PEARSON ETEXT ENGINEERING MECH & STATS
PEARSON ETEXT ENGINEERING MECH & STATS
15th Edition
ISBN: 9780137514724
Author: HIBBELER
Publisher: PEARSON
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Chapter 10, Problem 1FP

Determine the moment of inertia of the shaded area about the x axis.

Chapter 10, Problem 1FP, Determine the moment of inertia of the shaded area about the x axis.

Expert Solution & Answer
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To determine

The moment of inertia for the shaded area about the x axis.

Answer to Problem 1FP

The moment of inertia for the shaded area about the x axis is 0.1112m4_ .

Explanation of Solution

Given:

The height of the shaded area is 1m .

The width of the shaded area is 1m .

Show the area of the differential element parallel to the y axis as in Figure 1.

PEARSON ETEXT ENGINEERING MECH & STATS, Chapter 10, Problem 1FP

From Figure 1,

Compute the area of the differential element parallel to the y axis as shown with the shaded area from Figure 1.

dA=ydx

Here, the area of the differential element is dA and the centroid of the differential element along the y axis is y .

Express the moment of inertia of the differential element parallel to the x axis.

dIx=dI¯x+dAy˜2 (I)

Here, the first integral of the moment of inertia of the area about the centroidal axis is I¯x and the centroid of the differential element along the y axis is y˜ .

Substitute 112(dx)y3 for dI¯x , ydx for dA , and y/2 for y˜ in Equation (I).

dIx=112(dx)y3+ydx(y/2)2=[y312+yy24](dx)=[y312+y34](dx)

=4y312dx=y33dx (II)

Substitute x2 for y3 in Equation (II).

dIx=x23dx

Conclusion:

Express the moment of inertia for the shaded area about the x axis.

Ix=01mdIx (III)

Substitute x23dx for dIx in Equation (III).

Ix=01mx23dx=1301mx2dx=13[x33]01

=19(1)319(0)3=19m4=0.1112m4

Hence, the moment of inertia for the shaded area about the x axis is 0.1112m4_ .

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