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

= 2a 83b 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

10–19.

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 =

L

4 m

1 m

64

3x3 dx = ¢ -

323x2 ≤ `

1 m

4 m

= 10 m4

dIx =

13¢ 4

x≤3

dx =

643x3dxy =

4x

= 112

dxy3+ ydx¢y

2≤2

=

13

y3dx

dIx = dIx ¿+ dAy

'2dA = y dx

Ix = (Ix)1 + (Ix)2 = 9 + 1 = 10 m4

(Ix)2 =

112

(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¢ 4

y- 1≤dy =

L

4 m

1 ma4y - y2bdy

dA = ¢ 4y

- 1≤dyx =

4y

dA = (x - 1)dy

(Ix)2 =

112

bh3+ A2¢h

2≤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 =

112

(1)(33) + (1)(3)(2.52) = 21 m4

(Iy)1 =

LdIy =

L

4 m

1 m¢ 64

3y3 -

13≤dy = ¢ -

323y2 -

13

y≤ ` 4 m

1 m= 9 m4

dly = B13¢ 4

y≤3

-

13Rdy = ¢ 64

3y3 -

13≤dy

x =

4y

= a13

x3-

13≤dy

dIy = dIy¿

+ dAx'2

=

112

(dy)(x - 1)3+ (x - 1)dyB1

2(x + 1)R2

dA = (x - 1) dy

(Ix)2 =

112

bh3+ A2¢h

2≤2

Iy =

LAx2 dA =

L

4 m

1 mx2 a 4

xbdx =

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

10–46.

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¿

10–63.

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 -1

12(280)(360)3 = 511.36(10)6 mm4

150 mm

150 mm

yv

x

u

C

200 mm20 mm

20 mm

20

*10–64.

SOLUTIONMoment and Product of Inertia about x and y Axes: Since the shaded area issymmetrical about the x axis, Ixy = 0.

Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10-9 with, we have

Ans.

Ans.= 25.75 in4

=10.75 + 30.75

2-

10.75 - 30.752

cos 60° + 0(sin 60°)

Iv =Ix + Iy

2-

Ix - Iy

2cos 2u + Ixy sin 2u

= 15.75 in4

=10.75 + 30.75

2+

10.75 - 30.752

cos 60° - 0(sin 60°)

Iu =Ix + Iy

2+

Ix - Iy

2cos 2u - Ixy sin 2u

u = 30°

Iy =1

12(1)(43)+1(4)(2.5)2 +

112

(5)(13) = 30.75 in4

Ix =112

(1)(53) +112

(4)(13) = 10.75 in4

Determine the moments of inertia for the shaded area withrespect to the u and axes.v

y

x

vu

0.5 in.

0.5 in.0.5 in.

1 in. 4 in.

5 in.30

10–74.

SOLUTION

Ans.

Ans.Imin = 5.33 in4

Imax = 64.1 in4

=55.55 + 13.89

2;Ca55.55 - 13.89

2b2

+ (-20.73)2

Imax>min =Ix + Iy

2;CaIx - Iy

2b2

+ I2xy

= -20.73 in4

= -2[(1.813 + 0.1875)(3.813)(3.625)(0.375)] + 0

Ixy = ©xy A

= 13.89 in4

+112

(8)a38b3

Iy = 2 c 112a3

8b a4 -

38b3

+38a4 -

38b e a 4 - 3

8

2b +

316f2 d

= 55.55 in4

Ix = 2 c 112

(4)a38b3

+ 4a38b a4 -

316b2 d +

112a 3

8b a8 -

68b3

Determine the principal moments of inertia for the beam’scross-sectional area about the principal axes that have theirorigin located at the centroid C. Use the equationsdeveloped in Section 10.7. For the calculation, assume allcorners to be square.

x

4 in.

4 in.

4 in.

4 in.

y

C

in.38

in.38

10–75.

Solve Prob. 10–74 using Mohr’s circle.

SOLUTION

Center of circle:

Ans.

Ans.Imin = 34.72 - 29.39 = 5.33 in4

Imax = 34.72 + 29.39 = 64.1 in4

R = 2(55.55 - 34.72)2 + (-20.73)2 = 29.39 in4

Ix + Iy

2= 34.72 in4

= -20.73 in4

= -2[(1.813 + 0.1875)(3.813)(3.625)(0.375)] + 0

Ixy = ©xy A

+112

(8)a38b3

= 13.89 in4

Iy = 2 c 112a3

8b a4 -

38b3

+38a4 -

38b e a4 - 3

8

2b +

316f2 d

Ix = 2 c 112

(4)a38b3

+ 4a38b a4 -

316b2 d +

112a3

8b a8 -

68b3

= 55.55 in4

x

4 in.

4 in.

4 in.

4 in.

y

C

in.38

in.38

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 c 14

(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) - c 12

(0.1p)(0.12) + 0.1p(0.22) dIz = © AIz BG + md2