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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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Module - 6LECTURE - 34
Geosynthetics for reinforced soil retaining walls
Recap of previous lecture…..
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
+)a-L[{(wxa+)]αsin(x2w
+)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
+)a-L[{(wxa+)]αsin(x2w
+)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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. 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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Step 5: Calculation of factor of safety against overturning
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
Step 6: Calculation of factor of safety against sliding
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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)
Step 8: Check against bearing pressure
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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 5th layer (m) 4
width of the 6th layer (m) 3
width of the 7th layer (m) 3
width of the 8th layer (m) 2
width of the 9th layer (m) 2
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
)icos()cos()isin()sin(1)cos(cos
)(cosK
i = Backfill slope angle = Angle of friction between wall and soil α = Wall inclination with vertical = Soil friction angle
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Step 9: Calculation of factor of safety against overturning
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
Step 10: Calculation of factor of safety against sliding
Driving force (Fd) = Pa cos α
Resisting force (Fr)= (Wgabion + Ws + Wsoil - Pa sin α) Ci tan
(FOS)sliding = Fr / Fd > 1.5 (safe)
Step 11: Calculation of eccentricity (e)
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 12: Check against bearing pressure
Maximum base pressure developed (Pb) = (Wv / L) (1 + 6e/L) < qallowable (safe)
qallowable = Soil bearing pressure
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 17: Check for embedded length
At any layer,
The length of embedment past the wedge (Le) > Minimum embedded length (Lem) (OK)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. 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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 9: Calculation of factor of safety against overturning
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 10: Calculation of factor of safety against sliding
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 11: Calculation of eccentricity (e)
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Step 12: Check against bearing pressure
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)
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
Please let us hear from you
Any question?
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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