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CHAPTER 3
SOLUTION (3.1)
( a ) We obtain
" 4#
"x 4
= !12 pxy
Thus,
" 4#
"y 4
" 4#
"x 2"y 2
=0
= 6 pxy
# 4 " = !12 pxy + 2(6 pxy ) = 0
and the given stress field represents a possible solution.
= pxy 3 ! 2 px 3 y
" 2#
"x 2
(b)
Integrating twice
"=
px 3 y 3
6
!
px 5 y
10
+ f1 ( y ) x + f 2 ( y )
" 4 ! = 0 to obtain
=0
The above is substituted into
d 4 f1 ( y )
dy 4
d 4 f2 ( y )
x+
dy 4
This is possible only if
d 4 f1 ( y )
dy 4
d 4 f2 ( y )
=0
dy 4
=0
We find then
f 1 = c 4 y 3 + c5 y 2 + c 6 y + c 7
f 2 = c8 y 3 + c9 y 2 + c10 y + c11
Therefore,
"=
( c ) Edge y=0:
px 3 y 3
6
!
px 5 y
10
+ ( c4 y 3 + c5 y 2 + c6 y + c7 ) x + c8 y 3 + c9 y 2 + c10 y + c11
a
a
"a
a
"a
a
"a
"a
4
Vx = # ! xy tdx = # ( px2 + c3 )tdx =
pa5t
5
+ 2c3 at
Py = ! # y tdx = ! (0)tdx = 0
Edge y=b:
a
Vx = " (! 32 px 2b 2 + c1b 2 +
!a
= ! pa 3 (b 2 !
a2
5
px 4
2
+ c3 )tdx
)t + 2a ( c1b 2 + c3 )t
a
Py = " ( pxb3 ! 2 px 3b)tdx = 0
!a
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SOLUTION (3.2)
Edge
x = ±a :
" xy = 0 :
" xy = 0 :
Adding,
! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0
! 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0
( !3 pa 2 + 2c1 ) y 2 + pa 4 + 2c3 = 0
(CONT.)
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3.2 (CONT.)
c1 = 23 pa 2
Edge x = a :
"x = 0:
c3 = ! 12 pa 4
or
or
pa 3 y ! 2c1ay + c2 y = 0
c2 = 2 pa 3
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SOLUTION (3.3)
( a ) Equations (3.6) become
!# x
!x
+
!" xy
!y
!" xy
!x
=0
=0
Substituting the given stresses, we have
c 2 y ! 2 c3 y = 0
Thus
(b)
c 2 = 2 c3
c1 = arbitrary
# x = c1 y + c2 xy
Assume
" xy =
c2
2
(b 2 ! y 2 )
c1 > 0 and c2 > 0 .
y
" xy =
c2
2
2
2
(b ! y )
b
! x = c1 y
b
" xy =
c2
2
(b 2 ! y 2 )
x
! x = ( c1 + c2 a ) y
a
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SOLUTION (3.4)
Boundary conditions, Eq. (3.6):
!" x
!x
+
!# xy
!y
=0
!# xy
!x
+
!" y
!y
=0
( 2ab ! 2ab) x = 0
( !2ab + 2ab) y
or
are fulfilled.
However, equation of compatibility:
2
( !!x 2 +
!2
!y 2
=0
)(" x + " y ) = 0 or 4ab ! 0 is not satisfied.
Thus, the stress field given does not meet requirements for solution.
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SOLUTION (3.5)
It is readily shown that
" 4!1 = 0
4
" !2 = 0
is satisfied
is satisfied
(CONT.)
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3.5 (CONT.)
We have
" 2 #1
%x =
= 2 c,
"y 2
%y =
" 2 #1
"x 2
= 2a ,
Thus, stresses are uniform over the body.
Similarly, for ! 2 :
# x = 2cx + 6dy
# y = 6ax + 2by
2
$ xy = ! ""x#"y1 = !b
" xy = !2bx ! 2cy
Thus, stresses vary linearly with respect to x and y over the body.
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SOLUTION (3.6)
! z = 0 and ! y = 0 , we have plane stress in xy plane and plane strain
Note: Since
in xz plane, respectively.
