Finding elongation of bar and maximum tensile stress

[Blugga]

Homework Statement

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) σ_AB = (3P)/A
ε=(3P)/(AE)
δ_AB=(3PL)/(6AE) → δ_AB=(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...
σ_max=σ_AB+σ_CD
σ_max=[(-2P)/A]+[P/A]
solving for P and using 5000psi for σ_max 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.

 

[afreiden]

Don't add the stresses

 

[Blugga]

Don't add the stresses

So i should only do one of them? σ_max=σ_AB
or do the entire bar?
Can I get a better hint than that?

 

[Blugga]

I confirmed that I did part 4 right. I still need help with part 5. Anyone?

 

[Blugga]

I made a mistake on part 5. I plugged in the value of σ_BC (-2P/A) in place for σ_AB (3P/A) in σ_max=σ_AB+σ_CD

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

 

[Chestermiller]

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

 

[Blugga]

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 :)

 

[Chestermiller]

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.