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 3.5.13P

Two circular aluminum pipes of equal length L = 24 in. arc loaded by torsional moments T (sec figure). Pipe I has outside and inside diameters d2= 3 in. and L2, = 2.5 in., respectively. Pipe 2 has a constant outer diameter of d2along its entire length L and an inner diameter of d1but has an increased inner diameter of d3= 2.65 in. over the middle third.

Assume that E = 10,400 ksi, u = 0.33, and allowable shear stress ra= 6500 psi.

  1. Find the maximum acceptable torques that can be applied to Pipe 1; repeat for Pipe 2.

  • If the maximum twist e of Pipe 2 cannot exceed 5/4 of that of Pipe 1, what is the maximum acceptable length of the middle segment? Assume both pipes have total length L and the same applied torque T.
  • Find the new value of inner diameter d3of Pipe 2 if the maximum torque carried by Pipe 2 is to be 7/8 of that for Pipe L
  • If the maximum normal strain in each pipe is known to bemax = 811 x 10-6, what is the applied torque on each pipe? Also, what is the maximum twist of each pipe? Use the original properties and dimensions.
  •   Chapter 3, Problem 3.5.13P, Two circular aluminum pipes of equal length L = 24 in. arc loaded by torsional moments T (sec

    (a)

    Expert Solution
    Check Mark
    To determine

    The maximum acceptable torques for pipe (1).

    The maximum acceptable torques for pipe (2).

    Answer to Problem 3.5.13P

    Maximum acceptable torques for pipe (1) is = 17842lbin.

    Maximum acceptable torques for pipe (2) is = 13479.375lbin.

    Explanation of Solution

    Given information:

    The following figure shows the free body diagram of pipe 1:

           Figure-(1) shows the diagram of two pipes:

      Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.5.13P , additional homework tip  1

           Figure-(1)

    The following figure shows the free body diagram of pipe 2:

      Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.5.13P , additional homework tip  2

           Figure-(2)

    The length of pipe is 24in, outside diameter of pipe (1) is 3in, inside diameter of pipe (1) is 2.5in, the increased diameter of the pipe (2) is 2.65in, the young’s modulus is 10400ksi, Poisson ratio is 0.33and allowable shear stress is 6500psi.

    Write the expression for polar moment of inertia for pipe (1)

      Ip1=π32(d24d14)..... (I)

    Here, the polar moment of inertia is IP, the outside diameter is d2and inside diameter is d1.

    Write the expression for maximum torque for pipe (1)

      τa=Tad22Ip1...... (II)

    Here, acceptable shear stress is τa, the applied torque is T, polar moment of inertia is Ip1and outer diameter is d2.

    Write the expression for polar moment of inertia for pipe (2)

      Ip2=π32(d24d34)..... (III)

    Here, the polar moment of inertia is IP, the outside diameter is d2and inside diameter in the middle section is d3.

    Write the expression for maximum torque for pipe (2)

      τa=Tad22Ip2...... (IV)

    Here, acceptable shear stress is τa, the applied torque is T, polar moment of inertia is Ip1and outer diameter is d2.

    Calculation:

    Substitute, 3infor d2and 2.5infor d1in equation (I)

      Ip1=π32((3in)4(2.5in)4)=π32(81in439.0625in4)=4.1172in4

    Substitute 4.1172in4for Ip1, 6500psifor τaand 3infor d2in Equation (II).

      6500psi=Ta(3in)2(4.1172in4)6500psi=0.3643TaTa=6500psi0.3643Ta=17842lbin

    Substitute, 3infor d2and 2.65infor d3in equation (III)

      Ip2=π32((3in)4(2.65in)4)=π32(81in449.3155in4)=3.1106in4

    Substitute 3.1106in4for Ip1, 6500psifor τaand 3infor d2in Equation (II).

      6500psi=Ta(3in)2(3.1106in4)6500psi=0.482TaTa=6500psi0.482Ta=13479.375lbin

    Conclusion:

    Maximum acceptable torques for pipe (1) is = 17842lbin.

    Maximum acceptable torques for pipe (2) is = 13479.375lbin.

    (b)

    Expert Solution
    Check Mark
    To determine

    The maximum acceptable length of the middle segment.

    Answer to Problem 3.5.13P

    The maximum acceptable length of the middle segment.

    Explanation of Solution

    Given information:

    Maximum twist of pipe (2) cannot exceed 5/4of pipe (1).

    Write the expression for the angle of twist for pipe (1).

      ϕ1=GLTIp...... (V)

    Here the angle of twist is ϕand shear modulus is G.

    Write the expression for total angle of twist for pipe (2)

      ϕ2=(ϕ2)AB+(ϕ2)BC+(ϕ2)CD...... (VI)

    Here, the total angle of twist is ϕ2, the angle of twist in the section ABis (ϕ2)AB, the angle of twist in the section BCis (ϕ2)BC, ϕ2, the angle of twist in the section CDis (ϕ2)CD.

    Write the expression for the length of the segment ABand CDfor middle segment length as x.

      LAB=LCD=Lx2

    Write expression for the angle of twist in the section AB.

      (ϕ2)AB=G(Lx)T(Ip2)AB

    Write expression for the angle of twist in the section BC.

      (ϕ2)BC=G(x)T(Ip2)BC

    Write expression for the angle of twist in the section CD.

