.
.
10–3. Determine the moment of inertia of the area aboutthe axis.x
y
x
y2 � 2x
2 m
2 m
10–4. Determine the moment of inertia of the area aboutthe axis.y
y
x
y2 � 2x
2 m
2 m
10–5.
SOLUTION
Ans.Ix = 39.0 m4
Ix = 2B 2(15 y2 + 12(4)(y) + 8(4)2)2(4 - y)3
-105R4
0
= 2L4
0y224 - y dy
Ix = L4
0y2 dA = 2L
4
0y2 (x dy)
Determine the moment of inertia for the shaded area aboutthe x axis.
2 m
4 m
y 4 x2
x
y
2 m
10–6.
Determine the moment of inertia for the shaded area aboutthe y axis.
SOLUTION
Ans.Iy = 8.53 m4
= 2B4 x3
3-
x5
5R2
0
Iy = LA x2 dA = 2L
2
0x2 (4 - x2) dx
2 m
4 m
y � 4 � x2
x
y
2 m
10–7.
Determine the moment of inertia for the shaded area aboutthe x axis.
SOLUTION
Ans.
Also,
Ans.= 0.533 m4
= 2By3
3-
y5
5R1
-1
= L1
-12 y2 (1 - y2) dy
Ix = Ly2 dA
dA = x dy = 2(1 - y2) dy
= 0.533 m4
=23B 2
5(-0.5) (1 - 0.5x)5>2R2
0
= L2
0
23
(1 - 0.5 x)3>2 dx
Ix = Ld Ix
d Ix =112
dx (2y)3
y
y2 1 0.5x
x
2 m
1 m
1 m
10–8.
SOLUTION
Ans.
Also,
Ans.= 2.44 m4
= 2a83b By - y3 +
35
y5 -17
y7R1
0
= 2L1
0
83
(1 - y2)3 dy
= 2L1
0
13
x3 dy
Iy = Ld Iy
= 2.44 m4
= 2B 2(8 - 12(-0.5)x + 15(-0.5)2 x2)2(1 - 0.5x)3
105(-0.5)3
= L2
02 x2 (1 - 0.5x)1/2 dx
Iy = Lx2 dA
dA = 2y dx
Determine the moment of inertia for the shaded area aboutthe y axis.
y
y2 1 0.5x
x
2 m
1 m
1 m
R2
0
10–9.
Determine the moment of inertia for the shaded area aboutthe x axis.
SOLUTION
Ans.
Also,
Ans.=215
bh3
= cb3
y3 -b
5h2 y5 dh0
= Lh
0y2 (b -
b
h2 y2) dy
Ix = Ly2 dA
dA = (b - x) dy = (b -b
h2 y2) dy
=215
bh3
=13ah2
bb3>2
a25bx5>2 ]b
0
= Lb
0
y3
3dx = L
b
0
13ah2
bb3>2
x3>2 dx
Ix = Ld Ix
d Ix =13
y3 dx
bx
y
y2 —x
h
h2
b
10 10.
Determine the moment of inertia for the thin stripof area about the x axis.The strip is oriented at anangle � from the x axis. Assume that t << l.
Solution:
Ix Ay2����
d�0
lss2 sin2 �� �t
���
d�
A
Ix13
t l3 sin2 �� �� Ans.
–
10–11.
Determine the moment of inertia for the area about the x axis.
SOLUTION
Ans.
Also,
Ans.= 3.20 m4
= L2
0B 1
12 (2 - 22x)3 +
14
(2 - 22x)(2 + 22x)2R dx
Ix = L dIx
=112
(2 - 22x)3 dx +14
(2 - 22x)(2 + 22x)2 dx
=112dx(2 - 22x)3 + (2 - 22x)dx a2 - 22x
2+ 22xb2
dIx = dIx + dA y 2
dA = (2 - 22x)dx
= 3.20 m4
= B y5
10R2
0
= L2
0
y4
2dy
Ix = Ly2 dA
dA = x dy =y2
2dy
y
x2 m
2 my2 � 2x
10–12.
Determine the moment of inertia of the area about the y axis.
SOLUTION
Differential Element: The area of the differential element parallel to the y-axis is
Moment of Inertia: Applying Eq. 10–1 and performing the integration,
Ans. = 0.762 m4
= B23x3 -
2227x
72R `
0
2 m
= L2 m
0(2x2 - 22x
52)dx
Iy = LAx2dA = L
2 m
0x2(2 - 22x
12) dx
dA = (2 - y) dx = (2 - 22x12 )dx.
y
x2 m
2 my2 � 2x
10 13.
Determine the moment of inertia forthe shaded area about the x axis.
Given:
a 4m�
b 2m�
Solution:
Ix
0
b
yy2 a ayb
��
��
2��
���
���
�
����
d�
Ix 4.27 m4� Ans.
–
10 14.
Determine the moment of inertia for theshaded area about the y axis.
Given:
a 4m�
b 2m�
Solution:
Iy0
a
xx2 b�xa
�����
d�
Iy 36.6 m4� Ans.
–
10–15.
SOLUTIONArea of the differential element (shaded) where , hence,
.
Use Simpson’s rule to evaluate the integral: (to 500 intervals)
Ans.Iy = 0.628 m4
Iy = LA x2 dA = L1
0x2 (ex
2)dx
dA = ydx = ex2dx
y = ex2
dA = ydx
Determine the moment of inertia for the shaded area aboutthe y axis. Use Simpson’s rule to evaluate the integral.
y
x
y � ex2
1 m
1 m
10–16.
Determine the moment of inertia for the shaded area aboutthe x axis. Use Simpson’s rule to evaluate the integral.
SOLUTION
Ans.Ix =13L
1
0y3 dx =
13L
1
0(ex
2)3dx = 1.41 m4
=1
12 dxy3 + y dx ay
2b2
=13y3 dx
dIx = dIx + dAy2
y
x
y � ex2
1 m
1 m
10–17.
Determine the moment of inertia of the shaded area aboutthe x axis.
SOLUTION
Differential Element: The area of the rectangular differential element in Fig. ais . The moment of inertia of this element about the x axis is
Moment of Inertia: Performing the integration, we have
Ans. =4a4
9p
=a3
3b c - 1
3(p>a) cos ¢pa
x≤ d csin2 apa
xb + 2 d r `0
a
Ix = LdIx = La
0
a3
3 sin3 ¢p
a x≤dx
=13¢a sin
p
a x≤3 dx =
a3
3 sin3 ¢p
a x≤dx. dIx = dIx ¿ + dAy' 2 =
112
(dx)y3 + ydx¢y2≤2
=13y3 dx
dA = y dx
y
x
a
a––2
a––2
y � a sin xap
10–18.
Determine the moment of inertia of the shaded areaabout the y axis.
SOLUTION
Differential Element: The area of the rectangular differential element in Fig. a is
.
Moment of Inertia: Applying Eq. 10–1, we have
Ans. = ¢p2 - 4p3 ≤a4
= aB - ap¢x2 cos
p
ax≤ +
a2
p2 ¢2x sin p
ax≤ +
2a3
p3 cos p
axR ` a
0
Iy = LAx2dA = L
a
0x2¢a sin
p
ax≤dx
dA = y dx = a sin ¢pax≤dx
y
x
a
a––2
a––2
y � a sin xap
10–19.
Determine the moment of inertia of the shaded area aboutthe x axis.
SOLUTION
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
while the moment of inertia of segment (1) about the x axis
can be determined by applying Eq. 10–1. The area of the rectangular differential
element in Fig. a is . Here, . Thus, .
Applying Eq. 10–1 to segment (1) about the x axis
The moment of inertia of segment (2) about the x axis is
Thus, Ans.
The area of the rectangular differential element in Fig. b is .The moment inertia of this element about the x axis is
. Here, . Thus, .
Performing the integration, we have
Ans.Ix = Ldlx = L4 m
1 m
64
3x3 dx = ¢ - 323x2 ≤ `
1 m
4 m
= 10 m4
dIx =13¢ 4x≤3
dx =643x3dxy =
4x
= 1
12dxy3 + ydx¢y
2≤2
=13y3dx
dIx = dIx ¿ + dAy'2dA = y dx
Ix = (Ix)1 + (Ix)2 = 9 + 1 = 10 m4
(Ix)2 =1
12(3)(13) + (3)(1)(0.52) = 1 m4
= ¢2y2 -y3
3≤ ` 4 m
1 m= 9 m4
(Ix)1 = LAy2dA = L
4
m
1 my2¢ 4y
- 1≤dy = L4 m
1 ma4y - y2bdy
dA = ¢ 4y
- 1≤dyx =4y
dA = (x - 1)dy
(Ix)2 =1
12 bh3 + A2¢h2 ≤
2
,
4 m1 m
1 m
4 m
y
x
xy � 4
10–20.
Determine the moment of inertia of the shaded area aboutthe y axis.
SOLUTION
The area of the rectangular differential element in Fig. a is . Here, .
Thus, .
Applying Eq. 10–1,
Ans.
Here, the area must be divided into two segments as shown in Fig. b. The momentof inertia of segment (2) about the y axis can be determined using
, while the moment of inertia of segment (1) about the
x axis can be determined by computing the moment of inertia of the elementparallel to the x axis shown in Fig. b. The area of this element is and its moment of inertia about the y axis is
Here, . Thus,
Performing the integration, the moment of inertia of segment (1) about the y axis is
Ans.
