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Recap
Saint Venant Principle:(1)very small loading areacompared with the whole dimension. The affected area will
be much smaller than the unaffected area Aunaffected>>Aaffected. e.g in the tensile bar,
L>>a,in which the affected area will take roughly: z a.(2)Force replaced must be statically equivalent. The replacement must not changeeither the resultant force or resultant couple.
Combined shaft: As per Saint Venant (S-V) principle to define B.C.At z=0, z=L, Rr0 , 20 (flat end faces): ))(( drrddA
drrddrrdPR
zz
R
zz
2
0 0
2
0 0
))((
cos))((2
0 0
drrdrMMR
zzyy
0sin))((2
0 0
drrdrMR
zzxx
drdrQR
z
22
0 0
0zr
At r=R, 0 zL , 20 (cylindrical surface): 0 rzrrr
Bending BeamAt (front and back faces)
Lzh
yhb
x 0,22
,2
:
0 xzxyxx
At (top and bottom faces)
Lzh
yb
xb
0,2
,22
:
0 yzyxyy
At (side ends) Lorzh
yhb
xb
0,22
,22
: 0 zxzy ,
Moment:
2/
2/
2/
2/
h
h
b
b
zz Mdxdyy ,
2/
2/
2/
2/
0h
h
b
b
zzdxdyx ,
Resultant force
2/
2/
2/
2/
0
h
h
b
b
zzdxdy
Affected
zone
Affected
zoneunaffected zone
a
L
z z
z
y
x
dr
rdd
P
Q
MM
P
Q
y
x
MM
x
y
z
L
b
h
Symmetric line
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5.3 Thermal Stress AnalysisUse Di splacement method(u,v,w) to solve for equilibrium without mechanical body forces.
Step 1: Strain-Displacement relation
Mechanical strain mzzmyy
mxx ,,
Thermal strain: T
(Note that thermal strain does not affect the shear strains)
Total strain
mzzzz
myyyy
mxxxx
Tz
w
Ty
v
Tx
u
Tz
w
Ty
v
Tx
u
mzz
myy
mxx
Step 2: Use the Hookes law in mechanical strains:m
zzmyy
mxx ,,
mmxxzz
mmxxyy
mmxxxx
I
I
I
1
1
1
2
2
2
zxzx
yzyz
xyxy
2
2
2
where
T
z
wT
y
vT
x
uI
mzz
myy
mxx
m
1 .
Thusx-related components:
z
u
x
w
x
v
y
u
Tz
w
y
v
x
uT
x
u
xz
xy
xx
2
12
2
12
32
Step 3: Equilibrium Equation: 0
zyx
xzxyxx . Calculate different terms:
x
T
xz
w
xy
v
x
u
x
u
Tz
w
y
v
x
u
xT
x
u
xxTermst xx
322
321
22
2
2
2
2
yx
v
y
u
x
v
y
u
yyTermnd
xy
2
2
2
2
2
22
3z
u
zx
w
z
u
x
w
zzTermrd xz
So
2
222
2
2
22
2
2
2
2
322
zu
zxw
yxv
yu
x
T
xz
w
xy
v
x
u
x
u
zyxLHS xz
xyxx
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x
T
ux
I
x
T
z
u
y
u
x
u
z
w
y
v
x
u
xLHS
uI
32
32][][
21
2
2
2
2
2
2
21
)21()21)(1(
3)21(
)21)(1(3
)21)(1(
)21(
)21)(1(3
)1(2232
EEE
EEEE
Thus thermal Equilibrium Eqns:
0)21(
0)21(
0)21(
21
21
21
z
TEw
z
I
y
TEv
y
I
x
TEu
x
I
Compare with the Equilibrium equation (in terms of displacements) with body force.
0
0
0
21
21
21
z
y
x
fwz
I
fvy
I
fux
I
We can see that for the thermal deformation problems, the basic equations to be solved are
equivalent to a normal static problem with body force densities of
z
TE
f
y
TEf
x
TEf
z
y
x
)21(
)21(
)21(
Step 4: Boundary conditions
Displacement B.C. (if any): '',' wwvvuu and
Stress B.C.:
zzzyzxnz
yzyyyxny
xzxyxxnx
nmlnTE
nmlmTE
nmllTE
)21(
)21(
)21(
''
''
''
like a surface stress.
