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8.22.
p
rr
rr
A
r
A
p
r
A
uEEE
p
r
A
r
A
Brr
r
A
p
r
A
E
B
E
ruru
rr
r
r
A
p
r
A
E
up
r
A
r
A
p
r
A
r
A
rB
E
uB
r
rr
rr
rr
rr
2
1
2
2
2
2
2
1
)2(
2
2
)2(
2
1
)2(
)1(
221
2
2
)2(
2
1
)2(
)1(
1
)2(
1
)1(
2
2
)2(
2
2
1
)2(
2
2
)1(
1
11
1
)2(
1
)1(
1
2
2
)2(
2
2
)2(
2
2
)2(
2
2
)2(
2
)2(
)2(
2
2
)2(
2
)2(
)2(
)1(
1
11
)1()1()1()1(
)21(
)21(
0)21(
and core) rigidin n deformatio (no0 and, :problem For this
)()(
)21(
1)21)(1(
)()(
: @condition matchingWith
)21(
1
,,
:2 Material
)21)(1(
,
:1 Material
8.20 Exercise fromSolution
−+
−−
==
+−−−
===→
−−==
+−−−
+
=
−+
=
=
+−−−
+
=−−−=−−=
−+
===
8.23.
r
r
rr
pr
ppp
rr
prpr
rrr
pprr
+
−
=
==
−
−
+
−
−
−=
1
0 and withcase For the
1
)(
(8.4.3) from solution General
2
2
2
2
1
2
2
2
1
21
2
1
2
2
2
2
21
2
1
22
1
2
2
12
2
2
2
1
2
8.24.
Found Is Field Stress theand Determined Are Constants All
)(2
)log21()log21(
,
)/log(2
:Constantsfor Solving
02)log1()(,02)log1()( :ConditionsBoundary
2)log23(,2)log1(
:Solution Field Stress General
8
8
)0,()2,( :Condition Jump Cyclic
4
:MotionBody -Rigid without (8.3.9)Solution
3
22
22
23
22
22
1
2
2
1
32
2
1
3
2
2
1
32
2
1
3
33
32
−
+−+
=
−
=
=+++==+++=
+−+=+++=
==
=−
=
a
rr
rrrr
aa
rr
rrrr
a
a
r
a
rara
r
a
rar
a
r
a
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r
a
ra
E
ara
E
r
rruru
a
E
r
u
io
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8.25.
2
with sexpression same by thegiven is (2) problem osolution t The
2sin
23
1
2
2cos
3
1
2
1
2
2cos
43
1
2
1
2
(8.4.15) 7,-8 Example fromdirectly follows (1) problem osolution t The
,,
bygiven is problem loading biaxial theosolution t theshown, asion superposit Using
2
4
2
2
2
4
4
)1(
4
4
2
2
)1(
2
2
4
4
2
2
)1(
)2()1()2()1()2()1(
+→
+−−=
+−
+=
−++
−=
+=+=+=
r
r
rrrrrr
r
a
r
aT
r
aT
r
aT
r
a
r
aT
r
aT
=
+
(1)
(2)
8.26*.
r
r
r
a
r
aT
r
aT
r
rrrrrr
−++
−=
+=+=+=
2cos
1
2
1
2
2cos
43
1
2
1
2
(8.4.15) 7,-8 Example fromdirectly follows (1) problem osolution t The
,,
bygiven is problem loading biaxial theosolution t theshown, asion superposit Using
2
4
4
4
2
2
)1(
2
2
4
4
2
2
)1(
)2()1()2()1()2()1(
=
+
(1)
(2)
8.27.
,)21(
1)21)(1(
conditions matching gives now which and,let problem For this
)()(
)21(
1)21)(1(
)()(
: @ conditions Matching
)21(
1
,,
:2 Material
)21)(1(
,
:1 Material
:8.20 Exercise From
2
)2(
)1(
2
2
)2(
2
2
)1(
1
11
12
2
2
)2(
2
1
)2(
)1(
1
)2(
1
)1(
2
2
)2(
2
2
1
)2(
2
2
)1(
1
11
1
)2(
1
)1(
1
2
2
)2(
2
2
)2(
2
2
)2(
2
2
)2(
2
)2(
)2(
2
2
)2(
2
)2(
)2(
)1(
1
11
)1()1()1()1(
+=
−+−
+
=
−+
=→−→
−−==
+−−−
+
=
−+
=
=
+−−−
+
=−−−=−−=
−+
===
T
a
A
BT
a
A
E
B
E
arrTp
p
r
A
r
A
Brr
r
A
p
r
A
E
B
E
ruru
rr
r
r
A
p
r
A
E
u
r
A
p
r
A
r
A
p
r
A
rB
E
uB
rr
rr
rr
rr
8.28.
