Okay josephc68, I guess you don't want to tell me how you got your
numbers. We will just use your numbers and go ahead.
The maximum bending moment is:
M = Rx x L/2 = 6473 x 89.75 = 580,952 inlbs
The required section modulus is:
S = M/allowable stress = 580,952/(36,000 x .55) = 29.34 in^3
The allowable deflection is L/360 = 179.5/360 = 0.5 in
Using this deflection to calculate the required moment of inertia:
Where P = 2 x Rx = 12,946 lb
E is the modulus of elasticity of steel = 30,000,000 psi
I = PL^3/48ED = 12,946 (179.5^3)/48 (30,000,000)(0.5) = 104 in^4
The properties for a W6x25 are:
S = 16.7 in^3
I = 53.3 in^4
So, you see that the W6x25 is not adequate. It looks like your best
choice would be a W8x40 with S = 35.5 and I = 146.
Look this over and if you have something else you would like to try
please ask for a clarification.
Redhoss 
Clarification of Answer by
redhossga
on
30 Oct 2006 14:03 PST
I wish that you had started with the info that you state in your
clarification request. If you take your new numbers here is what we
have:
w (load in lbs. per foot) = 42 psf x 10 = 420 lb/ft
So, the end reactions would be:
Rx = 420 x 14.958 / 2 = 3,141 lbs.
This is less that half the 6,473 number that your "engineered lumber
program" gave you. I don't know what went wrong, but it yielded some
bogus numbers. Here is a recalculation using the correct numbers:
The maximum bending moment is:
M = Rx x L/2 = 3,141 x 89.75 = 281,905 inlbs
The required section modulus is:
S = M/allowable stress = 281,905/(36,000 x .55) = 14.24 in^3
The allowable deflection is L/360 = 179.5/360 = 0.5 in
Using this deflection to calculate the required moment of inertia:
Where P = 2 x Rx = 6,282 lb
E is the modulus of elasticity of steel = 30,000,000 psi
I = PL^3/48ED = 6,282 (179.5^3)/48 (30,000,000)(0.5) = 50.46 in^4
Now the W6x25 looks just fine. Good choice.
