10–3. Determine the moment of inertia of the area about
the x axis.
y
y2 ⫽ 2x
2m
x
2m
10–4. Determine the moment of inertia of the area about
the y axis.
y
y2 ⫽ 2x
2m
x
2m
10–23.
Determine the moment of inertia of the shaded area about
the x axis.
y
x2 ⫹ y2 ⫽ r02
a
––
2
SOLUTION
r0
Differential Element: The area of the differential element shown shaded in Fig. a is
dA = (rdu) dr.
Moment of Inertia: Applying Eq. 10–1, we have
a>2
Ix =
y2dA =
LA
a>2
=
L- a>2
a>2
=
L- a>2
However, sin2 u =
=
¢
r3 sin2 udrdu
r4 r0 2
≤ ` sin udu
4 0
r 04 2
sin udu
4
1
(1 - cos 2u). Thus,
2
a>2
Ix =
r2 sin2 u(rdu)dr
r0
L- a>2 L0
a>2
=
L- a>2 L0
r0
L- a>2
r0 4
(1 - cos 2u)du
8
a>2
r0 4
r0 4
1
=
(a - sin a)
B u - sin 2u R `
8
2
8
- a>2
Ans.
a
––
2
x
10–24.
Determine the moment of inertia of the shaded area about
the y axis.
y
x2 ⫹ y2 ⫽ r02
a
––
2
r0
SOLUTION
Differential Element: The area of the differential element shown shaded in Fig. a is
dA = (rdu)dr.
Moment of Inertia: Applying Eq. 10–1, we have
a>2
Iy =
x2dA =
LA
a>2
=
a>2
L- a>2
L- a>2 L0
r2 cos2 u(rdu)dr
r0
L- a>2 L0
=
r0
¢
r3 cos2 udrdu
r4 r0
≤ ` cos2 udu
4 0
a>2
=
r0 4
cos2 udu
L- a>2 4
1
( cos 2u + 1). Thus,
2
a>2 4
r0
Iy =
( cos 2u + 1)du
L- a>2 8
However, cos2 u =
=
a>2
r0 4 1
r0 4
=
( sin a + a)
B sin 2u + u R `
8 2
8
- a>2
Ans.
a
––
2
x
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
10–69.
Determine the moments of inertia Iu and Iv of the
shaded area.
y
20 mm
v
u
45°
SOLUTION
20 mm
x
20 mm
200 mm
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
1200214032 +
1402120032 = 27.7311062 mm4
12
12
Iy =
1
1
1402120032 + 4012002112022 +
1200214032
12
12
200 mm
40 mm
= 142.9311062 mm4
Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10–9 with
u = 45°, we have
Iu =
Ix + Iy
2
= a
+
Ix - Iy
2
cos 2u - Ixy sin 2u
27.73 + 142.93
27.73 - 142.93
+
cos 90° - 01sin 90°2b11062
2
2
= 85.311062 mm4
Iv =
Ix + Iy
= a
2
-
Ix - Iy
2
Ans.
cos 2u + Ixy sin 2u
27.73 - 142.93
27.73 + 142.93
cos 90° - 01sin 90°2b11062
2
2
= 85.3 106 mm4
Ans.
10–70.
y
Determine the moments of inertia and the product of
inertia of the beam’s cross sectional area with respect to the
u and v axes.
v
u
300 mm
SOLUTION
30⬚
Moments and product of Inertia with Respect to the x and y Axes: The
perpendicular distances measured from the centroid of the triangular segment to
the y axis are indicated in Fig. a.
Ix =
1
(400)(4503) = 1012.5(106) mm4
36
Iy = 2 B
Since the cross-sectional area is symmetrical about the y axis, Ixy = 0.
Moment and product of Inertia with Respect to the u and v Axes: Applying
Eq. 10–9 with u = 30°, we have
Iu =
Ix + Iy
2
= B
+
Ix - Iy
2
cos 2u - Ixy sin 2u
1012.5 + 600
1012.5 - 600
+ ¢
≤ cos 60° - 0 sin 60° R (106)
2
2
= 909.375(106) mm4 = 909(106) mm4
Iv =
Ix + Iy
2
= B
-
Ix - Iy
2
Iuv =
Ans.
cos 2u + Ixy sin 2u
1012.5 + 600
1012.5 - 600
- ¢
≤ cos 60° + 0 sin 60 R (106)
2
2
= 703.125(106) mm4 = 703(106) mm4
Ix - Iy
2
= B¢
Ans.
sin 2u + Ixy cos 2u
1012.5 - 600
≤ sin 60° + 0 cos 60° R (106)
2
= 178.62(106) mm4 = 179(106) mm4
150 mm
200 mm
1
1
(450)(2003) + (450)(200)(66.672) R = 600(106) mm4
36
2
Ans.
x
C
200 mm
10–71.
y
Solve Prob. 10–70 using Mohr’s circle. Hint: Once the circle
is established, rotate 2u = 60° counterclockwise from the
reference OA, then find the coordinates of the points that
