Question 14.4: A concrete gravity retaininvg wall is shown in Figure 14.13....
A concrete gravity retaininvg wall is shown in Figure 14.13. Determine:
a. The factor of safety against overturning
b. The factor of safety against sliding
c. The pressure on the soil at the toe and heel
(Note: Unit weight of concrete = \gamma_c = 150 lb/ft³.)
H^\prime =15+25=17.5 ft
K_a=\tan^2\left(45-\frac{\phi_1^\prime}{2}\right)=\tan^2\left(45-\frac{30}{2}\right)=\frac{1}{3}
P_a = \frac{1}{2} \gamma {(H^\prime)}^ 2 Ka=\frac{1}{2}(121)(17.5)^2\left(\frac{1}{3}\right)= 6176 lb/ft
= 6.176 kip/ft
Since α = 0
P_h = P_a = 6.176 kip/ft
P_v = 0
Part a: Factor of Safety Against Overturning
The following table can now be prepared to obtain \sum M_R.
| Area (from Figure 14.13) | Weight (kip) | Moment arm from C (ft) | Moment about C (kip.ft/ft) |
| 1 | \frac{1}{2}(0.8)(15)(\gamma_c)=0.9 | 1.25+\frac{2}{3}(0.8)=1.783 | 1.605 |
| 2 | (1.5)(15)(\gamma_c)=3.375 | 1.25+0.8+0.75=2.8 | 9.45 |
| 3 | \frac{1}{2}(5.25)(15)(\gamma_c)=5.906 | 1.25+0.8+1.5+\frac{25.5}{3}=5.3 | 31.30 |
| 4 | (10.3)(2.5)(\gamma_c)=3.863 | \frac{10.3}{2}=5.15 | 19.89 |
| 5 | \frac{1}{2}(5.25)(15)(0.121)=4.764 | 1.25+0.8+1.5+\frac{2}{3}(5.25)=7.05 | 33.59 |
| 6 | (1.5)(15)(0.121)=\frac{2.723}{21.531} | 1.25+0.8+1.5+5.25+0.75=9.55 | \frac{26.0}{121.84}=M_R |
The overturning moment
M_O = \frac{H^\prime}{3}P_a =\left(\frac{17.5}{3}\right)(6.176)= 36.03 kip/ft
FS_{(overturning)} = \frac{121.84}{36.03} = 3.38
Part b: Factor of Safety Against Sliding
From Eq. (14.10), with k_1 = k_2 = 2/3 and assuming that P_p = 0,
FS_{(sliding)} =\frac{\sum V \tan \left(\frac{2}{3}\right)\phi_2^\prime+B\left(\frac{2}{3}\right)c_2^\prime}{P_a}
=\frac{21.531 \tan \left(\frac{2\times 20}{3}\right)+10.3\left(\frac{2}{3}\right)(1.0)}{6.176}
=\frac{5.1+6.87}{6.176}=1.94
Part c: Pressure on the Soil at the Toe and Heel
e = \frac{B}{2}-\frac{ \sum M_R – \sum M_O}{\sum V}= \frac{10.3}{2}-\frac{ 121.84 – 36.03}{21.531}=5.15-3.99= 1.16 ft
q_{toe} =\frac{ \sum V}{B}\left[1+\frac{6e}{B}\right]=\frac{21.531}{10.3}\left[1+\frac{(6)(1.16)}{10.3}\right]= 3.5 kip/ft²
q_{heel} =\frac{ \sum V}{B}\left[1-\frac{6e}{B}\right]=\frac{21.531}{10.3}\left[1-\frac{(6)(1.16)}{10.3}\right]= 0.678 kip/ft²