1. “Design of Cantilever Retaining Wall”
GANDHINAGAR INSTITUTE OF TECHNOLOGY
“Design of Cantilever Retaining Wall”
Enrollment No Name
Jain Pritesh
Internal Guide: Prof. Jay Parmar Sanghvi Umang
Tiwari Yash
Pathan Mashira
Gandhinagar Institute of Technology: Department of Civil Engineering

2. Gandhinagar Institute of Technology : Department of Civil Engineering
Example :- Design of cantilever retaining wall to retain the earth of height 5.5m above lower ground level. Fix the basic dimension and carry out the stability checks of retaining wall. Design and detail all structural components.
Take SBC of soil = 175 Kpa
ø = 30˚, µ = 0.5
Unit weight of soil = 18 KN/m³
Use M20 grade of concrete and Fe 415 grade of steel.
Solution :-
1) Coefficient of earth pressure :-
Ø = 30˚ 𝑘 𝑎 = 1−𝑠𝑖𝑛∅ 1+𝑠𝑖𝑛∅ = 1−𝑠𝑖𝑛30° 1+𝑠𝑖𝑛30° = 1 3
𝑘 𝑝 = 1 𝑘 𝑎 =3
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2) Preliminary Dimensions :-
ϒ = 18 KN/m³
SBC of soil = 𝑞 0 = 175k 𝑝 𝑎 = 175 kN/m²
The minimum depth of foundation is
𝑑 𝑚𝑖𝑛 = 𝑞 0 ϒ 1−𝑠𝑖𝑛∅ 1+𝑠𝑖𝑛∅ 2 = =1.08 𝑚
Provide depth of foundation as 1.20 m.
Overall Height of retaining wall,
H = = 6.70 m
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4. Gandhinagar Institute of Technology : Department of Civil Engineering
Base Width (b)
For T shaped retaining wall, the minimum base width may be taken as :-
𝑏= 3𝑃 2𝛾 ⇒𝑏= 3× ×18 =3.35 𝑚
𝑃= 1 2 𝑘 𝑎 𝛾 𝐻 2 = 1 2 × 1 3 ×18× =134.67𝑘𝑁
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5. Gandhinagar Institute of Technology : Department of Civil Engineering
Consider toe width = 1 3 ×𝑏= 1 3 ×3.35=1.12 𝑚
Provide toe width = 1.20 m
Provide total base width = 4 m
Thickness of base slab :-
Thickness = 𝐻 12 𝑡𝑜 𝐻 15 = 𝑡𝑜 =0.558 𝑡𝑜 𝑚
Consider uniform thickness of base slab = 0.50 m
Thickness of stem at base :-
Pressure at the base of stem
𝑝 𝑎 = 𝑘 𝑎 𝛾ℎ h = = 6.20 m
Maximum moment at the base of stem
= 𝑘 𝑎 𝛾 ℎ 2 × ℎ 3 = 1 2 × 1 3 ×18× = 𝑘𝑁𝑚
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Factored moment = 1.5 × 238.3
𝑀 𝑢 = kNm
For Fe 415 steel 𝑀 𝑢 = 𝑓 𝑐𝑘 b 𝑑 2
357.48× = × 20 × 1000× 𝑑 2
d = mm
Consider 50 mm effective cover
Total d for stem required = = mm
Provide total depth of stem = 450 mm
Width of heel = 4.00 – 1.20 – 0.45 = 2.35 m
Provide top width of stem = 0.20 m
Provide shear key of size 0.45 m × 0.50 m below base to prevent sliding of the wall.
