How is the formula dA=A(1-m dL/L)^2-A derived?

[saravanan_n]

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
L=length

 

[FredGarvin]

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...

 

[saravanan_n]

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

R=PL/A

R=RESISTANCE
P=SPECIFIC RESISTANCE
L=LENGTH
A= AREA

Then,

dR=a(PdL+LdP) - P L da
-------------------
a^2

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

as

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)

 

[saravanan_n]

Please I need a answer How this formula is obtained?

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

please help me.

 

[FredGarvin]

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?

 

[saravanan_n]

Book reference

It is from "Transducer Engineering" by S.Renganathan

 

[saravanan_n]

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)

so

(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

saravanan_n@msn.com

 

[FredGarvin]

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

Start with the relationship for an electrical conductor:

$$R = \frac{\rho L}{A}$$

Where:
$$R$$ = Resistance
$$\rho$$ = Resistivity
$$A$$ = Cross sectional Area

Now differentiate it:

$$dR = \frac{A(L d\rho + \rho dL) - \rho L dA)}{A^2}$$

Now go back and divide the last equation by the first and have fun with the algebra. Remember the definition of Poisson's Ratio: $$\nu = \frac{dA/A}{dL/L}$$

 

[saravanan_n]

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?

 

[saravanan_n]

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)

 

[Astronuc]

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 $\pi$r², where r is radius.

Now looking in three dimensions, if l_x and l_y contract by $\nu\,(\frac{\Delta{l_z}}{l_z})$ then the new lengths are

l_x($1 - \nu (\frac{\Delta{l_z}}{l_z})$) and l_y($1 - \nu (\frac{\Delta{l_z}}{l_z})$),

and the Area is then given by the product. If A_o = l_x l_y, then the new area is

A = A_o * ($1 - \nu\,\frac{\Delta{l_z}}{l_z}$)²

and dA = A - A_o, which defines dA.

 

[saravanan_n]

Thanks for the reply.I am very much pleased.Thank you!