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View Full Version : Elongation of non-uniform (area) bar.


Starwatcher16
Jun1-09, 11:14 AM
Lets say I have a bar of uniform material where one end has a diameter given by R_1 and another end given by R_2. R_2 > R_1, R(x) is linear.

So I know now three equations:

A) s=\frac{E \Delta L}{L}

B)R_x=R_1+\frac{x}{L}*(R_2-R_1)

C)A_x=\pi*R_x^2

Therefore, I know:

\Delta L=\frac{P}{E}\int \frac{dx}{\pi*A_x^2}=\frac{P}{E} \int \frac{dx}{\pi*[R_1+\frac{X}{L}(R_2-R_1)]^2},

let u=R_1+\frac{x}{L}*(R_2-R_1),\frac{du}{dx}=\frac{R_2-R_1}{L}, so:

Delta L=\frac{P}{E}*\frac{L}{R_2-R_1} \int \frac{dx}{\pi*[R_1+\frac{X}{L}(R_2-R_1)]^2}=\frac{P}{E\pi}*\frac{L}{R_2-R_1}*[\frac{1}{R_1}-\frac{1}{R_2}]

=\frac{P*L}{\pi*E*R_1*R_2}

Is that right?

Mapes
Jun1-09, 04:21 PM
Lets say I have a bar of uniform material where one end has a diameter given by R_1

You mean radius here, right? Otherwise, looks good to me.

Starwatcher16
Jun2-09, 07:18 AM
You mean radius here, right? Otherwise, looks good to me.

Yes, I did. Thanks for the verification.