Lecture Retaining Wall Week 12

Transcription

1 Lecture Retaining Wall Week 12 Retaining walls which provide lateral support to earth fill embankment or any other form of material which they retain them in vertical position. These walls are also usually required to resist a combination of earth and hydrostatic loadings. The fundamental requirement is that the wall is capable of holding the retained material in place without undue movement arising from deflection, overturning or sliding. Types of Retaining Wall Concrete retaining walls may be considered in terms of three basic categories: (1) Gravity (2) Counterfort and (3) Cantilever Within these groups many common variations exist, for example cantilever walls may have additional supporting ties into the retained material. The structural action of each type is fundamentally different, but the techniques used in analysis, design and detailing are those normally used for concrete structures. (i) Gravity Walls These are usually constructed of mass concrete, with reinforcement included in the faces to restrict thermal and shrinkage cracking. As illustrated in Figure 1, reliance is placed on self weight to satisfy stability requirements both in respect of overturning and sliding. It is generally taken as a requirement that under working conditions the resultant of the self weight and overturning forces must lie within the middle third at the interface of the base and soil. This ensures that uplift is avoided at this interface. Friction effects which resist sliding are thus maintained across the entire base. Bending, shear, and deflections of such walls are usually insignificant in view of the large effective depth of Page 1 of

2 the section. Distribution steel to control thermal cracking is necessary, however, and great care must be taken to reduce hydration temperatures by mix design, construction procedure and curing techniques. Figure 1: Gravity wall (iii) Cantilever Walls These are designed as vertical cantilevers spanning from a large rigid base which often relies on the weight of backfill on the base to provide stability. Two forms of this construction are illustrated in Figure 2. In both cases, stability calculations follow similar procedures to those for gravity walls to ensure that the resultant force lies within the middle third of the base and that overturning and sliding requirements are met. Figure 2: Cantilever wall Page 2 of

3 (ii) Counterfort Walls This type of construction will probably be used where the overall height of wall is too large to be constructed economically either in mass concrete or as a cantilever. The basis of design of counterfort walls is that the earth pressures act on a thin wall which spans horizontally between the massive counterforts (Figure 3). These must be sufficiently large to provide the necessary dead load for stability requirements, possibly with the aid of the weight of backfill on an enlarged base. The counterforts must be designed with reinforcement to act as cantilevers to resist the considerable bending moments that are concentrated at these points. The spacing of counterforts will be governed by the above factors, coupled with the need to maintain a satisfactory span depth ratio on the wall slab, which must be designed for bending as a continuous slab. The advantage of this form of construction is that the volume of concrete involved is considerably reduced, thereby removing many of the problems of large pours, and reducing the quantities of excavation. Balanced against this must be considered the generally increased shuttering complication and the probable need for increased reinforcement. Figure 3: Counterfort Walls Page 3 of

4 Analysis and Design The design of retaining walls may be split into three fundamental stages: (1) Stability analysis ultimate limit state, (2) Bearing pressure analysis serviceability limit state, and (3) Member design and detailing ultimate and serviceability Limit states. (i) Stability Analysis Under the action of the loads corresponding to the ultimate limit state, a retaining wall must be stable in terms of resistance to overturning and sliding. This is demonstrated by the simple case of a gravity wall as shown in Figure 4. The critical conditions for stability are when a maximum horizontal force acts with a minimum vertical load. To guard against a stability failure, it is usual to apply conservative factors of safety to the force and loads. The values given in table 2.2 are appropriate to strength calculations but a value of y f = 1.6 or higher should be used for stability calculations. Page 4 of

5 If this force is predominantly hydrostatic and well defined, a factor of 1.4 may be used. A partial factor of safety of γ f = 1.0 is usually applied to the dead load G k. For resistance to overturning, moments would normally be taken about the toe of the base, point A on Figure 4, thus the requirement is that 1.0 G k x γ f H k y (Eqn 1) Figure 4: Forces and pressures on a gravity wall Resistance to sliding is provided by friction between the underside of the base and the ground, and thus is also related to total self weight Gk. Resistance provided by the passive earth pressure on the front face of the base may make some contribution, but since this material is often backfilled against the face, this resistance cannot be guaranteed and is usually ignored. Thus, if the coefficient of friction between base and soil is µ, the total friction force will be given by µ. G k for the length of wall of weight G k ; and the requirement is that 1.0 µ.g k y f H k (Eqn 2) Where H k is the horizontal force on this length of wall If this criterion is not met, a heel beam may be used, and the force due to the passive earth pressure over the face area of the heel may be included in resisting the sliding Page 5 of

6 force. The partial load factor γ f on the heel beam force should be taken as 1.0 to give the worst condition. To ensure the proper action of a heel beam, the front face must be cast directly against sound, undisturbed material, and it is important that this is not overlooked during construction. In considering cantilever walls, a considerable amount of backfill is often placed on top of the base, and this is taken into account in the stability analysis. The forces acting in this case are shown in Figure 5. In addition to G k and H k there is an additional vertical load V k due to the material above the base acting a distance q from the toe. The worst condition for stability will be when this is at a minimum; therefore, a partial load factor γ f = 1.0 is appropriate. The stability requirements then become 1.0 G k x V k q γ f H k y for overturning (Eqn 3) µ (1.0 G k V k ) γ f H k for sliding (Eqn 4) When a heel beam is provided the additional passive resistance of the earth must be included in equation 4. Stability analysis, as described here, will normally suffice. However, if there is doubt about the foundation material in the region of the wall or the reliability of loading values, it may be necessary to perform a full slip circle analysis, using techniques common to soil mechanics, or to use increased factors of safety. Page 6 of

7 Figure 5: Forces on a cantilever wall (ii) Bearing Pressure Analysis As with foundations, the bearing pressures underneath retaining walls are assessed on the basis of the serviceability limit state when determining the size of base that is required. The analysis will be similar to the combined effects of an eccentric vertical load, coupled with an overturning moment. Considering a unit length of the cantilever wall (figure 5) the resultant moment about the centroidal axis of the base is D D M f 1H k y f 2Gk ( x) f 3Vk ( q) (Eqn 5) 2 2 And the vertical load is N f 2Gk f 3 V (Eqn 6) k Where in the case of serviceability limit state the partial factor of safety are Page 7 of

8 γ f1 = γ f2 = γ f3 = 1.0 The distribution of bearing pressure will be as shown in the figure, provided the effective eccentricity lies within the middle third of the base, that is M N D 6 The maximum bearing pressure is then given by N M D p1 D I 2 Where I = D 3 / 12. Therefore, N 6M p1 (Eqn 7) 2 D D and N 6M p2 (Eqn 8) 2 D D Earth pressure on retaining walls (a) Active soil pressure Active soil pressures are given for the two extreme cases of a Cohesionless soil such as sand and a cohesive soil such as clay (Fig. 12.2). General formulae are available for intermediate cases. The formulae given apply to drained soils and reference should be made to textbooks on soil mechanics for pressure where the water table rises behind the wall. The soil pressures given are those due to a level backfill. If there is a surcharge of w kn/m2 on the soil behind the wall, this is equivalent to an additional soil depth of z = w/γ where γ is the density in kilonewtons per cubic meter. The textbooks give solutions for cases where there is sloping backfill. (i) Cohesionless soil, c = 0 (Fig. 12.2(a)) The pressure at any depth z is given by Where γ is the soil density and Φ is the angle of internal friction. The force on the wall of height H1 is Page 8 of

9 Fig. 12.2(a) (ii) Cohesive soil, ϕ = 0 (Fig. 12.2(b)) The pressure at any depth z is given theoretically by P = γz 2c where c is the cohesion at zero normal pressure. This expression gives negative values near the top of the wall. In practice, a value for the active earth pressure of not less than is used. (c) Vertical pressure under the base The vertical pressure under the base is calculated for service loads. For a cantilever wall a 1 m length of wall with base width b is considered. Then Area A = b m2 Page 9 of

10 Modulus Z = b 2 /6 m3 If ΣM is the sum of the moments of all vertical forces ΣW about the centre of the base and of the active pressure on the wall then ΣM = ΣW(x b/2) P 1 H 1 /3 The passive pressure in front of the base has been neglected again. The maximum pressure is This should not exceed the safe bearing pressure on the soil. Fig. 12.2(b) Page 10 of

