# Essay: Design a suitable counterfort retaining wall to support a leveled backfill of height 7.5m above ground level on the toe side

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• Subject area(s): Engineering essays
• Published on: December 27, 2019
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Design a suitable counterfort retaining wall to support a leveled backfill of height 7.5m above ground level on the toe side. Assume good soil for the foundation at a depth of 1.5m below ground level. The SBC of soil is 170kN/m2 with unit weight as 16kN/m3. The angle of internal friction is φ = 30o. The coefficient of friction between the soil and concrete is 0.5. Use M25 concrete and Fe415 steel.

Minimum depth of foundation =

Depth of foundation = 1.5m

Height of wall = 7.5 + 1.5 = 9m

Thickness of heel and stem = 5% of 9m = 0.45m ≈ 0.5m

Thickness of toe slab = 8% of 9m = 0.72m

Lmin = 1.5 x 3 = 4.5m

Thickness of counterfort = 6% of 9 = 0.54m

Stability conditions:

Earth pressure calculations:

Force ID

Force (kN)

Distance from heel (m)

Moment (kNm)

W1

16(7.5+1.5-0.5)(2.5) = 340

(3 – 0.5)/2 = 1.25

425

W2

25(0.5)(9 – 0.5) = 106.25

0.5/2 + 2.5 = 2.75

292.18

W3

25(0.5)(3) = 37.5

1.5

56.25

W4

25(1.5)(0.72) = 27

1.5/2 + 3 = 3.75

101.25

Total

W = 510.75

Mw = 874.69

Xw = 874.69/510.75 = 1.713m

FOS (overturning) = 0.9Mr/Mo

Where, Mo = Pa.h/3 = Ca.γe.h3/6 = 0.33x16x93/6 = 647.35kNm.

Mr = (L – Xw).W = 510.75(4.5 – 1.713) = 1423.6kNm.

FOS (overturning) = 1.98 > 1.4

Hence, section is safe against overturning.

Sliding:

FOS (sliding) = 0.9(μR)/PaCosθ

F = μR = 0.5 x 510.75 = 255.375kN

Pa = Ca.γe.h2/2 = 215.784

Base pressure calculation:

Where, LR = (M + Mo)/R, e = LR – L/2, where,

LR = (874.688 + 647.352)/510.75 = 2.98m

& e = LR – L/2 = 2.98 – (4.5/2) = 0.73 < L/6  (0.75m)

Since the maximum earth pressure is greater than SBC of soil, the length of base slab has to be increased preferably along the toe side. Increase the toe slab by 0.5m in length.

ΣW = 510.75 + 0.5 x 25 x 0.72 = 519.75kN

Moment = 0.5/2 + 4.5 = 4.75m

ΣM = 874.69 + 42.75 = 917.438kNm.

LR = (Mo + M) / R = (917.438 + 647.352)/519.75 = 3.011m

e = LR – L/2 = 3.011 – (5/2) = 0.511m < L/6 (0.833m)

FOS (Sliding) = 0.9(μR)/Pa = 0.9(0.5 x 519.75)/215.784 = 1.08 < 1.4.

Hence the section is not safe against sliding. Shear key is provided to resist sliding.

Assume shear key of size 300 x 300mm.

Pps = Cp.γe.(h22 – h12)/2 = 164.5kN/m

FOS (sliding) = 0.9(μR + Pps)/Pa = 1.77 > 1.4 [where, h1 = 1.2m, h2 = 1.2 + 0.3 + 1.39 = 2.88m]

Hence, section is safe in sliding with shear key 300 x 300mm.

Design of Toe Slab:

Effective cover = 75 + 20/2 = 85mm

Toe slab is designed similar to cantilever slab with maximum moment at front face of the stem and maximum shear at‘d’ from front face of stem.

d = 720 – 85 = 635mm.

M = 80.38×22/2 + ½ x 2 x 49.94 x 2/3 x 2 = 160.76 + 122.76 = 227.35kNm.

SF at 0.635m, = 49.94/2 x 0.635 = 15.606kN

Area of trapezoid = ½.h.(a + b) = ½(2 – 0.635)(130.32 + 95.98) = 154.44kN

Factored SF = 231.66kN; Factored Moment = 341.02kNm.

K = Mu/bd2 → Ast = 1551.25mm2 → Spacing = 1000ast/Ast → 16mm @125mm c/c.

Transverse reinforcement: = 0.12% of c/s

= 0.12/100 x 1000 x 720 = 864mm2

Provide 10mm @100mm c/c.

