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Design Of Counter Fort Retaining WallReference: Example 18.4, R.C.C. Design Vol. - I, B.C. Punamia
1 Design ConstantsHieght of cantilever wall from ground level = 11.5 m
= 18
Unit Weight of water (ϒw) = 10Angle of repose (φ) = 30
= 125Co-eff of friction (µ) = 0.5
= 25 MPa
= 415 MPaCover = 40 mmFoundation Depth = 1 mWidth of counterfort = 0.5 m
= 8.5
= 230Neuteral axis constant (k) = 0.289Lever arm constant (j) = 0.904Moment of resistance constant ® = 1.109
= 0.333
2 Dimension of Various PartsHeight of wall above base (H) = 11.5 +
= 12.5 mThe ratio of length of slabe (DE) to base width b is given by eq.
α = 1 -
= 1 -
α = 0.75Adopting α = 0.40 ….
The width of base is given by Eq.
b = 0.95 H x
b = 0.95 x
= 5.97Normal practice is to provide b between 0.7 to 0.8 H Taking maximum value of H = 0.7
b = 8.75 mWidth of toe slab = α x b
= 3.50 m
Provide toe slab = 3.50
Saturated weight of Earth (γsat ) kN/m3
kN/m3
SBC of Soil (qo) kN/m2
Unit Weight of Concrete (fck)
fy
σcbc N/mm2
σst N/mm2
Coefficient of earth pressure, Ka
Taking the uniform thickness of stem = 0.6 mHence width of heel slab = 4.65 mLet thickness of base slab = 0.8 m
Clear spacing of counter fort = 3.5 xH
= 3.20 m
Providing spacing of counterfort = 3.00 m
3 Stability of Wall
= Weight of rectangular portion of stem
= Weight of base slab
= Weight of soil on heel slab.
Detail
1 x 0.60
1 x 0.80
1 x 4.65
Total resisting moment = 7724 kNm
Earth pressure (p) == 158.33 kN
Moment == 4123.26 kNm
F.O.S. against overturning = 1.87 >
F.O.S. against sliding == 4.20 >
Pressure DistributionNet Moment (SM) = 3601 kNmDistance x of the point of application of resultant, from toe is
x == 2.71 m
e = b/2 - x= 1.67
b/6 = 1.46Hence un safe as e > b/6
= 1 +b
=1329.79
x8.75
= 325.69
γsat
Let w1
w2
w3
w1
w2
w3
(Ka x γ' x H)+(γw x H)
p*H2/6
µSw/p
SM/Sw
Pressure p1 at toeSW
kN/m2
Hence un safe as p1 > 125
= 1 -b
1329.79x
8.75
= -21.74
Hence un safe as p2 < 0
= 325.69 -
= 186.72
= 325.69 -
= 162.90
4 Design of Heel SlabClear spacing b/w counterforts = 3.00 mThe pressure distribution on the heel slab is shown in fig 2. Consider a strip 1 meter wide.
Upward pressure intensity (near outer edge, C) = -21.74
Down ward load due to weight of Earth. = 210.6
Down ward weight of slab per unit area = 15
= 203.86
== 152.89 kNm
Effective depth required (d) =BMRxb
= 371.39 mmProviding overall depth (D) = 500 mmEffective depth (d) = 460 mm
= 800
= 1598.94
= 1598.94
Dia of bar = 20 mmSpacing of bar calculated = 196 mmSpacing of bar provided = 150 mm
Shear force (V) = PL/2
Pressure p2 at heelSW
kN/m2
The Pressure intencity p1 under E,
p1
kN/m2
The Pressure intencity p2 under B,
p2
kN/m2
kN/m2
kN/m2
kN/m2
Hence net pressure intensities will be P kN/m2
M1 Pl2/12
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Ast
M=Ast σst j dmm2
Provide Ast1 mm2
= 305.79 kN
= 0.35
= 0.26Refernce: Pg. 84, Table 23, IS 456 : 2000
= V/(d x b)= 0.66 664.7609
Hence shear reinforcement is required as τv > τcHence depth required from shear point of veiw (d) = V/(τc x b)
= 1176.12 mmProviding overall depth (D) = 1200 mmHence effective depth (d) = 1160 mm
= 1920
= 634.06
= 1920
Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm
Distribution Steel = 1920Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm
Hence point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)
= 176.75 kN= 176746.62 N
Assuming that all the bars will avilable at point of contraflexure,
M == 462979286.112 Nmm
= 12φ or d, whichever is more= 1160 mm
= 45φ= 900 mm
= 3779.45 mmHence safe as M/V+Lo > Ld
= 1160 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
100Ast/bd
τc N/mm2
τv
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Ast
M=Ast σst j dmm2
Provide Ast1 mm2
mm2
Let us check this reinforcement for development length at point of contraflexure which is situated at distance of 0.211 x L. In this case, the slab is continuous, but we will assume the same position of contraflexure.
