Torsion of Prismatic Bars

Outline

• Elastic Cylinders with End Loading• Torsion of Cylinders: General formulation• Stress-Function Formulation• Displacement Formulation• Membrane Analogy• Solution: Boundary Equation Scheme• Solution: Fourier Method – Rectangular Section• Multiply Connected Cross-Sections• Hollow Sections

2

0.x y xyσ σ τ= = =

xyx

x yτσ ∂∂

+∂ ∂

xzxFz

τ∂+ +

∂0

xy y

x yτ σ

=

∂ ∂+

∂ ∂yz

yFzτ∂

+ +∂

0

yzxz zzFx y z

ττ σ

=

∂∂ ∂+ + +

∂ ∂ ∂

( )

( )2

2

0

0 ,

0 ,

0

xzxz xz

y

z

zyz yz

x yz

xz

z

y

τ τ τ

ττ

σ

τ

=

∂ = ⇒ = ∂∂⇒ = ⇒

∂=

∂

= ∂

Elastic Cylinders Subjected to End Loadings• Semi-Inverse Method• Zero lateral forces:• Let us guess the most general form for stresses

3

2xσ∇

2

2

11 x yx

σ σν

∂+ +

+ ∂ ( ) 21yx xz

z

FF FFx y z x

νσν

∂ ∂ ∂∂+ = − + + − − ∂ ∂ ∂ ∂

2

2

2 0z

y

x

σ

σ∂⇒ =

∂

∇2

2

11 x yy

σ σν

∂+ +

+ ∂ ( ) 21y yx z

z

F FF Fx y z y

νσν

∂ ∂ ∂ ∂+ = − + + − − ∂ ∂ ∂ ∂

2

2

2

0

xy

z

y

τ

σ=

∂

∇

∂⇒

211 x yx y

σ σν

∂+ +

+ ∂ ∂ ( )yx

z

FFy x

σ∂ ∂

+ = − + ∂ ∂ 2

22

2

0

11

,

z

z

x yz

x y

σ σ σν

σ

∂∇ + +

+ ∂

∂⇒ =

∂ ∂

( ) 21yx z z

z

FF F Fx y z z

νσν

∂ ∂ ∂ ∂+ = − + + − − ∂ ∂ ∂ ∂

satisfied

• Beltrami-Michell Equations

2 2 2 2

1 2 3 4 5 62 2 2 0z z z z

z C x C y C z C xz C yz Cx y z x yσ σ σ σ σ∂ ∂ ∂ ∂= = = = ⇒ = + + + + +∂ ∂ ∂ ∂ ∂

Elastic Cylinders Subjected to End Loadings

4

22 1

1xz x yx zτ σ σ

ν∂

∇ + ++ ∂ ∂ ( ) x zz

F Fz x

σ ∂ ∂ + = − + ∂ ∂

2 24

2 2

22

1

11

xz xz

yz x y

Cx y

y z

τ τν

τ σ σν

∂ ∂⇒ + = −

∂ ∂ +

∂∇ + +

+ ∂ ∂ ( )y z

z

F Fz y

σ∂ ∂

+ = − + ∂ ∂

2 25

2 2 1yz yz C

x yτ τ

ν

∂ ∂⇒ + = − ∂ ∂ +

Assumptions• Load Pz is applied at centroid of cross-

section so no bending effects• Using Saint-Venant Principle, exact end

tractions are replaced by statically equivalent uniform loading

• Thus assume stress σz is uniform over any cross-section throughout the solid

, 0zz xz yzPA

σ τ τ⇒ = = =

• Using stress results into Hooke’s law and combining with the strain-displacement relations gives

0,0,0

,,

=∂∂

+∂∂

=∂∂

+∂∂

=∂∂

+∂∂

=∂∂ν

−=∂∂ν

−=∂∂

zu

xw

yw

zv

xv

yu

AEP

zw

AEP

yv

AEP

xu zzz

Integrating and dropping rigid-body motion terms such that displacements vanish at origin zAE