The moment of inertia of segment (2) about the y axis is
Thus,
Ans.Iy = (Iy)1 + (Iy)2 = 9 + 21 = 30 m4
(Iy)2 =1
12 (1)(33) + (1)(3)(2.52) = 21 m4
(Iy)1 = LdIy = L4 m
1 m¢ 64
3y3 -13≤dy = ¢ - 32
3y2 -13y≤ ` 4 m
1 m= 9 m4
dly = B13¢ 4y≤3
-13Rdy = ¢ 64
3y3 -13≤dy
x =4y
= a13x3 -
13≤dy
dIy = dIy¿ + dAx'2 =1
12(dy)(x - 1)3
+ (x - 1)dyB12
(x + 1)R2
dA = (x - 1) dy
(Ix)2 =1
12bh3 + A2¢h2 ≤
2
Iy = LAx2 dA = L
4 m
1 mx2 a 4xbdx = L
4 m
1 m4x dx = a2x2b ` 4 m
1 m= 30 m4
dA =4x
dx
y =4x
dA = y dx
4 m1 m
1 m
4 m
y
x
xy � 4
10–21.
Determine the moment of inertia of the shaded area aboutthe x axis.
SOLUTION
Differential Element: The area of the rectangular differential element in Fig. a is
. Here, and .
Thus,
Moment of Inertia: Applying Eq. 10–1, we have
Ans. =ah3
28
= B ah1>2
¢27y7>2≤ -
a
h¢y4
4≤ R `
0
h
= Lh
0¢ ah1>2
y5>2 -a
hy3≤
dy
Ix = LAy2dA = L
h
0y2¢ ah1>2
y1>2 -a
hy≤ dy
dA = ¢ ah1>2
y1>2 -a
h y≤ dy.
x1 =a
hyx2 =
a
h1>2 y1>2dA = (x2 - x1) dy
y
xa
h
y � x2ha2
y � xha
10–22.
Determine the moment of inertia of the shaded area aboutthe y axis.
SOLUTION
Differential Element: The area of the rectangular differential element in Fig. a is
Moment of Inertia: Applying Eq. 10–1, we have
Ans. =a3h
20
= Bha¢x4
4≤ -
h
a2 ¢x5
5≤ R `
0
a
= La
0¢hax3 -
h
a2x4≤
dx
Iy = LAx2dA = L
a
0x2¢hax -
h
a2x2≤
dx
dA = (y2 - y1) dx = ¢hax -
h
a2x2≤dx.
y
xa
h
y � x2ha2
y � xha
10–23.
Determine the moment of inertia of the shaded area aboutthe x axis.
SOLUTION
Differential Element: The area of the differential element shown shaded in Fig. a is.
Moment of Inertia: Applying Eq. 10–1, we have
However, . Thus,
Ans. =r0
4
8Bu -
12
sin 2uR `-a>2a>2
=r0
4
8(a - sin a)
Ix = La>2
-a>2 r0
4
8(1 - cos 2u)du
sin2 u =12
(1 - cos 2u)
= La>2
-a>2 r 4
0
4 sin2 udu
= La>2
-a>2¢ r4
4≤ `
0
r0 sin2 udu
= La>2
-a>2Lr0
0r3 sin2 udrdu
Ix = LAy2dA = L
a>2
-a>2Lr0
0r2 sin2 u(rdu)dr
dA = (rdu) dr
y
x
x2 � y2 � r02
r0
––2a
––2a
10–24.
Determine the moment of inertia of the shaded area aboutthe y axis.
SOLUTION
Differential Element: The area of the differential element shown shaded in Fig. a is.
Moment of Inertia: Applying Eq. 10–1, we have
However, . Thus,
Ans. =r0
4
8B1
2 sin 2u + uR `
-a>2a>2
=r0
4
8( sin a + a)
Iy = La>2
-a>2r0
4
8( cos 2u + 1)du
cos2 u =12
( cos 2u + 1)
= La>2
-a>2r0
4
4 cos2 udu
= La>2
-a>2¢ r4
4≤ `
0
r0 cos2 udu
= La>2
-a>2Lr0
0r3 cos2 udrdu
Iy = LAx2dA = L
a>2
-a>2Lr0
0r2 cos2 u(rdu)dr
dA = (rdu)dr
y
x
x2 � y2 � r02
r0
––2a
––2a
10–25. Determine the moment of inertia of thecomposite area about the axis.x
y
x
150 mm
300 mm
150 mm
100 mm
100 mm
75 mm
10–26. Determine the moment of inertia of thecomposite area about the axis.y
y
x
150 mm
300 mm
150 mm
100 mm
100 mm
75 mm
10–27.
SOLUTION
Ans.kx = A9
3 = 103.5 mm
= 9 4
Ix =1
12( 0)( 0)3 + 2 B 1
12( 0)(200)3 + ( 0)(200)(1 0)2R
Determine the radius of gyration for the column’s cross-sectional area.
kx
200 mm200 mm
200 mm
200 mm0 mm
x
x
48 8 48 8
0.7543(10 )
0.7543(10 ) mm
70.4(10 )
40 mm
40 mm
8
a a
a a
a––2
y � ( � x)
y
x
10–28.
Determine the moment of intertia of the beam’s cross-sectionalarea about the x axis.
SOLUTION
Ans. =112
a4
Ix = 2B 136¢22 a≤ ¢ a
22≤3
+12a22 ab ¢ a
22≤ ¢1
3 a
22≤2R
.
.
10–30.
Determine the distance to the centroid of the beam’scross-sectional area: then find the moment of inertia aboutthe axis.y¿
x
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
Segment
1 160(80) 80 1.024(106)
2 40(80) 20 64.0(103)
16.0(103) 1.088(106)
Thus,
Ans.
Moment of Inertia: The moment of inertia about the axis for each segment can bedetermined using the parallel–axis theorem .
Segment
1 80(160) 12.0 1.8432(106) 29.150(106)
2 80(40) 48.0 7.3728(106) 7.799(106)
Thus,
Ans.Iy¿ = © AIy¿ B i = 36.949 A106 B mm4 = 36.9 A106 B mm4
112 (80)(403)
112 (80)(1603)
(Iy¿)i (mm4)(Ad 2x)i (mm4)(Iy ¿)i (mm4)(dx)i (mm)Ai (mm2)
Iy¿ = Iy¿ + Ad2x
y¿
x =©xA©A
=1.088(106)
16.0(103)= 68.0 mm
©
xA (mm3)x (mm)A (mm2)
C
40 mm120 mm
y y'
x'
x
40 mm
40 mm
40 mm
40 mm
–x
10–31.
SOLUTION
Moment of Inertia: The moment inertia for the rectangle about its centroidal axis
can be determined using the formula, , given on the inside back cover of
the textbook.
Ans.Ix¿ =112
(160) A1603 B - 112
(120) A803 B = 49.5 A106 B mm4
Ix¿ =112bh3
Determine the moment of inertia for the beam’s cross-sectional area about the axis.x¿
C
40 mm120 mm
y y'
x'
x
40 mm
40 mm
40 mm
40 mm
–x
10–32.
SOLUTION
Ans.=r4
24 (6u - 3 sin 2u - 4 cos u sin3 u)
=14r4au -
12
sin 2ub -1
18r4 cos u sin3 u -
19r4 cos u sin3 u
- 2 c 136
(r cos u)(r sin u)3 +12
(r cos u)(r sin u)a13r sin ub2 d
Ix = c14r4au -
12
sin 2ub d
Determine the moment of inertia of the shaded area aboutthe x axis.
y
xθ
θ
r
10–33.
Determine the moment of inertia of the shaded area aboutthe y axis.
SOLUTION
Ans. =r4
4 u +
12
sin 2u - 2 sin u cos3 u
=14
r4au +12
sin 2ub - c 118r4 sin u cos3u + r3 sin u cos3 u d
- c 136
(2r sin u)(r cos u)3 +12
(2r sin u)(r cos u)a23r cos ub2 d
Iy = c14
r4au +12
sin 2ub d
y
xθ
θ
r
C
x
y
x¿
_y
x¿
250 mm
50 mm
150 mm150 mm
25 mm25 mm
10–34.
Determine the moment of inertia of the beam’s cross-sectional area about the y axis.
SOLUTION
Moment of Inertia: The dimensions and location of centroid of each segment areshown in Fig. a. Since the y axis passes through the centroid of both segments, themoment of inertia about y axis for each segment is simply
Ans. = 115.10(106) mm4 = 115(106) mm4
Iy = g (Iy)i =1
12(50)(3003) +
112
(250)(503)
(Iy)i = (Iy¿)i.
10–35.
Determine which locates the centroidal axis for thecross-sectional area of the T-beam, and then find themoment of inertia about the x¿ axis.
x¿y,
SOLUTION
Ans.
