Design Of Counter Fort Retaining Wall Reference: Example 18.4, R.C.C. Design Vol. - I, B.C. Punamia 1 Design Constants Hieght of cantilever wall from ground level = Saturated weight of Earth (γsat ) =
11.5 18
Unit Weight of water (ϒw) Angle of repose (υ) SBC of Soil (qo) Co-eff of friction (µ) Unit Weight of Concrete (fck)
= = = = =
10 30 125 0.5 25
fy Cover Foundation Depth Width of counterfort σcbc
= = = =
415 40 1 0.5
= = = = = =
8.5 230 0.289 0.904 1.109 0.333
σst Neuteral axis constant (k) Lever arm constant (j) Moment of resistance constant ® Coefficient of earth pressure, Ka 2 Dimension of Various Parts Height of wall above base (H)
= = The ratio of length of slabe (DE) to base width b is given by eq. α
11.5 12.5
=
1
=
1
α Adopting α
= =
0.75 0.40
b
=
0.95 H
b
=
0.95
=
5.97
= = = =
0.7 8.75 αxb 3.50
= =
3.50 0.6
The width of base is given by Eq.
Normal practice is to provide b between 0.7 to 0.8 H Taking maximum value of H b Width of toe slab Provide toe slab Taking the uniform thickness of stem
Hence width of heel slab Let thickness of base slab
= =
4.65 0.8
Clear spacing of counter fort
=
3.5 x
=
3.20
=
3.00
Providing spacing of counterfort 3 Stability of Wall Let w1
=
Weight of rectangular portion of stem
w2
=
Weight of base slab
w3
=
Weight of soil on heel slab.
w1
1
Detail x
w2
1
x
w3
1
x
Total resisting moment Earth pressure (p) Moment F.O.S. against overturning F.O.S. against sliding
= = = = = = = =
Pressure Distribution Net Moment (SM) = Distance x of the point of application of resultant, from toe is x = = e = = b/6 = Pressure p1 at toe
= = =
Pressure p2 at heel
=
7724 (Ka x γ' x H)+(γw x H) 158.33 p*H2/6 4123.26 1.87 µSw/p 4.20
3601 SM/Sw 2.71 b/2 - x 1.67 1.46 Hence un safe as e > b/6 SW b 1329.79 8.75 325.69 Hence un safe as p1 > 125 SW
Pressure p2 at heel
=
=
b 1329.79 8.75 -21.74 Hence un safe as p2 < 0
The Pressure intencity p1 under E, p1
=
325.69
=
186.72
=
325.69
=
162.90
The Pressure intencity p2 under B, p2
4 Design of Heel Slab Clear spacing b/w counterforts = 3.00 The pressure distribution on the heel slab is shown in fig 2. Consider a strip 1 meter wide. Upward pressure intensity (near outer edge, C) Down ward load due to weight of Earth.
= =
-21.74 210.6
Down ward weight of slab per unit area Hence net pressure intensities will be P
=
15
=
203.86 2 Pl /12
M1
= =
= = =
152.90 BM Rxb 371.39 500 460
=
800
=
1598.94
Provide Ast1 Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
1598.94 20 196 150
Shear force (V)
= = =
PL/2 305.79 0.35
=
0.26
Effective depth required (d)
=
Providing overall depth (D) Effective depth (d) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Ast M=A st σ st j d
100Ast/bd τc Refernce: Pg. 84, Table 23, IS 456 : 2000
τv
Hence depth required from shear point of veiw (d) Providing overall depth (D) Hence effective depth (d) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Ast M=A st σ st j d
= = = = = =
V/(d x b) 0.66 Hence shear reinforcement is required as τv V/(τc x b) 1176.12 1200 1160
=
1920
=
634.06
Provide Ast1 Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
1920 20 163 150
Distribution Steel Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
1920 20 163 150
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. Hence point of contraflexure is at Shear force at this point is given
= 0.633 = P(L/2 - x) = 176.75 = 176746.62 Assuming that all the bars will avilable at point of contraflexure, Ast σst j d M = = 462979286.1 Lo = 12υ or d, whichever is more = 1160 Ld = 45υ = 900 M/V + Lo = 3779.45 Hence safe as M/V+Lo > Ld Cotinue these bars by a distance Lo = d = 1160 At the point of curtailment, length of each bar available = 1793 Hence safe These bars will be provide at the top face of heel slab. Maximum passive B.M. = PL2/16 3/4 x M1 =
Area of bottom steel (Ast2) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Provide Ast2 Dia of bar Spacing of bar calculated Spacing of bar provided
=
3/4 x Ast1
=
1440
=
1920
= = = =
1920 20 163 150
Let us check this reinforcement for development length crierion at point of contraflexur, Ast σst j d M = = 462979286.1 Lo = 12υ or d, whichever is more = 1160 Ld = 45υ = 900 M/V + Lo = 3779.45 Hence safe as M/V+Lo > Ld Cotinue these bars by a distance Lo = d = 1160 i.e. upto distance = -527 At this point half bars can be discontinued. Since this distance is quite small, it is better to continue these bar 5 Design of Toe Slab Providing counterfort over toe slab upto ground level. Assuming total depth of toe slab
