Vector Mechanics for Engineers: Statics, 11th Edition
Vector Mechanics for Engineers: Statics, 11th Edition
11th Edition
ISBN: 9780077687304
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 9.2, Problem 9.36P
To determine

Find the moment of inertia of the shaded area with respect to x axis (Ix) and y axis (Iy).

Expert Solution & Answer
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Answer to Problem 9.36P

The moment of inertia of the shaded area with respect to x axis (Ix) and y axis (Iy) are 69.9a4_ and 72a4_ respectively.

Explanation of Solution

Show the centroidal location of the given section as Figure 1.

Vector Mechanics for Engineers: Statics, 11th Edition, Chapter 9.2, Problem 9.36P

Consider x axis.

Consider the section 1.

Calculate the moment of inertia of the section 1 (I¯x1)1 using the formula:

(I¯x1)1=4a×(4a)312=256a412

The expression for the area of the section 1 (A) as follows;

A=4a×4a=16a2

The expression for the centroid of the section 1 (d) as follows;

d=4a2=2a

Calculate the moment of inertia of the section 1 (Ix)1 using the formula:

(Ix)1=(I¯x1)1+Ad2

Substitute 256a412 for (I¯x1)1 and 16a2 for A and 2a for d.

(Ix)1=256a412+16a2(2a)2=256a412+64a4=a4(25612+64)=256a43

The expression for the moment of inertia of the section AA (IAA)2 and section BB (IBB)3 as follows;

The section AA and BB compressed of semi-circle.

(IAA)2=(IBB)3=π8a4

The expression for the area of the semi-circle (A) as follows;

A=π2a2

The expression for the centroid of the semi-circle (d) as follows;

d=4a3π

Calculate the moment of inertia of the section 2 (I¯x2)2 and section 3 (I¯x3)3 using the formula:

Since both sections (2) and (3) are semi-circle which has same moment of inertia.

(IAA)2=(I¯x2)2+Ad2(I¯x3)3=(I¯x2)2=(IAA)2Ad2

Substitute π8a4 for (IAA)2, π2a2 for A and 4a3π for d.

(I¯x3)3=(I¯x2)2=π8a4π2a2(4a3π)2=π8a4πa22(16a29π2)=π8a48a49π=a4(π889π)

Consider the section 2.

The expression for the centroid of the section 2 (d) from neutral axis as follows;

d=52a+4a3π

Calculate the moment of inertia of the section 2 (Ix)2 using the formula:

(Ix)2=(I¯x2)2+Ad2

Substitute a4(π889π) for (I¯x2)2, π2a2 for A and (52a+4a3π) for d.

(Ix)2=a4(π889π)+[π2a2(52a+4a3π)2]=a4(π889π)+π2a2((52a)2+(4a3π)2+2(5a2)4a3π)=a4(103+13π4)

Consider the section 3.

The expression for the centroid of the section 3 (d) from neutral axis as follows;

d=32a4a3π

Calculate the moment of inertia of the section 3 (Ix)3 using the formula:

(Ix)3=(I¯x3)3+Ad2

Substitute a4(π889π) for (I¯x3)3, π2a2 for A and (32a4a3π) for d.

(Ix)3=a4(π889π)+[π2a2(32a4a3π)2]=a4(π889π)+π2a2((32a)2+(4a3π)22(32a)4a3π)=a4(2+5π4)

Calculate the total moment of inertia of the entire section (Ix) about x axis using the relation:

Ix=(Ix)1(Ix)2(Ix)3

Substitute 2563a4 for (Ix)1, a4(103+13π4) for (Ix)2 and a4(2+5π4) for (Ix)3.

Ix=2563a4a4(103+13π4)a4(2+5π4)=2563a410a4313a4π4+2a45a4π4=168a49πa42=69.9a4

Consider y axis.

Consider the section 1.

Calculate the moment of inertia of the section 1 (I¯y1)1 about y axis using the formula:

(I¯y1)1=4a×(4a)312=256a412

The expression for the area of the section 1 (A) as follows;

A=4a×4a=16a2

The expression for the centroid of the section 1 (d) as follows;

d=4a2=2a

Calculate the moment of inertia of the section 1 (Iy)1 using the formula:

(Iy)1=(I¯y1)1+Ad2

Substitute 256a412 for (I¯y1)1 and 16a2 for A and 2a for d.

(Iy)1=256a412+16a2(2a)2=256a412+64a4=a4(25612+64)=256a43

Consider the section 2.

Calculate the moment of inertia of the section 2 (I¯y2)2 about y axis using the relation:

(I¯y2)2=(IAA)2

Substitute π8a4 for (IAA)2.

(I¯y2)2=π8a4

The expression for the area of the semi-circle (A) as follows;

A=π2a2

The expression for the centroid of the semi-circle (d) as follows;

d=4a2=2a

Calculate the moment of inertia of the section 1 (Iy)1 using the formula:

(Iy)2=(I¯y2)2+Ad2

Substitute π8a4 for (I¯y2)2 and π2a2 for A and 2a for d.

(Iy)2=π8a4+π2a2(2a)2=π8a4+2πa4=17πa48

Since the section 2 and section 3 are same. Hence, (Iy)3=(Iy)2=17πa48.

Calculate the total moment of inertia of the entire section (Iy) about y axis using the relation:

Iy=(Iy)1(Iy)2(Iy)3

Substitute 256a43 for (Iy)1 and 17πa48 for (Iy)2 and 17πa48 for (Iy)3.

Iy=256a4317πa4817πa48=256a432(17πa48)=a4(256317π4)=72a4

Therefore, the moment of inertia of the shaded area with respect to x axis (Ix) and y axis (Iy) are 69.9a4_ and 72a4_ respectively.

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Chapter 9 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

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