chapter 8 two-dimensional problem solution

Chapter 8 Two-Dimensional Problem Solution

yxxy xyyx

2

2

2

2

2

,,

02 44

4

22

4

4

4

yyxx

Using Airy Stress Function approach, plane elasticity formulation with zero body forces reduces to a single governing biharmonic equation.

In Cartesian coordinates it is given by

and the stresses are related to the stress function by

We now explore solutions to several specific problems in both Cartesian and Polar coordinate systems

Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island

Cartesian Coordinate Solutions Using Polynomials

In Cartesian coordinates we choose Airy stress function solution of polynomial form

Method produces polynomial stress distributions, and thus would not satisfy general boundary conditions. However, using Saint-Venant’s principle we can replace a non-polynomial condition with a statically equivalent polynomial loading. This formulation is most useful for problems with rectangular domains, and is commonly based on inverse solution concept where we assume a polynomial solution form and then try to find what problem it will solve. Notice that the three lowest order terms with m + n 1 do not contribute to the stresses and will therefore be dropped. Second order terms will produce a constant stress field, third-order terms will give a linear distribution of stress, and so on for higher-order polynomials.Terms with m + n 3 will automatically satisfy biharmonic equation for any choice of constants Amn. However, for higher order terms, constants Amn will have to be related in order to have polynomial satisfy biharmonic equation.

Determined be toConstants,),(0 0

mn

m n

nmmn AyxAyx


Example 8.1 Uniaxial Tension of a Beam

x

y

T T

2l

2c

Boundary Conditions: 0),(),(

0),(,),(

cxyl

cxTyl

xyxy

yx

Since the boundary conditions specify constant stresses on all boundaries, try a second-order stress function of the form

202 yA 0,2 02 xyyx A

The first boundary condition implies that A02 = T/2, and all other boundary conditions are identically satisfied. Therefore the stress field solution is given by

0, xyyx T

Displacement Field (Plane Stress) Stress Field

E

T

Ee

y

vE

T

Ee

x

u

xyy

yxx

)(1

)(1

)(,)( xgyE

Tvyfx

E

Tu

0)()(02

xgyfex

v

y

u xyxy

oo

oo

vxxg

uyyf

)(

)(. . . Rigid-Body Motion

“Fixity conditions” needed to determine RBM terms0)()(0)0,0()0,0()0,0( xgyfvu z


Example 8.2 Pure Bending of a Beam

Boundary Conditions:

Expecting a linear bending stress distribution, try second-order stress function of the form

303 yA 0,6 03 xyyx yA

Moment boundary condition implies that A03 = -M/4c3, and all other boundary conditions are identically satisfied. Thus the stress field is

Stress Field

“Fixity conditions” to determine RBM terms:

x

y

M M

2l

2c

c

c x

c

c x

xyxyy

Mydyyldyyl

ylcxcx

),(,0),(

0),(),(,0),(

0,2

33

xyyx yc

M

)(4

3

2

3

)(2

3

2

3

233

33

xgyEc

Mvy

Ec

M

y

v

yfxyEc

Muy

Ec

M

x

u

0)()(2

30

3

xgyfxEc

M

x

v

y

u

oo

oo

vxxEc

Mxg

uyyf

234

3)(

)(

0)0,(and0)0,( lulv32 16/3,0 EcMlvu ooo

Displacement Field (Plane Stress)


Example 8.2 Pure Bending of a BeamSolution Comparison of Elasticity

with Elementary Mechanics of Materials

x

y

M M

2l

2c

]44[8

,

0,

222 lxyEI

Mv

EI

Mxyu

yI

Mxyyx

3/2 3cI

Elasticity Solution Mechanics of Materials SolutionUses Euler-Bernoulli beam theory to find bending stress and deflection of beam centerline

