retain Wall Counterfort design concept .ppt

Retain Wall Counterfort

1. 1
Summary Sheet
Session Number :
Date :
Subject Expert :

5

09.04.2007

Dr. M.C. NatarajaDr. M.C. Nataraja
Professor
Department of Civil Engineering,
Sri Jayachamarajendra College of
Engineering,
Mysore – 570 006.
Phone:[phone]
E-mail: [email protected]

2. 2
Design and Detailing of Counterfort
Retaining wall
Dr. M.C. NATARAJA

3. 3
•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
Counterfort Retaining wall
CRW
CF
Base Slab
Stem

4. 4
Parts of CRW
•Same as that of Cantilever Retaining wall Plus
Counterfort
Stem
Toe Heel
Base slab
Counterforts
Cross section Plan

5. 5
•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.
Design of Stem
BF
p=K
a
γh

6. 6
Maximum Bending moments for stem
Maximum +ve B.M= pl
2
/16
(occurring mid-way between counterforts)
and
Maximum -ve B.M= pl
2
/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
+
-

7. 7
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.
b
H

8. 8
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= pl
2
/16
(mid-way between counterforts)
And
Maximum -ve B.M= pl
2
/12
(occurring at counterforts)
Design of Heel Slab
BF

9. 9
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.
TC
d

10. 10
Behaviour of Counterfort RW
-M
-M
TOE
COUNTERFORT
+M
+M
STEM
HEEL SLAB
Important points
•Loads on Wall
•Deflected shape
•Nature of BMs
•Position of steel
•Counterfort details

11. 11
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/m
2
is available
at a depth of 1.25 m below the ground level. The
weight of earth is 18 kN/m
3
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.

12. 12
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:
f
ck = 20 N/mm
2
, f
y = 415N/mm
2
, H = 7 m above G.L,
Depth of footing below G.L. = 1.25 m, γ = 18 kN/m
3
,
μ = 0.58, f
b
=SBC= 220 kN/m
2

13. 13
a. Proportioning of Wall Components
Coefficient of active pressure = k
a
= 1/3
Coefficient of passive pressure= k
p = 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
H
b=5.5 m
1.25 m
h
1
=
7 m

14. 14
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
 Provide counterforts at 3 m c/c.
Assume width of counterfort = 400 mm
 clear spacing provided = l = 3 - 0.4 = 2.6 m
l

15. 15
4.05m
h=7.8 m
θ
d
250 mm
1.2 m
b=5.5 m
H=8.25 mh
1
=7 m
1.25m
CF: 3m c/c,
400 mm
T
Details of wall

16. 16
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
W
1
25x0.25x1x7.8
= 48.75
1.2 + 0.25/2
=1.325
64.59
2
Weight of base
slab W
2
25x5.5x1x0.45
= 61.88
5.5/2 =2.75 170.17
3
Weight of earth
over heel slab W
3
18x4.05x1x7.8
= 568.62
1.45 +4.05/2
= 3.475
1975.95
Total ΣW = 679.25
ΣW
=2210.71
b. Check Stability of Wall

17. 17
A B C D
H
8250
250 mm
1200 mm 4050 mm
450
D
f= 1250
ΣW
P
A
R
eX b/2
T
W3
W1
W2
P
A
Pressure distribution
Cross section of wall-Stability analysis
b/3
k
a
H
H/3
h
1
= 7000

18. 18
Stability of walls
Horizontal earth pressure on full height of wall
= P
h = k
aH
2
/2 =18 x 8.25
2
/(3 x 2) = 204.19 kN
Overturning moment = M
0

= P
h x H/3 = 204.19 x 8.25/3 = 561.52 kN.m.
Factor of safety against overturning
= ∑ M / M
0
= 2210.71/561.52 = 3.94 > 1.55
 safe.

19. 19
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 / P
h = 393.97/204.19
= 1.93 > 1.55 .
.
. safe.

20. 20
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
= p
A
= ∑W / b ( 1+6e/b) = 679.25/5.5 ( 1+ 6*0.32/5.5)
= 166.61 kN/m
2
< f
b
(i.e. SBC= 220 kN/m
2
)
Minimum pressure at heel
= p
D = 80.39 kN/m
2
compression.

