tion for V 1 , ft (m); F 1 and V 2 = vertical she'ar at critical section for stresses V 1 and V 2 , re-
spectively.
In accordance with the ACI Code, the critical section for V 1 is the surface GHJK, the
sides of which lie at a distance d/2 from the column faces. The critical section for V 2 is
plane LM, located at a distance d from the face of the column. The critical section for
bending stress and bond stress is plane EF through the face of the column. In calculating
V 2 , f, and u, no allowance is made for the effects of the orthogonal reinforcement.
DESIGN OF AN ISOLATED
SQUARE FOOTING
A 20-in (508-mm) square tied column reinforced with eight no. 9 bars carries a concentric
load of 380 kips (1690.2 kN). Design a square footing by the working-stress method us-
ing these values: the allowable soil pressure is 7000 lb/ft^2 (335.2 kPa);/c' = 3000 lb/in^2
(20,685 kPa); andfs - 20,000 lb/in^2 (137,900 kPa).
Calculation Procedure:- Record the allowable shear, bond, and bearing stresses
From the ACI Code table, V 1 = 110 lb/in^2 (758.5 kPa); V 2 = 60 lb/in^2 (413.7 kPa);/^ =
1125 lb/in^2 (7756.9 kPa); u = 4.8(/c')°-^5 /bar diameter - 264/bar diameter. - Check the bearing pressure on the footing
ThusX - 380/[20(2O)] = 0.95 kips/in^2 (7.258 MPa) < 1.125 kips/in^2 (7.7568 MPa). This
is acceptable. - Establish the length of footing
For this purpose, assume the footing weight is 6 percent of the column load. Then A =
1.06(380)77 = 57.5 ft^2 (5.34 in^2 ). Make L = 1 ft 8 in - 7.67 ft (2.338 m); A = 58.8 ft^2 (5.46
m^2 ). - Determine the effective depth as controlled by V 1
Apply
(4V 1 + p)f + h(4v{ + 2p)d = p(A - h^2 ) (49)Verify the result after applying this equation. Thus p = 380/58.8 = 6.46 kips/ft
2
(0.309
MPa); = 0.11(144) = 15.84 kips/ft^2 (0.758 MPa); 69.Sc/^2 + UlAd = 361.8; d =• 1.54 ft
(0.469 m). Checking in Fig. 24, we see GH = 1.67 + 1.54 = 3.21 ft (0.978 m); V 1 =
6.46(58.8 - 3.21
2
) = 313 kips (1392.2 kN); V 1 = V 1 I(^d) = 313/[4(3.2I)(1.54)] = 15.83
kips/ft^2 (0.758 MPa). This is acceptable.
- Establish the thickness and true depth of footing
Compare the weight of the footing with the assumed weight. Allowing 3 in (76.2 mm) for
insulation and assuming the use of no. 8 bars, we see that t = d + 4.5 in (114.3 mm). Then
t = 1.54(12) + 4.5 = 23.0 in (584.2 mm). Make t = 24 in (609.6 mm); d = 19.5 in = 1.63 ft
(0.496 m). The footing weight = 58.8(2)(0.150) = 17.64 kips (1384.082 kN). The as-
sumed weight = 0.06(380) = 22.8 kips (101.41 kN). This is acceptable. - Check V 2
In Fig. 24, AL = (7.67 - 1.67)72 - 1.63 = 1.37 ft (0.417 m); V 2 = 380(1.37/7.67) =