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===
−=−=
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Unit dimension in the direction perpendicular to the paper.
Stability Requirement
Settlement](https://image.slidesharecdn.com/retainingwall-170102190848/85/Retaining-wall-12-320.jpg)
![Design of Heel Cantilever
Wu = (1.7) (120) (8) + (1.4) [18 x 120 + 2 x 150 ]
= 5076 lb/ft
( )( )
ft.lb76.88947
92.576.5
2
1
2
W
M
2
u
2
=
=
=
Vu = Factored shear a joint of stem and heel
Solution
When the support reaction introduces compression into
the end region, then critical shear is at a distance d
from face of support. However, the support is not
producing compression, therefore, critical shear is at
joint of stem and heel.](https://image.slidesharecdn.com/retainingwall-170102190848/85/Retaining-wall-39-320.jpg)
![Now Wu=(1.7)(120)(8)+(1.4)[(17.5)(120)+(2.5)(150)] = 5097 lb/ft
.ft.lb75.89315)92.5)(5097(
2
1
M 2
u =
=
005.0
f
200
00337.0
f
mR2
11
m
1
psi23.131
)5.27)(12)(9.0(
1289315
bd
M
RquiredRe
y
min
y
n
u
n 22
==ρ
=
−−=ρ
=
×
=
φ
=
Solution
Design of Heel Cantilever](https://image.slidesharecdn.com/retainingwall-170102190848/85/Retaining-wall-41-320.jpg)
More Related Content
Retaining wall
- 1.
- 2. Retaining walls areused to hold back masses of earth or other loose material where conditions make it impossible to let those masses assume their natural slopes. Function of retaining wall Such conditions occur when the width of an excavation, cut, or embankment is restricted by conditions of ownership, use of the structure, or economy. For example, in railway or highway construction the width of the right of way is fixed, and the cut or embankment must be contained within that width. Similarly, the basement walls of the buildings must be located within the property and must retain the soil surrounding the base.
- 3. Free standing retainingwalls, as distinct from those that form parts of structures, such as basement walls, are of various types. Types of retaining walls The gravity retaining wall retains the earth entirely by its own weight and generally contains no reinforcement. It is used up to 10 ft height. The reinforced concrete cantilever retaining wall consists of the vertical arm that retains the earth and is held in position by a footing or base slab. In this case, the weight of the fill on top of the heel, in addition to the weight of the wall, contributes to the stability of the structure. Since the arm represents a vertical cantilever, its required thickness increase rapidly, with increasing height. It is used in the range of 10 to 20 ft height.
- 4. In the counterfortwall the stem and base slab are tied together by counterforts which are transverse walls spaced at intervals and act as tension ties to support the stem wall. Counterforts are of half or larger heights. Counterfort walls are economical for heights over 25 ft. Types of retaining walls Property rights or other restrictions sometimes make it necessary to place the wall at the forward edge of the base slab, i.e. to omit the toe. Whenever it is possible, toe extensions of one-third to one-fourth of the width of the base provide a more economical solution.
- 5. A buttress wallis similar to a counterfort wall except that the transverse support walls are located on the side of the stem opposite to the retained material and act as compression struts. Buttress, as compression elements, are more efficient than the tension counterforts and are economical in the same height range. Types of retaining walls A counterfort is more widely used than a buttress because the counterfort is hidden beneath the retained material, whereas the buttress occupies what may otherwise be usable space in front of the wall.
