design of rectangular water tank (1)

8/12/2019 Design of Rectangular Water Tank (1)

1/242

Capacity 80000 Litres (given)

Material M20 (given)

Fe 415 Grade HYSD reinforcement (given)

80400 Litres

L / B = 6 / 4 = 1.5 < 2 .

H / 4 = 3.5 / 4 = 0.875 m

2.5 x YW= 2.5 x 9.8

= 24.5 KN / m2

where Ywis unit weight of water = 9.8 KN / m3

6 m

6 m A E

F

3.5 m 24.5 KN / m2

D

34.3 KN / m2

Elevation Plan

MAB =

= 73.5 KNm

MAD =

= -32.66 KNm

- 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

Fixed end moments :-

Design of Rectangular water tank CASE-1 ( L / B < 2 )

Grade Concrete

Volume = 6 x 4 x 3.35 x 103

=

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

Solution :-

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

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.

bottom 1 m will be designed as cantilever.

Rotation factor at Joint A

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

Kani's Method :-

24.5 x 62/ 12 =

24.5 x 42/ 12 =

w x l2/ 12 =

w x l2/ 12 =

-3/10 -2/1040.84

2.5 m

1 m


2/242

Joint Member Relative

Stiffness( K ) K

Rotation Factor

u =(-1/2) k / K

AB I / 6 - 2 / 10

AD I / 4 - 3 / 10

MAF =

40.84 KNm

MAB=

=

= 57.16

MAD=

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

= -57.16

=

53.09 KNm

=

-8.16 KNm

= 49 KN

= 73.5 KN

M = 57.16 KNm

T = 49 KN

Q = 0.306

D = M / Q x b

= 57.16 x 106/ 0.306 x 1000

= 432.2 mm,

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

= 417 mm

Direct tension in long wall =

A 5 * I / 12

Sum of FEM

73.5-32.66

B.M. at centre of long span =

B.M. at centre of short span =

MABF+ 2 MAB' + MBA'

73.5 + 2 x (- 8.17 ) + 0

MADF+ 2 MAD' + MDA'

Direct tension in short wall = Yw( H - h ) x L / 2

24.5 x 6 / 2 =

Design of Long Walls :-

24.5 x 4 / 2 =

w x l2/ 8 - 57.16

w x l2/ 8 - 57.16

Yw( H - h ) x B / 2

24.5 x 42/ 8

- 57.16

24.5 x 62/ 8

- 57.16

Tension on liquid face.

Assuming d / D = 0.9

At support

From Table 9-6

From Table 9-5

Assuming d / D = 0.9


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M / stx j x d

=

= 1048 mm2

Ast2for direct tension = T / st

=

= 327 mm2

= 1375 mm2

=

= 146.15273 mm

Provide 16 mm O bar @ 130 mm C/Cmarked(a) = 1546 mm2/ m.

At centre

M = 53.09 KNm

T = 49 KN

e = M / T =

= 1.08 m

E = e + D / 2 - d b

=

= 888 mm

D

= 49 x 0.888d

= 43.51 KNm

d'

M / stx j x d

=

= 617 mm2


=

= 327 mm2

= 944 mm2

=

= 212.88136 mm

Total Ast1+ Ast2 = 617 + 327

Provide 16 mm O bar

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

200.96 x 1000 / 944

E = e + D / 2 - d

1080 + 450 / 2 - 417

modified moment

Ast1for moment =

43.51 x 106

/ 190 x 0.89 x 417

49 x 103

/ 150

Larger steel area is provided to match with the steel of short walls.

49 x 103

/ 150

Total Ast1+ Ast2 = 1048 + 327

i.e.tension is small

Line of action of forces lies outsi


57.16 x 106

/ 150 x 0.872 x 417

Ast1for moment =

53.09 / 49

200.96 x 1000 / 1375

Provide 16 mm O bar

tension on remote face


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Provide 16 mm O bar @ 200 mm C/Cmarked(b) = 1005 mm2

= 720 mm2

360 mm2

=

= 139.55556 mm

Provide 8 mm O bar @ 130 mm C/Cmarked(d) = 385 mm2

=

= 218.05556 mm

Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2

Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005 mm2

385 mm2

mm2

M = 57.16 KNm

T = 73.5 KN

M / stx j x d

== 1048 mm

2


=

= 490 mm2

= 1538 mm2

== 130.6632 mm

Provide 16 mm O bar @ 130 mm C/Cmarked(b) = 1546 mm2/ m.

At centre

M = 8.16 KNm

T = 73.5 KN

200.96 x 1000 / 1538

tension on liquid face

57.16 x 106

/ 150 x 0.872 x 417

73.5 x 103

/ 150

Total Ast1+ Ast2 = 1048 + 490

Provide 16 mm O bar


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

Design of short walls :-

At support

tension on liquid face

From Table 9-5

Ast1for moment =


50.24 x 1000 /360

Horizontal steel :-

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

Provide 10 mm O bar


78.50 x 1000 /360

Vertical Steel ( c)

Distribution steel =

From Table 9-3 minimum reinforcement 0.16 %

0.16 / 100 x 1000 x 450

On each face =

Provide 8 mm O bar


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M / stx j x d

=

= 150 mm2


=

= 490 mm2

= 640 mm2

=

= 176.625 mm

Provide 12 mm O bar @ 130 mm C/Cmarked(e) = 869 mm2/ m.

= 720 mm2

360 mm2

=

= 139.55556 mm

Provide 8 mm O bar @ 130 mm C/Cmarked(d) = 385 mm2

=

= 218.05556 mm

Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2.

Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869 mm2

385 mm2

mm2

M = OR Ywx H / 6 , whichever is greater.

= = 9.8 x 3.5 / 6

= 5.72 KNm = 5.72

78.50 x 1000 /360

Horizontal steel :-

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

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


50.24 x 1000 /360

Vertical Steel ( c)

Provide 10 mm O bar



Distribution steel = 0.16 / 100 x 1000 x 450

On each face =

Provide 8 mm O bar

Total Ast1+ Ast2 = 150 + 490

Provide 12 mm O bar


113.04 x 1000 / 640

From Table 9-5

Ast1for moment =

8.16 x 106

/ 150 x 0.872 x 417

73.5 x 103

/ 150

Bottom 1 m will be designed as cantilever

,tension on liquid face.

Cantilever moment : -

Ywx H x h2/ 6

9.8 x 3.5 x 1 / 6


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M / stx j x d

=

= 105 mm2

= 720 mm2

360 mm2

=

= 218 mm

Provide 10 mm O bar @ 200 mm C/Cmarked(c) = 392 mm2.

0.229%

=

= 344 mm2

=

= 292 mm

Ast = 346 mm2

lx= 4 + 0.15 = 4.15 say 4.5 m

ly= 6 + 0.15 = 6.15 say 6.5 m

3.75 KN / m2

1.0 KN / m2

1.5 KN / m2

6.25 KN / m2

PU=

= 9.38 KN / m

ly/ lx= 6.5 / 4.5

= 1.4


From Table 9-5

Astfor moment =

5.72 x 106

/ 150 x 0.872 x 417

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

78.50 x 1000 /360

each face

Distribution steel = 0.16 / 100 x 1000 x 450

On each face =

Provide 10 mm O bar

Base slab :-

Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.

From table 9-3

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

Minimum steel =

0.229 / 100 x 1000 x 150

Provide 8 mm O bar


50.24 x 1000 /172

Dead Load : self 0.15 x 25 =

floor finish =

1.5 x 6.25

,172 mm2bothway

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

Designed section,Elevation etc. are shown in fig.

Live load =

consider 1 m wide strip. Assume 150 mm thick slab.

For 1 m wide strip

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

Top slab : -


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Table 26 x = 0.085

y = 0.056

Mx= My=

= =

= 16.15 KNm = 10.64 KNm

M / Q x b

= 16.15 x 106/ 2.76 x 1000

= 76.50 mm

dshort=

= 130 > 76.50 mm

dlong= 130 - 10 = 120 mm

= 0.96

Pt=

fy/ fck

=

415 / 20

=

= 0.29%

= 377 mm2

=

= 208 mm

Provide 10 mm O bar @ 210 mm c/c = 374 mm2.

= 0.74

Pt=fy/ fck

=

415 / 20

(O.K.)

Larger depth is provided due to deflection check.

y x w x lx2

0.085 x 9.38 x 4.52

0.056 x 9.38 x 4.52

x x w x lx2

From Table 6-3 ,Q = 2.76

Mu/ b x d2(short) = 16.15 x 10

6/ 1000 x 130 x 130

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

Mu/ b x d2(long) = 10.64 x 10

6/ 1000 x 120 x 120

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

)

50 1-1-(4.6 / 20) x (0.74)

50 1-1-(4.6 / 20) x (0.96)

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

78.50 x 1000 /377

Ast(short) = 0.29 x 1000 x 130 / 100

Provide 10 mm O bar


drequired=

150 - 15(cover) - 5


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=

= 0.22%

= 264 mm2

=

= 190 mm

= 264 mm2.

