected to a pure torque of T =

75 N-m. Determine the minimum diameter required if the shear stress must not exceed 50 MPa and the

angle of twist must not exceed 4°. Report both the maximum shear stress and the angle of twist at

this minimum diameter.

Solution

Consider shear stress:


4

32J d

The elastic torsion formula can be rearranged to gather terms with d:

4 3

32 ( / 2) 16

d d T

d


316T

d

For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is:

32

16(75 N-m)(1,000 mm/m)19.70 mm

(50 N/mm )d

Consider angle of twist:

Rearrange the angle of twist equation:

4

32

TL TLJ d

JG G

and solve for the minimum diameter that will satisfy the angle of twist limitation:

4 42

32 32(75 N-m)(2,000 mm)(1,000 mm/m)22.46 mm

(0.069813 rad)(86,000 N/mm )

TLd

G

Therefore, the minimum diameter that could be used for the shaft is

min 22.5 mmd Ans.

The angle of twist for this shaft is = 0.069813 rad. To compute the shear stress in a 22.46-mm-

diameter shaft, first compute the polar moment of inertia:

4 4(22.46 mm) 24,983.625 mm32

J

The shear stress in the shaft is thus:

max 4

(75 N-m)(22.46 mm / 2)(1,000 mm/m)

24,983.6233

5.

m7 MP

m a Ans.




6.15 A hollow steel [G = 12,000 ksi] shaft with an outside diameter of 3.50 in. will be subjected to a

pure torque of T = 3,750 lb-ft. Determine the maximum inside diameter d that can be used if the shear

stress must not exceed 8,000 psi and the angle of twist must not exceed 3° in an 8-ft length of shaft.

Report both the maximum shear stress and the angle of twist for this maximum inside diameter.

Solution

Consider shear stress: The polar moment of inertia for a hollow shaft can be expressed as

4 4

32J D d

The elastic torsion formula can be rearranged to gather terms with D:

4 4

32 / 2

D d T

D

Rearrange this equation to isolate the inside diameter d term:

4 4 4 432 32

2 2

T D T DD d d D

From this equation, the unknown inside diameter of the hollow shaft can be expressed as

4

432

2

T Dd D

For the hollow steel shaft, the maximum diameter that will satisfy the allowable shear stress is:

44

32(3,750 lb-ft)(3.50 in.)(12 in./ft)(3.50 in.) 2.6564 in.

2 (8,000 psi)d

Consider angle of twist: Rearrange the angle of twist equation:

4 4

32

TL TLJ D d

JG G

Rearrange this equation to isolate the inside diameter d term:

4 4 32TL

d DG

and solve for the maximum inside diameter that will satisfy the angle of twist limitation:

2

4 444

32 32(3,750 lb-ft)(8 ft)(12 in./ft)(3.50 in.) 2.991 in.

(3 )( /180 )(12,000,000 psi)

TLd D

G

Therefore, the maximum inside diameter that could be used for the shaft is

max 2.66 in.d Ans.

The shear stress for this shaft is = 8,000 psi. To compute the angle of twist in an 8-ft length of the

hollow shaft, first compute the polar moment of inertia:

4 4 4 4 4(3.50 in.) (2.66 in.) 9.817318 in.32 32

J D d

The angle of twist in the shaft is thus:

2

4

(3,750 lb-ft)(8 ft)(12 in./ft)

(9.817318 in. )(12,000,000.0

0 psi367 rad 2.

)10

TL

JG Ans.




6.16 A compound steel [G = 80 GPa] shaft

(Fig. P6.16) consists of a solid 55-mm-

diameter segment (1) and a solid 40-mm-

diameter segment (2). The allowable shear

stress of the steel is 70 MPa, and the

maximum rotation angle at the free end of

the compound shaft must be limited to C ≤

3°. Determine the magnitude of the largest

torque TC that may be applied at C.

Fig. P6.16

Solution

Section properties:

4 4

1 (55 mm) 898,360.5 mm32

J

4 4

2 (40 mm) 251,327.4 mm32

J

Consider shear stress: The larger shear stress will occur in the smaller diameter segment; that is,

segment (2).

2 4

max

(70 N/mm )(251,327.4 mm )879,646 N-mm

40 mm/2

JT

c

Consider angle of twist: The rotation angle at C is the sum of the angles of twist in segments (1) and

(2):

1 1 2 21 2

1 1 2 2

C

T L T L

J G J G

Since T1 = T2 = TC and G1 = G2, this equation can be simplified to

max 1 2

1 2

3T L L

G J J

and solved for the maximum torque:

2

max

4 4

(3 )( /180 )(80,000 N/mm )739,395 N-mm

800 mm 1,200 mm

898,360.5 mm 251,327.4 mm

T

Therefore, the magnitude of the largest torque TC that may be applied at C is

,max 739,395 N-mm 739 N-mCT Ans.





aluminum segment (2). Segment (1) is a solid brass [G = 5,600 ksi]

shaft with an outside diameter of 1.75 in. and an allowable shear stress

of 9,000 psi. Segment (2) is a solid aluminum [G = 4,000 ksi] shaft

with an outside diameter of 1.25 in. and an allowable shear stress of

12,000 psi. The maximum rotation angle at the upper end of the

compound shaft must be limited to C ≤ 4°. Determine the magnitude

of the largest torque TC that may be applied at C.

Fig. P6.17

Solution

From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque TC. The

elastic torsion formula gives the relationship between shear stress and torque in a shaft.

Tc

J

Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments

(1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque.

An expression can be written for each shaft segment:

1 1 2 21 2

1 2

J JT T

c c

For shaft segment (1), the polar moment of inertia is:

4 4

1 (1.75 in.) 0.920772 in.32

J

Use this value along with the 9,000 psi allowable shear stress to determine the allowable torque T1:

4

1 11

1

(9,000 psi)(0.920772 in. )9,470.8 lb-in.

(1.75 in./2)

JT

c

(a)

For shaft segment (2), the polar moment of inertia is:

4 4

2 (1.25 in.) 0.239684 in.32

J

Use this value along with the 12,000 psi allowable shear stress to determine the allowable torque T2:

4

2 22

2

(12,000 psi)(0.239684 in. )4,601.9 lb-in.

(1.25 in./2)

JT

c

(b)

Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as:

1 1 2 21 2

1 1 2 2

T L T L

J G J G

The rotation angle at C is the sum of these two angles of twist:

1 1 2 21 2

1 1 2 2

C

T L T L

J G J G




and since T1 = T2 = TC:

1 2

1 1 2 2

C C

L LT

J G J G

Solving for TC gives:

1 2

1 1 2 2

4 4

rad(4 )

1803,308.4 lb-in.

12 in. 18 in.

(0.920772 in. )(5,600,000 psi) (0.239684 in. )(4,000,000 psi)

CCT

L L

J G J G

(c)

Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the

maximum torque that can be applied to the compound shaft at C is TC = 3,308.4 lb-in. = 276 lb-ft. Ans.





consists of solid 30-mm-diameter segments

(1) and (3) and tube segment (2), which has

an outside diameter of 60 mm and an inside

diameter of 50 mm (Fig. P6.18). Determine:

(a) the maximum shear stress in tube

segment (2).

(b) the angle of twist in tube segment (2).

(c) the rotation angle of gear D relative to

gear A.

Fig. P6.18

Solution

The torques in all shaft segments are equal; therefore, T1 = T2 = T3 = 240 N-m.

Polar moments of inertia in the shaft segments will be needed for this calculation. For segments (1) and

(3), which are solid 30-mm-diameter shafts, the polar moment of inertia is:

4 4

1 3(30 mm) 79,521.56 mm32

J J

For segment (2), which has a tube cross section, the polar moment of inertia is:

4 4 4

2 (60 mm) (50 mm) 658,752.71 mm32

J

Shear stress in tube segment (2): Use the elastic torsion formula to calculate the shear stress caused by

an internal torque of T2 = 240 N-m. Note that the radius term used in the elastic torsion formula for the

tube is the outside radius; that is, c2 = 60 mm/2 = 30 mm.

2 22 4

2

(240 N-m)(30 mm)(1,000 mm/m)

658,752.71 m10.93 MPa

m

T c

J Ans.

Angle of twist in tube segment (2): Apply the angle of twist equation to segment (2).

2 22 4 2

2 2

(240 N-m)(2,000 mm)(1,000 mm/m)0.009108 rad

(658,752.71 mm )(80,000 N/mm )0.00911 rad

T L

J G Ans.

Rotation angle of gear D relative to gear A: The angles of twist in segments (1) and (3) must be

calculated. Since both segments have the same shear modulus, polar moment of inertia, and length, they

will both have the same angle of twist:

1 11 34 2

1 1

(240 N-m)(300 mm)(1,000 mm/m)0.011318 rad

(79,521.56 mm )(80,000 N/mm )

T L

J G

Since gear A is the origin of the coordinate system for this problem, we will arbitrarily define the

rotation angle at gear A to be zero; that is, A = 0. The rotation angle of gear D relative to gear A is

found by adding the angles of twist for the three segments to A:

1 2 3

0 0.011318 rad 0.009108 rad 0.011318 rad

0.031743 0.03 rad 17 rad

D A

Ans.





consists of solid 40-mm-diameter segments

(1) and (3) and tube segment (2), which has

an outside diameter of 75 mm (Fig. P6.19).

If the rotation of gear D relative to gear A

must not exceed 0.01 rad, determine the

maximum inside diameter that may be used

for tube segment (2).

Fig. P6.19

Solution

The polar moment of inertia for shaft segments (1) and (3) is:

4 4

1 3 (40 mm) 251,327 mm32

J J

The rotation of gear D relative to gear A is equal to the sum of the angles of twist in segments (1), (2),

and (3):

1 2 3D

The angles of twist in segments (1) and (3) is

1 3 4 2

(240 N-m)(300 mm)(1,000 mm/m)0.003581 rad

(251,327 mm )(80,000 N/mm )

Therefore, the angle of twist that can allowed for segment (2) is

1 2 3 20.01 rad 0.01 rad 2(0.003581 rad) 0.002838 radD

The minimum polar moment of inertia required for segment (2) is thus:

2

42 22 2

2 2

(240 N-m)(2 m)(1,000 mm/m)0.002838 rad 2,114,165 mm

(80,000 N/mm )(0.002838 rad)

T LJ

J G

Since the outside diameter of segment (2) is D2 = 75 mm, the maximum inside diameter d2 is:

4 4 4 4 4

2 2 2

2

(75 mm

56.4 m

) 2,114,165 mm

m

32 32D d d

d

Ans.




6.20 The compound shaft shown in Fig. P6.20 consists of aluminum segment (1) and steel segment (2).

Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t1 = 0.25

in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2 =

2.50 in., a wall thickness of t2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaft

is subjected to torques applied at B and C, as shown in Fig. P6.20.

(a) Prepare a diagram that shows the internal

torque and the maximum shear stress in

segments (1), and (2) of the shaft. Use the sign

convention presented in Section 6-6.

(b) Determine the rotation angle of B with

respect to the support at A.

(c) Determine the rotation angle of C with

respect to the support at A.

Fig. P6.20

Solution

Section properties: Polar moments of inertia in the shaft segments will be needed for this calculation.

4 4 4 4 4

1 1 1 (4.00 in.) (3.50 in.) 10.4004 in.32 32

J D d

4 4 4 4 4

2 2 2 (2.50 in.) (2.25 in.) 1.3188 in.32 32

J D d

(a) Equilibrium

1

1

950 lb-ft 2,100 lb-ft 0

1,150 lb-ft

xM T

T

2

2

950 lb-ft 0

950 lb-ft

xM T

T

Shear stress:

1 11 4

1

(1,150 lb-ft)(4.00 in./2)(12 in./ft)2,653.75 psi

10.4004 in.2,650 psi

T c

J Ans.

2 22 4

2

( 950 lb-ft)(2.50 in./2)(12 in./ft)10,804.95 psi

1.3188 in.10,800 psi

T c

J

Ans.




(b) Rotation angle of B with respect to A:

1 0B A B B

Therefore,

2

1 11 4

1 1

(1,150 lb-ft)(9 ft)(12 in./ft)0.035826 rad

(10.4004 0.0358 rad

in. )(4,000,000 psi)B

T L

J G Ans.

(c) Rotation angle of C with respect to A:

2 2C B C B

The angle of twist in shaft (2) is

2

2 22 4

2 2

( 950 lb-ft)(6 ft)(12 in./ft)0.051864 rad

(1.3188 in. )(12,000,000 psi)

T L

J G

and thus, the rotation angle at C is

2 0.035826 rad ( 0.051864 rad) 0.016038 rad 0.01604 radC B Ans.




6.21 A solid 1.00-in.-diameter steel [G =

12,000 ksi] shaft is subjected to the torques

shown in Fig. P6.21.

(a) Prepare a diagram that shows the internal

torque and the maximum shear stress in

segments (1), (2), and (3) of the shaft. Use


(b) Determine the rotation angle of pulley C

with respect to the support at A.

(c) Determine the rotation angle of pulley D

with respect to the support at A.

Fig. P6.21

Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments

will be needed for this calculation.

4 4 4

1 1 2 3(1.00 in.) 0.098175 in.32 32

J d J J

(a) Equilibrium

1

1

240 lb-ft 280 lb-ft 130 lb-ft 0

90 lb-ft

xM T

T

2

2

280 lb-ft 130 lb-ft 0

150 lb-ft

xM T

T

3

3

130 lb-ft 0

130 lb-ft

xM T

T

Shear stress:

1 11 4

1

( 90 lb-ft)(1.00 in./2)(12 in./ft)5,500.4 psi 5,500 psi

0.098175 in.

T c

J

2 22 4

2

(150 lb-ft)(1.00 in./2)(12 in./ft)9,167.3 psi 9,170 psi

0.098175 in.

T c

J

3 33 4

3

( 130 lb-ft)(1.00 in./2)(12 in./ft)7,945.0 psi 7,950 psi

0.098175 in.

T c

J

The maximum shear stress in the entire shaft is max = 9,170 psi. Ans.




(b) Rotation angle of C with respect to A:

The angles of twist in the three shaft segments are:

1 11 4

1 1

( 90 lb-ft)(36 in.)(12 in./ft)0.033002 rad

(0.098175 in. )(12,000,000 psi)

T L

J G

2 22 4

2 2

(150 lb-ft)(36 in.)(12 in./ft)0.055004 rad

(0.098175 in. )(12,000,000 psi)

T L

J G

3 33 4

3 3

( 130 lb-ft)(36 in.)(12 in./ft)0.047670 rad

(0.098175 in. )(12,000,000 psi)

T L

J G

The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1)

and (2):

1 2 0.033002 rad 0.055004 rad 0.022002 rad 0.0220 radC Ans.

(c) Rotation angle of D with respect to A:

The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1),

(2), and (3):

1 2 3

0.033002 rad 0.055004 rad ( 0.047670 rad)

0.025 0.0257 ra669 rad d

D

Ans.




6.22 A compound shaft supports several

pulleys as shown in Fig. P6.22. Segments

(1) and (4) are solid 25-mm-diameter steel

[G = 80 GPa] shafts. Segments (2) and (3)

are solid 50-mm-diameter steel shafts. The

bearings shown allow the shaft to turn

freely.

(a) Determine the maximum shear stress in

the compound shaft.

(b) Determine the rotation angle of pulley D

with respect to pulley B.

(c) Determine the rotation angle of pulley E

with respect to pulley A.

Fig. P6.22

Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for this

calculation.

4 4 4

1 1 4(25 mm) 38,349.52 mm32 32

J d J

4 4 4

2 2 3(50 mm) 613,592.32 mm32 32

J d J

(a) Equilibrium

1

1

175 N-m 0

175 N-m

xM T

T

2

2

2,100 N-m 175 N-m 0

1,925 N-m

xM T

T

3

3

650 N-m 225 N-m 0

425 N-m

xM T

T

4

4

225 N-m 0

225 N-m

xM T

T




Shear stress:

1 11 4

1

( 175 N-m)(25 mm/2)(1,000 mm/m)57.041 MPa 57.0 MPa

38,349.52 mm

T c

J

2 22 4

2

(1,925 N-m)(50 mm/2)(1,000 mm/m)78.432 MPa 78.4 MPa

613,592.32 mm

T c

J

3 33 4

3

(425 N-m)(50 mm/2)(1,000 mm/m)17.316 MPa 17.32 MPa

613,592.32 mm

T c

J

4 44 4

4

( 225 N-m)(25 mm/2)(1,000 mm/m)73.339 MPa 73.3 MPa

38,349.52 mm

T c

J

The maximum shear stress in the entire shaft is max = 78.4 MPa. Ans.




(b) Angles of twist:

The angles of twist in the four shaft segments are:

1 11 4 2

1 1

( 175 N-m)(750 mm)(1,000 mm/m)0.042781 rad

(38,349.52 mm )(80,000 N/mm )

T L

J G

2 22 4 2

2 2

(1,925 N-m)(500 mm)(1,000 mm/m)0.019608 rad

(613,592.32 mm )(80,000 N/mm )

T L

J G

3 33 4 2

3 3

(425 N-m)(625 mm)(1,000 mm/m)0.005411 rad

(613,592.32 mm )(80,000 N/mm )

T L

J G

4 44 4 2

4 4

( 225 N-m)(550 mm)(1,000 mm/m)0.040336 rad

(38,349.52 mm )(80,000 N/mm )

T L

J G

Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is

found from the sum of the angles of twist in segments (2) and (3):

/ 2 3 0.019608 rad 0.005411 rad 0.025019 rad 0.0250 radD B Ans.

(c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is

found from the sum of the angles of twist in all four segments:

1 2 3 4

0.042781 rad 0.019608 rad 0.005411 rad ( 0.040336 rad)

0.05809 0.08 581 rar d da

E

Ans.




6.23 A solid steel [G = 80 GPa] shaft of variable

diameter is subjected to the torques shown in Fig.

P6.23. The diameter of the shaft in segments (1) and

(3) is 50 mm, and the diameter of the shaft in

segment (2) is 80 mm. The bearings shown allow

the shaft to turn freely.

(a) Determine the maximum shear stress in the

compound shaft.

(b) Determine the rotation angle of pulley D with

respect to pulley A.

Fig. P6.23

Solution

Section properties: The polar moments of inertia for the shaft segments will be needed for this

calculation.

4 4

1 3(50 mm) 613,592.32 mm32

J J

4 4

2 (80 mm) 4,021,238.60 mm32

J

(a) Equilibrium

1

1

1,200 N-m 0

1,200 N-m

xM T

T

2

2

1,200 N-m 4,500 N-m 0

3,300 N-m

xM T

T

3

3

500 N-m 0

500 N-m

xM T

T

Shear stress:

1 11 4

1

(1,200 N-m)(50 mm/2)(1,000 mm/m)48.892 MPa 48.9 MPa

613,592.32 mm

T c

J

2 22 4

2

( 3,300 N-m)(80 mm/2)(1,000 mm/m)32.826 MPa 32.8 MPa

4,021,238.60 mm

T c

J

3 33 4

3

( 500 N-m)(50 mm/2)(1,000 mm/m)20.372 MPa 20.4 MPa

613,592.32 mm

T c

J

The maximum shear stress in the compound shaft is max 1 48.9 MPa Ans.




Angles of twist:


2

1 11 4 2

1 1

(1,200 N-m)(0.7 m)(1,000 mm/m)0.017112 rad

(613,592.32 mm )(80,000 N/mm )

T L

J G

2

2 22 4 2

2 2

( 3,300 N-m)(1.8 m)(1,000 mm/m)0.018464 rad

(4,021,238.60 mm )(80,000 N/mm )

T L

J G

2

3 33 4 2

3 3

( 500 N-m)(0.7 m)(1,000 mm/m)0.007130 rad

(613,592.32 mm )(80,000 N/mm )

T L

J G

(b) Rotation angles:

1

2

3

0 rad

0 rad 0.017112 rad 0.017112 rad

0.017112 rad ( 0.018464 rad) 0.001352 rad

0.001352 rad ( 0.007130 rad 0.008482 ra) d

A

B A

C B

D C

Ans.




6.24 A compound shaft drives three gears,

as shown in Fig. P6.24. Segments (1) and

(2) of the compound shaft are hollow

aluminum [G = 4,000 ksi] tubes, which

have an outside diameter of 3.00 in. and a

wall thickness of 0.25 in. Segments (3) and

(4) are solid 2.00-in.-diameter steel [G =

12,000 ksi] shafts. The bearings shown



the compound shaft.

(b) Determine the rotation angle of flange

C with respect to flange A.

(c) Determine the rotation angle of gear E

with respect to flange A.

Fig. P6.24

Solution

Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments

will be needed for this calculation.

4 4 4 4 4

1 1 1 2(3.00 in.) (2.50 in.) 4.117204 in.32 32

J D d J

4 4 4

3 3 4(2.00 in.) 1.570796 in.32 32

J d J

(a) Equilibrium

1

1

14 kip-in. 42 kip-in.

35 kip-in. 0

7 kip-in.

xM

T

T

2

2

14 kip-in. 42 kip-in. 0

28 kip-in.

xM T

T

3

3

14 kip-in. 42 kip-in. 0

28 kip-in.

xM T

T

4

4

14 kip-in. 0

14 kip-in.

xM T

T




Shear stress:

1 11 4

1

( 7 kip-in.)(3.00 in./2)2.55 ksi

4.117204 in.

T c

J

2 22 4

2

( 28 kip-in.)(3.00 in./2)10.20 ksi

4.117204 in.

T c

J

3 33 4

3

( 28 kip-in.)(2.00 in./2)17.83 ksi

1.570796 in.

T c

J

4 44 4

4

(14 kip-in.)(2.00 in./2)8.91 ksi

1.570796 in.

T c

J

The maximum shear stress in the compound shaft is max 3 17.83 ksi Ans.

(b) Angles of Twist: The angles of twist in the four shaft segments are:

1 11 4

1 1

(7 kip-in.)(60 in.)0.025503 rad

(4.117204 in. )(4,000 ksi)

T L

J G

2 22 4

2 2

( 28 kip-in.)(6 in.)0.010201 rad

(4.117204 in. )(4,000 ksi)

T L

J G

3 33 4

3 3

( 28 kip-in.)(18 in.)0.026738 rad

(1.570796 in. )(12,000 ksi)

T L

J G

4 44 4

4 4

(14 kip-in.)(12 in.)0.008913 rad

(1.570796 in. )(12,000 ksi)

T L

J G

1

2

3

4

0 rad

0 rad 0.025503 rad 0.0255 rad

0.025503 rad ( 0.010201 rad) 0.01530 rad

0.015302 rad ( 0.026738 rad) 0.01144 rad

0.011436 rad 0.008913 rad 0.00252 rad

A

B A

C B

D C

E D

The rotation angle of flange C with respect to A is

0.01530 radC Ans.

(c) Rotation angle of E with respect to A:

0.00252 radE Ans.




6.25 A compound shaft drives several

pulleys, as shown in Fig. P6.25. Segments (1)

and (2) of the compound shaft are hollow

aluminum [G = 4,000 ksi] tubes, which have


thickness of 0.125 in. Segments (3) and (4)

are solid 1.50-in.-diameter steel [G = 12,000

ksi] shafts. The bearings shown allow the

shaft to turn freely.


the compound shaft.

(b) Determine the rotation angle of flange C


(c) Determine the rotation angle of pulley E


Fig. P6.25

Solution

(a) Equilibrium


3 395 lb-ft 270 lb-ft 0 175 lb-ftxM T T

Note: The internal torque in shaft segment (2) is the same as

in segment (3); therefore, T2 = –175 lb-ft.


Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2)

are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. – 2(0.125 in.) =

2.75 in. The polar moment of inertia of segments (1) and (2) is:

4 4 4 4 4

1 1 1 2(3.00 in.) (2.75 in.) 2.33740 in.32 32

J D d J

Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of:

4 4 4

3 3 4(1.50 in.) 0.49701 in.32 32

J d J




Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula:

1 11 4

1

(650 lb-ft)(1.50 in.)(12 in./ft)5,005.56 psi 5,010 psi

2.33740 in.

