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How to identify forces of compression or tension in simple truss?

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello. I have some troubles with a truss problem. In fact i believe i have correctly identified the direction of the forces involved in the rods, but i cannot identify which rod is in compression or which is in tension. Are there any tips/rules to identify only with a force diagram?

    https://www.physicsforums.com/attachment.php?attachmentid=9821&d=1177005673

    3. The attempt at a solution
    Well i have resolved forces at each joint and put a direction (of the force) in the rod so that the net force at any joint is zero (Since the system is in equilibrium).

    1. Direction of force at P : Horizontally to the left.
    2. Direction of force at R : Vertically upwards.
    3. Direction of force at Q : Vertically downwards.
    4. Direction of force at S : Horizontally to the left.

    However i cannot figure the least which rods are in compression or in tension.
     
  2. jcsd
  3. Feb 24, 2012 #2

    SammyS

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    Hello kirakun. Welcome to PF !

    attachment.php?attachmentid=9821&d=1177005673.jpg
    It depends upon what you mean by your statements. (I hope you're not being taught to actually state the forces that way for members of the truss.)

    If member, P, exerts a force on joint A of FP to the left, then member, P, is under tension and member, P, will also exert a force on the wall of FP to the right.

    However, if member, P, exerts a force on the wall of FP to the left, then member, P, is under compression.

    ... etc.
     
  4. Feb 24, 2012 #3
    If you remove or cut the member S, the triangle with PQ and R will rotate clockwise about the wall pin at the top. Thus B will move towards the wall, and you can conclude that S is in compression. Try similar arguments applied to the other members. On the other hand, there are also statics arguments with directions of arrows, but your question suggests that you have yet to grasp that idea. Hence my qualitative approach.
     
  5. Feb 27, 2012 #4
    I see thanks both of u. This has given me greater understanding of the problem.
     
  6. Mar 25, 2012 #5
    How would you identify the members Q and R using this logic? Do we need to consider any member to be rigid, consider gravity, or only consider what is given in the free body diagram?
     
  7. Mar 28, 2012 #6
    answer to skr999. If R is removed, how does the frame behave? A moves down and to the left with a centre at the wall support top left. Therefore AB gets smaller. Therefore R is in compression. If Q is 'removed', the quadrilateral remaining is a mechanism that will distort. Which way will it go? B will move to the right and down, because R and S together will try to be in compression in a shape that is 'pre-buckled'. Therefore length of Q gets longer. Therefore it is in tension.
     
  8. Jul 17, 2012 #7
    Well i found a simple way for this problem. Sry for bringing this post up, but maybe those having similar problems can refer to it.

    I'll just consider equilibrium of forces at point A (the rest is done similarly)

    1. You have a weight w downwards, so there must be an upward force to balance it somewhere provided by two rods P or R.

    2. Since O is horizontal, only R can do so. Therefore "rod R exerts a diagonal force upwards on the point A" According to Newton's 3rd law of motion, point A exerts an equal and opposite force on the rod R. Thus u have a force diagonally downwards acting on the rod. But a rod is either in compression or in tension, therefore the other end of rod R connected to point B experiences a diagonal force upwards. Thus R is in compression.

    Use the same logic to find identify compression or tension.
    :/ Many people don't seem to grasp this concept. Even after googling like for days, i did not see a convincing approach. I may be missing something, if so do correct me. I hope this will help people in the future.

    Of course, there exists mathematical approach or resolving forces, which i saw friends doing, but i found this method much more quicker and it is a qualitative analysis.
     
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