- 6. This is anfree standing wall category. A wall type bridge abutment acts similarly to a cantilever retaining wall except that the bridge deck provides an additional horizontal restraint at the top of the stem. Thus this abutment is designed as a beam fixed at the bottom and simply supported or partially restrained at the top. Types of retaining walls
- 7. w isunit weight of the soil Earth Pressure C0 is a constant known as the coefficient of earth pressure at rest According to Rankin, the coefficient for active and passive earth pressure are φ−δ+δ φ−δ−δ δ= 22 22 CosCosCos CosCosCos CosCa φ−δ−δ φ−δ+δ δ= 22 22 CosCosCos CosCosCos CosCp For the case of horizontal surface δ=0 φ+ φ− = sin1 sin1 Cah φ− φ+ = sin1 sin1 Cph For liquid P=wwh, ww is the unit weight of water. Soil retaining structure Ph=Cowh
- 8. Earth pressure forcommon condition of loading 2 whC 2 1 P 3 h y ah= = φ=φ=δ = = CosCFor Cawh 2 1 p 3 h y a 2
- 9. Earth pressure forcommon condition of loading )h2h(whC 2 1 P )h2h(3 hh3h y ah 2 ′+= ′+ ′+ =
- 10. 1. Individual partsshould be strong enough to resist the applied forces Stability Requirement 2. The wall as a whole should be stable against (i) Settlem ent (ii) Sliding (iii) Overtur ning
- 11. Stability Requirement It isnecessary to ensure that the pressure under the footing does not exceed the “permissible bearing pressure” for the particular soil. By the formula Settlement I MC A N q min max ±= If , compression will act throughout the section 3 a >
- 12. ( )[ ] [] [ ] v 21 2 v 22 v 1 2 v 2 vvv 2,1 3 v 2 v v 3 v 2 v v 2,1 3 v v 2,1 R qq 2 awhen 2a6 R q&a64 R q a63 RaR6R3R q 12 4 aR2R R 12 2 aR 4 R R q 12 2 Ra 2R q === −=−= −±= −±= − ±= − ±= − ±= Unit dimension in the direction perpendicular to the paper. Stability Requirement Settlement
- 13.
- 14. ( )( ) a3 R2 q a3q 2 1 R v v = = StabilityRequirement Settlement
- 15. Sliding F = µRv StabilityRequirement 5.1 P F h ≥ Overturning 2 momentgOverturnin momentgStabilizin ≥
- 16. 1. Lateral earthpressure will be considered to be live loads and a factor of 1.7 applied. Basis of Structural Design 2. In general, the reactive pressure of the soil under the structure at the service load stage will be taken equal to 1.7 times the soil pressure found for service load conditions in the stability analysis. 3. For cantilever retaining walls, the calculated dead load of the toe slab, which causes moments acting in the opposite sense to those produced by the upward soil reaction, will be multiplied by a factor of 0.9.
- 17. 4. For theheel slab, the required moment capacity will be based on the dead load of the heel slab itself, plus the earth directly above it, both multiplied by 1.4. 5. Surcharge, if present, will be treated as live load with load factor of 1.7. 6. The upward pressure of the soil under the heel slab will be taken equal to zero, recognizing that for severe over load stage a non linear pressure distribution will probably be obtained, with most of the reaction concentrated near the toe. Basis of Structural Design
- 18. Drainage Reduction inbearing capacity. Hydrostatic pressure. Ice pressure Drainage can be provided in various ways i. Weep holes, 6 to 8 in. 5 to 10 ft horizontally spaced. 1 ft3 stone at rear end at the bottom weep holes to facilitate drainage and to prevent clogging. ii. Longitudinal drains embedded in crushed stone or gravel, along rear face of the wall. iii. Continuous back drain consisting of a layer of gravel or crushed stone covering the entire rear face of the wall with discharge at the ends.
