EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 3.10, Problem 137P

A 4-m-long steel member has a W310 × 60 cross section. Knowing that G = 77.2 GPa and that the allowable shearing stress is 40 MPa, determine (a) the largest torque T that can be applied, (b) the corresponding angle of twist. Refer to Appendix C for the dimensions of the cross section and neglect the effect of stress concentrations. (Hint: consider the web and flanges separately and obtain a relation between the torques exerted on the web and a flange, respectively, by expressing that the resulting angles of twist are equal.)

Chapter 3.10, Problem 137P, A 4-m-long steel member has a W310  60 cross section. Knowing that G = 77.2 GPa and that the

Fig. P3.137

(a)

Expert Solution
Check Mark
To determine

Find the largest torque (T) that can be applied.

Answer to Problem 137P

The largest torque (T) is 1761.53Nm_.

Explanation of Solution

Given information:

The length of the steel member (L) is 4m.

The provided section of the member is W310×60.

The allowable shearing stress (τall) is 40MPa.

The modulus rigidity of the steel (G) is 77.2GPa.

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 203mm or 0.203m.

The thickness of the flange (b) is 13.1mm or 0.0131m.

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

Ratio=ab

Substitute 203mm for a and 13.1mm for b.

ab=20313.1=15.50

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 13.1mm for b and 203mm for a.

ba=13.1203=0.0645

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

c1=c2=13(10.630ba)

Substitute 0.0645 for ba.

c1=c2=13(10.630×0.0645)c1=c2=0.31979

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 0.31979 for c2, 0.203m for a, 0.0131m for b, 4m for L and 77.2GPa for G.

ϕf=Tf(4)0.31979×(0.203)×(0.0131)3×77.2×109=3.5503×104Tf

Consider web:

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

The thickness of the web (b) is 7.49mm or 0.00749m.

The depth of the member (D) is 302mm or 0.302m.

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

a=D2bf

Here, bf is the thickness of the flange.

Substitute 302mm for D and 13.1mm for bf.

a=3022(13.1)=275.8mm=275.8mm1000=0.2758m

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

Ratio=ab

Substitute 275.8mm for a and 7.49mm for b.

ab=275.87.49=36.822

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 7.49mm for b and 275.8mm for a.

ba=7.49275.8=0.02716

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

c1=c2=13(10.630ba)

Substitute 0.02716 for ba.

c1=c2=13(10.630×0.02716)c1=c2=0.32763

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

ϕw=TwLc2ab3G

Substitute 0.32763 for c2, 0.2758m for a, 0.00749m for b, 4m for L, and 77.2GPa for G.

ϕw=Tw(4)0.32763×(0.2758)×(0.00749)3×77.2×109=1.36464×103Tw

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

ϕf=ϕw

Substitute 3.5503×104Tf for ϕf and 1.36464×103Tw for ϕw.

3.5503×104Tf=1.36464×103TwTf=1.36464×103Tw3.5503×104

Tf=3.8437Tw (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 3.8437Tw for Tf.

2(3.8437Tw)+Tw=T7.6874Tw+Tw=T8.6874Tw=TTw=T8.6874

Tw=0.11511T (2)

Substitute 0.11511T for Tw in Equation (1).

Tf=3.8437(0.11511T)

Tf=0.44244T (3)

Calculate the torque in the flange (Tf) using the formula:

τmax=Tfc1ab2

Substitute 40MPa for τmax, 0.31979 for c1, 0.203m for a, and 0.0131m for b.

40×106=Tf0.31979×0.203×0.01312Tf=(40×106)×0.32763×0.203×0.01312Tf=445.62Nm

Substitute 445.62Nm for Tf in the equation (3).

445.62=0.44244TT=445.620.44244T=1,007.2Nm

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

τmax=Twc1ab2

Substitute 40MPa for τmax, 0.32763 for c1, 0.2758m for a, and 0.00749m for b.

40×106=Tw0.32763×0.2758×0.007492Tw=(40×106)×0.32763×0.2758×0.007492Tw=202.77Nm

Substitute 202.77Nm for Tw in the equation (2).

202.77=0.11511TT=202.770.11511T=1761.53Nm

Hence, take the lesser value from the torque produced in flange and web respectively.

Therefore, the largest torque (T) is 1761.53Nm_.

(b)

Expert Solution
Check Mark
To determine

Find the angle (ϕ) of twist.

Answer to Problem 137P

The angle (ϕ) of twist is 9.06°_.

Explanation of Solution

Given information:

The length of the steel member (L) is 4m.

The provided section of the member is W310×60.

The allowable shearing stress (τall) is 40MPa.

The modulus rigidity of the steel (G) is 77.2GPa.

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:

ϕf=3.5503×104Tf

Substitute 445.62Nm for Tf.

ϕf=3.5503×104(445.62)=0.15821radians=0.15821×180π=9.06°

Therefore, the angle of twist of the section is 9.06°_.

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Chapter 3 Solutions

EBK MECHANICS OF MATERIALS

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