Application Solutions of Plane Elasticity

Professor M. H. Sadd

Solutions to Plane ProblemsCartesian Coordinates

yxxy xyyx

2

2

2

2

2

,,

Airy Representation

02 44

4

22

4

4

4

yyxx

Biharmonic Governing Equation

),(,),( yxfTyxfT yyxx

Traction Boundary Conditions

RS

x

y

Uniaxial Tension of a Beam

x

y

TT

2l

2c

0),(),(

0),(,),(

ConditionsBoundary

cxyl

cxTyl

xyxy

yx

0,Try 202 xyyx TyA

)()(1

)()(1

ntsDisplaceme

xgyE

Tv

E

T

Ee

y

v

yfxE

Tu

E

T

Ee

x

u

xyy

yxx

0)()(02 xgyfex

v

y

u xyxy

oo

oo

vxxg

uyyf

)(

)(

00)0,0(

00)0,0(,00)0,0(

ConditionsBoundary nt Displaceme Overall

o

oo

x

v

vvuu

Pure Bending of a Beam

x

y

MM

2l

2c

c

c x

c

c x

xyxyy

Mydyyldyyl

ylcxcx

),(,0),(

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

CondtionsBoundary

0,2

33

303 xyyx y

c

MyA

)(4

3

2

3

)(2

3

2

3

233

33

xgyEc

Mvy

Ec

M

y

v

yfxyEc

Muy

Ec

M

x

u

oo

oo

vxxEc

Mxg

uyyf

234

3)(

)(

ntsDisplaceme

][2

,

0,

Elasticity ofTheory

222 lxyEI

Mv

EI

Mxyu

yI

Mxyyx

][2

)0,(

0,

Materialsof Strength

22 lxEI

Mxvv

yI

Mxyyx

0)0,(and0)0,( lulv

32 4/3,0 EcMlvu ooo

Note Integrated Boundary Conditions

Bending of a Beam by Uniform Transverse Loading

x

y

w

2c

2l

wl wl

c

c xy

c

c x

c

c x

yyxy

wldyylydyyldyyl

wcxcxcx

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

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

ConditionsBoundary

5233223

303

221

220 5

yA

yxAyAyxAxA

)(2

3

2

32

)53

()(2

Elasticity ofTheory

22

323

2322

ycxI

w

cycy

I

w

ycy

I

wyxl

I

w

xy

y

x

)(2

0

)(2

Materials ofStrength

22

22

ycxI

w

It

VQ

yxlI

w

I

My

xy

y

x

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

l/c = 2

l/c = 4

l/c = 3

Dimensionless Distance, y/c

Bending of a Beam by Uniform Transverse Loading

])25

4(

5

121[

24

5])

25

4(

2[

12

]562

)[(3

2

2122

)]3

2

3()

5

2

3

2()

3[(

2

ntsDisplaceme

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

5Materials ofStrength

4

max

For long beams l >>c, elasticity and strength of materials deflections will be approximately the same

Note that according to theory of elasticity, plane sections do not remain plane

x

y

w

2c

2l

wl wl

Cantilever Beam Problem

x

y

NP

L

2c

22

3

434

3y

c

N

c

xyxy

c

P

)1(4

3

022

3

2

2

3

c

y

c

P

c

N

c

Pxy

xy

y

x

)1(4

3

022

3

Materialsof Strength

2

2

3

c

y

c

P

c

N

c

Pxy

xy

y

x

PLydyyPdyyLNdyyL

ydyyPdyyNdyy

cxcx

c

c x

c

c xy

c

c x

c

c x

c

c xy

c

c x

xyy

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

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

0),(),(

Conditionsy Boundar

same! theare solutionsnt displaceme two theTherefore

)23(4

)23(6

)( Materialsof Strength From

)23(424

3

4)0,(

0 case For the24

0)0,(,2

0)0,(

4

30

4

30

)0,(

424

3

)3

(2

)1(3

424

3

)22

3()(

1,)

22

3(

1)(

1

3233

323

32333

3

3

2

3

3

3

3

3

3

3

2

3

2

3

3

3

2

2

3

3

3

3

2

33

LxLxEc

PLxLx

EI

Pxv

LxLxEc

P

Ec

PLx

Ec

PL

Ec

Pxxv

NEc

PLL

Ec

PLvLvL

Ec

NuLu

Ec

PL

Ec

PL

x

Lv

vxEc

Pxy

Ec

N

Ec

xyPv

uyc

yy

cE

P

Ec

yPx

Ec

N

Ec

yPxu

c

N

c

Pxy

EEy

v

c

N

c

Pxy

EEx

u

ooo

oo

oo

oo

xyyx

Stress Field Displacement Field

Cantilever Tapered Beam

x

y

L

p

A

B

px xy x y

y

x

co t

( co t )tan ( )( tan )

