RCC design B.C.Punmia
RETAINING WALL 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.
18.2 TYPE OF RETAINING WALLS 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) 1 2 3 4
Gravity walls Cantilever retaining walls Counterfort retainig walls. Buttresssed walls.
a. T- shaped
b. L- shaped
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.
The cantilever retaining wall resist the horizontal earth pressure as well as other vertical pressure by way of beending of varios com
vertical pressure by way of beending of varios components acting as cantilever s.A coomon form of cantilever retaining waal
DESIGN OF T SHAPED CANTILEVER RETAINING WALL with sloping back fill Hieght of cantilever wall from ground level = Unit weight of Earth = Angle of repose = Safe Bearing capacity of soil = Coffiecent of friction = Concrete M20 wt. of concrete sst Steel fe 415 N/mm2 2 scbc 7 m N/mm Nominal cover = b Surcharge angle Founadation depth = Stem thickness
2100
Key
m KN/m3 Degree KN/m3 3
N/m
N/mm2
8
At top Heel width
300
x
mm Degree
1000
200 mm 1200 mm 300 mm
12 12 12
mm F@ mm F@ mm F@
90 180 360
mm c/c mm c/c mm c/c
Distribution Tamprecture
8 8
mm F@ mm F@
160 300
mm c/c mm c/c
Main Distribution
12 8
mm F@ mm F@
120 170
mm c/c mm c/c
Main Distribution
10 8
mm F@ mm F@
210 170
mm c/c mm c/c
m 2.58 m 1.94
25% Reinforcement upto
m Top
12 mm F 180 @ c/c
540 200
200
12 mm F 120 @ c/c 1200 mm F 160 @ c/c 8F 300
300 2100 300
900 8 mm F 170 @ c/c
2.58
12 mm F 360 @ c/c
1.94
3730
12 mm F 180 @ c/c
TOE :-
12 90
HEEL :-
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mm F
90 @ c/c 10 mm F 210 @ c/c
730
m
8
50% Reinforcement upto
8 mm F 160 @ c/c
12
Reinforcement Summary STEM :Main 100% Reinforcement upto
mm F
300 @ c/c 3000 1.94 4000 8 mm F 300 @ c/c 3730 2.58 3000 3730 8 mm F 170 @ c/c
270
DESIGN SUMMARY At footing 0 mm Toe width 900 mm
Footing width
3.00 18 30 100 0.5 25000 230 13.33 30 16 1.00
b Surcharge angle 12 mm F 360 @ c/c
200
Out side
Earth side
mm F @ c/c
DESIGN OF T SHAPED CANTILEVER RETAINING WALL Hieght of cantilever wall from ground level Unit weight of Earth Angle of repose
q0
Safe Bearing capacity of soil Coffiecent of friction Concrete Steel Nominal cover Surcharge angle Founadation depth
m fe b
1 Design Constants:- For HYSD Bars
sst =
=
scbc = m
3 = 7 N/mm = 13.33
m*c+sst j=1-k/3 = 1 R=1/2xc x j x k = 0.5 2 Diamension of base:sin b Sin F
=
= = = = = = =
Cocrete M =
2 230 N/mm
m*c
k=
g
= = =
3.00 18 30 100 0.5
m kN/m3
M 415 30 16 1.00
20
-
0.289
x
7
= =
0.276 0.5
/
x 7
7 +
230
3
x 0.904
2 18000 N/m
Degree kN/m3 =
mm Degree m
2 = #### N/mm
= 0.289 = 0.904
x 0.289 = 0.913
cos b = Cos F =
0.96 0.87
tanb = 0.29
0.961 - 0.92 cos b - cos2 b -cos2 f = 0.96 x 0.961 + 0.92 cos b + cos2 b - cos2 f For surcharge wall, The ratio of length of slabe (DE) to base width q0 100 1 = 1 a = 2.7 y H 2.7 x 18 x 4.00 The base width is given by Eq. Ka cos b b = H x (1- a) x (1+3 a)
Ka
=
N/mm2
25
20
wt. of concrete
13.33 13.33 x
=
cos b
- 0.75 = 0.38 - 0.75 b is given by eq. =
0.49
Eq (1)
