Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
8.1.
033
042240
042240
4
4
04
22
22
4
40
=++
++=
AAA
yAyxAxA
8.2.
1
3
,0,
3
:Case 0
satisfied,
2
2
),(
yidenticall satisfied,0),(
:ConditionsBoundary
3
2
2
3
3
2
2
3
−−==−=
=
−=
=
+
−=
−−
y
P
Pxy
N
PLdy
c
c
PLydyyL
N
cx
y
N
xy
xy
P
xyyx
c
cx
c
c
y
y
L
P
8.3.
)23(
4
)23(
6
)( Materials ofStrength From
3
0 case For the
solution eVenant typ St. eapproximatour with 0),( and0),(
assuch conditions pointwise ensurecannot that weNote
24
0)0,(
2
0)0,(
4
3
0
4
3
0
)0,(
choose thusbeam; cantilever theconstrain properly
toconditionsboundary additional choosemust wesolution, thecomplete order toIn
4
2
4
3
)
3
(
2
)1(3
4
2
4
3
)
(
)1(3
)(,
)(
constant
4
3
)()1(
2
)1(3
)(
4
3
variables theseparating and gRearrangin
)1(
2
)1(3
)(
4
3
)(
4
3
)1(
2
)1(3)1(2
2
)()
2
4
3
()
2
2
3
()(
1
)()
2
4
3
(
1
)
2
2
3
(
1
)(
1
)1(
4
3
,0,
2
2
3
323
3
323
323
3
2
3
3
3
3
3
3
2
3
2
3
2
2
3
3
3
3
2
3
3
3
3
2
2
2
3
2
2
2
3
2
3
2
2
2
3
2
3
3
2
3
2
2
3
LxLx
Ec
P
LxLx
EI
P
xv
P
PL
PL
Px
N
yLvyLu
Ec
PL
L
Ec
PL
vLv
L
Ec
uLu
Ec
PL
Ec
PL
x
Lv
Ec
Px
Ec
N
Ec
xyP
uy
c
y
y
cE
P
Ec
yP
x
Ec
N
Ec
yPx
u
uy
y
y
P
yP
yfvx
Px
xg
Ec
yP
yf
c
y
Ec
P
xg
Ec
Px
c
y
Ec
P
xg
Ec
yP
yf
Ec
Px
c
y
Ec
P
E
e
x
v
y
u
xgy
c
N
c
Pxy
E
v
c
N
c
Pxy
EE
e
y
v
yfx
c
N
c
yPx
E
u
c
N
c
Pxy
EE
e
x
u
c
y
c
P
c
N
c
Pxy
oo
o
oo
oo
o
xyxy
xyy
yxx
xyyx
+−=+−=
=
==
=−−==
−==
−==+=
+−−
+
−
−+−=
+−−
+
−
−=++=
==
−
−−
+
−=
+−
−
+
−=
+
+
+−
−
+
−=
+
==
+
++−
−=+−
−=−==
++−=+−=−==
−−==+−=
8.4.
satisfied,000),0(
satisfied,0)206(0),0(
:0 end freeat Conditions
050),(
(same)150),(
2/5),(
150),(
:ConditionsBoundary
302
1022,)3020(6
50241200
3
43
3
421
2
42
3
421
2
42
2
42
2
3
421
2
2
23
43
2
2
4554
4
32
5
5
4
3
3
2
2
2
1
dyy
dyyCyCdyy
x
cCcCCcx
cCCcx
qcCcCCqcx
cCCcx
xyCxC
yx
yCyCC
x
yxyCyC
y
CCyCyC
yxCyCyCyxCxC
c
cxy
c
c
c
cx
y
xy
y
xy
xy
yx
==
=+=
=
=+−=−
==−
−=−+−=
==
+−=
−=
−+=
=−+=
=
−==+=
++++=
−
−−
8.5.