Pw

yAEPv

xAEPu

z

z

z

=

ν−=

ν−=

Extension of Cylinders

x

y

z Pz

ℓ

S

R

5

x

y

z T

ℓ

S

R

Guided by Observations from Strength of Materials:

Torsion of Cylinders

6

• Projection of each section on x-yplane rotates as rigid-body about central axis

• Amount of projected section rotation is linear function of axial coordinate

• Plane cross-sections will not remain plane after deformation thus leading to a warping displacement

sin , cosu r y v r xβ θ β β θ β= − = − = =

.zβ α=

, , ( , ).u yz v xz w w x yα α⇒ = − = =

Torsional Deformation• In-plane / projected displacements

• Angle of twist:• The warping displacement is assumed to be

a function of only the in-plane coordinates

7

• Now must show assumed displacement form will satisfy all elasticity field equations

0 0

12

( , )12

x y z xy x y z xy

xz xz

yz yz

u yzw wv xz y G yx x

w w x yw wx G xy y

ε ε ε ε σ σ σ τα

α ε α τ α

ε α τ α

= = = = = = = == −

∂ ∂ = ⇒ = − + ⇒ = − + ∂ ∂ = ∂ ∂

= + = + ∂ ∂

Stress Function Formulation• Strain and stress field

• Equilibrium equations result inzyzxz

x yσττ ∂∂∂

+ +∂ ∂ z

Fz

+∂

0

,xz yzy xψ ψτ τ

=

∂ ∂⇒ = = −

∂ ∂

• Two stress components satisfy

2 22

2 2

2

2

yzxz Gy x

Gx y

ττ α

ψ ψψ α

∂∂− = −

∂ ∂

∂ ∂⇒ ∇ = + = −

∂ ∂

• ψ = ψ(x,y)• Prandtl Stress Function 8

• Stress compatibility is automatically satisfied.

nx xT σ= x yxn τ+ y zx zn nτ+ 0 0 0ny xyT τ

= ⇒ =

= x yn σ+ y zy zn nτ+ 0 0 0n

z xz x yz y z zT n n nτ τ σ

= ⇒ =

= + +

i.e.

0

0 0, se t: 0.dy dx dy ds x ds ds

ψ ψ ψψ

= ⇒

∂ ∂ + − − = ⇒ = ∂ ∂ =

1 : 0 , 2 : 0

3 : 0

x xz y yzA A

z z

P dxdy P dxdy

P

τ τ

σ

= = = =

= =

∫∫ ∫∫, 4 : 0 x zA dxdy M yσ= =∫∫

5 : 0A

y z

dxdy

M xσ= =

∫∫( ), 6 : z yz xzA Adxdy T M x y dxdyτ τ= = −∫∫ ∫∫

Stress Function Formulation: BCs• On lateral surface

• On ends:• More interested in satisfying the resultant

end-loadings

, , 0n n nx xz y yz z zT T Tτ τ σ= ± = ± = =

,x ydy dxn nds ds

= = −

x

y

z T

ℓ

S

R n

( ) ( )

3 5 are automatically satisfied.

1 : 0 satisfied

2 : 0 satisfied

6 :

xzA A S S

yzA A S S

yz xz

y

A

x

A

dxdy dxdyy

d

dx n ds

dxdy dxdyx

xT x y dxdy x y dxdy

y n ds

x y x

ψτ ψ ψ

ψτ ψ ψ

ψψ ψτ τ

−∂

= = = =∂∂

= − = − = − =∂

∂ ∂ ∂= − = − + = − + ∂ ∂

−

∂

∫∫ ∫∫ ∫ ∫

∫∫ ∫∫ ∫ ∫

∫∫ ∫∫( )

( )

2A

S

ydxdy

y

y dx x dy

ψψ

ψ ψ

∂ − ∂

= − − +

∫∫

∫ 2 2A Adxdy T dxdyψ ψ+ ⇒ =∫∫ ∫∫

Stress Function Formulation: BCs• On ends (with a big help from the Green’s theorem)

• The assumed stress function yields a governing Poisson equation.• The stress function vanishes on the lateral boundary.• The overall torque is related to the integral of the stress Function.• The remaining end conditions are automatically satisfied.