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Solution of a thermal static deformation problem without body forces and surface stresses but
with a steady state field of temperature change, T, is equivalent to a solution of a statics
problem subjected to a set of body forces and a surface stresses
When a plane stress solution is available, the corresponding plane strain solution can be
obtained by replacing E, , by21
E,
1and )1( .
When a plane strain solution is available, the corresponding plane stress solution can be
obtained by replacing E, , by2)1(
)21(
E,
1and
21
1
.
Thermal fit of a hollow disk onto a shaftSimilar to interference fit, the thermal fit
first increases the temperature of hollow
disk allowing the diameter increase to
greater than the shaft diameter, and then
inserts the shaft into the hollow disk.
Assume: Ti= temperature change at the
inner surface and To= temperaturechange at the outer surface of the disk. Consider a plane stress problem in this thin disk case.
Step 1: Boundary condition of the hollow disk (without insertion of shaft yet)
At oi RandRr : 0 rrr
At iRr : iTT and At oRr : oTT
Step 2: Analysis:
If the temperature change Tis axisymmetric (only a function of r) )(rTT , the
displacement should be axisymmetric: 0),( vruu
Step 3: Heat conductionproblem: Heat conduction equation in polar system:
0
1
dr
Tdr
dr
d
r
Integrate twice: BrAT ln Constants A and B can be determined by the thermal B.C.
iRr , iTT BRAT ii ln
oRr , oTT BRAT oo ln
ioio RRATT lnln io
io
RR
TTA
lnln
iio
ioiii R
RR
TTTRATB ln
lnlnln
Thus the temperature change is calculated as
)/ln(
)/ln(
)/ln(
)/ln(
lnln
lnln
lnln
lnln
lnln
lnln
lnln
ln)ln(lnln
lnlnln
lnlnln
0RR
rRT
RR
rRTT
RR
RrT
RR
Rr
TRR
RrT
RR
RRRr
RRR
TTTr
RR
TTT
i
ioio
oioio
iiio
o
oio
ii
io
iio
iio
ioi
io
io
2RO2Ri
t
2Rs
Rf
r
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Step 4: thermal deformationAs discussed in Week 5, the equilibrium equation as per displacement in the polar system is
dr
Tdf
r
u
r
u
rr
ux
)()1(
122
2
Rewrite it as
dr
Tdru
dr
d
rdr
d
r
u
r
u
rr
u
rr
u
rr
u
r
u
rr
u )()1(
1122
2
Integrate once the equation above
')(
)1(1
Cdrdr
Tddrru
dr
d
rdr
d
')1(
1CTru
dr
d
r
rCrTrudr
d')1(
Integrate once again the above equation
drrCrTrudrd
')1(
DCrrdrTru 2
)1( (note that Tis a function of r)
Thus r
DCrrdrT
rru
r
Ri
1
)1()(
Step 5: Strain:
22
1)1()1(
1)1(
r
DCrdrT
rT
r
DCrrdrT
rrr
u r
R
r
R
rr
ii
22
)1(01
r
DCrdrT
rr
u
r
uv
r
r
Ri
Step 6: Hookes law with thermal strain
TE
rrrr
)1(1
2, T
Err
)1(
1 2
Step 7: Stresses
Dr
EC
ErdrT
rE
TE
r
DCrdrT
r
E
r
DCrdrT
rT
E
r
R
r
R
r
R
rr
i
i
i
22
2222
222
1
)1(1
1
)1(1
)1(1
1)1()1(
1
Dr
EC
ErdrT
rE
r
R
rr
i
22
1
)1(1
1
Similarly: Dr
E
C
E
TErdrTrE
r
Ri
22
1
11
1
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Step 8 Apply stress B.C. to determine constants C & D(At oi RandRr : 0 rrr ):
01
)1(1
1)(
22
D
R
EC
ErdrT
RERr
i
R
Ri
irr
i
i
which leads to DR
Ci2
1
1
1
. And plug it into next B.C. equation
01
)1(1
1)(
22
D
R
EC
ErdrT
RERr
o
R
Ro
orr
o
i
01
)1(
1
1
1
1
1222
D
R
ED
R
ErdrT
RE
oi
R
Ro
o
i
o
i
o
i
R
Rio
iR
Rio
rdrTRR
RDrdrT
RRC
22
2
22
)1(,
)1(
Step 8 Calculate the temperature by using fit condition: sii RRruR Thus the required temperature change for fit can be calculated.
