Simplified Design for Footings - Simplified Design of Reinforced Concrete Buildings - page 229

(4) Footing thickness

Footing projection

c = [(9.5 - 16/12)]/2 = 4.08 ft

P u 512

90.25 +

( )

h = 2.2c

A f +

4in. = 2.2 4.08 4 = 25.4 in. > 10 in. O.K.

Try h = 27 in. (2 ft-3 in.)
Check if the footing thickness is adequate for shear:

d 27- 4 = 23 in.

512
90.25

For wide-beam shear, use Fig. 7-3. With q u = = 5.7 ksf, read d/c 0.33. Therefore, the minimum d is

d = 0.33 4.08 = 1.35 ft = 16.2 in. < 23 in. O.K.

Use Fig. 7-4 for two-way shear:

A f 90.25

16 2

A c =

= 50.8

( )

/ 144

Interpolating between A f /A c = 45 and 60, read d/c 1 1.13 for q u = 5.7 ksf. The minimum d for two-way shear is:

d = 1.13 16 = 18.1 in. < 23 in. O.K.

Therefore, the 27 in. footing depth (d = 23 in.) is adequate for flexure and shear.

(5) Footing reinforcement
A s = 0.022 h = 0.022(27) = 0.59 in. 2 /ft

Try No.7 @ 12 in. (A s = 0.60 in. 2 /ft; see Table 3-7)
Determine the development length of the No.7 bars (see Fig. 7-6):

cover = 3 in. > 2d b = 2 0.875 = 1.8 in.

side cover = 3 in. > 2.5 0.875 = 2.2 in.

clear spacing = 12 - 0.875 = 11.1 in. > 5 x 0.875 = 4.4 in.4.4 in.

Since all of the cover and spacing criteria given in Fig. 7-6 are satisfied, Table 7-1 can be used to deter-
mine the minimum development length.
For › = 4000 psi: d = 42 in.
Check available development length:

L = 9.5 12 = 114 in. > (2 42) + 16 + 6 = 106 in. O.K.

Use 9 ft-6 in. 9 ft-6 in. square footing (L=118 in.)