GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE · 1 1 50 500 1 Solving, h r = 0.667 m = 66.7 cm Step 4: Saving in embankment thickness (%) With the use of geotextile saving in

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


Module - 8LECTURE - 42

Geosynthetics for embankments on soft foundations

Recap of previous lecture…..


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)

Step 4: Check for sliding failureEmbankment slides over the reinforcement after formationof crack in the embankment

Force diagram:


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


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


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


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


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)


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


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)


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


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 %


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 %


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

Geotextile base layer beneath the road embankment


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,


rut depth ≤ 50 mm

NC = 2.8 with no geotextile,


rut depth ≤50 mm

NC = 6 with geotextile,


rut depth ≥ 100 mm

Value of Nc for different rut depths (Stewart et al., 1977)


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


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


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


Substituting the values in the Boussinesq’s equation,

2/32

uh18.01

1150028Solving,hu = 0.91 m = 91cm

2/32h

hr1

11P


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


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


Please let us hear from you

Any question?


Prof. J. N. Mandal

Department of civil engineering, IIT Bombay, Powai , Mumbai 400076, India. Tel.022-25767328email: [email protected]


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GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE · 1 1 50 500 1 Solving, h r = 0.667 m = 66.7 cm Step 4: Saving in embankment thickness (%) With the use of geotextile saving in