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Theory of elasticity

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Problem 5: Derive the expressions for radial,hoop and shearstresses when a hollow cylinder, of internal radius a and

external radius b, is subjected only to internal pressure piand

external pressure po.

Solution to this problem was given by Lame in 1852.Hence, itis also known as Lames problem.

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Expressions for radial,circumferential and shear stress

components in polar co-ordinates (when body force is zero):

To have a possible stress distribution, stress function must

satisfy the compatibility condition, which is given by the

following equation for polar co-ordinates:

rrrrr

r

rrr

r

r

111

11

2

2

2

2

2

2

2

01111

2

2

22

2

2

2

22

24

rrrrrrrr

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Stress distribution symmetrical about and axis: If the stress

distribution is symmetrical with respect to an axis passing

through origin and perpendicular to the xy-plane, the stress

components do not depend on and are functions of r

only.From symmetry it follows also that the shear stress

component must vanish.

Thus,the equation of compatibility and stress components

will be as follows:

01121122

2

23

3

4

4

2

2

2

2

dr

d

rdr

d

rdr

d

rdr

d

dr

d

rdr

d

dr

d

rdr

d

0

1

2

2

r

r

r

rr

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The general solution of the above ordinary differential

equation is given below and it has 4 constants which must be

determined using boundary conditions:

The corresponding stress functions will be as follows:

For a circular plate with no hole at the origin of the

coordinates,constants A and B will vanish.And the plate will

be in uniform tension or compression in its plane.

DCrrBrrA 22 loglog

0

2)log23(

2)log21(1

22

2

2

r

r

CrBr

A

r

CrBrA

rr

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If there is a hole in the circular plate at origin then

constant B will be zero.(Proof that B must be zero

requires consideration of displacements.)

Thus,the equations for stresses will be as follows:

For the given problem, the boundary conditions are:

Substitution of these values in the equations of

stress gives:

CrA

Cr

Ar

2

2

2

2

iarr p)( obrr p)(

o

i

pCb

A

pCa

A

2

2

2

2

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On solving the above two equations, we get A

and B as :

Substitution of these constants in the

equation for stress gives:

22

22

22

22

2

)(

ab

bpapC

ab

ppbaA

oi

io

22

22

222

22

22

22

222

22

1)(

1)(

ab

bpap

rab

ppba

abbpap

rabppba

oiio

oiior

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Special Cases:

(i)A cylinder subjected to internal pressure only: In this case,

po= 0 andpi= p.

2

2

22

2

2

2

22

2

1

)(

1)(

r

b

ab

pa

rb

abpar

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Special Cases:

(ii)A cylinder subjected to external pressure only: In this

case,pi= 0 andpo= p.

2

2

22

2

2

2

22

2

1

1

r

a

ab

pb

ra

abpbr

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Problem 6: Define the plane polar components of stress, r,and rand assuming the relations:

between the Airys stress function and the Cartesiancomponents of stress prove that:

In the absence of body forces, show that:

for the solution of the problem of plane stress.

yxxy xyyx

2

2

2

2

2

,,

rrrrr

r

rrr

r

r

111

11

2

2

2

2

2

2

2

01111

2

2

22

2

2

2

22

24

rrrrrrrr

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Plane polar components of stress:

Consider the equilibrium of a small element cut out from a

plate by radial sections and cylindrical surfaces. The normal stress component in the radial direction is

denoted by r, the normal component in circumferential

direction by ,the shearing stress component by r.

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Derivation of expressions for stress components inpolar coordinates:

cosrx sinry22 yxr

x

y1tan

yxxyxyyx

2

2

2

2

2

,,

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Using chain rule:

Eqn 1-

Eqn 2-

Eqn 3-

rrxrx

r

x

sincos

rryry

r

y

cossin

rrrrrrrx

2

22

2

2

2

2

22

2

2 11cossin2

11sincos

rrrrrrry

2

22

2

2

2

2

22

2

2 11cossin211cossin

rrrrrrryx

2

2

22

2

2

22

22 11sincos

11cossin

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From figure one sees

that:

0

0

0

)(

)(

)(

xyr

y

xr

rrrrryx

rx

rrry

r

r

111

11

2

2

0

2

2

2

0

2

2

2

2

2

0

2

2

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Adding Eqn 1 and Eqn 2:

Thus, in the polar coordinate system Laplacian operator is:

In cartesian system we had derived the compatibility equation

as:

2

2

22

2

2

2

2

2 11

rrrryx

2

2

22

2

2

2

2

22 11

rrrryx

01111

0

0

2

2

22

2

2

2

22

2

2

2

2

2

2

2

2

2

22

rrrrrrrr

yxyx

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Main references:

-Theory of elasticity by Timonshenko andGoodier

-Advanced strength of materials by Volterraand Gaines

-Nptel web course by T.G.Sitharam &

L.GovindaRaju

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