geosynthetics engineering: in theory and practice · 1 1 50 500 1 solving, h r = 0.667 m = 66.7 cm...
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GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE
Prof. J. N. Mandal
Department of civil engineering, IIT Bombay, Powai , Mumbai 400076, India. Tel.022-25767328email: [email protected]
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Module - 8LECTURE - 42
Geosynthetics for embankments on soft foundations
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Recap of previous lecture…..
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Design of basal reinforced embankment (remaining)
Serviceability limit state
Placement of geosynthetics underneath embankment
Construction of basal reinforced embankment
Widening of existing roadway embankment
Design example (partly covered)
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Step 4: Check for sliding failureEmbankment slides over the reinforcement after formationof crack in the embankment
Force diagram:
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Total resisting force (Rg)
= Ws. tan e + Ca. Ls (Ca = 0 for granular soil)
= 0.5. γe. Ls. He. tan e + 0
= 0.5 x 17 x 8.75 x 3.5 x 0.8 x tan30°
= 120.233 kN/m
Factor of safety against sliding
= Rg / Pfill
= 120.233/34.36 = 3.5 > 2 (Safe)
Total driving force (Pfill) = 34.36 kN/m
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 5: Check for pullout strength
(Tg)design= τtop Le + τbottom Le
Le= Embedded length of the geosynthetic beyond slip line,τtop = σv tan δe, and
τbottom = Ca
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Now, For top layer (Tg)design = 200 kN/m
200 = (17 x 3.5 x 0.7 x tan 30° + 3.6) x Le
or, Le = 7.23 m
For both layers adopt Le = 7.23 m
Given that,Ci = Interaction coefficient = 0.7,
Ca = 40 % of Cu = 0.4 x 9 = 3.6 kPa in foundation
Hence, (Tg)design = σv tan δe Le + Ca Le
= γe He x Ci x tan e x Le + Ca Le
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 6: Required elastic strength of the geotextile
Considering 5% strain, εf = 0.05
Now, Tdesign = required tensile strength of the geotextile fortop layer = 200 kN/m.
Required Elastic Modulus of the geotextile (Erequired) for toplayer = 200/ 0.05 = 4000 kN/m
Similarly, Required Elastic Modulus of the geotextile(Erequired) for bottom layer = 100/ 0.05 = 2000 kN/m
f
reqdreqd
TE
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 7: Check for lateral squeezing
PA = 0.5.γf.Hf2 Ka - 2.cf. Hf Ka + qe1. Hf Ka
= 0.5. γf.Hf2 - 2.cf. Hf + qe1. Hf (qe1 = γe x He + qs)
PB = 0.5. γf.Hf2.Kp + 2.cf. Hf. Kp
= 0.5. γf.Hf2 + 2.cf. Hf (Kp = 1)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Shear force at the top of the foundation block in EFportion (Tt)= (Ca + vt tanf) x Ls
= Ca Ls
= 0.4 x 9 x 8.75 = 31.5 kN/m
Here, qs = 0
PA = 0.5 x 16 x 2.52 – 2 x 9 x 2.5 + (17 x 3.5) x 2.5 = 153.75 kN/m
PB = 0.5 x 16 x 2.52 + 2 x 9 x 2.5 = 95 kN/m
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Shear force at the bottom of the foundation block in EFportion (Tb)= (cf + vb tanf) x Ls
= cf Ls = 9 x 8.75 = 78.75 kN/m
Factor of safety against squeezing,
A
btBsq P
TTPFS
3.1335.175.153
75.785.3195FSsq
(Safe)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 8: Check for Drainage and Filtration
Required grain size distribution of sub-grade soil
Calculate :
1) Retention criteria (maximum apparent opening size of Geotextile)
2) Permeability criteria (Kg > Ks)
3) Clogging criteria (minimum apparent opening size of Geotextile)
Step 9: Check settlement and construction sequence
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 10: Required properties of geosynthetics
Ultimate tensile strength of Geotextile in the machine direction ≥ 200 kN/m (top layer )
Ultimate tensile strength of Geotextile in the machine direction ≥ 100 kN/m (bottom layer)
