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

Thus, the stress field given does not meet requirements for solution... Thus, Solution is not valid... On the basis of Eq... Stress field along the edges of the plate, as determined from

CHAPTER 3

SOLUTION (3.1)

( a ) We obtain

pxy

4 4

4 4

4

=

=

!

= """#

"

#

"

"

#

"

Thus, #4" = ! 12 pxy + 2 ( 6 pxy ) = 0

and the given stress field represents a possible solution

2

!

=

"

#

"

Integrating twice

) ( )

1 10 6

5 3 3

y f x y f y px y

=

"

The above is substituted into "4! = 0 to obtain

0

4 4 4

4dy f(y) +d dy f (y) =

This is possible only if

0

4 4

4dy f(y) = d dy f (y) =

d

We find then

7 6

2 5

3 4

11 10

2 9

3 8

Therefore,

11 10

2 9

3 8 7 6

2 5

3 4 10

c y c y c y c x c y c y c y c y px y

=

"

( c ) Edge y=0:

" "

!

!" = " =

a

a

a y

Edge y=b:

4

3

!

t c b c a t b

1 5

2

+ +

!

!

=

a

!

SOLUTION (3.2)

Edge x ± = a :

0 :

2 1 2 1 2 2 2

!

xy

"

0 :

2 1 2 1 2 2 2

!

xy

"

1

(CONT.)

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Solution Manual for Advanced Mechanics of Materials and Applied Elasticity

5th Edition by Ugural and Fenster

3.2 (CONT.)

3

2 2

3

Edge x = a :

0 2

:

= pa y c ay c y

x

"

or

3

2 2 pa

c =

SOLUTION (3.3)

( a ) Equations (3.6) become

0

=

!

!

x y

"

"

#

Substituting the given stresses, we have

0

2 3

2y ! c y =

c

Thus

=

2 2

xy

#

Assume c1 > 0 and c2 > 0

SOLUTION (3.4)

Boundary conditions, Eq (3.6):

0

= +!! !! !!

!

!

y x y

"

#

#

"

or ( 2 ab ! 2 ab ) x = 0 ( ! 2 ab + 2 ab ) y = 0

are fulfilled

However, equation of compatibility:

0 ) )(

2 2 2

= + + !!

!

!

y x y

x " " or 4 ! ab 0 is not satisfied

Thus, the stress field given does not meet requirements for solution

SOLUTION (3.5)

It is readily shown that

0 1

4! =

" is satisfied

0 2

4! =

" is satisfied

(CONT.)

y

) ( 2 2

22 b y

c

xy = !

"

y a c c

x = ( 1+ 2 )

!

x

) ( 2 2

22 b y

c

xy = !

"

y c

x = 1

!

b

b

a

3.5 (CONT.)

We have

b a

y

x = ""# = = ""# = = !""2#"1 = !

2 2 2

2

, 2 ,

%

Thus, stresses are uniform over the body

Similarly, for!2:

cy bx by

ax dy

x = 2 + 6 # = 6 + 2 " = ! 2 ! 2

#

Thus, stresses vary linearly with respect to x and y over the body

SOLUTION (3.6)

Note: Since !z = 0and!y = 0, we have plane stress in xy plane and plane strain

in xz plane, respectively

Equations of compatibility and equilibrium are satisfied by

0

!

x " " c "

"

0

=

=

= yz xz

xy ! !

!

We have

0

=

y

Stress-strain relations become

E y E

) ( )

( , " !"

!"

"

#

0 ,

) (

=

=

=

z x y " " "

# % $ $

Substituting Eqs (a,b) into Eqs (c), and solving

E z

y "!0 # "(1")!0

0

0

2 ) 1

!

y E

" #$

Then, Eqs (2.3) yield, after integrating:

0

0

2 ) 1

!

u #E" x

E z

w " + ( 1 " ) ! 0

=

SOLUTION (3.7)

Equations of equilibrium,

0 2

2 ,

= +!!

!

!#x x "y xy axy axy

0 ,

= + ""

"

" $y xy #x y ay ay

are satisfied Equation (3.12) gives

0 4 ) )(

2 2 2

!

"

= + + ##

#

Compatibility is violated; solution is not valid

(a)

(c)

SOLUTION (3.8)

We have

ay

x y

xy y

2 2

2

=

!

=

= "" " "

"

Equation of compatibility, Eq (3.8) is satisfied Stresses are

) (

)

1

y x E

) (

)

1

x y E

2 ) 1 (

xy

xy " !

Equations (3.6) become

0 )

2 3

1!aE2 x + xy + aE+ xy =

"

" "

0

2 1

2

1!aE2 y + !aE2 x =

"

"

These cannot be true for all values of x and y Thus, Solution is not valid

SOLUTION (3.9)

ax

y x

u

x = = ! 2 = "" = 2

"

#

0 2

!

= +

= ""

"

" x v cy cy

y

u xy

#

Thus

xy xy

y x E

% = 1&#2( + ) = 0 =

Ecx

x y

E

y =1! 2( # + "# ) = 2

$ "

Note that this is a state of pure bending

SOLUTION (3.10)

2

px pxy xy

y y

x = ""# = % = $ = !

%

Note that "4! = 0 is satisfied

( b )

( c ) Edge x = 0 : Vy = Px = 0

(CONT.)

y

x

pbx

y = 6

!

2

3px

xy =

!

0

=

x

!

2

3pa

xy =

!

0

=

y

! !xy = 3px2

0

=

x

!

0

=

xy

a

2a 2b

O

y

x

2Eac

2Eac

3.10 (CONT.)

!

=

!

=

0#

Edge y = b : Vx = pa3t !

!

=

SOLUTION (3.11)

( a ) We have #4" ! 0 is not satisfied

2 2 2

2 2

2 2 2

2 ) 4 ( )

xy a

xy x p x a

py x

%

( b )

6 2

0

Edge x = a :

!

=

0#

!

=

0#

Edge y = a :

!

=

0#

!

=

SOLUTION (3.12)

( a ) We have "4! = 0 is satisfied The stresses are

0 )

12 6

2 3

2 2

=

=

!

!

=

"

#

"

x y b

px y

$

) (

3

y x

xy = !"""# = !

$

(CONT.)

y

p

y =

!

0

=

x

!

2 2

2a

py

xy = !

"

) 4 1 (

p

xy = +

!

2p

) 4 (

2

2py a a y

!

) 1 ( a y

x = p +

!

0 ,

= xy

"

a a

3.12 (CONT.)

( b )

SOLUTION (3.13)

We have

2 2 2

[tan 1

y x y Py x y

x

xy x y P

"#

]

2 2 2 2 2 2

2

) ( 2 ) (

y x x y x y x y

x x

P

! + +

"

#

" = !$ +

The stresses are thus,

2 2 3 2

2

) (

2

y

x x

P y

x = !!" = # $ +

%

2 2 2 2 2

2

) (

2

y x xy P x

y = !!" = # $ +

%

2 2 2 2

) (

2

y x y x P y x

xy = #!!!" = # $ +

%

SOLUTION (3.14)

Various derivatives of ! are:

0 ),

2 2

3 2 0

"#

"

x h

y h

y

x $ y

) 1

( ,

2 0

2 2

2

4 h y h y y

x y

x = ""# = ! !

"

"

#

) 2

2

0 2

y = ! x ! + L +

"

#

" $

0 ,

4 4

4

=

= !!"

!

"

!

y x

It is clear that Eqs (a) satisfy Eq (3.17) On the basis of Eq (a) and (3.16), we obtain

(CONT.) (a)

y

) 12 6 (

b

pa

"

xy

!

x

xy

!

b

a

L

L

P !

2

P

x

!

xy

!

3.14 (CONT.)

0 ),

2 2

" #

) 1

2

4 h y h y

xy = !" ! !

"

From Eqs (b), we determine

Edge y = h : "y = 0 !xy = !0

Edge y = ! h : #y = 0 "xy = 0

2

4 h y h y xy

x L

x = # = " = !" ! !

It is observed from the above that boundary conditions are satisfied at y ± = h,

but not at x = L

SOLUTION (3.15)

( a ) For #4" = 0 , e = ! 5 dand a, b, c are arbitrary

( b ) The stresses:

) 3 2 ( 10

2 2

y x y d cy y

x = ""# = + !

3 10 2

2

2 2

dy by

a

x

y = ""# = + !

2 30 2

2

dxy bx

y x

xy = !"""# = ! !

Boundary conditions:

0

=

!

y p "

Equations (3), (4), and (5) give

p dh a

dh

0 0

h

h xy

h

h x

h

Equations (2), (4), and (7) yield

2

2dh

Similarly

0

Solution of Eqs (6), (8), and (9) results in

3

80 40

163

The stresses are therefore

) 3 2

8 20

3

h

p h

py

"

3 3

8 8

3

2 py h py h

p

y = ! ! !

"

) 1

2

8

3

h

y h

px

xy = !

"

(b)

SOLUTION (3.16)

We obtain

2 2 2 2

a y x p y

x = ""# = !

$

2 2 2 2

a

py x

y = !!" =

2

a

pxy y

x

xy = !"""# = !

$

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 Fx = 0

I

pxy

y xy =

!

!

Integrating,

) ( 1 2

2

x f I

pxy

xy = +

The boundary condition,

) ( 0

)

x f I

pxh h

y

xy = = = +

!

) ( 2 2

2px I h y

xy = ! !

Clearly, ( "xy)y !h = 0 is satisfied by Eq (b)

Then, the second of Eqs (3.6) with Fy = 0 results in

I y h p

y y (2 )

2

2 !

"

"

=

#

Integrating,

) ( )

2

2

x f h

I

p

Boundary condition, with t = 3 I 2 h2 ,

(CONT.)

p

y =

!

0

=

xy

!

2 2

2 a y

x = p

!

a x

xy = 2 p

!

2p

a y

xy = 2 p

!

) 2 1

2

a

y

x = p !

"

0 ,

= xy

"

a

a

p

y

2

a