Ast σst j d
Lo
Ld
M/V + Lo
Cotinue these bars by a distance Lo = d
At the point of curtailment, length of each bar availa = 1793 mmHence safe
These bars will be provide at the top face of heel slab.
Maximum passive B.M. =
=
=
= 1440
= 1920
= 1920
Dia of bar = 20 mmSpacing of bar calculated = 163 mmSpacing of bar provided = 150 mm
Let us check this reinforcement for development length crierion at point of contraflexur,
M == 462979286.112 Nmm
= 12φ or d, whichever is more= 1160 mm
= 45φ= 900 mm
= 3779.45 mmHence safe as M/V+Lo > Ld
= 1160 mm from the point of contraflexure.
i.e. upto distance = -527 mm from the centre of support. At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bars upto center of counterfors.
5 Design of Toe SlabProviding counterfort over toe slab upto ground level.Assuming total depth of toe slab = 500 mm
Total weight of toe slab = 12.5
Net upward intensity at D = 325.69-12.5
= 313.19
Net upward intensity at E = 186.72-12.5
= 174.22Cosidering strip of unit width at D.
=
= 234.8925 kNm
Effective depth required (d) =BMRxb
= 460.33 mmProviding overall depth (D) = 500 mm
PL2/16
3/4 x M1
Area of bottom steel (Ast2) 3/4 x Ast1
mm2
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Provide Ast2 mm2
Ast σst j d
Lo
Ld
M/V + Lo
Cotinue these bars by a distance Lo = d
kN/m2
kN/m2
kN/m2
kN/m2
kN/m2
Max. negative B.M. (M1) wL2/12
Effective depth (d) = 460 mm
= 800
= 2456.46
= 2456.46
Dia of bar = 20 mmSpacing of bar calculated = 127 mmSpacing of bar provided = 100 mm
Shear force (V) = PL/2= 469.78 kN
= 0.53
= 0.32Refernce: Pg. 84, Table 23, IS 456 : 2000
= V/(d x b)= 1.02
Hence shear reinforcement is required as τv > τcHence depth required from shear point of veiw (d) = V/(τc x b)
= 1468.08 mmProviding overall depth (D) = 1500 mmHence effective depth (d) = 1460 mm
= 2400
= 773.95
= 2400
Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm
Distribution Steel = 2400Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm
Let us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given = P(L/2 - x)
= 271.54 kN= 271535.73 N
M == 728394135.478 Nmm
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Ast
M=Ast σst j dmm2
Provide Ast1 mm2
100Ast/bd
τc N/mm2
τv
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Ast
M=Ast σst j dmm2
Provide Ast1 mm2
mm2
Ast σst j d
= 12φ or d, whichever is more= 1460 mm
= 45φ= 900 mm
= 4142.50Hence safe as M/V+Lo > Ld
= 1460 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
At the point of curtailment, length of each bar availa = 2093 mmHence safe
These bars will be provide at the top face of toe slab.
Maximum passive B.M. =
=
=
= 1800.00
= 2400
= 2400
Dia of bar = 20 mmSpacing of bar calculated = 130 mmSpacing of bar provided = 100 mm
Let us check this reinforcement for development length crierion at point of contraflexur,
M == 728394135.478 Nmm
= 12φ or d, whichever is more= 1460 mm
= 45φ= 900 mm
= 4142.50 mmHence safe as M/V+Lo > Ld
= 1460 mm from the point of contraflexure.
i.e. upto distance = -827 mm from the centre of support.
