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?
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
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]
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 = l^{2}, where l is side length, or a circle has area [itex]\pi[/itex]r^{2}, where r is radius.
Now looking in three dimensions, if l_{x} and l_{y} contract by [itex]\nu\,(\frac{\Delta{l_z}}{l_z})[/itex] then the new lengths are
l_{x}([itex]1 - \nu (\frac{\Delta{l_z}}{l_z})[/itex]) and l_{y}([itex]1 - \nu (\frac{\Delta{l_z}}{l_z})[/itex]),
and the Area is then given by the product. If A_{o} = l_{x} l_{y}, then the new area is
A = A_{o} * ([itex]1 - \nu\,\frac{\Delta{l_z}}{l_z}[/itex])^{2}
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