(Koerner, 1999) Design a 7m high geogrid-reinforced wall when the reinforcement vertical maximum spacing must be 1.0 m. The coverage ratio is 0.80 (Refer to Fig. Ex. 19.5). Given: T_{u}=156 kN / m , C_{r}=0.80, C_{i}=0.75. The other details are given in the figure.
Question 19.5: (Koerner, 1999) Design a 7m high geogrid-reinforced wall whe...
Internal Stability
From Eq. (19.14)
p_{h}=p_{a}+q_{h}=\left(\gamma z K_{A}+q_{h}\right) (9.14)
\begin{aligned}&p_{h}=\left(\gamma z K_{A}+q_{h}\right)=\gamma z K_{A}+q_{s} K_{A} \\&K_{A}=\tan ^{2}\left(45^{\circ}-\phi / 2\right)=\tan ^{2}\left(45^{\circ}-32 / 2\right)=0.31 \\&p_{h}=(18 \times z \times 0.31)+(15 \times 0.31)=5.58 z+4.65\end{aligned}
1. For geogrid vertical spacing.
Given T_{u}=156 kN/m
From Eq. (19.10) and Table 19.5, we have
T_{a}=T_{u}\left(\frac{1}{R F_{I D} \times R F_{C R} \times R F_{B D} \times R F_{C D}}\right) (19.10)
\begin{aligned}&T_{a}=T_{u}\left[\frac{1}{R F_{I D} \times R F_{C R} \times R F_{B D} \times R F_{C D}}\right] \\&T_{a}=156 \frac{1}{1.2 \times 2.5 \times 1.3 \times 1.0}=40 kN / m\end{aligned}
But use T_{ design }=28.6 kN / m \text { with } F_{ s }=1.4 \text { on } T_{a}
From Eq. (19.28)
h=\frac{T_{a} C_{r}}{p_{h}} (19.28)
T_{ design }=\frac{h p_{h}}{C_{r}}
28.6=h \frac{5.58 z+4.65}{0.8}
or h=\frac{22.9}{5.58 z+4.65}
Maximum depth for h = 1 m is
1.0=\frac{22.9}{5.58 z+4.65} \text { or } z=3.27 m
Maximum depth for h = 0.5m
0.5=\frac{22.9}{5.58 z+4.65} \text { or } z=7.37 m
The distribution of geogrid layers is shown in Fig. Ex. 19.5.
2. Embedment length of geogrid layers.
From Eqs (19.27) and (19.24)
T_{a}=T F_{s}=\left(\gamma z K_{A}+q_{h}\right) h F_{s} (19.24)
F_{R}=2 C_{i} C_{r} L_{e} p_{o} \tan \phi \geq T F_{s} (19.27)
2 C_{1} C_{r} L_{e} p_{o} \tan \phi=T_{H} F_{s}=p_{h} h F_{s}
Substituting known values
2 \times 0.75 \times 0.8 \times\left(L_{e}\right) \times 18 \times(z) \tan 32^{\circ}=h(5.58 z+4.65) 1.5
Simplfying L_{e}=\frac{(0.62 z+0.516) h}{z}
The equation for L_{R} is
\begin{aligned}L_{R} &=(H-z) \tan \left(45^{\circ}-\phi / 2\right)=(7-z) \tan \left(45^{\circ}-32 / 2\right) \\&=3.88-0.554(z)\end{aligned}
From the above relationships the spacing of geogrid layers and their lengths are given below.