11 (d) Resistance to sliding (Fig. 12.2) The resistance of the wall to sliding is as follows. (i) Cohesionless soil The friction R between the base and the soil is μσw where μ is the coefficient of friction between the base and the soil (μ= tanφ ). The passive earth pressure against the front of the wall from a depth H2 of soil is (ii) Cohesive soils The adhesion R between the base and the soil is βb where β is the adhesion in kilonewtons per square meter. The passive earth pressure is A nib can be added, as shown in Fig. 12.2, to increase the resistance to sliding through passive earth pressure. For the wall to be safe against sliding 1.4P 1 < P 2 +R where P 1 is the horizontal active earth pressure on the wall. (b) Wall stability Referring to Fig the vertical loads are made up of the weight of the wall and base and the weight of backfill on the base. Front fill on the outer base has been neglected. Surcharge would need to be included if present. If the centre of gravity of these loads is x from the toe of the wall, the stabilizing moment is ΣWx with a beneficial partial safety factory γ f =1.0. The overturning moment due to the active earth pressure is 1.4P 1 H 1 /3 with an adverse partial safety factor γ f =1.4. The stabilizing moment from passive earth pressure has been neglected. For the wall to satisfy the requirement of stability ΣWx 1.4 P 1 H 1 / 3 Page 11 of

12 Design procedure The steps in the design of a cantilever retaining wall are as follows. 1. Assume a breadth for the base. This is usually about 0.75 of the wall height. The preliminary thicknesses for the wall and base sections are chosen from experience. A nib is often required to increase resistance to sliding. 2. Calculate the horizontal earth pressure on the wall. Then considering all forces check stability against overturning and the vertical pressure under the base of the wall. Calculate the resistance to sliding and check that this is satisfactory. A partial safety factor of 1.4 is applied to the horizontal loads for the overturning and sliding check. The maximum vertical pressure is calculated using service loads and should not exceed the safe bearing pressure. 3. Reinforced concrete design for the wall is made for ultimate loads. The partial safety factors for the wall and earth pressure are each 1.4. Surcharge if present may be classed as either dead or imposed load depending on its nature. Referring to Fig the design consists of the following. (a) For the wall, calculate shear forces and moments caused by the horizontal earth pressure. Design the vertical moment steel for the inner face and check the shear stresses. Minimum secondary steel is provided in the horizontal direction for the inner face and both vertically and horizontally for the outer face. (b) The net moment due to earth pressure on the top and bottom faces of the inner footing causes tension in the top and reinforcement is designed for this position. (c) The moment due to earth pressure causes tension in the bottom face of the outer footing. The moment reinforcement is shown in Fig Page 12 of

13 Fig Page 13 of

14 Example 1: Specification Design a cantilever retaining wall to support a bank of earth 3.5 m high. The top surface is horizontal behind the wall but it is subjected to a dead load surcharge of 15 kn/m 2. The soil behind the wall is a well drained sand with the following properties: Density γ=1800 kg/m 3 =17.6 kn/m 3 angle of internal friction Φ=30 The material under the wall has a safe bearing pressure of 100 kn/m 2. The coefficient of friction between the base and the soil is 0.5. Design the wall using grade 30 concrete and grade 460 reinforcement. Answer: (a) Wall stability The proposed arrangement of the wall is shown in Fig The wall and base thickness are assumed to be 200 mm. A nib has been added under the wall to assist in the prevention of sliding. Consider 1 m length of wall. The surcharge is equivalent to an additional height of 15/17.6=0.85 m. The total equivalent height of soil is =4.6 m The horizontal pressure at depth y from the top of the surcharge is 17.6y (1 0.5)/ (1+0.5) = 5.87y kn/m 2 The horizontal pressure at the base is =27 kn/m 2 The weight of wall, base and earth and the moments for stability calculations are given in Table (i) Maximum soil pressure The base properties are Area, A=2.85 m2 Modulus, Z = 2.852/6 = 1.35 m 3 The maximum soil pressure at A calculated for service load is The maximum soil pressure is satisfactory. Page 14 of

