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 3, Problem 3.4.21P

A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T0/L at B. Segment BC has linearly distributed torque of intensity r(x) = T0/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC and the shear modulus of elasticity of the material is G.

  1. Find the reaction torque RA.

  • Find internal torsional moments T(x) in segments AB and BC.
  • Find the rotation t0
  • Find the maximum shear stress tmaxand its location along the bar, Draw the torsional moment diagram (TMD:T(x),0 < x < L).
  •   Chapter 3, Problem 3.4.21P, A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec

    (a)

    Expert Solution
    Check Mark
    To determine

    The reaction torque RA.

    Answer to Problem 3.4.21P

    The reaction torque RAis = To6.

    Explanation of Solution

    Given information:

    The intensity of torque at point (A) is 0, the intensity of torque at point (B) is ToL, intensity of torque in segment BCis To3L, the polar moment of inertia is twice in section ABas compared to section BC

    Write the expression for equilibrium torque equation.

      RA+12(ToL)(L2)12(To3L)(L2)=0..... (I).

    Here, the reaction torque is at point (A) is RA.and the maximum torque per distance is (ToL).

    Write the expression for the reaction torque is at point (A) from Equation (I)

      RA+(To4)(To12)=0RA+3ToTo12RA=To6

    Conclusion:

    The reaction torque RAis = To6.

    (b)

    Expert Solution
    Check Mark
    To determine

    The internal torsional moment in the segment AB.

    The internal torsional moment in the segment BC

    Answer to Problem 3.4.21P

    The internal torsional moment in the segment ABis To6x2ToL2.

    The internal torsional moment in the segment BCis [(Lx)2To3L2].

    Explanation of Solution

    Write the expression for internal torsional moment in section AB.

      TAB(x)=To6(xL2)(ToL)x2=To6x2ToL2

    Write the expression for internal torsional moment in section BC.

      TBC(x)=(Lx)(L2)(To3L)(12(Lx))=[(xLL)2(To3)]=[(Lx)2To3L2]

    Conclusion:

    The internal torsional moment in the segment ABis To6x2ToL2.

    The internal torsional moment in the segment BCis [(Lx)2To3L2].

    (c)

    Expert Solution
    Check Mark
    To determine

    The twist rotation at point (C).

    Answer to Problem 3.4.21P

    The twist rotation at point (C) is ToL144GIp.

    Explanation of Solution

    Write the expression angle of twist at point (C).

      ϕC=0LT(x)dxGIP(x)...... (II)

    Here, the angle of twist at point (C) is ϕC.

    Write the expression for polar moment of inertia of the bar.

      ϕC=ϕAB+ϕBC..... (III)

    Write the expression for polar moment of inertia in the section AB.

      ϕAB=0L2To6x2ToL22GIp..... (IV)

    Here, polar moment of inertia is in the section ABis ϕAB.

    Integrate Equation (IV).

      ϕAB=12GIp[To6xx3To3L2]0L2=12GIp[ToL12ToL24]=12GIp[2ToLToL24]=ToL48GIp..... (V)

    Write the expression for angle of twist in BC.

      ϕBC=L2LTBC(x)G(Ip)BCdx...... (VI)

    Here, angle of twist in section BCis ϕBC.

    Substitute [(Lx)2To3L2]for TBC(x)in Equation (VI).

      ϕBC=L2L[(Lx)2To3L2]G(Ip)dx=ToG(Ip)L2L(Lx)23L2dx=ToG(Ip)[(Lx)39L2]L/2L=ToL72G(Ip)

    Substitute ToL48GIpfor ϕABand ToL72G(Ip)for ϕBCin Equation (III)

      ϕC=ToL48GIp+ToL72G(Ip)=ToL144GIp

    Conclusion:

    The twist rotation at point (C) is ToL144GIp.

    (d)

    Expert Solution
    Check Mark
    To determine

    The maximum stress and the location of maximum stress in the bar.

    Answer to Problem 3.4.21P

    The maximum stress is 2.67ToπdAB3.

    The maximum stress in the bar is at point (A).

    Explanation of Solution

    Write the expression for the polar moment of inertia of section AB.

      2Ip=π32dAB4...... (VII)

    Here, the polar moment of inertia is Ipand the diameter of the section ABis dAB.

    Write the expression for the polar moment of inertia of section BC.

      Ip=π32dBC4..... (VIII)

    Here, the polar moment of inertia is Ipand the diameter of the section BCis dBC.

    Write the expression for the shear stress at point (A) from torsional equation.

      τA=TrAIp...... (IX)

    Here, the shear stress is τAand the radius of the shaft is r.

    Write the expression for the shear stress at point (A) from torsional equation.

      τB=TrBIp...... (X)

    Here, the shear stress is τBand the radius of the shaft is rB.

    Substitute To6for T, dAB2for rAand π32dAB4for Ipin Equation (IX).

      τA=(To6)(dAB2)r(π32dAB4)=8To3πdAB3(π32dAB4)=2.67ToπdAB3

    Substitute To12for T, dBC2for rBand π32dBC4for Ipin Equation (X).

      τB=(To12)(dBC2)(π32dBC4)=4To3πdBC3...... (XI)

    Substitute, (12)π32dAB4for Ipin Equation (VIII)

      (12)π32dAB4=π32dBC4(12)dAB4=dBC4dBC=(12)14dAB

    Substitute (12)14dABfor dBCin Equation

      τB=4To3π((12)14dAB)3=4To3π((12)34dAB3)=2.242ToπdAB3

    Conclusion:

    The maximum stress is= 2.67ToπdAB3.

    The maximum stress in the bar is at point (A).

    (d)

    Expert Solution
    Check Mark
    To determine

    The torsional moment diagrams.

    Explanation of Solution

    Write the torsional moment at point (A)

      RA=To6

    Write the torsional moment at point (B).

      RB=To12

    The following diagram shows the torsional moment diagram:

      Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.4.21P

           Figure-(1)

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    Chapter 3 Solutions

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