Retaining wall design_example

download Retaining wall design_example

of 8

  • date post

    30-May-2015
  • Category

    Design

  • view

    4.374
  • download

    0

Embed Size (px)

Transcript of Retaining wall design_example

  • 1. CE 437/537, Spring 2011 Retaining Wall Design Example1/8Design a reinforced concrete retaining wall for the psffollowing conditions. surcharge = qs = 400fc = 3000 psi Fill: = 32ofy = 60 ksiUnit wt = 100 pcfHT = 18 fttf wtoe tstem wheelNatural Soil: = 32oallowable bearing pressure = 5000psfDevelopment of Structural Design Equations. In this example, the structural design of thethree retaining wall components is performed by hand. Two equations are developed in thissection for determining the thickness & reinforcement required to resist the bending momentin the retaining wall components (stem, toe and heel).Equation to calculate effective depth, d: Three basic equations will be used to develop anequation for d.M u = M na M n = As f y d 2aM u = As f y d [ Eqn 1]2C = T , 0.85 f c a b = As f yf cAs = 0.85ab[ Eqn 2]fy 0.003 s + 0.003 a 0.003strain compatibility := ,=1 a / 1 d d s + 0.003Assuming 1 = 0.85,sa/d0.0050.319 0.00785 0.2350.0100.196and choosing a value for s in about the middle of the practical design range,a= 0.235, a = 0.235 d [ Eqn 3]d

