Chapter 8. Introduction to Geosynthetics Design
1. What are the major families of geosynthetics and the applications of each type of
geosynthetics in civil engineering?
Answers:
The major families of geosynthetics and their applications are briefly described below:
(1) Geotextiles: used as Separation, reinforcement, filtration, drainage, containment,
and protection, such as in retaining walls, filters and drains, pavement subgrade,
slope stabilization.
(2) Geomembrane: mainly used as a containment, such as landfill covers and bottom
liners, retention pond liners, facings of dams.
(3) Geogrid: mainly used as reinforcement and protection, such as in pavement
subgrade, railroad ballasts, slope stabilization, retaining walls.
(4) Geonet: mainly used in drainage, such as drainage in retaining walls, slopes,
landfills, embankments, and dams.
(5) Geosynthetic clay liner (GCL): mainly used as a containment, such as landfill
covers and bottom liners, retention pond liners.
(6) Geofoam: mainly used as a separation and lightweight backfill, such as
lightweight roadway fills, retaining wall/abutment backfills.
(7) Geocomposite: a combination of geosynthetics for combined functions such as
separation, reinforcement, filtration, drainage, containment, and protection. The
applications include landfill liners and drains, filter and drain and reinforcement
for retaining walls and embankments.
(8) Geocell: mainly used as reinforcement, filtration, protection, such as in erosion
control, slope protection, retaining walls, ground stabilization, temporary
floodwalls.
(9) Geopipe: mainly used as a drain, such as roadway drains, various pipelines.
2. What are the polymers used to manufacture geosynthetics? List the types of
geosynthetics manufactured from each of the polymers.
Answers:
 Polyethylene (PE): geotextile, geonet
 High-density polyethylene (HDPE): geomembrane, geogrid, geocell, geopipe
 Low-density polyethylene (LDPE): geomembrane
 Polypropylene (PP):geotextile, geomembrane, geopipe
 Polyvinyl chloride (PVC): geopipe
 Polyester (PET): geotextile, geocell
 Polyamide (PA): geotextile
 Polystyrene (PS): geofoam
3. For each of the following multiple-choice questions, select the correct answers.
The correct answers are marked by underline.
(1) The function of a geonet is primarily:
A. Reinforcement
B. Drainage
C. Filtration
D. Containment
E. Separation
(2) The following geosynthetics can be used for reinforcement purposes:
A. Geotextile
B. Geomembrane
C. Geogrid
D. Geonet
E. Geosynthetic clay liner
F. Geofoam
(3) The following geosynthetics can be used for containment purposes:
A. Geotextile
B. Geomembrane
C. Geonet
D. Geosynthetic clay liner
E. Geofoam
(4) The following geosynthetics can be used for both filtration and drainage purposes:
A. Geotextile
B. Geomembrane
C. Geogrid
D. Geonet
E. Geosynthetic clay liner
F. Geofoam
(5) The following geosynthetics can be used as the liner of a retention pond:
A. Geotextile
B. Geomembrane
C. Geogrid
D. Geonet
E. Geosynthetic clay liner
F. Compacted clay liner
(6) The advantages of MSE walls include:
A. Increased internal integrity due to the geosynthetics’ tensile strength and
the friction between the soil and the reinforcement.
B. Increased shear resistance to resist slope failure.
C. Rapid construction.
D. Flexible wall system can accommodate large differential settlement.
E. Suited for seismic region.
4. Briefly describe the geosynthetics market nowadays.
Answer: not provided.
5. Geotextile filtration is distinguished from drainage function. Describe these two
functions and how they are different. Two different parameters (permittivity and
transmissivity) are defined. What are the definitions of the two parameters? Which
parameter is used for which function?
Answer:
(1) Filtration is to prevent soil migration and loss, while drainage is to conduct water
away from a location or component.
(2) Permittivity is cross-plane permeability per unit thickness of geotextile:
[sec-1]
Transmissivity is in-plane permeability of geotextile that includes the effect of
geotextile thickness:
q×L
q = kp × t =
[cm2/sec]
Dh × w
(3) Permittivity is used to describe filtration; transmissivity is used to describe
drainage.
