GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE · allowable = Allowable bearing capacity of the subgrade soil Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay. Example:

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 - 6LECTURE - 34

Geosynthetics for reinforced soil retaining walls

Recap of previous lecture…..


Design of geotextile wrap-around-faced wall

Gabion walls

General

Design of gravity gabion wall (partly covered)

Step 2: Calculation of overturning momentOverturning moment (Mo) = Pa x hy /cos α

Step 3: Calculation of weight of Gabion (Wgabion)Wgabion = γg x (volume of wall per unit length)

γg = Gabion fill density

Step 4: Calculate the horizontal distance of point ofapplication of the weight of gabion wall from toe (hg )hg = (a . X) / A

a = Individual area of the gabions parallel to the slope,

X = distance of C.G. of the individual gabion from toe

A = Total area of the gabion wallProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

)wxa+wxL(

)]αsin(x)2w

+w(+)αcos(x}2a

+)a-L[{(wxa+)]αsin(x2w

+)αcos(x2L

[wLx=h g

)wxa+wxa+wxL(

)]αsin(x)2w

+w2(+)αcos(x}2a

+)a-L[{(wxa+)]αsin(x)2w

+w(+)αcos(x}2a


+)αcos(x2L

[wLx=hg

For two bottom gabions,

For three bottom gabions,

For four bottom gabions,

)wxa+wxa+wxL(

)]αsin(x)2w

+w3(+)αcos(x}2b

+)b-L[{(wxb+)]αsin(x)2w

+w2(+)αcos(x}2a

+)a-L[{(wxa

+)wxa+wxa+wxL(

)]αsin(x)2w

+w(+)αcos(x}2a


+)αcos(x2L

[wLx=h g

Similarly for more number of gabions, hg can be determined. Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay

w = thickness of each gabiona = width of the second and third gabion from the bottomb = width of the fourth and fifth gabion from the bottomc = width of the sixth and seventh gabion from the bottomd = width of the eighth and ninth gabion from the bottome = width of the tenth gabion from the bottom (top gabion asshown in the Figure)


Step 6: Calculation of factor of safety against overturning

Overturning moment (Mo) = Pa x hy /cos α

Resisting moment (Mr) = Wgabion x hg

(FOS)overturning = Mr/ Mo > 2 (safe)

Step 5: Calculation of factor of safety against sliding

Driving force (Fd) = Pa - Wgabion sin α

Resisting force (Fr) = Wgabion cos α x Ci tan

(FOS)sliding = Fr / Fd > 1.5 (safe)


Step 7: Calculation of eccentricity (e)

e = (L/2) – (Mr – Mo)/ Wg cos α

- L/6 < e < + L/6 (OK)

Step 8: Check against bearing pressure

Maximum base pressure developed (Pb) = (Wg cos α/ L) (1 + 6e/ L) < qallowable (safe)

qallowable = Allowable bearing capacity of the subgrade soil


Example:Design a gravity gabion wall with following information:

Wall height (H) = 10 m, Wall thickness (tg) = 1 m

Surcharge (q) = 0 kPa, Backfill slope angle (i) = 0°

Angle of friction between wall and soil () = 0°

Wall inclination with vertical (α) = -6°

Soil friction angle (s) = 32°, Soil density (γs) = 17 kN/m3

Gabion fill density (γg) = 25 kN/m3

Soil bearing pressure (qallowable) = 500 kPa

Scale correction factor (Ci) = 0.7

Maximum total base width (B) = 0.7 H = 0.7 x 10 = 7 mProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay


Solution:

Step 1: Calculation of earth pressure and its point ofapplication

The active earth pressure co-efficient = Ka

According to Coulombs’ derivation,

2

2

2

a

)icos()cos()isin()sin(1)cos(cos

)(cosK

Hence,

27.0

))6(0cos())6(0cos()032sin()032sin(1))6(0cos()6(cos

))6(32(cosK 2

2

2

a


Therefore, the total active thrust on the wall (Pa)= Ka (γsH2/2+qH)= 0.27 (17 x 102/2 + 0 x 10)= 229.5 kN/m

Vertical distance of the point of application of the resultantnormal force (Pa) from toe,

sinL

q2H

q3H

3Hh

sv

sv

vy

H = 10 m (Given)L = 0.7H = 0.7 x 10 = 7 mHv = H cos = 10 x cos(6) = 9.95 m

m6.2)6sin(7

170295.9

170395.9

395.9h y


Step 2: Calculation of overturning moment

Overturning moment (Mo)= Pa x hy /cos

= 229.5 x 2.6/ cos(6)