21. 21
Intensity of pressure at junction of stem with toe i.e. under
B
= p
B
= 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
=P
c
= 80.39 + (166.61 - 80.39) x 4.05/5.5 = 143.9 kN/m
2

22. 22
80.39
kN/m
2166.61
kN/m
2
143.9147.8153.9
A B C D
H
8250
250 mm
1200 mm
5500 mm
4050 mm
450
1250
ΣW
P
A
R
eX b/2
T

23. 23
b) Design of Toe slab
Max. BM
B = psf x (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.
M
u/bd
2
= 1.14 < 2.76, URS

24. 24
To find steel
p
t=0.34% <0.96%, A
st =1326 mm
2
,

# 16 @150
However, provide # 16 @110 from shear considerations.
Area provided =1827 mm
2
,

p
t=0.47%
Development length= 47 x 16=750 mm
Distribution steel = 0.12 x 1000 x 450/100 = 540 mm
2
Provide #12 mm at 200 mm c/c.
Area provided = 565 mm
2
b) Design of Toe slab- Contd.,

25. 25
Check for Shear
Critical section for shear: At distance d (= 390 mm)
from the face of the toe
p
E
= 80.39 + (166.61 - 80.39) (4.3 + 0.39)/5.5
= 153.9kN/m
2
Net vertical shear
= (166.61 + 153.9) x 0.81/2 - (25 x 0.45 x 0.81)
=120.7 kN.
Net ultimate shear = V
u.max = 1.5 x 120.7 =181.05
kN.
ζ
v= 181.05x 1000/1000x390 =0.46 MPa
p
t = 100 x 1827/ (1000 x 390) = 0.47 %
ζ
uc
= 0.36 + (0.48 - 0.36) x 0.22/0.25
= 0.47N/mm
2
> ζ
v
safe
d

26. 26
-M
-M
TOE
COUNTERFORT
+M
+M
STEM
HEEL SLAB
Counterfort RW

27. 27
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
M
u
= (psf) pl
2
/12 = 1.5 x 71.26 x 2.6
2
/12 = 60.2 kN-m (At the junction of CF)
(c) Design of Heel Slab

28. 28
71.26
kN/m
7.75
kN/m
DC
Forces on heel slab
80.39
kN/m
2166.61
kN/m
2
143.9147.8153.9
5500
mm

29. 29
To find steel
M
u/bd
2
=60.2x10
6
/(1000x390
2)=
0.39 < 2.76, URS
To find steel
p
t=0.114% <0.12%GA (Min. steel), <0.96%(p
t,lim.)
Provide 0.12% of GA
A
st= 0.12x1000x450/100 = 540 mm
2
Provide # 12 mm @ 200 mm c/c,
Area provided = 565 mm
2

p
t
= 100 x 565/ (1000 x 390) = 0.14 %

30. 30
Maximum shear = V
u,max
= 1.5 x 71.26 x 2.6/2 = 139 kN
For P
t
, = 0.14 % and M20 concrete, ζ
uc
= 0.28 N/mm
2

ζ
v= V
umax/bd =0.36 N/mm
2
,
ζ
uc
< ζ
v,
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.
Provide for 1m x 1m area as shown in figure
Check for shear (Heel slab)

31. 31
A B
C
1200
mm
4050
mm
450
1250
R
eX b/2
26003
0
0
0
4050
mm
C D
SFD
Net down force dia.
TOE
HEEL
139
71.28
kN/m
7.75
kN/m
x1
y1
Shear analysis and
Zone of shear steel
Area for
stirrups

32. 32
Area of steel for +ve moment
(Heel slab)
Maximum +ve ultimate moment = psf x pl
2
/16
= 3/4 M
u = 0.75 x 60.2= 45.15 kN-m.
M
u
/bd
2
=Very small and hence provide minimum steel.
A
st,min= 540 mm
2

Provide # 12 mm bars at 200 mm c/c.
Area provided = 565 mm
2
> 540 mm
2

33. 33
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
A
st
= 0.12 x 1000 x 450/100 = 540 mm
2