=
500
Total weight of toe slab
=
12.5
Net upward intensity at D
=
325.69-12.5
=
313.19
=
186.72-12.5
=
174.22
= =
= = =
wL2/12 234.8925 BM Rxb 460.33 500 460
=
800
=
2456.46
Net upward intensity at E Cosidering strip of unit width at D. Max. negative B.M. (M1) Effective depth required (d) Providing overall depth (D) Effective depth (d) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Ast M=A st σ st j d
=
Provide Ast1 Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
2456.46 20 127 100
Shear force (V)
= = =
PL/2 469.79 0.53
τc
=
0.32
τv
= =
100Ast/bd Refernce: Pg. 84, Table 23, IS 456 : 2000
Hence depth required from shear point of veiw (d) Providing overall depth (D) Hence effective depth (d) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Ast M=A st σ st j d
= = = =
V/(d x b) 1.02 Hence shear reinforcement is required as τv V/(τc x b) 1468.08 1500 1460
=
2400
=
773.95
Provide Ast1 Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
2400 20 130 100
Distribution Steel Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
2400 20 130 100
Let us check this reinforcement for development length crierion at point of contraflexur. Point of contraflexure is at = 0.633 Shear force at this point is given = P(L/2 - x) = 271.54 = 271535.73 Ast σst j d M = = 728394135.5 Lo = 12υ or d, whichever is more = 1460 Ld = 45υ = 900 M/V + Lo = 4142.50 Hence safe as M/V+Lo > Ld Cotinue these bars by a distance Lo = d = 1460
At the point of curtailment, length of each bar available These bars will be provide at the top face of toe slab. Maximum passive B.M. Area of bottom steel (Ast2) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Provide Ast2 Dia of bar Spacing of bar calculated Spacing of bar provided
=
2093 Hence safe
= =
PL /16 3/4 x M1
=
3/4 x Ast1
=
1800.00
=
2400
= = = =
2400 20 130 100
2
Let us check this reinforcement for development length crierion at point of contraflexur, Ast σst j d M = = 728394135.5 Lo = 12υ or d, whichever is more = 1460 Ld = 45υ = 900 M/V + Lo = 4142.50 Hence safe as M/V+Lo > Ld Cotinue these bars by a distance Lo = d = 1460 i.e. upto distance = -827 6 Design of Stem The stem acts as a continuous slab. Considred 1 m strip at B . The intencity of earth pressure is given by (ph) =
KaγH1
=
67.80
Negative B.M. in slab, (M1)
= = =
Effective depth required (d)
= = = =
11.30 phL2/12 50.85 BM Rxb 214.18 600 560
=
960
=
436.82
Hence revised H1
Providing overall depth (D) Effective depth (d) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Ast M=A st σ st j d
Provide Ast1 Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
960.00 16 209 200
Distribution Steel Dia of bar Spacing of bar calculated Spacing of bar provided
= = = =
960 16 209 200
Shear force (V)
= = =
PL/2 101.70 0.17
τc
=
0.20
τv
=
V/(d x b)
100Ast/bd Refernce: Pg. 84, Table 23, IS 456 : 2000
=
0.18 Hence no shear reinforcement is required as
Let us check this reinforcement for development length crierion at point of contraflexur. Point of contraflexure is at = 0.633 Shear force at this point is given (V) = V(L/2 - x) = 88.17 = 88173.90 Ast σst j d M = = 111753620.8 Lo = 12υ or d, whichever is more = 560 Ld = 45υ = 720 M/V + Lo = 1827.42 Hence safe as M/V+Lo > Ld Cotinue these bars by a distance Lo = d = 560 At the point of curtailment, length of each bar available = 1193 Hence safe These bars are to be provided at the inner face of the stem. Maximum passive B.M. = PL2/16 3/4 x M1 = Area of steel (Ast2) Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Provide Ast2 Dia of bar Spacing of bar calculated
=
3/4 x Ast1
=
720.00
=
960
= = =
960 16 209
Spacing of bar provided
=
200
Let us check this reinforcement for development length crierion at point of contraflexur, Ast σst j d M = = 111753620.8 Lo = 12υ or d, whichever is more = 560 Ld = 45υ = 720 M/V + Lo = 1827.42 Hence safe as M/V+Lo > Ld 7 Design of Main Counterfort Let us assuming thickness of counterforts is Spacing of counterforts At any section at depth h below the top A, the earth pressure acting on each counter forts will be
= =
500 350
=
Ka x γsat x h x L
Similarly, net down ward pressure on heel at C is
= =
21 (11.3 x 18) + (1200/1000) x 25 - -21.74
Similarly, net down ward pressure on heel at B is
= =
255.14 (11.3 x 18) + (1200/1000) x 25 - 162.9
= Hence reaction transferrred to each counterfort will be, At C At B Pressure intencity at h Shear force at F (Q) B.M.