5.4 Stress and Deformation due to Contact
There are many contact examples: gear tooth, tyres on road surface, train wheel and bearings.
When the contact area is much smaller than the characteristic dimension of a component, the
component can be considered to be semi-infinitebounded by a surface:
The governing equilibrium equations are as follows
.0
,0
,0
=f+w+z
I)+(
=f+v+y
I)+(
=f+u+x
I
)+(
z21
y21
x
21
The general solution to this equation without acceleration and body force was derived by
Papkovich in 1932 and Neuber in 1934:
dp=pdA
dpn
dpt
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
x
y
z
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)()1(4
1
)()1(4
1
)()1(4
1
0
0
0
zyxz
zyxy
zyxx
zyxz
w
zyxy
v
zyxx
u
For short )()1(4
10
Rd
grad
where: Twvu ),,(d , Tzyx ),,( are all harmonic functions that satisfy Laplace
equation 02 (2
2
2
2
2
22
zyx
)
How to select harmonic equations is skill-based. The textbook provides examples (P.190).
A Half-Space under a Normal Concentrated Load
Step 1: B.C.Displacement B.C.:
Axisymmetrical about z at 0,0 yxz : 0 vu
When R : 0 wvu (infinitely far field)
Stress B.C.:
Except at the origin, at z=0: 0 zyzxzz
0Pdxdyzz
(S-V principle)
Step 2: Select harmonic functions (refer to the textbook)
zRCR
Czyx ln,1
,0 201
Step 3: Displacement
RB
RR
zAw
zRR
yB
R
zyAv
zRR
xB
R
zxAu
1
)(
3
)(
)(
3
2
3
3
Step 4: Stresses
3
2
3
22
2
3
222
3
22
2
3
222
3
232
)()(23
2
)()(23
2
R
zBR
x
R
zA
zRR
y
zRR
zxB
R
y
R
zA
zRR
x
zRR
zyB
R
x
R
zA
zz
yy
xx
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3
2
3
3
2
3
235
232
232
)(
)2(2
6
R
xB
R
z
R
xA
R
yB
R
z
R
yA
zRR
RzxyB
R
xyzA
yz
yz
xy
Step 5: Apply B.C. to determine the constants A and B
Except at the origin, at z=0: 0 zyzxzz
0
4)2(4
BA
PBA
4
4
PB
PA
Step 6: Determine the stress and displacement functionsSubstitute them back to stress and displacement functions, one can determine the stress and
displacement completely.
R
P
R
zPw
zRR
yP
R
zyPv
zRR
xP
R
zxPu
1
4
)2(
4
)(44
)(44
3
2
3
3
5
2
5
2
235
5
3
22
2
3
222
3
22
2
3
222
3
2
3
2
3
)(
)2(
22
32
3
)()(23
2
)()(23
2
R
xzP
R
yzP
zRR
RzxyP
R
xyzR
zP
zRR
y
zRR
zxP
R
y
R
Pz
zRR
x
zRR
zyP
R
x
R
Pz
yz
yz
xy
zz
yy
xx
A Half-Space under a Tangential Concentrated Load
Step 1: B.C.
