Figure Ex. 19.4 shows a section of a retaining wall with geotextile reinforcement. The wall is backfilled with a granular soil having γ = 18 kN/m^3 and Φ = 34°. 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

Question

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

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 [latex]kN / m ^{2}[/latex]

Table 19.3 Peak soil-to-geotextile friction angles and efficiencies in selected cohesionless soils*
	Concrete sand	Rounded sand	Silty sand
Geotextile type	[latex]\left(\phi=30^{\circ} ight)[/latex]	[latex]\left(\phi=28^{\circ} ight)[/latex]	[latex]\left(\phi=26^{\circ} ight)[/latex]
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)

Accepted Answer

(a) The lateral pressure [latex]p_{h}[/latex] at any depth z is expressed as

[latex]p_{h}=p_{a}+q_{h}[/latex]

where [latex]p_{a}=\gamma z K_{A}, q_{h}=q K_{A}, K_{A}= an ^{2}\left(45^{\circ}-36 / 2 ight)=0.26[/latex]

Substituting

[latex]p_{h}=18 imes 0.26 z+0.26 imes 10=4.68 z+2.60[/latex]

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

[latex]T_{a}=T_{u}\left(\frac{1}{R F_{I D} imes R F_{C R} imes R F_{C D} imes R F_{B D}} ight)[/latex] (19.9)

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

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

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

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

where h = vertical spacing (lift thickness)

[latex]T_{a}=[/latex] allowable stress in the geotextile

[latex]p_{h}=[/latex] lateral earth pressure at depth z

[latex]F_{s}=[/latex] factor of safety = 1.4

Now substituting

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

At z = 6 m, [latex]h=\frac{13.2}{6.55 imes 6+3.64}=0.307 m ext { or say } 0.30 m[/latex]

At z = 33 m [latex]h=\frac{13.2}{6.55 imes 3.3+3.64}=0.52 m ext { or say } 0.50 m[/latex]

At z = 13 m [latex]h=\frac{13.2}{6.55 imes 1.3+3.64}=1.08[/latex] 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

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

[latex]L_{e}=\frac{T F_{s}}{2 \gamma z an \delta}=\frac{p_{h} h F_{s}}{2 \gamma z an \delta}=\frac{h(4.68 z+2.60) 1.4}{2 imes 18 z an 24^{\circ}}=L_{e}=\frac{h(6.55 z+3.64)}{16 z}[/latex]

From Fig. (19.15) the expression for [latex]L_{R}[/latex] is

[latex]L_{R}=(H-z) an \left(45^{\circ}-\phi / 2 ight)=(H-z) an \left(45^{\circ}-36 / 2 ight)=(6.0-z)(0.509)[/latex]

The total length L is

[latex]L=L_{R}+L_{e}[/latex]

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

Layer No	Depth z	Spacing h	[latex]L_{e}[/latex]	[latex]L_{e}[/latex] (min)	[latex]L_{R}[/latex]	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 [latex]L_{e}[/latex] 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 [latex]L_{o}[/latex]. The minimum value of [latex]L_{o}[/latex] is 1 .0 m. The equation for [latex]L_{o}[/latex] may be expressed as

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

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

[latex]L_{o}=\frac{0.65[4.68(0.65)+2.60] 1.4}{4 imes 18(0.65) an 24^{\circ}}=0.25 m[/latex]

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

(d) Check for external stability

The total active earth pressure [latex]P_{a}[/latex] is

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

[latex]F_{s}=\frac{ ext { Resisting moment } M_{R}}{ ext { 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)}[/latex]

where [latex]W_{1}=6 imes 2 imes 18=216 kN ext { and } l_{1}=2 / 2=1 m[/latex]

[latex]W_{2}=(6-2.1) imes(3-2)(18)=70.2 kN , ext { and } l_{2}=2.5 m[/latex]

[latex]W_{3}=(6-4.2)(4-3)(18)=32.4 kN ext { and } l_{3}=3.5 m[/latex]

[latex]F_{s}=\frac{213 imes(1)+70.2 imes(2.5)+32.4 imes(3.5)}{90.7 imes(2)}=2.78>2- OK[/latex]

Check for sliding

[latex]\begin{aligned}F_{s} &=\frac{ ext { Total resisting force } F_{R}}{ ext { Total driving force } F_{d}} \F_{R} &=\left(W_{1}+W_{2}+W_{3} ight) an \delta \&=(216+70.2+32.4) an 25.5^{\circ} \&=318.6 imes 0.477=152 kN \F_{d} &=P_{a}=90.7 kN\end{aligned}[/latex]

Hence [latex]F_{s}=\frac{152}{90.7}=1.68>1.5- OK[/latex]

Check for a foundation failure

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

[latex]q_{u}=\frac{1}{2} \gamma B N_{\gamma}[/latex]

Use Terzaghi's theory. For [latex]\phi=34^{\circ}, N_{\gamma}=38, ext { and } B=2 m[/latex]

[latex]q_{u}=\frac{1}{2} imes 18 imes 2 imes 38=684 kN / m ^{2}[/latex]

The actual load intensity on the base of the backfill

[latex]q( ext { actual })=18 imes 6+10=118 kN / m ^{2}[/latex]

[latex]F_{s}=\frac{684}{118}=5.8>3[/latex] which is acceptable

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

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