Page 261 - Handbook of Civil Engineering Calculations, Second Edition

2.46 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

4. 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 V/(Wd), or W V/(vd) 150/[8.64(2.63)] 6.60 ft (2.012 m). Set W
6 ft 8 in.(2.032 m).
5. 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/sq.ft. (0.236 MPa) < 5 kips/sq.ft. (0.239 MPa). This is
acceptable.
6. Check the punching shear
Thus, p 4.94 2.92(0.150) 4.50 kips/sq.ft. (0.215 MPa). At C1: b o 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/sq.in. (372.3 kPa) < 110 lb/sq.in. (758.5
kPa); this is acceptable.
2
At C2: b o 4(20 31.5) 206 in. (5232.4 mm); V 370 4.50(51.5) /144 287
kips (1276.6 kN); v 1 287,000/[206(31.5)] 44 lb/sq.in. (303.4 kPa). This is acceptable.
7. Design the longitudinal reinforcement for negative moment
2
Thus, M 851,400 ft·lb 10,217,000 in.·lb (1,154,316.6 N·m); M b 223(80)(31.5)
17,700,000 in.·lb (1,999,746.0 N·m). Therefore, the steel is stressed to capacity, and A s
2
10,217,000/[20,000(0.874)(31.5)] 18.6 sq.in. (120.01 cm ). Try 15 no. 10 bars with A s
2
19.1 sq.in. (123.2 cm ); o 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/sq.in. (944.6
0.5
kPa); u allow 3.4(3000) /1.25 149 lb/sq.in. (1027.4 kPa). This is acceptable.
8. Design the longitudinal reinforcement for positive moment
For simplicity, design for the maximum moment rather than the moment at the face of the
2
column. Then A s 158,400(12)/[20,000(0.874)(31.5)] 3.45 sq.in. (22.259 cm ). Try
2
six no. 7 bars with A s 3.60 sq.in. (23.227 cm ); o 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/sq.in. (1379.0
0.5
kPa); u allow 4.8(3000) /0.875 302 lb/sq.in. (2082.3 kPa). This is acceptable.
9. Design the transverse reinforcement under the interior column
For this purpose, consider member GNPH as an independent isolated footing. Then V ST
1
370(2.50/6.67) 138.8 kips (617.38 kN); M ST /2(138.8)(2.50)(12) 2082 in.·kips
(235.22 kN·m). Assume d 35 4.5 30.5 in. (774.7 mm); A s 2,082,000/
2
[20,000(0.874)(30.5)] 3.91 sq.in. (25.227 cm ). Try seven no. 7 bars; A s 4.20 sq.in.
2
(270.098 cm ); o 19.2 in. (487.68 mm); u 138,800/[19.2(0.874)(30.5)] 271
lb/sq.in. (1868.5 kPa); u allow 302 lb/sq.in. (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.
10. Design the transverse reinforcement under the exterior
column; disregard eccentricity
Thus, V UV 250(2.58/6.67) 96.8 kips (430.57 kN); M UV /2(96.8)(2.58)(12) 1498
1
2
in.·kips (169.3 kN·m); A s 2.72 sq.in. (17.549 cm ). Try five no. 7 bars; A s 3.00 sq.in.
2
(19.356 cm ); o 13.7 in. (347.98 mm); u 96,800/[13.7(0.874)(31.5)] 257
lb/sq.in. (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