15 (ii) Stability against overturning The stabilizing moment about the toe A of the wall for a partial safety factor γ f = 1.0 is ( ) = kn m The overturning moment for a partial safety factor γ f = 1.4 is = kn m The stability of the wall is adequate. (iii) Resistance to sliding The forces resisting sliding are the friction under the base and the passive resistance for a depth of earth of 850 mm to the top of the base: For the wall to be safe against sliding > = kn The resistance to sliding is satisfactory. (iv) Overall comment The wall section is satisfactory. The maximum soil pressure under the base controls the design. (b) Structural design The structural design is made for ultimate loads. The partial safety factor for each pressure and surcharge is γ f =1.4. Page 15 of

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17 (i) Wall reinforcement The pressure at the base of the wall is = 35.7 kn/m 2 The pressure at the top of the wall is =6.99 kn/m 2 Shear = ( ) + ( ) = = 74.8 kn Moment = ( ) + ( /3) = kn m The cover is 40 mm; assume 20 mm diameter bars. Then Provide 16 mm diameter bars at 140 mm centers to give a steel area of 1435 mm 2 /m. Determine the depth y 1 from the top where the 16 mm diameter bars can be reduced to a diameter of 12 mm. The depth y 1 is given by the equation 64=6.99(y 1 2 )/ (y 1 ) 3 /6 or y y =0 Solve to give y 1 =2.92 m. Referring to the anchorage requirements in BS8110: Part 1, clause , bars are to extend an anchorage length beyond the theoretical change point. The anchorage length from Table 3. of the code for grade 30 concrete is (section 5.2.1) = 595 mm Page 17 of

18 Stop bars off at = 2328 mm, say 2000 mm from the top of the wall. The shear stress at the base of the wall is The design shear stress in concrete is The shear stress is satisfactory. The deflection need not be checked. For control of cracking the bar spacing must not exceed 3 times the effective depth, i.e. 600 or 750 mm. The spacing at the bars in the wall is 140 mm. This is less than the 160 mm clear spacing given in Table 3.30 of the code for crack control. For distribution steel provide the minimum area of 0.13% from Table 3.27 of the code: A = /100 = 325 mm2/m Provide 10 mm diameter bars at 240 mm centers horizontally on the inner face. For crack control on the outer face provide 10 mm diameter bars at 240 mm centers each way. (iii) Inner footing Referring to Fig the shear and moment at the face of the wall are as follows: Provide 12 mm diameter bars at 120 mm centers to give 942 mm 2 /m. Page 18 of

19 This is satisfactory. For the distribution steel, provide 10 mm bars at 240 mm centers. (iv) Outer Footing Referring to Fig the shear and moment at the face of the wall are as follows: Shear Moment =1.4( / /2.85) =1.4( ) =80.74 kn = 1.4[( ) /3] =33.13 kn m Note that the sum of the moments at the bottom of the wall and at the face of the wall for the inner and outer footing is approximately zero. Reinforcement from the wall will be anchored in the outer footing and will provide the moment steel here. The anchorage length required is 592 mm and this will be provided by the bend and a straight length of bar along the outer footing. The radius of the bend is determined to limit the bearing stress to a safe value. The permissible bearing stress inside the bend is where a b is the bar spacing, 140 mm. The internal radius of the bend is Make the radius of the bend 150 mm: This is satisfactory. See the wall design below. The distribution steel is 10 mm diameter bars at 240 mm centers. (v) Nib Referring to Fig the shear and moment in the nib are as follows: Shear = 1.4( /2) = kn Moment = 1.4( ) = 8.65 kn m The minimum reinforcement is 0.13% or 325 mm 2 /m. Page 19 of

20 For crack control the maximum spacing is to be limited to 160 mm as specified in Table 3.30 of the code. Provide 10 mm diameter bars at 140 mm centers to lap onto the main wall steel. The distribution steel is 10 mm diameter bars at 240 mm centers. (v) Sketch of the wall reinforcement A sketch of the wall with the reinforcement designed above is shown in Fig Fig Page 20 of