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3) Stability calculation :-
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8. Consider 1 m length of wall
Horizontal active earth pressure at the base of the wall
= 𝑘 𝑎 𝛾𝐻 = 1 3 ×18×6.7=40.2 𝑘𝑁/ 𝑚 2
Horizontal Loads :-
Load Type
Horizontal Load (kN)
Perpendicular distance from A (m)
Moment about A
(kN.m)
Active earth Pressure
1 2 ×40.2×6.7= 𝑘𝑁
6.7 3 =2.23 𝑚
kN . m
Total 𝑃 𝑎ℎ
= kN←
𝑀 𝑜
= 𝑘𝑁𝑚
Sliding force, 𝑃 𝑎ℎ = 𝑘𝑁←
Overturning moment, 𝑀 𝑜 = 𝑘𝑁.𝑚
Gandhinagar Institute of Technology : Department of Civil Engineering

9. Vertical Load : density of concrete = 25 kN/ 𝑚 3
density of soil = 18 kN/ 𝑚 3
Load Type
Vertical Load (kN)
Perpendicular distance from A(m)
Moment about A (kN.m)
1. Stem - 𝑊 1
𝑊 1 = (6.20×0.20)×25
= 31 kN
= 1.55 m
48.05
2. Stem - 𝑊 2
𝑊 2 = 1 2 ×0.25×6.2 ×25
= kN
×0.25
= 1.37 m
26.54
3. Base slab
𝑊 3
𝑊 3 = 4.0 ×0.5 ×25
= 50 kN
4 2 =2.0 𝑚
100.00
4. Shear key
𝑊 4
𝑊 4 = 0.45×0.50 ×25
= kN
= m
8.02
5. Backfill
𝑊 5
𝑊 5 = 2.35×6.2 ×18
= kN
=2.825 𝑚
740.88
Total ∑W = kN
𝑀 𝑟 =923.49𝑘𝑁𝑚
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Total downward load, ∑W = kN↓
Resisting moment, 𝑀 𝑟 =923.49
Let, distance of C.G. of vertical
load from the face of the toe,
i.e. from point A is x.
x = Net moment at A (toe)
𝑥= − =1.69 𝑚
Hence, eccentricity e = 𝑏 2 −𝑥
= −1.69
=0.31 𝑚
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11. Gandhinagar Institute of Technology : Department of Civil Engineering
Maximum pressure at A (toe) :-
𝑝 𝑚𝑎𝑥 = ∑𝑊 𝑏 1+ 6𝑒 𝑏 = × = [ ]
= kN/ 𝑚 2 …… at toe
Minimum Pressure at B (Heel)
𝑝 𝑚𝑖𝑛 = ∑𝑊 𝑏 1− 6𝑒 𝑏 = × = [ ]
= kN/ 𝑚 2 …… at heel
𝑝 𝑚𝑎𝑥 = 𝑘𝑁 𝑚 2 <𝑆.𝐵.𝐶. 𝑜𝑓 𝑠𝑜𝑖𝑙=175 𝑘𝑁/ 𝑚 2 …Safe
𝑝 𝑚𝑖𝑛 =49.25 𝑘𝑁/ 𝑚 2 >0 i.e. not negative.
Therefore, no tension at base.
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Factor of safety against overturning :-
Restoring moment, 𝑀 𝑟 = kN.m
Overturning moment, 𝑀 𝑜 = kN.m
F.S. = 𝑀 𝑟 𝑀 𝑜 = =3.07>1.55……𝑆𝑎𝑓𝑒
Factor of safety against sliding :-
Sliding force = 𝑃 𝑎ℎ =134.67←
Frictional force = µ.∑W = 0.5 × =
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Passive Pressure under the base key
𝑝 𝑎 = 𝑘 𝑝 𝛾 ℎ 1 =3×18×1
{ ℎ 1 = } = 54 kN/ 𝑚 2
Total 𝑃 𝑝 = 1 2 ×54×1
= 27 kN →
Total restoring force = μ . ∑W + 𝑃 𝑝 =
= 𝑘𝑁 →
F.S. = restoring moment sliding force = =1.57 > 1.55 ………Safe
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14. Gandhinagar Institute of Technology : Department of Civil Engineering
4) Design of stem :-
Steam acts as a cantilever of height 6.20 m
Subjected to uniformly varying load of = 𝑘 𝑎 𝛾ℎ= 1 3 ×18×6.2
= 37.2 kN/ 𝑚 2
B.M. at the base of stem = 1 2 ×37.2×6.2× 1 3 ×6.2
= kN.m
Factored moment, 𝑀 𝑢 =1.5×238.32= 𝑘𝑁.𝑚
𝐴 𝑠𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑,
𝑀 𝑢 𝑏 𝑑 2 = × × = {d = mm}
𝑃 𝑡 =0.730 % 𝐴 𝑠𝑡 = 𝑃 𝑡 100 ×𝑏𝑑= ×1000×400
= 𝑚𝑚 2
Provide 20mm Ф 100mm c/c ( 𝐴 𝑠𝑡 =3142 𝑚𝑚 2 )
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Distribution Steel :-
Average thickness of stem = =325𝑚𝑚
Provide minimum 0.12% steel as distribution steel
𝐴 𝑠𝑡 = ×1000×325=390 𝑚𝑚 2
Provide mm c/c (390 𝑚𝑚 2 ) on tension face on
the outer face use 0.06% steel both ways.
Provide mm c/c both ways on the outer face.
Curtailment of Vertical bars :-
Bending moment at depth h from top of wall
𝑀= 𝑘 𝑎 𝛾 ℎ 2 × 1 3 ℎ= 1 6 𝑘 𝑎 𝛾 ℎ 3
Bending moment is proportional to ℎ 3 .
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𝐴 𝑠𝑡 required is proportional to B.M. / effective depth.
𝐴 𝑠𝑡 ∝ ℎ 3 𝑑 ⇒ 𝐴 𝑠𝑡1 𝐴 𝑠𝑡 = ℎ 1 3 / 𝑑 1 ℎ 3 /𝑑 ⇒ 𝐴 𝑠𝑡 1 𝐴 𝑠𝑡 = ℎ 1 3 ×𝑑 ℎ 3 × 𝑑 1
ℎ 1 =depth of any section from top of wall
h = depth from top of wall to the base of stem
𝑑 1 =effective depth of stem at depth ℎ 1 from top
d = effective depth of stem at base
𝑑 1 =effective depth of stem at depth ℎ 1 from top
= 𝑑 𝑡 + 𝑑− 𝑑 𝑡 ℎ × ℎ 1
= − × ℎ 1
= ℎ 1
𝑑 𝑡 =200−50
d =
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For 50% curtailment of steel,
𝐴 𝑠𝑡1 𝐴 𝑠𝑡 = 1 2 ⇒ 𝐴 𝑠𝑡 1 𝐴 𝑠𝑡 = ℎ 1 3 ×𝑑 ℎ 3 × 𝑑 1 ⇒ 1 2 = ℎ 1 3 × × ℎ 1
1.787× × ℎ 1 =400 ℎ 1 3
ℎ 1 3 −11.92× ℎ 1 −4.47× =0
Solving by calculator from 3-degree,
ℎ 1 = = 4.64 m from top of wall
Theoretical cutoff point from bottom of stem = 6.20 – 4.64
= 1.56 m
The bars should extend a development length( 𝐿 𝑑 ) from the theoretical cut off point.
For M20, Fe 415
𝐿 𝑑 =47 ф=47 ×20=940 𝑚𝑚 {Sp. 16, P. 184}
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Actual cut off point from bottom of stem,
= = 2.50 m
Curtail 50% of bars at 2.50 m from bottom of stem.
Similarly curtail 50% of remaining bars at 4.5 m from bottom of stem.
Check for shear :-
Pressure at base of stem = 𝑘 𝑎 𝛾ℎ= 1 3 ×18×6.2=37.2 𝑘𝑁 𝑚 2
Shear at base of stem = V = 1 2 ×37.2×6.2= 𝑘𝑁
𝑉 𝑢 =1.5 ×115.32= 𝑘𝑁
𝜏 𝑣 = 𝑉 𝑢 𝑏𝑑 = × ×400 =0.432 𝑁 𝑚𝑚 2
𝑃 𝑡 = 100 𝐴 𝑠𝑡 𝑏𝑑 = 100 × ×400 =0.785%
𝜏 𝑐 =0.57 𝑁/ 𝑚𝑚 2 >0.432𝑁/ 𝑚𝑚 2 ……….Safe
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5) Design of Heel Slab :-
Pressure at the junction of stem with heel slab is
4 m → 85.61
2.35 m → ? (50.29)
P =
= kN/ 𝑚 2 at D
Pressure distribution below base
Pressure at the junction of stem with toe slab is
4 m → 85.61
2.80 m → ? (59.93)
p = = kN/ 𝑚 2
Width of heel slab is 2.35 m.
