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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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b Surcharge anglem
mm F
Hieght of cantilever wall from ground level = 3.00 m @ c/c
Unit weight of Earth = 18 KN/m3 m
Angle of repose = 30 Degree mm F
Safe Bearing capacity of soil = 100 KN/m3 @ c/cCoffiecent of friction = 0.5 m
Concrete M- 20 25000 N/m3 mm F
cbc 7 N/mm2 m 13.33 @ c/cSteel fe 415 N/mm2 st 230 N/mm2 m
Nominal cover = 30 mm
Surcharge angle b 16 Degree mFounadation depth = 1.00 m Toe
Stem thickness At footing 310 mm At top 200 mm
Heel width 900 mm Toe width 1200 mm mm F
Footing width 2100 Key 300 x 300 mm @ c/c
Reinforcement Summary
STEM :- mm F @ c/c m
Main
2.58 12 mm F@ 90 mm c/c 8 F mm F
1.94 12 mm F@ 180 mm c/c @ c/c
Top 12 mm F@ 360 mm c/c
Distribution 8 mm F@ 160 mm c/c
Tamprecture 8 mm F@ 300 mm c/c mm F
TOE :- @ c/c
Main 12 mm F@ 120 mm c/c
Distribution 8 mm F@ 170 mm c/c mm F
HEEL :- @ c/c
Main 10 mm F@ 210 mm c/c
Distribution 8 mm F@ 170 mm c/c
wt. of concrete
3730
DESIGN OF T SHAPED CANTILEVER RETAINING WALL
with sloping back fill 8
3000
300 3000
170
180
3730
100% Reinforcement upto m
50% Reinforcement upto m
25% Reinforcement upto m
8
DESIGN SUMMARY
12
30
12
20
8
160
12
360
90
12
360
90
12
180
210
12
8
10
12
Heel
200270
1000
Out side
300
Earth side
2.58
1.94
9001200 300
8
170
540
730
160
2100
120
3730
200
3002.58
1.94
8
4000
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Hieght of cantilever wall from ground level = 3.00 m
Unit weight of Earth g = 18 kN/m = N/m
Angle of repose = 30 Degree
Safe Bearing capacity of soil q0 = 100 kN/m3
Coffiecent of friction m = 0.5 = 25 N/mmConcrete = M 20
Steel fe = 415
Nominal cover = 30 mm
Surcharge angle = 16 Degree
Founadation depth = 1.00 m
1 Design Constants:-For HYSD Bars = 20
st = = 230 N/mm = #### N/mm
cbc = = 7 N/mm3
m = 13.33x
13.33 x 7 + 230j=1-k/3 = 1 - 0.289 / 3 = 0.904
R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913
2 Diamension of base:-
sin b = 0.276 cos b = 0.96 tanb = 0.29Sin F = 0.5 Cos F = 0.87
cos b - cos2b -cos2f 0.961 - 0.92 - 0.75
cos b + cos2 b - cos2f 0.961 + 0.92 - 0.75
For surcharge wall, The ratio of length of slabe (DE) to base width b is given by eq.
q0
2.7 y H 2.7 x 18 x 4.00
The base width is given by Eq.
x
x x
( 1 - 0.49 )x( 1 + 1.46 )
x 4.00 x 0.38
( 1 - 0.49 )x 0.5
0.6 b = 0.60 x 4.00 = m
Hence Provided b = m
Width of toe slab = a x b = 0.49 x 2.40 = 1.17 m mLet the thickness of base be = H/12 = 4.00 / 12 = 0.33 or say = 0.30 m
= 2.40 - 0.30 - 1.20 = 0.90 m
3 Thickness of stem:-
= 4.00 - 0.30 = 3.70 m consider 1 m length of retaining wall
K x y x H12 0.38 x 18 x( 3.70 )
2
= 46.75 x 0.96 = 45 kN
46.75 Kn-m
=
Ka
Eq (1)
Hence the horizontal earth pressure is PH= P cos b
1.20
for design
purpose
=
The base width from the considration of sliding is given by Eq.
b
=
= 2.15
b
4.00
= 1 -
H
m
18000
100
=
= 0.289
2
0.7
=2
Hence width of heel slab
Heigth AB
Maximum Bending momentat B =
DESIGN OF T SHAPED CANTILEVER RETAINING WALL
Cocrete M
wt. of concrete
13.33 7m*c=k=
Provided toe slab =
m*c+sst
=
(1-a) m=
1
mb =
The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base .