21 Example 2 Design a T shaped cantilever retaining wall for retaining 5m high earth above the ground level. Specifications Unit weight of soil is γ s 15 kn/m 3 Angle of repose of soil is Φ = 30 Coefficient of friction at base μ = 0.5 Allowable bearing pressure of soil is q a = 150 kn/m 2 Grade of concrete = 35 MPa. Grade of steel = 460 MPa. Answer: i) Selection of retaining wall: Selection of retaining wall is proportioned as follows: Depth of foundation q a s 1 Sin 1 Sin Sin30 = 15 1 Sin30 = 1.11m 2 Consider that the base of foundation is located at a depth of 1.2m below which soil is not subjected to seasonal volume changes caused by alternate wetting and drying. Therefore, Total height of wall h a, = = 6.2m The selection of optimum dimensions of the retaining wall involves successive approximation. Reasonable dimensions are assumed based on approximate design and then various conditions of stability are checked. Necessary adjustments in cross sections are made to arrive at acceptable cross sectional dimensions and then the design and detailing for reinforcement are made, Base width of foundation based on safety against overturning ka B 0.955ha (1 )(1 3 ) Where 0.5 Page 21 of

22 1 Sin 1 Sin30 1 k a 1 Sin 1 Sin30 3 q 1 a = 2.2 h s a = Therefore 1 B = 2.96m ( )( ) Base width of foundation based on the consideration of safety against sliding, 0.707h B= a k a = =3.98m (1 ) 0.5( ) It will be economical to adopt base width of foundation on the consideration of safety against overturning and provide shear key for safety against sliding. Therefore, consider based width B = 3.25m width of toe slab α B = x 3.25 = m Consider width of toe slab = 0.9m Width of the heel slab = = 2.35m Consider thickness of the base slab = 0.07h a to 0.1h a = to = 434 to 620mm. A lower value of thickness of slab should be adopted when the height of retaining wall is large. Therefore, consider thickness of base slab equal to 450mm of the junction with the vertical wall it is reduced to 200mm of the free edges of toe and heel slabs. Consider gross d = 450mm of the base of wall which is reduced linearly to 200mm of the top ii) Stability Analysis: The checking for stability against overturning and sliding and foundation stability analysis is made as follows: a) Stability against overturning and sliding: Page 22 of

23 The stability against overturning and sliding are checked by neglecting earth on toe slab because the height of earth on toe slab is small which may not exist for some time during construction and may be eroded. This results in critical condition of stability. Factor of safety against overturning, StabilityMoment, M WA Fa OverturningMoment, M PA Where, M = moment of vertical load about -A PA WA M = moment of earth pressure about -A = P a h a /3 Where, P a = 0.5.k a γ s h a 2 = = 96.1 kn. Therefore, F a = / = > 2 (hence it is safe) Factor of safety against sliding, Horizontal Re sis tan ce, W Fs Horizontal Pr essure, Pa Where, W = total vertical load as computed in table-1 F s = / 96.1 = 1.34 < 1.55 (not safe) Hence it is unsafe (not safe) against sliding. It is considered to provide shear key to ensure that depth of shear key is 450mm below foundation slab as shown in fig. For the passive earth pressure the soil on the top of foundation slab is neglected because it may not exist for some times during construction and may be eroded. F s W P P a p Where, P p = Passive earth pressure = 0.5.k p γ s h p 2 1 Sin 1 Sin k p = 3 1 Sin 1 Sin h p = ( ) = 0.9m P p = = 9.5 kn Page 23 of

24 Therefore, F s = 1.65 > 1.55 (hence it is safe) Table-1: Vertical load & Moment calculation Loads (W kn) Distance from A Moment about A W 1 = =28.75 kn X 1 = = 1.0m M A1 = = 28.75kN m W 2 = =17.969kN X 2 = /3 = 1.183m M A2 = = kN m W 3 = 0.5 ( ) M A3 = = 3.712kNm X =7.313kN 3 = = m W 4 = =5.063kN X 4 = = 1.125m M A4 = = 5.696kNm W 5 = 0.5 ( ) X 5 = M A5 = = kNm = kN = m W 6 = =10.781kN X 6 = /3 = 1.266m M A6 = = kN m W 7 = 0.5 ( ) X 7 = M A7 = = = kN kn m = m W = kn M WA = kn m b) Foundation stability Analysis The foundation base should be under compression & the max upward soil press should be within it is permissible value. The soil pressure is determined as follows: W 6e P My W M W 6e q max., q min. = (1 ) => => => (1 ) B B A I A Z B 1 B 1 Where, e = 0.5B-(M WA -M PA )W = ( )/ = 0.42m < b/6 (3.25/6 = 0.54m) For unit length of retaining wall q max., q min. = (1 ) = kn/m 2 or 17.9 kn/m 2 < 150 kn/m Page 24 of

