Retaining Walls

1

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.

2

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 20ft height.

3

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.

4

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.

5

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

6

w is unit weight of the soil

Earth Pressure

C0 is a constant known as the coefficient of earth pressure at rest According to Rankine, 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

7

Earth pressure for common condition of loading

2whC21

P

3h

y

ah

CosCFor

Cawh21

p

3h

y

a

2

8

Earth pressure for common condition of loading

)h2h(whC21

P

)h2h(3hh3h

y

ah

2

9

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

10

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

11

v

vv

Rqqawhen

aR

qaR

q

21

2221

2

26&64

Stability Requirement Settlement

12

0&2

21 qR

q v

Stability Requirement Settlement

13

a

Rq v

3

2

Stability Requirement Settlement

14

Sliding

F = Rv

Stability Requirement

5.1PF

h

Overturning

2moment gOverturnin

moment gStabilizin

15

1. Lateral earth pressure will be considered to be live loads and a factor of 1.6 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 1.6 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.

16

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 1.2.

5. Surcharge, if resent, will be treated as live load with load factor of 1.6.

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

17

Drainage

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.

18

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

19

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.20

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 = 120 lb/ft3

Angle of internal friction = 35o

Coefficient of friction b/w concrete and soil = 0.4

fc’=3000 psi

fy=60,000 psi

Maximum soil pressure

= 5 k/ft221

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

22

ft148.8

82203820320

h2h3hh3h

y

lb2.11707

3620120271.05.0

822020120sin1sin1

21

)h2h(whC21

P

22

ah

Solution

23

Moments about point a

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

yP)2x

)(W(

)148.8)(2.11707()2x

)(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

24

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

'

'

25

Mu = (1.6) Py = (1.6) (9951.12) (7.412)

= 118004 lb.ft

Solution

),4(0203.0

005.0003.0

003.0

000,60

)3000)(85.0)(85.0(

005.0003.0

003.0)85.0( 1max

NilsonTableAOR

f

f

y

c

For economy and ease of bar placement,

008.04.0 max

26

Solution

"75.20

304129.0

12118004

ReRe

304

588.01

22

'

d

bd

M

R

Mquiredbdquired

psif

ffR

u

n

n

c

yyn

Total thickness = 20.75 + 0.5 + 2.5 = 23.75”

Try 24” thickness of base of stem and select 12” for top of the wall27

Shear at dd used now = 24-0.5-2.5=21”=1.75’ At 18’ – 1.75’ = 16.25’ from top

.inf,

20704

21123000285.0

2

13634

6.1

8521

8225.1625.16120271.02

1

)2(2

1

requiredisorcementreshearnoSoVVSince

lb

bdfV

lb

PV

lb

hhwhCP

uu

cu

u

ah

Solution

28

F.O.S Against Overturning

Component Force Arm Moment

W1 (5.667)(18)(120)=12239.7 3.67+2+5.667/2 104037.9

W2 (1)(18)(150)=1350.1 3.67+1+1/3 6750.6

W3 (18)(1)(150)=2700 3.67+0.5=4.17 11250

W4 (11.33)(2)(150)=3399.00 11.33/2 19267

W5 (18)(1)(120)=1080 3.67+1+2/3 4187.7

W6 (6.67)(8)(120)=6400 3.67+1+6.67/2 51200

Total 27170.1 198266.6

)2

1(

)2

1(

Solution

29

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 08.227.95390

6.198266

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

786.3

1.27170

27.953906.198266

Re

a

loadTotal

momentgOverturninmomentsistinga

Middle third=L/3= 3.77 ft, a>L/3,So resultant is within the middle third.

Solution

786.3633.114

33.11

1.27170

)64(

21

21

q

aR

q v

30

222 / 11)26( ftlbaR

q v

q1= 4783.7 lb/ft2 < 5 k/ft2

So O.K against bearing pressure.

Solution

31

Passive earth pressure against 2’ height of footing aphCwh21 2

Solution

.5.1966.02.11707

44210868...

