1.0 Check the stability of cantilever retaining wall against overturning, sliding and settlement.

2.0 Carry out cantilever retaining wall design by BS8110-1:1997.

3.0 Detail the cantilever retaining wall.

By completing this chapter, students shall be able to:

RETAINING WALLS Retaining walls are structures used to retain mainly earth but also other materials which not be able to stand vertically unsupported. The wall is subjected to overturning due to pressure of the retained material.

FUNCTION OF RETAINING WALLS

•  The most important consideration in proper design of retaining walls is that the retained material is attempting to move forward and down slope due to gravity. This creates a soil pressure behind the wall which depends on the angle of internal friction (Φ) and the cohesive strength (c) of the retained material, as well as the direction and magnitude of movement the retaining structure undergoes.

•  Lateral earth pressures are typically smallest at the top of the wall and increase toward the bottom. Earth pressures will push the wall forward or overturn it if not properly addressed. Also, any groundwater behind the wall that is not dissipated by a drainage system causes an additional horizontal hydraulic pressure on the wall.

TYPE OF RETAINING WALLS

•  Depend on their own weight and any soil resting on the concrete in resisting lateral earth forces. •  Generally economical up to 10 feet in height for cast concrete structures. •  Usually are sufficiently massive to be unreinforced. •  Monolithic cast walls are generally formed on site

TYPE OF RETAINING WALLS

•  Consist of a relatively thin stem and a base slab. The base is also divided into two parts, the heel and toe. The heel is the part of the base under the backfill. The toe is the other part of the base. •  Use much less concrete than monolithic gravity walls, but require more design and careful construction. •  Generally economical up to about 25 ft. in height. •  Can be precast in a factory or formed on site

TYPE OF RETAINING WALLS

•  These have some tension reinforcing steel included so as to minimize the thickness of the wall without requiring extensive reinforcement. •  They are a blend of the gravity wall and the cantilever wall designs.

TYPE OF RETAINING WALLS (cont) •  Similar to cantilever walls except they have thin vertical concrete webs at regular intervals along the backside of the wall. These webs are known as counterforts. •  The counterforts tie can reduce the shear forces and bending moments imposed on the wall by the soil. A secondary effect is to increase the weight of the wall from the added concrete. •  Can be precast or formed on site. •  Counterfort retaining walls are more economical than cantilever walls for heights above 25 ft.

ANALYSIS AND DESIGN

(a)  Active soil pressure – given for the two extreme cases of soil such as cohesionless soil a cohesive soil but formulae are available for intermediate cases ;

Pa = (γh + w) ka where ka = tan2 (45-φ/2) γ = unit weight in kN/m3

h = soil depth

w = load surcharge on the soil behind the wall in kN/m3 φ = angle of internal friction

The formulae given to drained soil and the soil pressures given are those due to a level backfill

ANALYSIS AND DESIGN

(b) Stability analysis – checking for ultimate limit state (i) Stability against over turning - Resisting Moment (Mr) > Over turning Moment (Mo) - The resisting moment made up of the weight of the wall stem and base and the weight of backfill on the base. - The over turning moment due to the active earth pressure. (ii) Resistance to sliding - Resisting Force (Fr) > Sliding Force (Fs) - Fr = µΣV +Pp - Fs = ΣHk where µ = coefficient of friction between the base and the soil Pp = passive earth pressure

ANALYSIS AND DESIGN

(c) Stability against Settlement (Vertical pressure under the base) - The vertical pressure under the base is calculated for service loads. - Allowable bearing pressure (σallow) > Safe bearing pressure on the soil (σactual)

(d) Main reinforcement – design for ultimate load (i) Wall - calculate shear forces and moment caused by horizontal pressure. Design the vertical moment steel for the inner face and check the shear stresses. (ii) Base - the net moment due to earth pressure on top and bottom faces of the inner footing causes tension in the top and reinforcement is design for this position.