]4[8

)0,(

0,

22 lxEI

Mxvv

yI

Mxyyx

Two solutions are identical, with the exception of the x-displacements


Example 8.3 Bending of a Beam by Uniform Transverse Loading

x

y

w

2c

2l

wl wl

Boundary Conditions: Stress Field

c

c xy

c

c x

c

c x

y

y

xy

wldyyl

ydyyl

dyyl

wcx

cx

cx

),(

0),(

0),(

),(

0),(

0),(

5233223

303

221

220 5

yA

yxAyAyxAxA

22321

3232120

322303

62

222

)3

2(66

xyAxA

yAyAA

yyxAyA

xy

y

x

23

33

3232

2

4

3

4

344

3

2

)3

2(

4

3

5

2

4

3

xyc

wx

c

w

yc

wy

c

ww

yyxc

wy

c

l

c

w

xy

y

x

BC’s


Example 8.3 Beam ProblemStress Solution Comparison of Elasticity with Elementary Mechanics of Materials

x

y

w

2c

2l

wl wl

Elasticity Solution Mechanics of Materials Solution

)(2

3

2

32

)53

()(2

22

323

2322

ycxI

w

cycy

I

w

ycy

I

wyxl

I

w

xy

y

x

)(2

0

)(2

22

22

ycxI

w

It

VQ

yxlI

w

I

My

xy

y

x

Shear stresses are identical, while normal stresses are not


Example 8.3 Beam ProblemNormal Stress Comparisons of Elasticity with Elementary Mechanics of Materials

y/w - Elasticity y/w - Strength of Materials

x/w - Elasticity x/w - Strength of Materials

l/c = 2

l/c = 4

l/c = 3

Maximum differences between two theories exist at top and bottom of beam, difference in stress is w/5. For most beam problems (l >> c), bending stresses will be much greater than w, and differences between elasticity and strength of materials will be relatively small.

Maximum difference between two theories is w and occurs at top of beam. Again this difference will be negligibly small for most beam problems where l >> c. These results are generally true for beam problems with other transverse loadings.

x – Stress at x=0 y - Stress


Example 8.3 Beam ProblemNormal Stress Distribution on Beam Ends

x

y

w

2c

2l

wl wl

End stress distribution does not vanish and is nonlinear but gives zero resultant force.

c

y

c

ywycy

I

wylx 5

1

3

1

2

3

53),(

3

323

wylx /),(


Example 8.3 Beam Problem

x

y

w

2c

2l

wl wl

)()]56

(2

)()3

2

212[(

2

)()]3

2

3()

5

2

3

2()

3[(

22242

223224

32

32332

xgycyy

xlycycy

EI

wv

yfc

ycy

xycy

xyx

xlEI

wu

oooo vxxclEI

wx

EI

wxguyyf 2224 ])

5

8([

424)(,)(

Choosing Fixity Conditions 0),(),0( ylvyu ])25

4(

5

121[

24

5,0

2

24

l

c

EI

wlvu ooo

])25

4(

5

121[

24

5])

25

4(

2[

12

]562

)[(3

2

2122

)]3

2

3()

5

2

3

2()

3[(

2

2

2422

24

224222

3224

32

32332

l

c

EI

wlxc

lx

ycyyxl

ycycy

EI

wv

cyc

yx

ycyxy

xxl

EI

wu

])25

4(

5

121[

24

5)0,0(

2

24

max l

c

EI

wlvv

EI

wlv

24

5 4

max Strength of Materials:

Good match for beams where l >> c



Cartesian Coordinate Solutions Using Fourier Methods

Fourier methods provides a more general solution scheme for biharmonic equation. Such techniques generally use separation of variables along with Fourier series or Fourier integrals.