T c

J

2 22 4

2

( 175 lb-ft)(1.50 in.)(12 in./ft)1,347.6 psi 1,348 psi

2.33740 in.

T c

J

3 33 4

3

( 175 lb-ft)(0.75 in.)(12 in./ft)3,169.0 psi 3,170 psi

0.49701 in.

T c

J

4 44 4

4

(95 lb-ft)(0.75 in.)(12 in./ft)1,720.3 psi 1,720 psi

0.49701 in.

T c

J

The maximum shear stress in the compound shaft is max 1 5,010 psi Ans.

Angles of twist:

1 11 4 2

1 1

(650 lb-ft)(72 in.)(12 in./ft)0.060067 rad

(2.33740 in. )(4,000,000 lb/in. )

T L

J G

2 22 4 2

2 2

( 175 lb-ft)(36 in.)(12 in./ft)0.008086 rad

(2.33740 in. )(4,000,000 lb/in. )

T L

J G

3 33 4 2

3 3

( 175 lb-ft)(54 in.)(12 in./ft)0.019014 rad

(0.49701 in. )(12,000,000 lb/in. )

T L

J G

4 44 4 2

4 4

(95 lb-ft)(54 in.)(12 in./ft)0.010322 rad

(0.49701 in. )(12,000,000 lb/in. )

T L

J G

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each

segment:

1 2 3 4B A C B D C E D

The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle

at pulley A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in

segment (1):

1

1 0 0.060067 rad 0.0601 rad

B A

B A

Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation

angle of pulley B:

2

2 0.060067 rad ( 0.008086 rad) 0.051981 rad 0.0520 rad

C B

C B

The rotation angle at D is:

3

3 0.051981 rad ( 0.019014 rad) 0.032967 rad 0.0330 rad

D C

D C

and the rotation angle at E is:

4

4 0.032967 rad 0.010322 rad 0.043289 rad 0.0433 rad

E D

E D




(b) Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the

system, the rotation angle of flange C with respect to pulley A is simply:

0.051981 ra 0.0520 radd C Ans.

(c) Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for the

system, the rotation angle of flange C with respect to pulley A is simply:

0.043289 ra 0.0433 radd E Ans.




6.26 A torque of TA = 460 lb-ft is applied to gear

A of the gear train shown in Fig. P6.26. The

bearings shown allow the shafts to rotate freely.

(a) Determine the torque TD required for

equilibrium of the system.

(b) Assume shafts (1) and (2) are solid 1.5-in.-

diameter steel shafts. Determine the magnitude

of the maximum shear stresses acting in each

shaft.

(c) Assume shafts (1) and (2) are solid steel

shafts, which have an allowable shear stress of

6,000 psi. Determine the minimum diameter

required for each shaft.

Fig. P6.26

Solution

(a) Torque TD required for equilibrium:

7 in. 7 in.

(460 lb-ft)4 in. 4 in.

805 lb-ftD AT T Ans.

(b) Shear stress magnitudes if shafts are solid 1.5-in.-diameter:

4 4 4(1.5 in.) 0.497010 in.32 32

J d

1 11 4

1

(460 lb-ft)(1.5 in./2)(12 in./ft)8,32 8,330 p9.82 psi

0.49si

7010 in.

T c

J Ans.

2 22 4

2

(805 lb-ft)(1.5 in./2)(12 in./ft)14,577.18 psi

0.497010 in14,580 p

.si

T c

J Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in

terms of the unknown diameter d:

4

332

/ 2 16

dTc J T

dJ c d

Using this expression, solve for the minimum acceptable diameter of each shaft:

3 311

allow

1

16 16(460 lb-ft)(12 in./ft)4.685522 in.

(6,000 psi)

1.673347 in. 1.673 in.

Td

d Ans.

3 322

allow

2

16 16(805 lb-ft)(12 in./ft)8.199663 in.

(6,000 psi)

2.016502 in 2.02 . in.

Td

d Ans.




6.27 A torque of TD = 450 N-m is applied to gear

D of the gear train shown in Fig. P6.27. The

bearings shown allow the shafts to rotate freely.

(a) Determine the torque TA required for

equilibrium of the system.

(b) Assume shafts (1) and (2) are solid 30-mm-

diameter steel shafts. Determine the magnitude

of the maximum shear stresses acting in each

shaft.

(c) Assume shafts (1) and (2) are solid steel

shafts, which have an allowable shear stress of

60 MPa. Determine the minimum diameter

required for each shaft.

Fig. P6.27

Solution

(a) Torque TA required for equilibrium:

90 mm 90 mm

(450 N 270 N--m)150 m

mm 150 mm

A DT T Ans.

(b) Shear stress magnitudes if shafts are solid 30-mm-diameter:

4 4 4(30 mm) 79,521.564 mm32 32

J d

1 11 4

1

(270 N-m)(30 mm/2)(1,000 mm/m)50.930 MPa

79,521.564 m50.9

mMPa

T c

J Ans.

2 22 4

2

(450 N-m)(30 mm/2)(1,000 mm/m)84.883 MPa

79,521.564 m84.9

mMPa

T c

J Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in


4

332

/ 2 16

dTc J T

dJ c d


3 311 2

allow

1

16 16(270 N-m)(1,000 mm/m)22,918.312 mm

(60 N/mm )

28.405 28.m 4 m mm

Td

d Ans.

3 322 2

allow

2

16 16(450 N-m)(1,000 mm/m)38,197.186 mm

(60 N/mm )

33.678 33.m 7 m mm

Td

d Ans.




6.28 The gear train system shown in Fig. P6.28

includes shafts (1) and (2), which are solid 1.375-

in.-diameter steel shafts. The allowable shear

stress of each shaft is 8,000 psi. The bearings

shown allow the shafts to rotate freely.

Determine the maximum torque TD that can be

applied to the system without exceeding the

allowable shear stress in either shaft.

Fig. P6.28

Solution

Section properties:

4 4

1 2(1.375 in.) 0.350922 in.32

J J

Maximum torque in either shaft:

4

allow 1max

1

(8,000 psi)(0.350922 in. )4,083.457 lb-in.

1.375 in./2

JT

c

Torque relationship:

1 21 2 2 1

10 in.or 1.667 controls

10 in. 6 in. 6 in.

T TT T T T

The torque in shaft (1) controls; therefore, T1 = 4,083.457 lb-in. Consequently, the maximum torque in

shaft (2) must be limited to:

2 1

6 in.0.6(4,083.457 lb-in.) 2,450.074 lb-in.

10 in.204 lb-ftT T Ans.




6.29 The gear train system shown in Fig. P6.29

includes shafts (1) and (2), which are solid 20-mm-

diameter steel shafts. The allowable shear stress of

each shaft is 50 MPa. The bearings shown allow the

shafts to rotate freely. Determine the maximum

torque TD that can be applied to the system without

exceeding the allowable shear stress in either shaft.

Fig. P6.29

Solution

Section properties:

4 4

1 2(20 mm) 15,707.963 mm32

J J

Maximum torque in either shaft:

2 4

allow 1max

1

(50 N/mm )(15,707.963 mm )78,539.816 N-mm

20 mm/2

JT

c


1 21 2 2 1

200 mmor 2.50 controls

200 mm 80 mm 80 mm

T TT T T T

The torque in shaft (1) controls; therefore, T1 = 78,539.816 N-mm. Consequently, the maximum torque

in shaft (2) must be limited to:

2 1

80 mm0.4(78,539.816 N-mm) 31,415.927 N-mm

200 m31.4 N-

mmT T Ans.




6.30 In the gear system shown in Fig. P6.30, the

motor applies a 210 lb-ft torque to the gear at A. A

torque of TC = 300 lb-ft is removed from the shaft

at gear C, and the remaining torque is removed at

gear D. Segments (1) and (2) are solid 1.5-in.-

diameter steel [G = 12,000 ksi] shafts, and the

bearings shown allow free rotation of the shaft.

Determine:

(a) the maximum shear stress in segments (1) and

(2) of the shaft.

(b) the rotation angle of gear D relative to gear B.

Fig. P6.30

Solution

Section properties:

4 4

1 2(1.5 in.) 0.497010 in.32

J J


10 in.

hence 2.50 2.50(210 lb-ft) 525 lb-ft4 in. 10 in. 4 in.

A BB A A

T TT T T

Shaft (1) has the same torque as gear B; therefore, T1 = 525 lb-ft. A torque of 300 lb-ft is removed from

the shaft at gear C; therefore, TC = 300 lb-ft. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

0

525 lb-ft 300 lb-ft 225 lb-ft

x C

C

M T T T

T T T

To summarize:

1 2525 lb-ft 225 lb-ftT T

(a) Maximum shear stress in shafts (1) and (2):

1 11 4

1

(525 lb-ft)(1.5 in./2)(12 in./ft)9,506 9,510 .856 psi

0.497010 in.psi

T c

J Ans.

2 22 4

2

(225 lb-ft)(1.5 in./2)(12 in./ft)4,074 4,070 .367 psi

0.497010 in.psi

T c

J Ans.

(b) Angles of twist in shafts (1) and (2):

1 11 4

1 1

(525 lb-ft)(60 in.)(12 in./ft)0.063379 rad

(0.497010 in. )(12,000,000 psi)

T L

J G

2 22 4

2 2

(225 lb-ft)(40 in.)(12 in./ft)0.018108 rad

(0.497010 in. )(12,000,000 psi)

T L

J G

Note: There is not enough information given to say whether these twist angles are positive or negative.

However, both twist angles will have the same sign.

1 2

say 0 rad (since this is our reference)

0 rad 0.063379 rad 0.018108 rad 0.0814 0.0887 rad 15 rad

B

D B

Ans.




6.31 In the gear system shown in Fig. P6.31, the motor applies a 360 lb-ft torque to the gear at A. A

torque of TC = 500 lb-ft is removed from the shaft at gear C, and the remaining torque is removed at gear

D. Segments (1) and (2) are solid steel [G = 12,000 ksi] shafts, and the bearings shown allow free

rotation of the shaft.

(a) Determine the minimum permissible

diameters for segments (1) and (2) of the

shaft if the maximum shear stress must not

exceed 6,000 psi.

(b) If the same diameter is to be used for

both segments (1) and (2), determine the

minimum permissible diameter that can be

used for the shaft if the maximum shear

stress must not exceed 6,000 psi and the

rotation angle of gear D relative to gear B

must not exceed 0.10 rad.

Fig. P6.31

Solution

Torque relationship: The motor applies a torque of 360 lb-ft to gear A. Since gear B is bigger than

gear A, the torque on gear B will be increased in proportion to the gear ratio:

10 in.

hence 2.50 2.50(360 lb-ft) 900 lb-ft4 in. 10 in. 4 in.

A BB A A

T TT T T

Let the positive x axis for shaft BCD extend from gear B toward gear D. In order for shaft BCD to be in

equilibrium with the torques as shown on gears C and D, the torque at B must act in the negative x

direction. Consequently, TB = −900 lb-ft.

Draw a free-body diagram that cuts through shaft (1) and includes gear B. From this FBD, the torque in

shaft (1) is:


From the problem statement, we are told that a torque

of 500 lb-ft is removed from the shaft at gear C. In

other words, TC = 500 lb-ft. Draw a FBD that cuts

through shaft (2) and includes both gears B and C.

From this FBD, the torque in shaft (2) is:

2

2

900 lb-ft 500 lb-ft 0

400 lb-ft

xM T

T

To summarize:


(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in


4

332

/ 2 16

dTc J T

dJ c d





3 311

allow

1

16 16(900 lb-ft)(12 in./ft)9.167325 in.

(6,000 psi)

2.092900 in 2.09 . in.

Td

d Ans.

3 322

allow

2

16 16(400 lb-ft)(12 in./ft)4.074367 in.

(6,000 psi)

1.597178 in. 1.597 in.

Td

d Ans.

(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft

segments (1) and (2):

1 2 0.10 radD B

Since the same solid steel shaft is to be used for both segments (1) and (2):

1 1 2 21 1 2 2

1 1 2 2

10.01 rad

T L T LT L T L

J G J G JG

or rearranging:

4 4(900 lb-ft)(60 in.)(12 in./ft) (400 lb-ft)(40 in.)(12 in./ft)1.883606 in.

32 (12,000,000 psi)(0.

1.63

1 rad

4 in.

)J d

d

Since the shaft has already been designed for stresses and since the diameter that is required to satisfy

the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required

for a constant diameter shaft between gears B and D is:

min 2.09 in.d Ans.





motor applies a torque of 220 N-m to the gear at

A. A torque of TC = 400 N-m is removed from the

shaft at gear C, and the remaining torque is

removed at gear D. Segments (1) and (2) are solid

40-mm-diameter steel [G = 80 GPa] shafts, and

the bearings shown allow free rotation of the

shaft.


segments (1) and (2) of the shaft.

(b) Determine the rotation angle of gear D

relative to gear B.

Fig. P6.32

Solution

Section properties:

4 4

1 2(40 mm) 251,327.41 mm32

J J


300 mm

hence 3 3(220 N-m) 660 N-m100 mm 300 mm 100 mm

A BB A A

T TT T T

Shaft (1) has the same torque as gear B; therefore, T1 = 660 N-m. A torque of TC = 400 N-m is removed

from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

0

660 N-m 400 N-m 260 N-m

x C

C

M T T T

T T T

To summarize:

1

2

660 N-m

260 N-m

T

T

(a) Maximum shear stress in shafts (1) and (2):

1 11 4

1

(660 N-m)(40 mm/2)(1,000 mm/m)52.521 MPa

251,327.41 m52.5

mMPa

T c

J Ans.

2 22 4

2

(260 N-m)(40 mm/2)(1,000 mm/m)20.690 MPa

251,327.41 m20.7

mMPa

T c

J Ans.





2

1 11 4 2

1 1

(660 N-m)(1.5 m)(1,000 mm/m)0.049239 rad

(251,327.41 mm )(80,000 N/mm )

T L

J G

2

2 22 4 2

2 2

(260 N-m)(1.0 m)(1,000 mm/m)0.012931 rad

(251,327.41 mm )(80,000 N/mm )

T L

J G

Note: There is not enough information given to say whether these twist angles are positive or negative.

However, both twist angles will have the same sign.

1 2

say 0 rad (since this is our reference)

0 rad 0.049239 rad 0.012931 rad 0.0621 0.0670 rad 22 rad

B

D B

Ans.





motor applies a torque of 400 N-m to the gear at

A. A torque of TC = 700 N-m is removed from the

shaft at gear C, and the remaining torque is

removed at gear D. Segments (1) and (2) are solid

steel [G = 80 GPa] shafts, and the bearings shown

allow free rotation of the shaft.

(a) Determine the minimum permissible

diameters for segments (1) and (2) of the shaft if

the maximum shear stress must not exceed 40

MPa.

(b) If the same diameter is to be used for

segments (1) and (2), determine the minimum

permissible diameter that can be used for the

shaft if the maximum shear stress must not

exceed 40 MPa and the rotation angle of gear D

relative to gear B must not exceed 3.0°.

Fig. P6.33

Solution


300 mm

hence 3 3(400 N-m) 1,200 N-m100 mm 300 mm 100 mm

A BB A A

T TT T T

Shaft (1) has the same torque as gear B; therefore, T1 = 1,200 N-m. A torque of TC = 700 N-m is

removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium:

1 2

2 1

0

1,200 N-m 700 N-m 500 N-m

x C

C

M T T T

T T T

To summarize:

1 21,200 N-m 500 N-mT T

(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in


4

332

/ 2 16

dTc J T

dJ c d


3 311 2

allow

1

16 16(1,200 N-m)(1,000 mm/m)152,788.745 mm

(40 N/mm )

53.460 53.m 5 m mm

Td

d Ans.

3 322 2

allow

2

16 16(500 N-m)(1,000 mm/m)63,661.977 mm

(40 N/mm )

39.929 39.m 9 m mm

Td

d Ans.




(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft

segments (1) and (2):

1 2 (3 )( /180 ) 0.052360 radD B

Since the same solid steel shaft is to be used for both segments (1) and (2):

1 1 2 21 1 2 2

1 1 2 2

10.052360 rad

T L T LT L T L

J G J G JG

or rearranging:

2 2

4 4

2

(1,200 N-m)(1.5 m)(1,000 mm/m) (500 N-m)(1.0 m)(1,000 mm/m)549,083.270 mm

32 (80,000 N/mm )(0.052360

48.

ra )

6 m

d

m

J d

d

Since the shaft has already been designed for stresses and since the diameter that is required to satisfy

the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required

for a constant diameter shaft between gears B and D is:

min 53.5 mmd Ans.





motor applies a torque of 250 N-m to the gear at A.

Shaft (1) is a solid 35-mm-diameter shaft, and

shaft (2) is a solid 50-mm-diameter shaft. The

bearings shown allow free rotation of the shafts.

(a) Determine the torque TE provided by the gear

system at gear E.

(b) Determine the maximum shear stresses in

shafts (1) and (2).

Fig. P6.34

Solution

(a) Torque relationship: The torque on gear B is:

72

3 3(250 N-m) 750 N-m24

B A AT T T

Shaft (1) has the same torque as gear B; therefore, T1 = 750 N-m and TC = 750 N-m. The torque on gear

D is found from:

60

2 2(750 N-m) 1,500 N-m30

D C CT T T

Shaft (2) has the same torque as gear D; therefore, T2 = 1,500 N-m and

1,500 N-mET Ans.

To summarize:

1 2750 N-m 1,500 N-mT T

(b) Section properties:

4 4

1

4 4

2

(35 mm) 147,323.51 mm32

(50 mm) 613,592.32 mm32

J

J

Maximum shear stress in shafts (1) and (2):

1 11 4

1

(750 N-m)(35 mm/2)(1,000 mm/m)89.090 MPa

147,323.51 m89.1

mMPa

T c

J Ans.

2 22 4

2

(1,500 N-m)(50 mm/2)(1,000 mm/m)61.115 MPa

613,592.32 m61.1

mMPa

T c

J Ans.





motor applies a torque of 600 N-m to the gear at A.

Shafts (1) and (2) are solid shafts, and the bearings

shown allow free rotation of the shafts.

(a) Determine the torque TE provided by the gear

system at gear E.

(b) If the allowable shear stress in each shaft must

be limited to 70 MPa, determine the minimum

permissible diameter for each shaft.

Fig. P6.35

Solution

(a) Torque relationship: The torque on gear B is:

72

3 3(600 N-m) 1,800 N-m24

B A AT T T

Shaft (1) has the same torque as gear B; therefore, T1 = 1,800 N-m and TC = 1,800 N-m. The torque on

gear D is found from:

60

2 2(1,800 N-m) 3,600 N-m30

D C CT T T

Shaft (2) has the same torque as gear D; therefore, T2 = 3,600 N-m and

3,600 N-mET Ans.

To summarize:

1 21,800 N-m 3,600 N-mT T

(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in


4

332

/ 2 16

dTc J T

dJ c d


3 311 2

allow

1

16 16(1,800 N-m)(1,000 mm/m)130,961.782 mm

(70 N/mm )

50.783 50.m 8 m mm

Td

d Ans.

3 322 2

allow

2

16 16(3,600 N-m)(1,000 mm/m)261,923.563 mm

(70 N/mm )

63.982 64.m 0 m mm

Td

d Ans.




6.36 In the gear system shown in Fig. P6.36, a

torque of TE = 1,500 lb-ft is delivered at gear E.

Shafts (1) and (2) are solid shafts, and the bearings

shown allow free rotation of the shafts.

(a) Determine the torque provided by the motor to

gear A.

(b) If the allowable shear stress in each shaft must

be limited to 4,000 psi, determine the minimum

permissible diameter for each shaft.

Fig. P6.36

Solution

(a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 1,500 lb-ft.

Accordingly, the torque applied to gear D is also equal to 1,500 lb-ft. The torque on gear C is found

with the gear ratio:

30

0.5 0.5(1,500 lb-ft) 750 lb-ft60

C D DT T T

Shaft (1) has the same torque as gear C; therefore, T1 = 750 lb-ft and TB = 750 N-m. The torque on gear

A is found from:

24 750 lb-ft

7250 l f

2 3b- t

3

CA B

TT T Ans.

To summarize:

1 2750 lb-ft 1,500 lb-ftT T

(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in


4

332

/ 2 16

dTc J T

dJ c d


3 311

allow

1

16 16(750 lb-ft)(12 in./ft)11.459156 in.

(4,000 psi)

2.255 in 2.26 in..

Td

d Ans.

3 322

allow

2

16 16(1,500 lb-ft)(12 in./ft)22.918313 in.

(4,

2.84

000 psi)

2.840 i in.n.

Td

d Ans.




6.37 In the gear system shown in Fig. P6.37, a

torque of TE = 720 lb-ft is delivered at gear E.

Shaft (1) is a solid 1.50-in.-diameter shaft, and

shaft (2) is a solid 2.00-in.-diameter shaft. The

bearings shown allow free rotation of the shafts.

(a) Determine the torque provided by the motor to

gear A.

(b) Determine the maximum shear stresses in

shafts (1) and (2).

Fig. P6.37

Solution

(a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 720 lb-ft.

Accordingly, the torque applied to gear D is also equal to 720 lb-ft. The torque on gear C is found with

the gear ratio:

30

0.5 0.5(720 lb-ft) 360 lb-ft60

C D DT T T

Shaft (1) has the same torque as gear C; therefore, T1 = 360 lb-ft and TB = 360 N-m. The torque on gear

A is found from:

24 360 lb-ft

7120 l f

2 3b- t

3

CA B

TT T Ans.

To summarize:


(b) Section properties:

4 4

1

4 4

2

(1.50 in.) 0.497010 in.32

(2.00 in.) 1.570796 in.32

J

J

Maximum shear stress in shafts (1) and (2):

1 11 4

1

(360 lb-ft)(1.50 in./2)(12 in./ft)6,51 6,520 p9.98 psi

0.49si

7010 in.

T c

J Ans.

2 22 4

2

(720 lb-ft)(2.00 in./2)(12 in./ft)5,500 5,500 p.396 psi

1si

.570796 in.

T c

J Ans.




6.38 Two solid 30-mm-diameter steel shafts are

connected by the gears shown in Fig. P6.38. The

shaft lengths are L1 = 300 mm and L2 = 500 mm.

Assume that the shear modulus of both shafts is

G = 80 GPa and that the bearings shown allow

free rotation of the shafts. If the torque applied at

gear D is TD = 160 N-m,

(a) determine the internal torques T1 and T2 in

the two shafts.

(b) determine the angles of twist 1 and 2.

(c) determine the rotation angles B and C of

gears B and C.

(d) determine the rotation angle of gear D.

Fig. P6.38

Solution

(a) Torque relationship: The torque on gear D creates a torque in shaft (2)

of T2 = 160 N-m.

2 20 160 N-mx D DM T T T T

Accordingly, the torque applied to gear C is also equal to 160 N-m. The

torque on gear B is found with the gear ratio:

54 teeth

1.8 1.8(160 N-m) 288 N-m30 teeth

B C CT T T

Shaft (1) has the same torque as gear B; therefore, T1 = −288 N-m. To summarize:

1 2288 N-m 160 N-mT T Ans.

Section properties:

4 4

1 2(30 mm) 79,521.56 mm32

J J


1 11 4 2

1 1

( 288 N-m)(300 mm)(1,000 mm/m)0.013581 rad

(79,521.56 mm )(80,00.01358 r

00 N mmd

/ )a

T L

J G Ans.

2 22 4 2

2 2

(160 N-m)(500 mm)(1,000 mm/m)0.012575 rad

(79,521.56 mm )(0.01258 ra

80,000 N/d

mm )

T L

J G Ans.

(c) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft

(1):

1 1 0 rad ( 0.013581 0.01358 rrad) adB A B A

Ans.

As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the

rotation is dictated by the gear ratio:

54 teeth 54 teeth

( 0.013581 rad) 0.024446 rad30 teeth 3

0.0244 r0 t

adeeth

C B Ans.

The rotation of gear D is found from:

2 2 0.024446 rad 0.012575 rad 0.0370 0.0370 rad21 radD C D C

Ans.




