Download - Design of Aqueduct
Design of Aqueduct
Input Datacanal dataFull Supply Discharge = 0.9825 cumecsFull Supply Level = 560.259 mCanal Bed Level = 559.499 mCanal Water Depth (D) = 0.76 mCanal Bed Width (B) = 0.80 mRugosity coff for concrete (n) = 0.016side slope = 1.5 :1bed slope = 1800free board = 0.4
Drainage DataHigh Flood Discharge (Q) = 18.00 cumecsHigh Flood Level = 557.00 mHigh Flood Depth = 2.00 mGeneral Ground Level = 555.00 m
Design of Drainage water way
Lacey`s regime Perimeter (P) == 20.15254
assume clear span b/w pier = 10 mthickness of pier = 1.5 mNo. of ways provided = 2no of piers = 1water way between abutments (L) = 21.5 m
Deisgn of canal water way
bed width of canal = 0.80 m
= 0.60 mAssumeSpaly in contraction = 2 :1splay in expansion = 3 :1
length of contraction transition = 0.2 mlength of expansion transition = 0.3 m
= 21.5 m
Head loss and Bed levels at different sections
At section 4-4
4.75*(Q1/2)
Based on perimeter assume clear span width and thickness of peir and accordingly no. of ways and no. of piers to provide. No. of
piers are one less than no. of ways.
let the cnala flumed to (B0)
length of flummed rectangular portion of the canal between abutments
In the transitions , the side slopes of the section will be warped from 1.5:1 to vertical.
Area of Section (A4) = (B+1.5D)*D
= 1.4744Velocity (V4) = Q/A
= 0.666373 m/s
Velocity Head == 0.022633 m
RL of Bed (given) = 559.499 mRL of Water Surface = 560.259 mRL of T.E.L. = 560.282 m
At section 3-3keeping the same water depth thoughout the channel
Area of Section (A3) =
= 0.456Velocity (V3) = Q/A
= 2.154605 m/s
m2
V2/2g
B0*D
m2
Velocity Head == 0.236612 m
loss of head in expnasion from section 3-3 to 4-4
== 0.064194
RL of T.E.L. = 560.346 mRL of Water Surface = 560.109 mRL of Bed (given) = 559.349 m
At section 2-2
Length of Trough Section = 21.5 m
Area of Trough = 0.456wetted perimeter = 2.12 mHydraulic mean depth = 0.215094 mvelocity in Trough = Q/A
= 2.154605 m/s
== 0.198263 m
RL of T.E.L. = 560.544 mRL of Water Surface = 560.307 mRL of Bed (given) = 559.547 m
At section 1-1
Loss of head in Contraction transition from 1-1 to 2-2
== 0.042796
RL of T.E.L. = 560.587 mRL of Water Surface = 560.564 mRL of Bed (given) = 559.804 m
Design of Transitions
V2/2g
0.3(V32-V4
2)/2g
m2
Froction loss b/w 2-2 to 3-3 (HL) (n2*V2*L)/(R4/3)
0.2(V32-V4
2)/2g
Contraction transitionBy Mitra`s Hyperbolic Transition equation
= 0.60 m
= 0.80 m
= 0.20 mx is the distance at which distance u want bed width
=
x = 0 0.05 0.1 0.15 0.2Bx = 0.6 0.64 0.685714 0.738462 0.8
Expansion transitionBy Mitra`s Hyperbolic Transition equation
= 0.60 m
= 0.80 m
= 0.30 m
=x = 0 0.05 0.15 0.2 0.25 0.3
Bx = 0.6 0.626087 0.685714 0.72 0.757895 0.8
Design of Trough
Tentative thickness of walls = 0.1 mBottom Slab of Trough = 0.15 mClear Width b/w Wall = 0.6 m
Bf
Bn
length of contaction tansition Lf
Bx Bn*Bf*Lf/Lf*Bn-x(Bn-Bf)
Bf
Bn
length of expansion tansition Lf
Bx Bn*Bf*Lf/Lf*Bn-x(Bn-Bf)
The trough will be of 0.6m and wall thickness will be 0.1m. A free board of 0.14m above FSD of 0.76m may be sufficient . Hence the Height of the Trough will therfore be kept equal to 0.76+0.14=0.9m.
The Entire Section will be constrcued in monolithic reinforced concrete & designed by usual strcutural methods.
1 2 3 4
0.80 0.60
0.2 21.5 0.3
TEL RL 560.587 560.544 560.346 560.282Water Surface RL 560.564 560.307 560.109 560.259
Bed Level RL 559.804 559.547 559.349 559.499
TroughContraction Tarnsition
Expansion Tarnsition
1 2 3 4
4321
Canal
AQUEDUCT TROUGH DESIGNNAME OF WORK:- PKN
CANAL DATA FOR AQUEDUCT (Vertical section)1 Discharge 0.9825 cumec 0.9825 cumec 270
2 Bed width 0.80 m 0.60 m
3 water Side slope 1.5 :1 vertical :1
4 F.S.D. 0.76 m 0.76 m
5 Free Board 0.40 m 0.40 m
6 Bed slope 1 in 1800 1 in 1800
7 C.B.L. 559.499 m 559.499 m
2008 F.S.L. 560.26 m 560.259
9 M.W.D. 1.16 m 1.16 m
10 Span 10.00 m 10000 mm
11 Concrete M- 20 wt. of concrete 25000
7 m 13 30012 Steel Inside Out side
150 190
13 Water wt 9800
14 Reinforcement (in wall) Main Vertical 10 90
15 Reinforcement (in Slab) Main 16 110
16 Reinforcement (in wall Beam) Main bottom 20 6
17 Distribution (in wall Beam) two lgd. Strrirps 8 300
18 Trough Wall thickness 270 mm or 0.27
19 Trough Slab thickness 300 mm or 0.30
kg/m3
scbc
sst sst
kg/m3
mm F @
mm F @
mm f
mm F @
pk_nandwana@yahoo,co,in
AQUEDUCT TROUGH DESIGNPKN
FOR AQUEDUCT (Vertical section)TBL 960.26 270600
F.B. 400
10 180
F.S.L. 560.259
8 mm 2 ledge stirrups@ 300
8 140
10 90
FSD 76016 220
16 110
Hench 8 130
10 300
6 20
CBL 559.499 0.12
720
200 200
mm c/c Disty. 8 140 mm c/c
mm c/c Disty. 8 130 mm c/c
Nos. 16 2 Nos.
mm c/c
mtr
mtr
2x 16 mm F top anchor bar
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
x Bars F
mm F @
mm F @
Top anchor mm Nos.
15 115 20 45 33 44 18 40 20 41 20 40 18
AQUEDUCT TROUGH DESIGN
NAME OF WORK:- PKN
CANAL DATA
Discharge 0.9825 cumec 0.9825 cumecBed width 0.80 m 0.60 mwater Side slope 1.5 :1 verticalF.S.D. 0.76 m 0.76 mFree Board 0.40 m 0.40 mBed slope 1800 1800C.B.L. 559.499 m 559.499 mF.S.L. 560.26 m 560.259 mM.W.D. 1.16 m 1.16 m
10.00 mNominal Cover 50 mmEffective cover 40 mm
1 Design Constants:- For HYSD Bars Concrete M- 20for water side force
= 150 wt. of concrete = 25000= 7 wt of water = 9800
m = 13m*c
=13 x 7
= 0.37813 x 7 + 150= 1 - 0.378 / 3 = 0.874= 0.5 x 7 x 0.87 x 0.378 = 1.155
for out side force
= 190 wt. of concrete = 25000= 7 wt of water = 9800
m = 13m*c
=13 x 7
= 0.32413 x 7 + 190= 1 - 0.324 / 3 = 0.892= 0.5 x 7 x 0.892 x 0.324 = 1.011
2 DESIGN OF VERTICAL WALL:-
The trough wall is to be designed as a beam having a span of = 10.00 mbetween supports Hence thickness should be equal to span/28
span=
10.00 x 1000= 360 mm say
28 28Max.depth of water = 1.16 m span = 10.00 m
B.M. = =9800 x 1.16 3
=6 6
Effective depth required =BM
=2549 x 1000
=Rxb 1.16 x 1000
Providing thickness "D"= 360 mm cover = 50 mm, Effective depth =
FOR QUEDUCT
Span (Proposed)
sst = N/mm2 N/m3
scbc = N/mm2 N/mm2
k=m*c+sst
j=1-k/3R=1/2xc x j x k
sst = N/mm2 N/m3
scbc = N/mm2 N/mm2
k=m*c+sst
j=1-k/3R=1/2xc x j x k
wh3
Steel required
Ast =BMx1000
=2549 x 1000
= 61150 x 0.892 x 310
using 10 mm bars = A = =3.14 x
4 x 100spacing =A/Ast = 78.50 x 1000 / 61.46 = 1277 mm
Hence Provided 10 mm bars @ 1270 mm c/c half the bars will be curtailed at
= 0.3 -0.1 ( 36 - 10 )
45 - 10
Area of distribution steel required = 0.23 % of x section area =0.23 x
100
Steel of Each face =813
= 4062
using 8 mm bars A = = 3.14 x4 x100
spacing =A/Ast = 50.24 x 1000 / 406.29 = 123.657 mmHence Provided 8 mm bars @ 120 mm c/c Each face
3 Design of Horizontal slabe :-
The trough slab having a span of of = 0.60 mbetween walls Hence thickness should be equal to span/20
span=
0.60 x 1000= 30 mm say
20 20Adopt effective thickness of slab "T" = 100 mm cover = 50 mm Total thickness
Effective span of slab = BW+ depth = 0.6 +Loading
Load of water column = mwd x 9800 = 1.16 x 9800 =Wt of slab = wt of concrete x area of slab = 25000 x 1.00 x 0.1 =
per meter length
Total water pressure on vertical wall= =9800 x 1.16 x 1.16
=2 2
\ Fixing moment at end of slab = 6593 x1.2
+3
Max. possible segging moment = =13868 x 0.96 x 0.96
=8 8
Net B.M. at center of span of slab= = 1598 - 3538 = -1941 kg-m The slab is design for this B.M.