25 Structural Design Vertical Wall Max ultimate moment, M up = 1.5 (Pa 1 ha 1 /3) Where, Pa 1 = 0.5 k a γ s ha 1 2 ha 1 = = 5.75m Pa 1 = (5.75) 2 = kn/m 2 Therefore, M up = 1.5 ( /3) = kn m/m Assume 20mm diameter bar used and cover 40mm. the effective depth of the base of the wall d = = 400mm k 6 M = f bd cu z = 0.94d = 376.6mm A s M = 0.87 f z y = mm Provide bar T20@200mm c/c with As (provided) == 1884mm 2 Check The max. spacing, 3d = = 1200mm or 450mm whichever is less. Hence, 200mm is ok. Consider that the theoretical point of curtailment is at a depth y from top where the area of steel required may be determined by As Moment of depth y- from top Mu.y Effective thickness of vertical slab of a depth y from top, dy Therefore, A s u /A s = y 3 d / h a 3.dy Where, dy = effective depth of wall of a height y from top Page 25 of

26 d d t d y dt y ha 1 Here, d t = effective depth (thickness) of the top of the wall = /2 = 150mm d y 150 y = y y y y y = 0 y = 4325 mm Distance of actual point of curtailment from the bottom of vertical wall = ( ) + development length of bar [Development length of bar, ] Here f bu = = 2.74 MPa fbu π Φ l = 0.87 fy π Φ 2 / l = 638mm Therefore, = 2063 mm Consider that the alternate bars are curtailed at height of 2200mm from the bottom of the vertical wall and remaining bar are extend up to the top. For secondary reinforcement, vertical wall is divided in to two zone of equal height Secondary reinforcement of 0.12 percent of gross section area is provided in each zone based on their average thickness As = (450+( )/2) 1000 = 465mm 2 /m = mm 2 /m on each face in horizontal direction Provide 8mm 200mm c/c (As = 251mm 2 /m on each face in the horizontal direction up to the height 5720/2 = 2875mm Secondary reinforcement in the upper zone of the vertical wall As = (200+( )/ = 315 mm2/m Page 26 of

27 = mm 2 /m on each face in the horizontal direction Provide 8mm 300mm c/c (As = 168mm 2 /m on each face in the horizontal direction in the upper zone of the vertical wall Foundation Slab Downward pressure on heel slab = (W 5 + W 7 ) / 1.95 = ( )/1.95 = 96.25kN/m 2 Downward pressure on toe slab = W 3 /0.9 = / 0.9 = kn/m kn/m kn/m kN/m kN/m 2 Upward and downward pressure on the foundation slab kN/m kN/m kN/m kN/m 2 Net pressure on the foundation slab Max. Ultimate bending moment on heel slab = 1.5( / ( ) / 3) = kn-m Assume 12mm bar in the slab with cover 40mm effective depth Page 27 of

28 d = = 404mm 6 M = 2 bd = A s bd = 0.13 As = 1090 mm2/m Provide 12mm 100mm c/c Check The max. spacing 3d = = 1212mm or 450mm whichever is less As = 1130mm 2 /m Max ultimate bending moment in toe slab Mu = 1.5 ( / ( ) ) = kn m Effective depth of the toe slab d = D 40 6 = 404mm A st = 540mm 2 /m Provide 12mm 200mm c/c Check The max. spacing 3d = = 1212mm or 450mm whichever is less As = 565mm 2 /m in heel & toe slab Secondary reinforcement for toe & heel slab Take 0.12% of gross sectional area As = / 100 = 540mm 2 /m Provide 12mm 200mm c/c Page 28 of

29 Shear key Max ultimate moment in shear key M u kN m 2 3 The moment in shear key is much less than the moment at the base of the vertical wall Therefore, extension of the reinforcement from the vertical wall into shear wkey shall be adequate Mu shear force in shear key V u = /2 = 27 kn Shear stress = N/mm Hence, safe < v c = 0.71 MPa Page of