0.38 0.961 x = m 2.15 ( 1 - 0.49 )x( 1 + 1.46 ) The base width from the considration of sliding is given by Eq. 0.7HKa 0.7 x 4.00 x 0.38 b = = = 4.13 m (1-a) m ( 1 - 0.49 )x 0.5 This width is excessive. Normal practice is to provide b between 0.4 to 0.6 H . Taking maximum value of H = 0.6 2.40 b = 0.60 x 4.00 = m 2.40 Hence Provided b = m The wall will be unsafe against sliding. This will be made safe by providing a shear Key at base . Width of toe slab = a x b = 0.49 x 2.40 = 1.17 m Provided toe slab = 1.20 m Let the thickness of base be = H/12 = 4.00 / 12 = 0.33 or say = 0.30 m for design purpose Hence width of heel slab = 2.40 - 0.30 - 1.20 = 0.90 m b
=
3 Thickness of stem:Heigth AB
=
4.00
x
4.00 - 0.30 Kxy x H12 Maximum Bending momentat B = 2 Hence the horizontal earth pressure is PH= P cos b
[email protected]
= =
3.70 m consider 1 m length of retaining wall 0.38
x
= 46.75 x
18 x( 3.70 )2 2 0.96 = 45
= kN
46.75
Kn-m
H1 3.70 = 45.00 x = 55.5 kN 3 3 55.50 x BM 10 6 Effective depth required = = = 247 mm Rxb 0.913 x 1000 Keep d = = 250 + 60 = 310 mm 250 mm and total thickness Assuming that 12 mm F bar will be used. a nominal cover of = 60 - 6 = 54 mm Reduce the total thickness = 200 mm at top so that effective depth of = 140 mm 45 x 1000 tv = = 0.18 N/mm2 > tc even at mimum steel 1000 x 250 =
PH
Length of heel slab
=
2.40 -
1.20 -
0.31 =
0.89 m
Height H2= H1+Ls tan b
=
3.70 +
0.89 x
0.29 =
3.96 m
Height H
=
3.96 +
=
4.26 m
B.M. at B
x
4 Stability of wall:-
Earth pressure p=
Ka x y x H2 2
0.30 0.38
=
x
18
x( 4.26 )2
2
Its horizontal and vertical component are PH = P cos b = 61.84 x 0.96
= 59.44
=
61.84
kN
..(2)
kN
= P sin b 61.84 x 0.28 = 17.04 kN P is acting on vertical face IG, at H/3 and hence Pv , will act the vertical line PV
Let W 1 w2 w3 w4
Full dimension wall is shown in fig 1a weight of rectangular portion of stem weight of triangular portion of stem weight of base slab weight of soil on heel slab. The calculation are arrenged in Table Detail 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
w5 Total resisting moment Over turning
=
215.89
force(kN) = 18.5 = 5.09 =
18
lever arm Moment about toe (KN-m)
1.41 1.255
26.085 6.38
1.2
21.6
18 = Pv =
62.01
1.95
120.91
17.04
40.90
Sw =
120.64
2.40 total MR
215.89
kN-m
..(1)
Over turning moment Mo =
61.8
x
4 3
= 87.7
kN-m
215.89 = 2.46 > 2 87.71 120.64 0.5 x mSw F.S. against Sliding = = = 1.01 < 1.5 PH 59.44 Hence not safe , To make safe against sliding will have to provide shear key \
F.S. against over turning
=
Pressure distribution net moment SM = 215.89 87.7 = 128.18 kN-m \ Distance x of the point of application of resultant, from toe is b SM = 128.18 = 1.06 m = 2.40 = 0.4 x = 120.64 6 6 Sw b 2.40 x = - 1.06 = 0.14 m < 0.4 Eccenticity e = 2 2 Pressure p1 at 6e 120.64 6x 0.14 67.55 < 100 SW = 1+ = x 1+ = 2.40 b b 2.40 toe kN -m2 Pressure p1 at 6e 120.64 6x 0.14 32.99 < 100 SW = 1= x 1= 2.40 Heel b b 2.40 kN -m2 Pressure p at the junction of stem with toe slab is 67.55 - 32.99 p = 67.55 x 1.20 = 50.27 kN-m2 2.40 Pressure p' at the junction of stem with Heel slab is
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Hence safe
