Engineering Mechanics - Statics
Chapter 10
Problem 10-1
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 2m
b = 4m
b
Solution:
⌠
y
2
Ix = 2 ⎮ y a 1 − d y
⎮
b
⌡
Ix = 39.0 m
4
0
Problem 10-2
Determine the moment of inertia for the shaded area about the y axis.
Given:
a = 2m
b = 4m
a
Solution:
⌠
2
⎮ 2 ⎡
⎛x⎞ ⎤
Iy = 2 ⎮ x b⎢1 − ⎜ ⎟ ⎥ dx
⎣ ⎝ a⎠ ⎦
⌡
Iy = 8.53 m
4
0
993
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Engineering Mechanics - Statics
Chapter 10
Problem 10-3
Determine the moment of inertia for the thin strip
of area about the x axis.The strip is oriented at an
angle θ from the x axis. Assume that t << l.
Solution:
l
⌠
⌠ 2 2
⎮ 2
Ix = ⎮ y d A = ⎮ s sin ( θ ) t ds
⌡0
⌡
A
1 3 2
t l sin ( θ )
3
Ix =
Problem 10-4
Determine the moment for inertia of the
shaded area about the x axis.
Given:
a = 4 in
b = 2 in
Solution:
⌠
⎮
Ix = ⎮
⎮
⌡
a
1 ⎡ ⎛x⎞
⎢b ⎜ ⎟
3 ⎣ ⎝ a⎠
3⎤
3
⎥ dx
⎦
0
Ix = 1.07 in
4
994
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Engineering Mechanics - Statics
Chapter 10
Problem 10-5
Determine the moment for inertia of the
shaded area about the y axis.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
3
⎮ 2 ⎛x⎞
I y = ⎮ x b ⎜ ⎟ dx
⎝ a⎠
⌡
0
Iy = 21.33 in
4
Problem 10-6
Determine the moment of inertia for the shaded area about the x axis.
Solution:
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
b
3
⎛ x⎞
⎜h ⎟
⎝ b ⎠ dx = 2 b h3
3
15
Ix =
2
3
bh
15
0
995
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Engineering Mechanics - Statics
Chapter 10
Alternatively
h
⌠
2
⎮ 2⎛
y ⎟⎞
2
3
⎜
Ix = ⎮ y b − b
bh
dy =
2⎟
⎜
15
⎮
h ⎠
⎝
⌡0
Ix =
2
3
bh
15
Ix =
ab
3( 1 + 3n)
Problem 10-7
Determine the moment of inertia for the
shaded area about the x axis.
Solution:
b
⌠
1⎤
⎮
⎡
⎢
⎮
n⎥
y⎞ ⎥
2⎢
⎛
⎮
Ix = A y a − a ⎜ ⎟
dy
⎮
b⎠ ⎦
⎣
⎝
⌡
3
0
Problem 10-8
Determine the moment of inertia for
the shaded area about the y axis.
996
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Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
⌠ 2
⎮ 2
I y = ⎮ x d A = ⎮ x y dx
⌡0
⌡
a
b ⌠ n+ 2
⎮ x
dx =
Iy =
n⌡
a 0
⎡⎢⎛ b ⎞ xn + 3 ⎤⎥
⎢⎜ an ⎟ n + 3⎥
⎣⎝ ⎠
⎦
a
0
3
ba
Iy =
n+3
Problem 10-9
Determine the moment of inertia for
the shaded area about the x axis.
Given:
a = 4 in
b = 2 in
Solution:
b
⌠
2
y⎞ ⎤
⎮ 2⎡
⎛
⎢
Ix = ⎮ y a − a ⎜ ⎟ ⎥ d y
⎣
⎝ b⎠ ⎦
⌡
0
Ix = 4.27 in
4
Problem 10-10
Determine the moment of inertia for the shaded area about the y axis.
997
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
2
Iy = ⎮ x b
⎮
⌡
x
dx
a
0
Iy = 36.6 in
4
Problem 10-11
Determine the moment of inertia for the shaded area about the x axis
Given:
a = 8 in
b = 2 in
Solution:
b
⌠
3
⎮ 2⎛
y ⎞⎟
⎜
Ix = ⎮ y a − a
dy
3⎟
⎜
⎮
b ⎠
⎝
⌡0
Ix = 10.67 in
4
Problem 10-12
Determine the moment of inertia for the shaded area about the x axis
998
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 2m
b = 1m
Solution:
b
⌠
⎮
2 ⎛
Ix = ⎮ y a⎜ 1 −
⎜
⎮
⎝
⌡− b
2⎞
y
b
⎟ dy
2⎟
⎠
Ix = 0.53 m
4
Problem 10-13
Determine the moment of inertia for the shaded area about the y axis
Given:
a = 2m
b = 1m
Solution:
a
⌠
x
2
Iy = ⎮ x 2b 1 − dx
⎮
a
⌡
Iy = 2.44 m
4
0
Problem 10-14
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 4 in
b = 4 in
999
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Engineering Mechanics - Statics
Chapter 10
Solution:
b
⌠
2
y⎞ ⎤
⎮ 2⎡
⎛
Ix = ⎮ y ⎢a − a ⎜ ⎟ ⎥ d y
⎣
⎝ b⎠ ⎦
⌡
0
Ix = 34.1 in
4
Problem 10-15
Determine the moment of inertia for the
shaded area about the y axis.
Given:
a = 4 in
b = 4 in
Solution:
a
⌠
2
Iy = ⎮ x b
⎮
⌡
x
dx
a
0
Iy = 73.1 in
4
1000
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Problem 10-16
Determine the moment of inertia of the
shaded area about the x axis.
Given:
a = 2 in
b = 4 in
Solution:
⌠
⎮
Ix = ⎮
⌡
a
3
1⎛
⎛ π x ⎞⎞
⎜ b cos ⎜ ⎟ ⎟ dx
3⎝
⎝ 2a ⎠ ⎠
−a
Ix = 36.2 in
4
Problem 10-17
Determine the moment of inertia for the shaded area
about the y axis.
Given:
a = 2 in
b = 4 in
Solution:
⌠
Iy = ⎮
⎮
⌡
a
2
⎛ π x ⎞ dx
⎟
⎝ 2a ⎠
x b cos ⎜
−a
Iy = 7.72 in
4
Problem 10-18
Determine the moment of inertia for the shaded area about the x axis.
1001
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Given:
a = 4 in
b = 2 in
Solution:
Solutiona
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
3
⎛
⎛ π x ⎞⎞
⎜b cos ⎜ ⎟⎟
⎝
⎝ 2a ⎠⎠ dx
3
−a
Ix = 9.05 in
4
Problem 10-19
Determine the moment of inertia for the shaded area about the y axis.
Given:
a = 4 in
b = 2 in
Solution:
⌠
Iy = ⎮
⎮
⌡
a
2
⎛ π x ⎞ dx
⎟
⎝ 2a ⎠
x b cos ⎜
−a
Iy = 30.9 in
4
Problem 10-20
Determine the moment for inertia of the shaded area about the x axis.
1002
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Engineering Mechanics - Statics
Chapter 10
Given:
a = 2 in
b = 4 in
c =
12 in
Solution:
⌠
⎮
Ix = ⎮
⎮
⌡
a+ b
2⎞
1 ⎛c
⎜
3⎝ x
3
⎟ dx
⎠
a
Ix = 64.0 in
4
Problem 10-21
Determine the moment of inertia of
the shaded area about the y axis.
Given:
a = 2 in
b = 4 in
c =
12 in
Solution:
⌠
⎮
Iy = ⎮
⌡
a+ b
2
2⎛ c ⎞
x
⎜ ⎟ dx
⎝x⎠
a
1003
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Engineering Mechanics - Statics
Iy = 192.00 in
Chapter 10
4
Problem 10-22
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 2m
b = 2m
Solution:
b
⌠
⎮ 2 ⎛ y2 ⎞
I x = ⎮ y a⎜ ⎟ d y
⎜ b2 ⎟
⎮
⎝ ⎠
⌡0
Ix = 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
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Engineering Mechanics - Statics
Chapter 10
Solution:
a
⌠
2
⎮
⎛x⎞
⎜a⎟
⎮
2
Iy = ⎮ x b e⎝ ⎠ dx
⌡0
Iy = 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:
⌠
⎮
⎮
⎮
Iy = ⎮
⎮
⌡
a
3
⎡⎢ ⎛ x ⎞ 2⎥⎤
⎜ ⎟
⎢b e⎝ a ⎠ ⎥
⎣
⎦ dx
3
Iy = 1.41 m
4
0
Problem 10-25
The polar moment of inertia for the area is IC about the z axis passing through the centroid C.
The moment of inertia about the x axis is Ix and the moment of inertia about the y' axis is Iy'.
Determine the area A.
Given:
IC = 28 in
4
1005
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Engineering Mechanics - Statics
Ix = 17 in
Chapter 10
4
Iy' = 56 in
4
a = 3 in
Solution:
IC = Ix + Iy
Iy = IC − Ix
2
Iy' = Iy + A a
A =
Iy' − Iy
a
A = 5.00 in
2
2
Problem 10-26
The polar moment of inertia for the area is Jcc about the z' axis passing through the centroid C. If
the moment of inertia about the y' axis is Iy' and the moment of inertia about the x axis is Ix.
Determine the area A.
