DESIGN AND DETAILING OF RETAINING WALLS

Learning Outcomes:g

•• AfterAfter thisthis presentpresent youyou willwill bebe ableable toto dodo thetheAfterAfter thisthis presentpresent youyou willwill bebe ableable toto dodo thethecompletecomplete designdesign andand detailingdetailing ofof differentdifferent typestypes ofofretainingretaining wallswalls..

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Cantilever Retaining wall with shear keywith shear key

Batter

Drainage HoleToeToe

2

Classification of Classification of Retaining wallsRetaining walls

• Cantilever retaining wall-RCC(Inverted T and L)(Inverted T and L)

• Counterfort retaining wall-RCC• Buttress wall-RCCButtress wall RCC

3

Classification of Retaining Classification of Retaining Classification of Retaining Classification of Retaining wallswalls

BackfillBackfill

Gravity RWT-Shaped RW

L-Shaped RWTile drain

Counterfort ButtressBackfill

Weep hole

4

Counterfort RW Buttress RW

Earth Pressure (P)Earth Pressure (P)

• Earth pressure is the pressure exerted by the retaining material on the retaining wall This GLon the retaining wall. This pressure tends to deflect the wall outward.

GL

• Types of earth pressure : Pa

• Active earth pressure or earth pressure (Pa) and

• Passive earth pressure (Pp).

• Active earth pressure tends to

Variation of Earth pressure

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• Active earth pressure tends to deflect the wall away from the backfill.

Factors affecting earth Factors affecting earth pressurepressure

• Earth pressure depends on type of backfill the height of wall and the backfill, the height of wall and the soil conditions

Soil conditions: The different soil conditions are

• Dry leveled back fill• Moist leveled backfill• Submerged leveled backfill• Leveled backfill with uniform

surcharge6

surcharge• Backfill with sloping surface

Analysis for dry back Analysis for dry back fillsfillsfillsfills

Maximum pressure at any height, p=kaγh GLMaximum pressure at any height, p kaγh Total pressure at any height from top,

pa=1/2[kaγh]h = [kaγh2]/2 h

GL

Bending moment at any height M=p xh/3= [k γh3]/6

Pa

HGL

M=paxh/3= [kaγh3]/6

∴ Total pressure, Pa= [kaγH2]/2

M

∴ Total pressure, Pa [kaγH ]/2 ∴Total Bending moment at bottom,

M = [kaγH3]/6

kaγH

H=stem height

7

• Where, ka= Coefficient of active earth pressure• = (1 sinφ)/(1+sinφ)=tan2φ• = (1-sinφ)/(1+sinφ)=tan2φ• = 1/kp, coefficient of passive earth

pressurepressure• φ= Angle of internal friction

γ=Unit weigh or density of backfillγ g y

• If φ= 30°, ka=1/3 and kp=3. Thus ka is 9 times kpp p

8

Backfill with sloping surface

• pa= ka γH at the bottom pa a γand is parallel to inclined surface of backfill

GL

• ka= ⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−

φθθφθθθ22

22 coscoscoscos

• Where θ=Angle of

⎥⎦⎢⎣ −+ φθθ 22 coscoscos

surcharge∴ Total pressure at bottom

9=Pa= ka γH2/2

Stability requirements of RWStability requirements of RW

• Following conditions must be f d f b l f llsatisfied for stability of wall

It h ld t t• It should not overturn• It should not slide

• i.e Max. pressure at the toe should not exceed the safe bearing not exceed the safe bearing capacity of the soil under working condition

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Check against overturningCheck against overturning

Factor of safety against overturning g

= MR / MO ≥ 1.55 (=1.4/0.9)Where,

M S bili i MR =Stabilising moment or restoring momentMO =overturning momentMO overturning moment

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Check against SlidingCheck against Sliding

• FOS against sliding • FOS against sliding • = Resisting force to

sliding/Horizontal force g/causing sliding

• = μ∑W/Pa ≥ 1.55 (=1.4/0.9)

Friction μ ∑W

SLIDING OF WALL

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SLIDING OF WALL

Design of Shear key• In case the wall is

unsafe against sliding

• pp= p tan2 (45 +φ/2) H

• = p kp• where pp= Unit

passive pressure il b

PAH+a

H

on soil above shearing plane AB

• p= Earth pressure at BCapp

RAC

at BC

• R=Total passive resistance=p xa

θ=45 + φ/2

app

Bθ

ΣμW kaγ(H+a)

