Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)
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Transcript of Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)
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RCC design B.C.Punmia
18.2 TYPE OF RETAINING WALLS
1 Gravity walls
2 Cantilever retaining walls a. T- shaped b. L- shaped
3 Counterfort retainig walls.
4 Buttresssed walls.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by w
A gravity retaining wall shown in fig 1 is the one in which the earth pressure exrted by the
back fill is resisted by dead weight of wall, which is either made of masonry or of mass concrete . The
stress devlop in the wall is very low ,These walls are no proportioned that no tension is devloped any
where, and the resultant of forces remain withen the middle third of the base.
A retaining wall or retaining structure is used for maintaining the ground surfgaces at defrent
elevations on either side of it. Whenever embankments are involed in construction ,retaining wall are
usually necessary. In the construction of buildins having basements, retaining walls are mandatory.
Similsrly in bridge work, the wing walls and abutments etc. are designed as retaining walls , to resist
earth pressure along with superimposed loads. The material retained or supported by a retaining wall
is called backfill lying above the horizontal plane at the elevation of the top of a wall is called the
surcharge , and its inclination to horizontal is called the surcharge angle b
In the design of retaining walls or other retaining structures, it is necessary to compute the
lateral earth pressure exerted bythe retaining mass of soil. The equation of finding out the lateral earth
pressure against retaining wall is one of the oldest in Civil Engineering field. The plastic state of
strees, when the failure is imminent, was invetigated by Rankine in1860. A Lot of theoretical
experiment work has been done in this field and many theory and hypothesis heve benn proposed.
RETAINING WALL
Retaining walls may be classified according to their mode of resisting the earth pressure,and
according to their shape. Following are some of commen types of retaining walls (Fig)
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y of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
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Hieght of cantilever wall from ground level = 7.00 mUnit weight of Earth = 18 KN/m
3
Angle of repose = 30 DegreeSafe Bearing capacity of soil = 180 KN/m
3
Coffiecent of friction = 0.5Concrete M- 20
wt. of concrete = 25000 N/m
3
Steel fe 415 N/mm sst 230 N/mm2m 13.33 scbc 7 N/mm2
Nominal cover = 30 mmFoundation depth = 1.00 m
Counter forts width = 0.50 m
Stem thickness At footing #REF! mm At top #REF! mm
Heel width 2000 mm Toe width 1700 mm
Footing width 4100 mm Key #REF! x #REF! mm
STEM : -
Main#REF! 10 mm F@ 90 mm c/c
#REF! 10 mm F@ #REF! mm c/c
Top 10 mm F@ #REF! mm c/c
Distribution 8 mm F@ #REF! mm c/c
Tamprecture 8 mm F@ #REF! mm c/c
TOE : -
Main 10 mm F@ 100 mm c/c
Distribution 8 mm F@ 0 mm c/c
HEEL : -
Main 10 mm F@ #REF! mm c/c
Distribution 8 mm F@ 180 mm c/c
DESIGN SUMMARY
100% Reinforcement upto m
50% Reinforcement upto m
25% Reinforcement upto m
DESIGN OF COUNTOR FORT RETAINING WALL with
horizontal back fill
Reinforcement Summary
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mm F
mm F @ c/c
@ c/c
mm F
mm F @ c/c
@ c/c
mm F
mm F @ c/c
@ c/c
mm F
@ c/c
mm F
@ c/c
mm F
@ c/c
mm F @ c/c mm F
@ c/c
## F mm F
@ c/c
mm F
@ c/c
mm F
@ c/c
1700
4100
####
####
#REF!
####
####
####
####
#REF!
####
1700
Toe
####
####
#### ####
Out side
####
####
Earth side
####
####
2000
400
10
####
####
####
10
####
10
####
####
####
####
####
Heel
2000
####
####
####
####
####
####
400
####
####
#REF!
####
####
####
####
#REF!
####
####
0
7000
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Hieght of cantilever wall from ground level = 7.00 m
Unit weight of Earth g = 18 kN/m = N/m
Angle of repose = 30 Degree
Safe Bearing capacity of soil q0 = 180 kN/m
Coffiecent of friction m = 0.5
Concrete = M 20Steel fe = 415
Nominal cover = 30 mm effective cover 40 mm
Foundation depth = 1.00 m
Counter forts width assume = 0.50 m 500 mm
1 Des ign Constan ts :For HYSD Bars = 20
sst = = 230 mm = #### N/mm2 = 25 kN/mm2scbc = = 7 mm
m = 13.33
x
13.33 x 7 + 230
j=1-k/3 = 1 - 0.289 / 3 = 0.904R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913
1 - 0.5 1
1 + 0.5 Ka
2Diamension of var ious parts : -
= 7.00 + 1.00 = 8.00 m
The ratio of length of slabe (DE) to base width b is given by eq.
q0
2.2 y H 2.2 x 18 x 8.00
Keep a = 0.43 Eq (1)The width of base is given by Eq.
( 1 - 0.43 )x( 1 + 1.29 )
x 8.00 x 0.33
( 1 - 0.43 )x 0.5
0.5 b = 0.50 x 8.00 = m
Hence Provided b = m
Width of toe slab = a x b = 0.43 x 4.00 = 1.72 m m
taking the uniform thickness of stem = 300 mm = 0.30 m
= 4.00 - 0.30 - 1.70 = 2.00 m
Let the thickness of base slab = 300 mm = 0.3 m
H 1/4 8.00 1/4
y 18
b
6.55
This width is excessive. Normal practice is to provide b between 0.5 to 0.6 H .
m
0.33
Provided toe slab =
3.50
The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .
1-sin F
1+sinF
=
for design purpose
Kp= =
180=
== 30 \ 3
for design purpose
3.84
.
m
1.70
m
3.5 x x
4.00
4.00
Hence width of heel slab
0.7HKa 0.7
=
Taking maximum value of H =
b =
2.86=
Ka
m*c+sst
=x
(1-a) m=
0.432
=
0.95 H
= 0.95
The base width from the considration of sliding is given by Eq.
b
m*c
=
Ka = =For soil, F
DESIGN OF COUNTERFORT RETAINING WALL
Cocrete M
wt. of concrete
13.33 7
k=
18000
= 0.289
a = 1 - = -1
8.00
(1- a)x(1+3 a)
0.333
Hence height of wall above base H
x
Clear spacing of counter fort =
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keep them at 3.00 m apart. Let us provide counterfort over toe slab, upto ground level at
3.00 m clear distance.
3
Full dimension wall is shown in fig 1a
Let w1 = weight of rectangular portion of stem
w2 = weight of base slab
w3 = weight of soil on heel slab.
The calculation are arrenged in Tableforce(kN) Moment about toe (KN-m)
w1 1 x 0.30 x 7.70 x 25 =
w2 1 x 0.30 x 4.00 x 25 =
w3 1 x 2.00 x 7.70 x 18 =
Sw =
Total resisting moment = kN-m ..(1)
0.33 x 18 x( 8.00 )2
2 2
8
3
mSw 0.50 x 364.95
PH
- 512 = kN-m
\ Distance x of the point of application of resultant, from toe is
SM 486.44 b 4.00
Sw 364.95 6 6
b 4.00
2 2SW 6 e 364.95 6x 0.67 183 > 180
b b 4.00
m, So that total wiodth is = 4.10
force(kN) Moment about toe (KN-m)
w1 1 x 0.30 x 7.70 x 25 =
w2 1 x 0.30 x 4.10 x 25 =
w3 1 x 2.00 x 7.70 x 18 =
Sw =
- = kN-m
\ Distance x of the point of application of resultant, from toe is
SM 523.00 b 4.10
Sw 365.70 6 6
b 4.10
2 2
SW 6 e 365.70 6x 0.62 170.1 < 180
b b 4.10
SW 6 e 365.70 6x 0.62 8.30 < 180
b b 4.10
1.50.95
kN -m2
==
=
Hence safe=4.10
Pressure p2 at Heel
The revised computations are arranged in table
Detail
1 + =
0.67
4.00
1.80
859
lever arm
58 1.95 113
31
277
1035.00
2.05 63
0.667
486.44
> 2Hence
safe
3.1
total MR365.70
x = = =
512.00 523.00
= 0.683
= Hence safe
Eccent ic i ty e =
kN -m2
< 0.683
Pressure p1 at toe= 1 +
Hence safe
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170.10 - 8.30
170.10 - 8.30
4
= 3.00 m
= 8.30 kN/m2
witch is minimum at C.
= 7.70 x 1 x 1.00 x 18 = kN-m2
= 0.30 x 1 x 1.00 x 25 = kN-m2
= 138.60 + 7.50 - 8.30 = N-m2
Maximum negative bending moment in heel slab.at counter fort is
Pl2 137.8 x 3.00
12
BM x 106
Rxb x
137.8 x 3.00
= 0.72 % tc = 0.33
V 207 x 1000
tc x b 0.33 x 1000
500 60 mm d = 500 - 60 = 440 mm
207 x 1000
1000 x 440
x
230 x 0.904 x 440
3.14xdia2 = 3.14 x 12 x 12 = 113 mm2
4 4
\ Spacing A x1000 / Ast = 113 x 1000 / 1130 = mm
Hence Provided 12 mm F bar, @ mm c/cLet us check this reinforcement for development length at point of contraflexure is situated at distance
of 0.211.L In over case, the slab is continuous, but we will assume the same position of contraflexure
i.e. at 0.211 x 3.00 = 0.63 m from the face of conunterforts.
pL l L l
2 2 2 2
3.00
2
Assuming that all the bars will avilable at point of contraflexure,
M = sst x Ast x j x d = 230 x 1130 x 0.904 x 440 = N-mmLo = witch ever is more = 440 mm
Ld = 45x F = 45 x 12 = 540 mm
M
V
440 mm beyond the point of contraflexure. After that, curtail
length of each bar available = 630 + 440 = 1070 > Ld = 540 m
PL2
3
16 4
3 3
4 4
M1
mm providing effective cover =
Hence depth required from shear point of veiw d
However keep =
=
These bars will be provide at the top face of heel slab. Maximum Passive B.M. =
- x
or d,
0.33
0.63
100
tv = N/mm2= 0.47 Shear reinforcement required
2'=
Effective depth required =
207 kN
N/mm2
= 626 mm=
1.80
119886
=
Design of Heel slab:-
The Pressure intencity p1 under E is p1
=
2
this is
excessive
103.35 kN-m
=
p1 =
For balance section , having P
mm0.913
>
=103.35
1000
Shear force V
336
Area of steel at supports is given by Ast
Shear force at this point is given
= 137.8
\ + Lo
99.07
138.6
87.20
=
The pressure distribution on the heel slab is shown in fig 1b .consider a strip 1 meter
170.10
137.8
170.10 -4.10
x
Down ward weight of slab per unit area
kN-m2
The Pressure intencity p2 under B is p2
p
Hence net pressure intensities will be P
- x
7.5
Clear spacing b etween co unter fort
kN-m2
wide.Near the outer edge C. The upward pressure intencity
Down ward load due to weight of Earth.
=M1
4.10=2.10=
x
Hence safe
N
=
- x + = p
=
=
12
-
100
mm210'
6103.35
=
=
103350000
+119886
103350000
using 12 mm bars
1130
A =
=
x 1130
540>1302
Hence safe
= mm2847
=
=
440
Cotinue these bars by a distance lo = d =
half bars, and continue the remaining half throughout the length. At the point of curtailment,
mm
\ Area of Bottom steel Ast2 = x Ast1
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3.14xdia2 = 3.14 x 12 x 12 = 113 mm2
4 4
\ Spacing A x1000 / Ast = 113 x 1000 / 847 = mm
Hence Provided 12 mm F bar, @ mm c/c
1000 x 113
Let us check this reinforcement for development length crierion at point of contraflexur,
M Where V = Shear at point of contraflexure= N
V = 0.63 mAssuming that all bars are available at point of contraflexure,
M = sst x Ast x j x d = 230 x 870 x 0.904 x 440 = N-mmLo = witch ever is more = 440 mm
Ld = 45x F = 45 x 12 = 540 mm as before
M
V
Thus continue all bottom bars to a point distance Lo = 440 mm from the point of contraflexure,
630 - 440 = 190 mm from the center of sports.