82.442

sin1

sin1)2)(120(

2

1

requirediskeySoSOF

lb

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

Frictional resistance = R

= (0.4) (27170.1)

= 10868 lb

32

The front of key is 4” in front of back face of the stem. This will permit anchoring the stem reinforcement in the key..

ftlbordinateTotal

ftlbx

x

/5.2538

/5.252733.11

)117.4783(

6

Frictional resistance between soil to soil = R

lb4.13773

)34.5(2

5.25387.4783)(tan

Solution

33

Frictional resistance between heel concrete to soil = R

lb4.3059

)6(2

115.2538)4.0(

lbh4.22169.3h12021

Cwh21

pressureearthPassive

22

ph2

Solution

F.O.S. against sliding = 1.5

fth

h

81.12.11707

4.2214.30594.137735.1

2

So use key of height = 2’34

Design of Heel CantileverWu = (1.6) (120) (8) + (1.2) [18 x 120 + 2 x 150 ]

= 4488 lb/ft

ftlb

WM u

.72055

67.544882

12

2

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.

35

Design of Heel CantileverVu = Factored shear a joint of stem

and heel= (5.67) (4488)

= 25432 lb

Solution

u

cc

V

bdfV

20211

5.20123000275.0

2

So depth is required to be increased.

"26)12)(3000)(2)(75.0(

25432

2

bf

Vd

c

u

Therefore heel thickness 30”, d = 26.5”36

Now Wu=(1.6)(120)(8)+(1.2)[(17.5)(120)+(2.5)(150)] = 4506 lb/ft

.. 72344)67.5)(4506(2

1 2 ftlbM u

0033.0200

002.02

111

53.2385.0

14.51)5.26)(12)(9.0(

1272344Re

min

'

22

y

y

n

c

y

un

f

f

mR

m

f

fm

psibd

MRquired

SolutionDesign of Heel Cantilever

37

As = min bd = 1.06 in2

Bar area of #8=3.14(1)^2/4=0.79, No of bars required=1.06/0.79=1.35, Take=2, S=12inch/No. of bars,

Use # 8 @ 6” c/c (As = 1.57 in2 )

Design of Toe Slab

33.11

66.7

7.4772

x

x = 3226.7

3226.7 + 11 = 3237.7

Wu=(6417.15)lb/ft

2

7.47837.3237)6.1(uW

Wu = 6417.15 lb/ft – 270 = 6147.15

Solution

Overload factor = 0.9{ d=24”-3.5”=20.5”

=1.71 ft

Self load=(0.9)(1×2×150)=270 lb/ft

38

109

5.20129.0

1241332Re

. 413322

)67.3)(15.6147(

2

22

22

bd

MRquired

ftlbW

M

un

uu

So min will control

As = (0.0033)(12)(20.5) = 0.82 in2

Use # 8 @ c/c (As = 1.26 in2) Table A-3 of Nilson21

7

At a distance d= 20.5” = 1.71’

3.67’ – 1.71’ = 1.96’

1.3958111.3947

1.394733.11

37.9

7.4772

x

x

Solution

39

Earth pressure lb 1.13707)6.1)(96.1(2

7.47831.3958

Vu = 13707.1 – 270 x 1.96 = 13177.9lb

u

cc

Vlb

bdfV

202115.20123000275.0

)2(

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

40

)57.1(/6@8#

26.12112005.0

005.0

000,60

27858.23211

58.23

1

53.23300085.0

000,60

85.0

211

1

278)21(129.0

12110271

.15.11027123.743.95326.1

2

2

22

inAccbarsUse

inbdA

f

fm

f

mR

m

psibd

MR

ftlbM

s

s

c

y

y

n

un

u

Reinforcement for stemSolution

41

Temperature & shrinkage reinforcement

Total amount of horizontal bars (h is average thickness)

ftinbhAs /43.02

241212002.0002.0 2

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 @ 8 in. c/c As=0.29 in2.ftinAs /28.03

2 2

Use # 3 @ 9 in. c/c As=0.15 in2. ftinAs / 14.03

1 2

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. 42

Solution

43