It's more like engineering, but whatever. Essentially, I'm doing a wee bit of studying of tensile strength and tensile stress. I figured you guys would know something useful about this, since I've never studied it until now.

The formula I've found for tensile stress is sigma = P/A where P is tensile force and A is the cross-section area. Where I'm hitting a wall is figuring out how to come up with the tensile force P.

Ejukate me!

P is usually the applied force so it is generally given in the problem, or solvable using the engineer's best friend: the free body diagram. On the other hand, sometimes you know the material being used, cross-section, and safety factor and you can work backwards to find the maximum appliable force.

What problem are you trying to solve?

Ah, see, I was thinking that P was some specific force that had a different formula as well.

I'm not trying to work out a problem, I'm trying to understand so I can apply it in some sort of demonstration.

Blow up a model bridge. That'll keep your students entertained.

I'm so glad people here are smart... lol I dont consider myself stupid, but I'm not a genius... unfortunately physics isnt my strong suite.

Fizziks? Hawt dang, you could finish Sith2 for us.

I couldn't even finish the series I started for MotS.

Another question about the sigma = P/A equation. If I understand this correctly, then the closer sigma gets to 1.0, the weaker the material, and values > 1.0 means it's busted right?

No because in SI units, sigma is measured in Pascals and many materials have yield stresses in the order of mega if not giga Pascals.

sigma is the stress applied on the part at that cross section. It is not a measure of when a part will break. Generally speaking, all parts of the same cross section with the same force applied on them will have the same stress on them.

sigma = P/A

P units are lbs (english) or N (SI)
A units are in^2 (english) or m^2 (SI)
Therefore
sigma units are lbs/in^2 aka psi (english) or N/m^2 aka Pascals (SI)

If you are using this in a demonstration, remember that this equation ONLY applies when the part is in tension like so...

P <-- |PART| --> P

To determine when a part will break, you need to look up the tensile yield strength and the ultimate tensile strength of the material in the part.

The tensile yield strength is the applied stress at which the part will begin to plastically deform (ie change shape or deflect and not return to its orginal form when the load is removed).

Ultimate tensile strength is the applied stress at which the part will fracture (aka break or fail catastrophically )

The Safety Factor (or Factor of Safety) for tensile strength is simply the ratio of yield strength over applied stress. If this number is below 1, the part should deform, above 1 and it should survive. (This is the simplified version; there are some other factors that may be needed depending on the geometry and material of the part that are to complex to describe here.)

You should be able to look up a lot of specific properties online, but as a general reference some typical values for yield strength are:

Lexan: 9,000 psi (what Nalgene water bottles are made of)
Commercial Aluminum: 5,000-22,000 psi depending on alloy
Aircraft Aluminum: 8,000-73,000 psi depending on alloy
Carbon Steel: 26,000-118,000 psi depending on carbon content
Titanium: 30,000-170,000 psi depending on alloy
Stainless Steel: 35,000-275,000 psi depending on alloy

Gah! Pounds of force are American not english!! All of Europe uses SI and loves it!

(But otherwise spot on)

Structural Steel comes in varying yield strengths from 275 up to about 390 (N/mm^2)