6. A geotextile is 50 mm in diameter and 0.50 mm in thickness. It is tested as a filter in
the lab. The following data are obtained using a constant-head cross-plane flow of
water through the geotextile. Calculate the geotextile’s average permittivity (s-1) and
cross-plane coefficient of permeability, kn (cm/s).
h (mm)
q (cm3/min)
45
450
91
978
180
1476
275
2530
Solution:
The geotextile is used as a filter, and the flow is cross-plane.
Use Darcy’s law:
kn =
q×t
Dh × A
Permittivity
Where: thickness t=0.50mm,
Cross sectional area A =
1 2 p 2
pd = 50 = 1962.5mm 2 =19.625cm 2
4
4
The calculation and results are given in the following table:
q (cm3/min)
q (cm3/sec)
Y (s-1)
4.5
450
7.5
0.0849
9.1
978
16.3
0.0913
18.0
1476
24.6
0.0698
27.5
2530
42.2
0.0782
(cm)
Calculate the average of Y:
ave = 0.081 s-1
Cross plane permeability kn = t=0.081 0.05=4.0510-3 cm/sec
7. A geotextile was tested as a drain in the lab. The geotextile is 2.0 mm thick, 300 mm
wide, and 600 mm long. The following constant-head data for planar flow of water in
the longitudinal direction in the geotextile were obtained. Calculate the geotextile’s
average transmissivity (m3/min-m) and the planar coefficient of permeability, kp
(cm/s).
h (mm)
q (cm3/min)
45
450
91
978
180
1476
275
2530
Solution:
The geotextile is used as a drain, the flow is in-plane.
Use Darcy’s law:
Dh
q = k × i × A = kp
(w × t)
L
q×L
so k p =
Dh × w × t
where L = 60 cm, w = 30 cm, t = 0.20 cm
q×L
Dh × w
The calculation and results are given in the following table:
(cm)
q (cm3/min)
 (cm2/min)
 (m3/min-m)
Transmissivity q = k p × t =
4.5
450
200
2.0010-2
9.1
978
215
2.1510-2
18.0
1476
164
1.6410-2
27.5
2530
184
1.8410-2
Calculate the average of :
ave = 1.9110-2 m3/min-m
In-plane permeability: k p =
q
t
=
1.91´10-2
= 9.55m / min =15.9cm / sec
2.0 ´10-3
8. Constant-head data for transmissivity measurements are given under different normal
pressures on the geotextile. The geotextile is 4-mm thick, 50-mm wide, and 60-mm
long. The tests were conducted at a constant head of 360 mm. Plot the transmissivity
(m3/min-m) versus applied normal pressure (kN/m2) response curve.
Pressure (kPa)
Flow rate, q (cm3/min)
25
1725
50
1240
75
1100
Solution: Not provided.
9. A non-woven, needle-punched geotextile is installed to accommodate cross-plane
flow, i.e., the flow is normal to the geotextile sheet. The geotextile is 3.0 mm thick,
1000 mm long, and 1000 mm wide. The total hydraulic head loss across the geotextile
is 5.0 mm, and the design flow rate is 2000 ml/min. Is the geotextile used as a filter or
a drain? Calculate the appropriate hydraulic properties of the geotextile, such as the
transmissivity or permittivity, in-plane permeability or cross-plane permeability,
whichever is applicable.
Solution:
The flow is cross-plane. So, the geotextile is used a filter.
q×t
Dh × A
Flow rate:
q = 2000 ml/min
Geotextile thickness:
t = 3.0 mm = 0.3 cm
Cross-sectional area for flow is: A = 100 cm  100 cm = 10000 cm2
Hydraulic head difference:
h = 5.0 mm = 0.5 cm
Cross-plane permeability is:
So, kn =
kn =
q ×t
2000 (ml/min) ´ 0.3(cm)
=
= 0.12 cm/min = 0.002 cm/sec
Dh × A
0.5(cm) ´10000(cm 2 )
Permittivity: Y =
kn 0.002
=
= 6.710-3 sec-1
t
0.3
10. Design a 5-m high wraparound woven geotextile reinforced soil wall carrying a road
consisting of 25-cm stone base ( = 21.2 kN/m3) and 12.5-cm asphalt ( = 24.0
kN/m3). The wall is to be backfilled with sand of  = 18.0 kN/m3,  = 35, and c = 0.
