10–5.
Determine the moment of inertia for the shaded area about
the x axis.
y
y
4
x2
4m
SOLUTION
x
4
Ix =
L0
4
y2 dA = 2
L0
2m
y2 (x dy)
4
= 2
L0
Ix = 2 B
y2 24 - y dy
2(15 y2 + 12(4)(y) + 8(4)2) 2(4 - y)3 4
R
- 105
0
Ix = 39.0 m4
Ans.
2m
10–6.
Determine the moment of inertia for the shaded area about
the y axis.
y
y ⫽ 4 ⫺ x2
4m
SOLUTION
Iy =
LA
= 2B
x
x2 dA = 2
3
4x
3
Iy = 8.53 m4
-
2
L0
2m
x2 (4 - x2) dx
5
x 2
R
5 0
Ans.
2m
10–7.
Determine the moment of inertia for the shaded area about
the x axis.
y
y2
1
0.5x
1m
x
1m
2m
SOLUTION
d Ix =
Ix =
1
dx (2y)3
12
L
d Ix
2
=
2
(1 - 0.5 x)3>2 dx
L0 3
=
2
2
2
(1 - 0.5x)5>2R
B
3 5(- 0.5)
0
= 0.533 m4
Ans.
Also,
dA = x dy = 2(1 - y2) dy
Ix =
L
y2 dA
1
=
L-1
= 2B
2 y2 (1 - y2) dy
y5 1
y3
R
3
5 -1
= 0.533 m4
Ans.
*10–8.
Determine the moment of inertia for the shaded area about
the y axis.
y
y2
1
0.5x
1m
x
1m
2m
SOLUTION
dA = 2y dx
Iy =
L
x2 dA
2
=
L0
= 2B
2 x2 (1 - 0.5x)1/2 dx
2(8 - 12(- 0.5)x + 15(- 0.5)2 x2)2(1 - 0.5x)3
105(-0.5)
3
= 2.44 m4
R
2
0
Ans.
Also,
Iy =
L
d Iy
1
= 2
1 3
x dy
L0 3
1
= 2
8
(1 - y2)3 dy
L0 3
1
8
3
1
= 2 a b B y - y3 + y5 - y7 R
3
5
7
0
= 2.44 m4
Ans.
10–19.
y
Determine the moment of inertia of the shaded area about
the x axis.
SOLUTION
xy ⫽ 4
4m
Here, the area must be divided into two segments as shown in Fig. a. The
moment of inertia of segment (2) about the x axis can be determined using
1 3
h 2
bh + A 2 ¢ ≤ , while the moment of inertia of segment (1) about the x axis
12
2
can be determined by applying Eq. 10–1. The area of the rectangular differential
1m
(Ix)2 =
element in Fig. a is dA = (x - 1)dy. Here, x =
4
4
. Thus, dA = ¢ - 1 ≤ dy.
y
y
Applying Eq. 10–1 to segment (1) about the x axis
4m
(Ix)1 =
y2dA =
LA
= ¢ 2y2 -
L1 m
y2 ¢
4m
4
a4y - y2 bdy
- 1 ≤ dy =
y
L1 m
y3 4 m
= 9 m4
≤`
3 1m
The moment of inertia of segment (2) about the x axis is
(Ix)2 =
1
(3)(13) + (3)(1)(0.52) = 1 m4
12
Thus, Ix = (Ix)1 + (Ix)2 = 9 + 1 = 10 m4
Ans.
The area of the rectangular differential element in Fig. b is dA = y dx.
'
The moment inertia of this element about the x axis is dIx = dIx ¿ + dAy 2
=
y 2
1
4
1 4 3
64
1
dxy3 + ydx ¢ ≤ = y3dx. Here, y = . Thus, dIx = ¢ ≤ dx =
dx.
x
12
2
3
3 x
3x3
Performing the integration, we have
4m
Ix =
L
dlx =
64
L1 m 3x
3
dx = ¢ -
32 4 m
= 10 m4
≤`
3x2 1 m
Ans.
x
1m
4m
*10–20.
y
Determine the moment of inertia of the shaded area about
the y axis.
SOLUTION
The area of the rectangular differential element in Fig. a is dA = y dx. Here, y =
Thus, dA =
4
dx.
x
4
.
x
xy ⫽ 4
4m
1m
Applying Eq. 10–1,
x
4m
4m
4
Ans.