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.
Reinforcement near B :- The c/c spacing of reinforcement near B may be increased, because P decrease
due to increase in upward soil reaction. Consider a strip 1 m wide near B
kN/m2 As found earlier.
weight of earth + weight of counterforts - upward soil reaction
\ Net downward load p = 138.60 +( 0.50 x 25 ) - 87.2 = 63.9 kN/m2
This is about = 63.90 / 137.8 = 0.464 of load intencity at C
= 100 / 0.464 = 200 mm c/c at the top face, near supports
Spacing of steel bars at the bottom face, at mid span= 133 / 0.464 = 300 mm c/c
0.12
100
P D2
3.14 x ( 12 )'2
41000 x 113
shear stress at C = tv = 0.47
100 x 1130
1000 x 440
0.26 % = 0.21 (See Table 3.1)
Safe if tv< tc Here 0.47 > 0.21 Hence shear reinforcement required
Vc = tc b x d= 0.21 x 1000 x 440 = #### N or 92.4 kN
= 92.4 kN
92.4 1.50 - x1
207
= 0.80 m on either side of each counterforts.
The requirement is there form a strip of unit width paassing through C,
such that shear force at the counterforts isd = 92.4 kN
Net down ward pressure at C = 137.80 kN/m2, 63.90 kN/m
2
Lt net down ward pressure at B1=w1 x 3/2=1.5w1 This is equal to = 92.4 kN
92.40
1.50
137.80 - 6461.6= 137.8 =y1 - 36.95 Y1w1= -137.80
2.00
1.50
Net down ward pressure at B =
w= = 61.6 kN/m2 However at Y1 from C,
0.80 mThe position is given by= =
Permissible shear stress for
Hence shear stirrups are required upto distance
steel provided tc
=
500
Shear reinforcement.
=
\ Spacing = = 188
\ +Lo>Ld
i.e. upto a distance =
119886+ 440
+Lo>LdInherent in criterion :
>
Distribution steel = x 1000
Hence spacing of steel bars
Net downward load p' =
x
=79531705
119886
Distance from face of supports
Hence safe5401103=
=130
=
79531705
870 mm2
using 12 mm bars A =
% of steel provided =
Actual Ast
mm2
133
130
or d,
600
Upward soil reaction at B is = 87.2
mm2113
N/mm2
or x1 -
Consider a section distance x1 from face of counterfort, where shear force is
=1.50
=
Using 12 mm F bars, Area ==
600
=
4
mm say = 180 mm c/c
0.70
0.26 %
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\ y1 = 2.10 m
Hence shear reinforcement is required in triangular portion on the other side of counterforts shown hatched in fig .
However, we will provide shear strirrups in reangular portion x1 x y1= 0.80 x 2.10 = 1.68 m on
either side of counterforts.
Let us provide 4 8 mm F wire
P D2
3.14 x ( 8 )'2
4
201 x 230 x 440
207 - 92.40 )x 1000Hence provided the 8 mm F 4 lgd strirrups @ 170 mm c/c either side of each counterforts.
5Design of toe slab :- Since the toe slab is also large, provide counterforts over it, upto ground level at
3.00 m clear distance face to face. The toe slab will thus bend like a contious slab.
= 500 mm or 0.5 m
= 0.50 x 1 x 1 x 25 = kN//m2
= 170.10 - 12.50 = kN//m2
= 99.07 - 12.50 = kN//m2
Cosidering strip of unit width at D.
wl2 157.60 x 3.00
2
12
BM x 10 6
Rxb x
157.6 x 3.00
= 0.30 N/mm2 assuming % steel 0.5 % table 3.1
236 x 1000
1000 x 0.300
500 60 mm d = 500 - 60 = 440 mm
x230 x 0.904 x 440
3.14xdia2 = 3.14 x 12 x 12
4 4
\ Spacing A x1000 / Ast = 113 x 1000 / 1292 = mm
Hence Provided 12 mm F bar, @ 80 mm c/c
1000 x 113
Let us check this reinforcement for development length crierion at point of contraflexur,
M
V
Hence shear force at the point of contraflexure is V =
w L 3.00
2 2
M = 230 x 1413 x 0.904 x 440 = NmmLo = witch ever is more = 440 mm
Ld = 45x F = 45 x 12 = 540 mm
M
V
Hence satisfied , continue these bars, at the bottom of toe slab, beyond the point of contraflexure
440.0 mm i.e. by a distance of 630 + 440.0 = 1070 mm
mm359.8
1292
mm2=
mm=
=
m
178Sv=Asv.ssv.d
V - Vc
=
= = 201 mm2
4
legged stirrups of
Using 8 mm F bars, Area =
1000
The depth of slab required from shear point of view is given by d= V / (b x tc)
d =
=
Total weight of toe slab
=
= 236.4 kN
Taking a permissible stress tc
kN/m2
Similarly Net upward pressure intencity at E
= 10'6
86.57
118.2 x
-
10'6
This is excessive ,However we will keep the same depth as that of heel
Effective depth required =
Max. negative B.M.
Shear force V =
Assume total depth of toe slab
Net upward pressure intencity at D 157.60
12.50
2
118.20
0.913
12=
mmArea of steel at supports, at bottom is Ast =
=
However keep =
using 12 mm bars A = 113
118.20
1413 mm2
mm providing effective cover =
and provide shear strirrups to take up excessive shearing stress.
