67.9 kips (302.02 kN); V 2 = V 2 I(Ld) = 67,900/[92(19.S)] = 38 lb/in
2
(262.0 kPa) < 60
lb/in^2 (413.7 kPa). This is acceptable.
- Design the reinforcement
In Fig. 24, EA = 3.00 ft (0.914 m); VEF= 380(3.00/7.67) = 148.6 kips (666.97 kN); MEF =
148.6(^1 / 2 )(3.00)(12) = 2675 in-kips (302.22 kN-m); A 5 = 2675/[20(0.874)(19.5)] = 7.85 in^2
(50.648 cm^2 ). Try 10 no. 8 bars each way. Then A 3 = 7.90 in^2 (50.971 cm^2 ); ^o = 31.4 in
(797.56 mm); u = VEFTZojd = 148,600/[31.4(0.874)(19.5)] = 278 lb/in^2 (1916.81 kPa);
"allow = 264/1 = 264 lb/in^2 (1820.3 kPa).
The bond stress at EF is slightly excessive. However, the ACI Code, in sections based
on ultimate-strength considerations, permits disregarding the local bond stress if the aver-
age bond stress across the length of embedment is less than 80 percent of the allowable
stress. Let Le denote this length. Then Le = EA - 3 = 33 in (838.2 mm); 0.80wallow = 211
lb/in^2 (1454.8 kPa); uav = Asfsl(L£o) = 0.79(20,000)/[33(3.1)] = 154 lb/in^2 (1061.8 kPa).
This is acceptable. - Design the dowels to comply with the Code
The function of the dowels is to transfer the compressive force in the column reinforcing
bars to the footing. Since this is a tied column, assume the stress in the bars is
0.85(20,000) = 17,000 lb/in^2 (117,215.0 kPa). Try eight no. 9 dowels withfy = 40,000
lb/in^2 (275,800.0 kPa). Then u = 2647(9/8) -
235 lb/in^2 (1620.3 kPa); Le = 1.00(17,00O)/
[235(3.5)] = 20.7 in (525.78 mm). Since the
footing can provide a 21-in (533.4-mm) em-
bedment length, the dowel selection is satis-
factory. Also, the length of lap = 20(9/8) =
22.5 in (571.5 mm); length of dowels = 20.7
- 22.5 = 43.2, say 44 in (1117.6 mm). The
footing is shown in Fig. 25.
COMBINED FOOTING DESIGN
An 18-in (457.2-mm) square exterior column
and a 20-in (508.0-mm) square interior col-
umn carry loads of 250 kips (1112 kN) and
370 kips (1645.8 kN), respectively. The col-
umn centers are 16 ft (4.9 m) apart, and the
footing cannot project beyond the face of the
exterior column. Design a combined rectan-
gular footing by the working-stress method,
using fi = 3000 lb/in^2 (20,685.0 kPa), /, =
20,000 lb/in^2 (137,900.0 kPa), and an allow- *I^UKE zs
able soil pressure of 5000 lb/in^2 (239.4 kPa).
Calculation Procedure:- Establish the length of footing, applying the criterion
of uniform soil pressure under total live and dead loads
In many instances, the exterior column of a building cannot be individually supported be-
each way
longdowels
long