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5.1.
p
40o
y
y
S
h
(d)
(c)
y
P
0),(),(
==
yavyau
0)0,()0,(
0),(,),(
==
==
xvxu
hxTShxT
yx
0),0(),0(
)30tan)(,(,0)30tan)(,(
0)0,()0,(
0)0,()0,(
==
=−=−
==
==
yvyu
PlxxTlxxT
xxT
xxT
o
y
o
x
yy
xyx
0),0(),0(
0),(),(
0),(),(
0),(),(
),(),(
0),(),(
0),(),(
==
=−−=−
=−−=−
==
−==
==
==
yvyu
hxhxT
hxhxT
hxhxT
ThxhxT
ylylT
ylylT
yy
xyx
yy
xyx
xyy
xx
5.1. Continued
x
a
b
(e)
x
y
w
(f)
kxbxTbxT
yaTyaT
yTyT
yx
yx
yx
−==
==
==
),(,0),(
0),(),(
0),0(),0(
0)0,()0,(
0),(,0),(
0),0(,),0(
==
==
==
xvxu
ywTywT
yTNyT
yx
yxx
5.2.
r1
r2
p2
(a)
0),(,),(
==
rTprT
S
(d)
0),(,),(
0),(),(,),(),(
0),(),(,0),(),(
:(2) and (1) materialsbetween bondingperfect Assume
2
)2(
2
)2(
1
)2(
1
)1(
1
)2(
1
)1(
1
)2(
1
)1(
1
)2(
1
)1(
=−=
===
====
rTprT
rrrr
rurururu
r
rrrr
rr
(c)
r2
p
NrTrT
rTSrT
r
r
==
=−=−
),(,0),(
0),(,),(
p
(b)
5.3.
5.4.
0),(,2),(
0),(,02),(
0),0(,02),0(
0)0,()0,((a)
=
+
=−=
+
+
=
=
+
==
+
+
=
=
+
==
+
+
=
==
x
v
y
u
axS
y
v
y
v
x
u
ax
x
v
y
u
ya
x
u
y
v
x
u
ya
x
v
y
u
y
x
u
y
v
x
u
y
xvxu
xyy
xyx
xyx
0)cos(2sin
0)tan)(,(
0)cos(sin2
0)tan)(,(
)0,(,02)0,(
0),0(),0((b)
=−
+
+
+
+
=+=−
=−
+
+
+
+
=+=−
=
+
==
+
+
=
==
y
v
y
v
x
u
x
v
y
u
nnaxxT
x
v
y
u
x
u
y
v
x
u
nnaxxT
S
x
v
y
u
x
y
v
y
v
x
u
x
yvyu
yyxxyy
yxyxxx
xyy
x
y
0,0
cos,sin:Surface Bottom
=+==+=
−==
yyxxyyyxyxxx
yx
nnTnnT
nn
(b)
x
y
S
a
S
(a)
a
a
x
y
5.5.
),(),(,),(),(
),(),(,),(),(
:Conditions Interface
0),(,0),(
0),0(,0),0(
0),(,),(
:(2) Material
0),(,0),(
0),0(,0),0(
0)0,()0,(
:(1) Material(a)
1
)2(
1
)1(
1
)2(
1
)1(
1
)2(
1
)1(
1
)2(
1
)1(
)2()2(
)2()2(
21
)2(
21
)2(
)1()1(
)1()1(
)1()1(
hxhxhxhx
hxvhxvhxuhxu
yaya
yy
hhxShhx
yaya
yy
xvxu
xyxyyy
xyx
xyx
xyy
xyx
xyx
==
==
==
==
=+−=+
==
==
==
),(),(,),(),(
),(),(,),(),(
:Conditions Interface
0),(,),(
:(2) Material
0at ntsdisplaceme & stresses Bounded
:(1) Material(b)
2
)2(
2
)1(
2
)2(
2
)1(
2
)1(
2
)1(
2
)2(
2
)1(
2
)2(
2
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==
==
=−=
=
rrrr
rurururu
rpr
r
rrrr
rr
rr
r1
r2
(1)
p
(2)
(b)
S
(a)
a
h1
x
y
h2
(1)
(2)
5.6.