Hence safe Hence safe Hence safe
p' =
67.55
-
67.55
- 32.99 2.40
x
0.90
= 54.59
kN-m2
5 Design of toe slab:The upward pressure distribution on the toe slab is shown in fig 1b .The weight of soil above the toe slab is neglicted . Thus two forces are acting on it (1) Up ward soil pressure (2) Down ward weight of slab Down ward weight of slab per unit area = 0.30 x 1 x 1.00 x 25 = 7.50 kN-m2 Hence net pressure intensities will be = 67.55 - 7.50 = 60.05 kN-m2 under D And at under E = 50.27 - 7.50 = 42.77 kN-m2 Total force = S.F. at E = 0.50 x( 60.05 + 42.77 ) x 1.20 = 61.69 kN 42.77 + 2.00 x 60.05 x 1.20 x from E = = 0.63 m 60.05 42.77 + 3 B.M. at E 39.09 kN-m = 61.69 x 0.63 = \ 6 39.09 x BM 10 Effective depth required = = = 207 mm Rxb 0.913 x 1000 Keep effective depth d = = 210 + 60 = 270 mm 210 mm and total thickness Reduce the total thickness to = say = 0.27 m 200 mm or 0.20 m at edge tv
=
61.69 x
1000
= 0.23 1000 x 270 39.09 x 10 6 230 x 0.904 x 210
N/mm2
<
tc even at mimum steel
BM x 106 = = 896 mm2 sst x j x D The reinforcement has to be provided at bottom face .If alternate bars of stem reiforcerment are
Ast =
are bent and continued in toe slab, area available = 1/2 628 mm2 (see step 7) x 1256 = 2 = 3.14 x 12 x 12 = 113 mm2 3.14xdia using 12 mm bars A = 4 4 120 Spacing A x1000 / Ast 113 x mm \ = 1000 / 896 = Hence Provided 120 mm c/c 12 mm F bar, @ 45 x 12 = 540 mm Let us check this reinforcement for development length Ld=45 F = Providing 30 mm clear side cover actual length available = 1200 30 = 1170 mm 1170 > Hence safe 540 270 + 200 0.12 Distribution steel = x 1000 x = 282 mm2 2 100 '2 2 3.14 x ( 8 ) Using 8 = 50 mm F bars, Area = P D = mm2 4 4 1000 x 50 mm c/c = 178 mm say = 170 \ Spacing = 282 6 Design of heel slab :Three force act on it 2 weight of heel slab 3.70 + 3.96 0.90 x
1. down ward weight of soil Total weight of soil over Heel Acting at
=
2
(H1+2xH2)xb
(H1+H2)x3
Total weight of heel slab
=
2 +
3.70 + 3.70
3 Down ward earth pressure 4 upward soil pressure
x
18
=
62
KN say =
0.90 x 3.96 = x 3.96 3 1 x 25 = x
0.455
63.00
kN
m from B
6.08 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 Similarly, Vertical component of earth pressure intencity at C =Ka.y.H2 tan b. sin b. =
0.90 x
0.27
….(I) ….(II)
Hence total force due to vertical component of earth pressure is (H1+H2) = Ka.y x b1 tan b x sin b 2 18 0.38 x = ( 3.70 + 3.96 )x 0.90 x 0.287 x
2
vertical earth pressure is =
[email protected]
1.86
kN
This Act at
0.45
0.28
m from B
Total upward soil pressure Acting at
= =
\ Total force 63 B.M. at B =(
1/2 x( 54.59 2 54.59 + 54.59 + S.F. at B 0.45 )+( 6 15.97 x 10'6
+ 32.99 )x 0.90 = x 32.99 0.90 x 32.99 = 63.00 + x 0.45 )+(
39.41
kN
= 0.41 m from B 3 6.08 - 39.41 = 29.67 kN 0.41 ) 1.86 x 0.45 )-( 39.41 x