788 mm
N
Inherent in criterion :
or d,
\ +Lo>Ld =129239020.21
by a distance of Lo=
> 540 Hence safe137112
+ 440 = 1383
+Lo>Ld
87.5
Actual Ast = =80
0.63=Where the point of contraflexure occure at
distance x rom supports
V =
129239020
137112.0- x 157.6= x 0.63 =
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from the face of counterforts
3 3
4 4
3 3
4 4
3.14xdia2 = 3.14 x 12 x 12 = 113 mm2
4 4
\ A x1000 / Ast = 113 x 1000 / 969 = mm
Hence Provided 12 mm F bar, @ mm c/c1000 x 113
Let us check this reinforcement for development length crierion at point of contraflexur,
M
V
Assuming that all bars provided at top face,are available at point of contraflexure,
M = sst x Ast x j x d = 230 x 1028 x 0.904 x 440 = N-mmLo = witch ever is more = 440 mm
Ld = 45x F = 45 x 12 = 540 mm
M
V
Thus continue all bottom bars to a point distance Lo = 440 mm from the point of contraflexure,
630 - 440 = 190 mm from the center of sports.
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.
Reinforcement at E :- At a section distance 1 meter from E,
170.1 - 8.30
\ Net upward pressure = 138.5 - 12.50 = kN/m2
This is about = 126.0 / 157.60 = 0.80 ofw at D
\ Spacing of bottom steel = 87.5 / 0.80 = mm say = 100 mm
Spacing of top steel = 117 / 0.80 = mm say = 140 mm
0.12100
P D2
3.14 x ( 12 )'2
4
1000 x 113
Shear reinforcement shear force at D = kN
236.40 x 1000
Beam Ht.x beam wt. 1000 x 440
100 x 1413 table
1000 x 440 3.1
0.32 % = 0.24 (See Table 3.1)
Safe if tv< tc Here 0.54 > 0.24 Shear reinforcement required
Vc = tc b x d= 0.24 x 1000 x 440 = N or 105.6 kN
= 105.6 kN
105.6 1.50 - x2
236.40
= 0.80 m on either side of each counterforts.
The requirement is there form a strip of unit width paassing through D, Let us consider a strip through E1,
distance y2 from D, such that shear force at the counterforts is kN. To find the position ofY2
consider the net pressure distribution below the toe.
= 969
> 540 Hence safe
as before
N
137112
or d,
93992015+ 440 = 1126
110
137112
236.4
146
600
x 0.80 =
x
170.1
mm2
93992015
mm2
N-m
mm2
Where V = Shear at point of contraflexure=
kN/m2
126.0
109
-4.10
Inherent in criterion : +Lo>Ld
i.e. upto a distance =
upward soil pressure =
117
110Actual Ast = = 1028
= x
\ +Lo>Ld =
Spacing =
using 12 mm bars A =
\ Area of Bottom steel Ast2 = x Ast1 = x 1292
106x 118.20 x 10'
6
0.70
Using 12 mm F bars, Area =
1.50=
Hence shear stirrups are required upto distance
N/mm2
N/mm2
The position is given by= = or x2 = 1.50
180
= x 500
-
4= 113
600=
Permissible shear stress tc for
% of steel provided =
Consider a section distance x1 from face of counterfort, where shear force is
105600
tc
shear force=
N/mm2
=
=
Shear stress tv =
\ Spacing = 188 mm say
=
Distribution steel
\ =
1000
steel provided tc
= 0.32 %
= 0.54
138.50
mm c/c
0.24
0.80 m
mm2
105.6
88.7Again, positive B.M. x M1 =
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= 12.50 Hence net pressure intencity below D an dE are
below D = 170.1 - 12.50 = 157.6 kN/m2,
and below E 99.1 - 12.50 = 86.6 kN/m
Let the net pressure intencity at E1 = w2 x 3/2 = 1.5 w2 kN/m2
\ Shear force at the counterforts at E1 = w2 x 3/2 =1.5m w2 kN/M2
This is equal to = 105.6 kN
105.60
1.50
157.60 - 86.6
= 157.60 - 39.46 Y2 = \ y2 = 2.20 mThis is > than DE DE = 1.80 m
= kN/m2
1.50 x 86.57 = kN/m2
Considered a section distence Z from the face of dounterforts (Point E), where S.F. is 105.6 kN
105.60 1.50 - Z
129.85
= 0.30 m only.
However, we will provide shear strirrups for whole of rectangular area (shown dotted),
for width DD 1 ,= x2 = 0.80 m and length DE = 0.30 m
Let us provide 8 8 mm F wire
P D2
3.14 x ( 8 )'2
4
402 x 230 x 440
236.4 - 105.60 )x 1000
Hence provided the 8 mm F 8 lgd strirrups @ 300 mm c/c either side of each counterforts.
6Design of stem (vert ical slab)
The stem acts as a continuous slab. Considred 1 m strip at B .
The intencity of earth pressure is given by. ph = KayH1 = 0.33 x 18 x 7.50 = 45.0 kN/m2
Hence revised H1= 8.00 - 0.50 = 7.50 m
L2 45.00 x( 3 )2
12
BMRxb x
60 mm, so total depth = 192 + 60 = 252 mm
300 mm and effective thickness = 300 - 60 = 240 mm
this increased thickness will keep the shear stress within limit so that additional shear
45.0 x 3 67.50 x 1000 N
1000 x 240 mm2
this is less than = tc = 0.3
230 x 0.904 x 240
Reinforcement corresponding to p = 0.50 %
0.50 x 1000 x 240
P D2
3.14 x ( 12 )'2
4
1000 x 113
113 100 x 1256
90 1000 x 240%1000 = = 0.52
0.28
N/mm2
at 0.5% reinfocement (see Table 3.1)
kN \ =
=
mm c/c
= mm2
=
=
=
=0.913
Shear force V =2
=
Effective depth required =
w2= = 70.4
reinforcement not required.
