Elongation of non-uniform (area) bar.

[Starwatcher16]

Lets say I have a bar of uniform material where one end has a diameter given by [tex]R_1[/tex] and another end given by [tex]R_2[/tex]. [tex]R_2 > R_1[/tex], R(x) is linear.

So I know now three equations:

A)[tex] s=\frac{E \Delta L}{L} [/tex]

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

C)[tex]A_x=\pi*R_x^2[/tex]

Therefore, I know:

[tex]\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}[/tex],

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

[tex] 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}][/tex]

[tex]=\frac{P*L}{\pi*E*R_1*R_2}[/tex]

Is that right?

 

[Mapes]

Lets say I have a bar of uniform material where one end has a diameter given by [tex]R_1[/tex]

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

 

[Starwatcher16]

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

Yes, I did. Thanks for the verification.