= 599.99 kN-m/m

Step 3: Calculation of weight of Gabion

Weight of gabion (Wgabion)= γg x (volume of wall per unit length)

= 25 x {1 x (7+5+5+4+4+3+3+2+2+1)}

= 900 kN/mProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Step 4: Calculation of horizontal distance from toe tothe point of application of Wgabion

hg = (a. X)/ A

a = Individual gabion area parallel to slope of 6,

X = distance of C.G. of the individual gabion from toe

A = Total area of the Gabion wall = 1x (7+5+5+4+4+3+3+2+2+1)

= 36 m2


36))6sin(5.3)6cos(5(1x4))6sin(5.2)6cos(5.4(1x5))6sin(5.1)6cos(5.4(1x5))6sin(5.0)6cos(x5.3(1x7hg

36))6sin(5.7)6cos(6(1x2))6sin(5.6)6cos(5.5(1x3))6sin(5.5)6cos(5.5(1x3))6sin(5.4)6cos(x5(1x4

36))6sin(5.9)6cos(5.6(1x1))6sin(5.8)6cos(6(1x2

Therefore,

or, hg = 5.17 m


Overturning moment (Mo) = 599.99 kN-m/m

Resisting moment (Mr)= Wgabion x hg = 900 x 5.17 = 4653 kN-m/m

(FOS)overturning = Mr/ Mo = 4653/ 599.99 = 7.76 > 2 (safe)Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay


Driving force (Fd)= Pa - Wgabion sin= 229.5 – 900 sin (6)= 135.424 kN/m

Resisting force (Fr)= Wgabion cos x Ci tanφ= 900 x 0.7 x tan (32)= 391.51 kN/m

(FOS)sliding = 391.51/ 135.424 = 2.89 > 1.5 (safe)


Step 7: Calculation of eccentricity

Eccentricity (e) = [(L/2) – (Mr – Mo)/ (Wg cos)]

Hence, e = (7/2) – (4653 – 599.99)/ (900 cos(6)) = -1.03

Now, L/6 = 7/ 6 = 1.17; Therefore, -1.17 < e < +1.17 (ok)


Maximum base pressure developed (Pb)= (Wg cos/ L) (1 + 6e/ L)

= (900 cos/ 7) {1 + (6 x (1.03)/7)}

= 240.76 kPa < (qallowable = 500 kPa) (safe)Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Design of Gabion wall in Excel

angle of internal friction of backfill soil (s) 32acute angle of back face slope of wall with vertical (α) -6wall base inclination with horizontal (α) -6angle of wall friction (δ) 0slope angle of backfill surface (i) 0Unit weight of backfill soil (γs) (kN/m3) 17Height of Gabion wall (H) (m) 10width of the wall (tg) (m) 1Gabion fill density(γg) (kN/m3) 25Maximum total base width (L) (m) 7surcharge load (q) (kPa) 0Scale correction factor (Ci) 0.7Soil bearing pressure (qallowable) (kPa) 500


width of the 1st layer (base layer) (m) 7

width of the 2nd layer (m) 5

width of the 3rd layer (m) 5

width of the 4th layer (m) 4






width of the 10th layer (m) 1Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Co-efficient of active earth pressure (Ka) 0.2687Active thrust on the wall (Pa) (kN/m) 228.4hy (m) 2.583hg (m) 5.165

Calculation:

Check for stability


Check for Stability

Weight of Gabion (Wgabion) (kN/m) 900

Overturning moment (kN-m/m)Mo 593.2

(FOS)overturning 7.837 > 2 (safe)Resisting moment (kN-m/m)Mr 4648.864

Driving force (kN/m)Fd 134.2988

(FOS)sliding 2.915 > 1.5 (safe)Resisting force (kN/m)Fr 391.5111

eccentricity (e) (m) -1.031086208 > - 1.166

Maximum base pressure (Pb) (kPa) 240.8745 < 500 (safe)

BACKProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Design of gabion wall with welded wire anchor mesh as horizontal tie-backs


Wall height vertically = Hg, Wall thickness = tgSurcharge = q, Backfill slope angle = i

Wall inclination with vertical = α

Soil friction angle =

Soil density = γs

Gabion fill density = γg

Soil bearing pressure = qallowable

Scale correction factor = Ci

Maximum base width (L) = 0.7 Hg

Ultimate tensile strength = Tult


Design Steps:External Stability:Step 1: Calculation of earth pressure and its point ofapplication

Total active thrust on the wall (Pa) = Ka (γsH2/2+qH)

