Need help with maximum stress in an L-shaped Beam

[Trackster]

Hello!

I'm trying to design a 60" long 3" x 3" x. 375" hot rolled steel L-beam (aka angle bar) used to support 3 loads for use in a framing platform. The beam will have 3 loads of 500lbs acting on it (already including a safety factor of 1.5) at 9.33", 45.33", and 53" from the left hand side of the L-beam. Both ends will be welded to other frames. I'm concerned if this L-beam will be able to support the weight and my math is a little rusty, so I thought I'd run it by some more experienced people . My questions are this:

a) what's the best way to account for these 3 forces when calculating stress?
b) Are the Z values I'm using below correct for this shape of a bar?
c) Does the beam hold the weight?

Right now, I've taken all 3 forces and put them as a single 1500lbs point force in the middle thinking that if the beam can support his, then it should be able to support the 3 separate loads located closer to the ends (and thus, closer to the supports). I'm using the equation Stress = (W*L)/(4*Z), where

W = Load = 1500lbs
L = Bar Length = 60"
Z = Section Modulus for given L-Bar = 1.982738 or 0.83298 (I believe this is corrrect for a 3"x3"x.375" L-bar?)
E (not used) = 29000ksi

This gives 11726.21psi or 27911.83psi, depending on which Z was used. Wanting to be safe, I went with the larger stress value. According to MatWeb and my book, the yield stress of HRS steel is ~29700psi. This is below that but I recall that the stress should be 30-50% lower, which this is not. However, again, this is a combined point force located in the center of the beam.

I was thinking about how to account for the multiple point forces by using a form of superposition but wasn't sure if that was the wrong method.

Also, I would like to thank everyone in advance! It's incredibly nice of you all to go out of your way to help others

Chris

[Cake of Doom]

The max and min Z values are to help you determine whether failure will occur in the tension or compression face. Consulting your local standards for steel design should solve your woes for this one.

I would recommend going through the moment distribution exercises to get the actual bending stress your angle will experience (which is going to be a lot less than a central max applied force option).

[Cake of Doom]

In addition to the above; I'm pretty sure the ultimate yield stress should also be factored by 0.6 or 0.66 or something.

29700*0.66 = 19602 psi

Yeah. Have a good look at your codes or wait for an American to turn up.

[Trackster]

Thank you very much Cake of Doom! That helped me out a lot

[Kelly_Bramble]

Thank you very much Cake of Doom! That helped me out a lot

[Cake of Doom]

Glad you could find something useful in all my ramblings.