PROBLEM 7.1
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
x
y =h1 + (h 2 − h1 ) ; dA =ydx
a
x

dI y = x 2 dA= x 2  h1 + (h 2 − h1 )  dx
a

a
h 2 − h1 3 
Iy
h1 x 2 +
x  dx
=

0
a


3
4
h 2 − h1 a
a
= h1
+
3
4
a
3
3
ba
h a
= 1 + 2
12
4
∫
=
Iy
a3
(h1 + 3h2 ) 
12
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PROBLEM 7.2
Determine by direct integration the moment of inertia of the shaded area with
respect to the y axis.
SOLUTION
y = kx1/3
For x = a :
b = ka1/3
k = b /a1/3
Thus:
y=
b 1/3
x
a1/3
2
=
dI y x=
dA x 2 ydx
b 7/3
2 b
=
dIy x=
x1/3 dx
x dx
1/3
a
a1/3
b a 7/3
b  3 10/3 
=
Iy =
dI y
x=
dx
 a 
1/3 0
a
a1/3  10

∫
∫
Iy =
3 3
ab 
10
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PROBLEM 7.3
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
 x x2
=
y 4h  − 2
a a



dA = ydx
 x x2
2
=
dI y x=
dA 4hx 2  − 2
a a
3
a x
x4 
=
I y 4h  − 2  dx
0
 a a 

 dx

∫
a
 x4
 a3 a3 
x5 
I y = 4h  − 2  = 4h  − 
5 
 4 a 5a  0
 4
1
I y = ha3 
5
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PROBLEM 7.4
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
At
Then
x=
a,
1
y=
y=
b
1
2
2
=
b ka
=
or k
y1 :
b
a2
y2 :
=
b ma
=
or m
b
a
y1 =
b 2
x
a2
y2 =
b
x
a
Now
 b
b
 
dI y = x 2 dA = x 2 [( y2 − y1 )dx = x 2  x − 2 x 2  dx 
a
a
 

Then
=
Iy
∫
=
dI y
∫
a
0
1
1

b  x3 − 2 x 4  dx
a
a

a
1
1

= b  x 4 − 2 x5 
5a
 4a
0
or
Iy =
1 3
ab 
20
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PROBLEM 7.5
Determine by direct integration the moment of inertia of the shaded area with
respect to the x axis.
SOLUTION
y =h1 + (h 2 − h1 )
Ix =
∫
dI y =
1
=
12
1
3
∫
x
a
1
dI x = y 3 dx
3
3
a
0
x
 h1 + (h 2 − h1 ) a  dx


a
x  a 

 h1 + (h 2 − h1 ) a   h − h 

  2 1 0
4
2
2
a
a (h 2 + h1 )(h 2 + h1 )(h 2 − h1 )
h 42 − h14 = ⋅
=
12(h 2 − h1 )
12
h 2 − h1
(
)
Ix =
a 2
(h1 + h 22 )(h1 + h 2 ) 
12
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PROBLEM 7.6
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
y=
b 1/3
x
a1/3
3
=
dI x
Ix
=
1 3
1  b 1/3 
1 b3
=
y dx
x
=
dx
xdx

3
3  a1/3
3 a

∫
dI x
=
∫
1 b3 a 2
b3
xdx
=
3 a
3 a 2
a1
0
Ix =
1 3
ab 
6
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PROBLEM 7.7
Determine by direct integration the moment of inertia of the shaded
area with respect to the x axis.
SOLUTION
 x x2 
=
y 4h  − 2 
a a 
1 3
1   x x2
dI x =
y dx
=
 4h  −
3
3   a a 2
Ix =
∫
dI x =
64h3
3
∫
3

  dx
 
a
0
x3
x4
x5 x 6 
 3 − 3 4 + 3 5 − 6  dx
a
a
a 
a
64h3 1 x 4 3 x5 1 x 6 1 x 7
=
−
+
−
3 4 a3 5 a 4 2 a5 7 a 6
=
a
0
64h  1 3 1 1  64h  3 
a − + =
a
− 

3
3
4 5 2 7
 420 
3
3
Ix =
16 3
ah 
105
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PROBLEM 7.8
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
At
Then
Now
Then
x=
a,
1
y=
y=
b
1
2
2
=
=
or k
y1 : b ka
b
a2
y2 : b ma
=
=
or m
b
a
y1 =
b 2
x
a2
y2 =
b
x
a
(
)
1 3
y2 − y13 dx
3
b3 
1  b3
=  3 x3 − 6 x 6  dx
3 a
a

=
dI x
=
Ix
=
∫
=
dI x
3
b
3
∫
a
0
b3  1 3 1 6 
x − 6 x  dx
3  a3
a

a
1 7
 1 4
 4a 3 x − 7 a 6 x 

0
or
Ix =
1
ab3 
28
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PROBLEM 7.9
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
x2 y 2
+
=
1
a 2 b2
=
x a 1−
y2
b2
dA = xdy
2
=
dI x y=
dA y 2 xdy
Ix
=
∫
∫
dI x
=
b
−b
2
xy=
dy a
∫
b
−b
y2 1 −
y2
dy
b2
=
y b=
sin θ
dy b cos θ dθ
Set:
Ix a
=
π /2
∫π
− /2
b 2 sin 2 θ 1 − sin 2 θ
π /2
∫π
b cos θ dθ
3
= ab
=
sin 2 θ cos 2 θ dθ ab3
− /2
1
= ab3
4
=
π /2
∫π
−
1 2
sin 2θ dθ
/2 4
π /2
1
1
1


(1 − cos 4θ )dθ = ab3 θ − sin 4θ 
−π /2 2
8
4

 −π /2
∫
π /2
1 3  π  π  π 2
− 
ab  −  =
ab
8
 2  2  8
1
I x = π ab3 
8
π /2
y2
dy =a
1 − sin 2 θ b cos θ dθ
2
−b
−b
−π /2
b
π /2
π /2 1
= ab
=
cos 2 θ dθ ab
(1 + cos 2θ ) dθ
−π /2
−π /2 2
∫
∫
+b
A = dA = xdy =a
∫
∫
b
1−
∫
∫
π /2
ab  π  π   1
1
= ab θ + sin 2θ =
π ab
 − − =

2
2  2  2   2
−π /2
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Problem 7.9 (Continued)
2
I=
k x2 A k=
x
x
Ix
=
A
1
π ab3
8
=
1
π ab
2
1 2
b
4
kx =
1
b 
2
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PROBLEM 9.10
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
At x 2=
a, y 0:
y1 : =
0 = c sin k (2a)
=
2ak π=
or k
At x a=
=
, y h:
y2 :
At x a=
, y 2h :
=
At x 2=
=
a, y 0:
h = c sin
π
2a
π
2a
(a) or
c=h
2h
= ma − b
=
0 m(2a) + b
Solving yields
2h
m=
− , b=
4h
a
Then
y1 =
h sin
Now
π 
 2h
dA= ( y2 − y1 )dx=  (− x + 2a ) − h sin
x dx
2a 
a
Then
=
A
π
2a
∫ dA= ∫
x
2a
a
2h
y2 =
x + 4h
−
a
2h
( − x + 2a )
=
a
π 
2
h  (− x + 2a ) − sin
x dx
2
a
a 

π 
2a
 1
= h  − ( − x + 2a ) 2 +
x
cos
π
2a  a
 a
2a
 2a  1

2

= h  −  + (−a + 2a)2 = ah 1 − 
π
π
a





= 0.36338ah
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SOLUTION Continued
Find:
We have
I x and k x
3
3
1 
1   2h
π  
1
 
=
dI
dx
(− x + 2a)  −  h sin
x  dx

x
 3 y2 − 3 y1 =
3   a
2a  


 
=
Then
Now
Then
=
Ix
h3  8
π 
(− x + 2a)3 − sin 3
x

3
3 a
2a 
∫
dI=
x
∫
2a
a
h3
3
8

3
3 π
 a3 (− x + 2a) − sin 2a x  dx


sin 3 θ =
sin θ (1 − cos 2 θ ) =
sin θ − sin θ cos 2 θ
=
Ix
h3
3
∫
π
π
π 
8

3
x − sin
x cos 2
x  dx
 3 (− x + 2a) −  sin
2a
2a
2a  

a
2a 
a
π
π 
h3  2
2a
2a
=
− 3 ( − x + 2a ) 4 +
x−
x
cos
cos3

π
3  a
2a
3π
2a  a
2a
=
=

h3  2 a 2 a  2
+
+ 3 ( − a + 2a ) 4 
 −

3  π 3π  a

2 3
2 
ah 1 −

3
 3π 
I x = 0.52520ah3
and
2
k=
x
I x 0.52520ah3
=
0.36338ah
A
or I x = 0.525ah3 
or
k x = 1.202h 
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PROBLEM 7.11
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the x axis.
SOLUTION
At =
x a1, y=
y=
b:
1
2
2
or k
y1: b ka
=
=
b
a2
b
2b − ca 2 or c =
y2 : b =
a2
Then
Now
Then
Now
=
y1
 
x2
dA =( y2 − y1 )dx =b  2 − 2
a
 
∫
∫
A = dA =
 b 2
2b 2
2
 − 2 x  dx = 2 (a − x )dx
a

 a
a
2b 2
2b 
1 
4
(a − x 2 )dx = 2  a 2 x − x3  = ab
2
3 0 3
a
a 
a
0

1 
1  
x2
1
dI x = y23 − y13  dx =  b  2 − 2
3 
3   
a
3

=
=
Then

b 2
x2 
=
−
x
y
b
2


2
a2
a 2 

3
   b 2  3 
  −  2 x   dx
 
   a

1 b3
(8a 6 − 12a 4 x 2 + 6a 2 x 4 − x 6 − x 6 )dx
3 a6
2 b3
(4a 6 − 6a 4 x 2 + 3a 2 x 4 − x 6 )dx
6
3a
∫
∫
I x = dI x =
a
0
2 b3
(4a 6 − 6a 4 x 2 + 3a 2 x 4 − x 6 )dx
6
3a
a
2 b3  6
3
1 
4a x − 2a 4 x 3 + a 6 x 5 − x 7 
6 
3a 
5
7 0
172 3
=
ab
105
ab3 43 2
I x 172
2
105
k=
=
=
b
x
4
A
ab
35
3
=
and
or
or
I x = 1.638ab3 
k x = 1.108b 
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consent of McGraw-Hill Education.
PROBLEM 7.12
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the x axis.
SOLUTION
2
=
y1 k=
y2 k2 x1/ 2
1x
For
=
x 0 and
y=
y=
b
1
2
2
b k=
b k2 a1/ 2
=
1a
k1
=
b
b
k2
=
2
1/ 2
a
a
Thus,
=
y1
b 2
b 1/ 2
=
x
y2
x
2
1/ 2
a
a
= ( y2 − y1 )dx
dA
=
A
∫
a
0
b 1/ 2 b 2 
 a1/ 2 x − a 2 x  dx


2 ba3/ 2 ba 3
− 2
3 a1/ 2
3a
1
A = ab
3
=
A
1 3
1
y2 dx − y13 dx
3
3
3
1 b
1 b3 6
3/ 2
=
x
dx
−
x dx
3 a3/ 2
3 a6
dI x
=
=
Ix
2
k=
x
∫
=
dI x
Ix
=
A
b3 a 3/ 2
b3 a 6
x
dx
−
x dx
3a3/ 2 0
3a 6 0
b3 a5/ 2 b3 a 7  2
1 
3 3
=3/ 2 5 − 6
=
−  ab3 I x =
ab 

35
3a
( 2 ) 3a 7  15 21 
(
∫
3
35
ab3
ab
b
)
∫
kx = b
9

35
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consent of McGraw-Hill Education.
PROBLEM 7.13
Determine the moment of inertia and the radius of gyration of the shaded area
shown with respect to the y axis.
SOLUTION
x2 y 2
+
=
1
a 2 b2
=
y b 1−
x2
a2
dA = 2 ydx
2
=
dI y x=
dA 2 x 2 ydx
=
Iy
Set:
∫
∫
=
dI y
a
2 x 2=
ydx 2b
0
∫
a
x2 1 −
0
x2
dx
a2
=
x a=
sin θ
dx a cos θ dθ
I y 2b
=
∫
π /2
0
∫
a 2 sin 2 θ 1 − sin 2 θ
a cos θ dθ
π /2
sin 2 θ cos 2 θ dθ 2a 3b
a 3b
= 2=
1
= a 3b
2
0
∫
π /2 1
0
π /2 1
0
4
sin 2 2θ dθ
1
1
(1 − cos 4θ )dθ = a 3b θ − sin 4θ
2
4
4
1 3 π
 π 3
a b =
ab
=
− 0
4
2
 8
From solution of Problem 7.9:
∫
π /2
0
1
I y = π a 3b 
8
1
A = π ab
2
Thus:
2
I=
k y2 A k=
y
y
Iy
=
A
1
π a 3b
8
=
1
π ab
2
1 2
a
4
ky =
1
a 
2
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consent of McGraw-Hill Education.
PROBLEM 7.14
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
At x 2=
a, y 0:
y1 : =
0 = c sin k (2a)
=
2ak π=
or k
At x a=
=
, y h:
y2 :
At x a=
=
, y 2h :
At x 2=
=
a, y 0:
π
h = c sin
2a
π
2a
(a ) or
c=h
2h
= ma − b
=
0 m(2a) + b
Solving yields
2h
m=
− , b=
4h
a
Then
y1 =
h sin
Now
π 
 2h
dA= ( y2 − y1 )dx=  (− x + 2a) − h sin
x dx
a
2
a 

Then
=
A
∫
π
2a
=
dA
∫
x
2a
a
2h
−
y2 =
x + 4h
a
2h
=
( − x + 2a )
a
π 
2
h  (− x + 2a) − sin
x dx
2a 
a
π 
2a
 1
= h  − ( − x + 2a ) 2 +
cos
x
π
2a  a
 a
2a
 2a  1

2

= h  −  + (−a + 2a)2 = ah 1 − 
 π
 π  a

= 0.36338ah
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consent of McGraw-Hill Education.
SOLUTION Continued
Find: I y and k y
  2h
π  
= x 2   (− x + 2a) − h sin
dI=
x 2 dA
x dx 
y
a
2
a  

π 
2
= h  (− x3 + 2ax 2 ) − x 2 sin
x dx
a
2
a 

We have
I=
y
Then
∫
dI=
y
∫
2a
a
π 
2
h  (− x3 + 2ax 2 ) − x 2 sin
x dx
a
a 
2

Now using integration by parts with
2
=
u x=
dv sin
du = 2 x dx
∫x
Then
2
sin
v= −
π
∫x
2
sin
π
2a
x dx
cos
π
π
2a
x
π   2a
π 
 2a
x dx = x 2  − cos
x  −  − cos
x (2 x dx)
2a
2a 
2a 
 π
 π
∫
=
du dx
=
v
Then
2a
2a
=
u x=
dv cos
Now let
π
2a
π
π
2a
sin
x dx
π
2a
x
π
π   2a
π  
2a 2
4a   2a
−
x dx =
x cos
x+
x
x −  sin
x dx 
sin
π
π   π
π
a  
2a
2a 
2

∫
2 1
2

I y = h   − x 4 + ax3 
a
4
3

 
2a
 2a 2
8a 2
16a3
π
π
π 
−  −
x cos
x + 2 x sin
x + 3 cos
x 
2a
2a
2a  
π
π
 π
a
 2  1
2
16a3 
 2a
=h   − (2a)4 + a(2a)3  −
(2a) 2 + 3 
3
π 
 π
 a  4
2
 2  1

2
 8a
− h   − ( a ) 4 + a ( a )3  − 2 ( a ) 
3
 π

 a  4
= 0.61345a3 h
and
2
k=
y
I y 0.61345a 3 h
=
0.36338ah
A
or
I y = 0.613a3 h 
or
k y = 1.299a 
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PROBLEM 7.15
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
At =
x a,
2
y=
y=
b: =
y1 : b ka
=
or k
1
2
b
a2
b
y2 : b =
2b − ca 2 or c =
a2
Then
Now
b 2
x
a2

x2
=
y2 b  2 − 2
a

y1 =



dA
= ( y2 − y1 ) dx
 

x2  b
= b  2 − 2  − 2 x 2  dx
a  a

 
2b
= 2 (a 2 − x 2 ) dx
a
Then
A
=
dA ∫
∫=
a
0
2b 2
(a − x 2 )dx
a2
a
2b  2
1 
a x − x3 
2 
3 0
a 
4
= ab
3
=
Now
 2b

2
dI y x=
dA x 2  2 (a 2 − x 2 ) dx 
=
a

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consent of McGraw-Hill Education.
SOLUTION Continued
Then
Iy
=
∫
dI y
=
∫
a
0
2b 2 2
x (a − x 2 )dx
2
a
a
2b  1
1 
= 2  a 2 x3 − x5 
5 0
a 3
or
and
2
k=
y
Iy
=
A
4 3
ab
15
=
4
ab
3
Iy =
4 3
ab 
15
1 2
a
5
or
ky =
a
5
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consent of McGraw-Hill Education.

