Okay skierbob, we have enough info to proceed. In each case the beam
supports 14' of the load. The rest being supported by the walls.
FIRST BEAM 20' SPAN:
The beam formulas for this beam loadings are:
M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
NOTE: Maximum deflection is limited to D = l/360 = 20 x 12 / 360 = 0.66 in
w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft
l (beam span) = 20 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 840 x 20^2 / 8 = 42,000 ft lb = 504,000 in lb
The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
= 19,800 psi
The section modulus of the required beam (S) = M/s = 504,000/19,800
= 25.45 in^3
Now we must calculate the required I (moment of inertia):
Solving for I in the above formula for deflection we get:
I = 5wl^4/384 ED = (5 x 840 x 20^4 / 384 x 30,000,000 x 0.66) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 152.73 in^4
It looks like a good choice for this beam would be:
W8x48 Depth = 8.5" Flange width = 8.117 Weight = 48 lb per ft
FIRST BEAM WITH CENTER SUPPORT (10 FT. SPANS):
The beam formulas for this beam loadings are:
M (maximum bending moment) = 49 wl^2/512
D (deflection @ center of span) = 0.0092 wl^4/EI
NOTE: Maximum deflection is limited to D = l/360 = 10 x 12 / 360 = 0.33 in
w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft
l (beam span) = 10 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 49 x 840 x 10^2 / 512 = 8,039 ft lb = 96,469 in lb
The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
= 19,800 psi
The section modulus of the required beam (S) = M/s = 96,469/19,800
= 4.87 in^3
Now we must calculate the required I (moment of inertia):
Solving for I in the above formula for deflection we get:
I = 0.0092 wl^4/ED = (0.0092 x 840 x 10^4 / 30,000,000 x 0.33) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 13.49 in^4
It looks like a good choice for this beam would be:
W5x16 Depth = 5.0" Flange width = 5.0 Weight = 16 lb per ft
SECOND BEAM:
The beam formulas for this beam loadings are:
M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
NOTE: Maximum deflection is limited to D = l/360 = 20 x 12 / 360 = 0.66 in
w (load per foot) = (10 + 5 + 20 + 10)psf x 14' = 630 lb per ft
l (beam span) = 20 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 630 x 20^2 / 8 = 31,500 ft lb = 378,000 in lb
The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
= 19,800 psi
The section modulus of the required beam (S) = M/s = 378,000/19,800
= 19.09 in^3
Now we must calculate the required I (moment of inertia):
Solving for I in the above formula for deflection we get:
I = 5wl^4/384 ED = (5 x 630 x 20^4 / 384 x 30,000,000 x 0.66) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 114.55 in^4
It looks like a good choice for this beam would be:
W8x35 Depth = 8.12" Flange width = 8.027 Weight = 35 lb per ft
If there is any of this you don't understand or you don't like the
beam choices, please ask for a clarification and I will try and
answer.
Good luck with your project, Redhoss |