4.13
This width is excessive. Normal practice is to provide b between 0.4 to 0.6 H .
Taking maximum value of H = 2.40
=0.7HKa
2.40
0.49-
Ka cos b
(1- a) x (1+3 a)
= cos b 0.96 0.38
0.38 0.961
= =x
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H1 3.70
3 3
BM 55.50 x 10 6
Rxb 0.913 x 1000
= 250 mm and total thickness = 250 + 60 = 310 mm
Assuming that 12 mm F bar will be used. a nominal cover of = 60 - 6 = 54 mm= 200 mm at top so that effective depth of = 140 mm
45 x 1000
1000 x 250
4
= 2.40 - 1.20 - 0.31 = 0.89 m
= 3.70 + 0.89 x 0.29 = 3.96 m
= 3.96 + 0.30 = 4.26 m
0.38 x 18 x( 4.26 )2
2 2
PH = P cos b = 61.84 x 0.96 = 59.44 kN
PV = P sin b 61.84 x 0.28 = 17.04 kN
Full dimension wall is shown in fig 1a
Let W1 = weight of rectangular portion of stem
w2 = weight of triangular portion of stem
w3 = weight of base slab
w4 = weight of soil on heel slab.
The calculation are arrenged in Table
force(kN) Moment about toe (KN-m)
w1 1 x 0.20 x 3.70 x 25 =
w2 1/2 x 0.11 x 3.70 x 25 =
w3 1 x 2.40 x 0.30 x 25 =
w4 1 x 0.90 x 3.83 x 18 =
w5 = 2.40
Sw =Total resisting moment = kN-m ..(1)
4
3
mSw 0.5 xPH
- 87.7 = kN-m
\ Distance x of the point of application of resultant, from toe is
SM 128.18 b 2.40Sw 120.64 6 6b 2.40
2 2
SW 6 e 120.64 6x 0.14 67.55 < 100b b 2.40
SW 6 e 120.64 6x 0.14 32.99 < 100b b 2.40
67.55 - 32.99
=
x =
Pressure p at the junction of stem with toe slab is
Pressure p1 at
toe
=59.44
x
x 1 +
1 -
=
Height H2= H1+Ls tan b
Height H
P is acting on vertical face IG, at H/3 and hence Pv , will act the vertical line
40.90Pv 17.04
Earth pressure p=
Its horizontal and vertical component are
26.085
lever arm
kN
N/mm2 > tc even at mimum steel
Length of heel slab
tv = = 0.18
PH x = 45.00 x = 55.5
2.401.20 50.27
=2.40 kN -m
2
Hence safe
kN-m2
Pressure p' at the junction of stem with Heel slab is
p = 67.55 -
1.06 m ==
1.01 2
..(2)
Over turning moment Mo = 87.7 kN-m
= 61.84
x =61.8
total MR120.64
18
62.01
1.2
1.95
1.41
247
215.89
=
1.255
Detail
mm
18.5
5.09
Reduce the total thickness
=Effective depth required
B.M. at B =
128.18
=Ka x y x H
2
-
Pressure p1 at
Heel= 1 -
Eccenticity e = x
F.S. against Sliding
215.89
Pressure distribution net moment SM =
Over turning
\ F.S. against over turning
Hence not safe , To make safe against sliding will have to provide shear key
0.4
Keep d
=
=
x=
-
=
=
= 1 + =
= Hence safe0.14 0.4
2.40=
m 540 Hence safe
0.12 270 + 200100
P D2 3.14 x ( 8 )'2
41000 x 50
6 Design of heel slab :-Three force act on it
2 weight of heel slab 3 Down ward earth pressure 4 upward soil pressure
3.70 + 3.96
3.70 + 2 x 3.96 0.90
+ 3
= 0.90 x 0.27 x 1 x 25 = kN
Acting at 0.45 m from B.