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Total downward pressure on heel slab
= weight of backfill + self weight of heel slab
= (6.2×18) + (0.5×25)
= kN/ 𝑚 2 ↓
Maximum S.F. at D = (24.56 × 2.35) ×2.35×50.29 =
= kN
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Factored S.F. 𝑉 𝑢 =1.5×116.82= 𝑘𝑁
Maximum B.M. at D = × × 2 3 ×2.35
= = kN.m
Factored B.M. 𝑀 𝑢 =1.5×160.41= 𝑘𝑁.𝑚
d = 500 – 50 = 450 mm
𝑀 𝑢 𝑏 𝑑 2 = × × 450 2
=1.188
𝑝 𝑡 =0.355%
𝐴 𝑠𝑡 = 𝑃 𝑡 100 ×𝑏𝑑
= ×1000×450
= 𝑚𝑚 2
Provide 16 mm 120mm c/c on top face of heel.
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22. Gandhinagar Institute of Technology : Department of Civil Engineering
Check for shear :-
𝜏 𝑣 = 𝑉 𝑢 𝑏𝑑 = × ×450 =0.389 𝑁/ 𝑚𝑚 2
𝑝 𝑡 = 100 𝐴 𝑠𝑡 𝑏𝑑 = 100× ×450 =0.372
From IS: , Table-19, page 73
0.25 → difference 𝜏 𝑐 =
0.122 → ? (0.058) = N/ 𝑚𝑚 2 >0.389𝑁/ 𝑚𝑚 2
Distribution Steel :-
Provide minimum 0.12 % steel of gross cross sectional area
𝐴 𝑠𝑡 = ×1000×500=600 𝑚𝑚 2
Provide mm c/c
Also provide mm c/c both ways on bottom face
for crack control
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Anchorage Length :-
Anchorage length required = 𝐿 𝑑 = 47 ф = 47 × 16 ≈ 800 mm
Extended main steel of heel for a length 800 mm from D to
the left side.
6) Design of toe slab :-
Toe slab designed as a cantilever slab.
Total downward pressure on toe
= Self weight of toe slab
= 0.5 × 25 = 12.5 kN/ 𝑚 2
Weight of soil above toe is neglected.
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Maximum B.M. at C
= (1.20 × 96.68) × ×1.20×25.68 × 2 3 ×1.20
= = kN.m
Factored B.M 𝑀 𝑢 =1.5×81.93= 𝑘𝑁.𝑚
𝑀 𝑢 𝑏 𝑑 2 = × × =0.61
𝑝 𝑡 =0.175% 𝐴 𝑠𝑡 = 𝑃 𝑡 100 ×𝑏𝑑= ×1000×450
= 𝑚𝑚 2
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Half the reinforcement of stem i.e mm c/c = =1571 𝑚𝑚 2 anchored in toe will serve as toe reinforcement.
Distribution Steel :-
Provide minimum 0.12% steel of bD.
𝐴 𝑠𝑡 = ×1000×500=600 𝑚𝑚 2
Provide mm c/c (604 𝑚𝑚 2 ).
7) Design of shear key :-
Size of shear key provided id 0.45 m × 0.50 m.
Provided minimum reinforcement in key
𝐴 𝑠𝑡 = ×1000×450=540 𝑚𝑚 2 (𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑)
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26. Gandhinagar Institute of Technology : Department of Civil Engineering
Half the main reinforcement of stem are anchored in key
= = 𝑚𝑚 2
Also extend temperature reinforcement of stem on outer face in the shear key.
This is mm c/c = 196 𝑚𝑚 2
Total 𝐴 𝑠𝑡 in key = = 𝑚𝑚 2 > 540 𝑚𝑚 2 …..O.K.
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8) Sketch :-
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THANK YOU
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