Ways to improve the stability of the retaining wall: 1. Drain the water behind the wall 2. Increase the thickness of the concrete 3. Embed the retaining wall below the surface 4. Different retaining wall dimensions

Introduction Structural design consists of the following

part: (a) Cantilever Wall – calculate V & M caused by

horizontal earth pressure. Design the vertical moment steel for the inner (earth side) face and check the shear stresses. Minimum secondary steel is provided in the horizontal direction for the inner face & both vertically & horizontally for the outer face.

(b) Inner Footing (heel slab) – The net moment due to vertical loads on the top and earth pressure on the bottom face causes tension in the top and reinforcement is designed for the position.

(c) Outer Footing (toe slab) – the moment due to earth pressure at the bottom face cause tension in the bottom face.

CANTILEVER WALL

Wall

Heel Toe

Horizontal Pressure

Net Pressure

Pressure

Example 1 A cantilever retaining wall is proposed as shown in Figure 1 to support a bank of earth 4.5m high. The top surface of earth is flat behind the wall and subjected to dead load surcharge,w of 10kN/m2. The retained earth is well-drained sand with unit weight γ, of 19kN/m3 and angle of internal friction, φ of 35o. Allowable bearing pressure, σallow is given of 150kN/m2. The friction factor (i.e., tan δ) for concrete to soil is 0.5. Check the stability of the retaining wall and design the wall with fcu = 30N/mm2, fy = 460N/mm2. Detail your design.

Figure 1

Solution:

Soil active pressure : γ = 19kN/m3; w = 10kN/m2

φ = 350

Pa = (γh + w) ka where ka = tan2 (45-φ/2) = 0.271

Depth 0m, Pa = [0+10] x 0.271 =

2.71kN/m2

Depth 5.0m, Pa = [(19x5) + 10] x 0.271 =

28.46kN/m2

Depth 5.4m, Pa = [(19x5.4) + 10] x 0.271

= 30.51kN/m2

Solution:

Solution: Part Loading (kN/m) Lever Arm

from A (m) Moment (kNm/m)

1. ½ x 0.15 x 5.0 x 24 = 9

0.25 x 5 x 24 = 30

0.9

1.075

8.1

32.25

2 0.4 x 3.8 x 24 = 36.48 1.9 69.31

3 Soil & surcharge

{(19 x 5 ) + 10} x 2.6 =273

2.5 682.5

4 Hk 2.71 x 5.4 x 1.4 = 20.49

0.5 x 27.8 x 5.4 x 1.4 = 105.08

2.7

1.8

55.32

189.14

Σ V = 348.48 Σ Hk = 125.57

792.16 244.46

Solution:

(a)  Stability against Overturning (Ultimate Load) Overturning Moment, Mo = 244.46kNm/m Resisting Moment, Mr = 792.16kNm/m Mr > Mo SAFE

(b) Stability against Sliding (Ultimate Load)

Sliding Force, Fs = ΣHk = 125.57kN/m Resisting Force, Fr = µΣV +Pp = 0.5 x 348.38 + 0 = 174.24 kN/m Fr > Fs SAFE

Solution:

(c)  Stability against Settlement (Service Load)

Overturning Moment, Mo = 244.46/1.4 = 174.61 kNm/m Resisting Moment, Mr = 792.16kNm/m ΔM = 792.16 – 174.61 = 617.55kNm/m

ΣV = 348.48kN/m x = ΔM / ΣV = 617.55/348.48 = 1.77m e = B/2 – x = 1.9 – 1.77 = 0.13m e < B/6 ok! (if > B/6 -ve pressure)

σactual = (V/B)(1 ± 6e/B) = [348.48/3.8][1 ± 6x0.13/3.8] = 110 kN/m2 & 72.4 kN/m2

σallow = 150kN/m2 > σactual SAFE

centerline

x e

ΣV

Solution: Soil Pressure Below Retaining Wall Base

+

Solution:

Final Soil Pressure Below Retaining Wall Base

Solution: Shear & Moment Analysis Va-a = 1.4[(101.4+93.3) x 0.8/2] = 109 kN/m Ma-a = 1.4 [(93.3x0.82/2)+(8.1x0.82/3)] = 44.2kNm/m

Vb-b = 1.4[(15.8+42.2) x 2.6/2] = 105.6 kN/m Mb-b = 1.4 [(15.8x2.62/2)+(26.4x2.62/3)]= 158 kNm/m