- 19. Problem A gravityretaining wall consisting of plain concrete w=144 lb/ft3 is shown in fig. The bank of supported earth is assumed to weigh 110 lb/ft3 ø=300 and to have a coefficient of friction against sliding on soil of 0.5. Solution 333.0 30sin1 30sin1 sin1 sin1 Ca = + − ⇒ φ+ φ− = 3 30sin1 30sin1 sin1 sin1 Cp = − + ⇒ φ− φ+ = lb36.2637)12)(110)(333.0( 2 1 whC 2 1 P 22 aah = == lb.ft44.10549 3 12 36.2637momentgOverturnin =×=
- 20. Restoring moment: Moment about toe Component Weights Force Moment arm Moment W1 (6)(1)(144) 864 x 3 2592 lb.ft W2 (1)(11)(144) 1584 x 1 1584 lb.ft W3 (4)(144)(11) 8976 lb.ft W4 (4)(11)(110) 10083.33 lb.ft W5 (0.5)(11)(110) 605 x 5.75 3478.75 lb.ft Rv = 8641 lb M=26714.08 lb.ft 2 1 ×+× 4 3 1 5.13168 Solution ×+× 4 3 2 5.12420 2 1 Safety factor against overturning = O.K253.2 44.10549 08.26714 >=
- 21. Distance ofresultant from the toe Thus resultant is out of kern distance The max. soil pressure will be )87.1)(3( )8641)(2( a3 R2 q v ⇒= q = 3080.57 lb/ft2 < 4000 psf Solution ft87.1 8641 44.1054908.26714 a = − =
- 22. Sliding Assuming thatsoil above footing toe has eroded and thus the passive pressure is only due to soil depth equal to footing thickness. Solution lb516)1)(110)(3( 2 1 whC 2 1 P 22 pph = == Friction force between footing concrete and soil. F.O.S. against sliding = O.K lb5.4320)8641)(5.0(RF v ==µ= 5.17.1 36.2637 1655.4320 >= +
- 23. Estimating size ofcantilever retaining wall The base of footing should be below frost penetration about 3’ or 4’. Height of Wall Stem is thickest at its base. They have thickness in the range of 8 to 12% of overall height of the retaining wall. The minimum thickness at the top is 10” but preferably 12”. Stem Thickness. Preferably, total thickness of base fall between 7 and 10% of the overall wall height. Minimum thickness is at least 10” to 12” used. Base Thickness
- 24. Estimating size ofcantilever retaining wall For preliminary estimates, the base length can be taken about 40 to 60% of the overall wall height. Base Length Another method refer to fig. W is assumed to be equal to weight of the material within area abcd. Take moments about toe and solve for x.
- 25. Problem Design acantilever retaining wall to support a bank of earth of 16 ft height above the final level of earth at the toe of the wall. The backfill is to be level, but a building is to be built on the fill. Assume that an 8’ surcharge will approximate the lateral earth pressure effect. Weight of retained material = 130 lb/ft3 Angle of internal friction = 35o Coefficient of friction b/w concrete and soil = 0.4 fc’=3000 psi fy=40,000 psi Maximum soil pressure = 5 k/ft2
- 26. Solution Allowing 4’for frost penetration to the bottom of footing in front of the wall, the total height becomes. Height of Wall h = 16 + 4 = 20 ft. At this stage, it may be assumed 7 to 10% of the overall height h. Thickness of Base Assume a uniform thickness t = 2’ ( 10% of h ) h = 20’ h’ = 8’ Base Length
- 27. ( )( )() ( )( )( )( )( ) ( ) ( ) ( )( )( ) ( ) ft148.8 82203 820320 h2h3 hh3h y lb2.11707 3620120271.05.0 822020120 sin1 sin1 2 1 )h2h(whC 2 1 P 22 ah = ×+ + = ′+ ′+ = = = ×+ φ+ φ− = ′+= Solution
- 28. Moments aboutpoint a W = (120)(x)(20+8) = 3360 x lb yP) 2 x )(W( ×= )148.8)(2.11707() 2 x )(x3360( = x = 7.54 ft So base length = 1.5 x x = 11.31 ft Use 11 ft 4” with x = 7’-8” and 3’-8’ toe length Solution
