- Compute the vertical shear at distance d from the column face
Establish the width of the footing. Thus V= 229.2 - 2.63(30.1) - 150.0 kips (667.2 kN);
v = VI(Wd), or W= VI(Vd) = 150/[8.64(2.63)] - 6.60 ft (2.012 m). Set W= 6 ft 8 in
(2.032 m). - Check the soil pressure
The footing weight - 20.67(6.67)(2.92)(0.150) = 60.4 kips (268.66 kN); p' = (620 +
60.4)/[(20.67)(6.67)] - 4.94 kips/ft^2 (0.236 MPa) < 5 kips/ft^2 (0.239 MPa). This is accept-
able. - Check the punching shear
Thus,/? = 4.94 - 2.92(0.150) = 4.50 kips/ft^2 (0.215 MPa). At Cl: b 0 = 18 + 31.5 + 2 (18 +
15.8) = 117 in (2971.8 mm); V= 250 - 4.50(49.5)(33.8)/144 = 198 kips (880.7 kN); V 1 =
198,000/[117(31.5)] = 54 lb/in^2 (372.3 kPa) < 110 lb/in^2 (758.5 kPa); this is acceptable.
At C2: b 0 = 4(20 + 31.5) - 206 in (5232.4 mm); V= 370 - 4.50(51.5)^2 /144 = 287 kips
(1276.6 kN); V 1 = 287,000/[206(31.5)] = 44 lb/in^2 (303.4 kPa). This is acceptable. - Design the longitudinal reinforcement for negative moment
Thus, M= 851,400 ft-lb = 10,217,000 in-lb (1,154,316.6 N-m);'Mb = 223(80)(31.5)^2 =
17,700,000 in-lb (1,999,746.0 N-m). Therefore, the steel is stressed to capacity, and As =
10,217,000/[20,000(0.874)(31.5)] - 18.6 in^2 (120.01 cm^2 ). Try 15 no. 10 bars with A 3 =
19.1 in^2 (123.2 cm^2 ); Zo = 59.9 in (1521.46 mm).
The bond stress is maximum at the point of contraflexure, where V= 15.81(30.1) -
250 = 225.9 kips (1004.80 kN); u = 225,900/[59.9(0.874)(31.5)] = 137 lb/in^2 (944.6 kPa);
"allow = 3.4(3000)°^5 /l.25 = 149 lb/in^2 (1027.4 kPa). This is acceptable. - Design the longitudinal reinforcement for positive moment
For simplicity, design for the maximum moment rather than the moment at the face of the
column. Then A 5 = 158,400(12)/[20,000(0.874)(31.5)] = 3.45 in^2 (22.259 cm^2 ). Try six
no. 7 bars with As = 3.60 in^2 (23.227 cm^2 ); So = 16.5 in (419.10 mm). Take LM as the
critical section for bond, and u = 90,800/[16.5(0.874)(31.5)] = 200 lb/in^2 (1379.0 kPa);
"allow = 4.8(3000)°-^5 /0.875 - 302 lb/in^2 (2082.3 kPa). This is acceptable. - Design the transverse reinforcement under the interior column
For this purpose, consider member GNPH as an independent isolated footing. Then VST =
370(2.50/6.67) = 138.8 kips (617.38 kN); MST =^1 / 2 (138.8)(2.50)(12) = 2082 hvkips
(235.22 kN-m). Assume d = 35 - 4.5 = 30.5 in (774.7 mm); A 8 = 2,082,000/
[20,000(0.874)(30.5)] = 3.91 in^2 (25.227 cm^2 ). Try seven no. 7 bars; A 5 = 4.20 in^2
(270.098 cm^2 ); Zo = 19.2 in (487.68 mm); u = 138,800/[19.2(0.874)(30.5)] = 271 lb/in^2
(1868.5 kPa); t/allow = 302 lb/in^2 (2082.3 kPa). This is acceptable.
Since the critical section for shear falls outside the footing, shearing stress is not a cri-
terion in this design. - Design the transverse reinforcement under the exterior
column; disregard eccentricity
Thus, Vuv = 250(2.58/6.67) = 96.8 kips (430.57 kN); M 1 Jv=^1 / 2 (96.8)(2.58)(12) = 1498
in-kips (169.3 kN-m); As = 2.72 in^2 (17.549 cm^2 ). Try five no. 7 bars; A 5 = 3.00 in^2
(19.356 cm^2 ); ^o - 13.7 in (347.98 mm); u = 96,800/[13.7 (0.874)(31.5)] - 257 lb/in^2
(1772.0 kPa). This is acceptable.
Cantilever Retaining Walls
Retaining walls having a height ranging from 10 to 20 ft (3.0 to 6.1 m) are generally built
as reinforced-concrete cantilever members. As shown in Fig. 28, a cantilever wall com-