satisfied,0
32
1
4
),(
satisfied,0),(&0),(
0),( that Note: end freeat Conditions
satisfied,
3
1
2
),0(
satisfied,0
1
4
),0(
satisfied,
3
1
2
),0(
:0 end fixedat Conditions
satisfied,0),(
satisfied,),(
satisfied,0),(
:ConditionsBoundary
3
11
4
32
1
4
,0
)(
3
1
2
6262
4
0
0
6262
4
0
32
1
4
0
4
4
2
2
2
2
,,
22
,
4
,
22
,
,
2
2
,
,,,
2
32
,
2
32
2
32
−−
−−
−−
−−
=
−−−=
==
==
=
+=
=
=
+=
=
=−
=
=
−
+−=
−−−=−===
−
+=
−−+==
=
=
−−+=
=
−−=
===
−−=
−−++=
c
c
c
cx
c
cx
x
c
c
c
cx
c
cxy
c
c
c
cx
xy
xy
y
xyxyxxy
yyx
yyyyyy
xxyyxy
xxxxxxxxxx
dy
c
y
c
ys
dyyl
ydyyldyyl
yllx
slcydy
c
y
c
sl
ydyy
dy
c
c
dyy
sldy
c
y
c
sl
dyy
x
cx
scx
cx
c
y
c
ys
c
y
c
ys
xl
c
y
c
s
c
xy
c
x
c
ly
c
ls
c
xy
c
x
c
ly
c
ls
c
y
c
ys
c
y
c
y
y
s
c
xy
c
xy
c
ly
c
ly
xy
s
8.6.
+−−+=
−+−=+−=
−===−=−=−=
=
=
=
=
=
=+=++=
=−−=−
−=++−=
++++−=
−=
++=
=++=
=
−==+=
+++++=
−
−
−
−
−
−
42222224
3
323
3
222
3
3
6
3
54321
2
6
41
2
53
4
6
2
41
3
532
3
532
4
6
22
5
2
4
2
31
2
3
532
2
2
3
6
3
54
2
2
5665
4
5
6
33
5
3
4
3
3
3
21
5
6
40
8
8
)32(
4
,)6105(
20
2
,
8
3
,
10
3
,
4
3
,
2
,
40
constantssix thedetermines end free and bottom top,on
conditonsboundary and equation aldifferenti from conditionssix theUsing
results previous using satisfied,
6
1
),(
results previous using satisfied,
2
1
),(
satisfiedly indentical,0),(
satisfiedly indentical,0),0(
20
6
satisfiedly indentical,0),0(
0
2
1
and0
4
1
2
1
0),(
0
3
2
0),(
/
3
2
/),(
:ConditionsBoundary
4
1
2
1
2
1
3
2
,
3
2
3
4
0680
209666
Lc
p
yycc
Lc
px
cyx
Lc
pxy
Lc
p
C
Lc
p
C
Lc
p
C
Lc
p
C
L
p
C
L
pc
C
pLydyyL
pLdyyL
dyyL
ydyy
dyy
cCCcCcCCcx
cCcCCcx
LpcCcCCLpxcx
yCyxCyCxCC
yx
xyCxyCxC
x
xyCyxCxyC
y
CCxyCxyC
xy
C
yx
C
xy
C
yx
C
x
CxyC
yx
c
cx
c
cxy
c
cx
c
cx
cxy
c
cx
xy
y
y
xy
yx
8.7.
2
2
31
2
876
2
876
3
8
2
765
2
876
2
3
8
2
765
2
2
2
8
2
7
3
4
2
321
2
2
84738743
4
32
8
22
7
2
6
2
5
5
4
4
3
3
2
2
1
020),0(
satisfiedly indentical,0),0(
020),0(
/642/),(
06420),(
022220),(
:ConditionsBoundary
642
2222
62201262
05 and030248120240
cccydyy
dyy
cccdyy
lccccclxcx
ccccccx
ccccccccx
xycxycxc
yx
ycycycc
x
yxcxcycycycc
y
ccccyccycc
yxcyxcyxcxcycycycyc
c
c
cxy
c
cx
ooxy
xy
y
xy
y
x
=+=
=
=+=
=++−=
=+−=−
=+=
−−−=
−=
+++=
=
+++++=
=
=+=+=+++=
+++++++=
−
−
8.8*.