)()(),( yYxXyx 024

4

22

4

4

4

yyxx

i

00

]cosh)(sinh)[(cos

]cosh)(sinh)[(sin

]cosh)(sinh)[(cos

]cosh)(sinh)[(sin

xxHFxxGEy

xxHFxxGEy

yyDByyCAx

yyDByyCAx

33

22100 xCxCxCC

29

287

36

2540 xyCyxCxyCyCyCyC

yx eYeX ,Choosing


Example 8.4 Beam with Sinusoidal Loading

x

y qosinπx/l

l

2c

qol/ qol/

/),(

/),0(

)/sin(),(

0),(

0),(

0),(),0(

lqdyyl

lqdyy

lxqcx

cx

cx

yly

c

c oxy

c

c oxy

oy

y

xy

xx

Boundary Conditions: ]cosh)(sinh)[(sin yyDByyCAx

)]cosh2sinh(sinh

)sinh2cosh(cosh[(cos

]cosh)(sinh)[(sin

)]sinh2cosh(cosh

)cosh2sinh(sinh[(sin

2

2

2

yyyDyB

yyyCyAx

yyDByyCAx

yyyDyB

yyyCyAx

xy

y

x

)1coth(

)1tanh(

ccCB

ccDA

l

c

l

c

l

c

l

l

cq

Co

coshsinh2

sinh

2

2

l

c

l

c

l

c

l

l

cq

Do

coshsinh2

sinh

2

2

Stress Field

l



x

y qosinπx/l

l

2c

qol/ qol/ Bending Stress

l

c

l

clc

l

yl

l

cc

l

yl

l

yy

l

c

l

clc

l

yl

l

cc

l

yl

l

yy

l

x

l

cq

lccCBccDA

l

c

l

c

l

c

l

l

cq

D

l

c

l

c

l

c

l

l

cq

C

yyyDyB

yyyCyAx

ox

oo

x

coshsinh

coshcothcosh2sinh

coshsinh

sinhtanhsinh2coshsinsinh

2

,)1coth(,)1tanh(

coshsinh2

cosh,

coshsinh2

sinh

)]sinh2cosh(cosh

)cosh2sinh(sinh[(sin

2

2

2

2

2

l

xy

c

lq

c

ylxlq

I

My

l

xy

c

lq

l

x

l

y

l

y

l

y

c

lq

BDACc

lqDcl

o

o

x

oox

o

sin2

3

3/2

sin:Theory Materials ofStrength

sin2

3sinsinhcosh

4

3

0,,0,4

3: case For the

23

2

3

2

2

23

2

33

3

53

5

2/lx



x

y qosinπx/l

l

2c

qol/ qol/

oo uyyyyD

yyyC

yByAxE

u

]}sinh2cosh)1[(

]cosh2sinh)1[(

cosh)1(sinh)1({cos

0)0,()0,0()0,0( lvvu ]2)1([,0 CBE

uv ooo

]tanh)1(2[sin)0,( ccxE

Dxv

]tanh2

11[sin

2

3)0,(

43

4

l

c

l

c

l

x

Ec

lqxv o

oo vyyyyD

yyyC

yByAxE

v

]}cosh)1(sinh)1[(

]sinh)1(cosh)1[(

sinh)1(cosh)1({sin

For the case l >> c 53

5

4

3

c

lqD o

Strength of Materials l

x

Ec

lqxv o

sin

2

3)0,(

43

4



Example 8.5 Rectangular Domain with Arbitrary Boundary Loading

Boundary Conditions

Must use series representation for Airy stress function to handle general boundary loading.

)(),(,0),(

0),(,0),(

xpbxbx

yaya

yxy

xyx

p(x)

x

y

a a

p(x)

b

b

20

1

1

]sinhcosh[cos

]sinhcosh[cos

xCxxGxFy

yyCyBx

mmmmmmm

nnnnnnn

1

2

1

2

1

2

01

2

1

2

1

2

)]sinhcosh(sinh[sin

)]sinhcosh(sinh[sin

)]cosh2sinh(cosh[cos

2]sinhcosh[cos

]sinhcosh[cos

)]cosh2sinh(cosh[cos

mmmmmmmmm

nnnnnnnnnxy

mmmmmmmmm

nnnnnnnny

mmmmmmmm

nnnnnnnnnx

xxxGxFy

yyyCyBx

xxxGxFy

CyyCyBx

xxGxFy

yyyCyBx

Using Fourier series theory to handle general boundary conditions, generates a doubly infinite set of equations to solve for unknown constants in stress function form. See text for details


Polar Coordinate FormulationAiry Stress Function Approach = (r,θ)

