me-603 aep stress function polar coordinate

8/10/2019 ME-603 AEP Stress Function Polar Coordinate

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ME 603: Applied Elasticity and Plasticity

STRESS FUNCTIONSTRESS FUNCTION

Prof. S.K.Sahoo


2/32

z dzzzz

zz

dzz

zrzr

dzz

zz

dzz

d

Static equilibrium equations in polar coordinates

Consider stresses on an infinitesimal element with

dimension dr, dz and d.

d

r

y

x

r

rr

rz

r

zz

zr

zDifferential stresses acting on an

element shown in positive

directions

drr

rzrz

dr

r

rr

drrrr

rr

d

rr

z

z


3/32

0rF

d

rdrd

z

drrdz

rzdzrddzddrrdr

r rz

rzrzrr

rrrr

022cossin

dzdrd

dr

rR

d

drdzdrz

d

drdz rr

r

Considering equilibrium in radial direction we have

Considering equilibrium in r direction

Let Ris the component of body force in radially outward direction.

12

cos&22

sin ddd

0....

dzddrrRdzdrdr

dzdrddzdrdrz

dzdrdr

rdzdrd rrzrrrr

01

Rzrrrr

rzrrrrr

Neglecting higher order terms and putting

we have,

or


4/32

Considering equilibrium in direction 0F

0..22

sin.2

cos

2..

dzdrddr

rd

drdzdr

drdzdr

drddr

rdzz

dzrddzddrrdrr

rr

r

zz

zrr

r

Let is the component of body force in tangential direction.

0....

zrrzrzrzrz

rzrr

rzr rzr

r

rr

012

rzrr

zrr

or

Putting

we have,


5/32

0..2

.

.2

dzddrdr

rZdrdzd

dzrddzddrrdrrdrd

dr

rdzz

zz

z

zrzr

zrzzzz

zz

dzdrddzdrdrdzdrdrdzdrdrdzdrddzdrdr zzrzzrzz

Considering equilibrium in Z-direction

or

0zF

WhereZ is the component of body force in z-direction.

0

dzrZdrd

rrrz

01 Zrrzr

zzrzzzr

or


6/32

0 zzrzz

01

2

01

rrr

Rrrr

rr

rrrrr

0 zzrzzr

Incase of plane stress system the equilibrium equations reduce to

Incase of plane strain systems

0zz

0

0

1

2

01

z

rrr

Rrrr

zz

rr

rrrrr


7/32

Strain and Displacement of an element in Polar Coordinates


8/32

Figure shows an element subjected to displacement as indicated. There

are three normal strains rr, , zz & three shearing strain r, z, zr

Considering the displacement in the rdirection, u, we see from fig (a) that

From fig(a) it is also clear that pure radial displacement yields a

strain in -direction, since the fibers of the element have elongated in

- direction. The length of element ab was originally rd , but after

r

u

dr

udrruur

)/(

Strain displacement relationships..

ra a sp acement, u, a ecome r+u .

The tangential strain due to this radial displacement is. Therefore,

From Fig(b), the tangential displacement v gives rise to a tangential

strain equal to,

So total tangential strain

r

u

rd

rddur

1

v

rrd

dv 1)/(2

21

rr

u 1


9/32

The normal strain in Z-direction is

The shearing strain is given by thedifference between the angle Cab &

Cab (fig c) values shows are in angle

The first term comes from the change in the radial displacement u in

the tangential direction; the second term comes from the change in

r

v

r

vu

rr

1

zz

as in rectangular

coordinates.

e angen a sp acemen v n e ra a rec on, an e as erm

appears since part of the scope change of the line accomes from the rotation of the element as a solid body about the axis

through O. Similarly other two shearing strains maybe obtained

If we assume that angles are equal to their tangents & that the rate of

change of displacement with length is small compared to unity.

rz

u

drrudr

drrw

zwdz

dzzuand

z

V

rzdz

dzzv

vrd

d

zr

z

11,

1

/1/1


10/32

Strain displacement relationships in

cylindrical coordinates

rz

u

z

v

rr

vu

rr

v

z

v

rr

u

r

u

zrzr

zr

,1

,1

,1

,

Where, u,v,w are the velocity/displacements in r, , z-directions

For two dimensional case

r

vu

rr

vv

rr

u

r

urr

1

,1

,


11/32


12/32

Compatibility equations in cylindrical coordinates (2D)

Differentiating equation (b) w.r.t . r and putting relationship of Eq. (a),

We get,

c

r

v

r

vu

r

bv

rr

ua

r

urr .....