The external friction angle between the soil and geotextile is  = 25. The geotextile
has an ultimate tensile strength of 50 kN/m. Use reduction factors for installation
damage, creep, and chemical/biological degradation. The factor of safety for tensile
strength is 1.5.
Determine:
(1) How many layers of geotextile are needed and the vertical spacing of each layer.
(2) Geotextile embedment length of each layer.
(3) Geotextile overlap length of each layer.
(4) The external stability in terms of overturning, sliding, and bearing capacity of the
MSE wall that you designed in (1) to (3). The foundation soil is ML-CL with  = 18.8
kN/m3,  = 15,  = 0.9, c= 500 psf, and ca = 0.90c.
Solution: Not provided.
11. A mechanically stabilized earth (MSE) wall is needed to retain a 10-meter high earth
fill. Geogrid is used as reinforcement. Lab testing shows the ultimate tensile strength
of the geogrid is 80 kN/m, and the external friction angle between the backfill and the
geogrid is 30. The earth fill material will be compacted at 95% of its maximum dry
unit weight (dmax = 21 kN/m3) at the optimum water content. Under typical field
conditions, the moisture content of the backfill is 5%. Direct shear tests indicate the
effective friction angle of the backfill is 30 degrees, and the effective cohesion is
zero. No surcharge is on the backfill. Use reduction factors of 1.5, 2.5, 1.2 for
installation damage, creep, and chemical/biological degradation. The foundation soil
beneath the MSE wall is silty clay; its moist unit weight in its natural condition is 19
kN/m3, the internal friction angle is 10 degrees, and the cohesion is 25 kN/m2.
Between the geogrid and the foundation soil, the external friction angle is 9 degrees,
and the adhesion is 22 kN/m2.
Determine:
(1) How many layers of geogrid are needed and the vertical spacing of each layer.
(2) Geotextile embedment length of each layer.
(3) Geotextile overlap length of each layer.
(4) The external stability in terms of overturning, sliding, and bearing capacity of the
MSE wall you designed in (1) to (3).
Solution:
Step 1: Determine wall dimension, external loads.
Step 2: Determine engineering properties of the foundation soil.
Step 3: Determine engineering properties of the reinforced backfill and retained backfill.
The wall dimensions, external loads, and soil engineering properties in the first three
steps are shown in Figure below.
Step 4: The following design factors of safety are required:
A. External stability:
o Sliding: FS  1.5
o Overturning: FS  2.0
o Bearing capacity: FS  2.5
B. Internal stability:
o Pullout resistance: FS  1.5
o Minimum embedment length: 1 m
o The allowable tensile strength of the reinforcement:
Tallowable =
Tult
RFID × RFCR × RFCBD
It is given: RFID = 1.5; RFCR = 2.5; RFCBD = 1.2.
So:
Tallowable =
Tult
80
=
= 17.8 kN / m
RFID × RFCR × RFCBD 1.5 ´ 2.5 ´1.2
Step 5: Determine wall embedment depth
The minimum embedment depth at the front of the wall is:
H1 = H/20 = 0.5 m. Use the minimum depth of 0.5 m.
Step 6: Internal stability design, including the following four major steps:
(1) Calculate the lateral earth pressure, h.
Only the lateral earth pressure caused by the backfill is considered.
s h = s hs
Rankine active earth pressure is: s hs = Kas 0¢
30 ö
æ
K a = tan 2 ç 45 - ÷ = 0.333
è
2ø
s 0¢ = g 1z = 21z
where:
s h = s hs = 0.333 ´ 21z = 7z
So:
(2) Calculate the vertical spacing, Sv, given FS = 1.5
Sv =
Tallowable
17.8
1.7
=
=
FS × s h 1.5 ´ (7z) z
Note:
 To be conservative, the chosen vertical spacing should be  the calculated
vertical spacing;
 The calculated reinforcement length varies with depth. For easy construction,
the reinforcement lengths can be grouped with one length in each group. The
selected length should be  the calculated length.
z = 10 m (bottom of the wall):
Sv = 0.17 m, use 0.15 m.
After 10 layers with Sv = 0.15 m, z = 8.5 m: Sv = 0.20 m, use 0.20 m.
