ANN A UNIVERSITY RAM ANATHA AN ATHAPURA PURAM-ROYAL M-ROYAL CIVIL DESIGN AND DETAIL DETAIL ING OF RETAINING RETAINING WALL S
Definition: Retaining walls are usually built to hold back soil mass. However, retaining walls can also be constructed for aesthetic landscaping purposes. Retaining walls are structures that are constructed to retail soil or any such materials which are unable to stand vertically by themselves. They are also provided to maintain the grounds at two different levels.
(a)
(b)
Figure. Cross section of cantilever retaining wall (a) and the finished retaining wall (b)
Figure. Types of retaining walls
Figure. Photo of concrete retaining wall
Figure. Types of gravity retaining walls
Classification of retaining walls:
Following are the different types of retaining walls, which is based on the shape and the mode of resisting the pressure. 1. Gravity wall-Masonry or Plain concrete 2. Cantilever retaining wall-RCC (Inverted T and L) 3. Counterfort retaining wall-RCC 4. Buttress wall-RCC
Backfill
Backfill
Tile drain Gravity RW
T-Shaped RW
L-Shaped RW
Back fill Counterfort
Counterfort RW
Buttress
Weep holes
Buttress RW Figure. Types of retaining walls
Earth Pressure (P)
Earth pressure is the pressure exerted by the retaining material on the reta ining wall. This pressure tends to deflect the wall outward. There are two types of earth pressure and they are; Active earth pressure or earth pressure (P a) and Passive earth pressure (P p). Active earth pressure tends to deflect the wall away from the backfill. Earth pressure depends on type of backfill, the height of wall and the soil conditions Soil conditions: The different soil conditions are Dry leveled back fill Moist leveled backfill Submerged leveled backfill Leveled backfill with uniform surcharge Backfill with sloping surface
Analysis for dry back fills
h H Pa
M Df kah Maximum pressure at any height, p=k ah Total pressure at any height from top, P=1/2[k ah]h = [k ah2]/2 Bending moment at any height = M=Pxh/3= [k ah3]/6
Total pressure at bottom, P a= [k aH2]/2 Total Bending moment at bottom, M = [k aH3]/6 Where, k a= Coefficient of active earth pressure= (1-sin )/(1+sin)=tan = 1/k p, coefficient of passive earth pressure = Angle of internal friction or angle of repose =Unit weigh or density of backfill
2
If = 30 , k a=1/3 and k p=3. Thus k a is 9 times k p Backfill with sloping surface
pa= k a H at the bottom and is parallel to inclined surface of backfill
cos cos 2 cos 2 k a= cos cos cos 2 cos2
Where =Angle of surcharge Total pressure at bottom=P a= k a H2/2 Stability requirements of RW:
Following conditions must be satisfied for stability of wall. 1. It should not overturn 2. It should not slide 3. It should not subside i.e Max. pressure at the toe should not exceed the safe bearing capacity of the soil under working condition Check against overturning
Factor of safety against overturning = M R / MO 1.55 (=1.4/0.9) Where, MR =Stabilising moment or restoring moment MO =overturning moment As per IS:456-2000, MR >1.2 M O, ch DL + 1.4 M O, ch IL 0.9 MR 1.4 MO, ch IL Check against Sliding
FOS = Resisting force to sliding/Horizontal force causing sliding = W/Pa 1.55 (=1.4/0.9) As per IS:456:2000 1.4 = ( 0.9 W)/Pa Design of Shear key :
If the wall is not safe against sliding, then a shear key is to be provided. It is provided either below the stem or at the end of heel. It should not be provided at the end of toe. If shear key is provided, then it should be designed taking the effect of passive pressure.
