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Summary SheetSession Number

:Date :

Subject Expert :

5 09.04.2007

Dr. M.C. NatarajaDr. M.C. Nataraja

Professor

Department of Civil Engineering,

Sri Jayachamarajendra College of Engineering,

Mysore – 570 006.

Phone:0821-2343521, 9880447742

E-mail: nataraja96@yahoo.com

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Design and Detailing of Counterfort Retaining wall

Dr. M.C. NATARAJA

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• When H exceeds about 6m,• Stem and heel thickness is more• More bending and more steel• Cantilever-T type-Uneconomical• Counterforts-Trapezoidal section• 1.5m -3m c/c

Counterfort Retaining wall

CRW

CF

Base Slab

Stem

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Parts of CRW• Same as that of Cantilever Retaining wall Plus

Counterfort

Stem

Toe Heel

Base slab

Counterforts

Cross section Plan

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

slab. • Earth pressure varies linearly over

the height• The slab deflects away from the

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

maximum at the base and reduces towards top.

• But the thickness of the wall is kept constant and only the area of steel is reduced.

Design of Stem

BF

p=Kaγh

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Maximum Bending moments for stem

Maximum +ve B.M= pl2/16 (occurring mid-way between counterforts)andMaximum -ve B.M= pl2/12 (occurring at inner face of counterforts)

Where ‘l’ is the clear distance between the counterforts and ‘p’ is the intensity of soil pressure

l

p+

-

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Design of Toe Slab

The base width=b =0.6 H to 0.7 HThe projection=1/3 to 1/4 of base width. The toe slab is subjected to an upward soil

reaction and is designed as a cantilever slab fixed at the front face of the stem.

Reinforcement is provided on earth face along the length of the toe slab.

In case the toe slab projection is large i.e. > b/3, front counterforts are provided above the toe slab and the slab is designed as a continuous horizontal slab spanning between the front counterforts.

b

H

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The heel slab is designed as a continuous slab spanning over the counterforts and is subjected to downward forces due to weight of soil plus self weight of slab and an upward force due to soil reaction.

Maximum +ve B.M= pl2/16 (mid-way between counterforts)AndMaximum -ve B.M= pl2/12 (occurring at counterforts)

Design of Heel Slab

BF

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Design of Counterforts• The counterforts are subjected to

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.

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

• The depth of the counterfort is measured perpendicular to the sloping side.

TC

d

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Behaviour of Counterfort RW

-M

-M

TOE

COUNTERFORT

+M

+M

STEM

HEEL SLAB

Important points

•Loads on Wall

•Deflected shape

•Nature of BMs

•Position of steel

•Counterfort details

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PROBLEM-Counterfort Retaining Wall

• A R.C.C. retaining wall with counterforts is required to support earth to a height of 7 m above the ground level. The top surface of the backfill is horizontal. The trial pit taken at the site indicates that soil of bearing capacity 220 kN/m2 is available at a depth of 1.25 m below the ground level. The weight of earth is 18 kN/m3 and angle of repose is 30°. The coefficient of friction between concrete and soil is 0.58. Use concrete M20 and steel grade Fe 415. Design the retaining wall.

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Draw the following:• Cross section of wall near the counterfort • Cross section of wall between the counterforts• L/s of stem at the base cutting the counterforts

Given:

fck = 20 N/mm2, fy = 415N/mm2, H = 7 m above G.L, Depth of footing below G.L. = 1.25 m, γ = 18 kN/m3,

μ = 0.58, fb =SBC= 220 kN/m2

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a. Proportioning of Wall Components

Coefficient of active pressure = ka = 1/3

Coefficient of passive pressure= kp = 3

The height of the wall above the base

= H = 7 + 1.25 = 8.25 m.

Base width = 0.6 H to 0.7 H

(4.95 m to 5.78 m), Say b = 5.5 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

H

b=5.5 m

1.25 m

h1= 7 m

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Spacing of counterforts

l = 3.5 (H/γ)0.25 = 3.5 (8.25/18)0.25 = 2.88 m c/c spacing = 2.88 + 0.40 = 3.28 m say 3 m

Provide counterforts at 3 m c/c.

