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Engineering Mechanics - Statics Chapter 10
Problem 10-16
Determine the moment of inertia of the
shaded area about the x axis.
Given:
a 2in=
b 4in=
Solution:
I
x
a−
a
x
1
3
bcos
π
x
2a













3




d=
I
x
36.2 in
4
=
Problem 10-17
Determine the moment of inertia for the shaded area
about the y axis.
Given:
a 2in=
b 4in=
Solution:
I
y
a−
a
xx
2
bcos
π
x
2a











d=
I
y
7.72 in
4
=
Problem 10-18
Determine the moment of inertia for the shaded area about the x axis.
1001
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Given:
a 4in=
b 2in=
Solution:
Solution
I
x
a−
a

x
bcos
π
x
2a












3
3






d=
I
x
9.05 in
4

=
Problem 10-19
Determine the moment of inertia for the shaded area about the y axis.
Given:
a 4in=
b 2in=
Solution:
I
y
a−
a
xx
2
bcos
π
x
2a










d=
I
y

30.9 in
4
=
Problem 10-20
Determine the moment for inertia of the shaded area about the x axis.
1002
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Given:
a 2in=
b 4in=
c 12 in=
Solution:
I
x
a
ab+
x
1
3
c
2
x







3





d=
I
x
64.0 in
4
=
Problem 10-21
Determine the moment of inertia of
the shaded area about the y axis.
Given:
a 2in=
b 4in=
c 12 in=
Solution:
I
y
a
ab+
xx
2
c
2
x











d=
1003
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
I
y
192.00 in
4
=
Problem 10-22
Determine the moment of inertia for the shaded area about the x axis.
Given:
a 2m=
b 2m=
Solution:
I
x
0

b
yy
2
a
y
2
b
2













d=
I
x
3.20 m
4
=
Problem 10-23
Determine the moment of inertia for the shaded area about the y axis. Use Simpson's rule to
evaluate the integral.

Given:
a 1m=
b 1m=
1004
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics Chapter 10
I
y
0
a
xx
2
be
x
a






2





d=

I
y
0.628 m
4
=
Problem 10-24
Determine the moment of inertia for the shaded area about the x axis. Use Simpson's rule to
evaluate the integral.
Given:
a 1m=
b 1m=
Solution:
I
y
0
a
x
be
x
a






2









3
3







d=
I
y
1.41 m
4
=
Problem 10-25
The polar moment of inertia for the area is I
C
about the z axis passing through the centroid C.
The moment of inertia about the x axis is I
x
and the moment of inertia about the y' axis is I
y'
.
Determine the area A.

Given:
I
C
28 in
4
=
1005
Solution:
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
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Engineering Mechanics - Statics Chapter 10
I
x
17 in
4
=
I
y'
56 in
4
=
a 3in=
Solution:
I
C
I
x
I
y

+=
I
y
I
C
I
x
−=
I
y'
I
y
Aa
2
+=
A
I
y'
I
y

a
2
= A 5.00 in
2
=
Problem 10-26
The polar moment of inertia for the area is J
cc
about the z' axis passing through the centroid C. If

the moment of inertia about the y' axis is I
y'
and the moment of inertia about the x axis is I
x
.
Determine the area A.
Given:
J
cc
548 10
6
× mm
4
=
I
y'
383 10
6
× mm
4
=
I
x
856 10
6
× mm
4
=
h 250 mm=
Solution:

I
x'
I
x
Ah
2
−=
J
cc
I
x'
I
y'
+=
J
cc
I
x
Ah
2
− I
y'
+=
1006
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
A
I

x
I
y'
+ J
cc

h
2
=
A 11.1 10
3
× mm
2
=
Problem 10-27
Determine the radius of gyration k
x
of the column’s
cross-sectional area.
Given:
a 100 mm=
b 75 mm=
c 90 mm=
d 65 mm=
Solution:
Cross-sectional area:
A 2b()2a() 2d()2c()−=
Moment of inertia about the x axis:
I
x

1
12
2b()2a()
3
1
12
2d()2c()
3
−=
Radius of gyration about the x axis:
k
x
I
x
A
= k
x
74.7 mm=
Problem 10-28

Determine the radius of gyration k
y
of the column’s cross-sectional area.
Given:
a 100 mm=
b 75 mm=
1007
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.