2. CE 437/537, Spring 2011Retaining Wall Design Example2/8Substituting Eqn. 2 into Eqn. 1:f aM u = 0.85 c ab f y d fy 2 And substituting Eqn. 3 into the above: f c 0.235 d M u = 0.850.235d b f y d fy2 0.883dInserting the material properties: fc = 3 ksi and fy = 60 ksi, and b = 12in (1-foot-widestrip of wall, in the direction out of the paper).kM u = 0.90(0.85) 3ksi (12 in )(0.235)(0.883)d 2M u = 5.71 in d 2Equation for area of reinforcement, As. The area of reinforcement required is calculatedfrom Eqn. 1:M u = As f y 0.883d = 0.90 As 60 ksi 0.883 d M u = 47.7 ksi As d Design Procedure (after Phil Ferguson, Univ. Texas)1. Determine HT. Usually, the top-of-wall elevation is determined by the client. Thebottom-of-wall elevation is determined by foundation conditions. HT = 18 feet.2. Estimate thickness of base. tf 7% to 10% HT (12" minimum) Tf = 0.07 (18 x 12"/) = 15.1" use tf = 16" 3. CE 437/537, Spring 2011Retaining Wall Design Example3/83. Design stem (tstem, Asstem). The stem is a vertical cantilever beam, acted on by thehorizontal earth pressure. ft inin/ft h = 8 16 /12ft h =16.67htf = 16inka h ka qs wtoe tstem wheelcalc. d:1Pfill = ( k a h ) h (1 ft out of page)21 sin 1 sin( 32 o )ka = == 0.311 + sin 1 + sin( 32 o )1Pfill = (0.31)(100 pcf )(16.67 ft ) 2 (1 ft ) = 4310 lb2Psur = k a q sur h (1 ft ) = 0.31 ( 400 psf )(16.67 ft )(1 ft ) = 2070 lb hhM u = ( Earth Pressure LoadFactor)( Pfill )( ) + ( Live LoadFactor)(Psur )( ) 32 16.67 ft16.67 ftM u = (1.6)(4310 lb )() + (1.6)(2070 lb )() = 65.9 k ft3 2 kMu =5.71 in d2 kin65.9 k ft (12 ) = 5.71 in d 2 , d = 11.8in ft 1t stem = 11.8in + 2 in cover + (1.0 in ) = 14.3in , ( assume #8 bars ) use t stem = 15in 2d = 15 2 0.5 = 12.5inininin 4. CE 437/537, Spring 2011Retaining Wall Design Example 4/8calc. As:M u = 47.7 ksi As din65.9 k ft (12 ) = 47.7 ksi As (12.5in ), As = 1.33 in 2 ftAs of one #8 bar = 0.79 in 2 in 20.79 barin in 212 = 7.13 ,use #8 @ 6 inin ft bar1.33 ft of wall4. Choose Heel Width, wheel Select wheel to prevent sliding. Use a key to force slidingfailure to occur in the soil (soil-to-soil has higher friction angle than soil-to-concrete). 12inNeglect soil resistance in front of thewall. Fresist 18ftset= FslidingFSFS = Factor of Safety = 1.5 for slidingFresist = (Vertical Force)(coefficient of friction)Fresist = WT (tan natural soil ) tf = 16intan natural soil = tan(32 o ) = 0.62WT = W fill + Wstem + W found15in lbW fill = (100 pcf )(16.67 ft )( wheel )(1 ft ) = 1670wheelft 12 in + 15in 1 ftWstem = (150 pcf )(16.67 ft )(in (1 ft ) = 2810 lb 2 12 ) 1615W found = (150 pcf )( ft )( wheel +ft + 3 ft )(1 ft ) = 200 plf wheel + 850 1212Fsliding = Pfill + Psur1Pfill = (0.31 100 pcf )(18 ft ) 2 (1 ft ) = 5020 lb2Psur = (0.31 400 psf )(18 ft )(1 ft ) = 2230 lbFsliding = 5020 lb + 2230 lb = 7250 lb 5. CE 437/537, Spring 2011Retaining Wall Design Example 5/8 lb lb lb lb 1670 ft wheel + 2810 + 200 ft wheel + 850 (0.62)7250lb = 1.5 1.5lb7250lb= 3660lb + 1870 wheel , wheel = 7.42 ft , use wheel = 7.5 ft 0.62ft5. Check Overturning. 12in 18 ft 18 ftM over = P fill ( ) + Psur ( ) 32M over = 5.02 k (6 ft ) + 2.23k (9 ft ) = 50.2 k ft 18 ft7.5 ft 15M resist = W fill (+ ft + 3 ft ), assume wtoe = 3 ft 212 1.25 fttf = 16in + Wstem (3 ft + )2 11.75 ft 3 15" 7.5 + W found ( )2M resist = (1.67 klf 7.5 ft )(8 ft ) + ( 2.81k )(3.625 ft ) + (0.20klf 7.5 ft + 0.85k )(5.875 ft ) 12.53k 2.35kM resist = 124.2 k ftM resist 124.2 k ft= = 2.47 > 2.0 = FSover , OKM over50.2 k ft6. Check Bearing.WTM L v at end of toe = + , equation is valid only if e < bL bL2 66WT = W fill + Wstem + W foundWT = 12.45 k + 2.81k + 2.35 k = 17.69 k7.5 ft 1.25 ftM = M over W fill (5.875 ft ) + Wstem (7.5 ft + 5.875 ft ) + W found (0)2 2M = 50.2 k ft 12.53k ( 2.125 ft ) + 2.81k ( 2.25 ft ) = 29.9 k ftCheck that e < L/6:m29.9 k ft L 11.75 ftL e== k = 1.68 ft , == 1.96 ft , e < , OKWT 17.69 666 6. CE 437/537, Spring 2011 Retaining Wall Design Example6/817.69 k29.9 k ftv = + = 2.80 ksf < 5.0 ksf = allowable bearing capacity, OK ft (1 )(11.75 ) ft 1 ft (1 )(11.75 ft ) 2 67. Heel Design.Max. load on heel is due to the weight of heel + fill + surcharge as the wall tries to tip over.Flexure:W = Wheel + W fill + Wsur16W = 1.2(150 pcf )(ft )(1 ft )wu12 Vu + 1.2(100 pcf )(16.67 ft )(1 ft )16 inMu + 1.6( 400 plf ) 7.5ftklfW = 2.88wu L2 2.88klf (7.5 ft ) 2Mu = == 81.0 k ft 22k 2 M u = 5.71d in ink81.0 k ft (12 ) = 5.71 d 2 , d = 13.0in for flexureftinShear: Vu = wu (7.5 ft ) = 2.88 klf (7.5 ft ) = 21.6 kVc = (0.75) 2 f c bw d = (0.75) 2 3000 psi (12 in ) dsetVu = Vc ,21,600lb = (0.75) 2 3000 psi (12 in ) d , d = 21.9 in for shear, controlsShear controls the thickness of the heel.1 t heel = 21.9 in + 2 in cover + in = 24.4 in (assume #8 bar ),use t heel = 21.5 in2Reinforcement in heel: M u = 47.7 ksi As d in81.0 k ft (12 ) = 47.7 ksi As (21.9 in ), As = 1.07in 2ft in 20.79 bar (12 in ) = 8.83 in , use #8 @ 8" in 2 ft1.07ft 7. CE 437/537, Spring 2011Retaining Wall Design Example7/88. Toe Design.Earth Pressure at Tip of Toe: WuMuv = bL 1 2 bL 6Wu = 1.2(W fill + Wstem + W found ) + 1.6 (Wsur )Wu = 1.2(12.53k + 2.81k + 2.35k ) + 1.6(0.4 ksf )(18 ft )(1 ft ) = 32.7 k , (did not recalc foundation wt b.c. neglible change)M u = 1.6 M over 1.2(Wsoil 2.125 ft + Wstem 1.0 ft )3 1.257.5 [ ]M u = 1.6(50.2 k ft ) 1.2 12.53k ( 2.125 ft ) + 2.81k (1 ft ) = 45.0k ft32.7 k45.0k ftv =+(1 ft )(11.75 ft ) 1 (1 ft )(11.75 ft ) 26 ksf vA = 2.78 +1.96ksf = 4.74 ksf A BC vC = 2.78ksf 1.96ksf= 0.82 ksf 4.74 ksf 0.82 ksf vB = 0.82 ksf +(8.75 ft ) = 3.74 ksf 11.75 ftd for flexure: 3 ft12M u = (3.74 ksf )(3 ft )(1 ft )() + (1.00 ksf )(3 ft )(1 ft )( 3 ft ) = 19.8k ft223 k 2M u = 5.71 din in k19.8k ft (12 ) = 5.71 d 2 , d = 6.5in for flexureft ind for shear:Assume theel = ttoe = 21.5inCritical section for shear occurs at "d" from face of stem, d = 21.5" 3"cover-1/2"=18"4.74 ksf 0.82 ksf18 vcritical sec tion = 0.82 ksf + ft(8.75 ft +ft ) = 4.24 ksf 11.75 12118Vu =( 4.74 ksf + 4.24 ksf )(3 ft ft )(1 ft ) = 6.74 k212Vc = (.75)2 3000 psi (12in )(18in ) = 17,750lb > Vu , OK , d for flexure controls 8. CE 437/537, Spring 2011Retaining Wall Design Example 8/8Reinforcement in toe:M u = 47.7 ksi As d in19.8k ft (12 ) = 47.7 ksi As (18in ), As = 0.28 in 2ft in 20.79 bar (12 in ) = 33in , try smaller bars , say #4 in 2 ft0.28ft in 20.20 bar (12 in ) = 8.6inuse #4 @ 8" in 2 ft0.28ft