Strength of materials- simple truss problem

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Jeff231
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Homework Statement



The bars in the truss each have a cross sectional area of 1.2in^2. If the maximum axial stress in any bar is not to exceed 25ksi, determine the maximum magnitude of P of the loads that can be applied to the truss. Determine the elongation of each member.

E=29,000 ksi


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Homework Equations



[tex] \sigma[/tex] = E* [tex]\epsilon[/tex]

[tex]\sigma[/tex] = P/A


[tex]\epsilon[/tex] = [tex]\delta[/tex] / L

Having trouble with Latex, but the top equation should read:
sigma=E*epsilon

the other two equations are stress=force/area and strain=elongation/lenth.

The Attempt at a Solution



I got an answer, but I'm not sure if it's correct and if I'm solving this correctly.

Since I know [tex]\sigma[/tex]=25ksi, and E=29,000 ksi I used [tex]\sigma[/tex]=P/A solving for P, P= [tex]\sigma[/tex]A=(25ksi)(1.2in^2)=30 kip.

I assumed this to be Pmax.

Using [tex]\sigma[/tex]=E*epsilon I solved for epsilon. epsilon= [tex]\sigma[/tex]/E=25ksi/29000ksi = 8.62E-4.

Then used epsilon=delta/L. Solved for elongation, delta. delta=epsilon*L.

Then I just used plugged in my solved value of epsilon, and then the length of each bar.

Here's an example for AC: delta=epsilon*L=(8.62E-4)(4ft)(12in/1ft)=0.0414in.


I'm not sure if this correct since it would mean every bar that is the same length would have the same elongation. Is this correct? Or do I need to solve the system for the forces and then do something? I'm confused. Thank you for the help.
 
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Axial stress is generated by a force perpendicular to the cross section, what is the direction of P in the diagram ?
 
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