Counter Forrt Ret Wall

Counterfort retaining wall Fardous Rababah Martha al- Rabadi 

When we use Counterfort Retaining wall??

• When the height of the retaining wall exceeds about 6 m, the thickness of the stem and heel slab works out to be sufficiently large and the design becomes uneconomical.so we use counterfort RW

Parts of counterfort RW

Proportioning

Stability • Before the design we must check stability as we learned previous.

Check for sliding Check for overturning •

Check for bearing capacity

Other two failures in counterfort • Tearing of wall slab • tearing of wall along with counterfort.

The main components that must be design:

Design of Stem Design of Toe Slab Design of Heel slab Design of Counterforts

Design of Stem • The stem acts as a continuous slab spanning longitudinally over the counterforts.

BF

The loads The horizontal active 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

p=Kaγh

Maximum Bending moments for stem Maximum +ve B.M= pl2/16 occurring mid-way between counterforts Maximum -ve B.M= pl2/12 (occurring at inner face of counterforts) Where( I) is the clear distance between the counterforts

and (p) is the intensity of soil pressure

The bending moment in the stem is maximum at the base and reduces towards top. Note :

• l • •

+

• p

Reinforcement of stem *The main reinforcement is provided horizontally along the length of the wall. *The ties are provided horizontally for the full value of reaction to prevent slab separating from counterforts.

Design of Toe Slab • designed as a cantilever slab fixed at the front face of the stem. H

The loads The toe slab is subjected to an upward soil reaction b

Reinforcement of Toe * Due to upward soil pressure, the tension develops on the earth face and the reinforcement is provided on earth face along the length of the toe slab.

Design of Heel Slab • The heel slab is designed as a continuous slab spanning over the counterforts, as in the case of stem.

The loads The heel slab is subjected to downward forces due to weight of soil plus self weight of slab and an upward force due to soil reaction.

BF

Maximum Bending moments for Heel

Maximum +ve B.M= pl2/16 mid-way between counterforts towards earth face Maximum -ve B.M= pl2/12 occurring at counterforts

• Where ( I) is the clear distance between the counterforts and (p) p is the net downward force

The downward force will be maximum at the edge of the slab where intensity of soil pressure is minimum.

Design of Counterforts • The compression zone with respect to the bending of the counterforts and hence the counterforts are designed as a T-beam or cantilever fixed at the base slab of varying depth.

The loads

*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.

C

T d

Reinforcement of counterfort *The main steel provided

along the sloping face shall be anchored properly at both ends. Note : The depth of the counterfort is measured perpendicular to the sloping side

In conclusion -M +M

COUNTERFORT STEM

-M

HEEL SLAB TOE

+M

example 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.

o i t u l so n

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 =

•

h 1= 7 m

• 1.25 m • b=5.5 m

•

H

• Spacing of counterforts  Provide counterforts at 3 m c/c. Assume width of counterfort = 400 mm  clear spacing provided = l = 3 - 0.4 = 2.6 m

l

Details of wall •

250 mm

•

Counter F @: 3m c/c, 400 mm

• •

•

h =7 m 1

h=7. 8m •

•

1.25 m

•

• T

•

b=5. 5m

H=8.2 5m

d •

1.2 m

•

• 4.05 m

θ

b. Check Stability of Wall •

• h 1= 7000

•

250 mm

•

W 1

• •

Σ W•

W 3

R • PA

• H • 8250 •

PA

D f= • • 4050 • A 1200 • B • C • D • H/ 1250 mm mm • 4 • W 3 • kaH 5• T • X 2 • e • b/ 0 • b/ • 2 Pressure distribution 3

• Cross section of wall-Stability analysis

b. Check Stability of Wall

Loads in kN

Dist. of e.g. from T in m

Moment about T in kN-m

Weight of stem W1

25x0.25x1x7.8 = 48.75

1.2 + 0.25/2 =1.325

64.59

2

Weight 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

#. No.

Description of loads

1

Total

ΣW = 679.25

ΣW =2210.71

Stability of walls Horizontal earth pressure on full height of wall = Ph = kaH2 /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 > 2  safe.

