chapter 8 two-dimensional problem solution.ppt
TRANSCRIPT
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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
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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
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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
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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)
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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
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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),(
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
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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
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
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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 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
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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 /),(
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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
Displacement Field (Plane Stress)
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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
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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
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
lcq
Do
coshsinh2
sinh
2
2
Stress Field
l
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Example 8.4 Beam Problem
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
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Example 8.4 Beam Problem
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
Displacement Field (Plane Stress)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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,(
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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
=
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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
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Stress Concentration Around Stress Free
Hole in Orthotropic MaterialChapter 11
Isotropic Case
Orthotropic Case Carbon/Epoxy
x(0,y)/S
SS
x
y
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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)
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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
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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
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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/,(
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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
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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
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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
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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
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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
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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
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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
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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
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Photoelastic Contact Stress Fields
(Uniform Loading)(Point Loading)
(Flat Punch Loading) (Cylinder Contact Loading)
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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<
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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
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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
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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
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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
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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
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Disk Under Diametrical Compression
+
P
P
D=
+
Flamant Solution (1)
Flamant Solution (2) Radial Tension Solution (3)
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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
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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
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Applications to Granular Media ModelingContact Load Transfer Between Idealized Grains
(Courtesy of URI Dynamic Photomechanics Lab)
P
P
P
P
Four-contact grain