PROBLEM 7.16
Determine the moment of inertia and the radius of gyration of the
shaded area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 7.12.
=
A
1
ab
3
a
2
=
dI y x=
dA x 2 ( y2 − y1 )dx
b
b
 b

= 1/ 2
x 2  1/ 2 x1/ 2 − 2 x 2  dx
a
a
a


I=
y
∫
Iy =
b b7/2
b a5  2 1  3
⋅
−
⋅
=  − a b
a1/ 2 ( 72 ) a 2 5  7 5 
2
k=
y
Iy
=
A
0
(
3
35
a 3b
ab
3
)
∫
a
0
x5/ 2 dx −
b
a2
∫
a
x 4 dx
0
Iy =
3 3
a b
35
ky = a
9

35
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consent of McGraw-Hill Education.
PROBLEM 7.17
Determine the polar moment of inertia and the polar radius of gyration
of the rectangle shown with respect to the midpoint of its (a) longer
sides, (b) shorter sides.
SOLUTION
1
dI x = a 3 dx
3
=
Ix
1 3 +a
a3
2 4
a=
dx
=
a
(2a )
∫
3 −a
3
3
2
=
dI y x=
dA x 2 (a dx)
+a
+a
=
I y a∫ =
x dx
2
−a
x3
2 4
=
a
3 −a 3
2
2
Jo = I x + I y = a4 + a4
3
3
4 4
a
J
2
3 = 2 a2
o
r=
=
o
A 2a 2 3
J=
ro2 A
o
4 4
a
3

ro = 0.816a

Jo =
 1  a 3  1 3
dI x 2=
a dx
=
   dx 
 3  2 
 12
2a 1
a3 2a
I x ∫=
a 3 dx
x
=
0 12
12 o
1 4
a3
(2a )
Ix =
a
=
12
6
2
d=
I y x=
dA x 2 (a dx)
2a
(2a )3
x3
=
I y a∫ =
x dx a= a
0
3 0
3
2a
2
8
I y = a4
3
1
8
 1 16 
Jo = I x + I y = a4 + a4 =  +  a4
6
3
6 6 
Jo
2
J=
ro2 A r=
=
o
o
A
17 4
a
6 = 17 a 2
2a 2
12
Jo =
17 4
a 
6
ro = 1.190a

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consent of McGraw-Hill Education.
PROBLEM 7.18
Determine the polar moment of inertia and the polar radius of gyration
of the rectangle shown with respect to one of its corners.
SOLUTION
MOMENT OF INERTIA ABOUT THE BASE OF A
RECTANGLE
1
I = bh3
3
1
=
Ix =
(2a )(a )3
3
1
(2a )3 (a )
=
Iy =
3
THUS:
2 4
a
3
8 4
a
3
J=
Ix + I y
o
J=
ro2 A
o
OR
Jo
2
r=
=
o
A
10 4
a
3 = 5 a2
2a 2
3
Jo =
10 4
a 
3
ro = 1.291a

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consent of McGraw-Hill Education.
PROBLEM 7.19
Determine the polar moment of inertia and the polar radius of gyration of the
shaded area shown with respect to Point P.
SOLUTION
a a y 1
x=+
= (a + y )
2 2a 2
=
dA x=
dy
1
A=
2
∫
1
Ix=
2
∫
a
dA =
0
y 2 dA=
∫
a
0
∫
1
1
y2
(a + y )dy = ay +
2
2
2
a
a
=
0
a1
1
=
x3 dy
0 3
3
∫
1
1
=
Iy
2
24
∫
a
1
1
(a + y )dy=
2
2
y2
0
1 y3 y 4
= a
+
2 3
4
1
=
Iy
2
1
(a + y )dy
2
∫
0
=
0
1  2 a2  3 2
a +
= a
2 
2  4
A=
Ix =
7 4
a 
12
Iy =
5 4
a 
16
JO =
43 4
a 
48
3
1

 2 (a + y )  dy


a
(a + y )3 dy
=
0
1 1
1 
15 4
4
(a + y )4=
(2a )4 − a =
a


24 4
96
96
0
1
5 4
Iy =
a
2
32
From Eq. (7.4):
3 2
a 
2
(ay 2 + y 3 )dy
11 1 4 1 7 4
+
a =
a
2  3 4 
2 12
a
0
∫
a
a
JO = I x + I y =
J=
kO2 A
O
7 4 5 4  28 + 15  4
a + a =
a
12
16
 48 
2
k=
O
JO
=
A
43 4
a
48
=
3 2
a
2
43 2
a
72
k=
a
O
43
72
kO = 0.773a 
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consent of McGraw-Hill Education.
PROBLEM 7.20
Determine the polar moment of inertia and the polar radius of
gyration of the shaded area shown with respect to Point P.
SOLUTION
By observation
First note
Now
Then
y=
b
x
a
dA
= ( y + 2b)dx
b
( x + 2a )dx
=
a
1 3
1 3

=
− ybottom
dI x  ytop
 dx
3
3

=
3

1  b 
 x  − (−2b)3  dx
3  a 

=
1 b3 3
( x + 8a 3 )dx
3 a3
=
Ix
∫
=
dI x
∫
1 b3 2
( x + 8a3 )dx
−a 3 a3
a
a
=
=
1 b3  1 4

x + 8a 3 x 
3 
3 a 4
 −a
1 b3
3 a3
 1 4
 1
  16 3
3
4
3
ab
(−a)  
  (a) + 8a (a)  −  (−a) + 8a=
 4
 3
 4
Also
b

2
dI y x=
dA x 2  ( x + 2a ) dx 
=
a

Then
=
Iy
∫
=
dI y
∫
a
−a
b 2
x ( x + 2a )dx
a
a
b 1 4 2 3
=
x + ax 
a  4
3
 −a
=
b  1 4 2
2
 4 3
3 1
4
a )3  
ab
  (a) + a (a )  −  (−a ) + a (−=
a  4
3
3
 4
 3
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consent of McGraw-Hill Education.
SOLUTION Continued
Now
JP = Ix + I y =
16 3 4 3
ab + a b
3
3
=
JP
and
2
k=
P
=
4
ab(a 2 + 4b 2 ) 
3
4
ab(a 2 + 4b 2 )
JP
3
=
A (2a )(3b) − 12 (2a )(2b)
1 2
(a + 4b 2 ) or
3
kP =
a 2 + 4b 2

3
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consent of McGraw-Hill Education.
PROBLEM 7.21
(a) Determine by direct integration the polar moment of inertia of the annular area
shown with respect to Point O. (b) Using the result of Part a, determine the
moment of inertia of the given area with respect to the x axis.
SOLUTION
dA = 2π u du
(a)
2
=
dJ O u=
dA u 2 (2π u du )
= 2π u 3 du
JO
=
dJ
∫=
O
2π
= 2π
(b)
From Eq. (9.4):
∫
R2
u 3 du
R1
1 4
u
4
R2
=
JO
π 4
R2 − R14  
=
Ix
π 4
R2 − R14  
R1
2

(Note by symmetry.) I x = I y
JO = I x + I y = 2I x
Ix
=
1
π 4
JO
R2 − R14 
=
2
4
4

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consent of McGraw-Hill Education.
PROBLEM 7.22
(a) Show that the polar radius of gyration k O of the annular area shown is
approximately equal to the mean radius R=
( R1 + R2 ) 2 for small values of
m
the thickness =
t R2 − R1. (b) Determine the percentage error introduced by
using R m in place of k O for the following values of t/R m : 1, 12 , and
1
10
.
SOLUTION
(a)
From Problem 9.23:
JO
=
π 4
4
 R2 − R1 
2
J O 2  R2 − R1 
=
A π R22 − R12
π
2
k=
O
4
(
=
kO2
1 2
R2 + R12 
2
=
Rm
1
( R1 + R2 )
2
4
)
Thickness:=
t R2 − R1
Mean radius:
Thus,
1
R1 =
Rm − t and
2
1
R2 =
Rm + t
2
2
2
1 
1  
1  
1
kO2 =  Rm + t  +  Rm − t   =Rm2 + t 2
2 
2  
2  
4
For t small compared to Rm : kO2 ≈ Rm2
(b)
Percentage error =
kO ≈ Rm 
(exact value) − (approximation value)
100
(exact value)
( )
( )
2
1 + 14 Rt
−1
Rm2 + 14 t 2 − Rm
m
P.E. =
(100) =
100
−
−
2
Rm2 + 14 t 2
t
1
1+ 4 R
m
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consent of McGraw-Hill Education.
SOLUTION Continued
For
t
= 1:
Rm
P.E. =
1+
1
4
−1
(100) = −10.56%

1 + 14 ( 12 ) − 1
−
−2.99%
P.E. =
(100) =
2
1 + 14 ( 12 )

1+
1
4
2
t
1
For
= :
Rm 2
t
1
For
= :
Rm 10
P.E. =
( ) −1
(100) =
2
1 + 14 ( 101 )
1+
1 1
4 10
2
−0.1250%