10

0 nz xz x yz y z zT n n nτ τ σ= = + +

( )2 2

For arbitrary n: ,

or equivalently:

2

x y

xy yxw wn nx y

w wG y n G

dy dx

x nx y

w

dwd

wy xx y

yn y xds ds

d x y

xn

n ds

α α

α

αα α α

α

α

+

∂ ∂ = − + + +

∂ ∂+

∂ ∂

∂ ∂

∂ ∂= = −

∂ ∂

= − =

⇒ = +

Displacement Formulation• Expressing the equilibrium equation with warping displacement

• BCs on the lateral surface:

,x ydy dxn nds ds

= = −

2G u∇ ( ) uGx x

λ ∂ ∂+ +∂ ∂

vy∂

+∂

wz

∂+∂ x

F

+

2

0 satisfied

G v

=

∇ ( ) uGy x

λ ∂ ∂+ +∂ ∂

vy∂

+∂

wz

∂+∂ y

F

+

( )2

0 satisfied

uG w Gz x

λ

=

∂ ∂∇ + +

∂ ∂vy∂

+∂

wz

∂+∂ z

F

+

2 2

2 20 0w w

x y∂ ∂

= ⇒ + =∂ ∂

11

1 : 0 , 2 : 0

3 : 0

x xz y yzA A

z z

P dxdy P dxdy

P

τ τ

σ

= = = =

= =

∫∫ ∫∫, 4 : 0 x zA dxdy M yσ= =∫∫

5 : 0A

y z

dxdy

M xσ= =

∫∫( )

( )

2 2

2 2

, 6 :

1 5 are automatically satisfied.

6 :

( )

z yz xzA A

z yz xzA A

A

dxdy T M x y dxdy

w wT M x y dxdy xG x yG y dxdyy x

w wT G x y x y dxdy Jy x

x w y wJ G x y dy x

τ τ

τ τ α α

α α

α α

= = −

−

∂ ∂ = = − = + − − + ∂ ∂ ∂ ∂

⇒ = + + − = ∂ ∂ ∂ ∂

= + + − ∂ ∂

∫∫ ∫∫

∫∫ ∫∫

∫∫

. . . Torsional RigidityA

xdy∫∫

• BCs on the ends

Displacement Formulation

12

2 22

2 2 2

0 Lateral Surface

2 EndsA

G Rx y

T dxdy

ψ ψψ α

ψ

ψ

∂ ∂∇ = + = − ∈

∂ ∂= ∈

= ∈∫∫( )

2 2

2 2

2 2

2 2

0

12

Lateral Surface

( )

Ends

A

w w Rx y

dw d x ydn ds

w wT G x y x y dxdyy x

α

α

∂ ∂+ = ∈

∂ ∂

= +

∈

∂ ∂= + + − ∂ ∂

∈

∫∫

• Stress Function Formulation • Displacement Formulation

• Relatively Simple Governing Equation

• Relatively Simple Boundary Conditions

• Simple Governing Equation• Complicated Boundary

Condition

Formulation Comparison

13

x

y

S

z

R

p

Deflected Membrane

Static Deflection of a Stretched Membrane

Ndy

Ndy

Ndx Ndx

Membrane Element

dx dy

x

z Ndy

Ndy xz

∂∂ dxx

zxz

2

2

∂∂

+∂∂

pdxdy

Equilibrium of Membrane Element

Membrane Analogy• Consider a thin elastic membrane stretched with uniform tension

over a closed frame and subjected to a uniform pressure.

