Handbook of Civil Engineering Calculations

FIGURE 27

her thus functions as a beam that overhangs one support. However, since the footing is
considerably wider than the columns, there is a transverse bending as well as longitudinal
bending in the vicinity of the columns. For simplicity, assume that the transverse bending
is confined to the regions bounded by planes AB and EF and by planes GH and NP, the
distance m being h/2 or d/2, whichever is smaller.
In Fig. 260, let Z denote the location of the resultant of the column loads. Then x =
370(16)7(250 + 370) - 9.55 ft (2.910 m). Since Z is to be the centroid of the footing, L =
2(0.75 + 9.55) = 20.60 ft (6.278 m). Set L = 20 ft 8 in (6.299 m), but use the value 20.60 ft
(6.278 m) in the stress calculations.

2. Construct the shear and bending-moment diagrams
   The net soil pressure per foot of length = 620/20.60 = 30.1 kips/lin ft (439.28 kN/m).
   Construct the diagrams as shown in Fig. 26.
   3. Establish the footing thickness
   Use

(Pv 2 + OMVL + Pp')d-Q.\7Pd^2 - VLp' (50)

where P = aggregate column load, kips (kN); V = maximum vertical shear at a column
face, kips (kN);/?' = gross soil pressure, kips/ft^2 (MPa).
Assume that the longitudinal steel is centered 3/4 in (88.9 mm) from the face of the
footing. Then P = 620 kips (2757.8 kN); V= 229.2 kips (1019.48 kN); V 2 = 0.06(144) =
8.64 kips/ft^2 (0.414 MPa); 926Od- 105Ad^2 = 23,608; d = 2.63 ft (0.801 m); t = 2.63 +
0.29 = 2.92 ft. Set t = 2 ft 11 in (0.889 m); d = 2 ft 71 X 2 in (0.800 m).