Solution manual for advanced mechanics of materials and applied elasticity, 5th edition ansel c. ugural saul k. fenster sample

1. CHAPTER 3
SOLUTION (3.1)
( a ) We obtain
pxypxy yxyx
6012 22
4
4
4
4
4
==!= ""
#"
"
#"
"
#"
Thus, 0)6(2124
=+!="# pxypxy
and the given stress field represents a possible solution.
( b ) ypxpxyx
33
22
2
!="
#"
Integrating twice
)()( 21106
533
yfxyfypxypx
++!="
The above is substituted into 04
=!" to obtain
04
2
4
4
1
4
)()(
=+ dy
yfd
dy
yfd
x
This is possible only if
00 4
2
4
4
1
4
)()(
== dy
yfd
dy
yfd
We find then
76
2
5
3
41 cycycycf +++=
1110
2
9
3
82 cycycycf +++=
Therefore,
1110
2
9
3
876
2
5
3
4106 )(
533
cycycycxcycycycypxypx
++++++++!="
( c ) Edge y=0:
4 5
3 32 5( ) 2
a a px pa t
x xya a
V tdx c tdx c at!
" "
= = + = +# #
!! ""
===
a
a
a
a
yy tdxtdxP 0)0(#
Edge y=b:
4
2 2 23
1 32 2( )
a px
x a
V px b c b c tdx
!
= ! + + +"
tcbcatbpa a
)(2)( 3
2
15
23 2
++!!=
3 3
( 2 ) 0
a
y a
P pxb px b tdx
!
= ! ="
______________________________________________________________________________________
SOLUTION (3.2)
Edge :ax ±=
0:0 3
4
2
12
1
22
2
3
=+++!= cpaycypaxy"
0:0 3
4
2
12
1
22
2
3
=+++!= cpaycypaxy"
Adding, 02)23( 3
42
1
2
=+++! cpaycpa
(CONT.)
______________________________________________________________________________________
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2. ______________________________________________________________________________________
3.2 (CONT.)
or
4
2
1
3
2
2
3
1 pacpac !==
Edge :ax =
02:0 21
3
=+!= ycaycypax"
or
3
2 2pac =
______________________________________________________________________________________
SOLUTION (3.3)
( a ) Equations (3.6) become
00 ==+ !
!
!
!
!
!
xyx
xyxyx
""#
Substituting the given stresses, we have
02 32 =! ycyc
Thus
== 132 2 ccc arbitrary
( b ) )( 22
221
2
ybxycyc c
xyx !=+= "#
Assume 01 >c and 02 >c .
______________________________________________________________________________________
SOLUTION (3.4)
Boundary conditions, Eq. (3.6):
00 =+=+ !
!
!
!
!
!
!
!
yxyx
yxyxyx
"##"
or 0)22(0)22( =+!=! yababxabab
are fulfilled.
However, equation of compatibility:
0))(( 2
2
2
2
=++ !
!
!
!
yxyx
"" or 04 !ab is not satisfied.
Thus, the stress field given does not meet requirements for solution.
______________________________________________________________________________________
SOLUTION (3.5)
It is readily shown that
01
4
=!" is satisfied
02
4
=!" is satisfied
(CONT.)
______________________________________________________________________________________
y
)( 22
2
2
ybc
xy !="
yaccx )( 21 +=!
x
)( 22
2
2
ybc
xy !="
ycx 1=!
b
b
a

3. ______________________________________________________________________________________
3.5 (CONT.)
We have
bac yxxyxyyx !=!===== ""
#"
"
#"
"
#" 1
2
2
1
2
2
1
2
,2,2 $%%
Thus, stresses are uniform over the body.
Similarly, for 2! :
cybxbyaxdycx xyyx 222662 !!=+=+= "##
Thus, stresses vary linearly with respect to x and y over the body.
______________________________________________________________________________________
SOLUTION (3.6)
Note: Since 0=z! and 0=y! , we have plane stress in xy plane and plane strain
in xz plane, respectively.
Equations of compatibility and equilibrium are satisfied by
00 =!=!= zyx c """"
0=== xzyzxy !!!
We have
0=y! (b)
Stress-strain relations become
EyEx
xyyx )()(
,
!""!""
##
$$
==
0,
)(
====
+!
xzyzxyEz
yx
"""#
$$%
Substituting Eqs. (a,b) into Eqs. (c), and solving
Ezy
0)1(
0
!""
#"!! +
=$=
00
2
)1(
=!= !
yEx "" #
$
Then, Eqs. (2.3) yield, after integrating:
00
2
)1(
=!= !
vu E
x"#
E
z
w 0)1( !"" +
=
______________________________________________________________________________________
SOLUTION (3.7)
Equations of equilibrium,
022,0 =+=+ !
!
!
!
axyaxyyx
xyx
"#
0,0 22
=!=+ "
"
"
"
ayayxy
yxy #$
are satisfied. Equation (3.12) gives
04))(( 2
2
2
2
!"=++ #
#
#
#
ayyxyx
$$
Compatibility is violated; solution is not valid.
______________________________________________________________________________________
(a)
(c)