Ans.Ix¿ = 222(106) mm4
+ c 112
(300)(50)3 + 50(300)(275 - 206.818)2 d
Ix¿ = c 112
(50)(250)3 + 50(250)(206.818 - 125)2 dy = 207 mm
= 206.818 mm
y =©yA©A
=125(250)(50) + (275)(50)(300)
250(50) + 50(300)
C
x
y
x¿
y
x¿
250 mm
50 mm
150 mm150 mm
25 mm25 mm
10–36. Determine the moment of inertia of the beam’scross-sectional area about the axis.x
y
50 mm 50 mm
15 mm115 mm
115 mm
7.5 mmx
15 mm
10–37. Determine the moment of inertia of the beam’scross-sectional area about the axis.y
y
50 mm 50 mm
15 mm115 mm
115 mm
7.5 mmx
15 mm
10–38. Determine the moment of inertia of the beam’scross-sectional area about the axis.x
y
100 mm12 mm
125 mm
75 mm12 mm
75 mmx
12 mm
25 mm
125 mm
12 mm
10–39. Determine the moment of inertia of the beam’scross-sectional area about the axis.y
y
100 mm12 mm
125 mm
75 mm12 mm
75 mmx
12 mm
25 mm
125 mm
12 mm
10–40. Determine the moment of inertia of the cross-sectional area about the axis.x
100 mm10 mm
10 mm
180 mm x
y¿y
C
100 mm
10 mm
x
10–41. Locate the centroid of the beam’s cross-sectional area, and then determine the moment of inertia ofthe area about the centroidal axis.y¿
x
100 mm10 mm
10 mm
180 mm x
y¿y
C
100 mm
10 mm
x
10–42.
Compute the moments of inertia and for the beam’scross-sectional area about the x and y axes.
IyIx
SOLUTION
Ans.Ix = 154(106) mm4
+1
12(100)(30)3 + 100(30)(185)2
+1
12(30)(170)3 + 30(170)(85)2
Ix =1
12(170)(30)3 + 170(30)(15)2
170 mm30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x
x¿
y¿
C
y
x
10–43.
Determine the moment of inertia of the beam’s cross-sectional area about the y axis.
SOLUTION
Ans. Iy = 91.3(106) mm4
+1
12(30)(100)3 + 30(100)(50)2
+1
12(170)(30)3 + 30(170)(15)2
Iy =1
12(30)(170)3 + 30(170)(115)2
170 mm30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x
x¿
y¿
_x
C
_y
10–44.
SOLUTION
Ans.
Ans.Ix¿ = 67.6(106) mm4
+1
12(100)(30)3 + 100(30)(185 - 80.68)2
+ c 112
(30)(170)3 + 30(170)(85 - 80.68)2 d
Ix¿ = c 112
(170)(30)3 + 170(30)(80.68 - 15)2 d= 80.68 = 80.7 mm
y =170(30)(15) + 170(30)(85) + 100(30)(185)
170(30) + 170(30) + 100(30)
Determine the distance to the centroid C of the beam’scross-sectional area and then compute the moment ofinertia about the axis.x¿Ix¿
y
170 mm30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x
x¿
y¿
C
y
x
10–45.
Determine the distance to the centroid C of the beam’scross-sectional area and then compute the moment ofinertia about the axis.y¿Iy¿
x
SOLUTION
Ans.
Ans.Iy¿ = 41.2(106) mm4
+112
(30)(100)3 + 100(30)(50 - 61.59)2
+ c 112
(170)(30)3 + 30(170)(15 - 61.59)2 d
Iy¿ = c 112
(30)(170)3 + 170(30)(115 - 61.59)2 d= 61.59 = 61.6 mm
x =170(30)(115) + 170(30)(15) + 100(30)(50)
170(30) + 170(30) + 100(30)170 mm30 mm
30 mm
70 mm
140 mm
30 mm
30 mm
y
x
x¿
y¿
C
y
x
10–46.
Determine the distance to the centroid of the beam’scross-sectional area; then determine the moment of inertiaabout the axis.x¿
y
x
x¿C
y
50 mm 50 mm75 mm
25 mm
25 mm
75 mm
100 mm
_y
25 mm
25 mm
100 mm
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
Segment A (mm2)1 50(100) 75 375(103)2 325(25) 12.5 10l.5625(103)3 25(100) –50 –125(103)
15.625(103) 351.5625(103)
Thus,
Ans.
Moment of Inertia: The moment of inertia about the axis for each segment can bedetermined using the parallel-axis theorem .
Segment
1 50(100) 52.5 13.781(106) 17.948(106)
2 325(25) 10 0.8125(106) 1.236(106)
3 25(100) 72.5 13.141(106) 15.224(106)
Thus,
Ans.Ix¿ = ©(Ix¿)i = 34.41 A106 B mm4 = 34.4 A106 B mm4
112 (25) (1003)
112 (325) (253)
112 (50) (1003)
AIx¿ B i (mm4)AAd2y B i (mm4)AIx–B i (mm4)Ady B i (mm)Ai (mm2)
Ix¿ + Ix¿ + Ad2y
x¿
y =©yA©A
=351.5625(103)
15.625(103)= 22.5 mm
©
yA (mm3)y (mm)
10–47.
Determine the moment of inertia of the beam’s cross-sectional area about the y axis.
x
x¿C
y
50 mm 50 mm75 mm
25 mm
25 mm
75 mm
100 mm
_y
25 mm
25 mm
100 mm
SOLUTIONMoment of Inertia: The moment of inertia about the axis for each segment can bedetermined using the parallel-axis theorem .
Segment
1 2[100(25)] 100 50.0(106) 50.260(106)
2 25(325) 0 0 71.517(106)
3 100(25) 0 0 0.130(106)
Thus,
Ans.Iy¿ = ©(Iy¿)i = 121.91 A106 B mm4 = 122 A106 B mm4
112 (100) (253)
112 (25) (3253)
2 C 112 (100) (253) D
AIy–B i (mm4)AAd2x B i (mm4)AIy–B i (mm4)Adx B i (mm)Ai (mm2)
Iy¿ = Iy¿ + Ad2x
y¿
10–48. Determine the beam’s moment of inertia aboutthe centroidal axis.x
Ix y
x50 mm
50 mm
100 mm
15 mm15 mm
10 mm
100 mm
C
10–49.the centroidal axis.y
Iy y
x50 mm
50 mm
100 mm
15 mm15 mm
10 mm
100 mm
C
Determine the beam’s moment of inertia about
Ans.
10–50.
Locate the centroid of the cross section and determine themoment of inertia of the section about the axis.x¿
y
SOLUTIONCentroid: The area of each segment and its respective centroid are tabulated below.
2 m
0.5 m
4 m
2 m 2 m 2 m0.3 m
x'–y
Thus,
Ans.
Moment of Inertia: The moment of inertia about the axis for each segment can bedetermined using the parallel-axis theorem Ix¿ = Ix¿ + Ady2.
x¿
y =©yA©A
=46.04225.5
= = 1.81 m
Thus,
Ans.Ix¿ = © Ix¿ i = 42.33 m4
Segment A (m2) y (m) yA (m3)
1 3(4) 2.5 30
2 12 142142 1.833 14.667
3 11(0.5) 0.25 1.375
© 25.5 46.042
Segment Ai (m2) (dy)i (m) (Ady2)i (m4) (Ix¿)i (m4)
1 3(4) 0.6944 112 1321432 0.5787 21.787
2 12 142142 0.02778 1
36 1421432 0.6173 7.117
3 11(0.5) 112 111210.532 1.3309
110-3213.423
(Ix¿)i (m4)
1.806 m
1.556
2 m 3 m 2 m
4 m
0.5 m 0.25 m
11 m
1.833 m
2.5 m
2 m 3 m 2 m 1.806 m
4 m
0.5 m
1.556 m 11 m 0.02778 m
0.6944 m
10–51.
SOLUTION
Ans.+ 2B 112(75)(20)3 + (75)(20)a275
2+ 10b2R = 162 A106 B mm4
Ix ¿ =112(15)(275)3 + 4B1.32 A106 B + 1.36 A103 B a275
2- 28b2R
Determine the moment of inertia for the beam’s cross-sectional area with respect to the centroidal axis. Neglectthe size of all the rivet heads. R, for the calculation.Handbook values for the area, moment of inertia, andlocation of the centroid C of one of the angles are listed inthe figure.
x¿
75mm
275 mm
20mm
x¿
xa¿Aa � 1.36(103) mm2
(Ia)xa¿ � 1.32(106) mm4
R
C
15 mm
28 mm
10–52.
SOLUTION
Ans.Ix¿ =112bh3 =
112(b)(a sin u)3 =
112a3b sin3u
h = a sin u
Determine the moment of inertia for the parallelogramabout the axis, which passes through the centroid C ofthe area.
x¿y
bx
Ca
y¿
x¿
u
10–53.
Determine the moment of inertia for the parallelogramabout the axis, which passes through the centroid C ofthe area.
y¿
SOLUTION
Ans.=ab sin u
12(b2 + a2 cos2 u)
+1
12(a sin u)(b - a cosu)3
Iy¿ = 2B 136
(a sin u)(a cos u)3 +12
(a sin u)(a cos u)ab2+a
2cos u -
23a cos ub2R
x = a cos u +b - a cos u
2=
12
(a cosu + b)
y
bx
Ca
y¿
x¿
u
u
y
x
l
t
10–54.
Determine the product of inertia of the thin strip of areawith respect to the and axes. The strip is oriented at anangle from the axis. Assume that .
SOLUTION
Ans.=16
l3t sin 2u-
lxy = LA xydA = Ll
0(s cos u)(s sin u)tds = sin u cos utL
1
0s2ds
t V lxuyx
10–55.
Determine the product of inertia for the shaded area withrespect to the x and y axes.
