Page 257 - Handbook of Civil Engineering Calculations, Second Edition

2.42 REINFORCED AND PRESTRESSED CONCRETE ENGINEERING AND DESIGN

ft (m); h side of column, in (mm); d effective depth of footing, ft (m); t thickness of
footing, ft (m); f b bearing stress at interface of column, lb/sq.in. (kPa); v 1 nominal
shearing stress under column, lb/sq.in. (kPa); v 2 nominal shearing stress caused by beam
action, lb/sq.in. (kPa); b o width of critical section for v 1 , ft (m); V 1 and V 2 vertical
shear at critical section for stresses v 1 and v 2 , respectively.
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/sq.ft. (335.2 kPa); f c
3000
lb/sq.in. (20,685 kPa); and f s 20,000 lb/sq.in. (137,900 kPa).

Calculation Procedure:

1. Record the allowable shear, bond, and bearing stresses
From the ACI Code table, v 1 110 lb/sq.in. (758.5 kPa); v 2 60 lb/sq.in. (413.7 kPa);
0.5
f b 1125 lb/sq.in. (7756.9 kPa); u 4.8( f c
) /bar diameter 264/bar diameter.
2. Check the bearing pressure on the footing
Thus, f b , 380/[20(20)] 0.95 kips/sq.in. (7.258 MPa) < 1.125 kips/sq.in. (7.7568
MPa). This is acceptable.
3. Establish the length of footing
For this purpose, assume the footing weight is 6 percent of the column load. Then A
1.06(380)/7 57.5 sq.ft. (5.34 sq.in.). Make L 7 ft 8 in. 7.67 ft (2.338 m); A 58.8
2
sq.ft. (5.46 m ).
4. Determine the effective depth as controlled by v 1
Apply
2
2
(4v 1 p)d h(4v 1 2p)d p(A h ) (49)
Verify the result after applying this equation. Thus p 380/58.8 6.46 kips/sq.ft. (0.309
2
MPa) 0.11(144) 15.84 kips/sq.ft. (0.758 MPa); 69.8d 127.1d 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
2
6.46(58.8 3.21 ) 313 kips (1392.2 kN); v 1 V 1 /(b o d) 313/[4(3.21)(1.54)] 15.83
kips/sq.ft. (0.758 MPa). This is acceptable.
5. 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 assumed weight 0.06(380) 22.8 kips (101.41 kN). This is acceptable.
6. Check v 2
In Fig. 24, AL (7.67 1.67)/2 1.63 1.37 ft (0.417 m); V 2 380(1.37/7.67)
67.9 kips (302.02 kN); v 2 V 2 /(Ld) 67,900/[92(19.5)] 38 lb/sq.in. (262.0 kPa) < 60
lb/sq.in. (413.7 kPa). This is acceptable.