define the diameter of the circle.
v
u
300 mm
SOLUTION
30⬚
Moments and product of Inertia with Respect to the x and y Axes: The
perpendicular distances measured from the centroid of the triangular segment to the
y axis are indicated in Fig. a.
Ix =
1
(400)(4503) = 1012.5(106) mm4
36
Iy = 2 B
Since the cross-sectional area is symmetrical about the y axis, Ixy = 0.
Construction of Mohr’s Circle: The center of C of the circle lies along the I axis at a
distance
Iavg =
Ix + Iy
2
= a
150 mm
200 mm
1
1
(450)(2003) + (450)(200)(66.672) R = 600(106) mm4
36
2
1012.5 + 600
b (106)mm4 = 806.25(106) mm4
2
The coordinates of the reference point A are [1012.5, 0](106) mm4. The circle can be
constructed as shown in Fig. b. The radius of the circle is
R = CA = (1012.5 - 806.25)(106) = 206.25(106) mm4
Moment and Product of Inertia with Respect to the u and v Axes: By referring to
the geometry of the circle, we obtain
Iu = (806.25 + 206.25 cos 60°)(106) = 909(106) mm4
Ans.
Iv = (806.25 - 206.25 cos 60°)(106) = 703(106) mm4
Ans.
Iuv = 206.25 sin 60° = 179(106) mm4
Ans.
x
C
200 mm
10–72. Locate the centroid Y of the beam’s cross-sectional
area and then determine the moments of inertia and the
product of inertia of this area with respect to the U and
V axes.
y
u
V
50 mm
450 mm
450 mm
50 mm
60
400 mm
x
C
50 mm
800 mm
y
Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s
cross – sectional area are indicated in Fig. a. Thus,
y =
ΣyC A
ΣA
=
1225(1000)(50) + 2[1000(400)(50)] + 600(12000)(100)
= 825 mm
1000(50) + 2(400)(50) + 1200(100)
Ans.
Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid
of each segment to the x and y axes are indicated in Fig. b. Using the parallel – axis theorem,
⎡1
⎡1
⎡1
⎤
⎤
⎤
Ix = ⎢ (1000)(50 3 ) + 1000(50)(400) 2 ⎥ + 2 ⎢ (50)(400 3 ) + 50(400)(175) 2 ⎥ + ⎢ (100)(1200 3 ) + 100(1200)(225) 2 ⎥
⎣ 12
⎣ 12
⎣ 12
⎦
⎦
⎦
8
4
)
mm
= 302.44 (10
Iy =
⎡1
⎤
1
1
(50)(10003) + 2 ⎢ (400)(50 3 ) + 400(50)(75) 2 ⎥ +
(1200)(1003)
12
12
⎣ 12
⎦
= 45 (108) mm4
Since the cross – sectional area is symmetrical about the y axis, Ixy = 0.
Moment and Product of Inertia with Respect to the u and v Axes: With ␪ = 60,
Iu =
Ix + Iy
2
+
Ix – Iy
2
cos 2␪ – Ixy sin 2␪
⎡ 302.44 + 45
⎤
302.44 – 45
+
cos 120° – 0 sin
n 120°⎥ (108)
= ⎢
2
2
⎣
⎦
= 109.36 (108) mm4 = 109 (108) mm4
Iv =
Ix + Iy
2
–
Ix – Iy
2
Ans.
cos 2␪ + Ixy sin 2␪
⎡ 302.44 + 45 302.44 – 45
⎤
–
cos 120° + 0 sin
n 120°⎥ (108)
= ⎢
2
2
⎣
⎦
= 238.08 (108) mm4 = 238 (108) mm4
Ans.
Iuv =
Ix – Iy
2
sin 2␪ + Ixy cos 2␪
⎡ 302.44 – 45
⎤
= ⎢
sin 120° + 0 cos 120°⎥ (108)
2
⎣
⎦
= 111.47 (108) mm4 = 111 (108) mm4
Ans.
500 mm
75 mm
500 mm
400 mm
75 mm
4000 mm
1225 mm
1000 mm
1000 mm
175 mm
50 mm
600 mm
1200 mm
50 mm
225 mm
y = 825 mm
10–73. Solve Prob. 10–72 using Mohr’s circle.
Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s cross
– sectional area are indicated in Fig. a. Thus,
ΣyA
1225(1000)(50) + 2[1000(400)(50)] + 600(12000)(100)
=
= 825 mm
ΣA
1000(50) + 2(400)(50) + 1200(100)
y =
Ans.
Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of
each segment to the x and y axes are indicated in Fig. b. Using the parallel – axis theorem,
⎡1
⎤
⎡1
⎤
⎡1
⎤
Ix = ⎢ (1000)(50 3 ) + 1000(50)(400) 2 ⎥ + 2 ⎢ (50)(400 3 ) + 50(400)(175) 2 ⎥ + ⎢ (100)(1200 3 ) + 100(1200)(225) 2 ⎥
12
12
12
⎣
⎦
⎣
⎦
⎣
⎦
= 302.44 (108) mm4
