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Minimum beam dimension for specific max stress

  1. Oct 12, 2012 #1
    1. The problem statement, all variables and given/known data
    A beam (note: part of a truss, I chose to use the beam in the truss with the most force and length for my calculations, which I assume is the correct thing to do) of length [itex]6m[/itex] with a force of [itex]9N[/itex] is being constructed out of a material with a Young's Modulus of [itex]E = 70 \times 10^9 N/m^2[/itex] and it's ultimate stress is [itex]σ_{u} = 110 \times 10^6 N/m^2[/itex]. What is the minimum dimension of the square beam (i.e. cross-section) so that the structure will not fail?


    2. Relevant equations
    Spring constant of beam: [itex]k_{beam} = \frac{A \times E}{L}[/itex]
    Hooke's Law: [itex] F = k \times ΔL[/itex]

    [itex]E = \frac{σ}{ε} = \frac{F/A}{ΔL/L}[/itex]


    3. The attempt at a solution
    I'm pretty lost on what to do here, we barely covered this stuff in lectures. My first thought was simply letting [itex]σ = F/A[/itex] but that doesn't make much sense because surely length is a factor and that method doesn't use the young's modulus at all. My second thought was to substitute [itex]k_{beam}[/itex] into the hooke's law equation, giving [itex] F = \frac{A \times E}{L} \times ΔL[/itex] => [itex]9 = \frac{A \times 70 \times 10^9}{6}[/itex] but I'm not sure where this gets me.


    P.S. Sorry if I didn't explain the question very well, I paraphrased a bit because the original question is part of a series of questions and doesn't make a lot of sense on it's own.
     
  2. jcsd
  3. Oct 12, 2012 #2
    Just so its clear in my mind: we have a beam spanning 6m with a force of 9N? do you have a sketch of what you are doing? Or perhaps scan in the question page. Just so we're all on the same page lol
     
  4. Oct 12, 2012 #3
    We have a truss with seven beams in total, this part of the question is asking for the minimum area of the square cross-section of an individual beam so that the structure will not fail (i.e. so that any individual beam will not fail). There is one beam that has the greatest length and force (6m and 9N) so I am assuming that is the one that should be used for calculations as it will be under the most stress (I think?).

    I think that the fact that the beam is part of a truss structure is largely irrelevant to this part of the question but I may be approaching this the wrong way.
     
  5. Oct 12, 2012 #4

    PhanthomJay

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    Well, it is relevant in that the 9N force is an axial force directed along the longitudinal axis of the member.
    Please check the units of the force...9 Newtons is barely any force at all.
    You must determine whether the member is in compression or tension.
    If in tension, then your F/A equation applies; if in compression, you must not only check F/A, but also buckling allowables. Length and modulus E is important when condsidering compressive loads, in addition to the properties of the cross section. For the tension case, typically you just need to know the area.
    You dont have to get into strain or stiffness calculations if you are not looking for deformations.
     
  6. Oct 12, 2012 #5
    Hi, sorry the beam is in compression I forgot to mention that. It is definitely 9N.
     
  7. Oct 12, 2012 #6

    PhanthomJay

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    Elements of a truss are typically not called 'beams', especially in this case where the loading is so slight that you have very small 'members'.
    Anyway, compare the compressive stress F/A versus the Euler buckling stress to see which controls the member size required.
     
  8. Oct 13, 2012 #7
    Woops sorry, the wording in the question seems to alternate between using "beam" and "member" to describe them. I'll give that a shot, thanks! Somehow I completely missed that when looking through my lecture notes.
     
  9. Oct 13, 2012 #8

    PhanthomJay

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    OK and please check again on the value of the force which you state is 9N. Are you assuming the members have no weight? The weight alone is probably more than that. Where does that number come from?
     
  10. Oct 13, 2012 #9
    They have given us the value of 9N. This is a very introductory class so I think they are just trying to teach us the basics with examples that are not necessarily practical or realistically possible. The weight of the members hasn't been mentioned at any point in anything we've covered involving trusses. Example questions involving calculating the force in a member done in our lectures have had similarly low forces (typically 2-10N) as the answer.
     
  11. Oct 15, 2012 #10
    Thanks PhanthomJay, I figured it and the rest of the questions out. Also my apologies, after all my stubbornness it was actually 9kN. I don't know how I managed to repeatedly misread that every time.
     
  12. Oct 15, 2012 #11

    PhanthomJay

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    Ok, good!
     
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