After 10 layers with Sv = 0.20 m, z = 6.5 m: Sv = 0.26 m, use 0.25 m.
After 10 layers with Sv = 0.25 m, z = 4.0 m: Sv = 0.425 m, use 0.40 m.
After 5 layers with Sv = 0.40 m, z = 2.0 m: Sv = 0.85 m, use 0.50 m.
From the bottom of the wall:
10 layers with Sv of 0.15 m +
10 layers with Sv of 0.20 m +
10 layers with Sv of 0.25 m +
5 layers with Sv of 0.40 m +
4 layers with Sv of 0.50 m
=1.5 + 2.0 + 2.5 + 2.0 + 2.0 = 10.0 m.
(3) Calculate the effective length, Le, and the non-reacting length, LNR, of the
geogrid.
Le =
Tallowable
2(ca + s v tan d )
Since c = 0 for backfill, ca = 0.
 = 30
s v = g z = 21z
So:
Le =
17.8
0.734
=
2(0 + 21z × tan 30)
z
The non-reacting length:
f¢ ö
30
æ
LNR = (H - z)× tan ç 45 - 1 ÷ = (10 - z)tan(45 - ) = 5.77 - 0.58z
è
2ø
2
Total embedment length: L = LNR + Le
(4) Overlap length:
Tallowable
0.367
L0 =
=
4 ´ (ca + s v tan d 1 )
z
zmin = 0.5 m, so L0 = 0.734 m.
Use minimum L0 = 1.0 m
The following table is developed to calculate the embedment length of each layer.
Layer
No.
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
Depth,
z
(m)
0.50
1.00
1.50
2.00
2.40
2.80
3.20
3.60
4.00
4.25
4.50
4.75
5.00
5.25
5.50
5.75
6.00
6.25
6.50
6.70
6.90
7.10
7.30
7.50
7.70
7.90
8.10
8.30
8.50
8.65
8.80
8.95
9.10
9.25
9.40
9.55
9.70
9.85
10.00
Spacing,
Sv
(m)
0.5
0.5
0.5
0.5
0.4
0.4
0.4
0.4
0.4
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.25
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.15
0.15
0.15
0.15
0.15
0.15
0.15
0.15
0.15
0.15
Le
Le(min)
LNR
Lcal
Lused
(m)
1.47
0.73
0.49
0.37
0.31
0.26
0.23
0.20
0.18
0.17
0.16
0.15
0.15
0.14
0.13
0.13
0.12
0.12
0.11
0.11
0.11
0.10
0.10
0.10
0.10
0.09
0.09
0.09
0.09
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.08
0.07
0.07
(m)
1.5
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
(m)
5.48
5.19
4.90
4.62
4.39
4.15
3.92
3.69
3.46
3.32
3.17
3.03
2.89
2.74
2.60
2.45
2.31
2.16
2.02
1.90
1.79
1.67
1.56
1.44
1.33
1.21
1.10
0.98
0.87
0.78
0.69
0.61
0.52
0.43
0.35
0.26
0.17
0.09
0.00
(m)
6.98
6.19
5.90
5.62
5.39
5.15
4.92
4.69
4.46
4.32
4.17
4.03
3.89
3.74
3.60
3.45
3.31
3.16
3.02
2.90
2.79
2.67
2.56
2.44
2.33
2.21
2.10
1.98
1.87
1.78
1.69
1.61
1.52
1.43
1.35
1.26
1.17
1.09
1.00
(m)
7.0
5.5
4.5
3.0
2.0
The configuration of the MSE wall and the reinforcement are shown in Figure below.
Step 7: The following external stability is checked.
(1) Overturning resistance, check with and without surcharge.
FSOT =
å (Resisting moments)
å (Overturning moments)
The overturning moment is solely caused by the horizontal component of the
active soil force, Pa.
Since the active soil pressure is: s h = s hs = 0.333 ´ 21z = 7z
the total resultant active soil force is:
1
1
Pa = g H 2 K a = ´ 21´10 2 ´ 0.333 = 350 kN/m
2
2
It is assumed that the lateral earth pressure is horizontal without vertical
component. This assumption is conservative.