Given:
6
4
Jcc = 548 × 10 mm
6
4
6
4
Iy' = 383 × 10 mm
Ix = 856 × 10 mm
h = 250 mm
Solution:
2
Ix' = Ix − A h
Jcc = Ix' + Iy'
2
Jcc = Ix − A h + Iy'
1006
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Engineering Mechanics - Statics
A =
Chapter 10
Ix + Iy' − Jcc
2
h
3
2
A = 11.1 × 10 mm
Problem 10-27
Determine the radius of gyration kx 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:
Ix =
1
1
3
3
( 2b) ( 2a) −
( 2d) ( 2c)
12
12
Radius of gyration about the x axis:
kx =
Ix
A
kx = 74.7 mm
Problem 10-28
Determine the radius of gyration ky of the column’s cross-sectional area.
Given:
a = 100 mm
b = 75 mm
1007
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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:
Iy =
1
12
1
3
( 2a) ( 2b) −
12
( 2c) ( 2d)
3
Radius of gyration about the y axis:
ky =
Iy
A
ky = 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:
IE =
1
12
⎡
( 6)
( 3)
2 ⎛ 275 mm
( 15 mm) ( 275 mm) + 4⎢1.32 10 mm + 1.36 10 mm
3
⎡1
+ 2⎢
⎣12
⎣
4
3
( 75 mm) ( 20 mm) + ( 75 mm) ( 20 mm)
⎜
⎝
2
2
⎛ 275 mm + 10mm⎞ ⎥⎤
⎜
⎟
⎝ 2
⎠⎦
− 28 mm⎟⎞
2⎤
⎥ ...
⎠⎦
1008
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
6
Chapter 10
4
IE = 162 × 10 mm
Problem 10-30
Locate the centroid yc of the cross-sectional area for the angle. Then find the moment
of inertia Ix' about the x' centroidal axis.
Given:
a = 2 in
b = 6 in
c = 6 in
d = 2 in
Solution:
⎛ c ⎞ + b d⎛ d ⎞
⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
a c⎜
yc =
Ix' =
yc = 2.00 in
ac + bd
1
12
a c + a c ⎛⎜
3
c
⎝2
2
− yc⎞⎟ +
⎠
1
12
b d + b d ⎛⎜ yc −
3
⎝
d⎞
⎟
2⎠
2
Ix' = 64.00 in
4
Problem 10-31
Locate the centroid xc of the cross-sectional area for the angle. Then find the moment
1009
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Engineering Mechanics - Statics
Chapter 10
of inertia Iy' about the centroidal y' axis.
Given:
a = 2 in
b = 6 in
c = 6 in
d = 2 in
Solution:
⎛ a ⎞ + b d⎛ a +
⎟
⎜
⎝ 2⎠
⎝
a c⎜
xc =
Iy' =
b⎞
⎟
2⎠
xc = 3.00 in
ac + bd
1
12
c a + c a ⎛⎜ xc −
3
⎝
a⎞
⎟
2⎠
2
+
1
12
d b + d b ⎛⎜ a +
3
⎝
b
2
− xc⎞⎟
⎠
2
Iy' = 136.00 in
Problem 10-32
Determine the distance xc to the centroid of the beam's cross-sectional area: then find
the moment of inertia about the y' axis.
1010
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4
Engineering Mechanics - Statics
Chapter 10
Given:
a = 40 mm
b = 120 mm
c = 40 mm
d = 40 mm
Solution:
⎛ a + b ⎞ + 2a d a
⎟
2
⎝ 2 ⎠
2( a + b)c⎜
xc =
xc = 68.00 mm
2( a + b)c + 2d a
⎡1
Iy' = 2⎢
⎣ 12
c ( a + b) + c( a + b) ⎛⎜
3
a+b
⎝ 2
6
− xc⎞⎟
2⎤
1
⎥ + 2d a3 + 2d a ⎛⎜xc −
⎠ ⎦ 12
⎝
a⎞
2
⎟
2⎠
4
Iy' = 36.9 × 10 mm
Problem 10-33
Determine the moment of inertia of the beam's cross-sectional area about the x' axis.
1011
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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:
1
Ix' =
12
3
( a + b) ( 2c + 2d) −
1
12
b ( 2d)
3
6
4
Ix' = 49.5 × 10 mm
Problem 10-34
Determine the moments of inertia for the shaded area about the x and y axes.
Given:
a = 3 in
b = 3 in
c = 6 in
d = 4 in
r = 2 in
Solution:
⎡ 1 3 1 ⎛ 2c ⎞ 2⎤ ⎛ π r4
2 2⎞
Ix = ( a + b) ( c + d) − ⎢ b c + b c ⎜ d + ⎟ ⎥ − ⎜
+ πr d ⎟
3
2 ⎝
3⎠⎦ ⎝ 4
⎣36
⎠
1
3
1012
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Ix = 1192 in
Iy =
Chapter 10
4
1
2b ⎞
3 ⎡1
3 1 ⎛
( c + d) ( a + b) − ⎢ c b + b c ⎜ a +
⎟
3
36
2
3⎠
⎣
⎝
Iy = 364.84 in
2⎤
⎛ π r4
2 2⎞
⎥−⎜
+ πr a ⎟
⎦ ⎝ 4
⎠
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 Cc, of Ix'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
Ac = 11.8 in
Ix'c = 349 in
2
4
Solution:
2A c d + a b⎛⎜ c +
yc =
⎝
b⎞
⎟
2⎠
2A c + a b
Ix' = ⎡⎣Ix'c + A c ( yc − d) ⎤⎦ 2 +
2
yc = 15.74 in
1
3
⎞
⎛ b
a b + a b ⎜ c + − yc⎟
12
⎝ 2
⎠
2
Ix' = 2158 in
4
Problem 10-36
Compute the moments of inertia Ix and Iy for the beam's cross-sectional area about
1013
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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:
Ix =
1
1
e⎞
3 1 3
3
⎛
a ( c + d + e) + b c +
g e + g e ⎜c + d + ⎟
3
3
12
2⎠
⎝
Iy =
1
3 1 3 1
3
c ( a + b ) + d f + c ( f + g)
3
3
3
2
6
6
Determine the distance yc to the centroid C of the
beam's cross-sectional area and then compute the
moment of inertia Icx' about the x' axis.
Given:
e = 30 mm
b = 170 mm
f = 30 mm
c = 30 mm
g = 70 mm
4
Iy = 91.3 × 10 mm
Problem 10-37
a = 30 mm
4
Ix = 154 × 10 mm
d = 140 mm
1014
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Engineering Mechanics - Statics
Chapter 10
Solution:
( a + b)c⎛⎜
⎛
⎟ + d f⎜ c +
⎝ 2⎠
⎝
yc =
c⎞
d⎞
⎛
⎟ + ( f + g)e⎜ c + d +
2⎠
⎝
e⎞
⎟
2⎠
( a + b)c + d f + ( f + g)e
yc = 80.7 mm
2
Ix' =
2
1
c⎞
1 3
3
⎞
⎛
⎛ d
( a + b) c + ( a + b)c ⎜ yc − ⎟ +
f d + f d ⎜ c + − yc⎟ ...
12
2⎠
12
⎝
⎝ 2
⎠
e
1
3
⎞
⎛
( f + g) e + ( f + g)e ⎜ c + d + − yc⎟
2
12
⎝
⎠
+
6
2
4
Ix' = 67.6 × 10 mm
Problem 10-38
Determine the distance xc to the centroid C of the beam's cross-sectional area and then compute the
moment of inertia Iy' 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:
b c⎛⎜
xc =
b
⎝2
+ a⎞⎟ + ( c + d) f⎛⎜
f⎞
f+g
⎟ + ( f + g)e
2
⎝ 2⎠
⎠
b c + b c + ( f + g)e
xc = 61.6 mm
1015
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
2
Iy' =
2
1
3
⎛ a + b − x ⎞ + 1 d f 3 + d f ⎛ x − f ⎞ ...
c ( a + b) + c( a + b) ⎜
⎜c
⎟
c⎟
12
2⎠
12
⎝ 2
⎠
⎝
f + g⎞
1
3
⎛
+
e ( f + g) + e( f + g) ⎜ xc −
⎟
2 ⎠
12
⎝
6
2
4
Iy' = 41.2 × 10 mm
Problem 10-39
Determine the location yc 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 =
f−c
2
e =
f−c
2
Solution:
( a + g) f⎛⎜
yc =
a + g⎞
⎛
⎟ + c b⎜ a + g +
⎝ 2 ⎠
⎝
b⎞
⎟
2⎠
( a + g) f + c b
yc = 48.25 mm
2
1
a + g⎞
1 3
3
⎞
⎛
⎛b
Ix' =
f ( a + g) + ( f) ( a + g) ⎜ yc −
⎟ + c b + c b ⎜ + a + g − yc⎟
12
2 ⎠
12
⎝
⎝2
⎠
6
2
4
Ix' = 15.1 × 10 mm
1016
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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 yc, which locates the centroidal axis x' for the cross-sectional area of the T-beam, and then
find the moments of inertia Ix' and Iy' .
Given:
a = 25 mm
b = 250 mm
c = 50 mm
d = 150 mm
Solutuion:
⎛ b ⎞ b2a + ⎛ b +
⎜ ⎟
⎜
2⎠
⎝
⎝
yc =
c⎞
⎟ 2d c
2⎠
b2a + c2d
yc = 207 mm
2
1
b⎞
1
3
3
⎞
⎛
⎛ c
Ix' =
2a b + 2a b ⎜ yc − ⎟ +
2d c + c2d ⎜ b + − yc⎟
12
2⎠
12
⎝
⎝ 2
⎠
6
2
4
Ix' = 222 × 10 mm
Iy' =
1
1
3
3
b ( 2a) +
c ( 2d)
12
12
6
4
Iy' = 115 × 10 mm
Problem 10-41
Determine the centroid y' for the beam’s cross-sectional area; then find Ix'.
