Structural Concrete - Hassoun

466 Chapter 13 Footings
The value of A s to be used must be greater than or equal to 9.26 in. 2 Use 22 no. 6 bars
(A s = 9.68 in. 2 ):
γ s = 2
β + 1 = 2
(12∕8.5)+1 = 0.83
The number of bars in an 8.5-ft band is 22(0.83) = 19 bars. The number of bars left on each side is
1
(22 − 19) ≈2 bars. Therefore, place 19 no. 6 bars within the 8.5-ft band; then place 2 no. 6 bars
2
(A s = 0.88 in. 2 ) within (12 − 8.5)/2 = 1.63 ft on each side of the band. The total number of bars is
23 no. 6 bars (A s = 10.12 in. 2 ). Details of reinforcement are shown in Fig. 13.19.
23 no. 6 10 no. 9
Figure 13.19
Example 13.3: Rectangular footing.
8. Check the bearing stress at the base of the column, as explained in the previous example. Use four
no. 8 dowel bars.
a. Bearing strength N 1 at the base of the column (A 1 = 18 × 18 in.) is
N 1 = φ(0.85f c ′ A 1)=0.65(0.85 × 4)(18 × 18) =716 K > P u = 614 K OK
b. Bearing strength N 2 of footing is
√
A
N 2 = N 2
1 ≤ 2N
A 1
1
A 1 = 18 × 18 = 324 in. 2
A 2 =(48 + 18 + 48)(42 + 18 + 42) =11,628 in. 2 (Fig. 13.20)
√ √
A 2 11,628
=
A 1 18 × 18 = 5.99 > 2