(Overturning moments) = Pa
𝐻
3
= 350 
10
3
= 1166.7 kNm/m
The reinforced soil of the MSE wall acts as one entity. Therefore, the resisting
moments are caused by the weight of the reinforced soil section, which can be
divided into three sections, as shown in the Figure.
Calculate the reinforced section weights per meter length of the wall (the unit
weight of backfill is 20 kN/m3):
W1 = 2.0  10.0  21 = 420 kN/m
W2 = 1.0  8.5  21 = 178.5 kN/m
W3 = 1.5  6.5  21 = 204.75 kN/m
W4 = 1.0  4  21 = 84 kN/m
W5 = 1.5  2  21 = 63 kN/m
So: (Resisting moments) =
moments caused by the reinforced soil weight +
= 420  1.0 + 178.5  2.5 + 204.75  3.75 + 84  5 + 63  6.25
= 2447.8 kNm/m
2447.8
So: FSOT = 1166.7 = 2.1 > 2.0, acceptable
(2) Sliding resistance between the first layer of reinforcement and the underlying
foundation soil. The contact width is B = 2.0 m.
FSS =
S(Resisting forces) [ca + s v tan d 2 ]B
=
S(Sliding forces)
P cosf ¢
a
1
(Resisting forces) = [ca + s v tan d2 ]B =(22 + 2110tan9)2=110.5 kN/m
(Sliding forces) = Pa = 350 kN/m
110.5
= 0.31 < 1.5 Not Acceptable
350
In order to increase the FS to 1.5, use the 10.0 m geogrid at the bottom of the
wall. So:
(Resisting forces) = [ca + s v tan d2 ]B =(22 + 2110tan9)10=552.5 kN/m
FS =
FS =
552.5
= 1.57 >1.5
350
Acceptable
(3) Bearing capacity of the foundation.
The entire soil wall exerts even pressure on the foundation soil, the pressure is
not eccentric. Use Terzaghi’s ultimate bearing capacity equation for wall
footing:
1
qult = c¢2N c + qN q + g 2 BNg
2
Use embedment depth of 0.5m, find q = 19  0.5 = 9.5 kN/m2
Note: B = 10.0 m
From 2 = 10, find Nc = 9.607, Nq = 2.694, N = 0.559
So:
kN/m2
qult = 25  9.607 + 9.5  2.694 + 0.5  19  10  0.559 = 318.9
The even pressure exerted by the MSE wall is:
v = 210 kN/m2.
FSbearing =
318.9
= 1.52 < 2.0
210
Not Acceptable.
Conclusion: the foundation soil is too weak. The foundation should be
improved to increase the shear strength before the construction of the MSE
wall.
12. An existing roadside embankment needs to be reinforced using geogrid sheets. The
embankment is 15-m high and the slope angle is 40. The in-situ soil will be used as
the reinforced backfill. The soil strength parameters are  = 220 and c = 15 kN/m2 in
both the embankment and foundation sections. The unit weight is 17.3 kN/m3. The
critical failure surface is assumed to be a toe circle, and the center is located at (+3 m,
+18 m) with respect to the toe at (0,0). Assume the groundwater table is far below the
foundation. The ultimate tensile strength of the geogrid is 75 kN/m. The vertical
spacing is chosen to be 40 cm. For the critical failure circle, how many layers of
geogrid are required to raise the factor of safety to 1.4?
Solution: Not provided.
13. Geosynthetic reinforced slope design. A 1-km long, 5-m high, 2.5H:1V side-slope
road embankment is to be widened to add one more lane. At least a 6-meter wide
extension is required to allow the additional lane plus shoulder improvements. A
1H:1V reinforced soil slope from the toe of the existing slope will provide a 7.5meter width to the alignment. The following design parameters have been provided
for the first three design steps.
Step 1: Establish the geometric, loading, and performance requirements for design.
(1) Geometric and loading requirements:
 Slope height, H = 5m
 Slope angle,  = 45
 Surcharge load, q = 10 kPa (for pavement)
 Temporary live load, q = 0
 Design seismic acceleration, Am = 0
(2) Performance requirements:
 External stability
o Sliding: FS  1.3
o Deep-seated (overall stability): FS  1.3
o Local bearing failure: FS  1.3
o Dynamic loading: no requirement
 Compound failure: FS  1.3
 Internal slope stability: FS  1.3
Step 2: Determine the engineering properties of the in situ soils.