SOLUTION
Ans.=316
b2 h2
=12
B a b2
h2>3b a 38by8>3Rh
0
= Lh
0
12a bh1>3b
2
y5>3 dy
Ixy = L d Ixy
d Ixy =x2y
2dy
dA = x dy
y' =
x
2x' =
y
y
x
h
b
y x3hb3
10–5 .6
.
10–57.
Determine the product of inertia for the shaded area withrespect to the x and y axes.
SOLUTION
Ans.=a2b2
4(n + 1)
= ¢ b2
2a2n ≤ ¢ 12n + 2
≤x2n+ 2 ` a0
=b2a2n+ 2
4(n + 1)a2n
Ixy = 0 +La
0(x)¢y
2≤(y dx) =
12L
a
0a b2
a2n bx2n+ 1 dx
dIxy = dIxy + x'y'dA
y
x
a
by � xn
anb
10–58.
Determine the product of inertia of the shaded area withrespect to the x and y axes, and then use the parallel-axistheorem to find the product of inertia of the area withrespect to the centroidal and axes.
SOLUTION
Differential Element: The area of the differential element parallel to the y axisshown shaded in Fig. a is . The coordinates of the centroid of
this element are Thus, the product of inertia of this
element with respect to the x and y axes is
Product of Inertia: Performing the integration, we have
Ans.
Using the information provided on the inside back cover of this book, the location of
the centroid of the parabolic area is at and
and its area is given by Thus,
Ans. Ix¿y¿ = 1.07 m4
10.67 = Ix¿y¿ + 5.333(2.4)(0.75)
Ixy = Ix¿y¿ + Adxdy
A =23
(4)(2) = 5.333 m2.
y =38
(2) = 0.75 mx = 4 -25
(4) = 2.4 m
Ixy = LdIxy = L4 m
0
12
x2dx = a16x3b 2
0
4 m
= 10.67 m4 = 10.7 m4
=12
x2dx
= 0 + Ax1>2 dx B(x)a12
x1>2b dIxy = dIx¿y¿ + dA~x~y
x'
= x and y'
=y
2=
12
x1>2dA = y dx = x1>2 dx
y¿x¿y2 � x
2 m
y y¿
x
4 m
Cx¿
10–59.
Determine the product of inertia for the shaded area withrespect to the x and y axes. Use Simpson’s rule to evaluatethe integral.
SOLUTION
Ans.= 0.511 m4
= 0.32 L1
0xe2x
2dx
= L1
0
12x(0.8 ex
2)2 dx
Ixy = L dIxy
dIxy =xy2
2dx
d A = ydx
y =y
2
x = xx
1 m
y
y 0.8ex2
10–60.
Determine the product of inertia for the area with respectto the x and y axes.
SOLUTION
Ans.= 0.333 m4
=12cx2
2-
16x3 d2
0
= L2
0
12
(x - 0.5x2) dx
Ixy = Ld Ixy
d Ixy =xy2
2dx
dA = y dx
y' =
y
2
x' = x
y
x
2 m
1 m
y2 1 0.5x
10–61.
Determine the product of inertia of the parallelogram withrespect to the x and y axes.
SOLUTIONProduct of Inertia of the Triangle: The product of inertia with respect to x and yaxes can be determined by integration. The area of the differential element parallel
to y rof diortnec eht fo The coordinates.])a(.giF[ si sixa
rof aitreni fo tcudorp eht nehT era tnemele siht
this element is
Performing the integration, we have
The product of inertia with respect to centroidal axes, and can be determinedby applying Eq. 10–8 [Fig. (b) or (c)].
Here, and Then,
Product of inertia of the parallelogram [Fig. (d)] with respect to centroidal and axes, is
The product of inertia of the parallelogram [Fig. (d)] about x and y axes is
Ans.=a2c sin2 u
124a cos u + 3c
=a3c sin2 u cos u
12+ 1a sin u21c2a c + a cos u
2b aa sin u
2b
Ixy = Ix¿y¿ + Adxdy
=a3c sin2 u cos u
12
Ix¿y¿ = 2Ba4 cos2 u sin2 u
72+
121a sin u21a cos u2a3c - a cos u
6b aa sin u
6b R
y¿x¿
Ix¿y¿ =a4 sin2 u cos2 u
72.h = a sin u.b = a cos u
Ix¿y¿ =b2h2
72
-b2h2
24= Ix¿y¿ +
12bh¢ -
b
3b ah
3b
Ixy = Ix¿y¿ + Adxdy
y¿,x¿
Ixy = L dIxy = 12 L
0
-b¢h2x +
h2
b2 x3 +
2h2
bx2≤dx =
b2h2
24
= 12¢h2x +
h2
b2 x3 +
2h2
bx2≤dx
= 0 + c ah +h
bxbdx d1x2B1
2ah +
h
bxb R
dIxy = dIx¿y¿ + dA x'y'
y' =
y
2=
12¢h +
h
bxb .x
' = x,
dA = ydx = ah +h
bxbdx
y
x
a
c
θ
-
.
10–63.
Determine the product of inertia for the beam’s cross-sectional area with respect to the u and axes.v
SOLUTIONMoments of inertia Ix and Iy
The section is symmetric about both x and y axes; therefore .
Ans.= 135(10)6 mm4
= a511.36 - 90.242
sin 40° + 0 cos 40°b 106
Iuv =Ix - Iy
2sin 2u + Ixy cos 2u
Ixy = 0
Iy = 2 c 112
(20)(300)3 d + 112
(360)(20)3 = 90.24(10)6 mm4
Ix =112
(300)(400)3 -112
(280)(360)3 = 511.36(10)6 mm4
150 mm
150 mm
yv
x
u
C
200 mm20 mm
20 mm
20
.
Determine the product of inertia for the beam's cross-sectional area with respect to the x and yaxes that have their origin located at the centroid C.
Given:
a �
b �
c �
Solution:
Ixy 2 b� a�a2
b2
���
��
� ca2
���
��
�� Ixy 1.10 (10 ) 4� �
10mm
50mm
50mm
mm6
10 65.
Ans.
–
10–66.
SOLUTIONProduct of Inertia: The area for each segment, its centroid and product of inertiawith respect to x and y axes are tabulated below.
Determine the product of inertia of the cross-sectional areawith respect to the x and y axes.
400 mm
100 mm
20 mm
20 mm
400 mm
x
y
100 mm
20 mm
C
Segment Ai (mm2) (dx)i (mm) (dy)i (mm) (Ixy)i (mm4)
1 100(20) 60 410 49.2110622 840(20) 0 0 0
3 100(20) -60 -410 49.211062Thus,
Ans.Ixy = ©1Ixy2i = 98.411062mm4
10–67. Determine the product of inertia of the beam’scross-sectional area with respect to the x and y axes.
x
y
300 mm
100 mm
10 mm
10 mm
10 mm
10–68.
Determine the distance to the centroid of the area andthen calculate the moments of inertia and for thechannel’s cross-sectional area. The u and axes have theirorigin at the centroid C. For the calculation, assume allcorners to be square.
vIvIu
y
150 mm150 mm
y
x
u
v
50 mm
10 mm
10 mm10 mm
y20C
SOLUTION
Ans.
Ans.
Ans.= 38.5(106) mm4
=0.9083(106) + 43.53(106)
2-
0.9083(106) - 43.53(106)
2cos 40°+0
Iv =Ix + Iy
2-Ix - Iy
2cos 2u + Ixy sin 2u
= 5.89(106) mm4
=0.9083(106) + 43.53(106)
2+
0.9083(106) - 43.53(106)
2cos 40° - 0
Iu =Ix + Iy
2+Ix - Iy
2cos 2u - Ixy sin 2u
Ixy = 0 (By symmetry)
= 43.53(106) mm4
Iy =112
(10)(300)3 + 2 c 112
(50)(10)3 + 50(10)(150 - 5)2 d= 0.9083(106) mm4
+ 2 c 112
(10)(50)3 + 10(50)(35 - 12.5)2 d
Ix = c 112
(300)(10)3 + 300(10)(12.5 - 5)2 d
y =300(10)(5) + 2[(50)(10)(35)]
300(10) + 2(50)(10)= 12.5 mm
10–69.
y
x
vu
20 mm
20 mm
20 mm
40 mm
200 mm
200 mm
45°
Determine the moments of inertia and of the shaded area.
IvIu
SOLUTIONMoment and Product of Inertia about x and y Axes: Since the shaded area issymmetrical about the x axis,
Moment of Inertia about the Inclined u and Axes: Applying Eq. 10–9 withwe have
Ans.
Ans.= 85.3 106 mm4
= a27.73 + 142.932
-27.73 - 142.93
2cos 90° - 01sin 90°2b11062
Iv =Ix + Iy
2-
Ix - Iy
2cos 2u + Ixy sin 2u
= 85.311062 mm4
= a27.73 + 142.932
+27.73 - 142.93
2cos 90° - 01sin 90°2b11062
Iu =Ix + Iy
2+
Ix - Iy
2cos 2u - Ixy sin 2u
u = 45°,v
= 142.9311062 mm4
Iy =1
121402120032 + 4012002112022 +
1121200214032
Ix =1
121200214032 +
1121402120032 = 27.7311062 mm4
Ixy = 0.
10–70.
Determine the moments of inertia and the product ofinertia of the beam’s cross sectional area with respect to theu and axes.
SOLUTION
Moments and product of Inertia with Respect to the x and y Axes: Theperpendicular distances measured from the centroid of the triangular segment tothe y axis are indicated in Fig. a.
