From Eq. (19.17a), the tension in a strip at depth z is
T=p_{h} \times h \times s / \text { strip }=\left(\gamma K_{A}+q_{h}\right) h \times s (19.17a)
T=\gamma z K_{A} s h \text { for } q_{h}=0
where \gamma=17.5 kN / m ^{3}, K_{A}=\tan ^{2}\left(45^{\circ}-36 / 2\right)=0.26, s=1 m ; h=1 m.
Substituting
T=17.5 \times 0.26(1)[1] z=4.55 z kN/strip.
L_{e}=\frac{F_{s} T}{2 \gamma z b \tan \delta}=\frac{1.5 \times 4.55 z}{2 \times 17.5 \times 0.1 \times 0.47 \times z}=4.14 m
This shows that the length L_{e}=4.14 m is a constant with depth. Fig. Ex. 19.2 shows the positions of L_{e} for strip numbers 1, 2 … 9. The first strip is located 0.5 m below the backfill surface and the 9th at 8.5 m below with spacings at 1 m apart. Tension in each of the strips may be obtained by using the equation T = 4.55 z. The total tension \Sigma T as computed is
\Sigma T=184.29 kN/m since s = 1 m.
As a check the total active earth pressure is
P_{a}=\frac{1}{2} \gamma H^{2} K_{A}=\frac{1}{2} 17.5 \times 9^{2} \times 0.26=184.28 kN / m =\Sigma T
The maximum tension is in the 9th strip, that is, at a depth of 8.5 m below the backfill surface. Hence
T=\gamma z K_{A} s h=17.5 \times 8.5 \times 0.26 \times 1 \times 1=38.68 kN / strip
The allowable tension is
T_{a}=f_{a} t b
where f_{a}=\frac{240 \times 10^{3}}{1.67}=143.7 \times 10^{3} kN / m ^{2}
Substituting T_{a}=143.7 \times 10^{3} \times 0.005 \times 0.1 » 72 kN >T- OK.
The total length of strip L at any depth z is
L=L_{R}+L_{e}=(H-z) \tan (45-\phi / 2)+4.14=0.51(9-z)+4.14 m
where H = 9 m.
The lengths as calculated have been shown in Fig. Ex. 19.2. It is sometimes convenient to use the same length L with depth or stepped in two or more blocks or use a linear variation as shown in the figure.
Check for External Stability
Check of bearing capacity
It is necessary to check the base of the wall with the backfill for the bearing capacity per unit length of the wall. The width of the wall may be taken as equal to 4.5 m (Fig. Ex. 19.2). The procedure as explained in Chapter 12 may be followed. For all practical purposes, the shape, depth, and inclination factors may be taken as equal to 1.
Check for sliding resistance
F_{s}=\frac{\text { Sliding resistance } F_{R}}{\text { Driving force } P_{a}}
where F_{R}=W \tan \delta=\frac{4.5+8.5}{2} \times 17.5 \times 9 \tan 36^{\circ}
= 1024 x 0.73 = 744 kN
where \delta=\phi=36^{\circ} for the foundation soil, and W – weight of the reinforced wall
\begin{aligned}&P_{a}=184.28 kN \\&F_{s}=\frac{744}{184.28}=4>1.5- OK\end{aligned}
Check for overturning
F_{s}=\frac{M_{R}}{M_{o}}
From Fig. Ex. 19.2 taking moments of all forces about O, we have
\begin{aligned}M_{R} &=4.5 \times 9 \times 17.5 \times \frac{4.5}{2}+\frac{1}{2} \times 9 \times(8.5-4.5)\left(4.5+\frac{4}{3}\right) \times 17.5 \\&=1595+1837=3432 kN – m \\M_{o} &=P_{a} \times \frac{H}{3}=184.28 \times \frac{9}{3}=553 kN – m \\F_{s} &=\frac{3432}{553}=6.2>2- OK\end{aligned}