- 29. Stem Thickness Priorcomputing stability factors, a more accurate knowledge of the concrete dimensions is necessary. Solution The thickness of the base of the stem is selected with the regard for bending and shear requirements. P for 18’ height and h’ = 8’ ( )( )( ) lb12.9951 )8218(18120271.0 2 1 )h2h(whC 2 1 P ah = ×+ = ′+ = ft412.7 )8218(3 )8)(18)(3()18( )h2h(3 hh3h y 22 ' ' = ×+ + = + + =
- 30. Mu =(1.7) Py = (1.7) (9951.12) (7.412) = 125388.09 lb.ft Solution 02784.075.0 03712.0 000,4087000 87000 000,40 )3000)(85.0)(85.0( f87000 87000 f f)85.0( bmax yy c b 1 =ρ=ρ = + = + ′β =ρ For adequate deflection control, choose 686.15 300085.0 000,40 f85.0 f m )m 2 1 1(fRthen c y yn = × = ′ = ρ−ρ= 01392.0 2 1 max =ρ=ρ
- 31. Solution ( )( ) 67.16d 65.3370 4969.0 1209.125388 R M R MquiredRe bdquiredRe 496 )686.1501392.0 2 1 1)(000,40)(01392.0( m 2 1 1fR n n n n2 yn ′′= = × = φ = = = ××−= ρ−ρ= Total thickness = 16.7 + 0.5 + 3 = 20.26” Try 21” thickness of base of stem and select 12” for top of the wall
- 32. Shear at d dused now = 17.5” = 1.458’ At 18’ – 1.458’ = 16.542’ from top ( )( )( )( ) .requiredisorcementinfreshearnoSo,VVSince lb7.19553 5.17123000285.0 bdf2V lb96.17879 P7.1V lb92.8752 82542.16542.16120271.0 2 1 )h2h(whC 2 1 P uu cu u ah >φ = ××××= ′×φ=φ = = = ×+ = ′+ = Solution
- 33. F.O.S Against Overturning ComponentForce Arm Moment W1 (5.92)(18)(120)=12787.2 107156.74 W2 (0.75)(18)(150)=1012.5 3.67+1+0.25=4.92 4981.50 W3 (18)(1)(150)=2700 3.67+0.5=4.17 11259 W4 (11.33)(2)(150)=3399.00 19255.34 W5 (18)(0.75)(120)=810 3.67+1+0.5=5.17 4187.7 W6 (6.67)(8)(120)=6403.2 51257.62 Total 27111.9 198097.9 ) 2 1 ( ) 2 1 ( 38.8 2 92.5 2167.3 " =++ 665.5 2 33.11 = 005.8 2 67.6 167.3 =++ Solution
- 34. P =11707.2 lb y= 8.148 ft Overturning Moment = 11707.2 x 8.148 = 95390.27 lb.ft F.O.S. against overturning O.K2077.2 27.95390 9.198097 >== Location of Resultant & Footing Soil Pressure Distance of the resultant from the front edge is 7883.3 9.27111 27.953909.198097 a )righting(loadTotal momentgOverturninmomentRighting a = − = − = Middle third = 3.7778 ft, So resultant is within the middle third. Solution ( ) ( )7883.3633.114 33.11 9.27111 q )a64( R q 21 v 21 ×−×= −=
- 35. 2 22 ft/lb74.14)2a6( R q v =−= q1=4771.12 lb/ft2 < 5 k/ft2 So O.K against bearing pressure. Solution
- 36. Passive earthpressure against 2’ height of footing aphCwh 2 1 2 = Solution .requirediskeySo5.1964.0 2.11707 44276.10844 .S.O.F lb82.442 sin1 sin1 )2)(120( 2 1 <= + = = φ− φ+ = F.O.S. against sliding force causing sliding = P = 11707.2 lb Frictional resistance = µR = (0.4) (27111.9) = 10844.76 lb
- 37. The frontof key is 4” in front of back face of the stem. This will permit anchoring the stem reinforcement in the key.. ft/lb58.2635ordinateTotal ft/lb84.2620x 33.11 38.4756 243.6 x = = Frictional resistance between soil to soil = µR lb17.13191 )087.5( 2 58.263512.4771 )(tan = + φ= Solution
- 38. Frictional resistancebetween heel concrete to soil = µR lb19.3309 )243.6( 2 74.1458.2635 )4.0( = + = ( )( ) ( ) lbh4.22169.3h120 2 1 Cwh 2 1 pressureearthPassive 22 ph 2 = = = Solution F.O.S. against sliding = 1.5 ft19.2h 2.11707 h4.22119.330917.13191 5.1 2 = ++ = So use key of height = 2’-3” = 2.25’
- 39. Design of HeelCantilever Wu = (1.7) (120) (8) + (1.4) [18 x 120 + 2 x 150 ] = 5076 lb/ft ( )( ) ft.lb76.88947 92.576.5 2 1 2 W M 2 u 2 = = = Vu = Factored shear a joint of stem and heel Solution When the support reaction introduces compression into the end region, then critical shear is at a distance d from face of support. However, the support is not producing compression, therefore, critical shear is at joint of stem and heel.