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consent of McGraw-Hill Education.
PROBLEM 7.23
Determine the moment of inertia of the shaded area with respect to the xaxis.
SOLUTION
y = a cos x
1
dI x = y 3 dx
3
1
= a 3 cos3 x dx
3
=
Ix
π
π
d I x 2∫ 2 dI x
∫=
0
2
π
−
2
π
1
2 3 π2
3
=
=
I x 2 ∫ 2 a 3 cos
x dx
a
(1 − sin x) cos x dx
0 3
3 ∫0
=
Set sin x u=
cos x dx du
1
2 3 1
2  u3 
2 3 2
Ix =
a ∫ (1 − u 2 )du = a 3 u −  =
a ( )
0
3
3 
3 0 3
3
Ix =
4 3
a
9

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consent of McGraw-Hill Education.
PROBLEM 7.24
Determine the moment of inertia of the shaded area with respect to the yaxis.
SOLUTION
y = a cos x
2
=
dI y x=
dA x 2 ( y=
dx) x 2 (a cos x)dx
=
Iy
π
π
dI = ∫ ax cos x dx ∫ 2 2ax
∫=
0
y
2
π
−
2
2
2
cos x dx
π
I y = ∫ 2 2ax 2 cos x dx
0
=
u 2=
ax 2 , du 4ax
=
dv cos
=
x dx, v sin x
We integrate by parts, with
π
π
I y =[u v ] − ∫ v du = 2ax sin x  − ∫ 2 (sin x)(4ax dx)
0
2
π
2
o
π
=
I y 2a ( ) − ∫ 2 4ax sin x dx
0
2
2
=
u 2=
ax, du 4a
dv = sin x dx, v = − cos x
Integrating again by parts, with
π
π
1 2
π a −  4ax(− cos x)  02 + ∫ 2 (− cos x)(4a dx)
0
2
π
π
1 2
1 2
Iy =
π a − 4a ∫ 2 cos x dx =
π a − 4a sin x  02
0
2
2
π2

1
I y = π 2 a − 4a =  − 4  a
2
 2

I y=
I y = 0.935a

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consent of McGraw-Hill Education.
PROBLEM 7.25
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24)(6) + (8)(48) + (48)(6)] mm 2
= (144 + 384 + 288) mm 2
= 816 mm 2
Now
I x = ( I x )1 + ( I x ) 2 + ( I x )3
where
1
(24 mm)(6 mm)3 + (144 mm 2 )(27 mm)2
12
= (432 + 104,976) mm 4
=
( I x )1
= 105, 408 mm 4
1
=
(8 mm)(48 mm)3 73, 728 mm 4
12
1
=
( I x )3
(48 mm)(6 mm)3 + (288 mm 2 )(27 mm)2
12
=
(864 + 209,952) mm 4 =
210,816 mm 4
=
( I x )2
Then
I x = (105, 408 + 73, 728 + 210,816) mm 4
= 389,952 mm 4
and
2
k=
x
I x 389,952 mm 4
=
A
816 mm 2
or =
I x 390 × 103 mm 4 
or
k x = 21.9 mm 
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consent of McGraw-Hill Education.
PROBLEM 7.26
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] in 2
= (30 − 8 − 4) in 2
= 18 in 2
Now
where
I x = ( I x )1 − ( I x ) 2 − ( I x )3
1
=
(5 in.)(6 in.)3 90 in 4
12
=
( I x )1
1
(4 in.)(2 in.)3 + (8 in 2 )(2 in.)2
12
2
= 34 in 4
3
=
( I x )2
=
( I x )3
1
3

(4 in.)(1 in.)3 + (4 in 2 )  in. 
12
2


1 4
= 9 in
3
Then
2
1

I x = 90 − 34 − 9  in 4
3
3

and
2
k=
x
I x 46.0 in 4
=
A
18 in 4
2
or
I x = 46.0 in 4 
or k x = 1.599 in. 
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consent of McGraw-Hill Education.
PROBLEM 7.27
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
A= (125)(250) −
=
Ix
π
2
2
(75)=
22 414 mm 2
1
π
π

(125)(250)3 −  (75) 4 + (75) 2 (125) 2 
3
2
8

= 500.56 ×106 mm 4
=
I x 501×106 mm 4 
2
r=
x
I x 500.56 ×106 mm 4
=
A
22 414 mm 2
=
rx2 22
=
332 mm 2
rx 149.4 mm 
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consent of McGraw-Hill Education.
PROBLEM 7.28
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the x axis.
SOLUTION
1 4 1 4
(6
) 432 in.4
=
a
=
3
3
1
11 4
Portion  I x
( I x for full circle)
=
=
 πr 
4
44

1
4
=
π (4=
) 16
=
π 50.3 in.4
16
Portion  =
Ix
For entire area:
I x = 432 − 50.3 =381.7
I x =382 in.4 
1
A =36 − π (42 ) =36 − 4π = 23.43 in.2
4
2
r=
x
I x 381.7
=
= 16.29 in.2
A 23.43
4.04 in. 
r=
x
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PROBLEM 7.29
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 + A2 + A3
= [(24 × 6) + (8)(48) + (48)(6)] mm 2
= (144 + 384 + 288) mm 2
= 816 mm 2
Now
I y = ( I y )1 + ( I y )2 + ( I y )3
where
1
(6 mm)(24 mm)3 6912 mm 4
=
12
1
( I y )2 =
(48 mm)(8 mm)3 2048 mm 4
=
12
1
( I y )3 =
(6 mm)(48 mm)3 55, 296 mm 4
=
12
( I y )1
=
Then
and
I y = (6912 + 2048 + 55, 296) mm 4 = 64, 256 mm 4
2
k=
y
I y 64, 256 mm 4
=
A
816 mm 2
or
=
I y 64.3 × 103 mm 4 
or
k y = 8.87 mm 
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PROBLEM 7.30
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
First note that
A = A1 − A2 − A3
= [(5)(6) − (4)(2) − (4)(1)] in 2
= (30 − 8 − 4) in 2
= 18 in 2
Now
I y = ( I y )1 − ( I y )2 − ( I y )3
where
=
( I y )1
1
=
(6 in.)(5 in.)3 62.5 in 4
12
=
( I y )2
1
2
=
(2 in.)(4 in.)3 10 in 4
12
3
=
( I y )3
1
1
=
(1 in.)(4 in.)3 5 in 4
12
3
Then
2
1

I y =  62.5 − 10 − 5  in 4
3
3

and
2
k=
y
I y 46.5 in 4
=
A
18 in 2
or
I y = 46.5 in 4 
or k y = 1.607 in. 
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PROBLEM 7.31
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
A= (125)(250) −
Iy
=
π
2
2
(75)=
22 414 mm 2
π
1
(250)(125)3 − (75) 4
3
8
= 150.335 ×106 mm 4
I y 150.3 ×106 mm 4 
=
I y 150.335 ×106 mm 4
r= =
= 6707.2 mm 2
2
A
22 414 mm
2
y
ry = 81.9 mm 
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PROBLEM 7.32
Determine the moment of inertia and the radius of gyration of the
shaded area with respect to the y axis.
SOLUTION
1 4 1 4
=
a
(6
=
) 432 in.4
3
3
1
11 4
Portion  I y '
=
=
( I y ' for full circle)
 πr 
4
44

1
4
) 16
=
π (4=
=
π 50.27 in.4
16
Portion  =
Iy
2
1
  16 
50.27 −  π 42  
I y " =I y ' − Ad =