Assume width of counterfort = 400 mm

clear spacing provided = l = 3 - 0.4 = 2.6 m

l

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4.05m

h=7.8 m

θ

d

250 mm

1.2 m

b=5.5 m

H=8.25 mh1=7 m

1.25m

CF: 3m c/c, 400 mm

T

Details of wall

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Sr.No.

Description of loads Loads in kN

Dist. ofe.g. from

T in m

Moment about

T in kN-m

1Weight of stem

W1

25x0.25x1x7.8= 48.75

1.2 + 0.25/2=1.325 64.59

2Weight of base

slab W2

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

3Weight of earth

over heel slab W3

18x4.05x1x7.8= 568.62

1.45 +4.05/2= 3.475

1975.95

Total ΣW = 679.25 ΣW =2210.71

b. Check Stability of Wall

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A B C D

H8250

250 mm

1200 mm 4050 mm

450

Df= 1250

ΣW

PA

R

eX b/2T

W3W1

W2

PA

Pressure distribution

Cross section of wall-Stability analysis

b/3

kaH

H/3

h1= 7000

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Stability of walls

Horizontal earth pressure on full height of wall

= Ph = kaH2 /2 =18 x 8.252/(3 x 2) = 204.19 kN

Overturning 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 > 1.55 safe.

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Check for sliding

Total horizontal force tending to slide the wall

= Ph = 204.19 kN

Resisting force = ∑µ.W = 0.58 x 679.25

= 393.97 kN

Factor of safety against sliding

= ∑µ.W / Ph = 393.97/204.19

= 1.93 > 1.55 ... safe.

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Check for pressure distribution at base

Let x be the distance of R from toe (T), x = ∑ M / ∑ W = 2210.71 -561.52 /679.25 = 2.43 m Eccentricity=e = b/2 - x = 5.5/2 - 2.43 = 0.32 < b/6 (0.91m)Whole base is under compression.

Maximum pressure at toe= pA = ∑W / b ( 1+6e/b) = 679.25/5.5 ( 1+ 6*0.32/5.5)= 166.61 kN/m2 < f b (i.e. SBC= 220 kN/m2)

Minimum pressure at heel= pD = 80.39 kN/m2 compression.

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Intensity of pressure at junction of stem with toe i.e. under B

= pB = 80.39 + (166.61 - 80.39) x 4.3/5.5 = 147.8kN/m2

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

=Pc= 80.39 + (166.61 - 80.39) x 4.05/5.5 = 143.9 kN/m2

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80.39kN/m2166.61

kN/m2143.9147.8153.9

A B C D

H8250

250 mm

1200 mm

5500 mm

4050 mm

450

1250

ΣW

PA

R

eX b/2T

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b) Design of Toe slab

Max. BMB = psf x (moment due to soil pressure - moment due to wt. of slab TB]

= 1.5 [147.8 x 1.22/2 + (166.61 - 147.8) x 1.2 (2/3 x 1.2)-(25x 1.2 x 0.45 x 1.2/2) =174.57 kN-m.

Mu/bd2= 1.14 < 2.76, URS

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To find steel

pt=0.34% <0.96%, A st =1326 mm2, # 16 @150 However, provide # 16 @110 from shear considerations.

Area provided =1827 mm2 , pt=0.47%Development length= 47 x 16=750 mmDistribution steel = 0.12 x 1000 x 450/100 = 540 mm2

Provide #12 mm at 200 mm c/c. Area provided = 565 mm2

b) Design of Toe slab- Contd.,

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Check for ShearCritical section for shear: At distance d (= 390 mm)

from the face of the toe pE = 80.39 + (166.61 - 80.39) (4.3 + 0.39)/5.5

= 153.9kN/m2

Net vertical shear = (166.61 + 153.9) x 0.81/2 - (25 x 0.45 x 0.81)

=120.7 kN. Net ultimate shear = Vu.max = 1.5 x 120.7 =181.05

kN.ζv= 181.05x 1000/1000x390 =0.46 MPapt = 100 x 1827/ (1000 x 390) = 0.47 %ζuc = 0.36 + (0.48 - 0.36) x 0.22/0.25

