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Retaining Walls

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Page 1: Retaining Wall

Retaining Walls

Page 2: Retaining Wall

Retaining walls are used 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.

Page 3: Retaining Wall

Free standing retaining walls, 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.

Page 4: Retaining Wall

In the counterfort wall 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.

Page 5: Retaining Wall

A buttress wall is 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.

Page 6: Retaining Wall

This is an free 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

Page 7: Retaining Wall

w is unit 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

22

22

CosCosCos

CosCosCosCosCa

22

22

CosCosCos

CosCosCosCosCp

For the case of horizontal surface =0

sin1sin1

Cah

sin1sin1

Cph

For liquid P=wwh, ww is the unit weight of water.Soil retaining structure Ph=Cowh

Page 8: Retaining Wall

Earth pressure for common condition of loading

2whC21

P

3h

y

ah

CosCFor

whC2

1p

3

hy

a

a2

Page 9: Retaining Wall

Earth pressure for common condition of loading

)h2h(whC21

P

)h2h(3hh3h

y

ah

2

Page 10: Retaining Wall

1. Individual parts should be strong enough to resist the applied forces

Stability Requirement

2. The wall as a whole should be stable against (i) Settlement (ii) Sliding (iii) Overturning

Page 11: Retaining Wall

Stability Requirement

It is necessary to ensure that the pressure under the footing does not exceed the “permissible bearing pressure” for the particular soil.

By the formula

Settlement

IMC

AN

qminmax

If , compression will act throughout the section3

a

Page 12: Retaining Wall

v21

2v

22v

1

2v

2vvv

2,1

3

v2

v

v3

v2

v

v2,1

3

vv

2,1

Rqq

2awhen

2a6R

q&a64R

q

a63RaR6R3R

q

12

4aR2R

R

12

2aR

4R

Rq

12

2Ra

2Rq

Unit dimension in the direction perpendicular to the paper.

Stability Requirement Settlement

Page 13: Retaining Wall

0q&R2

q

11Rq

2R

612

1R2

R6

2312

1Rq

12

2R

32Rq

2v

1

v2,1

v

v

v

v2,1

vv

2,1

33

3

Stability Requirement Settlement

Page 14: Retaining Wall

a3R2

q

a3q21

R

v

v

Stability Requirement Settlement

Page 15: Retaining Wall

Sliding

F = Rv

Stability Requirement

5.1PF

h

Overturning

2moment gOverturnin

moment gStabilizin

Page 16: Retaining Wall

1. Lateral earth pressure 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.

Page 17: Retaining Wall

4. For the heel 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 resent, 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

Page 18: Retaining Wall

Drainage

Reduction in bearing 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.

Page 19: Retaining Wall

Problem A gravity retaining 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

333.030sin130sin1

sin1sin1

Ca

330sin130sin1

sin1sin1

Cp

lb36.2637)12)(110)(333.0(21

whC21

P 22

aah

lb.ft 44.105493

1236.2637momentgOverturnin

Page 20: Retaining Wall

Restoring moment: Moment about toe

Component Weights

Force Moment arm Moment

W1 (6)(4)(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

21

4

31

5.13168

Solution

4

32

5.12420

21

Safety factor against overturning = O.K253.244.1054908.26714

Page 21: Retaining Wall

Distance of resultant from the toe

Thus resultant is out of kern distance

The max. soil pressure will be )87.1)(3()8641)(2(

a3R2

q v

q = 3080.57 lb/ft2 < 4000 psf

Solution

ft87.18641

44.1054908.26714a

Page 22: Retaining Wall

Sliding

Assuming that soil above footing toe has eroded and thus the passive pressure is only due to soil depth equal to footing thickness.

Solution

lb 516)1)(110)(3(21

whC21

P 22

pph

Friction force between footing concrete and soil.

F.O.S. against sliding = O.K

lb 5.4320)8641)(5.0(RF v

5.17.136.26371655.4320

Page 23: Retaining Wall

Estimating size of cantilever 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

Page 24: Retaining Wall

Estimating size of cantilever 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.

Page 25: Retaining Wall

Problem Design a cantilever 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

Page 26: Retaining Wall

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

Page 27: Retaining Wall

ft148.8

82203820320

h2h3hh3h

y

lb2.11707

3620120271.05.0

822020120sin1sin1

21

)h2h(whC21

P

22

ah

Solution

Page 28: Retaining Wall

Moments about point a

W = (120)(x)(20+8) = 3360 x lb

yP)2

)(W(

)148.8)(2.11707()2

)(3360(

x = 7.54 ft So base length = 1.5 x = 11.31 ft

Use 11 ft 4” with x = 7’-8” and 3’-8’ toe length

Solution

Page 29: Retaining Wall

Stem Thickness Prior computing 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(18120271.021

)h2h(whC21

P ah

ft412.7

)8218(3)8)(18)(3()18(

)h2h(3hh3h

y22

'

'

Page 30: Retaining Wall

Mu = (1.7) Py = (1.7) (9951.12) (7.412)

= 125388.09 lb.ft

Solution

02784.075.0

03712.0

000,408700087000

000,40)3000)(85.0)(85.0(

f8700087000

ff)85.0(

bmax

yy

cb

1

For adequate deflection control, choose

686.15

300085.0000,40

f85.0

fm

)m21

1(fRthen

c

y

yn

01392.021

max

Page 31: Retaining Wall

Solution

67.16d

65.3370

4969.01209.125388

R

M

RMquiredRe

bdquiredRe

496

)686.1501392.021

1)(000,40)(01392.0(

m21

1fR

n

n

n

n2

yn

Total thickness = 16.7 + 0.5 + 3 = 20.26”

Try 21” thickness of base of stem and select 12” for top of the wall

Page 32: Retaining Wall

Shear at dd used now = 17.5” = 1.458’ At 18’ – 1.458’ = 16.542’ from top

.requiredisorcementinfreshearnoSo,VVSince

lb27.18483

542.16123000285.0

bdf2V

lb96.17879

P7.1V

lb92.8752

82542.16542.16120271.02

1

)h2h(whC2

1P

uc

cc

u

ah

Solution

Page 33: Retaining Wall

F.O.S Against Overturning

Component Force 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

38.82

92.52167.3 "

665.5233.11

005.8267.6

167.3

Solution

2

1

Page 34: Retaining Wall

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.K2 077.227.953909.198097

Location of Resultant & Footing Soil Pressure Distance of the resultant from the front edge is

7883.39.27111

27.953909.198097a

)righting( load Totalmoment gOverturninmoment Righting

a

Middle third = 3.7778 ft, So resultant is within the middle third.

Solution

7883.3633.114

33.11

9.27111q

)a64(R

q

21

v21

Page 35: Retaining Wall

2

22 ft/lb 74.14)2a6(R

q v

q1= 4771.12 lb/ft2 < 5 k/ft2

So O.K against bearing pressure.

Solution

Page 36: Retaining Wall

Passive earth pressure against 2’ height of footing aphCwh21 2

Solution

.requirediskeySo5.1964.02.1170744276.10844

.S.O.F

lb82.442

sin1sin1

)2)(120(21

F.O.S. against sliding force causing sliding = P = 11707.2 lb

Frictional resistance = R

= (0.4) (27111.9)

= 10844.76 lb

Page 37: Retaining Wall

The front of 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/lb84.2620x33.11

38.4756243.6x

Frictional resistance between soil to soil = R

lb17.13191

)087.5(2

58.263512.4771)(tan

Solution

Page 38: Retaining Wall

Frictional resistance between heel concrete to soil = R

lb19.3309

)243.6(2

74.1458.2635)4.0(

lbh4.22169.3h12021

Cwh21

pressureearthPassive

22

ph2

Solution

F.O.S. against sliding = 1.5

ft19.2h2.11707

h4.22119.330917.131915.1

2

So use key of height = 2’-3” = 2.25’

Page 39: Retaining Wall

Design of Heel CantileverWu = (1.7) (120) (8) + (1.4) [18 x 120 + 2 x 150 ]

= 5076 lb/ft

ft.lb76.88947

92.576.5212

WM

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.

Page 40: Retaining Wall

Design of Heel CantileverVu = Factored shear a joint of stem

and heel= (5.92) (5076)

= 30049.92 lb

Solution

u

cc

V11.24023

5.21123000285.0

bdf2V

So depth is required to be increased.