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

Provide 8 mm O bar @ 190 mm c/c

Provide 8 mm O bar


50.24 x 1000 /264

Ast(long) = 0.22 x 1000 x 120 / 100


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B

4 m

C


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Q = M / bD2

Pt Q = M / bD2

Pt

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 0.258

0.9 0.314 0.335 0.306 0.246

Balanced Design Factors for members in bending

For M20 Grade Concrete Mix

Mild steel HYSD barsd / D

TABLE 9-5

TABLE 9-6

Members in bending ( Cracked condition )

Coefficients for balanced design


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Grade of

concrete

Grade of

steelcbc

N / mm2

st

N / mm2 k j Q

M20 Fe250 7 115 0.445 0.851 1.33

Fe415 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

D / 2

e = M / T

0.214

0.24

0.2290.217

0.206

0.194

0.183

0.171

300

350

400

0.3

0.2860.271

0.257

0.243

0.229

Thickness, mm

100

150200

250

TABLE 9-3

Minimum Reinforcement for Liquid Retaining Structures

% of reinforcement

Mild Steel HYSD bars

ide the section

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


12/242

0.2 0.16450 or more


13/242


14/242

3500

150

1 : 4 : 8 P.C.C.

150 450 6000 450 150

Elevation

1500

450

1000

4000

1000

450

1500 1500

6000

450 450

Section A-A

Table 6-3

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

1000

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 )


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250 415 500 550

15 2.22 2.07 2.00 1.94

20 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 Qlim, N / mm2

fy, N / mm2

fck N / mm2


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17/242


18/242


19/242

Pt,bal

1.36

0.98

1.2

0.61


20/242

(given)

Material M20 (given)

Fe 415 (given)

86400 Litres

L / B = 8 / 3.6 = 2.22 > 2 .

H / 4 = 3.0 / 4 = 0.75 m

=

=

= 44.1 KNm.

=

=

= 26.13 KNm.

==

= 19.60 KNm.

For bottom portion

M = OR

= =

= 4.90 KNm = 4.90 KNm

=

= 35.28 KN

=

= 19.6 KN

Volume = 3.6 x 8 x 3.0 x 103

=

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.

The long walls are designed as vertical cantilevers from the base.

bottom 1 m or H / 4 whichever is more is designed as cantilever.

bottom h = 1 m will be designed as cantilever.

Design of Rectangular water tank CASE-2 ( L / B 2 )

Grade Concrete

Solution :-

Size of tank : 3.6 m x 8.0 m x 3.0 m high

Grade HYSD reinforcement

Size of tank : 3.6 m x 8.0 m x 3.0 m high

Maximum B.M. in long walls at the base

(1 / 6 ) x Ywx H3

( 1 / 6 ) x 9.8 x 33

Moments and tensions :

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

Ywx ( H - h ) x B2/ 12

9.8 x ( 3 - 1 ) x 42/ 12

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

Direct tension in long wall = Yw x( H - h ) x B / 2

Direct tension in short wall= Yw( H - h ) x 1

Ywx ( H - h ) x B2

/ 169.8 x ( 3 - 1 ) x 4

2/ 16

Ywx H x h2/ 6

9.8 x 3.0 x 1 / 6

Ywx H / 6 , whichever is greater

9.8 x 3.0 / 6

9.8 x ( 3 - 1 ) x 1

9.8 x ( 3 - 1 ) x 3.6 / 2

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


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M ( - ) = 44.1 KNm

T = 35.28 KN

D = M / Q x b

=

= 379.6 mm,

Take D = 400 mm d =

= 367 mm

Ast =

=

= 918.68 mm2

=

= 218.75 mm

= 1005 mm2.

As=

= 684 mm2.

342 mm2.

= T / st

=

= 235 mm2.

== 146.9 mm

= 357 mm2


44.1 x 10 6/ 0.306 x 1000

400 - 25 - 8

M / stx j x d

Design of long walls : -

( water face )

( perpendicular to moment steel )

Assume d / D = 0.9 Q = 0.306

200.96 x 1000 / 918.68

( 0.171 / 100 ) x 1000 x 400

From Table 9-6

From Table 9-5 ,

Provide 16 mm O bar

spacing of bar =

Astfor Moment

Provide 8 mm O bar

Note : The design is made at the base. The moment reduces from base to top.For ec

reinforcement can be curtailed or the thickness of wall can also be reduced as we hav

cantilever retaining wall.

Distribution steel = 0.171 % for 400 mm depth

From Table 9-3

Steel required for direct tension

35.28 x 103/ 150

( 2 )

44.1 x 106

/ 150 x 0.872 x 367

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

From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.


50.24 x 1000 / 342

on each face = ( 1 )

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

on each face

Design of short walls :-


22/242

M = 26.13 KNm

T = 19.6 KN

=

= 544 mm2

T / st

=

= 131 mm2

= 675 mm2

=

= 167.47 mm= 706 mm

2.

1000

203.56

400 367

163.44

= 13.33

=

= 203.56 mm

D - x = 196.44 mm

d - x = 163.44 mmAT=

=

= 408705 mm2

Ixx=

Provide 12 mm O bar@160 mm c/c

checking :

At support

From Table 9-5

Ast1for moment =

113.04 x 1000 / 675

19.6 x 103

/ 150

M / stx j x d

26.13 x 106

/ 150 x 0.872 x 367

Ast2for direct tension =

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

Total Ast1+ Ast2 = 544 + 131

Provide 12 mm O bar

spacing of bar =

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

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

b x D + ( m - 1 ) x Ast

1000 x 400 + (13.33 - 1 ) x 706

modular ratio m = 280 / 3 x cbc

x =b x D

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

b x D + ( m - 1 ) x Ast

( 1 / 3 ) x b x ( x3+ ( D - x )

3) + ( m - 1 ) x Astx ( d - x )

2


23/242

=

= 5.34E+09 + 2.33E+08

= 5.57E+09 mm4

fct = T / AT

=

= 0.048 N / mm2

fcbt =

=

= 0.767 N / mm2

check :

1

0.4912 1

M = 19.6 KNm

T = 19.6 KN

=

= 408 mm2

T / st

=

= 131 mm2

= 539 mm2

=

= 209.72 mm

= 565 mm2.

As=

= 684 mm2.

342 mm2.

26.13 x 106x 163.44 / 5.57 x 10

9

( fct/ ct ) + ( fcbt/ cbt )( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) 1

( 1 / 3 ) x 1000 x ( 203.563+ 196.44

3) + ( 13.33 - 1 ) x 706 x 163.44

2

19.6 x 10 3/ 408705

M x ( d - x ) / Ixx

From Table 9-2

At centre :

From Table 9-5

Ast1for moment = M / stx j x d

0.04 + 0.4512 1

.. ( O. K. )

Provide 12 mm O bar


113.04 x 1000 / 539

19.6 x 106

/ 150 x 0.872 x 367

Ast2

for direct tension =

19.6 x 103

/ 150

Total Ast1+ Ast2 = 408 + 131

on each face = ( 1 )


From Table 9-3

Distribution steel = 0.171 % for 400 mm depth

( 0.171 / 100 ) x 1000 x 400


24/242

= T / st

=

= 131 mm2.

=

= 146.9 mm

= 357 mm2

M = 4.9 KNm

Ast =

=

= 102 mm2

= 357 mm2

0.229%

=

= 344 mm

2

=

= 292 mm

Ast = 346 mm2

lx= 3.6 + 0.4 = 4 say 4 mly= 8 + 0.15 = 8.15 say 8.5 m

3.75 KN / m2

1.0 KN / m2

1.5 KN / m2

6.25 KN / m2

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

Top slab : -

consider 1 m wide strip. Assume 150 mm thick slab.

Dead Load : self 0.15 x 25 =

floor finish =

Live load =

Provide 8 mm O bar


50.24 x 1000 /172

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

Designed section,Elevation etc. are shown in fig.

Minimum steel =

0.229 / 100 x 1000 x 150

,172 mm

2

bothway

Base slab :-

Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.

From table 9-3

Steel required for direct tension

19.6 x 103/ 150

50.24 x 1000 / 342

on each face

Bottom cantilever

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

( 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 faces

on each face

From Table 9-5

M / stx j x d

4.9 x 106

/ 150 x 0.872 x 367

Minimum steel = 342 mm2on each face.


25/242

PU=

= 9.38 KN / m

= 18.76 KNm

= 16.88 KN

drequired= M / Q x b

= 18.76 x 10 6/ 2.76 x 1000= 82.44 mm

dprovided=

=

= 1.13

Pt=

fy/ fck

=

415 / 20

== 0.34%

= 439 mm2

=

= 114 mm

= 457 mm2.