rr

r

rrr

r

r

1

11

2

2

2

2

2

01111

2

2

22

2

2

2

22

24

rrrrrrrr

RS

x

y

r

Airy Representation

Biharmonic Governing Equation

),(,),( rfTrfT rr

Traction Boundary Conditions

r

r


Polar Coordinate FormulationPlane Elasticity Problem

r

u

r

uu

re

uu

re

r

ue

rr

r

rr

1

2

1

1

0,2

)()(

2)(

2)(

rzzrr

rrz

r

rrr

e

ee

eee

eee

Strain Plane

0,1

)(1

)(

)(1

,)(1

rzzrr

rrz

rrr

eeE

e

eeE

e

Ee

Ee

Stress Plane

Strain-Displacement

Hooke’s Law


General Solutions in Polar CoordinatesMichell Solution

berfr )(),(

0)4(21212

4

22

3

2

2

2

fr

bbf

r

bf

r

bf

rf

Choosing the case where b = in, n = integer gives the general Michell solution

2

243

221

2

243

221

16153

1413

1211

16153

1413

1211

27

2654

23

2210

sin)(

cos)(

sin)loglog(

cos)loglog(

)loglog(

loglog

n

nn

nn

nn

nn

n

nn

nn

nn

nn

nrbrbrbrb

nrararara

rrbrbrbr

brrbrb

rrararar

arrara

rrararaa

rrararaa

Will use various terms from this general solution to solve several plane problems in polar coordinates

01111

2

2

22

2

2

2

22

24

rrrrrrrr


Axisymmetric Solutions

rrararaa loglog 23

2210

0

23log2

2log2

2321

3

2321

3

r

r

aar

ara

aar

ara

CrBAaE

ru

BA

rararraarE

ur

sincos4

cossin

)1(2)1(log)1(2)1(1

3

2331

Stress Function Approach: =(r) Navier Equation Approach: u=ur(r)er

(Plane Stress or Plane Strain)

011

22

2

rrr u

rdr

du

rdr

ud

rCrCur

121

Displacements - Plane Stress Case

Gives Stress Forms

0,,22

rr Br

AB

r

A

• a3 term leads to multivalued behavior, and is not found following the displacement formulation approach

• Could also have an axisymmetric elasticity problem using = a4 which gives r = = 0 and r = a4/r 0, see Exercise 8-15

Underlined terms represent rigid-body motion


Example 8.6 Thick-Walled Cylinder Under Uniform Boundary Pressure

r1

r2

p1

p2

Br

A

Br

Ar

2

2

Boundary Conditions General AxisymmetricStress Solution

2211 )(,)( prpr rr

21

22

22

212

1

21

22

122

22

1 )(

rr

prprB

rr

pprrA

21

22

22

212

122

12

2

122

22

1

21

22

22

212

122

12

2

122

22

1

1)(

1)(

rr

prpr

rrr

pprr

rr

prpr

rrr

pprrr

Using Strain Displacement Relations and Hooke’s Law for plane strain gives the radial displacement

rrr

prpr

rrr

pprr

E

r

ABr

Eur

21

22

22

212

12

12

2

122

22

1

2

)21(1)(1

])21[(1


Example 8.6 Cylinder Problem ResultsInternal Pressure Only

r1/r2 = 0.5

r/r2

r /p

θ /p

Dim

ensi

on

less

Str

ess

Dimensionless Distance, r/r2

Thin-Walled Tube Case:

pprrrr )3/5()/()()( 21

22

22

21max

t

pro112 rrt 2/)( 21 rrro Matches with Strength of Materials Theory

r1

r2

p


Special Cases of Example 8-6Pressurized Hole in an Infinite Medium

r1

p

22 and0 rp

0,,2

21

12

21

1 zr r

rp

r

rp

r

rp

Eur

2111

Stress Free Hole in an Infinite Medium Under Equal Biaxial Loading at Infinity

221 ,,0 rTpp

2

21

2

21 1,1

r

rT

r

rTr

Tr 2)()( 1maxmax

T

T

r1


Example 8.7 Infinite Medium with a Stress Free Hole Under Uniform Far Field

Loading

T a T

x

y

2sin2

),(

)2cos1(2

),(

)2cos1(2

),(

0),(),(

T

T

T

aa

r

r

rr

Boundary Conditions

2cos)(

loglog

242

234

222

21

23

2210

ararara

rrararaa

2sin)26

62(

2cos)6

122(2)log23(

2cos)46

2(2)log21(

224

4232

2221

4234

222121

23

224

423

2121

23

r

a

r

araa

r

araa

r

aara

r

a

r

aa

r

aara

r

r

Try Stress Function

2sin23

12

2cos3

12

12

2cos43

12

12

2

2

4

4

4

4

2

2

2

2

4

4

2

2

r

a

r

aT

r

aT

r

aT

r

a

r

aT

r

aT

r

r


Example 8.7 Stress Results

T a T

x

y

2sin23

12

2cos3

12

12

2cos43

12

12

2

2

4

4

4

4

2

2

2

2

4

4

2

2

r

a

r

aT

r

aT

r

aT

r

a

r

aT

r

aT

r

r

1

2

3

30

210

60

240

90

270

120

300

150

330

180 0

Ta /),(

Ta /),(

0)30,(,)0,(

)2cos21(),(

oaTa

Ta

Ta

r/)