Providing effective cover =
However provide total depth d =
12
Sv=
Actual shear forceat E =
=
(2)
Equating the two we get,
1.80Y2157.60However at Y2 from D, w2=
70.4
39.46157.6
mmV - Vc
Asv.ssv.d= =
= 402 mm2
4
129.8
The position of Z is given = 1.22
-
=1.50
=
- y2 =
1.50 -
is = pbd/100
100
=
mm say =
4
90
kN/m2
\
Hence shear stirrups are to be procided for a region DEE2D 1 , where EE2
B.M. ph x
=
legged stirrups of
311
=
mm2
Area of steel near conuterforts is
Actual AS provided= x 1256
Using 12 mm F bars, Area
1200
=
Spacing = = 94
m
N-mm
.(1)
= 0.30
= 677
113
mm2
= 1200
100As
bdmm
2and
=
Self weight of toe slab
105.6Hence shear force at E is more than
Using
or Z
8 mm F bars, Area
=
33750000
=
=
33750000
33750000 mm192
tc
1000
67.5
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Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion
M
V
0.211 x 3.00 = 0.63 m from the face of counterforts ,
shear force at this point given by V= pL/2(l/2-x)+(L/2)
pL l L l 3
2 2 2 2 2
= 58.7 kN
M = = 230 x 1256 x 0.904 x 240 = N-mmLo = 12 F or D , whichever is more = 240 mm
Ld = 45 F = 45 x 12 = 540 mm
M
V 58.70 x 1000
It is thus essencial to continue all the bars upto a point distance= 240 mm beyond
240 + 630 = 870 mm say = 900 mm from the
face of counterforts. These bars are to be provided at the inner face of stem slab.
3 x M1 3
4 4
3x Ast 3
4 4
P D2 3.14 x ( 12 )'2
4
1000 x 113
113 100 x 942
120 1000 x 942
Let us check these bars for devlopment length, near points of contraflexure, so as to satisfy the criterion
M
V
M = = 230 x 942 x 0.904 x 240 = N-mm
Lo = 12 F or D , whichever is more = 240 mm
Ld = 45 F = 45 x 12 = 540 mm V = 58.7 As before\ M
V 58.70 x 1000
The spacing of reinforcement at B , found above can be increased with height .
The pressurep h and hence the bending moment decreaases linearly with height.
Hence the spacing of bars can be increase gradually to say 300 mm c/c near top.
0.12
100
P D2
3.14 x ( 10 )'2
4
1000 x 79
7Design of main counterfort.
Let us assuming thickness of counterforts is = 500 mm. The counterfort will thus
be spaced @ 300 x 50 = 350 cm c/c. They will thus receive earth pressure from
3.5 3.5 m
At any section at depth h below the top A, the eerth pressure acting on each counter forts will be
1
3
62661343
1307 mm > 540 Hence safe+ Lo =62661343
sst xAst x jc xd
- x
240 =
= 67.5=
= + Lo > Ld
942
\ Ast
\ Spacing = = 120
540
=
mm2
+
For fixed beam or slab carrying U.D.L. , the point of
contraflexure is at a distance of 0.211 L
x - 0.63
+
Spacing =
= x
+46996007
a width of m and down ward reaction from heel slab for width of
240
79
1041
= 200
contraflexure point =
point of contraflexure, i.e. upto a point
Maximum positive B.M. = =
V = - x
Assuming that all the bars will be available at the point of contrflexure,
p x
N-mm
Area of steel = = = 1256 =
25312500x 33750000
942mm say = 120 mm c/c
Using 12 mm F bars, Area = = = 113 mm2
4
Actual AS provided= 1000 x = 942 mm2
and100As
= = 0.1 %bd
= + Lo > LdAssuming that all reinforcement is extended upto poin
of contraflexure.
mm2
4
sst xAst x jc xd 46996007
= 360
Using 10 mm F bars, Area = = =
mm2
Lo mm >=+ Hence safe
h
300
=
\ Distribution reinforcement = x 1000 x
mm c/c360
mm say= 218
x 3.5 = 21 h kN/m18.00 x h
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similarly, net down ward pressure on heel at c is
= 7.50 x 18 + 0.50 x 25 - 8.30 = kN/m2
and that at B = 7.50 x 18 + 0.50 x 25 - 87.2 = kN/m2
Hence reaction transferrred to each counterfort are will be
At C, = 139.2 x 3.50 = kN/m
At B, = 60.3 x 3.50 = kN/m
The variations of horizontal and vertical forces on counterfort are shown in fig.
The critical section for the counterfort will be F, since below this, enormous depth will be available to resist bending.
7.00 m is = 21 x 7.00 = 147 kN/m1
2
B.M. = 514.5 x 7 / 3 = kN-m or N-mm
Conterforts act as a T beam. However, even as a reactangular beam, depth required
BM
Rxb x
60 mm, so total depth = 1622 + 60 = 1682 mm
1700 mm and effective thickness = 1700 - 60 = 1640 mm
Angle F of faceAC is given by Tan F = 2.00 / 7.50 = 0.3 \ F = 15
\ sin F = 0.259 and cos F = 0.966
Depth F1 C1 = AF1 sin F = 7 x 0.259 = 1.82 m or mm
\ Depth FG = 1820 + 300 = 2120
Asssuming that the steel reinforcement is provided in 2 layer with mm space
between them and providing a nominal cover 30 mm and main bars of mm F dia
the effective depth will be = 2120 -( 30 + 20 + 12 + 10 = 2048 mm
230 x 0.904 x 2048
3.14xdia2 = 3.14 x 20 x 20
4 4
\ No. of bars 2820 / 314 = 9 2 layers
M d 2048
d' cos F 0.966
500 x 2120
100xAs 100 x 9 x 314
b xd x
the height h where half of the reinforcement can curtailed will be equal to HH = 8.00 = 2.8 m
below A, i.e. at point H. To locate the position of point of curtailmenton AC, draw Hl parallel to FG.
