A typical section of a retaining wall with the backfill reinforced with metal strips is shown in Fig. Ex. 19.2. The following data are available: Height H = 9 m; b = 100 mm; t = 5 mm fy = 240 MPa; Fs for steel = 1.67; Fs on soil friction = 1.5; Φ = 36°; γ = 17.5 kN/m^3; δ = 25°; h x s = 1 x 1 m.

Question

A typical section of a retaining wall with the backfill reinforced with metal strips is shown in Fig. Ex. 19.2. The following data are available: Height H = 9 m; b = 100 mm; t = 5 mm $f_{y}=240 MPa ; F_{s} \text { for steel }=1.67 ; F_{s}$ on soil friction $=1.5 ; \phi=36^{\circ} ; \gamma=17.5 kN / m ^{3} ; \delta=25^{\circ} ; h \times s=1 \times 1$ m.

Required:

(a) Lengths L and $L_{e}$ at varying depths.

(b) The largest tension Tin the strip.

(c) The allowable tension in the strip.

(d) Check for external stability.

Required:

(a) Lengths L and [latex]L_{e}[/latex] at varying depths.

(b) The largest tension Tin the strip.

(c) The allowable tension in the strip.

(d) Check for external stability.

Accepted Answer

From Eq. (19.17a), the tension in a strip at depth z is

[latex]T=p_{h} imes h imes s / ext { strip }=\left(\gamma K_{A}+q_{h} ight) h imes s[/latex] (19.17a)

[latex]T=\gamma z K_{A} s h ext { for } q_{h}=0[/latex]

where [latex]\gamma=17.5 kN / m ^{3}, K_{A}= an ^{2}\left(45^{\circ}-36 / 2 ight)=0.26, s=1 m ; h=1 m[/latex].

Substituting

[latex]T=17.5 imes 0.26(1)[1] z=4.55 z[/latex] kN/strip.

[latex]L_{e}=\frac{F_{s} T}{2 \gamma z b an \delta}=\frac{1.5 imes 4.55 z}{2 imes 17.5 imes 0.1 imes 0.47 imes z}=4.14 m[/latex]

This shows that the length [latex]L_{e}=4.14[/latex] m is a constant with depth. Fig. Ex. 19.2 shows the positions of [latex]L_{e}[/latex] 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 [latex]\Sigma T[/latex] as computed is

[latex]\Sigma T=184.29[/latex] kN/m since s = 1 m.

As a check the total active earth pressure is

[latex]P_{a}=\frac{1}{2} \gamma H^{2} K_{A}=\frac{1}{2} 17.5 imes 9^{2} imes 0.26=184.28 kN / m =\Sigma T[/latex]

The maximum tension is in the 9th strip, that is, at a depth of 8.5 m below the backfill surface. Hence

[latex]T=\gamma z K_{A} s h=17.5 imes 8.5 imes 0.26 imes 1 imes 1=38.68 kN / strip[/latex]

The allowable tension is

[latex]T_{a}=f_{a} t b[/latex]

where [latex]f_{a}=\frac{240 imes 10^{3}}{1.67}=143.7 imes 10^{3} kN / m ^{2}[/latex]

Substituting [latex]T_{a}=143.7 imes 10^{3} imes 0.005 imes 0.1 » 72 kN >T- OK[/latex].

The total length of strip L at any depth z is

[latex]L=L_{R}+L_{e}=(H-z) an (45-\phi / 2)+4.14=0.51(9-z)+4.14 m[/latex]

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

[latex]F_{s}=\frac{ ext { Sliding resistance } F_{R}}{ ext { Driving force } P_{a}}[/latex]

where [latex]F_{R}=W an \delta=\frac{4.5+8.5}{2} imes 17.5 imes 9 an 36^{\circ}[/latex]

= 1024 x 0.73 = 744 kN

where [latex]\delta=\phi=36^{\circ}[/latex] for the foundation soil, and W - weight of the reinforced wall

[latex]\begin{aligned}&P_{a}=184.28 kN \&F_{s}=\frac{744}{184.28}=4>1.5- OK\end{aligned}[/latex]

Check for overturning

[latex]F_{s}=\frac{M_{R}}{M_{o}}[/latex]

From Fig. Ex. 19.2 taking moments of all forces about O, we have

[latex]\begin{aligned}M_{R} &=4.5 imes 9 imes 17.5 imes \frac{4.5}{2}+\frac{1}{2} imes 9 imes(8.5-4.5)\left(4.5+\frac{4}{3} ight) imes 17.5 \&=1595+1837=3432 kN - m \M_{o} &=P_{a} imes \frac{H}{3}=184.28 imes \frac{9}{3}=553 kN - m \F_{s} &=\frac{3432}{553}=6.2>2- OK\end{aligned}[/latex]

Question 19.2: A typical section of a retaining wall with the backfill rein...

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