Since tension face is out side = 190 J = 0.892 ,
Effective depth required =BM
=-1941 x 1000
= 43.813 mmRxb 1.011 x 1000
Provided Effective depth 44 mm cover = 50 mm providing thickness
Steel required
Ast =-1940.89 x 1000
= -261190 x 0.89 x 44
using 16 mm bars = A = =3.14 x
mm2
sst x j x D
3.14xdia2
minimum steel to be provided for distribution
mm2
3.14xdia2
wH2
WL2
s st
BMx100/sstxjxD= mm2
3.14xdia2
using 16 mm bars = A =4 x 100
=
spacing =A/Ast = 201 x 1000 / -261 = -768.877 mmHence Provided 16 mm bars @ -760 mm c/c
Area of steel required at end (Near support) =3538 x 1000
=150 x 0.874 x 44
This is < than half the steel provided at the center of span,However, half the bars from the center of the span may be bent up at L/2 meter from supports.
Let us check whether this bending of half bars satisfies the enchorage and devlopments envisaged in
1x
1000 x 201x 190 x
2 -760= -0.98 x N-mm
=13868 x 0.96
= 6657 N2
= - x' - + = -2 2
= Length of support = 360 mm and x' = side cover =
M1=
-0.98 x+
360- 50 +
V 6657 2
= =F x 150
= 46.88 F4 x 0.8
= 46.88 x 16 = 750 mm
or 191 < 750 Thus the requirement is not satisfied
= 0.3 -0.1 ( 94 - 100 )
450 - 100
Area of distribution steel required = 0.30 % of x section area =0.30 x
100
Steel of Each face =283
= 1422
using 8 mm bars A = = 3.14 x4
spacing =A/Ast = 50.24 x 1000 / 141.55 = 355 cmHence Provided 8 mm bars @ 350 cm c/c Each face
4 Design of side wall as Beam :-
live load from slab = total load on slab x bw / 2 = 13868 x 0.6 /Self load = mwd x thick. x wt = 1.16 x 0.36 x
Total Load
Max. possible segging moment= =14600 x 10.00 x 10.00
=8 8
= 190 k = 0.324 J =
Effective depth required =BM
=182505 x 1000
=Rxb 1.011 x 360
Actual depth '= 1.16 + 0.10 = 1.26 or 1260 mmBut providing thickness = 1260 mm - (2 x cover = 80 )=
Steel required
equation M1/V + Lo > Ld
Where M1= Ast x sst x j x d=
10'6
V = shear force at the ends
Lo
ls 3 F 16 Fls
Where Ls
+ Lo
10'6
Ld
F sst4 t bd
minimum steel to be provided for distribution
mm2
3.14xdia2
WL2
using sst N/mm2
Ast =182505.0 x 1000
= 913190 x 0.892 x 1180
using 20 mm bars A = = 3.14 x4 x100
Nomber of Bars = Ast/A = 913 / 314 = 2.91Hence Provided 3 bars of 20
% of steel provided =3 x 314
x 100 = 0.22 %360 x 1180
Shear force =total load x span
=14600 x 10.0
=2 2
Shear stress =
shea force=
73002.0=
Beam Ht. x Beam Dt. 360 x 1180Permissible shear stress for 0.22 % = 0.2
Shear reinforcenment required if < Hence shear reinforcement not required
= 0.20 x 1180 x 360 == = 73002 - 84960 = -11958
= =190 x 1180
-11958
<2.175 x fy x Asy
<2.175 x
B
Hence = 2.51
Hence using 8 mm dia 2 Legged stirrups A = 100.5
= 2.51 x 100.5 = 252 mm subject to a max.
