GRA VITY RETAINING−WALL1. INTRODUCTIONRetaining walls are structures used to provide stability for earth or other material where conditions disallow the mass to assume its natural slope, and are commonly used to hold back or support soilbanks,coal or ore piles, and water.Retaining walls are classified, based on the method of achieving stability, into six principal types (Fig.1). The gravity-wall depends upon its weight, as the name implies, for stability. The cantilever wall is a reinforced-concrete wall that utilizes cantilever action to retain the mass behind the wall from assuming a natural slope. Stability of this wall is partially achieved from the weight of soil on the heel portion of the base slab. A counterfort retaining wall is similar to a cantilever retaining wall, except that it is used where the cantilever is long or for very high pressures behind wall and has counterforts, which tie the wall and base together, built at intervals along the wall to reduce the bending moments and sheers. As indicated in Fig.1c, the counterfort is behind the wall and subjected to tensile forces. A buttressed retaining wall is similar to a counterfort wall, except that the bracing is in front of the wall and is in compression instead of tension. Two other types of walls not considered further are crib walls, which are built-up members of pieces of precast concrete, metal, or timber and are supported by anchor pieces embedded in the soil for stability, and semigravity walls, which are walls intermediate between a true gravity and a cantilever wall.(a)(b)(e)(f)Approah slabFigure.1 Types of retaining walls: (a) Gravity walls of stone masonry, brick, or plainconcrete. Weight provides overturning and sliding stability; (b)cantilever wall; (c) counterfort, or buttressed wall. If backfill covers counterforts, the wall is termed a counterfort; (d) crib wall; (e) semigravity wall (small amount of steel reinforcement is used); (f) bridge abutment.Bridge abutments (Fig.1f) are often retaining walls with wing wall extensions to retain the approach fill and provide protection against erosion. They differ in two major respects from the usual retaining wall in:1. They carry end reaction from the bridge span.2. They are restrained at the top so that an active earth pressure is unlikely to develop. Foundation walls of buildings including residential construction are retaining walls whose function is to contain the earth out of basements.Retaining walls must be of adequate proportions to resist overturning for (or excessive tilting) and sliding as well as being structurally adequate.Terms used in retaining-wall design are shown in Fig.2. Note that the “toe ” is both the front base projection and the forward edge ; similarly for the “heel ”.Figure.2 Principal terms used with retaining walls.2. COMMON PROPORTIONS OF GRA VITY WALLRetaining-wall design proceeds with the selection of tentative dimensions, which are then analyzed for stability and structural requirements and are revised as required. Since this a trial process, several solutions to the problem may be obtained, all of which are satisfactory. A computer solution greatly simplifies the work in retaining-wall design and provides the only practical means to optimize the design.Slape changeto reduceconcrete (a)(b)Figure.3 (a)Tentative demensions for a gravity retaining wall; (b)broken back retaining wall.Gravity-wall dimensions may be may be taken as shown in Fig.3. Gravity walls, generally, are trapezoidal-shaped but also may be built with broken backs. The base other dimensions should be such that the resultant falls within the middle one-third of the base. The top width of the stem should be on the order of 0.30m.If the heel projection is only 100 to 150mm, the Coulomb equation may be used for evaluating the lateral earth pressure, with the surface of sliding taken along the back face of the wall. The Rankine solution may also be used on a section taken through the heel. Because of the massive proportion and resulting low concrete stresses, low-strength concrete can generally used for the wall construction.A critical section for analysis of tensile flexure stresses will occur through the junction of the toe portion at the front face of wall.°-α+δ)°-α+δ)ββsin a cos a (b)(a)Figure.4 Forces on a gravity wall.(a)Coulmomb analysis;(b)Rankine analysis.3. GRA VITY W ALL FORCEThe forces on a gravity wall are as indicated in Fig.4.The active earth pressure is computed by either the Rankine or Coulomb methods. If the Coulomb method is used, it is assumed that there is incipient sliding on the back face of the wall, and the pressure acts at the angle of wall friction δ to a normal with the wall. The Rankine solution applies to Pa acting at the angle β on a vertical plane through the heel. The vector can then be added to the weight vector of the wedge of soil W between the vertical plane and back of the wall to get the direction and magnitude of the resultant Pa on the wall. The vertical resultant R acting on the base is equal to the sum of the forces acting downward, and will have an eccentricity e with respect to the geometrical center of the base. Taking moments about the toe,=x sum of overturning moments(net)/RIf the