6.39 Two solid 1.75-in.-diameter steel shafts are

connected by the gears shown in Fig. P6.39. The

shaft lengths are L1 = 6 ft and L2 = 10 ft. Assume

that the shear modulus of both shafts is G =

12,000 ksi and that the bearings shown allow

free rotation of the shafts. If the torque applied at

gear D is TD = 225 lb-ft,

(a) determine the internal torques T1 and T2 in

the two shafts.

(b) determine the angles of twist 1 and 2.

(c) determine the rotation angles B and C of

gears B and C.

(d) determine the rotation angle of gear D.

Fig. P6.39

Solution

(a) Torque relationship: The torque on gear D creates a torque in shaft (2)

of T2 = 225 lb-ft.

2 20 225 lb-ftx D DM T T T T

Accordingly, the torque applied to gear C is also equal to 225 lb-ft. The

torque on gear B is found with the gear ratio:

54 teeth

1.8 1.8(225 lb-ft) 405 lb-ft30 teeth

B C CT T T

Shaft (1) has the same torque as gear B; therefore, T1 = −405 lb-ft. To summarize:

1 2405 lb-ft 225 lb-ftT T Ans.

Section properties:

4 4

1 2(1.75 in.) 0.920772 in.32

J J


2

1 11 4

1 1

( 405 lb-ft)(6 ft)(12 in./ft)0.031669 rad

(0.920772 in. )(12,000,0000.0317 r

s )ad

p i

T L

J G Ans.

2

2 22 4

2 2

(225 lb-ft)(10 ft)(12 in./ft)0.029323 rad

(0.920772 in. )(120.0293 r

,000,000 psi)ad

T L

J G Ans.

(c) Rotation angles of gears B and C: The rotation of gear B is found from the twist angle in shaft (1):

1 1 0 rad ( 0.031 0.0317 r669 r dd) aaB A B A

Ans.



54 teeth 54 teeth

( 0.031669 rad) 0.057004 rad30 teeth 3

0.0570 r0 t

adeeth

C B Ans.

(d) Rotation angle of gear D:The rotation of gear D is found from:


Ans.




6.40 Two solid steel shafts are connected by the

gears shown in Fig. P6.40. The design

requirements for the system require (1) that both

shafts must have the same diameter, (2) that the

maximum shear stress in each shaft must be less

than 6,000 psi, and (3) that the rotation angle of

gear D must not exceed 3°. Determine the

minimum required diameter of the shafts if the

torque applied at gear D is TD = 345 lb-ft. The

shaft lengths are L1 = 10 ft and L2 = 8 ft.


G = 12,000 ksi and that the bearings shown

allow free rotation of the shafts.

Fig. P6.40

Solution

Torque relationship: The torque on gear D creates a torque in shaft (2) of

T2 = 1,000 lb-ft.

2 20 345 lb-ft 4,140 lb-in.x D DM T T T T

Accordingly, the torque applied to gear C is also 4,140 lb-in. The torque on

gear B is found with the gear ratio:

48 teeth

0.6667 0.6667(4,140 lb-in.) 2,760 lb-in.72 teeth

B C CT T T

Shaft (1) has the same torque as gear B; therefore, T1 = −2,760 lb-in.

Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship

between shear stress and torque in a shaft.

Tc

J

The torques in both shafts have been determined, and the allowable shear stress is specified. Rearrange

the elastic torsion formula, putting the known terms on the right-hand side of the equation:

J T

c

Express the left-hand side of this equation in terms of the shaft diameter D:

4

332

/ 2 16

dT

dd

Solve for the minimum acceptable diameter in shaft (1):

3 311 2

1

16 16(2,760 lb-in.)2.342761 in.

(6,000 lb/in. )

1.328 in.

Td

d





3 322 2

2

16 16(4,140 lb-in.)3.514141 in.

(6,000 lb/in. )

1.520 in.

Td

d

Of these two values, d2 controls. Therefore, both shafts could have a diameter of 1.520 in. or more and

the shear stress constraint would be satisfied.

The angles of twists in shafts (1) and (2) can be expressed as:

1 1 2 21 2

1 1 2 2

T L T L

J G J G

The rotations of gears B and C are related since the arclengths turned by the two gears have the same

magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced.

C C B BR R

The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation

angle of gear C can be expressed in terms of T1.

1 1 1 11

1 1 1 1

B B B BC B

C C C C

R R R T L N T L

R R R J G N J G

The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft

(2):

2D C

and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:

1 1 2 2

1 1 2 2

3BD

C

N T L T L

N J G J G

Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 =

J2 = J and G1 = G2 = G. Factor these terms out to obtain:

1 1 2 2 1 1 2 2

1 13

(3 )

B B

C C

N NT L T L J T L T L

JG N G N

Express the polar moment of inertia in terms of diameter to obtain the following relationship:

4

1 1 2 2

4

32

(3 )

32 48 teeth( 2,760 lb-in.)(120 in.) (4,140 lb-in.)(96 in.)

rad 72 teeth(12,000,000 psi)(3 )180

10.022529 in.

1.779 in.

B

C

Nd T L T L

G N

d

Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements,

the rotation angle constraint controls.

Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 1.779 in. Ans.




6.41 Two solid steel shafts are connected by the

gears shown in Fig. P6.41. The design

requirements for the system require (1) that both

shafts must have the same diameter, (2) that the

maximum shear stress in each shaft must be less

than 50 MPa, and (3) that the rotation angle of

gear D must not exceed 3°. Determine the

minimum required diameter of the shafts if the

torque applied at gear D is TD = 750 N-m. The

shaft lengths are L1 = 2.5 m and L2 = 2.0 m.


G = 80 GPa and that the bearings shown allow

free rotation of the shafts. Fig. P6.41

Solution

Torque relationship: The torque on gear D creates a torque in shaft (2) of

T2 = 1,200 N-m.

2 20 750 N-mx D DM T T T T

Accordingly, the torque applied to gear C is also 750 N-m. The torque on

gear B is found with the gear ratio:

48 teeth

0.6667 0.6667(750 N-m) 500 N-m72 teeth

B C CT T T

Shaft (1) has the same torque as gear B; therefore, T1 = −500 N-m.

Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship

between shear stress and torque in a shaft. This equation can be rearranged in terms of the outside

diameter:

3

16

Tc Td

J


3 311 2

1

16 16(500 N-m)(1,000 mm/m)50,929.582 mm

(50 N/mm )

37.067 mm

Td

d


3 322 2

2

16 16(750 N-m)(1,000 mm/m)76,394.373 mm

(50 N/mm )

42.431 mm

Td

d

Of these two values, d2 controls. Therefore, both shafts could have a diameter of 42.431 mm or more

and the shear stress constraint would be satisfied.

The angles of twists in shafts (1) and (2) can be expressed as:

1 1 2 21 2

1 1 2 2

T L T L

J G J G




The rotations of gears B and C are related since the arclengths turned by the two gears have the same

magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced.

C C B BR R

The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation

angle of gear C can be expressed in terms of T1.

1 1 1 11

1 1 1 1

B B B BC B

C C C C

R R R T L N T L

R R R J G N J G

The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft

(2):

2D C

and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2:

1 1 2 2

1 1 2 2

3BD

C

N T L T L

N J G J G

Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 =

J2 = J and G1 = G2 = G. Factor these terms out to obtain:

1 1 2 2 1 1 2 2

1 13

(3 )

B B

C C

N NT L T L J T L T L

JG N G N

Express the polar moment of inertia in terms of diameter to obtain the following relationship:

4

1 1 2 2

2

2

4

32

(3 )

48 teeth32 ( 500 N-m)(2.5 m) (750 N-m)(2.0 m) (1,000 mm/m)

72 teeth

(80,000 N/mm )(3 )( / 180 )

5,673,986.284 mm

48.806 mm

B

C

Nd T L T L

G N

d

Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements,

the rotation angle constraint controls.

Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 48.8 mm. Ans.




6.42 The driveshaft of an automobile is being designed to transmit 180 hp at 3,500 rpm. Determine the

minimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not to

exceed 6,000 psi.

Solution

The torque in the driveshaft is:

550 lb-ft/s180 hp

1 hp270.109 lb-ft

3,500 rev 2 rad 1 min

min 1 rev 60 s

PT

The minimum diameter required for the shaft can be found from:

3 3

allow

(270.109 lb-ft)(12 in./ft)0.540217 in.

16 6,000 psi

1.401241 in. 1.401 in.

Td

d Ans.




6.43 A tubular steel shaft transmits 225 hp at 4,000 rpm. Determine the maximum shear stress produced

in the shaft if the outside diameter is D = 3.000 in. and the wall thickness is t = 0.125 in.

Solution

The torque in the tubular steel shaft is:

550 lb-ft/s225 hp

1 hp295.431 lb-ft


min 1 rev 60 s

PT

The polar moment of inertia of the shaft is:

4 4 4 4 4(3.000 in.) (2.750 in.) 2.337403 in.32 32

J D d

and the maximum shear stress in the shaft is

4

(295.431 lb-ft)(3.000 in./2)(12 in./ft)2,275 2,280 p.074 psi

2si

.337403 in.

Tc

J Ans.




6.44 A tubular steel shaft is being designed to transmit 225 kW at 1,700 rpm. The maximum shear stress

in the shaft must not exceed 30 MPa. If the outside diameter of the shaft is D = 75 mm, determine the

minimum wall thickness for the shaft.

Solution


1,000 N-m/s225 kW

1 kW1,263.877 N-m


min 1 rev 60 s

PT

The maximum shear stress in the shaft will be determined from the elastic torsion formula:

Tc

J where 4 4

32J D d

Rearrange this equation, grouping the diameter terms on the left-hand side of the equation:

4 4

32 / 2

D d T

D

Substitute the known values for T, , and D and solve for the inside diameter d:

4 4

2

4 4 4

4 4 4 4

(75 mm) (1,263.877 N-m)(1,000 mm/m)

32 75 mm / 2 30 N/mm

(75 mm) 16,092,181.76 mm

(75 mm) 16,092,181.76 mm 15,548,443.24 mm

62.7945 mm

d

d

d

d

The minimum wall thickness for the tubular steel shaft is thus

2

75 mm 62.7945 mm6.10275 mm

2 26.10 mm

D d t

D dt Ans.




6.45 A solid 20-mm-diameter bronze shaft transmits 11 kW at 25 Hz to the propeller of a small sailboat.

Determine the maximum shear stress produced in the shaft.

Solution


1,000 N-m/s11 kW

1 kW70.028175 N-m

25 rev 2 rad

s 1 rev

PT

The polar moment of inertia of the shaft is:

4 4(20 mm) 15,707.963 mm32

J

and the maximum shear stress in the shaft is

4

(70.028175 N-m)(20 mm/2)(1,000 mm/m)44.581 MPa

15,707.963 m44.6

mMPa

Tc

J Ans.




6.46 A steel propeller for a windmill transmits 5.5 kW at 65 rpm. If the allowable shear stress in the

shaft must be limited to 60 MPa, determine the minimum diameter required for a solid shaft.

Solution


1,000 N-m/s5.5 kW

1 kW808.017 N-m

65 rev 2 rad 1 min

min 1 rev 60 s

PT

The minimum diameter required for the shaft can be found from:

3 3

2

allow

(808.017 N-m)(1,000 mm/m)13,466.957 mm

16 60 N/mm

40.934 m 40.9 mmm

Td

d Ans.




6.47 A solid steel [G = 12,000 ksi] shaft with a 3-in. diameter must not twist more than 0.06 rad in a 16-

ft length. Determine the maximum horsepower that the shaft can transmit at 4 Hz.

Solution

Section properties: The polar moment of inertia of the shaft is:

4 4(3 in.) 7.952156 in.32

J

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque:

4(0.06 rad)(7.952156 in. )(12,000 ksi)

29.820585 kip-in. 2,485.05 lb-ft(16 ft)(12 in./ft)

JGT

L

Power transmission: The maximum power that can be transmitted at 4 Hz is:

4 rev 2 rad

(2,485.05 lb-ft) 62,456.088 lb-ft/ss 1 rev

P T

or in units of horsepower,

62,456.088 lb-ft/s

550 lb-ft/s

1

113.

hp

6 hpP Ans.




6.48 A tubular steel [G = 80 GPa] shaft with an outside diameter of D = 100 mm and a wall thickness of

t = 6 mm must not twist more than 0.05 rad in a 7-m length. Determine the maximum power that the

shaft can transmit at 375 rpm.

Solution


4 4 4 4 4(100 mm) (88 mm) 3,929,981.613 mm32 32

J D d

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque:

4 2(0.05 rad)(3,929,981.613 mm )(80,000 N/mm )

(7 m)(1,000 mm/m)

2,245,703.779 N-mm 2,245.704 N-m

JGT

L

Power transmission: The maximum power that can be transmitted at 375 rpm is:

375 rev 2 rad 1 min

(2,245.704 N-m) 88,188.581 N-m/smin 1 rev

88.2 kW60 s

P T Ans.




6.49 A solid 3-in.-diameter bronze [G = 6,000 ksi] shaft is 7-ft long. The allowable shear stress in the

shaft is 8 ksi and the angle of twist must not exceed 0.03 rad. Determine the maximum horsepower that

this shaft can deliver

(a) when rotating at 150 rpm.

(b) when rotating at 540 rpm.

Solution


4 4(3 in.) 7.952156 in.32

J

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not

exceed 8 ksi:

4(8 ksi)(7.952156 in. )

42.4115 kip-in.1.5 in.

JT

c (a)

Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed

0.03 rad:

4(0.03 rad)(7.952156 in. )(6,000 ksi)

17.0403 kip-in.(7 ft)(12 in./ft)

JGT

L (b)

Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft

is:

max 17.0403 kip-in. 1,420.028 lb-ftT

(a) Power transmission at 150 rpm: The maximum power that can be transmitted at 150 rpm is:

150 rev 2 rad 1 min

(1,420.028 lb-ft) 22,305.746 lb-ft/smin 1 rev 60 s

P T


22,305.746 lb-ft/s

550 lb-ft/s

1

40

hp

hp

.6P Ans.

(b) Power transmission at 540 rpm: The maximum power that can be transmitted at 540 rpm is:

540 rev 2 rad 1 min

(1,420.028 lb-ft) 80,300.684 lb-ft/smin 1 rev 60 s

P T


80,300.684 lb-ft/s

550 lb-ft/s

1

146.

hp

0 hpP Ans.




6.50 A hollow titanium [G = 43 GPa] shaft has an outside diameter of D = 50 mm and a wall thickness of t

= 1.25 mm. The maximum shear stress in the shaft must be limited to 150 MPa. Determine:

(a) the maximum power that can be transmitted by the shaft if the rotation speed must be limited to 20 Hz.

(b) the magnitude of the angle of twist in a 700-mm length of the shaft when 30 kW is being transmitted at

8 Hz.

Solution


4 4 4 4 4(50 mm) (47.5 mm) 113,817.540 mm32 32

J D d

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not

exceed 100 MPa:

2 4(150 N/mm )(113,817.540 mm )

682,905.237 N-mm 682.905 N-m50 mm/2

JT

c

(a) Power transmission: The maximum power that can be transmitted at 20 Hz is:

20 rev 2 rad

85(682.905 N-m) 85,816.403 N-m/ss 1 re

v

.8 kWP T Ans.

(b) Angle of twist: The torque in the shaft when 30 kW is being transmitted at 8 Hz is:

30,000 N-m/s

596.831 N-m8 rev 2 rad

s 1 rev

PT

The corresponding angle of twist is:

4 2

(596.831 N-m)(700 mm)(1,000 mm/m)

(113,817.540 mm )(43,000 N/mm0.0854 r

)ad

T L

JG Ans.




6.51 A tubular aluminum alloy [G = 4,000 ksi] shaft is being designed to transmit 400 hp at 1,500 rpm.

The maximum shear stress in the shaft must not exceed 6 ksi and the angle of twist is not to exceed 5° in

an 8-ft length. Determine the minimum permissible outside diameter if the inside diameter is to be three-

fourths of the outside diameter.

Solution

Torque from power transmission equation: The torque in the shaft when 400 hp is being transmitted

at 1,500 rpm is:

550 lb-ft/s(400 hp)

1 hp1,400.56 lb-ft


min 1 rev 60 s

PT

Polar moment of inertia: The inside diameter of the shaft is to be three-fourths of the outside

diameter; therefore, d = 0.75D. From this, the polar moment of inertia can be expressed as:

4 4

4 4 4 4 4 4 4(0.75 ) 1 (0.75) 0.683594 0.06711232 32 32 32

D DJ D d D D D (a)

Diameter based on shear stress: The maximum shear stress must not exceed 6 ksi; thus, from the

elastic torsion formula:

Tc J T

J c

Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of

the outside diameter D:

4

30.0671120.134224

/ 2

J DD

c D

Solve for the outside diameter D:

3 3(1,400.56 lb-ft)(12 in./ft)

0.134224 2.801120 in. 2.753175 in.6,000 psi

D D (b)

Diameter based on twist angle: The angle of twist is not to exceed 5° in an 8-ft length; thus, from the

angle of twist formula:

T L T L

JJG G

Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D:

2

4 4(1,400.56 lb-ft)(8 ft)(12 in./ft)0.067112 4.622180 in. 2.881972 in.

(5 )( /180 )(4,000,000 psi)D D (c)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is:

min 2.88 in.D Ans.




6.52 A tubular steel [G = 80 GPa] shaft is being designed to transmit 150 kW at 30 Hz. The maximum

shear stress in the shaft must not exceed 80 MPa and the angle of twist is not to exceed 6° in a 4-m

length. Determine the minimum permissible outside diameter if the ratio of the inside diameter to the

outside diameter is 0.80.

Solution

Torque from power transmission equation: The torque in the shaft when 150 kW is being transmitted

at 1,500 rpm is:

150,000 N-m/s

795.7747 N-m30 rev 2 rad

s 1 rev

PT

Polar moment of inertia: The ratio of inside diameter to outside diameter for the shaft is to be d/D =

0.80 or in other words d = 0.80D. From this, the polar moment of inertia can be expressed as:

4 4

4 4 4 4 4 4 4(0.80 ) 1 (0.80) 0.590400 0.05796232 32 32 32

D DJ D d D D D (a)

Diameter based on shear stress: The maximum shear stress must not exceed 80 MPa; thus, from the

elastic torsion formula:

Tc J T

J c

Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of

the outside diameter D:

4

30.0579620.115925

/ 2

J DD

c D

Solve for the outside diameter D:

3 3

2

(795.7747 N-m)(1,000 mm/m)0.115925 9,947.18375 mm 44.1070 mm

80 N/mmD D (b)

Diameter based on twist angle: The angle of twist is not to exceed 6° in a 4-m length; thus, from the

angle of twist formula:

T L T L

JJG G

Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D:

2

4 4

2

(795.7747 N-m)(4 m)(1,000 mm/m)0.057962 379,954.43 mm 50.5996 mm

(6 )( /180 )(80,000 N/mm )D D (c)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is:

min 50.6 mmD Ans.




6.53 An automobile engine supplies 180 hp at 4,200 rpm to a driveshaft. If the allowable shear stress in

the driveshaft must be limited to 5 ksi, determine:

(a) the minimum diameter required for a solid driveshaft.

(b) the maximum inside diameter permitted for a hollow driveshaft if the outside diameter is 2.00 in.

(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The

weight of a shaft is proportional to its cross-sectional area.)

Solution

Power transmission equation: The torque in the shaft is

550 lb-ft/s180 hp

1 hp225.0906 lb-ft


min 1 rev 60 s

PT

(a) Solid driveshaft: The minimum diameter required for a solid shaft can be found from:

3 3

allow

(225.0906 lb-ft)(12 in./ft)0.540217 in.

16 5,000 psi

1.401241 in. 1.401 in.

Td

d Ans.

(b) Hollow driveshaft: From the elastic torsion formula:

4 4

32 / 2

D dTc J T

J c D

Solve this equation for the maximum inside diameter d:

4 4

3

4 4 4

4 4 4 4

(2.00 in.) (225.0906 lb-ft)(12 in./ft)0.540217 in.

32 2.00 in. / 2 5,000 psi

(2.00 in.) 5.502605 in.

16 in. 5.502605 in. 10.497395 in.

1.800 in.

d

d

d

d Ans.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective

cross-sectional areas. The cross-sectional area of the solid shaft is

2 2

solid (1.401241 in.) 1.5421 in.4

A

and the cross-sectional area of the hollow shaft is

2 2 2

hollow (2.00 in.) (1.80 in.) 0.5969 in.4

A

The weight savings can be determined from

2 2

solid hollow

2

solid

( ) (1.5421 in. 0.5929 in. )weight savings (in percent)

1.5426

11.3

.%

in

A A

A Ans.




6.54 The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in

the impeller shaft must be limited to 80 MPa, determine:

(a) the minimum diameter required for a solid impeller shaft.

(b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm.

(c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The

weight of a shaft is proportional to its cross-sectional area.)

Solution

Power transmission equation: The torque in the shaft is

28,000 N-m/s

607.6825 N-m440 rev 2 rad 1 min

min 1 rev 60 s

PT

(a) Solid impeller shaft: The minimum diameter required for a solid shaft can be found from:

3 3

2

allow

(607.6825 N-m)(1,000 mm/m)7,596.0313 mm

16 80 N/mm

33.8209 m 3 8 mmm 3.

Td

d Ans.

(b) Hollow driveshaft: From the elastic torsion formula:

4 4

32 / 2

D dTc J T

J c D

Solve this equation for the maximum inside diameter:

4 4

3

2

4 4 4

4 4 4 4

(40 mm) (607.6825 N-m)(1,000 mm/m)7,596.0313 mm

32 40 mm / 2 80 N/mm

(40 mm) 1,547,450.779 mm

2,560,000 mm 1,547,450.779 mm 1,012,549.2

31.7 mm

21 mm

31.7215 mm

d

d

d

d Ans.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective

cross-sectional areas. The cross-sectional area of the solid shaft is

2 2

solid (33.8209 mm) 898.3803 mm4

A

and the cross-sectional area of the hollow shaft is

2 2 2

hollow (40 mm) (31.7215 mm) 466.3262 mm4

A

The weight savings can be determined from

2 2

solid hollow

2

solid

( ) (898.3803 mm48.1

466.3262 mm )weight savings (in percent)

898.3803 m%

m

A A

A Ans.




6.55 A motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Fig. P6.55. Gear B transfers 125

kW of power to operating machinery in the factory, and the remaining power in the shaft is transferred

by gear D. Shafts (1) and (2) are solid aluminum [G = 28 GPa] shafts that have the same diameter and an

allowable shear stress of = 40 MPa. Shaft (3) is a solid steel [G = 80 GPa] shaft with an allowable

shear stress of = 55 MPa. Determine:

(a) the minimum permissible diameter for

aluminum shafts (1) and (2).

(b) the minimum permissible diameter for

steel shaft (3).

(c) the rotation angle of gear D with respect

to flange A if the shafts have the minimum

permissible diameters as determined in (a)

and (b).

Fig. P6.55

Solution

Power transmission: The torque on flange A is

200,000 N-m/s

5,305.165 N-m6 rev 2 rad

s 1 rev

A

PT

thus, the torque in shaft segment (1) is T1 = 5,305.165 N-m. The torque transferred by gear B is

125,000 N-m/s

3,315.728 N-m6 rev 2 rad

s 1 rev

B

PT

therefore, the remaining torque in segments (2) and (3) of the shaft is

2 5,305.165 N-m 3,315.728 N-m 1,989.437 N-mA BT T T

(a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid

aluminum shaft can be found from:

3 311 2

1,allow

1

(5,305.165 N-m)(1,000 mm/m)132,629.125 mm

16 40 N/

87.

m

7 mm

m

Td

d Ans.

(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft

can be found from:

3 333 2

3,allow

3

(1,989.437 N-m)(1,000 mm/m)36,171.582 mm

16 55 N/mm

56.9 mmd

Td

Ans.