= 180 mm2

=


Provide 6 mm O bar


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

28.26 x 1000 /180

Maximum moment = 9.38 x 42/ 8

Maximum shear = 9.38 x 3.6 / 2

From Table 6-3 ,Q = 2.76

Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement

Provide 8 mm O bar


50.24 x 1000 /439

50 1-1-(4.6 / 20) x (1.13)

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

Ast = 0.34 x 1000 x 129 / 100

(O.K.)


Mu/ b x d2 = 18.76 x 10

6/ 1000 x 129 x 129

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

129 > 82.44

Design for flexure :

150 - 15(cover) - 6

1.5 x 6.25

For 1 m wide strip


26/242

= 157 mm

= 188 mm2.Provide 6 mm O bar @ 150 mm c/c


27/242


28/242


29/242


30/242


31/242

nomy, the

e done for


32/242

x


33/242

M15 1.1 1.5 1.5

M20 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 2.5

M40 1.7 2.4 2.7

Table 9-2

Grade of

concrete

Permissible stresses in N / mm2

Direct

tension ct

Tension due to

bending cbtShear stress

v = V / b j d

Permissible concrete stresses in calculations relating

to resistance to cracking


34/242

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

3000

Base details not

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

900

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

B

B

8 O @ 140 c/c

8 O @ 140 c/c

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

16 O @ 200 c/c ( chipiya )

8 O @ 140 c/c


35/242

150

150

150 400 3600 400 150

Section B- B

shown for clarity900 900


36/242


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( given )

2.5 KN / m2 ( given )

1 KN / m2 ( given )

( given )

( given )

= 25.2

= 158.7 mm

D =

= 178.7 mm180 mm

4.5 KN / m2

1.0 KN / m2

2.5 KN / m2

Total 8.0 KN / m2

1.5 x 8 = 12 KN / m

w x l2/ 8

= 12 x 42/ 8

= 24 KNm

w x l / 2

=

= 24 KN

d =

= 160 mm

= 0.94

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

fy/ fck

=

415 / 15

=

= 0.289%

M15 grade concrete

HYSD reinforcement grade Fe415

Design of simply supported one way slab

effective span = 4 m supported on masonry wall of 230 mm thickness

Live load =

( span / d ) ratio permissible = 1.26 x 20

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

for simply supported, basic span / effective depth ratio = 20

solution : -

Assume 0.4 % steel , a trial depth by deflection criteria

modification factor = 1.26

Floor finish =

material

DL = 0.18 x 25 =

Floor finish =

Live load =

Factored load =

drequired= 4000 / 25.2

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

Assume an overall depth =

Mu/ b x d2= 24 x 10

6/ 1000 x (160)

2

50 1-1-(4.6 / 15) x (0.94)

Maximum moment =

Maximum shear =

12 x 4 / 2

Design for flexure : -

Consider 1 m length of slab

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

180 - 15 - 5


51/242

= 462 mm2

== 170 mm

= 462 mm2.

= 0.128

= 216 mm2

=

= 233 mm

= 218 mm2.

Vu= 24 KN

=

= 0.150 N / mm2

d = 160 mm

As= 231 mm .

= 0.144

6 x =

= 0.8 x 15 / 6.89 x 0.144

= 12.1

6 x 12.1

= 0.277

N / mm2


78.50 x 1000 /462


Ast= 0.289 x 1000 x 160 / 100

Provide 10 mm O bar

> 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

Half the bars are bent at 0.1 l = 400 mm , and

remaining bars provide 231 mm2area

100 x As/ ( b x D ) = 100 x 231 / ( 1000 x 180 )

spacing of bar =

50.24 x 1000 /216


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

Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mm

Provide 8 mm O bar

100 x As/ b x d = 100 x 231 / 1000 x 160

For bars at support

Check for shear : -

Actual Shear stress = Vu/ b x d

24 x 103/ 1000 x 160

for Pt = 0.144 c= 0.28

IS 456-2000 Table 19 from table 7-1

< ( C)N / mm2 ( too small )

Design shear strength c= 0.85 0.8 x fck( 1 + 5 x - 1 )

0.8 x fck/ 6.89 Pt , but not less than 1.0

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


52/242

IS 456-2000 clause 40.2.1.1 -0.05

k = 1.24 ? -0.04

= 0.347 N / mm2 .( O.K.)

8 OAs= 231 mm

Mu1= OR

= { 1 - (415 x 231 / 15 x 1000 x 160 ) }

= 13.34 { 1- 0.0399 }

= 12.812 KNm

Vu = 24 KN

693.875 + 8 O 56 O48 O 693.88

O 14.46 mm .( O.K.)

20

Pt=

= 0.289IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement

= 28.4

= 25.00 < 28.4 .( O.K.)

= 3 x 160 = 480 mm170 mm

5 x effective depth of slab or

= 5 x 160

230 mm

Check for development length : -

Assuming L0=

for 180 mm slab depth

Design shear strength = 1.24 x 0.28

0.87 x 415 x 231 x 160

= 56 O ( from Table 7-6 )

Check for deflection : -

Basic ( span / d ) ratio =

100 x Ast/ b x d = 100 x 462 / 1000 x 160


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


Actual (span / d ) ratio = 4000 / 160

Main bars : maximum spacing permitted =

Distribution bars : maximum spacing permitted =

spacing provided =

25 difference

20 difference

(HYSD Fe415 steel ) For continuing bars

0.87 x fyx Astx d { 1 - ( fyx Ast/ fckx b x d ) }

The depth could be reduced

1.3 x ( 12.81 x 106/ 24 x 10

3) + 8 O 56 O

which gives

.( O.

= 800 mm

.( O.

spacing provided = < 300 mm

3 x effective depth of slab or 300 mm

or 300 m

Check for cracking : -

< 450 mm

IS 456-2000 , clause 26.3.3

Development length of bars Ld= O s/ 4 x bd1.3 x ( Mu1/ Vu ) + L0Ld


53/242

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


54/242

For mild steel minimum reinforcement 0.15 %


55/242

Pt=

=

= 0.14

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

we get , 0.49

Mu1=

= 12.54 KNm

160

180

450 mm whichever is small

0.49x 1000 x 1602

x 10-6

400

For checking development lmay be assumed as 8 O for

bars ( usually end anchorag

provided ) and 12 O for mild

U hook is provided usually

anchorage length is 16 O.

100 x As/ b x d

100 x 231 / 1000 x 160

From equation

Mu1/ b x d2=

4000

400

.)

or 450 mm i.e. 450 mm

.)

whichever is small

i.e. 300 mm

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


56/242

( 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

centre to centre distance shall be used.

where it forms the end of a continuous beamwhere the length to the centre of support shall

be taken.

( d )Frames-In the analysis of a continuous frame,

of the beam or slab or the clear span plus

half the width of the discontinuous support,

whichever is less;

3) In the case of spans with roller or rocket

bearings, the effective span shall always be

the distance between the centres of bearings.

other continuous or for intermediate spans,

the effective span shall be the clear span

between supports;

2) For end span with one end free and the other

continuous, the effective span shall be equal

to the clear span plus half the effective depth

shall be taken as under:

( b )Continuous Beam or Slab - In the case of

continuous 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

1) For end span with one end fixed and the

Effective Span

IS 456-2000 clause 22.2

( a ) Simply Supported Beam or Slab -

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.

supports are wider than I/12 of the clear span

or 600 mm whichever is less, the effective span

If ly/ lx 2 ,called one way slab provided thatit is supported on all four edges . Note that , if

all four edge is not supported and ly/ lx< 2 ,

then also it is one-way slab,If ly/ lx< 2 , called

two-way slab.provided that it is supported on

all four edges.


57/242


58/242

ngth , l0HYSD

is not

steel (

hose


59/242

material ( given )

( given )

( Balcony slab )

DL LL

For S2 3 0 KN / m2

1 0 KN / m2

0 2 KN / m2

Total 4 2 KN / m2

Pu= ( 6 + 3 ) KN / m2

DL LL

For S1 3 0 KN / m2

1 0 KN / m2

0 3 KN / m2

Total 4 3 KN / m2

Pu= (6 + 4.5) KN / m2

1.875 KN / m

Pu= 2.8 KN / m

9 KN/m 6 KN/m

A 3m B 1.2m C

9 KN/m 2.8 KN

A 3m B 1.2m C

considering fig (a)

wx l2/ 2

= 6 x 1.22/ 2

mild steel grade Fe250

1.5 ( 4 + 3 ) =

Weight of parapet 0.075 x 25 x 1 =

1.5 x 1.875 =

Consider 1 m long strip(1) To get maximum positive moment in slab S2only dead load on slab