TTTTTTTTar
rTT
r
a
E
uTT
r
a
TT
r
a
Tap
rr
rr
AarTpr
p
rr
rr
A
r
r
A
p
r
A
E
up
r
A
r
A
p
r
A
r
A
r
rr
rr
=+−−=−=+−==
−−
−
−
+
=+
−
−=+
−
=
−=
−+
−−
==−→→
−+
−−
=
+−−−
+
=−−−=−−=
2)21(,)1(2)21(:at Stresses
)21(
)21(
1
,
)21(
,
)21(
)21(
)21(
)21(
and,let problem For this
)21(
)21(
)21(
1
,,
8.22 Exercise fromSolution
2
2
2
2
2
2
2
2
2
2
2
1
2
2
2
2
2
1
)2(
12
2
1
2
2
2
2
2
1
)2(
2
2
)2(
2
2
)2(
2
2
2
2
)2(
2
)2(
2
2
)2(
2
)2(
8.29.
:Simplify andFirst Laplacian Do
2sinlog2cos
2
,sin2sinlogsin2
2sin
2
1
2coslog2sin,sin2sinlogsin2
sin2sin
2
1
logsinsincossinlogsin
2
22
2
,
22
,
22
2
22
2
oo
oo
r
oo
r
r
r
r
r
r
r
r
r
r
r
−
=++
=
−+
=−+
=
−+
=−+
=
8.30.
solved. is problem the thusand
)(4)2sin2(sin)2cos2(cos
)2cos2(cos)](2)2sin2[(sin
)(4)2sin2(sin)2cos2(cos
)](2)2sin2[(sin)2cos2(cos
gives and for relations twoprevious theSolving
)2cos2(cos)](2)2sin2(sin[sin),(
)2cos2(cos)](2)2sin2[(sincos),(
0,]sincos[
2
00),(),(),(),(
:ConditionsBoundary
]cossin[
1
]sincos[
1
]sin)2(cos)2[(
1
sin)log(cos)log(
:Solution Flamant General
222
15
222
15
1515
1515
1515
1515
1212
1212
1212
15121512
15121512
−−−+−
−−−+−
=
−−−+−
−+−+−
=
−+−+−−=−=
−+−+−=−=
==−=
======
−=
+=
−++=
+++=
YX
b
YX
a
ba
badaaY
abdaaX
ab
r
barrrr
ba
r
ba
r
abba
r
rbrrbrarra
r
r
rr
rr
r
r
8.31.
X
Aruy
ditionsfixity con
y
eK
KCrBA
E
X
r
E
X
E
X
u
BAr
E
X
E
X
u
KCrrgBA
E
X
f
Krgrrgdf
E
X
E
X
f
r
u
r
u
u
r
e
rgdfr
E
X
E
X
ufr
E
X
E
X
u
Er
X
E
u
rr
u
e
fr
E
X
u
Er
X
Er
u
e
r
X
r
r
r
r
r
r
r
rr
r
r
rr
)1(
0)
,( axis- theof
movement verticalzero enforce could we, Using
axis.- on the vanish strains all hence and stresses all that Note
0satisfy to0 case, loading vertical with theAs
sincossin
)1(
sinlog
2
cos
)1(
cossincoslog
2
sin
)1(
)(,cossinsin
)1(
)(
constant)()()(sin
2
sin
2
)(
0
11
2
)()(logsin
2
sin
2
)(logcos
2
cos
2
cos
2
)(
11
)(logcos
2
cos
2
)(
1
0,cos
2
−
==
==
++−+
+
+
+
−
−=
++
−
−
−=
+=++
−
−=
==
−=+
−
+
=
=−
+
=
+−
+
=−
+
=
=−=
+=
+
−=
−=−=
=
==
−=
8.32.
−=
=
−=
−
−=
+
+
−
−=
−=
−=
−=
−+
−=
−+
−=+−=
→
==
=
=
−−
→
→→
−−
2
2
2
1
22
1
11
sin
2
0
cossin
4
relationsation differenti standard thefrom follow stresses theand
)cossin(
M
tan
M
tan
P
),(),(
lim
),(),(
lim),(),(lim
is problemmoment for thefunction stress the,0 aslimit the takingandion superposit Using
tan),( where,),(tan
P
cos
P
bygiven isorigin at the acting force downward verticalsingle for thesolution The
r
M
r
M
x
y
yx
xy
x
y
x
x
d
x
d
x
F
M
d
yxFydxF
M
d
yxFydxF
Pdydxyx
d
x
yx
yxFyxPF
x
y
xr
r
r
d
dd
M
8.33.