H
H+a Pa
C A
R
p p a
B
W
k a(H+a)
=45 + /2
In case the wall is unsafe against sliding p p= p tan2 (45 + /2) = p k p where p p= Unit passive pressure on soil above shearing plane AB If W= Total vertical force acting at the key base = shearing angle of passive resistance R= Total passive force = p p x a PA=Active horizontal pressure at key base for H+a W=Total frictional force under flat base For equilibrium, R +
W =FOS x PA
FOS= (R + W)/ PA 1.55 Pressure below the wall
Consider the retaining wall as shown. All forces acting on the wall are shown. The moment of all forces at the end of toe is considered and the requirements of stability are to be established. For stability earth pressure at the end of the heel for the entire height of wall should be considered. The maximum and minimum pressure below the wall can be determined from the principles of static.
x1
h
x2
W1
W4
H
W
W2
Pa R
H/3
W3 T
e b/6
x
b
b/2
Pressure below the Retaining Wall Maximum pressure at the toe Let the resultant R due to W and P a lie at a distance x from the toe. X = M/W, M = sum of all moments about toe. Eccentricity of the load = e = (b/2-x) Minimum pressure at heel W 6e Pmin 1 b
b
This should not be less than zero to avoid tension at the base. From this e=b/6, resultant should cut the base within the middle third. Otherwise the wall tends to separate from the base due to tension. Maximum pressure at toe W 6e Pmax 1 b
b
This should not be greater than SBC of soil to avoid the subsidence of wall. Depth of foundation
Rankine’s formula: D f =
SBC
1 sin 1 sin
2
=
SBC
γ
k a
2
Preliminary Proportioning (T shaped wall)
Following guidelines are to be followed for initial proportioning of wall with out surcharge. For surcharge and other cases, good text books should be followed. Stem: Top width 200 mm to 400 mm Base slab width b= 0.4H to 0.6H, and 0.6H to 0.75H for surcharged wall Base slab thickness= H/10 to H/14 Toe projection= (1/3-1/4) Base width
200
H
tp= (1/3-1/4)b
H/10 –H/14
b= 0.4H to 0.6H Behaviour or structural action and design All the three elements namely stem, toe and heel acts as cantilever slabs and hence the design and detailing principles are same as that of conventional cantilever slabs.
Stem design: Mu=partial safety factor x (k a H /6) Determine the depth d from M u = Mu, lim=Qbd 2 Determine the steel based on balanced or under reinforced design. Provide enough development length at the junction for all bars. 3
Curtailment of steel
Maximum steel is needed at the base where the BM is maximum. As the BM decreases towards the top, steel can be suitably curtailed at one or two levels. Usually steel is curtailed at one level where the steel quantity is about 50% or 67% of the base steel. Effective depth is Proportional to h Bending moment is proportional to h 3 l l 2 Ast is α to BM/Eff. depth and is α to h i.e.
Ast 1 Ast 2
2
h1
2
h2
Distance From top
h1 Ast/2
Every alternate bar curtailed
h2 h1c Ldt
Ast
h2 Ast/2
Ast Provided
Ast
Distribution steel: 0.12% Gross area for HYSD bars, 0.15% for Mild steel bars Temperature steel: Provide this steel at the outer face which is same as t he distribution steel. Also provide suitable development lengths for all steel meeting at the junction. Provide suitable construction keys, drainage facilities, tile drains and weep holes as shown in the drawing. Sketch the drawings and detail as per the requirements. Retaining wall Design Design example-1
Design a cantilever retaining wall (T type) to retain earth for a height of 4m. the backfill is horizontal. The density of soil is 18kN/m 3. Safe bearing capacity of soil is 200 kN/m 2. Take the co-efficient of friction between concrete and soil as 0.6. The angle of repose is 30 degrees. Use M20 concrete and Fe415 steel. Solution
