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Calculate the total change in length of a brass bar

  1. Apr 23, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-4-23_21-15-3.png

    2. Relevant equations

    Stress = force/ cross sectional area

    elongation = force x length / modulus of elasticity x cross sectional area

    3. The attempt at a solution

    The given solution is 0.149mm

    I assumed that the bar needs to be solved in two sections because of the hole of diameter 20mm in the first 70mm of the bar. I will call this part A and the 90mm section part B.

    Part A:

    Stress = force/ cross sectional area
    Force = stress x cross sectional area
    = 125x10^6 x pi/4(.03-.02)^2
    = 9.8kN

    elongation = force x length / modulus of elasticity x cross sectional area
    = 9.8x10^3 x 0.07 / 100x10^9 x pi/4(.03-.02)^2
    = 8.73x10^-5mm

    Part B:

    Force = stress x cross sectional area
    = 125x10^6 x pi/4(.03)^2
    = 88.4kN

    elongation = force x length / modulus of elasticity x cross sectional area
    = 88.4x10^3 x 0.09 / 100x10^9 x pi/4(.03)^2
    = 1.126x10^-4mm

    now if I add the elongation from part A & B together I do not get 0.149mm.
    Please show me where I am going wrong.
     
  2. jcsd
  3. Apr 23, 2017 #2

    TSny

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    You did not calculate the cross-sectional area of the left section correctly.

    You have assumed that the stress is the same in both sections. But is that correct? Imagine that the stress is created by applying an overall force on the left end and an overall force on the right end. How do these two forces compare?
     
  4. Apr 23, 2017 #3
    Hi TSny

    I understand what you are trying to get me to imagine I just don't know how to apply it.
    I guess I need to find a ratio between the diameters and/or lengths? Then multiply this by the stress of 125MPa to find the stress induced on each end?
     
  5. Apr 24, 2017 #4

    haruspex

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    Did you understand from TSny's post that this is wrong:
    Having fixed that, consider the two lengths as separate bodies. What is the interaction between them?
     
  6. Apr 24, 2017 #5
    I have found the solution. I will post it when I get a minute.
    I realize that I need to use pi/4(.03^2-.02^2)
     
  7. Apr 24, 2017 #6
    upload_2017-4-24_12-54-23.png
    upload_2017-4-24_12-54-39.png
     

    Attached Files:

  8. Apr 24, 2017 #7

    haruspex

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    You calculate the force as about 49,000kN. Are you sure about the units?
     
  9. Apr 24, 2017 #8
    That is 49kN not 49000kN. I should have used a period not a comma
     
  10. Apr 24, 2017 #9

    haruspex

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    Sorry, I should have noticed you were using the decimal comma in the radii. All looks good,then.
     
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