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Homework Help: Strength of materials- simple truss problem

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    The bars in the truss each have a cross sectional area of 1.2in^2. If the maximum axial stress in any bar is not to exceed 25ksi, determine the maximum magnitude of P of the loads that can be applied to the truss. Determine the elongation of each member.

    E=29,000 ksi


    2. Relevant equations

    \sigma[/tex] = E* [tex] \epsilon[/tex]

    [tex]\sigma[/tex] = P/A

    [tex]\epsilon[/tex] = [tex]\delta[/tex] / L

    Having trouble with Latex, but the top equation should read:

    the other two equations are stress=force/area and strain=elongation/lenth.

    3. The attempt at a solution

    I got an answer, but I'm not sure if it's correct and if I'm solving this correctly.

    Since I know [tex]\sigma[/tex]=25ksi, and E=29,000 ksi I used [tex]\sigma[/tex]=P/A solving for P, P= [tex]\sigma[/tex]A=(25ksi)(1.2in^2)=30 kip.

    I assumed this to be Pmax.

    Using [tex]\sigma[/tex]=E*epsilon I solved for epsilon. epsilon= [tex]\sigma[/tex]/E=25ksi/29000ksi = 8.62E-4.

    Then used epsilon=delta/L. Solved for elongation, delta. delta=epsilon*L.

    Then I just used plugged in my solved value of epsilon, and then the length of each bar.

    Here's an example for AC: delta=epsilon*L=(8.62E-4)(4ft)(12in/1ft)=0.0414in.

    I'm not sure if this correct since it would mean every bar that is the same length would have the same elongation. Is this correct? Or do I need to solve the system for the forces and then do something? I'm confused. Thank you for the help.
  2. jcsd
  3. Aug 27, 2010 #2
    Axial stress is generated by a force perpendicular to the cross section, what is the direction of P in the diagram ?
  4. Aug 28, 2010 #3
    Can you determine the force in each member of the truss, as a function of P?
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