4. ______________________________________________________________________________________
SOLUTION (3.8)
We have
ayay yxxy
xyyx
220
2
2
2
2
2
=!== ""
"
"
"
"
" #$$
Equation of compatibility, Eq. (3.8) is satisfied. Stresses are
)()( 23
11 22 yxxaE
yx
E
x !!""# !!
+=+= $$
)()( 32
11 22 xyxaE
xy
E
y !!""# !!
+=+= $$
2
)1(2 xyG aE
xyxy !"# +==
Equations (3.6) become
0)23( 1
2
1 2 =++ +!
xyxyx aEaE
""
"
02
1
2
1 22 =+ !!
xy aEaE
""
These cannot be true for all values of x and y. Thus, Solution is not valid.
______________________________________________________________________________________
SOLUTION (3.9)
axcx y
v
yx
u
x 22 ==!== "
"
"
"
#$#
022 =+!=+= "
"
"
"
cycyx
v
y
u
xy#
Thus
xyxyyx
E
x G!"#$$% #
==+= &
0)(2
1
Ecxxy
E
y 2)(2
1
=+= !
"##$ "
Note that this is a state of pure bending.
______________________________________________________________________________________
SOLUTION (3.10)
( a )
2
36,02
2
pxpxy xyyyx !==== "
#"
$%%
Note that 04
=!" is satisfied.
( b )
( c ) Edge 0:0 === xy PVx
Edge 0: == xPax
(CONT.)
______________________________________________________________________________________
y
x
pbxy 6=!
2
3pxxy =!
0=x!
2
3paxy =!
0=y! 2
3pxxy =!
0=x!
0=xy!
b
a
2a
2b
O
y
x
2Eac
2Eac
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5. ______________________________________________________________________________________
3.10 (CONT.)
!== " btpatdyV
b
xyy
2
0
3#
Edge 0:0 == yPy
!== " tpatdxV
a
xyx
3
0
#
Edge !== tpaVby x
3
:
!== " btpatdxP
a
yy
2
0
3#
______________________________________________________________________________________
SOLUTION (3.11)
( a ) We have 04
!"# is not satisfied.
2
2
2
2
2
2
2
2
2
)4()(
,, a
yxyp
xya
xyxp
xa
py
xy
++
!
"!
#==== $%%
( b )
( c ) Edge
2
2
1
620
0: 0
a py
y xa
x V dy pat P= = = =!
Edge :ax =
!== " pattdyV
a
xyy 6
7
0
#
!== " pattdyP
a
xx 2
3
0
#
Edge 00:0 === yx PVy
Edge :ay =
!== " pattdxV
a
xyx 2
3
0
#
!== " patptdxP
a
yy
0
#
______________________________________________________________________________________
SOLUTION (3.12)
( a ) We have 04
=!" is satisfied. The stresses are
0)126( 2
2
32
2
==!!== "
#"
"
#"
xyb
px
yx yb $$
)(3
2 6
ybb
py
yxxy !=!= ""
#"
$
(CONT.)
______________________________________________________________________________________
y
py =!
0=x!
2
2
2a
py
xy !="
)41(2 a
xp
xy +=!
2p
p
x
)4(2
2
yaa
py
xy +=!
)1( a
y
x p +=!
0,0 == xyy !"
a
a

