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Chapter 9 Torsion of Thin-Walled Tubes Summary of Saint-Venant Torsion Theory Warping function, Ψ shear stress ( ) xy G z y Ψ τ = β , ( ) xz G y z ∂Ψ τ = β + compatibility relationship automatic equilibrium equations 2 0 Ψ= boundary conditions 0 xy xz dz dy ds ds τ −τ = 2 2 1 ( ) 2 dz dy d y z y ds z ds ds Ψ ∂Ψ = + torque ( ) xz xy A T y z dA = τ −τ ∫∫ ( ) A T G y z dA G J z y Ψ ∂Ψ = β + β ∫∫ Prandtl stress function, φ shear stress xy z φ τ = , xz y φ τ =− compatibility relationship 2 2 G ∇φ =− β equilibrium equations automatic boundary conditions 0 d φ = torque 2 A T dA = φ ∫∫
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### Transcript of Chapter 9, Torsion of Thin-Walled Tubes - ase.uc.edupnagy/ClassNotes/AEEM6001 Advanced Strength...

• Chapter 9 Torsion of Thin-Walled Tubes

Summary of Saint-Venant Torsion Theory

Warping function,

shear stress ( )xy G zy

=

, ( )xz G yz

= +

compatibility relationship automatic

equilibrium equations 2 0 =

boundary conditions 0xy xzdz dyds ds

=

2 21 ( )2

dz dy d y zy ds z ds ds

= +

torque ( )xz xy

AT y z dA=

( )A

T G y z dA G Jz y

= +

Prandtl stress function,

shear stress xy z

=

, xz y

=

compatibility relationship 2 2G =

equilibrium equations automatic

boundary conditions 0d =

torque 2

AT dA=

• Membrane Analogy

y

z

S dy

S dz

yx

S dzS dz

p dy dzy

dyy y y

+

(y,z)

y

z

S dy

S dz

yx

S dzS dz

p dy dzy

dyy y y

+

(y,z)

sin sin

sin sin 0

xdF S dz S dz dyy y y y

S dy S dy dz p dy dzz z z z

= + +

+ + + =

2 2

2 2 0S dz dy S dy dz p dy dzy z

+ + =

2 2

22 2

pSy z

= + =

compatibility relationship:

2 2G =

boundary condition: 0d =

• Examples of Solid Cross Sections

elliptical cross section rectangular cross section

triangular cross section I-beam cross section

• Discontinuous Boundary

y

z

yx (y,z)

= 0

= 1p

y

z

yx (y,z)

= 0

= 1p

elliptical cross section with hole rectangular cross section with hole

• Narrow Rectangular Cross Section

b >> t

b

t y

z

b

t y

z

equilibrium in x-direction

2

2 0xdF S dy dz p dy dzz

= + =

2

2pSz

=

0p zz S

= +

2

2p z CS

= +

2( ) 0

2 8t pz t C

S = = + =

2

8pC tS

=

2

22 4p t zS

=

22

4tG z

=

• torque

2/ 2 2

/ 22 2

4

t

A z t

tT dA G b z dz=

= =

/ 22 3

/ 2

24 3

t

t

t z zT G b

=

313

T G bt=

shear stress

2xy G zz

= =

, 0xz y

=

max 23TG tbt

= =

torsional stiffness

( )tTK JG

= =

31

3K bt=

torque contributions

bulk end( )xy xzA

T z y dA T T= + = +

/ 2 2

bulk/ 2

2t

xyA z t

T z dA G b z dz=

= =

3bulk

1 16 2

T G bt T= =

• End Correction

h

y = b/2 y = +b/2

hh

y = b/2 y = +b/2

h

1D approximation:

22 if and 0 else

4 2 2t b bG z y

= =

0 if and if2 2 2xzb b by y

y y

= = < < = =

2D tapered approximation:

2 22 2 2if and else

4 2 2 4

b yt b b tG z h y h G zh

= + =

end xzdT y dA=

2/ 2 / 2

2end

/ 2 / 24 2

t bxz

A z t b h

t b bT y dA G z dz y dyh=

=

3end

1 16 2

T G bt T= =

• General Thin-Walled Shapes

TKG

=

R

t

R

t

3 31 23 3

K bt R t=

b2

b 1

t1

t 2

bt = t(s)

s

b2

b 1

t1

t 2

b2

b 1

t1

t 2

bt = t(s)

s

bt = t(s)

s

31

13

ni i

iK b t

== 3

0

1 ( )3

bK t s ds=

• Single-Cell Thin-Walled Tube

y

L

zx

T

A

D

B

C

t

1 2A

D

B

C

dx

t1

t2t

P

d

ds

q ds

r

y

L

zx

T

A

D

B

C

ty

L

zx

T

A

D

B

C

t

1 2A

D

B

C

dx

t1

t2

1 2A

D

B

C

dx

t1t1

t2t2t

P

d

ds

q ds

rt

P

d

ds

q ds

r

1 1 2 2 0xdF t dx t dx= + =

constant shear flow q

1 1 2 2q t t= =

twisting moment T

dT dF r t ds r q ds r= = = 12

d ds r =

2dT q d=

2T q=

2Tq =

surf( ) 2T

t = =

• Rate of Twist, Torsional Stiffness

total strain energy over length L

20

12 2V

W T L U U dV Lt dsG

= = = =

2

212 8

T L dsT LtG

=

Bredts formula

24T ds

tG =

2q dsG t

=

Example: closed tube of circular cross section:

R

t

R

t

closed24

TG t

=

, 2 R= , 2R =

closed32

TG R t

=

closed 32K R t=

• Warping Displacement

z

y

x, uT

T

Q

axis of twist

z

y

x, uT

T

Qaxis of twist

closed cell open cell

z

y

x, uT

T

Q

axis of twist

z

y

x, uT

T

Qaxis of twist

closed cell open cell

closed 32K R t=

open 3 closed23

K Rt K=

• Multicell Tubes

1 2 0axial wF q L q L q L= + + = 1 2wq q q=

1 2 w

CBA ADC ACT q r ds q r ds q r ds= + +

1 1 2 22( ) 2( ) 2w w w wT q q q= + +

1 1 2 22 2T q q= +

For general n-cell cross section:

12

ni i

iT q

==

shear flow qi in the ith loop:

12 i ii i

qG dst

=

• Restraint of Warping

Example: I Sections

top view of the upper flange with differential element

axial view

( )x =

• bending of the upper flange: 2

2f fvM E I

x

=

3

12f

ft b

I =

2hv =

2

21 12 2f f f

M E I h E I hxx

= =

2

212

ff f

MV E I h

x x

= =

torque due to shear force in the flanges:

f fT V h=

total torque:

2

22

12f f

T V h K G E I h K Gx

= + = +

3 22

2 24f

ft b hhJ I = =

(called the sectorial area of inertia)

2

2T E J K Gx

= +

2

2K G TE J E Jx

=

2 2

22

k TkG Kx

=

where 2

2(1 )K G KkE J J

= =+

1 2( ) sinh coshTx C kx C kx G K = + +

• boundary conditions:

( 0) 0x = = (no warping at the center)

0x Lx =

=

(no bending moment at the end)

2TC G K= and 1 tanh

TC kLG K=

( )( ) tanh sinh cosh 1Tx kL kx kxG K = +

surf( ) G t = =

0

tanh( ) ( ) 1 ( )L T L kL T LL x dx kLG K kL G K

= = =

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 50Normalized Length, kL

Nor

mal

ized

Tw

ist, (

kL)

0

0.2

0.4

0.6

0.8

1

0

0.2

0.4

0.6

0.8

1

0 10 20 30 40 500 10 20 30 40 50Normalized Length, kL

Nor

mal

ized

Tw

ist, (

kL)

( )( ) tanh cosh sinh2f

fE I hT k

M x kL kx kxG K=

( )2

( ) tanh sinh cosh2f

fE I hT k

V x kL kx kxG K=