Given:
a = 25 mm
1017
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Engineering Mechanics - Statics
Chapter 10
b = 100 mm
c = 25 mm
d = 50 mm
e = 75 mm
Solution:
⎛ c ⎞ + 2a b⎛c + b ⎞
⎟
⎜
⎟
⎝ 2⎠
⎝ 2⎠
2( a + e + d)c⎜
yc =
Ix' =
yc = 37.50 mm
2( a + e + d)c + 2a b
2
12
( a + e + d) c + 2( a + e + d)c ⎛⎜ yc −
3
⎡1
+ 2⎢
⎣ 12
a b + a b ⎛⎜ c +
3
⎝
6
b
2
− yc⎞⎟
2⎤
⎝
c⎞
2
⎟ ...
2⎠
⎥
⎠⎦
4
Ix' = 16.3 × 10 mm
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
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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:
ly =
⎡1 3
1 3
a
3
2 ( a + d + e) c + 2⎢ b a + a b ⎛⎜ e + ⎞⎟
12
⎣12
⎝ 2⎠
6
2⎤
⎥
⎦
4
ly = 94.8 × 10 mm
Problem 10-43
Determine the moment for inertia Ix of the shaded area about the x axis.
Given:
a = 6 in
b = 6 in
c = 3 in
d = 6 in
Solution:
3
Ix =
ba
3
+
1
12
3
ca +
1
12
( b + c) d
3
Ix = 648 in
4
Problem 10-44
Determine the moment for inertia Iy of the shaded area about the y axis.
Given:
a = 6 in
b = 6 in
c = 3 in
d = 6 in
1019
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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:
3
2
2
c⎞
1
2( b + c) ⎤
3 1
⎛
⎡
Iy =
+
a c + a c ⎜b + ⎟ +
d ( b + c) + d( b + c) ⎢
⎥
3
36
2 ⎝
3⎠
36
2
⎣ 3 ⎦
ab
Iy = 1971 in
1
3
1
4
Problem 10-45
Locate the centroid yc 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 = 2 in
b = 12 in
c = 2 in
d = 4 in
Solution:
yc =
c
⎛ c + d ⎞ ( c + d)a
b c + 2⎜
⎟
2
⎝ 2 ⎠
b c + 2( c + d)a
yc = 2 in
2
2
c⎞
2
c+d
3
⎞
⎛
⎛
Ix =
b c + b c ⎜ yc − ⎟ +
a ( c + d) + 2a( c + d) ⎜
− yc⎟
12
2⎠
12
⎝
⎝ 2
⎠
1
Ix = 128 in
3
4
Problem 10-46
Determine the moments for inertia Ix and Iy of the shaded area.
Given:
r1 = 2 in
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
r2 = 6 in
Solution:
⎛ π r24 π r14 ⎞
⎟
Ix = ⎜
−
8 ⎠
⎝ 8
Ix = 503 in
4
⎛ π r24 π r14 ⎞
⎟
Iy = ⎜
−
8 ⎠
⎝ 8
Iy = 503 in
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:
h = ( a)sin ( θ )
Ixc =
Ixc =
1
12
3
bh =
1
12
b ⎡⎣( a)sin ( θ )⎤⎦ =
3
1 3
3
a b sin ( θ )
12
1 3
3
a b sin ( θ )
12
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
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Solution:
A = b( a) sin ( θ )
xc =
Iy' =
⎡⎡b( a) sin ( θ ) b − 1 ( a) cos ( θ ) ( a) sin ( θ ) ( a)cos ( θ )⎤
⎥
⎢⎢
3
2 2
b( a) sin ( θ ) ⎣
⎦
⎢ 1
( a)cos ( θ )⎤
⎡
⎢+ ( a) cos ( θ ) ( a) sin ( θ ) ⎢b +
⎥
3
⎣ 2
⎣
⎦
1
1
12
⎡1
+ −⎢
( a) sin ( θ ) ⎡⎣( a)cos ( θ )⎤⎦ +
b + ( a)cos ( θ )
⎥
⎥
⎦
2
2
− xc⎟⎞ ...
⎝2
⎠
b
3
( a) sin ( θ ) b + ( a)sin ( θ ) b ⎛⎜
⎤
...⎥ =
1
( a)cos ( θ )⎤
⎡
( a) sin ( θ ) ( a) cos ( θ ) ⎢xc −
⎥
2
3
⎣
⎦
2⎤
⎥ ...
⎦
2
(
)
1
( a)cos θ
3 1
⎤
⎡
(
)
(
)
(
)
(
)
+
( a) sin θ ⎡⎣( a)cos θ ⎤⎦ + ( a) sin θ ( a) cos θ ⎢b +
− xc⎥
36
3
2
⎣
⎦
⎣ 36
3
Simplifying we find.
Iy' =
(
)
ab 2
2
2
b + a cos ( θ ) sin ( θ )
12
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:
Ix' =
36
3
bh
2 1
⎛ b − a ⎞ 1 h ( b − a)
a h a + ⎜a +
⎟
3 2
3 ⎠2
⎝
xc =
=
1
1
h a + h ( b − a)
2
2
Iy' =
Iy' =
1
1
36
1
36
3
ha +
(2
1
2
h a ⎛⎜
b+a
⎝ 3
hb b − ab + a
2
−
2
3
2
a⎞⎟ +
⎠
1
36
b+a
3
3
h ( b − a) +
1
2
h( b − a) ⎛⎜ a +
⎝
b−a
3
−
b + a⎞
3
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:
Ix' =
1
12
⎡⎣2a ⎡⎣2( c sin ( θ ) + b)⎤⎦ 3⎤⎦ ...
⎡ 1 ( c cos ( θ ) ) ( c sin ( θ ) ) 3⎤ − 2⎡1 c4⎛ θ − 1 sin ( 2θ )⎞⎤
⎥ ⎢ ⎜
⎟⎥
2
⎣ 12
⎦ ⎣4 ⎝
⎠⎦
+ 4⎢
6
4
Ix' = 520 × 10 mm
Problem 10-51
Determine the moment of inertia of
the composite area about the x axis.
Given:
a = 2 in
b = 4 in
c = 1 in
d = 4 in
Solution:
d
⌠
3
⎮
4
2⎤⎤
⎞
⎡
⎛
⎡
1
π
c
1
x
3
2 2
⎢2a⎢1 − ⎛⎜ ⎟⎞ ⎥⎥ dx
Ix = ( a + b) ( 2a) − ⎜
+ πc a ⎟ + ⎮
⎮
4
3
3
⎝
⎠ ⌡ ⎣ ⎣ ⎝ d ⎠ ⎦⎦
0
Ix = 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 = 2 in
b = 4 in
c = 1 in
d = 4 in
Solution:
d
⌠
4
2
1
x⎞ ⎤
3 ⎛ πc
2 2⎞ ⎮ 2 ⎡
⎛
⎜
⎟
⎢
Iy = ( 2a) ( a + b) −
+ π c b + ⎮ x 2a 1 − ⎜ ⎟ ⎥ dx
3
⎝ 4
⎠ ⌡
⎣ ⎝ d⎠ ⎦
0
Iy = 271.1 in
4
Problem 10-53
Determine the radius of gyration kx for the column's
cross-sectional area.
Given:
a = 200 mm
b = 100 mm
Solution:
2
⎡1 3
a b⎞ ⎤
⎛
⎢
Ix =
( 2a + b) b + 2
ba + ba⎜ + ⎟ ⎥
12
⎣ 12
⎝ 2 2⎠ ⎦
1
kx =
3
Ix
b( 2a + b) + 2a b
kx = 109 mm
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.
Engineering Mechanics - Statics
Chapter 10
Problem 10-54
Determine the product of inertia for the shaded portion of the parabola with respect
to the x and y axes.
Given:
a = 2 in
b = 1 in
a
⌠
Ixy = ⎮
⎮
⌡− a
b
⌠
⎮
x y d y dx
⎮
2
⌡ ⎛x⎞
b
Ixy = 0.00 m
4
⎜a⎟
⎝ ⎠
Also because the area is symmetric about the y axis, the product of inertia must be zero.
Problem 10-55
Determine the product of inertia for the shaded
area with respect to the x and y axes.
Solution:
1
x
h ⎛⎜ ⎞⎟
b ⌠
⎮
⌠
⎝b⎠
⎮
⎮
Ixy = ⎮ ⎮
⌡ ⌡
0
0
3
x y d y dx =
3
16
2 2
Ixy =
b h
3 2 2
b h
16
1026
© 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 10-56
Determine the product of inertia of the shaded area of the ellipse with respect to the x
and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
2
⎮ ⎡⎢
x ⎞ ⎥⎤
⎛
⎮
2
⎢ b 1 − ⎜⎝ a ⎟⎠ ⎥
x⎞
⎛
⎮
Ixy =
x
⎥ b 1 − ⎜ a ⎟ dx
⎮ ⎢⎣
2
⎦
⎝ ⎠
⌡0
Ixy = 8.00 in
4
Problem 10-57
Determine the product of inertia of the parabolic area with respect to the x and y axes.
1027
© 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:
a
⌠
⎮ ⎛⎜ b x ⎟⎞
a
⎮
⎟b
Ixy = ⎮ x⎜
2 ⎠
⎝
⌡0
x
a
dx =
1
6
a
2
3 b
Ixy =
a
1 2 2
a b
6
Problem 10-58
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a = 8 in
b = 2 in
Solution:
a
⌠
1
⎮
3
1
⎮
⎛x⎞
⎮
b⎜ ⎟
3
⎝ a ⎠ b ⎛ x ⎞ dx
Ixy = ⎮ x
⎜ ⎟
⎮
2
⎝ a⎠
⌡0
Ixy = 48.00 in
4
Problem 10-59
Determine the product of inertia for the shaded parabolic area with respect to the x and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
b
Ixy = ⎮ x
⎮
2
⌡0
x
a
b
x
a
dx
1028
© 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
Ixy = 10.67 in
Chapter 10
4
Problem 10-60
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a = 2m
b = 1m
Solution:
a
⌠
⎛b
Ixy = ⎮ x⎜
⎮ ⎝2
⌡0
1−
x⎞
x
⎟ b 1 − dx
a⎠
a
Ixy = 0.333 m
4
Problem 10-61
Determine the product of inertia for the shaded area with respect to the x and y axes.