1.....

1.....

vvvu

v

rr

v

rr

u

rr

v

r

v

rr

r

ru

r

ur

r

rrrr

rr

1111111

11111

22

2

2

22

2

2

Ar

v

rrrr ...............................

2

2

22

2

22

22

22

2

11

11

r

vr

rr

r

u

rr

v

r

u

r

r

v

r

vr

r

v

r

u

r

u

rr

r

r

r

r

DifferentiationEq. (c) w.r.t. r


13/32

Differentiation Eq. (a) w.r.t. , we get,

rurr

2

We have,

This two expression gives:-

Differentiating (A) w.r.t. r, we get

2

22

r

vr

rr

r

ur

r

)......(2

2

Br

vr

rr r

rrr

2

3

2

2

2 r

v

rrrr

rr

Differentiating (B) w.r.t , we have,

Eliminating v from above two equations we have, Compatibility

equation in 2D:

2

32

2

2

r

vr

rr rrrr

rrrrrrrrrrrrrrr 2

2

2

2

22

2 11121


14/32

Since problems concerning cylindrical bodies, such as shaft, gun,

barrels, rotating, disk can be easily solved by means of stress function

equation, it is frequently, admissible to have these equations in the form

of cylindrical rather rectangular coordinates. Considering Z is the

cylinder axis, following equation holds good,

Stress function in cylindrical coordinates

x

y

x

yyxr 1222 tan,tan,

yrxr X

Y

yx

r

Considering stress

function as a

function of r & ,

we have,

rr

x

yrr

y

x

ryrx

cos,

sin

,,

22

Z

rryy

r

ry

rrxx

r

rx

cossin

sincos


15/32

1.........sin1

cossin2

sin1

cossin2

cos

sincos

sincos

2

2

2

22

22

2

2

2

2

2

rrrrrrr

rrrrx

2.........cos1cossin2cos1cossin2sin

cossin

cossin

22

22

22

2

2

rrrry

Repeating the above operations gives us the second derivation of

3...........11

2

2

22

2

2

2

2

2

rrrryx

rrrrrrr

4...........2

2

2

2

2

2

2

2

2

4

4

22

4

4

44

yxyxyyxx

5.................1111

2

2

22

2

2

2

22

24

rrrrrrrr

from equation (1) & (2)

Similarly, in cylindrical coordinates,

In Cartesian

coordinate,


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rrrrrr 22

222

2

11121

This indicate, for 2-D problem without body force, in Cylindrical

Coordinates, we will take airy stress function such that

it will identically satisfy the equilibrium and compatibility conditions

6..........01111

2

2

22

2

2

2

22

24

rrrrrrrr

021

,01

rrrrrrrrrrrrr

Equilibrium eqn.

Com atibilit e n.

(7)

8..................111

11

2

2

2

2

2

2

2

rrrrr

r

rrr

r

rr

When is bi-harmonic,

Stresses in cylindrical coordinate system will be,


17/32

y Y R

Let us consider that the only body

force acting is that in radial direction

orR. This will cover most cases of

practical interest.

The equilibrium equations are then

The body force R can be expressed in terms of the two body force

components X & Y so as:

Cylindrical coordinates with body force-Plane strain

0

021

01

z

rrr

Rrrr

z

rr

rrr

222 YXR

x

X

However, we

may write,

Since we assume that body force

R is a function of r only, does

not depend on. So,

22

yx

R

yY

xX

,

sin.

cos

ry

r

ry

rx

r

rx

22

2

2

sincosrr

R

The body force can be expressed by a

potential function , so that;

22 YXR


18/32

The Equilibrium equations

then become:

These equations are

identically satisfied by a

stress function defined as,

The compatibility equation

for two dimension in plane

strain case:

0,021

01

zrrr

rrrr

zrr

rrr

rrrrrr

r

r

rr

rrrr

r

r

111

,11

2

2

2

2

2

2

2

,01

21 24

Us ng cy n r ca unct on eqn 4 & eqn 5

In addition, there will be present a z stress acting in the axial directionwhich will be determined by the restrictions on z as given by the

following equations;

From Which,

11.......01

1

211111

1

212

2

2

2

22

2

2

2

22

224

rrrrrrrrrrr

tconsKE

rzz tan1

rz KE


19/32


20/32

Biharmonic Governing Equation

1111 22224

Summary: Solutions to Plane Problems in Polar Coordinates

If there is no body force,

Plane Stress

Plane Strain 01

1

2111112

2

2

2

22

2

2

2

22

2

rrrrrrrrrrr

01

11111

2

2

2

2

22

2

2

2

22

2

rrrrrrrrrr

),(,),( rfrfR r

Boundary Conditions

Airy Representation for stresses

rrrrrr

rr

1,,

112

2

2

2

2

222222

rrrrrrrr

R S

x

y

r


21/32

Conditions for these cases are:-

The ends are unrestricted i.e., Plane-stress condition.

It is not rotated or gravity is neglected i.e.,NO body force.Stress distribution will be a function of r

Stress distribution symmetrical about an axis

Equation of compatibility (without body force) becomes:-

011

.111

.11

32

2

23

3

22

2

4

4

2

2

2

2

dr

d

rdr

d

rdr

d

rdr

d

rdr

d

rdr

d

dr

d

dr

d

rdr

d

dr

d

rdr

d

01112111 232234

dddddddd

0zz

This is a homogenous differential equation that can be transformed

into a linear differential equation with constant coefficients by a

change of variableThis equation has, as a general solution, 4 constant of integration, which

must be determined from boundary conditions,

By substitution, the general solution is

32233222234 drrdrrdrrdrrdrrdrrdrrdr

011232

2

23

3

4

4

dr

d

rdr

d

rdr

d

rdr

d

0,,1

2

2

rrdr

d

dr

d

r

.ter

DCrrBrrA 22 loglog

Stress are:


22/32

So, the corresponding stress

components are:

If there is no hole at the origin of

coordinates i.e., if the cylinder is

solid then, when r =0,

Hence, for a plate without a hole at the origin & with no body forces,

only one case of stress distribution symmetrical w.r.t. the axis may exist,

0

2log23

2log211

22

2

2

r

r

CrB

r

A

r

CrBr

A

rr

&r

unless constants A & B vanish/zero.

name y, a

If there is a hole at the origin other solutions, than uniform tension or

compression can be derived, taking B as zero, for instance.It become,

r

C

r

A

Cr

Ar

2

2

2

2

& plate is in a condition of uniform tension or uniform compression in

all directions in its plane.


23/32

Application: Thick cylinder under uniform pressure

rr=- Pi at r = a ,rr=- P0 at r= b

If the ends are unrestrained,zz=0 i.e., Plane stress case

A.When it is a solid cylinder: a= 0

Consider a long, hollow, circular, cylinder having its

axis coincide with Z-axis subjected to internal pressure

Pi , external pressure Po , Internal radius = a, Externalradius = b.

,& rat r = 0

Boundary Condition are:

.

and if that solid cylinder is subjected to an

external pressure Po, We have Boundary Condition

at r = b from which we can say that,

throughout the body.

The sign isve since a positive pressure (as normally measured with

a gauge) will give rise to compression stress (-ve stress) in the body.

tconsCr tan2

or P

or P


24/32

Equilibrium equation is

we have,

Stress components are: 1..............0,,1

2

2

rrr

d

dr

d

r

B. When it is a hollow cylinder:

)2......(0rrdr

d rr asr=0 & stress is

a function ofronly.

3..........0,2log23

2log21

2

2

r

rr

CrBr

A

CrBr

A

In the above expression for stress compression we have three constants

of integration, whereas we have only two boundary conditions. For

determining the third constant, we have to examine the displacement:

The strain Components are: 4.............,r

u

dr

dur

The stress-strain

relationships are:

rArCrrrB

ruEor

Er

u

Edrdu

rr

rr

rrrr

1

112log123

.,

6.............1

5..........1


25/32

Integrating,

K is the integration

constant

In order that above two expression foru to be the same, we must have,

So that,

We have, rrrr Edrdu 1

Kr

ArCrrrrB

E

u 1

112log1231.1

Kr

ArCrrrrBuE 1

112log121.

