Design of Rectangular water tank.xls

Design of Rectangular water tank CASE-1 ( L / B < 2 )Capacity 80000 Litres (given)Material M20 Grade Concrete (given)

Fe 415 Grade HYSD reinforcement (given)Solution :-

Provide 6 m x 4 m x 3.5 m tank with free board of 150 mm.

80400 Litres

L / B = 6 / 4 = 1.5 < 2 .The top portion of side walls will be designed as a continuous frame.bottom 1 m or H / 4 whichever is more is designed as cantilever.

H / 4 = 3.5 / 4 = 0.875 mbottom 1 m will be designed as cantilever.

Water pressure at 3.5 - 1 = 2.5 m height from top = 2.5 x 9.8

= 24.5

6 m

6 m A E

F

3.5 m

D

Elevation Plan

Fixed end moments :-

= 73.5 KNm

= -32.66 KNm

Kani's Method :-

- 32.66 73.5

0 -12.25 -8.17 0

0 -12.25 -8.17 0

D 0 A 0 B-57.16 57.16

Volume = 6 x 4 x 3.35 x 10 3 =

2.5 x YW

=

KN / m2

where Yw is unit weight of water = 9.8 KN / m3

To find moment in side walls, moment distribution or kani's method is used. As the frame is symmetrical about both the axes, only one joint is solved

24.5 KN / m2

34.3 KN / m2

MAB

= w x l2 / 12 =

24.5 x 62 / 12 =

MAD

= w x l2 / 12 =

24.5 x 42 / 12 =

-3/10 -2/1040.84

2.5 m

1 m

Rotation factor at Joint A

Joint Member ∑ K

AAB I / 6

5 * I / 12- 2 / 10

AD I / 4 - 3 / 10

Sum of FEM

73.5-32.6640.84 KNm

= 73.5 + 2 x (- 8.17 ) + 0= 57.16

= (- 32.66 ) + 2 x (- 12.25 ) + 0= -57.16

B.M. at centre of long span =

=53.09 KNm

B.M. at centre of short span =

=-8.16 KNm

Direct tension in long wall = = 24.5 x 4 / 2 = 49 KN

Direct tension in short wall = = 24.5 x 6 / 2 = 73.5 KN

Design of Long Walls :-At support

M = 57.16 KNm

T = 49 KN Tension on liquid face.From Table 9-6 Q = 0.306 Assuming d / D = 0.9

D =

== 432.2 mm, Assuming d / D = 0.9

Take D = 450 mm d = 450 - 25 - 8

Relative Stiffness( K )

Rotation Factor u =(-1/2) k / ∑ K

MA

F =

MAB

= MAB

F + 2 MAB

' + MBA

'

MAD

= MAD

F + 2 MAD

' + MDA

'

w x l 2 / 8 - 57.16

24.5 x 62 / 8 - 57.16

w x l 2 / 8 - 57.16

24.5 x 42 / 8 - 57.16

Yw ( H - h ) x B / 2

Yw ( H - h ) x L / 2

√M / Q x b

√57.16 x 10 6 / 0.306 x 1000

= 417 mmFrom Table 9-5

=

= 1048

=

= 327

1048 + 327

= 1375Provide 16 mm O bar

spacing of bar == 200.96 x 1000 / 1375= 146.152727 mm

Provide 16 mm O bar @ 130 mm C/C…marked(a) = 1546Larger steel area is provided to match with the steel of short walls.At centre

M = 53.09 KNmT = 49 KN tension on remote facee = M / T = 53.09 / 49

= 1.08 m Line of action of forces lies outside the sectioni.e.tension is small

E = e + D / 2 - d b

= 1080 + 450 / 2 - 417 = 888 mm

D

modified moment = 49 x 0.888 d

= 43.51 KNmd'

E = e + D / 2 - d

=

= 617

=

= 327

617 + 327


Ast1

for moment = M / σst x j x d

57.16 x 10 6 / 150 x 0.872 x 417

mm2

Ast2

for direct tension = T / σst

49 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2

Area of one bar x 1000 / required area in m2 / m

mm2 / m.

Ast1


43.51 x 10 6 / 190 x 0.89 x 417

mm2

Ast2


49 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2


Provide 16 mm O bar @ 200 mm C/C…marked(b) = 1005From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450

= 720

On each face = 360Provide 8 mm O bar

spacing of bar == 50.24 x 1000 /360= 139.555556 mm

Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385Vertical Steel ( c)Provide 10 mm O bar


Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392Horizontal steel :-

Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005

Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385

Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392

Design of short walls :-At support M = 57.16 KNm

T = 73.5 KN tension on liquid faceFrom Table 9-5

=

= 1048

=

= 490

1048 + 490



Provide 16 mm O bar @ 130 mm C/C…marked(b) = 1546


mm2

mm2

mm2


mm2


mm2

mm2

mm2

mm2

Ast1


57.16 x 10 6 / 150 x 0.872 x 417

mm2

Ast2


73.5 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2


mm2 / m.

At centreM = 8.16 KNmT = 73.5 KN tension on liquid face

From Table 9-5

=

= 150

=

= 490

150 + 490



Provide 12 mm O bar @ 130 mm C/C…marked(e) = 869

From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450

= 720



Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385Vertical Steel ( c)Provide 10 mm O bar


Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392

Horizontal steel :-

Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869

Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385

Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392

Bottom 1 m will be designed as cantilever

Ast1


8.16 x 10 6 / 150 x 0.872 x 417

mm2

Ast2


73.5 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2


mm2 / m.

mm2

mm2


mm2


mm2 .

mm2

mm2

mm2

Cantilever moment : -

M = OR , whichever is greater.= 9.8 x 3.5 x 1 / 6 = 9.8 x 3.5 / 6= 5.72 KNm = 5.72 ,tension on liquid face.

From Table 9-5

=

= 105From Table 9-3 minimum reinforcement 0.16 %Distribution steel = 0.16 / 100 x 1000 x 450

= 720


spacing of bar == 78.50 x 1000 /360= 218 mm

Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392each face

Base slab :-Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.From table 9-3Minimum steel = 0.229%

= 0.229 / 100 x 1000 x 150



Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.

346Designed section,Elevation etc. are shown in fig.

Top slab : - consider 1 m wide strip. Assume 150 mm thick slab.

4 + 0.15 = 4.15 say 4.5 m

6 + 0.15 = 6.15 say 6.5 m

Dead Load : self 0.15 x 25 = 3.75

floor finish = 1.0

Live load = 1.5

6.25For 1 m wide strip

Yw x H x h2 / 6 Y

w x H / 6

Ast for moment = M / σ

st x j x d

5.72 x 10 6 / 150 x 0.872 x 417

mm2

mm2

mm2

Area of one bar x 1000 / required area in m2 /m

mm2 .

mm2 ,172 mm2 bothway


Ast = mm2

Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m2

lx =

ly =

KN / m2

KN / m2

KN / m2

KN / m2

1.5 x 6.25 = 9.38 KN / m

6.5 / 4.5= 1.4

AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.

Table 26 0.085

0.056

= == 16.15 KNm = 10.64 KNm

From Table 6-3 ,Q = 2.76

== 76.50 mm

150 - 15(cover) - 5= 130 > 76.50 mm …………(O.K.)

130 - 10 = 120 mmLarger depth is provided due to deflection check.

= 0.96

=415 / 20

= 50 [(1-0.88) x 20 / 415 ]= 0.29%

0.29 x 1000 x 130 / 100



Provide 10 mm O bar @ 210 mm c/c = 374

= 0.74

PU =

ly / l

x =

αx =

αy =

Mx = α

x x w x l

x2 M

y = α

y x w x l

x2

0.085 x 9.38 x 4.52 0.056 x 9.38 x 4.52

drequired

= √M / Q x b

√16.15 x 10 6 / 2.76 x 1000

dshort

=

dlong

=

Mu / b x d2 (short) = 16.15 x 106 / 1000 x 130 x 130

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

fy / f

ck

50 1-√1-(4.6 / 20) x (0.96)

Ast (short) =

mm2


mm2 .

Mu / b x d2 (long) = 10.64 x 106 / 1000 x 120 x 120

=415 / 20

= 50 [(1-0.91) x 20 / 415 ]= 0.22%

0.22 x 1000 x 120 / 100

= 264

Provide 8 mm O bar



Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

fy / f

ck

50 1-√1-(4.6 / 20) x (0.74)

Ast (long) =

mm2


mm2 .

B

4 m

C

TABLE 9-6Balanced Design Factors for members in bending

For M20 Grade Concrete Mix

d / DMild steel HYSD bars

0.75 0.3 0.4 0.295 0.289

0.8 0.305 0.37 0.299 0.272

0.85 0.31 0.355 0.302 0.2580.9 0.314 0.335 0.306 0.246

TABLE 9-5

Q = M / bD2 Pt Q = M / bD2 P

t

Members in bending ( Cracked condition )Coefficients for balanced design

k j Q

For members less than 225mm thickness and tension on liquid face

M20 Fe250 7 115 0.445 0.851 1.33Fe415 7 150 0.384 0.872 1.17

For members more than 225mm thickness and tension away from liquid face

M20 Fe250 7 125 0.427 0.858 1.28

Fe415 7 190 0.329 0.89 1.03

Line of action of forces lies outside the section

D / 2

e = M / T

TABLE 9-3Minimum Reinforcement for Liquid Retaining Structures

Thickness, mm% of reinforcement

Mild Steel HYSD bars

100 0.3 0.24

150 0.286 0.229

200 0.271 0.217250 0.257 0.206

Grade of concrete

Grade of steel

σcbc

N / mm2

σst

N / mm2

300 0.243 0.194350 0.229 0.183400 0.214 0.171

450 or more 0.2 0.16

3500

150

1 : 4 : 8 P.C.C.

150 450 6000 450 150

Elevation1500

450

1000

4000

1000

450

1500 1500

6000

450 450

Section A-A

A A15001500

150

150

8 O @ 290 c/c both ways top and bottom

10 O @ 200 c/c

10 O @ 200 c/c - shape

- shape

10 O @ 210 c/c 8 O @ 190 c/c

150 Free board

vv vv1000

16 O @ 130 c/c (a)

10 O @ 200 c/c both faces (c)

12 O @ 130 c/c (e)

16 O @ 200 c/c (b)

8 O @ 130 c/c (d)

( a ) ( a )( b ) ( d )

( c )

( d )

Table 6-3

250 415 500 550

15 2.22 2.07 2.00 1.9420 2.96 2.76 2.66 2.5825 3.70 3.45 3.33 3.23

30 4.44 4.14 3.99 3.87

Limiting Moment of resistance factor Q lim

, N / mm2

fck

N / mm2f

y, N / mm2

TABLE 9-5

Members in bending ( Cracked condition )Coefficients for balanced design

1.360.98

1.2

0.61

Pt,bal

Design of Rectangular water tank CASE-2 ( L / B ≥ 2 )Size of tank : 3.6 m x 8.0 m x 3.0 m high (given)Material M20 Grade Concrete (given)

Fe 415 Grade HYSD reinforcement (given)Solution :-Size of tank : 3.6 m x 8.0 m x 3.0 m high

86400 Litres

L / B = 8 / 3.6 = 2.22 > 2 .The long walls are designed as vertical cantilevers from the base. The short walls are designed as supported on long walls.If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.bottom 1 m or H / 4 whichever is more is designed as cantilever.

H / 4 = 3.0 / 4 = 0.75 mbottom h = 1 m will be designed as cantilever.

Moments and tensions :Maximum B.M. in long walls at the base

=

== 44.1 KNm.

Maximum ( - ve ) B.M. in short walls at support

=

== 26.13 KNm.

Maximum ( + ve ) B.M. in short walls at centre

=

== 19.60 KNm.

For bottom portion

M = OR = 9.8 x 3.0 x 1 / 6 = 9.8 x 3.0 / 6= 4.90 KNm = 4.90 KNm

Direct tension in long wall = = 9.8 x ( 3 - 1 ) x 3.6 / 2 = 35.28 KN

Direct tension in short wall= = 9.8 x ( 3 - 1 ) x 1= 19.6 KN

Volume = 3.6 x 8 x 3.0 x 10 3 =

(1 / 6 ) x Yw x H3

( 1 / 6 ) x 9.8 x 33

Yw x ( H - h ) x B2 / 12

9.8 x ( 3 - 1 ) x 42 / 12

Yw x ( H - h ) x B2 / 16

9.8 x ( 3 - 1 ) x 42 / 16

Yw x H x h2 / 6 Y

w x H / 6 , whichever is greater

Yw

x ( H - h ) x B / 2

Yw ( H - h ) x 1

It is assumed that end one metre width of long wall gives direct tension to short walls.

Design of long walls : -M ( - ) = 44.1 KNm ( water face )

T = 35.28 KN ( perpendicular to moment steel )From Table 9-6 Assume d / D = 0.9 Q = 0.306

D =

== 379.6 mm,

Take D = 400 mm d = 400 - 25 - 8= 367 mm

From Table 9-5 ,

=

= 918.68Provide 16 mm O bar

spacing of bar == 200.96 x 1000 / 918.68= 218.749 mm


From Table 9-3Distribution steel = 0.171 % for 400 mm depth

( 0.171 / 100 ) x 1000 x 400

= 684

on each face = 342 …………………… ( 1 )

Steel required for direct tension

=

=

= 235 …………………… ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O bar


Provide 8 mm O bar @ 140 mm c/c on each face = 357on each face

√M / Q x b

√44.1 x 10 6 / 0.306 x 1000

Ast for Moment

Ast = M / σ

st x j x d

44.1 x 10 6 / 150 x 0.872 x 367

mm2


mm2 .

Note : The design is made at the base. The moment reduces from base to top.For economy, the reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for cantilever retaining wall.

As =

mm2 .

mm2 .

T / σst

35.28 x 103 / 150

mm2 .


mm2

Design of short walls :-At support M = 26.13 KNm

T = 19.6 KNFrom Table 9-5

=

= 544

T / σst

=

= 131

544 + 131



Provide 12 mm O bar@160 mm c/c = 706

1000

203.56

400 367

163.44

checking :

modular ratio m = = 13.33

x =

=( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 )

= 203.56 mm

D - x = 196.44 mmd - x = 163.44 mm

= 1000 x 400 + (13.33 - 1 ) x 706

Ast1


26.13 x 10 6 / 150 x 0.872 x 367

mm2

Ast2

for direct tension =

19.6 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2


mm2.

280 / 3 x σcbc

b x D2 / 2 + Ast ( m - 1 ) x d

b x D + ( m - 1 ) x Ast

( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 )

AT = b x D + ( m - 1 ) x A

st

= 408705

== 5.34E+09 + 2.33E+08

= 5.57E+09

=

= 0.048

=

= 0.767check :

( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) ≤ 1 From Table 9-2

0.4912 ≤ 1 ………………….. ( O. K. )

At centre :M = 19.6 KNmT = 19.6 KN

From Table 9-5

=

= 408

T / σst

=

= 131

408 + 131




From Table 9-3Distribution steel = 0.171 % for 400 mm depth

mm2

Ixx

= ( 1 / 3 ) x b x ( x3 + ( D - x )3 ) + ( m - 1 ) x Ast x ( d - x )2

( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442

mm4

fct

= T / AT

19.6 x 10 3 / 408705

N / mm2

fcbt

= M x ( d - x ) / Ixx

26.13 x 106 x 163.44 / 5.57 x 10 9

N / mm2

( fct / σ

ct ) + ( f

cbt / σ

cbt ) ≤ 1

0.04 + 0.4512 ≤ 1

Ast1


19.6 x 10 6 / 150 x 0.872 x 367

mm2

Ast2

for direct tension =

19.6 x 10 3 / 150

mm2

Total Ast1

+ Ast2

=

mm2


mm2.

( 0.171 / 100 ) x 1000 x 400

= 684

on each face = 342 …………………… ( 1 )

Steel required for direct tension

=

=

= 131 …………………… ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O bar


Provide 8 mm O bar @ 140 mm c/c on each face = 357on each face

Bottom cantileverM = 4.9 KNm

From Table 9-5

=

= 102

Provide 8 mm O bar @ 140 mm c/c on each faces = 357on each face

Base slab :-Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.From table 9-3Minimum steel = 0.229%

= 0.229 / 100 x 1000 x 150



Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.

346Designed section,Elevation etc. are shown in fig.

Top slab : - consider 1 m wide strip. Assume 150 mm thick slab.

3.6 + 0.4 = 4 say 4 m

As =

mm2 .

mm2 .

T / σst

19.6 x 103 / 150

mm2 .


mm2

Ast = M / σ

st x j x d

4.9 x 10 6 / 150 x 0.872 x 367

mm2

Minimum steel = 342 mm2 on each face.

mm2

mm2 ,172 mm2 bothway


Ast = mm2

Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m2

lx =

8 + 0.15 = 8.15 say 8.5 m

Dead Load : self 0.15 x 25 = 3.75

floor finish = 1.0

Live load = 1.5

6.25For 1 m wide strip

1.5 x 6.25 = 9.38 KN / m

Maximum moment = = 18.76 KNm

Maximum shear = 9.38 x 3.6 / 2= 16.88 KN

From Table 6-3 ,Q = 2.76

== 82.44 mm

150 - 15(cover) - 6= 129 > 82.44 …………(O.K.)

Larger depth is provided due to deflection check.Design for flexure :

= 1.13

=415 / 20

= 50 [(1-0.86) x 20 / 415 ]= 0.34%

0.34 x 1000 x 129 / 100




ly =

KN / m2

KN / m2

KN / m2

KN / m2

PU =

9.38 x 42 / 8

drequired

= √M / Q x b

√18.76 x 10 6 / 2.76 x 1000

dprovided

=

Mu / b x d2 = 18.76 x 106 / 1000 x 129 x 129

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

fy / f

ck

50 1-√1-(4.6 / 20) x (1.13)

Ast =

mm2


mm2 .

Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcementDistribution steel = ( 0.12 / 100 ) x 1000 x 150

= 180

Provide 6 mm O bar



mm2


mm2 .

If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.

It is assumed that end one metre width of long wall gives direct tension to short walls.

Note : The design is made at the base. The moment reduces from base to top.For economy, the reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for

x

Table 9-2

M15 1.1 1.5 1.5M20 1.2 1.7 1.7

M25 1.3 1.8 1.9

M30 1.5 2.0 2.2

M35 1.6 2.2 2.5M40 1.7 2.4 2.7

) + ( 13.33 - 1 ) x 706 x 163.442

Permissible concrete stresses in calculations relating to resistance to cracking

Grade of concrete

Permissible stresses in N / mm2

Direct tension σct

Tension due to bending σcbt

Shear stress ح

v = V / b j d

150

3000

150

1 : 4 : 8 P.C.C. 150

150 400 8000 400

Section A-A

2000

400

900

3600

900

400

2000 2000

8000

400 Sectional plan 400

150

8 O @ 140 c/c (chipiya)

20002000

8 O @ 290 c/c both ways top and bottom

12 O @ 200 c/c

- shape

8 O @ 110 c/c 6 O @ 150 c/c

vv vv900

12 O @ 160 c/c(chipiya)16 O @ 200 c/c ( chipiya )

150

8 O @ 140 mm c/c

8 O @ 140 c/c

12 O @ 200 c/c A A

B

B

8 O @ 140 c/c

8 O @ 140 c/c

8 O @ 110 c/c 6 O @ 150 c/c

3000

150

150

150 400 3600 400 150

Section B- B

Base details not shown for clarity

16 O @ 200 c/c ( chipiya )

900 900

8 O @ 140 c/c

Design of simply supported one way slabeffective span = 4 m supported on masonry wall of 230 mm thickness ( given )

Live load = 2.5 ( given )

Floor finish = 1 ( given )material M15 grade concrete ( given )

HYSD reinforcement grade Fe415 ( given )solution : - Assume 0.4 % steel , a trial depth by deflection criteria IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.26 for simply supported, basic span / effective depth ratio = 20( span / d ) ratio permissible = 1.26 x 20

= 25.2

4000 / 25.2

= 158.7 mmD = 158.7 + 15 ( cover ) + 5 ( assume 10 O bar )

= 178.7 mmAssume an overall depth = 180 mmDL = 0.18 x 25 = 4.5

Floor finish = 1.0

Live load = 2.5

Total 8.0Factored load = 1.5 x 8 = 12 KN / m

Consider 1 m length of slabMaximum moment =

== 24 KNm

Maximum shear = w x l / 2= 12 x 4 / 2= 24 KN

Design for flexure : -d = 180 - 15 - 5

= 160 mm

= 0.94

=415 / 15

= 50 [(1-0.84) x 15 / 415 ]

KN / m2

KN / m2

drequired

=

KN / m2

KN / m2

KN / m2

KN / m2

w x l2 / 8

12 x 42 / 8

Mu / b x d2 = 24 x 10 6 / 1000 x (160)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.94)

= 0.289%

0.289 x 1000 x 160 / 100




100 x 231 / ( 1000 x 180 )

= 0.128 > 0.12 % ( minimum steel for Fe415)i.e. remaining bars provide minimum steel. Thus, half the bars may be bent up.

Distribution steel = ( 0.12 /100 ) x 1000 x 180 = 216

Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mmProvide 8 mm O bar


Provide 8 mm O bar @ 230 mm c/c = 218Check for shear : -

24 KN

Actual Shear stress =

=

= 0.150 ( too small )

For bars at support

d = 160 mm

231

100 x 231 / 1000 x 160= 0.144

6 x β

β == 0.8 x 15 / 6.89 x 0.144= 12.1

6 x 12.1= 0.277

IS 456-2000 Table 19 from table 7-1

Ast =

mm2


mm2 .

Half the bars are bent at 0.1 l = 400 mm , and remaining bars provide 231 mm2 area

100 x As / ( b x D ) =

mm2


mm2 .

Vu =

Vu / b x d

24 x 103 / 1000 x 160

N / mm2 ح ) >C )

N / mm2

As = mm2 .

100 x As / b x d =

Design shear strength حc = 0.85 √ 0.8 x f

ck ( √ 1 + 5 x β - 1 )

0.8 x fck

/ 6.89 Pt , but not less than 1.0

Design shear strength حc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.1 - 1 )

IS 456-2000 clause 40.2.1.1 25 difference -0.05k = 1.24 for 180 mm slab depth 20 difference ? -0.04

Design shear strength = 1.24 x 0.28

= 0.347 ……………….( O.K.)Check for development length : -

8 O (HYSD Fe415 steel ) For continuing bars

231

OR= 0.87 x 415 x 231 x 160 { 1 - (415 x 231 / 15 x 1000 x 160 ) }= 13.34 { 1- 0.0399 }= 12.812 KNm

24 KN

= 56 O ( from Table 7-6 )

693.875 + 8 O693.88

which gives 14.46 mm ……………….( O.K.)Check for deflection : -

Basic ( span / d ) ratio = 20

100 x 462 / 1000 x 160= 0.289

IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.42( span / d ) ratio permissible = 1.42 x 20

= 28.4Actual (span / d ) ratio = 4000 / 160

= 25.00 < 28.4 ……………….( O.K.)The depth could be reducedCheck for cracking : -IS 456-2000 , clause 26.3.3

Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 160 = 480 mm or 300 mm i.e. 300 mm

spacing provided = 170 mm < 300 mm ……………….( O.K.)Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small

= 5 x 160 = 800 mm spacing provided = 230 mm < 450 mm ……………….( O.K.)

for Pt = 0.144 حc = 0.28 N / mm2

N / mm2

Assuming L0 =

As = mm2

Mu1

= 0.87 x fy x A

st x d { 1 - ( f

y x A

st / f

ck x b x d ) }

Vu =

Development length of bars Ld = O σ

s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 12.81 x 106 / 24 x 103 ) + 8 O ≥ 56 O≥ 56 O

48 O ≤O ≤

Pt = 100 x A

st / b x d =

For tying the bent bars at top , provide 8 mm O @ 230 mm c/c

NOTE : -

NOTE : -

If clear span = 3.77 m , effective span = 3.77 + 0.23 = 4 m OR effective span = 3.77 + 0.16 ( effective depth ) = 3.93 m

0.6 % for mild steel reinforcement and 0.3 to 0.4 % for HYSD Fe415 grade reinforcement

For mild steel minimum reinforcement 0.15 %

= 100 x 231 / 1000 x 160= 0.14

From equation

we get , 0.49

= 12.54 KNm

160

180

400 400

4000

3 x effective depth of slab or 300 mm whichever is smallor 300 mm i.e. 300 mm

……………….( O.K.)5 x effective depth of slab or 450 mm whichever is small

or 450 mm i.e. 450 mm……………….( O.K.)

For checking development length , l0

may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

Mu1

/ b x d2 =

Mu1

= 0.49x 1000 x 1602 x 10-6

vv

vv

vvvv

vv

8 O @ 230 c/c10 O @ 170 c/c ( alternate bent )

IS 456-2000 clause 22.2Effective Span( a ) Simply Supported Beam or Slab -

( b )Continuous Beam or Slab - In the case ofcontinuous beam or slab, if the width of the

support is less than l/12 of the clear span, theeffective span shall be as in (a). If the

supports are wider than I/12 of the clear span

or 600 mm whichever is less, the effective span

shall be taken as under:

1) For end span with one end fixed and theother continuous or for intermediate spans,the effective span shall be the clear spanbetween supports;

2) For end span with one end free and the othercontinuous, the effective span shall be equalto the clear span plus half the effective depthof the beam or slab or the clear span plushalf the width of the discontinuous support,whichever is less;3) In the case of spans with roller or rocketbearings, the effective span shall always bethe distance between the centres of bearings.

( c )Cantilever-The effective length of a cantilever

shall be taken as its length to the face of the

support plus half the effective depth except

where it forms the end of a continuous beam

where the length to the centre of support shall

be taken.

( d )Frames-In the analysis of a continuous frame,centre to centre distance shall be used.

If ly / l

x ≥ 2 ,called one way slab provided that it

is supported on all four edges . Note that , if all four edge is not supported and l

y / l

x < 2 , then

also it is one-way slab,If ly / l

x < 2 , called two-

way slab.provided that it is supported on all four edges.

The effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre to centre of supports , whichever is less.



Design of Cantilever one way slabused for residential purpose

material M15 grade concrete ( given )mild steel grade Fe250 ( given )

Live Load As per IS 875Solution : -

( Balcony slab )Assume 120 mm thick slab DL LL

self load = 0.12 x 25 = 3 0

floor finish = 1 0

live load = 0 2

Total 4 2

1.5 ( 4 + 2 ) = ( 6 + 3 )DL LL

self load = 0.12 x 25 = 3 0

floor finish = 1 0

live load = 0 3

Total 4 3

1.5 ( 4 + 3 ) = (6 + 4.5)Weight of parapet 0.075 x 25 x 1 = 1.875 KN / m

1.5 x 1.875 = 2.8 KN / mConsider 1 m long strip

9 KN/m 6 KN/m

A 3m B 1.2m C

(a) Loads for maximum positive moment9 KN/m 10.5 KN/m 2.8 KN

A 3m B 1.2m C

considering fig (a)

cantilever moment =

at the free end of slab S1 ,concrete parapet of 75 mm thick and 1 m high.

For slab S2 live load = 2 KN /m2

For slab S1 live load = 3 KN /m2

ly = 6m

For S2 KN / m2

KN / m2

KN / m2

KN / m2

Pu = KN / m2

For S1 KN / m2

KN / m2

KN / m2

KN / m2

Pu = KN / m2

Pu =

(1) To get maximum positive moment in slab S2 only dead load on slab

S1 and total load on slab S

2 shall be considered

(b) Loads for maximum negative moment,maximum shear for cantilever span and maximum reaction at support B

wx l2 / 2

S2S2

S2

S1

S1

== 4.32 KNm

shear = Reaction at A = w x l / 2 - Moment @ B at distance 3 m= 9 x 3 / 2 - 4.32 / 3= 12.06 KN

Point of zero shear from A = 12.06 ( KN ) / 9 ( KN / m )= 1.34 m

Maximum positive moment =

== 8.08 KNm

(2) To get maximum negative moment and maximum shear at B,the slab is loaded with full loads as shown in fig (b)

Maximum negative moment =

== 10.92 KNm

Maximum shear at B

w x l / 2 + Moment @ B at distance 3 m= 9 x 3 / 2 + 10.92 / 3= 17.14 KN

w x l + 2.8= 10.5 x 1.2 + 2.8= 15.4 KN

Moment steel :Maximum moment = 10.92 KNm

From Table 6-3 Q = 2.22

== 70.14 mm,

120 - 15 - 6 ( assume 12 O bar )= 99 mm, ……………….( O.K.)

= 0.82

=250 / 15

= 50 [(1-0.865) x 15 / 250 ]= 0.405%

6 x 1.22 / 2

12.06 x 1.34 - W x l2 / 2

12.06 x 1.34 - 9 x 1.342 / 2

w x l + w x l2 / 2

2.8 x 1.2 + 10.5 x 1.22 / 2

Vu,BA

=

Vu,BC

=

drequired

= √M / Q x b

√10.92 x 10 6 / 2.22 x 1000

dprovided

=

Mu / b x d2 ( + ) = 8.08 x 10 6 / 1000 x (99)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.82)

0.405 x 1000 x 99 / 100



Provide 10 mm O bar@170 mm c/c = 462 ( alternate bent up )

= 1.11

=250 / 15

= 50 [(1-0.812) x 15 / 250 ]= 0.564%

0.564 x 1000 x 99 / 100


Provide 10 mm O bar@340 mm c/c = 231 ( bent bars extended )Area provided = Area of one bar x 1000 / spacing of bar in m

= 78.5 x 1000 / 340

= 231remaining area = 558 - 231


spacing of bar == 113.04 x 1000 / 327= 346 mm

Provide 12 mm O bar@340 mm c/c = 332 ( Extra )

For mild steel minimum reinforcement 0.15 %Distribution steel = ( 0.15 /100 ) x 1000 x 120



Ast ( + ) =

mm2


mm2.

Mu / b x d2 ( - ) = 10.92 x 10 6 / 1000 x (99)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (1.11)

Ast ( - ) =

mm2

mm2.

mm2

mm2


mm2.

Total 231 + 332 mm2 = 563 mm2 steel provided

Note that at simple support , the bars are bent at 0.1 l whereas at continuity of slab it is bent at 0.2 l

mm2.


Provide 6 mm O bar@150 mm c/c = 188For negative moment reinforcement

( from Table 7-6 )= O x 0.87 x 250 / 4 x 1= 54.4 O

54.4 x ( 10 + 12 ) / 2= 598 mm. say 600 mm

Check for development length : -

12 O (mild steel )

At A ,= 100 x 231 / 1000 x 99= 0.233

From equation OR

=

we get , 0.487 = 4.97401

= 4.78= 4.77 KNm

12.06 KN

= O x 0.87 x 250 / 4 x 1 =54.4 O

514.179 + 12 O514.179

which gives 12.13 mm ……………….( O.K.)

At B ,= 100 x 231 / 1000 x 99= 0.233

From equation OR

=

we get , 0.487 = 4.974

= 4.78= 4.77 KNm

Near point of contraflexure i.e. 0.15 x l from B

17.14 - ( 0.15 x 3 ) x 9 = 13.09 KN

= O x 0.87 x 250 / 4 x 1 =54.4 O

mm2.


s / 4 x ح

bd

Ld =

As a thumb rule, a bar shall be given an anchorage equal to the length of the cantilever.

Assuming L0 =

Pt = 100 x A

s / b x d Half bars bent = 462 / 2 = 231 mm2 )

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 0.487 x 1000 x 992 x 10-6

Vu =


s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 4.77 x 106 / 12.06 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O

42.4 O ≤O ≤

Pt = 100 x A

s / b x d Half bars bent = 462 / 2 = 231 mm2 )

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 0.487 x 1000 x 992 x 10-6

Vu =


s / 4 x ح

bd

473.72 + 12 O473.72

which gives 11.2 mm ……………….( O.K.)Check for shear : - Span AB :

12.06 KN ( for maximum loading )At B , shear at point of contraflexure = 13.09 KN

13.09 KN

=

= 0.132 ……………….( O.K.)

100 x 231 / 1000 x 99= 0.233

6 x β

β == 0.8 x 15 / 6.89 x 0.233= 7.47

6 x 7.47= 0.34


IS 456-2000 clause 40.2.1.1 0.1 difference 0.07k = 1.3 for 120 mm slab depth 0.02 difference ? 0.014


= 0.442 ……………….( O.K.)Span BC :

17.14 KN

=

= 0.173 ……………….( O.K.)

100 x 563 / 1000 x 99= 0.569

6 x β

β == 0.8 x 15 / 6.89 x 0.569

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 4.77 x 106 / 13.09 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O

42.4 O ≤O ≤

At A , Vu,AB

=

Use Vu =

Shear stress حv = V

u / b x d

13.09 x 10 3 / 1000 x 99

N / mm2 ح >c

Pt = 100 x A

s / b x d =


ck ( √ 1 + 5 x β - 1 )

0.8 x fck



for Pt = 0.233 حc = 0.34 N / mm2

N / mm2 ح <v

Vu =

Shear stress حv = V

u / b x d

17.14 x 10 3 / 1000 x 99

N / mm2 ح >c

Pt = 100 x A

s / b x d =


ck ( √ 1 + 5 x β - 1 )

0.8 x fck


= 3.06

6 x 3.06= 0.487


IS 456-2000 clause 40.2.1.1 0.25 difference 0.08k = 1.3 for 120 mm slab depth 0.181 difference ? 0.05792


= 0.624 ……………….( O.K.)Check for deflection : -For span AB :


100 x 462 / 1000 x 99= 0.467

IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 2( span / d ) ratio permissible = 2 x 20

= 40Actual (span / d ) ratio = 3000 / 99

= 30.30 < 40 ……………….( O.K.)For span BC :


100 x 563 / 1000 x 99= 0.569



= 12.12 < 12.6 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small

= 3 x 99 = 297 mm or 300 mm i.e. 297 mmspacing provided = 170 mm < 297 mm ……………….( O.K.)

(2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small= 5 x 99 = 495 mm

spacing provided = 150 mm < 450 mm ……………….( O.K.)


for Pt = 0.569 حc = 0.48 N / mm2

N / mm2 ح <v

Pt = 100 x A

st / b x d =

Pt = 100 x A

st / b x d =

1.2m

0.15m

0.15m

0.15m 0.15m

1.2m

ly = 6m

lx = 3m

S2 S1

S3

B1B2

B3

B4

1m high parapet

Column 300 x 300

( 1 - 0.0389 )

KNm

( 1 - 0.0389 )

KNm

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6

1200 1200

120 125

300 600

150 3000 150 1200



or 450 mm i.e. 450 mm……………….( O.K.)

10 O @ 340 c/c (bent)+ 12 O @ 340 c/c (extra)

10 O @ 170 c/c

6 O @ 150 c/c6 O @ 150 c/c

Design of Continuous Two-way slabBanking hallslab is restrained with edge beams

givenMaterial M15 grade concreteHYSD reinforcement of grade Fe415

Solution : - Assume 150 mm thick slab

Self load 0.15 x 25 = 3.75

floor finish = 1.00

Live Load = 3.00

Total 7.75

1.5 x 7.75

= 11.625

5 / 5= 1 < 2 i.e. two-way slab.

Middle strip : IS 456-2000 Table -26 Two adjacent edges Discontinuous

13.66 KNm

10.17 KNmFrom Table 6-3

For M15 mix and Fe415 steel, Q = 2.07

== 81.23 mm,

= 150 - 15 - 10 - 5 ( assume 10 O bar )= 120 mm, > 81.23 mm ……………….( O.K.)

= 150 - 15 - 5 = 130 mm, > 81.23 mm ……………….( O.K.)

= 0.706

=415 / 15

= 50 [(1-0.885) x 15 / 415 ]= 0.208%

KN / m2

KN / m2

KN / m2

KN / m2

Pu =

KN / m2

ly / l

x =

Mu1

, Mu3

( - ) = αx x w x l

x2 = 0.047 x 11.625 x 52 =

Mu2

( + ) = αy x w x l

x2 = 0.035 x 11.625 x 52 =

drequired

= √M / Q x b

√13.66 x 10 6 / 2.07 x 1000

dprovided

for positive moment reinforcement

dprovided

for negative moment reinforcement

Mu / b x d2 ( + ) = 10.17 x 10 6 / 1000 x (120)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.706)

0.208 x 1000 x 120 / 100



Provide 8 mm O bar@ 200 mm c/c = 251

= 0.81

=415 / 15

= 50 [(1-0.867) x 15 / 415 ]= 0.240%

0.24 x 1000 x 130 / 100



Provide 8 mm O bar@ 150 mm c/c = 335For HYSD Fe415

Minimum steel = ( 0.12 / 100 ) x 1000 x 150

= 180At discontinuous edges 4 and 5 , 50 % of the positive steel is required at top

= ( 1 / 2 ) x 251

= 126This is less than minimum , therefore , use minimum steel at location 4 and 5 .Provide 8 mm O bar


Provide 8 mm O bar@ 260 mm c/c = 193More steel is provided to match with the torsion reinforcement.In edge strip , minimum reinforcement is provided equal to 8 mm O @ 260 mm c/c.Torsion steel :-At corner A , steel required = ( 3/4 ) x 250

Ast =

mm2


mm2.

Mu / b x d2 ( - ) = 13.66 x 10 6 / 1000 x (130)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.81)

Ast =

mm2


mm2.

mm2

mm2


mm2.



Provide 8 mm O bar@ 260 mm c/c = 193This will be provided by minimum steel of edge strip,At corner B , steel required = ( 1/2 ) x 188



Provide 8 mm O bar@ 260 mm c/c = 193This will be provided by minimum steel .

Check for shear : -At point 1 or 3

S.F. = w x l / 2 + Moment @ point 1 or 3 in that span= 11.625 x 5 / 2 + 13.66 / 5= 31.795 KN

100 x 335 / ( 1000 x 130 )= 0.258

from table 7-1

0.25 differen 0.11IS 456-2000 clause 40.2.1.1 0.242 differe ? -0.1065

k = 1.3 for 150 mm slab depthDesign shear strength = 1.3 x 0.354

= 0.460 ……………….( O.K.)OR

6 x β

β == 6.75

6 x 6.75= 0.356

IS 456-2000 clause 40.2.1.1k = 1.3 for 150 mm slab depth


= 0.463 ……………….( O.K.)

Actual shear stress =

mm2.


mm2.

mm2.


mm2.

Note that positive reinforcements are not curtailed because if they are curtailed , the remaining bars do not provide minimum steel.

100 x As / b x d =

for Ptح , 0.258 =

c = 0.354 N / mm2

N / mm2


ck ( √ 1 + 5 x β - 1 )

0.8 x fck



N / mm2

Vu / b x d

=

= 0.245 ……………….( O.K.)At point 4 or 5

S.F. = w x l / 2 = 11.625 x 5 / 2 = 29.06 KN

100 x 251 / ( 1000 x 120 )= 0.209

from table 7-1

0.1 differenc 0.07IS 456-2000 clause 40.2.1.1 0.041 differe ? -0.0287


= 0.417OR

6 x β

β == 8.33

6 x 8.33= 0.326



= 0.424

Actual shear stress =

=

= 0.242 ……………….( O.K.)Check for development length : -This is critical at point 4 or 5.

At point 4 or 5 , 29.06 KN No bar is curtailed or bent up.

8 O (HYSD Fe415 steel )

= 100 x 251 / 1000 x 120= 0.209

From equation OR

=

we get , 0.711 = 10.8748

31.795 x 103 / ( 1000 x 130 )

N / mm2 ح > c

100 x As / b x d =

for Ptح , 0.209 =

c = 0.321 N / mm2

N / mm2


ck ( √ 1 + 5 x β - 1 )

0.8 x fck



N / mm2

Vu / b x d

29.06 x 103 / ( 1000 x 120 )

N / mm2 ح > c

Vu =

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6

Mu1

/ b x d2 =

= 10.246= 10.24 KNm

=54.3 O (From Table 7-6 )

458.355 + 8 O458.355


Short span 11.25 x ( 4.5 / 2 ) = 25.31 KN


= 100 x 462 / 1000 x 160= 0.289

From equation OR

=

we get , 0.9595 = 26.6888

= 24.5539= 24.56 KNm

=56 O (From Table 7-6 )

1261.48 + 8 O1261.48

which gives 26.28 mm ……………….( O.K.)Note that the bond is usually critical along long direction.Check for deflection : -


positive moment steel = 251actual d = 150 - 15 -8 - 4

= 123 mm.

100 x 251 / 1000 x 123 240.7= 0.204



= 40.65 < 42.9 ……………….( O.K.)Check for cracking : -IS 456-2000 , clause 26.3.3

Mu1

= 0.711 x 1000 x 1202 x 10-6

Development length of bars Ld = O σs / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 10.246 x 106 / 29.06 x 103 ) + 8 O ≥ 54.3 O≥ 54.3 O

46.3 O ≤O ≤

Vu =

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 0.9595 x 1000 x 1602 x 10-6


s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O

48 O ≤O ≤

mm2.

Pt = 100 x A

st / b x d =

(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 130 = 390 mm

spacing provided = 200 mm < 300 mm ……………….( O.K.)(2) Distribu. bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small

= 5 x 120 = 600 mmspacing provided = 260 mm < 450 mm ……………….( O.K.)

This is minimum (0.12 / 100) x 150 x 1000 = 180Provide 8 mm O bar


Provide 8 mm O bar@ 260 mm c/c = 193 for uniformity in spacing.For clarity , top and bottom reinforcements are shown separately.

625 3750 625

625

3750

625

150

625 3750 625

Section A-A

Note that the bottom reinforcements are both ways and therefore there is no necessity of secondary reinforcements.However , top reinforcement in edge strip requires the secondary steel for tying the bars.

mm2


mm2.

Mid

dle

Str

ip

S1

Edg

e s

trip

Ed

ge

stri

p

Edge strip Edge stripMiddle Strip

8 O

@ 2

60

c/c

8 O

@ 2

60

c/c

8 O

@ 2

00

c/c

8 O @ 260 c/c8 O @ 260 c/c8 O @ 200 c/c

A A

B

B

500

1500 1500

8 O @ 260 c/c

8 O @ 150 c/c

8 O @ 260 c/c

8 O @ 200 c/c

Design of Continuous Two-way slab 5 m

5 m

5 m

625 3750 625 5 m

ly/8 (3/4)l

yly/8

Middle Strip

S1

S1

Ed

ge s

trip

Ed

ge

str

ip

A B

B

1

2

3

4

5

0.0

35

-0.0

47

-0.047

0.035

( 1 - 0.0579 )

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6

KNm

( 1 - 0.0799 )

KNm

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6



or 450 mm i.e. 450 mm……………….( O.K.)

1000

625 3750 625

1000 625

3750

625

150

625 3750 625

Mid

dle

Str

ip

S1

Ed

ge s

trip

Ed

ge

str

ip

Edge strip Edge stripMiddle Strip

8 O

@ 2

40

c/c

8 O

@ 2

40

c/c

8 O

@ 1

40

c/c

8 O @ 240 c/c8 O @ 240 c/c8 O @ 140 c/c

500

1500 1500

8 O @ 240 c/c

8 O @ 140 c/c

8 O @ 240 c/c

8 O @ 240 c/c

500

15

001

500

1500 1500

8 O @ 180 c/c

3

-0.047

Design of Continuous One-way slabA five span continuous one-way slab used as an office floor.The centre-to-centre distance of supporting beams is 3 m

givenMaterial M15 grade concrete

HYSD reinforcement of grade Fe415Solution : -Try 120 mm thick slab DL LL

Dead load 0.12 x 25 = 3 0

floor finish = 1 0

live load = 0 3

Total 4 3factored load = 1.5 ( 4 + 3 )

=Consider 1 m wide strip of the slab.

3m 3m 3m 3m 3m

The factored moments at different points using the coefficients are as follows :

== 4.5 + 4.05= 8.55 KNm

== 3.38 + 3.38= 6.75 KNm

== 5.4 + 4.5= 9.9 KNm

== 4.50 + 4.50= 9 KNm

0.6 x w x l + 0.6 x w x l= 0.6 x 6 x 3 + 0.6 x 4.5 x 3= 18.9 KN

KNm

Live load 3 KN / m2 and floor finish 1 KN / m2

KN / m2

KN / m2

KN / m2

KN / m2

( 6 + 4.5 ) KN / m2

Mu1

( + ) = ( 1 / 12 ) x w ( DL ) x l2 + ( 1 / 10 ) x w ( LL ) x l2

( 1 / 12 ) x 6 x 32 + ( 1 / 10 ) x 4.5 x 32

Mu2

( + ) = ( 1 / 16 ) x w ( DL ) x l2 + ( 1 / 12 ) x w ( LL ) x l2

( 1 / 16 ) x 6 x 32 + ( 1 / 12 ) x 4.5 x 32

Mu3

( - ) = ( 1 / 10 ) x w ( DL ) x l2 + ( 1 / 9 ) x w ( LL ) x l2

( 1 / 10 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32

Mu4

( - ) = ( 1 / 12 ) x w ( DL ) x l2 + ( 1 / 9 ) x w ( LL ) x l2

( 1 / 12 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32

Maximum shear is Vu(BA)

=

Maximum moment is Mu3

( - ) = 9.9

A B C D E F

1 2 2 2 1

( 6 + 4.5 ) KN / m


== 69.2 mm,

Try 110 mm overall depth

110 - 15 - 5 ( assume 10 O bar )= 90 mm, ……………….( O.K.)

spacing of bar =Table

point Steel Provided

1 ( + ) 8.55 1.06 0.322 29010 mm O @ 440 c/c + 8 mm O @ 440 c/c

2 ( + ) 6.75 0.83 0.248 2238 mm O @ 220 c/c

3 ( - ) 9.9 1.22 0.38 34210 mm O @ 220 c/c

4 ( - ) 9.0 1.11 0.34 30610 mm O @ 220 c/c

For Main steel ,HYSD Fe415 reinforcementminimum steel area = ( 0.12 / 100 ) x 1000 x 110

= 132For Distribution steel , mild steel Fe250 reinforcement

minimum steel area = ( 0.15 / 100 ) x 1000 x 110

= 165Use 6 mm O

spacing of bar == 28.26 x 1000 /165= 171

Use 8 mm O

spacing of bar == 50.24 x 1000 /146= 344

drequired

= √M / Q x b

√9.9 x 10 6 / 2.07 x 1000

dprovided

=

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

Ast = P

t x b x d / 100


Factored moment

KNmMu/(b x d2 ) P

t

Ast

mm2

= 178 +114 = 292 mm2 (Half 10 O+half 8 O)

= 228 mm2

= 357 mm2

= 357 mm2

mm2

mm2


Use 6 mm O @ 160 mm c/c = 177 mm2.

Note that the positive bars cannot be curtailed as the remaining bars in the internal spans ( + ve moment ) will not provide minimum area.

Provide 50 % Ast

at end support top bars i.e. 292 / 2 = 146 mm2 .


Use 8 mm O @ 340 mm c/c = 148 mm2.

Check for shear : -Maximum shear = 18.9 KN


=

= 0.210 ( too small )

For bars at support

d = 90 mm

357

100 x 357 / 1000 x 90= 0.397

6 x β

β == 0.8 x 15 / 6.89 x 0.397= 4.4

6 x 4.4= 0.42


IS 456-2000 clause 40.2.1.1 0.25 difference 0.11k = 1.3 for 110 mm slab depth 0.103 difference ? 0.04532Design shear strength = 1.3 x 0.42

= 0.546 ……………….( O.K.)At point of contraflexure i.e. 0.15 x l from B

18.9 - 0.15 x 3 x 10.5= 14.18 KN


=

= 0.158 ( too small )

For bars at support

d = 90 mm

292

100 x 292 / 1000 x 90= 0.324

6 x β

β =

Vu / b x d

18.9 x 103 / 1000 x 90

N / mm2 ح ) >C )

N / mm2

As = mm2 .

100 x As / b x d =


ck ( √ 1 + 5 x β - 1 )

0.8 x fck



for Ptح 0.397 =

c = 0.42 N / mm2

N / mm2

with positive moment reinforcement ( 292 mm2 )

Vu =

Vu / b x d

14.18 x 103 / 1000 x 90

N / mm2 ح ) >C )

N / mm2

As = mm2 .

100 x As / b x d =


ck ( √ 1 + 5 x β - 1 )

0.8 x fck


= 0.8 x 15 / 6.89 x 0.324= 5.4

6 x 5.4OR = 0.39IS 456-2000 Table 19 from table 7-1

IS 456-2000 clause 40.2.1.1 0.25 difference 0.11k = 1.3 for 110 mm slab depth 0.176 difference ? 0.07744Design shear strength = 1.3 x 0.38

= 0.494 ……………….( O.K.)Check for development length : -Span AB is critical for checking this requirementAt support A

0.4 x w x l + 0.45 x w x l= 0.4 x 6 x 3 + 0.45 x 4.5 x 3 = 13.28 KN

At A ,= 100 x 292 / 1000 x 90= 0.324

From equation OR

=

we get , 1.064 = 9.48839

= 8.64= 8.62 KNm


= O x 0.67 x 415 / 4 x 1 =56.4 O

845.7831 + 8 O845.783

which gives 17.47 mm ……………….( O.K.)At support B, point of contraflexure is assumed at 0.15 x l from B

18.9 - 0.15 x 3 x 10.5= 14.18 KN

8.64 KNm as before

12 O

792.1016 + 12 O


for Ptح 0.324 =

c = 0.38 N / mm2

N / mm2

Vu =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 1.064 x 1000 x 902 x 10-6

Assuming L0 =


s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 8.64 x 106 / 13.28 x 103 ) + 8 O ≥ 56.4 O≥ 56.4 O

48.4 O ≤O ≤

Vu =

Mu1

=

L0 = ( actual anchorage is more than 12 O but L

0 is limited to 12 O or d ,

i.e. 90 mm whichever is greater )

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 8.64 x 106 / 14.18 x 103 ) + 12 O ≥ 56.4 O≥ 56.4 O

792.102which gives 17.84 mm ……………….( O.K.)

Check for deflection : -Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB


100 x 292 / 1000 x 90= 0.324



= 33.33 < 34.84 ……………….( O.K.)For span BC :


100 x 228 / 1000 x 90= 0.253




(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 90 = 270 mm or 300 mm i.e. 270 mm

spacing provided = 220 mm < 270 mm ……………….( O.K.)(2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small

= 5 x 90 = 450 mm spacing provided = 160 mm < 450 mm ……………….( O.K.)

44.4 O ≤O ≤

Pt = 100 x A

st / b x d =

Pt = 100 x A

st / b x d =

IS 456-2000 Clause -22.5( 22.5.1 ) Unless more exact estimates are made, forbeams of uniform cross-section which support

substantially uniformly distributed loads over three ormore spans which do not differ by more than 15 percentof the longest, the bending moments and shear forcesused in design may be obtained using the coefficientsgiven in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans

meet or in case where the spans are not equally loaded,

the average of the two values for the negative moment

at the support may be taken for design.Where coefficients given in Table 12 are used for

calculation of bending moments, redistribution referredto in 22.7 shall not be permitted. (22.5.2 ) Beams and Slabs Over Free End SupportsWhere a member is built into a masonry wall whichdevelops only partial restraint, the member shall bedesigned to resist a negative moment at the face of thesupport of Wl / 24 where W is the total design loadand I is the effective span, or such other restrainingmoment as may be shown to be applicable. For such a

condition shear coefficient given in Table 13 at the

end support may be increased by 0.05.Table 12 Bending Moment coefficients

Type of load

Span moments Support moments

+ 1 / 12 + 1 / 16 -1 / 10

+ 1 / 10 + 1 / 12 -1 / 9

Table 13 Shear Force coefficients

Type of loadAt support next to the end support

Outer side Inner side

0.4 0.6 0.55

0.45 0.6 0.6

Near middle of end span

At middle of interior span

At support next to the end support

Dead load and imposed load ( fixed )

imposed load ( not fixed )

NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span.

At end support

Dead load and imposed load ( fixed )

imposed load ( not fixed )

90110

450 900 900 900

3000 3000

10 O @ 220 c/c8 O @ 340 c/c

6 O @ 160 c/c8 O @ 220 c/c8 O @ 440 c/c

+ 10 O @ 440 c/c

( 0.3 l1 ) ( 0.3 l1 ) ( 0.3 l1 )( 0.15 l1 )

with positive moment reinforcement ( 292 mm2 )

( 1 - 0.0898 )

KNm

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6

Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB



or 450 mm i.e. 450 mm……………….( O.K.)

( 22.5.1 ) Unless more exact estimates are made, forbeams of uniform cross-section which support

substantially uniformly distributed loads over three ormore spans which do not differ by more than 15 percentof the longest, the bending moments and shear forcesused in design may be obtained using the coefficientsgiven in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans

meet or in case where the spans are not equally loaded,

the average of the two values for the negative moment

at the support may be taken for design.Where coefficients given in Table 12 are used for

calculation of bending moments, redistribution referredto in 22.7 shall not be permitted. (22.5.2 ) Beams and Slabs Over Free End SupportsWhere a member is built into a masonry wall whichdevelops only partial restraint, the member shall bedesigned to resist a negative moment at the face of thesupport of Wl / 24 where W is the total design loadand I is the effective span, or such other restrainingmoment as may be shown to be applicable. For such a

condition shear coefficient given in Table 13 at the

end support may be increased by 0.05.Table 12 Bending Moment coefficients

Support moments

-1 / 12

-1 / 9

Table 13 Shear Force coefficients

0.5

0.6

At other interior supports

NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total

At all other interior

supports

900

3000

10 O @ 220 c/c

( 0.3 l1 )

Design of Simply supported two way slabresidential building drawing room 4.3 m x 6.55 m

It is supported on 350 mm thick walls on all four sides.given

material M15 grade concreteHYSD reinforcement of grade Fe415

Solution :Consider 1 m wide strip. Assume 180 mm thick slab.

4.3 + 0.18 = 4.48 say 4.5 m

6.55+0.18 = 6.73 say 6.75 m

Dead load : self 0.18 x 25 = 4.5

floor finish = 1.0

Live load ( residence ) = 2.0

Total 7.5For 1 m wide strip

1.5 x 7.5= 11.25 KN / m

6.75 / 4.5 = 1.5

IS 456-2000 Table -27

23.7 KNm

10.48 KNmFrom Table 6-3 Q = 2.07

== 107 mm,

180 - 15 - 5 ( assume 10 O bar )= 160 mm, > 107 mm ……………….( O.K.)

160 - 10 = 150 mm, > 107 mm

Larger depth is provided due to deflection check.

= 0.926

lx =

ly =

KN / m2

KN / m2

KN / m2

KN / m2

Pu =

ly / l

x =

Mux

= αx x w x l

x2 = 0.104 x 11.25 x 4.52 =

Muy

= αy x w x l

x2 = 0.046 x 11.25 x 4.52 =

drequired

= √M / Q x b

√23.7 x 10 6 / 2.07 x 1000

dshort

provided

=

dlong

provided

=

Mu / b x d2 ( short ) = 23.7 x 10 6 / 1000 x (160)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

=

415 / 15

= 50 [(1-0.846) x 15 / 415 ]= 0.28%

0.28 x 1000 x 160 / 100



Provide 10 mm O bar@170 mm c/c = 462 ( short span )

= 0.466

=415 / 15

= 50 [(1-0.926) x 15 / 415 ]= 0.134%

0.134 x 1000 x 150 / 100

= 201For HYSD Fe415 minimum reinforcement 0.12 %

Minimum steel = ( 0.12 /100 ) x 1000 x 180



Provide 8 mm O bar@ 230 mm c/c = 218 ( long span )The bars cannot be bent or curtailed because if 50 % of long span bars are curtailed , the remaining bars will be less than minimum

50 1-√1-(4.6 / 15) x (0.926)

Ast ( short ) =

mm2


mm2.

Mu / b x d2 ( long ) = 10.48 x 10 6 / 1000 x (150)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.466

Ast ( long ) =

mm2

mm2.


mm2.

At top on support , provide 50 % of bars of respective span to take into account negative moment due to slab nature.


Long span 11.25 x ( 4.5 / 2 ) = 25.31 KN


= 100 x 218 / 1000 x 150= 0.145

From equation OR

=

we get , 0.5023 = 11.8063

= 11.3313= 11.30 KNm


580.403 + 8 O580.403


Short span 11.25 x ( 4.5 / 2 ) = 25.31 KN


= 100 x 462 / 1000 x 160= 0.289

From equation OR

=

we get , 0.9595 = 26.6888

= 24.5539= 24.56 KNm


1261.48 + 8 O1261.48

which gives 26.28 mm ……………….( O.K.)Note that the bond is usually critical along long direction.

Vu =

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 415 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 0.5023 x 1000 x 1502 x 10-6


s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 11.30 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O

48 O ≤O ≤

Vu =

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2) M

u1 = 0.87 x f

y x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

fy / f

ck 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6

Mu1

/ b x d2 =

Mu1

= 0.9595 x 1000 x 1602 x 10-6

Development length of bars Ld = O σs / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O≥ 56 O

48 O ≤O ≤

Check for shear : -This is critical along long span

25.31 KN

Shear stress =

=

= 0.169 ( too small )

100 x 218 / 1000 x 150= 0.145

from table 7-1

25 difference -0.05IS 456-2000 clause 40.2.1.1 20 difference ? -0.04


= 0.347 ……………….( O.K.)OR

6 x β

β == 12.0114

6 x 12.011= 0.278

IS 456-2000 clause 40.2.1.1 25 difference -0.05k = 1.24 for 180 mm slab depth 20 difference ? -0.04


= 0.345 ……………….( O.K.)Check for deflection : -This check shall be done along short span


100 x 462 / 1000 x 160= 0.289




Vu =

Vu / b x d

25.31 x 103 / 1000 x 150

N / mm2 ح ) >C )

N / mm2

100 x As / b x d =

for Ptح 0.145 =

c = 0.28 N / mm2

N / mm2


ck ( √ 1 + 5 x β - 1 )

0.8 x fck



N / mm2

Pt = 100 x A

st / b x d =

(1) Main bars : maximum spacing permitted for short span steel = 3 x effective depth of slab or 300 mm whichever is small= 3 x 160 = 480

spacing provided = 170 mm < 300 mm ……………….( O.K.)(2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth of slab or 450 mm whichever is small

= 5 x 150 = 750spacing provided = 230 mm < 450 mm ……………….( O.K.)

Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provision for torsion at corners

Case No. Type of Panel and Moments considered

1.0 1.1 1.2 1.3

10.032 0.037 0.043 0.047

Positive moment at mid-span 0.024 0.028 0.032 0.036

2 One Short Edge Discontinuous:

Negative moment at continuous edge 0.037 0.043 0.048 0.051


3 One long Edge Discontinuous:

Negative moment at continuous edge 0.037 0.044 0.052 0.057


4 Two Adjacent Edges Discontinuous:

Negative moment at continuous edge 0.047 0.053 0.060 0.065Positive moment at mid-span 0.035 0.040 0.045 0.049

5 Two Short Edges Discontinuous:Negative moment at continuous edge 0.045 0.049 0.052 0.056Positive moment at mid-span 0.035 0.037 0.040 0.043

6 Two Long Edges Discontinuous:

Negative moment at continuous edge - - - -Positive moment at mid-span 0.035 0.043 0.051 0.057

7Three Edges Discontinuous

(One Long Edge Continuous):Negative moment at continuous edge 0.057 0.064 0.071 0.076


8 Three Edges Discrmntinuous

(One Short Edge Continuous) :Negative moment at continuous edge - - - -Positive moment at mid-span 0.043 0.051 0.059 0.065

9 Four-Edges Discontinuous:Positive moment at mid-span 0.056 0.064 0.072 0.079

Table - 27 Bending moment coefficients for slabs spanning in two directions at right angles , simply supported on Four sides

short span coefficient αx

( Values of l

Interior Panels:Negative moment at continuous edge

1.0 1.1 1.2 1.3 1.4

0.062 0.074 0.084 0.093 0.099

0.062 0.061 0.059 0.055 0.051

ly / l

x

αx

αy

( 1 - 0.0402 )

KNm

( 1 - 0.0799 )

KNm

350 6550 350

350

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 415 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10 -6

0.87 x fy x A

st x d ( 1 - f

y x A

st / b x d x f

ck )

0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6

A

4300 4650

350

6900

PLAN460 460

180

350 4300 350

Section A-A

460

460

690 690

A

8 O @ 230 c/c

8 O @ 460 c/c

10 O @ 170 c/c

10 O @ 340 c/c

vv

vv

vv

vv

vv

vv

10 O @ 340 c/c

10 O @ 170 c/c

8 O @ 230 c/c

3 x effective depth of slab or 300 mm whichever is smallmm or 300 mm i.e. 300 mm


mm or 450 mm i.e. 450 mm……………….( O.K.)

Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provision for torsion at corners

1.4 1.5 1.75 2.0

0.051 0.053 0.060 0.065 0.0320.039 0.041 0.045 0.049 0.024

0.055 0.057 0.064 0.068 0.037

0.041 0.044 0.048 0.052 0.028

0.063 0.067 0.077 0.085 0.037

0.047 0.051 0.059 0.065 0.028

0.071 0.075 0.084 0.091 0.0470.053 0.056 0.063 0.069 0.035

0.059 0.060 0.065 0.069 -0.044 0.045 0.049 0.052 0.035

- - - - 0.0450.063 0.068 0.080 0.088 0.035

0.080 0.084 0.091 0.097 -

0.060 0.064 0.069 0.073 0.043

- - - - 0.0570.071 0.076 0.087 0.096 0.043

0.085 0.089 0.100 0.107 0.056

Table - 27 Bending moment coefficients for slabs spanning in two directions at right angles , simply supported on Four sides

short span coefficient αx Long span

coefficient αy for all

values of ly / lx

( Values of ly / l

x )

1.5 1.75 2.0 2.5 3.0

0.104 0.113 0.118 0.122 0.124

0.046 0.037 0.029 0.020 0.014

Design of Continuous BeamAn R.C.C. floor is used as a banking hall

The Slab thickness is 120 mm .

(Given )Rib size = 230 mm x 450 mmColumn = 300 mm x 300 mmMain beams = 300 mm x 570 mm overall .Material M15 grade concrete

HYSD reinforcement of grade Fe415 .Solution : -( a ) Load calculations and analysis :

Load on beam = 3 ( 4 + 3 ) = 12 + 9 KN / m Self wt. = 0.23 x 0.45 x 25 = 2.58 + 0 KN / m

Total 14.58 + 9 KN / mFactored Load = 1.5 ( 14.58 + 9 )

= 21.87 + 13.5say ( 22 + 14 ) KN /m .

Case ( a ) Maximum moment at B

Using three moment equation for span ABC

= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 162= 648 = 648

Design the beams B10

-B11

-B12

.

Live load = 3 KN / m2

Floor finish = 1 KN / m2

Slab 120 mm thick 0.12 x 25 = 3 + 0 KN / m2

Floor finish = 1 + 0 KN /m2

Live load = 0 + 3 KN / m2

Total 4 + 3 KN / m2

MA ( L

1 / I

1 ) + 2M

B ( L

1 / I

1 + L

2 / I

2 ) + M

C ( L

2 / I

2 ) = - 6 A

1a

1 / ( I

1 L

1 ) -6 A

2a

2 / ( I

2L

2)

A1 = ( 2 / 3 ) x Base x h

1A

2 = ( 2 / 3 ) x Base x h

1

a1 = 3 m a

2 = 3 m

MA ( 6 / I ) + 2M

B ( 6 / I + 6 / I ) + M

C ( 6 / I ) = - [ 6 x 648 x 3 / ( I

x

6 ) ] - [ 6 x 648 x 3 / ( I

x 6) ]

6MA + 24M

B + 6M

C = - [ 1944 ] - [ 1944 ]

B10

6 m

6 m

36 KN / m 36 KN / m 22 KN / m

6 m 6 m 6 mA B C D

36 KN / m

162

36 KN / m

162

A B B C

…………………….( 1 )Using three moment equation for span BCD

= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 648 = 396

…………………….( 2 )

…………………….( 1 )

…………………….( 2 )

6 m Distance 138 KNm3 m Distance ( ? ) 696 m Distance 96 KNm3 m Distance ( ? ) 486 m Distance 42 KNm

4MB + M

C = - 648 As M

A = 0

MB ( L

2 / I

2 ) + 2M

C ( L

2 / I

2 + L

3 / I

3 ) + M

D ( L

3 / I

3 ) = - 6 A

2a

2 / ( I

2 L

2 ) -6 A

3a

3 / ( I

3L

3)

A2 = ( 2 / 3 ) x Base x h

2A

3 = ( 2 / 3 ) x Base x h

3

a2 = 3 m a

3 = 3 m

MB ( 6 / I ) + 2M

C ( 6 / I + 6 / I ) + M

D ( 6 / I ) = - [ 6 x 648 x 3 / ( I

x

6 ) ] - [ 6 x 396 x 3 / ( I

x 6) ]

6MB + 24M

C + 6M

D = - [ 1944 ] - [ 1188 ]

MB + 4M

C = - 522 As M

D = 0

4MB + M

C = - 648

MB + 4M

C = - 522

By putting Value of MB from Equation ( 2 ) into Equation ( 1 )

4(- 522 - 4MC ) + M

C = - 648

- 2088 - 16 MC + M

C = - 648

15 MC = - 1440

MC = - 96 KNm

By putting Value of MC into Equation ( 1 )

4MB + ( - 96 ) = - 648

4 MB = - 552

MB = - 138 KNm

36 KN / m

162

22 KN / m

99

B C C D

36 KN / m 36 KN / m 22 KN / m

6 m 6 m 6 mA B C D138

96162162 99

( - )( + ) ( - ) ( + )

( - ) ( - )

( + ) ( + ) ( + )

93 5145

3 m Distance ( ? ) 21

X dist.

6 - X dist.

131 X =85 ( 6 - X )

216 X = 510

X = 2.36 m

Three cases are considered for getting maximum values of moments. Case ( a ) gives maximumnegative moment at B.The same moment shall be used at C also because of symmetry. Case ( b ) gives the maximum positive moment in span BC while the case ( c ) gives maximum positive moments in span AB and CD. Factored maximum moments :B or C , negative moment = 138 KNmAB ( + ) = CD ( + ) = 110 KNmBC ( + ) = 58 KNmBC ( minimum - ve ) = 5 KNmThe moment redistribution shall be now carried out. Maximum negative moment = 138 KNm.reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( increase ordecrease ) the moment at B or C = 110.4 KNm ( Hogging ).Maximum design moments :Support B or C = 110.4 KNm > 0.7 x 138 KNm ………………( O.K.) ( As per IS 456-2000 , Clause 37.1.1 )AB or CD ( + ) = 111.6 KNm > 0.7 x 110 KNm BC ( - ) = 11.6 KNmBC ( + ) = 51.6 KNm > 0.7 x 58 KNm.Note that after redistribution , the design positive moments also have been reduced.( b ) Design for flexure :Span AB or CD

The beam acts as a flanged beam

MB = - 138 KNm

VA x 6 -36 x 6 x 3 = -138

VA = 85 KN

VA + V

B = 36 x 6

VB = 216 - 85

VB = 131 KN

MC = - 96 KNm M

B = - 138 KNm

VD x 6 -22 x 6 x 3 = -96 50 x 12 + V

C x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -138

VD = 50 KN V

c = 183 KN

VC + V

D = 22 x 6 V

C + 82 = 183

VC = 132 - 50 V

C = 101 KN

VC = 82 KN M

C = - 96 KNm

85 x 12 + VB x 6 - 36 x 6 x 9 - 36 x 6 x 3 = - 96

VB = 246 KN

VB + 131 = 246

VB = 115 KN

Mu ( + ) = 111.6 KNm.

96138

85

131

2.36 m50

82

115

101

( As per IS456-2000 ,Clause 23.1.2 , Note )= 1650 mm > 3000 mm

Assuming one layer of 20 mm diameter barsd = 450 + 120 -25 - 10 = 535 mm .

( As per IS456-2000 ,Clause 26.5( a ) )

For bf /bw = 7 For bf /bw = 8

0.01 diff. 0.014 0.016

( As per SP:16 ,Table 58 ) 0.006 diff. ? ?

-0.0084 -0.0096

= 662.597 KNm > 111.6 KNm For 0.224 0.6566 0.743

1 diff 0.08640.83 diff ? 0.07171

= 535 - 120 / 2 = 535 - 60

= 651

Span BC :

= 301

Continue 3 - 12 O in span BC as required for flexure.Support B or C :

( From Table 6-3 )The section is under-reinforced.

415 / 15

= 0.549

603 135

For T-beams , bf = ( l

0 / 6 ) + b

w + 6 D

f

bf = ( 0.7 x 6000 / 6 ) + 230 + 6 x 120

Minimum Ast = ( 0.205 / 100 ) x 230 x 535 = 252 mm2

bf / b

w = 1650 / 230 = 7.17

Df / d = 120 / 535 = 0.224

Mu,lim

/ fck

bw d2 = 0.671

Mu.lim

= 0.671 x 15 x 230 x 5352 x 10-6

If Mu < M

u,lim

: design as under-reinforced section (singly reinforced beam) as explained below.

Ast = M

u / 0.87 x f

y x lever arm

where lever arm = d - Df / 2

Ast = 111.6 x 106 / 0.87 x 415 x ( 535 - 60 )

mm2 .

Provide 6 - 12 mm O = 678 mm2 .

Mu ( + ) = 51.6 KNm

Ast = 51.6 x 106 / 0.87 x 415 x ( 535 - 60 )

mm2 .

Provide 3 - 12 mm O = 339 mm2 .

In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ℓ ) from A and at 900 mm ( 0.15 ℓ ) from B.

Mu ( - ) = 110.4 KNm

Mu / bd2 = 110.4 x 106 / 230 x 5352 = 1.68 < 2.07.

Pt = 50

1 - √ 1 - ( 4.6 / fck

) x ( Mu / bd2 )

fy / f

ck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.68 )

Ast = ( 0.549 / 100 ) x 230 x 535 = 676 mm2

20 % of steel should be carried through the span = 0.2 x 676 = 135 mm2

Provide 2- 10 mm O anchor bars = 157 mm2 . At support , provide 3 - 16 mm O extra at top = 603 mm2 , one of which may be curtailed at 0.15 ℓ = 900 mm from centre of support B and remaining 2 - 16 mm Oat 0.25 ℓ = 1500 mm from B .

Case ( b ) Maximum positive moment in span BC


= ( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 162= 396 = 648

MA ( L

1 / I

1 ) + 2M

B ( L

1 / I

1 + L

2 / I

2 ) + M

C ( L

2 / I

2 ) = - 6 A

A1 = ( 2 / 3 ) x Base x h

1A

2 = ( 2 / 3 ) x Base x h

a1 = 3 m a

2 = 3 m

6 ) ] - [ 6 x 648 x 3 / ( I x 6) ] M

A ( 6 / I ) + 2M

B ( 6 / I + 6 / I ) + M

C ( 6 / I ) = - [ 6 x 396 x 3 / ( I

6MA + 24M

B + 6M

C = - [ 1188 ] - [ 1944 ]

B11

B12

6 m 6 m

3 m

3 m

22 KN / m 36 KN / m 22 KN / m

6 m 6 m 6 mA B C D

22 KN / m

99

36 KN / m

162

A B B

Using three moment equation for span BCD

= ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 648 = 396

…………………….( 1 )

…………………….( 2 )

( 162 - 69 = 93 KNm )

( 99 - 48 = 51 KNm )

4MB + M

C = - 522 As M

A = 0

MB ( L

2 / I

2 ) + 2M

C ( L

2 / I

2 + L

3 / I

3 ) + M

D ( L

3 / I

3 ) = - 6 A

A2 = ( 2 / 3 ) x Base x h

2A

3 = ( 2 / 3 ) x Base x h

a2 = 3 m a

3 = 3 m

6 ) ] - [ 6 x 396 x 3 / ( I x 6) ] M

B ( 6 / I ) + 2M

C ( 6 / I + 6 / I ) + M

D ( 6 / I ) = - [ 6 x 648 x 3 / ( I

6MB + 24M

C + 6M

D = - [ 1944 ] - [ 1188 ]

MB + 4M

C = - 522 As M

D = 0

4MB + M

C = - 522

MB + 4M

C = - 522


4(- 522 - 4MC ) + M

C = - 522

- 2088 - 16 MC + M

C = - 522

15 MC = - 1566

MC = - 104 KNm


4MB + ( - 104 ) = - 522

4 MB = - 418

MB = - 104 KNm

36 KN / m

162

22 KN / m

99

B C C

22 KN / m 36 KN / m 22 KN / m

6 m 6 m 6 mA B C D104 104

99162

99( - )

( - ) ( + )

( - ) ( - )

( + ) ( + ) ( + )

47 4758

( 162 - ( 96 + 21 ) = 45 KNm )

85

131

131 X =85 ( 6 - X )

216 X = 510

X = 2.36 m

Three cases are considered for getting maximum values of moments. Case ( a ) gives maximumnegative moment at B.The same moment shall be used at C also because of symmetry. Case ( b ) gives the maximum positive moment in span BC while the case ( c ) gives maximum positive

The moment redistribution shall be now carried out. Maximum negative moment = 138 KNm.reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( increase or

( As per IS 456-2000 , Clause 37.1.1 ) ( b ) The ultimate moment of resistance provided at any section of a member is not less than 70 percent of the moment at that sectionobtained from an elastic maximum moment diagram covering all appropriate combinations of loads.

MC = - 104 KNm M

B = - 104 KNm

VD x 6 -22 x 6 x 3 = -104 48.67 x 12 + V

C x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -104

VD = 48.67 KN V

c = 191.33 KN

VC + V

D = 22 x 6 V

C + 83.33 = 191.33

VC = 132 - 48.67 V

C = 108 KN

VC = 83.33 KN M

C = - 104 KNm

48.67 x 12 + VB x 6 - 36 x 6 x 3 - 22 x 6 x 9 = - 104

VB = 191.33 KN

VB + 83.33 = 191.33

VB = 108 KN

IS 456-2000 ,Clause 37.1.1 Redistribution of moments in Continuous Beams and Frames -

( c ) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30 % of the numerically largest moment given anywhere by the elastic maximum moments diagram for the particular member , covering all appropriate combination of loads.

104

48.67

83.33

2.21 m48.67

83.33

108

108

104

IS 456-2000 ,Clause 23.1.2 Effective width of flange - ( As per IS456-2000 ,Clause 23.1.2 , Note )

( As per IS456-2000 ,Clause 26.5( a ) )

For bf /bw = 8

Where ,

0.671 b = actual width of the flange.

IS 456-2000 Clause 26.5 Requirements of Reinforcement for

Structural Members

26.5.1 Beams26.5.1.1 Tension Reinforcement

shall not be less than that given by the following :

where ,

b = breadth of the beam or the breadth of the web of T- beam ,d = effective depth , and

Minimum steel %

For mild steel

For HYSD steel , Fe415 grade


In the absence of more accurate determination , the effective width of flange may be taken as the following but in no case greater than the breadth of the web plus half the sum of the clear distances to the adjacent beams on either side :

( a ) For T-beams , bf = ( l

0 / 6 ) + b

w + 6 D

f

( b ) For L-beams , bf = ( l

0 / 12 ) + b

w + 3 D

bf = effective width of flange ,

l0 = distance between points of zero moments in the beam ,

bw = breadth of the web ,

Df = thickness of flange , and

: design as under-reinforced section (singly reinforced beam) as explained below. NOTE - For continuous beams and frames , ' l0 ' may be assumed

as 0.7 times the effective span.

a ) Minimum reinforcement - The minimum area of tension reinforcement

As / b d = 0.85 / f

y

As = minimum area of tension reinforcement ,

fy = characteristic strength of reinforcement in N / mm2 .

b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.

100 As / b d = 100 x 0.85 / 250 = 0.34

100 As / b d = 100 x 0.85 / 415 = 0.205

100 As / b d = 100 x 0.85 / 500 = 0.17

Table 6-3

For singly reinforced rectangular sections

250 415 50015 2.22 2.07 2.0020 2.96 2.76 2.6625 3.70 3.45 3.3330 4.44 4.14 3.99

. At support , provide 3 - 16 mm O extra at top = 603 mm2 , = 900 mm from centre of support B and remaining 2 - 16 mm O


, N / mm

fck

N / mm2

fy, N / mm2

Case ( b ) Maximum positive moment in span BC Case ( c ) Maximum positive moment in span AB and CD


( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 162= 648

) + MC ( L

2 / I

2 ) = - 6 A

1a

1 / ( I

1 L

1 ) -6 A

2a

2 / ( I

2L

2) M

A ( L

1 / I

1 ) + 2M

B ( L

= ( 2 / 3 ) x Base x h1

A1 = ( 2 / 3 ) x Base x h

a1 = 3 m

( 6 / I ) = - [ 6 x 396 x 3 / ( I x

6 ) ] - [ 6 x 648 x 3 / ( I

x 6) ] M

A ( 6 / I ) + 2M

B ( 6 / I + 6 / I ) + M

6MA + 24M

B + 6M

C = - [ 1944 ] - [ 1188 ]

36 KN / m

C

36 KN / m

6 mA B

36 KN / m

162

A


( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 99= 396

…………………….( 2 )

…………………….( 1 )

…………………….( 2 )

6 m Distance 104 KNm3 m Distance ( ? ) 52 ( 99 - 52 = 47 KNm ) Moment in span BC = ( 162 - 104 ) = 58 KNm

4MB + M

C = - 522

) + MD ( L

3 / I

3 ) = - 6 A

2a

2 / ( I

2 L

2 ) -6 A

3a

3 / ( I

3L

3) M

B ( L

2 / I

2 ) + 2M

C ( L

= ( 2 / 3 ) x Base x h3

A2 = ( 2 / 3 ) x Base x h

a2 = 3 m

( 6 / I ) = - [ 6 x 648 x 3 / ( I x

6 ) ] - [ 6 x 396 x 3 / ( I

x 6) ] M

B ( 6 / I ) + 2M

C ( 6 / I + 6 / I ) + M

= - [ 1944 ] - [ 1188 ] 6MB + 24M

C + 6M

D = - [ 1188 ] - [ 1944 ]

MB + 4M

C = - 522

4MB + M

C = - 522

MB + 4M

C = - 522

from Equation ( 2 ) into Equation ( 1 ) By putting Value of M

4(- 522 - 4MC ) + M

- 2088 - 16 MC + M

15 MC = - 1566

MC = - 104.4 KNm

By putting Value of M

4MB + ( - 104.4 ) = - 522

4 MB = - 417.6

MB = - 104.4 KNm

22 KN / m

D

22 KN / m

99

B

36 KN / m

6 mA B104162 ( - )

( + )

( - )

( + )

110

X dist. 48.67

6 - X dist. 83.33

83.33 X = 48.67 ( 6 - X )

132 X = 292.02

X = 2.21 m

( b ) The ultimate moment of resistance provided at any section of a member is not less than 70 percent of the moment at that sectionobtained from an elastic maximum moment diagram covering all

MB = - 104 KNm

VA x 6 -22 x 6 x 3 = -104

VA = 48.67 KN

VA + V

B = 22 x 6

VB = 132 - 48.67

VB = 83.33 KN

MC = - 104 KNm

48.67 x 12 + VC x 6 - 36 x 6 x 3 - 22 x 6 x 9 = -104 V

D x 6 -36 x 6 x 3 = -104

VD = 90.67 KN

+ 83.33 = 191.33 VC + V

D = 36 x 6

VC = 216 - 90.67

VC = 125.33 KN

48.67 x 12 + VB x 6 - 36 x 6 x 3 - 22 x 6 x 9 = - 104

+ 83.33 = 191.33

IS 456-2000 ,Clause 37.1.1 Redistribution of moments in Continuous Beams and Frames -

( c ) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30 % of the numerically largest moment given anywhere by the elastic maximum moments diagram for the particular member , covering all appropriate combination of loads.

104

90.67

125.33

2.52 m

66

IS 456-2000 ,Clause 23.1.2 Effective width of flange -

IS 456-2000 Clause 26.5 Requirements of Reinforcement for

Structural Members


b = breadth of the beam or the breadth of the web of T- beam ,

In the absence of more accurate determination , the effective width of flange may be taken as the following but in no case greater than the breadth of the web plus half the sum of the clear distances to the adjacent beams on either side :

/ 6 ) + bw + 6 D

f

/ 12 ) + bw + 3 D

f

= distance between points of zero moments in the beam ,

NOTE - For continuous beams and frames , ' l0 ' may be assumed

as 0.7 times the effective span.


= minimum area of tension reinforcement ,

= characteristic strength of reinforcement in N / mm2 .


/ b d = 100 x 0.85 / 250 = 0.34

/ b d = 100 x 0.85 / 415 = 0.205

/ b d = 100 x 0.85 / 500 = 0.17

Table 6-3


5501.942.583.233.87


, N / mm2

, N / mm2

Case ( c ) Maximum positive moment in span AB and CD


( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99= 396

/ I1 ) + 2M

B ( L

1 / I

1 + L

2 / I

2 ) + M

C ( L

2 / I

2 ) = - 6 A

1a

1 / ( I

1 L

1 ) -6 A

2a

2 / ( I

2L

2)

= ( 2 / 3 ) x Base x h1

A2 = ( 2 / 3 ) x Base x h

1

a2 = 3 m

( 6 / I ) + 2MB ( 6 / I + 6 / I ) + M

C ( 6 / I ) = - [ 6 x 648 x 3 / ( I

x

6 ) ] - [ 6 x 396 x 3 / ( I

x 6) ]

+ 24MB + 6M

C = - [ 1944 ] - [ 1188 ]

22 KN / m 36 KN / m

6 m 6 mC D

36 KN / m 22 KN / m

99

B B C


( 2 / 3 ) x 6 x 99 = ( 2 / 3 ) x 6 x 162= 648

…………………….( 2 )

…………………….( 1 )

…………………….( 2 )

6 m Distance 104 KNm3 m Distance ( ? ) 52 ( 162 - 52 = 110 KNm )

Moment in span BC = ( 104 - 99 ) = 5 KNm

C = - 522 As M

A = 0

/ I2 ) + 2M

C ( L

2 / I

2 + L

3 / I

3 ) + M

D ( L

3 / I

3 ) = - 6 A

2a

2 / ( I

2 L

2 ) -6 A

3a

3 / ( I

3L

3)

= ( 2 / 3 ) x Base x h2

A3 = ( 2 / 3 ) x Base x h

3

a3 = 3 m

( 6 / I ) + 2MC ( 6 / I + 6 / I ) + M

D ( 6 / I ) = - [ 6 x 396 x 3 / ( I

x

6 ) ] - [ 6 x 648 x 3 / ( I

x 6) ]

+ 24MC + 6M

D = - [ 1188 ] - [ 1944 ]

C = - 522 As M

D = 0

C = - 522

C = - 522


(- 522 - 4MC ) + M

C = - 522

- 2088 - 16 MC + M

C = - 522

= - 104.4 KNm


+ ( - 104.4 ) = - 522

= - 104.4 KNm

22 KN / m 36 KN / m

162

C C D

22 KN / m 36 KN / m

6 m 6 mC D104 104

99

162

( - ) ( - )( + )

( - )

( + )

110

5.0

X dist. 90.67

6 - X dist. 125.33

125.33 X =90.67 ( 6 - X )

216 X = 544.02

X = 2.52 m

MB = - 104 KNm

VA x 6 -36 x 6 x 3 = -104

VA = 90.67 KN

VA + V

B = 36 x 6

VB = 216 - 90.67

VB = 125.33 KN

MB = - 104 KNm

x 6 -36 x 6 x 3 = -104 90.67 x 12 + VC x 6 - 22 x 6 x 3 - 36 x 6 x 9 = -104

Vc = 191.33 KN

VC + 125.33 = 191.33

VC = 66 KN

MC = - 104 KNm

90.67 x 12 + VB x 6 - 36 x 6 x 9 - 22 x 6 x 3 = - 104

VB = 191.33 KN

VB + 125.33 = 191.33

VB = 66 KN

104

125.33

90.67

125.33

66

( 162 - 52 = 110 KNm )

Design of Cantilever BeamSpan = 3 mCharacteristic U.D.L. = 12 KN /mAssume that sufficient safety against overturning is there

( Given )and reinforcement anchorages are also available.Material M15 grade concrete

mild steel reinforcement.Solution : -Assume an initial trial section of 230 mm x 600 mm overall depth to consider self-weight.The self-weight = 0.23 x 0.6 x 25 = 3.45 KN / m

Total load = 12 + 3.45 = 15.45 KN / mFactored load = 1.5 x 15.45 = 23.2 KN / mFactored S.F. =

Factored B.M. =

Depth required = √M / Q b ( From Table 6-3 )= 452 mm

Using one layer of 20 mm O bars and overall depth of 550 mmd = 550 - 25 - 10 = 515 mm OR

= 1.71 < 2.22 ( Table 6-3 ) = 135.424 KNm

The section is singly reinforced ( under-reinforced )

250 / 15= 0.93

Check for development length :

( From Table 7-6 )IS 456-2000 clause 26.2.1 ( From Table 7-5 )

== 47.71 O= 47.71 x 20= 954 mm

The bar shall extend into the support for a straight length of 954 mm . Provide anchorage of 1200 mm . If in some case the bars are to be bent e.g. anchored in column , the bearing stress around the bend has to be checked as discussed

in the next example . Check for shear :

w x ℓ = 23.2 x 3 = 69.6 KN

w x ℓ2 / 2 = 23.2 x 32 / 2 = 104.4 KNm

= √( 104.4 x 106 ) / ( 2.22 x 230 )

Mu / b d2 == 104.4 x 106 / ( 230 x 5152) M

u,lim = 2.22 x 230 x 5152 x 10-6

Mu < M

u,lim

Pt = 50

1 - √ 1 - ( 4.6 / fck

) x ( Mu / bd2 )

fy / f

ck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.71 )

Ast = ( 0.93 / 100 ) x 230 x 515 = 1102 mm2

Provide 4 - 20 mm O giving Ast = 4 x 314 = 1256 mm2 .

Ld = 1102 / 1256 = 55 O


s / 4 x ح

bd

O ( 0.87 x 250 x 1102 ) / ( 1256 x 4 x 1.0 )

104.4 KNm

=

= 0.588

( From IS 456-2000 , table 19 table 7-1 )= 1.06 0.25 difference

0.61 0.19 differenceProvide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups

== 0.87 x 56 x 250 / 0.4 x 230= 132.4 mm

spacing should not exceed( i ) 450 mm

( ii ) 0.75 d = 0.75 x 515 = 386 mm( iii ) ≤132.4 mm ( minimum )

( iv ) 569.2 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/cCheck for deflection :Basic span / d ratio = 7

modification factor = 1.4 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.4 = 9.8Actual span / d = 3000 / 515 = 5.83 < 9.8 ………………………( safe )Check for cracking (spacing of bars ) :Clear distance between bars

= ( 230 - 50 - 4 x 20 ) / 3 = 10 mmMaximum clear distance permitted

= 300 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe )The design beam is shown in fig.

Vu = 69.6 KN

Actual shear strength حv = V

u / bd

69.6 x 103 / ( 230 x 515 )

N / mm2

100 x As / b d = ( 100 x 1256 ) / ( 230 x 515 )

design ( permissible ) shear strength حc = N / mm2 ح <

v

0.87 Asv

fy / 0.4 b

100 Ast / b d = 100 x 1256 / 230 x 515 = 1.06

6 O @ 130 c/c

Fixed end support

1200 3000

2-12 O

4 -20 O

v v v v

vv

230

550

Stirrups 6 O @ 130 c/c

Design of Cantilever BeamA cantilever rectangular bracket projects from a column of size 230 mm x 500 mmin the direction of 500 mm for a length of 3 mFactored load of 20 KN / m inclusive of self - weightMaterial M15 grade concrete

HYSD reinforcement of grade Fe415 .Solution : -

( a ) Moment steel :Take size of Beam 230 mm x 550 mm overall.Assuming 16 mm diameter bars in one layerd = 550 - 25 ( Cover ) -8 OR

= 517 mm

= 127.256 KNm

= 1.46 < 2.07 ( Table 6-3 )The section is singly reinforced ( under-reinforced ) Beam

415 / 15= 0.464

Use 2- 10 mm O as bottom anchor bars .( b ) Check for development length :IS 456-2000 clause 26.2.1

= 0.87 x 415 x 552 / 603 = 330.5

IS 456-2000 clause 26.2.1 ( From Table 7-5 )

== 51.64 O= 51.64 x 16= 826 mm

The arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorageof 1127 mm. The bearing stress inside the bend is now checked .O = 16 mma = 82 mm for internal bar ( centre to centre distance between bar ) ( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm )

= 25 + 16 = 41 mm for external bar . ( = cover + dia of bar )

Vu = w x ℓ = 20 x 3 = 60 KN

Mu = w x ℓ2 / 2 = 20 x 32 / 2 = 90 KNm

Mu,lim

= 2.07 x 230 x 5172 x 10-6

Mu / b d2 =90 x 106 / ( 230 x 5172)

Mu < M

u,lim

Pt = 50

1 - √ 1 - ( 4.6 / fck

) x ( Mu / bd2 )

fy / f

ck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.46 )

Ast = ( 0.464 / 100 ) x 230 x 517 = 552 mm2

Provide 16 mm O - 3 No. = 603 mm2.


s / 4 x ح

bd

Stress in bar σs = 0.87 x f

y N / mm2

ح ) bd

= 1.6 x 1 = 1.6 N / mm2


s / 4 x ح

bd

O 330.5 / ( 4 x 1.6 )

33

The bearing stress is critical in external bar . Check for this stress for a = 41 mm.

= 1.5 x 15 / [ 1 + ( 2 x 16 / 41 ) ]= 22.5 / 1.78

= 12.6404At the centre of bend , the anchorage available = 279 + 147 = 426 mm

Stress in bar at centre of the bend = 330.5 x ( 826 - 426 ) / 826

= 160

=

= ……………………..( safe )The arrangement is thus satisfactory. ( c )Check for shea

=

= 0.505


0.46 0.243 differenceshear design is necessary .

At support , OR

= = 5.35= 60 - 54.6986= 5.3 KN

== 1188 mm

Provide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups

== 0.87 x 56 x 250 / 0.4 x 230= 132.4 mm

spacing should not exceed( i ) 450 mm ( ii ) 0.75 d = 0.75 x 517 = 388 mm

Design bearing stress = 1.5 x fck

/ [1 + ( 2O/a ) ]

N / mm2

N / mm2

Fbt = 160 x 201 x 10-3 = 32.16 KN ( area of 16 O bar = 201 mm2 )

Bearing stress = Fbt / r O

32.16 x 103 / 180 x 16

11.16 N / mm2 < 12.64 N /mm2

Vu = 60 KN


u / bd

60 x 103 / ( 230 x 517 )

N / mm2

100 x As / b d = ( 100 x 603 ) / ( 230 x 517 )

design ( permissible ) shear strength حc = N / mm2 ح >v

Vus

= Vuح -

c b d V

us = ( 0.505 - 0.46 ) x 230 x 517 x10-3

60 - 0.46 x 230 x 517 x 10-3

Using 6 mm O ( mild steel ) Two legged stirrups , Asv

= 28 x 2 = 56 mm2 .

fy = 250 N / mm2

Sv = 0.87 f

y A

sv d / V

us

0.87 x 250 x 56 x 517 / 5.3 x 103

0.87 Asv

fy / 0.4 b

550

33

( iii ) ≤132.4 mm ( minimum )( iv ) 1188 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/c( d ) Check for deflection :Basic span / d ratio = 7

modification factor = 1.2 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.2 = 8.4Actual span / d = 3000 / 517 = 5.8 < 8.4 ………………………( safe )( e ) Check for cracking (spacing of bars ) :Clear distance between bars

= ( 230 - 50 - 3 x 16 ) / 2 = 66 mmMinimum clear distance permitted

= 20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted


100 Ast / b d = 100 x 603 / 230 x 517 = 0.507

hagg

+ 5 mm =

279295

295

108

150

internal radiusr = 180

r

r

500 3000 mm

2-10 O

3-16 O50

550

Stirrups 6 mm O @ 130 mm c/c

Assume an initial trial section of 230 mm x 600 mm overall depth to consider self-weight.

IS 456-2000 , Table 19Table 7-1

Concrete grade

2.22 x 230 x 5152 x 10-6

Design shear strength of concrete , حC, N / mm2P

t = 100

x As

3 m

23.2 KN /m

69.6 KN

-104.4 KNm

M15 M20 M25 M30 M35

0.28 0.28 0.29 0.29 0.29

0.25 0.35 0.36 0.36 0.37 0.37

0.50 0.46 0.48 0.49 0.50 0.50

( From IS 456-2000 , table 19 table 7-1 ) 0.75 0.54 0.56 0.57 0.59 0.590.04 1.00 0.60 0.62 0.64 0.66 0.67

? -0.0304 1.25 0.64 0.67 0.70 0.71 0.731.50 0.68 0.72 0.74 0.76 0.781.75 0.71 0.75 0.78 0.80 0.822.00 0.71 0.79 0.82 0.84 0.86

2.25 0.71 0.81 0.85 0.88 0.902.50 0.71 0.82 0.88 0.91 0.932.75 0.71 0.82 0.90 0.94 0.963.00 0.71 0.82 0.92 0.96 0.99

The above given table is based on the following formula

6 x β

β =

IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement

………………………( safe )

x As

b x d

≤ 0.15


ck ( √ 1 + 5 x β - 1 )

0.8 x fck


v

v

4-20 O

2-12 O

( given )

The arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorage

( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm )

2.07 x 230 x 5172 x 10-6

33 3382 82

IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends

Where ,

r = internal radius of the bend , and

O = size of the bar or , in bundle , the size of bar of equivalent area

Where,

( From IS 456-2000 , table 19 table 7-1 )0.08

? -0.0778

KN

The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall be calculated as given below :

Bearing stress = Fbt / r O

Fbt = tensile force due to design loads in a bar or group of bars,

For Limit state method of design , this stress shall not exceed 1.5 f

fck

is the characteristic strength of concrete and

a is the centre to centre distance between bars or group of bars , for a bar or group of bars adjacent to the face of the member a shall be taken as the cover plus size of bar ( O )

= ( 0.505 - 0.46 ) x 230 x 517 x10-3

230

33 3382 82

16 25


………………………( safe )

230

2-10 O

3-16 O

IS 456-2000 , Table 19Table 7-1

Concrete grade

حC, N / mm2

M40

0.30

0.38

0.51

0.600.68

0.740.790.840.88

0.920.950.981.01

1 + 5 x β - 1 )

, but not less than 1.0

IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends

O = size of the bar or , in bundle , the size of bar of equivalent area

The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall be

= tensile force due to design loads in a bar or group of bars,

For Limit state method of design , this stress shall not exceed 1.5 fck

/ [ ( 1+ 2O/a ) ]

a is the centre to centre distance between bars or group of bars , for a bar or group of bars adjacent to the face of the member a shall be taken as the cover plus size of

Design of Simply supported Doubly reinforced BeamSpan = 5 m ( simply supported rectangular beam )super-imposed load = 40 KN / m Beam section = 230 mm x 600 mm ( Given )Material M15 grade concrete

HYSD reinforcement of grade Fe415Solution : -

DL of beam 0.23 x 0.60 x 25 = 3.45 KN / m super-imposed load = 40 KN / m

Total 43.45 KN / mFactored load = 1.5 x 43.45 = 65.18 KN / m

== 203.688 KNm

w x ℓ / 2= 65.18 x 5 / 2= 162.95 KN

( a ) Moment steel :Assuming 20 mm diameter bars in two layerd = 600 - 25 ( Cover ) - 20 - 10 OR

= 545 mm

= 141.414 KNm

= 2.98 > 2.07 ( Table 6-3 )The section is Doubly reinforced ( over-reinforced )

= 141.414 KNm

= 203.69 - 141.41 = 62.28 KNmLet the compression reinforcement be provided at an effective cover of 40 mm .

d' / d = 40 / 545= 0.07 consider d' / d = 0.1 .

=

= 349Corresponding tension steel

= 349 x 353 / ( 0.87 x 415 )

= 341.2

=

=

Mu = w x ℓ2 / 8

65.18 x 52 / 8

Vu =

Mu,lim

= 2.07 x 230 x 5452 x 10-6

Mu / b d2 = 203.69 x 106 / ( 230 x 5452)

Mu > M

u,lim

Mu,lim

= 2.07 x 230 x 5452 x 10-6

Mu2

= Mu - M

u,lim

Stress in compression steel , fsc

= 353 N / mm2 ( refer to table 6-6 )

Asc

= Mu2

/ ( fsc

x ( d - d' ) )

62.28 x 106 / 353 ( 545 - 40 )

mm2

Ast2

= Asc

fsc

/ 0.87 fy

mm2

Ast,lim

= Mu,lim

/ ( 0.87 fy ( d - 0.42 X

u,max ) )

141.41 x 106 / ( 0.87 x 415 ( 545 - 0.42 x 0.48 x 545 ) )

141.46 x 106 / 361.05 ( 435.13 )

= 900

=

= 1241.2Provide

2-16 O = 402

4-20 O = 1256 ( all straight )( b ) Check for development length :

1625.02 ≥ 44 OO ≤ 36.93 mm

( c ) Check for shear

162.95 KN

=

= 1.3

( From IS 456-2000 , table 19 table 7-1 )= 1.0

0.6As the ends are confined with compressive reaction , shear at distance d will be used forchecking shear at support . At 545 mm , shear is equal to

162.95 - 0.545 x 43.45 = 139.27 KN

=

== 307 mm

From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 8 mm O stirrups

== 0.87 x 100 x 415 / 0.4 x 230= 392.4 mm

spacing should not exceed( i ) 450 mm ( ii ) 0.75 d = 0.75 x 545 = 408 mm

mm2

Ast = A

st,lim + A

st2 .

900 + 341.2 mm2

mm2

Asc

= mm2

Ast = mm2

As all the bars are taken into support , Mu1

may be taken as Mu .

Assume L0 = 12 O L

d = 56 O ( From Table 7-6 )

1.3 x M1 / V + L

0 ≥ L

d

1.3 x 203.69 x 106 / 162.95 x 103 +12 O ≥ 56 O

Oprovided

= 20 mm ……………………..( safe )

Vu =


u / bd

162.95 x 103 / ( 230 x 545 )

N / mm2

100 x As / b d = ( 100 x 1256 ) / ( 230 x 545 )

design ( permissible ) shear strength حc = N / mm2 ح >v

Vu =

Vus

= Vuح -

c b d

139.27 - 0.6 x 230 x 545 x 10-3 = 64.06 KN

Assuming 8 mm O two - legged stirrups , Asv

= 100 mm2 , fy = 415 N / mm2 .

Sv = 0.87 f

y A

sv d / V

us

0.87 x 415 x 100 x 545 / 64.06 x 103

0.87 Asv

fy / 0.4 b

( iii ) ≤392.4 mm ( minimum )( iv ) 307 mm ( designed )Minimum shear reinforcement of 8 mm O @ 390 mm c/c will be used .At support provide 8 mm O @ 300 mm c/c .Shear resistance of beam with minimum shear reinforcement

=

== 50.45 + 75.21= 125.66 KN

Designed shear reinforcement is required from face of the support upto the distance equal to

162.95 - 125.66 / 43.45 = 0.858 mNo. of stirrups = ( 858 / 300 ) + 1 = 3.86 say 4 Provide 8 mm O two-legged stirrups @ 300 mm c/c upto 4 no. from face of the supportand 8 mm O @ 390 mm c/c in remaining central portion .( d ) Check for deflection :Basic span / d ratio = 20

modification factor = 0.97 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement

modification factor = 1.08 IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcementSpan / d permissible = 20 x 0.97 x 1.08 = 20.95Actual span / d = 5000 / 545 = 9.17 ………………………( safe )( d ) Check for cracking (spacing of bars ) :Clear distance between bars

= ( 230 - 50 - 4 x 20 ) / 3 = 10 mmMinimum clear distance permitted



( 0.87 fy A

sv d / S

vح +(

c b d

(0.87 x 415 x 100 x 545 x 10-3 / 390 ) + 0.6 x 230 x 545 x 10-3

100 Ast / b d = 100 x 1256 / 230 x 545 = 1.0

100 Asc

/ b d = 100 x 402 / 230 x 545 = 0.32

hagg

+ 5 mm =

230

600545

2-16 O

4- 20 O

600

5000 c/c

2-16 O

4- 20 O ( straight ) 600

DIANO.

SPA.

8 O 8 O 8 O4 4rest

300 300390

bearing

Table 6-6

d ' / d0.05 0.1 0.15 0.2

250 217 217 217 217415 355 353 342 329500 424 412 395 370

550 458 441 419 380

where ,

STRESS IN COMPRESSION REINFORCEMENT fsc , N / mm2 IN DOUBLY REINFORCED BEAMS

fy N / mm2

2.07 x 230 x 5452 x 10-6

If Mu > M

u,lim

: design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M

u2

needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. M

u2 = M

u - M

u,lim

as explained below .

Ast,lim

= Mu,lim

/ ( 0.87 fy ( d - 0.42 X

u,max ) )

Asc

= Mu2

/ ( fsc

x ( d - d' ) )A

st2 = A

sc f

sc / 0.87 f

y

Ast = A

st,lim + A

st2 .

If Xu < X

u,max the section is under-reinforced ( singly reinforced )

If Xu = X

u,max the section is balanced

If Xu > X

u,max the section is over-reinforced ( doubly reinforced )

Xu,max

= 0.53 x d ( for Fe250 mild steel )

Xu,max

= 0.48 x d ( for Fe415 HYSD steel )

Xu,max


Xu,max


Xu =( 0.87 f

y A

st ) / ( 0.36 f

ck b )

( From IS 456-2000 , table 19 table 7-1 )

As the ends are confined with compressive reaction , shear at distance d will be used for

Provide 8 mm O two-legged stirrups @ 300 mm c/c upto 4 no. from face of the support


IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcement

………………………( safe )


u2

needed is obtained by providing compression( top ) reinforcement and additional tensile as explained below .

the section is under-reinforced ( singly reinforced )

the section is over-reinforced ( doubly reinforced )

Design of Simply supported Singly reinforced BeamSpan = 6 m ( simply supported rectangular beam )characteristic load = 20 KN / m inclusive of its self-weightBeam section = 230 mm x 600 mmMaterial M15 grade concrete ( Given )

HYSD reinforcement of grade Fe415

The beam is resting on R.C.C. columns.Solution : -Factored load = 1.5 x 20 = 30 KN /m.

=

= 135 KNm

w x ℓ / 2

= 30 x 6 / 2

= 90 KN

( a ) Moment steel :

Assuming 20 mm diameter bars in one layer

d = 600 - 25 ( Cover ) -10 OR= 565 mm

= 151.983 KNm

= 1.84 < 2.07 ( Table 6-3 )

The section is singly reinforced ( under-reinforced )

415 / 15

= 0.614

As per IS 456-2000 clause 26.5.1.1 ( a )

From Table 6-4

Let 2 bars are bent at 1.25 D = 1.25 x 600 = 750 mm , from the face of the support . ( b ) Check for development length :

Mu = w x ℓ2 / 8

30 x 62 / 8

Vu =

Mu,lim

= 2.07 x 230 x 5652 x 10-6

Mu / b d2 = 135 x 106 / ( 230 x 5652)

Mu < M

u,lim

Pt = 50

1 - √ 1 - ( 4.6 / fck

) x ( Mu / bd2 )

fy / f

ck

Pt = 50

1 - √ 1 - ( 4.6 / 15 ) x ( 1.84 )

Ast = ( 0.614 / 100 ) x 230 x 565 = 798 mm2

Minimum steel , As = ( 0.205 / 100 ) x 230 x 565 = 266 mm2

Ast,lim

= 0.72 / 100 x 230 x 565 = 936 mm2.

Provide 16 mm O - 4 No. = 804 mm2.

( 1 ) A bar can be bent up at a distance greater than Ld = 56 O ( Table 7-6 )

From the centre of the support , i.e. 56 x 16 = 896 mm .

in this case , the distance is ( 3000 - 750 ) = 2250 mm ……………….( safe )

== 82.01 x 0.91= 74.99 KNm

90 KN ,

As the reinforcement is confined by compressive reaction

1083.19 ≥ 44 OO ≤ 24.62 mm

( c ) Check for shear :

As the ends of the reinforcement are confined with compressive reaction , shear at distance dwill be used for checking shear at support.

=

= 0.562


0.376 0.191 differenceshear design is necessary .

At support , OR

= = 24.17= 73.05 - 48.86= 24.19 KN

Capacity of bent bars to resist shear

== 102.6 KN

Bent bars share 50 % = 12.09 KNStirrups share 50 % = 12.09 KN

== 569.2 mm

( 2 ) For the remaining bars , Ast = 402 mm2

Mu1

= 0.87 fy A

st d ( 1 - [( f

y A

st ) / ( f

ck b d ) ] )

0.87 x 415 x 402 x 565 ( 1 - [ ( 415 x 402 ) / ( 15 x 230 x 565 ) ] ) x 10-6

Vu = L

0 = 12 O ( assume )

1.3 x M1 / V + L

0 ≥ L

d

1.3 x 74.99 x 106 / 90 x 103 +12 O ≥ 56 O

Oprovided

= 16 mm ……………………..( safe )

At support , Vu = 90 KN

Vu = 90 - 0.565 x w = 90 - 0.565 x 30 = 73.05 KN


u / bd

73.05 x 103 / ( 230 x 565 )

N / mm2

100 x As / b d = ( 100 x 402 ) / ( 230 x 565 )

design ( permissible ) shear strength حc = N / mm2 ح >

v

Vus

= Vuح -

c b d V

us = ( 0.562 - 0.376 ) x 230 x 565 x10-3

73.05 - 0.376 x 230 x 565 x 10-3

2 x 201 x 0.87 x 415 x sin 45º x 10-3 ( 0.87 fy A

sv sin α )

Using 6 mm O ( mild steel ) Two legged stirrups , Asv

= 28 x 2 = 56 mm2 .

Sv = 0.87 f

y A

sv d / V

us

0.87 x 250 x 56 x 565 / 12.09 x 103

From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups

== 0.87 x 56 x 250 / 0.4 x 230

= 132.4 mmspacing should not exceed

( i ) 450 mm ( ii ) 0.75 d = 0.75 x 565 = 423 mm( iii ) ≤132.4 mm ( minimum )( iv ) 569.2 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/cAt 1.25 D = 750 mm from face of the support where contribution of bent bars is not available

== 67.5 - 48.86= 18.64 KN

Provide minimum 6 mm O M.S. two -legged stirrups @ 130 mm c/c throught the beam.

( d ) Check for deflection :Basic span / d ratio = 20

modification factor = 1.1 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement

Span / d permissible = 20 x 1.1 = 22

Actual span / d = 6000 / 565 = 10.62 ………………………( safe )

( d ) Check for cracking (spacing of bars ) :Clear distance between bars

= 230 - 50 - 2 x 16 = 148 mm

Minimum clear distance permitted


= 180 mm ( cracking - table 8-1 , IS 456-200 , table 15 ) ………………………( safe )

The design beam is shown in fig.

0.87 Asv

fy / 0.4 b

Vu = 90 - 0.75 x w = 90 - 0.75 x 30 = 67.5 KN

Vus

= Vuح -

c b d

67.5 - 0.376 x 230 x 565 x 10-3

100 Ast / b d = 100 x 804 / 230 x 565 = 0.62

hagg

+ 5 mm =

230

600565

2-10 O

4- 16 O

750

6000 c/c

2-10 O

4- 16 O ( 2 straight +2 bent )

6 mm O @ 130 mm c/c

Consider width of the beam equal to 230 mm. The depth may be assumedas 1 / 10 to 1 / 8 of the span.To find steel area( 1 ) For a given ultimate moment ( also known as factored moment )and assumed width of section , find out d from equation

( 2 ) For a given factored moment ,width and depth of section .

where ,

Table 6-2

Limiting Moment of Resistance and Reinforcement

d = √ Mu / Q

lim b

This is a balanced section and steel area may be found out from table P

Obtain Mu,lim

= Qlim

bd2 .

If Mu < M

u,lim


Pt = 50

1 - √ 1 - ( 4.6 / fck

) x ( Mu / bd2 )

fy / f

ck

If Mu = M

u,lim

: design as balanced section as explained in ( 1 ).

If Mu > M

u,lim


u2

needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. M

u2 = M

u - M

u,lim

as explained below .2.07 x 230 x 5652 x 10-6

Ast,lim

= Mu,lim

/ ( 0.87 fy ( d - 0.42 X

u,max ) )

Asc

= Mu2

/ ( fsc

x ( d - d' ) )

Ast2

= Asc

fsc

/ 0.87 fy

Ast = A

st,lim + A

st2 .

If Xu < X

u,max the section is under-reinforced ( singly reinforced )

If Xu = X

u,max the section is balanced

If Xu > X

u,max the section is over-reinforced ( doubly reinforced )

Xu,max

= 0.53 x d ( for Fe250 mild steel )

Xu,max


Xu,max


Xu,max


Xu =( 0.87 f

y A

st ) / ( 0.36 f

ck b )

Index for Singly Reinforced Rectangular Sections

250 415 500 550

0.148 0.138 0.133 0.129

21.93 19.86 19.03 18.2

Table 6-3


250 415 500 550

15 2.22 2.07 2.00 1.9420 2.96 2.76 2.66 2.5825 3.70 3.45 3.33 3.23

30 4.44 4.14 3.99 3.87

Table 6-4As the ends of the reinforcement are confined with compressive reaction , shear at distance d


250 415 500 550

15 1.32 0.72 0.57 0.50

20 1.75 0.96 0.76 0.66

( From IS 456-2000 , table 19 table 7-1 ) 25 2.19 1.20 0.95 0.830.11 30 2.63 1.44 1.14 0.99

? -0.084IS 456-2000 Clause 26.5 Requirements of Reinforcement for

Structural Members

KN 26.5.1 Beams26.5.1.1 Tension Reinforcement


where ,

b = breadth of the beam or the breadth of the web of T- beam ,

d = effective depth , and

fy , N / mm2

Mu,lim

/ fck

b d2

Pt,lim

fy / f

ck


, N / mm2

fck

N / mm2

fy, N / mm2

Limiting Percentage of Reinforcement Pt,lim

fck

N / mm2

fy, N / mm2

= ( 0.562 - 0.376 ) x 230 x 565 x10-3


As / b d = 0.85 / f

y

As = minimum area of tension reinforcement ,

fy = characteristic strength of reinforcement in N / mm2 .


Minimum steel %For mild steel



At 1.25 D = 750 mm from face of the support where contribution of bent bars is not available

check for development lengthIS 456-2000 clause 26.2.1

IS 456-200 clause 26.2.1.1

Table 7-5


M15 M20 M25 M30 M35 M40

1.0 1.2 1.4 1.5 1.7 1.9

For mild steel Fe250

For Fe415IS 456-2000 clause 26.2.3.3

………………………( safe )

Table 7-6Development length for single mild steel bars

Tension bars Compression bars

M15 M20 M15 M20250 55 O 26 O 44 O 37 O415 56 O 47 O 45 O 38 O

100 As / b d = 100 x 0.85 / 250 = 0.34

100 As / b d = 100 x 0.85 / 415 = 0.205

100 As / b d = 100 x 0.85 / 500 = 0.17


may be assumed as 8 O for HYSD bars ( usually end anchorage is not

provided ) and 12 O for mild steel ( U hook is provided usually whose

anchorage length is 16 O.


s / 4 x ح

bd

Design bond stress (حbd

) for plain bars in tension

Concrete grade

ح ) bd

N / mm

Note-1 : حbd

shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ح

bd shall be increased by 60 %.

σs = 0.87 x f

y

σs = 0.67 x f

y

Ld ≤ M

1 / V + L

0

L0 = effective depth of the members or 12 O , whichever is greater

if ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )

fy N /

mm2

500 69 O 58 O 54 O 46 O

check for shearIS 456-2000 , Table 19

Table 7-1

Concrete gradeM15 M20 M25 M30 M35 M400.28 0.28 0.29 0.29 0.29 0.30

0.25 0.35 0.36 0.36 0.37 0.37 0.380.50 0.46 0.48 0.49 0.50 0.50 0.510.75 0.54 0.56 0.57 0.59 0.59 0.601.00 0.60 0.62 0.64 0.66 0.67 0.681.25 0.64 0.67 0.70 0.71 0.73 0.741.50 0.68 0.72 0.74 0.76 0.78 0.791.75 0.71 0.75 0.78 0.80 0.82 0.842.00 0.71 0.79 0.82 0.84 0.86 0.882.25 0.71 0.81 0.85 0.88 0.90 0.922.50 0.71 0.82 0.88 0.91 0.93 0.952.75 0.71 0.82 0.90 0.94 0.96 0.983.00 0.71 0.82 0.92 0.96 0.99 1.01


6 x β

β =

IS 456-2000 , Table 20Table 7-2

M15 M20 M25 M30 M35 M40

2.5 2.8 3.1 3.5 3.7 4.0

IS 456-2000 Clause 26.5.1.6 Minimum shear reinforcementMinimum shear reinforcement in the form of stirrups shall be provided

such that :where,

total cross-sectional area of stirrup legs effective in shear ,

stirrup spacing along the length of the member , b = breadth of the beam or breadth of the web of flanged beam , and

Design shear strength of concrete , حC, N / mm2

Pt = 100 x A

s

b x d≤ 0.15


ck ( √ 1 + 5 x β - 1 )

0.8 x fck


Maximum shear stress , حC, N / mm2

Concrete grade

ح )c )

max N/mm2

Asv

/ b sv ≥ 0.4 / 0.87 f

y

Asv

=

Sv =

fy = characteristic strength of the stirrup reinforcement in N / mm2

which shall not be taken greater than 415 N / mm2 .

check for deflectionBasic values of span to effective depth ratios for spans upto 10 m :

cantilever 7simply supported 20

continuous 26

check for crackingIS 456-2000 26.3.3 Maximum distance between bars in tension

% redistribution to or from section considered-30 -15 0 +15 +30

Clear distance between bars

mm mm mm mm mm250 215 260 300 300 300415 125 155 180 210 235500 105 130 150 175 195

For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.

Table 15 Clear distance Between Bars ( Clause 26.3.3 )

fy

N / mm2

Consider width of the beam equal to 230 mm. The depth may be assumedas 1 / 10 to 1 / 8 of the span.

( 1 ) For a given ultimate moment ( also known as factored moment )

This is a balanced section and steel area may be found out from table Pt,lim

, SP : 16 ,2.3


: design as balanced section as explained in ( 1 ).


u2

needed is obtained by providing compression( top ) reinforcement and additional tensile

The minimum area of tension reinforcement

The maximum area of tension

= effective depth of the members or 12 O , whichever is greater

Minimum shear reinforcement in the form of stirrups shall be provided

total cross-sectional area of stirrup legs effective in shear ,

breadth of the beam or breadth of the web of flanged beam , and

characteristic strength of the stirrup reinforcement in N / mm2

Design of slender ( Long ) columns ( with biaxial bending )Size of column 400 x 300 mm Column is restrained against sway.Concrete grade M 30

Unsupported length = 7.0 m

Factored load 1500kN

Factored.moment in the direction of larger dimension = 40 kNm at top and 22.5 KNm at bottom.

Factored.moment in the direction of shorter dimension = 30 kNm at top and 20 KNm at bottom.

Solution : -About x axis :

For Beam :

= 0.7 x 5000 / 6 + 230 + 6 x 120= 1533.33333

2.07Beam stiffness

=

= 2836699Column stiffness

= 175238.095

= 2 x 175238 / 2 ( 175238 + 2836699 )= 0.0582

as per IS 456-2000 fig. 27

1.035 < 1.2 ……………….consider 1.2

1.2 x unsupported length= 1.2 ( 7000 - 620 )= 7656 mm

7656/400= 19.14 > 12

The column is long about x direction.About Y axis :

Characteristic strength of reinforcement 415 N/mm2

Effective length for bending parallel to larger dimension ℓex

= 6.0 m

Effective length for bending parallel to shorter dimension ℓey

= 5.0 m

β1 and β

2 are the same

bf = l

0 / 6 + b

w + 6 D

f

bf / b

w = 1533.3 / 230 = 6.67

Df / D = 120 / 600 = 0.2

Kt from chart 88 , SP : 16 =

Kb = 1.5 x I

b / l

1.5 x ( 2.07 x ( 1 / 12 ) x 230 x 6203 ) / 5000

mm3

Kc = I

c / l = 1/12 x 230 x 4003 / 7000

mm3

β1 = β

2 = ∑ K

c / ( ∑K

c + ∑K

b )

lef / l =

lex

=

lex

/ D =

Beam stiffness

=

= 426006Column stiffness

= 57938.0952

= 2 x 57938.1 / 2 ( 57938.1 + 426006 )= 0.1197

as per IS 456-2000 fig. 27

1.06 < 1.2 ……………….consider 1.2

1.2 x unsupported length= 1.2 ( 7000 -420 )= 7896 mm

7896/300= 26.32 > 12

The column is long about Y direction.The column is bent in double curvature. Reinforcement will be distributed equally on four sides.

6000 / 400= 15.0 > 12

5000 / 300= 16.7 > 12

Therefore the column is slender about both the axes.Additional moments

== 67.5 KNm

== 62.75 KNm

The above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000but multiplication factors can be evaluated only if the reinforcement is known. For first trial , assume p = 3.0 ( with reinforcement equally on all the four sides ).

400 x 300

= 120000 400 x 300 - 120000*3/100

OR = 116400

= 2700 KN = 1571.4 +

= 2692 KN

Kb = 1.5 x I

b / l

1.5 x ( ( 1 / 12 ) x 230 x 4203 ) / 5000

mm3

Kc = I

c / l = 1/12 x 400 x 2303 / 7000

mm3

β1 = β

2 = ∑ K

c / ( ∑K

c + ∑K

b )

lef / l =

ley

=

ley

/ b =

ℓex

/ D =

ℓey

/ b =

Max

= ( Pu D / 2000 ) x ( ℓ

ex / D )2

( 1500 x 400 / 2000 ) x (15)2 x 10-3

May

= ( Pu b / 2000 ) x ( ℓ

ey / b )2

( 1500 x 300 / 2000 ) x (16.7)2 x 10-3

Ag = P

uz = 0.45 f

ck A

c + 0.75 f

mm2 Ac =

From chart 63 , puz

/ Ag = 22.5 N/mm2

mm2

Puz

= 22.5 x 120000 x 10-3 Puz

= 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3

Calculation of Pb :

Assuming 25 mm dia bars with 40 mm cover

d' / D ( about xx-axis ) = 52.5 / 400 = 0.13 ……………use d' / D = 0.15

d' / D ( about yy-axis ) = 52.5 / 300 = 0.18 ……………use d' / D = 0.2

From Table 60 , SP 16

= 779 KN

= 672 KN

= ( 2700 - 1500 ) / ( 2700 - 779 )= 1200 / 1921= 0.625

= ( 2700 - 1500 ) / ( 2700 - 672 )= 1200 / 2028= 0.592

The additional moments calculated earlier , will now be multiplied by the above values of k .

67.5 x 0.625 = 42.2 KNm

62.75 x 0.592 = 37.15 KNmThe additional moments due to slenderness effects should be added to the initial moments

after modifying the initial moments as follows ( see note 1 under 39.7.1 of the code IS 456-2000 )

( 0.6 x 40 - 0.4 x 22.5 ) = 15.0 KNm

( 0.6 x 30 - 0.4 x 20 ) = 10.0 KNmThe above actual moments should be compared with those calculated from minimum eccentricity

consideration ( see 25.4 of the code ) and greater value is to be taken as the initial moment for

adding the additional moments.

( ℓ / 500 ) + ( D / 30 ) = ( 7000 / 500 ) + ( 400 / 30) = 27.3333

( ℓ / 500 ) + ( b / 30 ) = ( 7000 / 500 ) + ( 300 / 30) = 24Moments due to minimum eccentricity :

Total moments for which the column is to be designed are :

41.0 + 42.2 = 83.2 KNm

36.0 + 37.15 = 73.15 KNmThe section is to be checked for biaxial bending

Pb ( about xx-axis ) = ( k

1 + k

2 p / f

ck ) f

ck b D

Pbx

= ( 0.196 + 0.203 x 3 /30 ) 30 x 300 x 400 x10-3

Pb ( about yy-axis ) = ( k

1 + k

2 p / f

ck ) f

ck b D

Pby

= ( 0.184 + 0.028 x 3 /30 ) 30 x 300 x 400 x10-3

Kx = ( P

uz - P

u ) / ( P

uz - P

bx )

Ky = ( P

uz - P

u ) / ( P

uz - P

by )

Max

=

May

=

Mux

=

Muy

=

ex =

ey =

Mux

= 1500 x 27.33 x 10-3 = 41.0 KNm > 15.0 KNm

Muy

= 1500 x 24 x 10-3 = 36.0 KNm > 10.0 KNm

Mux

=

Muy

=

Pu / f

ck b D = 1500 x 103 / ( 30 x 300 x 400 )

= 0.417

3 / 30 = 0.10

referring to chart 45 (d' / D = 0.15 ) ,

0.104

= 149.8 KNm

referring to chart 46 (d' / D = 0.2 ) ,

0.096

= 103.7 KNm

83.2 / 149.8 = 0.56

73.15 / 103.7 = 0.71

referring to chart 64 , the maximum allowable value of Mux / Mux1 corresponding to the above values of Muy / Muy1 and Pu / Puz is 0.58 which is slightly higher than the actual

value of 0.56 . The assumed reinforcement of 3.0 % is therefore satisfactory.

From , IS 456-2000 , Clause 39.6

1.602 0.2 difference 0.34

check : 0.04 difference ?

+

( 0.56 ) + ( 0.71 )

0.395 + 0.577

= 0.972 ……………………..( O.K.)

p x b x D / 100 = 3.0 x 300 x 400 / 100

= 3600

Ties : -

25 / 4= 6.25

Provide 8 mm O M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm

p / fck

=

Mu / f

ck b D2 =

Mux1

= 0.104 x 30 x 300 x 4002

Mu / f

ck b D2 =

Muy1

= 0.096 x 30 x 400 x 3002

Mux

/ Mux1

=

Muy

/ Muy1

=

Pu / P

uz = 1500 / 2700 = 0.56

for Pu / P

uz = 0.56 , α

n = 1.602

αn =

Mux

Muy ≤ 1M

ux1M

uy1

≤ 1

As =

mm2

Provide 25 mm diameter bar ( 3600 / 491 ) = 8 no. = 3928 mm2

O min

=

αn

αn

1.602 1.602

500

300

( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c

( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .

Design of slender ( Long ) columns ( with Uniaxial bending )Size of column 230 x 450 mm Column of a braced frameConcrete grade M 20

Unsupported length in both the direction = 5.0 m

The column is bent in double curvature and is slender about both the axis.

Solution : -Assume adjustment factor k = 0.8 for the first trial .Additional moments

== 39.2 KNm

== 28 KNm

About XX

1000 KN

= 0.6 x 80 - 0.4 x 60= 48 - 24= 24 KNm < 0.4 x 80 = 32 KNm

Take , 32 KNm

32 + 0.8 x 39.2= 63.36 KNm. < 80 KNm

Note that the distance between corner bars in one face is more than 48 Otr

Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.

Characteristic strength of reinforcement 415 N/mm2 HYSD reinforcement

Factored load Pu = 1000 kN

Factored.moment in the direction of larger dimension Muxx

= 80 kNm at top and 60 KNm at bottom.

Factored.moment in the direction of shorter dimension Muyy


The slenderness ratios ℓex

/ Ixx

and ℓey

/ Iyy are respectively 13.2 and 15.6

Assume that the moments due to minimum eccentricities about both the axes are less than applied moments.

Max

= ( Pu D / 2000 ) x ( ℓ

ex / D )2

( 1000 x 450 / 2000 ) x (13.2 )2 x 10-3

May

= ( Pu b / 2000 ) x ( ℓ

ey / b )2

( 1000 x 230 / 2000 ) x (15.6 )2 x 10-3

Pu =

Muxx

= Mi + k x M

ax

Mi = 0.6 M

u2 + 0.4 M

u1

Mi =

Note that Mu1

is considered negative as the column bends in double curvature .

Mu,xx

=

300

Take , 80 KNmAbout YY

1000 KN

= 0.6 x 40 - 0.4 x 30= 24 - 12= 12 KNm < 0.4 x 40 = 16 KNm

Take , 16 KNm

16 + 0.8 x 28= 38.4 KNm. < 40 KNm

Take , 40 KNmFinally design the column for

1000 KN

80 KNm

40 KNmFor the first trial , assume uniaxial bending about y axis for the following values .

1000 KN

( 230 / 450 ) ( 80 + 40 ) = 61.33 KNmd' / D = 50 / 230 = 0.22 Say 0.2

from chart 46 , SP : 16 = 0.162p = 0.162 x 20

= 3.24

( 3.24 /100 ) x 230 x 450

= 3353Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel are

Now check the assumed section as follows :

230 x 450 - 3220

= 100280

= 902.52 + 1002.225= 1904.7 KN

( p / 100 ) x b x D

( 3220 x 100 ) / ( 230 x 450 x 20 )

Mu,xx

=

Pu =

Muyy

= Mi + k x M

ay

Mi = 0.6 M

u2 + 0.4 M

u1

Mi =

Mu,yy

=

Mu,yy

=

Pu =

Mu,xx

=

Mu,yy

=

P'u =

M'uy

=

P'u / f

ck b D = 1000 x 103 / 20 x 230 x 450 = 0.48

M'uy

/ fck

b D2 = 61.33 x 106 / 20 x 450 x 2302 = 0.129p / f

ck =

Asc

=

mm2

usually conservative . Let us try 4 - 25 O + 4 - 20 O = 3220 mm2

Puz

= 0.45 fck

Ac + 0.75 f

y A

sc

Ac =

mm2

Puz

= 0.45 x 20 x 100280 x 10-3 + 0.75 x 415 x 3220 x 10-3

Pb = ( k

1 + k

2 p / f

ck ) f

ck b D

Asc

=

p / fck

=

= 0.156

= 0.181 x 2070= 373.8 KN

k == ( 1904.7 - 1000 ) / ( 1904.7 - 373.8 )= 904.7 / 1530.9= 0.59

Design for

1000 KN

32 + 0.59 x 39.2 = 55.13 KNm < 80

16 + 0.59 x 28 = 32.52 KNm < 40

the moment capacities can be found out as follows :About XXd' / D = 50 / 450 = 0.11 ≈ 0.15From chart 45 , SP : 16

= 135.07 KNmAbout YYd' / D = 50 / 230 = 0.22 ≈ 0.2From chart 46 , SP : 16

= 61.89 KNmCheck :

From , IS 456-2000 , Clause 39.6

1.542 0.2 difference 0.34check : 0.075 difference ?

+

80 + 40135.07 61.89

0.446 + 0.51

0.956 …………….( O.K.)

For d' / D = 0.2 , k1 = 0.184 and k

2 = -0.022 from table 60 ,SP : 16

Pb = ( 0.184 - 0.022 x 0.156 ) x 20 x 230 x 450 x 10-3

( Puz

- Pu ) / ( P

uz - P

b )

Pu =

Mux

= Take , Mux

= 80 KNm

Muy

= Take , Muy

= 40 KNm

For p / fck

= 0.162 and Pu / f

ck b D = 0.48 , the reinforcement being equally distributed ,

Mux1

/ fck

b D2 = 0.145

Mux1

= 0.145 x 20 x 230 x 4502 x 10-6

Muy1

/ fck

b D2 = 0.13

Muy1

= 0.13 x 20 x 450 x 2302 x 10-6

Pu / P

uz = 1000 / 1904.7 = 0.525

αn =

Mux

Muy ≤ 1M

ux1M

uy1

≤ 1

αn

αn

1.542 1.542

230

Provide 4 - 25 O + 4 - 20 O equally distributed.

= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 230 mm ( least lateral dimension )( ii ) 16 x 25 = 400 mm ( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm ( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 230 mm c/c

( 450 - 80 -25 = 345 < 48 x 8 = 384 ). Therefore two sets of Opened ties shall be used .

Ties : Minimum diameter Otr = 25 / 4

Note that the distance between corner bars in one face is less than 48 Otr

Note that if this distance would be more than ( 48 x 8 ) mm , closed ties for internal bars would be sufficient.

230

Design of slender ( Long ) columns ( with biaxial bending ) IS 456-2000 clause 25

( Given ) D = depth in respect of the major axis

Factored.moment in the direction of larger dimension = 40 kNm at top and 22.5 KNm at bottom.b = width of the member

Factored.moment in the direction of shorter dimension = 30 kNm at top and 20 KNm at bottom.

IS 456 : 2000Table 28 Effective Length of Compression Members ( Clause E-3 )

Symbol

Note :- A column may be considered as short when both slenderness ratios lex

/ D and ley

/ b ≤ 12 OR lex

/ ixx

< 40 where, For Circular column lex

/ D ≤ 10 .

lex

= effective length in respect of the major axis

ley

= effective length in respect of the minor axis

ixx

= radius of gyration in respect of the major axis.

iyy = radius of gyration in respect of the minor axis.

Degree of End Restraint of Compression members

Effectively held in position and restrained against rotation in both ends

Effectively held in position at both ends , restrained against rotation at one ends

Effectively held in position at both ends , but not restrained against rotation .

Effectively held in position and restrained against rotation at one end , and at the other end restrained against rotation but not held in position

Effectively held in position and restrained against rotation at one end , and at the other partially restrained against rotation but not held in position

Effectively held in position at one end but not restrained against rotation and at the other end restrained against rotation but not held in position

Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end.

NOTE - ℓ is the unsupported length of compression member.

The column is bent in double curvature. Reinforcement will be distributed equally on four sides.

The above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000

400 x 300 - 120000*3/100

1120.5

Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end.

Effective length of column ( ℓef ) : It is the distance between the points of zero moment (contraflexure ) along the

column height .

Ac + 0.75 f

y A

sc

= 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3

( i + 1 ) th floor

( i ) th floor

slab

Beam

ℓ

SP : 16 ,Table 60

Section d' / D0.05 0.10 0.15

Rectangular 0.219 0.207 0.196

Circular 0.172 0.160 0.149

Section d' / D0.05 0.10

250 -0.045 -0.045415 0.096 0.082

500 0.213 0.173250 0.215 0.146415 0.424 0.328500 0.545 0.425

The additional moments calculated earlier , will now be multiplied by the above values of k .

Circular

250 0.193 0.148

415 0.410 0.323

500 0.543 0.443The additional moments due to slenderness effects should be added to the initial moments IS 456-2000 clause 39.7.1

after modifying the initial moments as follows ( see note 1 under 39.7.1 of the code IS 456-2000 )

The above actual moments should be compared with those calculated from minimum eccentricity Where ,

consideration ( see 25.4 of the code ) and greater value is to be taken as the initial moment for axial load on the member,

effective length in respect of the major axis ,

mm > 20 mm effective length in respect of the minor axis ,

mm > 20 mm D =

b = width of the member

NOTES : - 1) A column may be considered braced in a given plane if lateral

stability to the structure as a whole is provided by walls or

bracing or buttressing designed to resist all lateral forces inthat plane. It should otherwise be considered unbraced.

2 ) In the case of a braced column without any transverse loads

Slender compression members- Values of P

Rectangular sections : Pb / f

ck b D = k

1 + k

2 . p / f

Circular sections : Pb / f

ck D2 = k

1 + k

2 . p / f

ck

Values of k1

Values of k2

fy N/ mm2

Rectangular ; equal reinforcement on two

opposite sides

Rectangular ; equal reinforcement on four

sides

The additional moments Max

and May

shall be calculated by the following formulae :

Max

= ( Pu D / 2000 ) x ( ℓ

ex / D )2

May

= ( Pu b / 2000 ) x ( ℓ

ey / b )2

Pu =

ℓex

=

ℓey

=

depth of the cross - section at right angles to the major axis , and

occurring in its height, the additional moment shall be added

end moment (assumed negative if the column is bent in double

curvature). In no case shall the initial moment be less than

shall be added to the end moments.

3 ) Unbraced compression members, at any given level or storey,subject to lateral load are usually constrained to deflect

equally. In such cases slenderness ratio for each column may

be taken as the average for all columns acting in the same

direction.IS 456-2000 clause 39.7.1.1The values given by equation 39.7.1 may be multiplied by the following factor :

where ,

axial load on compression member,

as defined in 39.6,

-0.068 axial load corresponding to the condition

of maximum compressive strain of

0.0035 in concrete and tensile strain of0.002 in outer most layer of tension steel.

to an initial moment equal to sum of 0.4 Mu1

, and 0.6 Mu2

,

where Mu2

is the larger end moment and Mu1

is the smaller

0.4 Mu2

nor the total moment including the initial moment be

less than Mu2

. For unbraced columns, the additional moment

K = ( Puz

- Pu ) / ( P

uz - P

b ) ≤ 1

Pu =

Puz

= Puz

= 0.45 fck

Ac + 0.75 f

Pb =

8-25 O


Design of slender ( Long ) columns ( with Uniaxial bending )

( Given )

Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be


= 40 kNm at top and 30 KNm at

Assume that the moments due to minimum eccentricities about both the axes are less than

8 O @300 c/c(two sets )

Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel are

-0.1275

450

4-25 O + 4 -20 O

( 16 times least longitudinal diameter of bar )

( 450 - 80 -25 = 345 < 48 x 8 = 384 ). Therefore two sets of Opened ties shall be used .

Note that if this distance would be more than ( 48 x 8 ) mm , closed ties for internal bars would be

8 O @ 230 mm c/c( three sets )

Table 28 Effective Length of Compression Members ( Clause E-3 )

Symbol

0.5 ℓ 0.65 ℓ

0.7 ℓ 0.8 ℓ

1.0 ℓ 1.0 ℓ

1.0 ℓ 1.2 ℓ

- 1.5 ℓ

2.0 ℓ 2.0 ℓ

2.0 ℓ 2.0 ℓ

Note :- A column may be considered as short when both slenderness ratios < 40 where, For Circular column l

ex / D ≤




= radius of gyration in respect of the minor axis.

Theoretical Value of effective length

Recommended Value of effective length

2.0 ℓ 2.0 ℓ

NOTE - ℓ is the unsupported length of compression member.

) : It is the distance between the points of zero moment (contraflexure ) along the

slab

Beam

SP : 16 ,Table 60

d' / D0.20

0.184

0.138

d' / D0.15 0.20

-0.045 -0.0450.046 -0.022

0.104 -0.0010.061 -0.0110.203 0.0280.256 0.0400.077 -0.020

0.201 0.036

0.291 0.056

effective length in respect of the major axis ,

effective length in respect of the minor axis ,

1) A column may be considered braced in a given plane if lateral

stability to the structure as a whole is provided by walls or

bracing or buttressing designed to resist all lateral forces inthat plane. It should otherwise be considered unbraced.

2 ) In the case of a braced column without any transverse loads

Slender compression members- Values of Pb

b D = k1 + k

2 . p / f

ck

= k1 + k

2 . p / f

ck

shall be calculated by the following formulae :

depth of the cross - section at right angles to

occurring in its height, the additional moment shall be added

end moment (assumed negative if the column is bent in double

curvature). In no case shall the initial moment be less than

shall be added to the end moments.

3 ) Unbraced compression members, at any given level or storey,subject to lateral load are usually constrained to deflect

equally. In such cases slenderness ratio for each column may

be taken as the average for all columns acting in the same

The values given by equation 39.7.1 may be multiplied by the following factor :

, and 0.6 Mu2

,

is the smaller

nor the total moment including the initial moment be

. For unbraced columns, the additional moment

= 0.45 fck

Ac + 0.75 f

y A

sc

Design of short eccentrically loaded square columns - Biaxial bending.Size 500 mm x 500 mmAxial factored load 1500 KN

( Given )

moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete

HYSD reinforcement of grade Fe415Solution :-

= 0.4

= 0.112d' = 40 + 10 = 50 mmd' / D = 50 / 500 = 0.1From chart 32 , SP-16

p = 0.078 x 15 = 1.17

= 2925

p = 3220 x 100 / ( 500 x 500 )p = 1.288

= 0.086The assumed section is now checked.

The reinforcement being equally distributed , the moment capacities from

Chart - 44 , SP-16

= 202.5 KNm

500 x 500 - 3220

= 246780

= 1665.77 + 1002.23

= 2667.99 KN

Factored moment Mux

= 90 KNm

Muy

= 120 Knm

Assume an axial load P'u of 1500 KN and a uniaxial moment M'

ux = 90 + 120 = 210 KNm .

P'u / ( f

ck x b x D ) = 1500 x 103 / ( 15 x 500 x 500 )

M'ux

/ ( fck

x b x D2 ) = 210 x 106 / ( 15 x 500 x 5002 )

p / fck

= 0.078

As = 1.17 x b x D = 1.17 x 500 x 500 / 100

mm2

Provide 4-25 mm O + 4-20 mm O = 3220 mm2 , equally distributed

p / fck

= 1.288 / 15

For p / fck

= 0.086 and Pu / ( f

ck x b x D ) = 0.4 ,

Mux1

/ ( fck

x b x D2 ) = Muy1

/ ( fck

x b x D2 ) = 0.108

Mux1 = Muy1 = 0.108 x 15 x 500 x 5002 x 10-6

Puz

= 0.45 fck

Ac + 0.75 f

y A

sc

Ac =

mm2

Puz

= 0.45 x 15 x 246780 x 10-3 + 0.75 x 415 x 3220 x 10-3

1500 / 2668= 0.56

From , IS 456-2000 , Clause 39.6

1.602 0.2 difference 0.34check : 0.04 difference ? -0.068

+

90+

120202.5 202.5

0.273 + 0.433

0.706

= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 500 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 350 mm c/c

Design of short eccentrically loaded rectangle columns - Biaxial bending.Size 300 mm x 500 mmAxial factored load 1500 KN

( Given )

moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete

HYSD reinforcement of grade Fe415Solution :-

= 0.67

= 0.107d' = 40 + 10 = 50 mmd' / D = 50 / 300 = 0.167 ………………..use 0.2From chart 34 , SP-16

p = 0.167 x 15 = 2.5

Pu / P

uz =

αn =

Mux

Muy ≤ 1M

ux1M

uy1

≤ 1Ties : Minimum diameter O

tr = 25 / 4

Factored moment Mux

= 60 KNm

Muy

= 60 KNm

Assume an axial load P'u of 1500 KN and a uniaxial moment M'

uy = 300 / 500 ( 60 + 60 ) = 72 KNm .

P'u / ( f

ck x b x D ) = 1500 x 103 / ( 15 x 300 x 500 )

M'u / ( f

ck x b x D2 ) = 72 x 106 / ( 15 x 500 x 3002 )

p / fck

= 0.167

αn

αn

1.602 1.602

500

500

4-25 O +4-20 O

8 O @350 c/c

= 3750

p = 3928 x 100 / ( 300 x 500 )p = 2.62

= 0.175The assumed section is now checked.About X

Chart - 44 , SP-16

= 146.25 KNmAbout Y

Chart - 46 , SP-16

= 69.53 KNmPure axial load capacity

300 x 500 - 3928

= 146072

= 985.986 + 1222.59= 2208.58 KN

1500 / 2208.6= 0.68

From , IS 456-2000 , Clause 39.6

1.802 0.2 difference 0.33check : 0.12 difference ? -0.198

+

60 + 60146.25 69.53

0.2 + 0.77

As = 2.5 x b x D = 2.5 x 300 x 500 / 100

mm2

Provide 8-25 mm O = 3928 mm2 , equally distributed

p / fck

= 2.62 / 15

d' / D = 50 / 500 = 0.1 , For p / fck

= 0.175 and Pu / ( f

ck x b x D ) = 0.67 ,

Mux1

/ ( fck

x b x D2 ) = 0.13

Mux1

= 0.13 x 15 x 300 x 5002 x 10-6

d' / D = 50 / 300 = 0.167 say 0.2 , For p / fck

= 0.175 and Pu / ( f

ck x b x D ) = 0.67 ,

Muy1

/ ( fck

x b x D2 ) = 0.103

Muy1

= 0.103 x 15 x 500 x 3002 x 10-6

Puz

= 0.45 fck

Ac + 0.75 f

y A

sc

Ac =

mm2

Puz

= 0.45 x 15 x 146072 x 10-3 + 0.75 x 415 x 3928 x 10-3

Pu / P

uz =

αn =

Mux

Muy ≤ 1M

ux1M

uy1

αn

αn

1.802 1.802

0.97

= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c


≤ 1Ties : Minimum diameter O

tr = 25 / 4



500

300

39.6 Members Subjected to Combined Axial Loadand Biaxial Bending where ,

+

= 90 + 120 = 210 KNm .

M

Mux

Muy ≤ 1M

ux1M

uy1M

ux1, M

uy1 =

αn is related to P

u/P

uz

where Puz

= 0.45 fck

Ac + 0.75 f

y A

sc

For values of Pu / P

uz = 0.2 to 0.8, ( 0.2 , 0.4 , 0.6 , 0.8 )the values of

linearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, α

αn

αn

( least lateral dimension )( 16 times least longitudinal diameter of bar )

( 48 times diameter of tie )

1 .O; for values greater than 0.8, αn is 2.0.

= 300 / 500 ( 60 + 60 ) = 72 KNm .

4-25 O +4-20 O

8 O @350 c/c




8-25 O

8 O @300 c/c(two sets )

moments about x and y axes

due to design loads,

maximum uniaxial moment

bending about x and y axes

respectively, and

Mux

, Muy

=

Mux1

, Muy1

=

capacity for an axial load of Pu,

= 0.2 to 0.8, ( 0.2 , 0.4 , 0.6 , 0.8 )the values of αn vary

linearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, αn is

Design of short eccentrically loaded columns - uniaxial bending.Size 300 mm x 600 mmAxial factored load 600 KNFactored moment 300 KNm

( Given )moment due to minimum eccentricity is less than the applied moment.Material M15 grade concrete

HYSD reinforcement of grade Fe415Solution :-Using 25 mm diameter bars with 40 mm clear coverd' = 40 + 12.5 = 52.5 mmd' / D = 52.5 / 600 = 0.088 ,say 0.1

= 0.222

= 0.185From chart 32 , SP : 16

p = 0.1 x 15 = 1.5

= 2700

As the distance between two opposite corner bars is more than 300 mm ,provide 2- 12 mm O at centre of long sides of the column, which may be tied by open ties.

= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/c ( two sets )


Pu / ( f

ck x b x D ) = 600 x 103 / ( 15 x 300 x 600 )

Mu / ( f

ck x b x D2 ) = 300 x 106 / ( 15 x 300 x 6002 )

p / fck

= 0.1

As = 1.5 x b x D = 1.5 x 300 x 600 / 100

mm2

Provide 22 mm O ( 2700 / 380 = 8 no.) = 3041 mm2

Provide 8- 25 O + 2 - 12 O = 4151 mm2

Ties : Minimum diameter Otr = 25 / 4



600

300

provide 2- 12 mm O at centre of long sides of the column, which may be tied by open ties.




8-25 O+ 2-12 O

8 mm O @300 mm c/c( two sets )

Design of short circular columnWorking load = 1200 KN

Given( a ) lateral ties & ( b ) helical reinforcementMaterial M20 grade concrete

HYSD steel Fe415For Lateral reinforcement mild steel Fe250Solution : -

Factored load = 1.5 x 1200 = 1800 KN

( a ) lateral ties : -

Assume 0.8 % minimum steel.

Then ,

=Substituting , we have

7.936 2.2244

10.1604

177165If D is the diameter of the column

177165D = √ 225688D = 475 mm

Use 475 mm diameter column.

0.008 x 177165

= 1417Minimum 6 bars shall be used.Provide 16 mm diameter bars ( 1417 / 201 ) = 8 no.

8 x 201

= 1608Use 6 mm O lateral ties , Spacing shall be lesser of( i ) 475 mm ( least lateral dimension )( ii ) 16 x 16 = 256 mm ( 16 times least longitudinal diameter of bar )( iii ) 48 x 6 = 288 mm ( 48 times diameter of tie )Provide 6 mm O lateral ties @ 250 mm c/c .( b ) Helical reinforcement : -The column with helical reinforcement can support 1.05 times the load of a similarmember with lateral ties. Therefore

Assume emin

< 0.05 D

Pu = 0.4 f

ck A

c + 0.67 f

y A

sc

Asc

= 0.008 Ag

Ac = A

g - A

sc

0.992 Ag

1800 x 103 = 0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 A

g

1800 x 103 = Ag + A

g

1800 x 103 = Ag

Ag = mm2

( ¶ / 4 ) x D2 =

Asc

=

mm2

Asc

=

mm2

Pu = 1.05 [ 0.4 f

ck A

c + 0.67 f

y A

sc ]

8.3328 2.33562

10.66842

168729If D is the diameter of the column

168729D = √ 214941D = 463 mm

Use 450 mm diameter column.

Then ,

= 158962.5

= 1335289 - 8.4 + 291.953

464711 = 283.55

1638.903provide 20 mm diameter bars ( 1638.9 / 314 =) 6 No.

6 x 314

= 1884Assume 8 mm O M.S. bars for helix at 40 mm clear cover .

450 - 40 - 40= 370 mm

= 50

== 0.36 x 0.4792 x 20 / 250= 0.013801

Now0.0138 = 4 x 50 / p x 370

p = 39.17 mm …………………….. ( 1 )As per IS 456-2000 clause 26.5.3.2 ( d )The pitch < 75 mm

> 25 mm> 3 x dia of helix bar = 3 x 8 = 24 mm ……………( 2 )

From ( 1 ) and ( 2 ) , provide 8 mm O helix @ 35 mm pitch.

1800 x 103 = 1.05 [0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 A

g ]

1800 x 103 = Ag + A

g

1800 x 103 = Ag

Ag = mm2

( ¶ / 4 ) x D2 =

Ag = ( ¶ / 4 ) x 4502

mm2

1800 x 103 = 1.05 [0.4 x 20 x (158963 - Asc

) + 0.67 x 415 x Asc

]

Asc

Asc

Asc

Asc

= mm2

Asc

=

mm2

Dc =

asp

= ( ¶ / 4 ) x 82

mm2

Minimum ρs = 0.36 ( (A

g / A

cr) - 1 ) f

ck / f

y

0.36 ( ( 4502 / 3702 ) - 1 ) x 20 / 250

ρs = 4 x a

sp / p x D

c

< Dc / 6 ( = 370 / 6 = 61.67 mm )

( 16 times least longitudinal diameter of bar )

450

6 - 20 O

6 - 20 O

35

8 mm O @35 mm c/c

Design of short columnFactored load = 1500 KN

GivenMaterial M15 grade concrete

mild steel Fe250

Solution : -

Therefore , size of column shall be minimum ( 20 / 0.05 = 400 ) 400 mm x 400 mm

Assume 0.8 % minimum steel.Then ,

=

=5.952 1.34

7.292

205705

If the column is to be a square , the side of column = √205705 = 453 mm

Adopt 450 mm x 450 mm size column .

Then ,

1215000 + 167.5

285000 = 161.5

1765Provide 20 mm diameter bars = ( 1765 / 314 ) = 6 No.

Note that the distance between the bars exceeds 300 mm on two parallel sides .the arrangement of reinforcement should be changed.Provide then 4 no. 20 mm diameter bars plus 4 no. 16 mm diameter bars giving

4 x 314 + 4 x 201= 1256 + 804

= 2060Lateral ties :Use 6 mm O lateral ties.Spacing should be lesser of :( i ) 450 mm ( least lateral dimension )( ii ) 16 x 16 = 256 mm ( 16 times least longitudinal diameter of bar )(iii) 48 x 6 = 288 mm . ( 48 times diameter of tie )Provide 6 mm O ties about 250 mm c/c .

Assume emin

< 0.05 D

Here , emin

< 0.05 D , But emin

= 20 mm

Asc

= 0.008 Ag

Ac = A

g - A

sc

0.992 Ag

Pu = 0.4 f

ck A

c + 0.67 f

y A

sc

0.4 x 15 x 0.992 Ag + 0.67 x 250 x 0.008 A

g

1500 x 103 = Ag + A

g

1500 x 103 = Ag

Ag = mm2

1500 x 103 = 0.4 x 15 x ( 450 x 450 - Asc

)+ 0.67 x 250 x Asc

1500 x 103 = - 6 Asc

Asc

Asc

Asc

= mm2

giving , Asc

= 6 x 314 = 1884 mm2

Asc

=

mm2


wrong arrangement correct arrangement



450

450

450

450

6 - 20 O4 - 20 O + 4 - 16 O

6 mm O @ 250 mm c/c(double ties )

IS 456-2000 clause 25

D = depth in respect of the major axis

b = width of the member

400 mm x 400 mm

IS 456-2000 clause 39.3

IS 456-2000 clause 26.5.3.1

NOTE - The use of 6 percent reinforcement may involve

practical difficulties in placing and compacting of concrete;hence lower percentage is recommended. Where bars fromthe columns below have to be lapped with those in thecolumn under consideration, the percentage of steel shall

usually not exceed 4 percent26.5.3 columns

26.5.3.1 Longitudinal reinforcementa) The cross-sectional area of longitudinalreinforcement, shall be not less than 0.8 percentnor more than 6 percent of the gross crosssectionalarea of the column.NOTE - The use of 6 percent reinforcement may involvepractical difficulties in placing and compacting of concrete;hence lower percentage is recommended. Where bars from

Note :- A column may be considered as short when both slenderness ratios lex

/ D and ley

/ b ≤ 12 OR lex

/ ixx

< 40 where,

lex


ley


ixx


iyy = radius of gyration in respect of the minor axis.

All columns shall be designed for minimum eccentricity equal to the unsupported length of column / 500 plus lateral dimension / 30 , subject to a minimum of 20 mm. For bi-axial bending this eccentricity exceeds the minimum about one axis at a time.

When emin

≤ 0.05 D , the column shall be designed by the following equation:

Pu = 0.4 f

ck A

c + 0.67 f

y A

sc

Ac = Area of concrete ,

Asc

= Area of longitudinal reinforcement for columns.

If emin

> 0.05 D , the column shall be designed for moment also.

The cross-sectional area of longitudinal reinforcement , shall be not less than 0.8 % nor more than 6 % of the gross cross-sectional area of the column.

the columns below have to be lapped with those in the( 450 - 80 -20 = 350 > 48 x 6 = 288 ). Therefore two sets of closed ties shall be used . column under consideration, the percentage of steel shall

usually not exceed 4 percent.b) In any column that has a larger cross-sectionalarea than that required to support the load,the minimum percentage of steel shall bebased upon the area of concrete required toresist the direct stress and not upon the actualarea.c)The minimum number of longitudinal barsprovided in a column shall be four in rectangularcolumns and six in circular columns.d )The bars shall not be less than 12 mm indiameter.e) A reinforced concrete column having helicalreinforcement shall have at least six bars oflongitudinal reinforcement within the helicalreinforcement.f )In a helically reinforced column, the longitudinalbars shall be in contact with the helicalreinforcement and equidistant around its innercircumference.g )Spacing of longitudinal bars measured alongthe periphery of the column shall not exceed300 mm.h )In case of pedestals in which the longitudinalreinforcement is not taken in account in strengthcalculations, nominal longitudinal reinforcementnot less than 0.15 percent of the cross-sectionalarea shall be provided.

4 - 20 O + 4 - 16 O

6 mm O @ 250 mm c/c(double ties )

Note :- A column may be considered as short when both slenderness ratios

All columns shall be designed for minimum eccentricity equal to the unsupported length of column / 500 plus lateral dimension / 30 , subject to a minimum of 20 mm. For bi-axial bending this eccentricity exceeds the

≤ 0.05 D , the column shall be designed by the following

The cross-sectional area of longitudinal reinforcement , shall be not less than 0.8 % nor more than 6 % of the gross cross-sectional area of the

Design of dog-legged staircase

230

UP 900

Floor A B 150 1950

900

230300 900 2250 900 230

Rise of step = 160 mm Rise 175 mm to 200 mm

Tread = 250 mm Tread 250 mm to 280 mmNosing is not provided ( given )Material M15 Grade concretemild steel reinforcement Fe250Solution : - Assume 150 mm thick waist slab. Both the landings can span on walls.Landing A or B

Self-load 0.15 x 25 = 3.75

Floor finish = 1.00

Live load ( residence ) = 3.00

Total 7.75

11.63Span = 1950 + 150 = 2100 i.e. 2.1 m

Consider 1 m length of slab

M =

== 6.41 KNm

Reinforcement will be in second layer, Assuming 12 mm O barsd = 150 - 15 (cover ) -12 - 6

= 117 mm

= 0.468

=

> 200 mm

< 230 mm

KN / m2

KN / m2

KN / m2

KN / m2

PU = 1.5 x 7.75 = KN / m2

w x l2 / 8

11.63 x 2.12 / 8

Mu / b x d2 = 6.41 x 106 / 1000 x 1172

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (0.468)

1 5 9 10

11151920

vvv

vv

250 / 15

= 50 [(1-0.93) x 15 / 250 ]= 0.224%

0.224 x 1000 x 117 / 100

= 262

Minimum steel = ( 0.15 / 100 ) x 1000 x 150



Provide 10 mm O bar @ 280 mm c/c = 280Maximum spacing = 3 x d = 3 x 117

= 351 mm ……………….( O.K.)Check for shear : -

w x l / 2= 11.63 x 2.1 / 2= 12.21 KN

Shear stress =

=

= 0.104 ( too small )

( from table 7-1 )……………….( O.K.)


12 O (mild steel )

= 100 x 280 / 1000 x 117= 0.239

From equation

0.499

= 6.83 KNm

12.21 KN

Ast =

mm2

According to IS 456-200 clause 26.5.2.1 Minimum steel 0.15 % for Fe250 and 0.12 % for Fe415

mm2


mm2 .

Vu =

Vu / b x d

12.21 x 103 / 1000 x 117

N / mm2 ح ) >C )

N / mm2

for Pt = 0.224 حc = 0.28 < 0.28

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

Mu1

/ b x d2 =

Mu1

= 0.499 x 1000 x 1172 x 10-6

Vu =

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

727.191 + 12 O

727.19which gives 16.91 mm ……………….( O.K.)

Check for deflection : -


100 x 280 / 1000 x 117= 0.239


modification factor = 2

( span / d ) ratio permissible = 2 x 20

= 40Actual (span / d ) ratio = 2100 / 117

= 17.95 < 40 ……………….( O.K.)

Design of flight : -

Loads :

Inclined length of waist slab for one step = = 296.8 mm

Assuming 150 mm thick waist slab

self-load in plan = ( 296.8 / 250 ) x 0.15 x 25 = 4.45Floor finish length for one step = 160 + 250

= 410 mm

floor finish = ( 410 / 250 ) x 1 = 1.64 250

Weight of step ( ( 0 + 160 ) / 2000 ) x 25 = 2.00 160 296.8

Live load = 3.00Total = 4.45 + 1.64 + 2.00 + 3.00

= 11.09

16.64IS 456-2000 , clause 33.2

16.64 KN / m

11.63 KN / m 11.63 KN / m

525 2250 525 525 = 900 / 2 +75

1.3 x ( 6.83 x 106 / 12.21 x 103 ) + 12 O ≥ 55 O≥ 55 O

43 O ≤O ≤

Pt = 100 x A

st / b x d =

Span and loading on both the flights are the same. Therefore same design will be adopted.

√ 2502 + 1602

KN / m2

KN / m2

KN / m2

KN / m2

KN / m2

Pu = 1.5 x 11.09 = KN / m2

Landing is spanning in transverse direction. The span of stair according to discussion made in art 13-1 and loading for 1 m width of stair is shown in fig below.

RA = 24.83 KN RB = 24.83 KN

3300

0.525 x 11.63 + 16.64 x 2.25 / 2 = 6.11 + 18.72= 24.83 KN

= 40.97 - 15.83 - 3.17= 21.97 KNm


== 99.48 mm,

150 - 15 - 6= 129 mm, ……………….( O.K.)

= 1.32

=250 / 15

= 50 [(1-0.77) x 15 / 250 ]= 0.687%

0.687 x 1000 x 129 / 100



Provide 12 mm O bar@125 mm c/c = 904Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement

Distribution steel = ( 0.15 / 100 ) x 1000 x 150




RA = R

B =

Mu = 24.83 x 3.3/2 - 11.63 x (3.3/2)2 / 2 - ( ( 16.64 - 11.63 ) x ( 2.25/2 )2 ) / 2

drequired

= √M / Q x b

√21.97 x 10 6 / 2.22 x 1000

dprovided

=

Mu / b x d2 = 21.97 x 10 6 / 1000 x (129)2

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

50 1-√1-(4.6 / 15) x (1.32)

Ast =

mm2


mm2.

mm2


mm2 .

Check for shear : -

24.83 KN

Shear stress =

=

= 0.192 ( too small )

100 x 904 / 1000 x 129= 0.70

from table 7-1



= 0.681 ……………….( O.K.)Check for development length : -

12 O (mild steel )

= 100 x 904 / 1000 x 129= 0.70

From equation

we get , 1.344

= 22.37 KNm

24.83 KN

= O x 0.87 x 250 / 4 x 1 =54.4 O

1171.2 + 12 O1171.2

which gives 27.62 mm ……………….( O.K.)From Crossing of bars, the bars must extend upto 54.4 x 12 = 653

say 700 mm.Check for deflection : -


100 x 904 / 1000 x 129= 0.70


Vu =

Vu / b x d

24.83 x 103 / 1000 x 129

N / mm2 ح ) >C )

N / mm2

100 x As / b x d =

for Pt = 0.7 حc = 0.524 N / mm2

N / mm2

Assuming L0 =

Pt = 100 x A

s / b x d

Pt = 50 1-√1-(4.6 / fck) x (M

u / b x d2)

fy / f

ck

Mu1

/ b x d2 =

Mu1

= 1.344 x 1000 x 1292 x 10-6

Vu =


s / 4 x ح

bd

1.3 x ( Mu1

/ Vu ) + L

0 ≥ L

d

1.3 x ( 22.37 x 106 / 24.83 x 103 ) + 12 O ≥ 54.4 O≥ 54.4 O

42.4 O ≤O ≤

Pt = 100 x A

st / b x d =


= 25.58 < 32.4 ……………….( O.K.)The depth could be reducedCheck for cracking : -IS 456-2000 , clause 26.3.3

Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small= 3 x 129 = 387 mm or 300 mm

spacing provided = 125 mm < 300 mm ……………….( O.K.)Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is small

= 5 x 129 = 645 mmspacing provided = 220 mm < 450 mm ……………….( O.K.)

175 mm to 200 mm

250 mm to 280 mm

check for shear

IS 456-2000 , Table 19Table 7-1

Concrete grade

M15 M20 M25 M30 M350.28 0.28 0.29 0.29 0.29

0.25 0.35 0.36 0.36 0.37 0.37

0.50 0.46 0.48 0.49 0.50 0.50

0.75 0.54 0.56 0.57 0.59 0.59

1.00 0.60 0.62 0.64 0.66 0.67

1.25 0.64 0.67 0.70 0.71 0.731.50 0.68 0.72 0.74 0.76 0.781.75 0.71 0.75 0.78 0.80 0.82

2.00 0.71 0.79 0.82 0.84 0.86

2.25 0.71 0.81 0.85 0.88 0.902.50 0.71 0.82 0.88 0.91 0.932.75 0.71 0.82 0.90 0.94 0.963.00 0.71 0.82 0.92 0.96 0.99


6 x β

β =

IS 456-2000 , Table 20

Table 7-2

Design shear strength of concrete , حC, N / mm2

Pt = 100 x A

s

b x d≤ 0.15


ck ( √ 1 + 5 x β - 1 )

0.8 x fck


Concrete grade M15 M20 M25 M30 M35

2.5 2.8 3.1 3.5 3.7check for development lengthIS 456-2000 clause 26.2.1

IS 456-200 clause 26.2.1.1Table 7-5

Concrete grade M15 M20 M25 M30 M35

1.0 1.2 1.4 1.5 1.7

For mild steel Fe250

For Fe415IS 456-2000 clause 26.2.3.3

Table 7-6Development length for single mild steel bars

Tension bars Compression bars

M15 M20 M15 M20

250 55 O 26 O 44 O 37 O

415 56 O 47 O 45 O 38 O500 69 O 58 O 54 O 46 O

check for deflectionBasic values of span to effective depth ratios for spans upto 10 m :

cantilever 7simply supported 20

continuous 26

check for cracking(a)The horizontal distance between parallel main

Maximum shear stress , حC, N / mm2

ح ) c )

max N / mm2


s / 4 x ح

bd

Design bond stress (حbd


ح ) bd

N / mm2

Note-1 : حbd

shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ح

bd shall be increased by 60 %.

σs = 0.87 x f

y

σs = 0.67 x f

y

Ld ≤ M1 / V + L

0

L0 = effective depth of the members or 12 O , whichever is greater

if ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )

fy N / mm2

For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.

reinforcement bars ( spacing )shall not be more than ttimes the effective depth of solid slab or

300 mm whichever is smaller.(b)The horizontal distance between parallelreinforcement bars ( spacing ) provided against

shrinkage and temperature (distribution bar) shall not be morethan five times the effective depth of a solidslab or 450 mm whichever is smaller.

IS 456-2000 Clause 40.2.1.1

275 250 225 200k 1.00 1.05 1.10 1.15 1.20

sketch :The reinforcements for both the flights are shown in fig.

700

700

Overall depth of slab , mm

300 or more

v vv v

v

10 O @ 280 c/c ( Landing )

Flight-B

Flight-A

v vv

vv

v vv

v vv

10 O @ 280 c/c ( Landing )

12 O @ 125 c/c150 mm thick waist slab

8 O @ 220 c/c

12 O @ 125 c/c

vv

vv

vv

vv

v

v

vv v v v v

8 O @ 220 c/c

12 O @ 125 c/c

10 O @ 280 c/c ( Landing )

150 mm thick waist slab



or 450 mm i.e. 450 mm……………….( O.K.)

IS 456-2000 , Table 19Table 7-1

Concrete grade

M400.300.38

0.51

0.60

0.68

0.740.790.84

0.88

0.920.950.981.01

IS 456-2000 , Table 20

Table 7-2

, N / mm2

1 + 5 x β - 1 )

, but not less than 1.0

M40

4.0

IS 456-200 clause 26.2.1.1Table 7-5

M40

1.9

&

Basic values of span to effective depth ratios for spans upto 10 m :

, N / mm2


shall be increased by 25 % for bars in compressionNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the

= effective depth of the members or 12 O , whichever is greater For checking development length , l

0


if ends of the reinforcement are confined by a compressive reaction M1 /

For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection

1751.25 1.30

150 or less

150

v vv

v vv

v v vvv v v

vv

vv

vv

vv

10 O @ 280 c/c ( Landing )

12 O @ 125 c/c

8 O @ 220 c/c

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