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

• 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.

• 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

250 mm

H 8250

ΣW

R PA

1250

A

1200 mm B

C

4050 mm

e

b/2

D

450

T

X

166.61 153.9 147.8 143.9 kN/m2 5500 mm

80.39 kN/m2

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.

To find steel A st =1326 mm2, # 16 @150 However, provide # 16 @110 from shear considerations. Area provided =1827 mm2 Distribution steel = 0.15 x 1000 x 450/100 = 540 mm2 Provide #12 mm at 200 mm c/c. Area provided = 565 mm2

Check for Shear for 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 1.2/2 - (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 MPa pt = 100 x 1827/ (1000 x 390) = 0.47 % ζuc = 0.36 + (0.48 - 0.36) x 0.22/0.25 = 0.47N/mm2 > ζvsafe

(c) Design of Heel Slab • Continuous slab. • 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

l e e h n o s e c For slab

166. 15 14 14 61 3.9 7.8 3.9 kN/ 5500 m2 mm

• • •

C

•

D

7.75 kN/m 71.26 kN/m

80.3 9 kN/ m2

To find steel Mu/j*d*b=60.2x106/(1000x390x0.8)=540 Ast= 540 mm2  Provide # 12 mm @ 200 mm c/c,  Area provided = 565 mm2

Check for shear (Heel slab)

•

Maximum shear = Vu,max = 71.26 x 2.6/2 = 139 kN For M20 concrete, ζuc= 0.28 N/mm2

• •

ζv= Vumax/jbd =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.

•



R 1250 45 0

1200 A B mm

C

X

e

4050 mm b/2

Area for stirrups

4050 mm C

3000

TOE

D 260 HEEL 0 x1

139 SFD

y1 7.75 kN/m

71.2 8 Net down force kN/m dia.

Shear analysis and Zone of shear steel

• Check the force at junction of heel slab with stem • The 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 = 0.15 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.

(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 = ph l2/12 = 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/jbd= 39.54x106/0.87x1000x390=1.1

• Ast=646 mm2, #12 mm @ 170 mm c/c, • However provide #12 mm @ 110 mm c/c, •

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

• Max.ult. shear = Vumax = 46.8 x 2.6/2 = 91.26 kN • For M20 concrete ζuc= 0.5 N/mm2 • ζv= Vumax/0.87bd =91.28 x1000/ (0.87x100X190)=0.48 N/mm 2,

(e) Design of Counterfort • 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. • Counterfort acts as a a cantilever beam

• The effective depth is taken at right angle to the reinforcement. • tan θ = 7.8/4.05 =1.93, θ = 62.5°,

h =7.8 m

•  d = 4050 sin θ - eff. cover

d

• = 3535 mm > > 1390 mm • Mu/jbd=2135.6x106/ (0.87x400x3535)

=Ast=1696mm2

4.0 5m

θ

Ast.min = 0.15xbxd  Provided 4- # 22 mm + 4 - # 22 mm,  Area provided = 3041 mm2 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.

Design of Horizontal Ties The 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 kN Sperating force=140.4 KN 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.

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 =213.78 x 103/(0.87 x 415) = 888 mm2 Using # 8 mm 2-legged stirrups , A st = 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

• DRAWING AND DETAILING

• Cross section between counterforts • • •

•

0200mm COUNTERFORT

•

7000 • •

•

250 mm STEM

•

#12@20 0 #12@20 0

• 1200 1250 • 45 mm 0 • • #16@12 • TOE 0

• #12@20 0

8250 mm

4050 mm •

#12@20 0

•

HEEL

Cross section through counterforts • 25 0 • #1m 2@m • #1 400 • 2@ #1 2@ 200

• 8- • 1.7 #2 7m • 82 2 • #8@11 •0-450, 850 •##8@17 VS 220-450, • 12 •110 12 HS • 4 50 00 • #1 5 m • #1 • #1 6@ 2@ 2@ 0 300

Section through stem at the junction of Base slab

• Backfill

• 0.2 5l

• 0.3 l •

With straight bars

Cross section of heel slab

u o y k n a Th 