chapter 8 two-dimensional problem solution.ppt

8/14/2019 Chapter 8 Two-Dimensional Problem Solution.ppt

1/51

yxxy xyyx

2

2

2

2

2

,,

02 44

4

22

4

4

4

yyxx

Using the Airy Stress Function approach, it was shown that the 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


2/51

Cartesian Coordinate Solutions

Using PolynomialsIn Cartesian coordinates we choose Airy stress function solution of polynomial form

whereAmnare constant coefficients to be determined. This method produces

polynomial stress distributions, and thus would not satisfy general boundary

conditions. However, we can modify such boundary conditions using Saint-Venants

principle and replace a non-polynomial condition with a statically equivalent loading.

This formulation is most useful for problems with rectangular domains, and is

commonly based on the inverse solution concept where we assume a polynomial

solution form and then try to find what problem it will solve.

Noted that the three lowest order terms with m + n 1 do not contribute to thestresses and will therefore be dropped. It should be noted that 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 the biharmonic equation for anychoice of constantsAmn. However, for higher order terms, constantsAmnwill have to

be related in order to have the polynomial satisfy the biharmonic equation.

0 0

),(m n

nm

mn yxAyx


3/51

Example 8.1 Uniaxial Tension of a Beam

x

y

TT

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

2

02yA 0,2

02

xyyx

A

The first boundary condition implies thatA02= 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

v

ET

Ee

xu

xyy

yxx

)(1

)(1

)(,)( xgyE

Tvyfx

E

Tu

0)()(02

xgyfe

xv

yu xy

xy

oo

oo

vxxg

uyyf

)(

)(. . . Rigid-Body Motion

Fixity conditionsneeded to determine RBM terms

0)()(0)0,0()0,0()0,0( xgyfvu z


4/51

Example 8.2 Pure Bending of a Beam

Boundary Conditions:

Expecting a linear bending stress distribution,

try second-order stress function of the form

3

03yA 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 conditionsto determine RBM terms:

x

y

MM

2l

2c

c

c x

c

c x

xyxyy

Mydyyldyyl

ylcxcx

),(,0),(

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

0,2

33

xyyx yc

M

)(4

3

2

3

)(23

23

2

33

33

xgyEc

Mvy

Ec

M

y

v

yfxyEcMuy

EcM

xu

0)()(2

30

3

xgyfxEc

M

x

v

y

u

oo

oo

vxxEc

Mxg

uyyf

2

34

3)(

)(

0)0,(and0)0,( lulv32

16/3,0 EcMlvu ooo

Displacement Field (Plane Stress)


5/51

Example 8.2 Pure Bending of a BeamSolution Comparison of Elasticity

with Elementary Mechanics of Materials

x

y

MM

2l

2c

]44[8,

0,

222

lxyEI

M

vEI

Mxy

u

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 thex-displacements


6/51

Example 8.3 Bending of a Beam by

Uniform Transverse Loading

x

y

w

2c

2l

wl wl


Stress Field

c

c xy

c

c x

c

c x

y

y

xy

wldyyl

ydyyl

dyyl

wcx

cx

cx

),(

0),(

0),(

),(

0),(

0),(

52332

23

3

03

2

21

2

205

yA

yxAyAyxAxA

22321

3

232120

32

2303

62

222

)3

2(66

xyAxA

yAyAA

yyxAyA

xy

y

x

2

3

3

3

32

32

2

4

3

4

3

44

3

2

)3

2(

4

3

5

2

4

3

xyc

w

xc

w

yc

wy

c

ww

yyxc

wy

c

l

c

w

xy

y

x

BCs


7/51

Example 8.3 Beam ProblemStress Solution Comparison of Elasticity


x

y

w

2c

2l

wl wl

Elasticity Solution Mechanics of Materials Solution

)(2

32

32

)53

()(2

22

32

3

2322

ycxI

w

cycyIw

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


8/51

Example 8.3 Beam ProblemNormal Stress Comparisons of Elasticity


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 the two theories exist

at top and bottom of beam, and actual difference in

stress values is w/5. For most beam problems where

l>> c, the bending stresses will be much greater than

w, and thus the differences between elasticity and

strength of materials will be relatively small.