Earth pressure intencity at b = Ka.y.H1 per unit inclined area, at b to horizontal,\ Earth pressure at B, on horizontal unitarea = Ka.y.H1.tan b
Vertical component of this, at B = Ka.y.H1 .tan b.sin b .(I)
.(II)
Hence total force due to vertical component of earth pressure is
0.38 x 18
= 1.86 kN m from B
63.00
= 0.455 m from B
=2
Ka.y(H1+H2)
2
(H1+2xH2)xb
(H1+H2)x3=
x
Total weight of heel slab
Acting at3.70
Total weight of
soil over Heel=
2xx0.90
1. down ward weight of soil
Using
12 mm bars
And at underE
x 0.90
282 mm2
2
Spacing =
8 mm F bars, Area
Distribution steel
sst x j x D
= x x
=
3.96
4mm say =
=
170
b1 tan b x sin b
=
120
120
50
mm c/c
mm2
x
18 kNKN say
6.08
The reinforcement has to be provided at bottom face .If alternate bars of stem reiforcerment are
= 62
using
are bent and continued in toe slab, area available (see step 7)628
1000
Reduce the total thickness to
=
Effective depth required =
BM x 106
Keep effective depth d
x from E
=
tv
896
= 0.23=
0.63
39.09mm
2
tc even at mimum steel
m
N/mm2 0.14
7 Reinforcement in the stem:-We had earliar assume the thickeness of heel slab as = 0.30 m
0.27 m only. Hence revised H1= 4.00 - 0.27 = 3.73 m
ka.y 0.38 x 18
2
S.F. x H1 47.52 x 3.73
= 250 mm and total thickness = 250 + 60 = 310 mm
= 200 mm or m at edge
x
230 x 0.904 x 250
P D2
3.14 x ( 12 )
'2
4
1000 x 113
113
90
Bend these bars into toe slab, to serve as reiforcement there. Sufficient devlopment length ia available.
+ 250 - 140
PH
3
)2=
2.0047.52
kN-m
x( 3.73
Hence Safe
S.F at B = pcos b = H12 = kN =
Spacing =
Using 12
The effective depth d' at section is =
=Actual AS provided =
Between A and B some of bars can be curtailed. Cosider a section at depth below the top of stem
1000 x
H140 x
90 mm c/c
(where h In meter)
1256 mm2
h
mm24
= 991137
=mm F bars, Area =
mm say =
106
= 1137 mm2
= 113
=
Keep effective depth d
Reduce the total thickness to 0.20
Ast = BMx100/sstxjxD=
=
0.14 N/mm2
%
While it has now been fixed as
= 0.15
steel provided tc N/mm2Permissible shear stress for
Shear stress tv
B.M. at B3
x 100
=
=shear force
=
If tc > tvhence safe
= 50
Nomber of Bars =
Hence Provided 10
% of steel provided =
Using 8 mm F bars, Area =
mm c/c
mm c/c170
210
mm2
mm
2
=
x2
= 288
288
=
= 174
4
= x 1000
mm saySpacing =
Hence provided these @
from a distance of
59.08
= 78.5
Spacing =
10Using
Distribution steel
=
=4
mm F bars, Area =
366= 214 mm say
Ast = 15.97 10
6
sst x j x DBM x 10
6
=
m from B
=
200
< tc even at mimum steel
29.67
mm2
mm2
39.41
0.41
59.08
tv
Acting at =54.59
Total upward soil pressure
S.F. at B
mm and D
This is much lessthan the B.M. on slab. However, we keep the same depth, as that toe slab,i.e.
mm, reducing it to
N/mm2
32.99
= = 0.14
= 366
N-mm2
x =
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250 - 140
H3
d'
h Ast' d'1/3
H1 st
where Ast' = reinforcement at depth h Ast = reinforcement at depth H1
d' = effective depthat depth h d = effective depthat depth H1
Ast' 1 h 1 d'1/3
Ast 2 H1 2 d
Subsituting d = 255 =( 140 + 29.5 x h ) we get
140 + 29.5 x h 1/3
x
140 x 29.5 x hx
h = 0.467 x ( 140 + 29.5 x h 1/3 ..(3)
h = 2.83 m 0.467 x ( 140 + 29.5 x h)
- h =
Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.
\ h = 2.83 - 0.25 = 2.58 m. Hence curtailed half bars at at height of2.58 m below the top . If we wish to curtailed half of the remaining bars so that remaining
Ast' 1Ast 4
h 1 x d' 140 + 29.5 x h 1/3
H1 4 d x
x 140 x 29.5 x h4 x
h = 0.371 x ( 140 + 29.5 x h 1/3 ..(4)
h = 2.19 m 0.371 x ( 140 + 29.5 x h) - h =
.Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mmOr d = 250 mm whichever is more beyond the point.
