elastic stress

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Namas ChandraAdvanced Mechanics of Materials Chapter 11-1

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CHAPTER 11The Thick Walled Cylinder

EGM 5653Advanced Mechanics of Materials




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Introduction

This chapter deals with the basic

relations for axisymmetric

deformation of a thick walled cylinder

In most applications cylinder wall

thickness is constant, and is subjectedto uniform internal pressure p1,

uniform external pressure p2, and a

temperature change ΔT




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Introduction contd.

Solutions are derived for open cylinders or for cylinders with

negligible “end cap” effects.

The solutions are axisymmetrical- function of only radial

coordinate r

Thick walled cylinders are used in industry as pressure vessels,pipes, gun tubes etc..




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11.1 Basic Relations

Equations of equilibrium derived neglecting the body force

( )rr rr rr

d d r or r

dr dr

Strain- Displacement Relations and Compatibility Condition

, ,rr zz

u u w

r r z

Three relations for extensional strain are

where u= u(r,z) and w= w(r,z) aredisplacement components in the r and z

directions




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11.1 Basic Relations Contd.

At sections far from the end shear

stress components = 0 and we

assume

εzz = constant. Therefore, byeliminating u = u(r)

( )rr rr

d d r r or

dr dr




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11.1 Basic Relations Contd.

Stress Strain Temperature Relations

The Cylinder material is assumed to be Isotropic and linearly elastic

The Stress- Strain temperature relations are:

1( )

1( )

1( ) constant

rr rr ZZ

rr ZZ

zz zz rr

T

E

T E

T E

Where, E is Modulus of Elasticity

ν is Poisson’s ratio

α is the Coefficient of linear thermal expansion

ΔT is the change in the temp. from the uniform reference temp.




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11.2 Closed End Cylinders Stress Components at sections far from the ends

.

The expressions for the stress components σrr,σθθ, σzz for a cylinder with closed ends and subjected to internal pressure p1, external

pressure p2, axial load P and temperature change ΔT.

From the equation of equilibrium, the strain compatibility equation and

the stress- strain temperature relations we get the differential

expression,0

1rr

d E T

dr




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11.2 Closed End Cylinders Stress Components at sections far from the ends contd.

Eliminating the stress component σθθ and integrating, we get

2

212 2 2

1(1 )

r

rr

a

C E aTrdr C

r r r

Using this in the previous expression and evaluating σθθ, we get

2

212 2 2

1(1 ) 1

r

a

C E E T aTrdr C

r r r




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11.2 Closed End Cylinders Stress Components at sections far from the ends contd.

The effects of temperature are self- equilibrating . The expression for εzz

at section far away from the closed ends of the cylinder can be written

in the form

2 2

1 2closed end 2 2 2 2 2 2

1 2 2

( )( ) ( ) ( )

b

zz a

P

p a p b Trdr E b a b a E b a

2 2

1 2closed end 2 2 2 2 2 22

( ) 1 (1 )( )

b

zz

a

p a p b P E T E Trdr b a b a b a

The expression for σzz at section far away from the ends of the cylinder can

be written in the form




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Open Cylinder

No axial loads applied on its ends.

The equilibrium equation of an axial portion of the

cylinder is:2 0

b

zz

a

r dz

2 2

2 1(open end) 2 2 2 2

2 ( ) 2

( ) ( )

b

zz

a

p b p aTrdr

b a E b a

The expression for εzz

and σzz

may be written

in the form




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11.3 Stress Components and Radial Displacementfor Constant Temperature

.

For a closed cylinder (with end caps) in the absence of temperature

change ΔT = 0 the stress components are obtained as

2 2 2 2

1 2

1 22 2 2 2 2

( )( )rr

p a p b a b p p

b a r b a

2 2 2 2

1 21 22 2 2 2 2

( )( )

p a p b a b p p

b a r b a

2 2

1 2

2 2 2 2 constant( ) zz

p a p b P

b a b a

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11.3.2 Radial Displacement for an Closed Cylinder

The radial displacement u for a point in a thick wall closed cylinder may

be written as

2 22 2

(closed end) 1 2 1 22 2 2

(1 )(1 2 )( ) ( )

( )

r a b P u p a p b p p

E b a r

11.3.3 Radial Displacement for an Open Cylinder

2 22 2

(open end) 1 2 1 22 2 2

(1 )(1 )( ) ( )

( )

r a bu p a p b p p

E b a r

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Example 11.2 Stresses and Deformations in Hollow cylinder

A thick walled closed-end cylinder is made of an aluminum alloy,

E = 72 GPa, ν = 0.33,Inner Dia. = 200mm, Outer Dia. = 800 mm,

Internal Pressure = 150 MPa.