Maximum difference between the two theories is w

and this occurs at the top of the 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.

xStress at x=0 y- Stress


9/51

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 /),(


10/51

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

22

3224

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

23

()5

23

2()3

[(2

2

2422

24

224222

3224

3

2

3233

2

l

c

EI

wlxc

lx

ycyyxl

ycycy

EI

wv

cycyxycyxyxxlEIwu

])25

4(

5

121[

24

5)0,0(

2

24

ma xl

c

EI

wlvv

EI

wlv

24

5 4

ma xStrength of Materials:

Good match for beams where l>> c



11/51

Cartesian Coordinate Solutions

Using Fourier MethodsA more general solution scheme for the biharmonic equation may be

found using Fourier methods. Such techniques generally use separation of

variablesalong with Fourier seriesor Fourier integrals.

)()(),( yYxXyx 02 44

22

4

4

4

yyxx

i

00

]cosh)(sinh)[(cos

]cosh)(sinh)[(sin

]cosh)(sinh)[(cos

]cosh)(sinh)[(sin

xxHFxxGEy

xxHFxxGEy

yyDByyCAx

yyDByyCAx

3

3

2

2100 xCxCxCC 2

9

2

87

3

6

2

540 xyCyxCxyCyCyCyC

yx eYeX ,Choosing


12/51

Example 8.4 Beam with Sinusoidal Loading

x

y qosinx/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


]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

lcq

Do

coshsinh2

sinh

2

2

Stress Field

l


13/51


x

y qosinx/l

l

2c

qol/ qol/Bending Stress

l

c

l

clc

l

yl

l

cc

l

yl

l

yy

l

c

l

clc

lyl

lcc

lyl

lyy

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

sinhtanhsinh2cosh

sinsinh2

,)1coth(,)1tanh(

coshsinh2

cosh

,

coshsinh2

sinh

)]sinh2cosh(cosh

)cosh2sinh(sinh[(sin

2

2

2

2

2

l

xy

c

lq

c

yl

xlq

I

My

l

xy

c

lq

l

x

l

y

l

y

l

y

c

lq

BDACc

lqD

cl

o

o

x

oox

o

sin

2

3

3/2

sin

sin2

3sinsinhcosh

4

3

0,,0,4

3

:

23

2

3

2

2

23

2

33

3

53

5

heoryaterialsft rength

aseheor

2/lx


14/51


x

y qosinx/l

l

2c

qol/ qol/

oo uyyyyD

yyyC

yByAxEu

]}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

yByAxEv

]}cosh)1(sinh)1[(

]sinh)1(cosh)1[(

sinh)1(cosh)1({sin

For the case l>> c53

5

4

3

c

lqD o

Strength of Materialsl

x

Ec

lqxv o

sin2

3)0,(

43

4



15/51

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),(

xpbx

bx

ya

ya

y

xy

xy

x

p(x)

x

y

a a

p(x)

b

b

2

0

1

1

]sinhcosh[cos

]sinhcosh[cos

xCxxGxFy

yyCyBx

m

mmmmmm

n

nnnnnn

1

2

1

2

1

2

0

1

2

1

2

1

2

)]sinhcosh(sinh[sin

)]sinhcosh(sinh[sin

)]cosh2sinh(cosh[cos

2]sinhcosh[cos

]sinhcosh[cos

)]cosh2sinh(cosh[cos

m

mmmmmmmm

n

nnnnnnnnxy

m

mmmmmmmm

n

nnnnnnny

m

mmmmmmm

n

nnnnnnnnx

xxxGxFy

yyyCyBx

xxxGxFy

CyyCyBx

xxGxFy

yyyCyBx

Use Fourier series theory to handle general

boundary conditions, and this generates a

doubly infinite set of equations to solve for

unknown constants in stress function form.