0,0 KB

rArC

Eu

1112

1

Cr

A

Cr

Arr

2

2

2

2

A & C now can be determined from the

boundary condition: -

irr P

orr P

at r =a,

at r=b

22

22

22

20

2

2

ab

PPbaA

ab

bPaPC

io

i

gives,


26/32

22

22

222

22

22

22

222

22

1.

1

abbPaP

rabPPba

ab

bPaP

rab

PPba

oiio

oiiorr

tcons

ab

bPaP oirr tan

222

22

0zz

Putting the value of 2C & A

we will get, Lame solution

Which Indicate that

If the ends of the cylinder are free then,

i.e., independent ofr.

rrzzE

So it also satisfy the Plane-Strain condition.

= Constant

We can use the equations to find force & displacement

in shrink fit condition.

urr &,


27/32

Most common problems to which the above equations apply is that of

the stresses in a gun barrel due to explosion pressure of change. For

the simple-tube barrel, PP00 ==00 && PPii 00 ; and the only pressure

involved is that inside barrel, or Pi. S o rand will be reduced to,

22

22

2

2

22

22

2

2

ab

br

r

aP

ab

br

r

aP

i

i

r

22

22

ab

aPi

22

22

ab

baP

i

iP

The maximum values of rand occur at r=a

Pi when b becomes very large or a becomes very small.

It indicates, no matter how much material can be added to gun barrel

the tangential stress can not be reduced less than internal pressure.

ii

ir

Peiab

baP

P

max22

22

max

max

.,.

riP

a


28/32

In order to decrease the common practice is to install shrink bands on the gun barrel.

They are fitted in hot & when cool they provide compressive Po atoutside of barrel.

One or more bands are used depending amount of internal pressure

similar stress pattern on outer fiber can be generated by proper heat

treatment to induce compressive stress.


29/32

When PPii==00 && PP00 00 ; ie, only external pressure

So rand

will be reduced to,

22

22

2

2

0

22

22

2

2

0

ab

ar

r

bP

ab

ar

r

bPr

r

22

2

0

2

ab

bP

22

22

0ab

abP

0P

0P

ba

When b= or b>>>>>a and PP00==00 && PPii 00 :

2

2

r

aPir


30/32

r

b

a

c

Shrink Fit

Let is the Fit allowance, ie,

internal radius of outer cylinder is

less than outer radius of inner

cylinder before fitting by an amount

of .

After fitting by heating and coolingthere will be shrink pressure, Let = Pc.

The contact pressure Pc acting on the outer surface of the inner

cylinder reducing its radius by u1

On the other hand, the contact pressure Pc acting on the inner surface ofthe outer cylinder increasing its radius by u2

The sum of these two quantities, ie, (-u1 + u2 ) =


31/32

We know,

Putting the value ofA & C, we have,

For inner cylinder: P0 = Pc , Pi =0 , b=c, a=a

r

ArCE

u 1

1121

rPP

ab

ba

Er

ab

bPaP

Eu ioi 022

22

22

22 11

22

22

1

1

22

2

1

11

11

ac

P

c

ca

E

c

ac

cP

E

u cc 212122

1

)1()1(

)(

ac

acE

cPc

For outer cylinder: P0 = 0 , Pi = Pc , b=b, a=c

22

22

2

2

22

2

2

22

11

cb

P

c

cb

Ec

cb

cP

Eu cc

2222222

)1()1()(

bccbE

cPc

As (-u1 + u2 ) = ,

we can get,

222

22

2

122

22

1

)(1)(1

cb

cb

Eac

ac

E

cPc


32/32

For a=0 ,ie, inner cylinder is solid one

Let E1= E2 , 1= 2ie, same material

222

2222

2

))((

abc

cbac

c

EPc

2

22

2

)(

b

cb

c

EPc

Stresses in inner cylinder: a< r

me-603 aep stress function polar coordinate

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