4
  3π 
= 50.27 − (12.57)(1.698) 2 = 50.27 − 36.24 = 14.03
2
14.03 + (12.57)(6 − 1.698) 2
I y =I y " + Ad 2 =
=14.03 + (12.57)(4.302) 2 =14.03 + 232.6 = 246.6 in.4
For entire area:
Iy =
432 − 246.6 =
185.4
Iy =
185.4 in.4 
A=
36 − 12.57 =
23.43 in.2
2
r=
y
I y 185.4
=
= 7.913 in.2
A 23.43
r=
2.81 in. 
y
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PROBLEM 7.33
Determine the shaded area and its moment of inertia with respect to the
centroidal axis parallel to AA′ knowing that d 1 = 25 mm and d 2 = 10 mm
and that its moments of inertia with respect to AA′ and BB′ are 2.2 × 106
mm4 and 4 × 106 mm4, respectively.
SOLUTION
I AA′ =
2.2 × 106 mm 4 =
I + A(25 mm)2
(1)
I BB′ =
4 × 106 mm 4 =
I + A(35 mm) 2
I BB′ − I AA′ =(4 − 2.2) × 106 =A(352 − 252 )
1.8 × 106 =
A(600)
Eq. (1):
2.2 × 106 =I + (3000)(25) 2
A = 3000 mm 2 
=
I 325 × 103 mm 4 
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PROBLEM 7.34
Knowing that the shaded area is equal to 6000 mm2 and that its moment of
inertia with respect to AA′ is 18 × 106 mm4, determine its moment of inertia
with respect to BB′ for d 1 = 50 mm and d 2 = 10 mm.
SOLUTION
Given:
A = 6000 mm 2
I AA′ =
I + Ad12 ; 18 × 106 mm 2 =
I + (6000 mm 2 )(50 mm) 2
I = 3 × 106 mm 4
3 × 106 mm 4 + (6000 mm 2 )(50 mm + 10 mm)2
I BB′ =
I + Ad 2 =
3 106 + 6000(60)2 =
24.6 × 106 mm 4
=×
I=
24.6 × 106 mm 4 
BB′
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PROBLEM 7.35
Determine the moments of inertia I x and I y of the area shown with respect
to centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate centroid C of the area.
Then
A, in 2
x , in.
1
5×8 =
40
2.5
2
−2 × 5 =−10
1.9
Σ
30
where
4
100
160
4.3
–19
–43
81
117
X = 2.70 in.
YΣA =
Σ yA: Y (30 in 2 ) =
117 in 3
Y = 3.90 in.
or
Now
yA, in 3
XΣA =
Σ xA: X (30 in 2 ) =
81 in 3
or
and
xA, in 3
y , in.
=
I x ( I x )1 − ( I x )2
1
(5 in.)(8 in.)3 + (40 in 2 )[(4 − 3.9) in.]2
12
= (213.33 + 0.4) in 4 = 213.73 in 4
1
( I x )=
(2 in.)(5 in.)3 + (10 in 2 )[(4.3 − 3.9) in.]2
2
12
= (20.83 + 1.60) = 22.43 in 4
(I =
x )1
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SOLUTION Continued
Then
Also
where
Then
=
I x (213.73 − 22.43) in 4
or I x = 191.3 in 4 
=
I y ( I y )1 − ( I y ) 2
1
(8 in.)(5 in.)3 + (40 in 2 )[(2.7 − 2.5) in.]2
12
= (83.333 + 1.6) in 4 = 84.933 in 4
1
( I y )=
(5 in.)(2 in.)3 + (10 in 2 )[(2.7 − 1.9) in.]2
2
12
=(3.333 + 6.4) in 4 =9.733 in 4
( I y )1=
=
I y (84.933 − 9.733) in 4
or
I y = 75.2 in 4 
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PROBLEM 7.36
Determine the moments of inertia I x and I y of the area shown with respect to
centroidal axes respectively parallel and perpendicular to side AB.
SOLUTION
First locate C of the area.
Symmetry implies X = 18 mm.
A, mm 2
1
2
Σ
Then
y , mm
yA, mm3
14
14,112
42
31,752
36 × 28 =
1008
1
× 36 × 42 =
756
2
Dimensions in mm
1764
45,864
Y ΣA =
Σ yA: Y (1764 mm 2 ) =45,864 mm3
or
Now
where
Y = 26.0 mm
=
I x ( I x )1 + ( I x ) 2
1
(36 mm)(28 mm)3 + (1008 mm 2 )[(26 − 14) mm]2
12
=
(65,856 + 145,152) mm 4 =
211.008 × 103 mm 4
( I=
x )1
1
(36 mm)(42 mm)3 + (756 mm 2 )[(42 − 26) mm]2
36
=
(74, 088 + 193,536) mm 4 =
267.624 × 103 mm 4
I x )2
(=
Then
I x = (211.008 + 267.624) × 103 mm 4
or
=
I x 479 × 103 mm 4 
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SOLUTION Continued
Also
where
=
I y ( I y )1 + ( I y )2
1
(28 mm)(36=
mm)3 108.864 × 103 mm 4
12
1

1

=
( I y ) 2 2  (42 mm)(18 mm)3 +  × 18 mm × 42 mm  (6 mm)2 
2

 36

=
( I y )1
=
2(6804 + 13, 608) mm 4 =
40.824 × 103 mm 4
[( I y ) 2 is obtained by dividing A2 into
Then
]
I y = (108.864 + 40.824 × 103 mm 4
I y 149.7 × 103 mm 4 
or =
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PROBLEM 7.37
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to side
AB.
SOLUTION
First locate centroid C of the area.
A, in 2
or
and
or
y , in.
xA, in 3
yA, in 3
1
3.6 × 0.5 =
1.8
1.8
0.25
3.24
0.45
2
0.5 × 3.8 =
1.9
0.25
2.4
0.475
4.56
3
1.3 × 1 =
1.3
0.65
4.8
0.845
6.24
4.560
11.25
Σ
Then
x , in.
5.0
XΣA =
Σ xA: X (5 in 2 ) =4.560 in 3
X = 0.912 in.
Y Σ A =Σ yA: Y (5 in 2 ) =
11.25 in 3
Y = 2.25 in.
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SOLUTION Continued
Now
where
I x = ( I x )1 + ( I x ) 2 + ( I x )3
1
(3.6 in.)(0.5 in.)3 + (1.8 in 2 )[(2.25 − 0.25) in.]2
12
=(0.0375 + 7.20) in 4 =7.2375 in 4
( I=
x )1
1
(0.5 in.)(3.8 in.)3 + (1.9 in 2 )[(2.4 − 2.25) in.]2
12
=
(2.2863 + 0.0428) in 4 =
2.3291 in 4
(=
I x )2
1
(1.3 in.)(1 in.)3 + (1.3 in 2 )[(4.8 − 2.25 in.)]2
12
=
(0.1083 + 8.4533) in 4 =
8.5616 in 4
( I x=
)3
Then
I x = (7.2375 + 2.3291 + 8.5616) in 4 = 18.1282 in 4
or
Also
where
I x = 18.13 in 4 
I y = ( I y )1 + ( I y )2 + ( I y )3
1
(0.5 in.)(3.6 in.)3 + (1.8 in 2 )[(1.8 − 0.912) in.]2
12
=
(1.9440 + 1.4194) in 4 =
3.3634 in 4
=
( I y )1
1
(3.8 in.)(0.5 in.)3 + (1.9 in 2 )[(0.912 − 0.25) in.]2
12
=
(0.0396 + 0.8327) in 4 =
0.8723 in 4
(I y =
)2
( I y )3 =
Then
1
(1 in.)(1.3 in.)3 + (1.3 in 2 )[(0.912 − 0.65) in.]2
12
=
(0.1831 + 0.0892) in 4 =
0.2723 in 4
I y = (3.3634 + 0.8723 + 0.2723)in 4 = 4.5080 in 4
or
I y = 4.51 in 4 
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PROBLEM 7.38
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
A, in 2
1
2
Σ
π
2
y , in.
8
(6) 2 = 56.5487
π
yA, in 3
= 2.5465
1
− (12)(4.5) =
−27
1.5
2
29.5487
144
–40.5
103.5
Y ΣA =Σ yA: Y (29.5487 in 2 ) =103.5 in 3
Then
Y = 3.5027 in.
or
=
J O ( J O )1 − ( J O ) 2
(a)
where
=
( J O )1
Now
=
( I x′ ) 2
and
π
=
(6 in.) 4 107.876 in 4
4
( J=
( I x′ ) 2 + ( I y ′ ) 2
O )2
1
=
(12 in.)(4.5 in.)3 91.125 in 4
12
1
3
4
( I y′ ) 2 2=
=
12 (4.5 in.)(6 in.)  162.0 in


[Note: ( I y′ ) 2 is obtained using
]
SOLUTION Continued
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Then
Finally
=
( J O )2 (91.125 + 162.0) in 4
= 253.125 in 4
JO =
(1017.876 − 253.125) in 4 =
764.751 in 4
or
J O = 765 in 4 
or
J C = 402 in 4 
J=
J C + AY 2
O
(b)
or
=
J C 764.751 in.4 − (29.5487 in 2 )(3.5027 in.) 2
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PROBLEM 7.39
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
Symmetry: X = 0
Dimensions in mm
Determination of centroid C of entire section
Y Σ A =Σ yA
2
=
Y (4000 mm
) 122.67 × 103 mm3
Y = 30.667 mm
Area mm2
1
2
Σ
(a)
1
(160)(80) = 6400
2
y , mm
yA, mm3
80
3
170.67 × 103
1
20
− (80)(60) =
−2400
2
−48 × 103
122.67 × 103
4000
Polar moment of inertia J O :
Section :
=
Ix
1
3
(160)(80)
=
6.8267 × 106 mm 4
12
For I y consider the following two triangles
1
3
6
4
=
=
I y 2  (80)(80)
 6.8267 × 10 mm
12


J O = I x + I y = (6.8267 + 6.8267)106 = 13.653 × 106 mm 4
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SOLUTION Continued
(b)
Section :
=
Ix
1
3
=
(80)(60)
1.44 × 106 mm 4
12
For I y consider the following two triangles
1
3
6
4
=
I y 2  (60)(40)
=
 0.640 × 10 mm
12


