
#1
Nov1909, 03:09 PM

P: 8

I'm trying to calculate the max load of a rack made of 3X3X1/4 steel tube. It is 8ft long and 4ft wide. I just want to calculate the max load (distributed load) in bending (prob for just the 8ft sides) and also in compression for the legs. i can't find the right formulas. can anyone help?
so far i have MOI (I)=3.495 in^4 Section modulus=2.330 in^3 MOE (E)=29,000,000 psi ( i think this is correct but i'm not sure. just want to use low grade steel) 



#2
Nov1909, 03:18 PM

Sci Advisor
PF Gold
P: 2,234

You just need to calculate the bending stress in the beam. This page should set you on the right track: http://en.wikipedia.org/wiki/Bending




#3
Nov1909, 03:47 PM

P: 8

I'm not sure if that is what i'm looking for. I need to know how much weight i can put on this rack (distributed load). i don't know if that that be yield strength or what??? i want to assume fix ends on an 8ft square tube (actually 2 8ft tubes parallel).




#4
Nov2009, 07:34 AM

Sci Advisor
P: 1,498

Calculating max load of square tube (steel)
No, that's what you're looking for. It's a textbook problem really. Calculate the maximum bending moment in your beam. Then apply the bending stress equation to find the maximum stress.
Via http://www.engineeringtoolbox.com/be...ond_1312.html the maximum stress in a simply supported distributed loaded beam is: [tex] \sigma = \frac{y q L^2}{8 I} [/tex] Where y is the perpendicular distance from the neutral axis, q is the load in force/length, L is the length of the beam, and I is the area moment of inertia of the cross section. Be careful though, depending on the application. Yield strength is the strength assuming one cycle. If this beam is to be loaded and unloaded, then you should calculate a fatigue life, and then still use a safety factor. Good luck, 



#5
Nov2009, 07:45 AM

P: 8

but what i want to know it q. right?
would be 1.5? 



#6
Nov2009, 07:51 AM

Sci Advisor
P: 1,498

Yes, find an allowable maximum stress and solve for q.




#7
Dec609, 07:24 PM

P: 1

just as an addition to the topic.....don't confuse the moment of inertia given with the body moment of inerita.....this is the area moment of inertia whose unit in SI units system is bisqaure meter( m ^ 4)....the other moment of inertia of the body whose unit in the SI unit is (N.m) isn't applicable here



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