2
1
),(,),(,0),(
satisfymust moments and forcesresultant theend, in-built theof location thebe to Choosing
cos,sinwhere,0)tan,(,0)tan,(
0)0,(,)0,(
:ConditionsBoundary
2
tantan2,tan2
0
)(
88
,
)(
16
,
)(
16
)cot1(2
cot
,)tan)((tan
0
tan
2
tan
tan
22
2
,
22
1
,
22
1
,
4
322
33
,
322
3
,
322
3
,
1222
L
pLydyyLpLdyyLdyyL
Lx
nnnnxxTnnxxT
xpx
yx
y
K
yx
xy
x
y
K
yx
xy
x
y
K
yx
yxxy
K
yx
yx
K
yx
xy
K
p
K
x
y
yxxyxK
L
x
L
xy
L
x
yxyyxxyyyxyxxx
xyy
xyxy
xxyyyx
xxyy
yyyyxxxx
−=−==
=
=−==+==+=
=−=
+
−=−=
+
+−−==
+
−−==
=
+
−
=
+
=
+
−
=
−
=
−+++−=
−−
−
x
y
L
p
A
B
n
y = xtan
pL
M=pL2/2
0 0.1 0.2 0.3 0.4 0.5
-10
-8
-6
-4
-2
0
2
8
10
Distance, y
Dimensionless Stress
Strength of Materials
0 0.1 0.2 0.3 0.4 0.5
-5
-4.5
-4
-3.5
-3
-2.5
-2
-0.5
0
Distance, y
Dimensionless Stress
Strength of Materials
Elasticity
8.8*. Continued
:Plots MATLAB
1
3
18
)3(
18
12/
2/)
4
(
2
1
183219
12/
)
2
(
2
1
1
2,
1
tan2
6
3
1
)(tan)cot1(
cot
2
3/130tan:6/30for 1at stresses of Evaluation
23
2
2
3
2
2
2
2
1
−−=
−−=
−−
==
−=
−=
−
==
+
−=
+
−−=
−
=
−
=
−
=
======
−
h
y
h
y
p
yLy
L
p
h
y
h
pL
It
VQ
h
y
pL
y
p
h
y
h
pL
I
yM
y
y
K
y
y
yK
ppp
K
LhLx
xy
xy
x
x
xyx
oo
8.9*.
8.-8 Exercise as same thebe wouldplots and conditionsOther
satisfies,0]sin2sin211[2]tan2sin2cos1[2),(),(
satisfies,0)0,()0,(
satisfies,0]tancoscos[sin2),(
satisfies,
)cot1(
)1cot(
]tan[
)cot1(2
cot
2]tan[2)0,(
:ConditionsBoundary
0]tan2cos2sin[2
11
]tan2cos2sin[2
]tan2sin2cos1[2,]tan2sin2cos[
]tancoscossin)[(2
]tancoscossin)([2
tan)2cos1(
2
2sin
2
)(]tancoscossin)([
:gives formpolar toCartesian from Converting
22
,r
,r
2
4
,
2
,,
2
22
,
r,
222
,
2
,
2
,
22
22222
=+−+−=++−=−=
=−=
=−=
−=
−
−
=−
−
=−=
=+−=++=
+−=
−=++−=++−=
=−+−=
−+−=
+−+−=−+−=
KrKrrr
rr
Kr
p
pp
Kr
K
r
r
rrK
KrrrrK
K
rrrK
rr
rKrrrK
r
r
rrr
r
rr
r
8.10.
2
1431
31311414
31143114
31311414
3114
31311414
31311414
31311414
2
4
1414
2
2
31311414
2
31311414
3
31311414
3
31311414
3
31311414
3
2
31311414
3
6
2
)3cos(cos3cos
)sin33(sin3sin
)3cos(cos3)3sin3)(sinsin33(sin
)3cos(cos3sin
2
)3sin3(sin3cos
6
3/
gives constantsfour for the Solving
3cos33sin3cossin0),(
3030)0,(
3sin3cossincos0),(
6/)0,(
:ConditionsBoundary
]3cos33sin3cossin[2)/(
]3sin3cossincos[6
]3sin63cos6sin2cos2[
11
0]sincos[8
11
]3cos33sin3cossin[3
]3sin813cos81sincos[
]3cos273sin27cossin[
]3sin93cos9sincos[
]3cos33sin3cossin[
0,][6,][6,][3
]3sin3cossincos[
kk
kk
bb
babar
bbbbr
babar
kaakrr
babarr
babar
babar
r
r
bar
r
r
babar
babar
babar
babar
babar
rr
babar
r
r
rr
rr
rr
rrr
r
rrrrrrrrrr
−−−
−−−−
−−−
=−=
−+−=
−==−−=
+++=
−=+−=
−+−=−=
+++==
−−+=+=
=+=++=
+−+−=
+++=
+−+−−=
+++−=
+−+−=
====
+++=
8.11*.