(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are:

4 4

1 2

4 4

3

(87.7 mm) 5,807,621 mm32

(56.9 mm) 1,029,080 mm32

J J

J





2

1 1

1 4 2

1 1

2

2 2

2 4 2

2 2

3 3

3

3 3

(5,305.165 N-m)(0.8 m)(1,000 mm/m)0.026100 rad

(5,807,621 mm )(28,000 N/mm )

(1,989.437 N-m)(0.4 m)(1,000 mm/m)0.004894 rad

(5,807,621 mm )(28,000 N/mm )

(1,989.437

T L

J G

T L

J G

T L

J G

2

4 2

N-m)(0.8 m)(1,000 mm/m)0.019332 rad

(1,029,080 mm )(80,000 N/mm )

Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft

segments:

1 2 3

0.026100 rad 0.004894 rad 0.019332 rad

0.050326 0.0503 radrad

D

Ans.




6.56 A motor supplies 150 hp at 520 rpm to flange A of the shaft shown in Fig. P6.56. Gear B transfers

90 hp of power to operating machinery in the factory, and the remaining power in the shaft is transferred

by gear D. Shafts (1) and (2) are solid aluminum [G = 4,000 ksi] shafts that have the same diameter and

an allowable shear stress of = 6 ksi. Shaft (3) is a solid steel [G = 12,000 ksi] shaft with an allowable

shear stress of = 8 ksi. Determine:

(a) the minimum permissible diameter for

aluminum shafts (1) and (2).

(b) the minimum permissible diameter for

steel shaft (3).

(c) the rotation angle of gear D with respect

to flange A if the shafts have the minimum

permissible diameters as determined in (a)

and (b).

Fig. P6.56

Solution

Power transmission: The torque on flange A is

550 lb-ft/s150 hp

1 hp1,515.033 lb-ft

520 rev 2 rad 1 min

min 1 rev 60 s

A

PT

thus, the torque in shaft segment (1) is T1 = 1,515.033 lb-ft. The torque transferred by gear B is

550 lb-ft/s90 hp

1 hp909.020 lb-ft

520 rev 2 rad 1 min

min 1 rev 60 s

B

PT

therefore, the remaining torque in segments (2) and (3) of the shaft is

2 1,515.033 lb-ft 909.020 lb-ft 606.013 lb-ftA BT T T

(a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid

aluminum shaft can be found from:

3 311

1,all

1

ow

(1,515.033 lb-ft)(12 in./ft)3.030066 in.

16 6,000 ps

2.49 in.

i

d

Td

Ans.

(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft

can be found from:

3 333

3,all

3

ow

(606.013 lb-ft)(12 in./ft)0.909020 in.

16 8,000

1.66

psi

7 in.d

Td

Ans.




(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are:

4 4

1 2

4 4

3

(2.49 in.) 3.773960 in.32

(1.667 in.) 0.758128 in.32

J J

J


2

1 1

1 4

1 1

2

2 2

2 4

2 2

3 3

3

3 3

(1,515.033 lb-ft)(2 ft)(12 in./ft)0.028904 rad

(3.773960 in. )(4,000,000 psi)

(606.013 lb-ft)(1 ft)(12 in./ft)0.005781 rad

(3.773960 in. )(4,000,000 psi)

(606.013 l

T L

J G

T L

J G

T L

J G

2

4

b-ft)(2 ft)(12 in./ft)0.019185 rad

(0.758128 in. )(12,000,000 psi)

Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft

segments:

1 2 3

0.028904 rad 0.005781 rad 0.019185 rad

0.053870 0.0539 radrad

D

Ans.




6.57 A motor supplies sufficient power to the

system shown in Fig. P6.57 so that gears C and

D provide torques of TC = 800 N-m and TD =

550 N-m, respectively, to machinery in a

factory. Power shaft segments (1) and (2) are

hollow steel tubes with an outside diameter of D

= 60 mm and an inside diameter of d = 50 mm.

If the power shaft [i.e., segments (1) and (2)]

rotates at 40 rpm, determine:

(a) the maximum shear stress in power shaft

segments (1) and (2).

(b) the power (in kW) that must be provided by

the motor as well as the rotation speed (in rpm).

(c) the torque applied to gear A by the motor.

Fig. P6.57

Solution

Equilibrium:

2

2

0

550 N-m

x D

D

M T T

T T

1

1

0

800 N-m 550 N-m

1,350 N-m

x C D

C D

M T T T

T T T

Section properties:

4 4 4

1 2(60 mm) (50 mm) 658,752.71 mm32

J J

(a) Shear stresses:

1 11 4

1

(1,350 N-m)(60 mm/2)(1,000 mm/m)61.480 MPa

658,752.71 m61.5

mMPa

T c

J Ans.

2 22 4

2

(550 N-m)(60 mm/2)(1,000 mm/m)25.047 MPa

658,752.71 m25.0

mMPa

T c

J Ans.

(b) Power transmission: Compute the power in shaft segment (1):

1

40 rev 2 rad 1 min(1,350 N-m) 5,654.867 N-m/s 5.65 kW

min 1 rev 60 sP T

This power in the shaft must be provided by the motor. Therefore:

motor 5.65 kWP Ans.




The rotation speed of the motor is found with the use of the gear ratio:

motor

72 teeth3 3(40 rpm)

24 teet12

h0 rpmA B B Ans.

(c) Motor torque: The motor must provide a torque of

24 teeth

(1,350 N-m 450 N-m)72 teeth

AA B

B

NT T

N Ans.




6.58 A motor supplies sufficient power to the

system shown in Fig. P6.58 so that gears C and

D provide torques of TC = 100 lb-ft and TD = 80

lb-ft, respectively, to machinery in a factory.

Power shaft segments (1) and (2) are hollow

steel tubes with an outside diameter of D = 1.75

in. and an inside diameter of d = 1.50 in. If the

power shaft [i.e., segments (1) and (2)] rotates

at 540 rpm, determine:

(a) the maximum shear stress in power shaft


(b) the power (in hp) that must be provided by

the motor as well as the rotation speed (in rpm).

(c) the torque applied to gear A by the motor.

Fig. P6.58

Solution

Equilibrium:

2

2

0

80 lb-ft

x D

D

M T T

T T

1

1

0

100 lb-ft 80 lb-ft

180 lb-ft

x C D

C D

M T T T

T T T

Section properties:

4 4 4

1 2(1.75 in.) (1.50 in.) 0.423762 in.32

J J

(a) Shear stresses:

1 11 4

1

(180 lb-ft)(1.75 in./2)(12 in./ft)4,460 4,460 p.049 psi

0si

.423762 in.

T c

J Ans.

2 22 4

2

(80 lb-ft)(1.75 in./2)(12 in./ft)1,982 1,982 .244 psi

0.423762 in.psi

T c

J Ans.

(b) Power transmission: Compute the power in shaft segment (1):

1

540 rev 2 rad 1 min(180 lb-ft) 10,178.760 lb-ft/s 18.507 hp

min 1 rev 60 sP T

This power in the shaft must be provided by the motor. Therefore:

motor 18.51 hpP Ans.




The rotation speed of the motor is found with the use of the gear ratio:

motor

72 teeth3 3(540 rpm 1,6)

24 t20 rpm

eethA B B Ans.

(c) Motor torque: The motor must provide a torque of

24 teeth

(180 lb-ft) 60 lb-ft72 t

720 lb-ine

.eth

AA B

B

NT T

N Ans.




6.59 A motor supplies 25 hp at 6 Hz to gear A

of the drive system shown in Fig. P6.59. Shaft

(1) is a solid 2.25-in.-diameter aluminum [G =

4,000 ksi] shaft with a length of L1 = 16 in.

Shaft (2) is a solid 1.5-in.-diameter steel [G =

12,000 ksi] shaft with a length of L2 = 12 in.

Shafts (1) and (2) are connected at flange C, and

the bearings shown permit free rotation of the

shaft. Determine:

(a) the maximum shear stress in shafts (1) and

(2).

(b) the rotation angle of gear D with respect to

gear B.

Fig. P6.59

Solution

Power transmission: The torque acting on gear A is:

550 lb-ft/s25 hp

1 hp364.730 lb-ft

6 rev 2 rad

s 1 rev

A

PT

Equilibrium: The torque applied to gear B is:

60 teeth

(364.730 lb-ft) 911.825 lb-ft24 teeth

BB A

A

NT T

N

The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B.

1 2 911.825 lb-ftD BT T T T

Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are:

4 4 4

1 1 (2.25 in.) 2.516112 in.32 32

J d 4 4 4

2 2 (1.5 in.) 0.497010 in.32 32

J d

(a) Maximum shear stresses:

1 11 4

1

(911.825 lb-ft)(2.25 in./2)(12 in./ft)4,892.326 psi

2.516112 in.4,890 psi

T c

J Ans.

2 22 4

2

(911.825 lb-ft)(1.5 in./2)(12 in./ft)16,511.601 psi

0.497010 in.16,510 psi

T c

J Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are:

1 1

1 4

1 1

2 2

2 4

2 2

(911.825 lb-ft)(16 in.)(12 in./ft)0.017395 rad

(2.516112 in. )(4,000,000 psi)

(911.825 lb-ft)(12 in.)(12 in./ft)0.022015 rad

(0.497010 in. )(12,000,000 psi)

T L

J G

T L

J G

Rotation angle of D relative to B:

1 2 0.017395 rad 0.022015 rad 0.039410 rad 0.0394 radD B

Ans.




6.60 A motor supplies 20 kW at 400 rpm to

gear A of the drive system shown in Fig. P6.60.

Shaft (1) is a solid 50-mm-diameter aluminum

[G = 28 GPa] shaft with a length of L1 = 1,200

mm. Shaft (2) is a solid 40-mm-diameter steel

[G = 80 GPa] shaft with a length of L2 = 750

mm. Shafts (1) and (2) are connected at flange

C, and the bearings shown permit free rotation

of the shaft. Determine:

(a) the maximum shear stress in shafts (1) and

(2).


gear B.

Fig. P6.60

Solution

Power transmission: The torque acting on gear A is:

20,000 N-m/s

477.465 N-m400 rev 2 rad 1 min

min 1 rev 60 s

A

PT

Equilibrium: The torque applied to gear B is:

60 teeth

(477.465 N-m) 1,193.662 N-m24 teeth

BB A

A

NT T

N

The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B.

1 2 1,193.662 N-mD BT T T T

Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are:

4 4 4

1 1 (50 mm) 613,592.32 mm32 32

J d 4 4 4

2 2 (40 mm) 251,327.41 mm32 32

J d

(a) Maximum shear stresses:

1 11 4

1

(1,193.662 N-m)(50 mm/2)(1,000 mm/m)48.634 MPa

613,592.32 mm48.6 MPa

T c

J Ans.

2 22 4

2

(1,193.662 N-m)(40 mm/2)(1,000 mm/m)94.989 MPa

251,327.41 mm95.0 MPa

T c

J Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are:

1 1

1 4 2

1 1

2 2

2 4 2

2 2

(1,193.662 N-m)(1,200 mm)(1,000 mm/m)0.083373 rad

(613,592.32 mm )(28,000 N/mm )

(1,193.662 N-m)(750 mm)(1,000 mm/m)0.044526 rad

(251,327.41 mm )(80,000 N/mm )

T L

J G

T L

J G

Rotation angle of D relative to B:

1 2 0.083373 rad 0.044526 rad 0.127899 rad 0.1279 radD B

Ans.




6.61 The motor shown in Fig. P6.61 supplies 10 hp

at 1,500 rpm at A. The bearings shown permit free

rotation of the shafts.

(a) Shaft (1) is a solid 0.875-in.-diameter steel

shaft. Determine the maximum shear stress

produced in shaft (1).

(b) If the shear stress in shaft (2) must be limited to

6,000 psi, determine the minimum acceptable

diameter for shaft (2) if a solid shaft is used.

Fig. P6.61

Solution

Power transmission: The torque acting in shaft (1) and on gear B is:

1

550 lb-ft/s10 hp

1 hp35.014 lb-ft


min 1 rev 60 s

B

PT T

Section properties: The polar moment of inertia for shaft (1) is:

4 4 4

1 1 (0.875 in.) 0.057548 in.32 32

J d

(a) Maximum shear stress in shaft (1):

1 11 4

1

(35.014 lb-ft)(0.875 in./2)(12 in./ft)3,194.258 psi

0.057548 in.3,194 psi

T c

J Ans.

(b) Equilibrium: The torque applied to gear C is:

30 teeth

(35.014 lb-ft) 21.884 lb-ft48 teeth

CC B

B

NT T

N

The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C.

2 21.884 lb-ftD CT T T

The minimum diameter required for solid shaft (2) can be found from:

3 32

2

2,allow

2

(21.884 lb-ft)(12 in./ft)0.043768 in.

16 6,000 psi

0.606328 in. 0.606 in.

Td

d Ans.




6.62 The motor shown in Fig. P6.62 supplies 12

kW at 15 Hz at A. The bearings shown permit free


(a) Shaft (2) is a solid 35-mm-diameter steel shaft.

Determine the maximum shear stress produced in

shaft (2).

(b) If the shear stress in shaft (1) must be limited to

40 MPa, determine the minimum acceptable

diameter for shaft (1) if a solid shaft is used.

Fig. P6.62

Solution

Power transmission: The torque acting in shaft (1) and on gear B is:

1

12,000 N-m/s127.324 N-m

15 rev 2 rad

s 1 rev

B

PT T

The torque applied to gear C is:

30 teeth

(127.324 N-m) 79.577 N-m48 teeth

CC B

B

NT T

N

The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C.

2 79.577 N-mD CT T T

Section properties: The polar moment of inertia for shaft (2) is:

4 4 4

2 2 (35 mm) 147,323.515 mm32 32

J d

(a) Maximum shear stress in shaft (2):

2 22 4

2

(79.577 N-m)(35 mm/2)(1,000 mm/m)9.453 MPa

147,323.515 m9.45

mMPa

T c

J Ans.

(b) Minimum diameter for solid shaft (1):

3 31

1 2

1,allow

1

(127.324 N-m)(1,000 mm/m)3,183.099 mm

16 40 N/mm

25.309 m 25.3 m mm

Td

d Ans.





kW at 15 Hz at A. Shafts (1) and (2) are each

solid 25-mm-diameter steel [G = 80 GPa] shafts

with lengths of L1 = 900 mm and L2 = 1,200 mm,

respectively. The bearings shown permit free

rotation of the shafts. Determine:

(a) the maximum shear stress produced in shafts

(1) and (2).


flange A.

Fig. P6.63

Solution

Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is:

1

9,000 N-m/s95.493 N-m

15 rev 2 rad

s 1 rev

B

PT T

The magnitude of the torque applied to gear C is:

54 teeth

(95.493 N-m) 143.239 N-m36 teeth

CC B

B

NT T

N

The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is

equal to the torque applied to gear C.

2 143.239 N-mD CT T T

To determine the proper signs for T1 and T2, consider equilibrium. First,

consider a FBD that cuts through shaft (2) and includes gear D.

2 20 143.239 N-mx D DM T T T T

Since T2 is positive, the torque in shaft (1) must be negative; therefore,

1 95.493 N-mT

Section properties: The polar moments of inertia for shafts (1) and (2) are equal:

4 4 4

1 1 2(25 mm) 38,349.520 mm32 32

J d J

(a) Maximum shear stress magnitudes:

1 11 4

1

(95.493 N-m)(25 mm/2)(1,000 mm/m)31.126 MPa

38,349.520 mm31.1 MPa

T c

J Ans.

2 22 4

2

(143.239 N-m)(25 mm/2)(1,000 mm/m)46.689 MPa

38,349.520 mm46.7 MPa

T c

J Ans.




Angles of twist in shafts (1) and (2):

1 11 4 2

1 1

( 95.493 N-m)(900 mm)(1,000 mm/m)0.028013 rad

(38,349.520 mm )(80,000 N/mm )

T L

J G

2 22 4 2

2 2

(143.239 N-m)(1,200 mm)(1,000 mm/m)0.056027 rad

(38,349.520 mm )(80,000 N/mm )

T L

J G

Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1):

1 1 0 rad ( 0.028013 rad) 0.028013 radB A B A



36 teeth

( 0.028013 rad) 0.018676 rad54 teeth

BC B

C

N

N

(b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from:


Ans.





hp at 1,800 rpm at A. Shafts (1) and (2) are each

solid 2-in.-diameter steel [G = 12,000 ksi] shafts

with lengths of L1 = 75 in. and L2 = 90 in.,

respectively. The bearings shown permit free


(a) Determine the maximum shear stress

produced in shafts (1) and (2).

(b) Determine the rotation angle of gear D with

respect to flange A.

Fig. P6.64

Solution

Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is:

1

550 lb-ft/s150 hp

1 hp437.676 lb-ft


min 1 rev 60 s

B

PT T

The magnitude of the torque applied to gear C is:

54 teeth

(437.676 lb-ft) 656.514 lb-ft36 teeth

CC B

B

NT T

N

The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is

equal to the torque applied to gear C.

2 656.514 lb-ftD CT T T

To determine the proper signs for T1 and T2, consider equilibrium. First,

consider a FBD that cuts through shaft (2) and includes gear D.

2 20 656.514 lb-ftx D DM T T T T

Since T2 is positive, the torque in shaft (1) must be negative; therefore,

1 437.676 lb-ftT

Section properties: The polar moments of inertia for shafts (1) and (2) are equal:

4 4 4

1 1 2(2 in.) 1.570796 in.32 32

J d J

(a) Maximum shear stress magnitudes:

1 11 4

1

(437.676 lb-ft)(2 in./2)(12 in./ft)3,343 3,340 p.599 psi

1si

.570796 in.

T c

J Ans.

2 22 4

2

(656.514 lb-ft)(2 in./2)(12 in./ft)5,015 5,020 p.399 psi

1si

.570796 in.

T c

J Ans.




Angles of twist in shafts (1) and (2):

1 11 4

1 1

( 437.676 lb-ft)(75 in.)(12 in./ft)0.020897 rad

(1.570796 in. )(12,000,000 psi)

T L

J G

2 22 4

2 2

(656.514 lb-ft)(90 in.)(12 in./ft)0.037615 rad

(1.570796 in. )(12,000,000 psi)

T L

J G

Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1):

1 1 0 rad ( 0.020897 rad) 0.020897 radB A B A



36 teeth

( 0.020897 rad) 0.013932 rad54 teeth

BC B

C

N

N

(b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from:


Ans.




6.65 The gear train shown in Fig. P6.65

transmits power from a motor to a machine at E.

The motor turns at a frequency of 50 Hz. The

diameter of solid shaft (1) is 25 mm, the

diameter of solid shaft (2) is 32 mm, and the

allowable shear stress for each shaft is 60 MPa.

Determine:

(a) the maximum power that can be transmitted

by the gear train.

(b) the torque provided at gear E.

(c) the rotation speed of gear E (in Hz).

Fig. P6.65

Solution

Section properties:

4 4 4

1 1

4 4 4

2 2

(25 mm) 38,349.52 mm32 32

(32 mm) 102,943.71 mm32 32

J d

J d

Allowable torques in the shafts:

2 4

allow 11,allow

1

(60 N/mm )(38,349.52 mm )184,078 N-mm

25 mm/2

JT

c

2 4

allow 22,allow

2

(60 N/mm )(102,943.71 mm )386,039 N-mm

32 mm/2

JT

c

Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by:

B DB A D C

A C

N NT T T T

N N (a)

Equilibrium: The magnitudes of the torques in the gears and shafts are related by:

1 2B C D ET T T T T T (b)

Equations (a) and (b) express the relationship between the torque in shafts (1) and (2):

2 1D

C

NT T

N (c)

Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque

for shaft (2):

1,allow 2,allowcheck

60 teeth(184,078 N-mm) 368,156 N-mm 386,039 N-m

30 teeth

D

C

NT T

N

O.K.




This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact,

the torques throughout the gear train can be determined:

1 2184,078 N-mm 368,156 N-mm

184,078 N-mm 368,156 N-mm

24 teeth(184,078 N-mm) 61,359 N-mm

72 teeth

B C D E

A

A B

B

T T

T T T T

NT T

N

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found

from the torque and rotation speed at A:

50 rev 2 rad

19.(61.359 N-m) 19,276.571 28 N-m/s ks 1 rev

WA AP T Ans.

(b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the

torque supplied by the motor at A and the torque available at gear E:

D D B D B DE D C B A A

C C A C A C

N N N N N NT T T T T T

N N N N N N (d)

The torque available at gear E is thus:

72 teeth 60 teeth

(61.359 N-m) 368.154 N-m24 teeth 30 teeth

368 N-mB DE A

A C

N NT T

N N Ans.

This result could also be found from the torque in shaft (2), as calculated previously.

(c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios,

the speed of gear E can be found from:

24 teeth 30 teeth

(50 Hz)72 teeth 60 te

8.3et

3 zh

HA CE A

B D

N N

N N Ans.




6.66 The gear train shown in Fig. P6.66

transmits power from a motor to a machine at

E. The motor turns at 2,250 rpm. The diameter

of solid shaft (1) is 1.50 in., the diameter of

solid shaft (2) is 2.00 in., and the allowable

shear stress for each shaft is 6 ksi. Determine:

(a) the maximum power that can be

transmitted by the gear train.

(b) the torque provided at gear E.

(c) the rotation speed of gear E (in Hz).

Fig. P6.66

Solution

Section properties:

4 4 4

1 1

4 4 4

2 2

(1.50 in.) 0.497010 in.32 32

(2.00 in.) 1.570796 in.32 32

J d

J d

Allowable torques in the shafts:

4

allow 11,allow

1

(6,000 psi)(0.497010 in. )3,976.1 lb-in.

1.50 in./2

JT

c

4

allow 22,allow

2

(6,000 psi)(1.570796 in. )9,424.8 lb-in.

2.00 in./2

JT

c

Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by:

B DB A D C

A C

N NT T T T

N N (a)

Equilibrium: The magnitudes of the torques in the gears and shafts are related by:

1 2B C D ET T T T T T (b)

Equations (a) and (b) express the relationship between the torque in shafts (1) and (2):

2 1D

C

NT T

N (c)

Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque

for shaft (2):

1,allow 2,allowcheck

60 teeth(3,976.1 lb-in.) 7,952.2 lb-in. 9,424.8 lb-in.

30 teeth

D

C

NT T

N

O.K.




This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact,

the torques throughout the gear train can be determined:

1 23,976.1 lb-in. 7,952.2 lb-in.

3,976.1 lb-in. 7,952.2 lb-in.

24 teeth(3,976.1 lb-in.) 1,325.4 lb-in. 110.45 lb-ft

72 teeth

B C D E

AA B

B

T T

T T T T

NT T

N

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found

from the torque and rotation speed at A:


(110.45 lb-ft) 26,023.381 lb-ft/smin 1 rev

47.3 hp60 s

A AP T Ans.

(b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the

torque supplied by the motor at A and the torque available at gear E:

D D B D B DE D C B A A

C C A C A C

N N N N N NT T T T T T

N N N N N N (d)

The torque available at gear E is thus:

72 teeth 60 teeth

(110.45 lb-ft) 662.7 lb-ft24 teeth 30 teet

663 lb- th

fB DE A

A C

N NT T

N N Ans.

This result could also be found from the torque in shaft (2), as calculated previously.

(c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios,

the speed of gear E can be found from:

24 teeth 30 teeth

(2,250 rpm)72 teeth 60 teeth

375 rpmA CE A

B D

N N

N N Ans.




6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in.

and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless

steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.67. The allowable shear stress of tube (1) is 27 ksi,

and the allowable shear stress of core (2) is 60 ksi. Determine:

(a) the allowable torque T that can be applied to

the tube-and-core assembly.

(b) the corresponding torques produced in tube (1)

and core (2).

(c) the angle of twist produced in a 10-in. length

of the assembly by the allowable torque T.

