VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)
12th Edition
ISBN: 9781260916942
Author: BEER
Publisher: MCG
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Chapter 9.1, Problem 9.18P
To determine

Find the moment of inertia and radius of gyration of the shaded area with respect to y axis by using direct integration.

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Answer to Problem 9.18P

The moment of inertia of the shaded area with respect to y axis by using direct integration is 18πa3b_.

The radius of gyration of the shaded area with respect to y axis by using direct integration is a2_.

Explanation of Solution

Given information:

The curve equation is x2a2+y2b2=1.

Calculation:

Sketch the vertical strip shaded portion as shown in Figure 1.

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA), Chapter 9.1, Problem 9.18P

Write the curve Equation as follows:

x2a2+y2b2=1 (1)

Modify Equation (1).

x2a2+y2b2=1y=b1(x2a2)

Determine the area of the strip element dA as shown in below:

dA=ydx

Determine the moment of inertia (dIx) with respect to x axis of a rectangular strip:

dIy=x2dA (2)

Substitute ydx for dA in Equation (2).

dIy=x2(ydx)Iy=aax2(ydx)=aax2(b1(x2a2))dx=baax2(1(x2a2))dx (3)

Consider x=asinθ

Differentiate both sides of the Equation.

dx=acosθdθ

Find the moment of inertia of the shaded area with respect to y axis by using direct integration.

Substitute asinθ for x and acosθdθ for dx and apply the limits in Equation (3).

Iy=baaa2sin2θ1(a2sin2θa2)acosθdθ=bπ2π2a2sin2θ1sin2θacosθdθ=a3bπ2π2sin2θcos2θdθ=a3bπ2π214sin22θdθ

=14a3bπ2π212(1cos4θ)dθ=18a3b[(θ14sin4θ)]π2π2=18a3b[(π214sin4(π2))(π214sin4(π2))]=18a3b(π2+π2)

=18a3b(2π2)Iy=18πa3b

Thus, the moment of inertia of the shaded area with respect to y axis by using direct integration is 18πa3b_.

Find the area of shaded portion (A) as shown below:

A=dA

Substitute 2xdy for dA in Equation (4).

A=2xdy=2a0b1(y2b2)dy

Substitute bsinθ for y and bcosθdθ for dy and apply the limit.

A=2xdy=2a0b1(b2sin2θb2)bcosθdθ=2ab0π2cos2θdθ=2ab0π212(1+cos2θ)dθ

A=ab[θ+12sin2θ]0π2=ab[π2+12sin2(π2)]=12πab

Find the radius of gyration of the shaded area with respect to y axis by using direct integration.

ky2=IyA (4)

Here, A is are of shaded portion and Iy is moment of inertia about y axis.

Substitute 12πab for A and 18πa3b for Iy in Equation (4).

ky2=18πa3b12πab=14a2ky=12a

Thus, the radius of gyration of the shaded area with respect to y axis by using direct integration is a2_.

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Chapter 9 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

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