Mechanical Engineering Chapter 10 Homework The area of the differential element

1030

10–21.

Determine the moment of inertia for the shaded area about

the x axis.

SOLUTION

Differential Element. Here

x2=y

and x1=

1

2

y2. The area of the differential element

y

x

2 m

y2  2x

y  x

Ans:

10–22.

Determine the moment of inertia for the shaded area about

the y axis.

SOLUTION

Differential Element. Here, y

2x

1

2 and

y1=x

. The area of the differential

Moment of Inertia. Perform the integration,

Iy=

x2dA =

x2

a2

2x1

2–x

b

dx

y

x

2 m

y2  2x

y  x

Ans:

10–23.

Determine the moment of inertia for the shaded area about

the x axis.

SOLUTION

Differential Element. Here x2=

a

b

1

2

y

1

2 and x1=

a

b

2y2. Thus, the area of the

differential element parallel to the x axis shown shaded in Fig. a is

dA =(x2–x1) dy

=

aa

b

1

2

y1

2–

a

b2y2

b

dy.

b

x

y

a

y2  —x

b2

a

y — x2

b

a2

1033

*10–24.

Determine the moment of inertia for the shaded area about

the y axis.

SOLUTION

Differential Element. Here, y2=

b

a

1

2

x1

2 and y1=

b

a

2

x2. Thus, the area of the

differential element parallel to the y axis shown shaded in Fig. a is

dA =(y2–y1)dx

=

a

b

a

1

2

x1

2–b

a2x2

b

dx

Moment of Inertia. Perform the integration,

b

x

y

a

y2  —x

b2

a

y — x2

b

a2

10–25.

Determine the moment of inertia of the composite area

about the xaxis.

SOLUTION

Composite Parts: The composite area can be subdivided into three segments as

shown in Fig. a.The perpendicular distance measured from the centroid of each

segment to the xaxis is also indicated.

y

x

6 in.

3 in.

10–26.

Determine

the moment of inertia of the composite area

about the

yaxis.

SOLUTION

Composite

Parts: The composite area can be subdivided into three segments as

shown

in Fig. a.The perpendicular distance measured from the centroid of each

segment to the

xaxis is also indicated.

y

x

6 in.

3 in.

1036

10–27.

The polar moment of inertia for the area is

JC

= 642 (106) mm4, about the z

′

axis passing through the

centroid C. The moment of inertia about the y

′

axis is

264 (106) mm4, and the moment of inertia about the x axis is

938 (106) mm4. Determine the area A.

SOLUTION

Applying the parallel-axis theorem with

d

y

=200 mm

and I

x=

938

(

10

6

)

mm

4

,

I

x=

I

x′

+Ad2

y

200 mm

Cx

¿

x

¿

1037

*10–28.

50 mm

x

–

y

50 mm

350 mm

250 mm

Determine the location of the centroid of the channel’s

cross-sectional area and then calculate the moment of

inertia of the area about this axis.

y

SOLUTION

Centroid: The area of each segment and its respective centroid are tabulated below.

Segment A(mm2)y

‘(mm) y

‘A(mm3)

1 100(250) 125 3.12511062

Ans:

1038

10–29.

Determine

y

, which locates the centroidal axis

x′

for the

cross-sectional area of the T-beam, and then find the

moments of inertia

Ix′

and

I

y

′

.

SOLUTION

Centroid. Referring to Fig. a, the areas of the segments and their respective centroids

are tabulated below.

Segment

A(mm2)

y

∼(mm)

y

∼A(mm3)

1 150(20) 10 30

(103)

Moment of Inertia. The moment of inertia about the

x′

axis for each segment can

be determined using the parallel axis theorem, Ix′=Ix′+Ad

2

y. Referring to Fig. b,

Segment

Ai(mm2)

(d

y

)

i

(mm)

(Ix′)i

(mm4)

(Ad

y

2)

i

(mm4)

(Ix′)i

(mm4)

Thus

75 mm

x¿

y¿

C

75 mm

150 mm

20 mm

y

10–30.

Determine the moment of inertia for the beam’s cross-

sectional area about the x axis.

SOLUTION

Segment A

i

(in

2

)

(d

y

)

i

(in.)

(

I

x′)i

(in4)

(Ady

2

)i

(in

4

) (I

x

)

i

(in

4

)

1 1(8) 4

1

12

(1)(83)128 170.67

1

12

8 in.

y

x

10 in.

3 in.

1 in.

SOLUTION

Moment of Inertia. The moments of inertia about the y axis for each segment can

be determined using the parallel axis theorem,

I

x=

I

x′

+Ad2

y

.

Referring to Fig. a,

Segment A

i

(in

2

)

(dx)i

(in.)