= x = N-mm2 This is much lessthan the B.M. on slab. However, we keep the same depth, as that toe slab,i.e. mm and D mm, reducing it to 200 mm at edge 270 d= 210 6 15.97 BM x 106 x 10 = Ast = = 366 mm2 sst x j x D 230 x 0.904 x 210 29.67 x 1000 tv = = 0.14 N/mm2 < tc even at mimum steel 1000 x 210 '2 2 3.14 x ( 10 ) Using 10 mm F bars, Area = P D = = 78.5 mm2 4 4 1000 x 78.5 mm c/c = 214 mm say = 210 \ Spacing = 366 Hence provided these @ 210 mm c/c at the top of keep slab. Take the reinforcement into toe from a distance of 45 450 mm to the left of B and end should x 10 = 45 x D = hook 210 + 270 0.12 Distribution steel = x 1000 x = 288 mm2 2 100 '2 3.14 x ( 8 ) P D2 Using 8 = = 50 mm F bars, Area = mm2 4 4 1000 x 50 mm c/c = 174 mm say = 170 \ Spacing = 288 Nomber of Bars = Ast/A = 366 / 79 = 4.66 say = 5 No. mm F at Bottom Hence Provided 5 bars of 10 5 x 79 % of steel provided = x 100 = 0.15 % 1000 x 270 shear force 29.67 x 1000 Shear stress t v = = = 0.14 N/mm2 Beam Ht.x beam wt. 1000 x 210 Permissible shear stress for 0.15 % steel provided t c = 0.18 N/mm2 (See Table 3.1) Hence Safe If tc > tvhence safe here 0.18 > 0.14
\
7 Reinforcement in the stem:We had earliar assume the thickeness of heel slab as = While it has now been fixed as 0.27 m only. Hence revised H1= 4.00 ka.y 2 S.F. x B.M. at B = 3 Keep effective depth d = Reduce the total thickness to = S.F at B = pcos b =
H12 H1
=
0.38 x 2.00
18
x( 3.73 )2= 47.52
0.30 m 0.27 kN
= 3.73 m =
PH
47.52 x 3.73 = 59.08 kN-m 3 = 250 + 60 = 310 mm 250 mm and total thickness 0.20 m at edge 200 mm or 59.08 x 10 6 Ast = BMx100/sstxjxD= = 1137 mm2 230 x 0.904 x 250 '2 3.14 x ( 12 ) P D2 Using 12 mm F bars, Area = = = 113 mm2 4 4 1000 x 113 = 99 mm say = 90 mm c/c \ Spacing = 1137 113 Actual AS provided = 1000 x = 1256 mm2 90 Bend these bars into toe slab, to serve as reiforcement there. Sufficient devlopment length ia available. Between A and B some of bars can be curtailed. Cosider a section at depth below the top of stem + 250 - 140 The effective depth d' at section is = 140 x h (where h In meter) H
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=
d' Now As
=
140
+ H3 d'
Ast
250
- 140 3.73 or
x
h
=( 140
+ 29.49 x h )
,,,'(1)
H =( Ast d )1/3
1/3
Hence where
Ast' =
Ast' d' h = Ast d H1 reinforcement at depth h
d' =
Ast = reinforcement at depth H1 d = effective depthat depth H1
effective depthat depth h Ast' 1 h 1 d' 1/3 if Ast = 1/2 Ast than = x \ Ast = 2 H1 2 d Subsituting d = 255 mm and d' =( 140 + 29.5 x h ) we get 140 + 29.5 x h 1/3 H1 h = x 2 255 x 1/3 140 x 29.5 x h h = 3.73 x 2 255 x h = 0.467 x ( 140 + 29.5 x h )1/3 ..(3) 1/3 h) h = = m 0.467 x ( 140 + 29.5 x 2.83 0.01 h Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mm Or d = 250 mm whichever is more beyond the point. h = 2.83 - 0.25 = 2.58 m. Hence curtailed half bars at at height of \ 2.58 m below the top . If we wish to curtailed half of the remaining bars so that remaining Ast' 1 remaining reinforcement is one forth of that provided ar B , we have = Hence from ….(2) Ast 4 1/3 1/3 h 1 d' x 140 + 29.5 x h H1 x = \ \ h = 4 255 x H1 4 d 1/3 x 140 x 29.5 x h h = 3.73 255 4 x h = 0.371 x ( 140 + 29.5 x h )1/3 ..(4) 1/3 2.19 m 0.00 0.371 x ( 140 + 29.5 x h) - h = h = This can be solved bytrial and error, Noting that if the effective thickness of stem w=are constant, .Howerver, the bars should be extented by a distance of 12 F = 12 x 12 = 144 mm Or d = 250 mm whichever is more beyond the point. h = 2.19 0.25 = 1.94 m. Hence stop half bars the remaining bars \ by 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 =