Since the cross-sectional area is symmetrical about the y axis,
Moment and product of Inertia with Respect to the u and v Axes: Applying Eq. 10–9 with we have
Ans.
Ans.
Ans. = 178.62(106) mm4 = 179(106) mm4
= B ¢1012.5 - 6002
≤ sin 60° + 0 cos 60°R(106)
Iuv =Ix - Iy
2 sin 2u + Ixy cos 2u
= 703.125(106) mm4 = 703(106) mm4
= B1012.5 + 6002
- ¢1012.5 - 6002
≤ cos 60° + 0 sin 60R(106)
Iv =Ix + Iy
2-Ix - Iy
2 cos 2u + Ixy sin 2u
= 909.375(106) mm4 = 909(106) mm4
= B1012.5 + 6002
+ ¢1012.5 - 6002
≤ cos 60° - 0 sin 60°R(106)
Iu =Ix + Iy
2+Ix - Iy
2 cos 2u - Ixy sin 2u
u = 30°,
Ixy = 0.
Iy = 2B 136
(450)(2003) +12
(450)(200)(66.672)R = 600(106) mm4
Ix =1
36(400)(4503) = 1012.5(106) mm4
v
150 mm
200 mm
u
v
x
y
200 mm
C
300 mm
30�
10–71.
Solve Prob. 10–70 using Mohr’s circle. Hint: Once the circleis established, rotate counterclockwise from thereference , then find the coordinates of the points thatdefine the diameter of the circle.
SOLUTION
Moments and product of Inertia with Respect to the x and y Axes: Theperpendicular distances measured from the centroid of the triangular segment to they axis are indicated in Fig. a.
Since the cross-sectional area is symmetrical about the y axis,
Construction of Mohr’s Circle: The center of of the circle lies along the axis at adistance
The coordinates of the reference point are .The circle can beconstructed as shown in Fig. b.The radius of the circle is
Moment and Product of Inertia with Respect to the and v Axes: By referring tothe geometry of the circle, we obtain
Ans.
Ans.
Ans.Iuv = 206.25 sin 60° = 179(106) mm4
Iv = (806.25 - 206.25 cos 60°)(106) = 703(106) mm4
Iu = (806.25 + 206.25 cos 60°)(106) = 909(106) mm4
u
R = CA = (1012.5 - 806.25)(106) = 206.25(106) mm4
[1012.5, 0](106) mm4A
Iavg =Ix + Iy
2= a1012.5 + 600
2b(106)mm4 = 806.25(106) mm4
IC
Ixy = 0.
Iy = 2B 136
(450)(2003) +12
(450)(200)(66.672)R = 600(106) mm4
Ix =1
36(400)(4503) = 1012.5(106) mm4
OA2u = 60°
150 mm
200 mm
u
v
x
y
200 mm
C
300 mm
30�
10–72. Locate the centroid of the beam’s cross-sectionalarea and then determine the moments of inertia and theproduct of inertia of this area with respect to the and
axes.
y
x
u
800 mm
400 mm
50 mm
50 mm
450 mm50 mm
y
450 mm
C
60
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 = Σ
Σy A
AC =
1225(1000)(50) + 2[1000(400)(50)] + 600(12000)(100)
1000(50) + 2(400)(50) + 1200(100) = 825 mm 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,
Ix = 1
12(1000)(50 ) + 1000(50)(400)3 2⎡
⎣⎢
⎤
⎦⎥ + 2
1
12(50)(400 ) + 50(400)(175)3 2⎡
⎣⎢
⎤
⎦⎥ +
1
12(100)(1200 ) + 100(1200)(225)3 2⎡
⎣⎢
⎤
⎦⎥
= 302.44 (108) mm4
Iy = 1
12(50)(10003) + 2
1
12(400)(50 ) + 400(50)(75)3 2⎡
⎣⎢
⎤
⎦⎥ +
1
12(1200)(1003)
= 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 = I Ix y+
2 +
I Ix y–
2 cos 2� – Ixy sin 2�
= 302.44 + 45
2+
302.44 – 45
2cos 120° – 0 sinn 120°
⎡
⎣⎢
⎤
⎦⎥ (108)
= 109.36 (108) mm4 = 109 (108) mm4 Ans.
Iv = I Ix y+
2 –
I Ix y–
2 cos 2� + Ixy sin 2�
= 302.44 + 45
2–
302.44 – 45
2cos 120° + 0 sinn 120°
⎡
⎣⎢
⎤
⎦⎥ (108)
= 238.08 (108) mm4 = 238 (108) mm4 Ans.
50 mm
50 mm
75 mm 75 mm500 mm
500 mm
1000
mm
600
mm 12
25 m
m
400
mm
1200
mm
1000 mm
175 mm
225 mm
4000 mm
y = 825 mm
Iuv = I Ix y–
2 sin 2� + Ixy cos 2�
= 302.44 – 45
2sin 120° + 0 cos 120°
⎡
⎣⎢
⎤
⎦⎥ (108)
= 111.47 (108) mm4 = 111 (108) mm4 Ans.
175 mm
10–73. Solve Prob. 10–72 using Mohr’s circle.
500 mm500 mm
400
mm
1200 mm
100 mm
50 mm
600
mm
1000
mm
1225
mm
50 mm
400 mm
75 mm75 mm
225 mm
y = 825 mm
175 mm (108 mm4)
(108 mm4)
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 = ΣΣyA
A =
1225(1000)(50) + 2[1000(400)(50)] + 600(12000)(100)
1000(50) + 2(400)(50) + 1200(100) = 825 mm 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,
Ix = 1
12(1000)(50 ) + 1000(50)(400)3 2⎡
⎣⎢
⎤
⎦⎥ + 2
1
12(50)(400 ) + 50(400)(175)3 2⎡
⎣⎢
⎤
⎦⎥
+
1
12(100)(1200 ) + 100(1200)(225)3 2⎡
⎣⎢
⎤
⎦⎥
= 302.44 (108) mm4
Iy = 1
12(50)(10003) + 2
1
12(400)(50 ) + 400(50)(75)3 2⎡
⎣⎢
⎤
⎦⎥ +
1
12(1200)(1003)
= 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 = I Ix y+
2 =
302.44 + 45
2
⎛⎝⎜
⎞⎠⎟
(108) = 173.72 (108) mm4
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.
Iuv = (128.72 sin 60°) (108) = 111 (108) mm4 Ans.
10–74. Locate the centroid of the beam’s cross-sectionalarea and then determine the moments of inertia of this areaand the product of inertia with respect to the and axes.The axes have their origin at the centroid C.
vu
y
200 mm
25 mm
y
u
Cx
y
60�
75 mm75 mm
25 mm25 mm v
10–75. Solve Prob. 10–7 using Mohr’s circle.4
10–76. Locate the centroid of the beam’s cross-sectionalarea and then determine the moments of inertia and theproduct of inertia of this area with respect to the and
axes. The axes have their origin at the centroid C.vu
x y
x
u
x
200 mm
200 mm
175 mm
20 mm
20 mm
20 mm
C
60�
v
10–77. Solve Prob. 10–7 using Mohr’s circle.6
10–78.
Determine the principal moments of inertia for the angle’scross-sectional area with respect to a set of principal axesthat have their origin located at the centroid C. Use theequation developed in Section 10.7. For the calculation,assume all corners to be square.
SOLUTION
Ans.
Ans.Imin = 1.36(106) mm4
Imax = 4.92(106) mm4
= 3.142(106) ; 20 + {( - 1.778)(106)}2
Imax/min =Ix + Iy
2;CaIx - Iy
2b2
+ I2xy
= - 1.778(106) mm4
= -(32.22 - 10)(50-32.22)(100)(20) - (60-32.22)(32.22-10)(80)(20)
Ixy = ©xy A
= 3.142(106) mm4
+ c 112
(20)(80)3 + 80(20)(60 - 32.22)2 d
Iy = c 112
(100)(20)3 + 100(20)(32.22 - 10)2 d= 3.142(106) mm4
+ c 112
(80)(20)3 + 80(20)(32.22 - 10)2 d
Ix = c 112
(20)(100)3 + 100(20)(50 - 32.22)2 d100 mm
100 mm
20 mm
20 mm
y
x
32.22 mm
32.22 mmC
10–79.
SOLUTION
Center of circle:
Ans.
Ans.Imin = 3.142(106) - 1.778(106) = 1.36(106) mm4
Imax = 3.142(106) + 1.778(106) = 4.92(106) mm4
R = 2(3.142 - 3.142)2 + (-1.778)2(106) = 1.778(106) mm4
Ix + Iy
2= 3.142(106) mm4
–1.778(10 6) mm4Ixy =
3.142(106) mm4Iy =
3.142(106) mm4Ix =
Solve Prob. 10–78 using Mohr’s circle.
100 mm
100 mm
20 mm
20 mm
y
x
32.22 mm
32.22 mmC
Solve Prob. 10–78.
10–80. Determine the orientation of the principal axes,which have their origin at centroid C of the beam’s cross-sectional area. Also, find the principal moments of inertia.
y
Cx
100 mm
100 mm
20 mm
20 mm
20 mm
150 mm
150 mm
10–81. Solve Prob. 10–8 using Mohr’s circle.0
10–82. Locate the centroid of the beam’s cross-sectionalarea and then determine the moments of inertia of this areaand the product of inertia with respect to the and axes.The axes have their origin at the centroid C.
vu
y
200 mm
20 mm
y
u
Cx
y
60�
20 mm20 mm v
80 mm mm80
2[100(200)(20)] 10(20)(120) 79.23 mm2(200)(20) 20(120)
+= =+
Ans.