1029
© 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:
h
⌠
2
1⎤
⎮
⎡
⎮
⎢
3⎥
1 ⎛ y⎞
3 2 2
⎮
⎢
Ixy =
y b ⎜ ⎟ ⎥ dy =
b h
⎮
2 ⎣ ⎝ h⎠ ⎦
16
⌡0
Ixy =
3
16
2 2
h b
Problem 10-62
Determine the product of inertia of the shaded area with
respect to the x and y axes.
Given:
a = 4 in
b = 2 in
Solution:
a
⌠
3
3
⎮ ⎛ b⎞ ⎛ x ⎞ ⎛ x ⎞
Ixy = ⎮ x⎜ ⎟ ⎜ ⎟ b ⎜ ⎟ dx
2 a
a
⌡0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
Ixy = 4.00 in
4
1030
© 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 10-63
Determine the product of inertia for the shaded area with respect to the x and y axes.
Solution:
a
⌠
⎮ ⎛⎜ b xn ⎞⎟ xn
Ixy = ⎮ x
b
dx
n
n
⎮ ⎜⎝ 2 a ⎟⎠ a
⌡
2 2
Ix =
a b
4( n + 1)
provided n ≠ −1
0
Problem 10-64
Determine the product of inertia for the
shaded area with respect to the x and y axes.
Given:
a = 4 ft
Solution:
a
⌠
( a − x) 2
⎮
Ixy = ⎮ x
2
⌡0
Ixy = 0.91 ft
(
a−
x) dx
2
4
1031
© 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 10-65
Determine the product of inertia for the shaded area with respect to the x and y axes. Use Simpson's
rule to evaluate the integral.
Given:
a = 1m
b = 0.8 m
Solution:
a
⌠
2
2
⎮
⎛x⎞ ⎛x⎞
⎮
⎜a⎟
b ⎜a⎟
Ixy = ⎮ x⎛⎜ ⎟⎞ e⎝ ⎠ b e⎝ ⎠ dx
⎮ ⎝ 2⎠
⌡0
Ixy =
Problem 10-66
Determine the product of inertia for the parabolic
area with respect to the x and y axes.
Given:
a = 1 in
b = 2 in
Solution:
Due to symmetry about y axis
Ixy = 0
⌠
Also ⎮
⎮
⎮
Ixy = ⎮
⎮
⌡
a
−a
2
b+b
x
x
2
a
2
2
⎛⎜
x ⎞
b − b ⎟ dx
2
⎜
a ⎟⎠
⎝
Ixy = 0.00 m
4
1032
© 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 10-67
Determine the product of inertia for the
cross-sectional area with respect to the x and y axes
that have their origin located at the centroid C.
Given:
a = 20 mm
b = 80 mm
c = 100 mm
Solution:
Ixy = 2b a
a⎞
⎜ − ⎟
2 ⎝ 2 2⎠
c⎛b
4
Ixy = 4800000.00 mm
Problem 10-68
Determine the product of inertia for the beam's cross-sectional area with respect to the x and y axes.
Given:
a = 12 in
b = 8 in
c = 1 in
d = 3 in
Solution:
Ixy =
⎛ c ⎞ ⎛ b ⎞ c b + ⎛ a ⎞ ⎛ c ⎞ ( a − 2c)c + d c⎛ a −
⎜ ⎟⎜ ⎟
⎜ ⎟⎜ ⎟
⎜
⎝ 2⎠⎝ 2 ⎠
⎝ 2 ⎠⎝ 2 ⎠
⎝
c ⎞⎛ d ⎞
⎟⎜ ⎟
2 ⎠⎝ 2 ⎠
Ixy = 97.75 in
4
Problem 10-69
Determine the location (xc, yc) of the centroid C of the angle’s cross-sectional area, and then
1033
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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
compute the product of inertia with respect to the x' and y' axes.
Given:
a = 18 mm
b = 150 mm
Solution:
⎛ a ⎞ a b + a( b − a) ⎛ a + b ⎞
⎜ ⎟
⎜
⎟
2⎠
2 ⎠
⎝
⎝
xc =
a b + a( b − a)
xc = 44.1 mm
⎛ b ⎞ a b + ⎛ a ⎞ a( b − a)
⎜ ⎟
⎜ ⎟
⎝ 2⎠
⎝ 2⎠
yc =
a b + a( b − a)
yc = 44.1 mm
⎛
⎝
Ix'y' = a b ⋅ −⎜ xc −
6
a⎞⎛ b
a⎞⎛ b a
⎞
⎞
⎛
⎟ ⎜ − yc⎟ + a( b − a) ⋅ −⎜ yc − ⎟ ⎜ + − xc⎟
2⎠⎝ 2
2
2
2
⎠
⎝
⎠⎝
⎠
4
Ix'y' = −6.26 × 10 mm
1034
© 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 10-70
Determine the product of inertia of the beam’s
cross-sectional area with respect to the x and y axes
that have their origin located at the centroid C.
Given:
a = 5 mm
b = 30 mm
c = 50 mm
Solution:
xc =
a + b⎞
⎛ a⎞
⎟ + c a⎜ ⎟
⎝ 2 ⎠
⎝ 2⎠
a( b − a) ⎛⎜
a( b − a) + a c
xc = 7.50 mm
⎛c⎞
⎟ + c a⎜ ⎟
⎝ 2⎠
⎝ 2⎠
a( b − a) ⎛⎜
yc =
a⎞
a( b − a) + c a
yc = 17.50 mm
⎛ a − y ⎞ ⎛ a + b − x ⎞ + a c⎛ a − x ⎞ ⎛ c − y ⎞
⎜
c⎟ ⎜
c⎟
c⎟ ⎜
c⎟
⎝2
⎠⎝ 2
⎠
⎝2
⎠⎝ 2
⎠
Ixy = ( b − a)a⎜
3
4
Ixy = −28.1 × 10 mm
1035
© 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 10-71
Determine the product of inertia for the shaded area
with respect to the x and y axes.
Given:
a = 2 in
b = 1 in
c = 2 in
d = 4 in
Solution:
⎛ c + d ⎞ − π b2 a d
⎟
⎝ 2 ⎠
lxy = 2a( c + d)a⎜
lxy = 119 in
4
Problem 10-72
Determine the product of inertia for the beam's cross-sectional area with respect to the x and y
axes that have their origin located at the centroid C.
Given:
a = 1 in
b = 5 in
c = 5 in
Solution:
⎛ a − b ⎞ ⎛c + a ⎞
⎟⎜
⎟
⎝ 2 2⎠⎝ 2 ⎠
Ixy = 2b a⎜
Ixy = −110 in
4
1036
© 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 10-73
Determine the product of inertia for the
cross-sec-tional area with respect to the x
and y axes.
Given:
a = 4 in
b = 1 in
c = 6 in
Solution:
⎛ a ⎞ ⎛ c + 3b ⎞ + c b⎛b + c ⎞ ⎛ b ⎞
⎟⎜
⎟
⎜
⎟⎜ ⎟
2⎠
⎝ 2 ⎠⎝
⎝ 2⎠⎝ 2 ⎠
lxy = b a⎜
lxy = 72 in
4
Problem 10-74
Determine the product of inertia for the beam's cross-sectional area with respect to the u and
v axes.
1037
© 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 = 150 mm
b = 200 mm
t = 20 mm
θ = 20 deg
Solution:
Moments of inertia Ix and Iy:
Ix =
1
1
3
3
2a ( 2b) −
( 2a − t) ( 2b − 2t)
12
12
Ix = 511.36 × 10 mm
Iy =
2
2
3
3
t ( 2a) +
( b − t) t
12
12
Iy = 90240000.00 mm
6
4
The section is symmetric about both x and y axes;
therefore Ixy = 0.
Iuv =
4
⎛ Ix − Iy ⎞
⎜
⎟ sin ( 2θ ) + Ixy cos ( 2θ )
⎝ 2 ⎠
4
Ixy = 0mm
6
4
Iuv = 135 × 10 mm
Problem 10-75
Determine the moments of inertia Iu and Iv and the product of inertia Iuv for the rectangular area.The
u and v axes pass through the centroid C.
Given:
a = 40 mm
b = 160 mm
θ = 30 deg
1038
© 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:
Ix =
1
3
ab
12
Iu =
Ix + Iy ⎛ Ix − Iy ⎞
+⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
2
⎝ 2 ⎠
Iy =
6
1
3
ba
12
4
Ixy = 0 mm
4
Iu = 10.5 × 10 mm
Iv =
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎜
⎟−⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
6
4
Iv = 4.05 × 10 mm
Iuv =
⎛ Ix − Iy ⎞
⎜
⎟ sin ( 2θ ) + Ixy cos ( 2θ )
⎝ 2 ⎠
6
4
Iuv = 5.54 × 10 mm
Problem 10-76
Determine the distance yc to the centroid of the area and then calculate the moments of inertia
Iu and Iv for the channel`s cross-sectional area. The u and v axes have their origin at the
centroid C. For the calculation, assume all corners to be square.