Except at the origin, at z=0: 0 zyzxzz
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0Qdxdyzx
Step 2: Select harmonic functions
zRC
zRR
xC
RC zyx
1
,)(
,0,1
3021
After derive the displacement and stress functions, apply B.C. to determine the three constants
,)1(8
)21(,
)1(8,
)1(8
)21( 2
QC
QB
QA
Step 3: Displacement
)()21(
4
)(
)21(
4
)()21(1
4
2
22
2
2
2
2
zR
x
R
xz
R
Qw
zR
xy
R
xy
R
Qv
zR
x
zR
R
R
x
R
Qu
Step 4: Stresses
5
2
5
222
2
2
5
2
2
222
2
222
2
2
2
3,
2
3,
2
)(
213
2
,2
3
23
)(
213
2
2
)(
213
2
R
Qzx
R
Qxyz
zR
RxxR
zRR
x
R
Qx
R
Qxz
zR
RxxR
zRR
y
R
Qx
zR
RyyR
zRR
x
R
Qx
yzyzxy
zz
yy
xx
Solutions to the Contact Problems
The frictionless contact can be solved by using approach to normal concentratedforce on a half-space
The frictional contact can be solved by using approach to tangential concentratedforce on a half-space
No penetration between the two contact bodyEffective Youngs Modulus:
1
2
22
1
21 11*
EEE
Two balls in contact
*)(4
3
21
213
E
P
RR
RR
a ,
2
21
213
*
)(
16
9
E
P
RR
RR
R22a
P
P
E1,1
E2,2
R1
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2
2
0 1a
rqq (distribution of contact stress)
where PERR
RRq
2
2
21
21
3
30 *
6
A sphere in contact with flat half space ( 2R )
*4
3
*)1/(4
3
*)//(4
31
1
1
2221
13
E
PR
E
P
R
R
E
P
RRRR
Ra
2
1
3
*
1
16
9
E
P
R
2
2
0 1a
rqq (distribution of contact stress)
where PER
q 2
2
13
30 *16
A sphere in contact with concave half space ( 22 RR )
*)(4
3
*)(4
3
12
21
21
213
E
P
RR
RR
E
P
RR
RRa
2
21
123
*
)(
16
9
E
P
RR
RR
2
2
0 1
a
rqq (distribution of contact stress)
where PERR
RRq
2
2
21
12
3
30 *
6
Two Parallel Cylinders
*)(
4
21
212
E
p
RR
RRb
407.0
2ln
1407.0
2ln
12 2
2
221
1
213
b
R
Eb
R
E
p
2
2
0 1bxqq (distribution of contact stress)
where pERR
RRq *
1
21
2120
Cylinders with flat half-space
*
41
2
E
pRb
2
2
0 1b
xqq where
1
20
*
R
pEq
R2
R1
E2,2
E1,1
pl
b b
qq0
P
E1,1
R2
R1
P
E1,1
R2
R1
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Chapter 6 Stress Function Method
6.1 Airy Stress Function
Basic equations: Review the basic equation in 2D plane stressproblems:
Equilibrium Eqn:
0
0
yyyyx
xxyxx
fyx
fyx
Strain-disp:
y
u
x
v
y
v
x
u
xy
yy
xx
2
1
Compatibilityyxyx
xyxxyy
2
2
2
2
2
Hookes law:
xyxyxy
xxyyyy
yyxxxx
EG
E
E
)1(21
1
1
xyxy
xxyyyy
yyxxxx
G
E
E
2
2
1
1
Stress B.C.
yyyxy
xxyxx
Fml
Fml
Displacement B.C.
*
*
vv
uu
Stress method to solve for plane stress problemuse stress as primary variableWhen 0 yx ff ,: Differentiate the 1
stequilibrium equation w.r.t.xand the 2ndw.r.t.y:
0
0
2
22
2
2
2
yyxyxy
xyxyxx
yyyxyyyx
xyxxxyxx
xyyx
xyyyxx
2
2
2
2
2
2
which is the compatibility in terms of stress. Using Hookes law leads to:
xyyyxxxxyy
EyxEyEx
1
211
2
2
2
2
2
0)1(2
2
2
2
2
2
2
2
2
2
yxyyxx
xyyyxxxxyy
0)1(2
2
2
2
2
2
2
2
2
2
2
2
yxyyxx
yyxxyyxxxxyy
0)(2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
yyxxyyxx
yyyyxxxxyyxxxxyy
0)()( 22
2
2
2
2
2
2
2
2
2
2
2
yyxxyyxxyyxxxxyy
yxyxyx
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Now the key equations need to be considered are:
0)(
0
0
2yyxx
yyyx
xyxx
yx
yx
To solve for the above differential equations, one can assume a solution to the stress, namely
stress function (or Airy stress function) such that it can satisfy the equilibrium equations. One
option can be:
2
2
yxx
,2
2
xyy
,yx
xy
2
Equilibrium equations
0
0
2
3
2
3
2
22
2
3
2
32
2
2
xyxyxyyxxyx
yxxyyxyyxyx
yyyx
xyxx
Compatibility: 02 224
4
22
4
4
4
2
2
2
22
yyxxxy
Now we need to solve for a bi-harmonicequation:
024
4
22
4
4
4
yyxx
Polar system: The stress functions can be written as:
2
2
2
11
rrrrr , 2
2
r
,
2
1
rrr
Bi-harmonic functions have been studied extensively. Tables 8.1 and 8.2 in the textbook
include some typical functions.