3 2 3 2
6 4
1 12 (20)(200 ) 20(200)(20.77) (120)(20 ) 120(20)(69.23)12 12
41.70(10 ) mm
xI = + + +
=
3 2 3
6 4
1 12 (200)(20 ) 200(20)(70) (20)(120 )12 12
42.35(10 ) mm
yI = + +
=
6
6 4
cos2 sin 22 2
41.70 42.35 41.70 42.35 cos( 120 ) 0sin( 120 ) (10 )2 2
42.2(10 ) mm
x y x yu xy
I I I II Iθ θ
+ −= + −
+ − = + − − −
= Ans.
6
6 4
cos2 sin 22 2
41.70 42.35 41.70 42.35 cos( 120 ) 0sin( 120 ) (10 )2 2
41.9(10 ) mm
x y x yxy
I I I II Iθ θ
+ −= + +
+ − = + − − −
= Ans.
v
6 4
sin 2 cos22
41.70 42.35 sin( 120 ) 0cos( 120 )2
0.28(10 ) mm
x yuv xy
I II Iθ θ
−= +
+= + − − −
= Ans.
20
20
20 10
60 60
70 mm 70 mm
20.77 mm
69.23 mm 79.23 mm
10–83. Solve Prob. 10–82 using Mohr’s circle.
2[100(200)(20)] 10(20)(120) 79.23 mm2(200)(20) 20(120)
+= =+
Ans.
1 13 2 3 22 (20)(200 ) 20(200)(20.77) (120)(20 ) 120(20)(69.23)12 12
6 441.70(10 ) mm
Ix = + + +
=
1 13 2 32 (200)(20 ) 200(20)(70) (20)(120 )12 12
6 442.25(10 ) mm
I y = + +
=
41.70 42.25 6 4 6 4(10 )mm 41.975(10 )mm2 2
I Ix yIavg+ + = = =
R = CA = (41.975 – 41.70)(106) = 0.275(106) mm4
6 6 4(41.975 0.275cos60 )(10 ) 41.86(10 )mm6 6 4(41.975 0.275cos60 )(10 ) 42.19(10 )mm
6 40.325sin 60 0.28(10 )mm
Iv
Iu
Iuv
= − =
= − =
= =
Ans.
Ans.
Ans.
20 20
10 20
60 60
70 mm 70 mm
20.77 mm
69.23 mm 79.23 mm
Iu
41.975
R = 0.275
Iv
42.25
41.70
10–84.
SOLUTION
Thus,
Ans.Iz = mR2
m = L2p
0r A R du = 2p r A R
Iz = L2p
0r A(R du)R2 = 2p r A R3
Determine the moment of inertia of the thin ring about thez axis. The ring has a mass m.
x
y
R
10–85.
Determine the moment of inertia of the semi-ellipsoid withrespect to the x axis and express the result in terms of themass m of the semiellipsoid. The material has a constantdensity r.
SOLUTION
Differential Disk Element: Here, .The mass of the differential disk element is
. The mass moment of inertia of this element is
.
Total Mass: Performing the integration, we have
Mass Moment of Inertia: Performing the integration, we have
The mass moment of inertia expressed in terms of the total mass is.
Ans.Ix =25a23rp ab2b b2 =
25mb2
=415
rp ab4
=rp b4
2a x5
5a4 -2x3
3a2 + xb ` a0
Ix = LdIx = La
0
rp b4
2ax4
a4 -2x2
a2 + 1b dx
=23rpab2
m = Lmdm = La
0rp b2 a1 -
x2
a2 bdx = rp b2 ax -x3
3a2 b `a
0
dIx =12dmy2 =
12crp b2 a1 -
x2
a2 b dx d cb2 a1 -x2
a2 b d =rp b4
2ax4
a4 -2x2
a2 + 1b dxdm = rdV = rp y2 dx = rp b2 a1 -
x2
a2 b dxy2 = b2 a1 -
x2
a2 b
y
x
b
a
� 1�b2y2
a2x2
.
10–87.
SOLUTION
Differential Disk Element: The mass of the differential disk element is
. The mass moment of inertia of this element
is .
Total Mass: Performing the integration, we have
Mass Moment of Inertia: Performing the integration, we have
The radius of gyration is
Ans.kx = AIxm=A
3.333(109)rp
1(106)rp= 57.7 mm
= 3.333(109)rp
=rp
2a2500x3
3b ` 200 mm
0
Ix = LdIx = L200 mm
0
rp
2(2500x2)dx
m = Lmdm = L200 mm
0rp(50x)dx = rp(25x2)|200 mm
0 = 1(106)rp
dIx =12dmy2 =
12
[rp(50x) dx](50x) =rp
2(2500x2)dx
dm = rdV = rpy2 dx = rp(50x) dx
Determine the radius of gyration of the paraboloid. Thedensity of the material is r = 5 Mg>m3.
kx y
x
100 mm
y2 50 x
200 mm
10–88.
Determine the moment of inertia of the ellipsoid with respectto the x axis and express the result in terms of the mass m ofthe ellipsoid.The material has a constant density r.
SOLUTION
Thus,
Ans.Ix =25mb2
Ix = La
-a
12rpb4 a1 -
x2
a2 b2
dx =815
prab4
m = LV rdV = La
-arp b2a1 -
x2
a2 b dx = 43prab2
dIx =y2dm
dm=Lpy2dx
2
y
x
a
b
1b2y2
a2x2
10–89.
Determine the moment of inertia of the homogenoustriangular prism with respect to the y axis. Express theresult in terms of the mass m of the prism. Hint: Forintegration, use thin plate elements parallel to the x-y planehaving a thickness of dz.
SOLUTION
Differential Thin Plate Element: Here, The mass of the
ssam ehT si tnemele etalp niht laitnereffid
moment of inertia of this element about y axis is
Total Mass: Performing the integration, we have
Mass Moment of Inertia: Performing the integration, we have
The mass moment of inertia expressed in terms of the total mass is
Ans.Iy =16
rabh
2a2 + h2 =
m
6a2 + h2
=rabh
121a2 + h22
=rab
3¢a2z +
a2
h2 z3 -
3a2
2hz2 -
a2
4h3 z4 + z3 -
3z4
4h≤ `
0
h
Iy = L dIy = Lh
0
rab
3¢a2 +
3a2
h2 z2 -
3a2
hz -
a2
h3 z3 + 3z2 -
3z3
hb dz
m = Lm dm = Lh
0raba1 -
z
hb dz = rab ¢z - z2
2h≤ `
0
h
=12rabh
=rab
3¢a2 +
3a2
h2 z2 -
3a2
hz -
a2
h3 z3 + 3z2 -
3z3
hb dz
= Ba2
3a1 -
z
hb2
+ z2R Braba1 -z
hb dzR
=13x2 dm + z2 dm
=1
12 dmx2 + dm ¢x2
4+ z2≤
dIy = dIG + dmr2
dm = rdV = rbxdz = raba1 -z
hb dz.
x = a a1 -z
hb . x
y
z
–ha (x – a)z =
h
ab
10–90.
Determine the mass moment of inertia of the solidformed by revolving the shaded area around the z axis.Thedensity of the materials is . Express the result in terms ofthe mass m of the solid.
SOLUTION
Differential Element: The mass of the disk element shown shaded in Fig. a
is . Here, . Thus,
. The mass moment of inertia of this element about the z axis is
.
Mass: The mass of the solid can be determined by integrating dm.Thus,
Mass Moment of Inertia: Integrating dIz, we obtain
From the result of the mass, we obtain .Thus, Iz can be written as
Ans.Iz =7
48(rpa2h)a2 =
748a8m
3ba2 =
718ma2
rpa2h =8m3
Iz = LdIz = Lh>2
0
rpa4
2a1 -
z
hb2
dz =rpa4
2c13a1 -
z
hb3
(-h) d 2h>2
0
=7rpa4h
48
m = Ldm = Lh>2
0rpa2a1 -
z
hbdz = rpa2az -
z2
2hb 2h>2
0
=3rpa2h
8
12arpr2dzbr2 =
rp
2r4dz =
rp
2aaA
1 -z
hb4
dz =rpa4
2a1 -
z
hb2
dzdIz =12dmr2 =
= rpa2a1 -z
hbdz
dm = rpaaA
1 -z
hb2
dzr = y = aA
1 -z
hdm = rdV = rpr2dz
r
Iz
x
z � (a2 � y2)h––a2
a
h––2
h––2
y
z
10–91.
Determine the moment of inertia of the sphere andexpress the result in terms of the total mass m of the sphere.The sphere has a constant density r.
Ix
SOLUTION
Thus,
Ans.Ix =25mr2
=43rpr3
m = Lr
-rrp(r2 - x2) dx
=8
15 prr5
Ix = Lr
-r
12rp(r2 - x2)2 dx
d Ix =12rp(r2 - x2)2 dx
dm = r dV = r(py2 dx) = rp(r2 - x2)dx
d Ix =y2 dm
2
x
y
r
x2 � y2 � r2
10–92. Determine the mass moment of inertia of the 2-kg bent rod about the z axis.
300 mm
300 mm
z
yx
10–93.