Given:
a = 150 mm
b = 10 mm
c = 50 mm
θ = 20 deg
1039
© 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:
2a b
yc =
Ix =
b
⎛ c⎞
+ 2c b⎜ b + ⎟
2
⎝ 2⎠
yc = 12.50 mm
2a b + 2c b
1
12
2a b + 2a b ⎛⎜ yc −
3
⎝
b⎞
⎟
2⎠
2
⎡1
+ 2⎢
⎣12
b c + b c ⎛⎜ b +
3
⎝
c
2
− yc⎟⎞
2⎤
⎥
⎠⎦
3
4
6
4
Ix = 908.3 × 10 mm
2
⎡1 3
b⎞ ⎤
⎛
⎢
Iy =
b ( 2a) + 2
c b + c b ⎜a − ⎟ ⎥
12
⎣12
⎝ 2⎠ ⎦
1
3
Iy = 43.53 × 10 mm
4
Ixy = 0 mm
(By symmetry)
Iu =
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎜
⎟+⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iv =
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎜
⎟−⎜
⎟ cos ( 2θ ) + Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
6
4
Iu = 5.89 × 10 mm
6
4
Iv = 38.5 × 10 mm
Problem 10-77
Determine the moments of inertia for the shaded area with respect to the u and v
axes.
Given:
a = 0.5 in
b = 4 in
c = 5 in
θ = 30 deg
1040
© 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:
Moment and Product of Inertia about x and y Axes: Since the
shaded area is symmetrical about the x axis,
1
Ix =
3
12
2a c +
1
12
b ( 2a)
3
2
b⎞
1
3
⎛
Iy =
2a b + 2a b ⎜ a + ⎟ +
c ( 2a)
12
12
⎝ 2⎠
1
3
Ixy = 0 in
4
Ix = 10.75 in
4
Iy = 30.75 in
4
Moment of Inertia about the Inclined u and v Axes
Iu =
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎜
⎟+⎜
⎟ cos ( 2θ ) − Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iu = 15.75 in
Iv =
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
⎜
⎟−⎜
⎟ cos ( 2θ ) + Ixy sin ( 2θ )
⎝ 2 ⎠ ⎝ 2 ⎠
Iv = 25.75 in
4
4
Problem 10-78
Determine the directions of the principal axes with origin
located at point O, and the principal moments of inertia for
the rectangular area about these axes.
Given:
a = 6 in
b = 3 in
Solution:
Ix =
1
Iy =
1
Ixy =
3
3
ba
ab
3
Ix = 216 in
3
Iy = 54 in
a b
2 2
ab
4
4
Ixy = 81 in
4
1041
© 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
−2Ixy
tan ( 2θ ) =
Imax =
Imin =
Ix − Iy
Ix + Iy
2
Ix + Iy
2
Chapter 10
θ =
Ixy ⎞
1
⎛
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = −22.5 deg
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
2
⎝
⎠
Imax = 250 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 20.4 in
4
Problem 10-79
Determine the moments of inertia Iu , Iv and the product of inertia Iuv for the beam's cross-sectional
area.
Given:
θ = 45 deg
a = 8 in
b = 2 in
c = 2 in
d = 16 in
Solution:
2
1
3
3
⎛ d⎞
I x = ( a + b) c +
2b d + 2b d ⎜ ⎟
3
12
⎝ 2⎠
Iy =
2
3
4
3
4
Ix = 5.515 × 10 in
1
1
3
3
[ 2( a + b) ] c +
( 2b) d
12
12
Iy = 1.419 × 10 in
Ixy = 0 in
4
Iu =
Ix + Iy Ix − Iy
+
cos ( 2θ ) − Ixy sin ( 2θ )
2
2
Iu = 3.47 × 10 in
Iv =
Ix + Iy Ix − Iy
−
cos ( 2θ ) + Ixy sin ( 2θ )
2
2
Iv = 3.47 × 10 in
3
4
3
4
1042
© 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
Iuv =
Ix − Iy
Chapter 10
sin ( 2θ ) + Ixy cos ( 2θ )
2
3
Iuv = 2.05 × 10 in
4
Problem 10-80
Determine the directions of the principal axes with
origin located at point O, and the principal moments
of inertia for the area about these axes.
Given:
a = 4 in
b = 2 in
c = 2 in
d = 2 in
r = 1 in
Solution:
Ix =
1
3 ⎛ πr
2 2⎞
( c + d) ( a + b) − ⎜
+ πr a ⎟
3
⎝ 4
⎠
Ix = 236.95 in
4
Iy =
1
3 ⎛ πr
2 2⎞
( a + b) ( c + d) − ⎜
+ πr d ⎟
3
⎝ 4
⎠
Iy = 114.65 in
4
Ixy =
⎛ a + b ⎞ ⎛ d + c ⎞ ( a + b) ( d + c) − d aπ r2
⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
Ixy = 118.87 in
4
4
tan ( 2 θ p ) =
−Ixy
Ix− Iy
θp =
Ixy ⎞
1
⎛
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
4
θ p = −31.39 deg
2
θ p1 = θ p
θ p1 = −31.39 deg
θ p2 = 90 deg + θ p1
θ p2 = 58.61 deg
Imax =
Ix + Iy
2
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
⎝ 2 ⎠
Imax = 309 in
4
1043
© 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
2
Ix + Iy
Imin =
−
2
⎛ Ix − Iy ⎞
2
⎜
⎟ + Ixy
⎝ 2 ⎠
Imin = 42.1 in
4
Problem 10-81
Determine the principal moments of inertia for the
beam's cross-sectional area about the principal
axes that have their origin located at the centroid C.
Use the equations developed in Section 10.7. For
the calculation, assume all corners to be square.
a = 4 in
Given:
b = 4 in
t =
3
in
8
Solution:
⎡1
I x = 2⎢
⎣12
3
Ix = 55.55 in
⎡1
I y = 2⎢
⎣12
⎛
⎝
a t + a t ⎜b −
t⎞
⎟
2⎠
2⎤
1
⎥ + t ( 2b − 2t) 3
⎦ 12
4
3
2
⎛ a − t + t ⎞ ⎤⎥ + 1 2b t3
⎟
2 ⎠ ⎦ 12
⎝ 2
t ( a − t) + t( a − t) ⎜
⎡a − t + ⎛ t ⎞⎤ ⎛b − t ⎞ t( a − t)
⎜ ⎟⎥ ⎜
⎟
⎣ 2
⎝ 2 ⎠⎦ ⎝ 2 ⎠
Ixy = −2⎢
Imax =
Imin =
Ix + Iy
2
Ix + Iy
2
Iy = 13.89 in
4
Ixy = −20.73 in
4
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
2
⎝
⎠
Imax = 64.1 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 5.33 in
4
1044
© 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 10-82
Determine the principal moments of inertia for
the angle's cross-sectional area with respect to a
set of principal axes that have their origin located
at the centroid C. Use the equation developed in
Section 10.7. For the calculation, assume all
corners to be square.
Given:
a = 100 mm
b = 100 mm
t = 20 mm
Solution:
tb
xc =
t
2
+ ( a − t)t⎛⎜ t +
⎝
a − t⎞
2
⎟
⎠
xc = 32.22 mm
t b + ( a − t)t
tb
yc =
b
+ ( a − t)t
t
2
2
t b + ( a − t)t
yc = 32.22 mm
2
Ix =
1 3
t⎞
1 3
⎞
⎛
⎛b
t ( a − t) + t( a − t) ⎜ xc − ⎟ +
t b + t b ⎜ − xc⎟
12
2
12
2
⎝
⎠
⎝
⎠
2
2
1 3
t⎞
1
3
⎛
⎛ a−t −x ⎞
Iy =
b t + b t ⎜ xc − ⎟ +
t ( a − t) + t( a − t) ⎜ t +
c⎟
12
2⎠
12
2
⎝
⎝
⎠
⎛
⎝
Ixy = −⎜ xc −
Imax =
Imin
t ⎞⎛ b
⎞
⎛ a − t + t − x ⎞ ⎛ y − t ⎞ ( a − t)t
⎟ ⎜ − yc⎟ b t − ⎜
⎟
c⎟ ⎜ c
2⎠⎝ 2
2⎠
⎠
⎝ 2
⎠⎝
⎛ Ix + Iy Ix − Iy ⎞
−
⎜
⎟ − Ixy
2 ⎠
⎝ 2
6
4
6
4
Ix = 3.142 × 10 mm
2
Iy = 3.142 × 10 mm
6
4
Ixy = −1.778 × 10 mm
6
4
6
4
Imax = 4.92 × 10 mm
Imin = 2.22 × 10 mm
⎛ Ix + Iy ⎞ ⎛ Ix − Iy ⎞
= ⎜
⎟+⎜
⎟ + Ixy
⎝ 2 ⎠ ⎝ 2 ⎠
1045
© 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 10-83
The area of the cross section of an airplane wing has the listed properties about the x and y axes
passing through the centroid C. Determine the orientation of the principal axes and the principal
moments of inertia.
Ix = 450 in
Given:
4
Iy = 1730 in
4
Ixy = 138 in
4
Solution:
tan ( 2θ ) =
Imax =
Imin =
−2Ixy
Ix − Iy
Ix + Iy
2
Ix + Iy
2
θ =
1
⎛ Ixy ⎞
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = 6.08 deg
2
⎛ Ix − Iy ⎞
2
+ ⎜
⎟ + Ixy
⎝ 2 ⎠
Imax = 1745 in
4
2
⎛ Ix − Iy ⎞
2
− ⎜
⎟ + Ixy
2
⎝
⎠
Imin = 435 in
4
Problem 10-84
Using Mohr’s circle, determine the principal moments of inertia for the triangular area and the
orientation of the principal axes of inertia having an origin at point O.