=
−=−=
−===
+
=−===
−=+=
+=
244
244
2
244
2
24
2
244
),(
20),(or0)0,(
:ConditionsBoundary
2cos2
)/(,0,
2sin4
11
2sin
M
b
M
aMrdrr
barr
r
ba
r
r
b
r
r
ba
r
rr
rrrrrr
8.34*.
,5,3,1 where),
(
as zero decay to Stresses
sin,][cos
00)0,( :CondtionBoundary
1sin
andcoscos
cos)0,( :CondtionBoundary
][cos
][cos
/
max
//
/
2
2
2
2
2
=
−==
→
−=+
−=
==+
−=
−+
−
=−=
=
=
=
−=−
−=
+−==
+=
−
−−
−
−
−
nye
p
y
L
n
r
ye
L
x
L
p
ey
L
x
L
p
Lp
DD
Lp
x
e
L
y
D
Lp
L
x
L
Lpp
B
LL
x
pxB
L
x
px
eDyBx
eDyBx
Ly
o
xy
Ly
o
xy
Ly
o
y
oo
xy
Ly
o
xyxy
oo
o
oy
y
xxy
y
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
y/l
|xy/po|
MATLAB Plot of Shear Stress
8.35.
satisfiedalready 0),( and ),(
2sin2
2cos
02cos220),(
2sin2
2sin2),(
:ConditionsBoundary
2sin2)/(
2cos22
2cos22
11
04
11
2cos4,2sin2
2cos22,2cos22
2cos
2212
2121
21
212
212
2
4
2
2
2
2
21
2
21
212212
2
21
2
2
=−−=−
−==+=
===
=−=
+==
−=+=
==++=
−=−=
+=+=
+=
rSr
S
aaar
S
aSaSr
ar
aa
aa
r
r
a
r
r
rara
aarara
rara
r
r
rr
rr
rr
rrr
rrr
8.36*.
)sin()sin()(2cos1
2
2
)2cos2cos2sin2(sin22
2
]2cos2[cos
2
)2sin2(sin
2
2
]2cos2[cos
2
)]2sin2(sin)(2[
2
)]2sin2(sin)(2[
2
122121
1212
2
12
2
12
2
2
max
12
1212
1212
−
=−
=−−
=
+−
=
−
+
−
−=
+
−
=
−
=
−−−
−=
−+−
−=
ppp
p
pp
p
p
p
xy
yx
xy
y
x
8.37.
1212
2
2
12
12
2
1
3
3
22232
2sin2sin)(2
2
cos
2
cos
2
2cos2cos
2
cossin
2
cossin
2
2cos2cos
sin
sin
log4
2
cossin
sin
cos2
sin
cos2
sin
cos2
sin/,But
sincos
2
cossin,sincos
2
sin,cos
2
cos
0,cos
2
: Since
2
1
2
1
2
1
2
1
−+−
−=
−=
−=
−
=
−=
−=
+−
=
−
−=
−=
−=
=
−==
−==
−==
==
−=
t
d
t
d
t
d
t
d
t
d
t
d
t
d
t
d
t
t
d
trddX
r
X
r
X
r
X
r
X
xy
xy
y
y
x
x
rxyryrx
rr
8.38.
ysx
ysx
yx
yx
ysx
sxsty
ysx
spy
ysx
ysxst
ysx
ysp
yx
Xxy
yx
Yy
ds
ysx
sxst
ds
ysx
sxspy
ds
ysx
sxst
ds
ysx
ysxsp
d
yx
Xx
yx
yYx
tdsXpdsY
xyxy
a
a
a
a
y
a
a
a
a
x
xx
−−
−−
+−
+−
+
+
+−
−
+−
+−
−
+−
+
+
+−
−
−
+−
−
−=
+−
−
−
+−
−
−=
+
−
+
−=
==
222
2
222
2
222
2
222
2
222
2
222
3
222
2
222
3
222
2
222
3
222
3
222
2
222
3
222
2
222
3
222
2
1
])[(
])[(
)(
)(
])[(
))((2
])[(
)(2
])[(
)()(2
])[(
)(2
)(
2
)(
2
])[(
))((2
])[(
))((2
])[(
)()(2
])[(
)()(2
)(
2
)(
2
, Using28,-8 Exercise and (8.4.36) From
