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?
#6
saravanan_n
12
0
Book reference
It is from "Transducer Engineering" by S.Renganathan
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]
#9
saravanan_n
12
0
Sorry for being impatient,Since I have test on this I have to hurry.
Now we get
DR/R=DL/L+DP/P-DA/A
Then we divide whole eqn by dl/l
(dr/r)/(dl/l)=1+(dp/p)/(dl/l)-m
so we can put
Gauge Factor=1+Piezoresistivity-Poisson ratio
Is this equation right?
#10
saravanan_n
12
0
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)
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}
and dA = A - A_{o}, which defines dA.
#12
saravanan_n
12
0
Thanks for the reply.I am very much pleased.Thank you!!!