B.C. for stress function
To this end, all the stress functions selected for a specific problem need to satisfy B.C. as well.
We need to see what stress function could meet the requirement of B.C.
6.2 Rectangular plate under bending
Pure Bending Problem
MM
h/2
h/2
L
x
y
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Step 1: B.C.
2,0 h
yLx : 0 yxyy (no surface stresses, but 0xx )
22,and0
h
y
h
Lxx : 0xy , (pure bending)
As per S-V principle: 02/
2/
h
h
xxdy , (no axial force) Mydyh
h
xx
2/
2/
(bending moment)
Step 2: Observe the B.C. and Select stress function
Now we need to choose proper stress function to satisfy the above B.C. From Table 8.3, we
can see stress function #8 leads to a linear xxat the two endsand all other stress
components are zero. Therefore, we choose3
ay
It will be found that this stress function satisfy all the B.C. well.
Step 3: Use B.C. to determine stress functionTo determine constant a, the bending moment B.C. can be used:
ayayyy
xx 6)( 3
2
2
2
2
Mhahh
adyyaydyayydy
h
h
h
h
h
h
xx
3332/
2/
22/
2/
2/
2/ 2883
1666
Thus:3
2
h
Ma
So the stress function becomes: 33
2y
h
M
yh
My
h
Mayxx 33
12266
,
, 032
2
2
2
ay
xxyy
, 03
22
ay
yxyxxy
Cantilever Beam Problem
Step 1: B.C.
2,0 h
yLx : 0 yxyy (no surface stresses, but 0xx )
22,0
hy
hx : 0 vu (fully-clamped)
22,
h
y
h
Lx : 0xx (no axial force) Vdy
h
hxy
2/
2/ (S-V principle)
Vh/2
h/2
L
x
y
Vh/2
h/2
L
x
y
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Step 2: Observe the B.C. and Select stress functionThe clamped end can be separated (F.B.D.). We need linear normal (but vanish in total) and a
uniform shear.
Uniform shear: #4: axyall around the plates #10: uniform shear in top/bottom but non-uniform at other two ends. Thus superposition can cancel the top/bottom shear. #10 also leads to a normal stress xx at right end (x=L) but B.C. requires only normal
at left end. We therefore need other function to cancel the right end normal stress.
#8 is just what we wantThus 33 cybxyaxy
Step 3: Calculate the stress
23322
2
2
33
2
2
2
2
3
0
66
byacybxyaxyyxyx
x
cybxycybxyaxyyy
xy
yy
xx
Step 4: Applied the B.C. to determine the unknowns in the stress function:
Vhbhahbhabyaydybyady
b
h
ab
h
a
h
ba
h
yLx
cbLcybLyh
yh
Lx
h
h
h
h
h
hxy
xy
xx
33
2/
2/3
2/
2/
22/
2/
222
82823
4
3
04
3
232,0
006622
,
I
Vha
8
2
,I
Vb
6 ,
I
VLc
6
Step 5: Express stress
Thus:
2
4
42,0,)( y
h
I
VyxL
I
V
xyyyxx
6.3 Stress concentration around a circular hole
Step 1: Transfer to Polar system:
Ta
b
T
8/12/2019 mechanics of solids week 7 lectures
15/15
Week 7 MECH3361
15
It is convenient to use polar system. To do so, a stress needs to be transferred:
2sin
2
1cossin)sin(cos2sincos)(
2cos12
1coscossin2sincos
22
222
TT
TT
xyxxyyr
xyyyxxrr
Step 2: B.C.:
At ar : 0 rrr
At br : 2sin2
12cos1
2
1TT rrr
The superposition can be applied to the second B.C at br .
2costan
2cos2
1
2
12cos1
2
1
toonalproportatitCons
rr TTT
In this boundary ( br ):
Part 1: Trr2
1 and 0r
Part 2: 2cos2
1Trr and 2sin
2
1Tr
Step 3: Select the stress functionFrom Table 8.4, the stress function is to sum #2, 4, 15, 16 and 17. Determine cifrom BC.
2cos)(ln 2
52
4312
1 rcrccrcrc
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
rr
When 12cos , reaches the maximum
Ta
aT
a
aTar 3cos
31
21
2 4
4
2
2
Thus T3max
Stress concentration factor k: nominalk max (k=3 in this small hole case)
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