 Subsurface exploration revealed that the slope and the foundation contain stiff
and low-plasticity silty clay with traces of sand.
 Strength parameters of the slope:
o cu = 100 kPa,  = 28, c = 0
o dry = 19 kN/m3, wopt = 15%
o The groundwater table is 45 m below the toe level
Step 3: Determine the properties of reinforced fill and the retained natural soil.
 Gradation and plasticity index.
 Compaction results: dry = 21 kN/m3, wopt = 15%
  = 33, c = 0
 Soil pH = 7.5
Follow the remaining steps (4 ~7) in this book to design the reinforcements and check
the external stability.
Solution: Not provided.
14. A geotextile is preferred as the filter that wraps around a perforated pipe for a
highway trench drain. The depth of the trench drain is 1.0m. Along the highway, three
types of soil are encountered. The sieve analyses of the soils are listed in the
following table. The permeability of the soils are measured as follows: ksoil A =
3.010-2 cm/sec, ksoil B = 3.610-3 cm/sec, ksoil C = 2.010-2 cm/sec. Select a geotextile
filter that can be used as the filter along the entire highway section.
Sieve number
Percent passing by mass (%)
Soil B
100
96
86
74
40
15
Soil A
95
90
78
55
10
1
4
12
20
40
100
200
Soil C
100
100
93
70
11
0
15. Geotextile filter design
100
90
% Finer by Weight
80
70
60
50
40
30
20
10
0
10
1
0.1
Soil Diameters (mm)
Figure 8.22 Grain size distribution in Problem 8.15
0.01
A geotextile is used as a filter behind a geonet drain in a reinforced slope. The grain size
distribution of the retained natural soil is shown in Figure 8.22 The soil’s permeability is
measured to be 4.810-4 cm/sec. Select a geotextile that meets the soil retention,
drainage, clogging resistance, and survivability criteria and its type (woven or
nonwoven).
Solution:
(1) Soil Retention Criteria
Assume the flow is steady state in stable soils, based on Equation (8.42):
AOS  B  D85
From Figure 8.22, the backfill soil retained on the No. 200 sieve (0.075 mm) =
80%, so the soil is coarse material. Find D10 = 0.03 mm, D30 = 0.14 mm, D60 =
0.54 mm, D85 = 1.7 mm.
Cu = D60/D10 = 18  8:
So: B = 1.0
AOS  B  D85 = 1.0  1.7 = 1.7 mm
(2) Drainage Criteria
(a) Permeability criteria:
It is judged that the filtration in the reinforced slope is a critical application
under severe conditions. Based on Equation (8.45):
kgeotextile  10 ksoil = 10  (4.8  10-4) cm/sec = 4.8  10-3 cm/sec
If students judge the problem is for less critical applications and less severe
conditions, that is fine. The:
kgeotextile  ksoil = 4.8  10-3 cm/sec
(b) Permittivity criteria:
The backfill soil passing the No. 200 sieve (0.075 mm): about 20% (15% to
50%), so,
Permittivity   0.2 sec-1
(c) Flow criteria
It is assumed the entire geotextile area is available for flow, so the optional
criteria are not checked.
(3) Long-term flow compatibility (clogging resistance):
It is judged that filtration in the reinforced slope is a critical application under
severe condition, so geotextile is first selected based on the above specifications
(AOS  1.7 mm, kgeotextile  4.8  10-3 cm/s, and   0.2 sec-1); then laboratory
filtration tests should be conducted using the selected geotextile and the in-situ
soil samples and hydraulic conditions, per ASTM D5101, “Measuring the SoilGeotextile System Clogging Potential by the Gradient Ratio” and the FHWA
publication “Geosynthetic Design & Construction Guidelines Reference Manual”
(FHWA 2008), so the geotextile’s clogging resistance can be evaluated.
(If students judge the problem is for less critical applications and less severe
conditions, that is fine. The criteria will vary accordingly)
(4) Survivability and durability criteria:
Nonwoven geotextile is generally used filter. Based on Table 8.6, the selected
geotextile should have the following minimum strength values for construction
survivability and durability.
Grab strength:
Sewn seam strength:
Tear strength:
Puncture strength:
700 N
630 N
250 N
1375 N