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Working out young's modulus and Poison ratio

  1. Mar 4, 2013 #1
    I have a question wondering if you guys could check over for me . I have a solution no sure on its accuracy!

    the question is in two parts

    a bar has a 25mm diameter and is subjected to a tensile force of 61Kn, the extension on a gauge length of 50mm is 0.1mm. calculate the youngs modulus in Gpa

    my attempt is youngs modulus = stress over strain

    stress = 61kn (61000N) divided by pi(12.5^2) = 124.3
    strain = 0.1 over 25 = 0.0004
    124.3 divided by 0.004 = 31075 which is 311 Gpa?

    the second part is: a bar has a 25mm diameter and is subjected to a tensile force of 61Kn,the decease in length is 0.013mm. the axial extension on a gauge length of 50mm is 0.1mm. calculate poissons ratio

    Poissons ratio = lateral strain over stress
    25mm divided by 0.1 x 0.013 = 19230
     
  2. jcsd
  3. Mar 4, 2013 #2

    SteamKing

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    The gauge length for the strain is 50 mm, not 25 mm.
     
  4. Mar 4, 2013 #3
    The Poisson's ratio is defined as the lateral compressive strain divided by the axial tensile strain, at least as measured in this type of deformation. It does not involve the stress.
     
  5. Mar 4, 2013 #4

    SteamKing

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    Poisson's ratio for steel is approximately 0.3
     
  6. Mar 5, 2013 #5
    so are my calculations correct bar using 25mm in poisons ratio?

    which would give me a value of 38461.5 (dimensionless number)
     
  7. Mar 5, 2013 #6
    No. As I said, the poisson's ratio is the lateral compressive strain divided by the tensile strain. What do you calculate for the lateral compressive strain? As SteamKing sort of implied, Poisson's ratio is going to be on the order of unity. Is your answer on the order of unity?
     
  8. Mar 5, 2013 #7
    so the youngs modulus part is correct but not the poisson ratio?

    and ah on so should it be the original length + the length extended by divided by the extension ? so

    25mm + 0.013mm divided by 0.1 = 250.13

    using a formula from another thread this :
    Elastic modulus in Tensile test
    E = L*F / A*△L

    this would change my first answer to : 50 x 61000 divided by (0.1x(12.5^2 x pi)) = 62 GPA
     
    Last edited: Mar 5, 2013
  9. Mar 5, 2013 #8

    SteamKing

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    In all of your calculations, you have been rather careless in using the units of the given quantities.

    For example, take the formula for E. E is to have units of GPa, which is to say 10^9 N/m^2. If the test sample is measured mm, then the dimensions must be converted to meters before calculating E.

    Since E is derived from a tensile test, the strain is calculated as the extension / original length of the sample, or 0.1 mm / 50 mm = 0.002
     
  10. Mar 5, 2013 #9
    Now, from what SteamKing just told you, what is the lateral strain in the direction perpendicular to the axis of the cylinder? (i.e., if the original diameter is 25 mm, and if the diameter decreases by 0.013 mm)
     
  11. Mar 6, 2013 #10
    would it be the force / area?

    so 61000/pi*((25+0.013/2)^2 = 124.1
     
  12. Mar 6, 2013 #11
    solved!!..i fink :P

    i did some looking in a material book and found :

    stress would = 61000 / ((pi/4)*(25^2) = 124.3
    strain = 0.002 so my answer would be 62Gpa for E

    my poisons ratio would be

    0.013 / (0.002 x 25) = 0.26
     
  13. Mar 6, 2013 #12
    These look OK now. Now please tell us more importantly whether you now understand the fundamentals of what you were doing.
     
  14. Mar 6, 2013 #13
    i know understand to get E i have to do stress over strain and I have to always look at my units i.e if its kn or mpa its in thousand not just 61 like i had previously used

    also that in the poisons ratio i am using the extension of the bar :)
     
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