mm
beyond l, i.e. extented upto l1. The location of H corresponding to l1 can be locate by drawing line l1H1
parallel FG. It should be noted that l1G should not less than 45 F = 900 mm similarly, other bars can be
curtailed, if desired,
Design of Hor izontal t ies:-
The vertical stem has a tendency to saprate outfrom the counterforts, and hence
should be tie to it by horizontalties. At any depth h below the top, force causing sepration
1
3
here h = 7 \ force = 18 x 7 = 126 kN/m
x 18 h kN/mh x 3.00 = 18
However provide total depth d =
Providing effective cover =
Effective shear force
60.3
139.2
487.20
211.05
Pressure intencity at h =
Shear force at F = x 147 x 7.00 = 514.5 kN
1820
1200.5 1200500000
Effective depth required = =1200500000
= 1622 mm0.913 500
20
20
Area of steel at supports, at bottom is Ast = 2820 mm
using 20 mm bars A = = 314
1200500000=
=2120 mm
mm2
No. provode these in
where d' = ==
361606.1321\ Effective shear force = 514500 -
-Q tan F
N
tv = N/mm2
2120x 0.27 =1200500000
\361606.1321
= 0.341138
= 0.3 %
240Thus half bars can be curtailed at l. However these should be extent by a distance 12 F =
area of steel =500 2120
thus the shear stress tv is more than permisssible shear stress tc. However, the vertical
and horizontal ties provided in counterforts will bear the excess shear stress.
\ tc = 0.21 N/mm2
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126 x 1000
2x3.14
4
1000 x 157
200 300 mm at top
Design of vert ical t ies:-
Similar to the stem slab, heel slab has also tendency to seprate out from counterforts,due to net down ward force, unless tied properly by vertical ties.
487.20 x 3.00
211.05 x 3.00
417.6 x 1000
2x3.14
4
1000 x 226
180.9 x 1000
2x3.14
4
1000 x 226
120 280 mm
8Design of front counter for ts :-
170.1 99.07 kN/m2
at E.
500 mm thick toe slab = 0.5 x 25 = kN/m2
= 170.1 - 12.5 =kN/m
= 99.07 - 12.5 = kN/m2
The center to center spacing of counterforts, 500 mm wide is 3.50 m.Hence upward force transmitted
to counterforts at D 157.6 x 3.50 = kN/m and at E x 3.50 = 303 kN/m
The counterforts bent up as cantilever about face FE. Hence DF will be in compression while D 1 E1 will be
in tension, and main reinforcement will be provide at bottom face D 1 E1
( 551.6 + 303.0 )x
303 + 2 x 551.6 1.80
+ 3
\ B.M. = 769 x 0.99 = KN-m OR N-m
500 x 0.913
80 mm, so total depth = 1289 + 80 = 1369 mm
1400 mm and effective thickness = 1400 - 80 = 1320 mm
Thus project the counterforts 400 mm above groud level,to point F1 as shown in fig 4
230 x 0.904 x 1320
P D2
3.14 x ( 25 )'2
4
2767 provide these in one layer and continue by a=
=Using 25 mm F bars, Area = = 491 mm2
4
Area of steel near conuterforts is =759000000
= 2767 mm2
Providing effective cover =
However provide total depth d =
759 759000000
\ d =
759000000
= 1289 mm
551.63030.99 m from Eacting at x = x =
and at E 86.6
551.6
1.80 = 769
Down ward weight of 12.5
hence net w at D 157.6
Refer fig 1 The upward pressure intencity varies from kN/m2
at D, to
)2= 226 mm
2
kN/m
86.6
2.0Total upward force
mm
mm c / c at C toThus the spacing of vertical tie can be increase gradually from
\ spacing of ties = = 287 mm say 280
legged ties, As = x( 12
787
using 12 mm f 2
mm
steel required at B = = 787 mm230
mm1816 say 120\ spacing of ties = = 124
12 )2= 226 mm
2
1816 mm230
using 12 mm f 2 legged ties, As
steel required at C =
= x(
3.50417.6The down wars force ar C will be
The down wars force ar B will be 180.9
Near end C, the heel slab is tied to counterforts with the help of main reinforcement of counterforts.
kN/m3.50
548=spacing
however provide mm c/c at bottom and gradualy increase to
= 286 mm
=\ steel required
x( 10using 10 mm f 2 legged ties, As =
230
)2= 157 mm
2
= mm548
kN/m
see fig.
=
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491
M 0.90
d' 1.80
759 0.90
1.32 1.80
482 x 1000
500 x 1320
100xAs 100 x 6 x 491
b xd xtv > tc shear reinforcement is required
2x3.14
4
Vc = tcxbxd = 0.280 x 500 x 1320 =
V1 = V - Vc = - = N
230 x 226 x 1320
230 mm c/c provide 2 x 12 mm f bars on top for holding.
9Fixing effect in stem, toe and heel :-
At the junction of stem, toe and heel slab fixing moment are included,which
normal reinforcement given below may be provided.
(I) In stem@ 0.8x0.3 =0.24% of cross section, to be provided at inner face,
in vertical direction,for a length 45 F
0.24
100
P D2
3.14 x ( 10 )'2
4
1000 x 79
Length embedment in stem, above heel slab = 45 x 10 = 450 mm
(II) In toe slab @ 0.12% to be provided at the lowae face
0.12100
P D2
3.14 x ( 10 )'2
4
1000 x 79
Length embedment in stem, above toe slab = 45 x 10 = 450 mm
(III) In heel slab @ 0.12% to be provided in upper face
0.12
100
P D2
3.14 x ( 10 )'2
4
1000 x 79
Length embedment in stem, above heel slab = 45 x 10 = 450 mm
Each of above reinforcement should anchored properly in adjoining slab, as shown in fig 5
10. Design o f shear key:-
However providethese @
and tv = =
\ svVs
ssv.Asv.d= =
296700
0.730
= x =769 - kN
N/mm2
481500
% \ tc
12
= 0.280area of steel = = 0.45
500 1320
2 = x(
since
using 12 mm f
N
481500 184800 296700
)2= 226 mm
2
184800
table
3.1
legged ties, As
231
distance of 45 F beyond E
Effective shear force = Q - tan F
\
V
mmsubject to a maxi.
