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BENDING STRESSES IN SHELLS

Page 1

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Bending Stresses in Shells - TopicsPage 2

Introduction (3)

Stress Resultants (4-8)

Force, Moment, & Displacement Relations (9-16) Compound Stresses (17)

Axisymmetrically Loaded Circular Cylindrical Shells

(18-28)

Bending of an Infinite Cylinder (29-42)

Uniform Load on Cylinder (43-47)

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IntroductionPage 3

Membrane theory: Cannot always provide solutions compatible with the actual

conditions of deformation.

Fails to predict the state of stress at boundaries & other

areas of shells. Bending theory: Provides a complete solution to shell behavior.

Considers membrane forces, shear forces, and momentsacting on the shell structure.

Is mathematically intricate. Will limit consideration to the most significant practical case

involving rotationally symmetric loading.

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Shell Stress ResultantsPage 4

Consider the shell infinitesimal element:

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Due to curvature, the arc lengths of an element

located a distance z from the midsurface are not

dsx and dsy:

( ) ( )y

yy

yy

x

xx

xx dsr

z

r

zrdsds

r

z

r

zrds

-=

-

-=

-1&1

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Now the resultant normal force per unit length on

the yz-plane is:

( )dzzN

dzr

zN

dzdsr

z

dsN

y

t

t

xx

y

t

t

xx

yy

t

txyx

ks

s

s

-=

-=

-=

-

-

-

1

1

1

2

2

2

2

2

2

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The complete set of stress resultants is:

)( )( )( )( )

( )

dz

z

z

z

z

z

z

Q

Q

N

N

N

N

t

t

xyz

yxz

xyx

yxy

xy

yx

y

x

yx

xy

y

x

-

-

-

-

--

-

=

2

2

1

1

1

1

1

1

kt

kt

kt

ktks

ks

( )( )( )( )

zdz

z

z

z

z

M

M

M

Mt

t

xyx

yxy

xy

yx

yx

xy

y

x

-

-

-

--

=

2

2 1

1

1

1

kt

kt

ksks

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Since rx is typically not equal to ry:

Nxy & Nyx are not generally equal.

Mxy & Myx are not generally equal.

However: For thin shells t

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Force, Moment & Displacement

RelationsPage 9

Hookes Law:

Assumption: z = 0.

[ ]

[ ]

1

1

2

2

xyxy

xyy

yxx

G

E

E

gt

neen

s

neen

s

=

+-

=

+-

=

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RelationsPage 10

Consider the deformations of the midsurface:

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RelationsPage 11

Definitions and assumptions:

Lines mn and mn are straight lines (shell assumption).

x0 = midsurface unit deformation

rx = radius of curvature after deformation

dsx = length of midsurface fiber

The unit elongation of a fiber is then

( ) fx

xxf

x

xf

f

fx

lr

zdsl

rzdsl

l

l

-

-+=D

-=

D=

'11

1erewh

0e

e

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RelationsPage 12

Thus, x is: ( )

-+

--

-=

-

+---+=

-

--

-+

=

xxx

x

xx

xx

x

xxx

xx

x

x

x

x

x

x

xx

x

rrr

r

z

z

r

z

r

z

r

z

r

z

r

zr

zds

r

zds

r

zds

1

'

1

'11

1

1''

1

1

1'

11

00

00

0

eee

eee

e

e

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RelationsPage 13

Assuming:

For t

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RelationsPage 14

And the strains are now:

Twist of midsurface = cxy.

xyxyxy

yyy

xxx

z

z

z

cggcee

cee

20

0

0

-=-=

-=

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RelationsPage 15

Stress-strain relationship becomes:

( )[ ]( )[ ]

( )xyxyxy

xyxyy

yxyxx

zG

zE

zE

cgt

nccneen

s

nccneen

s

2

1

1

0

002

002

-=

+-+-

=

+-+-

=

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RelationsPage 16

And introduced into resultant expressions:

( )( )

( )

( ) ( )( )

( )23

002

0

002

112

1

1

12

1

n

ncc

cnncc

neen

n

gnee

n

-=

+-=

--==+-=

+-

=

+==+

-=

EtD

DM

DMMDM

EtN

EtNN

EtN

xyy

xyyxxyyxx

xyy

xy

yxxyyxx

Flexural rigidity of shell (and plate)

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Compound Stresses in a ShellPage 17

Comparing previous sets of relationships (relatestresses to forces and moments):

First terms are membrane, second terms are bending.

Stresses are linear through thickness.

3

3

3

12

12

12

t

zM

t

N

t

zM

t

Nt

zM

t

N

xyxy

xy

yy

y

xxx

+=

+=

+=

t

s

s

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Axisymmetrically Loaded Circular

Cylindrical ShellsPage 18

Pipes, tanks, boilers are examples.

Due to symmetry:

Only have N , M , Nx

, Mx

, Qx

(5 unknowns).

