ance to sliding. The sides of the key are sloped to ensure that the surrounding soil will re-
main undisturbed during excavation.- Analyze the trial section for stability
The requirements are that there be a factor of safety (FS) against sliding and overturning
of at least 1.5 and that the soil pressure have a value lying between O and 4000 Ib/ft^2 (O
and 191.5 kPa). Using the equation developed later in this handbook gives h = sur-
charge/soil weight = 500/130 - 3.85 ft (1.173 m); sin 35° = 0.574; tan 35° = 0.700; Ca =
0.271; Cp = 3.69; Caw = 35.2 lb/ft^3 (5.53 kN/m^3 ); Cpw = 480 lb/ft^3 (75.40 kN/m^3 ); TAB =
(^1) /
2 (35.2)18(18 + 2 x 3.85) =^8140 Ib (36,206.7 N); MAB = (V6)35.2(18)
(^2) (18 + 3 x 3.85) =
56,200 ft Ib (76,207.2 N-m).
The critical condition with respect to stability is that in which the surcharge extends to
G. The moments of the stabilizing forces with respect to the toe are computed in Table 2.
In Fig. 29c, x = 81,030/21,180 = 3.83 ft (1.167 m); e = 5.50 - 3.83 = 1.67 ft (0.509 m).
The fact that the resultant strikes the base within the middle third attests to the absence of
uplift. By/= (PIA)(I ± 6CxIdx ± (^6) eyldy\pa = (21,180/1I)(I + 6 x 1.67/11) = 3680 lb/ft^2
(176.2 kPa); pb = (21,180/1I)(I - 6 x 1.67/11) - 171 lb/ft^2 (8.2 kPa). Check: x =
(11/3)(3680 + 2 x 171)7(3680 + 171) = 3.83 ft (1.167 m), as before. Also,/?c = 2723 lb/ft^2
(130.4 kPa); pd = 2244 lb/ft^2 (107.4 kPa); FS against overturning = 137,230/56,200 =
2.44. This is acceptable.
Lateral displacement of the wall produces sliding of earth on earth to the left of C and
of concrete on earth to the right of C. In calculating the passive pressure, the layer of earth
lying above the base is disregarded, since its effectiveness is unknown. The resistance to
sliding is as follows: friction, A to C (Fig. 29): ^(368O + 2723)(3)(0.700) = 6720 Ib
(29,890.6 N); friction, C to B: '/2(2723 + 171)(8)(0.5) = 5790 Ib (25,753.9 N); passive
earth pressure: '/2(480^2.7S)^2 = 1820 Ib (8095.4 N). The total resistance to sliding is the
sum of these three items, or 14,330 Ib (63,739.8 N). Thus, the FS against sliding is
14,330/8140 = 1.76. This is acceptable because it exceeds 1.5. Hence the trial section is
adequate with respect to stability.
- Calculate the soil pressures when the surcharge extends to H
Thus W 3 = 500(6.5) = 3250 Ib (14,456 N); ^W = 21,180 + 3250 = 24,430 Ib (108,664.6
N); M 0 = 81,030 + 3250(7.75) = 106,220 ft-lb (144,034.3 N-m); x = 106,220/24,430 =
4.35 ft (1.326 m); e = 1.15 ft (0.351 m);pa = 3613 lb/ft^2 (173 kPa);/?^ = 828 lb/ft^2 (39.6
kPa); pc = 2853 lb/ft^2 (136.6 kPa); pd = 2474 lb/ft^2 (118.5 kPa).
TABLE 2. Stability of Retaining Wall
Force, Ib (N) Arm, ft (m) Moment, ft-lb (N-m)
W 1 1.5(11)(150) = 2,480(11,031.0) 5.50(1.676) 13,640 (18,495.8)
W 2 0.67(16.5)(150) = 1,650 (7,339.2) 3.33(1.015) 5,500 (7,458.0)
W 3 0.5(0.83X16.5)(150) = 1,030 (4,581.4) 3.95(1.204) 4,070 (5,518.9)
W 4 1.25(1.13)(150) = 210 (934.1) 3.75(1.143) 790 (1,071.2)
W 5 0.5(0.83X16.5X130) = 890 (3,958.7) 4.23(1.289) 3,760 (5,098.6)
W 6 6.5(16.5)(130) = 13,940(62,005.1) 7.75(2.362) 108,000(146,448.0)
W 1 2.5(3)(130) = 980 (4,359.1) 1.50(0.457) 1,470 (1993.3)
Total 21,180 (94,208.6) 137,230 (186,083.8)
Overturning moment 56,200 (76,207.2)Net moment about A 81,030 (109,876.6)