300 mm
N/mm2
From fig 4 tanF =
481.5
=
are at right angles to their normal direction of bending. These moment are not determine , but
Using 10 =
= 720=Ast x 1000
500 =
\ mm2x 300
= x 1000 x
mm F bars, Area = mm2
4= 79
= 100Spacing = = 109720
mm say mm c/c
600 mm2
Using 10 mm F bars, Area = = =
\
79
Ast
79 mm2
4
\ Spacing = = 131 mm say
\ Ast = x
= =
mm c/c600
1000 x 500 = 600 mm2
130=
= 130Spacing = = 131 mm say
Using 10 mm F bars, Area =
mm c/c600
The wall is in unsafe in sliding, and hence shear key will have to be provided, as shown in fig. 6
mm2
4
\
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Pp = KpP = 3.00 x 99.07 = kN/m2
Pp x a = a
18
2.00
or PH = 3.00 x( 8 + a )2
= 4.10 ax 18 = 73.80 a
\ SW = 364.95 + 73.80 a Refer force calculation table=
mSw+Pp 0.5 x ( 364.95 + 73.80 a) 297.20 a
3 x( 8.00 + a2 )
0.5 x ( 364.95 + 73.80 a) 297.20 a
1.5
( 8 + a)2 = 182.48 + 36.9 a+ 297.20 a
4.5
( 8 + a)2 = 40.55 + 74.24 a
64 + 16 a + a2 = 74.24 a- 16 a + 40.55 - 64
a2 = 58.24 a -23.45
or a = a2 - 58.24 a +
or a = 0.405 m say = 410 mm
= 400 mm. and width of key 400 mm
Now size of key = 400 x 400 mm
PH = 3.00 x( 8.00 + a )2
= 3.00 x( 8.00 + 0.40 )2
PH = kN
= a = x
= kN Hence
SW = 364.95 + 73.80 a
= 364.95 + 73.80 x 0.40
= kN
Actual force to be resisted by the key at F.S. 1.5 is = 1.5PH - mSW= 1.5 x 211.7 - 0.5 x 394.5 = kN
120.29 x 1000
400 x 1000
120.29 x 200 x 1000
1/6 x 1000 x( 400 )2
= N/mm2 Hence safe
The details reinforcement shown in fig 7
297.20 0.40
211.68
Sliding force at level D1C1 =
Hence keep depth of key
23.45
297.20
120.29
= 0.30 N/mm2
=
x3
0.90
394.5
118.88
Bending stress =
\ shear stress =
.(2)
a )2
297.20
x( 8 +1.5
+
Hence equilibrium of wall, permitting F.S. 1.5 against sliding we have
8
=a)2
PH=1.5
x(
Weight of the soil between bottom of the base and D 1C1
x0.33
Let the depth of key =a intensity of passive pressure Pp devloped in front of key depend upon
the soil pressure P in front of the key
297.20
\ total passive pressure Pp =
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0.3
`
0.3
7.50
7.00
3.00
0.3
2.00 D1
D E B C 1.00
2.00 E1 E B B
4.10 D y2
y2
w2
x2
63.9
0
1.80
FIG. 2
DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill
170.1
99.0
7
1.70
1.80
x2
8.3
0
FIG. 1
0.50
4.00
87.2
0
86.5
7
157.6
0
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0.30 A
F1
h
G.Level F
21 h kn/m
8.00 D F E7.50
G D1 E1
G1
F 147.00 kN/m
F1 F2 551.6
1.00 E B kN/m
C
168.00 kN/m
211.05 487.2kN/m 300
D E
mm F
mm c/c 450 D1
Pp
10 mm F
130 mm c/c
2.00
7.00
0.5
DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill
1.80
1
0.40
0.50
a
4.10FIG. 3
FIG. 4
0.51.00
0.50
FIG. 6FIG. 5
170.1
0
99.0
7
10
100
450
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300 300
12 mm F 10 mm F
300 mm c/c 300 mm c/c
12 mm F 1 10 mm F
240 mm c/c 265 mm c/c
12 mm F 10 mm F
180 mm c/c 230 mm c/c
8000
12 mm F 10 mm F
120 mm c/c 200 mm c/c
12 mm F 0 mm F
110 mm c/c 12 mm F 0 mm c/c
3 900 180 mm c/c 2x12 mm F
2 2
12 mm F2 lgd
230 mm c/c
3 12 mm F 12 mm F
12 mm F 130 mm c/c 80 mm c/c
130 mm c/c 400 1 25 mm F 400
0 mm F 6 No.Bars 0 mm
0 mm c/c 0 mm
400
DESIGN OF COUNTOR FORT RETAINING WALL with horizontal back fill
Cross -section mid way between co
2000
400
Holding bars
Cross -section mid way between counterfoorts
5001800
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mm
mm
Fron
tcounterforts
mm
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M-10 M-15 M-20 M-25 M-30 M-35 M-40
(N/mm2) Kg/m2
(N/mm2) Kg/m2
M 10 3.0 300 2.5 250
M 15 5.0 500 4.0 400
M 20 7.0 700 5.0 500
M 25 8.5 850 6.0 600
M 30 10.0 1000 8.0 800
M 35 11.5 1150 9.0 900
M 40 13.0 1300 10.0 1000
M 45 14.5 1450 11.0 1100M 50 16.0 1600 12.0 1200
M-10 M-15 M-20 M-25 M-30 M-35 M-40
Grade of concrete M-15 M-20 M-25 M-30 M-35 M-40
Modular Ratio 18.67 13.33 10.98 9.33 8.11 7.18
scbc N/mm2 5 7 8.5 10 11.5 13
