Finding elongation of bar and maximum tensile stress

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Discussion Overview

The discussion revolves around calculating the elongation of a bar and determining the maximum tensile stress in a structural problem involving three sections of a bar under tension. Participants are working through a homework problem that includes applying relevant equations and concepts from mechanics of materials.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the equations for stress and strain, calculating elongation based on given parameters.
  • Another participant suggests not to add the stresses from different sections of the bar when determining maximum tensile stress.
  • There is confusion about whether to consider only the stress from one section or the entire bar when calculating maximum tensile stress.
  • A participant acknowledges a mistake in their calculations and revises their value for maximum tensile stress.
  • Another participant confirms that the maximum tensile stress occurs in section AB and supports the idea of not adding stresses from other sections.

Areas of Agreement / Disagreement

Participants generally agree that the maximum tensile stress occurs at section AB and should not involve adding stresses from other sections. However, there is some uncertainty regarding the calculations and the approach to determining maximum stress.

Contextual Notes

There are unresolved aspects related to the calculations of stress and elongation, including potential errors in earlier steps and the dependence on specific assumptions about the bar's behavior under load.

Who May Find This Useful

Students and practitioners interested in mechanics of materials, structural engineering, and those working on similar homework problems may find this discussion relevant.

Blugga
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Homework Statement



wk1xkz.jpg

L=52 in
A=2.76 in^2
E=10.4*10^6 psi

Homework Equations



σ=F/A
ε=σ/E
δ=εL

The Attempt at a Solution



4) σAB = (3P)/A
ε=(3P)/(AE)
δAB=(3PL)/(6AE) → δAB=(PL)/(2AE)
solving for P
P=[0.17*2*2.76*(10.4*106)]/52 → P=187680 lb → P=187.7 kip

5) Because AB and CD are in tension i did this...
σmaxABCD
σmax=[(-2P)/A]+[P/A]
solving for P and using 5000psi for σmax i get
P=-5000*2.76 → P=13800 lb → P=13.8kip

I tried looking for an example in the book to follow, but they were completely different. I hope i didn't mess up too bad.
 
Last edited:
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Don't add the stresses
 
afreiden said:
Don't add the stresses

So i should only do one of them? σmaxAB
or do the entire bar?
Can I get a better hint than that?
 
I confirmed that I did part 4 right. I still need help with part 5. Anyone?
 
I made a mistake on part 5. I plugged in the value of σBC (-2P/A) in place for σAB (3P/A) in σmaxABCD

Now i get σmax=3450 lb or 3.45 kip. But still don't know if it's right.
 
Part 5:

You correctly determined the stresses in the 3 sections of the bar.

AB = 3P/A
BC = -2P/A
CD = P/A

So, you already know that the maximum tensile stress in the bar is 3P/A. This cannot exceed 5000 psi = 5 ksi

chet
 
So what I'm getting from this is that the maximum tension occurs at AB so I only set σmax=σAB and don't add them with the other member in tension.

Thanks :)
 
Blugga said:
So what I'm getting from this is that the maximum tension occurs at AB so I only set σmax=σAB and don't add them with the other member in tension.

Thanks :)

Yes. That's right. The other sections will be less prone to failure.
 

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