Given:
a = 30 mm
b = 40 mm
1046
© 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:
Moment of inertia Ix and Iy:
Ix =
1
3
ba
12
Ix = 90 × 10 mm
Iy =
1
3
ab
12
Iy = 160 × 10 mm
3
4
3
4
Product of inertia Ixy:
⌠
⎮
Ixy = ⎮
⌡
b
2
x⎛
a ⎞
⎜a − x⎟ dx
2⎝
b ⎠
3
4
Ixy = 60 × 10 mm
0
Mohr's circle :
2
OA =
⎞
⎛ Ix + Iy
2
− Ix⎟ + Ixy
⎜
2
⎝
⎠
3
4
OA = 69.462 × 10 mm
Imax =
⎞
⎛ Ix + Iy
+ OA⎟
⎜
⎝ 2
⎠
3
4
Imax = 194.462 × 10 mm
Imin =
⎞
⎛ Ix + Iy
− OA⎟
⎜
⎝ 2
⎠
3
4
Imin = 55.5 × 10 mm
tan ( 2θ ) =
Ixy
Ix + Iy
− Ix
2
θ =
1
⎛ Ixy ⎞
atan ⎜ 2
⎟
2
⎝ −Ix + Iy ⎠
θ = 29.9 deg
1047
© 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 10-85
Determine the directions of the principal axes with origin
located at point O, and the principal moments of inertia
for the rectangular area about these axes.
Solve using Mohr's circle.
Given:
a = 6 in
b = 3 in
Solution:
Ix =
1 3
ba
3
Ix = 216 in
Iy =
1 3
ab
3
Iy = 54 in
Ixy =
a b
ab
2 2
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
⎣ ⎝ 2 ⎠⎦
4
4
Ixy = 81 in
4
2
Imax =
Imin =
θ p1 =
Ix + Iy
+R
2
Ix + Iy
2
−R
−1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p2 = θ p1 + 90 deg
R = 114.55 in
4
Imax = 250 in
4
Imin = 20.4 in
4
θ p1 = −22.50 deg
θ p2 = 67.50 deg
1048
© 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 10-86
Determine the principal moments of inertia for the
beam's cross-sectional area about the principal
axes that have their origin located at the centroid
C. For the calculation, assume all corners to be
square. Solve using Mohr's circle.
Given:
a = 4 in
b = 4 in
t =
3
in
8
Solution:
⎡1
I x = 2⎢
⎣12
⎡1
I y = 2⎢
⎣12
⎛
⎝
3
a t + a t ⎜b −
t⎞
⎟
2⎠
2
⎛ a − t + t ⎞ ⎤⎥ + 1 2b t3
⎟
2 ⎠ ⎦ 12
⎝ 2
3
⎡a − t + ⎛ t ⎞⎤ ⎛b −
⎜ ⎟⎥ ⎜
⎣ 2
⎝ 2 ⎠⎦ ⎝
t⎞
⎟ t( a − t)
2⎠
Ix = 55.55 in
4
Iy = 13.89 in
4
Ixy = −20.73 in
2
Ix + Iy ⎞
⎛
2
⎜Ix −
⎟ + Ixy
2 ⎠
⎝
Imax =
Imin =
1
⎥ + t ( 2b − 2t) 3
⎦ 12
t ( a − t) + t( a − t) ⎜
Ixy = −2⎢
R =
2⎤
R = 29.39 in
4
Ix + Iy
+R
2
Imax = 64.1 in
Ix + Iy
Imin = 20.45 in
2
−R
4
4
4
Problem 10-87
Determine the principal moments of inertia for the angle's cross-sectional area with respect to a set
of principal axes that have their origin located at the centroid C. For the calculation, assume all
corners to be square. Solve using Mohr's ciricle.
1049
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
a = 100 mm
Given:
b = 100 mm
t = 20 mm
Solution:
⎛ a − t⎞
⎟ + ( a − t)t⎜ t +
⎟
2 ⎠
⎝ 2⎠
⎝
t b⎛⎜
xc =
xc = 32.22 mm
t b + ( a − t)t
⎛t⎞
⎟ + ( a − t)t⎜ ⎟
⎝ 2⎠
⎝ 2⎠
t b⎛⎜
yc =
t⎞
b⎞
yc = 32.22 mm
t b + ( a − t)t
2
1 3
t⎞
1 3
⎞
⎛
⎛b
Ix =
t ( a − t) + t( a − t) ⎜ xc − ⎟ +
t b + t b ⎜ − xc⎟
12
2⎠
12
⎝
⎝2
⎠
2
2
Iy =
1 3
t⎞
1
3
⎛
⎛ a−t −x ⎞
b t + b t ⎜ xc − ⎟ +
t ( a − t) + t( a − t) ⎜ t +
c⎟
12
2⎠
12
2
⎝
⎝
⎠
6
4
6
4
Ix = 3.142 × 10 mm
2
Iy = 3.142 × 10 mm
⎛
⎝
Ixy = −⎜ xc −
R =
t ⎞⎛ b
⎞
⎛ a − t + t − x ⎞ ⎛ y − t ⎞ ( a − t)t
⎟ ⎜ − yc⎟ b t − ⎜
⎟
c⎟ ⎜ c
2⎠⎝ 2
2⎠
⎠
⎝ 2
⎠⎝
2
Ix + Iy ⎞
⎛
2
⎜Ix −
⎟ + Ixy
2 ⎠
⎝
6
4
Ixy = −1.778 × 10 mm
6
4
R = 1.78 × 10 mm
1050
© 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
Imax =
Ix + Iy
+R
2
Imax = 4.92 × 10 mm
Imin =
Ix + Iy
−R
2
Imin = 1364444.44 mm
6
4
Problem 10-88
Determine the directions of the principal axes with
origin located at point O, and the principal moments
of inertia for the area about these axes. Solve using
Mohr's circle
Given:
a = 4 in
b = 2 in
c = 2 in
d = 2 in
r = 1 in
Solution:
Ix =
1
3 ⎛ πr
2 2⎞
( c + d) ( a + b) − ⎜
+ πr a ⎟
3
⎝ 4
⎠
Ix = 236.95 in
4
Iy =
1
3 ⎛ πr
2 2⎞
( a + b) ( c + d) − ⎜
+ πr d ⎟
3
⎝ 4
⎠
Iy = 114.65 in
4
Ixy =
⎛ a + b ⎞ ⎛ d + c ⎞ ( a + b) ( d + c) − d aπ r2
⎜
⎟⎜
⎟
⎝ 2 ⎠⎝ 2 ⎠
Ixy = 118.87 in
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
⎣ ⎝ 2 ⎠⎦
4
4
4
2
Imax =
Imin =
Ix + Iy
+R
2
Ix + Iy
2
4
−R
R = 133.67 in
4
Imax = 309 in
4
Imin = 42.1 in
4
1051
© 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
θ p1 =
Chapter 10
−1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p2 = θ p1 +
θ p1 = −31.39 deg
π
θ p2 = 58.61 deg
2
Problem 10-89
The area of the cross section of an airplane wing has the listed properties about the x and y axes
passing through the centroid C. Determine the orientation of the principal axes and the principal
moments of inertia. Solve using Mohr's circle.
Ix = 450 in
Given:
4
Iy = 1730 in
Ixy = 138 in
4
4
Solution:
2
R =
⎡ ⎛ Ix + Iy ⎞⎤
2
⎢Ix − ⎜
⎟⎥ + Ixy
2
⎣ ⎝
⎠⎦
R = 654.71 in
4
Imax =
⎞
⎛ Ix + Iy
+ R⎟
⎜
⎝ 2
⎠
Imax = 1.74 × 10 in
Imin =
⎞
⎛ Ix + Iy
− R⎟
⎜
⎝ 2
⎠
Imin = 435 in
3
4
4
1052
© 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
θ p1 =
Chapter 10
1
⎛ Ixy ⎞
asin ⎜ ⎟
2
⎝R⎠
θ p1 = 6.08 deg
θ p2 = θ p1 + 90 deg
θ p2 = 96.08 deg
Problem 10-90
The right circular cone is formed by revolving the shaded area
around the x axis. Determine the moment of inertia lx and
express the result in terms of the total mass m of the cone. The
cone has a constant density ρ.
Solution:
h
⌠
2
rx⎞
1
⎮
2
⎛
m = ⎮ ρ π ⎜ ⎟ dx =
hρ π r
3
⎝h⎠
⌡
0
⌠
3m ⎮
lx =
2⎮
πh r ⌡
h
4
1 ⎛ rx⎞
3
2
π ⎜ ⎟ dx =
mr
2 ⎝h⎠
10
0
lx =
3
2
mr
10
Problem 10-91
Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.
1053
© 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:
m = ρ 2π R
2π
⌠
I=⎮
⎮
⌡0
ρ =
m
2π R
⎛ m ⎞ R 2 R dθ = m R 2
⎜
⎟
⎝ 2π R ⎠
I=mR
2
Problem 10-92
The solid is formed by revolving the shaded area
around the y axis. Determine the radius of gyration
ky. The specific weight of the material is γ.
Given:
a = 3 in
b = 3 in
lb
γ = 380
ft
3
Solution:
b
⌠
2
⎮
⎡ ⎛ y ⎞ 3⎤
m = ⎮ γ π ⎢a ⎜ ⎟ ⎥ d y
⎮
⎣ ⎝ b⎠ ⎦
⌡
m = 2.66 lb
0
b
⌠
2
2
⎮
⎡ ⎛ y ⎞ 3⎤ 1 ⎡ ⎛ y ⎞ 3⎤
⎢a ⎜ ⎟ ⎥ d y
Iy = ⎮ γ π ⎢a ⎜ ⎟ ⎥
⎮
b⎠ ⎦ 2 ⎣ ⎝ b⎠ ⎦
⎣
⎝
⌡
Iy = 6.46 lb⋅ in
2
0
ky =
Iy
ky = 1.56 in
m
1054
© 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 10-93
Determine the moment of inertia Ix for the sphere and express the result in terms of
the total mass m of the sphere. The sphere has a constant density ρ.