\ h = 2.19 - 0.25 = 1.94 m. Hence stop half bars the remaining barsby 1.94 m below the top of the stem . Continue rest of the bars to the top of the stem
Check for shear:-
Shear force = 182
47.52 x 1000
1000 x 250
Ast/A = 1137 / 113 = 10.06 say = 11 No.
11 bars of mm F at Bottom11 x 113
1000 x 250
0.50 % = 0.3 (See Table 3.1)
here 0.30 > 0.19
Distribution and temprechure reinforcement:-
= 310 + 200
0.12
100
P D2 3.14 x ( 8 )'2
4
1000 x 50
= 8 = 300 mm c/cboth way in outer face
0.01
255
= H1
Hence
d' =( 140
=
Now As )1/3
=
Ast dor
x h
Ast H =(
140 x h )29.49++3.73
,,,'(1)
Hence Provided 12
255
x
h = 3.73
=\
2 255
mm and d'
x
h =
if Ast = than
H1
1/2 Ast
2
=
x
remaining reinforcement is one forth of that provided arB, we have
% of steel provided = x 100 = 0.50 %
Hence from .(2)=
Permissible shear stress for steel provided tc N/mm2
\ h
This can be solved bytrial and error, Noting that if the effective thickness of stem w=are constant,
x
0.00
\
h = 3.73
=4 255
= 0.38p = kNxkayH
2
23.73
2
=x
\ tc
If tc > tvhence safe Hence Safe
47.52
= 0.19 N/mm2 < (see table 3.1)
Average thickness of stem2
= 255
=tv
Nomber of Bars =
mm
\ Distribution reinforcement = x 1000 x 255 = 306
= = = 50.24 mm2
4
mm bars
mm c/c at the inner face ofwall,along its length
mm2
Using 8 mm F bars, Area
306= 164 mm say = 160
for tempreture reinforcement provide
\ spacing
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8 Design of shear key:-
300 x 310 Let Pp be the intensity of passive pressure devloped
in front of key this intencity Pp depend upon the soil pressure P in front of the key
Pp = KpP = 1/Ka= 1/ 0.38 = 2.64 x 50.27 = kN/m
Pp x a = x = 39.74 kN
18 `
2.00
or PH = 3.42 x( 4.53 )2x 0.961 = 67.22 kN
= 2.40 x 18 x 0.30 = 12.96 kN
\ W = 120.64 + 12.96 = kN Refer force calculation table=
mSw+Pp 0.5 x 133.60 + 39.74
= 300 mm. Keep width of key 310 mm (equal to stem width)
F where (45 + F/2) =2 shearing angle of passive resistance
\ a1 = 0.3 x ( 2.64 )1/2
a1 = 0.487 m = DE = 1.20 m
Hence satisfactory.
Now size of key = 300 x 310 mm
Actual force to be resisted by the key at F.S. 1.5 is = 1.5PH - mSW= 1.5 x 67.22 - 0.5 x
= kN
34.03 x 1000
300 x 1000
34.03 x 150 x 1000
1/6 x 1000 x( 300 )2
= N/mm2 Hence safe
Since concrete can take this much of tensile stress, no special reinforcement is necessary for the shear key
> 1.5 Hence safe
133.60
67.22= 1.58
Let u sprovide ashear key
0.30
The wall is in unsafe in sliding, and hence shear key will be provided below the stem as shown in fig.
132.46
\ total passive pressure Pp = 132.46
.(2)
Weight of the soil between bottom of the base and GJ
0.38 x x 4.53Sliding force at level GJ = cos bx
Hence equilibrium of wall, permitting F.S. 1.5 against sliding we have
PH=1.5 =
133.60
34.03
= 0.113 N/mm2
=
Actual length of the slab available
a kp45
Bending stress =
\ shear stress =
+
0.34
a tan F = a tana1 x
it should be noted that passive pressure taken into account above will be devloped only when length
a1 given below is avilable in front of key ;
However, provided minimum value of a
=
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b Surcharge angle b Surcharge angle b SurchA A A
m H1 m m H1= m m
toe heel
D E B C D E B C D E B
Toe Toe a1
m m
D1 e
p = pp
3.70
1.00
1.20
a b
W1
2.40
0.20
3.00 3.70
4.00H=
0.20
3.00
0.