:Plot and CodeSolution MATLAB
][
2
, :Solution Fieldnt Displaceme 2,-8 Example 222 lxy
EI
M
v
EI
Mxy
u−+=−=
clc;clear all;clf
nu=0.3;L=10;c=2;
hold on
axis equal
% line(X,Y,'color','k','linestyle','--')
8.12*.
shown. ison distributi thisofplot MATALBA
itself. not vanish doesclearly but forceresultant zero gives stress This
5
1
3
1
2
3
53
),(, endeach At
53
)(
2
bygiven wasstress normal the3,-8 Example From
3
323
23
22
−=
−==
−+−=
c
y
c
ywycy
I
w
yllx
ycy
I
w
yxl
I
w
x
x
-10 -5 0 5 10
-8
-4
-2
4
6
8.13*.
shown isplot MATALB the,3/ case For the
same theare sprediction elasticity and materials of strength thusand
sin
2
3
3/2
sin
:Theory Materialsof Strength
sin
2
sinsinhcosh
4
coshsinh
coshcothcosh2sinh
coshsinh
sinhtanhsinh2cosh
sinsinh
2
,)1coth(,)1tanh(
coshsinh2
cosh
,
coshsinh2
sinh
)]sinh2cosh(cosh)cosh2sinh(sinh[(sin
:4-8 Examplefrom Results
23
2
3
2
2
23
33
2
2
2
2
2
=
−=
−=−=
−
+
−
+
+
−
+
+
+
+
−
+
−=
=+−=+−=
−
=
+
−
=
+++++=
cl
l
x
y
c
lq
c
y
l
x
lq
I
My
l
y
c
l
l
l
l
c
c
c
lc
l
y
l
l
c
c
l
y
l
l
y
y
l
c
l
c
lc
l
y
l
l
c
c
l
y
l
l
y
y
l
x
l
c
q
l
ccCBccDA
l
c
l
c
l
c
l
l
c
q
D
l
c
l
c
l
c
l
l
c
q
C
yyyDyByyyCyAx
o
o
x
x
o
x
oo
x
8.14.
CrBAa
E
r
u
BArararraa
rE
u
D
DCrrgBAf
rgdfrgrf
Er
u
r
u
u
r
e
rgdfa
E
u
fa
E
r
uaa
r
a
raaa
r
a
ra
E
r
u
aa
r
a
raaa
r
a
ra
E
E
u
u
r
e
frararraa
rE
frara
r
a
rrrarara
r
a
rrra
E
u
aa
r
a
raaa
r
a
ra
EEr
u
e
aa
r
a
raaa
r
a
ra
r
r
r
r
r
rr
r
r
r
r
rr
+−+
=
++
−++−−+
+
−=
+=+=
=−+
+
=
+
=
−
+
=
+−
=
−=−
+++−++−=
+++−++−=
−=
+=
+
−++−−+
+
−=
+
+++−−++−−=
++−−+++=−=
=
=++−=+++=
sincos
4
cossin)1(2)1(log)1(2
)1(1
thusand nt,displaceme l tangentiaconsistentfor dropped bemust that Note
)( andcossin)(
0)()()()(0
11
2
1
)()(
)(
4
2log223log2
2log223log2
1
)(
11
)()1(2)1(log)1(2
)1(1
)(23)log(22)log(2
1
23log22log2
1
)(
1
stress planefor law s Hooke'andnt displaceme-strain Using
0,23log2,2log2
by given werestresses thecase, icaxisymmetr For the
3
2331
3
323
2
1
323
2
1
3
23
2
1
323
2
1
3
2331
23
1
323
1
3
23
2
1
323
2
1
3
23
2
1
323
2
1
3
8.15.
−
+
=
+
==+
+
−=
+
=
+
=−
−
+
=
+
=
+
=
=====
===
=
−==
==
+
==
r
r
r
E
r
u
r
E
r
CruCr
rE
r
u
r
E
r
r
u
dr
d
r
E
r
r
u
dr
du
r
u
r
u
u
r
r
r
EE
e
uruu
dr
du
e
r
r
rar
r
a
rr
rr
rr
a
r
rr
rrr
r
r
rr
rr
1
)1(
1
)1(
0)(but ,
1
)1(
:Solution
1
)1(2
1
)1(2
1
2
111
0,0)(but ,constant0 :ntsDisplaceme
)( :ConditionBoundary
1
,0,0
11
2
1
2
2
2
1
2
2
1
2
2
3
2
2
2
2
2
2
2
2
1
2
2
2
2
242
2
4
2
2
2
2
2
4
8.16.