Solution:
3
4π r
m=ρ
3
ρ =
r
3m
3
4π r
⌠ 1 3m
⎞ π r2 − x2 r2 − x2 dx = 2 m r2
⎛
Ix = ⎮
⎜
⎟
3
5
⎮ 2 4π r
⎠
⌡− r ⎝
(
)(
)
Ix =
2
2
mr
5
Problem 10-94
Determine the radius of gyration kx of the paraboloid. The density of the material is ρ.
Units Used:
Given:
Mg = 1000 kg
ρ = 5
Mg
m
3
a = 200 mm
b = 100 mm
1055
© 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:
a
⌠
⎛ b2 x ⎞
⎮
⎟ dx
mp = ⎮ ρ π ⎜
a ⎠
⎝
⌡
mp = 15.71 kg
0
⌠
⎮
Ix = ⎮
⌡
a
⎛ b x⎞⎛ b x⎞
1
⎟ ⎜ ⎟ dx
ρ π⎜
2
⎝ a ⎠⎝ a ⎠
2
2
Ix = 52.36 × 10
−3
kg⋅ m
2
0
kx =
Ix
kx = 57.7 mm
mp
Problem 10-95
Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and
express the result in terms of the mass m of the semiellipsoid. The material has a
constant density ρ.
Solution:
a
⌠
⎮
2⎛
m = ⎮ ρ π b ⎜1 −
⎜
⎮
⎝
⌡0
a
2⎞
⎟ dx = 2 a ρ π b2
3
a ⎠
x
ρ =
2⎟
⌠
⎮ 1 ⎛ 3m ⎞ 2 ⎛
Ix = ⎮
π b ⎜1 −
⎜
⎟
2
⎜
⎮ 2 ⎝ 2π a b ⎠
⎝
⌡0
2⎞
2
⎟ b2 ⎛⎜1 − x ⎟⎞ dx = 2 m b2
2⎟ ⎜
2⎟
5
a ⎠ ⎝
a ⎠
x
3m
2π a b
2
Ix =
2
2
mb
5
1056
© 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 10-96
Determine the radius of gyration kx of the body. The specific weight of the material is γ.
Given:
lb
γ = 380
ft
3
a = 8 in
b = 2 in
Solution:
a
⌠
2
⎮
⎮
3
2⎛x⎞
⎮
mb =
γ π b ⎜ ⎟ dx
⎮
⎝ a⎠
⌡
mb = 13.26 lb
0
⌠
⎮
⎮
Ix = ⎮
⎮
⌡
a
2
1
2 x
γ π b ⎛⎜ ⎟⎞
2
⎝ a⎠
3
2
2⎛x⎞
b
3
⎜ ⎟ dx
⎝ a⎠
Ix = 0.59 slug⋅ in
2
0
kx =
Ix
mb
kx = 1.20 in
Problem 10-97
Determine the moment of inertia for the ellipsoid with respect to the x axis and express the result in
terms of the mass m of the ellipsoid. The material has a constant density ρ.
1057
© 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:
a
⌠
⎮
2⎛
m = ⎮ ρ π b ⎜1 −
⎜
⎮
⎝
⌡− a
2⎞
⎟ dx = 4 a ρ π b2
2⎟
3
a ⎠
x
a
⌠
⎮
1 3m
2⎛
Ix = ⎮
π b ⎜1 −
2
⎜
2
⎮
4π a b
⎝
⌡− a
ρ =
3m
4π a b
2
2⎞
2
⎟ b2⎛⎜ 1 − x ⎟⎞ dx = 2 m b2
2⎟ ⎜
2⎟
5
a ⎠ ⎝
a ⎠
x
Ix =
2
2
mb
5
Problem 10-98
Determine the moment of inertia of
the homogeneous pyramid of mass m
with respect to the z axis. The density
of the material is ρ. Suggestion: Use a
rectangular plate element having a
volume of dV = (2x)(2y) dz.
Solution:
⌠
⎮
V=⎮
⌡
h
⎡⎛
⎢a⎜1 −
⎣⎝
2
z ⎞⎤
1 2
⎟⎥ dz = h a
h ⎠⎦
3
ρ =
0
⌠
3m ⎮
Iz =
2 ⎮
a h⌡
m
3m
=
2
V
a h
h
4
1⎡ ⎛
z ⎞⎤
1
2
⎢a⎜ 1 − ⎟⎥ dz = m a
6⎣ ⎝
h ⎠⎦
10
Iz =
1
2
ma
10
0
1058
© 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 10-99
The concrete shape is formed by rotating the shaded area about the y axis. Determine the moment of
inertia Iy. The specific weight of concrete is γ.
Given:
γ = 150
lb
ft
3
a = 6 in
b = 4 in
c = 8 in
Solution:
c
⌠
2
2
1
2
2 ⎮ 1 ⎛ a y⎞ a y
⎟
Iy = γ π ( a + b) c ( a + b) − ⎮
γ ⎜π
dy
2
2 ⎝ c ⎠ c
⌡
Iy = 2.25 slug⋅ ft
2
0
Problem 10-100
Determine the moment of inertia of the thin plate about an axis perpendicular to the page and passing
through the pin at O. The plate has a hole in its center. Its thickness is c, and the material has a
density of ρ
Given:
a = 1.40 m
c = 50 mm
b = 150 mm
ρ = 50
kg
m
3
Solution:
IG =
(
)
1
1
2 2
2
2 2
ρa c a + a − ρ πb c b
12
2
IG = 1.60 kg⋅ m
I0 = IG + m d
2
2
1059
© 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
2
Chapter 10
2
m = ρa c − ρ πb c
m = 4.7233 kg
I0 = IG + m ( a sin ( 45 deg) )
I0 = 6.23 kg⋅ m
2
2
Problem 10-101
Determine the moment of inertia Iz of the frustum of the cone which has a conical depression. The
material has a density ρ.
Given:
kg
ρ = 200
m
3
a = 0.4 m
b = 0.2 m
c = 0.6 m
d = 0.8 m
Solution:
h =
da
a−b
Iz =
3 ⎡ ⎛ 1 2 ⎞⎤ 2
3 ⎡ ⎛ 1 2 ⎞⎤ 2
3 ⎡ ⎡1 2
⎤⎤ 2
⎢ρ ⎜ π a h⎟⎥ a − ⎢ρ ⎜ π a c⎟⎥ a − ⎢ρ ⎢ π b ( h − d)⎥⎥ b
10 ⎣ ⎝ 3
10 ⎣ ⎝ 3
10 ⎣ ⎣ 3
⎠⎦
⎠⎦
⎦⎦
Iz = 1.53 kg⋅ m
2
Problem 10-102
Determine the moment of inertia for the assembly about an axis which is perpendicular to the
page and passes through the center of mass G. The material has a specific weight γ.
Given:
a = 0.5 ft
d = 0.25 ft
1060
© 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
b = 2 ft
c = 1 ft
Chapter 10
e = 1 ft
γ = 90
lb
ft
3
Solution:
IG =
1
2
2 1
2
2 1
2 2
γ π ( a + b) e ( a + b) − γ π b ( e − d) b − γ π c d c
2
2
2
IG = 118 slug⋅ ft
2
Problem 10-103
Determine the moment of inertia for the assembly about an axis which is perpendicular to the page and
passes through point O. The material has a specific weight γ .
Given:
a = 0.5 ft
d = 0.25 ft
b = 2 ft
e = 1 ft
c = 1 ft
γ = 90
lb
ft
3
Solution:
IG =
1
2
2 1
2
2 1
2 2
γ π ( a + b) e ( a + b) − γ π b ( e − d) b − γ π c d c
2
2
2
IG = 118 slug⋅ ft
2
2
2
2
M = γ π ( a + b ) e − γ π b ( e − d) − γ π c d
I O = I G + M ( a + b)
2
M = 848.23 lb
IO = 283 slug⋅ ft
2
1061
© 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 10-104
The wheel consists of a thin ring having a mass M1 and four spokes made from slender rods, each
having a mass M2. Determine the wheel’s moment of inertia about an axis perpendicular to the page
and passing through point A.
Given:
M1 = 10 kg
M2 = 2 kg
a = 500 mm
Solution:
1
2
2
IG = M1 a + 4 M2 a
3
IA = IG + ( M1 + 4M2 ) a
2
IA = 7.67 kg⋅ m
2
Problem 10-105
The slender rods have a weight density γ. Determine the moment of inertia for the assembly
about an axis perpendicular to the page and passing through point A.
Given:
γ = 3
lb
ft
a = 1.5 ft
b = 1 ft
c = 2 ft
Solution:
I =
1
1
2
2
2
γ ( b + c) ( b + c) +
γ 2a ( 2a) + γ 2a c
12
3
I = 2.17 slug⋅ ft
2
1062
© 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 10-106
Each of the three rods has a mass m. Determine the
moment of inertia for the assembly about an axis which is
perpendicular to the page and passes through the center
point O.