2,(

r/a

, /

T

Ta 3)2/,(max


Superposition of Example 8.7Biaxial Loading Cases

= +

T1

T2

T1

T2

Equal Biaxial Tension CaseT1 = T2 = T

Tension/Compression CaseT1 = T , T2 = -T

2sin23

1

2cos3

1

2cos43

1

2

2

4

4

4

4

2

2

4

4

r

a

r

aT

r

aT

r

a

r

aT

r

r

2

21

2

21 1,1

r

rT

r

rTr

Tr 2)()( 1maxmax

TaaTaa 4)2/3,()2/,(,4),()0,(


Review Stress Concentration FactorsAround Stress Free Holes

T

T

r1

T a T

x

y

T T

T

T

T T

T T

45o

(a) Biaxial Loading (b) Shear Loading

K = 2 K = 3

K = 4

=


Stress Concentration Around Stress Free Elliptical Hole – Chapter 10

x

y

Sx

b a

a

bS 21

max

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10

Eccentricity Parameter, b/a

Str

es

s C

on

ce

ntr

ati

on

Fa

cto

r

Circular Case

()max/S

Maximum Stress Field


Stress Concentration Around Stress Free Hole in Orthotropic Material – Chapter 11

Isotropic Case

Orthotropic Case Carbon/Epoxy

x(0,y)/S

S S

x

y


2-D Thermoelastic Stress Concentration Problem Uniform Heat Flow Around

Stress Free Insulated Hole – Chapter 12

x

y

q

a

cos2

1

sin2

1

sin2

1

3

3

3

3

3

3

r

a

r

a

k

qaE

r

a

r

a

k

qaE

r

a

r

a

k

qaE

r

r

sin),(max k

qaEa

Stress Field

kqaEa /)2/,(max

Maximum compressive stress on hot side of holeMaximum tensile stress on cold side

2/2/

Steel Plate: E = 30Mpsi (200GPa) and = 6.5in/in/oF (11.7m/m/oC), qa/k = 100oF (37.7oC), the maximum stress becomes 19.5ksi (88.2MPa)


Nonhomogeneous Stress Concentration Around Stress Free Hole in a Plane Under Uniform Biaxial Loading

with Radial Gradation of Young’s Modulus – Chapter 14

n = 0 (homogeneous case)

n = 0.2

n = 0.4

n = 0.6

b/a = 20 = 0.25

n = -0.2

n

o a

rErE

)(

-0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.41

1.5

2

2.5

3

3.5

Power Law Exponent, n

Str

ess

Con

cent

ratio

n F

acto

r, K

homogeneous case

b/a = 20 = 0.25


Three Dimensional Stress Concentration Problem – Chapter 13

x

y

z

a

S

S

Normal Stress on the x,y-plane (z = 0)

5

5

3

3

)57(2

9

)57(2

541)0,(

r

a

r

aSrz

Sa zz )57(2

1527)()0,( max

04.2

)(3.0 max

Sz

1.9

1.95

2

2.05

2.1

2.15

2.2

0 0.1 0.2 0.3 0.4 0.5

Poisson's Ratio

Str

ess

Co

nce

ntr

atio

n F

act

or

1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

2

2.5

3

3.5

Dimensionless Distance, r/a

Nor

mal

ized

Str

ess

in L

oadi

ng D

irect

ion

Two Dimensional Case: (r,/2)/S

Three Dimensional Case: z(r,0)/S , = 0.3


Wedge Domain Problems

)2sin2cos( 2121622 baaar

2sin22cos2

2sin22cos222

2sin22cos222

21216

212162

212162

aba

baaa

baaa

r

r

x

y

r

Use general stress function solution to include terms that are bounded at origin and give uniform stresses on the boundaries