- 40. Design of HeelCantilever Vu = Factored shear a joint of stem and heel = (5.92) (5076) = 30049.92 lb Solution u cc V11.24023 5.21123000285.0 bdf2V <= ××××= ′φ=φ So depth is required to be increased. 98.26 )12)(3000)(2)(85.0( 92.30049 bf2 V d c u ′′== ′φ = Therefore heel thickness 30”, d = 27.5”
- 41. Now Wu=(1.7)(120)(8)+(1.4)[(17.5)(120)+(2.5)(150)] =5097 lb/ft .ft.lb75.89315)92.5)(5097( 2 1 M 2 u = = 005.0 f 200 00337.0 f mR2 11 m 1 psi23.131 )5.27)(12)(9.0( 1289315 bd M RquiredRe y min y n u n 22 ==ρ = −−=ρ = × = φ = Solution Design of Heel Cantilever
- 42. As =ρmin bd = 1.8 in2 Use # 8 @ 5” c/c (As = 1.88 in2 ) Dev. Length required = 23” top bars so 23 × 1.3 = 29.9” Available = 5.92’-3” = 68.04” O.K Design of Toe Slab 33.11 66.7 38.4756 x = x = 3215.7 3215.7 + 14.74 = 3230.44 (6801.33)lb/ft + = 2 12.477144.3230 )7.1(Wu Wu = 6801.33 lb/ft – 270 = 6531.33 Solution Overload factor = 0.9 {d=24”-3.5”=20.5” =1.71 ft Self load=(0.9)(1×2×150)=270 lb/ft
- 43. ( )( )() 3.116 5.20129.0 1292.43984 bd M RquiredRe ft.lb92.43984 2 )67.3)(33.6531( 2 W M 22 u n u u 22 = × = φ = === So ρmin will control As = (0.005)(12)(20.5) = 1.23 in2 Use # 8 @ c/c (As = 1.26 in2 ) 2 1 7 Available dev. Length = 3.76’ – 3” = 41.01” Required = 23” At a distance d= 20.5” = 1.71’ 3.67’ – 1.71’ = 1.96’ 3.394874.1456.3933 56.3933x 33.11 37.9 38.4756 x =+ = = Solution
- 44. Earth pressure lb55.14526)7.1)(96.1( 2 12.47713.3948 = + = Vu = 14526.55 – 270 x 1.96 = 13997.37 lb u cc Vlb76.229055.20123000285.0 bd)f2(V >=××××= ′φ=φ So no shear reinforcement is required Reinforcement for stem Solution ( )( )( )( ) ( ) ( ) 32.7 5.100 25.726 825.173 85.173)5.17( h2h3 hh3h y lb43.9532 825.175.17120271.0 2 1 )h2h(whC 2 1 P 22 ah ′== ×+ ××+ = ′+ ′+ = = ×+ = ′+=
- 45. ( )( )() )in67.2A(c/c 2 1 4@bars9#Use in457.25.17120117.0bdA 0117.0 000,40 08.425686.152 11 686.15 1 686.15 300085.0 000,40 f85.0 f m f mR2 11 m 1 psi08.425 )5.17(129.0 121.117163 bd M R ft.lb1.11716323.743.95327.1M 2 s 2 s c y y n u n u 22 = ″ =××=ρ= = ×× −−=ρ = × = ′ = −−=ρ = ×× × = φ = == Reinforcement for stem Solution
- 46. At 5’ fromtop P=1707.3lb y=2.302’ Mu=6681.34 lb.ft =6.68 k.ft At 10’ from top P=4227.6 lb y=4.36’ Mu=31334.97 lb.ft =31.33 k.ft At 15’ from top P=7560.9 lb y=6.29’ Mu=80848.7 lb.ft =80.85 k.ft Solution ft.k16.117 in.lb2.1405957ft.lb1.117163baseatMu = =
- 47. With Full Reinforcement C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb T = Asfy = 2.67 x 40000 = 106800 a = 3.49 in At top of wall d = 8.5” ft.k11.54.in.lb6.649290 2 49.3 5.8)106800)(9.0( 2 a dfA9.0M ysn == −= −=φ At base of stem d=17.5” Solution ft.k2.126.in.lb6.1514370 2 49.3 5.17)106800)(9.0(Mn == −=φ
- 48. With half Reinforcement C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb T = Asfy = 1.335 × 40,000 = 53400 for #9@9” c/c(As=1.33 in2 ) a = 1.745 in At top of wall d = 8.5” ft.k55.30.in.lb65.366577 2 745.1 5.8)53400)(9.0(Mn == −=φ At base of stem d=17.5” Solution ft.k59.66.in.lb65.799117 2 745.1 5.17)53400)(9.0(Mn == −=φ
- 49. With one-fourth Reinforcement C = 0.85fc’ba = 0.85 × 3000 × 12 × a = 30600a lb T = Asfy = 0.67 × 40,000 = 26800 for #9@18” c/c(As=0.67 in2 ) a = 0.88 in At top of wall d = 8.5” ft.k2.16.in.lb2.194407 2 88.0 5.8)26800)(9.0(Mn == −=φ At base of stem d=17.5” Solution ft.k29.34.in.lb2.411487 2 88.0 5.17)26800)(9.0(Mn == −=φ
- 50. The actualtermination point is found by extending beyond the intersection of capacity moment line with the factored moment diagram a distance of either the effective depth d or 12 bar diameters, whichever is greater. Solution Thus half bars should be cut at 4’-8” distance from bottom and further half bars should be cut 8’-8” theoretically from bottom.
