Aluminum I beam size required
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Re: Aluminum I beam size required
Re: Aluminum I beam size required -- KaelFarmer	Post Reply	Top of thread	Forum

Posted by: acroduster1 ® 06/03/2004, 09:15:27 Author Profile Mail author Edit

I'll give you the basic equations to work with because the results heavily depend on what aluminum alloy you use. For strength: sigma = M/S = w*l^2/(8*S) = 3000/(10*12)*(10*12)^2/(8*S) = 45000/S where S is the section modulus of the beam you use in in^3, sigma is the stress on the beam in psi, M is the moment in the beam, w is the distributed load on the beam (in this case 3000lb/10ft or 300lb/ft, 25lb/in), and l is the length of the span. The sigma that you get needs to be less than the allowable yield stress of the aluminum alloy you pick which will be on the order of 60% of the actual yield stress. For deflection: delta = 5*w*l^4/(384*E*I) = 5*3000/(10*12)*(10*12)^4/(384*10000000*I)         = 6.75/I where E is the modulus of elasticity for aluminum (taken to be 10,000,000 psi - check against the alloy you get), I is the moment of inertia of the beam you use in in^4, and w and l as described above. You want the deflection to be pretty low, around l/240 to l/360 to be comfortable and not very noticeable.  This translates to 10*12/240 = 0.5" to 10*12/360 = 0.33" in the middle of the span under full load.  Set this equal to the 6.75/I to see if the I you have is good enough. Please note that these formulas are for an evenly distributed load.  If the load is concentrated at the center, the stresses and deflections get much worse... Let me know how it turns out. Acro

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