6 Design of StemThe stem acts as a continuous slab. Considred 1 m strip at B .
=
= 67.80
= 11.30 m
== 50.85 kNm
Effective depth required (d) =BM
Lo
Ld
M/V + Lo
Cotinue these bars by a distance Lo = d
PL2/16
3/4 x M1
Area of bottom steel (Ast2) 3/4 x Ast1
mm2
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Provide Ast2 mm2
Ast σst j d
Lo
Ld
M/V + Lo
Cotinue these bars by a distance Lo = d
The intencity of earth pressure is given by (ph) KaγH1
kN/m2
Hence revised H1
Negative B.M. in slab, (M1) phL2/12
Effective depth required (d) =Rxb
= 214.18 mmProviding overall depth (D) = 600 mmEffective depth (d) = 560 mm
= 960
= 436.82
= 960.00
Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm
Distribution Steel = 960Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm
Shear force (V) = PL/2= 101.70 kN
= 0.17
= 0.20Refernce: Pg. 84, Table 23, IS 456 : 2000
= V/(d x b)
= 0.18Hence no shear reinforcement is required as τv < τc
Let us check this reinforcement for development length crierion at point of contraflexur.Point of contraflexure is at = 0.633 mmShear force at this point is given (V) = V(L/2 - x)
= 88.17 kN= 88173.90 N
M == 111753620.786 Nmm
= 12φ or d, whichever is more= 560 mm
= 45φ= 720 mm
= 1827.42Hence safe as M/V+Lo > Ld
= 560 mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
At the point of curtailment, length of each bar availa = 1193 mmHence safe
These bars are to be provided at the inner face of the stem.
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Ast
M=Ast σst j dmm2
Provide Ast1 mm2
mm2
100Ast/bd
τc N/mm2
τv
N/mm2
Ast σst j d
Lo
Ld
M/V + Lo
Cotinue these bars by a distance Lo = d
Maximum passive B.M. =
=
=
= 720.00
= 960
= 960
Dia of bar = 16 mmSpacing of bar calculated = 209 mmSpacing of bar provided = 200 mm
Let us check this reinforcement for development length crierion at point of contraflexur,
M == 111753620.786 Nmm
= 12φ or d, whichever is more= 560 mm
= 45φ= 720 mm
= 1827.42 mmHence safe as M/V+Lo > Ld
7 Design of Main CounterfortLet us assuming thickness of counterforts is = 500 mmSpacing of counterforts = 350 cm c/c
=
= 21 h kN/mSimilarly, net down ward pressure on heel at C is = (11.3 x 18) + (1200/1000) x 25 - -21.74
= 255.14Similarly, net down ward pressure on heel at B is = (11.3 x 18) + (1200/1000) x 25 - 162.9
= 70.5Hence reaction transferrred to each counterfort will be,At C = 255.14 x (350/100)
= 892.99 kN/mAt B = 70.5 x (350/100)
= 246.75 kN/mPressure intencity at h = 11.5 m
= 241.5 kN/mShear force at F (Q) = 0.5 x 241.5 x 11.5
= 1388.63 kNB.M. = h/3 x 1388.625
= 5323.06 kNm= 5323062500 Nmm
PL2/16
3/4 x M1
Area of steel (Ast2) 3/4 x Ast1
mm2
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Provide Ast2 mm2
Ast σst j d
Lo
Ld
M/V + Lo
At any section at depth h below the top A, the earth pressure acting on each counter forts will be Ka x γsat x h x L
kN/m2
kN/m2
Effective depth required (d) =BMRxb
= 3099.03 mmProviding total depth (D) = 3150 mmEffective depth (d) = 3110 mmAngle θ of the face AC is given by, =
tan θ = 4.65/11.3tan θ = 0.41150442478
θ = 22.37 DegreesSin θ = 0.3805Cos θ = 0.9248
== 4.38 m= 4376.26 mm
FG = 4977.00 mmAsssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and 20 mm dia bar.Effective depth (d) = 4977 - (40 + 20 + 10)
= 4907.00 mm
= 5218.47
Using dia. of bar = 20 mm
= 314.16No. of bars = 17 barsProvide the bars in two layers.Effective shear force = Q-(M/d')tanθ
d' = d/cosθ= 5306.23 mm
Effective shear force = 975815.19 N= V/(d x b)
= 0.37
= 0.20
= 0.21Hence shear reinforcement is required as τv > τc
However, the vertical and horizontal ties provided in counterforts will bear the excess shear stress.