2 2

2 2

2 2

2 2

0 zz z z z z zF Ndy dx Ndy Ndx dy Ndx pdxdyx x x y y y

z z px y N

∂ ∂ ∂ ∂ ∂ ∂ = = + − + + − + ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂⇒ + = −

∂ ∂

∑

• The equilibrium of a membrane element requires

14

• The membrane is stretched over the boundary of the frame

Sz on0=• The volume enclosed by the deflected membrane and the x-y plane

AV zdxdy= ∫∫

2 2

2 2

0 on

A

z z px y N

z S

V zdxdy

∂ ∂+ = −

∂ ∂=

= ∫∫

• Membrane Equations2 2

2 2 2

0 on

2A

Gx y

S

T dxdy

ψ ψ α

ψ

ψ

∂ ∂+ = −

∂ ∂=

= ∫∫

• Torsion Equations

• Equations are same with: ψ = z , p/N = 2Gα , T = 2V

Membrane Analogy

15

Membrane Analogy• Along any contour line, the resultant shear stress must be tangent

16

cos sin 0 0 0 cos sin 0sin cos 0 0 0 sin cos 00 0 1 0 0 0 1

0 0 cos sin

n nt zn xz

nt t zt yz

zn zt z xz yz

xz yz

σ τ τ θ θ τ θ θτ σ τ θ θ τ θ θτ τ σ τ τ

τ θ τ θ

− = −

+

= 0 0 sin coscos sin

xz yz

xz yz

τ θ τ θτ θ τ θ

− ++ sin cos 0xz yzτ θ τ θ

− +

x

y

θ

( )( )

, ,

0 0

x y xz yz

yz y xz x zn

zt xz y yz x y x

dy dxn nds ds y x

dz d dx dy n nds ds x ds y ds

d dzn n n ny x dn dn

ψ ψτ τ

ψ ψ ψ τ τ τ

ψ ψ ψτ τ τ τ

∂ ∂= = − = = −

∂ ∂

∂ ∂= = = + = − − + = =

∂ ∂

∂ ∂ ⇒ = = − + = − + − = − = − ∂ ∂

• The shear stress at any point in the cross-section is given by the negative of the slope of the membrane in the direction normal to the contour line through the point.

• The maximum shear stress appears always to occur on the boundary where the largest slope of the membrane occurs.

• Using membrane visualizations, a useful qualitative picture of the stress function distribution can be determined and approximate solutions can be constructed.

• The torque is given as twice the volume under the membrane.

Membrane Analogy

x

y

θ

17

Solutions Derived from Boundary Equation

x

y

R

S

• If boundary is expressed by relation f(x,y) = 0, this suggests possible simple solution scheme of expressing stress function as ψ = Kf(x,y) where K is an arbitrary constant.

• This form satisfies boundary condition on S, and for some simple geometric shapes it will also satisfy the governing equation with appropriate choice of K.

• Unfortunately this is not a general solution method and works only for special cross-sections of simple geometry.

2 2

2 2

Boundary-Value Problem

2

0

2 EndsA

G Rx y

S

T dxdy

ψ ψ α

ψ

ψ

∂ ∂+ = − ∈

∂ ∂= ∈

= ∈∫∫

0),( =yxf

18

Elliptical Section2 2

2 2 1x ya b

+ =

• Look for Stress Function Solution:2 2

2 2 1 .x yKa b

ψ

= + −

• ψ satisfies the BC and will satisfy the governing equation if2 2 2 2 2 2

2 2 2 2 2 2 1 .a b G a b G x yKa b a b a b

α αψ

= − ⇒ = − + − + + • Since the governing equation and the BC are satisfied, we have

found the solution.

19

2 22 2

2 2 2 2

2 2 3 3 3 3

2 2 2 2 2 2

2 1 12

2 1 1 4 4

A A A A

a b GT dxdy x dxdy y dxdy dxdya b a b

a b G a b a b a b Gaba b a b a b

αψ

α π π π απ

= = − + − +

= − + − = + +

∫∫ ∫∫ ∫∫ ∫∫

• Relation to the carrying torque

2 2 2 2

3 3 2 2 1a b T x yTa b G ab a b

α ψπ π

+= ⇒ = − + −

• Angle of twist per unit length

2 22 2

3 3 4 4

2 2 2,xz yz xz yzTy Tx T x y

y ab x a b ab a bψ ψτ τ τ τ τ

π π π∂ ∂

= − = − = ⇒ = + = +∂ ∂

＝

• Stress field (two-fold symmetry)

• The maximum shear stress (Strength of Materials vs. Membrane analogy)

( )max 220, Tbab

τ τπ

= ± =

Elliptical Section

20

( )

3

3

2 2

3 3

2 1

2

xz

xz

wG yx

w y dxGTy y dxab GT xy

ab G

T b axy

a b G

τ α

τ α

απ

απ

π

∂ = − + ∂ ⇒

= +

= − +

= − +

−=

∫

∫

• Warping displacement (two-fold symmetry)

Elliptical Section

• If a = b, the results degenerate to torsion of circular shafts.

