Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Statically Indeterminate Stepped Bar Problem

  1. Nov 25, 2006 #1
    I've been having difficulty with this problem for the past 5 hours. I am understanding some part, but I can't get the final answer.

    http://img223.imageshack.us/img223/1103/1015zz1.th.jpg [Broken]

    This is what I got so far:

    EFx = Ra + Rb - P = 0
    Ra + Rb = P

    Sab = Sac + Scb = [(-Ra)(a)/AE] + [(Rb)(b)/AE] = 0
    -Ra(a) + Rb(b) = 0
    -Ra(a) + (P - Ra)b = 0
    Ra = (b/a+b)*P
    Rb = (a/a+b)*P

    Ra = (8in/18in)*100kips = 44.44 kips
    Rb = (10in/18in)*100kips = 55.56 kips

    I found the reactions but I cannot find the stress of each segment. The fact that the problem isn't giving me the modulus of elasticity is throwing me off.

    I know stress = P/A where P = force and A = cross-sectional area

    I figured then stress for segment AC = 44.44kips/6in^2 but it doesn't equal the answer in the back of the book. I also figured that would be the same for segment CB but it's not.

    Please help because I feel like I've done everything I can to this problem

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Nov 25, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You don't need to know the modulus of elasticity since it is given that both materials have the same modulus.
    It sure looks like the fat piece will be in compression and the skinny piece will be in tension. To get the force in each, you must take note that the deflections of each must be the same. The fat piece compresses by the same amount that the skinny piece elongates. What is the formula for axial deformation? Apply that to solve for the ratio of the loads in each piece, and thence the load in each; then stress is just P/A for each piece.
    EDIT: Oh, sorry, you seem to be on the right track, however, you assumed that A is the same for each section. It is not. Plug in the values of A for each and I think you've got it.
    Last edited by a moderator: May 2, 2017
  4. Nov 26, 2006 #3
    I didn't assume the cross-sectional areas were the same. The fat piece is 6 in^2 and the skinnier piece is 3 in^2.

    In my thinking:
    Stress of AC should = 44.44 kips/6 in^2 = 7.41
    Stress of BC should = 55.56 kips/3 in^2 = 18.52

    However those are not the answers. In the back of the book, AC is 10.xx and BC is 12.xx

    Any ideas?

  5. Nov 26, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Go back to your equation with Sab etc you will see that you cancelled out the AE term; this is where you assumed the x-section was the same...you can cancel out the E but not the the A's, they are different...
  6. Nov 26, 2006 #5
    Ah excellent! Thank you very much. The equations for Ra and Rb were a little more complicated but I got the answers the back of the book has.

    I'm surprised I didn't see that I was cancelling out both A's when they clearly aren't equal to each other.

    Thank you very much. This forum has always been helpful.
  7. Dec 2, 2006 #6
    Once you've found your reactions from the applications of Statics and your knowledge of Deformations, the determination of the stresses in a bar section reduces to


    where P is the resultant force acting on each section. For section AC, P = FBx - 100. Simlarly for the other section.

    You have the hard part figured out. Little details always hold you back.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook