The active pressure coefficient Ka is given by Coulomb theory as
where 0 = angle of slope of back wall to horizontal, degrees ф’ = effective angle of internal friction, degrees 8 = angle of wall friction, degrees P = angle of back slope, degrees
Refer to Figure 8.12 for the force diagram. The resultant horizontal earth force is to be determined for a design case wherein the following assumptions apply:
Design assumptions
ф’ = 34°
8 = 25°
P = 0°
0 = 90°
Y = 125 lb/ft3 (19.6 kN/m3)
H = height of wall = 20 ft (6.1 m)
Soil type = 1 (see Table 8.1)
Computations
sin (0 + ф’) = sin (90° + 34°) = sin 124° = 0.8290 sin2 (0 + ф’) = sin2 (90° + 34°) = sin2 124° = 0.6873 sin (ф’ + 8) = sin (34° + 25°) = sin 59° = 0.8572 sin (ф’ — P) = sin (34° — 0°) = sin 34° = 0.5592 sin (0-8) = sin (90° — 25°) = sin 65° = 0.9063 sin (0 + P) = sin (90° + 0°) = sin 90° = 1.0000 sin2 (0) = sin2 (90) = 1.0000
ka = horizontal active pressure = Ka8’H = 0.2542(125)20 = 635.5 lb/ft2 (U. S. Customary units)
= 0.2542 (19.6) 6.1 = 30.4 kPa (SI units)
Pa = force resultant due to horizontal active pressure
Alternate calculation. Figure 8.10 gives the horizontal and vertical components of active earth pressure, kh and kv, for the five soil types listed in Table 8.1. The pressures are given in terms of the ratio H1/H, when H1 is the surcharge height and H is the height of the fill from the base, both as defined by the sketches in Fig. 8.10.
From Fig. 8.10, soil is type 1, H1/H = 0, kh = 30 lb/ft2/ft. Use 35 lb/ft2/ft (5.50 kN/m2/m), per note 3, Fig. 8.10.
Pa = /2khH2 = /2 X 35(20)2 = 7000 lb/ft (U. S. Customary units)
= /2 X 5.50(6.1)2 = 102 kN/m (SI units)