98.26)12)(3000)(2)(85.0(

92.30049bf2

Vd

c

u

Therefore heel thickness 30”, d = 27.5”

Page 41: Retaining Wall

Now Wu=(1.7)(120)(8)+(1.4)[(17.5)(120)+(2.5)(150)] = 5097 lb/ft

.ft.lb 75.89315)92.5)(5097(21

M 2

u

005.0f

200

00337.0f

mR211

m1

psi 23.131)5.27)(12)(9.0(

1289315bdM

RquiredRe

ymin

y

n

un 22

SolutionDesign of Heel Cantilever

Page 42: Retaining Wall

As = min 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

33.1166.7

38.4756x

x = 3215.7

3215.7 + 14.74 = 3230.44

(6801.33)lb/ft

212.477144.3230

)7.1(Wu

Wu = 3801.33 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

Page 43: Retaining Wall

3.116

5.20129.0

1292.43984bdM

RquiredRe

ft.lb 92.439842

)67.3)(33.6531(2

WM

22u

n

uu

22

So min will control

As = (0.005)(12)(20.5) = 1.23 in2

Use # 8 @ c/c (As = 1.26 in2)21

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

56.3933x33.1137.9

38.4756x

Solution

Page 44: Retaining Wall

Earth pressure lb 55.14526)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

So no shear reinforcement is required

Reinforcement for stem

Solution

32.75.10025.726

825.17385.173)5.17(

h2h3hh3h

y

lb43.9532

825.175.17120271.021

)h2h(whC21

P

22

ah

Page 45: Retaining Wall

)in67.2A(c/c21

4@bars9#Use

in457.25.17120117.0bdA

0117.0

000,4008.425686.152

11686.151

686.15300085.0

000,40f85.0

fm

fmR2

11m1

psi 08.425)5.17(129.0

121.117163bdM

R

ft.lb1.11716323.743.95327.1M

2s

2s

c

y

y

n

un

u

22

Reinforcement for stemSolution

Page 46: Retaining Wall

At 5’ from top

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

Page 47: Retaining Wall

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

249.3

5.8)106800)(9.0(2a

dfA9.0M ysn

At base of stem d=17.5”

Solution

ft.k2.126.in.lb6.1514370

249.3

5.17)106800)(9.0(Mn

Page 48: Retaining Wall

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

2745.1

5.8)53400)(9.0(Mn

At base of stem d=17.5”

Solution

ft.k59.66.in.lb65.799117

2745.1

5.17)53400)(9.0(Mn

Page 49: Retaining Wall

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

288.0

5.8)26800)(9.0(Mn

At base of stem d=17.5”

Solution

ft.k29.34.in.lb2.411487

288.0

5.17)26800)(9.0(Mn

Page 50: Retaining Wall

The actual termination 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.

Page 51: Retaining Wall

Solution

Page 52: Retaining Wall

36.6y5.17

9x

125.179

y12x

d at 4.6’ from bottom (y=12.9’)

= 8.5 + 6.63 = 15.13”

15”

85.4y5.17

9x

d at 8.6’ from bottom (y=8.9’)

= 8.5 + 4.58 = 13.08”

13”

Solution

Page 53: Retaining Wall

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

Solution

Page 54: Retaining Wall

kips 68.9V32

kips 14.531000

113123000285.0V

.satisfiedis)1(Condition

kips31.10

10001

h2hwhC21

7.1V

kips 11.17V32

kips 16.761000

115123000285.0V

c

c

ahu

c

c

Solution

Page 55: Retaining Wall

kips13.6

10001

829.89.8120271.021

7.1

10001

h2hwhC21

7.1V ahu

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

kips56.191000

15.17123000285.0Vc

Since øVc > Vu so no need of shear reinforcement.

Page 56: Retaining Wall

The used 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.

yf200

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

005.0f

2000077.0

y

= 0.0045 ≈ 0.005 Therefore the above condition is satisfied.

Page 57: Retaining Wall

Another requirement is 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

Page 58: Retaining Wall

Temperature & shrinkage reinforcement

Total amount of horizontal bars (h is average thickness)

ft/in39.02

50.201212002.0bh002.0A 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/in26.0A32 2

s

Use # 3 @ 10 in. c/c As=0.13 in2. ft/in 13.0A31 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.

Page 59: Retaining Wall

Solution