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14.1.
( )
( )
0
11
22
1
0
1
2
111
22
1
0
11
12
1
1
22
11
2
1
0
11
12
11
:constant and)( of case special For the
1.-7 Table conversion use and (14.1.5)relation strain plane start withsimply also could We
0
1
2
11
givesfunction stressAiry usual thegIntroducin
0
1
2
11
give would this(7.2.2)relation using stress planefor and
2 isequation ity compatibil governing The
2
2
2
2
2
2
2
2
2
2
4
4
22
4
4
4
2
3
2
2
2
2
2
2
2
2
3
3
4
4
22
4
4
4
2
3
22
4
2
2
2
2
2
3
22
4
2
2
2
2
3
3
4
4
4
4
2
3
22
4
22
4
4
4
2
2
2
2
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=
−
+
+
+
+
+
=
+
−
+
+
+
+
=
+
++
−
−
−
+
+
+
=
+
++
−
+
−
==
=
+
+
−
+
−
=
+
−
−
+
−
=
+
yx
E
xyx
xEx
yyxx
E
yx
Ex
y
E
xx
E
xx
Ex
yyxx
E
yx
Ex
yx
E
y
E
x
yx
Ex
yx
E
x
E
xx
Ex
yx
E
yx
Ex
yx
E
yx
E
y
E
y
E
x
E
x
xEE
yxEyx
x
E
y
E
yy
E
x
E
x
EyxEE
y
EE
x
yx
e
y
e
x
e
xyyxxy
xy
x
y
14.2.
02)1(
0
022)1(
0)2(
0
:equations mequilibriuin sHooke' of form new theUsing
)()(,)1()(,
)2()2(
)2()2(
2
2
2
2
2
2
2
2
2
2
=
+
+
+
+
+
+
=
+
=
++
+
+
+
+
=
+
+
+
+
=
+
=+=
+
=
+
+=
+
+=
+
+=
+
+=
x
v
y
u
a
yx
u
x
v
y
v
y
kax
yx
x
u
ka
yx
v
y
u
x
u
x
kax
x
v
y
u
yy
v
k
x
u
k
x
yx
xkxaxx
x
v
y
u
x
u
k
y
v
k
x
u
y
vy
v
k
x
u
k
y
v
x
u
oo
yxy
oo
xy
x
oxy
y
x
14.3.
( ) ( )
0)1()(2
)(
)1(2
()(
1
)(
1
11
)1(2)(
1
)(
1
)(
1
1
2)(
1
)(
1
2:Relationity Compatibil
11
)(
1
)(
1
)(
1
)(
1
11
)( constant,
2
2
22224
,,
,,,,,,,,,,
,
2
,,,
2
,,,,
,
2
,,
2
2
,,
2
2
2
2
2
2
2
,
,,
,,
2
=
+−+
−
−
+
−=
−−−−−−−+−
−+−=
−
−−
+−
+
−
=
−
+
−
=
+
+
−=
+
=
−=−=
−=−=
=
=
==
==
y
aa
x
a
a
E
a
E
a
a
EE
x
E
E
Ex
E
E
ExE
EyxE
x
E
y
yx
e
x
e
y
e
EE
e
EE
e
EE
e
E
a
x
E
E
a
x
E
E
aEeaE
x
E
eExE
xyyxxyy
yyxxyyxxxxyyxxxxyyxxxxxxxxyyyyyy
xyyxxyyyyxxyyxxxxxxyyyyyy
xyyyxxxxyy
xyy
x
xyxyxy
yyxxxyy
xxyyyxx
ax
o
ax
o
14.5.*
x
0 0.5 1
0
1
x
0 0.5 1
0
1
x
y
0
0.5
x
y
0
0.5
14.6.