width of the base is B, the eccentricity of the base can then be computed asx Be -=2ββ=P sin a v a =P h cos W s =weight of abcdW c =weight of concrete of entire wall systemF=F r /P h ≥1.5F R ′+c ′+P p B P p=12γH p ′p K Figure.5 Forces involvedin the sliding stability of a retaining wall.4. STABILITY OF GRA VITY W ALLRetaining walls must provide adequate stability against sliding, as shown in Fig.5. The soil in front of the wall provides a passive-earth-pressure resistance, as the wall tends to slide into it. If the soil is excavated or eroded after the wall is built, the passive-pressure component is not available and sliding instability may occur. If there is certainty of no loss of the toe soil, the designer may use the passive pressure in this zone as part of the sliding resistance.Additional sliding stability may be derived from the use of a key beneath the base. Unless the key is quite deep, however, the sliding zone (Fig.6) may bridge over the key in taking that path of least resistance. A key into firm soil or rock may be quite advantageous, since the resistance is now the force necessary to sheer the key from the base slab.γ1p=P 2pK H p(b)(a)(c)this inclined planeFigure.6 Stability against sliding using a base key.(a)Base key near stem so that stem steel maybe run into the key;but(b)the sliding surface may develop as shown here where little aid is gained from using the key;(c)heel key which presents two possible modes of failure(passive and slip along the plane).The best key location is at the heel as indicated in Fig.6. This location creates aslightly lager sliding-resistance distance L, as well as an additional component of forcefrom the upward-sloping plane. The lesser of the two values1. Passive pressure developed to the bottom of the key.2. Sliding resistance up plane ab . is used in computing sliding stability.The sliding resistance along the base is taken as fR , where R includes all the vertical forces, including the vertical component of Pa, acting on the base.The coefficient of friction between the base and the soil may be taken as f =tg φ to 0.67 tgand base cohesion cas c=0.5c to 0.75cThe base soil is usually compacted prior to pouring the base slab; however, the wet concrete will always attach to the ground such that f =tg φ is obtained. The cohesion may be considerably destroyed from water and remolding; thus values of 0.5 to 0.75c are more appropriate.The safety factor against sliding should be at least 1.5 for cohesionless backfill and about 2.0 for cohesive backfill computed as follows:sliding F =sum resisting forces /sum deriving forcesThe usual safety factor against overturning with respect to the toe is 1.5, with a value of 2.0 suggested for cohesive soil:g overturnin F =sum of moments to resist overturning /sum of overturning moments (d)The safety factor can be computed in several ways depending on the interpretation of what goes in the numerator or denominator of Eqs. (a) and (b).-α+δ)-α+δ)°°Figure.7 Design of a gravity retaining wall with criticalpoints indicated.f=V/t5. DESIGN OF GRA VITY W ALLThe design of gravity wall will be illustrated by an example.The first step in the design of a gravity wall is to select proportions. Figure 3 is used as a guide for selecting initial wall dimensions. Figure.7 indicates critical sections and the method of computing concrete stresses. Also shown are the allowable ACI (USD) Code concrete stresses in Fps and SI units.(a)(b)Figure.8Example Design a solidg.8a. gravity wall to retain a 5.5m embankment. The general wall geometry is shown on FiSoil data:Mpa f c21,c t f f 42.0 ,65.0=φ c c f v 16.0 ,85.0=φ SOLUTIONStep1 Find the later wall force using the Rankine Ka:Ka==-)245(02ϕtg =-)23245(02tg 0.321 Pa=KN Ka H 142321.01.75.17212122=⨯⨯⨯=γKN P h 13910cos 1420==,KN P V 2510sin 1420==Step2 Computer wall stability. Neglect soil toe and not use passive pressure.*Using appropriciation of centroid 2.27/3 from shear plane. (a) The overturning safety factor is >1.5 O.K.(b) The sliding factor of safety (and neglecting any passive pressure) is Take cohesion c=Take cohesion >1.5 also O.K.Step3 Locate resultant on base and eccentricity:m x B e 50.020.17.12 <6L O.K.Step4 Computer actual soil pressure:(max)253]4.35.061[14.3457)61(kpa L e A P q<275kpa O.K. =16(min)Step5 Check shear and tensile bending stresses in toe at 0.15m from edge; refer to Fig.8b.(a) Shear check: q=253-69.7xV=27.69253)7.69253(2x x dx x xat x=0.15m and V=37 KN.For load factor=2, d=D for no rebars, andkpa bd V LF v a 2.829.01372=⨯⨯=⨯=kpa v c 6231021)85.0(16.03=⨯=>82kap O.K. (b) Tension check:67.692253320x x Vdx M xat x=0.15m and M=2.81m KNFor LF=2 and 62bh S x =Actual kpa bh M LF f t 429.0181.226)(622=⨯⨯⨯==O.K. Step6 Approximate check t f at 1/2 wall height (3.5m from top; ≃3.75 at slope):Approx. M= cos 212y Ka HM=m KN 6.4810cos 375.3321.075.35.172102Find wall h at 3.5m by proportion:8.56.25.3h m h 57.1 and h=1.57+0.5=2.07m Actual (approx.)kpa bh M LF f t 13607.216.4826)(622=⨯⨯⨯==≪1250kpa O.K. Note this check is very conservative as it neglects both wall batter and the weight of stem above 3.5m level to reduce t f . Wall proportions are adequate. Might consider (1) using boulder filler (2) using c f =14 or 17 Mpa concrete.。