# How to calculate the amount an object can take from PSI

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?

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.

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 $$S_p$$

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