Using # 12 mm bars, spacing = 1000 x 113/468 = 241 mm.
Provide # 12 mm at 200 mm c/c.
Area provided = 565 mm
2

34. 34
(d) Design of Stem (Vertical Slab).
Continuous slab spanning between the counterforts and
subjected to earth pressure.
The intensity of earth pressure
= p
h = k
a γh =18 x 7.8/3=46.8 kN/m
2
Area of steel on earth side near counterforts :
Maximum -ve ultimate moment,
M
u = 1.5 x p
h l
2
/12 = 1.5 x 46.8 x 2.6
2
/12 = 39.54 kN.m.
Required d = √ (39.54 x 106/(2.76 x 1000)) = 119 mm
However provide total depth = 250 mm
M
u/bd
2
= 39.54x10
6
/1000x390
2
=1.1 < 2.76, URS

35. 35
To find steel: P
t
=0.34% <0.96%,
A
st=646 mm
2,
#12 mm @ 170 mm c/c,
However provide #12 mm @ 110 mm c/c,
Area provided = 1027.27 mm
2,
P
t
= 0.54 %.
As the earth pressure decreases towards the top, the spacing of
the bars is increased with decrease in height.
Max.ult. shear = V
umax
= 1.5 x 46.8 x 2.6/2 = 91.26 kN
For P
t, = 0.54 % and M20 concrete ζ
uc= 0.5 N/mm
2

ζ
v
= V
umax
/bd =91.28 x1000/(100X190)=0.48 N/mm
2
,
Shear steel is not needed and hence safe.

36. 36
At any section at any depth h below the top, the total
horizontal earth pressure acting on the counterfort
= 1/2 k
a
y h
2
x c/c distance between counterfort
= 18 x h
2
x 3 x 1/6 = 9 h
2

B.M. at any depth h = 9h
2
xh/3 = 3h
3
B.M. at the base at C= 3 x 7.8
3
= 1423.7 kN.m.
Ultimate moment = M
u
= 1.5 x 1423.7 = 2135.60 kN.m.
Counterfort acts as a T-beam.
Even assuming rectangular section,
d =√(2135.6 x 10
6
(2.76 x 400)) = 1390 mm
(e) Design of Counterfort

37. 37
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
M
u/bd
2
=2135.6x10
6
/(400x3535
2
)
=0.427, p
t=0.12%, A
st=1696mm
2
Check for minimum steel 4.05m
h =7.8 m
θ
d

38. 38
A
st.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
p
t
= 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.

39. 39
Design of Horizontal Ties
The direct pull by the wall on counterfort for 1 m height at
base
= k
a
γh 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 10
3
/(0.87 x 415) = 583 mm
2
per m height.
Using # 8 mm 2-legged stirrups, A
st
= 100 mm
2
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.

40. 40
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 A
st = 1.5 x 213.78 x 10
3
/(0.87 x 415) = 888 mm
2
Using # 8 mm 2-legged stirrups , A
st = 100 mm
2
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

41. 41
DRAWING AND DETAILING
COUNTERFORT RETAINING WALL

42. 42
250 mm
1200 mm 4050 mm
450
1250
#16@120#12@200 #12@200
#12@200
#12@200
8250 mm
7000
Cross section between counterforts
0-200mm
STEM COUNTERFORT
TOE HEEL

43. 43
250 mm
1200 mm
450
1250
#16@120 #12@200
8 - # 22
#12@200
#8@170-450, HS
#8@110-450, VS
#12@
110-300
#12@200
#12@400
8-#22
1.77m
8250
Cross section through counterforts

44. 44
Section through stem at the junction of Base slab.
Backfill
With straight bars
0.3l
0.25 l
Backfill
With cranked
bars
STEM
STRAIGHT
BARS

45. 45
Cross section of heel slab
Backfill
Backfill

46. 46
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)

47. 47
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: h
1
=3.2 m, µ=0.5, γ=18 kN/m
2
, ө=30º, SBC= 120
kN/m
2
,

M20 Concrete and Fe 415 steel
Find H= h
1
+ D
f

48. 48