Iy =
x2 dA =
x2 a bdx =
4x dx = a2x2 b `
= 30 m4
x
L1 m
LA
L1 m
1m
Here, the area must be divided into two segments as shown in Fig. b. The moment
of inertia of segment (2) about the y axis can be determined using
4m
h 2
1 3
bh + A 2 ¢ ≤ , while the moment of inertia of segment (1) about the
12
2
x axis can be determined by computing the moment of inertia of the element
parallel to the x axis shown in Fig. b. The area of this element is dA = (x - 1) dy
and its moment of inertia about the y axis is
(Ix)2 =
2
1
'2
3
(dy)(x - 1) + (x - 1)dy B 1 (x + 1) R
dIy = dIy¿ + dAx =
12
2
1
1
= a x3 - ≤ dy
3
3
Here, x =
4
. Thus,
y
1 4 3
1
64
1
dly = B ¢ ≤ - R dy = ¢ 3 - ≤ dy
3 y
3
3
3y
Performing the integration, the moment of inertia of segment (1) about the y axis is
4m
(Iy)1 =
L
dIy =
L1 m
¢
4m
64
1
32
1
- ≤ dy = ¢ - 2 - y ≤ `
= 9 m4
3
3
3
3y
3y
1m
Ans.
The moment of inertia of segment (2) about the y axis is
(Iy)2 =
1
(1)(33) + (1)(3)(2.52) = 21 m4
12
Thus,
Iy = (Iy)1 + (Iy)2 = 9 + 21 = 30 m4
Ans.
1m
4m
10–46.
y
25 mm
Determine the distance y to the centroid of the beam’s
cross-sectional area; then determine the moment of inertia
about the x¿ axis.
25 mm
100 mm
C
x¿
_
y
25 mm
x
50 mm
100 mm
75 mm
75 mm
SOLUTION
Centroid: The area of each segment and its respective centroid are tabulated below.
Segment
1
2
3
A (mm2)
50(100)
325(25)
25(100)
©
15.625(103)
yA (mm3)
375(103)
10l.5625(103)
–125(103)
y (mm)
75
12.5
–50
351.5625(103)
Thus,
y =
351.5625(103)
©yA
=
= 22.5 mm
©A
15.625(103)
Ans.
Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be
determined using the parallel-axis theorem Ix¿ + Ix¿ + Ad2y.
Segment
1
2
3
Ai (mm2)
50(100)
325(25)
25(100)
A dy B i (mm) A Ix–B i (mm4)
52.5
10
72.5
1
12
1
12
1
12
A Ad2y B i (mm4)
A Ix¿ B i (mm4)
3
13.781(106)
17.948(106)
3
0.8125(106)
1.236(106)
3
13.141(106)
15.224(106)
(50) (100 )
(325) (25 )
(25) (100 )
Thus,
Ix¿ = ©(Ix¿)i = 34.41 A 106 B mm4 = 34.4 A 106 B mm4
Ans.
25 mm
50 mm
10–47.
Determine the moment of inertia of the beam’s crosssectional area about the y axis.
y
25 mm
25 mm
100 mm
C
x¿
_
y
50 mm
100 mm
Moment of Inertia: The moment of inertia about the y¿ axis for each segment can be
determined using the parallel-axis theorem Iy¿ = Iy¿ + Ad2x.
Ai (mm2)
75 mm
75 mm
25 mm
SOLUTION
Segment
25 mm
x
A dx B i (mm)
1
2[100(25)]
100
2
25(325)
0
3
100(25)
0
2
C
A Iy–B i (mm4)
1
3
12 (100) (25 )
1
3
12 (25) (325 )
1
3
12 (100) (25 )
D
A Ad2x B i (mm4) A Iy–B i (mm4)
50.0(106)
50.260(106)
0
71.517(106)
0.130(106)
0
Thus,
Iy¿ = ©(Iy¿)i = 121.91 A 106 B mm4 = 122 A 106 B mm4
Ans.
50 mm
10–63.
Determine the product of inertia for the beam’s crosssectional area with respect to the u and v axes.
y
v
150 mm
150 mm
u
20
SOLUTION
Moments of inertia Ix and Iy
20 mm
1
1
(300)(400)3 (280)(360)3 = 511.36(10)6 mm4
Ix =
12
12
Iy = 2 c
20 mm
1
1
(20)(300)3 d +
(360)(20)3 = 90.24(10)6 mm4
12
12
The section is symmetric about both x and y axes; therefore Ixy = 0.
Iuv =
Ix - I y
= a
2
sin 2u + Ixy cos 2u
511.36 - 90.24
sin 40° + 0 cos 40°b 106
2
= 135(10)6 mm4
x
C
Ans.
200 mm
*10–64.
Determine the moments of inertia for the shaded area with
respect to the u and v axes.
y
0.5 in.
v
u
SOLUTION
Moment and Product of Inertia about x and y Axes: Since the shaded area is
symmetrical about the x axis, Ixy = 0.
Ix =
1
1
(1)(53) +
(4)(13) = 10.75 in4
12
12
1 in.
Iy =
1
1
(1)(43)+ 1(4)(2.5)2 +
(5)(13) = 30.75 in4
12
12
Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10-9 with
u = 30°, we have
Iu =
=
Ix + Iy
2
+
I x - Iy
2
cos 2u - Ixy sin 2u
10.75 - 30.75
10.75 + 30.75
+
cos 60° - 0(sin 60°)
2
2
= 15.75 in4
Iv =
=
Ix + Iy
2
Ans.
-
Ix - I y
2
cos 2u + Ixy sin 2u
10.75 - 30.75
10.75 + 30.75
cos 60° + 0(sin 60°)
2
2
= 25.75 in4
Ans.
0.5 in.
x
0.5 in.
30
5 in.
4 in.
10–74.
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.
y
4 in.
3 in.
8
SOLUTION
Ix = 2 c
4 in.
3 in.
8
x
C
3 3
3 2
6 3
1
3
1 3
(4)a b + 4a b a4 b d +
a b a8 - b
12
8
8
16
12 8
8
4 in.
= 55.55 in4
4 in.
4 - 38
3 3
3
3 2
1 3
3
b +
f d
Iy = 2 c a b a 4 - b + a4 - b e a
12 8
8
8
8
2
16
+
3 3
1
(8) a b
12
8
= 13.89 in4
Ixy = ©xy A
= - 2[(1.813 + 0.1875)(3.813)(3.625)(0.375)] + 0
= -20.73 in4
Imax>min =
=
Ix + Iy
2
;
C
a
Ix - Iy
2
2
b + I2xy
55.55 + 13.89
55.55 - 13.89 2
a
;
b + (- 20.73)2
2
C
2
Imax = 64.1 in4
Ans.
Imin = 5.33 in4
Ans.
10–75.
Solve Prob. 10–74 using Mohr’s circle.
y
4 in.
3 in.
8
SOLUTION
Ix = 2 c
3 3
3 2
6 3
1
3
1 3
(4)a b + 4a b a 4 b d +
a b a8 - b = 55.55 in4
12
8
8
16
12 8
8
Iy = 2 c
4 - 38
3 3
3
3 2
1 3
3
a b a4 - b + a4 - b e a
b +
f d
12 8
8
8
8
2
16
+
4 in.
4 in.
1
3 3
(8)a b = 13.89 in4
12
8
= - 2[(1.813 + 0.1875)(3.813)(3.625)(0.375)] + 0
= - 20.73 in4
Center of circle:
2
x
C
Ixy = ©xy A
Ix + Iy
4 in.
3 in.
8
= 34.72 in4
R = 2(55.55 - 34.72)2 + (-20.73)2 = 29.39 in4
Imax = 34.72 + 29.39 = 64.1 in4
Ans.
Imin = 34.72 - 29.39 = 5.33 in4
Ans.
10–106.
The thin plate has a mass per unit area of 10 kg>m2.
Determine its mass moment of inertia about the y axis.
z
200 mm
200 mm
100 mm
200 mm
SOLUTION
100 mm
Composite Parts: The thin plate can be subdivided into segments as shown in Fig. a.
Since the segments labeled (2) are both holes, the y should be considered as
negative parts.
200 mm
200 mm
x
Mass moment of Inertia: The mass of segments (1) and (2) are
m1 = 0.4(0.4)(10) = 1.6 kg and m2 = p(0.12)(10) = 0.1p kg. The perpendicular
distances measured from the centroid of each segment to the y axis are indicated in
Fig. a. The mass moment of inertia of each segment about the y axis can be
determined using the parallel-axis theorem.
Iy = © A Iy B G + md2
= 2c
1
1
(1.6)(0.42) + 1.6(0.22) d - 2c (0.1p)(0.12) + 0.1p(0.22) d
12
4
= 0.144 kg # m2
Ans.
200 mm
200 mm
200 mm
y
10–107.
The thin plate has a mass per unit area of 10 kg>m2.
Determine its mass moment of inertia about the z axis.
z
200 mm
200 mm
100 mm
200 mm
SOLUTION
Composite Parts: The thin plate can be subdivided into four segments as shown in
Fig. a. Since segments (3) and (4) are both holes, the y should be considered as
negative parts.
100 mm
200 mm
200 mm
x
200 mm
200 mm
Mass moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) are
m1 = m2 = 0.4(0.4)(10) = 1.6 kg and m3 = m4 = p(0.12)(10) = 0.1p kg. The mass
moment of inertia of each segment about the z axis can be determined using the
parallel-axis theorem.
Iz = © A Iz B G + md2
=
1
1
1
1
(1.6)(0.42) + c (1.6)(0.42 + 0.42) + 1.6(0.22) d - (0.1p)(0.12) - c (0.1p)(0.12) + 0.1p(0.22) d
12
12
4
2
= 0.113 kg # m2
Ans.
200 mm
y