Structural Concrete - Hassoun

478 Chapter 13 Footings
b. Reinforcement under the interior column II:
Bandwidth = 20 + 31.5 + 31.5 = 83 in. = 6.91 ft
Use 7 ft(84 in.).
The steel percentage, ρ = 0.0008 is less than minimum ρ for shrinkage reinforcement ratio of
0.0018:
Min A s (shrinkage) =(0.0018)(7 × 12)(36) =5.44 in. 2
Use seven no. 8 bars placed within the bandwidth of 84 in. under column II, as shown in
Figs. 13.26 and 13.27. The development length l d of no. 8 bars in the short direction is 48 in.
Though not required by code, it is recommended to provide in the shorter direction at the
top of the footing no. 5 bars at 10 in. center-to-center to prevent possible shrinkage cracks and
to hold the reinforcement in the other direction.
13.7 FOOTINGS UNDER ECCENTRIC COLUMN LOADS
When a column transmits axial loads only, the footing can be designed such that the load acts at the
centroid of the footing, producing uniform pressure under the footing. However, in some cases, the
column transmits an axial load and a bending moment, as in the case of the footings of fixed-end
frames. The pressure q that develops on the soil will not be uniform and can be evaluated from the
following equation:
q = P A ± Mc ≥ 0 (13.20)
I
where A and I are the area and moment of inertia of the footing, respectively. Different soil conditions
exist, depending on the magnitudes of P and M and allowable soil pressure. The different
design conditions are shown in Fig. 13.28 and are summarized as follows:
1. When e = M/P < L/6, the soil pressure is trapezoidal:
q max = P A + Mc
I
q min = P A − Mc
I
2. When e = M/P = L/6, the soil pressure is triangular:
q max = P
LB + 6M
BL = 2P
2 LB
q min = 0 = P
LB − 6M or
BL 2
3. When e > L/6, the soil pressure is triangular:
x = L − y
3
P = q max
( 3x
2
q max = 2P
3xB =
= P LB + 6M
BL 2 (13.21)
= P LB − 6M
BL 2 (13.22)
= L 2 − e
(13.23)
P
LB = 6M
BL 2 (13.24)
)
B (13.25)
4P
3B(L − 2e)