Retain Wall Counter Fort

8/13/2019 Retain Wall Counter Fort

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Summary SheetSession Number :

Date :

Subject Expert :

5

09.04.2007

Dr. M.C. Nataraja

Professor

Department of Civil Engineering,

Sri Jayachamarajendra College of

Engineering,

Mysore 570 006.

Phone:0821-2343521, 9880447742

E-mail: [email protected]


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Design and Detailingof 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 PlusCounterfort

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 ismaximum at the base and reduces

towards top. But the thickness of the wall is kept

constant and only the area of steelis reduced.

Design of Stem

BF

p=Kah


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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 thecounterfortsand 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 H

The projection=1/3 to 1/4 of base width.

The toe slab is subjected to an upwardsoil reaction and is designed as a

cantilever slab fixed at the front face ofthe stem.

Reinforcement is provided on earth facealong the length of the toe slab.

In case the toe slab projection is large i.e.

> b/3, front counterforts are providedabove the toe slab and the slab isdesigned as a continuous horizontalslab spanning between the frontcounterforts.

b

H


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The heel slab is designed as a continuous slabspanning over the counterforts and issubjected to downward forces due to weight ofsoil 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 tooutward reaction from the stem. This produces tension along the

outer sloping face of the counterforts.

The inner face supporting the stem isin compression. Thus counterfortsare designed as a T-beam of varyingdepth.

The main steel provided along thesloping face shall be anchored

properly at both ends. The depth of the counterfort is

measured perpendicular to thesloping 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 isrequired to support earth to a height of 7 m above

the ground level. The top surface of the backfill ishorizontal. The trial pit taken at the site indicatesthat soil of bearing capacity 220 kN/m2is availableat a depth of 1.25 m below the ground level. Theweight of earth is 18 kN/m3and angle of repose is30. The coefficient of friction between concreteand soil is 0.58. Use concrete M20 and steelgrade 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/mm

2, H = 7 m above G.L,

Depth of footing below G.L. = 1.25 m, = 18 kN/m3,

= 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 mAssume 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/)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

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 ofloads

Loads in kNDist. of

e.g. fromT in m

Momentabout

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

3

Weight of earth

over heel slab W3

18x4.05x1x7.8

= 568.62

1.45+4.05/2

= 3.475 1975.95

Total W = 679.25 W=2210.71

b. Check Stability of Wall


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A B C D

H

8250

250 mm

1200 mm 4050 mm

450

Df= 1250

W

PA

R

eX b/2T

W3W1

W2

PA

Pressure distribution

Cross section of wall-Stability analysis

b/3kaH

H/3

h1= 7000


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Stability of wallsHorizontal earth pressure on full height of wall

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

Overturning moment = M0

= Phx 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 > 1.55safe.


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Check for sliding

Total horizontal force tending to slide the wall

= Ph= 204.19 kN

Resisting force = .W = 0.58 x 679.25= 393.97 kN

Factor of safety against sliding

= .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 mEccentricity=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/m2compression.


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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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80.39

kN/m2166.61

kN/m2143.9147.8153.9

A B C D

H

8250

250 mm

1200 mm

5500 mm

4050 mm

450

1250

W

PA

R

eX b/2T


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b) Design of Toe slab

Max. BMB= psf x (moment due to soil pressure - momentdue 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%


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Check for Shear

Critical 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.v= 181.05x 1000/1000x390 =0.46 MPapt= 100 x 1827/ (1000 x 390) = 0.47 %

uc= 0.36 + (0.48 - 0.36) x 0.22/0.25= 0.47N/mm2 > vsafe

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 are

Downward wt. of soil=18x7.8xl= 140.4 kN/m

Downward wt. of heel slab = 25 x 0.45 x 1= 11.25 kN/m

Upward 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

80.39

kN/m2166.61

kN/m2143.9147.8153.9

5500mm


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To find steel

Mu/bd2=60.2x106/(1000x3902)=0.39 < 2.76, URS

To find steel

pt=0.114%


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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, uc= 0.28 N/mm2

v= Vumax/bd =0.36 N/mm2,

uc< v, 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 C1200mm

4050mm

450

1250

R

eX b/2

26003000

4050

mm

C D

SFD

Net down force dia.

TOE

HEEL

139

71.28

kN/m7.75

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/bd

2=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 stem

The intensity of downward force decreases due toincreases in upward soil reaction. Consider m width ofthe 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 andsubjected to earth pressure.

The intensity of earth pressure

= ph= ka h =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 phl2/12 = 1.5 x 46.8 x 2.62/12 = 39.54 kN.m.

Required d = (39.54 x 106/(2.76 x 1000)) = 119 mm

However 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%


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

= 1/2 kay h2x c/c distance between counterfort

= 18 x h2x 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 rightangle to the reinforcement.

tan = 7.8/4.05 =1.93, = 62.5,

d = 4050 sin - eff. cover= 3535 mm > > 1390 mm

Mu/bd2=2135.6x106/(400x35352)

=0.427, pt=0.12%, Ast=1696mm2

Check for minimum steel4.05m

h =7.8 m

d


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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 becurtailed 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 =1.77mfrom top.


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Design of Horizontal Ties

The direct pull by the wall on counterfort for 1 m height atbase

= kah 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 mm2per m height.Using # 8 mm 2-legged stirrups, Ast = 100 mm

2

spacing = 1000 x 100/583 = 170 mm c/c.

Provide # 8 at 170 mm c/c.

Since the horizontal pressure decreases with h, thespacing of stirrups can be increased from 170 mm c/c to450 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 slabwhere 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 to450 mm c/c towards the end C


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250 mm

1200 mm 4050 mm

450

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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Section through stem at the junction of Base slab.

Backfill

With straight bars

0.3l

0.25 l

Backfill

With cranked

bars

STEM

STRAIGHT

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 earthembankment 3.2 m high above the ground level. Theunit weight of the earth is 18 kN/m2 and its angle ofrepose is 30 degrees. The embankment is horizontalat 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 thefollowing to a suitable scale:

1. Section of the retaining wall

2. Reinforcement details at the inner face of the stem.

60 Marks

Data: h1=3.2 m, =0.5, =18 kN/m2, =30, SBC= 120kN/m2,

M20 Concrete and Fe 415 steel

Find H= h1+ Df


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Retain Wall Counter Fort