Solution:
⎡1
I O = 3⎢
⎣12
IO =
2
2
⎛ a sin ( 60 deg) ⎞ ⎥⎤
⎟
3
⎝
⎠⎦
ma + m⎜
1
2
ma
2
Problem 10-107
The slender rods have weight density γ. Determine the moment of inertia for the assembly
about an axis perpendicular to the page and passing through point A
Given:
γ = 3
lb
ft
a = 1.5 ft
b = 2 ft
Solution:
IA =
1
1
2
2
2
γb b +
γ 2a ( 2a) + γ ( 2a) b
3
12
IA = 1.58 slug⋅ ft
2
Problem 10-108
The pendulum consists of a plate having weight Wp and a slender rod having weight Wr .
Determine the radius of gyration of the pendulum about an axis perpendicular to the page and
passing through point O.
1063
© 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
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be reproduced, in any form or by any means, without permission in writing from the publisher.
Engineering Mechanics - Statics
Chapter 10
Given:
Wp = 12 lb
a = 1 ft
Wr = 4 lb
b = 1 ft
c = 3 ft
d = 2 ft
Solution:
(
2
)
1
2
⎛ c + d − c⎞ + 1 W a2 + b2 + W ⎛c + b ⎞
I0 =
W r ( c + d) + W r ⎜
⎟
⎟
p
p⎜
12
12
⎝ 2
⎠
⎝ 2⎠
k0 =
I0
Wp + Wr
2
k0 = 3.15 ft
Problem 10-109
Determine the moment of inertia for the overhung crank about the x axis. The material is steel
having density ρ.
Units Used:
Mg = 1000 kg
Given:
ρ = 7.85
Mg
m
3
a = 20 mm
b = 20 mm
c = 50 mm
d = 90 mm
e = 30 mm
Solution:
2
⎛ a⎞ c
⎟
⎝ 2⎠
m = ρπ⎜
m = 0.12 kg
M = ρ 2d b e
M = 0.85 kg
1064
© 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
⎡1 ⎛ a ⎞
1 ⎡
2⎤
2
2
Ix = 2⎢ m ⎜ ⎟ + m ( d − e) ⎥ +
M⎣( 2d) + e ⎤⎦
12
2
2
⎣ ⎝ ⎠
⎦
2
Ix = 3.25 × 10
−3
kg⋅ m
2
Problem 10-110
Determine the moment of inertia for the overhung crank about the x' axis. The material is steel
having density ρ.
Units used:
Mg = 1000 kg
Given:
ρ = 7.85
Mg
m
3
a = 20 mm
b = 20 mm
c = 50 mm
d = 90 mm
e = 30 mm
Solution:
2
⎛ a⎞
m = ρπ⎜ ⎟ c
⎝ 2⎠
m = 0.12 kg
M = ρ 2d b e
M = 0.85 kg
⎡1 ⎛ a ⎞
1 ⎡
2⎤
2
2
Ix = 2⎢ m ⎜ ⎟ + m ( d − e) ⎥ +
M⎣( 2d) + e ⎤⎦
12
2
2
⎣ ⎝ ⎠
⎦
2
Ix' = Ix + ( M + 2m) ( d − e)
2
Ix' = 7.19 × 10
−3
kg⋅ m
2
1065
© 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 10-111
Determine the moment of inertia for the solid steel assembly about the x axis. Steel has a specific
weight γst.
Given:
a = 2 ft
b = 3 ft
c = 0.5 ft
d = 0.25 ft
lb
γ st = 490
ft
3
Solution:
h =
ca
c−d
⎡ 2 ⎛ c2 ⎞ π 2 ⎛ 3c2 ⎞ π 2
⎛ 3d2 ⎞⎤
⎢
⎜
⎟
⎜
⎟
⎟⎥
Ix = γ st π c b
+ c h
− d ( h − a) ⎜
⎣
⎝ 2 ⎠ 3 ⎝ 10 ⎠ 3
⎝ 10 ⎠⎦
Ix = 5.64 slug⋅ ft
2
Problem 10-112
The pendulum consists of two slender rods AB and OC which have a
mass density ρr. The thin plate has a mass density ρp. Determine the
location yc of the center of mass G of the pendulum, then calculate
the moment of inertia of the pendulum about an axis perpendicular to
the page and passing through G.
Given:
kg
m
ρr = 3
ρ s = 12
kg
m
2
a = 0.4 m
b = 1.5 m
1066
© 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 = 0.1 m
d = 0.3 m
Solution:
b
bρ r
2
yc =
IG =
2
2
+ π d ρ s( b + d) − π c ρ s( b + d)
2
yc = 0.888 m
2
bρ r + π d ρ s − π c ρ s + ρ r2a
1
1
2
2
2
2aρ r ( 2a) + 2aρ r yc +
bρ r b ...
12
12
2
1 2
2
2
2
⎞
⎛b
+ bρ r ⎜ − yc⎟ + π d ρ s d + π d ρ s ( b + d − yc) ...
2
⎝2
⎠
+
1
2
π c ρ s c − π c ρ s ( b + d − yc)
2
IG = 5.61 kg⋅ m
2
2
2
2
Problem 10-113
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 2 in
b = 8 in
Solution:
b
⌠
1
⎮
⎮
3
2 ⎛ y⎞
Ix = ⎮ y a ⎜ ⎟ d y
⎮
⎝ b⎠
⌡
0
Ix = 307 in
4
1067
© 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 10-114
Determine the moment of inertia for the shaded area about the y axis.
Given:
a = 2 in
b = 8 in
Solution:
a
⌠
3
⎮ 2⎡
⎛x⎞ ⎤
Iy = ⎮ x ⎢b − b ⎜ ⎟ ⎥ dx
⎣
⎝ a⎠ ⎦
⌡
0
Iy = 10.67 in
4
Problem 10-115
Determine the mass moment of inertia Ix of the body and express the result in terms of the total
mass m of the body. The density is constant.
Solution:
a
⌠
2
bx
7
⎮
2
⎞
⎛
m = ⎮ ρπ⎜
+ b⎟ dx = a ρ π b
3
⎝a
⎠
⌡
ρ =
0
3m
7π a b
2
1068
© 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
⌠
2
2
⎮ 1 ⎛ 3m ⎞ ⎛ b x
bx
93
2
⎞
⎞
⎛
Ix = ⎮
π⎜
+ b⎟ ⎜
+ b⎟ dx =
mb
⎜
⎟
2
70
⎠ ⎝a
⎠
⎮ 2 ⎝ 7π a b ⎠ ⎝ a
⌡0
Ix =
93 2
mb
70
Problem 10-116
Determine the product of inertia for the shaded area with respect to the x and y axes.
Given:
a = 1m
b = 1m
Solution:
⌠
⎮
⎮
Ixy = ⎮
⎮
⌡
b
1
1 ⎛ y⎞
ya⎜ ⎟
2 ⎝ b⎠
3
1
3
⎛ y⎞ dy
⎟
⎝ b⎠
a⎜
Ixy = 0.1875 m
4
0
Problem 10-117
Determine the area moments of inertia Iu and Iv and
the product of inertia Iuv for the semicircular area.
Given:
r = 60 mm
θ = 30 deg
1069
© 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:
4
Ix =
πr
Iy = Ix
8
4
Ixy = 0 mm
Iu =
Iv =
Iuv =
Ix + Iy
Ix − Iy
cos ( 2θ ) − Ixy sin ( 2θ )
Iu = 5.09 × 10 mm
Ix + Iy Ix − Iy
−
cos ( 2θ ) − Ixy sin ( 2θ )
2
2
Iv = 5.09 × 10 mm
2
+
2
Ix − Iy
sin ( 2θ ) + Ixy cos ( 2θ )
2
Iuv = 0 m
6
4
6
4
4
Problem 10-118
Determine the moment of inertia for the shaded area about the x axis.
Given:
a = 3 in
b = 9 in
Solution:
b
⌠
y
2
Ix = ⎮ y a 1 − d y
⎮
b
⌡
0
Ix = 333 in
4
Problem 10-119
Determine the moment of inertia for the shaded area about the y axis.
1070
© 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 = 3 in
b = 9 in
Solution:
a
⌠
⎮ 2 ⎛
Iy = ⎮ x b⎜ 1 −
⎜
⎮
⎝
⌡0
Iy = 32.4 in
2⎞
x
a
⎟ dx
2⎟
⎠
4
Problem 10-120
Determine the area moment of inertia of the area about the x axis. Then, using the parallel-axis
theorem, find the area moment of inertia about the x' axis that passes through the centroid C of the
area.
Given:
a = 200 mm
b = 200 mm
Solution:
b
⌠
2
Ix = ⎮ y 2a
⎮
⌡
y
dy
b
6
4
3
2
Ix = 914 × 10 mm
0
Find the area and the distance to the centroid
b
⌠
A = ⎮ 2a
⎮
⌡
y
dy
b
A = 53.3 × 10 mm
0
1071
© 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
1 ⌠
⎮ y2a
yc =
A⎮
⌡0
Ix' = Ix − A yc
y
b
yc = 120.0 mm
dy
6
2
4
Ix' = 146 × 10 mm
Problem 10-121
Determine the area moment of inertia for the triangular area about (a) the x axis, and
(b) the centroidal x' axis.
Solution:
h
⌠ 2b
1 3
Ix = ⎮ y ( h − y) d y =
⋅h ⋅b
⎮
h
12
⌡0
3
Ix' =
bh
12
−
1
2
b h ⎛⎜
h⎞
2
Ix =
1
3
⎟ = ⋅h ⋅b
3
36
⎝ ⎠
Ix' =
1
12
1
36
bh
3
3
bh
Problem 10-122
Determine the product of inertia of the shaded area with respect to the x and y axes.
Given:
a = 2 in
b = 1 in
1072
© 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:
b
⌠ a
Ixy = ⎮
⎮ 2
⌡0
Ixy = 0.667 in
y
b
ya
y
b
dy
4
1073
© 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.