⎡1
⎤
1
1
(50)(10003) + 2 ⎢ (400)(50 3 ) + 400(50)(75) 2 ⎥ +
(1200)(1003)
12
12
12
⎣
⎦
Iy =
= 45 (108) mm4
Ixy = 60(5)(–14.35)(13.15) + 55(15)(15.65)(–14.35)
= –11.837 (104) mm4
Since the cross – sectional area is symmetrical about the y axis, Ixy = 0.
Construction of Mohr’s Circle: The center C of the circle lies along the u axis at a distance
Iavg =
Ix + Iy
2
⎛ 302.44 + 45 ⎞
8
8
4
= ⎜
⎟⎠ (10 ) = 173.72 (10 ) mm
2
⎝
The coordinates of the reference point A are (302.44, 0) (108) mm4. The circle can be constructed as shown in Fig. c. The radius of
the circle is
R = CA = (302.44 – 173.72) (108) = 128.72 (108) mm4
Moment and Product of Inertia with Respect to the u and v Axes: By referring to the geometry of the circle,
Iu = (173.72 – 128.72 cos 60°) (108) = 109 (108) mm4
Ans.
Iv = (173.72 + 128.72 cos 60°) (108) = 238 (108) mm4
Ans.
8
8
4
Iuv = (128.72 sin 60°) (10 ) = 111 (10 ) mm
Ans.
(108 mm4)
500 mm
75 mm
500 mm
400 mm
50 mm
100 mm
1225 mm
1000 mm
50 mm
600 mm
1200 mm
400 mm
75 mm
175 mm
225 mm
y = 825 mm
(108 mm4)
10–74. Locate the centroid y of the beam’s cross-sectional
area and then determine the moments of inertia of this area
and the product of inertia with respect to the u and v axes.
The axes have their origin at the centroid C.
y
25 mm
200 mm
v
25 mm
x
C
60⬚
25 mm
75 mm 75 mm
u
y
10–75. Solve Prob. 10–7 4using Mohr’s circle.
10–76. Locate the centroid x of the beam’s cross-sectional
area and then determine the moments of inertia and the
product of inertia of this area with respect to the u and
v axes. The axes have their origin at the centroid C.
y
x
20 mm
v
200 mm
C
x
60⬚
200 mm
20 mm
20 mm
175 mm
u
10–77. Solve Prob. 10–76 using Mohr’s circle.
10–78.
y
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.
20 mm
32.22 mm
100 mm
32.22 mm
C
x
SOLUTION
Ix = c
20 mm
100 mm
1
(20)(100)3 + 100(20)(50 - 32.22)2 d
12
+ c
1
(80)(20)3 + 80(20)(32.22 - 10)2 d
12
= 3.142(106) mm4
Iy = c
1
(100)(20)3 + 100(20)(32.22 - 10)2 d
12
+ c
1
(20)(80)3 + 80(20)(60 - 32.22)2 d
12
= 3.142(106) mm4
Ixy = ©xy A
= - (32.22 - 10)(50-32.22)(100)(20) - (60 - 32.22)(32.22 - 10)(80)(20)
= - 1.778(106) mm4
Imax/min =
Ix + Iy
2
;
C
a
Ix - Iy
2
2
b + I2xy
= 3.142(106) ; 20 + {( - 1.778)(106)}2
Imax = 4.92(106) mm4
Ans.
Imin = 1.36(106) mm4
Ans.
10–79.
Solve Prob. 10–78 using Mohr’s circle.
y
20 mm
32.22 mm
100 mm
SOLUTION
32.22 mm
C
x
Solve Prob. 10–78.
20 mm
Ix = 3.142(106) mm4
100 mm
Iy = 3.142(106) mm4
Ixy = –1.778(10 6) mm4
Center of circle:
Ix + I y
2
= 3.142(106) mm4
R = 2(3.142 - 3.142)2 + ( -1.778)2(106) = 1.778(106) mm4
Imax = 3.142(106) + 1.778(106) = 4.92(106) mm4
Ans.
Imin = 3.142(106) - 1.778(106) = 1.36(106) mm4
Ans.
10–80. Determine the orientation of the principal axes,
which have their origin at centroid C of the beam’s crosssectional area. Also, find the principal moments of inertia.
y
100 mm
20 mm
20 mm
150 mm
x
C
150 mm
100 mm
20 mm
10–81.
Solve Prob. 10–80 using Mohr’s circle.
10–82. Locate the centroid y of the beam’s cross-sectional
area and then determine the moments of inertia of this area
and the product of inertia with respect to the u and v axes.
The axes have their origin at the centroid C.
y
20 mm
v
20 mm
x
C
200 mm
60⬚
20 mm
8 0 mm 8 0 mm
=
2[100(200)(20)] + 10(20)(120)
= 79.23 mm
2(200)(20) + 20(120)
Ans.
1
 1