Quarter Plane Example ( = 0 and = /2)

x

y

S

r

0)0,()0,( rr r

)2sin2

2cos1(2

)2sin2cos2

22

(2

)2sin2cos2

22

(2

S

S

S

r

r

Sr

r

r

)2/,(

0)2/,(


Half-Space ExamplesUniform Normal Stress Over x 0

x

y

T

r Try Airy Stress Function

2sin221

26 rbra

2cos2

2sin22

216

216

ba

ba

r

Trr

rr

r

r

),(,0),(

0)0,()0,(Boundary Conditions

)2cos1(2

)22(sin2

)22(sin2

T

T

T

r

r

Use BC’s To Determine Stress Solution


Half-Space Under Concentrated Surface Force System (Flamant Problem)

x

y

Y

X

r

C

Try Airy Stress Function

Boundary Conditions

Use BC’s To Determine Stress Solution

sin)log(

cos)log(

1512

1512

rbrrb

rarra

21 eeForces YX

rr

rr

C

r

r

0),(,0),(

0)0,()0,(

0

]sincos[2

r

r YXr


Flamant Solution Stress ResultsNormal Force Case

Dimensionless Distance, x/a

y/(Y/a)

xy/(Y/a)

Dim

en

sio

nles

s S

tre

ss

x

y

Y

r = constant

or in Cartesian components

222

2

222

32

222

22

)(

2cossin

)(

2sin

)(

2cos

yx

Yxy

yx

Yy

yx

yYx

rxy

ry

rx

0

sin2

r

r r

Y

y = a

aYy /2


Flamant Solution Displacement ResultsNormal Force Case

011

sin2

)(11

sin2

)(1

rr

r

rr

rr

r

r

u

r

uu

r

Er

Y

E

u

rr

uEr

Y

Er

u

]cos)1(coslog2sin)2

)(1([

]sinlog2cos)2

)(1[(

rE

Yu

rE

Yur

)1(2

),()0,( E

Yruru rr

]log2)1[(),()0,( rE

Yruru

On Free Surface y = 0

-0.5 0 0.5

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0

0.1

Y

Note unpleasant feature of 2-D model that displacements become unbounded as r


Comparison of Flamant Results with 3-D Theory - Boussinesq’s Problem

x

y z

P

0

)1(24

)21(

4

2

2

2

u

R

z

R

Pu

zR

r

R

rz

R

Pu

z

r

5

2

5

3

2

3

2

2

2

3,

2

3

2

)21(

)21(3

2

R

rzP

R

Pz

zR

R

R

z

R

P

zR

R

R

zr

R

P

rzz

r

5

2

5

2

2325

3

2

2

3

2

2

2

2

3

2

2

2

2

22

2

3,

2

3

)(

)2)(21(3

2,

2

3

)(

)2()21(

3

2

)(

)2()21(

3

2

)1(24

,21

4,

21

4

R

Pxz

R

Pyz

zRR

xyzR

R

xyz

R

P

R

Pz

zRR

zRy

zR

R

R

z

R

zy

R

P

zRR

zRx

zR

R

R

z

R

zx

R

P

R

z

R

Pw

zRR

z

R

Pyv

zRR

z

R

Pxu

xzyz

xyz

y

x

Cartesian Solution

Cylindrical Solution

Free Surface Displacements

R

PRuz

2

)1()0,(

Corresponding 2-D Results

]log2)1[()0,( rE

Pru

3-D Solution eliminates the unbounded far-field behavior


Half-Space Under Uniform Normal Loading Over –a x a

p

x

y

a a 1 2

cossin2

cossin

sin2

sin

cossin2

cos

2

32

22

r

Yr

Yr

Y

rxy

ry

rx

dp

d

dp

d

dp

d

xy

y

x

cossin2

sin2

cos2

2

2

]2cos2[cos2

cossin2

)]2sin2(sin)(2[2

sin2

)]2sin2(sin)(2[2

cos2

12

12122

12122

2

1

2

1

2

1

pd

p

pd

p

pd

p

xy

y

x

dx

r

d

dY = pdx = prd /sin


Half-Space Under Uniform Normal Loading - Results

Dimensionless Distance, x/a

Dim

ensi

onle

ss S

tres

s

y/p

xy /p

0 1 2 3 4 5 6 7 8 9 100

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Dimensionless Distance, y /

Dim

ensi

onle

ss M

axim

um S

hear

Str

ess

a

Distributed Loading max/p

Concentrated Loading max/(Y/a)