= h m
= 3.54 m below A, i.e. at point H.To locate the position of point of curtailmenton AC, drawing Hl parallel to FG.
= 240
= 900 mm
Design of Horizontal Ties
F1G1 AF1 x sinθ
Area of steel at supports, at bottom (Ast ) M=Ast σst j d mm2
Aφ mm2
τv
N/mm2
100Ast/bd
τc N/mm2
The height h where half of the reinforcement can curtailed (H)
Thus half bars can be curtailed at l. However these should be extent by a distance 12φ
mm beyond I, i.e. extented upto I1
The location of H corresponding to I1 can be locate by drawing line I1H1 parallel FG
It should be noted that I1G should not less than 45φ
At any depth h below the top, force causing sepratio == 207 kN/m
= Force/Stress
= 900Using 2 legged ties of dia. = 10 mm
=
= 157.08Spacing required = 174.53 mmSpacing provided = 150 mm
Design of Vertical TiesThe downward force at C = (892.99 x 3)/(350/100)
= 765.42 kN/mThe downward force at B = (246.75 x 3)/(350/100)
= 211.5 kN/m
= Force/Stress
= 3327.91Using 2 legged ties of dia. = 20 mm
=
= 628.32Spacing required = 188.80 mmSpacing provided = 150 mm
= Force/Stress
= 919.57Using 2 legged ties of dia. = 10 mm
=
= 157.08Spacing required = 170.82 mmSpacing provided = 150 mm
8 Design of Front Counterfort
The upward pressure intensity varies from 325.69 186.72
Downward weight of toe slab = 37.5
Net weight at D = 288.19
Net weight at E = 149.22The center to center spacing of counterforts = 3.5 mHence upward force transmitted to counterforts at D = 288.19 x 3.5 kN/m
= 1008.67 kN/mUpward force transmitted to counterforts at E = 149.22 x 3.5 kN/m
= 522.27 kN/mTotal upward force = 1/2 x (1008.665 + 522.27) x 3.5
= 2679.14 kNForce will be acting at x = ((522.27 + 2 x 1008.665)/(1008.665 + 522.27)) x (3.5/3)
Ka x γsat x h x L
Ast required
mm2
Aφ 2Π/4 x D2
mm2
Ast required at C
mm2
Aφ 2Π/4 x D2
mm2
Ast required at B
mm2
Aφ 2Π/4 x D2
mm2
kN/m2 at D to
kN/m2
kN/m2
kN/m2
= 1.94 m from EB.M. = 2679.14 x 1.94 kNm
= 5197.53 kNm
Effective depth required (d) =BMRxb
= 3062.27Providing total depth (D) = 4000 mmEffective depth (d) = 3960 mmThus height of counterfort above ground level = 3000 mm
= 6313.93
= 6400
= 6400.00
Dia of bar = 32 mm
= 804.25No. of bars = 8 barsEffective shear force (V) = Q-(M/d)tanθ
tan θ = 1.14Effective shear force (V) = 1179130.88 N
= V/(d x b)
= 0.60
= 0.11
= 0.19Hence shear reinforcement is required as τv > τc
Using 2 legged ties of dia. = 10 mm
=
= 157.08
== 376200 N
== 802930.88 N
== 178.18
= 150 mm c/c Provide 2 x 10 mm holding bar at top
9 Fixing Effect in Stem, Toe and HeelAt the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.
(i)
i.e. upto point F3.