Hence provideed 8 mm Dia 2 legged shear stirrus @
Provide
BMx100/sstxjxD= mm2
3.14xdia2
mm F at Bottom
steel provided tc N/mm2
TV Tc
Vc = shear resistance of concrete = tc.b.d
Vs V - Vc
Spacing of strirrups is given by
Sv
sst .d.AsvAsv
Vs
While maximum permissible spacing of shear stirip is
Sv Asv
mm2
Sv
2 x 12 mm F hoilding bars at the top.
40 16 34 16 34 26
AQUEDUCT TROUGH DESIGN
PKN
for water side force
K = 0.378
J = 0.874R = 1.155
for out side force
K = 0.324
J = 0.892R = 1.011
mm say 360 mm
2549 N-m2549463n-mm
47 mm
mm, Effective depth = 310 mm
N/m3
N/mm2
N/m3
N/mm2
10 x 10= 78.5
4
0.78 m from base
= 0.23 %
360 x 1000= 813
100
8 x 8= 50.2
4
mm say 100 mm
Total thickness = 150 mm0.36 = 0.96 m
11368 N2500 N
13868 N
6593
0.3= 3538 N-m
2
1598 N-m
The slab is design for this B.M.
R = 1.011
mm
providing thickness = 94 mm
16 x 16= 201
mm2
mm2
mm2
mm2
mm2
mm2
4= 201
616
than half the steel provided at the center of span,However, half the bars from the center of the span may be bent up at L/2 meter from supports.
Let us check whether this bending of half bars satisfies the enchorage and devlopments envisaged in
0.892 x 44
x' +
side cover = 50 mm
13 x 16 = 190.5 mm
See table Concrete 3.4 M 20
Thus the requirement is not satisfied
= 0.30 %
94 x 1000= 283
100
8 x 8= 50.2
4
2 = 4160 kg-m25000 = 10440 kg-m
Total Load = 14600 kg-m
182505 Kg-m
0.892 R = 1.011
708 mm
mm1180 mm
mm2
mm2
13 F
mm2
mm2
20 x 20= 314
4say = 3 No.
73002 kg.
0.17
(See Table 3.1)Hence shear reinforcement not required
84960 NN
= -18.75
415 x Asv< 2.51 Asv
360
subject to a max. = 300 mm300 mm c/c
mm2
mm2
N/mm2
N/mm2
Asv Asv
NAME OF WORK:- PKN
270 TBL 758.58 2705000
F.B. 500
10 180
F.S.L. 258.58
8 mm 2 ledge stirrups@ 300
8 140
10 90
FSD 200016 220
16 110
8 130
200 10 300
6 20
CBL 256.58 1.00
300 720
200 200
2x 16 mm F top anchor bar
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
mm F bars@
x Bars F
Grade of co M-10 M-15 M-20 M-25 M-30 M-35 M-40 bd
1.2 2.0 2.8 3.2 3.6 4.0 4.40.250.500.751.001.25
(N/mm2) (N/mm2) (N/mm2) 1.50M 10 3.0 300 2.5 250 -- -- 1.75M 15 5.0 500 4.0 400 0.6 60 2.00M 20 7.0 700 5.0 500 0.8 80 2.25M 25 8.5 850 6.0 600 0.9 90 2.50M 30 10.0 1000 8.0 800 1.0 100 2.75M 35 11.5 1150 9.0 900 1.1 110 3.00 and above
M 40 13.0 1300 10.0 1000 1.2 120M 45 14.5 1450 11.0 1100 1.3 130M 50 16.0 1600 12.0 1200 1.4 140
Over all depth of slab
k
Grade of co M-10 M-15 M-20 M-25 M-30 M-35 M-40Modular ra
Grade of concrete
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 Grade of concrete M5 7 8.5 10 11.5 13
93.33 93.33 93.33 93.33 93.33 93.330.4 0.4 0.4 0.4 0.4 0.4
0.867 0.867 0.867 0.867 0.867 0.8670.867 1.214 1.474 1.734 1.994 2.2540.714 1 1.214 1.429 1.643 1.857