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resistance=ppxa

Design of Shear keyDesign of Shear key--Contd.,Contd.,

• If ∑W= Total vertical force acting at the key base

,,

baseφ= shearing angle of passive resistanceR T t l i f • R= Total passive force = pp x a

• PA=Active horizontal pressure at key base for H+aH+aμ∑W=Total frictional force under flat base

• For equilibrium, R + μ∑W =FOS x PA

14• FOS= (R + μ∑W)/ PA ≥ 1.55

M i h M i h Maximum pressure at the toeMaximum pressure at the toe

W Hx1

x

W1

W2

W4

PΣW

Hh

x2

W3

H/3

Pa

R

TW3

b/2b/6e

xb

Pressure below the Retaining WallPmax

Pmin.

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Retaining Wall

• Let the resultant R due to ∑W and Pa• lie at a distance x from the toe.e at a d sta ce o t e toe• X = ∑M/∑W, ∑M = sum of all moments about toe.

• Eccentricity of the load = e = (b/2-x) < b/6

• Minimum pressure at heel= >Zero. ⎥⎦

⎤⎢⎣⎡ −

∑=

be

bW 61Pmin

• For zero pressure, e=b/6, resultant should cut the base within the middle thirdthe base within the middle third.

• Maximum pressure at toe= < SBC of soil.⎤⎡∑ eW 6

16⎥⎦⎤

⎢⎣⎡ +

∑=

be

bW 61Pmax

Depth of foundationDepth of foundationDepth of foundationDepth of foundation

R ki ’ f l • Rankine’s formula: 2

sin1 ⎤⎡ − φSBC• Df = sin1sin1

⎥⎦

⎤⎢⎣

⎡+ φ

φγSBC

•• =

2akγ

SBC Df

•γ

17

Preliminary Proportioning Preliminary Proportioning (T shaped wall)(T shaped wall)(T shaped wall)(T shaped wall)

• Stem: Top width 200 mm 200

to 400 mm• Base slab width b= 0.4H to

f0.6H, 0.6H to 0.75H for surcharged wallB l b thi k H/10

H

H/10 –• Base slab thickness= H/10 to H/14

• Toe projection= (1/3 1/4) b 0 4H t 0 6H

tp= (1/3-1/4)b H/14

• Toe projection= (1/3-1/4) Base width

b= 0.4H to 0.6H

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Behaviour or structural action

• Behaviour or • Behaviour or structural action and design of stem, heel and toe slabs are same as that of any that of any cantilever slab.

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DistDist.fromtop Every

alternateh1

alternate bar cutAst/2 h2

L

h1c

AA /2h2

Ast

Ldt

AstProvided

Ast/2

Ast

C ti C t il t20

Cross section Curtailment curve

Design of Heel and ToeDesign of Heel and ToeDesign of Heel and ToeDesign of Heel and Toe

1 Heel slab and toe slab should also be1. Heel slab and toe slab should also bedesigned as cantilever. and determinethe maximum bending moments at thethe maximum bending moments at thejunction.

2 D t i th i f t 2. Determine the reinforcement. 3. Also check for shear at the junction. 4. Provide enough development length.

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Design of stem

• As(min)=14/fy bd

g

• As(min)=14/fy bd• 1.34As =• Development length (Stem • Development length (Stem

steel)

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•Check for shear

x x

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x1 W1

W

W4

ΣW

Hh

x2W2

WH/3

Pa

R

TW3

b/2b/6e

xb Forces acting

Pmax

Pmin.

0.75m 0.45m 1.8m

30.16 T/m2

Forces acting on the wall

and the 120.6 T/m2

T/m224.1

97.99 22.6

P b l th R t i i W ll

pressure below the wall

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Pressure below the Retaining Wall

Design of heelDesign of heel

x

x

120.6 T/m2

30.16 T/m2

25

T/m

Ch k f h t d f j ti

Design of toe slabDesign of toe slab--Contd.,Contd.,

• Check for shear: at d from junction (at xx as wall is in compression)

xd

xd

Ldt

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Other deatailsOther deatails

• Construction joint• DrainageDrainage

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Design and DetailingDesign and Detailing of of g gg gCounterfort Retaining Counterfort Retaining

llllwallwall

Counterfort Retaining wall

• When H exceeds about 6m,

g

When H exceeds about 6m,• Stem and heel thickness is more• More bending and more steel

Cantilever T type Uneconomical

CF

• Cantilever-T type-Uneconomical• Counterforts-Trapezoidal section• 1.5m -3m c/c Stem

Base Slab

CRW

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Parts of CRWParts of CRWSame as that of Cantilever Retaining wall • Same as that of Cantilever Retaining wall Plus Counterfort

Stem Counterforts

Toe Heel ToeBase slab

C ti Pl31

Cross section Plan

Design of Stem• The stem acts as a continuous slab • Soil pressure acts as the load on the

slab. BF• Earth pressure varies linearly over the

height• The slab deflects away from the earth

f b t th t f t

BF

face between the counterforts• The bending moment in the stem is

maximum at the base and reduces towards top towards top.

p=Kaγh

32

Maximum Bending moments for Maximum Bending moments for stemstem

• Maximum + B.M= pl2/16 • (occurring mid-way between

counterforts)• and l• and• Maximum - B.M= pl2/12 • (occurring at inner face of

f )-

counterforts)

• Where ‘l’ is the clear distance

p+

• Where l is the clear distance between the counterforts

• and ‘p’ is the intensity of soil

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pressure

Design of Toe SlabDesign of Toe Slab• The base width=b =0.6 H to 0.7 H• The projection=1/3 to 1/4 of base

width.• The toe slab is subjected to an upward

soil reaction and is designed as acantilever slab fixed at the front face ofthe stem

H

the stem.• Reinforcement is provided on earth

face along the length of the toe slab.In case the toe slab projection is large• In case the toe slab projection is largei.e. > b/3, front counterforts areprovided above the toe slab and theslab is designed as a continuous bslab is designed as a continuoushorizontal slab spanning between thefront counterforts.

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Design of Heel Slab

• The heel slab is designed as a continuousslab spanning over the counterforts and issubjected to downward forces due to weight ofsubjected to downward forces due to weight ofsoil plus self weight of slab and an upward forcedue to soil reaction.

• Maximum +ve B.M= pl2/16• (mid-way between counterforts)• And

BF• And• Maximum -ve B.M= pl2/12• (occurring at counterforts)

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Design of CounterfortsDesign of Counterforts• The counterforts are subjected to j

outward reaction from the stem. • This produces tension along the outer

sloping face of the counterforts. • The inner face supporting the stem is

in compression. Thus counterforts are designed as a T-beam of varying depth depth.

• The main steel provided along the sloping face shall be anchored properly at both ends

TC

at both ends. • The depth of the counterfort is

measured perpendicular to the sloping side.

d

side.

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Behaviour of Counterfort Behaviour of Counterfort Behaviour of Counterfort Behaviour of Counterfort RWRW

-M

+M Important points

COUNTERFORT

•Loads on Wall

•Deflected shapeCOUNTERFORTSTEM

•Deflected shape

•Nature of BMs

-M•Position of steel

•Counterfort details

TOE +MHEEL SLAB

Counterfort details

37

a. Proportioning of Wall a. Proportioning of Wall p gp gComponentsComponents

• Coefficient of active pressure = ka = 1/3• Coefficient of passive pressure= kp = 3Coefficient of passive pressure kp 3• The height of the wall above the base • = H = 7 + 1.25 = 8.25 m. Hh1=

7 m• Base width = 0.6 H to 0.7 H • (4.95 m to 5.78 m), Say b = 5.5 m

1 25 m

7 m

• Toe projection = b/4 = 5.5/4 = say 1 .2 m • Assume thickness of vertical wall = 250 mm • Thickness of base slab = 450 mm

1.25 m

• Thickness of base slab = 450 mm b=5.5 m

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S i f fS i f fSpacing of counterfortsSpacing of counterforts

l = l = 33..5 5 (H/γ)(H/γ)00..2525 = = 33..5 5 ((88..2525//1818))00..2525 = = 22..88 88 m m

