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Mechanical principles - Stess and modulus of elasticity

  1. Dec 1, 2012 #1
    Mechanical principles - Stress and modulus of elasticity

    1. The problem statement, all variables and given/known data
    A 2.2m long steel towing bar of solid circular section diameter of 45mm is expected to carry a maximum load of 210 kN. The safety factor is 4; and for the steel the UTS is 540 MNm-2 and the modulus of elasticity is 200 GNm-2

    i) Calculate the actual towing capacity of the bar and state whether the proposed limit is acceptable
    ii) Determine the extension of the bar under the maximum proposed load

    2. Relevant equations

    i) σ = Load / Area
    Proposed limit acceptable= SF = UTS / σ

    ii) Change in length = (σ / Modulus of elasticity) / Length

    3. The attempt at a solution

    Hi I have worked out what I think is right but I have trouble writing the correct units, I will put what i have worked out could someone check if I am right please, thank you very much:-

    i) σ = Load / Area = 210kN / ((∏(45 x 10-3) / 4)

    =

    840x10-3 / 6.362x10-3 = 132.04 (now I think this is MNm2)

    Proposed limit acceptable = SF = 540 / 132.04 = 4.09

    So the proposed limit in this case is acceptable as it is over the required Safety Factor of 4

    ii) Change in length = (132.04 / 200) x 2.2

    = 0.66 x 2.2 = 1.45mm (??)
     
    Last edited: Dec 1, 2012
  2. jcsd
  3. Dec 1, 2012 #2

    SteamKing

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    If the load is in newtons and the area is in m^2, what are the units of load/area?
    Hint: it's not MNm^2.

    If P/A is in MNm^-2 and E is in GNm^-2, then what are the units of (P/A)/E?
    Do MN and GN cancel without using any other factors?
     
  4. Dec 1, 2012 #3
    Hi steamking this is where i draw a mental block, i dont know why but im working on it
    1st point putting it like that is it Nm^2?

    2nd point do you mean that both ^-2 cancel each other out leaving MNm and GNm?

    Thanks, I keep telling myself i will get this!
     
  5. Dec 1, 2012 #4

    SteamKing

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    My point is, if you divide a force by an area, a la P/A, the units must be N/m^2 or alternately, Nm^-2 (You do know there is a difference between a positive and a negative exponent, don't you?)

    On the second point, what is the difference between 1 MN and 1 GN? How many MN are in 1 GN? In other words, your calculation of the change in length omits a factor of 1000, although the final answer is correct.
     
  6. Dec 1, 2012 #5
    Thankyou very much for taking the time in showing me where i am going wrong SteamKing, that definately makes more sense, im going to go and do some more examples to make sure its sinked in whilst its still fresh! Thanks again
     
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