RETAINING WALL DESIGN
EXAMPLE - 1
Prof. Hassan Irtaza, Department of Civil Engineering,
A.M.U., Aligarh – 202002, India
Design Example - Cantilever Retaining Wall
# Design a cantilever retaining wall (i.e. T-type) to
retain earth for a height of 4m. The backfill is
horizontal. The density of soil is 18 kN/m3. Safe
bearing capacity of soil is 150 kN/m2. Take the co-
efficient of friction between concrete and soil as
0.6. The angle of repose of earth is 30o. Use M25
grade concrete and Fe500 TMT steel.
# Solution
Data: 150
h1 =4m
Safe Bearing Capacity of soil =
150 kN/m2
γ = 18 kN/m3, μ = 0.6, ø = 30o
Height of the retaining wall
H = h1 + Df
2
SBC 1-sin
γ 1+sin
Df =
= 0.926 m say 1.0 m
Therefore H = 5.0 m
PROPORTIONING OF WALL
Thickness of the base slab = 1/10th
to 1/14th of H [0.50 m to 0.357 m]
say 400 mm
Width of the base slab = 0.5 to 0.6
of H [2.5 m to 3.0 m] say 3 m
Toe projection = 1/3 to 1/4 of B
[1m to 0.75 m] say 0.85 m
Provide 300 mm thickness for the
stem at the base and 150 mm at the
top of the stem.
Pressure Below the Retaining Wall
STABILITY ANALYSIS
Load Magnitude Distance from A BM about A
(kN) (m) (kN-m)
Stem W1 0.15 x 4.6 x 1 x 25 0.85 + 0.15 + 0.1 = 18.975
= 17.25 1.1
Stem W2 ½ x 0.15 x 4.6 x 1 0.85 +2/3 x 0.15 = 8.194
x 25 = 8.625 0.95
Base Slab W3 3.0 x 0.40 x 1 x 25 1.5 45.0
= 30.0
Back fill W4 1.85 x 4.60 x 1 x 2.075 317.85
18 = 153.18
Total ΣW = 209.06 ΣMR = 390.02
Earth Pressure = Ph Ph= 0.333 x 18 x H/3 = 5.0/3 Mo= 125
5.02/2 = 75
Stability Checks
Check for overturning
FOS = ΣMR/Mo = 390.02/125 = 3.12 > 1.55 safe
Check for sliding
FOS = μ ΣW/PH = (0.6x209.06)/75 = 1.67 > 1.55 safe
Check for subsidence
x = ΣM/ΣW = 390.02/209.06 = 1.866 m
and e = 1.866 – 3/2 = 1.866 -1.5 = 0.366< b/6
Pressure below the base slab
W 6e W 6e
Pmax = 1 + , Pmin = 1 −
bl b bl b
where, b = 3 m, l = 1 m
Pressure Below the Base Slab
Pmax = 120.70 kN/m2 < SBC, safe
Pmin = 18.68 kN/m2 > zero,
No tension or separation, safe
DESIGN OF STEM
1 - sin 1 - sin30o 1
Ca = = =
1 + sin 1+sin30 o 3
Pa = 1 .Ca .γ s .h 2 = 1 × 1 ×18 × 4.62 = 63.48 kN
2 2 3
M = Pa . h = 63.48 4.6 = 97.336 kN-m
3 3
M u = 1.5×M = 146.004 kN-m
Taking 1m width of the vertical of
wall, its thickness is given by
B.M. = 0.133 σ ck bd 2
146.004 106 = 0.133×25×1000×d 2
or, d = 210 mm
`
Adopt a 270 mm effective depth and 300 mm overall thickness
of the stem at base and 150 mm thickness at the top of the stem.
Area of tension steel is given by
0.36σ ck bx m 0.36×25×1000×0.46×270
At = =
0.87σ y 0.87×500
= 2570 mm 2
Use 20 mm diameter TMT bars, spacing = 122 mm c/c
Let us provide 20 mm ϕ bar @ 110 mm c/c [* less than 300 mm
and 3d, o.k.]
314×1000 110
A t provided = ×100 = 1.056% > 0.12%
1000×270
CURTAILMENT OF BARS
Curtail 50% steel from top where
the moment is 50% of base
moment. Position where the B.M.
is half from top = 3.16m
Actual point of cut off = 3.16 - Ld
= 3.16 - 49φ bar h
= 3.16 - 0.98
= 2.18 m
say 2.2.0 m from top
spacing of bars = 220 mm c/c
< 300 mm and 3d o.k.
Distribution and Secondary (Temperature) Steel
Distribution steel for the stem
(300+150) 1000
0.12% GA = 0.12× ×
2 100
= 270 mm2
Provide 10 bar @ 200 mm c/c Temperature steel for stem at front both ways
< 450 mm and 5d o.k. Distribution steel
[Note: Actaul spacing 291 mm c/c] = 0.12% GA = 0.12×225×1000 100
= 2700 mm 2
Provide 10 mm bar @ 200 mm c/c
in both the direction
< 450 mm and 5d o.k.
