Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 5, Problem 5.6.5P

A cantilever beanie B is loaded by a uniform load q and a concentrated load P, as shown in the figure.

  1. Select the most economical steel C shape from Table F-3(a) in Appendix F; use q = 20 lb/ft and P = 300 lb (assume allowable normal stress is cra= IS ksi).

  • Select the most economical steel S shape from Table F-2(a) in Appendix F; use q = 45 lb/ft and P = 2000 lb (assume allowable normal stress is tra= 20 ksi),
  • Select the most economical steel W shape from Table F-1(a) in Appendix F; use q = 45 lb/ft and P = 2000 lb (assume allowable normal stress is (T = 20 ksi). However, assume that the design requires that the W shape must be used in weak axis bending, i.e., it must bend about the 2-2 (or v) axis of the cross section.
  • Note: For parts (a), (b), and (c), revise your initial beam selection as needed to include the distributed weight of the beam in addition to uniform load q.

      Chapter 5, Problem 5.6.5P, A cantilever beanie B is loaded by a uniform load q and a concentrated load P, as shown in the

    (a)

    Expert Solution
    Check Mark
    To determine

    The most economical steel Cshape.

    Answer to Problem 5.6.5P

    The most economical steel Cshape is C15×33.9.

    Explanation of Solution

    Given information:

    The uniform distributed load is 20lb/ft, the load is 300lb, maximum normal stress is 18ksi.

    The following figure shows the free body diagram:

    Mechanics of Materials (MindTap Course List), Chapter 5, Problem 5.6.5P

    Figure-(1)

    Write the expression for the maximum moment of beam.

    Mmax=qL22+P(6)…… (I)

    Here, the load is P, the uniform load on beam is q,and the length of the beam is L.

    Write the expression for the section modulus.

    S=Mmaxσallow…… (II)

    Here, the maximum stress is σallow, and the maximum moment is Mmax,and the section modulus is S.

    Write the expression for the maximum stress.

    σmax=MmaxS……(III)

    Here, the maximum stress is σmax.

    Calculation:

    Substitute 20lb/ftfor q, and 300lbfor Pand 10ftfor Lin Equation (I).

      Mmax=20lb/ft(102ft2)2+(300lb)(6ft)=(2800lbft)(12in1ft)=33600lbin.

    Substitute 33600lbin.for Mmax, and 18ksifor σallowin Equation (II).

    S=33600lbin18ksi=(33600lbin)(18ksi)(1000psi1ksi)=(33600lbin)(18psi)(1lb/in21psi)=1.87in3

    Refer to the table (a)of “Appendix F” to get the value of weight intensity w=30lbft, and the section modulus S=2.05in3.

    Substitute 20lb/ftfor q, and 300lbfor Pand 10ftfor L, 30lbftfor win Equation (I).

    Mmax=((20+30)lbft)×102ft22+(300lb×6ft)=((20+30)lbft)×102ft22+(1800lbft)=2150lbft

    Substitute 2150lbftfor Mmax, 2.05in.3for Sin Equation (III).

    σmax=2150lbft2.05in.3=(2150lbft)(12in1ft)2.05in.3(1psi1lb/in2)=(25797psi)(103ksi1psi)=25.797ksi

    Here, the maximum stress is greater than the calculated stress, so we neglect this section.

    Refer to the table (a)of “Appendix F” to get the value of weight intensity w=33.9lbft, and the section modulus S=3.09in3.

    Substitute 20lb/ftfor q, and 300lbfor Pand 10ftfor L, 33.9lbftfor win Equation (I).

    Mmax=((20+33.9)lbft)×102ft22+(300lb×6ft)=((20+33.9)lbft)×102ft22+(1800lbft)=3595lbft

    Substitute 3595lbftfor Mmax, 3.09in.3for Sin Equation (III).

    σmax=3595lbft3.09in.3=(3595lbft)(12in1ft)3.09in.3(1psi1lb/in2)=(13961.16psi)(103ksi1psi)=13.961ksi

    Hence, from the table “Appendix F” we will use the value C15×33.9.

    Conclusion:

    The most economical steel Cshape is C15×33.9.

    (b)

    Expert Solution
    Check Mark
    To determine

    The most economical steel Sshape.

    Answer to Problem 5.6.5P

    The most economical steel Sshape is S8×18.4.