max - Contours


Generalized Superposition MethodHalf-Space Loading Problems

x

y

a a

t(s)

p(s)

dsysx

sxstys

ysx

sxspy

dsysx

sxstyds

ysx

spy

dsysx

sxstds

ysx

sxspy

a

a

a

axy

a

a

a

ay

a

a

a

ax

222

2

222

2

222

2

222

3

222

3

222

2

])[(

))((2

])[(

))((2

])[(

))((2

])[(

)(2

])[(

))((2

])[(

))((2


Photoelastic Contact Stress Fields

(Uniform Loading) (Point Loading)

(Flat Punch Loading) (Cylinder Contact Loading)


Notch/Crack Problem

y

= 2 -

r

Stress Free Faces x

])2cos()2sin(cossin[ DCBAr

])2sin()2()2cos()2(sincos[)1(

])2cos()2sin(cossin[)1(2

2

DCBAr

DCBAr

r

0)2,()2,()0,()0,( rrrr rrBoundary Conditions:

,2,1,0,12

0)1(2sin nn

At Crack Tip r 0:

Try Stress Function:

)(,)( 12 rOrO ntDisplaceme Stress

Finite Displacements and Singular Stresses at Crack Tip 1< <2 = 3/2


Notch/Crack Problem Results

)2

sin3

1

2

3(sin)

2cos

2

3(cos

1

4

3

)2

cos2

3(cos)

2sin3

2

3(sin

1

4

3

)2

cos3

5

2

3(cos)

2sin5

2

3(sin

1

4

3

BAr

BAr

BAr

r

r

)cos31(2

cos2

)cos1(2

sin2

3

)cos1(2

sin2

3)cos1(

2cos

2

3

)cos31(2

sin2

)cos3(2

cos2

3

r

B

r

Ar

B

r

Ar

B

r

A

r

r

y

= 2 -

r

Stress Free Faces x

Transform to Variable

• Note special singular behavior of stress field O(1/r)• A and B coefficients are related to stress intensity factors and are useful in fracture

mechanics theory• A terms give symmetric stress fields – Opening or Mode I behavior• B terms give antisymmetric stress fields – Shearing or Mode II behavior


Crack Problem ResultsContours of Maximum Shear Stress

Mode I (Maximum shear stress contours) Mode II (Maximum shear stress contours)

Experimental Photoelastic Isochromatics Courtesy of URI Dynamic Photomechanics Laboratory


Mode III Crack Problem – Exercise 8-41

y

r

x

●

z

Contours for Mode III Crack Problem

),(,0 yxwwvu

011

2

2

22

22

w

rr

w

rr

ww

2sin

2,

2cos

2,

2sin

r

A

r

ArAw zrz

Anti-Plane Strain Case

2/1rOStresses Again

z - Stress Contours


Curved Beam Under End Moments

b

a

b

a

rr

rr

Mrdr

dr

ba

ba

0

0)()(

0)()(

a

b

r

M M

0

])log()log()log([4

)]log()log()log([4

22222

22

222

22

r

r

abr

aa

b

rb

a

b

r

ba

N

M

r

aa

b

rb

a

b

r

ba

N

M Dimensionless Distance, r/a

Dim

en

sio

nle

ss

Str

es

s,

a2 /M

Theory of Elasticity Strength of Materials

b/a = 4

rrararaa loglog 23

2210


Curved Cantilever BeamP

ab

r

cos)(

sin)3(

sin)(

22

3

22

22

3

22

22

3

22

r

ba

r

bar

N

P

r

ba

r

bar

N

P

r

ba

r

bar

N

P

r

r

Dimensionless Distance, r/aD

imen

sio

nle

ss S

tres

s,

a/P

Theory of Elasticity Strength of Materials

= /2 b/a = 4

b

a r

b

a

b

a

b

a

b

a

b

a r

rr

rr

drr

baPrdrr

Pdrr

rdrrdrr

Pdrr

ba

ba

0)2/,(

2/)()2/,(

)2/,(

0)0,()0,(

)0,(

0),(),(

0),(),(

sin)log( 3 rDrCrr

BAr


Disk Under Diametrical Compression

+

P

P

D=

+

Flamant Solution (1)

