Strength of materials- simple truss problem

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SUMMARY

The discussion focuses on solving a simple truss problem involving axial stress and elongation of truss members. Given a cross-sectional area of 1.2 in² and a maximum axial stress of 25 ksi, the maximum load (P) that can be applied to the truss is calculated to be 30 kip using the formula P = σA. The modulus of elasticity (E) is specified as 29,000 ksi, leading to a calculated strain (ε) of 8.62E-4. The elongation (δ) for each member is determined to be 0.0414 inches, assuming uniform length across members.

PREREQUISITES
  • Understanding of axial stress and strain concepts
  • Familiarity with the equations of state: σ = P/A and ε = δ/L
  • Knowledge of material properties, specifically modulus of elasticity (E)
  • Basic principles of truss analysis and member forces
NEXT STEPS
  • Study the method of joints for analyzing forces in truss members
  • Learn about the effects of different loading conditions on truss stability
  • Explore advanced topics in structural analysis, such as the finite element method
  • Investigate the impact of varying cross-sectional areas on stress distribution
USEFUL FOR

This discussion is beneficial for civil engineers, structural engineers, and students studying mechanics of materials or structural analysis, particularly those working with truss systems.

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


4930827627_46098ef021.jpg


Homework Equations



<br /> \sigma = E* \epsilon

\sigma = P/A


\epsilon = \delta / 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 \sigma=25ksi, and E=29,000 ksi I used \sigma=P/A solving for P, P= \sigmaA=(25ksi)(1.2in^2)=30 kip.

I assumed this to be Pmax.

Using \sigma=E*epsilon I solved for epsilon. epsilon= \sigma/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.
 
Physics news on Phys.org
Axial stress is generated by a force perpendicular to the cross section, what is the direction of P in the diagram ?
 
Can you determine the force in each member of the truss, as a function of P?
 

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