Ka = active earth pressure co-efficient

2

2

2

a


)(cosK

i = Backfill slope angle = Angle of friction between wall and soil α = Wall inclination with vertical = Soil friction angle


When surcharge is applied over the backfill, the verticaldistance of point of application of the resultant normalforce (Pa) from base = hy

)q2H(

)q3H(x

3H

h

sg

sg

gy

γs = Soil density

Hg = Wall height


Step 2: Calculation of overturning moment about toe

Overturning moment (Mo) = Pa cos α x hy + Pa sin α x (tg + hy tan α)

Step 3: Calculation of weight of Gabion (Wgabion)

Weight of gabion (Wgabion)= ½ x (tg + tg) x Hg x γg

= Hg x tg x γg

γg = Gabion fill density


Step 4: Calculation of horizontal distance from toe to thepoint of application of Wgabion

hg = tg/ 2 + (Hg/ 2) tan α

tg = Wall thickness, Hg = Wall heightα = Wall inclination with vertical

Step 5: Calculation of weight of surcharge (Ws)

Weight of surcharge (Ws) = q x l

l = L – tg – Hg tan α (L = base width = 0.7 Hg)

q = surcharge over the backfill surface


Step 6: Horizontal distance from toe to the point ofapplication of Ws

Horizontal distance of the weight of surcharge from thetoe of the wall = hs

hs = tg + Hg tanα + l/2

l = L – tg – Hg tanα

Step 7: Calculation of weight of Backfill soil (Wsoil)

Wsoil = (½ x Hg tanα x Hg + l x Hg) γs

γs = Density of backfill soilProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Step 8: Horizontal distance from toe to the point ofapplication of Wsoil

hsoil = [(Hg2 tanα){tg + (Hg/ 3)tanα} + (Hg x l){tg + Hg tanα + l/2}] x (γs/ Wsoil)

Wsoil = Weight of backfill soil, l = L – tg – Hg tanα


Overturning moment (Mo)

= Pa cos α x hy + Pa sin α x (tg + hy tanα)

Resisting moment (Mr)

= Wgabion x hg + Ws x hs + Wsoil x hsoil

(FOS)overturning = Mr / Mo > 2 (safe)Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay


Driving force (Fd) = Pa cos α

Resisting force (Fr)= (Wgabion + Ws + Wsoil - Pa sin α) Ci tan

(FOS)sliding = Fr / Fd > 1.5 (safe)


e = (L/ 2) – (Mr – Mo)/ Wv -L/6 < e < L/6 (ok)

Wv = Total vertical downward force over the sub-grade soil = Wgabion + Ws + Wsoil - Pa sinα



Maximum base pressure developed (Pb) = (Wv / L) (1 + 6e/L) < qallowable (safe)

qallowable = Soil bearing pressure


Step 13: Calculate spacing and tensile force at each layer

The vertical pressure at any layer,σz = γs x z + q

γs = Soil densityz = depth of the layer from the top of the wallq = surcharge

Therefore, tensile strength at any layer,Tcalculated = σz x sv x Ka

sv = vertical spacing of reinforcementsKa = co-efficient of active earth pressure


Provided ultimate tensile strength = Tultimate

Hence, Tallowable = Tultimate/ Factor of safety (FS)

Using Tallowable = σz x sv x Ka, determine the maximumspacing required at the bottom.

Getting an idea, assume suitable spacing for the layersand calculate tensile strength (Tcalculated) at any layer

Step 14: Check tensile strength at each layer

Tcalculated < Tallowable (OK)

Where, at any layer, Tcalculated = σz x sv x Ka


Step 15: Calculation of minimum embedded length (Lem)Minimum embedded length (Lem)

= FS x Tcalculated/ (2 x σz x Ci x tan) Ci = scale correction factorϕ = soil - to - soil friction angle

Step 16: Calculation of actual embedded length (Le)At the top of the wall, distance to the wedge failure planefrom the back of the wall,La = Hg tan (45⁰ - /2) – Hg tanα

At any layer at a depth z, the length of embedment past theWedge, Le = L – tg – La x (Hg – z)/ Hg


Step 17: Check for embedded length

At any layer,

The length of embedment past the wedge (Le) > Minimum embedded length (Lem) (OK)


Example:Design a Gabion wall with welded wire Anchor mesh ashorizontal tie-backs for soil reinforcement (MSE Walls).

