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I'm pretty sure this is a very straight forward question. But here goes.

So 6061 Aluminum has a yield point of 37,600 PSI. Which is Lbs force per square inch. How do I determine the yield point if I have .25" x .25" piece of 6061 Aluminum?

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Are you asking what is the yield force/load on a 1/4 inch squares bar ? If so, the bar is 0.0625 square inch X 37,600 yield = 2350 pounds. With a few assumptions like the aluminum has homogenous properties. And there is no affect of grips. And axial load with no bending.

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It will yield (T or C) under the axial force, $Fy$ = $f_y$ x $A$, and

the extreme fibers (T & C) will reach yield stress under the flexural moment, $My$ = $f_y$ x $S_x$, or

the entire cross-section will yield (T & C) under the plastic moment $M_p$ = $f_y$ x $S_p$

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Stress is defined as:

$$\sigma = \frac{F}{A}$$

where:

  • A is the cross-section.
  • F is the force on the cross-sectional area

Solving for F you can obtain:

$$F = \sigma\cdot A$$

In the case that you are considering:

  • $\sigma =\sigma_y = 37600$ PSI the stress at yield
  • $A= 0.0625 \;in^2$

Therefore, $ F = 2350 \text{ lb}$

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