    Explanation of Solution

    Given Information:

    The uniform distributed load is 45lb/ft, the load is 2000lb, maximum normal stress is 20ksi.

    Write the expression for the maximum moment of beam.

    Mmax=qL22+P(AC)…… (IV)

    Write the expression for the section modulus.

    S=Mmaxσallow…… (V)

    Write the expression for the maximum stress.

    σmax=Mmax20lb/ft

    S…… (VI)

    Calculation:

    Substitute for q, and 300lbfor Pand 10ftfor Lin Equation (IV).

    Mmax=45lb/ft(102ft2)2+(2000lb)(6ft)=(14250lbft)(12in1ft)=171000lbin.

    Substitute 171000lbin.for Mmax, and 20ksifor σallowin Equation (V).

    S=171000lbin.(20ksi)(103psi1ksi)=171000lbin.(20000psi)(1lb/in21psi)=8.55in.3

    Refer to the table (a)of “Appendix F” to get the value of weight intensity w=17.2lbft, and the section modulus S=8.74in3.

    Substitute 45lb/ftfor q, and 2000lbfor Pand 10ftfor L, 17.2lbftfor win Equation (I).

    Mmax=((45+17.2)lbft)×102ft22+(2000lb×6ft)==((45+17.2)lbft)×102ft22+(12000lbft)=18220lbft

    Substitute 18220lbftfor Mmax, 8.74in.3for Sin Equation (VI).

    σmax=(18220lbft)(12in1ft)8.74in.3=(25016.01lb/in2)(1psi1lb/in2)=(25016.01psi)(103ksi1psi)=25.016ksi

    Here, the maximum stress is greater than the calculated stress, so we neglect this section.

    Refer to the table (a)of “Appendix F” to get the value of weight intensity w=18.4lbft, and the section modulus S=14.4in3.

    Substitute 45lb/ftfor q, and 2000lbfor Pand 10ftfor L, 18.4lbftfor win Equation (I).

    Substitute 18340lbftfor Mmax, 14.4in.3for Sin Equation (VI).

    σmax=(15283.33lb/in2)(1psi1lb/in2)=(15283.33psi)(103ksi1psi)=15.283ksi=12.64ksi

    Hence, from the table “Appendix F” we will use the value S8×18.4.

    Conclusion:

    The maximum value of load Pis S8×18.4.

    (c)

    Expert Solution
    Check Mark
    To determine

    The most economical steel Wshape.

    Answer to Problem 5.6.5P

    The most economical steel Wshape is W8×35.

    Explanation of Solution

    Given Information:

    The uniform distributed load is 45lb/ft, the load is 2000lb, maximum normal stress is 20ksi.

    Write the expression for the maximum moment of beam.

    Mmax=qL22+P(AC)…… (VII)

    Write the expression for the section modulus.

    S=Mmaxσallow…… (VIII)

    Write the expression for the maximum stress.

    σmax=MmaxS…… (IX)

    Calculation:

    Substitute 20lb/ftfor q, and 300lbfor Pand 10ftfor Lin Equation (V)

    Mmax=45lb/ft(102ft2)2+(2000lb)(6ft)=(14250lbft)(12in1ft)=171000lbin.

    Substitute 171000lbin.for Mmax, and 20ksifor σallowin Equation (VI)

    S=171000lbin.(20ksi)(103psi1ksi)=171000lbin.(20000psi)(1lb/in21psi)=8.55in.3

    Refer to the table (a)of “Appendix F” to get the value of weight intensity w=35lbft, and the section modulus S=10.6in3.

    Substitute 45lb/ftfor q, and 2000lbfor Pand 10ftfor L, 35lbftfor win Equation (I).

    Mmax=((45+35)lbft)×102ft22+(2000lb×6ft)=((45+35)lbft)×102ft22+(12000lbft)=10000lbft

    Substitute 18220lbftfor Mmax, 8.74in.3for Sin Equation (VI).

    σmax=(10000lbft)(12in1ft)10.6in.3=(10000lbft)(12in1ft)10.6in.3(1psi1lb/in2)=11320.75psi(103ksi1psi)=11.320ksi

    Hence, from the table “Appendix F” we will use the value W8×35.

    Conclusion:

    The maximum value of load Pis W8×35.

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