N & M do not vary with .

Displacement v vanishes only have u & w.

Only 3 of 6 equilibrium equations remain to be

satisfied.

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Stress resultants on an element of axisymmetrically

loaded circular cylindrical shell:

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Write equilibrium expressions:

0:0

0:0

0:0

=-=

=++=

=+=

dxadQaddx

dx

dMM

dxadpddxNaddxdx

dQF

dxadpaddxdx

dN

F

xx

y

rx

z

x

x

x

qq

qqq

qq

q

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Rewriting:

Note that there are 5 unknowns and only 3 equations.

Need more equations.

0

01

0

=-

=++

=+

xx

rx

xx

Qdx

dM

pNadx

dQ

pdx

dN

q

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Examine the midsurface displacements:

n Recall that v = 0 from symmetry:

( )

( )

a

wN

Etdx

dua

w

dx

duEtEtN

a

w

ad

addwa

dx

du

x

xx

x

nn

n

n

nee

n

qqq

e

e

q

q

+-

=

-

-

=+

-

=

-=--

=

=

2

22

111

First governing

displacement condition

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Consider N :

Returning to 3rd equilibrium equation:

( )

xx

x

MMdx

wdDM

dy

wd

dy

dw

dx

du

a

wEtEtN

n

q

nn

neen

q

qq

=-=

==

--

-=+-

=

&

)withvaries(Nothing0

11

2

2

2

2

22

2

2

0dx

Md

dx

dQQ

dx

dM xxx

x ==-

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Substituting expressions into 2nd equilibrium equation:

01

0

1

1

1

01

1

01

4

4

2

22

2

2

2

22

2

=+

+-+-

=+

+

-

---+

-

=+

--

-+

=++

rx

rx

rx

rx

pNa

wEt

adx

wdD

pa

w

NEta

wEt

adx

wd

Ddx

d

pdx

du

a

wEt

adx

Md

pN

ady

dQ

n

n

n

nn

nn

q

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Rewriting:

( )22

2

2

4

4

4

4

24

4

24

4

13

4

4

0

taDa

Et

D

pN

aDw

dx

wd

D

pN

aDw

Da

Et

dx

wd

pNa

wa

Et

dx

wdD

rx

rx

rx

nb

nb

n

n

-==

=-+

=-+

=--+

Second governing

displacement condition

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If there is no axial load (Nx = 0), the two displacement

equations simplify further:

Procedure:

n Find u from direct integration of first equation.

n Find w from the second equation (ordinary differential

equation with constant coefficients).

D

pw

dx

wd

a

w

dx

du

r=+

=4

4

4

4b

n

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Solution to differential equation:

( )

( )

( ) ( )( )xfw

ececeececew

im

mm

mm

m

mmmm

ececececw

p

xixixxixix

h

xmxmxmxm

h

=

+++=

=

=+

=-+

=+

+++=

--- bbbbbb

b

bb

bb

b

4321

22

22222

444321

4321

1

22

042

04

ofrootsare,,,where

4321

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Function f(x) represents the particular solution wp.

The results of membrane theory can always be

considered as the particular solution of the equations of

bending. Can rewrite expression for w in a different form:

( ) ( ) ( )xfxCxCexCxCew xx ++++= - bbbb bb sincossincos

4321

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A Typical Case of the Axisymmetrically

Loaded Cylindrical ShellPage 30

Since there is no pressure distributed over surface of

the shell: pr = 0. Also Nx = f(x) =0.

As x approaches infinity, the deflection & all

derivatives with respect to x must vanish:

( ) ( )xCxCexCxCewxx

bbbb

bb

sincossincos 4321 +++=

-

( )xCxCewCC

x bbb sincos0

21

43

+=== -

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From first displacement condition with Nx = 0:

Substituting into expression for N from Hookes Law:

a

w

dx

du

a

w

dx

duEtNx nn

n

==

-

-

= 0

1

2

a

EtwN

a

w

a

wEt

dx

du

a

wEtN

-=

-

-

-=

-

-

-=

q

q n

n

n

n

2

22

11

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Writing expressions for moment and shear, and

observing that due to loading:

n Half of load is carried by each side.

n The slope is zero at x = 0 due to symmetry.

0

2

&

3

3

2

2

2

2

=

-=-==

-=-=

dx

dw

P

dx

wdD

dx

dMQ

dx

wdDM

dx

wdDM

xx

x nq

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Take first derivative of w to evaluate boundary

condition:

( )( )

[ ] [ ]( )

210

21

21

21

0

cossincossin

cossin

sincos

CCdx

dw

xxCxxCedx

dw

xCxCe

xCxCedx

dw

x

x

x

x

==

-++-=

+-++-=

=

-

-

-

bbbbb

bbbbbbb

b

b

b

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Need third derivative to evaluate other boundary

condition:

[ ]

[ ] [ ]

[ ]xxCedx

wd

xCexCedx

wd

xCedx

dw

xxeCw

x

xx

x

x

bbb

bbbb

bb

bb

b

bb

b

b

cossin2

cos2sin2

sin2

)sin(cos

1

2

2

2

1

2

1

2

2

2

1

1

-=

-=

-=

+=

-

--

-

-

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Need third derivative to evaluate other boundary

condition (continued):

[ ] [ ]

2311303

3

1

3

3

3

1

3

1

3

3

3

8242

cos4

sincos2cossin2

CD

PCD

PCD

P

dx

wd

xCedx

wd

xxCexxCedx

wd

x

x

xx

====

=

++--=

=

-

--

bb

bb

bbbbbb

b

bb

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Can now write the final expressions for the

displacement w and membrane and bending terms:

( ) ( )

( )

( ) ( )

( )xfP

dx

wdDM

xfP

xxeP

dx

wdDM

xfDa

EtPw

a

EtN

xfD

P

xxeD

P

w

xx

x

bb

nn

bbbbb

bb

bbbbb

q

b

q

b

32

2

32

2

13

133

4

4sincos

4

8

8cossin8

=-=

=-=-=

-=-=

=+=

-

-

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Can now write the final expressions for the

displacement w and membrane and bending terms

(continued):

n Recalling that b is a function of geometry and material

properties, can evaluate functions f1(bx), f3(bx), etc. as afunction of x.

( )xfPxePdxwdDQx

x bbb 433

2cos

2 -=-=-=-

( )22

2

2

4 13

4 taDa

Et nb

-==

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Maximum displacement at x=0:

Maximum moments at x=0:

Maximum axial stress:

Et

Paw

Da

Et

D

Pw

2

4

8

2

max2

4

3max

bb

b===

bn

b q 44

max,max,

PM

PMx ==

23

max,

max,2

312

2,0

t

P

t

zMtzx

x

x bs ==

==

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Maximum circumferential stress:

+-=+-=

=+-=

+=

==

222max,

2

4

23max,

3

max,

max,

3

22

3

2

4

2

3

8

12

2,0

tt

aP

t

P

t

aP

Da

Et

t

P

Dta

EtP

t

zM

t

Ntzx

b

nb

b

nbs

bb

nb

s

s

q

q

qqq

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Function values:

These functions are typically supplied in tables, shownon next slide.

( ) ( )

( )

( ) ( )

( ) '''13

''

22

'

34

'

12

'

23

'

12

1

4

1

2

1

2

1cos

2

11sincos

2

1sin

sincos

fffxexf

ffxxexf

fxexf

xxexf

x

x

x

x

bbbbb

bbbbb

bbb

bbb

b

b

b

b

=-=-==

-==-=

-==

+=

-

-

-

-

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Function values:

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It is observed that the f-functions all decrease with

increasing bx.

Thus, in most engineering applications, the effect of the

concentrated loads may be neglected at locations:

Therefore, bending is of a local character.

n A shell of length L = 2p/b loaded at mid-length willexperience maximum deflection and bending moment nearly

identical with those associated with a long shell.

bp

>x

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Example:

Long cylinder of radius a.

Uniform load p over L of its length.

Find w at arbitrary point O within length L.

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Know displacement Dw at point O owing to portion of

load Px = pdx:

Thus, displacement at O produced by entire load:( )xf

D

pdxw b

b 138=D

( ) ( )dxxfD

pdxxf

D

pw

cb

+=0

13

0

13

88

b

b

b

b

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Evaluating integrals:

( ) ( )

( )[ ]( )

( )

( ) xedxxf

xxexdxe

xxe

xxexdxe

xxexf

x

xx

x

xx

x

bb

b

bbbb

bbb

bbbbbb

bbb

b

bb

b

bb

b

cos1

cossin2

1cos

cossin2

1

cossin2

1

sin

cossin

1

2

1

-

--

-

--

-

-=

-=

+-=

--=

+=

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Substituting into expression for w:

[ ]

[ ]

( ) ( )[ ]cfbfEt

paw

cebeEt

paw

Da

Et

cebeD

pw

ce

D

pbe

D

pw

cb

cb

cb

bb

bbb

bbb

b

b

bbb

b

bb

bb

bb

bb

44

2

2

2

4

4

33

22

coscos22

4

coscos28

1cos

1

8

1cos

1

8

--=

--==

--=

+-+

+-=

--

--

--

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Maximum deflection occurs when b = c:

If b & c are large:

( )[ ]bfEt

paw b4

2

max 1-=

( )Et

pawbf

2

max4 0 =b

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Homework Problem 19Page 48

A long steel pipe of 0.75 m in diameter and 10 mmthickness is subjected to loads P uniformly distributedalong two circular sections 0.05 m apart. Assume n =0.3.

For the mid-length between the loads, obtain the radialcontraction.

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10 Bending