Hello J. R., you have done the hardest part of the calculations by
providing the load applied to the beam. I can help you with the rest.
The beam formulas for this loading 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 = 14 x 12 / 360 = 0.46 in
w (load per foot) = 1200 lb per ft
l (beam span) = 14 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 1200 x 14^2 / 8 = 29,400 ft lb = 352,800 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 = 352,800/19,800
= 17.8 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 1200 x 14^4 / 384 x 30,000,000 x 0.46) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 75.2 in^4
Now we can look for a beam with these minimum properties and a maximum
depth of 12 inches.
A good choice would be a 10.25 inch deep Wide Flange beam weighing 19 #
per ft (W10x19)
S = 18.8 in^3
I = 96.3 in^4
Of course, you can use a heavier beam if there is one more available.
Please ask for a clarification if there is anything you don't
understand and I will try to explain.
Good luck with your kitchen project, Redhoss