Data: h' = 4m, SBC= 200 kN/m
2
, γ= 18 kN/m3, μ=0.6, φ=30°
To fix the height of retaining wall, H H= h' +Df Depth of foundation Rankine’s formula: D f =
SBC
1 sin 1 sin
2
=
1.23m say 1.2m , therefore H= 5.2m
SBC
γ
k a
2
Proportioning of wall Thickness of base slab= (1/10 to 1 /14) H, 0.52m to 0.43m, say 450 mm Width of base slab=b = (0.5 to 0.6) H, 2.6m to 3.12m say 3m Toe projection= pj= ( 1/3 to ¼)H, 1m to 0.75m say 0.75m Provide 450 mm thickness for the stem at the base and 200 mm at the top Design of stem To find Maximum bending moment at the junction 2
Ph= ½ x 1/3 x 18 x 4.75 =67.68 kN M= Ph h/3 = 0.333 x 18 x 4.75 3/6 = 107.1 kN-m Mu= 1.5 x M = 160.6 kN-m Taking 1m length of wall, 2 Mu/bd = 1.004 < 2.76, URS (Here d=450- effective cover=450-50=400 mm) To find steel Pt=0.295% <0.96% 2 Ast= 0.295x1000x400/100 = 1180 mm #12 @ 90 < 300 mm and 3d ok 2 Ast provided= 1266mm Development length Ld =47 φ bar =47 x 12 = 564 mm Curtailment of bars Curtail 50% steel from top 2 (h1/h) = ½ 2
(h1/4.75) = ½, h1 = 3.36m Actual point of cutoff= 3.36-L d =3.36-47 φ bar = 3.36-0.564 = 2.74m from top. Spacing of bars = 180 mm c/c < 300 mm and 3d ok Distribution steel 2 = 0.12% GA = 0.12x450 x 1000/100 = 540 mm #10 @ 140 < 450 mm and 5d ok
Secondary steel for stem at front (Temperature steel) 2 0.12% GA = 0.12x450 x 1000/100 = 540 mm #10 @ 140 < 450 mm and 5d ok Check for shear
Max. SF at Junction = P h=67.68 kN Ultimate SF= Vu=1.5 x 67.68 = 101.52 kN Nominal shear stress =τv=Vu/bd = 101.52 x 1000 / 1000x400 = 0.25 MPa To find τc : 100Ast/bd = 0.32%, From IS:456- 2000, τc= 0.38 MPa
τv< τc Hence safe in shear. Stability analysis
x1
h
x2
W1
W4
H
W
W2
Pa H/3 W3
T
e b/6
x
b/2
b 0.75m 0.45m
1.8m 30.16 kN/m 2
120.6 kN/m
2
24.1 22.6
97.99
Pressure below the Retaining Wall
Load
Magnitude, kN
Stem W1
0.2x4.75x1x25
= 23.75
Stem W2
½ x0.25x4.75x1x25
= 14.84
Base slab W3 3.0x0.45x1x25 = 33.75 Back fill, W4 1.8x4.75x1x18 = 153.9 ΣW= 226.24 total 2 Hori. earth PH =0.333x18x5.2 /2 pressure =P H =81.04 kN
Stability checks:
Check for overturning: FOS = ΣMR / MO= 2.94 >1.55 safe
Distance from A, m
1.1 0.75 + 2/3x0.25=0.316 1.5 2.1 H/3 =5.2/3
Bending moment about A kN-m 26.13
13.60 50.63 323.20 ΣMR=413.55 MO=140.05
Check for Sliding: FOS = µ ΣW/ PH= 2.94 >1.55 safe Check for subsidence: Let the resultant cut the base at x from toe T, x= ΣM/ ΣW= 1.20 m > b/3 e= b/2 –x = 3/2 – 1.2 = 0.3m < b/6 Pressure below the base slab Max. pressure= Pmax
W 6e 1 b b
120.66 kN/m2 < SBC, safe Min. pressure = Pmin
W 6e 1 b b
30.16 kN/m 2 > zero, No tension or separation, safe Design of Heel
To fine the maximum bending moment
153.9 0.45x1.8x25 = 27.25
Distance from C, m 0.9 0.9
BM, MC, kN-m 138.51 18.23
30.16 x 1.8 =54.29
0.9
-48.86
½ x 24.1 x1.8=21.69
1/3x1.8
-13.01
105.17
Total BM at unction
ΣMC=94.86
Load
Magnitude, kN
Backfill Heel slab Pressure distribution, rectangle Pressure distribution, triangle Total Load at junction
Mu= 1.5 x 94.86 =142.3 kNm Mu/bd 2= 0.89 < 2.76, URS Pt=0.264% < 0.96% Ast= 0.264x1000x400/100 = 1056 mm 2 #16@ 190 < 300 mm and 3d ok 2 Ast provided= 1058mm Development length Ld =47 φ bar =47 x 16 = 752mm Distribution steel Same, #10 @ 140 < 450 mm and 5d ok Check for shear at junction (Tension) Net downward force causing shear = 142.3kN. Critical section for shear is at the face as it is subjected to tension. Maximum shear =V=105.17 kN, V U, max= 157.76 kN, τv =0.39 MPa
pt=100x1058/(1000x400)=0.27% τuc =0.37 MPa Allowable shear force= 0.37x 1000 x 400 =148kN, slightly less than V U, max. May be ok Design of toe To find the maximum bending moment
0.75x0.45x25=8.44
Distance from C, m 0.75/2
BM, MC, kN-m -3.164
97.99x0.75=73.49
0.75/2
27.60
2/3x1=0.75
4.24
