Mechanics of Materials, Ch01-Axial Load
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Transcript of Mechanics of Materials, Ch01-Axial Load
Russell C. Hibbeler
Chapter 4: Axial Load
Elastic Deformation of an Axially Loaded Member
� Using Hooke’s law and the definitions of stress and
strain, we are able to develop the elastic deformation of a member subjected to axial loads.of a member subjected to axial loads.
� Suppose an element subjected to loads,
( )( )
and P x dδ
εA x dx E
σσ = = =
( )( )∫=
L
ExA
dxxP
0
δ
= small displacementδ
( )( )
P x dxd
A x Eδ =
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
= small displacement
L = original length
P(x) = internal axial force
A(x) = cross-sectional area
E = modulus of elasticity
δ
Constant Load and CrossConstant Load and Cross--Sectional AreaSectional Area
� When a constant external force is applied at each end of the member, end of the member,
Sign ConventionSign Convention
� Force and displacement is positive when tension and
AE
PL=δ
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
elongation and negative will be compression and contraction.
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
For small elements, small deflection approximation assumes that higher order terms can be ignored:
2 2 2 2, , , ,dx dy dw dz dwdz
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
Example 4.4A member is made from a material that has a specific weight
and modulus of and elasticity E. If it is formed into a cone,
find how far its end is displaced due to gravity when it is
suspended in the vertical position.
γ
suspended in the vertical position.
Solution:Radius x of the cone as a function of y is determined by proportion,
yL
rx
L
r
y
x oo == ;
The volume of a cone having a base of radius x and height y is
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
The volume of a cone having a base of radius x and height y is
3
2
2
2
33y
L
ryxV oππ
==
Since , the internal force at the section becomes
Solution:
( ) 3
2
2
3 ;0 y
L
ryPF o
y
γπ==↑+ ∑
VW γ=
3L
( ) 2
2
2
2y
L
rxyA oπ
π ==
The area of the cross section is also a function of position y,
Between the limits of y =0 and L yields
( ) ( )[ ]( )[ ] (Ans)
3222
LdyLrdyyPL
o
L γγπδ === ∫∫
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
( )( )
( )[ ]( )[ ] (Ans)
6
3
0
22
0E
L
ELr
dyLr
EyA
dyyP
o
o γγπ
γπδ === ∫∫
Principle of Superposition
� Principle of superposition is to simplify stress and
displacement problems by subdividing the loading into components and adding the results.
� A member is statically indeterminate when equations
of equilibrium are not sufficient to determine the reactions on a member.
Statically Indeterminate Axially Loaded Member
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
0, 0x B AF F F P= + − =∑
Statically Indeterminate Axially Loaded Member
Equilibrium equation:
/ 0A Bδ =
A ACAC
F L
AEδ = B CB
CB
F L
AEδ
−=
Method of solving 1: Compatibility
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
0A AC B CBF L F L
AE AE− =
CB ACA B
L LF P and F P
L L
= =
Method of solving 2: Superposition
0AC BPL F L
AE AE− =
Displacement without support at B
Reaction at support to compress bar to original length: remove P
ACB p
PL
AEδ δ= =AF P=
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
ACB
LF P
L
=
0ACA
LP F P
L
+ − =
ACB ABB
PLF L
AE AEδ
−− = = −
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
Thermal Stress
� Change in temperature cause a material to change
its dimensions.
Since the material is homogeneous and isotropic,� Since the material is homogeneous and isotropic,
TLT ∆−= αδ
linear coefficient of thermal expansionα
0
L
T Tdxδ α= ∆∫Td T dxδ α= ∆
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
= linear coefficient of thermal expansion, property of the material
= algebraic change in temperature of the member
= original length of the member
= algebraic change in length of the member
αT∆T
Tδ
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
*Inelastic Axial Deformation and Residual stresses
� Member may be designed to yield and permanently
deform; this is referred to as being elastic perfectly
plastic or elastoplastic.plastic or elastoplastic.
AdAP Y
A
Yp σσ == ∫
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
� The yield stress and A is the bar’s cross-sectional
area at section a–a.
CB ACA B
L LF P and F P
L L
= =
Since AC yielded re-calculate FA
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
P applied in reverse direction
Force P removed:
Residual Stress:
Permanent displacement:
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition
Alternative way:
© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd
Chapter 4: Axial LoadChapter 4: Axial Load
Mechanics of Material 7Mechanics of Material 7thth EditionEdition