Fig. P6.67

Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(1.75 in.) (1.25 in.) 0.681087 in.32

(1.25 in.) 0.239684 in.32

J

J

Equilibrium:

1 2 0xM T T T (a)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the

angles of twist in both members must be equal; therefore,

1 2 (b)

Torque-Twist Relationships:

1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of

deformation relationship [Eq. (b)] to obtain the compatibility equation:

1 1 2 2

1 1 2 2

T L T L

J G J G (d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to

rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as:

Tc T

J J c

which allows Eq. (d) to be rewritten as:

1 1 2 2

1 1 2 2

L L

G c G c

(e)

Solve Eq. (e) for 1:

2 1 11 2 2 2

1 2 2

6,500 ksi 1.75 in./20.728

12,500 ksi 1.25 in./2

L G c

L G c

(f)




Assume the core controls: If the shear stress in core (2) reaches its allowable value of 60 ksi (in other

words, if the core controls), then the shear stress in tube (1) will be:

1 0.728(60 ksi) 43.68 ksi 27 ksi N.G.

which is larger than the 27-ksi allowable shear stress for the tube. Therefore, the shear stress in the tube

controls.

Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given

that the shear stress in tube (1) is at its 27-ksi allowable value:

2

27 ksi37.088 ksi 60 ksi

0.728 O.K.

Now that the maximum shear stresses in the tube and the core are known, the torques in each component

can be computed:

4

1 11

1

4

2 22

2

(27 ksi)(0.681087 in. )21.016 kip-in.

1.75 in./2

(37.088 ksi)(0.239684 in. )14.223 kip-in.

1.25 in./2

JT

c

JT

c

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed:

max 1 2 21.016 kip-in. 14.223 kip-in. 35.239 kip-in. 35.2 kip-in.T T T Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is:

1 21.0 kip-in.T Ans.

and the torque in core (2) is:


(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the

same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of

the entire assembly.

1 11 4

1 1

(21.016 kip-in.)(10 in.)0.047473 rad

(0.681087 in. )(0.0475 ra

6,500 ksi)d

T L

J G Ans.




6.68 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in.

and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless

steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.68. If a torque of T = 20 kip-in. is applied to the

tube-and-core assembly, determine:

(a) the torques produced in tube (1) and core (2).

(b) the maximum shear stress in the bronze tube

and in the stainless steel core.

(c) the angle of twist produced in a 10-in. length

of the assembly.

Fig. P6.68

Solution

Section Properties: For tube (1) and core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(1.75 in.) (1.25 in.) 0.681087 in.32

(1.25 in.) 0.239684 in.32

J

J

Equilibrium:

1 2 20 kip-in. 0xM T T (a)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the


1 2 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

T L T L

J G J G (d)

Solve the Equations: Solve Eq. (d) for T1:

4

2 1 11 2 2 24

1 2 2

0.681087 in. 6,500 ksi1.477632

0.239684 in. 12,500 ksi

L J GT T T T

L J G

and substitute this result into Eq. (a) to compute the torque T2 in core (2):

1 2 2 2 2

2

1.477632 2.477632 20 kip-in.

8.072 kip-in. 8.07 kip-in.

T T T T T

T

Ans.

The torque in tube (1) is therefore:

1 220 kip-in. 20 kip-in. 8.072 kip-in. 11.928 ki 11.93 kip-in.p-in.T T Ans.




(b) Maximum Shear Stress: The maximum shear stress in tube (1) is:

1 11 4

1

(11.928 kip-in.)(1.75 in. / 2)

0.681087 in15.32 ksi

.

T c

J Ans.

The maximum shear stress in core (2) is:

2 22 4

2

(8.072 kip-in.)(1.25 in. / 2)

0.239684 in.21.0 ksi

T c

J Ans.

(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the

same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of

the entire assembly.

1 11 4

1 1

(11.928 kip-in.)(10 in.)0.026943 rad

(0.681087 in. )(0.0269 ra

6,500 ksi)d

T L

J G Ans.




6.69 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the

ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.69a. The cross-sectional dimensions of

the assembly are shown in Fig. P6.69b. If a torque of T = 1,100 N-m is applied to the composite

assembly, determine:

(a) the maximum shear stress in the aluminum tube and in the steel core.

(b) the rotation angle of end B relative to end A.

Fig. P6.69a Tube-and-Core Composite Shaft Fig. P6.69b Cross-Sectional Dimensions

Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(50 mm) (40 mm) 362,265 mm32

(20 mm) 15,708.0 mm32

J

J

Equilibrium:

1 2 1,100 N-m 0xM T T (a)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the


1 2 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

T L T L

J G J G (d)


4

2 1 11 2 2 24

1 2 2

362,265 mm 28 GPa8.071858

15,708.0 mm 80 GPa

L J GT T T T

L J G

and substitute this result into Eq. (a) to compute the torque T2 in steel core (2):

1 2 2 2 2

2

8.071858 9.071858 1,100 N-m

121.254 N-m

T T T T T

T




The torque in aluminum tube (1) is therefore:

1 21,100 N-m 1,100 N-m 121.254 N-m 978.746 N-mT T

(a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is:

1 11 4

1

(978.746 N-m)(50 mm / 2)(1,000 mm/m)

362,265 mm67.5 MPa

T c

J Ans.

The maximum shear stress in steel core (2) is:

2 22 4

2

(121.254 N-m)(20 mm / 2)(1,000 mm/m)

15,708.0 m77.2 MPa

m

T c

J Ans.

(b) Rotation Angle of B relative to A: Since both the tube and the core twist exactly the same amount

[i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire

assembly.

1 11 4 2

1 1

(978.746 N-m)(300 mm)(1,000 mm/m)0.028947 rad

(362,265 mm )(0.028

28,000 N/r

m )9 ad

m

T L

J G Ans.




6.70 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the

ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.70a. The cross-sectional dimensions of

the assembly are shown in Fig. P6.70b. The allowable shear stress of aluminum tube (1) is 90 MPa, and

the allowable shear stress of steel core (2) is 130 MPa. Determine:

(a) the allowable torque T that can be applied to the composite shaft.

(b) the corresponding torques produced in tube (1) and core (2).

(c) the angle of twist produced by the allowable torque T.

Fig. P6.70a Tube-and-Core Composite Shaft Fig. P6.70b Cross-Sectional Dimensions

Solution

Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(50 mm) (40 mm) 362,265 mm32

(20 mm) 15,708.0 mm32

J

J

Equilibrium:

1 2 0xM T T T (a)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the


1 2 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

T L T L

J G J G (d)



Tc T

J J c





1 1 2 2

1 1 2 2

L L

G c G c

(e)


2 1 11 2 2 2

1 2 2

28 GPa 50 mm/20.875

80 GPa 20 mm/2

L G c

L G c

(f)

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 130 MPa (in

other words, if the core controls), then the shear stress in tube (1) will be:

1 0.875(130 MPa) 113.75 MPa 90 MPa N.G.

which is larger than the 90 MPa allowable shear stress for the tube. Therefore, the shear stress in the

tube controls.

Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given

that the shear stress in tube (1) is at its 90 MPa allowable value:

2

90 MPa102.857 MPa 130 MPa

0.875 O.K.

Now that the maximum shear stresses in the tube and the core are known, the torques in each component

can be computed:

2 4

1 11

1

4

2 22

2

(90 N/mm )(362,265 mm )1,304,154 N-mm

50 mm/2

(102.857 MPa)(15,708.0 mm )161,568 N-mm

20 mm/2

JT

c

JT

c


max 1 2 1,304,154 N-mm 161,568 N-mm 1,465,721 N-mm 1,466 N-mT T T Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is:

1 1,304 N-mT Ans.


2 161.6 N-mT Ans.

(c) Angle of Twist: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either

torque-twist relationship can be used to compute the angle of twist of the entire assembly.

1 11 4 2

1 1

(1,304,154 N-mm)(300 mm)0.038571 rad

(362,265 mm )(28,0.0386 ra

000 N/d

mm )

T L

J G Ans.




6.71 The composite shaft shown in Fig. P6.71 consists of a bronze sleeve (1) securely bonded to an inner

steel core (2). The bronze sleeve has an outside diameter of 35 mm, an inside diameter of 25 mm, and a

shear modulus of G1 = 45 GPa. The solid steel core has a diameter of 25 mm and a shear modulus of G2

= 80 GPa. The allowable shear stress of sleeve (1) is 180 MPa, and the allowable shear stress of core (2)

is 150 MPa. Determine:

(a) the allowable torque T that can be applied to the

composite shaft.

(b) the corresponding torques produced in sleeve (1) and

core (2).

(c) the rotation angle of end B relative to end A that is

produced by the allowable torque T.

Fig. P6.71

Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(35 mm) (25 mm) 108,974 mm32

(25 mm) 38,349.5 mm32

J

J

Equilibrium:

1 2 0xM T T T (a)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the


1 2 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

T L T L

J G J G (d)



Tc T

J J c


1 1 2 2

1 1 2 2

L L

G c G c

(e)





2 1 11 2 2 2

1 2 2

45 GPa (35 mm/2)0.7875

80 GPa (25 mm/2)

L G c

L G c

(f)

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 150 MPa (in

other words, if the core controls), then the shear stress in sleeve (1) will be:

1 0.7875(150 MPa) 118.125 MPa 180 MPa O.K.

This calculation shows that the shear stress in the core does in fact controls.

Now that the maximum shear stresses in the sleeve and the core are known, the torques in each

component can be computed:

2 4

1 11

1

4

2 22

2

(118.125 N/mm )(108,974 mm )735,574 N-mm

35 mm/2

(150 MPa)(38,349.5 mm )460,194 N-mm

25 mm/2

JT

c

JT

c


max 1 2 735,574 N-mm 460,194 N-mm 1,195,769 N-mm 1,196 N-mT T T Ans.

(b) Torques in each component: As computed previously, the torque in sleeve (1) is:

1 736 N-mT Ans.


2 460 N-mT Ans.

(c) Angle of Twist: Since both the sleeve and the core twist exactly the same amount [i.e., Eq. (b)],

either torque-twist relationship can be used to compute the angle of twist of the entire assembly.

1 11 4 2

1 1

(735,574 N-mm)(360 mm)0.054000 rad

(108,974 mm )(45,00.0540 ra

00 N )d

/mm

T L

J G Ans.




6.72 The composite shaft shown in Fig. P6.72 consists

of a bronze sleeve (1) securely bonded to an inner steel

core (2). The bronze sleeve has an outside diameter of

35 mm, an inside diameter of 25 mm, and a shear

modulus of G1 = 45 GPa. The solid steel core has a

diameter of 25 mm and a shear modulus of G2 = 80 GPa.

The composite shaft is subjected to a torque of T = 900

N-m. Determine:

(a) the maximum shear stresses in the bronze sleeve and

the steel core.

(b) the rotation angle of end B relative to end A.

Fig. P6.72

Solution

Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are:

4 4 4

1

4 4

2

(35 mm) (25 mm) 108,974 mm32

(25 mm) 38,349.5 mm32

J

J

Equilibrium:

1 2 900 N-m 0xM T T (a)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the


1 2 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

T L T L

J G J G (d)


4

2 1 11 2 2 24

1 2 2

108,974 mm 45 GPa1.598401

38,349.5 mm 80 GPa

L J GT T T T

L J G

and substitute this result into Eq. (a) to compute the torque T2 in steel core (2):

1 2 2 2 2

2

1.598401 2.598401 900 N-m

346.367 N-m

T T T T T

T

The torque in bronze sleeve (1) is therefore:

1 2900 N-m 900 N-m 346.367 N-m 553.633 N-mT T




(a) Maximum Shear Stress: The maximum shear stress in bronze sleeve (1) is:

1 11 4

1

(553.633 N-m)(35 mm / 2)(1,000 mm/m)

108,974 mm88.9 MPa

T c

J Ans.

The maximum shear stress in steel core (2) is:

2 22 4

2

(346.367 N-m)(25 mm / 2)(1,000 mm/m)

38,349.5 mm112.9 MPa

T c

J Ans.

(b) Rotation Angle of B relative to A: Since both the sleeve and the core twist exactly the same

amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the

entire assembly.

1 11 4 2

1 1

(553.633 N-m)(360 mm)(1,000 mm/m)0.040643 rad

(108,974 mm )(0.040

45,000 N/r

m )6 ad

m

T L

J G Ans.




6.73 The composite shaft shown in Fig. P6.73

consists of two steel pipes that are connected at

flange B and securely attached to rigid walls at A

and C. Steel pipe (1) has an outside diameter of

168 mm and a wall thickness of 7 mm. Steel pipe

(2) has an outside diameter of 114 mm and a wall

thickness of 6 mm. Both pipes are 3-m long and

have a shear modulus of 80 GPa. If a concentrated

torque of 20 kN-m is applied to flange B,

determine:

(a) the maximum shear stress magnitudes in pipes

(1) and (2).

(b) the rotation angle of flange B relative to

support A.

Fig. P6.73

Solution

Section Properties: The polar moments of inertia for the two steel pipes are:

4 4 4

1

4 4 4

2

(168 mm) (154 mm) 22,987,183 mm32

(114 mm) (102 mm) 5,954,575 mm32

J

J

Equilibrium:

1 2 20 kN-m 0xM T T (a)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed

supports at A and C, the sum of the angles of twist in the two pipes must equal zero:

1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)


2 1 1

1 2

1 2 2

4

2 4

4

2 24

3 m 22,987,183 mm 80 GPa

3 m 5,954,575 mm 80 GPa

22,987,183 mm3.860424

5,954,575 mm

L J GT T

L J G

T

T T




and substitute this result into Eq. (a) to compute the torque T2:

1 2 2 2 2

2

3.860424 4.860424 20 kN-m

4.1149 kN-m

T T T T T

T

The torque in member (1) is therefore:

1 2 20 kN-m 4.1149 kN-m 20 kN-m 15.8851 kN-mT T

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:

2

1 11 4

1

(15.8851 kN-m)(168 mm / 2)(1,000)

22,987,183 mm58.0 MPa

Tc

J Ans.

The maximum shear stress magnitude in member (2) is:

2

2 22 4

2

(4.1149 kN-m)(114 mm / 2)(1,000)

5,954,575 mm39.4 MPa

T c

J Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be

defined by the difference in rotation angles at the two ends; hence,

1 B A

Since joint A is restrained from rotating, A = 0 and thus

1 B

The rotation angle at B can be determined by computing the angle of twist in member (1):

1 11

1 1

4 2

(15.8851 kN-m)(3,000 mm)(1,000 mm/m)(1,000 N/kN)

(22,987,183 mm )(80,000 N/mm )

0.025914 ra 0.0259 rd ad

T L

J G

Ans.




6.74 The composite shaft shown in Fig. P6.74 consists of two steel pipes that are connected at flange B

and securely attached to rigid walls at A and C. Steel pipe (1) has an outside diameter of 8.625 in., a wall

thickness of 0.322 in., and a length of L1 = 15 ft. Steel pipe (2) has an outside diameter of 6.625 in., a

wall thickness of 0.280 in., and a length of L2 = 25 ft. Both pipes have a shear modulus of 12,000 ksi.

If a concentrated torque of 36 kip-ft is

applied to flange B, determine:

(a) the maximum shear stress magnitudes

in pipes (1) and (2).

(b) the rotation angle of flange B relative

to support A.

Fig. P6.74

Solution


4 4 4

1

4 4 4

2

(8.625 in.) (7.981 in.) 144.978 in.32

(6.625 in.) (6.065 in.) 56.284 in.32

J

J

Equilibrium:

1 2 36 kip-ft 0xM T T (a)



1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)


2 1 11 2

1 2 2

4

2 24

25 ft 144.978 in. 12,000 ksi4.293050

15 ft 56.284 in. 12,000 ksi

L J GT T

L J G

T T





1 2 2 2 2

2

4.293050 5.293050 36 kip-ft

6.801 kip-ft

T T T T T

T


1 2 36 kip-ft 6.801 kip-ft 36 kip-ft 29.199 kip-ftT T


1 11 4

1

(29.199 kip-ft)(8.625 in. / 2)(12 in./ft)

144.978 in.10.42 ksi

T c

J Ans.


2 22 4

2

(6.801 kip-ft)(6.625 in. / 2)(12 in./ft)

56.284 in4.80 ksi

.

T c

J Ans.



1 B A


1 B


1 1

1

1 1

2

4

(29.199 kip-ft)(15 ft)(12 in./ft)

(144.978 in. )(12,000 ksi)

0.036252 rad 0.0363 rad

T L

J G

Ans.




6.75 The composite shaft shown in Fig. P6.75 consists of a solid brass segment (1) and a solid aluminum

segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Brass

segment (1) has a diameter of 1.00 in., a length of L1 = 15 in., a shear modulus of 5,600 ksi, and an

allowable shear stress of 8 ksi. Aluminum segment (2) has a diameter of 0.75 in., a length of L2 = 20 in.,

a shear modulus of 4,000 ksi, and an allowable shear stress of 6 ksi. Determine:

(a) the allowable torque TB that can be

applied to the composite shaft at flange B.

(b) the magnitudes of the internal torques in


(c) the rotation angle of flange B that is

produced by the allowable torque TB.

Fig. P6.75

Solution

Section Properties: The polar moments of inertia for the two shafts are:

4 4

1

4 4

2

(1.00 in.) 0.098175 in.32

(0.75 in.) 0.031063 in.32

J

J

Equilibrium:

1 2 0x BM T T T (a)

Geometry of Deformation Relationship: Since the two shafts are securely attached to fixed supports

at A and C, the sum of the angles of twist in the two members must equal zero:

1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)



Tc T

J J c





1 1 2 2

1 1 2 2

L L

G c G c

(e)


2 1 11 2 2 2

1 2 2

20 in. 5,600 ksi 1.00 in./22.488889

15 in. 4,000 ksi 0.75 in./2

L G c

L G c

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 6 ksi,

then the shear stress magnitude in shaft (1) will be:

1 2.488889(6 ksi) 14.9333 ksi 8 ksi N.G.

which is larger than the 8 ksi allowable shear stress magnitude for shaft (1). Therefore, the shear stress

magnitude in shaft (1) must control.

Shaft (1) actually controls: Rearrange Eq. (f) to solve for the shear stress magnitude in shaft (2) given

that the shear stress magnitude in shaft (1) is at its 8 ksi allowable magnitude:

2

8 ksi3.214 ksi 6 ksi

2.488889

O.K.

Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in

each component can be computed:

4

1 11

1

4

2 22

2

(8 ksi)(0.098175 in. )1.571 kip-in.

1.00 in./2

(3.214 ksi)(0.031063 in. )0.266 kip-in.

0.75 in./2

JT

c

JT

c

Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated

equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving

it a negative value. Therefore, by inspection

2 0.266 kip-in.T

(a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed:

,max 1 2 1.571 kip-in. ( 0.266 kip-in.) 1.837 kip-in.BT T T Ans.

(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is:


and the torque magnitude in shaft (2) is:


(c) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be


1 B A


1 B


1 11 4

1 1

(1.571 kip-in.)(15 in.)0.042857 rad

(0.098175 in. )(0.0429 ra

5,600 ksi)d

T L

J G Ans.





consists of a solid brass segment (1) and a solid

aluminum segment (2) that are connected at flange

B and securely attached to rigid walls at A and C.

Brass segment (1) has a diameter of 18 mm, a

length of L1 = 235 mm, and a shear modulus of 39

GPa. Aluminum segment (2) has a diameter of 24

mm, a length of L2 = 165 mm, and a shear modulus

of 28 GPa. If a concentrated torque of 270 N-m is

applied to flange B, determine:

(a) the maximum shear stress magnitudes in



support A.

Fig. P6.76

Solution


4 4

1

4 4

2

(18 mm) 10,306.0 mm32

(24 mm) 32,572.0 mm32

J

J

Equilibrium:

1 2 270 N-m 0xM T T (a)



1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)


2 1 11 2

1 2 2

4

2 24

165 mm 10,306.0 mm 39 GPa0.309434

235 mm 32,572.0 mm 28 GPa

L J GT T

L J G

T T





1 2 2 2 2

2

0.309434 1.309434 270 N-m

206.1959 N-m

T T T T T

T


1 2 270 N-m 206.1959 N-m 270 N-m 63.8041 N-mT T


1 11 4

1

(63.8041 N-m)(18 mm / 2)(1,000 mm/m)

10,306.0 m55.7 MPa

m

T c

J Ans.


2 22 4

2

(206.1959 N-m)(24 mm / 2)(1,000 mm/m)

32,572.0 m76.0 MPa

m

T c

J Ans.



1 B A


1 B


1 11

1 1

4 2

(63.8041 N-m)(235 mm)(1,000 mm/m)

(10,306.0 mm )(39,000 N/mm )

0.037305 rad 0.0373 rad

T L

J G

Ans.





consists of a stainless steel tube (1) and a brass

tube (2) that are connected at flange B and securely

attached to rigid supports at A and C. Stainless

steel tube (1) has an outside diameter of 2.25 in., a

wall thickness of 0.250 in., a length of L1 = 40 in.,

and a shear modulus of 12,500 ksi. Brass tube (2)

has an outside diameter of 3.500 in., a wall

thickness of 0.219 in., a length of L2 = 20 in., and a

shear modulus of 5,600 ksi. If a concentrated

torque of TB = 42 kip-in. is applied to flange B,

determine:

(a) the maximum shear stress magnitudes in tubes

(1) and (2).


support A.

Fig. P6.77

Solution

Section Properties: The polar moments of inertia for the two tubes are:

4 4 4

1

4 4 4

2

(2.250 in.) (1.750 in.) 1.595340 in.32

(3.500 in.) (3.062 in.) 6.102156 in.32

J

J

Equilibrium:

1 2 42 kip-in. 0xM T T (a)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at

A and C, the sum of the angles of twist in the two tubes must equal zero:

1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)


2 1 11 2

1 2 2

4

2 24

20 in. 1.595340 in. 12,500 ksi0.291784

40 in. 6.102156 in. 5,600 ksi

L J GT T

L J G

T T





1 2 2 2 2

2

0.291784 1.291784 42 kip-in.

32.5132 kip-in.

T T T T T

T


1 2 42 kip-in. 32.5132 kip-in. 42 kip-in. 9.4868 kip-in.T T


1 11 4

1

(9.4868 kip-in.)(2.250 in. / 2)

1.595340 in.6.69 ksi

T c

J Ans.


2 22 4

2

(32.5132 kip-in.)(3.500 in. / 2)

6.102156 in.9.32 ksi

T c

J Ans.



1 B A


1 B


1 11

1 1

4

(9.4868 kip-in.)(40 in.)

(1.595340 in. )(12,500 ksi)

0.019029 rad 0.01903 rad

T L

J G

Ans.





consists of a stainless steel tube (1) and a brass

tube (2) that are connected at flange B and securely

attached to rigid supports at A and C. Stainless

steel tube (1) has an outside diameter of 2.25 in., a

wall thickness of 0.250 in., a length of L1 = 40 in.,

and a shear modulus of 12,500 ksi. Brass tube (2)

has an outside diameter of 3.500 in., a wall

thickness of 0.219 in., a length of L2 = 20 in., and a

shear modulus of 5,600 ksi. The allowable shear

stress in the stainless steel is 50 ksi, and the

allowable shear stress in the brass is 18 ksi.

Determine:

(a) the allowable torque TB that can be applied to

the composite shaft at flange B.

(b) the rotation angle of flange B that is produced

by the allowable torque TB.

Fig. P6.78

Solution

Section Properties: The polar moments of inertia for the two tubes are:

4 4 4

1

4 4 4

2

(2.250 in.) (1.750 in.) 1.595340 in.32

(3.500 in.) (3.062 in.) 6.102156 in.32

J

J

Equilibrium:

1 2 0x BM T T T (a)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at

A and C, the sum of the angles of twist in the two tubes must equal zero:

1 2 0 (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0T L T L

J G J G (d)



Tc T

J J c





1 1 2 2

1 1 2 2

L L

G c G c

(e)


2 1 11 2 2 2

1 2 2

20 in. 12,500 ksi 2.250 in./20.717474

40 in. 5,600 ksi 3.500 in./2

L G c

L G c

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 18 ksi,

then the shear stress magnitude in shaft (1) will be:

1 0.717474(18 ksi) 12.914541 ksi 50 ksi O.K.

This calculation shows that the shear stress in shaft (2) does in fact control.



4

1 11

1

4

2 22

2

(12.914541 ksi)(1.595340 in. )18.3139 kip-in.

2.250 in./2

(18 ksi)(6.102156 in. )62.7650 kip-in.