6. ______________________________________________________________________________________
3.12 (CONT.)
( b )
______________________________________________________________________________________
SOLUTION (3.13)
We have
2222 ],[tan 1
yx
yPy
xyx
xy
x
yP
y +
!
"
#"
+
!
"
#"
!=+!= $$
][ 222
222
222
2
)(
2)(
yx
xyxyx
yx
xP
y +
!+
+"
#"
+!= $
The stresses are thus,
222
3
2
2
)(
2
yx
xP
yx +!
"!
#== $%
222
2
2
2
)(
2
yx
xyP
xy +!
"!
#== $%
222
22
)(
2
yx
yxP
yxxy +!!
"!
#=#= $%
______________________________________________________________________________________
SOLUTION (3.14)
Various derivatives of ! are:
0),( 2
2
2
32
0
4 =!!= "
#"
"
#"
xh
y
h
y
x y$
)1(,0 2
2
0
2
22
4 32
4 h
y
h
y
yxyx
!!== ""
#"
""
#" $
)22( 2
0
2
2 66
4 h
Ly
h
xy
hy
Lx ++!!="
#" $
0,0 4
4
4
4
== !
"!
!
"!
yx
It is clear that Eqs. (a) satisfy Eq. (3.17). On the basis of Eq. (a) and (3.16), we obtain
(CONT.)
______________________________________________________________________________________
(a)
y
)126(3 ybb
pa
x !!="
xy!
x
xy!b
a
L
LP !2
P
x!
xy!

7. ______________________________________________________________________________________
3.14 (CONT.)
0),22( 66
4
0
=++!!= yh
Ly
h
xy
hx Lx "" #
)1( 2
2
0 32
4 h
y
h
y
xy !!!= "
"
From Eqs. (b), we determine
Edge 00: !!" === xyyhy
Edge 00: ==!= xyyhy "#
Edge )1(,0: 2
2
0 32
4 h
y
h
y
xyxLx !!!=== "
"#
It is observed from the above that boundary conditions are satisfied at hy ±= ,
but not at Lx = .
______________________________________________________________________________________
SOLUTION (3.15)
( a ) For de 5,04
!=="# and a, b, c are arbitrary.
Thus )5( 325322
yxydcyybxax !+++=" (1)
( b ) The stresses:
)32(106 23
2
2
yxydcyyx !+== "
#"
$ (2)
3
10222
2
dybyaxy !+== "
#"
$ (3)
2
302
2
dxybxyxxy !!=!= ""
#"
$ (4)
Boundary conditions:
0=!= xyy p "# (at y=h) (5)
Equations (3), (4), and (5) give
pdhadhb !=!!= 32
40215 (6)
000 === !!! """
h
h
xy
h
h
x
h
h
x dydyydy #$$ (at x=0) (7)
Equations (2), (4), and (7) yield
2
2dhc != (8)
Similarly
00 == xyy !" (at y=-h)
give
3
20dha = (9)
Solution of Eqs. (6), (8), and (9) results in
33
16804016
3
4 h
p
h
p
h
p
h
pp
edcba !==!=!=!= (10)
The stresses are therefore
)32( 23
820
3
3 yxyh
p
h
py
x !+!="
3
3
88
3
2 h
py
h
pyp
y !!!="
)1( 2
2
8
3
h
y
h
px
xy !="
______________________________________________________________________________________
(b)

8. ______________________________________________________________________________________
SOLUTION (3.16)
We obtain
2
22
2
2 )2(
a
yxp
yx
!
"
#"
==$
2
2
2
2
a
py
xy == !
"!
# (a)
2
2 2
a
pxy
yxxy !=!= ""
#"
$
Taking higher derivatives of ! , it is seen that Eq. (3.17) is not satisfied.
Stress field along the edges of the plate, as determined from Eqs. (a),
is sketched bellow.
______________________________________________________________________________________
SOLUTION (3.17)
The first of Eqs. (3.6) with 0=xF
I
pxy
y
xy
=!
!"
Integrating,
)(12
2
xfI
pxy
xy +=! (a)
The boundary condition,
)(0)( 12
2
xfI
pxh
hyxy +===!
gives .2)( 2
1 Ipxhxf != Equation (a) becomes
)( 22
2 yhI
px
xy !!=" (b)
Clearly, 0)( =!= hyxy" is satisfied by Eq. (b).
Then, the second of Eqs. (3.6) with 0=yF results in
I
yhp
y
y
2
)( 22
!
"
"
=
#
Integrating,
)()( 23
2
2
2
xfhy y
I
p
y +!=" (c)
Boundary condition, with
2
23 hIt = ,
(CONT.)
______________________________________________________________________________________
py =!
0=xy!
2
2
2 a
y
x p=!
a
x
xy p2=!
2p
p
x
a
y
xy p2=!
)21( 2
2
a
y
x p !="
0,0 == xyy !"
a
a
p
y
2
a
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