Statics Allowable load problem.

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SUMMARY

The discussion centers on calculating the allowable load on a steel tube with specific parameters: length (L) of 14 inches, cross-sectional area (A) of 11 square inches, Young's modulus (Es) of 30 x 10^6 psi, and a strain (δ) of 0.008. The user initially calculated the load (P) using the formula δ=PL/EA, resulting in P=188,571.4 lb, and then applied the formula θ=P/A, yielding P=264,000 lb. However, the instructor indicated the correct allowable load is P=2,181.8 lb, prompting confusion regarding the axial tensile load and the calculations performed.

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Homework Statement


Determine allowable load on steel tube

L= 14 in
A= 11 in ^2
Es= 30 x10^6
θ=24,000 psi
δ= .008



Homework Equations


δ=PL/EA
θ=P/A


The Attempt at a Solution



.008=P(14 in)/(30 x 10^6)(11 in)^2

P=188571.4 lb


θ=P/A
24,000 psi=P/11 in^2

P=264,000?

My teach is telling me that the answer is P=2181.8 lbs i don't understand how or what i did wrong
 
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I don't understand either if the load is axial tensile. Is it?
 

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