21

Equilateral Triangular Section• For stress function try product

form of each boundary line equation

( )( )( )3 2 3 2K x y a x y a x aψ = − + + + −• ψ satisfies boundary condition

and will satisfy governing equation

• Since governing equation and boundary condition are satisfied, we have found the solution.

2 2

2 2 2 , if .6GG K

x y aψ ψ αα∂ ∂+ = − = −

∂ ∂

22

Equilateral Triangular Section

2 2

max 3

( ) ,

( 2 )2

3 5 3( ,0)2 18

xz

yz

yz

G x a yy a

G x ax yx a

Ta G aa

ψ ατ

ψ ατ

τ τ α

∂= = −∂∂

= − = + −∂

⇒ = = =

• Stress field (three-fold symmetry)

( )2 236xzw y dx w y x y

G aτ αα = + ⇒ = − ∫

• Warping displacement (three-fold symmetry)

23

( )

4

2 2 44

27 3255 3

5 3 5 , 3 327 3

pA

p Ap

T dxdy G a G I

T T I x y dxdy aGa GI

ψ α α

α

= = =

= = = + =

∫∫

∫∫

• Relation to the loading torque

Additional Examples Using Boundary Equation Scheme

Section with Higher Order Polynomial Boundary

Circular Shaft with Circular Keyway

r = b

r = 2acosθ

r

θx

y

.

( )2 2

max

max

2 cos12

( ) 2As / 0 : 2( )

Stress Concentration

keyway

shaft

G ab rr

G ab aG a

α θψ

τ ατ α

= − −

→ → =

∴

x

y

22 cxay +−=

22 cyax +−=

22 cyax +=

22 cxay +=

a

a

2 2 2 2 2 2

2

max

( )( )

, 3 84 (1 2)

( ,0) (0, ) 2

K a x cy a cx yGK c

a

a a G a

ψα

τ τ τ α

= − + + −

= − = −−

= ± = ± =

24

When Boundary Equation Scheme Does Not Work?

x = a

y = m1x

y = -m2x

x

y

General Triangular Section Rectangular Section

• Trying the previous scheme of products of the boundary lines does not create a stress function that can satisfy the governing equation.

25

Rectangular Section – Fourier Method Solution• Previous boundary equation scheme will not create

a stress function that satisfies the governing equation. Thus we must use a more fundamental solution technique - Fourier method. Thus look for stress function solution of the standard form:

2 2with ( , ) ( )h p p x y G a xψ ψ ψ ψ α= + = −

2 2 20 , ( , ) 0 , ( , ) ( )h h ha y x b G a xψ ψ ψ α∇ = ± = ± = − −• The homogeneous solution must then satisfy

2 2

2

2

( , ) ( ) ( ) : 0 0

sin00

h hX Yx y X x Y y X Y XYX Y

X A xX XY Y

ψ ψ λ

λλ

λ

′′ ′′′′ ′′= ∇ = ⇒ + = ⇒ = − = −

=′′ + =⇒ ⇒′′ − =

cos

sinh

B x

Y C y

λ

λ

+

= cosh

1( , ) 0 , 1,3,5,2h

D y

a y a n n

λ

ψ λ π

+

± = ⇒ = =

• Separation of Variables Method

26

1

2 2

1

( , ) cos cosh cos cosh ,2 2

( , ) ( ) cosh cos ,2

1,3,5,

1,3,52

,

h nn

h nn

n x n yx y BD x y Ba an b n xx

n

nb G a x Ba a

π πψ λ λ

π πψ α

∞

=

∞

=

= =

⇒ ± = − − =

=

=

∑

∑

Rectangular Section – Fourier Method Solution• Homogeneous solution

• By Fourier cosine series

( ) ( )