,
relations,nt displaceme-strain theUsing
equation aldifferenti thegIntegratin
0)(
1
or 0
1
0
1
)2(
0)(
1
)()(
1
0)(:forcebody no with Equations sNavier'
)(
1
)(
)(
1
)(,)(
1
,
5,-1 Example from and , :Case icAxisymmetr
2
1
2
1
2
1
22
2
22
2
2
2
22
222
+==−==
+=
=
=−+
=
−++
=
++
−
=++
−=
−==
===
=
rr
r
r
r
rrr
rrr
r
rr
rrr
rr
rr
r
C
C
u
e
C
C
du
e
r
C
rCu
ru
dr
d
rdr
d
r
u
dr
du
r
dr
ud
r
u
dr
du
r
dr
ud
ru
dr
d
rdr
d
r
u
dr
du
r
dr
d
r
)(
r
u
dr
du
r
dr
d
r
r
u
uu
ru
dr
d
rdr
d
ru
dr
d
rdr
d
u
rr
rrr
rr
r
ee
uu
eeeu
euue
eu
8.18.
2
3
,
,
gives constantssix for the Solving
0330),(
0330),(
4
23,
2
2)2cos1(
2
sin),(
023,020),(
2sin)33(2)/(
2cos)6122(2
2cos)462(2
11
0)(
1
)(
1
)(
42cos)412(
11
2cos)(4
2sin)(2
2cos)6122(2
2cos)242(2
2cos)(log
6
1
2
2
4
1
6
2
21
2
2
2
2
2
2
1
1
224
223
222212
2
124
4
123
2
122211
2
4
2
2
2
124
4
123212
2
111
2
24
4
23
2
2221
4
23
2
22212
2
1
2
24
4
23212
2
1
2
2
2
224
2
2
24
2
22
2
2
24
2
23
4
22
2
21
24
2
23
4
22
2
21
4
23
2
22212
2
1
3
23
3
22212
1
1
24
2
23
4
22
2
21
2
21
rrrr
p
a
r
p
a
rr
p
a
rararaar
rararaar
p
raraa
p
ara
p
pr
raraaarar
rararaar
raraaara
raraaara
r
r
r
r
arara
r
r
ararara
ararara
raraaara
rarararara
ararararara
ooo
r
r
ooo
r
rr
rr
rr
rrr
rrr
rr
r
+−−
−=
−=
=
=−−+=
=−−+=
−=++−=+−−=−=
=++=+=
−−+=−=
++++−==
++−+=+=
=++=
+−=++=
+++−=
+++−=
++++−=
−+++=
+++++=
−−
−−−
−−−
−−
−−
−−−
−
−
−
−−
−−
−
8.19.
+−=
−=
−+
+
−+
+
=
−+−
+
=
−==+=
−+−
+
=
=
−+−
+
=−−
+
=
−+
+
=−−
+
=
+−=+=
2
2
22
2
2
2
1
2
2
1
2
2
2
1
2
2
2
21
1
1
1
2
2
2
2
2
2
2
22
11
,
11
)21(
1
)21(
1
gives and constants for the Solving
)21(
1
)(
00)(
:ConditionsBoundary
)21(
1
:relationnt displaceme-Strain
)21(
1
])1[(
1
)21(
1
])1[(
1
:Law sHooke'Strain Plane
, :Solution icAxisymmetr General
rr
A
rr
A
rr
r
E
rr
rr
E
BA
Br
r
A
E
ru
r
A
BB
r
A
r
Br
r
A
E
u
r
u
e
B
r
A
EE
e
B
r
A
EE
e
B
r
A
B
r
A
r
r
r
r
r
r
rr
r
8.20.
solution. complete to and for relationscondition matching Solve
)()(
1)21)(1(
: @ conditions Matching
)21(
1
,
)(conditon Boundary
,,
:2 Material
)21)(1(
,
0 0at stresses finitefor but
,,
:1 Material
)21(
1
,,
(8.4.5) and (8.4.1) relationsby given assolution w icaxisymmetrstrain plane General
)2()1(
2
2
)2(
2
1
)2(
)1(
1
)2(
1
)1(
2
)2(
1
)2(
2
2
)1(
1
11
)2(
)1(
1
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(
)1(
1
11
)1()1()1()1(
)1(
)1(
2
)1(
)1()1(
2
)1(
)1(
22
AB
p
r
A
r
A
Brr
r
A
r
A
E
E
rr
r
r
A
p
r
A
E
up
r
A
r
A
r
A
pBpB
r
A
pr
B
r
A
B
r
A
rB
E
uB
Ar
B
r
A
B
r
A
Br
r
A
E
uB
r
A
B
r
A
rr
rr
r
r
rr
r
rr
−−==
+−−−
+
−+
=
+−−−
+
=−−=
−−=−=+−=
+−=+=
−+
===
==
+−=+=
−+−
+
=+−=+=