∴∴ c/c spacing = c/c spacing = 22 88 88 + + 00 40 40 = = 33 28 28 m m l

∴∴ c/c spacing = c/c spacing = 22..88 88 + + 00..40 40 = = 33..28 28 m m say say 3 3 mm

∴∴ Provide counterforts at Provide counterforts at 3 3 m c/c. m c/c. Assume width of counterfort = Assume width of counterfort = 400 400 mmmm∴∴ clear spacing provided = l = clear spacing provided = l = 3 3 -- 00..4 4 = =

22..6 6 m m

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250

Details of wall250 mm

CF 3 /

h 7 8

CF: 3m c/c, 400 mm

h=7.8 mH=8.25 mh1=7 m

θ

d

4.05m θ1.2 m

b=5 5 m

1.25m

T40

b=5.5 mT

Di f M

b. Check Stability of Wall

Sr.No.

Description of loads Loads in kN

Dist. ofe.g. from

T in m

Moment about

T in kN-m

1 Weight of stemW1

25x0.25x1x7.8= 48.75

1.2 + 0.25/2=1.325 64.59

2 Weight of baseslab W2

25x5.5x1x0.45= 61.88 5.5/2 =2.75 170.17

3 Weight of earthover heel slab W3

18x4.05x1x7.8= 568.62

1.45 +4.05/2= 3.475 1975.95

Total ΣW = 679.25 ΣW =2210.71

41

250 mm

W1

H8250

W3W1

h1= 70008250ΣW

PA

R

A B C D1200 mm 4050 mmDf= 1250

PA PA

H/3450

eX b/2TW2

b/3

kaγH

Pressure distribution

Cross section of wall-Stability analysis

b/3

42

Cross section of wall Stability analysis

Stability of Stability of Stability of Stability of wallswalls

• Horizontal earth pressure on full heightf llof wall

• = Ph = kaγH2 /2 =18 x 8.252/(3 x 2) =204 19 kN204.19 kN

• Overturning moment = M0Overturning moment M0• = Ph x H/3 = 204.19 x 8.25/3 = 561.52

kN.m.• Factor of safety against overturning• = ∑ M / M0 = 2210.71/561.52 = 3.94 >

1 5543

1.55∴ safe.

CheckCheck forfor slidingsliding

TotalTotal horizontalhorizontal forceforce tendingtending toto slideslide thethewallwallwallwall

== PPhh == 204204..1919 kNkNResistingResisting forceforce == ∑µ∑µ WW == 00 5858 xx 679679 2525ResistingResisting forceforce == ∑µ∑µ..WW == 00..5858 xx 679679..2525== 393393..9797 kNkN∴∴FactorFactor ofof safetysafety againstagainst slidingsliding∴∴FactorFactor ofof safetysafety againstagainst slidingsliding== ∑µ∑µ..WW // PPhh == 393393..9797//204204..1919== 11..9393 >> 11..5555 ...... safesafe..

44

CheckCheck forfor pressurepressure distributiondistribution atat basebase

LetLet xx bebe thethe distancedistance ofof RR fromfrom toetoe (T),(T),∴∴ xx == ∑∑ MM // ∑∑ WW∑∑ // ∑∑Eccentricity=eEccentricity=e == b/b/22 -- xx << b/b/66∴∴WholeWhole basebase isis underunder compressioncompression..

MaximumMaximum pressurepressure atat toetoe== pp == ∑W∑W // bb (( 11++66e/b)e/b) ==== ppAA == ∑W∑W // bb (( 11++66e/b)e/b) ==

MinimumMinimum pressurepressure atat heelheeluu p essu ep essu e atat eeee

45

I t it f t j ti f t ith t I t it f t j ti f t ith t Intensity of pressure at junction of stem with toe Intensity of pressure at junction of stem with toe i.e. under Bi.e. under B

Intensity of pressure at junction of stem with heel Intensity of pressure at junction of stem with heel i.e. under Ci.e. under C

46

250 mm

H8250ΣW R

PA

R

A B C D1200 mm 4050 mm

450

1250

T

80.39kN/ 2166 61 143 9147 8153 9

eX b/2T

kN/m2166.61kN/m2

143.9147.8153.9

475500 mm

b) Design of Toe slabb) Design of Toe slab• Max. BMB = psf x (moment due to soil pressure -

moment due to wt. of slab TB]