0),(),(,),(),(
),(),(
:Conditions Interface
0),(,0),(
0),0(,0),0(
0),(,),(
:(2) Material
0),(,0),(
0),0(,0),0(
0)0,()0,(
:(1) Material(a)
1
)2(
1
)1(
1
)2(
1
)1(
1
)2(
1
)1(
)2()2(
)2()2(
21
)2(
21
)2(
)1()1(
)1()1(
)1()1(
===
=
==
==
=+−=+
==
==
==
hxhxhxhx
hxvhxv
yaya
yy
hhxShhx
yaya
yy
xvxu
xyxyyy
xyx
xyx
xyy
xyx
xyx
0),(),(,),(),(
),(),(
:Conditions Interface
:(2) Material
0at ntsdisplaceme & stresses Bounded
:(1) Material(b)
2
)2(
2
)1(
2
)2(
2
)1(
2
)2(
2
)1(
2
)2(
2
)2(
===
=
=
rrrr
ruru
r
rrrr
rr
rr
5.7.
),0(,0),0(,),0( (d)
30cos2/),0(,30cos/),0(,0),0( (c)
)0,(,0)0,(,)0,( (b)
22
000
2
30tan
30tan
30tan
000
TlhydyyTdyyTTldyyT
PlydyyTPldyyTdyyT
SwhxdxxTdxxTSwdxxT
yx
w
yy
w
yx
w
x
h
hx
h
hy
h
hx
o
lx
o
ly
lx
w
y
w
y
w
x
ooo
===
=−==
==−=
−−−
−−−
r2
(1)
p
(b)
S
(a)
a
h1
x
y
h2
(1)
(2)
5.8.
ijjikkijijkkkkij
iikkii
ijjiijkkmmijkkkkij
ijmmijkkmmijmmjimmijmmijkkmm
jkikmmikjkmmkkijmmijkkmmikjkjkikijkkkkij
ikjkjkikijkkkkij
FFF
Fji
FF
eeeelk
,,,,,
,,
,,,,,
,,,,,,
,,,,,,,,
2
,,,,
11
1
result desired theyieldsstatement ity compatibil
theintoback result thisusing and ,
1
1
givesrelation above the, case For the
11
1
)(
1
)3(
1
)(
1
relations theseinto (5.1.4) law sHooke' Using
0by given are with (5.1.2) relationsity compatibil The
−−
−
−=
+
+
−
+
−==
−−
+
=
+
+
+
+
=−−+
+
=
−−+
+
=−−+
=−−+=
5.9.
0)(
0)()(
0 :equations mequilibriuin Using
)()( :(5.4.1)Equaton
,,
,,,
,
,,,
=+++
=++++
=+
+++=
ikikkki
iijjjjikik
ijij
ijjiijkkij
Fuu
Fuuu
F
uuu
5.10.
0
21
1
01
)21(
2
0
)1(2
)1(2)21)(1(
00)(
:forcesbody zero with gives (5.4.3)Relation
,,
,,
,,,,
=
−
+
=
+
−
+
=
+
+
+
−+
+
=
+
+=++
kikkki
kikkki
i
kikkkikikkki
uu
uu
u
E
EE
u
uuuu
5.11.