3.500 in./2

JT

c

JT

c

Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated

equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving

it a negative value. Therefore, by inspection

2 62.7650 kip-in.T

(a) Allowable Torque TB: From Eq. (a), the total torque acting at flange B must not exceed:

,max 1 2 18.3138 kip-in. ( 62.7650 kip-in.) 81.0788 kip-in 81.1 kip- n. i .BT T T Ans.



1 B A


1 B


1 11 4

1 1

(18.3139 kip-in.)(40 in.)0.036735 rad

(1.595340 in. )(10.0367 ra

2,500 ksi)d

T L

J G Ans.




6.79 The torsional assembly of Fig. P6.79

consists of a cold-rolled stainless steel tube

connected to a solid cold-rolled brass

segment at flange C. The assembly is

securely fastened to rigid supports at A and

D. Stainless steel tube (1) and (2) has an

outside diameter of 3.50 in., a wall thickness

of 0.120 in., and a shear modulus of G =

12,500 ksi. The solid brass segment (3) has a

diameter of 2.00 in. and a shear modulus of

G = 5,600 ksi. A concentrated torque of TB =

6 kip-ft is applied to the stainless steel pipe

at B. Determine:

(a) the maximum shear stress magnitude in

the stainless steel tube.

(b) the maximum shear stress magnitude in

brass segment (3).

(c) the rotation angle of flange C.

Fig. P6.79

Solution

Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass

segment are:

4 4 4

1 2

4 4

3

(3.500 in.) (3.260 in.) 3.643915 in.32

(2.0 in.) 1.570796 in.32

J J

J

Equilibrium: Consider a free-body diagram cut around joint

B, where the external torque TB is applied:

1 2 0x BM T T T (a)

and also consider a FBD cut around joint C. Although there is

not an external torque applied at joint C, the section properties

of the torsion structure change at C.

2 3 2 30xM T T T T (b)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely

attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero:

1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (d)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of

deformation relationship [Eq. (c)] to obtain the compatibility equation:

1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (e)




Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1

and T3 with equivalent expressions involving T2. From Eq. (a):

1 2 BT T T (f)

Substitute Eqs. (b) and (f) into Eq. (e) and simplify to derive an expression for T2:

2 1 2 2 2 3

1 1 2 2 3 3

1 2 3 12

1 1 2 2 3 3 1 1

1

1 12

1 2 3

1 1 2 2 3 3

( )0B

B

B

T T L T L T L

J G J G J G

L L L T LT

J G J G J G J G

T L

J GT

L L L

J G J G J G

(e)

Note that G1 = G2 and J1 = J2. Compute T2:

4

2

4 4

(6 kip-ft)(42 in.)(12 in./ft)

(3.643915 in. )(12,500 ksi)

42 in. 30 in. 24 in.

(3.643915 in. )(12,500 ksi) (1.570796 in. )(5,600 ksi)

15.4070 kip-in.

T

Substitute this result into Eq. (f) to compute internal torque T1:

1 2 15.4070 kip-in. (6 kip-ft)(12 in./ft) 56.5930 kip-in.BT T T

and from Eq. (b):

3 2 15.4070 kip-in.T T

Maximum Shear Stress: The maximum shear stress magnitudes in the three members are:

1 11 4

1

(56.5930 kip-in.)(3.500 in. / 2)27.2 ksi

3.643915 in.

T c

J

2 22 4

2

(15.4070 kip-in.)(3.500 in. / 2)7.40 ksi

3.643915 in.

T c

J

3 33 4

3

(15.4070 kip-in.)(2.000 in. / 2)9.81 ksi

1.570796 in.

T c

J

(a) Maximum Shear Stress Magnitude in Stainless Steel Tube:

1 27.2 ksi Ans.

(b) Maximum Shear Stress Magnitude in Brass Shaft:

3 9.81 ksi Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be

defined by the difference in rotation angles at the two ends:

1 1B A B A (f)

Similarly, the angle of twist in member (2) can be defined by:

2 2C B C B (g)




To derive an expression for C, substitute Eq. (f) into Eq. (g), and note that joint A is restrained from

rotating; therefore, A = 0.

1 2C

The angle of twist in member (1) is:

1 11 4

1 1

(56.5930 kip-in.)(42 in.)0.052184 rad

(3.643915 in. )(12,500 ksi)

T L

J G


2 22 4

2 2

( 15.4070 kip-in.)(30 in.)0.010148 rad

(3.643915 in. )(12,500 ksi)

T L

J G

The rotation angle of flange C is thus:

1 2 0.052184 rad ( 0.010148 rad) 0.042036 rad 0.0420 radC Ans.




6.80 The torsional assembly of Fig. P6.80

consists of a cold-rolled stainless steel tube

connected to a solid cold-rolled brass

segment at flange C. The assembly is

securely fastened to rigid supports at A and

D. Stainless steel tubes (1) and (2) have an

outside diameter of 3.50 in., a wall thickness

of 0.120 in., a shear modulus of G = 12,500

ksi, and an allowable shear stress of 30 ksi.

The solid brass segment (3) has a diameter

of 2.00 in., a shear modulus of G = 5,600

ksi, and an allowable shear stress of 18 ksi.

Determine the maximum permissible

magnitude for the concentrated torque TB. Fig. P6.80

Solution

Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment

are:

4 4 4

1 2

4 4

3

(3.500 in.) (3.260 in.) 3.643915 in.32

(2.0 in.) 1.570796 in.32

J J

J

Equilibrium: Consider a free-body diagram cut around joint

B, where the external torque TB is applied:

1 2 0x BM T T T (a)

and also consider a FBD cut around joint C. Although there is

not an external torque applied at joint C, the section properties

of the torsion structure change at C.

2 3 2 30xM T T T T (b)



1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (d)



1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (e)


rewrite Eq. (e) in terms of stresses. In general, the elastic torsion formula can be rearranged as:




Tc T

J J c

which allows Eq. (e) to be rewritten as:

1 1 2 2 3 3

1 1 2 2 3 3

0L L L

G c G c G c

(f)

Also using the elastic torsion formula, Eq. (b) can be expressed in terms of stress:

2 2 3 3 3 2 2

2 3 3 2 3

J J J

c c c c J

(g)

Substitute Eq. (g) into Eq. (f) to obtain

1 1 2 2 3 2 2

1 1 2 2 3 2 3

0L L L J

G c G c G c J

Simplify:

2 3 2

1 2 3 2 2 2 3 2 31 2 1 2

11 1 2 2 3 2 3

1 1

L L J

L L L J G c G c J

LG c G c G c J

G c

Substitute values:

4

4

1 2

2

30 in. 24 in. 3.643915 in.

(12,500 ksi)(3.50 in./2) (5,600 ksi)(3.50 in./2) 1.570796 in.

42 in.

(12,500 ksi)(3.50 in./2)

3.6732

This calculation demonstrates that the shear stress in segment (1) of the stainless steel tube is much

larger than the shear stress in segment (2). If the shear stress magnitude in segment (1) is 30 ksi, then

the shear stress magnitude in segment (2) will be:

12 8.1673 ksi

3.6732

(h)

Next, we need to check the corresponding shear stress in brass shaft (3). From Eq. (g):

3 23 2

2 3

c J

c J

Substitute the magnitude obtained for 2 in Eq. (h) into this expression and calculate the corresponding

shear stress magnitude in shaft (3):

4

3 4

2.00 in./2 3.643915 in.(8.1673 ksi) 10.8265 ksi 18 ksi

3.50 in./2 1.570796 in.

O.K.

Now that the maximum shear stresses in the three shaft segments are known, the torques in each

component can be computed:

4

1 1

1

1

4

2 2

2

2

4

3 3

3

3

(30 ksi)(3.643915 in. )62.4671 kip-in.

3.50 in./2

( 8.1673 ksi)(3.643915 in. )17.0062 kip-in.

3.50 in./2

( 10.8265 ksi)(1.570796 in. )17.0062 kip-in.

2.00 in./2

JT

c

JT

c

JT

c




Maximum Permissible Torque TB: From Eq. (a), the torque TB acting on the assembly must not

exceed:





6.81 The torsional assembly of Fig. P6.81a consists of a solid 75-mm-diameter bronze [G = 45 GPa]

segment (1) securely connected at flange B to solid 75-mm-diameter stainless steel [G = 86 GPa]

segments (2) and (3). The flange at B is secured by four 14-mm-diameter bolts, which are each located

on a 120-mm-diameter bolt circle (Fig. P6.81b). The allowable shear stress of the bolts is 90 MPa, and

friction effects in the flange can be neglected for this analysis. Determine:

(a) the allowable torque TC that can be applied to the assembly at C without exceeding the capacity of

the bolted flange connection.

(b) the maximum shear stress magnitude in bronze segment (1).

(c) the maximum shear stress magnitude in stainless steel segments (2) and (3).

Fig. P6.81a Fig. P6.81b Flange B Bolts

Solution

Section Properties: For bronze segment (1) and stainless steel segments (2) and (3), the polar moments

of inertia are identical:

4 4

1 2 3 (75 mm) 3,106,311 mm32

J J J

Equilibrium: Consider a free-body diagram cut around

flange B:

1 2 1 20xM T T T T (a)

and also consider a FBD cut around joint C, where the

external torque TC is applied:

2 3 0x CM T T T (b)

Capacity of Flange B: The flange at B is secured by four 14-mm-diameter bolts, which are each

located on a 120-mm-diameter bolt circle. The allowable shear stress of the bolts is 90 MPa, and each

bolt acts in single shear. The cross-sectional area of a single bolt is:

2 2

bolt (14 mm) 153.9380 mm4

A

The shear force that can be provided by each bolt is:

2 2

bolt (90 N/mm )(153.9380 mm ) 13,854.4236 NV

The total torque that can resisted by the bolts is thus:

,max (4 bolts)(13,954.4236 N/bolt)(120 mm/2) 3,325,062 N-mmBT




The internal torques in segments (1) and (2) cannot exceed this magnitude; therefore, based on Eq. (a):

1 2 3,325,062 N-mmT T (c)

From Eq. (b), we observe that the external torque applied to the shaft is related to internal torques T2 and

T3. We must determine the relationship between these two internal torques by developing a

compatibility equation, which is based on the geometry of deformations for this configuration.


attached to fixed supports at A and D, the sum of the angles of twist in the three segments must equal

zero:

1 2 3 0 (d)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (e)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (e)] into the geometry of

deformation relationship [Eq. (d)] to obtain the compatibility equation:

1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (f)

Solve the Equations: Solve Eq. (f) for T3, noting that J1 = J2 = J3, G2 = G3, and T1 = T2:

1 1 2 2 3 3 1 3 23 1

1 1 2 2 3 3 1 3

T L T L G J L G LT T

J G J G L L G L

(g)

Substitute the values determined for T1 and T2 in Eq. (c) into Eq. (g) to compute T3:

3

400 mm 86 GPa 750 mm(3,325,062 N-mm)

400 mm 45 GPa 400 mm

(3,325,062 N-mm)(3.786111)

12,589,054 N-mm

T

(a) Allowable Torque TC: From equilibrium equation (b),

2 3

3,325,062 N-mm ( 12,589,054 N-mm)

15,914,116 N-mm 15.91 kN-m

CT T T

Ans.

(b) Shear Stress Magnitude in Bronze Segment (1):

1 11 4

1

(3,325,062 N-mm)(75 mm / 2)

3,106,311 mm40.1 MPa

T c

J Ans.

(c) Shear Stress Magnitude in Stainless Steel Segments (2) and (3):

3 33 4

3

(12,589,054 N-mm)(75 mm / 2)

3,106,311 mm152.0 MPa

T c

J Ans.




6.82 The torsional assembly shown in Fig. P6.82

consists of solid 2.50-in.-diameter aluminum [G =

4,000 ksi] segments (1) and (3) and a central solid

3.00-in.-diameter bronze [G = 6,500 ksi] segment

(2). Concentrated torques of TB = T0 and TC = 2T0

are applied to the assembly at B and C,

respectively. If T0 = 20 kip-in., determine:


aluminum segments (1) and (3).


bronze segment (2).

(c) the rotation angle of joint C.

Fig. P6.82

Solution

Section Properties: The polar moments of inertia for the aluminum and bronze segments are:

4 4

1 3

4 4

2

(2.50 in.) 3.834952 in.32

(3.00 in.) 7.952156 in.32

J J

J


joint B, where the external torque TB is applied:

1 2 0 0xM T T T (a)



2 3 02 0xM T T T (b)



1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (d)



1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (e)



1 2 0T T T (f)

and from Eq. (b):

3 2 02T T T (g)




Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2:

2 0 1 2 2 2 0 3

1 1 2 2 3 3

1 2 3 0 3 0 12

1 1 2 2 3 3 3 3 1 1

0 3 0 1

3 3 1 12

1 2 3

1 1 2 2 3 3

( ) ( 2 )0

2

2

T T L T L T T L

J G J G J G

L L L T L T LT

J G J G J G J G J G

T L T L

J G J GT

L L L

J G J G J G

(h)

Since G1 = G3 and J1 = J3, Eq. (h) can be further simplified and T2 can be computed as:

3 102

1 3 21 1

1 1 2 2

4

4 4

2

20 kip-in. 2(12 in.) 12 in.

12 in. 12 in. 24 in.(3.834952 in. )(4,000 ksi)

(3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi)

7.711461 kip-in.

L LTT

L L LJ G

J G J G

Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3:

1 7.711461 kip-in. 20 kip-in. 27.711461 kip-in.T

3 7.711461 kip-in. 2(20 kip-in.) 32.288539 kip-in.T

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are:

1 11 4

1

(27.711461 kip-in.)(2.500 in. / 2)9.03 ksi

3.834952 in.

T c

J

2 22 4

2

(7.711461 kip-in.)(3.000 in. / 2)1.455 ksi

7.952156 in.

T c

J

3 33 4

3

(32.288539 kip-in.)(2.500 in. / 2)10.52 ksi

3.834952 in.

T c

J

(a) Maximum Shear Stress Magnitude in Aluminum Segments:

3 10.52 ksi Ans.

(b) Maximum Shear Stress Magnitude in Bronze Segment:

2 1.455 ksi Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be


1 1B A B A (i)


2 2C B C B (j)

To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from





1 2C


1 11 4

1 1

(27.711461 kip-in.)(12 in.)0.021678 rad

(3.834952 in. )(4,000 ksi)

T L

J G


2 22 4

2 2

(7.711461 kip-in.)(24 in.)0.003581 rad

(7.952156 in. )(6,500 ksi)

T L

J G


1 2 0.021678 rad 0.003581 rad 0.025259 ra 0.0253 rad dC Ans.





consists of solid 2.50-in.-diameter aluminum [G =

4,000 ksi] segments (1) and (3) and a central solid

3.00-in.-diameter bronze [G = 6,500 ksi] segment

(2). Concentrated torques of TB = T0 and TC = 2T0

are applied to the assembly at B and C,

respectively. If the rotation angle at joint C must

not exceed 3°, determine:

(a) the maximum magnitude of T0 that may be

applied to the assembly.


aluminum segments (1) and (3).

(c) the maximum shear stress magnitude in

bronze segment (2).

Fig. P6.83

Solution


4 4

1 3

4 4

2

(2.50 in.) 3.834952 in.32

(3.00 in.) 7.952156 in.32

J J

J



1 2 0 0xM T T T (a)



2 3 02 0xM T T T (b)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G

Torque in Segment (3): From the problem statement, the rotation angle at C must not exceed 3°;

consequently, the angle of twist in segment (3) must be the opposite of this value:

3 3 0.052360 rad

From the torque-twist relationships, the internal torque in segment (3) can be computed:

4

3 3 3 3 33 3

3 3 3

( 0.052360 rad)(3.834952 in. )(4,000 ksi)66.9325 kip-in.

12 in.

T L J GT

J G L

Consider Rotation Angle of Flange C Relative to Support A: The rotation angle at C is given by the

sum of the angles of twist in segments (1) and (2). This rotation angle must not exceed 3°.

1 1 2 21 2

1 1 2 2

3 0.052360 radC

T L T L

J G J G (c)

from Eq. (b):

2 3 02T T T (d)




and from Eq. (a):

1 2 0 3 0 0 3 0( 2 ) 3T T T T T T T T (e)

Substitute the expressions derived for T1 and T2 in Eqs. (d) and (e) into Eq. (c):

3 0 1 3 0 2

1 1 2 2

( 3 ) ( 2 )0.052360 rad

T T L T T L

J G J G

Simplify

1 2 1 23 0

1 1 2 2 1 1 2 2

3 20.052360 rad

L L L LT T

J G J G J G J G

Solve for T0:

1 23

1 1 2 2

01 2

1 1 2 2

0.052360 rad

3 2

L LT

J G J GT

L L

J G J G

and compute:

4 4

0

4 4

12 in. 24 in.0.052360 rad ( 66.9325 kip-in.)

(3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi)

3(12 in.) 2(24 in.)

(3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi)

41.4590 kip-in.

T

Backsubstitute this value into Eq. (d) to compute T2:

2 3 02 66.9325 kip-in. 2(41.4590 kip-in.) 15.9855 kip-in.T T T

and backsubstitute this value into Eq. (e) to compute T1:

1 3 03 66.9325 kip-in. 3(41.4590 kip-in.) 57.4445 kip-in.T T T

(a) Maximum magnitude of T0:

0,max 41.5 kip-in.T Ans.

Maximum Shear Stress Magnitudes: The maximum shear stress magnitudes are:

1 11 4

1

(57.4445 kip-in.)(2.500 in. / 2)18.72 ksi

3.834952 in.

T c

J

2 22 4

2

(15.9855 kip-in.)(3.000 in. / 2)3.02 ksi

7.952156 in.

T c

J

3 33 4

3

(66.9325 kip-in.)(2.500 in. / 2)21.8 ksi

3.834952 in.

T c

J

(b) Maximum Shear Stress Magnitude in Aluminum Segments:

3 21.8 ksi Ans.

(c) Maximum Shear Stress Magnitude in Bronze Segment:

2 3.02 ksi Ans.





consists of a solid 60-mm-diameter aluminum [G

= 28 GPa] segment (2) and two bronze [G = 45

GPa] tube segments (1) and (3), which have an

outside diameter of 75 mm and a wall thickness

of 5 mm. If concentrated torques of TB = 9 kN-m

and TC = 9 kN-m are applied in the directions

shown, determine:


bronze tube segments (1) and (3).


aluminum segment (2).


Fig. P6.84

Solution


4 4 4

1 3

4 4

2

(75 mm) (65 mm) 1,353,830 mm32

(60 mm) 1,272,345 mm32

J J

J



1 2 0x BM T T T (a)



2 3 0x CM T T T (b)



1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (d)



1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (e)






1 2 BT T T (f)

and from Eq. (b):

3 2 CT T T (g)


2 1 2 2 2 3

1 1 2 2 3 3

1 2 3 1 32

1 1 2 2 3 3 1 1 3 3

1 3

1 1 3 32

1 2 3

1 1 2 2 3 3

( ) ( )0B C

B C

B C

T T L T L T T L

J G J G J G

L L L T L T LT

J G J G J G J G J G

T L T L

J G J GT

L L L

J G J G J G

(h)

Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as:

1 3

21 3 21 1

1 1 2 2

4 2

4 4

1

1 (9,000,000 N-mm)(325 mm) (9,000,000 N-mm)(325 mm)

325 mm 325 mm 400 mm(1,353,830 mm )(45,000 N/mm )

(1,353,830 mm )(45,000) (1,272,345 mm )(28,000)

B CT L T LT

L L LJ G

J G J G

4,385,216 N-mm


1 4,385,216 N-mm 9,000,000 N-mm 4,614,784 N-mmT

3 4,385,216 N-mm 9,000,000 N-mm 4,614,784 N-mmT


1 11 4

1

(4,614,784 N-mm)(75 mm / 2)127.8 MPa

1,353,830 mm

T c

J

2 22 4

2

(4,385,216 N-mm)(60 mm / 2)103.4 MPa

1,272,345 mm

T c

J

3 33 4

3

(4,614,784 N-mm)(75 mm / 2)127.8 MPa

1,353,830 mm

T c

J

(a) Maximum Shear Stress Magnitude in Bronze Segments:

1 127.8 MPa Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment:

2 103.4 MPa Ans.




(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be


1 1B A B A (i)


2 2C B C B (j)



1 2C


1 11 4 2

1 1

(4,614,784 N-mm)(325 mm)0.024618 rad

(1,353,830 mm )(45,000 N/mm )

T L

J G


2 22 4 2

2 2

( 4,385,216 N-mm)(400 mm)0.049237 rad

(1,272,345 mm )(28,000 N/mm )

T L

J G







consists of a solid 60-mm-diameter aluminum [G

= 28 GPa] segment (2) and two bronze [G = 45

GPa] tube segments (1) and (3), which have an

outside diameter of 75 mm and a wall thickness

of 5 mm. If concentrated torques of TB = 6 kN-m

and TC = 10 kN-m are applied in the directions

shown, determine:


bronze tube segments (1) and (3).


aluminum segment (2).


Fig. P6.85

Solution


4 4 4

1 3

4 4

2

(75 mm) (65 mm) 1,353,830 mm32

(60 mm) 1,272,345 mm32

J J

J



1 2 0x BM T T T (a)



2 3 0x CM T T T (b)



1 2 3 0 (c)


1 1 2 2 3 31 2 3

1 1 2 2 3 3

T L T L T L

J G J G J G (d)



1 1 2 2 3 3

1 1 2 2 3 3

0T L T L T L

J G J G J G (e)






1 2 BT T T (f)

and from Eq. (b):

3 2 CT T T (g)


2 1 2 2 2 3

1 1 2 2 3 3

1 2 3 1 32

1 1 2 2 3 3 1 1 3 3

1 3

1 1 3 32

1 2 3

1 1 2 2 3 3

( ) ( )0B C

B C

B C

T T L T L T T L

J G J G J G

L L L T L T LT

J G J G J G J G J G

T L T L

J G J GT

L L L

J G J G J G

(h)

Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as:

1 3

21 3 21 1

1 1 2 2

4 2

4 4

1

1 (6,000,000 N-mm)(325 mm) (10,000,000 N-mm)(325 mm)

325 mm 325 mm 400 mm(1,353,830 mm )(45,000 N/mm )

(1,353,830 mm )(45,000) (1,272,345 mm )(28,000)

B CT L T LT

L L LJ G

J G J G

3,897,970 N-mm


1 3,897,970 N-mm 6,000,000 N-mm 2,102,030 N-mmT

3 3,897,970 N-mm 10,000,000 N-mm 6,102,030 N-mmT


1 11 4

1

(2,102,030 N-mm)(75 mm / 2)58.2 MPa

1,353,830 mm

T c

J

2 22 4

2

(3,897,970 N-mm)(60 mm / 2)91.9 MPa

1,272,345 mm

T c

J

3 33 4

3

(6,102,030 N-mm)(75 mm / 2)169.0 MPa

1,353,830 mm

T c

J

(a) Maximum Shear Stress Magnitude in Bronze Segments:

3 169.0 MPa Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment:

2 91.9 MPa Ans.




(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be


1 1B A B A (i)


2 2C B C B (j)



1 2C


1 11 4 2

1 1

(2,102,030 N-mm)(325 mm)0.011214 rad

(1,353,830 mm )(45,000 N/mm )

T L

J G


2 22 4 2

2 2

( 3,897,970 N-mm)(400 mm)0.043766 rad

(1,272,345 mm )(28,000 N/mm )

T L

J G






6.86 A solid 1.50-in.-diameter brass [G = 5,600 ksi] shaft [segments (1), (2), and (3)] has been stiffened

between B and C by the addition of a cold-rolled stainless steel tube (4) (Fig. P6.86a). The tube (Fig.

P6.86b) has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and a shear modulus of G =

12,500 ksi. The tube is attached to the brass shaft by means of rigid flanges welded to the tube and to the

shaft. (The thickness of the flanges can be neglected for this analysis.) If a torque of 400 lb-ft is applied

to the shaft as shown in Fig. P6.86a, determine:

(a) the maximum shear stress magnitude in segment (1) of the brass shaft.

