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Elastic deformation of axially loaded members

  1. Mar 11, 2014 #1
    Hi all,

    I have a text book question which I am stuck on and should be quite basic.

    There is a bar that has a cross-sectional area of 1750mm2, and E = 220GPa.

    http://imgur.com/gU29J1q
    Edit: okay link is not working, it is http://imgur.com/gU29J1q

    I am asked to find the displacement with the loading w=500x^1/3 N/m.

    In class we have been using the formula displacement = PL/AE, where

    P=load
    L=length of member
    A=cross sectional area tangent to load
    E=Young's or elastic modulus

    So far I have this

    500(1.5m)^1/3 N/m x 1.5m
    0.00175m^2 x 2.2x10^11 N/m^2

    So it would seem all the units would cancel out apart from the m^1/3

    However to me this didn't make much sense as the question is only asking for the answer to three significant figures and I got 1.467x10^-6 m^1/3

    Can someone tell me if I am doing this correctly?

    Thanks
     
  2. jcsd
  3. Mar 11, 2014 #2

    SteamKing

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    You have a bar where the axial load is distributed longitudinally. Exactly at what location are you supposed to calculate the axial deflection of the bar?
     
  4. Mar 11, 2014 #3
    There is a point A at the end of the 1.5m long bar
     
  5. Mar 11, 2014 #4
    I think you made a good attempt with trying to cancel out the units. The fact you came out with a strange unit for length should tell you did something wrong some where.

    The mistake you made was using the wrong formula. The formula used is for a constant axial load. You have a distributed axial load here.

    To solve this problem you need to develop your own formula by looking at how PL/AE is derived. Once you set this up and do the integral calculus you should arrive at a more sensible answer.
     
  6. Mar 11, 2014 #5
    Yes thank you, I just found the right formula to use. It states the displacement is = integral between 0 and L of P(x)/A(x)E(x)

    However I still don't know how to do it unfortunately. They give me w in the diagram, is this supposed to be P in the equation? And how would you substitute x in? Do just make it 500*1.5m^1/3 N/m? Or do I actually leave x in the equation?
     
  7. Mar 11, 2014 #6
    Yes thank you, I just found the right formula to use. It states the displacement is = integral between 0 and L of P(x)/A(x)E(x)

    However I still don't know how to do it unfortunately. They give me w in the diagram, is this supposed to be P in the equation? And how would you substitute x in? Do just make it 500*1.5m^1/3 N/m? Or do I actually leave x in the equation?
     
  8. Mar 11, 2014 #7
    You're trying to substitute in x = L = 1.5m too soon. x does not equal 1.5m as it is variable that ranges from 0 to 1.5m.

    But otherwise you guessed right, P(x)= w = 500x^1/3. Stick that in there and do the integration.
     
  9. Mar 11, 2014 #8
    Duplicate.
     
    Last edited: Mar 11, 2014
  10. Mar 11, 2014 #9
    Okay thanks. And for the A(x) and E(x) am I just using the area multiplied by x, and the elastic modulus multiplied by x?
     
  11. Mar 11, 2014 #10
    This is what I am putting into Wolfram Alpha - 'definite integral of (500x^(1/3))/((0.00175x)(2.2x10^(11)x)) between 0 and 1.5'

    However I don't know how to interpret the answer...
     
  12. Mar 11, 2014 #11
    This is what I am putting into Wolfram Alpha - 'definite integral of (500x^(1/3))/((0.00175x)(2.2x10^(11)x)) between 0 and 1.5'

    However I don't know how to interpret the answer...
     
  13. Mar 11, 2014 #12

    SteamKing

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    No. A(x) and E(x) represent the cross-sectional area and the modulus of elasticity of the beam as functions of the length variable, x. For a prismatic beam made of a single material, A(x) and E(x) will both be constants.
     
  14. Mar 11, 2014 #13
    So what do I type into Wolfram Alpha? What does the formula look like when written out fully, before simplifying etc?
     
  15. Mar 11, 2014 #14
    Put in the same formula but drop the extra x's that you put in for A(x) and E(x).

    Generally, when someone defines a function F(x) this means the value of F depends on the variable x without having to state exactly what the function is. In this specific case P(x) = 500x^1/3 but A(x) = A and E(x) = E since these are constant along the length of the beam.
     
  16. Mar 11, 2014 #15
    Duplicate.
     
  17. Mar 12, 2014 #16
    And what about the metres and Newtons in the formula?
     
  18. Mar 12, 2014 #17
    Oh sorry they cancel out, no worries
     
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