Check for Shear
Grade of Concrete = 25 MPa
Grade of steel = 500 TMT bars
Percentage of steel in the vertical stem = 1.056
τcm =3.1 N/mm 2
τc =0.644 N/mm 2
Corresponding to grade of concrete and percentage of steel provided.
The critical section for sheer strength is taken at a distance d from the
bottom of the vertical stem.
Now h = 4.6-0.27 = 4.33m
Ph = 1 .Ca .γs .h 2 = 1 × 1 ×18×4.332 = 56.25 kN
2 2 3
Ultimate SF = Vu = 1.5 x 56.25 = 84.375 kN
Check for Shear
Nominal shear stress =
84.375 1000
τ v = Vu bd =
1000 270
= 0.3125 MPa
To find τ c : 100Ast bd = 1.056%
From IS: 456 - 2000 corresponding
to M25
τ c = 0.644 MPa
τ v < τ c , hence safe in shear
Design of Base Slab
Load Magnitude Distance BM, Mc
(kN) from C (m) (kN-m)
Back fill 1.85 x 4.6 x 18 0.925 141.69
=153.18
Heel Slab 0.40 x 1.85 x 0.925 17.11
25 = 18.5
Upward Pressure 18.68 x 1.85 = 0.925 -31.97
(Rectangular portion) - 34.56
Upward Pressure ½ x 62.91 x 1/3 x 1.85 -35.88
(Triangular portion) 1.85 = - 58.19 = 0.6167
Total Load Σ =78.93 ΣMc= 90.95
Design of Heel Slab
W 6e W 6e
Pmax = 1 + , Pmin = 1−
bl l bl l
where, b = 3 m, l = 3 m
Mu = 1.5 x 90.95 = 136.425 kN-m
S.F. = 78.93 kN
Vu = 1.5 x 78.93 = 118.4 kN
Effective depth of heel is given as
136.425 106
d= = 202.6mm
0.133 25 1000
Adopt an effective depth of 370 mm
and overall depth of 400 mm.
y At
M u = 0.87 y At d −
ck b
500 At
136.425 10 = 0.87 500 At (350 −
6
)
25 1000
At = 947.5mm 2
Spacing = 212 mm c/c
Provide 16 mm ϕTMT bars @ 175 -200 mm c/c < 300 mm and
3d o.k.
Percentage of steel Pt = 0.328% for spacing of 175 mm c/c and
0.287% for spacing of 200 mm c/c.
Development length Ld = 49 φbar
= 49 x16 = 784 mm
Bars must be embedded at least by
784 mm into the slab to transfer the
tensile force through bond into the
concrete.
Provide 10 mm ϕ distributed steel
@ 150 mm c/c [%age steel = 0.13 %
> 0.12%
< 450mm and 5d o.k.
Check for Shear at Junction ( Tension )
Maximum shear force = 78.93 kN
Vu, max= 118.395 kN
Nominal shear stress
τ v = Vu bd =118.395×1000 1000×350
= 0.34 MPa
τc corresponding to percent steel = 0.328%
& grade of concrete M25
τc = 0.42MPa
τ v < τc
Hence safe in shear.
Design of Toe Slab
Load Magnitude Distance from D Bending Moment,
(kN) (m) M (kN-m)
Weight of Toe Slab 0.85 x 0.40 x 1 x 0.85/2 = 0.425 - 3.6125
25 = 8.5
(downward)
Pressure distribution, 91.79 x 0.85 x 1 = 0.85/2 = 0.425 33.16
rectangular 78.02
Pressure distribution, ½ x 28.91 x 1 x 2/3 x 0.85 = 0.567 6.96
Triangular 0.85 = 12.29
Total load at 81.81 kN Total B.M. ΣM=36.51
Junction at junction
Design of Toe Slab
M u =1.5 x 36.51 = 54.765 kN-m
Area of steel corresponding to M u =367 mm 2
Min m %age of steel = 0.12% GA
Ast = 480 mm 2
Hence provide 10 @150 mm c/c
< 300 mm and 3d o.k.
Development length:
Ld = 49φ bar = 49 x 10 = 490 mm
Check for shear
The toe is treated as a cantilever beam
with critical section for shear at a
distance d from the front face of the wall.
Neglecting the earth on the toe, the net
shear force at the junction
V = 103.684 0.48+ 1 (120.7 − 103.684) ) 0.48
2
= 53.85
Vu = 1.5 × 53.85 = 80.775 kN
τ v = 80.775x1000/(1000x350) = 0.231 MPa
Percentage of steel Pt =
(
78.5 1000 )
150 100 = 0.23%,
1000 350
from IS:456 - 2000
Corresponding to steel of 0.23 and M25 grade of concrete
τc = 0.35 MPa
τ v < τc
Hence safe in shear
OTHER DETAILS
1. Construction joint
2. A key 200 mm wide x 50 mm deep with nominal steel
#10@250, 600 mm length in two rows.
3. Drainage:
100 mm dia. pipes as weep holes at 3 m c/c at bottom.
Also provide 200 mm gravel blanket at the back of the
stem for back drain.
Drawing and Detailing
THANKS