Determine the mass moment of inertia of the solidformed by revolving the shaded area around the axis. Thedensity of the material is . Express the result in terms ofthe mass of the solid.m
r
yIy
SOLUTIONDifferential Element: The mass of the disk element shown shaded in Fig. a is
. Here, . Thus, .
The mass moment of inertia of this element about the y axis is
.
Mass: The mass of the solid can be determined by integrating dm. Thus,
Mass Moment of Inertia: Integrating dIy,
From the result of the mass, we obtain . Thus, Iy can be written as
Ans.Iy =19a5m
2b =
518m
pr =5m2
=rp
512¢y9
9≤ `
2m
0
=pr
9
Iy = L dIy = L2 m
0
rp
512y8dy
m = Ldm = L2 m
0
rp
16y4dy =
rp
16 ¢y5
5≤ `
2m
0
=25rp
12
(rpr2dy)r2 =12rpr4dy =
12rpa1
4y2b4
dy =rp
512y8dy
dIy =12dmr2 =
dm = rpa14y2b2
dy =rp
16y4dyr = z =
14y2dm = rdV = rpr2dy
z y2
x
y
z
14
2 m
1 m
10–94.
SOLUTIONDifferential Element: The mass of the disk element shown shaded in Fig. a is
. Here, .Thus, .
The mass moment of inertia of this element about the y axis is
.
Mass: The mass of the solid can be determined by integrating dm. Thus,
The mass of the solid is . Thus,
Mass Moment of Inertia: Integrating ,
Substituting into Iy,
Ans.Iy =32p
7a375pb = 1.71(103)kg # m2
r =375p
kg>m3
Iy = LdIy = L4m
0
rp
512y6dy =
rp
512 ¢y7
7≤ ` 4m
0=
32p7r
dIy
1500 = 4pr r =375p
kg>m3
m = 1500 kg
m = Ldm = L4m
0
rp
16y3dy =
rp
16¢y4
4≤ ` 4 m
0
= 4pr
dIy =12dmr2 =
12Arpr2dy Br2 =
rp
2r4dy =
rp
2a1
4y3>2b4
dy =rp
512y6dy
dm = rpa14y3>2b2
dy =rp
16y3dyr = z =
14y3>2dm = rdV = rpr2dy
Determine the mass moment of inertia of the solidformed by revolving the shaded area around the axis. Thetotal mass of the solid is .1500 kg
yIy
y
x
z
4 m
2 mz2 y31––16
O
.
.
10–96.
SOLUTIONLocation of Centroid: This problem requires .
Ans.
Mass Moment of Inertia About an Axis Through Point O: The mass moment of inertiaof each rod segment and disk about an axis passing through the center of mass can be
determine using and . Applying Eq. 10–15, we have
Ans.= 53.2 kg # m2
+12
(6) A0.22 B + 6 A12 B
+112
[6.39(2)] A6.392 B + [6.39(2)] A0.52 B
=112
[1.3(2)] A1.32 B + [1.3(2)] A0.152 BIO = ©(IG)i + mid2
i
(IG)i =12mr2(IG)i =
112ml2
L = 6.39 m
0.5 =1.5(6) + 0.65[1.3(2)] + 0[L(2)]
6 + 1.3(2) + L(2)
x =©xm©m
x = 0.5 m
The pendulum consists of a disk having a mass of 6 kg andslender rods AB and DC which have a mass of .Determine the length L of DC so that the center of mass is at the bearing O . What is the moment of inertia of the assembly about an axis perpendicular to the page andpassing through point O?
2 kg >m
O
0.2 mL
A B
C
D0.8 m 0.5 m
10–97.
SOLUTION
Ans.
Ans.= 4.45 kg # m2
=112
(3)(2)2 + 3(1.781 - 1)2 +112
(5)(0.52 + 12) + 5(2.25 - 1.781)2
IG = ©IG¿ + md2
y =©ym©m
=1(3) + 2.25(5)
3 + 5= 1.781 m = 1.78 m
The pendulum consists of the 3-kg slender rod and the 5-kgthin plate. Determine the location of the center of mass Gof the pendulum; then find the mass moment of inertia ofthe pendulum about an axis perpendicular to the page andpassing through G.
y
G
2 m
1 m
0.5 m
y
O
10–98.
SOLUTIONLocation of Centroid:
Ans.
Mass Moment of Inertia About an Axis Through Point G: The mass moment ofinertia of a rectangular block and a semicylinder about an axis passing through thecenter of mass perpendicular to the page can be determined using
.ylevitcepser dna
Applying Eq. 10–16, we have
Ans.= 0.230 kg # m2
+ C0.319915210.222 + 510.0880822 D= c 1
12 13210.32 + 0.422 + 310.146822 d
IG = ©1Iz2Gi + midi2
1Iz2G =12mr2 - ma 4r
3pb2
= 0.3199mr21Iz2G =1
12 m1a2 + b22
y =©ym©m
=350132 + 115.12152
3 + 5= 203.20 mm = 203 mm
Determine the location of of the center of mass G of theassembly and then calculate the moment of inertia about anaxis perpendicular to the page and passing through G. Theblock has a mass of 3 kg and the mass of the semicylinder is 5 kg.
y 400 mm
200 mmy–
300 mm
G
•10–109. If the large ring, small ring and each of the spokesweigh 100 lb, 15 lb, and 20 lb, respectively, determine the massmoment of inertia of the wheel about an axis perpendicularto the page and passing through point A.
A
O
1 ft
4 ft
0.3 m
1.2 m
1.2 m0.3 m
1.2 m
0.75 m
10–99.spokes weigh 500 N, 75 N, and 100 N, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A.
Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of
(1.2 – 0.3) = 0.9 m and a center of mass located at a distance of 0.3 +0.9
2
⎛⎝⎜
⎞⎠⎟
m = 0.75 m from point O can be grouped
as segment (2).
Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O.
IO = 500
9.81
⎛⎝⎜
⎞⎠⎟
(1.22) + 81
12
100
9.81(0.9 ) +
100
9.81(0.752⎛
⎝⎜⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
22 )⎡
⎣⎢⎢
⎤
⎦⎥⎥
+ 75
9.81
⎛⎝⎜
⎞⎠⎟
(0.32)
= 125.5 kg · m2
The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found
using the parallel – axis theorem IA = IO + md2, where m = 500
9.81 + 8
100
9.81
⎛⎝⎜
⎞⎠⎟
+ 75
9.81 = 140.16 kg and d = 1.2 m. Thus,
IA = 125.5 + 140.16 (1.22) = 327.3 kg · m2 Ans.
If the large ring, small ring and each of the
10–100.
Determine the mass moment of inertia of the assemblyabout the z axis. The density of the material is 7.85 Mg m3.
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
Mass: The mass of each segment is calculated as
Mass Moment of Inertia: Since the z axis is parallel to the axis of the cone and thehemisphere and passes through their center of mass, the mass moment of inertia can be
computed from and .Thus,
Ans. = 29.4 kg # m2
=3
10(158.9625p)(0.32) +
25
(141.3p)(0.32) -3
10(5.8875p)(0.12)
Iz = ©(Iz)i
310
m3r2
3(Iz)1 =3
10 m1r
21 , (Iz)2 =
25
m2r2
2 ,
m3 = rV3 = ra13pr2hb = 7.85(103) c1
3p(0.12)(0.225) d = 5.8875p kg
m2 = rV2 = ra23pr3b = 7.85(103) c2
3p(0.33) d = 141.3p kg
m1 = rV1 = ra13pr2hb = 7.85(103) c1
3p(0.32)(0.675) d = 158.9625p kg
z
0.45 + z=
0.10.3
z = 0.225m
>z
yx
450 mm
300 mm
300 mm
100 mm
10–101.
Determine the moment of inertia of the frustum of thecone which has a conical depression. The material has adensity of 200 kg>m3.
Iz
SOLUTION
Ans.Iz = 1.53 kg # m2
-310
c13p(0.4)2 (0.6)(200) d(0.4)2
-310c13p(0.2)2(0.8)(200) d(0.2)2
Iz =310c13p (0.4)2 (1.6)(200) d(0.4)2
z
0.8 m0.6 m
0.2 m
0.4 m
10 102.
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.
Given:
Wp 60kg� a 1m�
Wr 20kg� b 1m�
c 3m�
d 2m�
Solution:
I0112
Wr� c d�( )2 Wrc d�
2c��
� ��
2��
112
Wp� a2 b2�� �� Wp c
b2
���
��
2���
k0I0
Wp Wr�� k0 3.15 m� Ans.
–
10 10
The slender rods have mass of 3 . Determine the moment of inertia for the assemblyabout an axis perpendicular to the page and passing through point A.
Given:
� 3kgm
�
a 1.5m�
b 2m�
Solution:
IA13� b� b2
�112
�� 2� a� 2 a�( )2� �� 2 a�( )� b2
��� IA 50.75 kg m2��
3.
Ans.
kg/m
–
a
10–104.
Determine the moment of inertia of the frustrum of thecone which has a conical depression. The material has adensity of 2000 kg>m3.
Iz
SOLUTION
Mass Moment of Inertia About z Axis: From similar triangles,
The mass moment of inertia of each cone about z axis can be
determined using
Ans.= 342 kg # m2
-3
10 cp
310.22210.62120002 d10.222
-3
10 cp
310.22210.3332120002 d10.222
Iz = ©1Iz2i = 310
cp310.82211.3332120002 d10.822
Iz =3
10mr2.
z = 0.333 m.
z
0.2=z + 1
0.8,
z
200 mm
800 mm
600 mm
400 mm
10–105.