++−=+=
======
+−−=+−=
==
−=
+=
+
+=
==
+
+
+
−=++
−=+−=−=
+
+
=+==
===
+=+=
=
===
=
+
+
−
+
−
22
,)1(
0 0)0,0(;0 0)0,0( ;0 0)0,0( :ConditionsFixity
2
)(;)(
constant)()(0)()(02
)(
2
)()1()(
)(
)1(
)(
0,by given 1,-14 Example as same theare stresses The
constants are and where,)1/(or
1
twicegIntegratin
0
1
)6.1.14(, 2/ and ,constant ,)( case For the
0
1
2
11
)6.1.14( stress, planeFor
22
2
2
x
2
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
xy
Ky
E
T
vxKy
E
T
u
vvuu
vx
x
E
TK
xguyyf
yfxgx
E
TK
xgyfx
E
TK
e
x
v
y
u
xg
y
Ky
E
T
xgdyKy
E
T
xgdy
E
T
v
E
T
y
v
yfx
E
KyT
yfdx
E
T
u
E
T
x
u
T
KEKyEEBAy
E
E
dy
d
TyyEE
yxEyx
x
E
y
E
yy
E
x
E
x
oo
ozoo
oo
o
oo
o
oo
xy
xy
oo
o
xyy
oo
14.7.
0,
/
become stresses theso and
nintegratio of constantsarbitrary are and where,
)/(
)/(
)/(
solution with 0
)(
2
dy
dY
becomes ODE previous Letting
02)(0)(
)( and ,constant ,)/(1)( with case For the
0
1
2
11
)6.1.14( stress, planeFor
2
1
2
2
212
1
2
2
2
1
3
3
2
1
3
3
3
4
2
2
22
2
2
2
2
2
2
2
2
2
2
2
2
==+
+
−=
=
+
+
−=
+
=
+
==
+
+
=
=
+
+=
+
==+==
=
+
+
−
+
−
xyyx C
ABy
C
dy
d
CCC
ABy
C
dy
d
ABy
C
dy
d
ABy
C
YY
BAy
A
dy
d
Y
dy
A
dy
BAy
y
BAy
dy
yBAyyEE
yxEyx
x
E
y
E
yy
E
x
E
x
14.8.
0)(
1
)(
1
)(
)(
1
0)((8.4.76)
11
))((
)(
1
11
))((
)(
1
stress plane Law sHooke'
2
2
2
2
2
2
2
2
=
−
−
−
=−
−
=−=
−
=−=
dr
d
r
dr
d
rrEdr
d
r
dr
d
r
rEdr
d
ere
dr
d
dr
d
r
dr
d
E
r
rE
e
dr
d
dr
d
rE
r
rE
e
r
r
rr
14.9.
( )
( )
( )
( )
−+−
=
−++−
−
−=
−++−
−
−
=
−+−−
=
++−
−+
−
++−
−
=
−
=
++−
+
−+
−=
−
=
++−
+
−+
−=
+−
−+
+
−
−
=
+
−
=
+
+−
+
+
+
−
−
=
+
−
=
+=
++
+−−−−
++−−+−−+−++−
++−++−
++−−+−−+−++−
++−−+−−+−++−
−+−−+−
−+−++−
−+−++−
−+−++−
−+−++−
+−+−
)2()(
(
gives realtions stress theinto ngsubstitutiBack
)2(
2
1
)2(
2
2
2
22
2
1
and for equations two theSolving
1
2
2
2
2
)(
1
2
2
2
2
)(
:ConditionsBoundary
2
)(
1
2
)(
1
11
2
)(
2
)(
11
gives stresses for the relations theUsing
:bygiven werentsdisplaceme theproblemcylinder hollow For the
2/)2(2/2/2/)2(
2/)2(2/2/
2/)2(2/)2(2/)2(2/)2(
2/)2(2/)2(
2
2/)2(2/)2(2/)2(2/)2(
2/)2(2/)2(2/)2(2/)2(
2/)2(2/)2(
2
2
2/)2(2/)2(
2
2/)2(2/)2(
2/)2(2/)2(
22
2/)2(2/)2(
22
2/)(2/)(
nkrpbapba
rba
rbpbarpbapba
ab
babakn
bpap
E
B
babakn
E
knkn
apbp
kn
E
A
BA
p
E
b
kn
Bb
kn
Apb
p
E
a
kn
Ba
kn
Apa
r
kn
Br
kn
A
E
dr
du
r
uE
r
kn
Br
kn
A
E
r
u
dr
duE
BrAru
k
o
kn
i
nk
kk
nknn
i
o
o
kk
r
knknknkn
kn
i
kn
o
knknknkn
knknknkn
kn
o
kn
i
o
knkn
or
i
knkn
ir
knkn
knkn
r
knkn
14.10.