S1and total load on slab S2shall be considered

(b) Loads for maximum negative

moment,maximum shear for cantilever span

and maximum reaction at support B

cantilever moment =

Solution : -

For slab S2live load = 2 KN /m2

For slab S1live load = 3 KN /m2

Assume 120 mm thick slab

self load = 0.12 x 25 =

floor finish =

live load =

1.5 ( 4 + 2 ) =

self load = 0.12 x 25 =

Live Load As per IS 875

ly=

(a) Loads for maximum positive moment

Design of Cantilever one way slab

floor finish =

live load =

10.5 KN/m

M15 grade concrete

used for residential purpose

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

S S2

S2 S1

S1


60/242

= 4.32 KNm

shear =

=

= 12.06 KN

= 1.34 m

=

= 8.08 KNm

=

= 10.92 KNm

Vu,BA=

== 17.14 KN

Vu,BC=

=

= 15.4 KN

10.92 KNm

Q = 2.22

=

= 70.14 mm,

= 99 mm,

= 0.82


=

250 / 15

=

= 0.405%

= 401 mm2

Moment steel :

50 1-1-(4.6 / 15) x (0.82)

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

10.5 x 1.2 + 2.8

Maximum moment =

Mu/ b x d2( + ) =

120 - 15 - 6 ( assume 12 O bar )

Ast( + ) = 0.405 x 1000 x 99 / 100

.( O.K.)8.08 x 10

6/ 1000 x (99)

2

From Table 6-3

dprovided=

10.92 x 10 6/ 2.22 x 1000

Maximum shear at B

9 x 3 / 2 + 10.92 / 3

w x l / 2 + Moment @ B at distance 3 m

w x l + 2.8

the slab is loaded with full loads as shown in fig (b)

Maximum positive moment = 12.06 x 1.34 - W x l2/ 2

12.06 x 1.34 - 9 x 1.342/ 2

M / Q x b

9 x 3 / 2 - 4.32 / 3

w x l / 2 - Moment @ B at distance 3 mReaction at A =

2.8 x 1.2 + 10.5 x 1.22/ 2

Maximum negative moment = w x l + w x l2/ 2

(2) To get maximum negative moment and maximum shear at B,

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

drequired=


61/242

=

= 195.76 mm

= 462 mm2.

= 1.11


=

250 / 15

=

= 0.564%

= 558 mm2

= 231 mm2. ( bent bars extended )

=

= 231 mm2

remaining area = 558 - 231

= 327 mm2

=

= 346 mm

= 332 mm2. ( Extra )

= 180 mm2.

=

= 157 mm

= 188 mm2.

For negative moment reinforcement

Total 231 + 332 mm2= 563 mm

2steel provided

( from Table 7-6 )

28.26 x 1000 / 180


Development length of bars Ld= O s/ 4 x bd

Distribution steel = ( 0.15 /100 ) x 1000 x 120

Provide 6 mm O bar


For mild steel minimum reinforcement 0.15 %

Provide 12 mm O bar



Area provided = Area of one bar x 1000 / spacing of bar in m

Provide 10 mm O bar

Mu/ b x d2( - ) = 10.92 x 106

/ 1000 x (99)2

78.5 x 1000 / 401

113.04 x 1000 / 327

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

( alternate bent up )

78.5 x 1000 / 340


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

Ast( - ) = 0.564 x 1000 x 99 / 100


Provide 10 mm O bar


50 1-1-(4.6 / 15) x (1.11)


62/242

=

= 54.4 O

Ld= 54.4 x ( 10 + 12 ) / 2

= 598 mm.

12 O (mild steel )

At A , Pt=

=

= 0.233

OR

Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=fy/ fck =

we get , 0.487 = 4.974

Mu1= = 4.78= 4.77 KNm

Vu = 12.06 KN

=54.4 O

514.18 + 12 O 54.4 O42.4 O 514.18

O 12.13 mm

At B , Pt=

== 0.233

OR


we get , 0.487 = 4.974

Mu1= = 4.78

= 4.77 KNm

Vu = 13.09 KN

=54.4 O

473.72 + 12 O 54.4 O42.4 O 473.72

O 11.2 mmwhich gives

Mu1/ b x d2=

.( O.K.)

0.487 x 1000 x 992

x 10-6

Development length of bars Ld= O s/ 4 x bd = O x 0.87 x 250 / 4 x 1

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

1.3 x ( Mu1/ Vu ) + L0Ld

1.3 x ( 4.77 x 106/ 13.09 x 10

3) + 12 O 54.4 O

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

100 x 231 / 1000 x 99

From equation

0.87 x fy

0.87 x 25

which gives .( O.K.)

Half bars bent = 462 / 2 = 231 mm2)

0.87 x fy

0.87 x 25


1.3 x ( Mu1/ Vu ) + L0Ld

100 x 231 / 1000 x 99

From equation

Mu1/ b x d2=

0.487 x 1000 x 992

x 10-6

1.3 x ( 4.77 x 106/ 12.06 x 10

3) + 12 O 54.4 O


Assuming L0=

100 x As/ b x d

O x 0.87 x 250 / 4 x 1

say 600 mmAs a thumb rule, a bar shall be given an anchorage equal to the length of

the cantilever.

17.14 - ( 0.15 x 3 ) x 9 =


63/242

13.09 KN

=

= 0.132 N / mm < c

Pt=

= 0.233

6 x =

=

= 7.47

6 x 7.47

= 0.34

N / mm2

IS 456-2000 clause 40.2.1.1 0.07

k = 1.3 ? 0.014

= 0.442 N / mm2

> v

Vu=

=

= 0.173 N / mm2

< c

Pt=

= 0.569

6 x =

=

= 3.06

6 x 3.06

= 0.487

N / mm2

Span AB :

Check for shear : -

Use Vu= 13.09 KN

Shear stress v= Vu/ b x d

At A , Vu,AB= 12.06 KN ( for maximum loading )

At B , shear at point of contraflexure =




13.09 x 103/ 1000 x 99

100 x As/ b x d = 100 x 231 / 1000 x 99


.( O.K.)

0.8 x 15 / 6.89 x 0.233

.( O.K.)

for Pt = 0.233 c= 0.34

0.1 difference

for 120 mm slab depth 0.02 difference

Span BC :

17.14 KNShear stress v= Vu/ b x d




0.8 x 15 / 6.89 x 0.569

17.14 x 103/ 1000 x 99

.( O.K.)

100 x As/ b x d = 100 x 563 / 1000 x 99



for Pt = 0.569 c= 0.48


64/242

IS 456-2000 clause 40.2.1.1 0.08

k = 1.3 ? 0.0579

= 0.624 N / mm2

> v

20

Pt=

= 0.467


= 40

= 30.30 < 40 .( O.K.)

7

Pt=

= 0.569


= 12.6

= 12.12 < 12.6 .( O.K.)

= 3 x 99 = 297 mm170 mm


= 5 x 99

150 mm

0.25 difference



modification factor = 2

( span / d ) ratio permissible = 2 x 20


.( O.K.)Check for deflection : -


100 x Ast/ b x d = 100 x 462 / 1000 x 99

For span AB :

(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm



For span BC :


100 x Ast/ b x d = 100 x 563 / 1000 x 99


spacing provided = < 450 mm .( O.K

or 300 mspacing provided = < 297 mm .( O.K

(2 )Distribution bars : maximum spacing permitted == 495 mm


IS 456-2000 , clause 26.3.3


65/242

1.2m

0.15m

0.15m

0.15m 0.15m

1.2mlx= 3m

6m S2 S1

S3

B1B2

B3

B4

1m high parapet

Column 300 x 300


66/242


67/242


68/242

KNm

KNm

( 1 - 0.0389 )

Astx d ( 1 - fyx Ast/ b x d x fck)

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


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

( 1 - 0.0389 )


69/242


70/242

1200 1200

150 3000 150 1200


hichever is small

.)

i.e. 297 mm .)

120 125

600300

or 450 mm i.e. 450 mm

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

10 O @ 170 c/c

6 O @ 150 c/c

6 O @ 150 c/c


71/242


72/242


73/242


74/242


75/242


76/242


77/242


78/242

= 250 mm2

=

= 200.96 mm

= 251 mm2.

= 0.81


=

415 / 15

== 0.240%

= 312 mm2

=

= 161 mm

= 335 mm2.

For HYSD Fe415

= 180 mm2

=

= 126 mm2

=

= 279.11 mm

= 193 mm

2

.

= 188 mm2.

Provide 8 mm O bar@ 260 mm c/cMore 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

Provide 8 mm O bar

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

This is less than minimum , therefore , use minimum steel at location 4 and 5 .

Provide 8 mm O bar


50.24 x 1000 / 180


50.24 x 1000 / 312

Provide 8 mm O bar@ 150 mm c/c

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

50.24 x 1000 / 250


Mu/ b x d2( - ) = 13.66 x 10

6/ 1000 x (130)

2

( 1 / 2 ) x 251

50 1-1-(4.6 / 15) x (0.81)

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

Ast = 0.24 x 1000 x 130 / 100

Provide 8 mm O bar

Provide 8 mm O bar

Ast = 0.208 x 1000 x 120 / 100



79/242

=

= 267.23 mm

= 193 mm2.

= 94 mm2.

=

= 279.11 mm

= 193 mm2.

