Elongation of non-uniform (area) bar.

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SUMMARY

The discussion focuses on the elongation of a non-uniform bar with varying diameters, specifically where one end has a radius R_1 and the other R_2, with R_2 > R_1. The equations provided include the elongation formula ΔL, derived from the relationship between pressure (P), Young's modulus (E), and the cross-sectional area (A_x) of the bar. The final expression for elongation is confirmed as ΔL = (P*L) / (π*E*R_1*R_2), validating the calculations and assumptions made regarding the bar's geometry and material properties.

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  • Understanding of material mechanics, specifically Young's modulus (E)
  • Familiarity with calculus, particularly integration techniques
  • Knowledge of geometric properties of circles, including area calculations
  • Basic principles of stress and strain in materials
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  • Explore advanced integration techniques for solving elongation problems
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Mechanical engineers, materials scientists, and students studying mechanics of materials who are interested in understanding the behavior of non-uniform bars under load.

Starwatcher16
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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?
 
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Starwatcher16 said:
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.
 
Mapes said:
You mean radius here, right? Otherwise, looks good to me.

Yes, I did. Thanks for the verification.
 

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