−+
−−
+
+−
−+
−
=
−+
++−−
+
+−
−+
−
=
−
−
−=−
−
−=
=
−+
++−+−
+
+−
−+−
−
=
−++−
−
−=
+−−++−
−+
++
+−−−−
+−−++−
−+
+
+−−−−
++
++
+−−−−
++++
+−−−−
2/)2(2/)2(
2/)2(
2/)(2/2/2/2/)2(
2/)2(2/2/
2/)2(2/)2(
2/)2(
2/2/)2(
2/)2(2/2/
2/)(2/)(2/2/
2/)2(2/2/2/)2(
2/)2(2/2/
2/2/)2(2/)2(2/2/)2(2/
2/)2(2/2/
2
2
2
2
2
)2)((
2
)2()(
((
0 :only pressure internal of case special For the
2
)2)((
2
)2()(
(
:solution general theFrom
nkknk
kk
nk
i
i
nkkkk
i
nk
kk
nknn
nkknk
kk
nk
i
kk
i
nk
kk
nknn
r
o
o
nk
i
nkkk
k
o
kn
i
nk
kk
nknn
kk
i
nkk
o
kn
o
knk
kk
nknn
r
rb
nk
nk
r
nk
nk
ab
ap
nk
nkpabba
nk
nkrpba
ab
rba
rbr
ab
ap
rbpba
ab
rba
p
nk
nkbpapabba
nk
nkrpbapba
ab
rba
rbpbarpbapba
ab
rba
14.11.
)0ith equality w(1
2
2
1
2/)(2
2
22
)0ith equality w(02
)0ith equality w(02
242224
244 0,1/20 with case For the
)1(444With
22
22
22
222
=
+−
−+
+−
−+
+−
−+
−
=+−−
=++−
+−+−++−+−+
−++
−+=−+=
n
nk
nk
nk
nk
nk
nk
nk
nnk
nnk
nnnnknn
nnknn
nnknnk
14.12.*
2/)2(
2/)2(2/)2(
2/)2(2/)2(
2/)2(
2/)2(
2/)2(2/)2(2/)2(2/)2(
2/)2(
2
2
2
2
2
2
(
let andonly pressure internal of case for thesolution special theUse
nk
i
nknk
i
nkknk
kk
nk
i
nk
i
nknk
i
nkknk
kk
nk
i
r
r
a
nk
nk
pr
nk
nk
ap
rb
nk
r
nk
ab
r
a
praprbr
ab
ap
b
−+
+−−−+
+−−++−
−+
−+
+−−−++−−++−
−+
−+
−−
=
−+
−−
→
−+
+
+−
−
=
−=−→−
−
−=
→
1 2 3 4 5 6 7 8 9 10
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
r/a
r/pi
= 0.25
n = 0
n = 1/2
n = 1
n = 2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
/pi
= 0.25
n = 0
n = 1
n = 1/2
n = 2
14.13.
0)0(
)(
:(14.3.6) from followssolution nt Displaceme
,0
(14.3.11) relationsby given isSolution
13 3 and0 case For the
32
32
2/)313(2/)(
22
=
++=
==
===
−+−
o
o
kn
r
E
a
u
E
a
BrArru
r
nkn
14.14*.
MATLAB plot of hoop stress in rotating disk problem
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Dimensionless Distance, r/a
Dimensionless Stress, /2a2
n=0
n=0.5
n=-0.5
n=1 n=2
for the case with n < 0, while for n > 0, the hoop stress goes to zero at the origin.
14.15.
)(
2
1
)(1
)(
and
2
0)(conditon boundary theUsing
)(1
)(
:soluton thegives ngIntergrati
)(1
)(
)(1
)(
)(
)(1
)(
law sHooke' Using
soluton nt displaceme thegives ngIntergrati rdr
d
(14.3.2)relation , with case For the
222
1
22
2
2
22
1
2
1
2
1
2
1
ra
Cr
rE
a
Ca
Cr
r
rEC
r
rEC
dr
d
rdr
d
r
rEC
r
u
r
dr
du
r
rE
rCu
uu
r
r
r
r
r
r
−=
+
==
+−=
+
=
+
−
+
=
+
−
=
=
==
14.16.