= 0.47N/mm2 > ζvsafe

d

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-M

-M

TOE

COUNTERFORT

+M

+M

STEM

HEEL SLAB

Counterfort RW

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Continuous slab. Consider 1 m wide strip near the outer edge D The forces acting near the edge areDownward wt. of soil=18x7.8xl= 140.4 kN/mDownward wt. of heel slab = 25 x 0.45 x 1= 11.25 kN/mUpward soil pressure 80.39 kN/m2= 80.39 x 1= 80.39 kN/m Net down force at D= 140.4 + 11.25 - 80.39 = 71.26 kN/m Also net down force at C = 140.4 + 11.25 - 143.9 = 7.75 kN/m

Negative Bending Moment for heel at junction of counterfort

Mu= (psf) pl2 /12 = 1.5 x 71.26 x 2.62/12 = 60.2 kN-m (At the junction of CF)

(c) Design of Heel Slab

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71.26 kN/m

7.75 kN/m

DC

Forces on heel slab

80.39kN/m2166.61

kN/m2143.9147.8153.9

5500 mm

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To find steel

Mu/bd2=60.2x106/(1000x3902)= 0.39 < 2.76, URS To find steelpt=0.114% <0.12%GA (Min. steel), <0.96%(pt,lim.) Provide 0.12% of GA

Ast= 0.12x1000x450/100 = 540 mm2

Provide # 12 mm @ 200 mm c/c, Area provided = 565 mm2

pt= 100 x 565/ (1000 x 390) = 0.14 %

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Maximum shear = Vu,max = 1.5 x 71.26 x 2.6/2 = 139 kNFor Pt, = 0.14 % and M20 concrete, ζuc= 0.28 N/mm2

ζv= Vumax/bd =0.36 N/mm2 , ζuc < ζv, Unsafe, Hence shear steel is needed

Using #8 mm 2-legged stirrups,Spacing=0.87x415x100/[(0.36-0.28)x1000] = 452 mm < (0.75 x 390 = 290 mm or 300 mm ) Provide #8 mm 2-legged stirrups at 290 mm c/c.

Provide for 1m x 1m area as shown in figure

Check for shear (Heel slab)

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A B C1200 mm

4050 mm

450

1250

R

eX b/2

26003000

4050 mm

C D

SFD

Net down force dia.

TOE

HEEL

139

71.28 kN/m

7.75 kN/m

x1

y1Shear analysis and Zone of shear steel

Area for stirrups

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Area of steel for +ve moment(Heel slab)

Maximum +ve ultimate moment = psf x pl2/16= 3/4 Mu = 0.75 x 60.2= 45.15 kN-m.

Mu/bd2=Very small and hence provide minimum steel.

Ast,min= 540 mm2

Provide # 12 mm bars at 200 mm c/c. Area provided = 565 mm2 > 540 mm2

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Check the force at junction of heel slab with stemThe intensity of downward force decreases due to

increases in upward soil reaction. Consider m width of the slab at C

Net downward force= 18 x 7.8 +25 x 0.45 - 143.9 = 7.75 kN/m. Provide only minimum reinforcement.

Distribution steel

Ast = 0.12 x 1000 x 450/100 = 540 mm2 Using # 12 mm bars, spacing = 1000 x 113/468 = 241 mm. Provide # 12 mm at 200 mm c/c.Area provided = 565 mm2

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(d) Design of Stem (Vertical Slab).Continuous slab spanning between the counterforts and

subjected to earth pressure.The intensity of earth pressure

= ph = ka γh =18 x 7.8/3=46.8 kN/m2

Area of steel on earth side near counterforts : Maximum -ve ultimate moment,

Mu = 1.5 x ph l2/12 = 1.5 x 46.8 x 2.62/12 = 39.54 kN.m. Required d = √ (39.54 x 106/(2.76 x 1000)) = 119 mmHowever provide total depth = 250 mm

Mu/bd2= 39.54x106/1000x3902=1.1 < 2.76, URS

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To find steel: Pt=0.34% <0.96%, Ast=646 mm2, #12 mm @ 170 mm c/c, However provide #12 mm @ 110 mm c/c, Area provided = 1027.27 mm2,Pt= 0.54 %.

As the earth pressure decreases towards the top, the spacing of the bars is increased with decrease in height.

Max.ult. shear = Vumax = 1.5 x 46.8 x 2.6/2 = 91.26 kNFor Pt, = 0.54 % and M20 concrete ζuc= 0.5 N/mm2

ζv= Vumax/bd =91.28 x1000/(100X190)=0.48 N/mm2,

Shear steel is not needed and hence safe.