I x = 2  (20)(2003 ) + 20(200)(20.77) 2  +  (120)(203 ) + 120(20)(69.23) 2 
12
 12

= 41.70(106 ) mm 4
1
 1
I y = 2  (200)(203 ) + 200(20)(70)2  + (20)(1203 )
12
 12
= 42.35(106 ) mm 4
Ix + I y
Ix − I y
cos 2θ − I xy sin 2θ
2
2
 41.70 + 42.35  41.70 − 42.35 

 
6
=
+
 cos(−120 ) − 0sin(−120 )  (10 )
2
2




Iu =
+
= 42.2(106 ) mm 4
Ix + I y
Ans.
Ix − I y
cos 2θ + I xy sin 2θ
2
2
 41.70 + 42.35  41.70 − 42.35 

 
6
=
+
 cos(−120 ) − 0sin(−120 )  (10 )
2
2




Iv =
+
= 41.9(106 ) mm 4
Ans.
u
y
Ix − I y
sin 2θ + I xy cos 2θ
2
41.70 + 42.35
=
+ sin(−120 ) − 0cos(−120 )
2
= 0.28(106 ) mm 4
I uv =
20
Ans.
20
70 mm
70 mm
20.77 mm
20
10
60
60
79.23 mm
69.23 mm
10–83. Solve Prob. 10–82 using Mohr’s circle.
=
2[100(200)(20)] + 10(20)(120)
= 79.23 mm
2(200)(20) + 20(120)
Ans.
1
 1

I x = 2  (20)(2003 ) + 20(200)(20.77)2  +  (120)(203 ) + 120(20)(69.23)2 
12
 12

= 41.70(106 ) mm 4
1
 1
I y = 2  (200)(203 ) + 200(20)(70)2  + (20)(1203 )
12
 12
= 42.25(106 ) mm 4
I avg =
Ix + I y
2
 41.70 + 42.25  6
4
6
4
=
 (10 )mm = 41.975(10 )mm
2