| Layer | Depth | Spacing | L_{e} | L_{e} (min) | L_{R} | L (cal) | L (required) |
| No. | (m) | h (m) | (m) | (m) | (m) | (m) | (m) |
| 1 | 0.75 | 0.75 | 0.98 | 1 | 3.46 | 4.46 | 5 |
| 2 | 1.75 | 1 | 0.92 | 1 | 2.91 | 3.91 | 5 |
| 3 | 2.75 | 1 | 0.81 | 1 | 2.36 | 3.36 | 5 |
| 4 | 3.25 | 0.5 | 0.39 | 1 | 2.08 | 3.08 | 5 |
| 5 | 3.75 | 0.5 | 0.38 | 1 | 1.8 | 2.8 | 5 |
| 6 | 4.25 | 0.5 | 0.37 | 1 | 1.52 | 2.52 | 5 |
| 7 | 4.75 | 0.5 | 0.36 | 1 | 1.25 | 2.25 | 5 |
| 8 | 5.25 | 0.5 | 0.36 | 1 | 0.97 | 1.97 | 5 |
| 9 | 5.75 | 0.5 | 0.36 | 1 | 0.69 | 1.69 | 5 |
| 10 | 6.25 | 0.5 | 0.35 | 1 | 0.42 | 1.42 | 5 |
| 11 | 6.75 | 0.5 | 0.35 | 1 | 0.14 | 1.14 | 5 |
External Stability
(a) Pressure distribution
\begin{aligned}&P_{a}=\frac{1}{2} \gamma H^{2} K_{A}=\frac{1}{2} \times 17 \times 7^{2} \tan ^{2}\left(45^{\circ}-30 / 2\right)=138.8 kN / m \\&P_{q}=q_{s} K_{A} H=15 \times 0.33 \times 7=34.7 kN / m \\&\text { Total } \approx 173.5 kN / m\end{aligned}
1. Check for sliding (neglecting effect of surcharge)
\begin{aligned}&F_{R}=W \tan \delta=\gamma \times H \times L \tan 25^{\circ}=18 \times 7 \times 5.0 \times 0.47=293.8 kN / m \\&P=P_{a}+P_{q}=173.5 kN / m \\&F_{s}=\frac{293.8}{173.5}=1.69>1.5 OK\end{aligned}
2. Check for overturning
Resisting moment M_{R}=W \times \frac{L}{2}=18 \times 7 \times 5 \times \frac{5}{2}=1575 kN – m
Overturning moment M_{O}=P_{a} \times \frac{H}{3}+P_{q} \times \frac{H}{2}
or M_{O}=138.8 \times \frac{7}{3}+34.7 \times \frac{7}{2}=445.3 kN – m
F_{s}=\frac{1575}{445.3}=3.54>2.0 \quad OK
3. Check for bearing capacity
\text { Eccentricity } e=\frac{M_{O}}{W+q_{s} L}=\frac{445.3}{18 \times 7 \times 5+15 \times 5}=0.63
e=0.63<\frac{L}{6}=\frac{5}{6}=0.83 Ok
Effective length =L-2 e=5-2 \times 0.63=3.74 m
Bearing pressure =[18 \times 7+15]\left(\frac{5}{3.74}\right)=189 kN / m ^{2}
F_{s}=\frac{600}{189}=3.17>3.0 OK
| Table 19.5 Recommended reduction factor values for use in Eq. (19.10) for determining allowable tensile strength of geogrids | ||||
| Application Area | R F_{ID} | R F_{CR} | R F_{C D} | RF _{B D} |
| Unpaved roads | 1.1 to1.6 | 1.5 to 2.5 | 1.0 to 1.5 | 1.0 to 1.1 |
| Paved roads | 1.2 to 1.5 | 1.5 to 2.5 | 1.1 to 1.6 | 1.0 to 1.1 |
| Embankments | 1.1 to 1.4 | 2.0 to 3.0 | 1.1 to 1.4 | 1.0 to 1.2 |
| Slopes | 1.1 to 1.4 | 2.0 to 3.0 | 1.1 to 1.4 | 1.0 to 1.2 |
| Walls | 1.1 to 1.4 | 2.0 to 3.0 | 1.1 to 1.4 | 1.0 to 1.2 |
| Bearing capacity | 1.2 to 1.5 | 2.0 to 3.0 | 1.1 to 1.6 | 1.0 to 1.2 |
Related Answered Questions
Check for Rankine's condition
FromEq. (19.1b)
&nbs...
From Eq. (19.17a), the tension in a strip at depth...
FromEq. (19.17a)
T=p_{h} \times h \t...
(a) The lateral pressure p_{h} at a...