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Help to find force given a diameter reduction (Poissons Ratio)

  1. Jul 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Mild steel bar 40 mm diameter and 100 mm long, subjected to a tensile force along its axis (through central length)
    Youngs Modulus = 200GN m^-2
    Poisson's ratio = 0.3

    Calculate the force required to reduce the diameter to 39.99mm

    (there is also a picture saying Use the x- y co-ordinate system above). I shall upload that momentarily if needed?

    2. Relevant equations

    Poissons ratio = Transverse strain = - εt / εl

    Diameter = 0.04m
    Lo = 0.1m
    E = 200GN m^-2 = 200x10^9 Pascals (Pa)
    V = 0.3

    Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter

    Axial strain = Transverse strain / V

    Δd = εtd

    ε = εt / -V

    ε = σ / E

    σ = F/A


    3. The attempt at a solution

    So here's what I did and hoping it is right?

    Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter
    (0.3999 - 0.4000) / 0.4000 = -0.00025 Transverse strain (εt)

    Axial strain = Transverse strain / V
    -0.00025 / 0.3 = .00083333333 Axial strain

    I confirmed these by dividing Transverse strain by Axial strain = -0.3

    Since decrease in diameter = transverse strain x Original Diameter then:
    Δd = εtd = -0.00025 x .40 = 0.1mm which is correct.

    As εt = -Vε Then:
    -0.00025 = -0.3 x ε

    ε = εt / -V
    ε = 0.00083333333

    ε = σ / E
    σ is unknown ∴
    σ = ε x E
    σ = 16666666 Pa (stress value)

    σ = F / A

    Force = σ x A (Area)
    = 16666666 Pa x 0.00125663706
    = 20943.95019 Nm

    I think here I can bring the figure down a thousand to 20.944 KNm?

    Thank guys for ANY help! This took me ages!!
     
  2. jcsd
  3. Jul 31, 2013 #2

    SteamKing

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    1. What are the derived units for a Pascal?
    2. Since when did force in SI become newton-meters?
     
  4. Jul 31, 2013 #3
    Oh, every time one of us Brit's ask a question regarding GNm, we get told to use the proper term GPa (I think that's the equivalent). So I swapped the GNm to Gpa. Yes just checked online and 1 Newtom metre = 1 Pascal.

    Force SI should be Newtons? As in N or KN or MN but not NM?

    Thanks!

    I'm worried my force looks too low? .01mm should surely take more than my final figure?

    Thanks

    Lloyd
     
  5. Jul 31, 2013 #4
    Just went through double checking figures as "felt" force too low to make a .01mm impact on diameter.....

    Found σ was missing a unit, one 6 short!

    New final figure: 209439.51N
    Or simplified to 209.43951KN
    Or to 0.20943951 GN?
     
  6. Jul 31, 2013 #5

    SteamKing

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    Not even close.

    The pascal is the SI unit for pressure or stress, which implies a force divided by an area.

    1 Pascal = 1 Newton / m^2
     
  7. Jul 31, 2013 #6

    SteamKing

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    F = 209.44 kN looks like a good result.
     
  8. Sep 23, 2013 #7
    ignore this.
     
    Last edited: Sep 23, 2013
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