- #1

- 12

- 0

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

where m= Poisson's ratio

A=area of cross section

L=length

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter saravanan_n
- Start date

- #1

- 12

- 0

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

where m= Poisson's ratio

A=area of cross section

L=length

- #2

FredGarvin

Science Advisor

- 5,066

- 9

- #3

- 12

- 0

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

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)

- #4

- 12

- 0

Please I need a answer How this formula is obtained?

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

please help me.

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

please help me.

- #5

FredGarvin

Science Advisor

- 5,066

- 9

- #6

- 12

- 0

It is from "Transducer Engineering" by S.Renganathan

- #7

- 12

- 0

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

[email protected]

- #8

FredGarvin

Science Advisor

- 5,066

- 9

Start with the relationship for an electrical conductor:

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

Where:

[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]

- #9

- 12

- 0

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

- 12

- 0

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:

- #11

Astronuc

Staff Emeritus

Science Advisor

- 19,173

- 2,617

But A is proportional to the square of the length, i.e. a square has area, A = l

Now looking in three dimensions, if l

l

and the Area is then given by the product. If A

A = A

and dA = A - A

- #12

- 12

- 0

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

Share:

- Replies
- 5

- Views
- 386