10–3. determine the moment of inertia of the area...

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.


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.


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.


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.


SOLUTION


. Here, and .

Thus,


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.


SOLUTION



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.


SOLUTION

Differential Element: The area of the differential element shown shaded in Fig. a is.


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.


SOLUTION

Differential Element: The area of the differential element shown shaded in Fig. a is.


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


y

xθ

θ

r

10–33.


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.


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


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.


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


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


=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


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

10–78.

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

10–82. 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 .



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



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.

10–3. determine the moment of inertia of the area...

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