Effective depth required (d)
= = = = = = = = = = = =
Providing total depth (D) Effective depth (d) Angle θ of the face AC is given by, tan θ tan θ θ Sin θ Cos θ F1G1
= = = = = = = = = =
70.5 255.14 x (350/100) 892.99 70.5 x (350/100) 246.75 11.5 241.5 0.5 x 241.5 x 11.5 1388.63 h/3 x 1388.625 5323.06 5323062500 BM Rxb 3099.03 3150 3110 4.65/11.3 0.411504425 22.37 0.3805 0.9248 AF1 x sinθ
= 4.38 = 4376.26 FG = 4977.00 Asssuming that the steel reinforcement is provided in two layers and providing a nominal cover of 40 mm and Effective depth (d) = 4977 - (40 + 20 + 10) = 4907.00 Area of steel at supports, at bottom (Ast ) = 5218.47 M=Ast σst j d Using dia. of bar = 20 Aυ = 314.16 No. of bars = 17 Provide the bars in two layers. Effective shear force = Q-(M/d')tanθ d' = d/cosθ = 5306.23 Effective shear force = 975815.19 τv = V/(d x b) = =
100Ast/bd τc
0.37 0.20
0.21 Hence shear reinforcement is required as τv However, the vertical and horizontal ties provided in counterforts will bear the excess shear stress. The height h where half of the reinforcement can curtailed (H)
=
=
h
= 3.54 To locate the position of point of curtailmenton AC, drawing Hl parallel to FG. Thus half bars can be curtailed at l. However these = 240 should be extent by a distance 12υ 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 45υ
=
900
= = =
Ka x γsat x h x L 207 Force/Stress
= = =
900 10 2Π/4 x D2
Spacing required Spacing provided
= = =
157.08 174.53 150
Design of Vertical Ties The downward force at C
=
Design of Horizontal Ties At any depth h below the top, force causing sepration Ast required Using 2 legged ties of dia. Aυ
(892.99 x 3)/(350/100)
= = = =
The downward force at B Ast required at C Using 2 legged ties of dia. Aυ Spacing required Spacing provided Ast required at B Using 2 legged ties of dia. Aυ Spacing required Spacing provided
765.42 (246.75 x 3)/(350/100) 211.5 Force/Stress
= = =
3327.91 20 2Π/4 x D2
= = = =
628.32 188.80 150 Force/Stress
= = =
919.57 10 2Π/4 x D2
= = =
157.08 170.82 150
8 Design of Front Counterfort The upward pressure intensity varies from Downward weight of toe slab
325.69 =
Net weight at D
=
Net weight at E The center to center spacing of counterforts Hence upward force transmitted to counterforts at D
= = = = = = = = = = = =
Upward force transmitted to counterforts at E Total upward force Force will be acting at x B.M. Effective depth required (d) Providing total depth (D) Effective depth (d) Thus height of counterfort above ground level i.e. upto point F3. Area of steel at supports, at bottom (Ast ) M=Ast σst j d
= = = = =
=
2
kN/m at D to 37.5 288.19
149.22 3.5 288.19 x 3.5 1008.67 149.22 x 3.5 522.27 1/2 x (1008.665 + 522.27) x 3.5 2679.14 ((522.27 + 2 x 1008.665)/(1008.665 + 522.27 1.94 2679.14 x 1.94 5197.53 BM Rxb 3062.27 4000 3960 3000
6313.93
Minimum steel required 0.2% of b D ― further 20% [IS :3370,7.1] Provide Ast1 Dia of bar Aυ No. of bars Effective shear force (V) tan θ Effective shear force (V) τv 100Ast/bd
=
6400
= = = = = = = =
6400.00 32 804.25 8 Q-(M/d)tanθ 1.14 1179130.88 V/(d x b)
= =
0.60 0.11
τc
=
Aυ
= =
0.19 Hence shear reinforcement is required as τv 10 2Π/4 x D2
= = = = = = = = 10
157.08 τc x b x d 376200 V - Vc 802930.88 (σst x Aυ x d)/Vs 178.18 150 mm holding bar at top
Using 2 legged ties of dia.
