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

10–23.

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

10–46.

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¿

10–69.

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

10–94.

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

10–100.

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

10–106.

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