)()(2
*)(22)( :component-for Likewise
21
* where
*)(2
21
2
221
2
2
2
2
,222)(
=
+
==
+
=++=
−
=
+
=
−
+
=
−
+
=
+
=
+
=+
=
+
+
=++=
z
x
w
z
u
ze
z
w
zeeez
x
u
z
x
u
x
u
x
u
z
w
x
u
x
u
z
w
x
u
x
u
eee
xzxz
zxzz
xzxx
14.17.
+
==
+=
+=
+=+
++
=+++=
+
+
=
+
++
=+++=
+
=
−
+
=
−
+
=
+
=
++
=+
=
+
++
=+++=
r
u
z
u
e
z
r
u
r
u
r
u
r
u
z
u
r
u
r
u
eeee
z
u
z
u
z
u
z
u
z
u
r
u
r
u
eeee
z
r
u
z
z
r
u
z
z
r
u
r
u
z
u
r
u
r
u
r
u
r
u
z
u
r
u
r
u
eeee
zr
zrzr
rr
rrzrr
zr
zz
zzzrr
zzrz
rrrr
zrrrrzrr
rzrr
2
)(*2
2
2
222)(
2
222)(
)(*2
)(21
)(
2
2
)(21
)(2
2
2
2
,222)(
14.18.
( )
1
1
1
1
by through Divide
0)(
2
1
2
1
)()(,)()( :Variables of Separation
0)(
2
1
2
1
0)(
1
)(
1
2
1
2
1
0
22
0
2
1
2
1
00
(14.5.4)equation governing into (14.5.5)ation transformUsing
2
2
222222
222
22
2
2
2
2
=
−
−
+
+
+
==
=
−−+
+
=
+
−+
+
+
=
−
+
−
=
−
+
−
=
+
=
+
q
p
q
p
Y
X
pqXY
Wqpqpqpqp
pq
YXYXpq
yqxpyYxXW
WW
y
W
x
W
y
W
y
W
yx
W
x
W
x
y
W
y
W
yx
W
x
W
x
W
yy
W
xxy
w
yx
w
x
o
oooo
o
o
o
yyxxyx
yyxxyx
14.19.
( )
oo
oo
yx
bb
dy
dq
qdy
qd
q
dx
dp
pdx
pd
p
eyqexp
=
=−
=
−
==
44
1
2
1
4
1
2
1
44
1
2
1
4
1
2
1
)(,)( , functions theUsing
2
2
2
2
2
2
2
2
2
2
2
2
14.20.
dr
dr
r
dr
r
xx
dr
dr
dr
d
rdr
d
dr
d
rdr
d
dr
d
dr
dy
dr
d
r
y
dr
dx
dr
d
r
x
dr
d
rdr
d
r
yx
dr
d
r
y
dr
d
r
y
dr
d
r
x
dr
d
r
x
yyxx
dr
d
r
y
y
r
dr
d
ydr
d
r
x
x
r
dr
d
x
r
r
eeeeee
rzuuu
yzxzz
zrrzzr
zrrzzr
zr
−=
+
−=
+
−=+−=
−=
−=
+
+
−=
+
+
+
−=
+
−=
+
=
=
=
=
======
======
===
2222
2
2
22
2
1
2
1111
2
1111
2
11
2
11
,
:case icaxisymmetr For the
,0 (A.8) using calculated becan then stress The 2
,0 as (A.2)relation from follow then strains The
,0body rigid a as rotatessection
theandnt displaceme warpingno y,axisymmetr have wecase,section circular For the
14.21.
=
==
=
−=
=
===
+
+−=
+−=
+−=
−=
aa
z
zz
a
r
a
drrrdrrTJ
rr
dr
d
dr
rdrrCaC
Cdr
r
r
Crdrr
r
C
r
dr
d
Cr
dr
dr
r
dr
dr
dr
d
0
3
0
2
21
21
1
1
2
)(2
2
/
)(
)()( issolution
)(0)(condition boundary and,0 in text, discussed As
)(
)(
gIntegratin
2(14.6.9)relation From
14.22.
r
a
oo
a
r
m
o
a
r
m
o
n
a
a
d
a
r
ma
r
a
n
rar
m
d
a
n
dr
r
a
n
r
+
−=
+
=
−=
−
+−
=
=
+==
+=
1log
1
)(
gives case,1 for the while
)(
3
)(
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