      (ϕ2)CD=G(Lx)T(Ip2)CD

    Substitute G(Lx)T(Ip2)ABfor (ϕ2)AB, G(x)T(Ip2)BCfor (ϕ2)BCand G(Lx)T(Ip2)CDfor (ϕ2)CDin Equation (VI).

      ϕ2=G(Lx)T(Ip2)AB+G(x)T(Ip2)BC+G(Lx)T(Ip2)CD=GT[(Lx)2(Ip2)AB+(x)(Ip2)BC+(Lx)2(Ip2)CD]...... (VII)

    Write the expression for relation between ϕ1and ϕ2.

      ϕ2=54ϕ1..... (VIII)

    Calculation:

    Substitute 4.1172in4for Ip, 24infor Lin equation (V)

      ϕ1=G(24in)T(4.1172in4)

    Substitute 24infor L

      4.1172in4for (Ip2)ABand 3.1106in4for (Ip2)BCin Equation (VII).

      ϕ2=GT[(24inx)24.1172in4+(x)3.1106in4+(24inx)24.1172in4]=GT[(24inx)4.1172in4+(x)3.1106in4]

    Substitute GT[(24inx)4.1172in4+(x)3.1106in4]for ϕ2and G(24in)T(4.1172in4)for ϕ1in Equation (VIII).

      GT[(24inx)4.1172in4+(x)3.1106in4]=54G(24in)T(4.1172in4)[(24inx)4.1172in4+(x)3.1106in4]=54(24in)(4.1172in4)x3.1106in4x4.1172in4=(24in)(4.1172in4)(541)0.7859x=1.4573

      x=18.5412in

    Conclusion:

    The maximum acceptable length of the middle segment is = 18.5412in.

    (c)

    Expert Solution
    Check Mark
    To determine

    The inner diameter for the given parameters.

    Answer to Problem 3.5.13P

    The inner diameter for the given parameters is = 2.58in.

    Explanation of Solution

    Given information:

    the maximum torque carried by pipe (2) is 7/8of pipe (1).

    Write the expression for allowable torque.

      Tallow=78Ta...... (IX)

    Here, the allowable torque is Tallow

    Substitute 78Tafor Taand π32((3in)4d34)for Ip2in Equation (IV).

      τa=78Tad22π32((d2)4d34)..... (X)

    Calculation:

    Substitute 6500psifor τa, 17810psifor Taand 3infor d2in Equation (X).

      6500psi=78(17810psi)(3in)2π32((3in)4d34)((3in)4d34)=78(17810psi)(3in)2π326500psi((3in)4d34)=36.63in4d34=81in436.63in4

      d34=44.369in4d3=2.58in

    Conclusion:

    The inner diameter for the given parameters is 2.58in.

    (c)

    Expert Solution
    Check Mark
    To determine

    Applied torque on pipe (1)

    Maximum twist of pipe (1)

    Answer to Problem 3.5.13P

    Applied torque on pipe (1) is = 17406.56lbin.

    Applied torque on pipe (2) is = 13148.35lbin.

    Maximum twist of pipe (1) is = 1.487°.

    Explanation of Solution

    Write the expression for maximum shear strain.

      γmax=2εmax..... (XI)

    Here, maximum shear strain is γmaxand the allowable tensile strain is εmax.

    Write the expression for maximum shear stress.

      τmax=Gγmax...... (XII)

    Here, the maximum shear stress is τmax, the shear modulus of the shaft is Gand maximum shear strain is γmax.

    Write the expression for shear modulus of elasticity.

      G=E2(1+v)..... (XIII)

    Write the expression for maximum torque for pipe (1)

      τmax=Tmaxd22Ip1...... (XIV)

    Write the expression for maximum angle of twist for pipe (1)

      ϕ1=Tmax×LGIp...... (XV)

    Maximum angle of twist in pipe (1) is ϕ1.

    Calculation:

    Substitute 10400ksifor Eand 0.33for v, 4.1172in4for Ip1and 24infor Lin Equation (XIII).

      G=10400ksi2(1+0.33)G=10400ksi2.66G=3909.77psi

    Substitute 811×106for εmaxin Equation (XI).

      γmax=2(811×106)=1.622×103

    Substitute 1.622×103for γmaxand 3909.77psifor Gin Equation (XII).

      τmax=(3909.77ksi)(1.622×103)=6341.65psi

    Substitute 6341.65psifor τmax, 4.1172in4for Ip1, 3infor d2in equation (XIV).

      6341.65psi=Tmax(3in)2(4.1172in4)Tmax=(6341.65psi)(2(4.1172in4))3inTmax=17406.56lbin

    Substitute 17406.56lbinfor Tmax

      24infor L, 4.1172in4for Ip

      3909.77psifor Gin equation (XV).

      ϕ1=(17406.56lbin)×24in(3909.77psi)(4.1172in4)=427740lbin242818880lbin2=(0.02594rad)(180°πrad)=1.487°

    Substitute 6341.65psifor τmax, 3.11in4for Ip1, 3infor d2in equation (XIV).

      6341.65psi=Tmax(3in)2(3.11in4)Tmax=(6341.65psi)(2(3.11in4))3inTmax=13148.35lbin

    Conclusion:

    Applied torque on pipe (1) is = 17406.56lbin.

    Applied torque on pipe (2) is = 13148.35lbin.

    Maximum twist of pipe (1) is = 1.487°.

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    Chapter 3 Solutions

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