J O = I x + I y = (1.44 + 0.640)106 = 2.08 × 106 mm 4
Entire section
J O= ( J O )1 − ( J O )2= 13.653 × 106 − 2.08 × 106
=
J O 11.573 × 106 mm 4
(c)
=
J O 11.57 × 106 mm 4 
Polar moment of inertia J O of intire area
J=
J C + AY 2
O
11.573 × 106 = J C + (4000)(30.667) 2
=
J C 7.811 × 106 mm 4
=
J C 7.81 × 106 mm 4 
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PROBLEM 7.40
Determine the polar moment of inertia of the area shown with respect to (a) Point O,
(b) the centroid of the area.
SOLUTION
Area, in2
x , in.
(4.5) 2 = 15.904
1.9099
30.375
1.2732
–9.00
Section
1
2
π
4
−
π
4
(3) 2 =
−7.069
Σ
Then
8.835
21.375
XA=
Σ xA: X (8.835 in 2 ) =
21.375 in 3
X = 2.419 in.
or
Then
xA, in 3
JO =
=
OC
π
8
(4.5 in.) 4 −
=
2X
π
8
(3 in.) 4 = 129.22 in 4
J O = 129.2 in 4 
2(2.419=
in.) 3.421 in.
J=
J C + A(OC ) 2 :
O
4
129.22 in=
J C + (8.835 in.)(3.421 in.) 2
J C = 25.8 in 4 
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PROBLEM 7.41
Two channels are welded to a rolled W section as shown. Determine
the moments of inertia and the radii of gyration of the combined
section with respect to the centroidal x and y axes.
SOLUTION
W section:
A = 9.12 in 2
I x = 110 in 4
I y = 37.1 in 4
Channel:
A = 3.37 in 2
I x = 1.31 in 4
I y = 32.5 in 4
Atotal
= AW + 2Achan
=
9.12 + 2(3.37) =
15.86 in 2
Now
where
=
I x ( I x ) W + 2( I x )chan
( I x )=
I xchan + Ad 2
chan
1.31 in 4 + (3.37 in 2 )(4.572 in.)2 =
71.754 in 4
=
Then
4
I=
(110 + 2 × 71.754) in=
253.51 in 4
x
Ix
253.51 in 4
=
Atotal 15.86 in 2
and
=
k x2
Also
=
I y ( I y ) W + 2( I y )chan
4
= (37.1 + 2 × 32.5) in=
102.1 in 4
and
=
k y2
Iy
102.1 in 4
=
Atotal 15.86 in 2
I x = 254 in 4 
k x = 4.00 in. 
I y = 102.1 in 4 
k y = 2.54 in. 
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PROBLEM 7.42
Two L6 × 4 × 12 -in. angles are welded together to form the section shown.
Determine the moments of inertia and the radii of gyration of the combined
section with respect to the centroidal x and y axes.
SOLUTION
From Figure 9.13A
From Appendix. B:
Area =A= 4.75 in2
I x′ = 17.3 in 4
I y′ = 6.22 in 4
I x =2[ I x′ + AyO2 ] =2[17.3 in 4 + (4.75 in 2 )(1.02 in.) 2 ] =44.484 in 4
I x = 44.5 in 4 
Total area = 2(4.75) = 9.50 in2
=
k x2
Ix
44.484 in 4
=
= 4.6825 in 2
area
9.50 in 2
k x = 2.16 in. 
I y = 2[ I y′ + AxO2 ] = 2[6.22 in 4 + (4.75 in 2 )(1.269 in.) 2 ]
= 27.738 in 4
I y = 27.7 in 4 
Total area = 2(4.75) = 9.50 in2
=
k y2
Iy
27.738 in 4
=
= 2.9198
area
9.50 in 2
k y = 1.709 in 
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PROBLEM 7.43
Two channels and two plates are used to form the column
section shown. For b = 200 mm, determine the moments of
inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
From Figure 9.13B
For C250 × 22.8
A = 2890 mm 2
I x 28.0 × 106 mm 4
=
I y 0.945 × 106 mm 4
=
Total area
A =+
2[2890 mm 4 (10 mm)(375 mm)] =
13.28 × 103 mm 2
Given b= 200 mm:
1

2[28.0 × 106 mm 4 ] + 2  (375 mm)(10 mm)3 + (375 mm)(10 mm)(132 mm)2 
Ix =
12


= 186.743 × 106
2
k=
x
I x 186.743 × 106
=
= 14.0620 × 103 mm 2
A
13.28 × 103
=
I x 186.7 × 106 mm 4 
k x = 118.6 mm 
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consent of McGraw-Hill Education.
SOLUTION Continued
Channel
Plate
1

Iy =
Σ( I y + Ad 2 ) =
2[ I y + Ad 2 ] + 2  (10 mm)(375 mm)3 
12


2

 200 mm
 
= 2 0.945 × 106 mm 4 + (2890 mm 2 ) 
+ 16.10 mm   + 87.891 × 106 mm 4
2

 

= 2[0.945 × 106 + 38.955 × 106 ] + 87.891 × 106
= 167.691 × 106 mm 4
2
k=
y
I y 167.691 × 106
=
= 12.6273 × 103
3
A
13.28 × 10
=
I y 167.7 × 106 mm 4 
k y = 112.4 mm 
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PROBLEM 7.44
Two 20-mm steel plates are welded to a rolled S section as shown. Determine
the moments of inertia and the radii of gyration of the combined section with
respect to the centroidal x and y axes.
SOLUTION
A = 6010 mm 2
S section:
=
I x 90.3 × 106 mm 4
=
I y 3.88 × 106 mm 4
Atotal
= AS + 2 Aplate
Note:
= 6010 mm 2 + 2(160 mm)(20 mm)
= 12, 410 mm 2
=
I x ( I x )S + 2( I x ) plate
Now
( I x )=
I xplate + Ad 2
plate
where
1
(160 mm)(20 mm)3 + (3200 mm 2 )[(152.5 + 10) mm]2
12
= 84.6067 × 106 mm 4
=
I=
(90.3 + 2 × 84.6067) × 106 mm 4
x
Then
= 259.5134 × 106 mm 4
=
k x2
and
Also
Ix
259.5134 × 106 mm 4
=
Atotal
12410 mm 2
or =
I x 260 × 106 mm 4 
or
k x = 144.6 mm 
=
I y ( I y )S + 2( I y ) plate
1

=
3.88 × 106 mm 4 + 2  (20 mm)(160 mm)3 
12

= 17.5333 × 106 mm 4
and
=
k y2
Iy
17.5333 × 106 mm 4
=
Atotal
12, 410 mm 2
I y 17.53 × 106 mm 4 
or =
or
k y = 37.6 mm 
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PROBLEM 7.45
Two L4 × 4 × 12 -in. angles are welded to a steel plate as shown. Determine
the moments of inertia of the combined section with respect to centroidal
axes respectively parallel and perpendicular to the plate.
SOLUTION
1
For 4 × 4 × -in. angle:
2
=
A 3.75 in 2 , I=
I=
5.52 in 4
x
y
YA = Σ y A
Y (12.5 in 2 ) = 33.85 in 3
Y = 2.708 in.
Area, in2
y in.
yA, in 3
Plate
(0.5)(10) = 5
5
25
Two angles
2(3.75) = 7.5
1.18
8.85
Section
Σ
12.5
33.85
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SOLUTION Continued
Entire section:
1
3
2
2
Ix =
Σ( I x′ + Ad 2 ) =
12 (0.5)(10) + (0.5)(10)(2.292)  + 2[5.52 + (3.75)(1.528) ]