Area of steel at supports, at bottom (Ast ) M=Ast σst j d mm2
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1]
mm2
Provide Ast1 mm2
Aφ mm2
τv
N/mm2
100Ast/bd
τc N/mm2
Aφ 2Π/4 x D2
mm2
Vc τc x b x d
Vs V - Vc
Spacing required (sv) (σst x Aφ x d)/Vs
Spacing required (sv)
In stem @0.3% of cross section, to be provided at inner face, in vertical direction,for a length 58φ
= (0.3/100) x 1000 x 600
= 1800Dia. of bar = 16 mm
= 201.06Spacing required = (201.06 x 1000)/1800
= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm
(ii) In toe slab @0.15% to be provided at the lowar face
= (0.15/100) x 1000 x 1500
= 2250Dia. of bar = 20 mm
= 314.16Spacing required = (314.16 x 1000)/2250
= 139.63 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 1160 mm
(iii) In heel slab @ 0.15% to be provided in upper face
= (0.15/100) x 1000 x 1200
= 1800Dia. of bar = 16 mm
= 201.06Spacing required = (201.06 x 1000)/1800
= 111.70 mmSpacing provided = 100 mmLength of embedment in slab above heel slab = 928 mm
Ast
mm2
Aφ mm2
Ast
mm2
Aφ mm2
Ast
mm2
Aφ mm2
1
2.2 y H125
2.2 x 18 x 12.50
Eq (1)
(1-α) x (1+3α)
12.500.333
( 1 - 0.40 )x( 1
q0
Ka
Weight of rectangular portion of stem
Weight of soil on heel slab.
Detail force(kN) lever arm
x 11.70 x 25 = 175.5 3.8
x 8.75 x 25 = 175 4.38
x 11.70 x 18 = 979.29 6.43
= 1329.79
1.5 Safe against overturning
1.5 Safe against sliding
6 e
b
1 +6 x 1.67
8.75
1/4
Sw Total MR
6 e
b
1 +6 x 1.67
8.75
325.69 - -21.74x 3.50
8.75
325.69 - -21.74x 4.10
8.75
0.66Hence shear reinforcement is required as τv > τc
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
mm from the point of contraflexure.
mm from the centre of support. At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bars upto center of counterfors.
Hence shear reinforcement is required as τv > τc
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
mm from the point of contraflexure.
mm from the centre of support.
Hence no shear reinforcement is required as τv < τc
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
12φ or d, whichever is more
Hence safe as M/V+Lo > Ld
(11.3 x 18) + (1200/1000) x 25 - -21.74
(11.3 x 18) + (1200/1000) x 25 - 162.9
Asssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and 20 mm dia bar.
Hence shear reinforcement is required as τv > τcHowever, the vertical and horizontal ties provided in counterforts will bear the excess shear stress.
m below A, i.e. at point H.
mm beyond I, i.e. extented upto I1
1/2 x (1008.665 + 522.27) x 3.5
((522.27 + 2 x 1008.665)/(1008.665 + 522.27)) x (3.5/3)
kN/m2 at E.
m from E
Hence shear reinforcement is required as τv > τc
At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.
0.333+ 1.20 )
A 0.6
θ
II1
H1
H12.50
11.5 G
3.00F
3.50 1.00 4.65 D E B C
8.75
FIGURE 1Moment about toe (KN-m)
666.90
765.63
6291.94
325.
697724
FIGURE 2
F3 G1
F1 F2
kN/m
2
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
mm beyond the point of contraflexure. After that, curtail half bars, and continue the remaining half throughout the length.
At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given below may be provided.
16 mm φ
0.928100 mm c/c
0.928
20 mm φ 20 mm φ100 mm c/c 100 mm c/c
1.16
1.50
-21.
74kN
/m2
M-15 M-20 M-25 M-30 M-35 M-40
0.15 0.18 0.18 0.19 0.20 0.20 0.20 0.210.25 0.22 0.22 0.23 0.23 0.23 0.23 0.260.50 0.29 0.30 0.31 0.31 0.31 0.32 0.320.75 0.34 0.35 0.36 0.37 0.37 0.381.00 0.37 0.39 0.40 0.41 0.42 0.421.25 0.40 0.42 0.44 0.45 0.45 0.461.50 0.42 0.45 0.46 0.48 0.49 0.491.75 0.44 0.47 0.49 0.50 0.52 0.522.00 0.44 0.49 0.51 0.53 0.54 0.552.25 0.44 0.51 0.53 0.55 0.56 0.572.50 0.44 0.51 0.55 0.57 0.58 0.60