(Iy‘)i

(in

4

)(Adx

2

)

i

(in

4

) (Iy)i

(in

4

)

1 8(1) 9.5

1

12

(8)(1

3)722 722.67

1

12

1

12

10–31.

Determine the moment of inertia for the beam’s cross-

sectional area about the y axis.

8 in.

y

x

10 in.

3 in.

1 in.

SOLUTION

Moment of Inertia. The moment of inertia about the x axis for each segment can

be determined using the parallel axis theorem,

Ix=Ix′+Ad2y.

Referring to Fig. a

Segment A

i

(mm

2

)

(d

y

)

i

(mm)

(I

x′

)

i

(mm

4

) (Ady)

2

i

(mm

4

) (I

x

)

i

(mm

4

)

1 200(300) 150

(200)(3003)1.35(10

9

) 1.80(10

9

)

*10–32.

Determine the moment of inertia Ix of the shaded area

about the x axis.

Ox

150 mm

100 mm 100 mm

75 mm

150 mm

y

Ans:

1042

SOLUTION

Moment of Inertia. The moment of inertia about the y axis for each segment can be

determined using the parallel-axis theorem,

I

y=

I

y′

+Ad2

x

.

Referring to Fig. a

Segment A

i

(mm

2

)

(dx)i

(mm)

Iy

′

(mm

4

) (Ad

2

x

)

i

(mm

4

) (Iy)i

(mm

4

)

10–33.

Determine the moment of inertia Ix of the shaded area

about the x axis.

Ox

150 mm

100 mm 100 mm

75 mm

150 mm

y

10–34.

C

y

x¿

250 mm

50 mm

150 mm

Determine

the moment of inertia of the beam’s cross-

sectional area about the

yaxis.

SOLUTION

M

oment of Inertia: The dimensions and location of centroid of each segment are

shown

in Fig. a.Since the yaxis passes through the centroid of both segments, the

moment of inertia about

yaxis for each segment is simply

(Iy)i=(Iy¿)i.

10–35.

Determine which locates the centroidal axis for the

cross–sectional area of the T-beam, and then find the

moment of inertia about the x¿ axis.

x

¿

y

,

SOLUTION

y=

©yA

250(50) +50(300)

C

y

x¿

250 mm

50 mm

150 mm

SOLUTION

Moment of Inertia. Since the x axis passes through the centroids of the two segments,

Fig. a,

*10–36.

Determine the moment of inertia about the x axis.

150 mm

y

x

C

200 mm

20 mm

10–37.

Determine the moment of inertia about the y axis.

SOLUTION

Moment of Inertia. Since the y axis passes through the centroid of the two segments,

Fig. a,

150 mm

y

x

C

200 mm

20 mm

SOLUTION

Moment of Inertia. The moment of inertia about the x axis for each segment can

be determined using the parallel-axis theorem,

I

x=

I

x′

+Ad

y

2.

Referring to Fig. a,

Segment A

i

(in

2

)

(d

y

)

i

(in.)

(

I

x′)i

(in4)

(Ady

2

)i

(in

4

) (I

x

)

i

(in

4

)

1 6(6) 3

1

12

(6)(63)324 432.0

10–38.

Determine the moment of inertia of the shaded area about

the x axis.

x

6 in.

3 in.

6 in.

y

6 in.

SOLUTION

Moment of Inertia. The moment of inertia about the y axis for each segment can

be determined using the parallel-axis theorem,

I

y=

I

y′

+Ad

x

2.

Referring to Fig. a,

Segment A

i

(in

2

)

(dx)i

(in.)

(Iy′)i

(in

4

)(Ad

x

2

)

i

(in

4

) (Iy)i

(in

4

)

1 6(6) 3

(6)(63)324 432.0

Thus,

Iy=Σ(Iy)i=2376 in

4

Ans.

10–39.

Determine the moment of inertia of the shaded area about

the y axis.

x

6 in.

3 in.

6 in.

y

6 in.

1049

SOLUTION

Centroid. Referring to Fig. a, the areas of the segments and their respective centroids

are tabulated below.

Segment

A(in2)

y

∼(in.)

y

∼A(in3)

Σ

14.0 19.00

Segment A

i(in2)

(d

y

)

i

(in.)

(

I

x′)i

(in4)

(

Ady

2)

i

(in4)

(

I

x′)i

(in4)

*10–40.

Determine the distance

y

to the centroid of the beam’s

cross-sectional area; then find the moment of inertia about

the centroidal

x′

axis.

y

x

3 in. 1 in.

1 in.

4 in.

1 in.

y¿

x ¿

C

y

3 in.

Ans: