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Poisson ratio

  1. Nov 19, 2006 #1
    My question is a bout this question below (Bear with me people)

    A tie bar 25mm in diameter and 1m long extends in length by 1.2mm, when subjected to an axial tensile force of 80KN. If the diameter decreases by 0.007mm, determine the values of poisson's ration, and E,G and K.

    My question is how do you solve the Poissions ratio if it is equal to lateral strain divided by Axial strain. Can someone help me to determine the lateral strain and do I need to covert 80KN into strain.

    How can I figure out lateral strain that what I really need to know?
     
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 20, 2006 #2

    PhanthomJay

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    axial strain (e) is 1.2mm/1000mm = .0012; lateral strain is .007mm/25mm = .00028; Then poissons ratio is, as you noted, .00028/.0012 = .23.
    The 80kN force creates an axial stress of P/A = 80,000N/((pi)(.025)^2/4) = 163MPa; the since stress is eE, solve E = 136GPa, and then G is .23(E) = 37.5MPa.
     
  4. Nov 21, 2006 #3

    Your gonna have to clarify this bit, I am at a loss where you got the figures to solve E. Your answer of 136GPa is right though, I want to clarify to solve E isn't it stress/strain. How did you get this answer.
     
  5. Nov 21, 2006 #4

    PhanthomJay

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    There are 2 'e''s, big E and little e. Big E usually denotes the modulus of elasticity, in units of force/length^2, and little e denotes the strain, in dimensionless units of length/length. E is the slope of the stress strain curve, that is Stress/Strain = E, as you noted. Since axial stress is Force/Area, calculated above, and e is delta L/L, as was calculated above, solve for E.
     
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