0.20
3.00
2.40
W1
W2W2 0.90 0.00
2.40
P=
P=
0.30 0.30
b =b =
50.2
7
W1
2.40
1.20W2 0.90
67.5
5
50.2
7
54.5
9
32.9
9
67.5
5
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b Surcharge angleEarth side Face A Outer side face
mm F@ mm F `
@ c/c
m
mm F@
mm F@ c/c
mm F@
mm F@ c/c
mm FN.S.L. @
mm F mm F
@ c/c @
Heel Toe
Reinforcement Detail Reinforcemen
Foundation level
mm F mm F@ c/c @ c/c `
12
8
310
C/C
0.20
Outer sideEarth side Face
360
C/C
C/C
8
300 12
4.00H=
170
3.70
8
8
160
12
90
1.9
4
2.5
8
12
210 C/C
8
160 C/C
200
300
310
200
8
170120
180
9001200
1000
300
10
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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 13m 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
(d) sst =275
N/mm2
(Fe 500)
(c ) sst =
230N/mm2
(Fe 415)
(b) sst =190
N/mm2
Table 2.1. VALUES OF DESIGN CONSTANTS
(a) sst =140
N/mm2
(Fe 250)
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)
1.4 140
Table 1.18. MODULAR RATIO
1.1 110
1.2 120
1.3 130
0.8 80
0.9 90
1.0 100
2.8 3.2
-- --
0.6 60
4.4
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
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS
Table 1.16.. Permissible stress in concrete (IS : 456-2000)
Grade of concrete
Tensile stress N/mm2 1.2 2.0 3.6 4.0
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M-15 M-20 M-25 M-30 M-35 M-40 < %
0.15 % 0.18 0.18 0.19 0.20 0.20 0.20 %
0.25 % 0.22 0.22 0.23 0.23 0.23 0.23 %
0.50 % 0.29 0.30 0.31 0.31 0.31 0.32 %
0.75 % 0.34 0.35 0.36 0.37 0.37 0.38 %
1.00 % 0.37 0.39 0.40 0.41 0.42 0.42 %
1.25 % 0.40 0.42 0.44 0.45 0.45 0.46 %
1.50 % 0.42 0.45 0.46 0.48 0.49 0.49 %
1.75 % 0.44 0.47 0.49 0.50 0.52 0.52 %
2.00 % 0.44 0.49 0.51 0.53 0.54 0.55 %
2.25 % 0.44 0.51 0.53 0.55 0.56 0.57 %
2.50 % 0.44 0.51 0.55 0.57 0.58 0.60 %
2.75 % 0.44 0.51 0.56 0.58 0.60 0.62
3.00 and above % 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
100As
bd
Table 3.4. Permissible Bond stress Table bd in concrete (IS : 456-2000)
1.4 25 2.24 26
1.3 27 2.081.2 29 1.92 30
28
331.1 32 1.76
40
1 35 1.6 36
0.9 39 1.44
60
0.8 44 1.28 45
0.6 58 0.96
Grade of concrete
tc.max
Table 3.5. Development Length in tension
Grade of
concrete
Plain M.S. Bars H.Y.S.D. Bars
kd = LdF kd = LdF
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)
Table 3.2. Facor k
Over all depth of slab
k
Table 3.1. Permissible shear stress Table c in concrete (IS : 456-2000)
Permissible shear stress in concrete tc N/mm2
7/29/2019 Reatining WAll With SLOPING Bach Fill
16/16
Degree sin cos tan
10 0.174 0.985 0.176
15 0.259 0.966 0.268
16 0.276 0.961 0.287
17 0.292 0.956 0.306
18 0.309 0.951 0.325
19 0.326 0.946 0.344
20 0.342 0.940 0.364
21 0.358 0.934 0.384
22 0.375 0.927 0.404
23 0.391 0.921 0.424
24 0.407 0.924 0.445
25 0.422 0.906 0.466
30 0.500 0.866 0.577
35 0.573 0.819 0.700
40 0.643 0.766 0.839
45 0.707 0.707 1.000
50 0.766 0.643 1.192
55 0.819 0.574 1.428
60 0.866 0.500 1.732
65 0.906 0.423 2.145
Value of angle