Ultimate tensile strength of Geotextile in the cross-machine direction ≥ 60 kN/m (for both layers)
Seam strength of Geotextile ≥ 60 kN/m (for both layers)
Limit strain = 5 %
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Summary of required geotextile properties:
Ultimate tensile strength of geotextile in the machinedirection ≥ 200 kN/m (top layer)
Ultimate tensile strength of geotextile in the machinedirection ≥ 100 kN/m (bottom layer)
Ultimate tensile strength of geotextile in the cross-machine direction ≥ 60 kN/m (for both layers)
Seam strength of Geotextile ≥ 60 kN/m (for both layers)
Limit strain = 5 %
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Example:
Determine the minimum height of a low embankment (i.e.unpaved roadways) without and with geotextile toprevent foundation failure due to wheel load. Alsocompare the results. Use the following given details:
Cohesion of the sub grade soil is 10 kPa,Axle load = 102 kN,
Wheel load = 51 kN,
Allowable tire pressure (P) = 500 kPa,
Rut depth = 5 cm, and
Number of passages = 1000Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Geotextile base layer beneath the road embankment
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 1: Given DataAxle load =102 kN
Allowable tyre pressure (P) = 500 kpa
Rut depth = 50 mm
Cu=10 kPa
NC = 3.3 with no geotextile,
No. of passes = 100, and
rut depth > 100 mm
NC = 5 with geotextile,
No. of passes = 1000, and
rut depth ≤ 50 mm
NC = 2.8 with no geotextile,
No. of passes = 1000, and
rut depth ≤50 mm
NC = 6 with geotextile,
No. of passes = 100, and
rut depth ≥ 100 mm
Value of Nc for different rut depths (Stewart et al., 1977)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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The vertical stress (h) at a depth ‘h’ is given by Boussinesq (1883) for a loaded circular area.
Vertical stress due to circular loaded area
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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2/1
Pwr
= {51/( x 500)}1/2 = 0.18 m
2/32h
hr1
11P
P = allowable tire pressure = 500 kPar = Radial horizontal distance
Axle load =102 kN
w = single wheel load = 102/ 2 = 51 kN
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 2: Calculation of the depth of the embankment (hu) without geotextile.
For rut depth = 50 mm and number of passes =1000, from Stewart et al. (1977),
Nc = 2.8 (without geotextile)
hu = Nc x Cu = 2.8 x Cu = 2.8 x 10 = 28 kPa
hu = bearing capacity of the foundation soil without geotextile
cu = cohesion of the foundation soil
hu = thickness of embankment without geotextile
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Substituting the values in the Boussinesq’s equation,
2/32
uh18.01
1150028Solving,hu = 0.91 m = 91cm
2/32h
hr1
11P
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Step 3: Calculation of the depth of the embankmentwith geotextile (hr)
For rut depth = 50 mm and number of passes =1000, from Stewart et al. (1977),
Nc = 5 (with geotextile)
hr = Nc x Cu = 5 x Cu = 5 x 10 = 50 kPa
hr = bearing capacity of the foundation soil with geotextile
hr = thickness of embankment with geotextile
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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2/32
rh18.01
1150050
Solving, hr = 0.667 m = 66.7 cm
Step 4: Saving in embankment thickness (%)
With the use of geotextile saving in thickness= 91cm - 66.7 cm = 24.3 cmPercentage saving = (24.3/ 91) x 100 = 26.7 %
Substituting the values in the Boussinesq’s equation,
2/32h
hr1
11P
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Please let us hear from you
Any question?
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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Prof. J. N. Mandal
Department of civil engineering, IIT Bombay, Powai , Mumbai 400076, India. Tel.022-25767328email: [email protected]
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay