1 Length of cantilever roof each side 3.00 mtr bw 400 mm

3 Clear span 6.00 mtr 6000 mm

4 Super imposed Load 2.00 kN/m2

5 Conrete M - m - 20 Grade 25000 N/m3

scbc 7 N/mm2 m 13.33

6 Steel fy 415 N/mm2 230 N/mm2

7 Nominal cover 15 mm 25 mm

8 Reinforcement Beam

Bottom Main reinforcement 30 mm F 5 Nos bars

Top holding bars 12 mm F 3 Nos bars

2 lgd Stirrups 8 mm F 220 mm c/c

Reinforcement slab

main bars at supports 10 mm F 140 mm c/c

main bars at end 10 mm F 280 mm c/c

Distrbution at supports 8 mm F 140 mm c/c

Distrbution at ends 8 mm F 210 mm c/c

3 nos. holding bars

30 mm f 12 mm f 30 mm f

2 Nos bars 2 Nos bars

mm mm

860 30 mm f main 5 Nos bars 860

(A) Section at beam

10 mm f @ 10 mm f @ 12 mm f @

280 mmc/c 140 mm c/c 3 Nos bars

3000 3000

100

290 mm

8 mmf @ 8 mmf @ 8 mm f @

mm c/c 140 mm c/c 220 mm c/c 2 lgd stirrups

500

mm 30 mm f @

5 Nos bars

pk_nandwana@yahoo.co.in

210

220220

Tensile stress

(B) Section at mid - span

400

6000

Design of Cycle stand shade

pkn

Effective cover

Unit weight concrete

Name of work:-

`

Name of work:-

1 Length of cantilever roof each side 3.00 mtr 3000 mm bw 400 mm

3 Clear span 6.00 mm 6000 mm

4 Super imposed load 2.00 kN/m 2000 N/m

5 Concrete m 20 Unit weight concrete N/m3

scbc 7 N/mm2

m 13.3

6 Steel fy 415 N/mm2

Tensile stess = 230 N/mm2

7 Nominal cover 15 mm Effective cover = 25 mm

Solution :-

For m 20 concrete we have

scbc = 7 N/mm2

m = 13.3 sst = 230 N/mm2,

mc x

mc + t 13.33 x 7 + 230

j = 1 - k/3 = 1 - 0.29 / 3 = 0.904

R=1/2xc x j x k = 0.5 x 7 x 0.904 x 0.289 = 0.913

(A) Design of cantilever slab

= 7

20 concrete =0.44% . Assuming pt =0.25

1.6 7 x 1.6 = 11.2

span 3000

11.2 11.2

100 mm at ends. Mean thickness of slab( 290 + 100 )/ 2 = 195 mm

15 mm nominal cover and using 10 mm f bars,

available d = 290 - 15 - 10 / 2 = 270 mm at supports

and d = 100 - 15 - 10 / 2 = 80 mm at free ends

0.195 x 1.0 x 1 x 25000 = N

= N/m2

\ = N/m2

wL2 6875 x 3.000 x 3.000

2

V = wL = 6875 x 3.000 = N/m

width considering b=1 meter =1000 mm

BM = 30938 x 1000

Rxb 0.913 x 1000

x

230 x 0.904 x 270

3.14xdia2

3.14 x 10 x 10

4 x100 4 x

1000 x 79 / 551 = 142 mm

Hence Provided 10 mm F bar, @ 140 mm c/c

a f 1000 x 78.5

s

1000 x 270

Let us curtail alternate bars at half the cantilever length.

wx2 6875 x 1.5

2 2

pk_nandwana@yahoo.co.in

pkn

Design of Cycle stand shade

Spacing of Bars =

Actual area of stell provided =1000 =

x 10'67.73=

A*1000/Ast =

A = = =

N-mm

sst x jx D

using 10 mm bars 78.5 mm2

100

551

= mm184

140= 561 mm

2

mm2

N-mm/m

20625

Effective depth required =

Ast =BM

=30938.00 1000

=

M = =2

= 30938

keeping

Dead load of slabper sq meter = 4875

super imposed load 2000

Total load /m2 6875

= 290 mmmm keep total depth of 268 + 15

= =13.33 7

25000

=kc 0.29=

and reduce it to

For stiffness the ratio of span to effective depth of cantilever

For a balanced design, the percentage reinforcement for m =

\ span/d=We get, from fig 1. modification factore, using fe 415 steel. =

\ Effective depth = = = 268

561

k

1000

=pt =

= =B.M. at the point 2x

x %100 0.21

270 + 80

x

230 x 0.904 x 175

1

2

Check for shear V

bxd 1000 x 270

0.12 x 1000 x 290

3.14xdia2

3.14 x 8 x 8

4 x100 4 x

1000 x 50 / 348 = 144 mm

Hence Provided 8 mm F bar, @ 140 mm c/c

At the half the cantilever length, avrage depth '= 195 mm

0.12 x 1000 x 195

3.14xdia2

3.14 x 8 x 8

4 x100 4 x

1000 x 50 / 234 = 215 mm

Hence Provided 8 mm F bar, @ 210 mm c/c

(B) Design of beam :-

The beam will be designed as T-beam with tapered flanges of uniform average thickness.

Width of web, bw = L / 10 or width of column which is heigher

bw = 3.00 / 10 = 0.30 m width of column = 0.40 m here mtr

290 + 100

Total width of flange = b = 2 x 3 + 0.40 = 6.40 mm

= lo = 2 x 3 + 0.40 = 6.40 mtr or mm

For an isolated T-Beam,the width of the flange is

lo 6400

b 6400

400 mm, equal the bw, for the purpose of calculating the dead load

(I) Dead load of slab = 0.195 x 1 x 6 x #### = N

(II) Dead load of beam = 0.40 x 0.40 x 1 x #### = N

(III) Super imposed load = 2000 x 6.40 x 1 = N

= N

wlo2 46050 x 6.4 x 6.4 x 1000

8

Effective depth of beam:-

2x 0.289 d- 195

2 x 0.289 d- 195

0.45 x 1680 x 7.00 x

390 d - = 44553 + or

= / or mm

keep D = 500 mm 30 8 mm f bars, pk_nandwana@yahoo.co.in

=Asd 100

= mm2

mm2

100

Hence O.K.mm2

=20625

N/mm2 which is quite small= 0.08

561 =x

Ast =

Actual Ast available =

Effective depth of slab

BM =

= = mm2

175

10'6

=

Ast =0.12xbD

280 213

213 mm2

>

7.73

sst x jx D

348

Ast = mm2

Asd 100

A*1000/Ast =

tv =

using 8

Distribution reinforcement

234

=

0.12xbD=

A = =

Spacing of Bars =

= = 50.2using 8

50.2mm bars

mm bars A

=

2= 195=Avrage depth of flange, Df

mm2

100

Spacing of Bars = A*1000/Ast =

=

=

6400

mm = 0.195 mtr

Let effective span of beam

bf =+ 4

lo

+ 4

12800

46050Total w

M = = =8

Assume over all depthof beam

29250

4000

+ 400 = 1680+ bw6400

=

x 1680 x

for balance section, we have

or 0.45

M

= 7

keeping

195

mm nominal cover and using

675.66

131753

176306

176306

390 452

=

235776000

235776000=0.289

2kcd-Df

kc

235776000

0.45bf.c.Df

x

0.40

mm

6400mtr or

N-mm

available d = 500 - 30 - 8 - 30 / 2 = 447 mm

= 0.50 x 0.40 x 1 x #### = N

= 29250 + 5000 + 12800 = N

wlo2 47050 x 6.4 x 6.4 x 1000

8

Df 195

2 2

230 x 0.904 x 350

3.14xdia2

3.14 x 30 x 30

4 x100 4 x

= 3311 x 707 = 5 No.

= 5 No. 30 mm F bar, = 3535 mm2

Determination N.A.and actual stresses.

1680 13.33 x 3535 x( 447 - n)

2

\ n2 = 25075 - 56.1 n or n

2+ n - = 0

or n = 56.10 + 3147 -4x 1.00 x

2 a 2 x 1

or n = 56.10 +( 3147 - )1/2

2

or n = 56.10 +( )1/2

2 x 1

or n = 56.10 + 322

2

or n = 266 / 2 = 133 mm

n 133

3 3

m

Ast xa 3535 x 403

t n 169 N

m d - n 13.3 447 - 133 mm2

Shear reinforcement:-

wl 47050 x 6.00 V = N

2 2 bxd 400 x 447 mm2

100As 100 x 2121

bxd 400 x 447

Hence from Table permissible shear (tc)for M 20 concrete, for 1.19 % steel = 0.41 N/mm2

> tc Shear reinforcement necessary

20 Concrete, tcmax. = 1.80 N/mm2 < than Tc section O.K.

Shear reinfocement will be provided both in the form of inclined bars as well as in the form of vertical stirrups.

Out of 5 30 mm f bars, let us bend 2 bars at distance of √2 a = √ 2 x 403 mm

= 1.414 x 403 = 570 mm from the edge of the supports.

3.14xdia2

3.14 x 30 x 30

4 x

= 2 x 707 = 1414 mm2

\ Shear resistance of 2 bent up bars = v1 sin 45

v1 = ssv x Asv x sina = 230 x 1414 x 0.71 = N

more than the S.F. it self. Through shear esistance of concrete (v2 = tc. bd )

is additionally available, we will have to provide nominal shear stirrups.

3.14 x 8 x 8

4 x

using 8 mm 2 lgd stirrups area = 2 x 50.2 = mm2

pk_nandwana@yahoo.co.in

100.4

V=

4 x100bend up bars 2Area of cross -section of

Since tv

mm2

100

Also from table for M- Since tv

= =

M = =

Revised Dead load of beam

8

Total weight w

240896000=

5000

N-mm

47050

350 mm= d = 447-

Ast =BM

=sst x jx D

- =

mm bars A

= 3311

Approximate L.A., a

mm2240896000

mm2

100

Noumber of Bars

25075

= = = 707using 30

-25075

Hence Provided Actual steel provide

(n)2=

56.10

Assuming that N.A. falls in the flange, we have N.A. =

447

-100301.585

103448.47

= 403 mm

=240896000

x

-

n =b +b

2-4.a.c

d -L.A.= =

230<

1337

corresponding stress In

concrete is given by =

= 169Avrage stress in steel =

<

= 141150141150

O.K.

= x = 5.4

N/mm2

%

=N tv =

= 1.19bars will be bent up available

using 8

0.79

707

Area of cross -section of

229931

This is

Assuming that at the ends, 2

mm2mm bars area

4 x100

3.14xdia2

= = = 50.2100

2.18 x 100 x 415

Hence Provided 8 220 860 mm from edge of supports

141150

3.00

= 0.41 x 400 x 447 = N

Hence shear to be rasisted by stirrups is given by Vs =V-tc*bw*d = - =

ssv.Asv.d 230 x 100.4 x 447

227 mm. Hence provide 8 mm f stirrups @ 220 mm

3 x 12

check for development length at the end.

1.3xM1 L0

V

M1 = moment of resistance of section, assuming all reinforcement stress at sst

M1 "= Ast x s st x jc x d

2121 x 230 x 0.904 x 447

V = 141150 N Ls = 0 Alternatively, Ld = 45 F = 45 x 30 = 1350 mm

400 mm and a side cover x' 30 mm,

And Provide 900 bend at the ends

Ls 400

2 2

f .sst x

4.tbd 4 x 0.8 x 2

1.3 xM1 197.1 x 10 6

V

= 2075 > 1350

pk_nandwana@yahoo.co.in

Vs=

27379

100687 N0.86 )=

But maximum spacing can not exceed

100687 73308 27379

= 377 mmThe spacing of 8 mm f 2 lgd stirrups is, sv =

At

73308shear resistance of concrete =tc.bw.d

mm F 2 lgd stirrups, @ mm c/c upto

860 mm shear force x( 3.00 -

the maximum spacing of stirrups is given by 227

30 + 90

mm2.175 x Asv.x fy

b=

400=

we have, Lo =( - x'

Ld

mm

>

f = 45230

mm

+ L0 = 1.3 x + 260

+ 3 x f

=141150

mm2075

= 260= -

Hence Code requirement are satisfied

The code requires that + 2121

Ld = = = 45 1350

Let us assume the width of supports equal to

N-mm1000000

= 197.1 x 10 6

mm f holding bars at topProvidethrough out the length of beam.

f x 30 =

mm2Ast available at end.

=

#

mm F 2 - lgd strirrups @ mm c/c

3 - mm F holding bars 2 - mm F bars

mm mm

- mm F bars

mm f @ mm f @ 3 - mm F holding bars

Nos bars mm c/c

mm

mmf @ mmf @

mm c/c mm c/c mm F 2 - lgd strirrups @ mm c/c

- mm F bars

pk_nandwana@yahoo.co.in

mm

500

mm

500

1500

3000

860

100

290

220 220

5

400

1500

30 860

10

280

10

140

140

8 8

210

pkn

2208

12 30

5 30

12

8 220

M-15 M-20 M-25 M-30 M-35 M-40 Grade of concrete

18.67 13.33 10.98 9.33 8.11 7.18 tbd (N / mm2)

5 7 8.5 10 11.5 13

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 0.904

Rc 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

M-15 M-20 M-25 M-30 M-35 M-40

0.18 0.18 0.19 0.2 0.2 0.2

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

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

tc.max

2.50

2.753.00 and above

(3) Maximum shear stress tc.max in concrete (IS : 456-2000)

2.00

2.25

1.50

1.75

1.00

1.25

0.50

0.75

0.25

bd

< 0.15

(2) Permissible shear stress Table tv in concrete (IS : 456-2000)

100As Permissible shear stress in concrete tv N/mm2

(d) sst =

275

N/mm2

(Fe 500)

(c ) sst =

230

N/mm2

(Fe 415)

(b) sst =

190

N/mm2

(a) sst =

140

N/mm2

(Fe 250)

(1) VALUES OF DESIGN CONSTANTS

Grade of concrete

Modular Ratio

scbc N/mm2

m scbc

100As 100As

bd bd Degree sin cos tan

0.15 0.18 0.18 0.15 10 0.17 0.98 0.18

0.16 0.18 0.19 0.18 11 0.19 0.98 0.19

0.17 0.18 0.2 0.21 12 0.21 0.98 0.21

0.18 0.19 0.21 0.24 13 0.23 0.97 0.23

0.19 0.19 0.22 0.27 14 0.24 0.97 0.25

0.2 0.19 0.23 0.3 15 0.26 0.97 0.27

0.21 0.2 0.24 0.32 16 0.28 0.96 0.29

0.22 0.2 0.25 0.35 17 0.29 0.96 0.31

0.23 0.2 0.26 0.38 18 0.31 0.95 0.32

0.24 0.21 0.27 0.41 19 0.33 0.95 0.34

0.25 0.21 0.28 0.44 20 0.34 0.94 0.36

0.26 0.21 0.29 0.47 21 0.36 0.93 0.38

0.27 0.22 0.30 0.5 22 0.37 0.93 0.40

0.28 0.22 0.31 0.55 23 0.39 0.92 0.42

0.29 0.22 0.32 0.6 24 0.41 0.92 0.45

0.3 0.23 0.33 0.65 25 0.42 0.91 0.47

0.31 0.23 0.34 0.7 30 0.50 0.87 0.58

0.32 0.24 0.35 0.75 35 0.57 0.82 0.70

0.33 0.24 0.36 0.82 40 0.64 0.77 0.84

0.34 0.24 0.37 0.88 45 0.71 0.71 1.00

0.35 0.25 0.38 0.94 50 0.77 0.64 1.19

0.36 0.25 0.39 1.00 55 0.82 0.57 1.43

0.37 0.25 0.4 1.08 60 0.87 0.50 1.73

0.38 0.26 0.41 1.16 65 0.91 0.42 2.14

0.39 0.26 0.42 1.25

0.4 0.26 0.43 1.33

0.41 0.27 0.44 1.41

0.42 0.27 0.45 1.50

0.43 0.27 0.46 1.63

0.44 0.28 0.46 1.64

0.45 0.28 0.47 1.75

0.46 0.28 0.48 1.88

0.47 0.29 0.49 2.00

0.48 0.29 0.50 2.13

0.49 0.29 0.51 2.25

0.5 0.30

0.51 0.30

0.52 0.30 % fy 200 250 328 415

0.53 0.30 0.0

0.54 0.30 0.05

0.55 0.31 0.10

0.56 0.31 0.15 1.90

0.57 0.31 0.20 1.80

0.58 0.31 0.25 2 1.70

0.59 0.31 0.30 1.85 1.60

0.6 0.32 0.35 1.75 1.50

0.61 0.32 0.4 1.65 1.40

0.62 0.32 0.5 2.0 1.5 1.30

modification factore Table

Value of angle

Shear stress tc Reiforcement %

M-20 M-20

0.63 0.32 0.6 1.75 1.4 1.20

0.64 0.32 0.7 1.90 1.65 1.35 1.15

0.65 0.33 0.8 1.80 1.55 1.30 1.05

0.66 0.33 0.9 1.70 1.5 1.25 1.02

0.67 0.33 1.0 1.60 1.45 1.2 1.20

0.68 0.33 1.1 1.55 1.4 1.16 0.98

0.69 0.33 1.2 1.50 1.35 1.13 0.96

0.7 0.34 1.3 1.50 1.3 1.1 0.94

0.71 0.34 1.4 1.45 1.3 1.1 0.92

0.72 0.34 1.5 1.40 1.25 1.07 0.91

0.73 0.34 1.6 1.35 1.2 1.05 0.90

0.74 0.34 1.7 1.35 1.2 1.03 0.89

0.75 0.35 1.8 1.30 1.18 1.01 0.86

0.76 0.35 1.9 1.30 1.16 1.0 0.86

0.77 0.35 2.0 1.25 1.14 0.99 0.85

0.78 0.35 2.1 1.25 1.13 0.97 0.84

0.79 0.35 2.2 1.20 1.12 0.96 0.83

0.8 0.35 2.3 1.18 1.1 0.95 0.83

0.81 0.35 2.4 1.17 1.1 0.94 0.82

0.82 0.36 2.5 1.16 1.08 0.93 0.82

0.83 0.36 2.6 1.15 1.06 0.92 0.81

0.84 0.36 2.7 1.14 1.05 0.92 0.81

0.85 0.36 2.8 1.13 1.04 0.91 0.81

0.86 0.36 2.9 1.12 1.03 0.91 0.81

0.87 0.36 3.0 1.11 1.02 0.90 0.81

0.88 0.37 3.1 1.11 1.01 0.87 0.81

0.89 0.37 3.2 1.11 1.00 0.86 0.81

0.9 0.37

0.91 0.37

0.92 0.37

0.93 0.37

0.94 0.38

0.95 0.38

0.96 0.38

0.97 0.38

0.98 0.38

0.99 0.38

1.00 0.39

1.01 0.39

1.02 0.39

1.03 0.39

1.04 0.39

1.05 0.39

1.06 0.39

1.07 0.39

1.08 0.4

1.09 0.4

1.10 0.4

1.11 0.4

1.12 0.4

1.13 0.4

1.14 0.4

1.15 0.4

1.16 0.41

1.17 0.41

1.18 0.41

1.19 0.41

1.20 0.41

1.21 0.41

1.22 0.41

1.23 0.41

1.24 0.41

1.25 0.42

1.26 0.42

1.27 0.42

1.28 0.42

1.29 0.42

1.30 0.42

1.31 0.42

1.32 0.42

1.33 0.43

1.34 0.43

1.35 0.43

1.36 0.43

1.37 0.43

1.38 0.43

1.39 0.43

1.40 0.43

1.41 0.44

1.42 0.44

1.43 0.44

1.44 0.44

1.45 0.44

1.46 0.44

1.47 0.44

1.48 0.44

1.49 0.44

1.50 0.45

1.51 0.45

1.52 0.45

1.53 0.45

1.54 0.45

1.55 0.45

1.56 0.45

1.57 0.45

1.58 0.45

1.59 0.45

1.60 0.45

1.61 0.45

1.62 0.45

1.63 0.46

1.64 0.46

1.65 0.46

1.66 0.46

1.67 0.46

1.68 0.46

1.69 0.46

1.70 0.46

1.71 0.46

1.72 0.46

1.73 0.46

1.74 0.46

1.75 0.47

1.76 0.47

1.77 0.47

1.78 0.47

1.79 0.47

1.80 0.47

1.81 0.47

1.82 0.47

1.83 0.47

1.84 0.47

1.85 0.47

1.86 0.47

1.87 0.47

1.88 0.48

1.89 0.48

1.90 0.48

1.91 0.48

1.92 0.48

1.93 0.48

1.94 0.48

1.95 0.48

1.96 0.48

1.97 0.48

1.98 0.48

1.99 0.48

2.00 0.49

2.01 0.49

2.02 0.49

2.03 0.49

2.04 0.49

2.05 0.49

2.06 0.49

2.07 0.49

2.08 0.49

2.09 0.49

2.10 0.49

2.11 0.49

2.12 0.49

2.13 0.50

2.14 0.50

2.15 0.50

2.16 0.50

2.17 0.50

2.18 0.50

2.19 0.50

2.20 0.50

2.21 0.50

2.22 0.50

2.23 0.50

2.24 0.50

2.25 0.51

2.26 0.51

2.27 0.51

2.28 0.51

2.29 0.51

2.30 0.51

2.31 0.51

2.32 0.51

2.33 0.51

2.34 0.51

2.35 0.51

2.36 0.51

2.37 0.51

2.38 0.51

2.39 0.51

2.40 0.51

2.41 0.51

2.42 0.51

2.43 0.51

2.44 0.51

2.45 0.51

2.46 0.51

2.47 0.51

2.48 0.51

2.49 0.51

2.50 0.51

2.51 0.51

2.52 0.51

2.53 0.51

2.54 0.51

2.55 0.51

2.56 0.51

2.57 0.51

2.58 0.51

2.59 0.51

2.60 0.51

2.61 0.51

2.62 0.51

2.63 0.51

2.64 0.51

2.65 0.51

2.66 0.51

2.67 0.51

2.68 0.51

2.69 0.51

2.70 0.51

2.71 0.51

2.72 0.51

2.73 0.51

2.74 0.51

2.75 0.51

2.76 0.51

2.77 0.51

2.78 0.51

2.79 0.51

2.80 0.51

2.81 0.51

2.82 0.51

2.83 0.51

2.84 0.51

2.85 0.51

2.86 0.51

2.87 0.51

2.88 0.51

2.89 0.51

2.90 0.51

2.91 0.51

2.92 0.51

2.93 0.51

2.94 0.51

2.95 0.51

2.96 0.51

2.97 0.51

2.98 0.51

2.99 0.51

3.00 0.51

3.01 0.51

3.02 0.51

3.03 0.51

3.04 0.51

3.05 0.51

3.06 0.51

3.07 0.51

3.08 0.51

3.09 0.51

3.10 0.51

3.11 0.51

3.12 0.51

3.13 0.51

3.14 0.51

3.15 0.51

Grade of concreteM-10 M-15 M-20 M-25 M-30 M-35 M-40 M-45

tbd (N / mm2) -- 0.6 0.8 0.9 1 1.1 1.2 1.3

M 15

M 20

M 25

M 30

M 35

M 40

M 45

M 50

(N/mm2) Kg/m2 (N/mm2) Kg/m

2

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 1100

M 50 16.0 1600 12.0 1200 1.4 140

1.2 120

1.3 130

1.0 100

1.1 110

0.8 80

0.9 90

-- --

0.6 60

(N/mm2) in kg/m2

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)

1.3 27 2.08

Permissible stress in concrete (IS : 456-2000)

1.4 25 2.24 26

1.1 32 1.76 33

1.2 29 1.92 30

28

39 1.44 40

1 35 1.6 36

0.6 58 0.96 60

0.8 44 1.28 45

0.9

Development Length in tension

Grade of

concrete

Plain M.S. Bars H.Y.S.D. Bars

tbd (N / mm2) kd = Ld F tbd (N / mm2) kd = Ld F

Permissible Bond stress Table tbd in concrete (IS : 456-2000)

tan Degree sin cos

0.18 10 0.17 0.98

0.19 11 0.19 0.98

0.21 12 0.21 0.98

0.23 13 0.23 0.97

0.25 14 0.24 0.97

0.27 15 0.26 0.97

0.29 16 0.28 0.96

0.31 17 0.29 0.96

0.32 18 0.31 0.95

0.34 19 0.33 0.95

0.36 20 0.34 0.94

0.38 21 0.36 0.93

0.40 22 0.37 0.93

0.42 23 0.39 0.92

0.45 24 0.41 0.92

0.47 25 0.42 0.91

0.58 30 0.50 0.87

0.70 35 0.57 0.82

0.84 40 0.64 0.77

1.00 45 0.71 0.71

1.19 50 0.77 0.64

1.43 55 0.82 0.57

1.73 60 0.87 0.50

2.14 65 0.91 0.42

500

2.00

1.80

1.65

1.50

1.40

1.35

1.30

1.20

1.16

1.08

modification factore Table

Value of angle

1.00

0.95

0.90

0.86

0.84

0.82

0.81

0.80

0.79

0.78

0.77

0.76

0.75

0.74

0.73

0.72

0.72

0.72

0.71

0.71

0.71

0.70

0.70

0.69

0.69

0.68

0.68

M-50

1.4

0.0

Percentage of tension reinforcement

2 2.4 2.80.4 0.8 1.2 1.6

2.0

0.4

1.4

1.2

0.8

Modific

ation facto

rePermissible Bond stress Table tbd in concrete (IS : 456-2000)