Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 4.10, Problem 187P
To determine

The expression of R for a circular cross section.

Expert Solution & Answer
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Answer to Problem 187P

The expression of R for a circular cross section is 12(r¯+r¯2c2)_.

Explanation of Solution

Consider w be the width as a function of β, w=2csinβ

Sketch the cross section for the polar coordinate as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 4.10, Problem 187P

Refer to Figure 1.

The distance r=r¯ccosβ

Differentiate both sides with respect to β as shown below.

drdβ=0c(sinβ)dr=csinβdβ

Calculate the area of the strip (dA) as shown below.

dA=wdr

Substitute 2csinβ for w and csinβdβ for dr.

dA=2csinβ×csinβdβdAr=2c2sin2βdβr

Substitute r¯ccosβ for r.

dAr=2c2sin2βdβr¯ccosβ

Integrate both sides of the Equation.

dAr=0π2c2sin2βdβr¯ccosβ=20πc2(1cos2β)dβr¯ccosβ=20π(c2c2cos2β+r¯2r¯2)dβr¯ccosβ=20π(r¯2c2cos2β(r¯2c2))dβr¯ccosβ

=20π(r¯2c2cos2β)dβr¯ccosβ20π(r¯2c2)dβr¯ccosβ=20π(r¯ccosβ)(r¯+ccosβ)dβr¯ccosβ2(r¯2c2)0πdβr¯ccosβ=20π(r¯+ccosβ)dβ2(r¯2c2)0πdβr¯ccosβ=2(r¯β+csinβ)0π2(r¯2c2)(2r¯2c2tan1r¯2c2tan12βr¯+c)0π

={2(r¯π+csinπ0)4r¯2c2{(tan1r¯2c2tanπ2r¯+c)(tan1r¯2c2tan0r¯+c)}}={2(r¯π)4r¯2c2(π20)}=2πr¯2πr¯2c2

Calculate the area of the circle (A) as shown below.

A=πc2

Calculate the expression of R as shown below.

R=AdAr

Substitute πc2 for A and 2πr¯2πr¯2c2 for dAr.

R=πc22πr¯2πr¯2c2=12π×πc2r¯r¯2c2=12×c2r¯r¯2c2×r¯+r¯2c2r¯+r¯2c2=12×c2(r¯+r¯2c2)(r¯2(r¯2c2)2)

=12×c2(r¯+r¯2c2)(r¯2r¯2+c2)=12×c2(r¯+r¯2c2)c2=12(r¯+r¯2c2)

Therefore, the expression of R for a circular cross section is 12(r¯+r¯2c2)_.

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Chapter 4 Solutions

Mechanics of Materials, 7th Edition

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