Essay: Determine suitable dimensions of a cantilever retaining wall

Let us determine suitable dimensions of a cantilever retaining wall, which is required to support a bank of earth 4.0m high above ground level on the toe side of the wall. Consider the backfill surface to be inclined at an angle of 15o with the horizontal. Assume good soil for foundation at a depth of 1.25m below ground level with SBC of 160kN/m2. Further assume the backfill to comprise of granular soil with unit weight of 16kN/m3 and an angle of shearing resistance of 30o. Assume the coefficient of friction between soil and concrete to be 0.5.
Given:
h = 4.0 + 1.25m
θ = 15o
ф = 30o
γe = 16kN/m3
qa = 160kN/m2
μ = 0.5
Solution :
Earth pressure coefficient,
Preliminary proportioning:
Thickness of footing base slab = 0.08h = 0.08 x 5.25 = 0.42m
Provide a base thickness of 420mm for base slab.
Assume stem thickness of 450mm at base of stem tapering to 150mm at top of wall.
For economical proportioning of length ‘L’, assume vertical reaction R at the footing base to be in line with front face of the stem.
= 2.0m [where 0.4m is assumed as height above wall]
Assuming a triangular base pressure distribution,
L = 1.5X = 3.0m
Preliminary proportions are shown in figure.
For the assumed proportions, the retaining wall is checked for stability against overturning and sliding.
Stability against overturning:
Force ID
Force (kN)
Distance from heel (m)
Moment (kNm)
W1
16(1.85)(5.25 – 0.42) = 143.0
0.925
132.3
W2
16(1.85)(0.5×0.536)=7.9 [2tan150=0.536]
0.617
4.9
W3
25(0.15)(5.25-0.42) = 18.1
1.925
34.8
W4
(25-16)(4.83)(0.5×0.3) = 6.5
1.75
11.4
W5
25(3)(0.42) = 31.5
1.50
47.2
PaSinθ
25.9
Total
W = 232.9
Mw = 230.6
Pa = Active pressure exerted by retained earth on wall [both wall and earth move in same direction]
Pp = Passive pressure exerted by wall on retained earth [both move in opposite direction]
Ca -> same for dry and submerged condition, since ф for granular soil does not change significantly.
Force due to active pressure,
Pa = Ca. γe.h’2/2
Where, h’ = h + X tanθ
= 5250 + 2000 tan150 = 5786mm
Pa = 0.373(16)(5.786)2/2 = 99.9kN [per m length of wall]
Therefore, FOS(overturning) =
Overturning moment, Mo = (PaCosθ)h’/3 = 96.5(5.786/3) = 186.1kNm
To find the distance of resultant vertical force from heel,
Distance of resultant vertical force from heel,
Xw = Mw/W = 230.6/232.9 = 0.99m
Stabilising moment (about toe),
Mr = W(L – Xw) + PaSinθ(L) = 232.9(3 –0.99) + 77.6
= 468.1kNm [per m length of wall]
Soil pressure at footing base:
Resultant vertical reaction, R = W = 232.9kN [per m length of wall]
Distance of R from heel, LR = (Mw + Mo) / R = (230.6 + 186.1)/232.9 = 1.789m
Eccentricity, e = LR – L/2 = 1.789 – 3/2 = 0.289m < L/6 ->[0.5m]
Hence the resultant lies within the middle third of the base, which is desirable.
Maximum pressure as base,
Stability against sliding:
Sliding force, PaCosθ = 96.5kN
Resisting force, F = μR = 0.5 x 232.9 = 116.4kN [Ignoring passive pressure on toe side]
Hence a shear key may be provided.
Assume a shear key 300mm x 300mm at a distance of 1300mm from toe as shown in figure.
h2 = 950 + 300 + 1300 tan300 = 2001mm
Design of toe slab:
Assuming a clear cover of 75mm and 16mm Ф used,
d = 420 –75 –8 = 337mm
Mu = 1.5{(81.9 x 12/2) + (112 – 81.9) x 1/2 x 12 x 2/3}
= 76.48 kNm/m length
Nominal shear stress,
Using M20 concrete,
For a c = 0.29N/mm2, pt (required) = 0.2% [Page 178, SP-16]
For pt = 0.2%, Ast = 0.2/100 x 1000 x 337 = 674 mm2 / m
Spacing = = 298mm
Provide 16mm φ @ 290mm c/c at bottom of toe slab
=16 x 47 = 752mm, beyond face of stem.
Since length available is 1m, no curtailment is sorted.
Design of heel slab:
Mu = 1.5{(82.54 x 1.552/2) + (128.6 – 82.54) x 1/2 x 1.552 x 2/3}
= 203.96 kNm/m length
Nominal shear stress,
Using M20 concrete,
For a c = 0.39N/mm2, pt (required) = 0.3% [Page 178, SP-16]
For pt = 0.565%, Ast = 0.565/100 x 1000 x 337 = 1904.05 mm2 / m
Spacing = = 105.61mm
Provide 16mm φ @ 100mm c/c at bottom of toe slab
Since length available is 1.55m, no curtailment is sorted.
Design of vertical stem:
Height of cantilever above base = 5250 – 420 = 4830mm
Assume a clear cover of 50mm and 16mmФ bar,
dat base = 450 – 50 – 16/2 = 392mm
Mu = 1.5(Ca.γe.h3/6) = 1.5(1/3)(16 x 4.923/6) = 150.24kNm.
pt = 0.3%, Ast = 0.295/100 x 1000 x 392 = 1200mm2.
Spacing = 1000 x 201/1200 = 160mm
Provide 16mm φ @ 160mm c/c in the stem, extending into the shear key upto 47Ф = 752mm.
Check for shear: [at ‘d’ from base]
Vu (stem) = 1.5(Ca.γe.Z2/2) = 1.5(1/3 x 16 x (4.83 – 0.392)2/2) = 53.83kN
Hence, SAFE.
Curtailment of bars:
Curtailments of bars in stem are done in two stages:
At 1/3rd and 2/3rd height of the stem above base.
Temperature and shrinkage reinforcement:
Ast = 0.12/100 x 10 x 450 = 540mm2
For I 1/3rd height, provide 2/3rd of bar near front face (exposed to weather) and 1/3rd near rear face.
For II 1/3rd height, provide 1/2 the above.
For III 1/3rd height, provide 1/3rd of I case.
Provide nominal bars of 10mm @ 300mm c/c vertically near front face.