( )

0 01

2 ( 1)/22 2

0 3 3

2 ( 1)/22 2

331,3,5

1 2cos , cos ,2

2 32 ( 1)cosh ( ) cos2 2 cosh

232 ( 1)( ) cos cosh

2 2cos

0,1,2,

h2

l

n nn

na

n n

n

h pn

n x nf x a a a f dl l l

n b n G aB G a d B n ba

n

a a na

G a n x n yG a x n b a ana

π πξξ ξ

π πξ αα ξ ξ ππ

α π πψ ψ ψ α ππ

∞

=

−

−∞

=

= + =

−= − − ⇒ = −

−⇒ = + = − −

=∑ ∫

∫

∑

27

( 1)/2

221,3,5

( 1)/2

221,3,5

max 221,3,5

16 ( 1) cos sinh2 2cosh

216 ( 1)2 sin cosh

2 2cosh2

16 1( ,0) 2cosh

2

n

xzn

n

yzn

yzn

G a n x n yn by a an

aG a n x n yG x n bx a an

aG aa G a n bn

a

ψ α π πτ ππ

ψ α π πτ α ππ

ατ τ α ππ

−∞

=

−∞

=

∞

=

∂ −= = −∂

∂ −= − = −

∂

= = −

∑

∑

∑

3 4

5 51,3,5

16 1024 12 tanh 3 2A n

G a b G a n bT dxdyn a

αψ αα ππ

∞

=

= = − ⇒ =∑∫∫

2 ( 1)/2

331,3,5

32 ( 1) sin sinh2 2cosh

2

nxz

n

a n x n yw y dx w xy n bG a ana

τ α π πα α ππ

−∞

=

− = + ⇒ = −

∑∫

• Stress field

• Relation to the loading torque

• Warping displacement

Rectangular Section – Fourier Method Solution

28

Rectangular Section – Fourier Method Solution

29

Thin Rectangles (a

Multiply Connected Cross-Sections

31

A

0 Const t an .idds

ψψ == ⇒• On all lateral boundaries:• Value of ψi may be arbitrarily chosen only on

one boundary, commonly taken as zero on So .• The single-valuedness of warping displacement

requires:

( ) ( )

( ) ( ) ( )

10

1 22 ,

,

xz yz S

i

xz

S

S

z

S

S

y

S

w wdw dx dyx y

ydxG

s ds G

w wG y G xx y

dx dy

s dG

A As

xdyα

α

τ τ

ατ

τ

τ

α τ α

∂ ∂= = + = + ∂ ∂

−

−

∂ ∂ = − + = +

= + ⇒

∂ ∂

=

+

∫ ∫ ∫∫

∫ ∫

• where A is the area enclosed by an arbitrary closed-path S that is encircling an inner boundary Si.

• The value of ψi on the inner boundary Si must be chosen so that the above relation is satisfied.

Multiply Connected Cross-Sections

• For multiple holes:

1 : 0 , 2 : 0

3 : 0

n nx x y yA A

nz z

P T dxdy P T dxdy

P T

= = = =

= =

∫∫ ∫∫, 4 : 0 nx zA dxdy M y T= =∫∫

5 : 0A

ny z

dxdy

M x T= =

∫∫, 6 : ( )n nz y xA Adxdy T M xT yT dxdy= = −∫∫ ∫∫

• The BCs on cylinder ends require

( )

( ) ( ) ( )

( )

1

11 1 1 1

1 1

, ,

1 5 are automatically satisfied.

