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Homework Help: What is the maximum load the concrete can support?

  1. Jan 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A certain type of concrete has a tensile breaking stress of 3.1 MN/m^2, a compressive breaking stress of 37.7 MN/m^2 and a shear breaking stress of 9.4 MN/m^2. A circular pillar of this concrete has a radius of 0.6 m. What is the maximum load it can support assuming that it is evenly distributed over the top of the pillar?

    2. Relevant equations
    Normal stress, [PLAIN]http://upload.wikimedia.org/math/9/d/4/9d43cb8bbcb702e9d5943de477f099e2.png=F/A [Broken]
    Shear stress, [PLAIN]http://upload.wikimedia.org/math/d/9/5/d95fd1519e587418ebe3da8fb081701f.png=F/A [Broken]

    3. The attempt at a solution
    I haven't dealt with this sort of problem where the object has more than one type of stress applied to it. I tried finding the maximum load (or force) by adding up all the stresses and multiplying by the area which in this case is πr^2 but this doesn't give me the correct answer. I'm running out of ideas, could Pythagoras' theorem be considered? Please help!
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 19, 2015 #2


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    Homework Helper

    You added a tension stress to a compression stress? is the pillar going to be compressed, or will it be stretched, if it is supporting some "load"?
    Is the load above it, below it, or beside it?
  4. Jan 19, 2015 #3


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    Staff: Mentor

    Which of those stresses would apply to an upright pillar supporting a load from above?
  5. Jan 19, 2015 #4
    Ah of course, how silly of me! Even though there are multiple stresses in the problem, only one is required for solving. Thank you!
  6. Jan 20, 2015 #5


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    Education Advisor
    Gold Member

    Well, for completeness you need to check two not just one. The "compressive breaking stress" is just another way of referring to the normal stress that will break the pillar. But this limit is bigger than the other limit. So the check for this one is that it's stronger than the other. So if the other one is ok this one is too. And if the other one fails this one does not matter.

    But it tells you that you can predict the method of failure of the pillar as you increase the load to the point it does fail.
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