See text for details


16/51

Polar Coordinate Formulation

Airy Stress Function Approach = (r,)

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 rrTraction Boundary Conditionsr

r


17/51

Polar Coordinate Formulation

Plane Elasticity Problem

ru

ruu

re

uu

re

r

ue

rr

r

rr

121

1

0,2

)()(

2)(2)(

rzzrr

rrz

r

rrr

e

ee

eeeeee

Strainlane

0,1

)(1

)(

)(1,)(1

rzzrr

rrz

rrr

eeE

e

eeE

e

Ee

Ee

Stresslane

Strain-Displacement

Hookes Law


18/51

General Solutions in Polar Coordinates

Michell 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

2

43

2

21

2

2

43

2

21

16153

1413

1211

1615

3

1413

1211

2

7

2

654

2

3

2

210

sin)(

cos)(

sin)loglog(

cos)loglog(

)loglog(

loglog

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

n

nrbrbrbrb

nrararara

rrbrbrbr

brrbrb

rrararar

arrara

rrararaa

rrararaa

We will use various

terms from this general

solution to solve

several plane problems

in polar coordinates

011112

2

22

2

2

2

22

24

rrrrrrrr


19/51

Axisymmetric Solutions

rrararaa loglog 232

210

0

23log2

2log2

232

13

232

13

r

r

aa

r

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

a3term leads to multivalued behavior, and is not found following the

displacement formulation approach

Could also have an axisymmetric elasticity problem using = a4which gives

r=

= 0 and

r

= a

4/r 0, see Exercise 8-14

Underlined terms representrigid-body motion


20/51

Example 8.6 Thick-Walled Cylinder

Under Uniform Boundary Pressure

r1

r2

p1

p2

BrA

Br

Ar

2

2

Boundary ConditionsGeneral Axisymmetric

Stress Solution2211 )(,)( prpr rr

2

1

2

2

2

2

21

2

1

2

1

2

2

12

2

2

2

1 )(

rr

prprB

rr

pprrA

2

1

2

2

2

2

21

2

1

22

1

2

2

12

2

2

2

1

2

1

2

2

2

2

21

2

1

22

1

2

2

12

2

2

2

1

1)(

1)(

rr

prpr

rrr

pprr

rr

prpr

rrr

pprrr

Using Strain Displacement

Relations and Hookes Law

for plane strain gives the

radial displacement

rrr

prpr

rrr

pprr

E

r

ABr

Eur

2

1

2

2

2

2

21

2

1

2

1

2

2

12

2

2

2

1

2

)21(1)(1

])21[(1


21/51

Example 8.6 Cylinder Problem Results

Internal Pressure Only

r1/r2= 0.5

r/ r2

r/p

/p

DimensionlessStress

Dimensionless Distance, r/r2

Thin-Walled Tube Case:

pprrrr )3/5()/()()( 2

1

2

2

2

2

2

1max

t

pro112 rrt 2/)( 21 rrro Matches with Strength

of Materials Theory

r1

r2

p


22/51

Special Cases of Example 8-6

Pressurized Hole in an Infinite Medium

r1

p

22 and0 rp

0,,2

2

112

2

11 zr

r

rp

r

rp

r

rp

Eur

2

111

Stress Free Hole in an Infinite Medium

Under Equal Biaxial Loading at Infinity

221 ,,0 rTpp

2

2

1

2

2

1 1,1r

rT

r

rTr

Tr 2)()( 1maxmax

T

T

r1


23/51

Example 8.7 Infinite Medium with a Stress

Free Hole Under Uniform Far Field Loading

TaT

x

y

2sin2),(

)2cos1(2

),(

)2cos1(2

),(

0),(),(

T

T

T

aa

r

r

rr

Boundary Conditions

2cos)(

loglog

24

2

23

4

22

2

21

2

3

2

210

ararara

rrararaa

2sin)26

62(

2cos)6

122(2)log23(

2cos)46

2(2)log21(

2

24

4

232

2221

4

234

22212

123

224423212123

r

a

r

araa

r

araa

r

aara

r

a

r

aar

aara

r

r

Try Stress Function

2sin23

12

2cos3

12

12

2cos

43

1212

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


24/51

Example 8.7 Stress Results

TaT

x

y

2sin23

12

2cos3

12

12

2cos

43

1212

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

,

Ta 3)2/,(ma x


25/51

Superposition of Example 8.7

Biaxial Loading Cases

= +

T1

T2

T1

T2

Equal Biaxial Tension Case

T1= T2= T

Tension/Compression Case

T1= 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

2

1

2

2

1 1,1

r

rT

r

rTr

Tr 2)()( 1maxmax

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


26/51

Review Stress Concentration Factors

Around Stress Free Holes

T

T

r1 TaT

x

y

TT

T

T

TT

T T

45o

K = 2 K = 3

K = 4

=


27/51

Stress Concentration Around

Stress Free Elliptical HoleChapter 10

x

ySx

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

Stres

sConcentrationFactor

Circular Case

()max/S

Maximum Stress Field


28/51

Stress Concentration Around Stress Free

Hole in Orthotropic MaterialChapter 11

Isotropic Case

Orthotropic Case Carbon/Epoxy

x(0,y)/S

SS

x

y


29/51

2-D Thermoelastic Stress Concentration

Problem Uniform Heat Flow Around

Stress Free Insulation HoleChapter 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),(ma xk

qaEa

Stress Field

kqaEa /)2/,(max

Maximum compressive stress on hot side of hole

Maximum 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)


30/51

Nonhomogeneous Stress Concentration Around Stress

Free Hole in a Plane Under Uniform Biaxial Loading

with Radial Gradation of Youngs Modulus Chapter 14

n= 0 (homogeneous case)

n= 0.2

n= 0.4

n= 0.6

b/a= 20

= 0.25n= -0.2

n

oa

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

StressConcentrationFactor,K

homogeneous case

b/a= 20

= 0.25


31/51

Three Dimensional Stress Concentration

ProblemChapter 13

x

y

z

a

S

S

Normal Stress on thex,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,( ma x

04.2

)(3.0 max

S

z

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

StressConcentrationFactor

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

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


32/51

Wedge Domain Problems

)2sin2cos( 2121622 baaar

2sin22cos22sin22cos222

2sin22cos222

21216

212162

212162

ababaaa

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/,(


33/51

Half-Space ExamplesUniform Normal Stress Overx0

x

y

T

r Try Airy Stress Function

2sin22126 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 BCs To Determine Stress Solution


34/51

Half-Space Under Concentrated Surface

Force System (Flamant Problem)

x

y

Y

X

r

C

Try Airy Stress Function

Boundary Conditions

Use BCs 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


35/51

Flamant Solution Stress Results

Normal Force Case

Dimensionless Distance,x/a

y/(Y/a)

xy/(Y/a)

