Download - 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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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 /),(
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Displacement Field (Plane Stress)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
Example 8.4 Beam Problem
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
Example 8.4 Beam Problem
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
Displacement Field (Plane Stress)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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,(
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
=
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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/,(
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
Photoelastic Contact Stress Fields
(Uniform Loading) (Point Loading)
(Flat Punch Loading) (Cylinder Contact Loading)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
Disk Under Diametrical Compression
+
P
P
D=
+
Flamant Solution (1)
Flamant Solution (2) Radial Tension Solution (3)
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
Applications to Granular Media ModelingContact Load Transfer Between Idealized Grains
(Courtesy of URI Dynamic Photomechanics Lab)
P
P
P
P
Four-Contact Grain
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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
Elasticity Theory, Applications and NumericsM.H. Sadd , University of Rhode Island
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