Flamant Solution (2) Radial Tension Solution (3)


Disk Problem – Superposition of Stresses

P

P

2

y

x

1 r1

r2

112

1

)1(

13

1

)1(

12

11

)1(

sincos2

cos2

sincos2

r

P

r

P

r

P

xy

y

x

0,2 )3()3()3(

xyyx D

P

42

2

41

2

42

3

41

3

42

2

41

2

)()(2

1)()(2

1)()(2

r

xyR

r

xyRP

Dr

yR

r

yRP

Dr

xyR

r

xyRP

xy

y

x

222,1 )( yRxr

222

2

)2(

23

2

)2(

22

22

)2(

sincos2

cos2

sincos2

r

P

r

P

r

P

xy

y

x


Disk Problem – Results

0)0,(

1)4(

42)0,(

4

42)0,(

222

4

2

22

22

x

xD

D

D

Px

xD

xD

D

Px

xy

y

x

0),0(

1

2

2

2

22),0(

2),0(

y

DyDyD

Py

D

Py

xy

y

x Constant

(Theoretical max Contours) (Photoelastic Contours) (Courtesy of URI Dynamic Photomechanics Lab)

x-axis (y = 0) y-axis (x = 0)

P

P

2

y

x

1 r1

r2


Applications to Granular Media ModelingContact Load Transfer Between Idealized Grains

(Courtesy of URI Dynamic Photomechanics Lab)

P

P

P

P

Four-Contact Grain


Contact Between Two Elastic Solids

Creates Complicated Nonlinear Boundary Condition:Boundary Condition Changing With Deformation; i.e. w and pc Depend on Deformation, Load, Elastic Moduli, Interfacial Friction Characteristics

w

pc


Generates:- Contact Area (w) - Interface Tractions (pc) - Local Stresses in Each Body

2-D Elastic Half-Space Subjected to a Rigid Indenter

x

y

Rigid Indenter

aa

yu

constantsmotion body rigid are and

)()(2

1log)(

E

2

log)(E

2 )()(

2

1

21

2

1

aa

adsstdsstE

dssxspu

adssxstdsspdsspE

u

a

x

x

-a

a

-ay

a

-a

a

x

x

-ax

Local stresses and deformation determined from Flamant solutionSee Section 8.4.9 and Exercise 8.38


Frictionless Case (t = 0)

a

-a

y

x

dssx

sp

dx

ud

xpEdx

ud

)(

E

2

)(1

2-D Elastic Half-Space Subjected Frictionless Flat Rigid Indenter


constant oyy uu

a

-ads

sx

sp0

)(

22)(

xa

Pxp

x

y

Rigid Indenter

aa

yu

P

axua

x

a

xu

axaxPE

u

oyy

x

,1logE

2

,)/(sin1

2/1

2

2

1

dsysxsa

sxPy

dsysxsa

Py

dsysxsa

sxPy

a

axy

a

ay

a

ax

222222

2

222222

3

22222

2

2

])[(

)(2

])[(

12

])[(

)(2

Frictionless Rigid Punch Loading on a Half-Space

22)(

xa

Pxp

x

y

Max Shear Stress Contours

Solution

Unbounded Stresses at Edges of Indenter


2-D Elastic Half-Space Subjected Frictionless Cylindrical Rigid Indenter

x

y

Rigid Indenter

a

R

yu

a

P

Rxu y 2/ toalproportion 2 xR

dssx

spa

-a 2

E)(

222

2)( xa

a

Pxp

E

PRa

42

dsysx

sxsa

a

Py

dsysx

sa

a

Py

dsysx

sxsa

a

Py

a

axy

a

ay

a

ax

222

22

22

2

222

22

22

3

222

222

22

])[(

)(4

])[(

4

])[(

)(4

Max Shear Stress Contours

Solution

Elliptical Distributed Normal Loading on a Half-Space

222

2)( xa

a

Pxp

x

y

a -a

chapter 8 two-dimensional problem solution

Documents

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