The area A 1 , in 2 (mm 2 ), required for a base plate under a column supported by concrete should be taken as the larger of the values calculated from the equation cited earlier, with R taken as the total column load, kip (kN), or
A 1 = R / 0.70 f ‘ c
Unless the projections of the plate beyond the column are small, the plate may be designed as a cantilever assumed to be fixed at the edges of a rectangle with sides equal to 0.80b and 0.95d, where b is the column flange width, in (mm), and d is the column depth, in (mm).
To minimize material requirements, the plate projections should be nearly equal. For this purpose, the plate length N, in (mm) (in the direction of d), may be taken as
N = ( A 1 ) ½ + 0.5 ( 0.95d – 0.80b )
The width B, in (mm), of the plate then may be calculated by dividing A1 by N. Both B and N may be selected in full inches (millimeters) so that BN ³ A 1 . In that case, the bearing pressure f p , ksi (MPa), may be determined from the preceding equation. Thickness of plate, determined by cantilever bending, is given by
t = 2p (f p / F y ) ½
where F y = minimum specified yield strength, ksi (MPa), of plate; and p = larger of 0.5 (N – 0.95d) and 0.5 ( B – 0.80b).
When the plate projections are small, the area A 2 should be taken as the maximum area of the portion of the supporting surface that is geometrically similar to and concentric with the loaded area. Thus, for an H-shaped column, the column load may be assumed distributed to the concrete over an H-shaped area with flange thickness L, in (mm), and web thickness 2L:
L = ( 1 / 4 ) (d + b) – 1 / 4 ( (d + b) 2 – 4R / F P ) ½
where F p = allowable bearing pressure, ksi (MPa), on support. (If L is an imaginary number, the loaded portion of the supporting surface may be assumed rectangular as discussed earlier.) Thickness of the base plate should be taken as the larger
of the values calculated from the preceding equation and
t = L (¬3 f p / F b ) ½