0.329 0.329 0.329 0.329 0.329 0.329 M 15
0.89 0.89 0.89 0.89 0.89 0.89 M 200.732 1.025 1.244 1.464 1.684 1.903 M 250.433 0.606 0.736 0.866 0.997 1.127 M 300.289 0.289 0.289 0.289 0.289 0.289 M 350.904 0.904 0.904 0.904 0.904 0.904 M 400.653 0.914 1.11 1.306 1.502 1.698 M 450.314 0.44 0.534 0.628 0.722 0.816 M 50
Table 1.15. PERMISSIBLE DIRECT TENSILE STRESSTable 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)
100A s
Tensile stress N/mm2
< 0.15
Table 1.16.. Permissible stress in concrete (IS : 456-2000)
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)
Kg/m2 Kg/m2 in kg/m2
Table 3.2. Facor k
Table 1.18. MODULAR RATIO
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)31 (31.11)
19 (18.67)
13 (13.33)
11 (10.98)
9 (9.33)
8 (8.11)
7 (7.18)
tc.max Table 2.1. VALUES OF DESIGN CONSTANTS
Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)
scbc N/mm2 tbd (N / mm2)
m scbc
(a) sst = 140
N/mm2 (Fe 250)
kcTable 3.5. Development Length in tension
jc
Rc Grade of concretePc (%)
(b) sst = 190
N/mm2
kc
jc
Rc
Pc (%)
(c ) sst = 230 N/mm2 (Fe 415)
kc
jc
Rc
Pc (%)
Reiforcement %
M-20 M-20bd bd
0.15 0.18 0.18 0.15
0.16 0.18 0.19 0.18
0.17 0.18 0.2 0.21
0.18 0.19 0.21 0.24
0.19 0.19 0.22 0.270.2 0.19 0.23 0.3
0.21 0.2 0.24 0.32
0.22 0.2 0.25 0.350.23 0.2 0.26 0.38
0.24 0.21 0.27 0.410.25 0.21 0.28 0.44
0.26 0.21 0.29 0.470.27 0.22 0.30 0.5
0.28 0.22 0.31 0.550.29 0.22 0.32 0.6
0.3 0.23 0.33 0.650.31 0.23 0.34 0.7
0.32 0.24 0.35 0.750.33 0.24 0.36 0.82
0.34 0.24 0.37 0.88
0.35 0.25 0.38 0.94
0.36 0.25 0.39 1.000.37 0.25 0.4 1.080.38 0.26 0.41 1.160.39 0.26 0.42 1.250.4 0.26 0.43 1.33
0.41 0.27 0.44 1.410.42 0.27 0.45 1.500.43 0.27 0.46 1.630.44 0.28 0.46 1.640.45 0.28 0.47 1.750.46 0.28 0.48 1.880.47 0.29 0.49 2.000.48 0.29 0.50 2.130.49 0.29 0.51 2.250.5 0.30
0.51 0.300.52 0.300.53 0.300.54 0.300.55 0.31
Shear stress tc
100A s 100A s
0.56 0.310.57 0.310.58 0.310.59 0.310.6 0.32
0.61 0.320.62 0.320.63 0.320.64 0.320.65 0.330.66 0.330.67 0.330.68 0.330.69 0.330.7 0.34
0.71 0.340.72 0.340.73 0.340.74 0.340.75 0.350.76 0.350.77 0.350.78 0.350.79 0.350.8 0.35
0.81 0.350.82 0.360.83 0.360.84 0.360.85 0.360.86 0.360.87 0.360.88 0.370.89 0.370.9 0.37
0.91 0.370.92 0.370.93 0.370.94 0.380.95 0.380.96 0.380.97 0.380.98 0.380.99 0.381.00 0.391.01 0.391.02 0.391.03 0.39
1.04 0.391.05 0.391.06 0.391.07 0.391.08 0.41.09 0.41.10 0.41.11 0.41.12 0.41.13 0.41.14 0.41.15 0.41.16 0.411.17 0.411.18 0.411.19 0.411.20 0.411.21 0.411.22 0.411.23 0.411.24 0.411.25 0.421.26 0.421.27 0.421.28 0.421.29 0.421.30 0.421.31 0.421.32 0.421.33 0.431.34 0.431.35 0.431.36 0.431.37 0.431.38 0.431.39 0.431.40 0.431.41 0.441.42 0.441.43 0.441.44 0.441.45 0.441.46 0.441.47 0.441.48 0.441.49 0.441.50 0.451.51 0.45