Equations of compatibility and equilibrium are satisfied by
" y = !c
" x = !" 0
"z = 0
! xy = ! yz = ! xz = 0
We have
!y = 0
(a)
(b)
Stress-strain relations become
#x =
(" x $!" y )
E
#z =
!% ($ x +$ y )
E
,
#y =
,
(" y $!" x )
E
(c)
" xy = " yz = " xz = 0
Substituting Eqs. (a,b) into Eqs. (c), and solving
! y = $"! 0
2
" x = ! (1#!0$E )
# z = " (1+E" )! 0
"y = 0
Then, Eqs. (2.3) yield, after integrating:
2
u = ! (1!# E)" 0 x
v=0
w = " (1+"E )! 0 z
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SOLUTION (3.7)
Equations of equilibrium,
!# x
!x
+
!" xy
!y
= 0,
2axy + 2axy = 0
"$ xy
"y
+
"# y
"x
= 0,
ay 2 ! ay 2 = 0
are satisfied. Equation (3.12) gives
2
( ##x 2 +
#2
#y 2
)($ x + $ y ) = "4ay ! 0
Compatibility is violated; solution is not valid.
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SOLUTION (3.8)
We have
" 2$ x
" 2$ y
=0
"y 2
"x 2
" 2# xy
"x"y
= !2ay
= 2ay
Equation of compatibility, Eq. (3.8) is satisfied. Stresses are
E
1$! 2
#x =
(" x + !" y ) =
aE
1$! 2
( x 3 + !x 2 y )
# y = 1$E! 2 (" y + !" x ) = 1$aE! 2 ( x 2 y + !x 3 )
# xy = G" xy =
aE
2 (1+! )
xy 2
Equations (3.6) become
aE
1!" 2
(3x 2 + 2"xy ) + 1aE
+" xy = 0
aE
1!" 2
y 2 + 1!aE" 2 x 2 = 0
These cannot be true for all values of x and y. Thus, Solution is not valid.
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SOLUTION (3.9)
#x =
"u
"x
"u
"y
# xy =
Thus
= !2$cx
+
"v
"x
#y =
y
= 2ax
"v
"y
= !2cy + 2cy = 0
% x = 1&E# 2 ($ x + #$ y ) = 0
" xy = G! xy
2a
$ y = 1!E" 2 (# y + "# x ) = 2 Ecx
O
2Eac
x
2b
2Eac
Note that this is a state of pure bending.
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SOLUTION (3.10)
(a)
%x =
" 2#
"y 2
Note that
% y = 6 pxy
= 0,
$ xy = !3 px 2
" 4 ! = 0 is satisfied.
(b)
! y = 6 pbx
y
!x = 0
b
! xy = 0
! xy = 3px 2
!x = 0
a
!y =0
( c ) Edge
x = 0:
V y = Px = 0
Edge
x=a:
Px = 0
! xy = 3pa 2
x
! xy = 3px 2
(CONT.)
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______________________________________________________________________________________
3.10 (CONT.)
b
V y = " # xy tdy = 3 pa 2 bt !
Edge
y = 0:
0
Py = 0
a
V x = " # xy tdx = pa 3t !
0
Edge
V x = pa 3t !
y = b:
a
Py = " # y tdx = 3 pa 2 bt !
0
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SOLUTION (3.11)
( a ) We have
# 4 " ! 0 is not satisfied.
2
2
% y = !!x"2 = pya 2 ,
%x =
p ( x 2 + xy )
a2
y
(b)
! x = p(1 + ay )
py 2 a
" xy = ! 2a
2
a
" y = 0, ! xy = 0 p
Edge
x=a:
Vy = !
a py 2
2
0 2a
a
dy = 16 pat
V y = " # xy tdy =
7
6
pat !
Px = " # x tdy =
3
2
pat !
0
a
0
y = 0:
Edge y = a :
Vx = 0
Edge
a
V x = " # xy tdx =
0
a
p ( 4 xy + y 2 )
2a2
2p
!x = 0
x = 0:
$ xy = #
! xy = 2p (1 + 4 ax )
!y = p
( c ) Edge
,
! xy =
py
2a2
( 4a + y )
x
Px = 0
Py = 0
3
2
pat !
Py = " # y ptdx = pat !
0
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SOLUTION (3.12)
( a ) We have
" 4 ! = 0 is satisfied. The stresses are
2
$ x = ""y#2 = ! bpx3 (6b ! 12 y )
$y =
2
$ xy = ! ""x"#y =
6 py
b3
" 2#
"x 2
=0
(b ! y )
(CONT.)
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3.12 (CONT.)
(b)
y
"x = !
! xy
b
! xy
pa
b3
(6b ! 12 y )
x
a
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SOLUTION (3.13)
We have
"#
"y
= ! $P [tan !1 xy +
" 2#
"y 2
= ! $P [ x 2 +x y 2 +
xy
x2 + y2
],
"#
"x
( x 2 + y 2 ) x !2 y 2 x
( x 2 + y 2 )2
= ! Py
$
!y
x2 + y2
]
The stresses are thus,
%x =
! 2"
!y 2
= # 2$P
x3
( x 2 + y 2 )2
%y =
! 2"
!x 2
= # 2$P
xy 2
( x 2 + y 2 )2
2
% xy = # !!x!"y = # 2$P
x2 y
( x 2 + y 2 )2
P
!x
2 P !L
! xy
L
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SOLUTION (3.14)
! are:
2
3
( y ! yh ! hy2 ),
Various derivatives of
"#
"x
=
" 4#
"x 2"y 2
" 2#
"y 2
4
! "
!x 4
=
$0
4
" 2#
"x"y
= 0,
$0
4h
= 0,
( !2 x !
4
! "
!y 4
6 xy
h
" 2#
"x 2
= $40 (1 !
2y
h
+ 2L +
6 Ly
h2
=0
2
! 3hy2 )
)
(a)
=0
It is clear that Eqs. (a) satisfy Eq. (3.17). On the basis of Eq. (a) and (3.16), we obtain
(CONT.)
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3.14 (CONT.)
"x =
#0
4h
( !2 x ! 6hxy + 2 L +
6 Ly
h
),
"y =0
(b)
2
" xy = ! "40 (1 ! 2hy ! 3hy2 )
From Eqs. (b), we determine
"y =
Edge y = h :
Edge
y = !h :
Edge
x = L:
0
! xy = ! 0
#y =0
" xy = 0
2
" xy = ! "40 (1 ! 2hy ! 3hy2 )
# x = 0,
It is observed from the above that boundary conditions are satisfied at
y = ±h ,
but not at x = L .
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SOLUTION (3.15)
(a)
# 4 " = 0, e = !5d and a, b, c are arbitrary.
For
" = ax 2 + bx 2 y + cy 3 + d ( y 5 ! 5 x 2 y 3 )
Thus
( b ) The stresses:
(1)
= 6cy + 10d ( 2 y 3 ! 3x 2 y )
(2)
= 2a + 2by ! 10dy 3
(3)
$ xy = ! ""x"#y = !2bx ! 30dxy 2
(4)
$x =
" 2#
"y 2
$y =
2
" #
"x 2
2
Boundary conditions:
# y = !p
" xy = 0
(at y=h)
(5)
Equations (3), (4), and (5) give
b = !15dh 2
!
h
"h
2a ! 40dh 3 = ! p
$ x dy = 0
!
h
"h
y$ x dy = 0
Equations (2), (4), and (7) yield
Similarly
!
h
"h
# xy dy = 0
(at x=0)
c = !2dh 2
"y =0
give
(6)
(8)
! xy = 0
(at y=-h)
a = 20dh 3
(9)
Solution of Eqs. (6), (8), and (9) results in
a = ! 4p
(7)
b = ! 163ph
The stresses are therefore
c = ! 40ph
d=
p
80 h 3
e = ! 16ph3
(10)
" x = ! 320pyh + 8hp3 ( 2 y 3 ! 3x 2 y )
3
" y = ! 2p ! 38pyh ! 8pyh3
" xy =
3 px
8h
(1 !
y2
h2
)
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SOLUTION (3.16)
We obtain
$x =
" 2#
"y 2
=
p ( x 2 !2 y 2 )
a2
#y =
! 2"
!x 2
=
py 2
a2
(a)
2
$ xy = ! ""x"#y = !
2 pxy
a2
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.
y
!y = p
! xy = 2 p ax
p
2p
! xy = 2 p ay
! xy = 0
a
2
! x = 2 p ay2
2
" x = p(1 ! 2 ay2 )
a
2
a
" y = 0, ! xy = 0
p
x
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SOLUTION (3.17)
Fx = 0
The first of Eqs. (3.6) with
!" xy
!y
=
pxy
I
! xy =
pxy 2
2I
Integrating,
+ f1 ( x )
(a)
The boundary condition,
(! xy ) y =h = 0 =
gives
pxh 2
2I
+ f1 ( x )
f1 ( x ) = ! pxh 2 2 I . Equation (a) becomes
" xy = ! 2pxI ( h 2 ! y 2 )
Clearly,
(b)
(" xy ) y = ! h = 0 is satisfied by Eq. (b).
Then, the second of Eqs. (3.6) with
"# y
"y
=
2
Fy = 0 results in
2
p(h ! y )
2I
Integrating,
"y =
p
2I
y (h 2 !
Boundary condition, with
y2
3
) + f 2 ( x)
(c)
2
t = 3I 2h ,
(CONT.)
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