Total BM at unction
ΣM=28.67kNm
Load
Magnitude, kN
Toe slab Pressure distribution, rectangle Pressure distribution, triangle
½ x22.6 x1x0.75=8.48
Total Load at junction
73.53
Mu= 1.5 x 28.67 =43 kNm 2 Mu/bd = 0.27< 2.76, URS Pt=0.085% Very small, provide 0.12%GA 2 Ast= 540 mm #10 @ 140 < 300 mm and 3d ok Development length: Ld =47 φ bar =47 x 10 = 470 mm Check for shear: Since the soil pressure introduces compression in the wall, the critical section is taken at a distance d from junction. Net shear force at the section= (120.6+110.04)/2 x 0.35 -0.45x0.35x25=75.45kN V=75.46 kN, VU,max=75.45x1.5=113.18 kN τv=113.17x1000/(1000x400)=0.28 MPa pt=0.25% τuc =0.37 MPa V,allowable = 0.37x 1000 x 400 =148 kN > V U,max, ok Construction joint A key 200 mm wide x 50 mm deep with nominal steel #10 @ 250, 600 mm length in two rows Drainage: 100 mm dia. pipes as weep holes at 3m c/c at bottom Also provide 200 mm gravel blanket at the back of the stem for back drain. Sketch
Following section will be asked in the examination. 1. Cross section of wall
2. Longitudinal section of wall for about 2m 3. Sectional plan of the base slab 4. Longitudinal section of stem near the base slab
#12 @ 180
#10 @ 140
#12 @ 90 #16 @ 190
#10 @ 140 Cross section of wall
Longitudinal section of wall
Note
Sectional plan of base slab
Adopt a suitable scale such as 1:20 Show all the details and do neat drawing Show the development length for all bars at the junction Name the different parts such as stem, toe, heel, backfill, weep holes, blanket, etc., Show the dimensions of all parts Detail the steel in all the drawings Lines with double headed arrows represents the development lengths in the cross section
DESIGN AND DETAILING OF COUNTERFORT RETAINING WAL L
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. In such a case counterforts having t rapezoidal section fixed at the base slab are provided at intervals of 1.5 m to 3 m. The counterforts support the heel slab and the vertical stem. The design principles for different components of the wall are discussed as under. Design of Stem The stem acts as a continuous slab spanning longitudinally over the counterforts. The horizontal active soil pressure acts as the load on the slab. Since the earth pressure varies linearly over the height of the stem, the slab deflects away from the earth face between the counterforts and hence the main steel is pr ovided at the outer face of the stem a nd at the inner face near the supporting 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.
If I is the clear distance between the counterforts and p is the intensity of soil pressure, the slab is designed for bending moment as under: Maximum +ve B.M= pl 2/16 (occurring mid-way between counterforts) and 2 Maximum -ve B.M= pl /12 (occurring at inner face of counterforts) 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 The base width is approximately taken equal to 0.6 H to 0.7 H, where H is the overall height of the wall. The projection of toe slab is approximately taken between 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. 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. In case the toe slab projection is large i.e. > b/3, front counterforts are provided above the toe slab (normally up to the ground level) and the slab is designed as a co ntinuous horizontal slab spanning between the front cou nterforts. Design of Heel Slab The heel slab is designed as a continuous slab spanning over the counterforts, as in the case of stem. 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. The net force acts downward producing tension towards the earth face between the counterforts and negative moment develops at the support provided by counterforts.
If p is the net downward force and I is the clear span between the counterforts the B.M. is given by:
Maximum +ve B.M= pl 2/16 (mid-way between counterforts towards earth face) Maximum -ve B.M. = pl 2/12 (occurring at counterforts) Design of Counterforts Since the active earth pressure on stem acts outward and stem is considered to be fixed at 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, the stem lies in the compression zone with respect to the bending of the counterforts and hence the counterforts are designed as a T-beam of varying depth. The main steel provided along the sloping face shall be anchored p roperly at both ends. The depth of the counterfort is measured perpendicular to t he sloping side.
In order that the counterfort and stem should act as one unit, it is joined firmly to the stem by providing ties in the horizontal plane. The base is tied with vertical ties to prevent its tendency to separate out under the action of net downward force. The provision of ties ensures transfer of forces to the counterforts. The net forces acting on the different components of the counterforts, position of main steel and horizontal and vertical ties are schematically shown. PROBLEM: 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 2 pit taken at the site indicates that soil of bearing capacity 220 kN/m is available at a depth of 3 1.25 m below the ground level. The weight of earth is 18 kN/m 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. 2
2
Given: f ck = 20 N/mm , f y = 415N/mm , H = 7 m above G.L, Depth of footing below G.L. = 1.25 m, γ = 18 kN/m3, µ = 0.58, f b = 220 kN/m 2 Required : Design the counterfort retaining wall. Design constants 2 Q=2.76 N/mm Ld = (0.87 f y / 4 τ bd ) φ = 0.87x415/4x(1.2x1.6)φ = 47φ bar
For φ = 30°
Coefficient of active pressure = k a = (1 - sin φ)/(l + sin φ) = 1/3 Coefficient of passive pressure = k p = (1 + sin φ)/(l - sinφ) = 3 a. Proportioning of Wall Components
The height of the wall above the base = H = 7 + 1.25 = 8.25 m. Base width = 0.6 H to 0.7 H i.e. between 4.95 m to 5.78 m. Assume base width 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
clear spacing between counterforts is given by : L = 3.5 (H/γ) 0.25 = 3.5 (8.25/18) 0.25 = 2.88 m c/c spacing = 2.88 + 0.25 = 3.13 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 The preliminary dimensions of the components of the wall are shown in Figure. 250 mm
CF: 3m c/c, 400 mm h=7.8 m
1
h =7 m
H=8.25 m
d
1.25m
1.2 m
4.05m
θ
b=5.5 m
b. Check Stability of Wall
The stability of the wall will be checked at working load because safe bearing capacity of soil is in the working load condition.
Sr. No.
Description of loads
Loads in kN
Dist. of e.g. from T in m
Moment about T in kN-m
1
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
Total
ΣW = 679.25
Horizontal earth pressure on full height of wall 2 2 = Ph = γh k a /2 =18 x 8.25 /(3 x 2) = 204.19 kN
ΣW =2210.71
Overturning moment = M 0 = 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 = 3.937 > 1.55 . . safe. Check for sliding Total horizontal force tending to slide the wall = P h = 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.
250 mm
H 8250
ΣW
R
PA 1250
A
1200 mm
C
B
4050 mm
D
450 e
X
166.61 2 kN/m
153.9
147.8
b/2
143.9
5500 mm Check for pressure distribution at base Net moment = M = 2210.71 -561.52 = 1649.19 kN.m. Let x be the distance from the toe where the resultant R acts, x = M / ∑W = 1649.19/679.25 = 2.43 m Eccentricity = e = b/2 - x = 5.5/2 - 2.43 = 0.32 < b/6 (= 0.91 m)
Whole base is under compression. Maximum pressure at toe = p A = ∑W / b ( 1+6e/b) = 679.25/5.5 ( 1+ 6*0.32/5.5) 2 2 = 166.61 kN/m < f b (= 220 kN/m )
80.39 kN/m2
Minimum pressure at heel = p D = ∑W / b (1-6e/b) = 679.25/5.5 ( 1- 6 *0.32/5.5) 2 = 80.39 kN/m compression. The distribution of stresses under the base is shown in Fig Intensity of pressure at junction of stem with toe i.e. under B 2 = p B = 80.39 + (166.61 - 80.39) x 4.3/5.5 = 147.8kN/m Intensity of pressure at junction of stem with heel i.e. under C 2 =P c= 80.39 + (166.61 - 80.39) x 4.05/5.5 = 143.9 kN/m (b) Design of Toe slab Since the projection of the toe is small, it is designed as a cantilever fixed at the stem. 2 Intensity of pressure at B = 147.8 kN/m Neglecting the weight of soil above the toe slab, the forces acting on the toe slab are : (i) downward force due to weight of toe slab TB (ii) upward soil pressure on length AB. Ultimate moment at B, MB = L.F (moment due to soil pressure - moment due to wt. of slab TB = 1.5 [147.8 x 1.2 2/2 + (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. d =√ (174.57 x 106) / ( 2.76 x1000) = 251.49mm < d (=390mm ) . .. o.k. 2 2 Mu/bd =1.15, p t=0.343, A st= 1336mm Using # 16 mm bars, spacing = 1000 x 201/1335 = 150 mm However, the spacing is limited to 110 mm c/c from shear considerations. Provide #16 mm @ 110 mm c/c, Area provided =1827 mm 2, pt=0.47%
The bars shall be extended beyond the front face of the wall for a distance equal to development length of 750 mm (= 47 x 16) Distribution steel = 0.12 x 1000 x 450/100 = 540 2 mm 2 Provide #12 mm at 200 mm c/c. Area provided = 565 mm Check for Shear Since the soil pressure induces compression in the wall the critical section for shear is taken at a distance d from the face of the stem. Intensity of pressure at distance d (= 390 mm) from the face of the toe. pE = 80.39 + (166.61 - 80.39) (4.3 + 0.39)/5.5 = 153.9 kN/m 2 2
2
Net vertical shear = Shear due to pressure varying from 166.61 kN/m to 153.9 kN/m - Shear due to downward force of slab in length of 0.81 m (= 1.2 - 0.39) = (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= 181.05x10 6/1000x390=0.46 MPa ζc =0.47MPa for p t=0.47%, ... safe.
(c) Design of Heel Slab The heel slab is designed as a continuous slab supported on counterforts. The downward force will be maximum at the edge of the slab where intensity of soil pressure is minimum.
Consider 1 m wide strip near the outer edge D The forces acting near the edge are (a) Downward wt. of soil of height 7 . 8 m = 1 8 x7 . 8 x l = 140.4 kN/m (b) Downward wt. of heel slab = 25 x 0.45 x 1 = 11.25 kN/m 2 (c) Upward soil pressure of intensity 80.39 kN/m = 80.39 x 1 = 80.39 kN/m Net downward force at D = p = 140.4 + 11.25 - 80.39 = 71.26 kN/m Also net downward force at C = 140.4 + 11.25 - 143.9 = 7.75 kN/m Let the width of the counterfort = 400 mm Clear spacing between counterforts = l = 2.6 m Maximum -ve ultimate moment in heel slab at counterfort 2 2 = Mu= (L.F.) pl / 12 = 1.5 x 71.26 x 2.6 /12 = 60.2 kN.m. 2 6 2 Mu/bd =60.2x10 /(1000x390 )= 0.4, pt=0.114, provide 0.12%GA 2 ( Ast ) min = 0.12 x 1000 x 450/100 = 540 mm Provide # 12 mm @ 200 mm c/c, Area provided = 565 mm 2 Pt, = 100 x 565/ (1000 x 390) = 0.14 % Check for shear
Maximum shear = V umax = 1.5 x 71.26 x 2.6/2 = 139 kN
ζv= 139x10 3/1000x390=0.36 MPa ζc=0.28MPa. Unsafe and hence shear steel i s 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 ) Spacing = 290 mm Provide #8 mm 2-legged stirrups at 290 mm c/c. Since shear force varies linearly along the span of 2.6 m, the zone of design shear reinforcement can be determined. Let x1 be the distance from the counterfort where S.F. = 109.2 kN then x1 = 1.30 x (139-109.2) / 139= 0.28m Further in the transverse direction the S .F. decreases due to increase in t he soil pressure. Let the net down ward ultimate force/m at a distance y 1 from C be equal to w 1 Then ultimate S.F. at y1 = w1 x 2.60/2 = 1.30 w 1 . and this must be equal to V uc i.e. 1.30 w1 = Vu c = 109.2 . . w1= 84 kN
R 1250
A
1200 mm B
C
X
e
4050 mm
450 b/2 4050 mm
TOE
C
D 0 0 0 3
2600
HEEL
x1 139 SFD y1 71.28 kN/m
7.75 kN/m Net down force dia.
net downward working load = w' 1 = 84/1.5 = 56 kN/m. Now, variation of net downward force is linear having value of 71.26 kN/m. at D and 7.75 kN/m at C. Let y2 be the distance from C where net downward force is 56 kN/m.. . . y2 ./ (56 - 7 .75 ) = . . 4.05 / (17.26- 7 .75) . . y2 = 3.08 m from C. y1 = 4.05 - 3.08 = say 1 m from end D Provide # 8 mm 2-legged stirrups in heel slab at 290 mm c/c for a distance of 0.340 m on either side of the counterfort and for a length of 1 m along the length of the counterfort in the triangular portion. But from practical considerations provide the stirrups in the rectangular portion of (0.34 m x 1 m). Check for development length For a continuous slab the check for development length satisfying the curtailments rules as per SP34. Area of steel for +ve moment 2 Maximum +ve ultimate moment = L.F. x p/ /16 = 3/4 Mu = 0.75 x 60.2 = 45.15 kN.m.
Mu/bd 2=Very small Too small and hence provide minimum steel. A stmin(= 540 mm 2) Provide # 12 mm bars at 200 mm c/c., Area provided = 565 mm 2 > 540 mm2 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. Provide # 12 mm bars at 200 mm c/c. Distribution steel 2 Ast = 0.12 x 1000 x 450/100 = 540 mm Using # 12 mm bars, spacing = 1000 x 113/468 = 241 mm. Provide # 12 mm at 200 mm c/c. , Area provided = 565 mm 2 (d) Design of Stem (Vertical Slab) The stem acts as a continuous slab spanning between the counterforts. It is subjected to linearly varying earth pressure having maximum intensity at bo ttom. Consider 1 m wide strip at bottom of stem at C. The intensity of earth pressure 2
= PH = k a γ h =18
x 7.8/3=46.8 kN/m Area of steel on earth side near counterforts : Maximum -ve ultimate moment, 2
2
Mu = 1.5 x ph 1 /12 = 1.5 x 46.8 x 2.6 /12 = 39.54 kN.m. 6 Required d = √ (39.54 x 10 /(2.76 x 1000)) = 119 mm However provide total depth = 250 mm Assuming effective cover = 60 mm , d = 250 - 60 = 190 mm 2
2
Mu/bd =1.1, p t=0.33, Ast=627mm Provide #12 mm @ 180 mm c/c,
However provide #12 mm @ 110 mm c/c from shar. Area provided = 1000 x 113/110 = 1027.27 mm 2 P t= 100 x 1027.27/(100 x 190) = 0.54 % , As the earth pressure decreases towards the top, the spacing of the bars is increased with decrease in height. Distribution steel 2 Asl = 0.12 x 1000 x 250/100 = 300 mm Area of steel on each face = 300/2 = 150 mm 2 Provide # 8 mm @ 300 mm on each face in the vertical direction. 2 Area provided = 1000 x 50/300 = 1 67 mm
On the front face provided nominal steel φ 8 mm at 300 mm c/c to support the vertical bars.
