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4.1.
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123112231212123312221211
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313131233112313331223111
233123232312233323222311
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113111231112113311221111
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result theimplies relations two theseComparing
222
222
222
222
222
222
(4.2.3) from While
222
222
222
222
222
222
(4.2.1)relation From
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ij
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4.2.
2
*
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1
(4.2.4) is which thusand , and in symmetric isit of definition thefrombut
)(
2
1
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1
2
12
1
2
1
(4.2.4) iswhich
,
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4.3.
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2
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4.4.
)(
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and
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Since
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3
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Since
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(4.2.6) From
jlikjkilklijijkl
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4.5.
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if truebe will thisand
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and
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providing)10.2.4( form equal willThis
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4.6.
)21)(1(2
1
2
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2
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)(2
)1(2
)22(2
)(
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3,-4 exercise From
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E
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4.7.
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2 and imply that then materials isotropicfor results These
strain. for the directions and valuesprincipal theare and where0)(
as expressed becan strain for the problem valueprincipal theHowever,
2
1
0)(
2
1
0)2()(
thusand2 materials, isotropicFor
0)( then stress, for the directions and valuesprincipal theare and If
ij
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4.8.
18.52MPa 2
64.16MPa 2)(
64.16MPa- 2)(
80.2GPa) 111GPa,( steelfor moduli elastic with law sHooke' Using
vanish. toassumed are componentsstrain other All
10
3
200
,10400,10400 :strains for the Solving
2
3
4
3
4
1
10100
120cos120sin2120sin120cos
10400
90cos90sin290sin90cos
2
3
4
3
4
1
10300
60cos60sin260sin60cos
666
6
22
6
22
6
22
==
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4.9.
plane in the beamr rectangula a of bending pure toscorrespond stress of state thisThus
0 2
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components stress plane-in following thegives law sHooke'
0,0
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2
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22
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−
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4.10.
( )
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0
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inequality erestrictiv more thechoose weif and ,
211
that Noting
2
1
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1
then ,0and0 if so and213
1,-4 FromTable
−+
=
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EE
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4.11.
( )( )
( )
unbounded. become and 1/2,For
2/1,
4/1,3/2
0,3/
213
2/1,3/
4/1,5/2
0,2/
2/1,
4/1,5/2
0,0
211
k
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4.12.
( )( )
3/3232
ateindetermin becomes thusand 0 termthe,2 From
211
and
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)21(3
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==
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4.13.
3-
3-
3-
3-
3-
3-
3-
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10
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10
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00
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8.71,7.25,34.0,9.68 :Aluminum (a)
−
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4.14.
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1
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=−=+−+=
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4.15.
−−
=
−
−=+−=
====
2
2
is curvestrain -stress uniaxial theof slope theand
2
2
2
2
2
11
1
gives law sHooke',0,2With
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4.16.*
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1
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1
:Confined
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2)(
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:Unconfined
=
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−
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=−=+−=
=
+
+
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+
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+
−
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−=++=+++=
−=++=+++=
4.16.*
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0
2
4
6
8
10
12
14
16
18
Poisson's Ratio,
E*/E
4.17.
mmelz
mmely
mmelx
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GPaE
zz
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00028.0)100.7)(4(
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1
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107.1110)20(29.030
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)(
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104.510)30(29.020
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)(
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5
5
5
53
53
53
−=−==
===
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==
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4.18.
0)0)(4(
062.0)101.3)(200(
093.0)101.3)(300(
207
1
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101.310)50(29.050
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4
4
3
43
43
===
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−=−==
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4.19.
TT
E
e
TTTET
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e
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x
+=++−=
+=+=++−=
===
)1()(
1
)1()(
1
1
gives then (4.4.4) law sHooke'
0 stresses e transversno and 0strain axial No
x
zy
4.20.
zxzxyzyzxyxyxy
oyxzz
oxzyy
ozyxx
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+
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==
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=
−+−+++=
−+−+++=
−+−+=
1
,
1
,
1
2
)(
)21(
)]()1[(
)21)(1(
)(
)21(
)]()1[(
)21)(1(
Similarly
)(
)21(
)]()1[(
)21)(1(
1-4 Table Using
)()23()()2(
)()23(2)(
)()23(2