Determine the Principal stresses, Maximum shear stress at the inner

radius (r= a = 100 mm), and the increase in the inside diameter caused

by the internal pressure

Solution:2 p 0 and r a

2 2

1 12 2150rr

a b p p MPa

b a

2 2 2 2

1 2 2 2 2

100 400

150 170400 100

a b

p MPab a

2 2

1 2 2 2 2

100150 10

400 100 zz

a p MPa

b a

The Principal stresses for the conditions that

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Example 11.2 Stresses and Deformations in Hollow cylinder contd.

max minmax

170 ( 150)160

2 2 MPa

2 0and r a p P

The maximum shear stress is given by the equation

The increase in the inner diameter caused by the internal pressure is

equal to twice the radial displacement for the conditions

2 21( ) 2 2

2 2

2 2

(1 2 ) (1 )( )

150(100)1 0.66 100 (1 0.33)400

72,000(400 100 )0.3003

r a

p au a b

E b a

mm

The increase in the internal pressure caused by the internal pressure

is 0.6006 mm

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11.4 Criteria of Failure

Recap

Maximum Principal stress criterion – Design of Brittle isotropic materials

– If the principal stress of largest magnitude is the tensile stress

Maximum shear stress or the Octahedral shear- stress criterion

– Design of Ductile isotropic materials

11.4.1 Failure of Brittle Materials

Maximum principal stress = Ultimate tensile Strength σu

Condition for Failure

At sections far removed from the ends

Maximum Principal stress = Circumferential stress σθθ(r=a)

or axial stress σzz

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11.4.2 Failure of Ductile Materials

.

General Yielding Failure

Yielding at sections other than the points of stressconcentration

Thick walled cylinders occasionally subjected to static

loads or peak loads

Member has yielded over a considerable region as with

fully plastic loads

Fatigue failure

Subjected to repeated pressurizations (loading and unloading)

Found predominantly around the region of stress

concentration Maximum shear stress or maximum octahedral shear stress to

be determined at the regions of stress concentration

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11.4.3 Material Response Data for Design

General Yielding

Property : Yield stressCriteria : Maximum shear stress or Octahedral shear stress

Fatigue Failure

Property : Fatigue strength

Criteria : Maximum shear stress and Octahedral shear stress in conjunction

Values obtained from tests of either a tension specimen or hollow thin

walled cylinder in torsion

The hollow thin walled cylinder specimen values led to more accurate

prediction of thick walled cylinders than the tension specimen

The critical state of stress is usually at the inner wall , for a pressure

loading it is only pure shear in addition to hydrostatic state of stress

Values from tensile test specimen and hollow thin-wall tube

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11.4.3 Material Response Data for Design contd.

Since for most materials the hydrostatic stress does not affect

yielding yielding is caused by the pure shear

Hence maximum shear- stress criterion and octahedral shear-stress

criterion predict with errors of <1% in comparison with 15.5 %

11.4.4 Ideal Residual Stress Distributions for

Composite Open Cylinders

Stress distributions in a closed cylinder at initiation of yielding (b=2a)

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11.4.4 Ideal Residual Stress Distributions forComposite Open Cylinders contd.

Stress distributions in composite cylinder made of brittle material that fails

at inner radius of both cylinders simultaneously

(a) Residual stress distributions

(b) Total stress distributions

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11.4.4 Ideal Residual Stress Distributions forComposite Open Cylinders contd.

Stress distributions in composite cylinder made of ductile material

that fails at inner radius of both cylinders simultaneously

(a) Residual stress distributions

(b) Total Stress distributions

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Example 11.5Yield of a Composite Thick Wall Cylinder

Problem: Consider a Composite cylinder,

Inner cylinder: Inner Radii a =10 mm and outer radii ci=25.072 mm

Outer cylinder: Inner Radii co =25 mm and outer radii b= 50 mm

Ductile steel ( E=200 GPa and ν=0.29)

Determine minimum yield stress for a factor of safety SF=1.75Solution:

It is necessary to consider the initiation of yielding for theinside of both the cylinders.σzz =0 for both the cylinders

At the inside of the inner cylinder, the radial and circumferential

stresses for a pressure (SF)p1 are,(1.75)(300) 525

(1.75)(325) 450.2 118.6

rr MPa

MPa

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Example 11.5Yield of a Composite Thick Wall Cylinder contd.

2 2 21(118.6 525) (525) (118.6) 593.3

2Y MPa

At the inside of the outer cylinder, the radial and circumferential

stresses for a pressure (SF)p1 are,

(1.75)(37.5) 189.1 245.7

(1.75)(62.5) 315.1 424.5

rr MPa

MPa

2 2 21(424.5 254.7) (254.7) (424.5)

2

594.3 593.3

Y

MPa MPa

In an ideal design, the required yield stress will be the same for the

inner and the outer cylinders.

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