1.52 0.451.53 0.451.54 0.451.55 0.451.56 0.451.57 0.451.58 0.451.59 0.451.60 0.451.61 0.451.62 0.451.63 0.461.64 0.461.65 0.461.66 0.461.67 0.461.68 0.461.69 0.461.70 0.461.71 0.461.72 0.461.73 0.461.74 0.461.75 0.471.76 0.471.77 0.471.78 0.471.79 0.471.80 0.471.81 0.471.82 0.471.83 0.471.84 0.471.85 0.471.86 0.471.87 0.471.88 0.481.89 0.481.90 0.481.91 0.481.92 0.481.93 0.481.94 0.481.95 0.481.96 0.481.97 0.481.98 0.481.99 0.48
2.00 0.492.01 0.492.02 0.492.03 0.492.04 0.492.05 0.492.06 0.492.07 0.492.08 0.492.09 0.492.10 0.492.11 0.492.12 0.492.13 0.502.14 0.502.15 0.502.16 0.502.17 0.502.18 0.502.19 0.502.20 0.502.21 0.502.22 0.502.23 0.502.24 0.502.25 0.512.26 0.512.27 0.512.28 0.512.29 0.512.30 0.512.31 0.512.32 0.512.33 0.512.34 0.512.35 0.512.36 0.512.37 0.512.38 0.512.39 0.512.40 0.512.41 0.512.42 0.512.43 0.512.44 0.512.45 0.512.46 0.512.47 0.51
2.48 0.512.49 0.512.50 0.512.51 0.512.52 0.512.53 0.512.54 0.512.55 0.512.56 0.512.57 0.512.58 0.512.59 0.512.60 0.512.61 0.512.62 0.512.63 0.512.64 0.512.65 0.512.66 0.512.67 0.512.68 0.512.69 0.512.70 0.512.71 0.512.72 0.512.73 0.512.74 0.512.75 0.512.76 0.512.77 0.512.78 0.512.79 0.512.80 0.512.81 0.512.82 0.512.83 0.512.84 0.512.85 0.512.86 0.512.87 0.512.88 0.512.89 0.512.90 0.512.91 0.512.92 0.512.93 0.512.94 0.512.95 0.51
2.96 0.512.97 0.512.98 0.512.99 0.513.00 0.513.01 0.513.02 0.513.03 0.513.04 0.513.05 0.513.06 0.513.07 0.513.08 0.513.09 0.513.10 0.513.11 0.513.12 0.513.13 0.513.14 0.513.15 0.51
bd M-15 M-20 M-25 M-30 M-35 M-400.18 0.18 0.19 0.2 0.2 0.2
0.25 0.22 0.22 0.23 0.23 0.23 0.230.50 0.29 0.30 0.31 0.31 0.31 0.320.75 0.34 0.35 0.36 0.37 0.37 0.381.00 0.37 0.39 0.40 0.41 0.42 0.421.25 0.40 0.42 0.44 0.45 0.45 0.461.50 0.42 0.45 0.46 0.48 0.49 0.491.75 0.44 0.47 0.49 0.50 0.52 0.522.00 0.44 0.49 0.51 0.53 0.54 0.552.25 0.44 0.51 0.53 0.55 0.56 0.572.50 0.44 0.51 0.55 0.57 0.58 0.602.75 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
Over all depth of slab 300 or more 275 250 225 200 175 150 or lessk 1.00 1.05 1.10 1.15 1.20 1.25 1.30
Grade of concrete M-15 M-20 M-25 M-30 M-35 M-401.6 1.8 1.9 2.2 2.3 2.5
10 15 20 25 30 35 40 45 50-- 0.6 0.8 0.9 1 1.1 1.2 1.3 1.4
Plain M.S. Bars H.Y.S.D. Bars
0.6 58 0.96 600.8 44 1.28 450.9 39 1.44 401 35 1.6 36
1.1 32 1.76 331.2 29 1.92 301.3 27 2.08 281.4 25 2.24 26
Table 3.1. Permissible shear stress Table tc in concrete (IS : 456-2000)100A s Permissible shear stress in concrete tc N/mm2
< 0.15
Table 3.2. Facor k
Table 3.3. Maximum shear stress tc.max in concrete (IS : 456-2000)
tc.max
Table 3.4. Permissible Bond stress Table tbd in concrete (IS : 456-2000)
Table 3.5. Development Length in tension
tbd (N / mm2) kd = Ld F tbd (N / mm2) kd = Ld F