(b) the maximum shear stress magnitude in segment (2) of the brass shaft (i.e., between flanges B and C)

(c) the maximum shear stress magnitude in the stainless steel tube (4).

(d) the rotation angle of end D relative to end A.

Fig. P6.86a Fig. P6.86b Cross Section Through Tube

Solution

Section Properties: The polar moments of inertia for the brass shaft and the stainless steel tube are:

4 4

1 2 3

4 4 4

4

(1.50 in.) 0.497010 in.32

(3.50 in.) (3.26 in.) 3.643915 in.32

J J J

J

(a) Shear Stress Magnitude in Brass Segment (1):

1 11 4

1

(400 lb-ft)(1.500 in. / 2)(12 in./ft)7,243.3 psi

0.497010 in.7,240 psi

T c

J Ans.

Statically Indeterminate Portion of the Shaft: The portion of the shaft between B and C is statically

indeterminate.

Equilibrium: Consider a free-body diagram that cuts

through the shaft between B and C and includes the

free end of the shaft at A:

2 4 400 lb-ft 0xM T T (a)

Geometry of Deformation Relationship: Since the tube and shaft are securely connected by the rigid

flanges, the angles of twist in both members must be equal; therefore,

2 4 (b)


2 2 4 42 4

2 2 4 4

T L T L

J G J G (c)






2 2 4 4

2 2 4 4

T L T L

J G J G (d)


4

4 2 22 4 4 44

2 4 4

20 in. 0.497010 in. 5,600 ksi0.0611047

20 in. 3.643915 in. 12,500 ksi

L J GT T T T

L J G

and substitute this result into Eq. (a) to compute the torque T4 in the stainless steel tube (4):

2 4 4 4 4

4

0.0611047 1.0611047 400 lb-ft

376.9656 lb-ft

T T T T T

T

The torque in brass shaft segment (2) is therefore:

2 4400 lb-ft 400 lb-ft ( 376.9656 lb-ft) 23.0344 lb-ftT T

(b) Maximum Shear Stress in Brass Segment (2): The maximum shear stress in segment (2) is:

2 22 4

2

(23.0344 lb-ft)(1.50 in. / 2)(12 in./ft)417.1 psi

0.497010 i417 ps

.i

n

T c

J Ans.

(c) Maximum Shear Stress in Stainless Steel Tube (4): The maximum shear stress in tube (4) is:

4 44 4

4

(376.9656 lb-ft)(3.50 in. / 2)(12 in./ft)2,172.5 psi

3.643915 in2,1 0

.7 psi

T c

J Ans.

(d) Rotation Angle of End D Relative to End A: The angle of twist of segment (1) is:

1 11 4

1 1

( 400 lb-ft)(12 in.)(12 in./ft)0.020695 rad

(0.497010 in. )(5,600,000 psi)

T L

J G

The angle of twist in brass segment (3) has the same value:

3 0.020695 rad

The angle of twist in brass segment (2) is:

2 22 4

2 2

( 23.0344 lb-ft)(20 in.)(12 in./ft)0.0019863 rad

(0.497010 in. )(5,600,000 psi)

T L

J G

The rotation angle of end D relative to end A is the sum of these three angles of twist:

1 2 3

0.020695 rad ( 0.0019863 rad) ( 0.020695

0.0434 r

rad)

0.0433763 ra add

D

Ans.




6.87 A solid 60-mm-diameter cold-rolled brass [G = 39 GPa] shaft that is 1.25-m long extends through

and is completely bonded to a hollow aluminum [G = 28 GPa] tube, as shown in Fig. P6.87. Aluminum

tube (1) has an outside diameter of 90 mm, an inside diameter of 60 mm, and a length of 0.75 m. Both

the brass shaft and the aluminum tube are securely attached to the wall support at A. When the two

torques shown are applied to the composite shaft, determine:


aluminum tube (1).

(b) the maximum shear stress magnitude in brass

shaft segment (2).

(c) the maximum shear stress magnitude in brass

shaft segment (3).

(d) the rotation angle of joint B.

(e) the rotation angle of end C.

Fig. P6.87

Solution

Section Properties: The polar moments of inertia for tube (1) and brass shafts (2) and (3) are:

4 4 4

1

4 4

2 3

(90 mm) (60 mm) 5,168,902 mm32

(60 mm) 1,272,345 mm32

J

J J

Equilibrium: Consider a free-body diagram cut

around end C through segment (3):

3 38 kN-m 0 8 kN-mxM T T (a)

and also consider a FBD cut around joint B:

1 2 3 20 kN-m 0xM T T T

1 2 3 20 kN-m 12 kN-mT T T (b)

Eq. (b) reveals that the portion of the shaft between A and B is statically indeterminate. The five-step

solution procedure will be used to determine T1 and T2 in this portion of the shaft.

Geometry of Deformation Relationship: Since the aluminum tube and the brass shaft are securely

bonded together, the angles of twist in both members must be equal; therefore,

1 2 (c)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (d)






1 1 2 2

1 1 2 2

T L T L

J G J G (e)

Solve the Equations: Solve Eq. (e) for T1:

4

2 1 11 2 2 24

1 2 2

5,168,902 mm 28 GPa2.916667

1,272,345 mm 39 GPa

L J GT T T T

L J G

and substitute this result into Eq. (b) to compute the torque T2 in brass shaft (2):

1 2 2 2 2

2

2.916667 3.916667 12,000 N-m

3,063.830 N-m

T T T T T

T

The torque in aluminum tube (1) is therefore:

1 212,000 N-m 12,000 N-m 3,063.830 N-m 8,936.170 N-mT T

(a) Maximum Shear Stress in Aluminum Tube: The maximum shear stress in aluminum tube (1) is:

1 11 4

1

(8,936.170 N-m)(90 mm / 2)(1,000 mm/m)

5,168,902 m77.8 MPa

m

T c

J Ans.

(b) Maximum Shear Stress Magnitude in Brass Shaft Segment (2):

2 22 4

2

(3,063.830 N-m)(60 mm / 2)(1,000 mm/m)

1,272,345 m72.2 MPa

m

T c

J Ans.

(c) Maximum Shear Stress Magnitude in Brass Shaft Segment (3):

3 33 4

3

(8,000 N-m)(60 mm / 2)(1,000 mm/m)

1,272,345 mm188.6 MPa

T c

J Ans.

(d) Rotation Angle of Joint B:

2

1 1

1 4 2

1 1

2

2 2

2 4 2

2 2

0.04

(8,936.170 N-m)(0.75 m)(1,000 mm/m)0.046308 rad

(5,168,902 mm )(28,000 N/mm )

or

(3,063.830 N-m)(0.75 m)(1,000 mm/m)0.046308 rad

(1,272,345 mm )(39,0

63 rad

00 N/mm )

B

T L

J G

T L

J G

Ans.

(e) Rotation Angle of End C:

2

3 33 4 2

3 3

3

( 8,000 N-m)(0.5 m)(1,000 mm/m)0.080610 rad

(1,272,345 mm )(39,000 N/mm )

0.046308 rad ( 0.080 0.0343 r610 rad) adC B

T L

J G

Ans.




6.88 The gear assembly shown in Fig. P6.88 is

subjected to a torque of TC = 140 N-m. Shafts (1)

and (2) are solid 20-mm-diameter steel shafts, and

shaft (3) is a solid 25-mm-diameter steel shaft.

Assume L = 400 mm and G = 80 GPa. Determine:

(a) the maximum shear stress magnitude in shaft

(1).

(b) the maximum shear stress magnitude in shaft

segment (3).

(c) the rotation angle of gear E.

(d) the rotation angle of gear C.

Fig. P6.88

Solution

Section Properties: The polar moments of inertia for the shafts are:

4 4

1 2

4 4

3

(20 mm) 15,708.0 mm32

(25 mm) 38,349.5 mm32

J J

J

Equilibrium: Consider a free-body diagram cut through

shaft (2) and around gear C:

2 20 140 N-mx C CM T T T T (a)

Next, consider a FBD cut around gear B through shafts (1)

and (2). The teeth of gear C exert a force F on the teeth of

gear B. This force F opposes the rotation of gear B. The

radius of gear B will be denoted by RB for now (even though

the gear radius is not given explicitly).

2 1 0x BM T T F R (b)

Finally, consider a FBD cut around gear E through shaft (3).

The teeth of gear B exert an equal magnitude force F on the

teeth of gear C, acting opposite to the direction assumed in

the previous FBD. The radius of gear E will be denoted by

RE for now.

33 0x E

E

TM T F R F

R (c)

The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give

1 3140 N-m B

E

RT T

R

The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of

gear teeth NB and NE:

1 3 3 3

24 teeth140 N-m 140 N-m 140 N-m 0.4

60 teeth

B

E

NT T T T

N

(d)




Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns

in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the

problem, an additional equation must be developed. This second equation will be derived from the

relationship between the angles of twist in shafts (1) and (3).

Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1):

1B

and the rotation of gear E is equal to the angle of twist in shaft (3):

3E

However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The

arclengths associated with the respective rotations must be equal, but the gears turn in opposite

directions. The relationship between gear rotations can be stated as:

B B E ER R

where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to

the shaft angles of twist, this relationship can be expressed as:

1 3B ER R (e)


1 1 3 31 3

1 1 3 3

T L T L

J G J G (f)

Compatibility Equation:

Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)]

to obtain:

1 1 3 3

1 1 3 3

B E

T L T LR R

J G J G

which can be rearranged and expressed in terms of the gear ratio NB /NE:

1 1 3 3

1 1 3 3

B

E

N T L T L

N J G J G (g)

Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved

simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3).

Solve the Equations: Solve for internal torque T3 in Eq. (g):

1 3 33 1

3 1 1

B

E

N L J GT T

N L J G

and substitute this result into Eq. (d):

1 3

1 3 31

3 1 1

4

1 4

1

140 N-m 0.4

140 N-m 0.4

24 teeth 38,349.5 mm140 N-m 0.4

60 teeth 15,708.0 mm

140 N-m 0.390624

B

E

T T

N L J GT

N L J G

T

T




Group the T1 terms to obtain:

1

140 N-m100.6742 N-m

1.390624T

Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3):

1 3

13

140 N-m 0.4

140 N-m 100.6742 N-m 140 N-m98.3146 N-m

0.4 0.4

T T

TT

(a) Maximum Shear Stress Magnitude in Shaft (1):

1 11 4

1

(100.6742 N-m)(20 mm / 2)(1,000 mm/m)

15,708.0 m64.1 MPa

m

T c

J Ans.

(b) Maximum Shear Stress Magnitude in Shaft (3):

3 33 4

3

(98.3146 N-m)(25 mm / 2)(1,000 mm/m)

38,349.5 m32.0 MPa

m

T c

J Ans.

(c) Rotation Angle of Gear E:

3 33 4 2

3 3

3

( 98.3146 N-m)(1.25)(400 mm)(1,000 mm/m)0.016023 rad

(38,349.5 mm )(80

0.01602 r

,000 N

a

/ )

d

mm

E

T L

J G

Ans.

(d) Rotation Angle of Gear C:

1 1

1 4 2

1 1

2 2

2 4 2

2 2

2 1 2

(100.6742 N-m)(1.25)(400 mm)(1,000 mm/m)0.040057 rad

(15,708.0 mm )(80,000 N/mm )

(140 N-m)(400 mm)(1,000 mm/m)0.044563 rad

(15,708.0 mm )(80,000 N/mm )

0.040057 C B

T L

J G

T L

J G

rad 0.044563 r 0.0846a ad r d Ans.




6.89 The gear assembly shown in Fig. P6.89 is

subjected to a torque of TC = 1,100 lb-ft. Shafts

(1) and (2) are solid 1.625-in.-diameter aluminum

shafts and shaft (3) is a solid 2.00-in.-diameter

aluminum shaft. Assume L = 20 in. and G = 4,000

ksi. Determine:

(a) the maximum shear stress magnitude in shaft

(1).

(b) the maximum shear stress magnitude in shaft

segment (3).


(d) the rotation angle of gear C. Fig. P6.89

Solution


4 4

1 2

4 4

3

(1.625 in.) 0.684563 in.32

(2.00 in.) 1.570796 in.32

J J

J



2 20 1,100 lb-ftx C CM T T T T (a)


and (2). The teeth of gear C exert a force F on the teeth of




2 1 0x BM T T F R (b)





RE for now.

33 0x E

E

TM T F R F

R (c)


1 31,100 lb-ft B

E

RT T

R



1 3 3 3

24 teeth1,100 lb-ft 1,100 lb-ft 1,100 lb-ft 0.4

60 teeth

B

E

NT T T T

N

(d)









1B


3E




B B E ER R



1 3B ER R (e)


1 1 3 31 3

1 1 3 3

T L T L

J G J G (f)



to obtain:

1 1 3 3

1 1 3 3

B E

T L T LR R

J G J G


1 1 3 3

1 1 3 3

B

E

N T L T L

N J G J G (g)




1 3 33 1

3 1 1

B

E

N L J GT T

N L J G


1 3

1 3 31

3 1 1

4

1 4

1

1,100 lb-ft 0.4

1,100 lb-ft 0.4

24 teeth 1.570796 in.1,100 lb-ft 0.4

60 teeth 0.684563 in.

1,100 lb-ft 0.367135

B

E

T T

N L J GT

N L J G

T

T





1

1,100 lb-ft804.602 lb-ft

1.367135T


1 3

13

1,100 lb-ft 0.4

1,100 lb-ft 804.602 lb-ft 1,100 lb-ft738.495 lb-ft

0.4 0.4

T T

TT


1 11 4

1

(804.602 lb-ft)(1.625 in. / 2)(12 in./ft)11,459.677 psi

0.684563 in.11.46 ksi

T c

J Ans.


3 33 4

3

(738.495 lb-ft)(2.00 in. / 2)(12 in./ft)5,641.687 psi

1.570796 in.5.64 ksi

T c

J Ans.


3 33 4

3 3

3

( 738.495 lb-ft)(1.25)(20 in.)(12 in./ft)0.035261 rad

(1.570

0.0353 ra

796 in. )(4,000,000 psi)

d

E

T L

J G

Ans.


1 1

1 4

1 1

2 2

2 4

2 2

2 1 2

(804.602 lb-ft)(1.25)(20 in.)(12 in./ft)0.088151 rad

(0.684563 in. )(4,000,000 psi)

(1,100 lb-ft)(20 in.)(12 in./ft)0.096412 rad

(0.684563 in. )(4,000,000 psi)

0.C B

T L

J G

T L

J G

088151 rad 0.09641 0.182 r 46ad rad Ans.




6.90 A torque of TC = 460 N-m acts on gear C of

the assembly shown in Fig. P6.90. Shafts (1)

and (2) are solid 35-mm-diameter aluminum

shafts and shaft (3) is a solid 25-mm-diameter

aluminum shaft. Assume L = 200 mm and G =

28 GPa. Determine:


shaft (1).


shaft segment (3).



Solution


4 4

1 2

4 4

3

(35 mm) 147,323.5 mm32

(25 mm) 38,349.5 mm32

J J

J



2 20 460 N-mx C CM T T T T (a)


and (2). The teeth of gear E exert a force F on the teeth of




2 1 0x BM T T F R (b)





RE for now.

33 0x E

E

TM T F R F

R (c)


1 3460 N-m B

E

RT T

R



1 3 3 3

54 teeth460 N-m 460 N-m 460 N-m 1.285714

42 teeth

B

E

NT T T T

N

(d)









1B


3E




B B E ER R



1 3B ER R (e)


1 1 3 31 3

1 1 3 3

T L T L

J G J G (f)



to obtain:

1 1 3 3

1 1 3 3

B E

T L T LR R

J G J G


1 1 3 3

1 1 3 3

B

E

N T L T L

N J G J G (g)




1 3 33 1

3 1 1

B

E

N L J GT T

N L J G


1 3

1 3 31

3 1 1

4

1 4

1

460 N-m 1.285714

460 N-m 1.285714

54 teeth 38,349.5 mm460 N-m 1.285714

42 teeth 147,323.5 mm

460 N-m 0.430305

B

E

T T

N L J GT

N L J G

T

T





1

460 N-m321.610 N-m

1.430305T


1 3

13

460 N-m 1.285714

460 N-m 321.610 N-m 460 N-m107.637 N-m

1.285714 1.285714

T T

TT


1 11 4

1

(321.610 N-m)(35 mm / 2)(1,000 mm/m)

147,323.5 m38.2 MPa

m

T c

J Ans.


3 33 4

3

(107.637 N-m)(25 mm / 2)(1,000 mm/m)

38,349.5 m35.1 MPa

m

T c

J Ans.


3 33 4 2

3 3

3

( 107.637 N-m)(2)(200 mm)(1,000 mm/m)0.040096 rad

(38,349.5 mm )(28,

0.0401 r

000 N m

a

)

d

/ m

E

T L

J G

Ans.


1 1

1 4 2

1 1

2 2

2 4 2

2 2

2 1 2

(321.610 N-m)(2)(200 mm)(1,000 mm/m)0.031186 rad

(147,323.5 mm )(28,000 N/mm )

(460 N-m)(200 mm)(1,000 mm/m)0.022303 rad

(147,323.5 mm )(28,000 N/mm )

0.031186 raC B

T L

J G

T L

J G

d 0.022303 ra 0.0535 ad r d Ans.




6.91 A torque of TC = 40 kip-in. acts on gear C

of the assembly shown in Fig. P6.91. Shafts

(1) and (2) are solid 2.00-in.-diameter stainless

steel shafts and shaft (3) is a solid 1.75-in.-

diameter stainless steel shaft. Assume L = 8 in.

and G = 12,500 ksi. Determine:


shaft (1).


shaft segment (3).



Solution


4 4

1 2

4 4

3

(2.00 in.) 1.570796 in.32

(1.75 in.) 0.920772 in.32

J J

J



2 20 40 kip-in.x C CM T T T T (a)


and (2). The teeth of gear E exert a force F on the teeth of




2 1 0x BM T T F R (b)





RE for now.

33 0x E

E

TM T F R F

R (c)


1 340 kip-in. B

E

RT T

R



1 3 3 3

54 teeth40 kip-in. 40 kip-in. 40 kip-in. 1.285714

42 teeth

B

E

NT T T T

N

(d)









1B


3E




B B E ER R



1 3B ER R (e)


1 1 3 31 3

1 1 3 3

T L T L

J G J G (f)



to obtain:

1 1 3 3

1 1 3 3

B E

T L T LR R

J G J G


1 1 3 3

1 1 3 3

B

E

N T L T L

N J G J G (g)




1 3 33 1

3 1 1

B

E

N L J GT T

N L J G


1 3

1 3 31

3 1 1

4

1 4

1

40 kip-in. 1.285714

40 kip-in. 1.285714

54 teeth 0.920772 in.40 kip-in. 1.285714

42 teeth 1.570796 in.

40 kip-in. 0.968994

B

E

T T

N L J GT

N L J G

T

T





1

40 kip-in.20.3149 kip-in.

1.968994T


1 3

13

40 kip-in. 1.285714

40 kip-in. 20.3149 kip-in. 40 kip-in.15.3106 kip-in.

1.285714 1.285714

T T

TT


1 11 4

1

(20.3149 kip-in.)(2.00 in. / 2)

1.570796 in12.93 ksi

.

T c

J Ans.


3 33 4

3

(15.3106 kip-in.)(1.75 in. / 2)

0.920772 in14.55 ksi

.

T c

J Ans.


3 33 4

3 3

3

( 15.3106 kip-in.)(2)(8 in.)0.021284 rad

(0.920772 in. )(12,

0.0213

500

d

k

r

si)

aE

T L

J G

Ans.


1 1

1 4

1 1

2 2

2 4

2 2

2 1 2

(20.3149 kip-in.)(2)(8 in.)0.016554 rad

(1.570796 in. )(12,500 ksi)

(40 kip-in.)(8 in.)0.016297 rad

(1.570796 in. )(12,500 ksi)

0.016554 rad 0.0 0.16297 ra 9 d 032 rC B

T L

J G

T L

J G

ad Ans.




6.92 The steel [G = 12,000 ksi] pipe shown in Fig. P6.92 is fixed to the wall support at C. The bolt holes

in the flange at A were supposed to align with mating holes in the wall support; however, an angular

misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation

torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside

diameter of the pipe is 3.50 in. and its wall thickness is 0.216 in.

(a) Determine the temporary installation

torque T′B that must be applied at B to align

the bolt holes at A.

(b) Determine the maximum shear stress

initial in the pipe after the bolts are

connected and the temporary installation

torque at B is removed.

(c) If the maximum shear stress in the pipe

shaft must not exceed 12 ksi, determine the

maximum external torque TB that can be

applied at B after the bolts are connected.

Fig. P6.92

Solution

Section Properties: The polar moment of inertia for the pipe is:

4 4 4

1 2 (3.50 in.) (3.068 in.) 6.034313 in.32

J J

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated

4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or

0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2)

2 22

2 2

T L

J G

the temporary torque T′B is:

4

2 2 2

2

(0.069813 rad)(6.034313 in. )(12,000 ksi)

(15 ft)(12 in./ft)28.1 kip-in.B

J GT

L

Ans.

(b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle

of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to

the 4° misfit is thus:

4

misfitinitial

1 2

(0.069813 rad)(6.034313 in. )(12,000 ksi)16.851 kip-in.

( ) (25 ft)(12 in./ft)

JGT

L L

and the maximum shear stress magnitude in the pipe due to this internal torque is:

initialinitial 4

(16.851 kip-in.)(3.50 in. / 2)

6.034313 in4.89 ksi

.

T c

J Ans.




After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate

torsion structure.

Equilibrium:

1 2 0x BM T T T (a)

Note: By inspection, the internal torque T2 will end up having

a negative value.

Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A

and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit

angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then

the pipe must twist in a negative sense to reach a zero rotation angle at C.

1 2 0.069813 rad (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0.069813 radT L T L

J G J G (d)



Tc T

J J c


1 1 2 2

1 1 2 2

0.069813 radL L

G c G c

(e)


1 1 2 1 1

1 2

1 1 2 2

2

2

( 0.069813 rad)

(12,000 ksi)(3.50 in./2) 15 ft( 0.069813 rad)

(10 ft)(12 in./ft) 10 ft

12.217305 ksi 1.5

G R L G c

L L G c

(f)

Assume the shaft (2) controls: By inspection of the FBD, the internal torque T2 should have a negative

value. Therefore, we can assume that the maximum shear stress in shaft (2) will be −12 ksi. If the shear

stress in shaft (2) reaches this value, then the shear stress in shaft (1) will be:

1 12.217305 ksi 1.5( 12 ksi) 5.7827 ksi 12 ksi O.K.

This calculation demonstrates that the shear stress magnitude in shaft (2) must control.






4

1 11

1

4

2 22

2

(5.7827 ksi)(6.034313 in. )19.9398 kip-in.

3.50 in./2

( 12 ksi)(6.034313 in. )41.3781 kip-in.

3.50 in./2

JT

c

JT

c

(c) Maximum Allowable Torque: From Eq. (a), the total torque acting at B must not exceed:





6.93 The steel [G = 12,000 ksi] pipe shown in Fig. P6.93 is fixed to the wall support at C. The bolt holes

in the flange at A were supposed to align with mating holes in the wall support; however, an angular

misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation

torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside

diameter of the pipe is 2.875 in. and its wall thickness is 0.203 in.

(a) Determine the temporary installation

torque T′B that must be applied at B to align

the bolt holes at A.

(b) Determine the maximum shear stress

initial in the pipe after the bolts are

connected and the temporary installation

torque at B is removed.

(c) Determine the magnitude of the

maximum shear stress in segments (1) and

(2) if an external torque of TB = 80 kip-in. is

applied at B after the bolts are connected.

Fig. P6.93

Solution

Section Properties: The polar moment of inertia for the pipe is:

4 4 4

1 2 (2.875 in.) (2.469 in.) 3.059108 in.32

J J

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated

4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or

0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2)

2 22

2 2

T L

J G

the temporary torque T′B is:

4

2 2 2

2

(0.069813 rad)(3.059108 in. )(12,000 ksi)

(15 ft)(12 in./f14

t).24 kip-in.B

J GT

L

Ans.

