Structural Concrete - Hassoun

508 Chapter 14 Retaining Walls
Figure 14.13
Example 14.3: Footing details.
sliding. Another function of the key is to provide sufficient development length for the dowels of
the stem. The key is therefore placed such that its face is about 6 in. from the back face of the stem
(Fig. 14.13). In the calculation of the passive pressure, the top foot of the earth at the toe side is
usually neglected, leaving a height of 2 ft in this example. Assume a key depth of t = 1.5 ft and a
width of b = 1.5 ft.
1 + sin φ
C p =
1 − sin φ = 1 = 1
C a 0.271 = 3.69
H p = 1 2 C p w(h′ + t) 2 = 1 2 × 3.69 × 110(2 + 1.5)2 = 2486 lb
The sliding may occur now on the surfaces AC, CD, andEF (Fig. 14.13). The sliding surface
AC lies within the soil layers with a coefficient of internal friction = tan φ = tan 35 ∘ = 0.7, whereas
the surfaces CD and EF are those between concrete and soil with a coefficient of internal friction
of 0.5, as given in this example. The frictional resistance is F = μ 1 R 1 + μ 2 R 2 .
( )
3.13 + 1.96
R 1 = reaction of AC =
× 4.5 = 11.44 K
2
R 2 = R − R 1 = 18.44 − 11.44 = 7.0K
( )
1.96 + 0.39
= reaction of CDF =
× 6 = 7.05 K
2
F = 0.7(11.44)+0.5(7.00) =11.50 K