0
),(),(
0
)(constant a equalmust also )( thusand
)(
),(
)()(
),(
)(
0
),(),(
0
)()(),(,)()(),(
)(
),(),(
0
),(
2
),(
),(
0,, :fieldStrain
43
21
2
2
zyf
E
gx
x
yxh
z
u
x
w
e
CzCzHzH
CzCzF
yxh
zFzH
yxh
xzF
gy
y
yxh
z
zxg
E
gy
y
w
z
v
e
zHxzFzxgzGyzFzyf
zF
x
zxg
y
zyf
x
v
y
u
e
yxh
E
gz
w
E
gz
z
w
e
zxg
E
gzy
v
E
gz
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v
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zyf
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gzx
u
E
gz
x
u
e
eee
E
gz
e
E
gz
ee
zx
yz
xy
z
y
x
xzyzxyzyx
=
+
−
=
+
=
+=
+=
=
−=
+
−
−
=
+
+
−=
+
=
+−=+=
=
−=
=
+
=
+
=
=
=
+
−=
−=
=
+
−=
−=
=
===
=
−==
5.12.
theory.SOMth exactly wimatch results elasticity the,
2
,
2
, constantsWith
)(
2
,
theory materials ofstrength case, For this figure. in theshown as loading endunder andright the
at supported problem bending beam cantilever arepresent toused be could field that thesuggests This
beam of bottom and on top vanish willstressshear then the, if that Note
),(,0),(
),(,),(
),0(,0),0(
:conditionsboundary following thesatisfying while
domain, beam in theon distributi stressshear quadratic a and stress bending aryinglinearly v a gives
field stress the, with 0domain r rectangula heconsider t weIf
satisfied0)0()(
:forcesbody no with (7.2.7) Equationsity Compatibil
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2/020
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2
22
xy
2
2
2
2
2
2
2
2
2
2
I
P
C
I
Ph
B
I
P
A
yh
I
It
I
I
ChB
ChBhxhx
CyBylAlyyl
CyByy
hlhyh,lx
Axy
yx
yx
ACCyAy
yx
CyBAxy
x
xyy
xyx
xyx
yx
yxy
xy
x
xyyx
−===
−====−==
−=
+==
+==
+==
−
=+
+
=+
==
+
−==+=
+
+===
x
l
h
h
P
5.13.
solution. elasticityproper a not are thusand ity,compatibilnot but mequilibriusatisfy stresses
check)not (does0)1(40)])(1([)]([)1(
0)()1(
:check equationity compatibil Michell-Beltrami
(checks)00000
(checks)00220
(checks)00220
:forces)body (nocheck equation mEquilibriu
ity.compatibil and mequilibriu bothsatisfy must field stress problem, elasticity tosolutiona be To
0,0,2
)(,)]([,)]([
22
2
2
2222
2
2
2
22222222
=+=++
+−++
=++
++
=++=
+
+
=++−=
+
+
=+−=
+
+
==−=
+=−+=−+=
cyxc
x
yxyc
x
zyx
ycyc
zyx
xcxc
zyx
cxyc
yxcxyxcyxyc
zyxx
z
yz
xz
zyyxy
zx
yx
x
zxyzxy
zyx
5.14*.
( )
belowshown are100 and10 of cases specific for the results MATALB following The
identical. are (b) and (a) problems from stresses theloading, thefrom distances largeat Thus
.components stressother for the made becan arguementssimilar and ,
)(
2
])[(
)(
)(])[(
)(
)(
:stress normal horizontal theingInvestigat
)( and)(, loadings, thefromaway far pointsAt
])[(
)(
)(
])[()(
])[(
)(
)(
:(b) problemfor field Stress
)(
2
,
)(
2
,
)(
2
:(a) problemfor field Stress
)(
222
2
222
2
222
2
222
2
222
2
)(
222
2
222
2
)(
222
3
222
3
)(
222
2
222
2
)(
222
2
)(
222
3
)(
222
2
)(
ayay
yx
yPx
yx
yxP
yx
yPx
yax
yaxP
yx
yPx
yayxaxyx
yax
yaxP
yx
Pxy
yax
Py
yx
Py
yax
yaxP
yx
yPx
yx
Pxy
yx
Py
yx
yPx
a
x
b
x
b
xy
b
y
b
x
a
xy
a
y
a
x
==
+
−
+
+
+
−
++
+
+
+
−=
++→
++
+
+
+
−=
++
+
+
−=
++
+
+
+
−=
+
−=
+
−=
+
−=
y = 10a
y = 100a