Vc Vs Spacing required (sv) Spacing required (sv) Provide 2 x
9 Fixing Effect in Stem, Toe and Heel At the junction of stem, toe and heel slab fixing moment are induced, which are at right angles to their normal In stem @0.3% of cross section, to be provided at (i) inner face, in vertical direction,for a length 58υ Ast = (0.3/100) x 1000 x 600 Dia. of bar Aυ Spacing required Spacing provided Length of embedment in slab above heel slab (ii) In toe slab @0.15% to be provided at the lowar face Ast Dia. of bar
= = = = = = =
1800 16 201.06 (201.06 x 1000)/1800 111.70 100 928
=
(0.15/100) x 1000 x 1500
= =
2250 20
Aυ Spacing required Spacing provided Length of embedment in slab above heel slab (iii) In heel slab @ 0.15% to be provided in upper face Ast Dia. of bar Aυ Spacing required Spacing provided Length of embedment in slab above heel slab
= = = = =
314.16 (314.16 x 1000)/2250 139.63 100 1160
=
(0.15/100) x 1000 x 1200
= = = = = = =
1800 16 201.06 (201.06 x 1000)/1800 111.70 100 928
m kN/m3 kN/m3 kN/m2 MPa MPa mm m m 2 N/mm N/mm2
+ m -
1
q0 2.2 y H 2.2
125 18
x
x
12.50
…. Eq (1) Ka (1-α) x (1+3α)
x x
m m m
12.50
(
1
-
0.40
)x(
0.333 1
m m H γsat m
1/4
m
Weight of rectangular portion of stem Weight of base slab Weight of soil on heel slab. Detail
force(kN)
lever arm
0.60
x
11.70
x
25
=
175.5
3.8
0.80
x
8.75
x
25
=
175
4.38
4.65
x
11.70
x
18
=
979.29
=
1329.79
Sw kNm (Ka x γ' x H)+(γw x H) kN kNm >
1.5
Safe against overturning
>
1.5
Safe against sliding
kNm
m
Hence un safe as e > b/6 1+
6e b
x
1+
kN/m2 Hence un safe as p1 > 125 1-
6e
6x
1.67 8.75
6.43 Total MR
1-
b
x
1+
6x
1.67 8.75
2
kN/m Hence un safe as p2 < 0
-
325.69
8.75
-21.74
325.69
8.75
-21.74
x
3.50
x
4.10
kN/m2 kN/m2
m nsider a strip 1 meter wide. kN/m2 kN/m2 2
kN/m
kN/m2 kNm
mm mm mm mm2 mm2 mm2 mm mm mm
kN N/mm2
664.761 0.66 Hence shear reinforcement is required as τv > τc mm mm mm mm2 mm2 mm2 mm mm mm 2
mm mm mm mm
mm kN N
Nmm 12υ or d, whichever is more mm
mm mm 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 leng mm
mm2 mm2 mm2 mm mm mm
n at point of contraflexur, Nmm 12υ or d, whichever is more mm
mm mm Hence safe as M/V+Lo > Ld mm from the point of contraflexure. mm from the centre of support. is quite small, it is better to continue these bars upto center of counterfors.