2 12 2 2 1

B o u n d a ry C o n d itio n s

y xy x y

xL

xL

xyL

x p x T x x T x x

L y d y L y yd y p L L y d y p L

( , ) , ( , ) , ( , tan ) ( , tan )

( , ) , ( , ) , ( , )ta n ta n ta n

0 0 0 0

01

20 02

0

22

2

,221

221 2,tantan2,tan2

yx

yK

yx

xy

x

yK

yx

xy

x

yK xyxyyx

Stress Field

x = L

x = L

Solutions to Plane ProblemsPolar Coordinates

RS

),(,),( rfTrfT rr

Traction Boundary Conditions

Airy Representation

rrrrrr rr

1,,

112

2

2

2

2

Biharmonic Governing Equation

01111

2

2

22

2

2

2

22

24

rrrrrrrr

x

y

r

General Solutions in Polar Coordinates

011

2

2

2

22

24

rrrr

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

rrararaa loglog

Case icAxisymmetr 2

32

210

Thick-Walled Cylinder Under Uniform Boundary Pressure

r1

r2

p1

p2

Br

A

Br

Ar

2

2

2211 )(,)(

ConditionsBoundary

prpr rr

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

r1/r2 = 0.5

r/r2

r /p

/p

Dimensionless Distance, r/r2

Internal Pressure Case

Stress Free Hole in an Infinite Medium Under Uniform Uniaxial Loading at Infinity

TaT

x

y

2sin2

),(

)2cos1(2

),(

)2cos1(2

),(

0),(),(

CondtionsBoundary

T

T

T

aa

r

r

rr

2cos)(

loglog

242

234

222

21

23

2210

ararara

rrararaa

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

Ta 3)2/,(max

r/a

Ta

r/)

2,(

1

2

3

30

210

60

240

90

270

120

300

150

330

180 0

Ta /),(

Ta /),(

Stress Concentrations for Other Loading Cases

TT

T

TBiaxial Loading

TT

Unaxial Loading

TT

T

Biaxial Loading

K=3 K=4K=2

Stress Concentration Around Elliptical Hole

x

y

Sx

ba

0

5

10

15

20

25

0 1 2 3 4 5 6 7 8 9 10

Eccentricity Parameter, b/a

Str

ess

Co

nce

ntr

atio

n F

acto

rCircular Case (K=3)

()max/S

a

bSbx 21),0(

max

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

x

y

YX

r

C

sin)log(cos)log( 15121512 rbrrbrarra

150

150

sin),(

cos),(

adaaY

bdaaX

r

r

Conditions Boundary

]cossin[1

]sincos[1

]sin)2(cos)2[(1

1212

1212

15121512

bar

bar

abbar

r

r

0

]sincos[2

r

r YXr

0

sin2

r

r r

Y

222

2

222

32

222

22

)(

2cossin

)(

2sin

)(

2cos

yx

Yxy

yx

Yy

yx

yYx

rxy

ry

rx

Dimensionless Distance, x/a

y/(Y/a)

xy/(Y/a)

Normal Loading Case (X=0)

y = a

Notch-Crack Problems y

= 2 -

r

x

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

2

3

tsdisplacmen finite stresses,singular 0lim

0)0,(,0)0,(

r

rr 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

Contours of Maximum Shear Stress

Two-Dimensional FEA Code MATLAB PDE Toolbox

- Simple Application Package For Two-Dimensional Analysis Initiated by Typing “pdetool” in Main MATLAB Window

- Includes a Graphical User Interface (GUI) to: - Select Problem Type - Select Material Constants - Draw Geometry - Input Boundary Conditions - Mesh Domain Under Study - Solve Problem - Output Selected Results

FEA Notch-Crack Problem

(vonMises Stress Contours)

Curved Beam ProblemP

ab

r

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),(),(

ConditionsBoundary

sin)log( 3 rDrCrr

BAr

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

)log()( 2222

a

bbabaN

Dimensionless Distance, r/a

a/P

Theory of Elasticity

Strength of Materials

= /2 b/a = 4

Disk Under Diametrical Compression

+

P

P

D =

+

Flamant Solution (1)

Flamant Solution (2) Radial Tension Solution (3)

Disk Under Diametrical Compression

112

1

)1(

13

1

)1(

12

11

)1(

sincos2

cos2

sincos2

r

P

r

P

r

P

xy

y

x

0

2

2

)3(

)3(

)3(

xy

y

x

D

PD

P

222

2

)2(

23

2

)2(

22

22

)2(

sincos2

cos2

sincos2

r

P

r

P

r

P

xy

y

x

+ +

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

P

P

2

y

x

1r1

r2

Disk ResultsTheoretical, Experimental, Numerical

Photoelastic Contours(Courtesy of Dynamic Photomechanics Laboratory, University of Rhode Island)

Theoretical Contours of Maximum Shear Stress

D

Pyx

2),0(

Finite Element Model(Distributed Loading)