41.70
R = CA = (41.975 – 41.70)(106) = 0.275(106) mm4
I v = (41.975 − 0.275cos60 )(106 ) = 41.86(106 )mm 4
Ans.
Iu = (41.975 − 0.275cos60 )(106 ) = 42.19(106 )mm 4
Ans.
Iuv = 0.325sin 60 = 0.28(106 )mm 4
Ans.
Iu
20
41.975
20
70 mm
70 mm
20.77 mm
20
10
60
60
79.23 mm
69.23 mm
R = 0.275
Iv
42.25
10–94.
Determine the mass moment of inertia Iy of the solid
formed by revolving the shaded area around the y axis. The
total mass of the solid is 1500 kg.
z
4m
z2
O
SOLUTION
x
Differential Element: The mass of the disk element shown shaded in
2
1
1
dm = rdV = rpr2dy. Here, r = z = y3>2.Thus, dm = rpa y3>2 b dy =
4
4
The mass moment of inertia of this element about the y
rp 4
rp 1 3>2 4
rp 6
1
1
r dy =
a y b dy =
y dy.
dIy = dmr2 = A rpr2dy B r2 =
2
2
2
2 4
512
Fig. a is
rp 3
y dy.
16
axis is
Mass: The mass of the solid can be determined by integrating dm. Thus,
4m
m =
L
dm =
L0
4
4m
rp 3
rp y
= 4 pr
y dy =
¢ ≤`
16
16 4 0
The mass of the solid is m = 1500 kg. Thus,
1500 = 4pr
r =
375
kg>m3
p
Mass Moment of Inertia: Integrating dIy,
4m
Iy =
L
Substituting r =
dIy =
L0
rp 6
rp y7 4 m
32p
y dy =
r
=
¢ ≤`
512
512 7 0
7
375
kg>m3 into Iy,
p
Iy =
32p 375
a
b = 1.71(103) kg # m2
p
7
Ans.
1 y3
––
16
2m
y
10–100.
z
Determine the mass moment of inertia of the assembly
about the z axis. The density of the material is 7.85 Mg> m3.
100 mm
SOLUTION
Composite Parts: The assembly can be subdivided into two circular cone segments (1)
and (3) and a hemispherical segment (2) as shown in Fig. a. Since segment (3) is a hole,
it should be considered as a negative part. From the similar triangles, we obtain
0.1
z
=
0.45 + z
0.3
450 mm
300 mm
z = 0.225m
Mass: The mass of each segment is calculated as
300 mm
1
1
m1 = rV1 = r a pr2h b = 7.85(103) c p(0.32)(0.675) d = 158.9625p kg
3
3
2
2
m2 = rV2 = ra pr3 b = 7.85(103) c p(0.33) d = 141.3p kg
3
3
1
1
m3 = rV3 = ra pr2h b = 7.85(103) c p(0.12)(0.225) d = 5.8875p kg
3
3
Mass Moment of Inertia: Since the z axis is parallel to the axis of the cone and the
hemisphere and passes through their center of mass, the mass moment of inertia can be
3
2
3
computed from (Iz)1 =
m r12, (Iz)2 = m2r22, and
m r32. Thus,
10 1
5
10 3
Iz = ©(Iz)i
=
3
2
3
(158.9625p)(0.32) + (141.3p)(0.32) (5.8875p)(0.12)
10
5
10
= 29.4 kg # m2
Ans.
x
y
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 - 2 c (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
10–108. Determine the mass moment of inertia of the
overhung crank about the x axis. The material is steel
having a density of r = 7.85 Mg>m3.
20 mm
30 mm
90 mm
50 mm
x
180 mm
20 mm
x¿
30 mm
20 mm
2
= 0.00325 kg · m = 3.25 g · m
2
.
50 mm
30 mm
10–109.
Determine the mass moment of inertia of the
overhung crank about the x¿ axis. The material is steel
having a density of r = 7.85 Mg>m3.
20 mm
30 mm
90 mm
50 mm
x
180 mm
20 mm
x¿
30 mm
20 mm
kg · m2
50 mm
30 mm