48

Check for ShearCheck for ShearCritical section for shear: At distance d ( • Critical section for shear: At distance d (= 390 mm) from the face of the toe

d

49

-MCounterfort RW+M

COUNTERFORTSTEMSTEM

-M

TOE +MHEEL SLAB

50

(c) Design of Heel Slab( ) g

80.39kN/m2166.61

kN/m2143.9147.8153.9

kN/m

5500 mm

71.26

7.75 kN/m Forces on heel slab71.26 kN/m

DC

51

A B C1200 4050 1250

R

A B Cmm mm

450

eX b/2 Area for4050 mm

C DTOE

Area for stirrups

26003000

TOE

HEEL

SFD139

x1

1

Net down force dia

71.28 kN/m

7.75 kN/m

y1 Shear analysis and Zone of shear steel

52

Net down force dia.

(d) Design of Stem (Vertical (d) Design of Stem (Vertical (d) Design of Stem (Vertical (d) Design of Stem (Vertical Slab).Slab).

C ti l b i b t th • Continuous slab spanning between the counterforts and subjected to earth pressure.

• The intensity of earth pressure• The intensity of earth pressure• = ph = ka γh• Area of steel on earth side near counterforts : Area of steel on earth side near counterforts : • Maximum -ve ultimate moment,• Mu = ph l2/12 u ph /

• Required d

53

( ) D i f C t f tAt any section at any depth h below the top, the At any section at any depth h below the top, the

(e) Design of CounterfortAt any section at any depth h below the top, the At any section at any depth h below the top, the

total horizontal earth pressure acting on the total horizontal earth pressure acting on the counterfortcounterfort

= = 11//2 2 kk yy hh22x c/c distance between counterfortx c/c distance between counterfort= = 11//2 2 kkaayy hh22x c/c distance between counterfortx c/c distance between counterfort

Counterfort acts as a TCounterfort acts as a T--beam. beam.

Even assuming rectangular section,Even assuming rectangular section,

54

The effective depth is taken at The effective depth is taken at right angle to the reinforcement.

h =7.8 m

d

4.05m θ

55

Design of Horizontal TiesDesign of Horizontal TiesTh di t ll b th ll t f t f 1 • The direct pull by the wall on counterfort for 1 m height at base

• = k γh x c/c distance =1/3x18 x 7 8 x 3 = 140 4 • = kaγh x c/c distance =1/3x18 x 7.8 x 3 = 140.4 kN

• Area of steel required to resist the direct pullq p• = 140.4 x 103/(0.5fy) = 583 mm2 per m height.• Using 8 mm 2-legged stirrups, Ast = 100 mm2

• spacing = 1000 x 100/583 = 170 mm c/c.• Since the horizontal pressure decreases with h,

th i f ti b i dthe spacing of stirrups can be increased

56

Design of Vertical TiesDesign of Vertical TiesDesign of Vertical TiesDesign of Vertical Ties

• The maximum pull will be exerted at the end of heel slab where the net downward force = 71.26 kN/m.kN/m.

• Total downward force at D• = 71.26 x c/c distance. • Required Ast = 1.5 x 213.78 x 103/(0.87 x 415) =

888 mm2888 mm• Using 8 mm 2-legged stirrups , Ast = 100 mm2

• spacing = 1000 x 100/888 = 110 mm c/c.p g / /

• Increase the spacing of vertical stirrups towards the end C

57

the end C

250 mm 0-200mm

STEM COUNTERFORT

7000

STEM COUNTERFORT

12@200 8250 mm

12@200

1200 mm 4050 mm450

1250

16@120 12@200 12@200

Cross section between counterforts

TOE HEEL

58

Cross section between counterforts

250 mm

12@400 8-22 1.77m

8 - # 22

#8@110-450, VS#12@

#12@200 8250

8 # 22

#8@170-450, HS

@110-300

1200 mm450

1250

#16@120 #12@200 #12@200

Cross section through counterforts59

Cross section through counterforts

STEMSTRAIGHT BARS STEMBARS

Backfill B kfillBackfill

0.3l

Backfill

0.25 l

Section through stem at the junction of Base slab

With straight barsWith cranked

bars

60

Section through stem at the junction of Base slab.

Lateral pressure against retaining wall d t h l d due to surcharge loads

61

62

63

64

65

66

67

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