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Calculating the maximum load that can be carried & extension

  1. Jan 26, 2015 #1
    Ive been given a question which i'm stuck on and cannot answer. Ive only been able to calculate the maximum allowable working stress and other then that im stuck on how to answer the following question:

    A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm). On testing this specimen it was found that the ultimate tensile stress was 420 N/mm^2. Calculate:

    i) the maximum load that can be carried using a Factor of Safety of 2.5
    ii) the extension under this load

    So what i currently have worked out is the maximum allowable working stress which is 420/2.5=168 KN/m^2 (is this right?) and the cross section area which is presume is 10.5mm x12.5mm = 131.25mm^2 (is this right too?)

    Thank you in advance
     
  2. jcsd
  3. Jan 26, 2015 #2

    SteamKing

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    The maximum allowable working stress, with the factor of safety, has units of N/mm2, instead of kN/m2, going by the original text.
    The area of the cross section is correct.
     
  4. Jan 26, 2015 #3
    Thank you but what do i do after is what im stuck on
     
  5. Jan 26, 2015 #4

    SteamKing

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    Well, what's your understanding of how to calculate the stress in the sample, if a load P were applied to it?

    You have calculated the maximum allowable stress using the factor of safety. Now you need to determine the load P which produces this stress.
     
  6. Jan 26, 2015 #5

    DEvens

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    The hint about the stress value for your sample comes from the units of the ultimate tensile stress. Notice that it is N (for Newton) per square mm. Or, to be a bit more obvious: If you had two identical wires, how much force could they hold in combination compared to what one wire could hold alone?

    I think that in order to calculate the extension under this load you need more data. The ultimate tensile stress does not tell you what the extension will be at stress much less than that. Do you maybe have something like Young's modulus for the steel?
     
  7. Jan 26, 2015 #6
    The young's modulus from my previous question was 192.31 Kn/mm^2
    To find out the load i suppose i would need to maximum tensile force right? is this done by dividing the maximum allowable stress by the cross sectional area so 168/131.25=1.28 Kn? im not sure whether this is right
     
  8. Jan 26, 2015 #7
    I don't know whether I'm allowed to bump this thread but my deadline is on Friday and I really need to get this done am I on track or far off? Continuing from my previous post
     
  9. Jan 26, 2015 #8

    haruspex

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    As has been suggested, consider the units. The max stress is given in units of N/mm2. The area can be calculated in mm2. If you divide the first by the second, what will the units be?
     
  10. Jan 27, 2015 #9
    I think it's either N/mm^2 or Kn/mm^2? - I'm sorry for being so slow by the way guys, I'm pretty new to this field so I hope you guys can bear with me
     
  11. Jan 27, 2015 #10

    haruspex

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    Treat N and mm as unknowns in an algebraic expression. You have a = N/mm2 and b = mm2. You want a result of N. Will you get that by dividing a by b? If not, what operation will give the desired result?
     
  12. Jan 27, 2015 #11
    I gave it another try and got the following:

    Cross section = 10.5mm x 12.5mm = 131.25mm^2
    0.9m length =900mm
    Ultimate tensile stress = 420/2.5=168N/mm^2

    168N/mm^2=168,000,000 Pa
    168,000,000 x 0.00013125m^2 (131.25mm^2) = 22050N = 22.05 Kn
    Force = 22.05 Kn
    At this moment im assuming i did this correctly i acquired the force, but is the force "the maximum load that can be carried" as the question is asking? or do i do something else from here?

    radius= 22.05/pie x 168= 0.4177817256
    0.4177817256 square rooted = 0.2044m

    Diameter = 2x0.2044=0.4088m=408.8mm
    408.8m/2=204.4
    pie x (204.4^2) = 131253.7304

    extension = E = FxL / A x change in Length
    so Change in length = 22.05 x 900/ 131253,7304 x 192.31 (this is the youngs modulus i got from my previous question)
    = 0.000786mm extension

    Now i feel as ive gone wrong somewhere here but im not sure
     
  13. Jan 27, 2015 #12

    SteamKing

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    The SI symbol for the newton is 'N'. The force should be written as 22.05 kN.

    Wait a minute here! When did the specimen cross section become circular? According to the OP, the cross section was 10.5 mm x 12.5 mm, which suggests a rectangular cross section. Did you omit some information from your problem statement?
    It's not clear what units were used for your value for Young's modulus.

    Always try to include units in your calculations so that the numbers are intelligible to someone else trying to follow your work.
     
  14. Jan 27, 2015 #13
    Thank you for that, so is 22.05 kN (the force) the maximum load that can be applied? Im not too familiar with this subject
    The young's modulus calculation was stress/strain so 0.2/1.04 x10 to the power of -3 = 192.307692 = 192.31 kN/mm^2 (this was the previous question which ive managed to complete)

    And how did you determine that the specimen was of rectangular shape from the cross section? My lecturer told me that to find the diameter of a mild steel bar is to find the radius and then times the radius by 2 to get the diameter?
     
  15. Jan 27, 2015 #14

    SteamKing

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    Instead of writing E = 192.31 kN/mm2, write E = 192.31 MPa instead. You might want to check your calculation of the extension of the specimen.
    I think you are taking what your instructor said out of context. Not all cross sections are circular unless specified as such.

    This is the text from your OP:
    "A specimen of the same material that was used in the above test (mild steel), had a cross section of 10.5mm x 12.5mm and a length of 0.9m (900mm)." [emphasis added]

    This suggests to me that the cross section is rectangular (two different dimensions given). If the cross section were indeed circular, all you would need to say is that the specimen had a diameter (or radius) of so many mm.
     
  16. Jan 27, 2015 #15
    What is the formula to find the diameter of a rectangular specimen?
     
  17. Jan 27, 2015 #16
    Disregard my previous question, what about this for the extension:

    22.05 x 900/131,25 x 192.31 (young's modulus from before) = 0.786mm extension?
     
  18. Jan 27, 2015 #17

    SteamKing

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    Yes, this appears correct.

    I should correct my previous post. Young's modulus of 192.31 kN/mm2 should be equivalent to 192.31 GPa, instead of 192.31 MPa.
     
  19. Jan 27, 2015 #18
    Thank you very much for your and everybody that posed on this thread's help!
    Would you say that i've finished this question with the answers ive provided?
     
  20. Jan 27, 2015 #19
    Have I completed this question or do I still need to work out the maximum load?
     
  21. Jan 28, 2015 #20

    SteamKing

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    You worked out the maximum load to answer part i) of the question.
     
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