2 18 kayH 2 47.52 kN = 0.38 x x 3.73 2 2 = 47.52 x 1000 tv tc (see table 3.1) = = 0.19 < \ N/mm2 1000 x 250 Nomber of Bars = Ast/A = 1137 / 113 = 10.06 say = 11 No. mm F at Bottom Hence Provided 11 bars of 12 11 x 113 % of steel provided = x 100 = 0.50 % 1000 x 250 Permissible shear stress for 0.50 % steel provided t c = 0.3 N/mm2 (See Table 3.1) Hence Safe If tc > tvhence safe here 0.30 > 0.19 Distribution and temprechure reinforcement:Average thickness of stem = 310 + 200 = 255 mm 2 0.12 Distribution reinforcement = x 1000 x 255 = 306 \ mm2 100 '2 3.14 x ( 8 ) P D2 Using 8 = = 50.24 mm2 mm F bars, Area = 4 4 1000 x 50 \ spacing = = 164 mm say = 160 mm c/c at the inner face of 306 wall,along its length for tempreture reinforcement provide = mm bars = 300 mm c/c both way in outer face 8
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p=
8 Design of shear key:The wall is in unsafe in sliding, and hence shear key will be provided below the stem as shown in fig. Let u sprovide ashear 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 = 0.38 = 2.64 x 50.27 KpP = 1/Ka= 1/ 132.46 kN/m3 = \ total passive pressure Pp = Pp x a = 132.46 0.30 x = 39.74 kN 18 ` Sliding force at level GJ = 0.38 x x 4.53 x cos b 2.00 or PH ….(2) = 3.42 x( 4.53 )2x 0.961 = 67.22 kN Weight of the soil between bottom of the base and GJ = 2.40 x 18 x 0.30 = 12.96 kN 133.60 120.64 + 12.96 = kN Refer force calculation table \ SW = against sliding we have Hence equilibrium of wall, permitting F.S. = 1.5 m Sw+Pp 0.5 x 133.60 + 39.74 1.5 = = = 1.58 > 1.5 Hence safe PH 67.22 However, provided minimum value of a = 300 mm. Keep width of key 310 mm (equal to stem width) 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 ; F where (45 + F/2) = a1 = a tan F = a tan x 45 + = a kp shearing angle of passive resistance 2 a1 = 0.3 x ( 2.64 )1/2 \ a1 = Actual length of the slab available = DE = 1.20 m 0.487 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 133.60 = 34.03 kN 34.03 x 1000 = 0.113 N/mm2 \ shear stress = 300 x 1000 34.03 x 150 x 1000 Bending stress = 1/6 x 1000 x( 300 )2 0.34 = Hence safe N/mm2 Since concrete can take this much of tensile stress, no special reinforcement is necessary for the shear key
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0.20
0.20
0.20
b Surcharge angle
b Surcharge angle
A
b Surcharge angle
A
A
H= 4.00
H= 4.00
H1=3.70 m
3.00 m
H1=
3.00 m
W1
W2
2.40 0.00
0.90
E
heel C
B
D
W1
W2
2.40 1.20
0.90
E
B
0.30
C
D
E
0.90 B
a1 m 32.99
2.40 P= 54.59
P= 50.27
b= 67.55
m
C 0.30
Toe 2.40
W2
0.30
Toe b=
3.70 m
W1
67.55
toe 1.00 D
3.00 m
Kay(H+a)
a D1
e Pp = 50.27
ab 1.20
3.70 m
Kpp
C1
0.20 b Surcharge angle A Outer side face
Earth side Face 8 mm F @ 300 C/C
` 8 mm F @ 300 c/c
4.00 m
1.94
H=
12 mm F @ 360 c/c 8 mm F @ 160 C/C 2.58
8 mm F 3.70 @ 160 c/c
8 mm F @ 300 c/c
12 mm F @ 180 C/C
mm F @ 300 c/c
8
12 mm F @ 90 C/C
N.S.L.
8
mm F
10 mm F @ 210 C/C
@ 170 c/c #REF!