= 41.667 + 26.266 + 1604 + 17.511= 96.48 in 4
=
Iy
I x = 96.5 in 4 
1
(10)(0.5)3 + 2[5.52 + (3.75)(1.43) 2 ]
12
= 0.104 + 11.04 + 15.367 = 26.51 in 4
I y = 26.5 in 4 
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PROBLEM 7.46
A channel and a plate are welded together as shown to form a
section that is symmetrical with respect to the y axis.
Determine the moments of inertia of the combined section
with respect to its centroidal x and y axes.
SOLUTION
From Figure 9.13B
B = 3.37 in 2
For C8 × 11.5:
I y = 1.31 in 4
(Note change of axes)
I y = 32.5 in 4
Location of centroid
YΣ A=
Σ y A:
Y [3.7 in 2 + (12 in.)(0.5 in.)] =
(3.37 in 2 )(1.938 in.)
Y = 0.69702 in.
Moment of inertia with respect to x axis.
1
Ix =
I x + AY 2 = (12 in.)(0.5 in.)3 + (12 in.)(0.5 in.)(0.69702 in.) 2
12
=0.125 + 2.9150 =3.0400 in 4
Plate:
Ix =
I x′ + AY02 =
1.31 in 4 + (3.37 in.)(1.938 in. − 0.69702 in.)2
Channel:
=
1.31 + 5.1899 =
6.4999 in 4
Entire section:
I x = 3.0400 in 4 + 6.4999 in 4 = 9.5399 in 4
I x = 9.54 in 4 
Moment of inertia with respect to y axis.
Plate:
=
Iy
1
=
(0.5 in.)(12 in.)3 72 in 4
12
Channel:
I y = 32.5 in 4
Entire section:
Iy =
72 in 4 + 32.5 in 4 =
104.5 in 4
I y = 104.5 in 4 
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PROBLEM 7.47
Two L76 × 76 × 6.4-mm angles are welded to a C250 × 22.8 channel.
Determine the moments of inertia of the combined section with respect to
centroidal axes respectively parallel and perpendicular to the web of the
channel.
SOLUTION
A = 929 mm 2
Angle:
I=
I=
0.512 × 106 mm 4
x
y
A = 2890 mm 2
Channel:
=
I x 0.945 × 106 mm 4
=
I y 28.0 × 106 mm 4
First locate centroid C of the section
Then
A, mm 2
y , mm
yA, mm3
Angle
2(929) = 1858
21.2
39,389.6
Channel
2890
−16.1
−46,529
Σ
4748
−7139.4
Y ΣA=
Σ y A: Y (4748 mm 2 ) =
−7139.4 mm3
Y = −1.50366 mm
or
Now
=
I x 2( I x ) L + ( I x )C
where
(I x )L =
I x + Ad 2 =
0.512 × 106 mm 4 + (929 mm 2 )[(21.2 + 1.50366) mm]2
= 0.990859 × 106 mm 4
( I x )C =
0.949 × 106 mm 4 + (2890 mm 2 )[(16.1 − 1.50366) mm]2
I x + Ad 2 =
= 1.56472 × 106 mm 4
Then
=
I x [2(0.990859) + 1.56472 × 106 ] mm 4
or
=
I x 3.55 × 106 mm 4 
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SOLUTION Continued
Also
=
I y 2( I y ) L + ( I y )C
where
(I y )L =
I y + Ad 2 =
0.512 × 106 mm 4 + (929 mm 2 )[(127 − 21.2) mm]2
= 10.9109 × 106 mm 4
( I y )C = I y
Then
=
I y [2(10.9109) + 28.0] × 106 mm 4
or
=
I y 49.8 × 106 mm 4 
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PROBLEM 7.48
The strength of the rolled W section shown is increased by welding a channel to
its upper flange. Determine the moments of inertia of the combined section with
respect to its centroidal x and y axes.
SOLUTION
A = 14, 400 mm 2
W section:
=
I x 554 × 106 mm 4
=
I y 63.3 × 106 mm 4
A = 2890 mm 2
Channel:
=
I x 0.945 × 106 mm 4
=
I y 28.0 × 106 mm 4
First locate centroid C of the section.
W Section
Channel
Σ
Then
A, mm2
y , mm
yA, mm3
14,400
−231
−33,26,400
2,890
49.9
1,44,211
17,290
YΣ A=
Σ y A: Y (17, 290 mm 2 ) =
−3,182,189 mm3
Y = −184.047 mm
or
Now
where
−31,82,189
=
I x ( I x ) W + ( I x )C
( I x ) W= I x + Ad 2
=×
554 106 mm 4 + (14, 400 mm 2 )(231 − 184.047) 2 mm 2
= 585.75 × 106 mm 4
( I x )C= I x − Ad 2
=
0.945 × 106 mm 4 + (2,890 mm 2 )(49.9 + 184.047)2 mm 2
= 159.12 × 106 mm 4
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SOLUTION Continued
Then
I x = (585.75 + 159.12) × 106 mm 4
or
also
=
I x 745 × 106 mm 4 
=
I y ( I y ) W + ( I y )C
= (63.3 + 28.0) × 106 mm 4
I y 91.3 × 106 mm 4 
or =
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PROBLEM 7.49
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
By observation
Now
y=
h
x
b
=
dI y x 2 =
dA x 2 [(h − y )dx]
x

= hx 2 1 −  dx
 b
Then
=
Iy
dI
∫=
∫
y
b
0
x

hx 2 1 −  dx
 b
b
1
x4 
= h  x3 − 
4b  0
3
or
Iy =
1 3
b h
12
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PROBLEM 7.50
Determine by direct integration the moment of inertia of the shaded area
with respect to the x axis.
SOLUTION
By observation
y=
h
x
b
or
x=
b
y
h
Now
2
=
dI x y=
dA y 2 ( x dy )
b

= y 2  y dy 
h

b
= y 3 dy
h
Then
=
Ix
dI
∫=
∫
x
h
0
b 3
y dy
h
h
=
b 1 4 
y
h  4  0
or
Ix =
1 3
bh 
4
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PROBLEM 7.51
Determine by direct integration the moment of inertia of the shaded area
with respect to the y axis.
SOLUTION
h = k ( 2a )
At x 2=
a, y h :
=
2
or
k=
h
4a 2
Then
y=
h 2
x
4a 2
=
dA ydx
=
Now
=
A
dA
∫=
h
4a 2
∫
2a
h 2
x dx
4a 2
x 2 dx
a
2a
3
3
a3  7
 h  x 
 h   8a
=
=
−
A  2  =
ah




3  12
 4a   3  a  4a 2   3
=
dI x
1 3
1 h  6
=
y dx

 x dx
3
3  64a 6 
=
Ix
Then
∫
=
dI x
h3
192a 6
∫
2a
x 6 dx
0
2a
 h3  x 7
h3  128a 7 − a 7 
=
I x =


6
6 
7
 192a  7 a 192a 

=
Ix
127
=
ah3 0.094494ah3
7
192
( )( )
I x = 0.0945ah3 
2
k x=
I x 0.094494ah3
=
= 0.161983h 2
7
A
ah
12
k x = 0.402h 
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PROBLEM 7.52
Determine the moment of inertia and the radius of gyration of the shaded
area shown with respect to the y axis.
SOLUTION
See figure of solution on Problem 9.16
=
y
h 2
7
 h

2
x =
A
ah =
dA ydx =
dI y x=
dA x 2  2 x 2  dx
2
12
4a
 4a

h 2a 4
h
=
Iy =
x dx
2 0
4a
4a 2
∫
=
Iy
h
4a 2
2
k=
y
Iy
=
A
 x5

 5
2a


a
 32a5 a5  31 3
−=
ha


5  20
 5
(
)
ha3
93 2
=
a
7
35
ah
12
31
20
Iy =
31 3
ha 
20
ky = a
93

35
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PROBLEM 7.53
Determine the polar moment of inertia and the polar radius of gyration of
the isosceles triangle shown with respect to Point O.
SOLUTION
By observation:
y=
h
or
x=
b
y
2h
Now
b
2
x
 b
=
dA xdy
= 
 2h
and
2
=
dI x y=
dA
Then
=
Ix
∫

y  dy

b 3
y dy
2h
=
dI x 2
∫
h
0
b 3
y dy
2h
h
1 3
b y4
= =
bh
h 4 0 4
From above:
Now
y=
2h
x
b
2h 

(h − y )dx =
dA =
 h − b x  dx


=
and
2
=
dI y x=
dA x 2
h
(b − 2 x)dx
b
h
(b − 2 x)dx
b
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SOLUTION Continued
Then
=
Iy
2∫
dI
∫=
y
b/2
0
h 2
x (b − 2 x)dx
b
b/2
h 1
1 
= 2  bx3 − x 4 
b 3
2 0
3
4
h b  b  1  b   1 3
= 2    −  =
bh

b  3  2  2  2   48
Now
1
1
J O = I x + I y = bh3 + b3 h
4
48
and
2
k=
O
JO
=
A
bh
48
(12h 2 + b 2 ) 1
(12h 2 + b 2 )
=
1
24
bh
2
or J O
=
or
bh
(12h 2 + b 2 ) 
48
kO =
12h 2 + b 2

24
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PROBLEM 7.54
Determine the moments of inertia of the shaded area shown with respect to the x
and y axes when a = 20 mm.
SOLUTION
We have
where
=
I x ( I x )1 + 2( I x ) 2
1
(40 mm)(40 mm)3
12
= 213.33 × 103 mm 4
( I x )1 =
2
π
π
 