Design of heel slab:

The heel slab is designed as an one way continuous slab with moment wl2/12 at the support and wl2/16 at the midspan.

The maximum shear at the support is w(l/2 – d).

The maximum pressure at the heel slab is considered for the design.

Moment at the support, Msup = wl2/12 = 106.92 x 2.52/12 = 55.688kNm.

Moment at the midspan, Mmid = wl2/16 = 41.76kNm

The maximum pressure acting on the heel slab is taken as ‘w’ for which the Ast required at midspan and support are found.

Factored Msup = 83.53kNm → Ast = 570.7mm2

Factored Mmid = 62.64kNm → Ast = 425.4mm2

Using 16mm φ bar, Spacing = 1000ast / Ast = 110.02mm → Provide 16mm @ 110mm c/c

At midspan, spacing = 156.72mm → Provide 16mm @ 150mm c/c

Transverse reinforcement = 0.12% of c/s = 0.12/100 x 1000 x 500 = 600mm2

For 8mm bar, Spacing = 83.775mm → Provide 8mm @80mm c/c.

Check for shear:

Maximum shear = w (l/2 – d) = 107 (2.5/2 – 0.415) = 89.345kN

Factored shear force = 134.0175kN

ζv = 0.33N/mm2, ζc = 0.29N/mm2, ζcmax = 3.1N/mm2 Depth has to be increased.

Design of stem:

The stem is also designed as one way continuous slab with support moment wl2/12 and midspan moment wl2/16. For the negative moment at the support, reinforcement is provided at the rear side and for positive moment at midspan, reinforcement is provided at front face of the stem.

The maximum moment varies from a base intensity of Ka.γe.h = 1/3x16x(9–0.5) = 45.33kN/m

Msup = wl2/12 = 1.5 x 45.33 x 3.542/12 = 71kNm

Mmid = wl2/16 = 1.5 x 45.33 x 3.542/16 = 53.26kNm

Effective depth = 500 – (50 + 20/2) = 440mm

Ast at support = 1058mm2, For 16mm φ, Spacing = 190mm. Provide 16mm @ 190mm c/c

Ast at midspan =718mm2, For 16mm φ, Spacing = 280mm. Provide [email protected] c/c

Max. SF = w (l/2 –d) = 60.29kN, Factored SF = 90.44kN

Transverse reinforcement = 0.12% of c/s → 8mm @ 80mmc/c

ζv = 0.188N/mm2, ζc = 0.65N/mm2, ζcmax = 3.1N/mm2 Safe in Shear.

Design of Counterfort:

The counterfort is designed as a cantilever beam whose depth is equal to the length of the heel slab at the base and reduces to the thickness of the stem at the top. Maximum moment at the base of counterfort, Mmax = Ka.γe.h3/6 x Le

Where, Le  c/c distance from counterfort

Mmax = 1932.5kNm, Factored Mmax = 2898.75kNm

Ast = 2755.5mm2, Assume 25mmφ bar, No. of bars required = 2755.5/491.5 = 5.61 ~ 6

The main reinforcement is provided along the slanting face of the counterfort.

Curtailment of reinforcement:

Not all the 6 bars need to be taken to the free end. Three bars are taken straight to the entire span of the beam. One bar is cut at a distance of,

,where, n is the total number of bars and h1 is the distance from top.

When n = 6, h1 = 7.75m [from bottom]

The second part is cut at a distance of,

,h2 = 6.94m [from bottom]

The third part is cut at a distance of,

,h3 = 6.01m [from bottom]

Vertical ties and horizontal ties are provided to connect the counterfort with the vertical stem and the heel slab.

Design of horizontal ties:

Closed stirrups are provided to the vertical stem and the counterfort. Considering 1m strip,the tension resisted by reinforcement is given by lateral pressure on the wall multiplied by contributing area.

T = 1/3 x 16 x (9 – 0.5) x 3.54 = 160.48kN

Factored force, T = 1.5 x 160.48kN

Ast = 666.72mm2. For 10mm φ, Spacing = 110mm.

Provide [email protected] c/c closed stirrups as horizontal ties.

Design of vertical ties:

The vertical stirrup connects the counterfort and the heel slab. Considering 1m strip, the tensile force is the product of the average downward pressure and the spacing between the counterforts.

T = Avg(43.56 & 107) x Le = 266.49 kN

Factored T = 399.74 kN

Ast = 1107.15mm2. For 10mm φ, Spacing = 70.93mm.

Provide 10mm @ 70mm c/c.

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