DimensionlessStress

x

y

Y

r= constant

or in Cartesian

components

222

2

222

32

222

2

2

)(

2cossin

)(

2sin

)(

2cos

yx

Yxy

yx

Yy

yx

yYx

rxy

ry

rx

0

sin2

r

rr

Y

y= a

aYy /2


36/51

Flamant Solution Displacement Results

Normal Force Case

011

sin2

)(11

sin2

)(1

rr

r

rr

rr

r

r

u

r

uu

r

Er

Y

E

u

rr

u

Er

Y

Er

u

]cos)1(coslog2sin)2

)(1([

]sinlog2cos)2

)(1[(

rE

Yu

rE

Yur

)1(2),()0,( EY

ruru rr

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

Yruru

On Free Surface y= 0

Y

Note unpleasant feature of 2-D model that

displacements become unbounded as r


37/51

Comparison of Flamant Results with

3-D Theory - BoussinesqsProblem

x

yz

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


38/51

Half-Space Under Uniform Normal

Loading Overaxap

x

y

aa 12

cossin2

cossin

sin2

sin

cossin2

cos

2

32

22

r

Y

r

Y

r

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

1212

2

1212

2

2

1

2

1

2

1

pd

p

pd

p

pd

p

xy

y

x

dx

r

d

dY=pdx=prd/sin


39/51

Half-Space Under Uniform Normal

Loading - Results

Dimensionless Distance, x/a

Dim

ensionlessStress

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 /

DimensionlessMaximumS

hearStress

a

Distributed Loading

max/p

Concentrated Loading

max/(Y/a)

max- Contours


40/51

Generalized Superposition Method

Half-Space Loading Problems

x

y

aa

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

22

222

2

222

3

222

3

222

2

])[(

))((2

])[(

))((2

])[(

))((2

])[(

)(2

])[(

))((2

])[(

))((2


41/51

Photoelastic Contact Stress Fields

(Uniform Loading)(Point Loading)

(Flat Punch Loading) (Cylinder Contact Loading)


42/51

Notch/Crack Problemy

= 2-

r

Stress Free Facesx

])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 r0:

Try Stress Function:

)(,)( 12 rOrO ntDisplacemeStress

Finite Displacements and Singular Stresses at Crack Tip 1<


43/51

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(2sin2)cos3(2cos2

3

r

B

r

A

r

B

r

A

r

B

r

A

r

r

y

= 2-

r

Stress Free Facesx

Transform to Variable

Note special singular behavior of stress field O(1/r) Aand Bcoefficients are related to stress intensity factors and are useful in fracture

mechanics theory

Aterms give symmetric stress fieldsOpening or Mode I behavior

Bterms give antisymmetric stress fieldsShearing or Mode II behavior

C k P bl R lt


44/51

Crack Problem Results

Contours 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


45/51

Mode III Crack ProblemExercise 8-32

y

r

x

),(,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


46/51

Curved Beam Under End Moments

b

a

b

a

rr

rr

Mrdr

dr

ba

ba

0

0)()(

0)()(

a

b

r

MM

0

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

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

2222

2

22

22

2

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

DimensionlessStress

a2

MTheory of Elasticity

Strength of Materials

b/a = 4

rrararaa loglog 232

210


47/51

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/a

Dimen

sionlessStress,

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


48/51

Disk Under Diametrical Compression

+

P

P

D=

+

Flamant Solution (1)

Flamant Solution (2) Radial Tension Solution (3)


49/51

Disk ProblemSuperposition of Stresses

P

P

2

y

x

1 r1

r211

2

1

)1(

1

3

1

)1(

12

1

1

)1(

sincos2

cos2

sincos2

r

P

r

P

rP

xy

y

x

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

xyyxD

P

4

2

2

4

1

2

4

2

3

4

1

3

4

2

2

4

1

2

)()(2

1)()(2

1)()(2

r

xyR

r

xyRP

DryR

ryRP

Dr

xyR

r

xyRP

xy

y

x

22

2,1 )( yRxr

22

2

2

)2(

2

3

2

)2(

2

2

2

2

)2(

sincos2

cos2

sincos2

r

P

r

P

r

P

xy

y

x


50/51

Disk ProblemResults

0)0,(

1)4(

42)0,(

442)0,(

222

4

2

22

22

x

xD

D

D

Px

xDxD

DPx

xy

y

x

0),0(

1

2

2

2

22),0(

2),0(

y

DyDyD

Py

DPy

xy

y

x Constant

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

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

P

P

2

y

x

1 r1

r2


51/51

Applications to Granular Media ModelingContact Load Transfer Between Idealized Grains

(Courtesy of URI Dynamic Photomechanics Lab)

P

P

P

P

Four-contact grain

chapter 8 two-dimensional problem solution.ppt

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