Download solution manual for advanced mechanics of materials and applied elasticity 5th edition by ugural and fenster

Solution Manual for Advanced Mechanics of Materials and Applied Elasticity
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5th Edition by Ugural and Fenster

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

______________________________________________________________________________________
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

______________________________________________________________________________________
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
______________________________________________________________________________________
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.
______________________________________________________________________________________
SOLUTION (3.5)
It is readily shown that

" 4!1 = 0
4

" !2 = 0

is satisfied
is satisfied

(CONT.)
______________________________________________________________________________________


______________________________________________________________________________________
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.
______________________________________________________________________________________
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
______________________________________________________________________________________
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.
______________________________________________________________________________________


______________________________________________________________________________________
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.
______________________________________________________________________________________
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.
______________________________________________________________________________________
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

______________________________________________________________________________________
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

______________________________________________________________________________________
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.)

______________________________________________________________________________________


______________________________________________________________________________________
3.12 (CONT.)
(b)

y

"x = !

! xy

b

! xy

pa
b3

(6b ! 12 y )

x

a
______________________________________________________________________________________
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

______________________________________________________________________________________
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.)
______________________________________________________________________________________


______________________________________________________________________________________
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 .
______________________________________________________________________________________
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

)

______________________________________________________________________________________


______________________________________________________________________________________
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

______________________________________________________________________________________
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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