310 Heel
Toe
200
Earth side Face Reinforcement Detail
200 Foundation level 12 mm F @ 120 c/c 1200
300
310
8 mm F @ 170 c/c 900
`
Outer side face Reinforcement Detail
mm F c/c
mm F c/c
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESS Grade of concrete
Tensile stress N/mm2
M-10
M-15
M-20
M-25
M-30
M-35
M-40
1.2
2.0
2.8
3.2
3.6
4.0
4.4
Table 1.16.. Permissible stress in concrete (IS : 456-2000) Permission stress in compression (N/mm2) Permissible stress in bond (Average) for 2 Bending acbc plain bars in tention (N/mm ) Direct (acc)
Grade of concrete M M M M M M M M M
10 15 20 25 30 35 40 45 50
(N/mm2) 3.0 5.0 7.0 8.5 10.0 11.5 13.0 14.5 16.0
Kg/m2 300 500 700 850 1000 1150 1300 1450 1600
(N/mm2) 2.5 4.0 5.0 6.0 8.0 9.0 10.0 11.0 12.0
Kg/m2 250 400 500 600 800 900 1000 1100 1200
(N/mm2) -0.6 0.8 0.9 1.0 1.1 1.2 1.3 1.4
2
in kg/m -60 80 90 100 110 120 130 140
Table 1.18. MODULAR RATIO Grade of concrete
Modular ratio m
M-10 31 (31.11)
M-15 19 (18.67)
M-20 13 (13.33)
M-25 11 (10.98)
M-30 9 (9.33)
M-35 8 (8.11)
Table 2.1. VALUES OF DESIGN CONSTANTS Grade of concrete Modular Ratio
scbc N/mm2 m scbc kc jc Rc Pc (%) kc (b) sst = jc 190 Rc N/mm2 Pc (%) (c ) sst = kc jc 230 N/mm2 Rc (Fe 415) Pc (%) kc (d) sst = jc 275 N/mm2 Rc (Fe 500) Pc (%) (a) sst = 140 N/mm2 (Fe 250)
M-15 18.67 5 93.33 0.4 0.867 0.867 0.714 0.329 0.89 0.732 0.433 0.289 0.904 0.653 0.314 0.253 0.916 0.579 0.23
M-20 13.33 7 93.33 0.4 0.867 1.214 1 0.329 0.89 1.025 0.606 0.289 0.904 0.914 0.44 0.253 0.916 0.811 0.322
M-25 10.98 8.5 93.33 0.4 0.867 1.474 1.214 0.329 0.89 1.244 0.736 0.289 0.904 1.11 0.534 0.253 0.916 0.985 0.391
M-30 9.33 10 93.33 0.4 0.867 1.734 1.429 0.329 0.89 1.464 0.866 0.289 0.904 1.306 0.628 0.253 0.914 1.159 0.46
M-35 8.11 11.5 93.33 0.4 0.867 1.994 1.643 0.329 0.89 1.684 0.997 0.289 0.904 1.502 0.722 0.253 0.916 1.332 0.53
M-40 7.18 13 93.33 0.4 0.867 2.254 1.857 0.329 0.89 1.903 1.127 0.289 0.904 1.698 0.816 0.253 0.916 1.506 0.599
M-40 7 (7.18)
Table 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000) 2
100As bd
Permissible shear stress in concrete tc N/mm 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
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. Facor k Over all depth of slab
300 or more
k
1.00
275 1.05
250 1.10
225 1.15
200 1.20
175 150 or less 1.25 1.30
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000) Grade of concrete
M-15 1.6
tc.max
M-20 1.8
M-25 1.9
M-30 2.2
M-35 2.3
M-40 2.5
Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000) Grade of concrete M-10 -tbd (N / mm2)
M-15 0.6
M-20 0.8
M-25 0.9
M-30 1
M-35 1.1
M-40 1.2
M-45 1.3
Table 3.5. Development Length in tension Grade of concrete M 15 M 20 M 25 M 30 M 35 M 40 M 45 M 50
Plain M.S. Bars kd = Ld F tbd (N / mm2) 0.6 58 0.8 44 0.9 39 1 35 1.1 32 1.2 29 1.3 27 1.4 25
H.Y.S.D. Bars kd = Ld F tbd (N / mm2) 0.96 60 1.28 45 1.44 40 1.6 36 1.76 33 1.92 30 2.08 28 2.24 26
M-50 1.4
Value of angle 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
