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Poisson's ratio

  1. Aug 5, 2005 #1
    please help on this formula how it has been arrived?

    dA=A(1-m dL/L)^2-A

    where m= Poisson's ratio
    A=area of cross section
  2. jcsd
  3. Aug 5, 2005 #2


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    In what context have you seen this? I can't recall seeing this form. I'll try to work out a proof. It looks like it's in regards to a tensile test...
  4. Aug 6, 2005 #3
    gauge factor

    While deriving the formulae for Gauge factor of materials(resitors) in strain gauges it comes.In between they have given this formula.

    First they told


    A= AREA


    dR=a(PdL+LdP) - P L da

    Divide L.H.S by R and R.H.S by PL/A we get

    dR dL dP dA
    -- = --- + ---- - ----- (eqn 1)
    R L P A

    And suddenly they give this formula


    dA= A(1-m dL/L)^2 -A

    m=poisson's ratio

    they substituted this in equation (1)

    and finally arrived at gauge factor (dR/R)/(dL/L)
  5. Aug 9, 2005 #4
    Please I need a answer How this formula is obtained?

    da=a(1-m dl/l)^2 - a

    please help me.
  6. Aug 9, 2005 #5


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    I can not find anything in my references that has that particular form. I am trying to derive it but not having much luck right now. I am still trying though. BTW...who is "they"? What book/reference states this?
  7. Aug 10, 2005 #6
    Book reference

    It is from "Transducer Engineering" by S.Renganathan
  8. Aug 11, 2005 #7
    please provide me the proof for this formulae:

    da=a(1-m dl/l)^2-a
    i.e., da/a=(1-m dl/l)^2 -1
    =1+m^2 (dl/l)^2-2m dl/l -1
    =(m dl/l)^2 - 2m dl/l

    so da/a= (m dl/l)^2 - 2m dl/l
    =m dl/l(m dl/l -2)


    (da/a)/(dl/l)=m^2 (dl/l)-2m

    now atleast you can say me how we arrive at this formula.Because I don't know any of the mechanical concepts and also about Poisson ratio.If you find any material regarding these please send me.My e-mail Id is

  9. Aug 11, 2005 #8


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    Pretty impatient people these days...Maybe "at least" I can tell you this:

    Start with the relationship for an electrical conductor:

    [tex]R = \frac{\rho L}{A}[/tex]

    [tex]R[/tex] = Resistance
    [tex]\rho[/tex] = Resistivity
    [tex]A[/tex] = Cross sectional Area

    Now differentiate it:

    [tex]dR = \frac{A(L d\rho + \rho dL) - \rho L dA)}{A^2}[/tex]

    Now go back and divide the last equation by the first and have fun with the algebra. Remember the definition of Poisson's Ratio: [tex]\nu = \frac{dA/A}{dL/L}[/tex]
  10. Aug 11, 2005 #9
    Sorry for being impatient,Since I have test on this I have to hurry.
    Now we get

    Then we divide whole eqn by dl/l


    so we can put
    Gauge Factor=1+Piezoresistivity-Poisson ratio

    Is this equation right?
  11. Aug 17, 2005 #10
    Equation Not Agrreable

    If you say Poisson Ratio =(da/a)/(dl/l)

    then do you agree with this equation also

    da=a(1-((poisson ratio) dl/l)^2)-a
    (in whatever context it comes)
    Last edited: Aug 17, 2005
  12. Aug 17, 2005 #11


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    Poisson's ratio is the lateral contraction per unit breadth divided by the longitudinal extension per unit length.

    But A is proportional to the square of the length, i.e. a square has area, A = l2, where l is side length, or a circle has area [itex]\pi[/itex]r2, where r is radius.

    Now looking in three dimensions, if lx and ly contract by [itex]\nu\,(\frac{\Delta{l_z}}{l_z})[/itex] then the new lengths are

    lx([itex]1 - \nu (\frac{\Delta{l_z}}{l_z})[/itex]) and ly([itex]1 - \nu (\frac{\Delta{l_z}}{l_z})[/itex]),

    and the Area is then given by the product. If Ao = lx ly, then the new area is

    A = Ao * ([itex]1 - \nu\,\frac{\Delta{l_z}}{l_z}[/itex])2

    and dA = A - Ao, which defines dA.
  13. Aug 18, 2005 #12
    Thanks for the reply.I am very much pleased.Thank you!!!
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