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 3.10, Problem 138P

(a)

To determine

Find the maximum shearing stress (τmax) along the line a-a.

(a)

Expert Solution
Check Mark

Answer to Problem 138P

The maximum shearing stress (τmax) along the line a-a is 4.55ksi_.

Explanation of Solution

Given information:

The length of the steel member (L) is 8ft.

The provided section of the member is W8×31.

The torque in the member (T) is 5kipin..

The modulus rigidity of the steel (G) is 11.2×106psi.

Assume the angle of twist in flange and web is equal.

Calculation:

Consider flange:

Refer Appendix C, “Properties of Rolled-Steel shapes”.

The width of the flange (a) is 8in.

The thickness of the flange (b) is 0.435in.

Calculate the ratio of width to thickness of the steel ab.

Ratio=ab

Substitute 8in. for a and 0.435in. for b.

ab=80.435=18.391

Hence, the ratio of ab is greater than 5. Therefore, use the below formula for calculating the coefficient of rectangular bar.

Calculate the ratio of thickness to width of the steel ba.

Ratio=ba

Substitute 0.435in. for b and 8in. for a.

ba=0.4358=0.0544

Calculate the coefficient for rectangular bar (c1) and (c2) using the formula:

c1=c2=13(10.630ba)

Substitute 0.0544 for ba.

c1=c2=13(10.630×0.0544)c1=c2=0.32191

Calculate the angle of twist in flange (ϕf) using the formula:

ϕf=TfLc2ab3G

Here, Tf is the torque in flange, L is the length of the member and G is the modulus of rigidity.

Substitute 0.32191 for c2, 8in. for a, 0.435in. for b, 8ft for L and 11.2×106psi for G.

ϕf=Tf(8×12)0.32191×(8ft)×(0.435ft)3×11.2×106=4.044×105Tf

Consider web:

Refer Appendix C, “Properties of Rolled-Steel shapes”.

The thickness of the web (b) is 0.285in..

The depth of the member (D) is 8in.

Calculate the width of the web (a) using the formula:

a=D2bf

Here, bf is the thickness of the flange.

Substitute 8in. for D and 0.435in. for bf.

a=82(0.435)=7.13in.

Calculate the ratio of width to thickness of the steel ab.

Ratio=ab

Substitute 7.13in. for a and 0.285in. for b.

ab=7.130.285=25.0175

Hence, the ratio of ab is greater than 5. Therefore, use the below formula for calculating the coefficient of rectangular bar.

Calculate the ratio of thickness to width of the steel ba.

Ratio=ba

Substitute 0.285in. for b and 7.13in. for a.

ba=0.2857.13=0.039972

Calculate the coefficient for rectangular bar (c1) and (c2) using the formula:

c1=c2=13(10.630ba)

Substitute 0.039972 for ba.

c1=c2=13(10.630×0.039972)c1=c2=0.32494

Calculate the angle of twist in web (ϕw) using the formula:

ϕw=TwLc2ab3G

Substitute 0.32494 for c2, 7.13in. for a, 0.285in. for b, 8ft for L and 11.2×106psi for G.

ϕw=Tw(8×12)0.32494×(7.13)×(0.285)3×11.2×106=1.5982×104Tw

Since the angle of twist in flange and web is equal, therefore,

ϕf=ϕw

Substitute 4.044×105Tf for ϕf and 1.5982×104Tw for ϕw.

4.044×105Tf=1.5982×104TwTf=1.5982×104Tw4.044×105

Tf=3.952Tw (1)

By taking the sum of torque exerted on two flanges and web in the member is equal to the total torque T applied to member. Therefore,

2Tf+Tw=T

Substitute 3.952Tw for Tf and 5kipin. for T.

2(3.952Tw)+Tw=5×1037.904Tw+Tw=5,0008.904Tw=5,000

Tw=5,0008.904Tw=561.5lbin.

Substitute 561.5lbin. for Tw in Equation (1).

Tf=3.952(561.5)=2,219.05lbin.

Calculate the maximum shearing stress (τmax) along line a-a using the formula:

τmax=Tfc1ab2

Substitute 2,219.05lbin. for Tf, 0.32191 for c1, 8in. for a, and 0.435in. for b.

τmax=2,219.050.32191×8×0.4352=4,553psi=4.55ksi

Therefore, maximum shearing stress (τmax) along the line a-a is 4.55ksi_.

(b)

To determine

Find the maximum shearing stress (τmax) along the line b-b.

(b)

Expert Solution
Check Mark

Answer to Problem 138P

The maximum shearing stress (τmax) along the line b-b is 2.98ksi_.

Explanation of Solution

Given information:

The length of the steel member (L) is 8ft.

The provided section of the member is W8×31.

The torque in the member (T) is 5kipin..

The modulus rigidity of the steel (G) is 11.2×106psi.

Assume the angle of twist in flange and web is equal.

Calculation:

Calculate the torque in the web (Tw) using the formula:

τmax=Twc1ab2

Substitute 561.5lbin. for Tw, 0.32494 for c1, 7.13in. for a, and 0.285in. for b.

τmax=561.50.32494×7.13×0.2852=2983.8psi=2.98ksi

The maximum shearing stress (τmax) along the line b-b is 2.98ksi_.

(c)

To determine

Find the angle (ϕ) of twist.

(c)

Expert Solution
Check Mark

Answer to Problem 138P

The angle (ϕ) of twist is 5.14°_.

Explanation of Solution

Given information:

The length of the steel member (L) is 8ft.

The provided section of the member is W8×31.

The torque in the member (T) is 5kipin..

The modulus rigidity of the steel (G) is 11.2×106psi.

Assume the angle of twist in flange and web is equal.

Calculation:

From the above calculation of angle of twist, take the critical angle to compute the angle of twist.

Calculate the angle of twist (ϕf) using the relation:

Consider the torque equation,

2Tf+Tw=5×103

Substitute ϕc2ab3GL for Tf and ϕc2ab3GL for Tw.

2(ϕc2ab3GL)+(ϕc2ab3GL)=5×103ϕGL[2(c2ab3)f+(c2ab3)w]=5×103ϕGL[2Kf+Kw]=5,000

ϕ=5,000LG(2Kf+Kw) (2)

Assume Kf=(c2ab3)f and Kw=(c2ab3)w.

Calculate the value of Kf.

Kf=(c2ab3)f

Substitute 0.32191 for c2, 8in. for a, and 0.435in. for b.

Kf=0.32191×8×0.4353=0.21198in.4

Calculate the value of Kw.

Kw=(c2ab3)w

Substitute 0.32494 for c1, 7.13in. for a, and 0.285in. for b.

Kw=0.32494×7.13×0.2853=0.053632in.4

Find the angle of twist:

Substitute 11.2×106psi for G, 8ft for L, 0.21198in.4 for Kf and 0.053632in.4 for Kw.

ϕ=5000(8×12)in.11.2×106[2(0.21198)+0.053632]=0.0897rad×180π=5.14°

Therefore, the angle of twist of the section is 5.14°_.

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Chapter 3 Solutions

Mechanics of Materials, 7th Edition

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