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Question 19.4: Figure Ex. 19.4 shows a section of a retaining wall with geo...

Figure Ex. 19.4 shows a section of a retaining wall with geotextile reinforcement. The wall is backfilled with a granular soil having \gamma=18 kN / m ^{3} \text { and } \phi=34^{\circ}.

A woven slit-film geotextile with warp (machine) direction ultimate wide-width strength of 50 kN/m and having δ = 24° (Table 19.3) is intended to be used in its construction.

The orientation of the geotextile is perpendicular to the wall face and the edges are to be overlapped to handle the weft direction. A factor of safety of 1.4 is to be used along with sitespecific reduction factors (Table 19.4).

Required:

(a) Spacing of the individual layers of geotextile.

(b) Determination of the length of the fabric layers.

(c) Check the overlap.

(d) Check for external stability.

The backfill surface carries a uniform surcharge dead load of 10 kN / m ^{2}

 

Table 19.3 Peak soil-to-geotextile friction angles and efficiencies in selected cohesionless soils*
Concrete sand Rounded sand Silty sand
Geotextile type \left(\phi=30^{\circ}\right) \left(\phi=28^{\circ}\right) \left(\phi=26^{\circ}\right)
Woven, monofilament 26° (84 %) – –
Woven, slit-film 24° (77%) 24° (84 %) 23° (87 %)
Nonwoven, heat-bonded 26° (84 %) – –
Nonwoven, needle-punched 30° (100%) 26° (92 %) 25° (96 %)
* Numbers in parentheses are the efficiencies. Values such as these should not be used in final design. Site
specific geotextiles and soils must be individually tested and evaluated in accordance with the particular project conditions: saturation, type of liquid, normal stress, consolidation time, shear rate, displacement amount, and so on. (Koerner, 1999)

 

Table 19.4 Recommended reduction factor values for use in [Eq. (19.9)]
Range of Reduction Factors
Application Installation Chemical Biological
Area Damage Creep* Degradation Degradation
Separation 1.1 to 2.5 1.5 to 2.5 1.0 to 1.5 1.0 to 1.2
Cushioning 1.1 to 2.0 1.2 to 1.5 1.0 to 2.0 1.0 to 1.2
Unpaved roads 1.1 to 2.0 1.5 to 2.5 1.0 to 1.5 1.0 to 1.2
Walls 1.1 to 2.0 2.0 to 4.0 1.0 to 1.5 1.0 to 1.3
Embankments 1.1 to 2.0 2.0 to 3.5 1.0 to 1.5 1.0 to 1.3
Bearing capacity 1.1 to 2.0 2.0 to 4.0 1.0 to 1.5 1.0 to 1.3
Slope stabilization 1.1 to 1.5 2.0 to 3.0 1.0 to 1.5 1.0 to 1.3
Pavement overlays 1.1 to 1.5 1.0 to 2.0 1.0 to 1.5 1.0 to 1.1
Railroads (filter/sep.) 1.5 to 3.0 1.0 to 1.5 1.5 to 2.0 1.0 to 1.2
Flexible forms 1.1 to 1.5 1.5 to 3.0 1.0 to 1.5 1.0 to 1.1
Silt fences 1.1 to 1.5 1.5 to 2.5 1.0 to 1.5 1.0 to 1.1
* The low end of the range refers to applications which have relatively short service lifetimes and / or situations where creep deformations are not critical to the overall system performance. (Koerner, 1999)
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(a) The lateral pressure p_{h} at any depth z is expressed as

 

p_{h}=p_{a}+q_{h}

 

where p_{a}=\gamma z K_{A}, q_{h}=q K_{A}, K_{A}=\tan ^{2}\left(45^{\circ}-36 / 2\right)=0.26

 

Substituting

 

p_{h}=18 \times 0.26 z+0.26 \times 10=4.68 z+2.60

 

From Eq. (19.9), the allowable geotextile strength is

 

T_{a}=T_{u}\left(\frac{1}{R F_{I D} \times R F_{C R} \times R F_{C D} \times R F_{B D}}\right) (19.9)

 

\begin{aligned}T_{a} &=T_{u} \frac{1}{R F_{I D} \times R F_{C R} \times R F_{C D} \times R F_{B D}} \\&=50 \frac{1}{1.2 \times 2.5 \times 1.15 \times 1.1}=13.2 kN / m\end{aligned}

 

From Eq. (19.17a), the expression for allowable stress in the geotextile at any depth z may be expressed as

 

T=p_{h} \times h \times s / \text { strip }=\left(\gamma K_{A}+q_{h}\right) h \times s (19.17a)

 

\begin{aligned}&T=T_{a}=p_{h} h F_{s} \\&h=\frac{T_{a}}{p_{h} F_{s}}\end{aligned}

 

where h = vertical spacing (lift thickness)

T_{a}= allowable stress in the geotextile

p_{h}= lateral earth pressure at depth z

F_{s}= factor of safety = 1.4

 

Now substituting

 

h=\frac{13.2}{[4.68(z)+2.60] 1.4}=\frac{13.2}{6.55(z)+3.64}

 

At z = 6 m, h=\frac{13.2}{6.55 \times 6+3.64}=0.307 m \text { or say } 0.30 m

 

At z = 33 m h=\frac{13.2}{6.55 \times 3.3+3.64}=0.52 m \text { or say } 0.50 m

 

At z = 13 m h=\frac{13.2}{6.55 \times 1.3+3.64}=1.08 m,  but use 0.65 m for a suitable distribution

 

The depth 3.3 m or 1.3 m are used just as a trial and error process to determine suitable spacings. Figure Ex. 19.4 shows the calculated spacings of the geotextiles.

