Summary Sheet Session Number : 5 Date : 09.04.2007 Subject Expert :
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] 1
Design and Detailing of Counterfort
Retaining wall Dr. M.C. NATARAJA
2
Counterfort Retaining wall • • • • • •
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
CF
Stem Base Slab
CRW 3
Parts of CRW • Same as that of Cantilever Retaining wall Plus Counterfort
Stem
Counterforts
Heel
Toe Base slab
Cross section
Plan
4
Design of Stem • 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 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.
BF
p=Kaγh
5
Maximum Bending moments for stem Maximum +ve B.M= pl2/16 (occurring mid-way between counterforts) and Maximum -ve B.M= pl2/12 (occurring at inner face of counterforts) Where ‘l’ is the clear distance between the counterforts and ‘p’ is the intensity of soil pressure
l
+
p
6
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 upward soil reaction and is designed as a cantilever slab fixed at the front face of the stem. 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 and the slab is designed as a continuous horizontal slab spanning between the front counterforts.
H
b
7
Design of Heel Slab The heel slab is designed as a continuous slab spanning over the counterforts and is subjected to downward forces due to weight of soil plus self weight of slab and an upward force due to soil reaction. Maximum +ve B.M= pl2/16 (mid-way between counterforts) And Maximum -ve B.M= pl2/12 (occurring at counterforts)
BF
8
Design of 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 counterforts are designed as a T-beam of varying depth. • The main steel provided along the sloping face shall be anchored properly at both ends. • The depth of the counterfort is measured perpendicular to the sloping side.
C
T d
9
Behaviour of Counterfort RW -M
Important points
+M
•Loads on Wall COUNTERFORT STEM
•Deflected shape •Nature of BMs •Position of steel
-M
•Counterfort details
HEEL SLAB TOE
+M
10
PROBLEM -Counterfort Retaining Wall • 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 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. 11
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/mm2, H = 7 m above G.L, Depth of footing below G.L. = 1.25 m, γ = 18 kN/m 3, μ = 0.58, fb =SBC= 220 kN/m2
12
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 m Assume thickness of vertical wall = 250 mm Thickness of base slab = 450 mm
h1= 7m
H
1.25 m
b=5.5 m
13
Spacing of counterforts l = 3.5 (H/γ)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
l
Provide counterforts at 3 m c/c. Assume width of counterfort = 400 mm clear spacing provided = l = 3 - 0.4 = 2.6 m
14
Details of wall 250 mm CF: 3m c/c, 400 mm h =7 m 1
h=7.8 m
H=8.25 m
d 1.25m
T
1.2 m
4.05m
θ
b=5.5 m 15
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
Sr. No.
Description of loads
1
Total
ΣW = 679.25
ΣW =2210.71 16
250 mm W1
W3 H 8250
h1= 7000 ΣW R PA Df= 1250
A
1200 mm
B
C
450
4050 mm
PA D
H/3
W2 T
X
e b/3
kaH b/2 Pressure distribution
Cross section of wall-Stability analysis 17
Stability of walls Horizontal earth pressure on full height of wall = Ph = kaH2 /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 > 1.55 safe.
18
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 = 1.93 > 1.55 ... safe.
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. 20
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/m 2 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/m 2
21
250 mm
H 8250
ΣW
R PA
1250
A
1200 mm B
C
4050 mm
D
450
T
X
e
166.61 153.9 147.8 143.9 kN/m2
b/2 80.39 kN/m2
5500 mm 22
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. Mu/bd2= 1.14 < 2.76, URS
23
b) Design of Toe slab- Contd., To find steel pt=0.34% ζvsafe
25
Counterfort RW
-M
+M
COUNTERFORT STEM
-M HEEL SLAB TOE
+M 26
(c) Design of Heel Slab 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/m 2= 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)
27
166.61 153.9 147.8 143.9 kN/m2
80.39 kN/m2
5500 mm
7.75 kN/m 71.26 kN/m C
Forces on heel slab
D 28
To find steel Mu/bd2=60.2x106/(1000x3902)= 0.39 < 2.76, URS To find steel pt=0.114% 1390 mm Mu/bd2=2135.6x106/(400x35352) =0.427, pt=0.12%, Ast=1696mm2 Check for minimum steel
h =7.8 m
d 4.05m θ
37
Ast.min = 0.85 bd/fy = 0.85 x 400 x 3535/415 = 2896 mm 2 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 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.
38
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 Area of steel required to resist the direct pull = 1.5 x 140.4 x 103/(0.87 x 415) = 583 mm2 per m height. 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. 39
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 = 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 to 450 mm c/c towards the end C 40
DRAWING AND DETAILING COUNTERFORT RETAINING WALL
41
250 mm STEM
0-200mm COUNTERFORT
7000 #12@200
8250 mm
#12@200
1250
1200 mm
4050 mm
450 TOE #16@120 #12@200
#12@200
HEEL
Cross section between counterforts 42
250 mm
#12@400
8-#22
#12@200
1.77m
#8@110-450, VS
#12@ 110-300
8250
8 - # 22 #8@170-450, HS
1250
1200 mm 450 #16@120
#12@200
#12@200
Cross section through counterforts 43
STRAIGHT BARS
Backfill
STEM
Backfill
0.3l 0.25 l
With straight bars
With cranked bars
Section through stem at the junction of Base slab. 44
Backfill
Backfill
Cross section of heel slab 45
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) 46
Exam Problem (Dec. 2006) Design a T shaped cantilever retaining wall to retain earth embankment 3.2 m high above the ground level. The unit weight of the earth is 18 kN/m2 and its angle of repose is 30 degrees. The embankment is horizontal at it top. The SBC of soil is 120 kN/m 2 and the co-efficient of friction between soil and concrete is 0.5. Use M20 concrete and Fe 415 steel. Draw the following 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, µ=0.5, γ=18 kN/m 2, ө=30º, SBC= 120 kN/m2, M20 Concrete and Fe 415 steel Find H= h1 + Df 47
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