mm 2 kN/m kN/m2 2
kN/m
kN/m2 kN/m2
kNm
mm mm mm mm2 mm2
mm2 mm mm mm
kN N/mm2
Hence shear reinforcement is required as τv > τc mm mm mm 2
mm
2
mm
mm2 mm mm mm mm2 mm mm mm
n at point of contraflexur. mm kN N Nmm 12υ or d, whichever is more mm mm
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 leng
mm
mm2 mm2 mm2 mm mm mm
n at point of contraflexur, Nmm 12υ or d, whichever is more mm mm mm Hence safe as M/V+Lo > Ld mm from the point of contraflexure. mm from the centre of support.
kN/m2 m kNm
mm mm mm mm2 mm2
mm2 mm mm mm mm2 mm mm mm
kN N/mm2
N/mm2 Hence no shear reinforcement is required as τv < τc
n at point of contraflexur. mm kN N Nmm 12υ or d, whichever is more mm mm
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 leng mm
mm2 2
mm
mm2 mm mm
mm
n at point of contraflexur, Nmm 12υ or d, whichever is more mm mm mm Hence safe as M/V+Lo > Ld
mm cm c/c
h kN/m (11.3 x 18) + (1200/1000) x 25 - -21.74 kN/m2 (11.3 x 18) + (1200/1000) x 25 - 162.9 2 kN/m 255.14 x (350/100) kN/m 70.5 x (350/100) kN/m m kN/m 0.5 x 241.5 x 11.5 kN kNm Nmm
mm mm mm
Degrees
m mm mm s and providing a nominal cover of 40 mm and 20 mm dia bar. 4977 - (40 + 20 + 10) mm 2
mm
mm mm2 bars
mm N N/mm2
N/mm2 Hence shear reinforcement is required as τv > τc rts will bear the excess shear stress. m m below A, i.e. at point H. mm beyond I, i.e. extented upto I1 line I1H1 parallel FG mm
kN/m mm2 mm 2
mm mm mm
(892.99 x 3)/(350/100)
kN/m (246.75 x 3)/(350/100) kN/m mm2 mm mm2 mm mm mm2 mm 2
mm mm mm
2
186.72 kN/m at E. kN/m2 kN/m2 kN/m2 m kN/m kN/m kN/m kN/m 1/2 x (1008.665 + 522.27) x 3.5 kN ((522.27 + 2 x 1008.665)/(1008.665 + 522.27)) x (3.5/3) m from E kNm kNm
mm mm mm 2
mm
mm2 2
mm mm
mm2 bars
N N/mm2 2
N/mm
Hence shear reinforcement is required as τv > τc mm mm2 N N (σst x Aυ x d)/Vs mm c/c mm holding bar at top
duced, which are at right angles to their normal direction of bending. These moment are not determine , but normal reinforcement given bel
(0.3/100) x 1000 x 600 mm2 mm mm2 (201.06 x 1000)/1800 mm mm mm
(0.15/100) x 1000 x 1500 mm2 mm
mm2 (314.16 x 1000)/2250 mm mm mm
(0.15/100) x 1000 x 1200 mm2 mm 2 mm (201.06 x 1000)/1800 mm mm mm
0.333 +
1.20
) A
0.6
θ I1 H1 H
I
11.5
G F3 F
3.50 D
E
G1
3.00 1.00
F1 B
4.65
8.75
FIGURE 1 Moment about toe (KN-m) 666.90
kN/m2
6291.94 7724
325.69
765.63
FIGURE 2
F2 C
the remaining half throughout the length.
the remaining half throughout the length.
the remaining half throughout the length.
ne , but normal reinforcement given below may be provided.
0.928
16 mm υ 100 mm c/c 0.928
20 mm υ 100 mm c/c 1.16
20 mm υ 100 mm c/c
12.50
kN/m2
-21.74
12.50
1.50
0.15
0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50
M-15
M-20
M-25
M-30
M-35
M-40
0.18 0.22 0.29 0.34 0.37 0.40 0.42 0.44 0.44 0.44 0.44
0.18 0.22 0.30 0.35 0.39 0.42 0.45 0.47 0.49 0.51 0.51
0.19 0.23 0.31 0.36 0.40 0.44 0.46 0.49 0.51 0.53 0.55
0.20 0.23 0.31 0.37 0.41 0.45 0.48 0.50 0.53 0.55 0.57
0.20 0.23 0.31 0.37 0.42 0.45 0.49 0.52 0.54 0.56 0.58
0.20 0.23 0.32 0.38 0.42 0.46 0.49 0.52 0.55 0.57 0.60
0.21 0.26 0.32
