FIGURE 14. Grillage under column.mum bending moment occurs at the center of the span; its value is M = P(A - a)/8 =fS;
therefore, O 1 =A- SfSIP.
At the toe of the fillet, the load P is distributed across a distance a + 2k. Then/^ = Pl(a
+ 2k)tw; therefore, a 2 = P/fyw - 2k. Try four beams; then P = 2790/4 = 697.5 kips (3102.5
kN);/= 24 kips/in^2 (165.5 MPa); fb = 27 kips/in^2 (186.1 MPa). Upon substitution, the
foregoing equations reduce to a} = 60 - 0.2755; a 2 = 25.8/*w - 2k.
Select the trial beam sizes shown in the accompanying table, and calculate the corre-
sponding values of Or 1 and a 2.Size S, in^3 (cm^3 ) tw, in (mm) k, in (mm) ^ 1 , in (mm) a 2 , in (mm)
818x54.7 88.4(1448.6) 0.460(11.68) 1.375(34.93) 35.7(906.8) 53.3(1353.8)
S18 x 70 101.9 (1669.8) 0.711 (18.06) 1.375 (34.93) 32.0 (812.8) 33.6 (853.4)
820x65.4 116.9(1915.7) 0.500(12.7) 1.563(39.70) 27.9(708.7) 48.5(1231.9)
820 x 75 126.3 (2069.7) 0.641 (16.28) 1.563 (39.70) 25.3 (642.6) 37.1 (942.3)Try S18 x 70, with a = 34 in (863.6 mm). The flange width is 6.25 in (158.8 mm). The
maximum vertical shear occurs at the edge of the plate; its magnitude is V = P(A - d)(2A)
= 697.5(60 - 34)/[2(60)] = 151.1 kips (672.1 kN); v = 151.1/[18(0.71I)] = 11.8 < 14.5
kips/in
2
(99.9 MPa), which is acceptable.
- Design the base plate
Refer to the second previous calculation procedure. To permit the deposition of concrete,
allow a minimum space of 2 in (50.8 mm) between the beam flanges. The minimum value
of b is therefore b = 4(6.25) + 3(2) = 31 in (787.4 mm).
The dimensions of the effective bearing area under the column are 0.95(16.81 + 2 x
- = 18.82 in (478.0 mm); 0.80(20) = 16 in (406.4 mm). The projection of the plate are
(34 - 18.82)/2 = 7.59 in (192.8 mm); (31 - 16)/2 = 7.5 in (190.5 mm).
Therefore, keep 6 = 31 in (787 mm), because this results in a well-proportioned plate.
Separators