(e) Design of Counterfort Width of counterfort = 400 mm. The counterforts are provided at 3 m c/c. They are subjected to earth pressure and downward reaction from the heel slab. At any section at any depth h below the top E the total horizontal earth pressure acting on the counterfort. = 1/2 y h 2 k x c/c distance between counterfort 2 2 = 18 x h x 3 x 1/6 = 9 h B.M. at any depth h = 9h 2xh/3 = 3h3 3 B.M. at the base at C= 3 x 7.8 = 1423-7 kN.m. Ultimate moment = Mu= 1.5 x 1423.7 = 2135.60 kN.m. Net downward pressure on heel slab at D = wt. due to earth pressure + wt. of heel slab - upward soil pressure = 18 x 7.8 + 25 x 0.45 - 80.39 = 71.26 kN/m Net downward pressure on heel slab at C 2 = 18 x 7.8 + 25 x 0.45 - 143.9 = 7.75 kN/m . Total down ward force at D = 71.26 x c/c distance = 71.26 x 3 = 213.78 kN.m. Total down ward force at C = 7.75 x c/c distance = 7.75 x 3 = 23.25 KN.m. . As mentioned earlier the counterfort acts as a T-b eam. As can be seen that the depth available is much more than required from B.M. considerations. Even assuming rectangular section, d =√(2135.6 x 106(2.76 x 400)) = 1390 mm The available depth is obtained as under :
7.8 m
d 4.05m
θ
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/bd 2=2135.6x106/(400x35352) =0.427, p t=0.12%, Ast=1696mm2
Ast.min = 0.85 bd/f y = 0.85 x 400 x 3535/415 = 2896 mm 2 Provided 4- # 22 mm + 4 - # 22 mm, Area provided = 3041 mm 2 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-L dt from top i.e, 2.79-1.03 =1.77m from top . Design of Horizontal Ties Due to horizontal earth pressure, the vertical stem has a tendency of separating out from the counterforts, Hence it should be tied to it by horizontal ties. The direct pull by the wall on counterfort for 1 m height at base = γhk ax c/c distance =18 x 7.8 x 3 x 1/3 = 140.4 kN Area of steel required to resist the direct pull 3 2 = 1.5 x 140.4 x 10 /(0.87 x 415) = 583 mm per meter height. 2 2 Using # 8 mm 2-legged stirrups, A st = 2 x π x 8 /4 = 100 mm spacing = 1000 x 100/583 = 170 m m c/c. Provide # 8 mm 2-legged stirrups 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 Due to net vertical down ward force acting on the base slab, it has a tendency to separate out from the counterfort. This is prevented b y providing vertical ties. The maximum pull will be exerted at the end of heel slab where the net downward force = 71.26 kN/m. Consider one meter strip Total downward force at D = 71.26 x c/c distance between counterforts = 71.28 x 3 = 213.78 kN. 3 2 Required A st = 1.5 x 213.78 x 10 /(0.87 x 415) = 888 mm 2 Using # 8 mm 2-legged stirrups , A st = 100 mm spacing = 1000 x 100/888 = 110 mm c/c. Provide # 8 mm 2-legged stirrups at 110 mm c/c. Total downward force at C= 3 x 7.75 = 23.25 kN Required A st = 1.5 x 23.25 x l0 3/(0.87 x 415) = 96.6mm 2 very less. Increase the spacing of vertical stirrups from 110 mm c/c to 450 mm c/c towards the end C.
250 mm
8-#22
#12@400
#12@200
1.77m
#8@110-450, VS
#12@ 110-300
8 - # 22 #8@170-450, HS
1250
1200 mm 450 #16@120
#12@200
#12@200
Cross sectional details of wall through the counterfort
250 mm
7000 8250 mm
#12@200 #12@200
1200 mm
1250
4050 mm
450 #16@120
#12@200
#12@200
Cross section between counterforts
8250
Backfill
With straight bars
Backfill
With cranked bars
Section through stem at the junction of Base slab.