S.F. =

=

= 31.795 KN

= 0.258

N / mm2

0.25 differen 0.11

IS 456-2000 clause 40.2.1.1 0.242 differe ? -0.1065

k = 1.3

= 0.460 N / mm2

OR

6 x =

= 6.75

6 x 6.75

= 0.356

IS 456-2000 clause 40.2.1.1

k = 1.3

= 0.463 N / mm2

=

= 0.245 N / mm2

< c .( O.K.)


.( O.K.)Actual shear stress = Vu/ b x d

31.795 x 103/ ( 1000 x 130 )





from table 7-1

for Pt= 0.258 , c= 0.354



.( O.K.)

Note that positive reinforcements are not curtailed because if they are curtailed , the

remaining bars do not provide minimum steel.

Check for shear : -

At point 1 or 3w x l / 2 + Moment @ point 1 or 3 in that span

11.625 x 5 / 2 + 13.66 / 5

100 x As/ b x d = 100 x 335 / ( 1000 x 130 )

Provide 8 mm O bar


50.24 x 1000 / 180


This will be provided by minimum steel .


50.24 x 1000 / 188


This will be provided by minimum steel of edge strip,

At corner B , steel required = ( 1/2 ) x 188


80/242

S.F. = w x l / 2

=

= 29.06 KN

= 0.209

N / mm2

0.1 differenc 0.07

IS 456-2000 clause 40.2.1.1 0.041 differe ? -0.0287

k = 1.3

= 0.417 N / mm2

OR

6 x =

= 8.33

6 x 8.33

= 0.326

IS 456-2000 clause 40.2.1.1

k = 1.3

= 0.424 N / mm2

=

= 0.242 N / mm < c

Vu= 29.06 KN

8 O (HYSD Fe415 steel )

Pt=

=

= 0.209

OR

Pt= 50 1-1-(4.6 / fck) x (Mu/ b x d2) Mu1=

fy/ fck =

we get , 0.711 = 10.875

Mu1= = 10.246

= 10.24 KNm

=54.3 O

This is critical at point 4 or 5.

No bar is curtailed or bent up.

Mu1/ b x d2=

0.711 x 1000 x 1202

x 10-6

Development length of bars Ld= O s/ 4 x bd (From Table 7-6 )

Assuming L0=

0.87 x fy

Actual shear stress = Vu/ b x d

29.06 x 103/ ( 1000 x 120 )

.( O.K.)

0.87 x 41

0.85 0.8 x 15 ( 1 + 5 x 8.33 - 1 )



100 x As/ b x d

100 x 251 / 1000 x 120

From equation


At point 4 or 5 ,

1.3 x 0.321



Design shear strength c=

At point 4 or 5

11.625 x 5 / 2

100 x As/ b x d = 100 x 251 / ( 1000 x 120 )

from table 7-1

for Pt= 0.209 , c= 0.321


Design shear strength =


81/242

458.36 + 8 O 54.3 O46.3 O 458.36

O 9.90 mm

Vu= = 25.31 KN


Pt=

=

= 0.289

OR


we get , 0.9595 = 26.689

Mu1

= = 24.554

= 24.56 KNm

=56 O

1261.5 + 8 O 56 O48 O 1261.5

O 26.28 mm

26

251 mm2.

= 123 mm.

Pt= 240.7

= 0.204


= 42.9

= 40.65 < 42.9

= 3 x 130 = 390 mm

200 mm

= 5 x 120 = 600 mm

< 300 mm .( O.(2) Distribu. bars : maximum spacing permitted = 5 x effective depth of slab or 4

.( O.K.)

IS 456-2000 , clause 26.3.3Check for cracking : -

3 x effective depth of slab or 3(1) Main bars : maximum spacing permitted =

spacing provided =

1.3 x ( Mu1/ Vu ) + L0Ld

1.3 x ( 24.56 x 106/ 25.31 x 10

3) + 8 O 56 O

which gives

Note that the bond is usually critical along long direction.


.( O.K.)

0.9595 x 1000 x 1602

x 10-6

Development length of bars Ld= (From Table 7-6 )

.( O.K.)

Short span 11.25 x ( 4.5 / 2 )

0.87 x fy

0.87 x 41

1.3 x ( Mu1/ Vu ) + L0Ld

1.3 x ( 10.246 x 106/ 29.06 x 10

3) + 8 O 54.3 O

which gives

O s/ 4 x bd



Mu1/ b x d2=

Assuming L0=

100 x As/ b x d

100 x 462 / 1000 x 160

From equation

Check for deflection : -

100 x Ast/ b x d = 100 x 251 / 1000 x 123

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



82/242

260 mm

This is 180 mm2

=

= 279.11 mm

= 193 mm2.

625 3750 625

150

625 3750 625

Section A-A

50.24 x 1000 / 180

Provide 8 mm O bar@ 260 mm c/c for uniformity in spacing.

For clarity , top and bottom reinforcements are shown separately.

3750

625

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.

minimum (0.12 / 100) x 150 x 1000 =

Provide 8 mm O barspacing of bar = Area of one bar x 1000 / required area in m

2/ m

spacing provided = < 450 mm .( O.

625

MiddleStrip

S1

Edgestrip

Edgestrip

Edge strip Edge stripMiddle Strip

8O@2

60c/c

8O@2

60c/c

8O@200c/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


83/242

5 m

625 3750 625 5 m

ly/8 (3/4)ly ly/8

5 m

5 m

Middle Strip

S1

S1

Edgestrip

Edgestrip

A B

B

1

2

3

4

50.0

35

-0.0

47

-

0.035


84/242


85/242


86/242

KNm

( 1 - 0.0579 )


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


87/242

KNm

or 300 mm i.e. 300 mm

or 450 mm i.e. 450 mm

.) 50 mm whichever is small


( 1 - 0.0799 )


5 x 462 x 160 ( 1 - 415 x 46


88/242

1000

625 3750 625

150

625 3750 625

1000

.)

625

3750

625

MiddleStrip

S1

Edgestrip

Edgestrip

Edge strip Edge stripMiddle Strip

8O@2

40c/c

8O@2

40c/c

8O@140c/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

1500

1500

1500 1500

8 O @ 180 c/c


89/242

0.047


90/242

given

Material

DL LL

3 0 KN / m2

1 0 KN / m2

0 3 KN / m2

Total 4 3 KN / m2

=

3m 3m 3m 3m 3m

=

= 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

=

= 18.9 KN

KNm

Q = 2.07

0.6 x 6 x 3 + 0.6 x 4.5 x 3

Maximum moment is Mu3( - ) = 9.9

From Table 6-3

( 1 / 10 ) x 6 x 32+ ( 1 / 9 ) x 4.5 x 3

2

Mu4( - ) = ( 1 / 12 ) x w ( DL ) x l2+ ( 1 / 9 ) x w ( LL ) x l

2

( 1 / 12 ) x 6 x 32+ ( 1 / 9 ) x 4.5 x 3

2

Maximum shear is Vu(BA)= 0.6 x w x l + 0.6 x w x l

Mu3( - ) = ( 1 / 10 ) x w ( DL ) x l2+ ( 1 / 9 ) x w ( LL ) x l

2

( 1 / 12 ) x 6 x 32+ ( 1 / 10 ) x 4.5 x 3

2

Mu2( + ) = ( 1 / 16 ) x w ( DL ) x l2+ ( 1 / 12 ) x w ( LL ) x l

2

( 1 / 16 ) x 6 x 32+ ( 1 / 12 ) x 4.5 x 3

2

Design of Continuous One-way slab

A five span continuous one-way slab used as an office floor.

The centre-to-centre distance of supporting beams is 3 m

Live load 3 KN / m2and floor finish 1 KN / m

2

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

Mu1( + ) = ( 1 / 12 ) x w ( DL ) x l2+ ( 1 / 10 ) x w ( LL ) x l

2

Dead load 0.12 x 25 =

floor finish =

live load =

factored load =

HYSD reinforcement of grade Fe415

M15 grade concrete

Solution : -

Try 120 mm thick slab

1.5 ( 4 + 3 )

( 6 + 4.5 ) KN / m2

Consider 1 m wide strip of the slab.

A B C D E F

1 2 2 2 1

( 6 + 4.5 ) KN / m


91/242

=

= 69.2 mm,

= 90 mm,


= 132 mm2

= 165 mm2

=

= 171

=

= 344

306

Table

8 mm O @ 220 c/c

= 228 mm2

10 mm O @ 220 c/c

= 357 mm2

10 mm O @ 220 c/c

= 357 mm24 ( - ) 9.0 1.11 0.34

223

3 ( - ) 9.9 1.22 0.38 342

2 ( + ) 6.75

10 mm O @ 440 c/c + 8 mm O @ 440 c/c

= 178 +114 = 292 mm2(Half 10 O+half 8 O)290

0.83 0.248

1 ( + ) 8.55 1.06 0.322

Factored

moment

KNmpoint Mu/(b x d2) Pt

Ast

mm2

Steel Provided

dprovided= 110 - 15 - 5 ( assume 10 O bar ).( O.K.)

Try 110 mm overall depth


Ast = Ptx b x d / 100

drequired= M / Q x b9.9 x 10 6/ 2.07 x 1000

For Main steel ,HYSD Fe415 reinforcement

minimum steel area = ( 0.12 / 100 ) x 1000 x 110

For Distribution steel , mild steel Fe250 reinforcement

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

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

Use 6 mm O


28.26 x 1000 /165

Check for shear : -


50.24 x 1000 /146

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

Maximum shear = 18.9 KN


92/242

=

= 0.210 N / mm2

d = 90 mm

As= 357 mm2.

= 0.397

6 x =

=

= 4.4

6 x 4.4

= 0.42

N / mm2

IS 456-2000 clause 40.2.1.1 0.11

k = 1.3 ? 0.0453

= 0.546 N / mm2

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

Vu=

= 14.18 KN

=

= 0.158 N / mm2

d = 90 mm

As= 292 mm2.

= 0.324

6 x =

=

= 5.4


18.9 x 103/ 1000 x 90




Design shear strength c=0.85 0.8 x 15 ( 1 + 5 x 4.4 - 1 )

0.8 x 15 / 6.89 x 0.397


for Pt= 0.397 c= 0.42

< ( C)N / mm2

0.25 difference

( too small )

For bars at support

100 x As/ b x d = 100 x 357 / 1000 x 90

.( O.K.)

18.9 - 0.15 x 3 x 10.5



100 x As/ b x d = 100 x 292 / 1000 x 90


14.18 x 103/ 1000 x 90

< ( C)N / mm2 ( too small )

For bars at support

with positive moment reinforcement ( 29


0.8 x 15 / 6.89 x 0.324


93/242

6 x 5.4

OR = 0.39

N / mm2

IS 456-2000 clause 40.2.1.1 0.11k = 1.3 ? 0.0774

= 0.494 N / mm2

Vu=

= 13.28 KN

At A , Pt=

=

= 0.324

OR


we get , 1.064 = 9.4884

Mu1= = 8.64

= 8.62 KNm


=56.4 O

845.783 + 8 O 56.4 O48.4 O 845.78

O 17.47 mmAt support B, point of contraflexure is assumed at 0.15 x l from B

Vu=

= 14.18 KN

Mu1= 8.64 KNm

L0= 12 O

792.102 + 12 O 56.4 O44.4 O 792.1

O 17.84 mm


for Pt= 0.324 c= 0.38

0.25 differencefor 110 mm slab depth 0.176 difference


0.4 x 6 x 3 + 0.45 x 4.5 x 3 =

100 x 292 / 1000 x 90

From equation

1.064 x 1000 x 902

x 10-6


.( O.K.)

0.87 x fy

0.87 x 41

Mu1/ b x d2=


Assuming L0=

100 x As/ b x d

Span AB is critical for checking this requirement

At support A

0.4 x w x l + 0.45 x w x l

as before

( actual anchorage is more than 12 O but L0is limited to 12 O or d

, i.e. 90 mm whichever is greater )

1.3 x ( Mu1/ Vu ) + L0Ld


1.3 x ( Mu1/ Vu ) + L0Ld1.3 x ( 8.64 x 10

6/ 13.28 x 10

3) + 8 O 56.4 O


1.3 x ( 8.64 x 106/ 14.18 x 10

3) + 12 O 56.4 O

18.9 - 0.15 x 3 x 10.5

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


94/242

26

Pt=

= 0.324


= 34.84

= 33.33 < 34.84 .( O.K.)

26

Pt=

= 0.253


= 41.6

= 33.33 < 41.6 .( O.K.)

= 3 x 90 = 270 mm

220 mm


= 5 x 90

160 mm

.( O.


modification factor = 1.34( span / d ) ratio permissible = 1.34 x 26

or 300 m

IS 456-2000 , clause 26.3.3

(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm

100 x Ast/ b x d = 100 x 292 / 1000 x 90

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


100 x Ast/ b x d = 100 x 228 / 1000 x 90




For span BC :




(2 )Distribution bars : maximum spacing permitted == 450 mm



95/242

Outer side Inner side

IS 456-2000 Clause -22.5

( 22.5.1 ) Unless more exact estimates are made, for

beams of uniform cross-section which support

substantially uniformly distributed loads over three or

more spans which do not differ by more than 15 percent

of the longest, the bending moments and shear forcesused in design may be obtained using the coefficients

Where coefficients given in Table 12 are used for

calculation of bending moments, redistribution referred

to in 22.7 shall not be permitted.

(22.5.2 ) Beams and Slabs Over Free End Supports

given in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans

the average of the two values for the negative moment

at the support may be taken for design.

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

and I is the effective span, or such other restraining

moment 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.

Where a member is built into a masonry wall whichdevelops only partial restraint, the member shall be

designed to resist a negative moment at the face of the

support of Wl / 24 where W is the total design load

+ 1 / 12

+ 1 / 10

Table 12 Bending Moment coefficients

Near middle

of end span

At middle of

interior span

Support m

At support next to

the end support

Type of load

Span moments

NOTE -For obtaining the bending moment, the coefficient shall be multiplie

design load and effective span.

Table 13 Shear Force coefficients

+ 1 / 16

+ 1 / 12

-1 / 10

-1 / 9

Dead load and imposed

load ( fixed )

imposed load (

not fixed )

Type of load

Dead load and imposed

load ( fixed )

imposed load (

not fixed )

At support next to the end supportAt end

support

0.45 0.6 0.6

0.4 0.6 0.55


96/242

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 )


97/242

2 mm2)


98/242

KNm


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

( 1 - 0.0898 )


99/242


.) i.e. 270 mm

hichever is small

B

.)or 450 mm i.e. 450 mm


100/242

oments

At other interior

supports

-1 / 12

-1 / 9

by the total

0.6

At all other

interior

supports

0.5


101/242

900

3000

10 O @ 220 c/c

( 0.3 l1 )


102/242


103/242


104/242


105/242


106/242


107/242


108/242


109/242

material

lx= say 4.5 m

ly = say 6.75 m

4.5 KN / m2

1.0 KN / m2

2.0 KN / m2

Total 7.5 KN / m

2

Pu= 1.5 x 7.5

= 11.25 KN / m

ly/ lx=

= 1.5

Mux = 23.7 KNm

Muy= 10.48 KNm

Q = 2.07

=

= 107 mm,

= 160 mm,

160 - 10

= 150 mm,

= 0.926

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

fy/ fck

=

Solution :

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

Design of Simply supported two way slab

residential building drawing room 4.3 m x 6.55 m

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

M15 grade concrete

6.75 / 4.5

4.3 + 0.18 = 4.48

6.55+0.18 = 6.73

Live load ( residence ) =

For 1 m wide strip

Dead load : self 0.18 x 25 =

floor finish =

HYSD reinforcement of grade Fe415

given

IS 456-2000 Table -27

dshortprovided= 180 - 15 - 5 ( assume 10 O bar )

.( O.K.)> 107 mm

dlongprovided=

> 107 mm

xx w x lx2

=

yx w x lx2

= 0.046 x 11.25 x 4.52 =

0.104 x 11.25 x 4.52

=

From Table 6-3

drequired= M / Q x b23.7 x 10 6/ 2.07 x 1000


Mu/ b x d2( short ) = 23.7 x 10

6/ 1000 x (160)

2

50 1-1-(4.6 / 15) x (0.926)


110/242

415 / 15

=

= 0.28%

= 448 mm2

=

= 175.22 mm

= 462 mm2.

= 0.466

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

fy/ fck

=

415 / 15

=

= 0.134%

= 201 mm2

= 216 mm2.

=

= 232.59 mm

= 218 mm2.

Vu= = 25.31 KN


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

Ast( short ) = 0.28 x 1000 x 160 / 100

Provide 10 mm O bar

Mu/ b x d2( long ) = 10.48 x 10

6/ 1000 x (150)

2

50 1-1-(4.6 / 15) x (0.466

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


78.5 x 1000 / 448

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

Ast( long ) = 0.134 x 1000 x 150 / 100

For HYSD Fe415 minimum reinforcement 0.12 %

Provide 8 mm O bar

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


Provide 8 mm O bar@ 230 mm c/c ( long span )

50.24 x 1000 / 216

The bars cannot be bent or curtailed because if 50 % of long span bars are curtailed ,the remaining bars will be less than minimum

At top on support , provide 50 % of bars of respective span to take into account

negative moment due to slab nature.


Assuming L0=

Long span 11.25 x ( 4.5 / 2 )


111/242

Pt=

=

= 0.145

OR

Pt= 50 1- 1-(4.6 / fck) x (M

u/ b x d

2) M

u1=

fy/ fck =

we get , 0.5023 = 11.806

Mu1= = 11.331

= 11.30 KNm

=56 O

580.4 + 8 O 56 O

48 O 580.403O 12.09 mm

Vu= = 25.31 KN


Pt=

=

= 0.289

OR

Pt= 50 1- 1-(4.6 / fck) x (Mu/ b x d2) Mu1=

fy/ fck =

we get , 0.9595 = 26.689

Mu1= = 24.554

= 24.56 KNm

=56 O

1261.5 + 8 O 56 O48 O 1261.48

O 26.28 mm

Vu= 25.31 KN

=

From equation

0.87 x fy

1.3 x ( Mu1/ Vu ) + L0Ld

0.9595 x 1000 x 1602x 10

-6

Check for shear : -

Shear stress = Vu/ b x d

Note that the bond is usually critical along long direction.

Development length of bars Ld= O s/ 4 x bd (From Table 7-6 )

Mu1/ b x d2=

0.87 x 41

Mu1/ b x d2=

0.5023 x 1000 x 1502x 10

-6

1.3 x ( 24.56 x 106/ 25.31 x 10

3) + 8 O 56 O


From equation

0.87 x fy

0.87 x 41

100 x 462 / 1000 x 160

Short span 11.25 x ( 4.5 / 2 )

(From Table 7-6 )

100 x As/ b x d

100 x 218 / 1000 x 150

Development length of bars Ld= O s/ 4 x bd1.3 x ( Mu1/ Vu ) + L0Ld

1.3 x ( 11.30 x 106/ 25.31 x 10

3) + 8 O 56 O


Assuming L0=

100 x As/ b x d

25.31 x 103/ 1000 x 150

This is critical along long span


112/242

= 0.169 N / mm2

= 0.145

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

k = 1.24

= 0.347 N / mm2

OR

6 x =

= 12.011

6 x 12.011

= 0.278

IS 456-2000 clause 40.2.1.1 25 difference -0.05

k = 1.24 20 difference ? -0.04

= 0.345 N / mm2

20

Pt=

= 0.289


= 28.8

= 28.00 < 28.8

= 3 x 160 = 480

170 mm

= 5 x 150 = 750

230 mm



< ( C)N / mm2 ( too small )100 x As/ b x d = 100 x 218 / 1000 x 150




.( O.K.)


from table 7-1

for Pt= 0.145 c= 0.28

IS 456-2000 , clause 26.3.3


100 x Ast/ b x d = 100 x 462 / 1000 x 160



.( O.K.)Check for deflection : -

This check shall be done along short span




.( O.K.)

.( O.

3 x effective depth


(1) Main bars : maximum spacing permitted for short span steel =

(2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth



113/242


114/242

1.0 1.1 1.2 1.3Interior Panels:

Negative moment at continuous edge 0.032 0.037 0.043 0.047

Positive moment at mid-span 0.024 0.028 0.032 0.036

One Short Edge Discontinuous:



One long Edge Discontinuous:


Positive moment at mid-span 0.028 0.033 0.039 0.044Two Adjacent Edges Discontinuous:



Two Short Edges Discontinuous:



Two Long Edges Discontinuous:

Negative moment at continuous edge - - - -


Three Edges Discontinuous

(One Long Edge Continuous):



Three Edges Discrmntinuous

(One Short Edge Continuous) :

Negative moment at continuous edge - - - -


Four-Edges Discontinuous:


ly/ lx 1.0 1.1 1.2 1.3 1.4

Table - 26 Bending moment coefficients for rectangular panels supported on four sid

short span c

( Values

3

4

5

Type of Panel and Moments

consideredCase No.

1

2

6

7

8

9

Table - 27 Bending moment coefficients for slabs spanning in two directions at right a


115/242

x 0.062 0.074 0.084 0.093 0.099

y 0.062 0.061 0.059 0.055 0.051


116/242

KNm

KNm

350 6550 350

350

4300 4650


( 1 - 0.0799 )

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

Astx d ( 1 - f

yx A

st/ b x d x f

ck)

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

( 1 - 0.0402 )

460

690 690

A

8 O @ 230 c/c

10 O @ 170 c/c

10 O @ 340 c/c


117/242


118/242


119/242

1.4 1.5 1.75 2.0

0.051 0.053 0.060 0.065 0.032

0.039 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 0.047

0.053 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

- - - - 0.045

0.063 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

- - - - 0.057

0.071 0.076 0.087 0.096 0.043

0.085 0.089 0.100 0.107 0.056

1.5 1.75 2.0 2.5 3.0

s with provision for torsion at corners

oefficient x

of ly/ lx)

Long span

coefficient yfor

all values of ly/ lx

ngles , simply supported on Four sides


120/242

0.104 0.113 0.118 0.122 0.124

0.046 0.037 0.029 0.020 0.014


121/242

Material

say

= == 648 = 648

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

6MA+ 24MB+ 6MC= - [ 1944 ] - [ 1944 ]

MA( 6 / I ) + 2MB( 6 / I + 6 / I ) + MC( 6 / I ) = - [ 6 x 648 x 3 / ( Ix 6 ) ] - [ 6 x 648 x 3 / ( Ix

4MB+ MC= - 648 As MA= 0 .( 1 )

Using three moment equation for span ABC

MA( L1/ I1) + 2MB( L1/ I1+ L2/ I2) + MC( L2/ I2) = - 6 A1a1/ ( I1L1) -6 A2a2/ ( I2L2)

A1= ( 2 / 3 ) x Base x h1

a1= 3 m

A2= ( 2 / 3 ) x Base x h1

a2= 3 m

Live load = 3 KN / m2

Floor finish = 1 KN / m2

Rib size = 230 mm x 450 mm

Main beams = 300 mm x 570 mm overall .

Design of Continuous Beam

An R.C.C. floor is used as a banking hall

Design the beams B10-B11-B12.

The Slab thickness is 120 mm .

Floor finish = 1 + 0 KN /m2

Total 4 + 3 KN / m2

Load on beam = 3 ( 4 + 3 ) = 12 + 9 KN / m

(Given )

M15 grade concrete

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

HYSD reinforcement of grade Fe415 .

Column = 300 mm x 300 mm

Solution : -

( a ) Load calculations and analysis :

Self wt. = 0.23 x 0.45 x 25 = 2.58 + 0 KN / m

Total 14.58 + 9 KN / m

Factored Load = 1.5 ( 14.58 + 9 )

= 21.87 + 13.5

Live load = 0 + 3 KN / m2

( 22 + 14 ) KN /m .

Case ( a ) Maximum moment at B

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


122/242

= =

= 648 = 396

6 m Distance 138 KNm3 m Distance ( ? ) 69

6 m Distance 96 KNm

3 m Distance ( ? ) 48

6 m Distance 42 KNm

3 m Distance ( ? ) 21

- 2088 - 16 MC+ MC= - 648

15 MC= - 1440

MC= - 96 KNm

4MB+ ( - 96 ) = - 648

4 MB= - 552MB= - 138 KNm

By putting Value of MCinto Equation ( 1 )

4MB+ MC = - 648MB+ 4MC = - 522

.( 1 ).( 2 )

By putting Value of MBfrom Equation ( 2 ) into Equation ( 1 )

4(- 522 - 4MC) + MC= - 648

MB+ 4MC= - 522 As MD= 0 .( 2 )

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

a2= 3 m a3= 3 m

MB( 6 / I ) + 2MC( 6 / I + 6 / I ) + MD( 6 / I ) = - [ 6 x 648 x 3 / ( Ix 6 ) ] - [ 6 x 396 x 3 / ( Ix

6MB+ 24MC+ 6MD= - [ 1944 ] - [ 1188 ]

Using three moment equation for span BCD

MB( L2/ I2) + 2MC( L2/ I2+ L3/ I3) + MD( L3/ I3) = - 6 A2a2/ ( I2L2) -6 A3a3/ ( I3L3)

A2= ( 2 / 3 ) x Base x h2 A3= ( 2 / 3 ) x Base x h3

36 KN / m

162

22 KN / m

99B

CC D

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

6 m 6 m 6 mA B C D138

96162

162 99( - )

( + )( - ) ( + )

96

( - ) ( - )

( + ) ( + ) ( + )

93 5145

138


123/242

X dist.

6 - X dist.

Span AB or CD

= 1650 mm > 3000 mm

BC ( + ) = 58 KNmBC ( minimum - ve ) = 5 KNm

Factored maximum moments :

B or C , negative moment = 138 KNm

AB ( + ) = CD ( + ) = 110 KNm

Three cases are considered for getting maximum values of moments. Case ( a ) gives ma

negative moment at B.The same moment shall be used at C also because of symmetry. C

gives the maximum positive moment in span BC while the case ( c ) gives maximum positi

moments in span AB and CD.

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

VC= 101 KN

50 x 12 + VCx 6 - 36 x 6 x 3 - 22 x 6 x 9 = -138

VB= 246 KN

VB+ 131 = 246

VB= 115 KN

VC= 132 - 50

VC= 82 KN MC= - 96 KNm

VDx 6 -22 x 6 x 3 = -96

VD= 50 KN

VC+ VD= 22 x 6

VB= 216 - 85VB= 131 KN

MB= - 138 KNm

Vc= 183 KN

VC+ 82 = 183

MB= - 138 KNm

VAx 6 -36 x 6 x 3 = -138

VA= 85 KN

MC= - 96 KNm

VA+ VB= 36 x 6 131 X =8

216 X =X = 2.36

BC ( - ) = 11.6 KNm

BC ( + ) = 51.6 KNm > 0.7 x 58 KNm.

The moment redistribution shall be now carried out. Maximum negative moment = 138 KN

reduce it by 20 % . Then Mu = 0.8 x 138 = 110.4 KNm. For all the cases , redistribute ( inc

decrease ) the moment at B or C = 110.4 KNm ( Hogging ).

Maximum design moments :

The beam acts as a flanged beamFor T-beams , bf= ( l0/ 6 ) + bw+ 6 Dfbf= ( 0.7 x 6000 / 6 ) + 230 + 6 x 120

( As per I

( As per IS456-2000 ,Clause 23.1.2 , Note

Note that after redistribution , the design positive moments also have been reduced.

( b ) Design for flexure :

Mu( + ) = 111.6 KNm.

Support B or C = 110.4 KNm > 0.7 x 138 KNm ( O.K.)AB or CD ( + ) = 111.6 KNm > 0.7 x 110 KNm

85

131

2.36 m50

82

115

101


124/242

For bf /bw = 7

0.01 diff. 0.014 0.016

0.006 diff. ? ?

-0.0084 -0.0096

= 662.6 For 0.224 0.6566 0.743

1 diff 0.0864

0.83 diff ? 0.0717

= 651 mm2.

Span BC :

= 301 mm .

fy/ fck

415 / 15

= 0.549

Ast= 603 135

Assuming one layer of 20 mm diameter bars

d = 450 + 120 -25 - 10 = 535 mm .

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

( As per IS456-2000 ,Cl

If Mu< Mu,lim : design as under-reinforced section (singly reinforced beam) as explaine

Ast= Mu/ 0.87 x fyx lever arm

KNm > 111.6 KNm

bf/ bw= 1650 / 230 = 7.17

Df/ d = 120 / 535 = 0.224

Mu,lim/ fckbwd2= 0.671 ( As per SP:16 ,Table 58 )

For bf /

Mu.lim= 0.671 x 15 x 230 x 5352x 10

-6

Mu( + ) = 51.6 KNm

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

Provide 3 - 12 mm O = 339 mm2.

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

where lever arm = d - D f/ 2 = 535 - 120 / 2 = 535 - 60

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

Provide 6 - 12 mm O = 678 mm2.

Continue 3 - 12 O in span BC as required for flexure.

Support B or C :Mu( - ) = 110.4 KNm

Mu/ bd2= 110.4 x 10

6/ 230 x 535

2= 1.68 < 2.07. ( From Table 6-3 )

The section is under-reinforced.

Pt= 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )

Provide 2- 10 mm O anchor bars = 157 mm2. At support , provide 3 - 16 mm O extra at to

one of which may be curtailed at 0.15 = 900 mm from centre of support B and remainingat 0.25 = 1500 mm from B .

Pt= 501 - 1 - ( 4.6 / 15 ) x ( 1.68)

( 0.549 / 100 ) x 230 x 535 = 676 mm

2

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


125/242


126/242


127/242


128/242

= == 396 = 648

6MA+ 24MB+ 6MC= - [ 1188 ] - [ 1944 ]

4MB+ MC= - 522 As MA= 0

A2= ( 2 / 3 ) x Bas

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

a1= 3 m a2= 3 m

MA( 6 / I ) + 2MB( 6 / I + 6 / I ) + MC( 6 / I ) = -6) ]

Using three moment equation for span ABC

MA( L1/ I1) + 2MB( L1/ I1+ L2/ I2) + MC( L2/ I

A1= ( 2 / 3 ) x Base x h1

Case ( b ) Maximum positive moment in span B

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 K

162

A B B


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= =

= 648 = 396

MB= - 104 KNm

MC= - 104 KNm

MB+ 4MC = - 522 .

By putting Value of MCinto Equation ( 1 )

4MB+ ( - 104 ) = - 522

4 MB= - 418

By putting Value of MBfrom Equation ( 2 ) into

4(- 522 - 4MC) + MC = - 522

- 2088 - 16 MC+ MC= - 522

15 MC= - 1566

MB( 6 / I ) + 2MC( 6 / I + 6 / I ) + MD( 6 / I ) = -

6MB+ 24MC+ 6MD= - [ 1944 ] - [ 1188 ]

MB+ 4MC= - 522 As MD= 0

4MB+ MC = - 522 .

A2= ( 2 / 3 ) x Base x h2 A3= ( 2 / 3 ) x Bas

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

a2= 3 m a3= 3 m

Using three moment equation for span BCD

MB( L2/ I2) + 2MC( L2/ I2+ L3/ I3) + MD( L3/ I

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

( 99 - 48 = 51 KNm )

( 162 - 69 = 93 KNm )

6) ]

36 KN / m

162

22 K

99B

CC

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

6 m 6 m 6 mA B C D104 104

99162

99( - )

( - ) ( + )

( - ) ( - )

( + ) ( + ) ( + )

47 4758

104 104


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85

131

imum

ase ( b )

ve

VC= 132 - 48.67 VC= 108 KN

VC= 83.33 KN MC= - 104 KNm

VD= 48.67 KN Vc= 191.33 KN

VB= 191.33 KN

VB= 108 KN

48.67 x 12 + VBx

VB+ 83.33 = 191.3

VC+ VD= 22 x 6 VC+ 83.33 = 191.

MC= - 104 KNm MB= - 104 KNm

VDx 6 -22 x 6 x 3 = -104 48.67 x 12 + VCx

5 ( 6 - X )

10 m

m.

rease or

appropriate combinations of loads.

( c ) The elastic moment at any sec

particular combination of loads shallthan 30 % of the numerically largest

the elastic maximum moments di

member , covering all appropriate co

IS 456-2000 ,Clause 23.1.2 EffectiveIn the absence of more accurate dete

width of flange may be taken as the f

IS 456-2000 ,Clause 37.1.1 Redistri

Continuous Beams and Frames -

( b ) The ultimate moment of resistan

member is not less than 70 percent of

obtained from an elastic maximum m

S 456-2000 , Clause 37.1.1 )

)

48.67

83.33

2.21 m48.67

83.33

108

108


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Where ,

0.671

26.5.1 Beams

26.5.1.1 Tension Reinforcement

where ,

For mild steel

greater than the breadth of the web pl

distances to the adjacent beams on ei

( a ) For T-beams , bf= ( l0/ 6 ) + bw+

( b ) For L-beams , bf= ( l0/ 12 ) + bw

NOTE - For continuous beams and fr

assumed as 0.7 times the effective sp

bf= effective width of flange ,

l0= distance between points of zero

bw= breadth of the web ,

Df= thickness of flange , and

b = actual width of the flange.

a ) Minimum reinforcement -The mi

shall not be less than that given by th

As/ b d = 0.85 / fy

lause 26.5( a ) )

IS 456-2000 Clause 26.5 Requireme

below.

b ) Maximum reinforcement -The m

reinforcement shall not exceed 0.04 b

Minimum steel %

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

For HYSD steel , Fe415 grade

As= minimum area of tension reinforc

b = breadth of the beam or the breadt

d = effective depth , andfy= characteristic strength of reinforce

Structural Members

bw = 8

For singly reinforced rectangular sections

100 As/ b d = 100 x 0.85 / 415 = 0.20

For HYSD steel , Fe500 grade100 As/ b d = 100 x 0.85 / 500 = 0.17

Table 6-3

Limiting Moment of resistance factor Q lim, N /

p = 603 mm2,

2 - 16 mm O


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250 415 500

15 2.22 2.07 2.00

20 2.96 2.76 2.66

25 3.70 3.45 3.33

30 4.44 4.14 3.99

fck

N / mm2

fy, N / mm2


133/242


134/242


135/242

== 648

Using three mome

MA( L1/ I1) + 2MB

A1= ( 2 / 3 ) x Bas

( 2 / 3 ) x

.( 1 )

x h1

6 x 162

[ 6 x 396 x 3 / ( Ix 6 ) ] - [ 6 x 648 x 3 / ( Ix 6) ]

Case ( c ) Maximu

2) = - 6 A1a1/ ( I1L1) -6 A2a2/ ( I2L2)

6MA+ 24MB+ 6MC

4MB+ MC = - 522

MA( 6 / I ) + 2MB(

C

a1= 3 m

N / m

C

36 KN

design of rectangular water tank (1)

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