Engineering Mechanics - Statics Chapter 10
c 90 mm=
d 65 mm=
Solution:
Cross-sectional area:
A 2b()2a() 2d()2c()−=
Moment of inertia about the y axis:
I
y
1
12
2a()2b()
3
1
12
2c()2d()
3
−=
Radius of gyration about the y axis:
k
y
I
y
A
= k
y
59.4 mm=
Problem 10-29
Determine the moment of
inertia for the beam's

cross-sectional area with
respect to the x' centroidal axis.
Neglect the size of all the rivet
heads, R, for the calculation.
Handbook values for the area,
moment of inertia, and location
of the centroid C of one of the
angles are listed in the figure.
Solution:
I
E
1
12
15 mm( ) 275 mm()
3
4 1.32 10
6
()
mm
4
1.36 10
3
()
mm
2
275 mm
2
28 mm−







2
+






+
2
1
12
75 mm( ) 20 mm()
3
75 mm( ) 20 mm()
275 mm
2
10mm+






2
+







+
=
1008
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
I
E
162 10
6
× mm
4
=
Problem 10-30
Locate the centroid y
c
of the cross-sectional area for the angle. Then find the moment
of inertia I
x'
about the x' centroidal axis.
Given:
a 2in=
b 6in=

c 6in=
d 2in=
Solution:
y
c
ac
c
2






bd
d
2






+
ac bd+
= y
c
2.00 in=
I
x'

1
12
ac
3
ac
c
2
y
c







2
+
1
12
bd
3
+ bd y
c
d
2








2
+= I
x'
64.00 in
4
=
Problem 10-31
Locate the centroid x
c
of the cross-sectional area for the angle. Then find the moment
1009
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
of inertia I
y'
about the centroidal y' axis.
Given:
a 2in=
b 6in=
c 6in=
d 2in=
Solution:
x
c
ac

a
2






bd a
b
2
+






+
ac bd+
= x
c
3.00 in=
I
y'
1
12
ca
3
ca x

c
a
2







2
+
1
12
db
3
+ db a
b
2
+ x
c







2
+= I

y'
136.00 in
4
=
Problem 10-32
Determine the distance x
c
to the centroid of the beam's cross-sectional area: then find
the moment of inertia about the y' axis.
1010
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Given:
a 40 mm=
b 120 mm=
c 40 mm=
d 40 mm=
Solution:
x
c
2 ab+()c
ab+
2







2ad
a
2
+
2 ab+()c 2da+
= x
c
68.00 mm=
I
y'
2
1
12
ca b+()
3
ca b+()
ab+
2
x
c







2
+







1
12
2da
3
+ 2da x
c
a
2







2
+=
I
y'
36.9 10
6
× mm
4
=

Problem 10-33
Determine the moment of inertia of the beam's cross-sectional area about the x' axis.
1011
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Given:
a 40 mm=
b 120 mm=
c 40 mm=
d 40 mm=
Solution:
I
x'
1
12
ab+()2c 2d+()
3
1
12
b 2d()
3
−= I
x'
49.5 10
6
× mm
4
=

Problem 10-34
Determine the moments of inertia for the shaded area about the x and y axes.
Given:
a 3in=
b 3in=
c 6in=
d 4in=
r 2in=
Solution:
I
x
1
3
ab+()cd+()
3
1
36
bc
3
1
2
bc d
2c
3
+







2
+







π
r
4
4
π
r
2
d
2
+






−=
1012
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may

be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
I
x
1192in
4
=
I
y
1
3
cd+()ab+()
3
1
36
cb
3
1
2
bc a
2b
3
+






2

+







π
r
4
4
π
r
2
a
2
+






−=
I
y
364.84 in
4
=

Problem 10-35
Determine the location of the centroid y' of the beam constructed from the two channels and the
cover plate. If each channel has a cross-sectional area A
c
and a moment of inertia about a horizontal
axis passing through its own centroid C
c
, of I
x'c ,
determine the moment of inertia of the beam’s
cross-sectional area about the x' axis.

Given:
a 18 in=
b 1.5 in=
c 20 in=
d 10 in=
A
c
11.8 in
2
=
I
x'c
349 in
4
=
Solution:
y
c

2A
c
dabc
b
2
+






+
2A
c
ab+
= y
c
15.74 in=
I
x'
I
x'c
A
c
y
c
d−
()
2

+




2
1
12
ab
3
+ ab c
b
2
+ y
c







2
+= I
x'
2158in
4
=
Problem 10-36
Compute the moments of inertia I

x
and I
y
for the beam's cross-sectional area about
1013
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
the x and y axes.
Given:
a 30 mm=
b 170 mm=
c 30 mm=
d 140 mm=
e 30 mm=
f 30 mm=
g 70 mm=
Solution:
I
x
1
3
ac d+ e+()
3
1
3
bc
3
+

1
12
ge
3
+ ge c d+
e
2
+






2
+= I
x
154 10
6
× mm
4
=
I
y
1
3
ca b+()
3
1
3

df
3
+
1
3
cf g+()
3
+= I
y
91.3 10
6
× mm
4
=
Problem 10-37
Determine the distance y
c
to the centroid C of the
beam's cross-sectional area and then compute the
moment of inertia I
cx'
about the x' axis.
Given:
a 30 mm= e 30 mm=
b 170 mm= f 30 mm=
c 30 mm= g 70 mm=
d 140 mm=
1014
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may

be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
y
c
ab+()c
c
2






df c
d
2
+






+ fg+()ec d+
e
2
+







+
ab+()cdf+ fg+()e+
=
y
c
80.7 mm=
I
x'
1
12
ab+()c
3
ab+()cy
c
c
2







2
+
1
12

fd
3
+ fd c
d
2
+ y
c







2
+
1
12
fg+()e
3
fg+()ec d+
e
2
+ y
c








2
++
=
I
x'
67.6 10
6
× mm
4
=
Problem 10-38
Determine the distance x
c
to the centroid C of the beam's cross-sectional area and then compute the
moment of inertia I
y'
about the y' axis.
Given:
a 30 mm=
b 170 mm=
c 30 mm=
d 140 mm=
e 30 mm=
f 30 mm=
g 70 mm=
Solution:
x
c

bc
b
2
a+






cd+()f
f
2






+ fg+()e
fg+
2
+
bc bc+ fg+()e+
=
x
c
61.6 mm=
1015
Solution:

© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
I
y'
1
12
ca b+()
3
ca b+()
ab+
2
x
c







2
+
1
12
df
3
+ df x
c

f
2







2
+
1
12
ef g+()
3
ef g+()x
c
fg+
2







2
++
=
I

y'
41.2 10
6
× mm
4
=
Problem 10-39
Determine the location y
c
of the centroid C of the beam’s cross-sectional area. Then compute
the moment of inertia of the area about the x' axis
Given:
a 20 mm=
b 125 mm=
c 20 mm=
f 120 mm=
g 20 mm=
d
fc−
2
=
e
fc−
2
=
Solution:
y
c
ag+()f
ag+

2






cb a g+
b
2
+






+
ag+()fcb+
=
y
c
48.25 mm=
I
x'
1
12
fa g+()
3
f()ag+()y

c
ag+
2







2
+
1
12
cb
3
+ cb
b
2
a+ g+ y
c







2
+=

I
x'
15.1 10
6
× mm
4
=
1016
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Problem 10-40
Determine y
c
, which locates the centroidal axis x' for the cross-sectional area of the T-beam, and then
find the moments of inertia I
x'
and I
y'
.
Given:
a 25 mm=
b 250 mm=
c 50 mm=
d 150 mm=
Solutuion:
y
c
b

2






b2ab
c
2
+






2dc+
b2ac2d+
=
y
c
207mm=
I
x'
1
12
2ab
3
2ab y

c
b
2







2
+
1
12
2dc
3
+ c2db
c
2
+ y
c







2
+=

I
x'
222 10
6
× mm
4
=
I
y'
1
12
b 2a()
3
1
12
c 2d()
3
+=
I
y'
115 10
6
× mm
4
=
Problem 10-41
Determine the centroid y' for the beam’s cross-sectional area; then find I
x'
.
Given:

a 25 mm=
1017
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
b 100 mm=
c 25 mm=
d 50 mm=
e 75 mm=
Solution:
y
c
2 ae+ d+()c
c
2






2ab c
b
2
+







+
2 ae+ d+()c 2ab+
= y
c
37.50 mm=
I
x'
2
12
ae+ d+()c
3
2 ae+ d+()cy
c
c
2







2
+
2
1
12
ab

3
ab c
b
2
+ y
c







2
+






+
=
I
x'
16.3 10
6
× mm
4
=

Problem 10-42
Determine the moment of inertia for the beam's cross-sectional area about the y axis.
Given:
a 25 mm=
b 100 mm=
c 25 mm=
d 50 mm=
e 75 mm=
1018
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Solution:
l
y
1
12
2
3
ad+ e+()
3
c 2
1
12
ba
3
ab e
a
2

+






2
+






+=
l
y
94.8 10
6
× mm
4
=
Problem 10-43
Determine the moment for inertia I
x
of the shaded area about the x axis.
Given:
a 6in=
b 6in=

c 3in=
d 6in=
Solution:
I
x
ba
3
3
1
12
ca
3
+
1
12
bc+()d
3
+= I
x
648in
4
=
Problem 10-44
Determine the moment for inertia I
y
of the shaded area about the y axis.
Given:
a 6in=
b 6in=
c 3in=

d 6in=
1019
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Solution:
I
y
ab
3
3
1
36
ac
3
+
1
2
ac b
c
3
+