6 :

2 2

2 2 2

2 2

o

n ny xA A

S C S

S

A A

A A

A

T xT yT dxdy x y dxdyx y

x ydxdy y dx x dy dxdy

x y

y dx x dy dxdy A dxdy

T dxdy A

ψ ψ

ψ ψψ ψ ψ ψ

ψ ψ ψ ψ ψ

ψ ψ

−

∂ ∂= − = − + ∂ ∂ ∂ ∂

= − + − = − + ∂ ∂

= − − + = +

⇒ = +

∫∫ ∫∫

∫∫ ∫ ∫∫

∫ ∫∫ ∫∫

∫∫

2 2 i iAi

T dxdy Aψ ψ= + ∑∫∫32

Hollow Elliptical Section

x

y

122

2

2

=+by

ax1

)()( 22

2

2

=+kby

kax• For this case lines of constant shear stress coincide

with both inner and outer boundaries, and so no stress will act on these lateral surfaces. Therefore, hollow section solution is found by simply removing inner core from solid solution. This gives same stress function and stress distribution in remaining material.

( )2 2

22 2 1 .i

a b G ka b

αψ = −+

2 2 2 2

2 2 2 2 1a b G x ya b a b

αψ

= + − + • Constant value of stress function on inner boundary is:

• Load carrying capacity is determined by subtracting load carried by the removed inner cylinder from the torque relation for solid section

3 3 3 33 3 4

2 2 2 2 2 2

( ) ( ) (1 )( ) ( )

a b G ka kb G GT a b ka b ka kb a bπ α π α π α

= − = −+ + +

• Maximum stress still occurs at x = 0 and y = ±b: max 2 42 1 .

1T

ab kτ

π=

−33

Closed Thin-Walled Tubes

( ) ( )1 .s t sτ ψ=

where Sc = length of tube centerline, Ac = area enclosed by tube centerline.

oψ 1ψ

( ) ( ) ( )1 21

2 2 4 cSc c c

T T Ts dsA A t s GA t s

ψ τ α= = = ∫

34

• Thin wall thickness implies that the membrane slope BC can be approximated by a straight line.

• Since the membrane slope equals resultant shear stress:

• Determine the value of ψ1:( )

( )1

22 ,1cc

ccS

S

G As ds G Ads

t s

ατ α ψ= ⇒ =∫∫

• Load carrying relation:

where A = section area, A1 = area enclosed by S1.

11 1 1 1 12 2 2 2 2 .2 cA

T dxdy A A A Aψψ ψ ψ ψ = + = + = ∫∫

• Combining these relations:

• τmax occurs across the narrowest wall.

Open vs. Closed Thin-Walled Tubes• Cut creates an open tube and produces

significant changes to stress function, stress field and load carrying capacity.

• Open tube solution can be approximately determined using results from thin rectangular solution.

• Stresses for open and closed tubes can be compared and for identical applied torques, the following relation can be established

Open Tube Open Tube

Closed Tube Closed Tube

Open Tube Closed Tube

3

6 , but since 1

2

Stresses are higher in open tube and thus closed tube is stronger.

cc

c

TAAt A AT A

A t

τ ττ τ

τ τ

≈ = >> ⇒ >>

⇒ >>

∴

Cut

35

Outline

• Elastic Cylinders with End Loading• Torsion of Cylinders: General formulation• Stress-Function Formulation• Displacement Formulation• Membrane Analogy• Solution: Boundary Equation Scheme• Solution: Fourier Method – Rectangular Section• Multiply Connected Cross-Sections• Hollow Sections

36

幻灯片编号 1Outline幻灯片编号 3幻灯片编号 4幻灯片编号 5幻灯片编号 6幻灯片编号 7幻灯片编号 8幻灯片编号 9幻灯片编号 10幻灯片编号 11幻灯片编号 12幻灯片编号 13幻灯片编号 14幻灯片编号 15幻灯片编号 16幻灯片编号 17Solutions Derived from Boundary EquationElliptical SectionElliptical SectionElliptical SectionEquilateral Triangular SectionEquilateral Triangular SectionAdditional Examples Using Boundary Equation SchemeWhen Boundary Equation Scheme Does Not Work?Rectangular Section – Fourier Method SolutionRectangular Section – Fourier Method SolutionRectangular Section – Fourier Method SolutionRectangular Section – Fourier Method SolutionThin Rectangles (a