SOLUTION
Mass Moment of Inertia About an Axis Through Point O: The mass for each wiretnemges hcae fo aitreni fo tnemom ssam ehT si tnemges
about an axis passing through the center of mass can be determined using
Applying Eq. 10–16, we have
Ans.
Location of Centroid:
Ans.
Mass Moment of Inertia About an Axis Through Point G: Using the resultand and applying Eq. 10–16, we have
Ans.IG = 0.150 10-3 kg # m2
0.450110-32 = IG + 310.03210.0577422IO = IG + md2
d = y = 0.05774 mIO = 0.450110-32 kg # m2
= 0.05774 m = 57.7 mm
y =©ym©m
=230.05 sin 60°10.0324 + 0.1 sin 60°10.032
310.032
= 0.450110-32 kg # m2
+1
12 10.03210.122 + 0.0310.1 sin 60°22
= 2 c 11210.03210.122 + 0.0310.0522 d
IO = ©1IG2i + mi di21IG2i = 1
12ml2.
mi = 0.310.12 = 0.03 kg.
Determine the moment of inertia of the wire triangle aboutan axis perpendicular to the page and passing throughpoint O.Also, locate the mass center G and determine themoment of inertia about an axis perpendicular to the pageand passing through point G.The wire has a mass of 0.3 kg/m.Neglect the size of the ring at O.
100 mm
y
O
BA
G60° 60°
–
10–106.
SOLUTIONComposite 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 asnegative parts.
Mass moment of Inertia: The mass of segments (1) and (2) areand . The perpendicular
distances measured from the centroid of each segment to the y axis are indicated inFig. a. The mass moment of inertia of each segment about the y axis can bedetermined using the parallel-axis theorem.
Ans.= 0.144 kg # m2
= 2 c 112
(1.6)(0.42) + 1.6(0.22) d - 2 c14
(0.1p)(0.12) + 0.1p(0.22) dIy = © AIy BG + md2
m2 = p(0.12)(10) = 0.1p kgm1 = 0.4(0.4)(10) = 1.6 kg
The thin plate has a mass per unit area of .Determine its mass moment of inertia about the y axis.
10 kg>m2
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
z
yx
100 mm
100 mm
10–107.
The thin plate has a mass per unit area of .Determine its mass moment of inertia about the z axis.
10 kg>m2
SOLUTIONComposite Parts: The thin plate can be subdivided into four segments as shown inFig. a. Since segments (3) and (4) are both holes, the y should be considered asnegative parts.
Mass moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) areand .The mass
moment of inertia of each segment about the z axis can be determined using theparallel-axis theorem.
m3 = m4 = p(0.12)(10) = 0.1p kgm1 = m2 = 0.4(0.4)(10) = 1.6 kg
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
200 mm
z
yx
100 mm
100 mm
Ans.= 0.113 kg # m2
=112
(1.6)(0.42) + c 112
(1.6)(0.42 + 0.42) + 1.6(0.22) d -14
(0.1p)(0.12) - c12
(0.1p)(0.12) + 0.1p(0.22) dIz = © AIz BG + md2
10–108. Determine the mass moment of inertia of theoverhung crank about the x axis. The material is steelhaving a density of .r = 7.85 Mg>m3
90 mm
50 mm
20 mm
20 mm
20 mm
x
x¿
50 mm30 mm
30 mm
30 mm
180 mm
.= 0.00325 kg · m2 = 3.25 g · m2
10–109.overhung crank about the axis. The material is steelhaving a density of .r = 7.85 Mg>m3
x¿
90 mm
50 mm
20 mm
20 mm
20 mm
x
x¿
50 mm30 mm
30 mm
30 mm
180 mm
Determine the mass moment of inertia of the
kg · m2
.
10–111.
SOLUTIONDifferential Element: Here, . The area of the differential element
parallel to the x axis is .
Moment of Inertia: Applying Eq. 10–1 and performing the integration, we have
Ans.
The moment of inertia about the axis can be determined using the parallel–axis
theorem. The area is
Ans.Ix¿ = 146 A106 B mm4
= 914.29 A106 B = Ix¿ + 53.33 A103 B A1202 BIx = Ix¿ + Ad2
y
A = LA dA = L200 mm
022200y
12dy = 53.33 A103 B mm2
x¿
= 914.29 A106 B mm4 = 914 A106 B mm4
= 22200a27y
72b 2 200 mm
0
Ix = LA y2 dA = L200 mm
0y2a22200y
12 dyb
dA = 2xdy = 22200y12 dy
x = 2200y12
Determine the area moment of inertia for the area aboutthe x axis. Then, using the parallel-axis theorem, find thearea moment of inertia about the axis that passesthrough the centroid C of the area. y = 120 mm.
x¿200 mm
200 mm
y
x
x¿–y
C1
200y � x2
10–112.
SOLUTIONDifferential Element: Here, . The area of the differential element parallel tothe x axis is .The coordinates of the centroid for this element are
, . Then the product of inertia for this element is
Product of Inertia: Performing the integration,we have
Ans.Ixy = LdIxy = L1 m
0
12y33 dy =
316y35 2 1 m
0= 0.1875m4
=12y33 dy
= 0 + Ay13 dy B a12y13b(y)
dIxy = dIx¿y¿ + dAx y
y = yx =x
2=12y13
dA = xdy = y13 dyx = y
13
Determine the product of inertia of the shaded area with respect to the x and y axes.
y x 3
y
1 m
1 m
x
10–113.
Determine the area moment of inertia for the triangulararea about (a) the x axis, and (b) the centroidal axis.x¿
SOLUTION
(a)
Ans.
(b)
Ans.Ix¿ =bh3
36
bh3
12= Ix¿ +
12bhah
3b2
Ix = Ix¿ + A d2
=bh3
12
= Lh
0y2 c b
h(h - y) ddy
Ix = L y2 dA
dA = s dy = c bh(h - y) d dy
s =b
h(h - y)
s
h - y=
b
h
y
x
x¿h
b
Ch–3
10–114.
Determine the mass moment of inertia for the body andexpress the result in terms of the total mass m of the body.The density is constant.
Ix
SOLUTION
Ans.Ix =9370mb2
m = Lm dm = rp La
0ab2
a2x2 +
2b2
ax + b2b dx =
73rpab2
=3110
rpab4
Ix = L dIx = 12 rp L
a
0ab4
a4 x4 +
4b4
a3x3 +
6b4
a2 x2 +
4b4
ax + b4b dx
dIx =12rp ab4
a4 x4 +
4b4
a3x3 +
6b4
a2 x2 +
4b4
ax + b4b dx
dIx = 12 dmy
2 = 12 rpy
4 dx
dm = rdV = rpy2 dx = rp ab2
a2 x2 +
2b2
ax + b2b dx
y
x
2b
a
z
b
y ab
x b
10 115.
Determine the moment of inertia for the shaded area about the x axis.
Given:
a 4m�
b 2m�
Solution
Solution
Ixa�
a
xb cos
� x�2 a�
��
��
���
��
3
3
������
d� Ix 9.05 m4� Ans.
–
10 116.
Determine the moment of inertia for the shaded area about the y axis.
Given :
a 4m�
b 2m�
Solution :
Iya�
a
xx2 b� cos� x�2 a�
��
��
�����
d� Iy 30.9 m4� Ans.
–
10–117.
Determine the area moments of inertia and and theproduct of inertia for the semicircular area.Iuv
IvIu
SOLUTION
Ans.
Ans.
Ans.Iuv = 0
= 0 + 0
Iuv =Ix - Iy
2sin2u + Ixy cos 2u
Iv = 5.09(106) mm4
=5 089 380.1 + 5 089 380.1
2- 0 + 0
Iv =Ix + Iy
2-
Ix - Iy
2cos2u + Ixy sin 2u
Iu = 5.09(106) mm4
=5 089 380.1 + 5 089 380.1
2+ 0 - 0
Iu =Ix + Iy
2+
Ix - Iy
2cos2u - Ixy sin2u
Ixy = 0 (Due to symmetry)
Ix = Iy =18p(60)4 = 5 089 380.1 mm4
x
y
v
u
30�
30�60 mm
10–118.
SOLUTIONMoment of Inertia:The moment of inertia about the x axis for the composite beam’scross section can be determined using the parallel-axis theorem .
Ans.= 0.0954d4
Iy = c 112
(d) Ad3 B + 0 d + 4B 136
(0.2887d)ad2b3 +
12
(0.2887d)ad2b ad
6b2R
Ix = © AIx + Ad2y B i
Cx
y
d2
d2
d2
d2 60
60
Determine the area moment of inertia of the beam’s cross-sectional area about the x axis which passes through the centroid C.
10–119.
SOLUTIONMoment of Inertia: The moment of inertia about y axis for the composite beam’scross section can be determined using the parallel-axis theorem .
Ans.= 0.187d4
Iy = c 112
(d) Ad3 B + 0 d + 2 c 136
(d)(0.2887d)3 +12
(d)(0.2887d)(0.5962d)2 dIy = © AIy + Ad2
x B i
Cx
y
d2
d2
d2
d2 60
60
Determine the area moment of inertia of the beam’s cross-sectional area about the y axis which passes through the centroid C.