m scbc 93.33 93.33 93.33 93.33 93.33 93.33
kc 0.4 0.4 0.4 0.4 0.4 0.4
jc 0.867 0.867 0.867 0.867 0.867 0.867
Rc 0.867 1.214 1.474 1.734 1.994 2.254
Pc (%) 0.714 1 1.214 1.429 1.643 1.857
kc 0.329 0.329 0.329 0.329 0.329 0.329
jc 0.89 0.89 0.89 0.89 0.89 0.89
Rc 0.732 1.025 1.244 1.464 1.684 1.903
Pc (%) 0.433 0.606 0.736 0.866 0.997 1.127
kc 0.289 0.289 0.289 0.289 0.289 0.289
jc 0.904 0.904 0.904 0.904 0.904 0.904Rc 0.653 0.914 1.11 1.306 1.502 1.698
Pc (%) 0.314 0.44 0.534 0.628 0.722 0.816
kc 0.253 0.253 0.253 0.253 0.253 0.253
jc 0.916 0.916 0.916 0.914 0.916 0.916
Rc 0.579 0.811 0.985 1.159 1.332 1.506
Pc (%) 0.23 0.322 0.391 0.46 0.53 0.599
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS
Grade of concrete
2.8 3.2 3.6 4.0 4.4
Table 1.16.. Permissible stress in concrete (IS : 456-2000)
Tensile stress N/mm2 1.2 2.0
Grade of
concrete
Permission stress in compression (N/mm2) Permissible stress in bond (Average) for
plain bars in tention (N/mm2)Bending acbc Direct (acc)
(N/mm2) in kg/m2
-- --
0.6 60
0.8 80
0.9 90
1.0 100
1.1 110
1.2 120
1.3 1301.4 140
Table 1.18. MODULAR RATIO
Grade of concrete
Modular ratio m31
(31.11)
19
(18.67)
13
(13.33)
11
(10.98)
9
(9.33)
8
(8.11)
7
(7.18)
Table 2.1. VALUES OF DESIGN CONSTANTS
(a) sst =
140
N/mm2
(Fe 250)
(b) sst =
190
N/mm2
(d) sst =
275
N/mm2
(Fe 500)
(c ) sst =
230N/mm2
(Fe 415)
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M-15 M-20 M-25 M-30 M-35 M-40
0.18 0.18 0.19 0.20 0.20 0.20
0.22 0.22 0.23 0.23 0.23 0.23
0.29 0.30 0.31 0.31 0.31 0.32
0.34 0.35 0.36 0.37 0.37 0.38
0.37 0.39 0.40 0.41 0.42 0.42
0.40 0.42 0.44 0.45 0.45 0.46
0.42 0.45 0.46 0.48 0.49 0.49
0.44 0.47 0.49 0.50 0.52 0.52
0.44 0.49 0.51 0.53 0.54 0.55
0.44 0.51 0.53 0.55 0.56 0.57
0.44 0.51 0.55 0.57 0.58 0.60
0.44 0.51 0.56 0.58 0.60 0.62
0.44 0.51 0.57 0.6 0.62 0.63
300 or more 275 250 225 200 175 50 or less
1.00 1.05 1.10 1.15 1.20 1.25 1.30
M-15 M-20 M-25 M-30 M-35 M-40
1.6 1.8 1.9 2.2 2.3 2.5
Grade of concrete M-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45 M-50
tbd (N / mm2) -- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4
tbd (N / mm2) tbd (N / mm2)
M 15
M 20
M 25
M 30
M 35
M 40M 45
M 50
Table 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)100As Permissible shear stress in concrete tc N/mm
2
bd
< 0.15
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00 and above
Table 3.2. Facork
Over all depth of slab
k
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)
Grade of concrete
tc.max
Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)
Table 3.5. Development Length in tension
Grade of
concrete
Plain M.S. Bars H.Y.S.D. Bars
kd = LdF kd = LdF
0.6 58 0.96
0.9 39 1.44
1 35 1.6
60
0.8 44 1.28 45
40
36
33
1.2 29 1.92 30
1.1 32 1.76
28
1.4 25 2.24 26
1.3 27 2.08
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7/28/2019 Counter+Fort++Reatining+WAll+With+HORZONTAL+Bach+Fill (1)
35/35
Degree sin cos tan tan Degree sin cos
10 0.17 0.98 0.18 0.18 10 0.17 0.98
11 0.19 0.98 0.19 0.19 11 0.19 0.98
12 0.21 0.98 0.21 0.21 12 0.21 0.98
13 0.23 0.97 0.23 0.23 13 0.23 0.97
14 0.24 0.97 0.25 0.25 14 0.24 0.97
15 0.26 0.97 0.27 0.27 15 0.26 0.97
16 0.28 0.96 0.29 0.29 16 0.28 0.96
17 0.29 0.96 0.31 0.31 17 0.29 0.96
18 0.31 0.95 0.32 0.32 18 0.31 0.95
19 0.33 0.95 0.34 0.34 19 0.33 0.95
20 0.34 0.94 0.36 0.36 20 0.34 0.94
21 0.36 0.93 0.38 0.38 21 0.36 0.93
22 0.37 0.93 0.40 0.40 22 0.37 0.93
23 0.39 0.92 0.42 0.42 23 0.39 0.92
24 0.41 0.92 0.45 0.45 24 0.41 0.92
25 0.42 0.91 0.47 0.47 25 0.42 0.91
30 0.50 0.87 0.58 0.58 30 0.50 0.87
35 0.57 0.82 0.70 0.70 35 0.57 0.82
40 0.64 0.77 0.84 0.84 40 0.64 0.77
45 0.71 0.71 1.00 1.00 45 0.71 0.71
50 0.77 0.64 1.19 1.19 50 0.77 0.64
55 0.82 0.57 1.43 1.43 55 0.82 0.57
60 0.87 0.50 1.73 1.73 60 0.87 0.50
65 0.91 0.42 2.14 2.14 65 0.91 0.42
Value of angle Value of angle