2

+
1
36
db c+()
3
+
1
2
db c+()
2 bc+()
3






2
+=
I
y
1971in
4
=
Problem 10-45
Locate the centroid y
c
of the channel's cross-sectional area, and then determine the moment of
inertia with respect to the x' axis passing through the centroid.
Given:

a 2in=
b 12 in=
c 2in=
d 4in=
Solution:
y
c
c
2
bc 2
cd+
2






cd+()a+
bc 2 cd+()a+
=
y
c
2in=
I
x
1
12
bc
3

bc y
c
c
2







2
+
2
12
ac d+()
3
+ 2ac d+()
cd+
2
y
c







2

+=
I
x
128in
4
=
Problem 10-46
Determine the moments for inertia I
x
and I
y
of the shaded area.
Given:
r
1
2in=
1020
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
r
2
6in=
Solution:
I
x
π
r
2

4
8
π
r
1
4
8







= I
x
503in
4
=
I
y
π
r
2
4
8
π
r
1
4

8







= I
y
503in
4
=
Problem 10-47
Determine the moment of inertia for the parallelogram about the x' axis, which passes through
the centroid C of the area.
Solution:
ha( )sin
θ
()
=
I
xc
1
12
bh
3
=
1
12

ba( )sin
θ
()




3
=
1
12
a
3
bsin
θ
()
3
=
I
xc
1
12
a
3
bsin
θ
()
3
=
Problem 10-48

Determine the moment of inertia for the parallelogram about the y
'
axis, which passes through
the centroid C of the area.
1021
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Solution:
Aba( ) sin
θ
()
=
x
c
1
ba( ) sin
θ
()
ba( ) sin
θ
()
b
2
1
2
a( ) cos
θ
()

a( ) sin
θ
()
a( )cos
θ
()
3







1
2
a( ) cos
θ
()
a( ) sin
θ
()
b
a( )cos
θ
()
3
+







+











=
ba( )cos
θ
()
+
2
=
I
y'
1
12
a( ) sin
θ
()

b
3
a( )sin
θ
()
b
b
2
x
c







2
+
1
36
a( ) sin
θ
()
a( )cos
θ
()





3
1
2
a( ) sin
θ
()
a( ) cos
θ
()
x
c
a( )cos
θ
()
3







2
+







−+

1
36
a( ) sin
θ
()
a( )cos
θ
()




3
1
2
a( ) sin
θ
()
a( ) cos
θ
()
b
a( )cos
θ
()
3
+ x

c







2
++

=
Simplifying we find.
I
y'
ab
12
b
2
a
2
cos
θ
()
2
+
()
sin
θ
()

=
Problem 10-49
Determine the moments of inertia for the triangular area about the x' and y' axes, which pass
through the centroid C of the area.
1022
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Solution:
I
x'
1
36
bh
3
=
x
c
2
3
a
1
2
ha a
ba−
3
+







1
2
hb a−()+
1
2
ha
1
2
hb a−()+
=
ba+
3
=
I
y'
1
36
ha
3
1
2
ha
ba+
3
2
3

a−






2
+
1
36
hb a−()
3
+
1
2
hb a−()a
ba−
3
+
ba+
3







2

+=
I
y'
1
36
hb b
2
ab− a
2
+
()
=
Problem 10-50
Determine the moment of inertia for the
beam’s cross-sectional area about the x'
axis passing through the centroid C of
the cross section.
Given:
a 100 mm=
b 25 mm=
c 200 mm=
θ
45 deg=
1023
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Solution:
I

x'
1
12
2a 2 csin
θ
()
b+
()




3




4
1
12
ccos
θ
()()
csin
θ
()()
3







2
1
4
c
4
θ
1
2
sin 2
θ
()













−+
=
I

x'
520 10
6
× mm
4
=
Problem 10-51
Determine the moment of inertia of
the composite area about the x axis.
Given:
a 2in=
b 4in=
c 1in=
d 4in=
Solution:
I
x
1
3
ab+()2a()
3
π
c
4
4
π
c
2
a
2

+







0
d
x
1
3
2a 1
x
d






2














3





d+=
I
x
153.7 in
4
=
1024
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics Chapter 10
Problem 10-52
Determine the moment of inertia of
the composite area about the y axis.
Given:
a 2in=
b 4in=
c 1in=

d 4in=
Solution:
I
y
1
3
2a()ab+()
3
π
c
4
4
π
c
2
b
2
+







0
d
xx
2
2a 1

x
d






2











d+=
I
y
271.1 in
4
=
Problem 10-53
Determine the radius of gyration k
x
for the column's

cross-sectional area.
Given:
a 200 mm=
b 100 mm=
Solution:
I
x
1
12
2ab+()b
3
2
1
12
ba
3
ba
a
2
b
2
+






2
+







+=
k
x
I
x
b 2ab+()2ab+
= k
x
109mm=
1025
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may
be reproduced, in any form or by any means, without permission in writing from the publisher.

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