Va-b = 1.4[(2.71+28.46) x 5/2] = 109 kN/m Ma-b = 1.4 [(27.1x52/2)+(25.75x52/6)] = 197.6kNm/m

0.8m 2.6m a b

Solution:

Main Reinforcement (a) Wall (a-b)

M = 197.6kNm/m; d= 400-40-10 = 350mm K = M/fcubd2 = 197.6x106/(30x1000x3502) = 0.054 < 0.156 z = d {0.5+(0.25-K/0.9)0.5} = 0.94d = 329mm As = M/0.95fyz = 197.6x106/(0.95x460x329) = 1375mm2/m

Table 3.25: Asmin = 0.13x1000x400/100 = 520mm2/m

Provide T20-200c/c (1571mm2/m) 100As/bd = (100x1571)/(1000x350) = 0.45

Solution: Main Reinforcement (cont) (b) Inner Base (b-b)

M = 158kNm/m; d= 400-40-10 = 350mm K = M/fcubd2 = 158x106/(30x1000x3502) = 0.043 < 0.156 z = d {0.5+(0.25-K/0.9)0.5} = 0.95d = 333mm As = M/0.95fyz = 158x106/(0.95x460x333) = 1086mm2/m

Table 3.25: Asmin = 0.13x1000x400/100 = 520mm2/m Provide T20-225c/c (1396mm2/m)

100As/bd = (100x1396)/(1000x350) = 0.4

(c) Outer Base (a-a)

M = 44.2kNm/m; d= 400-40-6 = 354mm K = M/fcubd2 = 44.2x106/(30x1000x3542) = 0.012 < 0.156 z = d {0.5+(0.25-K/0.9)0.5} = 0.99d > 0.95d = 336mm As = M/0.95fyz = 44.2x106/(0.95x460x336) = 301mm2/m

Table 3.25: Asmin = 0.13x1000x400/100 = 520mm2/m Provide T16-300c/c (670mm2/m)

100As/bd = (100x670)/(1000x354) = 0.19

Solution:

Shear Since the shear value for three section is almost the same,

take the smaller 100As/bd for checking 0.19

V = 109 kN/m

v = V/bd = 109x103/(1000x354) = 0.31N/mm2

vc = (0.79/1.25)(100As/bd)1/3(400/d)0.25(fcu/25)1/3 = (0.79/1.25)(0.19)1/3(400/354)0.25(30/25)1/3 = 0.40 N/mm2 > v

OK!

Solution:

Cracking - cl 3.12.11.2.7 no further check is required on bar spacing if either. 1) grade 250 steel is used and the slab depth does not exceed 250 mm; or 2) grade 460 steel is used and the slab depth does not exceed 200 mm; or 3) the reinforcement percentage (100As/bd) is less than 0.3 %.

a-b: 100 As/bd = 0.45 > 0.3 Allowable bar spacing = 155/0.45 = 344.4mm Actual bar spacing = 200 – 20 = 180 < 344.4mm ok!

b-b: 100 As/bd = 0.4 > 0.3 Allowable bar spacing = 155/0.4 = 388mm Actual bar spacing = 225 – 20 = 205 < 388mm ok!

a-a: 100 As/bd = 0.19 < 0.3 Allowable bar spacing = 3d or 750mm which is lesser Actual bar spacing = 300 – 16 = 284 < 750mm ok!

Solution:

Distribution of Bar Transverse/horizontal bar and outer vertical bar minimum As Asmin = 0.13x1000x400/100 = 520mm2/m Provide T16-300 (670mm2/m)

Anchorage length ,La pass through critical section Table 3.27: La = 40x diameter bar a-b & b-b : 40x20 = 800mm a-a : 40x16 = 640mm a-b : not enough space, bent up the bar b-b & a-a: enough space, bent up bar not required.

Curtailment (based on cantilever beam rules) T20-200 provided up to 5000x2 or 45 diameter = 900 from the face

of the base. The remainder will use the reduction of 50% As

(T20-200) = 1571/2 provide T16-200 (1006mm2/m)

Detailing