(b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle

of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to

the 4° misfit is thus:

4

misfitinitial

1 2

(0.069813 rad)(3.059108 in. )(12,000 ksi)8.5426 kip-in.

( ) (25 ft)(12 in./ft)

JGT

L L

and the maximum shear stress magnitude in the pipe due to this internal torque is:

initialinitial 4

(8.5426 kip-in.)(2.875 in. / 2)

3.059108 in4.01 ksi

.

T c

J Ans.




After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate

torsion structure.

Equilibrium:

1 2 0x BM T T T (a)

Note: By inspection, the internal torque T2 will end up having

a negative value.

Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A

and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit

angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then

the pipe must twist in a negative sense to reach a zero rotation angle at C.

1 2 0.069813 rad (b)


1 1 2 21 2

1 1 2 2

T L T L

J G J G (c)



1 1 2 2

1 1 2 2

0.069813 radT L T L

J G J G (d)


1 1 2 1 11 2

1 1 2 2

4

2

2

( 0.069813 rad)

(3.059108 in. )(12,000 ksi) 15 ft( 0.069813 rad)

(10 ft)(12 in./ft) 10 ft

21.356600 kip-in. 1.5

J G L J GT T

L L J G

T

T


1 2 2 2 2

2

21.356600 kip-in. 1.5 21.356600 kip-in. 2.5 80 kip-in.

101.3566 kip-in.40.5426 kip-in.

2.5

T T T T T

T


1 2 80 kip-in. 40.5426 kip-in. 80 kip-in. 39.4574 kip-in.T T

(c) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is:

1 11 4

1

(39.4574 kip-in.)(2.875 in. / 2)

3.059108 in18.54 ksi

.

T c

J Ans.


2 22 4

2

(40.5426 kip-in.)(2.875 in. / 2)

3.059108 in19.05 ksi

.

T c

J Ans.




6.94 A stepped shaft with a major diameter of D = 1.375 in. and a minor diameter of d = 1.00 in. is

subjected to a torque of 500 lb-in. A fillet with a radius of r = 3/16 in. is used to transition from the

major diameter to the minor diameter. Determine the maximum shear stress in the shaft.

Solution

Fillet: From Fig. 6.17b:

1.375 in. 0.1875 in.

1.375 0.1875 1.221.0 in. 1.0 in.

D rK

d d

Section Properties of Minor Diameter Section:

4 4(1.00 in.) 0.098175 in.32

J

Maximum Shear Stress:

max 4

(500 lb-in.)(1.00 in./2)(1.22) 3,106.7 psi

0.098175 in3,110 ps

.i

TcK

J Ans.




6.95 A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mm is

subjected to a torque of 25 N-m. A full quarter-circular fillet having a radius of r = 2 mm is used to

transition from the major diameter to the minor diameter. Determine the maximum shear stress in the

shaft.

Solution


20 mm 2 mm

1.25 0.125 1.3016 mm 16 mm

D rK

d d


4 4(16 mm) 6,434.0 mm32

J

Maximum Shear Stress:

4

(25 N-m)(16 mm/2)(1,000 mm/m)(1.30)

6,434.0 mm40.4 MPa

TcK

J Ans.




6.96 A fillet with a radius of 1/2 in. is used at the junction in a stepped shaft where the diameter is

reduced from 8.00 in. to 6.00 in. Determine the maximum torque that the shaft can transmit if the

maximum shear stress in the fillet must be limited to 5 ksi.

Solution


8.0 in. 0.50 in.

1.33 0.0833 1.416.0 in. 6.0 in.

D rK

d d


4 4(6.00 in.) 127.2345 in.32

J

Maximum Torque:

4

allow

max

(5 ksi)(127.2345 in. )

(1.41)(6.00 in./2)150.4 kip-in.

TcK

J

JT

Kc Ans.




6.97 A stepped shaft with a major diameter of D = 2.50 in. and a minor diameter of d = 1.25 in. is

subjected to a torque of 1,200 lb-in. If the maximum shear stress must not exceed 4,000 psi, determine

the minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet

radius must be chosen as a multiple of 0.05 in.

Solution


4 4(1.25 in.) 0.239684 in.32

J

Fillet Requirement: From Fig. 6.17b:

4(4,000 psi)(0.239684 in. )1.28

(1,200 lb-in.)(1.25 in. / 2)

2.5 in.2.00

1.25 in.

0.17

Tc JK K

J Tc

D

d

r

d

Minimum Fillet Radius:

min min(0.17)(1.25 in.) 0.213 in. say 0.25 in.r

r d rd

Ans.




6.98 A fillet with a radius of 16 mm is used at the junction in a stepped shaft where the diameter is

reduced from 200 mm to 150 mm. Determine the maximum torque that the shaft can transmit if the

maximum shear stress in the fillet must be limited to 55 MPa.

Solution


200 mm 16 mm

1.33 0.107 1.34150 mm 150 mm

D rK

d d


4 4(150 mm) 49,700,978 mm32

J

Maximum Torque:

2 4

6allow

max

(55 N/mm )(49,700,978 mm )27.1995 10 N-mm

(1.34)(150 mm/227.2 kN-m

)

TcK

J

JT

Kc Ans.




6.99 A stepped shaft with a major diameter of D = 50 mm and a minor diameter of d = 32 mm is

subjected to a torque of 210 N-m. If the maximum shear stress must not exceed 40 MPa, determine the

minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet

radius must be chosen as a multiple of 1 mm.

Solution


4 4(32 mm) 102,943.7 mm32

J

Fillet Requirement: From Fig. 6.17b:

2 4(40 N/mm )(102,943.7 mm )1.23

(210 N-m)(32 mm / 2)(1,000 mm/m)

50 mm1.56

32 mm

0.19

Tc JK K

J Tc

D

d

r

d

Minimum Fillet Radius:

min min(0.19)(32 mm) 6.08 mm say 7 m mr

r d rd

Ans.




6.100 A stepped shaft has a major diameter of D = 2.00 in. and a minor diameter of d = 1.50 in. A fillet

with a 0.25-in. radius is used to transition between the two shaft segments. The maximum shear stress

in the shaft must be limited to 9,000 psi. If the shaft rotates at a constant angular speed of 800 rpm,

determine the maximum power that may be delivered by the shaft.

Solution


2.00 in. 0.25 in.

1.33 0.167 1.241.50 in. 1.50 in.

D rK

d d


4 4(1.50 in.) 0.497010 in.32

J

Maximum Torque:

allow

2 4

allow

max

(9,000 lb/in. )(0.497010 in. )4,809.77 lb-in. 400.8143 lb-ft

(1.24)(1.50 in./2)

TcK

J

JT

Kc


max max

800 rev 1 min 2 rad(400.8143 lb-ft) 33,578.54 lb-ft/s

min 60 s 1 revP T


max

33,578.54 lb-ft/s

550 lb-ft/s

1

61

hp

hp

.1P Ans.




6.101 A stepped shaft has a major diameter of D = 100 mm and a minor diameter of d = 75 mm. A fillet

with a 10-mm radius is used to transition between the two shaft segments. The maximum shear stress in

the shaft must be limited to 60 MPa. If the shaft rotates at a constant angular speed of 500 rpm,

determine the maximum power that may be delivered by the shaft.

Solution


100 mm 10 mm1.33 0.133 1.29

75 mm 75 mm

D rK

d d


4 4(75 mm) 3,106,311.1 mm32

J

Maximum Torque:

allow

2 4

allow

max

(60 N/mm )(3,106,311.1 mm )3,852,789 N-mm 3,852.789 N-m

(1.29)(75 mm/2)

TcK

J

JT

Kc


max max

500 rev 1 min 2 rad(3,852.789 N-m)

min 60 s 1 rev

202 kW

201,731.6 N-m/s

P T

Ans.




6.102 A 2-in.-diameter shaft contains a ½-in.-deep U-shaped groove that has a ¼-in. radius at the bottom

of the groove. The shaft must transmit a torque of T = 720 lb-in. Determine the maximum shear stress in

the shaft.

Solution

Groove: From Fig. 6.17a:

2.0 in. 2(0.5 in.) 1.0 in.

2.0 in. 0.25 in.2.00 0.25 1.29

1.0 in. 1.0 in.

d

D rK

d d

Section Properties at Minimum Diameter Section:

4 4(1.00 in.) 0.098175 in.32

J

Shaft Shear Stress:

max 4

(720 lb-in.)(1.00 in./2)(1.29) 4,730.34 psi

0.098175 in.4,730 psi

TcK

J Ans.




6.103 A semicircular groove with a 6-mm radius is required in a 50-mm-diameter shaft. If the maximum

allowable shear stress in the shaft must be limited to 40 MPa, determine the maximum torque that can be

transmitted by the shaft.

Solution


50 mm 2(6 mm) 38 mm

50 mm 6 mm1.32 0.158 1.39

38 mm 38 mm

d

D rK

d d


4 4(38 mm) 204,707.75 mm32

J

Maximum Torque:

2 4

allow

max

(40 N/mm )(204,707.75 mm )310,045.8 N-mm

(1.39)(38 m310 N-

mm

/2)

TcK

J

JT

Kc Ans.




6.104 A 40-mm-diameter shaft contains a 10-mm-deep U-shaped groove that has a 6-mm radius at the

bottom of the groove. The maximum shear stress in the shaft must be limited to 60 MPa. If the shaft

rotates at a constant angular speed of 22 Hz, determine the maximum power that may be delivered by

the shaft.

Solution


40 mm 2(10 mm) 20 mm

40 mm 6 mm2.00 0.30 1.25

20 mm 20 mm

d

D rK

d d


4 4(20 mm) 15,707.96 mm32

J

Maximum Torque:

2 4

allow

max

(60 N/mm )(15,707.96 mm )75.3982 N-m

(1.25)(20 mm/2)

TcK

J

JT

Kc


max max

22 rev 2 rad(75.3982 N-m)

s 1 rev

10,42

1

2.3 N-m/

0.42

s

kW

P T

Ans.




6.105 A 1.25-in. -diameter shaft contains a 0.25-in.-deep U-shaped groove that has a 1/8-in. radius at the

bottom of the groove. The maximum shear stress in the shaft must be limited to 12,000 psi. If the shaft

rotates at a constant angular speed of 6 Hz, determine the maximum power that may be delivered by the

shaft.

Solution


1.25 in. 2(0.25 in.) 0.75 in.

1.25 in. 0.125 in.1.67 0.167 1.39

0.75 in. 0.75 in.

d

D rK

d d


4 4(0.75 in.) 0.031063 in.32

J

Maximum Torque:

4

allow

max

(12,000 psi)(0.031063 in. )715.1219 lb-in. 59.5935 lb-ft

(1.39)(0.75 in./2)

TcK

J

JT

Kc


max max

6 rev 2 rad(59.5935 lb-ft)

s 1 rev

2,246.6219 lb-ft/s

P T


max

2,246.6219 lb-ft/s

550 lb-ft/s

1

4.

hp

hp

08P Ans.




6.106 A torque of magnitude T = 1.5

kip-in. is applied to each of the bars

shown in Fig. P6.106. If the allowable

shear stress is specified as allow = 8 ksi,

determine the minimum required

dimension b for each bar.

Fig. P6.106

Solution

(a) Circular Section


4 3

which can be simplified to 32 ( / 2) 16

d T d T

d


316T

d

To support a torque of T = 1.5 kip-in. without exceeding the maximum shear stress of 8 ksi, the solid

shaft must have a diameter (i.e., dimension b shown in the problem statement) of

3 3min

16 16(1.5 kip-in.)

(8 ksi)0.985 in.

Tb

Ans.

(b) Square Section

From Table 6.1,

aspect ratio 1 0.208b

a ba

The maximum shear stress in a rectangular section is given by Eq. (6.22):

max 2

T

a b

For a square section where a = b,

3 3

min

max

1.5 kip-in.0.901442 in.

(0.208)(8 ks0.966 in.

i)

Tb b

Ans.

(c) Rectangular Section

From Table 6.1,

2


b

For a rectangular section where a = 2b,

3 3

min

max

1.5 kip-in.0.381098 in.

2 2(0.246)(8 k0.725 in.

si)

Tb b

Ans.




6.107 A torque of magnitude T = 270 N-

m is applied to each of the bars shown in

Fig. P6.107. If the allowable shear stress

is specified as allow = 70 MPa, determine

the minimum required dimension b for

each bar.

Fig. P6.107

Solution

(a) Circular Section


4 3

which can be simplified to 32 ( / 2) 16

d T d T

d


316T

d

To support a torque of T = 270 N-m without exceeding the maximum shear stress of 70 MPa, the solid

shaft must have a diameter (i.e., dimension b shown in the problem statement) of

3 3min 2

16 16(270 N-m)(1,000 mm/m)

(70 N/mm )27.0 mm

Tb

Ans.

(b) Square Section

From Table 6.1,


a ba

The maximum shear stress in a rectangular section is given by Eq. (6.22):

max 2

T

a b

For a square section where a = b,

3 3

min2

max

(270 N-m)(1,000 mm/m)18,543.946 mm

(0.208)(726.5 m

0 N/mm )m

Tb b

Ans.

(c) Rectangular Section

From Table 6.1,

2


b

For a rectangular section where a = 2b,

3 3

min2

max

(270 N-m)(1,000 mm/m)7,839.721 mm

2 2(0.246)(19.87 mm

70 N/mm )

Tb b

Ans.




6.108 The bars shown in Fig. P6.108

have equal cross-sectional areas and they

are each subjected to a torque of T = 160

N-m. Determine:

(a) the maximum shear stress in each

bar.

(b) the rotation angle at the free end if

each bar has a length of 300 mm.

Assume G = 28 GPa.

Fig. P6.108

Solution

Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22):

max 2

T

a b

and the angle of twist is given by Eq. (6.23)

3

TL

a bG

Rectangular Section 15 mm by 50 mm

From linear interpolation of the values given in Table 6.1,

50 mm

aspect ratio 3.333 0.272 0.26915 mm

Shear stress:

max 2 2

(160 N-m)(1,000 mm/m)

(0.272)(15 mm) (50 mm)52.3 MPa

T

a b

Ans.

Angle of twist:

3 3 2

(160 N-m)(300 mm)(1,000 mm/m)

(0.269)(15 mm) (50 mm)(28,000 N/mm )0.0378 rad

TL

a bG

Ans.


From Table 6.1,

30 mm

aspect ratio 1.2 0.219 0.16625 mm

Shear stress:

max 2 2

(160 N-m)(1,000 mm/m)

(0.219)(25 mm) (30 mm)39.0 MPa

T

a b

Ans.

Angle of twist:

3 3 2

(160 N-m)(300 mm)(1,000 mm/m)

(0.166)(25 mm) (30 mm)(28,000 N/mm )0.0220 rad

TL

a bG

Ans.




6.109 The allowable shear stress for each

bar shown in Fig. P6.109 is 75 MPa.

Determine:

(a) the largest torque T that may be

applied to each bar.

(b) the corresponding rotation angle at the

free end if each bar has a length of 300

mm. Assume G = 28 GPa.

Fig. P6.109

Solution

Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22):

max 2

T

a b

and the angle of twist is given by Eq. (6.23)

3

TL

a bG


From linear interpolation of the values given in Table 6.1,

50 mm

aspect ratio 3.333 0.272 0.26915 mm

Maximum torque:

2 2 2

max (75 N/mm )(0.272)(15 mm) (50 mm) 229,500 N-mm 230 N-mT a b Ans.

Angle of twist:

3 3 2

(229,500 N-mm)(300 mm)

(0.269)(15 mm) (50 mm)(28,000 N/mm )0.0542 rad

TL

a bG

Ans.


From Table 6.1,

30 mm

aspect ratio 1.2 0.219 0.16625 mm

Maximum torque:

2 2 2

max (75 N/mm )(0.219)(25 mm) (30 mm) 307,979 N-mm 308 N-mT a b Ans.

Angle of twist:

3 3 2

(307,979 N-mm)(300 mm)

(0.166)(25 mm) (30 mm)(28,000 N/mm )0.0424 rad

TL

a bG

Ans.




6.110 A solid circular rod having

diameter D is to be replaced by a

rectangular tube having cross-sectional

dimensions D × 2D (which are

measured to the wall centerlines of the

cross section shown in Fig. P6.110).

Determine the required minimum

thickness tmin of the tube so that the

maximum shear stress in the tube will

not exceed the maximum shear stress

in the solid bar.

Fig. P6.110

Solution

For the solid circular rod, the maximum shear stress is given by

max 34

( / 2) 16

32

Tc T D T

J DD

(a)

For the rectangular tube, the area enclosed by the median line is

2( )(2 ) 2mA D D D

The maximum shear stress for the thin-walled section is given by Eq. (6.25)

max 2 22 2(2 ) 4m

T T T

A t D t D t (b)

Equate Eqs. (a) and (b)

3 2

16

4

T T

D D t

and solve for t:

min

64t

D Ans.




6.111 A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet is to be formed into a hollow section

by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a cross-

sectional medial length of 24 in. (no stretching of the sheet due to bending). If the maximum shear stress

must be limited to 12 ksi, determine the maximum torque that can be carried by the hollow section if

(a) the shape of the section is a circle.

(b) the shape of the section is an equilateral triangle.

(c) the shape of the section is a square.

(d) the shape of the section is an 8 × 4 in. rectangle.

Solution

The maximum shear stress for a thin-walled section is given by Eq. (6.25)

max2 m

T

A t

and thus, the maximum torque that can be carried by the hollow section is

max max2 mT A t

(a) Circle:

2 2

2

max max

24 in. 7.639437 in.

(7.639437 in.) 45.8366 in.4

2 2(12 ksi)(45.8366 in. )(0. 110.0 kip-i100 in n.) .

m m

m

m

d d

A

T A t

Ans.

(b) Equilateral triangle:

2

2

max max

triangle sides are each 24 in./3 8 in.

1 1(8 in.)(8 in.)sin 60 27.7128 in.

2 2

2 2(12 ksi)(27.7128 in. 66.)( 5 ki0.10 p-i0 in .) n.

m

m

A bh

T A t

Ans.

(c) Square:

2

2

max max

sides of the square are each 24 in./4 6 in.

(6 in.)(6 in.) 36 in.

2 2(12 ksi)(36 in. )(0. 86.4 ki1 p-in0 n.) .0 i

m

m

A bh

T A t

Ans.

(d) 8 × 4 in. rectangle:

2

2

max max

(8 in.)(4 in.) 32 in.

2 2(12 ksi)(32 in. )(0.100 in.) 76.8 kip-in.

m

m

A bh

T A t

Ans.




6.112 A 500 mm wide by 3 mm thick by 2 m long aluminum sheet is to be formed into a hollow section

by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a cross-

sectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear

stress must be limited to 75 MPa, determine the maximum torque that can be carried by the hollow

section if

(a) the shape of the section is a circle.

(b) the shape of the section is an equilateral triangle.

(c) the shape of the section is a square.

(d) the shape of the section is a 150 × 100 mm rectangle.

Solution


max2 m

T

A t

and thus, the maximum torque that can be carried by the hollow section is

max max2 mT A t

(a) Circle:

2 2

2 2

max max

5

8

00 mm 159.155 mm

(159.155 mm) 19,894.382 mm4

2 2(75 N/mm )(19,894.382 mm )(3 mm) 8,952,472 N-mm .95 kN-m

m m

m

m

d d

A

T A t

Ans.

(b) Equilateral triangle:

2

2 2

max max

triangle sides are each 500 mm/3 166.667 mm

1 1(166.667 mm)(166.667 mm)sin 60 12,028.131 mm

2 2

2 2(75 N/mm )(12,028.131 mm )(3 mm) 5,412,65 5.41 kN-m9 N-mm

m

m

A bh

T A t

Ans.

(c) Square:

2

2 2

max max

sides of the square are each 500 mm/4 125 mm

(125 mm)(125 mm) 15,625 mm

2 2(75 N/mm )(15,625 mm )(3 mm) 7,031,2 7.03 kN50 N-mm -m

m

m

A bh

T A t

Ans.

(d) 150 × 100 mm rectangle:

2

2 2

max max

(150 mm)(100 mm) 15,000 mm

2 2(75 N/mm )(15,000 mm )(3 mm) 6,750,000 6.75 kNN- -mmm

m

m

A bh

T A t

Ans.




6.113 A torque of T = 150 kip-in. will be applied to

the hollow, thin-walled aluminum alloy section

shown in Fig. P6.113. If the maximum shear stress

must be limited to 10 ksi, determine the minimum

thickness required for the section. (Note: The

dimensions shown are measured to the wall

centerline.)

Fig. P6.113

Solution


max2 m

T

A t

For the aluminum alloy section,

2 2(6 in.)(8 in.) (6 in.) 76.274 in.4

mA

The minimum thickness required for the section if the maximum shear stress must be limited to 10 ksi is

thus:

min 2

max

150 kip-in.

2 2(10 ksi)(76.274 in. )0.0983 in.

m

Tt

A Ans.




6.114 A torque of T = 2.5 kN-m will be applied to the hollow,

thin-walled aluminum alloy section shown in Fig. P6.114. If

the maximum shear stress must be limited to 50 MPa,

determine the minimum thickness required for the section.

(Note: The dimensions shown are measured to the wall

centerline.)

Fig. P6.114

Solution


max2 m

T

A t


2 21(100 mm) 7,853.982 mm

4mA

The minimum thickness required for the section if the maximum shear stress must be limited to 50 MPa

is thus:

min 2 2

max

(2.5 kN-m)(1,000 N/kN)(1,000 mm/m)

2 2(50 N/mm )(7,853.982 mm3.18 mm

)m

Tt

A Ans.




6.115 A torque of T = 100 kip-in. will be applied to


shown in Fig. P6.115. If the section has a uniform

thickness of 0.100 in., determine the magnitude of

the maximum shear stress developed in the section.

(Note: The dimensions shown are measured to the

wall centerline.)

Fig. P6.115

Solution


max2 m

T

A t


2 2(10 in.) 39.2699 in.8

mA

The maximum shear stress developed in the section is:

max 2

100 kip-in.

2 2(39.2699 in. )(0.100 in.1

)2.73 ksi

m

T

A t Ans.




6.116 A torque of T = 2.75 kN-m will be applied to


shown in Fig. P6.116. If the section has a uniform

thickness of 4 mm, determine the magnitude of the

maximum shear stress developed in the section.


wall centerline.)

Fig. P6.116

Solution


max2 m

T

A t


2 2(150 mm)(50 mm) (25 mm) 8,481.748 mm2

mA


max 2

(2.75 kN-m)(1,000 N/kN)(1,000 mm/m)

2 2(8,481.748 mm )(4 mm)40.5 MPa

m

T

A t Ans.




6.117 A cross section of the leading edge of an

airplane wing is shown in Fig. P6.117. The enclosed

area is 82 in.2. Sheet thicknesses are shown on the

diagram. For an applied torque of T = 100 kip-in.,

determine the magnitude of the maximum shear

stress developed in the section. (Note: The

dimensions shown are measured to the wall

centerline.)

Fig. P6.117

Solution


max2 m

T

A t


max 2

100 kip-in.

2 2(82 in. )(0.050 in.)12.20 ksi

m

T

A t Ans.




6.118 A cross section of an airplane fuselage made

of aluminum alloy is shown in Fig. P6.118. For an

applied torque of T = 1,250 kip-in. and an allowable

shear stress of = 7.5 ksi, determine the minimum

thickness of the sheet (which must be constant for

the entire periphery) required to resist the torque.


wall centerline.)

Fig. P6.118

Solution


max2 m

T

A t

For the fuselage,

2 2(30 in.)(20 in.) (15 in.) 1,306.858 in.mA

The minimum thickness required for the sheet if the maximum shear stress must be limited to 7.5 ksi is

thus:

min 2

max

1,250 kip-in.

2 2(7.5 ksi)(1,306.858 in. )0.0638 in.

m

Tt

A Ans.

Philpot MoM 2nd Ch01-06 ISM

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Transcript of Philpot MoM 2nd Ch01-06 ISM