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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 - 6LECTURE - 34
Geosynthetics for reinforced soil retaining walls
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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)
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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
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)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
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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
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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
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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
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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
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Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Design of gabion wall with welded wire anchor mesh as horizontal tie-backs
Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Prof. J. N. Mandal, Department of Civil Engineering, IIT Bombay
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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