4
2  4 × 20
=
( I x )2  (20 mm) − (20 mm) 
mm  
2
 3π
 
 8
+
 4 × 20

π

+ 20  mm 
(20 mm)2 
2

 3π

2
= 527.49 × 103 mm 4
Then
Ix =
[213.33 + 2(527.49)] × 103 mm 4
or =
I x 1.268 × 106 mm 4 
Also
where
Then
=
I y ( I y )1 + 2( I y )2
=
( I y )1
1
(40 mm)(40 =
mm)3 213.33 × 103 mm 4
12
(=
I y )2
π
4
(20 mm)
62.83 × 103 mm 4
=
8
I y =[213.33 + 2(62.83)] × 103 mm 4
or
=
I y 339 × 103 mm 4 
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PROBLEM 7.55
Determine the moments of inertia I x and I y of the area shown
with respect to centroidal axes respectively parallel and
perpendicular to side AB.
SOLUTION
Dimensions in mm
First locate centroid C of the area.
Symmetry implies Y = 30 mm.
1
Then
or
Now
where
A, mm 2
x , mm
108 × 60 =
6480
54
349,920
46
–59,616
2
1
− × 72 × 36 =−1296
2
Σ
5184
xA, mm3
290,304
X ΣA =
Σ xA : X (5184 mm 2 ) =290,304 mm3
X = 56.0 mm
=
I x ( I x )1 − ( I x )2
1
3
=
(108 mm)(60 mm)
1.944 × 106 mm 4
12
1

1

=
( I x ) 2 2  (72 mm)(18 mm)3 +  × 72 mm × 18 mm  (6 mm)2 
2

 36

I x )1
(=
=
2(11, 664 + 23,328) mm 4 =
69.984 × 103 mm 4
[( I x ) 2 is obtained by dividing A2 into
Then
]
Ix =
(1.944 − 0.069984) × 106 mm 4
or
=
I x 1.874 × 106 mm 4 
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SOLUTION Continued
Also
where
=
I y ( I y )1 − ( I y ) 2
1
(60 mm)(108 mm)3 + (6480 mm 2 )[(56.54) mm]2
12
=
(6, 298,560 + 25,920) mm 4 =
6.324 × 106 mm 4
=
( I y )1
1
(36 mm)(72 mm)3 + (1296 mm 2 )[(56 − 46) mm]2
36
=
(373, 248 + 129, 600) mm 4 =
0.502 × 106 mm 4
( I=
y )2
Then
I y (6.324 − 0.502)106 mm 4
=
or =
I y 5.82 × 106 mm 4 
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PROBLEM 7.56
Determine the moments of inertia I x and I y of the area shown with
respect to centroidal axes respectively parallel and perpendicular to
side AB.
SOLUTION
First locate C of the area:
Symmetry implies X = 12 mm.
Then
A, mm 2
y , mm
yA, mm3
1
12 × 22 =
264
11
2904
2
1
(24)(18) = 216
2
28
6048
Σ
480
8952
Y ΣA =Σ yA: Y (480 mm 2 ) =8952 mm3
Y = 18.65 mm
Now
=
I x ( I x )1 + ( I x ) 2
where
( I x )=
1
1
(12 mm)(22 mm)3 + (264 mm 2 )[(18.65 − 11) mm]2
12
= 26, 098 mm 4
1
(24 mm)(18 mm)3 + (216 mm 2 )[(28 − 18.65) mm]2
36
= 22, 771 mm 4
(=
I x )2
Then
I x = (26.098 + 22.771) × 103 mm 4
or
=
I x 48.9 × 103 mm 4 
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SOLUTION Continued
Also
where
I y ( I y )1 + ( I y )2
=
1
=
(22 mm)(12 mm)3 3168 mm 4
12
1

1

=
( I y ) 2 2  (18 mm)(12 mm)3 +  × 18 mm × 12 mm  (4 mm)2 
2

 36

=
( I y )1
= 5184 mm 4
[( I y ) 2 is obtained by dividing A2 into
Then
]
=
I y (3168 + 5184) mm 4
I y 8.35 × 103 mm 4 
or =
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PROBLEM 7.57
The shaded area is equal to 50 in 2 . Determine its centroidal moments
of inertia I x and I y , knowing that I y = 2 I x and that the polar moment
of inertia of the area about Point A is JA = 2250 in 4 .
SOLUTION
Given:
2
=
I y 2=
I x , J A 2250 in 4
A = 50 in
J=
J C + A(6 in.)2
A
4
2250 in=
J C + (50 in 2 )(6 in.) 2
J C = 450 in 4
JC =
I x + I y with
4
450 in=
I x + 2I x
Iy =
2I x
I x = 150.0 in 4 
=
I y 2=
I x 300 in 4 
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PROBLEM 7.58
Determine the polar moment of inertia of the area shown with respect
to (a) Point O, (b) the centroid of the area.
SOLUTION
First locate centroid C of the figure.
Note that symmetry implies Y = 0.
Dimensions in mm
A, mm 2
π
1
2
2
−
3
π
2
−
4
Σ
Then
2
π
56
π
−2290.22
(54)(27) =
π
112
(42)2 = 2770.88
−
72
π
36
(27)2 =
−1145.11
π
=
−35.6507
XA, mm3
−197,568
= 17.8254
49,392
=
−22.9183
52,488
= 11.4592
4877.32
–13,122
–108,810
X ΣA =Σ xA: X (4877.32 mm 2 ) =−108,810 mm3
X = −22.3094
or
(a)
−
(84)(42) = 5541.77
π
2
X , mm
J O = ( J O )1 + ( J O ) 2 − ( J O )3 − ( J O )4
where
=
( J O )1
π
(84 mm)(42 mm)[(84 mm)2 + (42 mm)2 ]
8
= 12.21960 × 106 mm 4
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SOLUTION Continued
( J O )2 =
π
(42 mm)4
4
= 2.44392 × 106 mm 4
=
( J O )3
π
(54 mm)(27 mm)[(54 mm) 2 + (27 mm) 2 ]
8
= 2.08696 × 106 mm 4
( J O )4 =
π
(27 mm) 4
4
= 0.41739 × 106 mm 4
Then
J O = (12.21960 + 2.44392 − 2.08696 − 0.41739) × 106 mm 4
= 12.15917 × 106 mm 4
or =
J O 12.16 × 106 mm 4 
J=
J C + AX 2
O
(b)
or
J C = 12.15917 × 106 mm 4 − (4877.32 mm 2 )(−22.3094 mm)2
or
=
J C 9.73 × 106 mm 4 
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PROBLEM 7.59
Determine the polar moment of inertia of the area shown with respect to
(a) Point O, (b) the centroid of the area.
SOLUTION
Determination of centroid C of entire section:
Area, in2
Section
1
π
4
(4)2 = 4π
2
1
(4)(4) = 8
2
3
1
(4)(4) = 8
2
Σ
28.566
x , in.
xA, in 3
16
3π
21.333
−
4
3
4
3
−10.6667
10.6667
21.333
X ΣA =
Σ xA: X (28.566 in 2 ) =21.333 in 3
X= 0.74680 in.; by symmetry, Y= X= 0.74680 in.
(a)
For section :
=
Ix
For sections ,:
(b)
=
Ix
11 4 1
=
πr  =
π (4)4 50.265 in 4
4  4
 16
1 3 1
3
=
bh
=
(4)(4)
21.333 in 4
12
12
For total area,
Ix =
50.265 + 2(21.333) =
92.931 in 4
By symmetry,
I y = I x ; J O = I x + I y = 185.862 in 4
Parallel axis theorem:
JC =
J O − A(OC )2 =
J O − A( X 2 + Y 2 )
JC =
185.862 − (28.566)(0.746802 + 0.746802 )
J O = 185.9 in 4 
J C = 154.0 in 4 
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PROBLEM 7.60
Two L6 × 4 × 12 -in. angles are welded together to form the section
shown. Determine the moments of inertia and the radii of gyration
of the combined section with respect to the centroidal x and y axes.
SOLUTION
=
A 9.12
=
in 2 I x 110
=
in 4 I y 37.1 in 4
W section:
Angle:
=
A 1.44
=
=
in 2 I x 1.23
in 4 I y 1.23 in 4
= AW + 4AA
Atotal
=
9.12 + 4(1.44) =
14.880 in 2
Now
where
=
I x ( I x ) W + 4( I x )A
I xA + Ad 2
( I x )=
A
=
1.23 in 4 + (1.44 in 2 )(4.00 in. + 0.836 in.)2 =
34.907 in 4
Then
4
I=
(110 + 4 × 34.907) in=
249.63 in 4
x
Ix
249.63 in 4
=
Atotal 14.880 in 2
and
=
k x2
Also
=
I y ( I y ) W + 4( I y )A
where
(Iy =
)A
I x = 250 in 4 
k x = 4.10 in. 
I yA + Ad 2
=(1.23 + 1.44 ( 5 − 0.836 ) ) in 4 =26.198 in 4
2
4
I=
(37.1 + 4 × 26.198) in=
141.892 in 4 I y = 141.9 in 4 
y
=
k y2
Iy
141.892 in 4
=
Atotal 14.880 in 2
k y = 3.09 in. 
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