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At any section at any depth h below the top, the total horizontal earth pressure acting on the counterfort

= 1/2 kay h2x c/c distance between counterfort= 18 x h2 x 3 x 1/6 = 9 h2

B.M. at any depth h = 9h2xh/3 = 3h3

B.M. at the base at C= 3 x 7.83 = 1423.7 kN.m. Ultimate moment = Mu= 1.5 x 1423.7 = 2135.60 kN.m.

Counterfort acts as a T-beam. Even assuming rectangular section,d =√(2135.6 x 106(2.76 x 400)) = 1390 mm

(e) Design of Counterfort

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

tan θ = 7.8/4.05 =1.93, θ = 62.5°,

d = 4050 sin θ - eff. cover

= 3535 mm > > 1390 mm

Mu/bd2=2135.6x106/(400x35352) =0.427, pt=0.12%, Ast=1696mm2

Check for minimum steel 4.05m

h =7.8 m

θd

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Ast.min = 0.85 bd/fy = 0.85 x 400 x 3535/415 = 2896 mm2

Provided 4- # 22 mm + 4 - # 22 mm, Area provided = 3041 mm2

pt = 100 x 3041/(400 x 3535) = 0.21 %

The height h where half of the reinforcement can be curtailed is approximately equal to √H= √7.8=2.79 m

Curtail 4 bars at 2.79-Ldt from top i.e, 2.79-1.03 =1.77m from top.

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Design of Horizontal TiesThe direct pull by the wall on counterfort for 1 m height at

base

= kaγh x c/c distance =1/3x18 x 7.8 x 3 = 140.4 kNArea of steel required to resist the direct pull= 1.5 x 140.4 x 103/(0.87 x 415) = 583 mm2 per m height.

Using # 8 mm 2-legged stirrups, Ast = 100 mm2

spacing = 1000 x 100/583 = 170 mm c/c. Provide # 8 at 170 mm c/c.Since the horizontal pressure decreases with h, the spacing

of stirrups can be increased from 170 mm c/c to 450 mm c/c towards the top.

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Design of Vertical Ties

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

Total downward force at D= 71.26 x c/c distance bet. CFs = 71.28 x 3 = 213.78 kN.

Required Ast = 1.5 x 213.78 x 103/(0.87 x 415) = 888 mm2

Using # 8 mm 2-legged stirrups , Ast = 100 mm2

spacing = 1000 x 100/888 = 110 mm c/c.

Provide # 8 mm 2-legged stirrups at 110 mm c/c.Increase the spacing of vertical stirrups from 110 mm c/c to

450 mm c/c towards the end C

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DRAWING AND DETAILING

COUNTERFORT RETAINING WALL

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250 mm

1200 mm 4050 mm450

1250

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

#12@200

#12@200

8250 mm

7000

Cross section between counterforts

0-200mm

STEM COUNTERFORT

TOE HEEL

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250 mm

1200 mm450

1250

#16@120 #12@200

8 - # 22

#12@200

#8@170-450, HS

#8@110-450, VS

#12@ 110-300

#12@200

#12@4008-#22 1.77m

8250

Cross section through counterforts

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Section through stem at the junction of Base slab.

Backfill

With straight bars

0.3l

0.25 l

Backfill

With cranked bars

STEMSTRAIGHT BARS

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Cross section of heel slab

BackfillBackfill

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Examination Problems

July 2006

• Single bay Fixed Portal Frame

• Combined footing (Beam and slab type)

December 2006

• T-shaped Cantilever Retaining wall

• Combined footing (Type not mentioned)

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Exam Problem (Dec. 2006)Design a T shaped cantilever retaining wall to retain earth

embankment 3.2 m high above the ground level. The unit weight of the earth is 18 kN/m2 and its angle of repose is 30 degrees. The embankment is horizontal at it top. The SBC of soil is 120 kN/m2 and the co-efficient of friction between soil and concrete is 0.5. Use M20 concrete and Fe 415 steel. Draw the following to a suitable scale:

1. Section of the retaining wall2. Reinforcement details at the inner face of the stem.

60 MarksData: h1=3.2 m, µ=0.5, γ=18 kN/m2, ө=30º, SBC= 120

kN/m2,

M20 Concrete and Fe 415 steelFind H= h1 + Df

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