# Elongation of non-uniform (area) bar.

1. Jun 1, 2009

### Starwatcher16

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?

Last edited: Jun 1, 2009
2. Jun 1, 2009

### Mapes

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

3. Jun 2, 2009

### Starwatcher16

Yes, I did. Thanks for the verification.