- 51.
- 52. 36.6y 5.17 9 x 125.17 9 y12 x ′′=×= × = × d at 4.6’from bottom (y=12.9’) = 8.5 + 6.63 = 15.13” 15”≅ 85.4y 5.17 9 x ′′=×= d at 8.6’ from bottom (y=8.9’) = 8.5 + 4.58 = 13.08” 13” Solution ≅
- 53. bar used# 9 of diameter = 1.128” 12 db = 13.54” = 14” Therefore half bars should be terminated actually at 4’-8”+15”=15’-11” from bottom and further half bars should be terminated at 8’-8”+14” =9’10” For tension bars to be terminated in the tension zone, one of the following condition must be satisfied. 2. Continuing bars must provide at last twice the area required for bending moment at the cut off point. 1. Vu at the cut-off point must not exceed two-thirds of the shear strength øVn. 3. Excess shear reinforcement is provided. at 12.9’ from topbd)f2(V cc ′φ=φ Solution
- 54.
- 55. ( ) ( ) kips13.6 1000 1 829.89.8120271.0 2 1 7.1 1000 1 h2hwhC 2 1 7.1Vahu = × ×+××××= × ′+= Condition (1) satisfied so bars can be terminated. The above condition are imposed as a check stress concentration. Shear at bottom = Vu = 1.7 x 9.53243 = 16.21 kips Solution kips56.19 1000 1 5.17123000285.0Vc =×××××=φ Since øVc > Vu so no need of shear reinforcement.
- 56. The ρused should not be less than at any point. This minimum limit, strictly speaking, does not apply to retaining walls. However, because the integrity of retaining wall depends absolutely on the vertical walls, it appear prudent to use this limit un such cases. yf 200 First termination point is 5’-11” from bottom where d=14.46” As =1.335 in2 Second termination point is 9’-10” form bottom where d=12.44” As=0.6675 in2 Solution 005.0 f 200 0077.0 y =>=ρ ρ = 0.0045 ≈ 0.005 Therefore the above condition is satisfied.
- 57. Another requirementis that maximum spacing of the primary flexural reinforcement exceed neither 3 times the wall thickness nor 18 in. These restrictions are satisfied as well. For splices of deformed bars in tension, at sections where the ratio of steel provided to steel required is less than 2 and where no more than 50% of the steel is spliced, the ACI code requires a class-B splice of length 1.3 ld. ld for # 9 bars =29” Splice length = 1.3 × 29 = 37.7” or 3’-2” O.K Solution
- 58. Temperature & shrinkagereinforcement Total amount of horizontal bars (h is average thickness) ft/in39.0 2 50.2012 12002.0bh002.0A 2 s = + ××== Since front face is more exposed for temperature changes therefore two third of this amount is placed in front face and one third in rear face. Solution Accordingly # 4 @ 9 in. c/c As=0.26 in.ft/in26.0A 3 2 2 s = Use # 3 @ 10 in. c/c As=0.13 in2 .ft/in13.0A 3 1 2 s = For vertical reinforcement on the front face, use any nominal amount. Use # 3 @ 18 in. c/c Since base is not subjected to extreme temperature changes, therefore # 4@ 12” c/c just for spacers will be sufficient.
- 59.