 

(b) Length of the Fabric Layers

From Eq. (19.26) we may write

 

F_{R}=2\left[(\gamma z+\Delta q) L_{1}+\gamma z L_{z}\right] \tan \delta \geq T_{a} F_{s} (19.26)

 

L_{e}=\frac{T F_{s}}{2 \gamma z \tan \delta}=\frac{p_{h} h F_{s}}{2 \gamma z \tan \delta}=\frac{h(4.68 z+2.60) 1.4}{2 \times 18 z \tan 24^{\circ}}=L_{e}=\frac{h(6.55 z+3.64)}{16 z}

 

From Fig. (19.15) the expression for L_{R} is

 

L_{R}=(H-z) \tan \left(45^{\circ}-\phi / 2\right)=(H-z) \tan \left(45^{\circ}-36 / 2\right)=(6.0-z)(0.509)

 

The total length L is

 

L=L_{R}+L_{e}

 

The computed L and suggested L are given in a tabular form below.

 

Layer No Depth z Spacing h L_{e} L_{e} (min) L_{R} L (cal) L (suggested)
(m) (m) (m) (m) (m) (m) (m)
1 0.65 0.65 0.49 1 2.72 3.72 4
2 1.3 0.65 0.38 1 2.39 3.39 –
3 1.8 0.5 0.27 1 2.14 3.14 –
4 2.3 0.5 0.26 1 1.88 2.88 3
5 2.8 0.5 0.25 1 1.63 2.63 –
6 3.3 0.5 0.24 1 1.37 2.37 –
7 3.6 0.3 0.14 1 1.22 2.22 –
8 3.9 0.3 0.14 1 1.07 2.07 –
9 4.2 0.3 0.14 1 0.92 1.92 2
10 4.5 0.3 0.14 1 0.76 1.76 –
11 4.8 0.3 0.14 1 0.61 1.61 –
12 5.1 0.3 0.14 1 0.46 1.46 –
13 5.4 0.3 0.14 1 0.31 1.31 –
14 5.7 0.3 0.14 1 0.15 1.15 –
15 6 0.3 0.13 1 0 1 –

 

It may be noted here that the calculated values of L_{e} are very small and a minimum value of 1.0 m should be used.

 

(c) Check for the overlap

When the fabric layers are laid perpendicular to the wall, the adjacent fabric should overlap a length L_{o}. The minimum value of L_{o} is 1 .0 m. The equation for L_{o} may be expressed as

 

L_{o}=\frac{h p_{h} F_{s}}{2 \times 2 \gamma z \tan \delta}=\frac{h[4.68(z)+2.60] 1.4}{4 \times 18(z) \tan 24^{\circ}}

 

The maximum value of L_{o} is at the upper layer at z = 0.65. Substituting for z we have

 

L_{o}=\frac{0.65[4.68(0.65)+2.60] 1.4}{4 \times 18(0.65) \tan 24^{\circ}}=0.25 m

 

Since this value of L_{o} calculated is quite low, use L_{o}=1.0 m for all the layers.

 

(d) Check for external stability

The total active earth pressure P_{a} is

 

P_{a}=\frac{1}{2} \gamma H^{2} K_{A}=\frac{1}{2} \times 18 \times 6^{2} \times 0.28=90.7 kN / m

 

F_{s}=\frac{\text { Resisting moment } M_{R}}{\text { Driving moment } M _{ o }}=\frac{W_{1} l_{1}+W_{2} l_{2}+W_{3} l_{3}+P_{v} l_{4}}{P_{a}(H / 3)}

 

where W_{1}=6 \times 2 \times 18=216 kN \text { and } l_{1}=2 / 2=1 m

 

W_{2}=(6-2.1) \times(3-2)(18)=70.2 kN , \text { and } l_{2}=2.5 m

 

W_{3}=(6-4.2)(4-3)(18)=32.4 kN \text { and } l_{3}=3.5 m

 

F_{s}=\frac{213 \times(1)+70.2 \times(2.5)+32.4 \times(3.5)}{90.7 \times(2)}=2.78>2- OK

 

Check for sliding

 

\begin{aligned}F_{s} &=\frac{\text { Total resisting force } F_{R}}{\text { Total driving force } F_{d}} \\F_{R} &=\left(W_{1}+W_{2}+W_{3}\right) \tan \delta \\&=(216+70.2+32.4) \tan 25.5^{\circ} \\&=318.6 \times 0.477=152 kN \\F_{d} &=P_{a}=90.7 kN\end{aligned}

 

Hence F_{s}=\frac{152}{90.7}=1.68>1.5- OK

 

Check for a foundation failure

Consider the wall as a surface foundation with D_{f}=0. Since the foundation soil is cohesionless, we may write

 

q_{u}=\frac{1}{2} \gamma B N_{\gamma}

 

Use Terzaghi’s theory. For \phi=34^{\circ}, N_{\gamma}=38, \text { and } B=2 m

 

q_{u}=\frac{1}{2} \times 18 \times 2 \times 38=684 kN / m ^{2}

 

The actual load intensity on the base of the backfill

 

q(\text { actual })=18 \times 6+10=118 kN / m ^{2}

 

F_{s}=\frac{684}{118}=5.8>3 which is acceptable

19.3.

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