Sign Discrepancy in Plane Stress Transformation Solutions?

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SUMMARY

The discussion focuses on a sign discrepancy encountered in plane stress transformation solutions, specifically regarding the force balance in the x and y directions. The participant reports achieving correct magnitudes for the forces but incorrect signs, suspecting an issue with the shear stress direction on their free body diagram. The stress tensor is defined as ##\vec{\sigma}=\frac{P}{(0.05)(0.08)}\vec{i}_x\vec{i}_x=250 P\vec{i}_x\vec{i}_x## Pa, leading to critical load calculations of $$P<3896\ N$$ and $$P<6266\ N$$, with the critical load determined to be 3896 N, aligning with results from the BEAST tool.

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theBEAST
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Homework Statement


Here is the problem with my attempt at the solution:
8Ezm0.jpg


The magnitude of my answers are correct, HOWEVER I am getting the wrong signs. For the force balance in the x direction I get a negative P but for the force balance in the y direction I get a positive P. Does anyone know why?

I was thinking it was because I have the wrong direction for shear stress on my free body diagram... But I thought this wouldn't matter since the math will always work itself out?
 
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The stress tensor in the members is ##\vec{\sigma}=\frac{P}{(0.05)(0.08)}\vec{i}_x\vec{i}_x=250 P\vec{i}_x\vec{i}_x## Pa. The unit normal to the joint is ##\cos {25}\vec{i}_x+\sin{25}i_y##. From the Cauchy stress relationship, the stress vector acting on the joint is ##250 P\cos {25}\vec{i}_x##. The component of this stress vector normal to the joint is ##250 P\cos^2 {25}##. The unit tangent to the joint is ##\cos{25} \vec{i}y-\sin{25}\vec{i_x}##. The component of the stress vector tangent to the joint is ##250 P\sin{25}\cos {25}##. So, for the joint not to fail,
$$250 P \cos^2 {25}< 800000$$and $$250 P\sin{25}\cos {25}<600000$$So,$$P<3896\ N$$and $$P<6266\ N$$So the critical load is 3896 N.

This result is basically the same as the result obtained by theBEAST, and thus confirms his answer.
 

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