A section of a retaining wall with a reinforced backfill is shown in Fig. Ex. 19.3. The backfill surface is subjected to a surcharge of 30 $kN / m ^{2}$ . Required:

(a) The reinforcement distribution.

(b) The maximum tension in the strip.

(c) Check for external stability.

Given: b = 100 mm, t = 5 mm, $f_{a}=143.7 MPa , c=0, \phi=36^{\circ}, \delta=25^{\circ}, \gamma=17.5 kN / m ^{3}$ , s = 0.5 m, and h = 0.5 m.

Question

A section of a retaining wall with a reinforced backfill is shown in Fig. Ex. 19.3. The backfill surface is subjected to a surcharge of 30 [latex]kN / m ^{2}[/latex]. Required:

(a) The reinforcement distribution.

(b) The maximum tension in the strip.

(c) Check for external stability.

Given: b = 100 mm, t = 5 mm, [latex]f_{a}=143.7 MPa , c=0, \phi=36^{\circ}, \delta=25^{\circ}, \gamma=17.5 kN / m ^{3}[/latex], s = 0.5 m, and h = 0.5 m.

Accepted Answer

FromEq. (19.17a)

[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=\left(\gamma z K_{A}+q_{h} ight) h imes s=\left(p_{o}+q_{h} ight) A_{c}[/latex]

where [latex]\gamma=17.5 kN / m ^{3}, K_{A}=0.26, A_{c}=h imes s=(0.5 imes 0.5) m ^{2}[/latex]

FromEq. (19.13a)

[latex]q_{h}=\frac{2 q_{s}}{\pi}(\beta-\sin \beta \cos 2 \alpha)[/latex] (19.13a)

[latex]q_{h}=\frac{2 q_{s}}{\pi}[\beta-\sin \beta \cos 2 \alpha][/latex]

Refer to Fig. Ex. 19.3 for the definition of [latex]\alpha ext { and } \beta[/latex].

[latex]q_{s}=30 kN / m ^{2}[/latex]

The procedure for calculating length L of the strip for one depth z = 1.75 m (strip number 4) is explained below. The same method is valid for the other strips.

Strip No. 4. Depth z = 1.75 m

[latex]p_{a}=\gamma z K_{A}=17.5 imes 1.75 imes 0.26=7.96 kN / m ^{2}[/latex]

From Fig. Ex. 19.3, [latex]\begin{aligned}&\beta=19.07^{\circ}=0.3327 ext { radians } \&\alpha=29.74^{\circ} \&q_{s}=30 kN / m ^{2}\end{aligned}[/latex]

[latex]q_{h}=\frac{2 imes 30}{3.14}\left[0.3327-\sin 19.07^{\circ} \cos 59.5^{\circ} ight]=3.19 kN / m ^{2}[/latex]

Figure Ex. 19.3 shows the surcharge distribution at a 2 (vertical) to 1 (horizontal) slope. Per the figure at depth z = 1.75 m, [latex]L_{1}=1.475[/latex] m from the failure line and [latex]L_{R}=(H-z) an \left(45^{\circ}-\phi / 2 ight)[/latex] [latex]=2.75 an \left(45^{\circ}-36^{\circ} / 2 ight)=1.4[/latex] 1.4m from the wall to the failure line. It is now necessary to determine [latex]L_{2}[/latex] (Refer to Fig. 19.15a).

Now [latex]T=(7.96+3.19) imes 0.5 imes 0.5=2.79 kN / strip[/latex].

The equation for the frictional resistance per strip is

[latex]F_{R}=2 b(\gamma z+\Delta q) L_{1} an \delta+\left(\gamma z L_{2} an \delta ight) 2 b[/latex]

From the 2:1 distribution [latex]\Delta q ext { at } z=1.75[/latex] m is

[latex]\Delta q=\frac{Q}{B+z}=\frac{30 imes 1}{1+1.75}=10.9 kN / m ^{2}[/latex]

[latex]p_{o}=17.5 imes 1.75=30.63 kN / m ^{2}[/latex]

Hence [latex]\overline{p_{o}}=10.9+30.63=41.53 kN / m ^{2}[/latex]

Now equating frictional resistance [latex]F_{R}[/latex] to tension in the strip with [latex]F_{s}=1.5[/latex], we have [latex]F_{R}=1.5[/latex] Given b = 100 mm. Now from Eq. (19.20)

[latex]F_{R}=2 b\left[\bar{p}_{o} L_{1}+p_{o} L_{2} ight] an \delta \leq T F_{s}[/latex] (19.20)

[latex]F_{R}=2 b an \delta\left(\bar{p}_{o} L_{1}+p_{o} L_{2} ight)=1.5 T[/latex]

Substituting and taking [latex]\delta=25^{\circ}[/latex], we have

[latex]2 imes 0.1 imes 0.47\left[41.53 imes 1.475+30.63 L_{2} ight]=1.5 imes 2.79[/latex]

Simplifying

Hence [latex]\begin{aligned}&L_{2}=-0.546 m \approx 0 \&L_{e}=L_{1}+0=1.475 m \&L=L_{R}+L_{e}=1.4+1.475=2.875 m\end{aligned}[/latex]

L can be calculated in the same way at other depths.

Maximum tension T

The maximum tension is in strip number 9 at depth z = 4.25m

Allowable [latex]T_{a}=f_{a} b t=143.7 imes 10^{3} imes 0.1 imes 0.005=71.85 kN[/latex]

[latex]T=\left(\gamma z K_{A}+q_{h} ight) s h[/latex]

where [latex]\gamma z K_{A}=17.50 imes 4.25 imes 0.26=19.34 kN / m ^{2}[/latex]

[latex]q_{h}=0.89 kN / m ^{2} ext { from equation for } q_{h} ext { at depth } z=4.25 m[/latex].

Hence [latex]T=(19.34+0.89) imes 1 / 2 imes 1 / 2=5.05 kN / strip <71.85 kN - OK[/latex]

Question 19.3: A section of a retaining wall with a reinforced backfill is ...

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