Wall height (Hg) = 10 m, Wall thickness (tg) = 1 m

Surcharge (q) = 39 kPa, Backfill slope angle (i) = 0°

Wall inclination with vertical (α) = -6°

Soil friction angle (ϕs) = 32°, Soil density (γs) = 18 kN/m3

Gabion fill density (γg) = 17 kN/m3

Soil bearing pressure (qallowable) = 500 kPa

Scale correction factor (Ci) = 0.7, Tult = 60 kN/m

Maximum total base width (L) = 0.7 Hg = 0.7 x 10 = 7 mProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay


Solution:

Step 1: Calculation of earth pressure and its point ofapplication

The active earth pressure co-efficient = Ka

According to Coulombs’ derivation,

2

2

2

a


)(cosK

Hence, 27.0

))6(0cos())6(0cos()032sin()032sin(1))6(0cos()6(cos

))6(32(cosK 2

2

2

a


Therefore, the total active thrust on the wall (Pa)= Ka (γsHg

2/2+qHg)= 0.27 (18 x 102/2+ 39 x 10)= 346.6 kN/m

Vertical distance of the point of application of the resultantnormal force (Pa) from base,

Hg = 10 m (Given) q = 39 kPa, γs = 18 kN/m3

sg

sg

gy q2H

q3Hx

3H

h

m84.3

1839x210

1839x310

x3

10h y


Step 2: Calculation of overturning moment

Overturning moment (Mo)= Pa cosα x hy + Pa sinα x (tg + hy tanα)

= 346.6 x cos(6) x 3.84 + 346.6 x sin(6) x {1 + 3.84 x tan(6)}

=1374.50 kN-m/m

Step 3: Calculation of weight of Gabion

Weight of gabion (Wgabion)= ½ x (tg + tg) x Hg x γg

= Hg x tg x γg

= 10 x 1 x 17 = 170 kN/mProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Step 4: Calculation of horizontal distance from toe tothe point of application of Wgabion

hg = tg/2 + (Hg/2) tan (α) = 1/2 + (10/2) tan(6) = 1.026 m

Step 5: Calculation of weight of surcharge

Weight of surcharge (Ws) = q x l

l = L – tg – Hg tanα = 7 – 1 - 10 x tan 6 = 4.95 m

Therefore, Ws = 39 x 4.95 = 193 kN/m

Step 6: Horizontal distance from toe to Ws

hs = tg + Hg tanα + l/2 = 1+ 10 x tan (6) + 4.95/2 = 4.53 mProf. J. N. Mandal, Department of Civil Engineering, IIT Bombay

Step 7: Calculation of weight of Backfill soil (Wsoil)

Wsoil = (½ x Hg tanα x Hg + l x Hg) γs

= (½ x 10 x tan (6) x 10 + 4.95 x 10) x18

= 985.41kN/m

Step 8: Horizontal distance from toe to Wsoil

hsoil = [(Hg2 tanα) {tg + (Hg/3)tanα}

+ (Hg x l){ tg + Hg tanα + l/2}] x (γs/ Wsoil)= [(102 tan6) {1 + (10/3)tan6}

+ (10 x 4.95) {1 + 10 tan6 + 4.95/2}] x (18 / 985.41)= 3.5 m



Overturning moment (Mo) = 1374.50 kN-m/m

Resisting moment (Mr)= Wgabion x hg + Ws x hs + Wsoil x hsoil

= 170 x 1.026 + 193 x 4.53 + 985.41 x 3.5

= 4492 kN/m

(FOS)overturning = Mr/ Mo = 4492/ 1374.50 = 3.27 > 2 (safe)



Driving force (Fd)= Pa cos α

= 346.6 x cos(6) = 344.69 kN/m

Resisting force (Fr)= (Wgabion + Ws + Wsoil - Pa sin α) x Ci x tan

= (170 + 193 + 985.41-36.23) x 0.7 x tan (32)

= 573.96 kN/m

(FOS)sliding = 573.96 / 344.69 = 1.67 > 1.5 (safe)



e = (L/2) – (Mr – Mo)/ Wv

Wv = Wgabion + Ws + Wsoil - Pa sinα = (170 + 193 + 985.41- 36.23) = 1312.18 kN/m

Hence, e = (7/2) – (4492 – 1374.50)/ 1312.18 = 1.124

Now, L/ 6 = 7/ 6 = 1.17

Therefore, e = 1.124 < L/ 6 (ok)



Maximum base pressure developed (Pb) = (Wv/ L) (1 + 6e/L)

= (1312.18 / 7) {1 + (6 x 1.096/7)}

= 363.65 kPa < 500 kPa (qallowable) (safe)


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]


Documents

GEOSYNTHETICS ENGINEERING: IN THEORY AND PRACTICE · allowable = Allowable bearing capacity of the subgrade soil Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay. Example: