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. Volume = 6 x 4 x 3.35 x 10 3 = 80400 Litres 6 / 4 = 1.5 <2. L/B= 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 m bottom 1 m will be designed as cantilever. 2.5 x YW = 2.5 x 9.8 Water pressure at 3.5 - 1 = 2.5 m height from top = =

24.5

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 6m

A

F

2.5 m

3.5 m

6m E

24.5 KN / m

2

1m D 34.3 KN / m

2

Elevation

Plan

Fixed end moments :MAB = w x l2 / 12 =

MAD =

= 24.5 x 62 / 12 =

73.5

KNm

w x l2 / 12 = = 24.5 x 42 / 12 =

-32.66

KNm

Kani's Method :-

0 0

D

- 32.66 -12.25 -12.25 0 -57.16

73.5 -3/10 40.84 -2/10 -8.17 -8.17 A 0 57.16

0 0

B

Rotation factor at Joint A Joint

Member

Relative Stiffness( K )

∑K

A

AB AD

I/6 I/4

5 * I / 12

Sum of FEM MAF = 73.5-32.66 40.84

Rotation Factor u =(-1/2) k / ∑ K

- 2 / 10 - 3 / 10

KNm

MAB =

MABF + 2 MAB' + MBA' = 73.5 + 2 x (- 8.17 ) + 0 = 57.16

MAD =

MADF + 2 MAD' + MDA' = (- 32.66 ) + 2 x (- 12.25 ) + 0 = -57.16

B.M. at centre of long span =

w x l 2 / 8 - 57.16 = 24.5 x 62 / 8 - 57.16 53.09 KNm

B.M. at centre of short span =

w x l 2 / 8 - 57.16 = 24.5 x 42 / 8 - 57.16 -8.16 KNm

=

Yw ( H - h ) x B / 2 24.5 x 4 / 2 =

49

KN

=

Yw ( H - h ) x L / 2 24.5 x 6 / 2 =

73.5

KN

Direct tension in long wall =

Direct tension in short wall =

Design of Long Walls :At support M= 57.16 T= From Table 9-6 D=

KNm

49 KN Tension on liquid face. Q = 0.306 Assuming d / D = 0.9

√M / Q x b = √57.16 x 10 =

Take D = 450 mm

432.2

/ 0.306 x 1000 mm, Assuming d / D = 0.9

6

d = 450 - 25 - 8

=

417 mm

From Table 9-5 Ast1 for moment = M / σst x j x d = =

57.16 x 10 6 / 150 x 0.872 x 417 1048 mm2

Ast2 for direct tension = T / σst = = Total Ast1 + Ast2 =

49 x 10 3 / 150 327 mm2

1048 + 327 = 1375 mm2

Provide 16 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 200.96 x 1000 / 1375 = 146.152727 mm Provide 16 mm O bar @ 130 mm C/C…marked(a) = 1546 mm2 / m. Larger steel area is provided to match with the steel of short walls. At centre M= 53.09 KNm T= 49 KN tension on remote face e= M/T= 53.09 / 49 = 1.08 m Line of action of forces lies outside the section i.e.tension is small E= e+D/2-d b = 1080 + 450 / 2 - 417 = 888 mm D

modified moment = 49 x 0.888 = 43.51

d

KNm d'

Ast1 for moment = M / σst x j x d = =

E=e+D/2-d

43.51 x 10 6 / 190 x 0.89 x 417 617

mm2

Ast2 for direct tension = T / σst = = Total Ast1 + Ast2 = Provide 16 mm O bar

49 x 10 3 / 150 327 mm2

617 + 327 = 944

mm2

spacing of bar =

Area of one bar x 1000 / required area in m 2 / m = 200.96 x 1000 / 944 = 212.881356 mm Provide 16 mm O bar @ 200 mm C/C…marked(b) = 1005 mm2 From Table 9-3 minimum reinforcement 0.16 % Distribution steel = 0.16 / 100 x 1000 x 450 = 720 mm2 On each face = 360 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /360 = 139.555556 mm Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385 mm2 Vertical Steel ( c) Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 78.50 x 1000 /360 = 218.055556 mm Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392 mm2 Horizontal steel :Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005 mm2 Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385 mm2 Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392 Design of short walls :At support M= T= From Table 9-5 Ast1 for moment = M / σst x j x d = =

57.16 73.5

KNm KN

tension on liquid face

57.16 x 10 6 / 150 x 0.872 x 417 1048 mm2

Ast2 for direct tension = T / σst = = Total Ast1 + Ast2 =

73.5 x 10 3 / 150 490 mm2

1048 + 490 = 1538 mm2

Provide 16 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 200.96 x 1000 / 1538 = 130.663199 mm Provide 16 mm O bar @ 130 mm C/C…marked(b) = 1546 mm2 / m.

mm2

At centre M= 8.16 T= 73.5 From Table 9-5 Ast1 for moment = M / σst x j x d = =

KNm KN

tension on liquid face

8.16 x 10 6 / 150 x 0.872 x 417 150 mm2

Ast2 for direct tension = T / σst = = Total Ast1 + Ast2 =

73.5 x 10 3 / 150 490 mm2

150 + 490 = 640

mm2

Provide 12 mm O bar spacing of bar =

Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 640 = 176.625 mm Provide 12 mm O bar @ 130 mm C/C…marked(e) = 869 mm2 / m. From Table 9-3 minimum reinforcement 0.16 % Distribution steel = 0.16 / 100 x 1000 x 450 = 720 mm2 On each face = 360 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /360 = 139.555556 mm Provide 8 mm O bar @ 130 mm C/C…marked(d) = 385 mm2 Vertical Steel ( c) Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 78.50 x 1000 /360 = 218.055556 mm Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392 mm2 . Horizontal steel :Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869 mm2 Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = 385 mm2 Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392 Bottom 1 m will be designed as cantilever

mm2

Cantilever moment : Yw x H x h2 / 6 M= = 9.8 x 3.5 x 1 / 6 = 5.72 KNm From Table 9-5 Ast for moment =

OR

Yw x H / 6 , whichever is greater. = 9.8 x 3.5 / 6 = 5.72 ,tension on liquid face.

M / σst x j x d = =

5.72 x 10 6 / 150 x 0.872 x 417 105 mm2 From Table 9-3 minimum reinforcement 0.16 % Distribution steel = 0.16 / 100 x 1000 x 450 = 720 mm2 On each face = 360 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 78.50 x 1000 /360 = 218 mm Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392 mm2 . each face Base slab :Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab. From table 9-3 Minimum steel = 0.229% = 0.229 / 100 x 1000 x 150 = 344 mm2 ,172 mm2 bothway Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /172 = 292 mm Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. Ast = 346 mm2 Designed section,Elevation etc. are shown in fig. Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m 2 Top slab : consider 1 m wide strip. Assume 150 mm thick slab. lx = 4 + 0.15 = 4.15 say 4.5 m ly =

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 6.25 For 1 m wide strip

KN / m2 KN / m2 KN / m2 KN / m2

PU =

1.5 x 6.25 = 9.38 KN / m

ly / l x =

6.5 / 4.5 = 1.4 AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span. αx = Table 26 0.085 αy = Mx =

αx x w x l

0.056 My =

2 x

= 0.085 x 9.38 x 4.52 = 16.15 KNm From Table 6-3 ,Q = 2.76

√M / Q x b = √16.15 x 10

αy x w x lx2 = 0.056 x 9.38 x 4.52 = 10.64 KNm

drequired =

=

76.50

6

/ 2.76 x 1000 mm

dshort = 150 - 15(cover) - 5 = 130 > 76.50 mm …………(O.K.) dlong = 130 - 10 = 120 mm Larger depth is provided due to deflection check. Mu / b x d2 (short) = 16.15 x 106 / 1000 x 130 x 130 = 0.96 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck =

50 1-√1-(4.6 / 20) x (0.96) 415 / 20

= 50 [(1-0.88) x 20 / 415 ] = 0.29% Ast (short) = 0.29 x 1000 x 130 / 100 = 377 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 78.50 x 1000 /377 = 208 mm Provide 10 mm O bar @ 210 mm c/c = 374 mm2 . Mu / b x d2 (long) = 10.64 x 106 / 1000 x 120 x 120 = 0.74

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

50 1-√1-(4.6 / 20) x (0.74) 415 / 20

= 50 [(1-0.91) x 20 / 415 ] = 0.22% Ast (long) = 0.22 x 1000 x 120 / 100 = 264 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /264 = 190 mm Provide 8 mm O bar @ 190 mm c/c = 264 mm2 .

B

4m

C

TABLE 9-6 Balanced Design Factors for members in bending For M20 Grade Concrete Mix Mild steel HYSD bars d/D Pt Pt Q = M / bD2 Q = M / bD2 0.75 0.3 0.4 0.295 0.289 0.8

0.305

0.37

0.299

0.272

0.85 0.9

0.31 0.314

0.355 0.335

0.302 0.306

0.258 0.246

TABLE 9-5

Members in bending ( Cracked condition ) Coefficients for balanced design σst Grade of Grade of σcbc 2 concrete steel k j N / mm N / mm2 For members less than 225mm thickness and tension on liquid face M20 Fe250 7 115 0.445 0.851 Fe415 7 150 0.384 0.872

Q 1.33 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

s lies outside the section

D/2

e=M/T

TABLE 9-3 Minimum Reinforcement for Liquid Retaining Structures Thickness, mm

% of reinforcement

100

Mild Steel 0.3

HYSD bars 0.24

150 200 250

0.286 0.271 0.257

0.229 0.217 0.206

300 350 400 450 or more

0.243 0.229 0.214 0.2

0.194 0.183 0.171 0.16

8 O @ 190 c/c

10 O @ 210 c/c 150

150 Free board

3500

A

10 O @ 200 c/c -

shape

10 O @ 200 c/c -

shape

1500

150 150

8 O @ 290 c/c both ways top and bottom 1500

A

1 : 4 : 8 P.C.C. 150 450

6000

450 150

Elevation 1500 450 1000

16 O @ 130 c/c (a) 16 O @ 200 c/c (b) 10 O @ 200 c/c both faces (c) 8 O @ 130 c/c (d) 12 O @ 130 c/c (e)

4000

v 1000 v

(a)

(b)

(d)

(c) (d)

( a ) 1000

vv

450 1500

1500 6000

450

450

Section A-A

Table 6-3 Limiting Moment of resistance factor Q lim, N / mm2 fck N / mm2

fy, N / mm2

15 20 25

250 2.22 2.96 3.70

415 2.07 2.76 3.45

500 2.00 2.66 3.33

550 1.94 2.58 3.23

30

4.44

4.14

3.99

3.87

Pt,bal 1.36 0.98 1.2 0.61

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 Volume = 3.6 x 8 x 3.0 x 10 3 = 86400 Litres 8 / 3.6 = 2.22 >2. L/B= 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 m bottom h = 1 m will be designed as cantilever. Moments and tensions : Maximum B.M. in long walls at the base 3 = (1 / 6 ) x Yw x H = ( 1 / 6 ) x 9.8 x 33 = 44.1 KNm. Maximum ( - ve ) B.M. in short walls at support 2 = Yw x ( H - h ) x B / 12 = 9.8 x ( 3 - 1 ) x 42 / 12 = 26.13 KNm. Maximum ( + ve ) B.M. in short walls at centre 2 = Yw x ( H - h ) x B / 16 = 9.8 x ( 3 - 1 ) x 42 / 16 = 19.60 KNm. For bottom portion Yw x H x h2 / 6 M= = 9.8 x 3.0 x 1 / 6 = 4.90 KNm Direct tension in long wall =

OR

Yw x H / 6 , whichever is greater = 9.8 x 3.0 / 6 = 4.90 KNm

Yw x ( H - h ) x B / 2 = 9.8 x ( 3 - 1 ) x 3.6 / 2 = 35.28 KN

Direct tension in short wall= = =

Yw ( H - h ) x 1 9.8 x ( 3 - 1 ) x 1 19.6 KN

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 T= 35.28 KN From Table 9-6 Assume d / D = 0.9

√M / Q x b = √44.1 x 10

( water face ) ( perpendicular to moment steel ) Q = 0.306

D=

= Take D =

379.6

6

/ 0.306 x 1000

mm,

400 mm

d = 400 - 25 - 8 = 367 mm

Ast for Moment From Table 9-5 , Ast = M / σst x j x d = 44.1 x 10 6 / 150 x 0.872 x 367 = 918.68

mm2

Provide 16 mm O bar spacing of bar =

Area of one bar x 1000 / required area in m 2 / m = 200.96 x 1000 / 918.68 = 218.749 mm Provide 16 mm O bar @ 200 mm c/c = 1005 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. From Table 9-3 Distribution steel = 0.171 % for 400 mm depth As = ( 0.171 / 100 ) x 1000 x 400 = 684 mm2 . on each face = 342

mm2 .

…………………… ( 1 )

Steel required for direct tension = T / σst = 35.28 x 103 / 150 = 235 mm2 . …………………… ( 2 ) From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension. Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 342 = 146.901 mm Provide 8 mm O bar @ 140 mm c/c on each face = 357 mm2 on each face

Design of short walls :At support From Table 9-5 Ast1 for moment =

M= T=

26.13 19.6

KNm KN

M / σst x j x d = 26.13 x 10 6 / 150 x 0.872 x 367 = 544 mm2

Ast2 for direct tension = T / σst = 19.6 x 10 3 / 150 = 131 mm2 Total Ast1 + Ast2 = 544 + 131 =

675

mm2

Provide 12 mm O bar spacing of bar =

Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 675 = 167.467 mm Provide 12 mm O bar@160 mm c/c = 706 mm2. 1000 203.56 400 367 163.44

checking : modular ratio m =

x=

280 / 3 x σcbc = 13.33

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

=

= 203.56

b x D + ( m - 1 ) x Ast ( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 ) ( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 ) mm

D-x= 196.44 mm d-x= 163.44 mm AT = b x D + ( m - 1 ) x Ast = 1000 x 400 + (13.33 - 1 ) x 706

=

408705 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 = 5.34E+09 + 2.33E+08 4 = 5.57E+09 mm fct = T / AT = 19.6 x 10 3 / 408705 = 0.048 N / mm2 fcbt = M x ( d - x ) / Ixx = 26.13 x 106 x 163.44 / 5.57 x 10 9 = 0.767 N / mm2 check : ( fct / σct ) + ( fcbt / σcbt ) ≤1 ( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) ≤ 1 From Table 9-2 0.04 + 0.4512 ≤ 1 0.4912 ≤ 1 ………………….. ( O. K. ) At centre : M= T= From Table 9-5 Ast1 for moment =

19.6 19.6

KNm KN

M / σst x j x d = 19.6 x 10 6 / 150 x 0.872 x 367 = 408 mm2

Ast2 for direct tension = T / σst = 19.6 x 10 3 / 150 = 131 mm2 Total Ast1 + Ast2 = 408 + 131 =

539

Provide 12 mm O bar spacing of bar =

mm2

Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 539 = 209.722 mm Provide 12 mm O bar @ 200 mm c/c = 565 mm2.

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

As = ( 0.171 / 100 ) x 1000 x 400 = 684 mm2 . on each face = 342 mm2 .

…………………… ( 1 )

Steel required for direct tension = T / σst = 19.6 x 103 / 150 = 131 mm2 . …………………… ( 2 ) From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension. Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 342 = 146.901 mm Provide 8 mm O bar @ 140 mm c/c on each face = 357 mm2 on each face Bottom cantilever M = 4.9 KNm From Table 9-5 Ast = M / σst x j x d = 4.9 x 10 6 / 150 x 0.872 x 367 = 102 mm2 Minimum steel = 342 mm2 on each face. Provide 8 mm O bar @ 140 mm c/c on each faces = 357 mm2 on each face Base slab :Base slab is resting on ground. For a water head 3 m, provide 150 mm thick slab. From table 9-3 Minimum steel = 0.229% = 0.229 / 100 x 1000 x 150 = 344 mm2 ,172 mm2 bothway Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /172 = 292 mm Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. Ast = 346 mm2 Designed section,Elevation etc. are shown in fig. Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m 2 Top slab : consider 1 m wide strip. Assume 150 mm thick slab. lx = 3.6 + 0.4 = 4 say 4 m

ly =

8 + 0.15 = 8.15 say 8.5 m Dead Load : self 0.15 x 25 = floor finish = Live load =

3.75 1.0 1.5 6.25

KN / m2 KN / m2 KN / m2 KN / m2

For 1 m wide strip PU =

1.5 x 6.25 = 9.38

KN / m

Maximum moment = 9.38 x 42 / 8 = 18.76 KNm Maximum shear = 9.38 x 3.6 / 2 = 16.88 KN From Table 6-3 ,Q = 2.76

√M / Q x b = √18.76 x 10

drequired =

=

82.44

6

/ 2.76 x 1000 mm

dprovided =

150 - 15(cover) - 6 = 129 > 82.44 …………(O.K.) Larger depth is provided due to deflection check. Design for flexure : Mu / b x d2 = 18.76 x 106 / 1000 x 129 x 129 = 1.13 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck =

50 1-√1-(4.6 / 20) x (1.13) 415 / 20

= 50 [(1-0.86) x 20 / 415 ] = 0.34% Ast = 0.34 x 1000 x 129 / 100 = 439 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 50.24 x 1000 /439 = 114 mm Provide 8 mm O bar @ 110 mm c/c = 457 mm2 .

Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement Distribution steel = ( 0.12 / 100 ) x 1000 x 150 = 180 mm2 Provide 6 mm O bar spacing of bar =

Area of one bar x 1000 / required area in m 2 / m = 28.26 x 1000 /180 = 157 mm Provide 6 mm O bar @ 150 mm c/c = 188 mm2 .

4.0 m.

hort walls.

op.For economy, the as we have done for

x

42

Table 9-2 Permissible concrete stresses in calculations relating to resistance to cracking Grade of concrete M15 M20 M25 M30 M35 M40

Permissible stresses in N / mm2 Direct Tension due to Shear stress ‫ﺡ‬v = V / b j d tension σct bending σcbt 1.1 1.5 1.5 1.2 1.7 1.7 1.3 1.5 1.6 1.7

1.8 2.0 2.2 2.4

1.9 2.2 2.5 2.7

6 O @ 150 c/c

8 O @ 110 c/c 150

12 O @ 200 c/c 8 O @ 140 c/c shape 8 O @ 140 c/c 8 O @ 290 c/c both ways top and bottom 2000

3000

150 150

2000

150

1 : 4 : 8 P.C.C. 150 400

8000

400

Section A-A

B

2000

8 O @ 140 mm c/c

400 900

16 O @ 200 c/c ( chipiya )

12 O @ 160 c/c(chipiya) 8 O @ 140 c/c (chipiya)

3600

8 O @ 140 c/c

A

900 400

A

12 O @ 200 c/c 900

v v 2000

vv

2000

B 8000

400

Sectional plan

8 O @ 110 c/c 150

400

6 O @ 150 c/c

16 O @ 200 c/c ( chipiya ) 8 O @ 140 c/c

3000

900

900

Base details not shown for clarity

150 150 150 400

3600 Section B- B

400 150

Design of simply supported one way slab effective span = 4 m supported on masonry wall of 230 mm thickness Live load = 2.5 KN / m2 ( given ) Floor finish = 1 KN / m2 ( 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 reinforcement modification factor = 1.26 for simply supported, basic span / effective depth ratio = 20 ( span / d ) ratio permissible = 1.26 x 20 = 25.2 drequired = 4000 / 25.2 = 158.7 mm D = 158.7 + 15 ( cover ) + 5 ( assume 10 O bar ) = 178.7 mm Assume an overall depth = 180 mm 2 DL = 0.18 x 25 = 4.5 KN / m Floor finish = 1.0 KN / m2 Live load = 2.5 KN / m2 Total 8.0 KN / m2 Factored load = 1.5 x 8 = 12 KN / m Consider 1 m length of slab Maximum moment = w x l2 / 8 = 12 x 42 / 8 = 24 KNm Maximum shear = w x l / 2 = 12 x 4 / 2 = 24 KN Design for flexure : d = 180 - 15 - 5 = 160 mm Mu / b x d2 = 24 x 10 6 / 1000 x (160)2 = 0.94 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.94) 415 / 15 = 50 [(1-0.84) x 15 / 415 ]

( given )

= 0.289% Ast = 0.289 x 1000 x 160 / 100 = 462 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 78.50 x 1000 /462 = 170 mm Provide 10 mm O bar @ 170 mm c/c = 462 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 ) = 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 mm2 Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mm Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 50.24 x 1000 /216 = 233 mm Provide 8 mm O bar @ 230 mm c/c = 218 mm2 . Check for shear : Vu = 24 KN Actual Shear stress =

Vu / b x d

= 24 x 103 / 1000 x 160 = 0.150

N / mm2

< (‫ﺡ‬

C

) N / mm2 ( too small )

For bars at support d = 160 mm As = 231 mm2 . 100 x As / b x d = 100 x 231 / 1000 x 160 = 0.144 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 0.8 x 15 / 6.89 x 0.144 = 12.1 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.1 - 1 ) 6 x 12.1 = 0.277 IS 456-2000 Table 19 from table 7-1

for Pt = 0.144 ‫ﺡ‬c = 0.28 N / mm2 IS 456-2000 clause 40.2.1.1 25 difference -0.05 k = 1.24 for 180 mm slab depth 20 difference ? Design shear strength = 1.24 x 0.28 = 0.347 N / mm2 ……………….( O.K.) Check for development length : Assuming L0 = 8 O (HYSD Fe415 steel ) For continuing bars Mu1 = = = =

As = 231 mm2 0.87 x fy x Ast x d { 1 - ( fy x Ast / fck x b x d ) } 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

-0.04

OR

Vu = 24

KN Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd

= 56 O ( from Table 7-6 )

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

1.3 x ( 12.81 x 106 / 24 x 103 ) + 8 O ≥ 56 O 693.875 + 8 O ≥ 56 O 48 O ≤ 693.88 which gives O ≤ 14.46 mm ……………….( O.K.) Check for deflection : Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 462 / 1000 x 160 = 0.289 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.42 ( span / d ) ratio permissible = 1.42 x 20 = 28.4 Actual (span / d ) ratio = 4000 / 160 = 25.00 < 28.4 ……………….( O.K.) The depth could be reduced Check 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 sm = 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 whiche = 5 x 160 = 800 mm spacing provided = 230 mm < 450 mm ……………….( O.K.)

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

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

OR

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

For mild steel minimum reinforcement 0.15 %

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 As / b x d = 100 x 231 / 1000 x 160 = 0.14 From equation Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck Mu1 / b x d2 = 0.49 Mu1 = 0.49x 1000 x 1602 x 10-6

we get ,

= 12.54

KNm

160 180 400

r 300 mm whichever is small or 300 mm i.e. 300 mm ……….( O.K.) h of slab or 450 mm whichever is small or 450 mm i.e. 450 mm ……….( O.K.)

v v

v 8 O @ 230 c/c v 10 O @ 170 c/c ( alternate bent ) 4000

v v 400

v v

If ly / lx ≥ 2 ,called one way slab provided that it 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 twoway slab.provided that it is supported on all four edges. IS 456-2000 clause 22.2 Effective Span ( 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. ( 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, the effective 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 the 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 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. ( 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.

elopment length , l0 as 8 O for HYSD anchorage is not O for mild steel ( U usually whose is 16 O.

Design of Cantilever one way slab used for residential purpose at the free end of slab S1 ,concrete parapet of 75 mm thick and 1 m high. material M15 grade concrete mild steel grade Fe250 Live Load As per IS 875 Solution : For slab S2 live load = 2 KN /m2

( given ) ( given )

For slab S1 live load = 3 KN /m2

( Balcony slab ) DL LL

Assume 120 mm thick slab For S2 self load = 0.12 x 25 = 3 floor finish = 1 live load = 0 Total 4 Pu = 1.5 ( 4 + 2 ) = ( 6 + 3 ) DL For S1 self load = 0.12 x 25 = 3 floor finish = 1 live load = 0 Total 4

0 0 2 2

ly = 6m

KN / m2 KN / m2 KN / m2 KN / m2

KN / m2 LL 0 0 3 3

KN / m2 KN / m2 KN / m2 KN / m2

Pu = 1.5 ( 4 + 3 ) = (6 + 4.5) KN / m2 Weight of parapet 0.075 x 25 x 1 = 1.875 KN / m Pu = 1.5 x 1.875 = 2.8 KN / m Consider 1 m long strip (1) To get maximum positive moment in slab S2 only dead load on slab S1 and total load on slab S2 shall be considered 9 KN/m

6 KN/m

SS22 A

3m

S1 B

1.2m

C

(a) Loads for maximum positive moment 9 KN/m

10.5 KN/m

S2 A

3m

2.8 KN

S1 B

1.2m

C

(b) Loads for maximum negative moment,maximum shear for cantilever span and maximum reaction at support B considering fig (a) cantilever moment = wx l2 / 2

= 6 x 1.22 / 2 = 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 = 12.06 x 1.34 - W x l2 / 2 = 12.06 x 1.34 - 9 x 1.342 / 2 = 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 = w x l + w x l2 / 2 = 2.8 x 1.2 + 10.5 x 1.22 / 2 = 10.92 KNm Maximum shear at B Vu,BA = w x l / 2 + Moment @ B at distance 3 m = 9 x 3 / 2 + 10.92 / 3 = 17.14 KN Vu,BC = 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

√M / Q x b = √10.92 x 10

drequired =

= dprovided = = Mu / b x d2 ( + ) = =

70.14

6

/ 2.22 x 1000

mm,

120 - 15 - 6 ( assume 12 O bar ) 99 mm, ……………….( O.K.) 8.08 x 10 6 / 1000 x (99)2 0.82

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.82) 250 / 15 = 50 [(1-0.865) x 15 / 250 ] = 0.405%

Ast ( + ) = 0.405 x 1000 x 99 / 100 = 401 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 78.5 x 1000 / 401 = 195.761 mm Provide 10 mm O bar@170 mm c/c = 462 mm2. ( alternate bent up ) Mu / b x d2 ( - ) = 10.92 x 10 6 / 1000 x (99)2 = 1.11 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (1.11) 250 / 15 = 50 [(1-0.812) x 15 / 250 ] = 0.564% Ast ( - ) = 0.564 x 1000 x 99 / 100 = 558 mm2 Provide 10 mm O bar Provide 10 mm O bar@340 mm c/c = 231 mm2. ( bent bars extended ) Area provided = Area of one bar x 1000 / spacing of bar in m = 78.5 x 1000 / 340 = 231 mm2 remaining area = 558 - 231 = 327 mm2 Provide 12 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 327 = 346 mm Provide 12 mm O bar@340 mm c/c = 332 mm2. ( Extra ) 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 For mild steel minimum reinforcement 0.15 % Distribution steel = ( 0.15 /100 ) x 1000 x 120 = 180 mm2. Provide 6 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 28.26 x 1000 / 180 = 157 mm

Provide 6 mm O bar@150 mm c/c = 188 mm2. For negative moment reinforcement Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd ( from Table 7-6 ) = O x 0.87 x 250 / 4 x 1 = 54.4 O Ld = 54.4 x ( 10 + 12 ) / 2 = 598 mm. say 600 mm As a thumb rule, a bar shall be given an anchorage equal to the length of the cantilever. Check for development length : Assuming L0 = 12 O (mild steel ) Pt = 100 x As / b x d = 100 x 231 / 1000 x 99 = 0.233 From equation At A ,

Half bars bent = 462 / 2 = 231 mm2 )

OR

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

Mu1 = 0.87 x fy x Ast x d ( 1 - fy x

fy / fck we get ,

= 0.87 x 250 x 231 x 99 ( 1

Mu1 / b x d2 = 0.487 Mu1 = 0.487 x 1000 x 992 x 10-6 = 4.77 KNm Vu = 12.06 KN

Development length of bars Ld =

O σs / 4 x ‫ﺡ‬bd

= 4.97401 = 4.78

= O x 0.87 x 250 / 4 x 1

=54.4 O

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 4.77 x 106 / 12.06 x 103 ) + 12 O ≥ 54.4 O 514.179 + 12 O ≥ 54.4 O 42.4 O ≤ 514.179 which gives O ≤ 12.13 mm ……………….( O.K.) Pt = 100 x As / b x d At B , Half bars bent = 462 / 2 = 231 mm2 ) = 100 x 231 / 1000 x 99 = 0.233 From equation OR Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

Mu1 = 0.87 x fy x Ast x d ( 1 - fy x

fy / fck

= 0.87 x 250 x 231 x 99 ( 1

Mu1 / b x d2 = 0.487 Mu1 = 0.487 x 1000 x 992 x 10-6 = 4.77 KNm Near point of contraflexure i.e. 0.15 x l from B Vu = 17.14 - ( 0.15 x 3 ) x 9 = we get ,

Development length of bars Ld =

O σs / 4 x ‫ﺡ‬bd

=

4.974

= 4.78

13.09

KN

= O x 0.87 x 250 / 4 x 1

=54.4 O

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 4.77 x 106 / 13.09 x 103 ) + 12 O ≥ 54.4 O 473.72 + 12 O ≥ 54.4 O 42.4 O ≤ 473.72 which gives O ≤ 11.2 mm Check for shear : Span AB : At A , Vu,AB = 12.06 KN ( for maximum loading ) At B , shear at point of contraflexure = 13.09 Use Vu = 13.09 KN

……………….( O.K.)

KN

Shear stress ‫ﺡ‬v = Vu / b x d = 13.09 x 10 3 / 1000 x 99 = 0.132 N / mm2 < ‫ﺡ‬c

……………….( O.K.)

100 x As / b x d = 100 x 231 / 1000 x 99 = 0.233 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 0.8 x 15 / 6.89 x 0.233 = 7.47 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 7.47 - 1 ) 6 x 7.47 = 0.34 IS 456-2000 Table 19 from table 7-1 for Pt = 0.233 ‫ﺡ‬c = 0.34 N / mm2 IS 456-2000 clause 40.2.1.1 0.1 difference 0.07 k = 1.3 for 120 mm slab depth 0.02 difference ? 0.014 Design shear strength = 1.3 x 0.34 = 0.442 N / mm2 > ‫ﺡ‬v ……………….( O.K.) Span BC : Vu = 17.14 KN Pt =

Shear stress ‫ﺡ‬v = Vu / b x d = 17.14 x 10 3 / 1000 x 99 = 0.173 N / mm2 < ‫ﺡ‬c

……………….( O.K.)

100 x As / b x d = 100 x 563 / 1000 x 99 = 0.569 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 0.8 x 15 / 6.89 x 0.569 Pt =

= 3.06 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 3.06 - 1 ) 6 x 3.06 = 0.487 IS 456-2000 Table 19 from table 7-1 for Pt = 0.569 ‫ﺡ‬c = 0.48 N / mm2 IS 456-2000 clause 40.2.1.1 0.25 difference 0.08 k = 1.3 for 120 mm slab depth 0.181 difference ? 0.05792 Design shear strength = 1.3 x 0.48 = 0.624 N / mm2 > ‫ﺡ‬v ……………….( O.K.) Check for deflection : For span AB : Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 462 / 1000 x 99 = 0.467 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 2 ( span / d ) ratio permissible = 2 x 20 = 40 Actual (span / d ) ratio = 3000 / 99 = 30.30 < 40 ……………….( O.K.) For span BC : Basic ( span / d ) ratio = 7 Pt = 100 x Ast / b x d = 100 x 563 / 1000 x 99 = 0.569 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.8 ( span / d ) ratio permissible = 1.8 x 7 = 12.6 Actual (span / d ) ratio = 1200 / 99 = 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 sma = 3 x 99 = 297 mm or 300 mm i.e. 297 mm spacing provided = 170 mm < 297 mm ……………….( O.K.) (2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whiche = 5 x 99 = 495 mm spacing provided = 150 mm < 450 mm ……………….( O.K.)

S3

1.2m 0.15m

ly = 6m

0.15m

Column 300 x 300

B3

B1

B2 S2

S1

B4

0.15m

0.15m lx = 3m

1.2m

1m high parapet

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6 ( 1 - 0.0389 ) KNm

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 250 x 231 x 99 ( 1 - 250 x 231 / 1000 x 99 x 15 ) x 10 -6 ( 1 - 0.0389 ) KNm

10 O @ 340 c/c (bent) + 12 O @ 340 c/c (extra) 1200

1200

120

125 10 O @ 170 c/c

6 O @ 150 c/c

6 O @ 150 c/c 300 150

r 300 mm whichever is small or 300 mm i.e. 297 mm ……….( O.K.) of slab or 450 mm whichever is small or 450 mm i.e. 450 mm ……….( O.K.)

600 3000

150

1200

Design of Continuous Two-way slab Banking hall slab is restrained with edge beams Material M15 grade concrete HYSD reinforcement of grade Fe415 Solution : Assume 150 mm thick slab Self load 0.15 x 25 = 3.75 KN / m2 floor finish = 1.00 KN / m2 Live Load = 3.00 KN / m2 Total 7.75 KN / m2 Pu = 1.5 x 7.75

given

= 11.625 KN / m2 ly / lx = 5 / 5 =1 < 2 i.e. two-way slab. Middle strip : IS 456-2000 Table -26 Two adjacent edges Discontinuous Mu1 , Mu3 ( - ) = αx x w x lx2 = 0.047 x 11.625 x 52 = Mu2 ( + ) = αy x w x l = 0.035 x 11.625 x 52 = From Table 6-3 For M15 mix and Fe415 steel, Q = 2.07 2 x

√M / Q x b = √13.66 x 10

13.66

KNm

10.17

KNm

drequired =

6

/ 2.07 x 1000

= 81.23 mm, dprovided for positive moment reinforcement = 150 - 15 - 10 - 5 ( assume 10 O bar ) = 120 mm, > 81.23 mm ……………….( O.K.) dprovided for negative moment reinforcement = 150 - 15 - 5 = 130 mm, > 81.23 mm ……………….( O.K.) Mu / b x d2 ( + ) = 10.17 x 10 6 / 1000 x (120)2 = 0.706 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.706) 415 / 15 = 50 [(1-0.885) x 15 / 415 ] = 0.208%

Ast = 0.208 x 1000 x 120 / 100 = 250 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 250 = 200.96 mm Provide 8 mm O bar@ 200 mm c/c = 251 mm2. Mu / b x d2 ( - ) = 13.66 x 10 6 / 1000 x (130)2 = 0.81 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.81) 415 / 15 = 50 [(1-0.867) x 15 / 415 ] = 0.240% Ast = 0.24 x 1000 x 130 / 100 = 312 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 312 = 161 mm Provide 8 mm O bar@ 150 mm c/c = 335 mm2. For HYSD Fe415 Minimum steel = ( 0.12 / 100 ) x 1000 x 150 = 180 mm2 At discontinuous edges 4 and 5 , 50 % of the positive steel is required at top = ( 1 / 2 ) x 251 = 126 mm2 This is less than minimum , therefore , use minimum steel at location 4 and 5 . Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 180 = 279.111 mm Provide 8 mm O bar@ 260 mm c/c = 193 mm2. More 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

= 188

mm2.

Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 188 = 267.234 mm Provide 8 mm O bar@ 260 mm c/c = 193 mm2. This will be provided by minimum steel of edge strip, At corner B , steel required = ( 1/2 ) x 188 = 94 mm2. Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 180 = 279.111 mm Provide 8 mm O bar@ 260 mm c/c = 193 mm2. This will be provided by minimum steel . 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 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 As / b x d = 100 x 335 / ( 1000 x 130 ) = 0.258 from table 7-1 for Pt = 0.258 , ‫ﺡ‬c = 0.354 N / mm2 0.11 0.25 differen IS 456-2000 clause 40.2.1.1 ? -0.1065 0.242 differe k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.354 = 0.460 N / mm2 ……………….( O.K.) OR Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 6.75 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 6.75 - 1 ) 6 x 6.75 = 0.356 IS 456-2000 clause 40.2.1.1 k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.356 = 0.463 N / mm2 ……………….( O.K.) Actual shear stress = Vu / b x d

= 31.795 x 103 / ( 1000 x 130 ) = 0.245 N / mm2 < ‫ﺡ‬c ……………….( O.K.) At point 4 or 5 S.F. = w x l / 2 = 11.625 x 5 / 2 = 29.06 KN 100 x As / b x d = 100 x 251 / ( 1000 x 120 ) = 0.209 from table 7-1 for Pt = 0.209 , ‫ﺡ‬c = 0.321 N / mm2 0.1 differenc IS 456-2000 clause 40.2.1.1 0.041 differe k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.321 = 0.417 N / mm2 OR Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 8.33 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 8.33 - 1 ) 6 x 8.33 = 0.326 IS 456-2000 clause 40.2.1.1 k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.326 = 0.424 N / mm2 Actual shear stress = Vu / b x d

0.07 ?

-0.0287

= 29.06 x 103 / ( 1000 x 120 ) = 0.242 N / mm2 < ‫ﺡ‬c ……………….( O.K.) Check for development length : This is critical at point 4 or 5. Vu = 29.06 At point 4 or 5 , KN No bar is curtailed or bent up. Assuming L0 = 8 O (HYSD Fe415 steel ) Pt = 100 x As / b x d = 100 x 251 / 1000 x 120 = 0.209 From equation Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck we get ,

Mu1 / b x d2 = 0.711

OR Mu1 = 0.87 x fy x Ast x d ( 1 - fy = 0.87 x 415 x 251 x 120 = 10.8748

Mu1 = 0.711 x 1000 x 1202 x 10-6 = 10.246 = 10.24 KNm Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd =54.3 O (From Table 7-6 ) 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 10.246 x 106 / 29.06 x 103 ) + 8 O ≥ 54.3 O 458.355 + 8 O ≥ 54.3 O 46.3 O ≤ 458.355 which gives O ≤ 9.90 mm ……………….( O.K.) Vu = 11.25 x ( 4.5 / 2 ) = 25.31 KN Short span Assuming L0 = 8 O (HYSD Fe415 steel ) Pt = 100 x As / b x d = 100 x 462 / 1000 x 160 = 0.289 From equation

OR

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d ) 2

fy / fck

Mu1 = 0.87 x fy x Ast x d ( 1 - fy = 0.87 x 415 x 462 x 160

Mu1 / b x d = 0.9595 = 26.6888 Mu1 = 0.9595 x 1000 x 1602 x 10-6 = 24.5539 = 24.56 KNm Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd =56 O (From Table 7-6 ) 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld we get ,

2

1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 1261.48 + 8 O ≥ 56 O 48 O ≤ 1261.48 which gives O ≤ 26.28 mm ……………….( O.K.) Note that the bond is usually critical along long direction. Check for deflection : Basic ( span / d ) ratio = 26 positive moment steel = 251 mm2. actual d = 150 - 15 -8 - 4 = 123 mm. Pt = 100 x Ast / b x d = 100 x 251 / 1000 x 123 240.7 = 0.204 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.65 ( span / d ) ratio permissible = 1.65 x 26 = 42.9 Actual (span / d ) ratio = 5000 / 123 = 40.65 < 42.9 ……………….( 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 whiche = 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 whiche = 5 x 120 = 600 mm spacing provided = 260 mm < 450 mm ……………….( O.K.) 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.

B 625

3750

625

625 A 3750

S1

625

Edge strip 8 O @ 260 c/c

Middle Strip 8 O @ 200 c/c

Edge strip 8 O @ 260 c/c

B 500

8 O @ 200 c/c

8 O @ 150 c/c 8 O @ 260 c/c

150

8 O @ 260 c/c 1500

1500 625

3750

Section A-A

625

8 O @ 200 c/c

Edge strip Middle Strip Edge strip 8 O @ 260 c/c 8 O @ 260 c/c

This is minimum (0.12 / 100) x 150 x 1000 = 180 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 180 = 279.111 mm Provide 8 mm O bar@ 260 mm c/c = 193 mm2. for uniformity in spacing. For clarity , top and bottom reinforcements are shown separately.

A

5m

5m

-0.047

S1 1

S1 Middle Strip

5

3750

625

ly/8

(3/4)ly

ly/8

3

2

A 625

-0.047

0.035

Edge strip

5m

Edge strip

B

0.035

4 5m

B

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 ( 1 - 0.0579 )

-6

KNm

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 ( 1 - 0.0799 ) KNm

-6

1000 625

625

625

1500

1500

500 3750

1500

S1

625

1500

1000

3750

Edge strip Middle Strip Edge strip 8 O @ 240 c/c 8 O @ 240 c/c 8 O @ 140 c/c

of slab or 300 mm whichever is small or 300 mm i.e. 300 mm ……….( O.K.) of slab or 450 mm whichever is small or 450 mm i.e. 450 mm ……….( O.K.)

Edge strip 8 O @ 240 c/c

Middle Strip 8 O @ 140 c/c

8 O @ 240 c/c

Edge strip 8 O @ 240 c/c

8 O @ 140 c/c 500

150 8 O @ 180 c/c 1500

8 O @ 240 c/c 1500 625

3750

625

8 O @ 240 c/c

-0.047 3

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 / m2 and floor finish 1 KN / m2 Material M15 grade concrete HYSD reinforcement of grade Fe415 Solution : Try 120 mm thick slab DL LL Dead load 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 factored load = 1.5 ( 4 + 3 ) = ( 6 + 4.5 ) KN / m2 Consider 1 m wide strip of the slab.

given

( 6 + 4.5 ) KN / m

1 A

3m

2 B

3m C

2

3m

2 D

3m

1 E

3m

F

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 l2 = = = Mu2 ( + ) =

( 1 / 12 ) x 6 x 32 + ( 1 / 10 ) x 4.5 x 32 4.5 + 4.05 8.55 KNm

= = = Mu4 ( - ) =

( 1 / 10 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32 5.4 + 4.5 9.9 KNm

( 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 = 3.38 + 3.38 = 6.75 KNm Mu3 ( - ) = ( 1 / 10 ) x w ( DL ) x l2 + ( 1 / 9 ) x w ( LL ) x l2

( 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 = 4.50 + 4.50 =9 KNm Maximum shear is Vu(BA) = 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 Maximum moment is Mu3 ( - ) = 9.9 KNm

From Table 6-3

Q = 2.07

√M / Q x b = √9.9 x 10

drequired =

/ 2.07 x 1000 mm, 6

= 69.2 Try 110 mm overall depth dprovided = 110 - 15 - 5 ( assume 10 O bar ) = 90 mm, ……………….( O.K.) Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck Ast = Pt x b x d / 100 spacing of bar = Area of one bar x 1000 / required area in m 2 / m Table Factored Ast Pt moment Mu/(b x d ) point Steel Provided KNm mm2 2

1(+)

8.55

1.06

0.322

290

2(+)

6.75

0.83

0.248

223

3(-)

9.9

1.22

0.38

342

4(-)

9.0

1.11

0.34

306

10 mm O @ 440 c/c + 8 mm O @ 440 c/c = 178 +114 = 292 mm2 (Half 10 O+half 8 O)

8 mm O @ 220 c/c = 228 mm2 10 mm O @ 220 c/c = 357 mm2 10 mm O @ 220 c/c = 357 mm2

For Main steel ,HYSD Fe415 reinforcement minimum steel area = ( 0.12 / 100 ) x 1000 x 110 = 132 mm2 For Distribution steel , mild steel Fe250 reinforcement minimum steel area = ( 0.15 / 100 ) x 1000 x 110 = 165 mm2 Use 6 mm O spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 28.26 x 1000 /165 = 171 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 mm 2 . Use 8 mm O spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 /146 = 344 Use 8 mm O @ 340 mm c/c = 148 mm2.

Check for shear : Maximum shear = 18.9 KN Actual Shear stress = Vu / b x d = 18.9 x 103 / 1000 x 90 = 0.210

N / mm2

< (‫ﺡ‬

C

) N / mm2 ( too small )

For bars at support d = 90

mm

As = 357 mm2 . 100 x As / b x d = 100 x 357 / 1000 x 90 = 0.397 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 0.8 x 15 / 6.89 x 0.397 = 4.4 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 4.4 - 1 ) 6 x 4.4 = 0.42 IS 456-2000 Table 19 from table 7-1 for Pt = 0.397 ‫ﺡ‬c = 0.42 N / mm2 IS 456-2000 clause 40.2.1.1 0.25 difference 0.11 k = 1.3 for 110 mm slab depth 0.103 difference ? 0.04532 Design shear strength = 1.3 x 0.42 = 0.546 N / mm2 ……………….( O.K.) At point of contraflexure i.e. 0.15 x l from B with positive moment reinforcement ( 292 mm2 ) Vu = 18.9 - 0.15 x 3 x 10.5 = 14.18 KN Actual Shear stress = Vu / b x d = 14.18 x 103 / 1000 x 90 = 0.158

N / mm2

< (‫ﺡ‬

C

) N / mm2 ( too small )

For bars at support d = 90

mm

As = 292 mm2 . 100 x As / b x d = 100 x 292 / 1000 x 90 = 0.324 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0

= 0.8 x 15 / 6.89 x 0.324 = 5.4 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 5.4 - 1 ) 6 x 5.4 OR = 0.39 IS 456-2000 Table 19 from table 7-1 for Pt = 0.324 ‫ﺡ‬c = 0.38 N / mm2 IS 456-2000 clause 40.2.1.1 0.25 difference 0.11 k = 1.3 for 110 mm slab depth 0.176 difference ? Design shear strength = 1.3 x 0.38 = 0.494 N / mm2 ……………….( O.K.) Check for development length : Span AB is critical for checking this requirement At support A Vu = 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 Pt = 100 x As / b x d At A , = 100 x 292 / 1000 x 90 = 0.324 From equation OR Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

Mu1 = 0.87 x fy x Ast x d ( 1 - fy

fy / fck

= 0.87 x 415 x 292 x 90 (

Mu1 / b x d = 1.064 Mu1 = 1.064 x 1000 x 902 x 10-6 = 8.62 KNm Assuming L0 = 8 O (HYSD Fe415 steel )

we get ,

2

Development length of bars Ld =

O σs / 4 x ‫ﺡ‬bd

0.07744

= O x 0.67 x 415 / 4 x 1

= 9.48839 = 8.64

=56.4 O

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 8.64 x 106 / 13.28 x 103 ) + 8 O ≥ 56.4 O 845.7831 + 8 O ≥ 56.4 O 48.4 O ≤ 845.783 which gives O ≤ 17.47 mm ……………….( O.K.) At support B, point of contraflexure is assumed at 0.15 x l from B Vu = 18.9 - 0.15 x 3 x 10.5 = 14.18 KN Mu1 = 8.64 KNm as before L0 =

12 O ( 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 ) + L0 ≥ Ld 1.3 x ( 8.64 x 106 / 14.18 x 103 ) + 12 O ≥ 56.4 O 792.1016 + 12 O ≥ 56.4 O

44.4 O ≤ 792.102 which gives O ≤ 17.84 mm ……………….( O.K.) Check for deflection : Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB Basic ( span / d ) ratio = 26 Pt = 100 x Ast / b x d = 100 x 292 / 1000 x 90 = 0.324 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.34 ( span / d ) ratio permissible = 1.34 x 26 = 34.84 Actual (span / d ) ratio = 3000 / 90 = 33.33 < 34.84 ……………….( O.K.) For span BC : Basic ( span / d ) ratio = 26 Pt = 100 x Ast / b x d = 100 x 228 / 1000 x 90 = 0.253 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.6 ( span / d ) ratio permissible = 1.6 x 26 = 41.6 Actual (span / d ) ratio = 3000 / 90 = 33.33 < 41.6 ……………….( 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 sm = 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 which = 5 x 90 = 450 mm spacing provided = 160 mm < 450 mm ……………….( O.K.)

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 forces used in design may be obtained using the coefficients given 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 referred to in 22.7 shall not be permitted. (22.5.2 ) Beams and Slabs Over Free End Supports Where a member is built into a masonry wall which develops 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 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. Table 12 Bending Moment coefficients Span moments Support moments Type of load Dead load and imposed load ( fixed ) imposed load ( not fixed )

Near middle At middle of At support next to of end span interior span the end support + 1 / 12

+ 1 / 16

-1 / 10

+ 1 / 10

+ 1 / 12

-1 / 9

NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span. Table 13 Shear Force coefficients Type of load Dead load and imposed load ( fixed ) imposed load ( not fixed )

At end support

At support next to the end support

Outer side

Inner side

0.4

0.6

0.55

0.45

0.6

0.6

( 0.15 l1 )

( 0.3 l1 ) 900

450 8 O @ 340 c/c 110

( 0.3 l1 ) 900

( 0.3 l1 ) 900 10 O @ 220 c/c

90 6 O @ 160 c/c 8 O @ 440 c/c + 10 O @ 440 c/c 3000

8 O @ 220 c/c

3000

cement ( 292 mm2 )

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6 ( 1 - 0.0898 ) KNm

in span AB

r 300 mm whichever is small or 300 mm i.e. 270 mm ……….( O.K.) of slab or 450 mm whichever is small or 450 mm i.e. 450 mm ……….( O.K.)

ts Support moments At other interior supports

-1 / 12 -1 / 9

e multiplied by the total

At all other interior supports 0.5 0.6

( 0.3 l1 ) 900 10 O @ 220 c/c

3000

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. material M15 grade concrete HYSD reinforcement of grade Fe415

given

Solution : Consider 1 m wide strip. Assume 180 mm thick slab. lx = 4.3 + 0.18 = 4.48 say 4.5 m ly = 6.55+0.18 = 6.73 say 6.75 m Dead load : self 0.18 x 25 = 4.5

KN / m2

floor finish = 1.0

KN / m2

Live load ( residence ) = 2.0

KN / m2

Total

7.5

KN / m2

For 1 m wide strip Pu = 1.5 x 7.5 = 11.25 KN / m ly / lx = 6.75 / 4.5 = 1.5 IS 456-2000 Table -27 Mux = αx x w x lx2 = Muy =

αy x w x lx2 =

From Table 6-3

0.104 x 11.25 x 4.52 =

23.7

KNm

0.046 x 11.25 x 4.52 = Q = 2.07

10.48

KNm

√M / Q x b = √23.7 x 10

drequired =

6

/ 2.07 x 1000

= 107 mm, dshort provided = 180 - 15 - 5 ( assume 10 O bar ) = 160 mm, > 107 mm dlong provided = 160 - 10 = 150 mm, > 107 mm Larger depth is provided due to deflection check. Mu / b x d2 ( short ) =

23.7 x 10 6 / 1000 x (160)2 = 0.926

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

……………….( O.K.)

= 50 1-√1-(4.6 / 15) x (0.926) 415 / 15 = 50 [(1-0.846) x 15 / 415 ] = 0.28% Ast ( short ) = 0.28 x 1000 x 160 / 100 = 448 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 78.5 x 1000 / 448 = 175.223 mm Provide 10 mm O bar@170 mm c/c = 462

mm2.

( short span )

Mu / b x d ( long ) = 2

10.48 x 10 6 / 1000 x (150)2 = 0.466

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.466 415 / 15 = 50 [(1-0.926) x 15 / 415 ] = 0.134% Ast ( long ) = 0.134 x 1000 x 150 / 100 = 201 mm2 For HYSD Fe415 minimum reinforcement 0.12 % Minimum steel = ( 0.12 /100 ) x 1000 x 180 = 216 mm2. Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 50.24 x 1000 / 216 = 232.593 mm Provide 8 mm O bar@ 230 mm c/c = 218 mm2. ( 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 At top on support , provide 50 % of bars of respective span to take into account negative moment due to slab nature.

Check for development length : Vu = 11.25 x ( 4.5 / 2 ) Long span

= 25.31 KN

Assuming L0 = 8 O (HYSD Fe415 steel ) Pt = 100 x As / b x d = 100 x 218 / 1000 x 150 = 0.145 From equation

OR

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

Mu1 = 0.87 x fy x Ast x d ( 1 - fy x

fy / fck

= 0.87 x 415 x 218 x 150 (

Mu1 / b x d2 = 0.5023 Mu1 = 0.5023 x 1000 x 1502 x 10-6 = 11.30 KNm Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd =56 O (From Table 7-6 ) 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld we get ,

= 11.8063 = 11.3313

1.3 x ( 11.30 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 580.403 + 8 O ≥ 56 O 48 O ≤ 580.403 which gives O ≤ 12.09 mm ……………….( O.K.) Vu = 11.25 x ( 4.5 / 2 ) = 25.31 KN Short span Assuming L0 = 8 O (HYSD Fe415 steel ) Pt = 100 x As / b x d = 100 x 462 / 1000 x 160 = 0.289 From equation Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

OR Mu1 = 0.87 x fy x Ast x d ( 1 - fy x

fy / fck Mu1 / b x d = 0.9595 Mu1 = 0.9595 x 1000 x 1602 x 10-6 = 24.56 KNm Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd =56 O (From Table 7-6 ) 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld we get ,

2

= 0.87 x 415 x 462 x 160 ( = 26.6888 = 24.5539

1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 1261.48 + 8 O ≥ 56 O 48 O ≤ 1261.48 which gives O ≤ 26.28 mm ……………….( O.K.) Note that the bond is usually critical along long direction.

Check for shear : This is critical along long span Vu = 25.31 KN Shear stress = Vu / b x d = 25.31 x 103 / 1000 x 150 2 = 0.169 N / mm2 < ( ‫ﺡ‬C ) N / mm ( too small ) 100 x As / b x d = 100 x 218 / 1000 x 150 = 0.145 from table 7-1 for Pt = 0.145 ‫ﺡ‬c = 0.28 N / mm2 -0.05 25 difference IS 456-2000 clause 40.2.1.1 ? 20 difference k = 1.24 for 180 mm slab depth Design shear strength = 1.24 x 0.28 = 0.347 N / mm2 ……………….( O.K.) OR Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 = 12.0114 Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.01 - 1 ) 6 x 12.011 = 0.278 IS 456-2000 clause 40.2.1.1 -0.05 25 difference k = 1.24 for 180 mm slab depth ? 20 difference Design shear strength = 1.24 x 0.278 = 0.345 N / mm2 ……………….( O.K.) Check for deflection : This check shall be done along short span Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 462 / 1000 x 160 = 0.289 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.44 ( span / d ) ratio permissible = 1.44 x 20 = 28.8 Actual (span / d ) ratio = 4480 / 160 = 28.00 < 28.8 ……………….( O.K.) Check for cracking : IS 456-2000 , clause 26.3.3

-0.04

-0.04

(1) Main bars : maximum spacing permitted for short span steel = = spacing provided = 170 mm < 300 mm (2) Distribu. bars : maximum spacing permitted for long span steel = = spacing provided = 230 mm < 450 mm

3 x effective depth of slab or 300 m 3 x 160 = 480 ……………….( O.K.) 5 x effective depth of slab or 450 m 5 x 150 = 750 ……………….( O.K.)

Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provis

short span coefficient α ( Values of l

Case No. Type of Panel and Moments considered 1.0

1.1

1.2

1.3

0.032 0.024

0.037 0.028

0.043 0.032

0.047 0.036

Negative moment at continuous edge

0.037

0.043

0.048

0.051

Positive moment at mid-span

0.028

0.032

0.036

0.039

Negative moment at continuous edge

0.037

0.044

0.052

0.057

Positive moment at mid-span Two Adjacent Edges Discontinuous:

0.028

0.033

0.039

0.044

Negative moment at continuous edge Positive moment at mid-span

0.047 0.035

0.053 0.040

0.060 0.045

0.065 0.049

Two Short Edges Discontinuous: Negative moment at continuous edge Positive moment at mid-span

0.045 0.035

0.049 0.037

0.052 0.040

0.056 0.043

0.035

0.043

0.051

0.057

0.057

0.064

0.071

0.076

0.043

0.048

0.053

0.057

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

0.043

0.051

0.059

0.065

Four-Edges Discontinuous: Positive moment at mid-span

0.056

0.064

0.072

0.079

1

Interior Panels: Negative moment at continuous edge Positive moment at mid-span

2

One Short Edge Discontinuous:

3

4

5

6

One long Edge Discontinuous:

Two Long Edges Discontinuous: Negative moment at continuous edge Positive moment at mid-span

7

Three Edges Discontinuous (One Long Edge Continuous): Negative moment at continuous edge

8

9

Positive moment at mid-span Three Edges Discrmntinuous

Table - 27 Bending moment coefficients for slabs spanning in two directions at right angles , simp

ly / lx

1.0

1.1

1.2

1.3

1.4

αx

0.062

0.074

0.084

0.093

0.099

αy

0.062

0.061

0.059

0.055

0.051

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 218 x 150 ( 1 - 415 x 218 / 1000 x 150 x 15 ) x 10

-6

( 1 - 0.0402 ) KNm

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10

-6

( 1 - 0.0799 ) KNm

A 350 350

6550

350

460

10 O @ 340 c/c 10 O @ 170 c/c

690

4300

8 O @ 230 c/c

690

4650

8 O @ 460 c/c

460 350 6900 A

PLAN 46010 O @ 340 c/c

460

180 v v v v 10 O @ 170 v v v c/c v 8 O @ 230 c/c 350

4300

Section A-A

v v

v v

350

tive depth of slab or 300 mm whichever is small mm or 300 mm i.e. 300 mm ……….( O.K.) tive depth of slab or 450 mm whichever is small mm or 450 mm i.e. 450 mm ……….( O.K.)

four sides with provision for torsion at corners

ort span coefficient αx ( Values of ly / lx )

Long span coefficient αy for all values of ly / lx

1.4

1.5

1.75

2.0

0.051 0.039

0.053 0.041

0.060 0.045

0.065 0.049

0.032 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 0.053

0.075 0.056

0.084 0.063

0.091 0.069

0.047 0.035

0.059 0.044

0.060 0.045

0.065 0.049

0.069 0.052

0.035

0.063

0.068

0.080

0.088

0.045 0.035

0.080

0.084

0.091

0.097

-

0.060

0.064

0.069

0.073

0.043

0.071

0.076

0.087

0.096

0.057 0.043

0.085

0.089

0.100

0.107

0.056

at right angles , simply supported on Four sides

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 Beam An R.C.C. floor is used as a banking hall Design the beams B10-B11-B12. The Slab thickness is 120 mm . Live load = 3 KN / m2 6m 2 (Given ) Floor finish = 1 KN / m Rib size = 230 mm x 450 mm Column = 300 mm x 300 mm Main beams = 300 mm x 570 mm overall . Material M15 grade concrete HYSD reinforcement of grade Fe415 . Solution : ( a ) Load calculations and analysis : Slab 120 mm thick 0.12 x 25 = 3 + 0 KN / m 2 Floor finish = 1 + 0 KN /m2 Live load = 0 + 3 KN / m2

B10

6m

Total 4 + 3 KN / m2 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 / m Factored Load = 1.5 ( 14.58 + 9 ) = 21.87 + 13.5 say ( 22 + 14 ) KN /m . Case ( a ) Maximum moment at B 36 KN / m

6m

A

22 KN / m

36 KN / m

B

6m

C

6m

D

Using three moment equation for span ABC MA ( L1 / I1 ) + 2MB ( L1 / I1 + L2 / I2 ) + MC ( L2 / I2 ) = - 6 A1a1 / ( I1 L1 ) -6 A2a2 / ( I2L2) 36 KN / m

B

A

36 KN / m

B

162

C 162

A1 = ( 2 / 3 ) x Base x h1 A2 = ( 2 / 3 ) x Base x h1 = ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 162 = 648 = 648 a1 = 3 m a2 = 3 m MA ( 6 / I ) + 2MB ( 6 / I + 6 / I ) + MC ( 6 / I ) = - [ 6 x 648 x 3 / ( I x 6 ) ] - [ 6 x 648 x 3 / ( I x 6) ] 6MA + 24MB + 6MC = - [ 1944 ] - [ 1944 ]

4MB + MC = - 648 As MA = 0 …………………….( 1 ) Using three moment equation for span BCD MB ( L2 / I2 ) + 2MC ( L2 / I2 + L3 / I3 ) + MD ( L3 / I3 ) = - 6 A2a2 / ( I2 L2 ) -6 A3a3 / ( I3L3) 36 KN / m

22 KN / m

C

B

C

D

162

99

A2 = ( 2 / 3 ) x Base x h2 A3 = ( 2 / 3 ) x Base x h3 = ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99 = 648 = 396 a2 = 3 m a3 = 3 m MB ( 6 / I ) + 2MC ( 6 / I + 6 / I ) + MD ( 6 / I ) = - [ 6 x 648 x 3 / ( I x 6 ) ] - [ 6 x 396 x 3 / ( I x 6) ] 6MB + 24MC + 6MD = - [ 1944 ] - [ 1188 ] MB + 4MC = - 522

As MD = 0

4MB + MC = - 648

…………………….( 1 )

…………………….( 2 )

MB + 4MC = - 522

…………………….( 2 ) By putting Value of MB from Equation ( 2 ) into Equation ( 1 ) 4(- 522 - 4MC ) + MC = - 648 - 2088 - 16 MC + MC = - 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

A

6m 162 (+)

22 KN / m

36 KN / m

B 138 (-)

6m

(-)

(-)

D

99 (+)

51

45 (+)

(+)

6m

96 162

93

C

(+) (-)

6 m Distance 3 m Distance 6 m Distance 3 m Distance 6 m Distance

138 KNm (?) 96 KNm (?) 42 KNm

69 48

3 m Distance

96

(?)

21

138 85

MB = - 138 KNm

115 82

2.36 m 50 131

MC = - 96 KNm

101

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

X dist.

VA = 85 KN

6 - X dist.

VA + VB = 36 x 6

131 X =85 ( 6 - X )

VB = 216 - 85

216 X = 510

VB = 131 KN

X = 2.36 m

MB = - 138 KNm

VD x 6 -22 x 6 x 3 = -96

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

VD = 50 KN

Vc = 183 KN

VC + VD = 22 x 6

VC + 82 = 183

VC = 132 - 50

VC = 101 KN

VC = 82 KN

MC = - 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 Three cases are considered for getting maximum values of moments. Case ( a ) gives maximum negative 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 KNm AB ( + ) = CD ( + ) = 110 KNm BC ( + ) = 58 KNm BC ( minimum - ve ) = 5 KNm 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 decrease ) 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 AB or CD ( + ) = 111.6 KNm > 0.7 x 110 KNm BC ( - ) = 11.6 KNm BC ( + ) = 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 Mu ( + ) = 111.6 KNm. The beam acts as a flanged beam

For T-beams , bf = ( l0 / 6 ) + bw + 6 Df bf = ( 0.7 x 6000 / 6 ) + 230 + 6 x 120 ( As per IS456-2000 ,Clause 23.1.2 , Note ) = 1650 mm > 3000 mm 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 mm 2 ( As per IS456-2000 ,Clause 26.5( bf / bw = 1650 / 230 = 7.17 Df / d = 120 / 535 = 0.224

For bf /bw = 7

0.01 diff.

Mu,lim / fck bw d = 0.671 2

( As per SP:16 ,Table 58 ) 0.006 diff. Mu.lim = 0.671 x 15 x 230 x 5352 x 10-6 = 662.597 KNm > 111.6 KNm

If Mu < Mu,lim :

For bf /bw = 8

0.014

0.016

?

?

For 0.224

-0.0084 -0.0096 0.6566

0.743

1 diff 0.0864 0.83 diff ? 0.07171 design as under-reinforced section (singly reinforced beam) as explained below.

Ast = Mu / 0.87 x fy x lever arm where lever arm = d - Df / 2 = 535 - 120 / 2 = 535 - 60 Ast = 111.6 x 106 / 0.87 x 415 x ( 535 - 60 ) = 651

mm2 .

Provide 6 - 12 mm O = 678 mm2 . Span BC : Mu ( + ) = 51.6 KNm Ast = 51.6 x 106 / 0.87 x 415 x ( 535 - 60 ) = 301 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. Continue 3 - 12 O in span BC as required for flexure. Support B or C : Mu ( - ) = 110.4 KNm Mu / bd2 = 110.4 x 106 / 230 x 5352 = 1.68 < 2.07. The section is under-reinforced. Pt = 50

( From Table 6-3 )

1 - √ 1 - ( 4.6 / fck) x ( Mu / bd2 ) fy / fck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.68 ) 415 / 15 = 0.549 Ast = ( 0.549 / 100 ) x 230 x 535 = 676 mm 2 20 % of steel should be carried through the span = 0.2 x 676 = 135 mm 2

603

135

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

3m B11

B12 3m

6m

6m

Case ( b ) Maximum positive moment in span BC 22 KN / m

6m

A

22 KN / m

36 KN / m

B

6m

C

6m

D

Using three moment equation for span ABC MA ( L1 / I1 ) + 2MB ( L1 / I1 + L2 / I2 ) + MC ( L2 / I2 ) = - 6 A 22 KN / m

B

A

36 KN / m

B 162

99

A1 = ( 2 / 3 ) x Base x h1 = ( 2 / 3 ) x 6 x 99 = 396 a1 = 3 m x 3 / ( I x 6) ]

A2 = ( 2 / 3 ) x Base x h = ( 2 / 3 ) x 6 x 162 = 648 a2 = 3 m

MA ( 6 / I ) + 2MB ( 6 / I + 6 / I ) + MC ( 6 / I ) = - [ 6 x 396 x 3 / 6MA + 24MB + 6MC = - [ 1188 ] - [ 1944 ]

4MB + MC = - 522 As MA = 0 Using three moment equation for span BCD MB ( L2 / I2 ) + 2MC ( L2 / I2 + L3 / I3 ) + MD ( L3 / I3 ) = - 6 A 36 KN / m

22 KN / m

C

B

C

162

99

A2 = ( 2 / 3 ) x Base x h2 A3 = ( 2 / 3 ) x Base x h = ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99 = 648 = 396 a2 = 3 m a3 = 3 m x 3 / ( I x 6) ]

MB ( 6 / I ) + 2MC ( 6 / I + 6 / I ) + MD ( 6 / I ) = - [ 6 x 648 x 3 / 6MB + 24MC + 6MD = - [ 1944 ] - [ 1188 ] MB + 4MC = - 522

As MD = 0

4MB + MC = - 522

…………………….( 1 )

MB + 4MC = - 522

…………………….( 2 ) By putting Value of MB from Equation ( 2 ) into Equation ( 1 ) 4(- 522 - 4MC ) + MC = - 522 - 2088 - 16 MC + MC = - 522 15 MC = - 1566 MC = - 104 KNm By putting Value of MC into Equation ( 1 ) 4MB + ( - 104 ) = - 522 4 MB = - 418 MB = - 104 KNm 22 KN / m

6m

A 99

B 104 (-)

6m

162

( 162 - 69 = 93 KNm ) ( 99 - 48 = 51 KNm )

22 KN / m

36 KN / m

C104 (-)

6m 99 (+)

47

58

47

(+)

(+)

(+)

(-)

(-)

D

( 162 - ( 96 + 21 ) = 45 KNm ) 104

85

104

108

48.67

83.33

131 131 X =85 ( 6 - X )

2.21 m 48.67

216 X = 510 X = 2.36 m

83.33

MC = - 104 KNm

108

MB = - 104 KNm

VD x 6 -22 x 6 x 3 = -104

48.67 x 12 + VC x 6 - 36 x 6 x 3

VD = 48.67 KN

Vc = 191.33 KN

VC + VD = 22 x 6

VC + 83.33 = 191.33

VC = 132 - 48.67

VC = 108 KN

VC = 83.33 KN

MC = - 104 KNm

48.67 x 12 + VB x 6 - 36 x 6 x 3 VB = 191.33 KN VB + 83.33 = 191.33 VB = 108 KN

gives maximum mmetry. Case ( b ) mum positive

= 138 KNm. ibute ( increase or

( As per IS 456-2000 , Clause 37.1.1 )

IS 456-2000 ,Clause 37.1.1 Redistribution of mom Continuous Beams and Frames ( b ) The ultimate moment of resistance provided at member is not less than 70 percent of the moment obtained from an elastic maximum moment diagram appropriate combinations of loads.

( c ) The elastic moment at any section in a me particular combination of loads shall not be reduce 30 % of the numerically largest moment given an elastic maximum moments diagram for the partic covering all appropriate combination of loads.

1.2 , Note )

6-2000 ,Clause 26.5( a ) ) For bf /bw = 8

0.671 explained below.

IS 456-2000 ,Clause 23.1.2 Effective width of flan In the absence of more accurate determination , the width of flange may be taken as the following but in greater than the breadth of the web plus half the su distances to the adjacent beams on either side : ( a ) For T-beams , bf = ( l0 / 6 ) + bw + 6 Df ( b ) For L-beams , bf = ( l0 / 12 ) + bw + 3 D Where , bf = effective width of flange ,

l0 = distance between points of zero moments in the bw = breadth of the web , Df = thickness of flange , and b = actual width of the flange.

NOTE - For continuous beams and frames , ' l0 ' ma as 0.7 times the effective span.

IS 456-2000 Clause 26.5 Requirements of Reinfo Structural Members 26.5.1 Beams 26.5.1.1 Tension Reinforcement

a ) Minimum reinforcement - The minimum area o shall not be less than that given by the following : As / b d = 0.85 / fy

where , As = minimum area of tension reinforcement , b = breadth of the beam or the breadth of the web o d = effective depth , and fy = characteristic strength of reinforcement in N / m

b ) Maximum reinforcement - The maximum area reinforcement shall not exceed 0.04 b D. Minimum steel % For mild steel 100 As / b d = 100 x 0.85 / 250 = 0.34 For HYSD steel , Fe415 grade 100 As / b d = 100 x 0.85 / 415 = 0.205 For HYSD steel , Fe500 grade 100 As / b d = 100 x 0.85 / 500 = 0.17

extra at top = 603 mm 2 , remaining 2 - 16 mm O

Table 6-3 Limiting Moment of resistance factor Q lim, N / mm For singly reinforced rectangular sections

fck N 2 / mm

15 20 25 30

fy, N / mm2 250 2.22 2.96 3.70 4.44

415 2.07 2.76 3.45 4.14

500 2.00 2.66 3.33 3.99

t in span BC

Case ( c ) Maximum positive mome 36 KN / m

6m

A

MC ( L2 / I2 ) = - 6 A1a1 / ( I1 L1 ) -6 A2a2 / ( I2L2)

Using three moment equation for s MA ( L1 / I1 ) + 2MB ( L

36 KN / m

C

B

36 KN / m

A 162

/ 3 ) x Base x h1 ( 2 / 3 ) x 6 x 162

A1 = ( 2 / 3 ) x Base x h = ( 2 / 3 ) x 6 x 162 = 648 a1 = 3 m

6 / I ) = - [ 6 x 396 x 3 / ( I x 6 ) ] - [ 6 x 648 x 3 / ( I x 6) ]

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

6MA + 24MB + 6MC = - [ 1944 ] - [ 1

…………………….( 1 )

MD ( L3 / I3 ) = - 6 A2a2 / ( I2 L2 ) -6 A3a3 / ( I3L3)

4MB + MC = - 522 Using three moment equation for s MB ( L2 / I2 ) + 2MC ( L

22 KN / m

22 KN / m

D

B 99

/ 3 ) x Base x h3 ( 2 / 3 ) x 6 x 99

A2 = ( 2 / 3 ) x Base x h = ( 2 / 3 ) x 6 x 99 = 396 a2 = 3 m

6 / I ) = - [ 6 x 648 x 3 / ( I x 6 ) ] - [ 6 x 396 x 3 / ( I x 6) ]

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

88 ]

6MB + 24MC + 6MD = - [ 1188 ] - [ 1 …………………….( 2 )

MB + 4MC = - 522

…………….( 1 )

4MB + MC = - 522

…………….( 2 ) ( 2 ) into Equation ( 1 )

MB + 4MC = - 522 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 36 KN / m

A

6m 162 (+)

6 m Distance 104 KNm 3 m Distance (?) 52 ( 99 - 52 = 47 KNm ) Moment in span BC = ( 162 - 104 ) = 58 KNm

B 104 (-)

110 (+) (-)

104

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

X dist.

48.67

VA = 48.67 KN

6 - X dist.

83.33

VA + VB = 22 x 6

83.33 X = 48.67 ( 6 - X )

VB = 132 - 48.67

132 X = 292.02

VB = 83.33 KN

X = 2.21 m

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

33 = 191.33

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

33 = 191.33

1 Redistribution of moments in rames of resistance provided at any section of a percent of the moment at that section aximum moment diagram covering all

at any section in a member due to a ads shall not be reduced by more than rgest moment given anywhere by the diagram for the particular member , mbination of loads.

90.67 66

2.52 m

125.33

MC = - 104 KNm VD x 6 -36 x 6 x 3 = -104 VD = 90.67 KN VC + VD = 36 x 6 VC = 216 - 90.67 VC = 125.33 KN

2 Effective width of flange urate determination , the effective n as the following but in no case the web plus half the sum of the clear eams on either side : 6 ) + bw + 6 Df 12 ) + bw + 3 Df

s of zero moments in the beam ,

ms and frames , ' l0 ' may be assumed pan.

equirements of Reinforcement for

nt - The minimum area of tension reinforcement

iven by the following :

on reinforcement , the breadth of the web of T- beam ,

of reinforcement in N / mm 2 .

nt - The maximum area of tension eed 0.04 b D.

50 = 0.34

15 = 0.205

00 = 0.17

3

tor Q lim, N / mm2

sections

/ mm2 550 1.94 2.58 3.23 3.87

) Maximum positive moment in span AB and CD 36 KN / m

22 KN / m

6m

C

6m

D

ree moment equation for span ABC I1 ) + 2MB ( L1 / I1 + L2 / I2 ) + MC ( L2 / I2 ) = - 6 A1a1 / ( I1 L1 ) -6 A2a2 / ( I2L2) 36 KN / m

B

22 KN / m

B

C 99

/ 3 ) x Base x h1 A2 = ( 2 / 3 ) x Base x h1 ( 2 / 3 ) x 6 x 162 = ( 2 / 3 ) x 6 x 99 = 396 a2 = 3 m ) + 2MB ( 6 / I + 6 / I ) + MC ( 6 / I ) = - [ 6 x 648 x 3 / ( I x 6 ) ] - [ 6 x 396 x 3 / ( I x 6) ]

4MB + 6MC = - [ 1944 ] - [ 1188 ]

= - 522 As MA = 0 …………………….( 1 ) ree moment equation for span BCD I2 ) + 2MC ( L2 / I2 + L3 / I3 ) + MD ( L3 / I3 ) = - 6 A2a2 / ( I2 L2 ) -6 A3a3 / ( I3L3)

C

22 KN / m

36 KN / m

C

C

D 162

/ 3 ) x Base x h2 ( 2 / 3 ) x 6 x 99

A3 = ( 2 / 3 ) x Base x h3 = ( 2 / 3 ) x 6 x 162 = 648 a3 = 3 m

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

4MC + 6MD = - [ 1188 ] - [ 1944 ]

C

= - 522

As MD = 0

C

= - 522

…………………….( 1 )

…………………….( 2 )

= - 522

…………………….( 2 ) ng Value of MB from Equation ( 2 ) into Equation ( 1 )

C

4MC ) + MC = - 522 16 MC + MC = - 522

104.4 KNm

ng Value of MC into Equation ( 1 )

- 104.4 ) = - 522

04.4 KNm 36 KN / m

22 KN / m

6m 104 (-)

99

C 104

6m

162 D (+)

(-)

110 (+) 5.0

(-)

6 m Distance 104 KNm 3 m Distance (?) 52 ( 162 - 52 = 110 KNm ) Moment in span BC = ( 104 - 99 ) = 5 KNm

104 125.33

MB = - 104 KNm

66 90.67 125.33

36 x 6 x 3 = -104

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

X dist.

VA = 90.67 KN

6 - X dist.

VA + VB = 36 x 6

125.33 X =90.67 ( 6 - X )

VB = 216 - 90.67

216 X = 544.02

VB = 125.33 KN

X = 2.52 m

MB = - 104 KNm 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

90.67 125.33

2 = 110 KNm )

Design of Cantilever Beam Span = 3 m Characteristic U.D.L. = 12 KN /m Assume 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 / m Factored load = 1.5 x 15.45 = 23.2 KN / m Factored S.F. = w x ℓ = 23.2 x 3 = 69.6 KN Factored B.M. = w x ℓ2 / 2 = 23.2 x 32 / 2 = 104.4 KNm = √( 104.4 x 106 ) / ( 2.22 x 230 ) ( From Table 6-3 ) = 452 mm Using one layer of 20 mm O bars and overall depth of 550 mm 104.4 KNm d = 550 - 25 - 10 = 515 mm OR Mu / b d2 == 104.4 x 106 / ( 230 x 5152) Mu,lim = 2.22 x 230 x 5152 x 10-6 = 1.71 < 2.22 ( Table 6-3 ) = 135.424 KNm Mu < Mu,lim The section is singly reinforced ( under-reinforced ) Depth required = √M / Q b

Pt = 50

1 - √ 1 - ( 4.6 / fck) x ( Mu / bd2 ) fy / fck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.71 ) 250 / 15 = 0.93 Ast = ( 0.93 / 100 ) x 230 x 515 = 1102 mm2 Provide 4 - 20 mm O giving Ast = 4 x 314 = 1256 mm2 . Check for development length : Ld = 1102 / 1256 = 55 O ( From Table 7-6 ) IS 456-2000 clause 26.2.1 ( From Table 7-5 ) Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd = O ( 0.87 x 250 x 1102 ) / ( 1256 x 4 x 1.0 ) = 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 :

Vu = 69.6 KN Actual shear strength ‫ﺡ‬v = Vu / bd = 69.6 x 103 / ( 230 x 515 ) = 0.588 N / mm2 100 x As / b d = ( 100 x 1256 ) / ( 230 x 515 ) ( From IS 456-2000 , table 19 table 7-1 ) = 1.06 0.25 difference design ( permissible ) shear strength ‫ﺡ‬c = 0.61 N / mm2 > ‫ﺡ‬v 0.19 difference Provide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 O stirrups = 0.87 Asv fy / 0.4 b = 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/c Check for deflection : Basic span / d ratio = 7 100 Ast / b d = 100 x 1256 / 230 x 515 = 1.06 modification factor = 1.4 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement Span / d permissible = 7 x 1.4 = 9.8 Actual 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 mm Maximum clear distance permitted = 300 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe ) The design beam is shown in fig. 4 -20 O v v v v 550 Fixed end support 1200 6 O @ 130 c/c

2-12 O 3000 Stirrups 6 O @ 130 c/c

v

v 230

Design of Cantilever Beam A cantilever rectangular bracket projects from a column of size 230 mm x 500 mm in the direction of 500 mm for a length of 3 m Factored load of 20 KN / m inclusive of self - weight Material M15 grade concrete HYSD reinforcement of grade Fe415 . Solution : Vu = w x ℓ = 20 x 3 = 60 KN Mu = w x ℓ2 / 2 = 20 x 32 / 2 = 90 KNm ( a ) Moment steel : Take size of Beam 230 mm x 550 mm overall. Assuming 16 mm diameter bars in one layer d = 550 - 25 ( Cover ) -8 OR = 517 mm 2 Mu / b d =90 x 106 / ( 230 x 5172)

Mu,lim =

2.07 x 230 x 5172 x 10-6

= 127.256 KNm

Mu < Mu,lim = 1.46 < 2.07 ( Table 6-3 ) The section is singly reinforced ( under-reinforced ) Beam Pt = 50

1 - √ 1 - ( 4.6 / fck) x ( Mu / bd2 ) fy / fck

Pt = 50 1 - √ 1 - ( 4.6 / 15 ) x ( 1.46 ) 415 / 15 = 0.464 Ast = ( 0.464 / 100 ) x 230 x 517 = 552 mm 2 Provide 16 mm O - 3 No. = 603 mm2. Use 2- 10 mm O as bottom anchor bars . ( b ) Check for development length : IS 456-2000 clause 26.2.1 Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd Stress in bar σs =

0.87 x fy = 0.87 x 415 x 552 / 603 =

330.5

N / mm2

‫ﺡ‬bd = 1.6 x 1 = 1.6 N / mm2 IS 456-2000 clause 26.2.1 ( From Table 7-5 ) Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd = O 330.5 / ( 4 x 1.6 ) = 51.64 O = 51.64 x 16 = 826 mm The arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorage of 1127 mm. The bearing stress inside the bend is now checked . O = 16 mm a = 82 mm for internal bar ( centre to centre distance between bar ) ( = ( 230 - 50 - 3 x 16 ) / 2 = 25 + 16 = 41 mm for external bar . ( = cover + dia of bar ) (

33

The bearing stress is critical in external bar . Check for this stress for a = 41 mm. Design bearing stress = 1.5 x fck / [1 + ( 2O/a ) ] = 1.5 x 15 / [ 1 + ( 2 x 16 / 41 ) ] 550 = 22.5 / 1.78

33

= 12.6404 N / mm2 At 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 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 The arrangement is thus satisfactory. ( c )Check for shea Vu = 60 KN

……………………..( safe )

Actual shear strength ‫ﺡ‬v = Vu / bd = 60 x 103 / ( 230 x 517 ) = 0.505 N / mm2 100 x As / b d = ( 100 x 603 ) / ( 230 x 517 ) ( From IS 456-2000 , table 19 table 7-1 ) = 0.507 0.25 difference design ( permissible ) shear strength ‫ﺡ‬c = 0.46 N / mm2 < ‫ﺡ‬v 0.243 difference shear design is necessary . Vus = Vu - ‫ﺡ‬c b d Vus = ( 0.505 - 0.46 ) x 230 x 517 x10 -3 At support , OR = 60 - 0.46 x 230 x 517 x 10-3 = 5.35 = 60 54.6986 = 5.3 KN Using 6 mm O ( mild steel ) Two legged stirrups , Asv = 28 x 2 = 56 mm2 . fy = 250 N / mm2 Sv = 0.87 fy Asv d / Vus = 0.87 x 250 x 56 x 517 / 5.3 x 103 = 1188 mm Provide minimum shear reinforcement . For 230 mm wide beam , From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 O stirrups = 0.87 Asv fy / 0.4 b = 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

( 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 = 7 100 Ast / b d = 100 x 603 / 230 x 517 = 0.507 modification factor = 1.2 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement Span / d permissible = 7 x 1.2 = 8.4 Actual 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 mm Minimum clear distance permitted = hagg + 5 mm = 20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm . Maximum clear distance permitted = 180 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe ) The design beam is shown in fig.

295

279 r

108 r

50

3-16 O

internal radius r = 180 550

295 150 500 Stirrups

2-10 O 3000 mm 6 mm O @ 130 mm c/c

er self-weight. 23.2 KN /m

3m

69.6 KN

104.4 KNm

-

230 x 5152 x 10-6

IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Pt = 100 Concrete grade x As

19 table 7-1 ) 0.04 ?

-0.0304

x As bxd

M15

M20

M25

M30

M35

≤ 0.15 0.25 0.50

0.28 0.35 0.46

0.28 0.36 0.48

0.29 0.36 0.49

0.29 0.37 0.50

0.29 0.37 0.50

0.75 1.00

0.54 0.60

0.56 0.62

0.57 0.64

0.59 0.66

0.59 0.67

1.25 1.50 1.75 2.00

0.64 0.68 0.71 0.71

0.67 0.72 0.75 0.79

0.70 0.74 0.78 0.82

0.71 0.76 0.80 0.84

0.73 0.78 0.82 0.86

2.25 0.71 0.81 0.85 0.88 0.90 2.50 0.71 0.82 0.88 0.91 0.93 2.75 0.71 0.82 0.90 0.94 0.96 3.00 0.71 0.82 0.92 0.96 0.99 The above given table is based on the following formula Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0

nforcement

………………( safe )

4-20 O v

v

2-12 O

( given )

230 x 5172 x 10-6

an anchorage

( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm ) 33 82 82 33

33 82 82 33

IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends

16 25

The bearing stress in concrete for bends and hooks described in IS : 2502 need not be checked . The bearing stress inside a bend in any other bend calculated as given below : Bearing stress = Fbt / r O Where , Fbt = tensile force due to design loads in a bar or group of bars,

230

r = internal radius of the bend , and O = size of the bar or , in bundle , the size of bar of equivalent area For Limit state method of design , this stress shall not exceed 1.5 f Where, fck is the characteristic strength of concrete and

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

19 table 7-1 ) 0.08 ?

-0.0778

0.505 - 0.46 ) x 230 x 517 x10 -3

KN

nforcement

………………( safe )

3-16 O

2-10 O 230

‫ ﺡ‬, N / mm2 C

M40 0.30 0.38 0.51 0.60 0.68 0.74 0.79 0.84 0.88 0.92 0.95 0.98 1.01 1+5xβ-1)

ss than 1.0

s at bends

ooks described in IS : 2502-1963 de a bend in any other bend shall be

or group of bars,

ar of equivalent area hall not exceed 1.5 fck / [ ( 1+ 2O/a ) ]

or group of bars , for a bar or group hall be taken as the cover plus size of

Design of Simply supported Doubly reinforced Beam Span = 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 Fe415 Solution : DL of beam 0.23 x 0.60 x 25 = 3.45 KN / m super-imposed load = 40 KN / m Total 43.45 KN / m Factored load = 1.5 x 43.45 = 65.18 KN / m Mu = w x ℓ2 / 8 = 65.18 x 52 / 8 = 203.688 KNm Vu = w x ℓ / 2 = 65.18 x 5 / 2 = 162.95 KN ( a ) Moment steel : Assuming 20 mm diameter bars in two layer d = 600 - 25 ( Cover ) - 20 - 10 = 545 mm 2 Mu / b d = 203.69 x 106 / ( 230 x 5452)

OR Mu,lim =

2.07 x 230 x 5452 x 10-6

= 141.414 KNm

Mu > Mu,lim = 2.98 > 2.07 ( Table 6-3 ) The section is Doubly reinforced ( over-reinforced ) Mu,lim = 2.07 x 230 x 5452 x 10-6 = 141.414 KNm Mu2 = Mu - Mu,lim = 203.69 - 141.41 = 62.28 KNm Let the compression reinforcement be provided at an effective cover of 40 mm . d' / d = 40 / 545 = 0.07 consider d' / d = 0.1 . 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 ) = 349 mm2 Corresponding tension steel Ast2 = Asc fsc / 0.87 fy = 349 x 353 / ( 0.87 x 415 ) = 341.2 mm2 Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,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 Ast = Ast,lim + Ast2 .

mm2

= 900 + 341.2 mm2 = 1241.2 mm2 Provide Asc = 2-16 O = 402 mm2 Ast = 4-20 O = 1256 mm2 ( all straight ) ( b ) Check for development length : As all the bars are taken into support , Mu1 may be taken as Mu . Assume L0 = 12 O

Ld = 56 O ( From Table 7-6 )

1.3 x M1 / V + L0 ≥ Ld 1.3 x 203.69 x 106 / 162.95 x 103 +12 O ≥ 56 O 1625.02 ≥ 44 O O ≤ 36.93 mm Oprovided = 20 mm ……………………..( safe ) ( c ) Check for shear Vu = 162.95 KN Actual shear strength ‫ﺡ‬v = Vu / bd = 162.95 x 103 / ( 230 x 545 ) = 1.3 N / mm2 100 x As / b d = ( 100 x 1256 ) / ( 230 x 545 ) ( From IS 456-2000 , table 19 table 7-1 ) = 1.0 design ( permissible ) shear strength ‫ﺡ‬c = 0.6 N / mm2 < ‫ﺡ‬v As the ends are confined with compressive reaction , shear at distance d will be used for checking shear at support . At 545 mm , shear is equal to Vu = 162.95 - 0.545 x 43.45 = 139.27 KN Vus = Vu - ‫ﺡ‬c 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 fy Asv d / Vus = 0.87 x 415 x 100 x 545 / 64.06 x 103 = 307 mm From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 8 mm O stirrups = 0.87 Asv fy / 0.4 b = 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

( 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 = ( 0.87 fy Asv d / Sv )+ ‫ﺡ‬c b d = (0.87 x 415 x 100 x 545 x 10-3 / 390 ) + 0.6 x 230 x 545 x 10-3 = 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 m No. 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 support and 8 mm O @ 390 mm c/c in remaining central portion . ( d ) Check for deflection : Basic span / d ratio = 20 100 Ast / b d = 100 x 1256 / 230 x 545 = 1.0 modification factor = 0.97 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement 100 Asc / b d = 100 x 402 / 230 x 545 = 0.32 modification factor = 1.08 IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcement Span / d permissible = 20 x 0.97 x 1.08 = 20.95 Actual 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 mm Minimum clear distance permitted = 20 + 5 = 25 mm or 20 mm ( O of bar ) i.e. 25 mm . = hagg + 5 mm Maximum clear distance permitted = 180 mm ( cracking - table 8-1 , IS 456-2000 , table 15 ) ………………………( safe ) The design beam is shown in fig. 2-16 O

2-16 O

600

545 4- 20 O ( straight )

600

230

5000 c/c

DIA NO. SPA.

8O 4 300

8O rest 390

600 bearing

8O 4 300

4- 20 O

Table 6-6 STRESS IN COMPRESSION REINFORCEMENT fsc , N / mm2 IN DOUBLY REINFORCED BEAMS d'/d fy N / mm2 0.05 0.1 0.15 0.2 250 217 217 217 217 415 355 353 342 329 500 424 412 395 370 230 x 5452 x 10-6

550

458

441

419

380

If Mu > Mu,lim : design the section either increasing the dimensions of se Over reinforced section ( doubly-reinforced beam ).The additional momen needed is obtained by providing compression( top ) reinforcement and ad reinforcement. Mu2 = Mu - Mu,lim as explained below . Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) ) Asc = Mu2 / ( fsc x ( d - d' ) ) Ast2 = Asc fsc / 0.87 fy Ast = Ast,lim + Ast2 . If Xu < Xu,max the section is under-reinforced ( singly reinforced ) If Xu = Xu,max the section is balanced If Xu > Xu,max the section is over-reinforced ( doubly reinforced ) where , Xu,max = 0.53 x d ( for Fe250 mild steel ) Xu,max = 0.48 x d ( for Fe415 HYSD steel ) Xu,max = 0.46 x d ( for Fe500 HYSD steel ) Xu,max = 0.44 x d ( for Fe550 HYSD steel ) Xu =( 0.87 fy Ast ) / ( 0.36 fck b )

19 table 7-1 )

l be used for

he support

nforcement

on reinforcement

………………( safe )

easing the dimensions of section or deign as am ).The additional moment of resistance M u2 top ) reinforcement and additional tensile low .

ngly reinforced )

ubly reinforced )

Design of Simply supported Singly reinforced Beam Span = 6 m ( simply supported rectangular beam ) characteristic load = 20 KN / m inclusive of its self-weight Beam section = 230 mm x 600 mm Material 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. Mu = w x ℓ2 / 8 = 30 x 62 / 8 = 135 Vu = w x ℓ / 2

KNm

= 30 x 6 / 2 = 90

KN

( a ) Moment steel : Assuming 20 mm diameter bars in one layer d = 600 - 25 ( Cover ) -10 = 565 mm

OR

Mu / b d2 = 135 x 106 / ( 230 x 5652) = 1.84

< 2.07 ( Table 6-3 )

Mu,lim =

2.07 x 230 x 5652 x 10-6

= 151.983 KNm Mu < Mu,lim

The section is singly reinforced ( under-reinforced ) Pt = 50 Pt = 50

1 - √ 1 - ( 4.6 / fck) x ( Mu / bd2 ) fy / fck 1 - √ 1 - ( 4.6 / 15 ) x ( 1.84 ) 415 / 15

= 0.614 Ast = ( 0.614 / 100 ) x 230 x 565 = 798 mm 2 As per IS 456-2000 clause 26.5.1.1 ( a ) Minimum steel , As = ( 0.205 / 100 ) x 230 x 565 = 266 mm 2 From Table 6-4 Ast,lim = 0.72 / 100 x 230 x 565 = 936 mm2. Provide 16 mm O - 4 No. = 804 mm2. 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 : ( 1 ) A bar can be bent up at a distance greater than L d = 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 ) ( 2 ) For the remaining bars , Ast = 402 mm2 Mu1 = 0.87 fy Ast d ( 1 - [( fy Ast ) / ( fck b d ) ] ) = = = Vu =

0.87 x 415 x 402 x 565 ( 1 - [ ( 415 x 402 ) / ( 15 x 230 x 565 ) ] ) x 10 -6 82.01 x 0.91 74.99 KNm L0 = 12 O ( assume ) 90 KN ,

As the reinforcement is confined by compressive reaction 1.3 x M1 / V + L0 ≥ Ld 1.3 x 74.99 x 106 / 90 x 103 +12 O ≥ 56 O 1083.19 ≥ 44 O O ≤ 24.62 mm Oprovided = 16 mm ……………………..( safe ) ( c ) Check for shear : At support , Vu = 90 KN As the ends of the reinforcement are confined with compressive reaction , shear at distance d will be used for checking shear at support. Vu = 90 - 0.565 x w = 90 - 0.565 x 30 = 73.05 KN Actual shear strength ‫ﺡ‬v = Vu / bd = 73.05 x 103 / ( 230 x 565 ) = 0.562 N / mm2 100 x As / b d = ( 100 x 402 ) / ( 230 x 565 ) ( From IS 456-2000 , table 19 table 7-1 ) = 0.309 0.25 difference design ( permissible ) shear strength ‫ = ﺡ‬0.376 N / mm2 < ‫ﺡ‬v 0.191 difference c

shear design is necessary . Vus = Vu - ‫ﺡ‬c b d At support ,

OR

= 73.05 - 0.376 x 230 x 565 x 10-3 = 73.05 48.86 = 24.19 KN Capacity of bent bars to resist shear ( 0.87 fy Asv sin α ) = 2 x 201 x 0.87 x 415 x sin 45º x 10-3 = 102.6 KN Bent bars share 50 % = 12.09 KN Stirrups share 50 % = 12.09 KN Using 6 mm O ( mild steel ) Two legged stirrups , Asv = 28 x 2 = 56 mm2 . Sv = 0.87 fy Asv d / Vus = 0.87 x 250 x 56 x 565 / 12.09 x 103 = 569.2 mm

Vus = ( 0.562 - 0.376 ) x 230 x 565

= 24.17

From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 O stirrups = 0.87 Asv fy / 0.4 b = 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 565 = 423 mm ( iii ) ≤132.4 mm ( minimum ) ( iv ) 569.2 mm ( designed ) Provide 6 mm O two-legged stirrups @ 130 mm c/c At 1.25 D = 750 mm from face of the support where contribution of bent bars is not available Vu = 90 - 0.75 x w = 90 - 0.75 x 30 = 67.5 KN Vus = Vu - ‫ﺡ‬c b d = 67.5 - 0.376 x 230 x 565 x 10-3 = 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 = 20 100 Ast / b d = 100 x 804 / 230 x 565 = 0.62 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 = hagg + 5 mm = 20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm . Maximum clear distance permitted = 180 mm

( cracking - table 8-1 , IS 456-200 , table 15 )

………………………( safe )

The design beam is shown in fig. 2-10 O

2-10 O

600

565 4- 16 O ( 2 straight + 2 bent )

750

230 6000 c/c

4- 16 O

6 mm O @ 130 mm c/c

Consider width of the beam equal to 230 mm. The depth may be assumed as 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 d = √ Mu / Qlim b This is a balanced section and steel area may be found out from table P

( 2 ) For a given factored moment ,width and depth of section . Obtain Mu,lim = Qlim bd2 . If Mu < Mu,lim :

design as under-reinforced section (singly reinforced beam) as exp

Pt = 50 If Mu = Mu,lim :

230 x 5652 x 10-6

1 - √ 1 - ( 4.6 / fck) x ( Mu / bd2 ) fy / fck

design as balanced section as explained in ( 1 ).

If Mu > Mu,lim : design the section either increasing the dimensions of section or d Over reinforced section ( doubly-reinforced beam ).The additional moment of resist needed is obtained by providing compression( top ) reinforcement and additional te reinforcement. Mu2 = Mu - Mu,lim as explained below . Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) ) Asc = Mu2 / ( fsc x ( d - d' ) ) Ast2 = Asc fsc / 0.87 fy Ast = Ast,lim + Ast2 . If Xu < Xu,max the section is under-reinforced ( singly reinforced ) If Xu = Xu,max the section is balanced If Xu > Xu,max the section is over-reinforced ( doubly reinforced ) where , Xu,max = 0.53 x d ( for Fe250 mild steel ) Xu,max = 0.48 x d ( for Fe415 HYSD steel ) Xu,max = 0.46 x d ( for Fe500 HYSD steel ) Xu,max = 0.44 x d ( for Fe550 HYSD steel ) Xu =( 0.87 fy Ast ) / ( 0.36 fck b ) Table 6-2 Limiting Moment of Resistance and Reinforcement

Index for Singly Reinforced Rectangular Sections

fy , N / mm2

250

415

500

550

Mu,lim / fck b d2

0.148

0.138

0.133

0.129

Pt,lim fy / fck

21.93

19.86

19.03

18.2

Table 6-3 Limiting Moment of resistance factor Q lim, N / mm2 For singly reinforced rectangular sections

fck N / mm2

fy, N / mm2

15 20 25

250 2.22 2.96 3.70

415 2.07 2.76 3.45

500 2.00 2.66 3.33

550 1.94 2.58 3.23

30

4.44

4.14

3.99

3.87

Table 6-4 Limiting Percentage of Reinforcement Pt,lim

at distance d

For singly reinforced rectangular sections fck N 2 / mm

19 table 7-1 ) 0.11 ?

fy, N / mm2

15 20

250 1.32 1.75

415 0.72 0.96

500 0.57 0.76

550 0.50 0.66

25 30

2.19 2.63

1.20 1.44

0.95 1.14

0.83 0.99

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

0.562 - 0.376 ) x 230 x 565 x10

KN

-3

Structural Members 26.5.1 Beams 26.5.1.1 Tension Reinforcement a ) Minimum reinforcement - The minimum area of tension reinforcement shall not be less than that given by the following : As / b d = 0.85 / fy where , As = minimum area of tension reinforcement , b = breadth of the beam or the breadth of the web of T- beam , d = effective depth , and fy = characteristic strength of reinforcement in N / mm 2 . b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.

not available

Minimum steel % For mild steel 100 As / b d = 100 x 0.85 / 250 = 0.34 For HYSD steel , Fe415 grade 100 As / b d = 100 x 0.85 / 415 = 0.205 For HYSD steel , Fe500 grade 100 As / b d = 100 x 0.85 / 500 = 0.17

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.

check for development length IS 456-2000 clause 26.2.1 Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd IS 456-200 clause 26.2.1.1 Table 7-5 Design bond stress (‫ﺡ‬bd ) for plain bars in tension

orcement

………………( safe )

Concrete grade (

M15

M20

M25

M30

M35

M40

‫ﺡ‬bd N / mm 1.0

1.2 1.4 1.5 1.7 1.9 Note-1 : ‫ﺡ‬bd shall be increased by 25 % for bars in compression Note-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ‫ﺡ‬bd shall be increased by 60 %. For mild steel Fe250

σs = 0.87 x fy

σs = 0.67 x fy For Fe415 IS 456-2000 clause 26.2.3.3 Ld ≤ M1 / V + L0 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 ) Table 7-6 Development length for single mild steel bars

fy N / mm2 250 415

Tension bars M15 M20 55 O 26 O 56 O 47 O

Compression bars

M15 44 O 45 O

M20 37 O 38 O

500

69 O

58 O

54 O

46 O

check for shear IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Concrete grade M15 M20 M25 M30 M35 M40 ≤ 0.15 0.28 0.28 0.29 0.29 0.29 0.30 0.25 0.35 0.36 0.36 0.37 0.37 0.38 0.50 0.46 0.48 0.49 0.50 0.50 0.51 0.75 0.54 0.56 0.57 0.59 0.59 0.60 1.00 0.60 0.62 0.64 0.66 0.67 0.68 1.25 0.64 0.67 0.70 0.71 0.73 0.74 1.50 0.68 0.72 0.74 0.76 0.78 0.79 1.75 0.71 0.75 0.78 0.80 0.82 0.84 2.00 0.71 0.79 0.82 0.84 0.86 0.88 2.25 0.71 0.81 0.85 0.88 0.90 0.92 2.50 0.71 0.82 0.88 0.91 0.93 0.95 2.75 0.71 0.82 0.90 0.94 0.96 0.98 3.00 0.71 0.82 0.92 0.96 0.99 1.01 The above given table is based on the following formula Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0

Pt = 100 x As bxd

IS 456-2000 , Table 20 Table 7-2 Maximum shear stress , ‫ﺡ‬C, N / mm2 Concrete grade

( ‫ﺡ‬c )max N/mm

2

M15 2.5

M20 2.8

M25 3.1

M30 3.5

M35 3.7

M40 4.0

IS 456-2000 Clause 26.5.1.6 Minimum shear reinforcement Minimum shear reinforcement in the form of stirrups shall be provided Asv / b sv ≥ 0.4 / 0.87 fy such that : where, Asv = total cross-sectional area of stirrup legs effective in shear , Sv = stirrup spacing along the length of the member , b = breadth of the beam or breadth of the web of flanged beam , and fy = characteristic strength of the stirrup reinforcement in N / mm 2 which shall not be taken greater than 415 N / mm 2 .

check for deflection Basic values of span to effective depth ratios for spans upto 10 m : cantilever 7 simply supported 20 continuous 26 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. check for cracking IS 456-2000 26.3.3 Maximum distance between bars in tension

Table 15 Clear distance Between Bars ( Clause 26.3.3 ) % redistribution to or from section considered

fy

-30

N / mm2 250 415 500

mm 215 125 105

-15 0 +15 +30 Clear distance between bars mm 260 155 130

mm 300 180 150

mm 300 210 175

mm 300 235 195

may be assumed

ed moment )

out from table P t,lim , SP : 16 ,2.3

gly reinforced beam) as explained below.

d in ( 1 ).

dimensions of section or deign as additional moment of resistance M u2 orcement and additional tensile

sion reinforcement

nsion

is greater

ll be provided

ve in shear ,

anged beam , and

ent in N / mm 2

Design of slender ( Long ) columns ( with biaxial bending ) Size of column 400 x 300 mm Column is restrained against sway. Concrete grade M 30 Characteristic strength of reinforcement 415 N/mm 2 Effective length for bending parallel to larger dimension ℓ ex = 6.0 m Effective length for bending parallel to shorter dimension ℓ ey = 5.0 m 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 : β1 and β2 are the same For Beam : bf = l 0 / 6 + b w + 6 D f = 0.7 x 5000 / 6 + 230 + 6 x 120 = 1533.33333 bf / bw = 1533.3 / 230 = 6.67 Df / D = 120 / 600 = 0.2 Kt from chart 88 , SP : 16 = 2.07 Beam stiffness Kb = 1.5 x Ib / l = 1.5 x ( 2.07 x ( 1 / 12 ) x 230 x 6203 ) / 5000 = 2836699 mm3 Column stiffness Kc = Ic / l = 1/12 x 230 x 4003 / 7000 = 175238.095 mm3 β1 = β2 = ∑ Kc / ( ∑Kc + ∑Kb ) = 2 x 175238 / 2 ( 175238 + 2836699 ) = 0.0582 as per IS 456-2000 fig. 27 lef / l = 1.035 < 1.2 ……………….consider 1.2 lex = 1.2 x unsupported length = 1.2 ( 7000 - 620 ) = 7656 mm lex / D = 7656/400 = 19.14 > 12 The column is long about x direction. About Y axis :

Beam stiffness Kb = 1.5 x Ib / l = 1.5 x ( ( 1 / 12 ) x 230 x 4203 ) / 5000 = 426006 mm3 Column stiffness Kc = Ic / l = 1/12 x 400 x 2303 / 7000 = 57938.0952 mm3 β1 = β2 = ∑ Kc / ( ∑Kc + ∑Kb ) = 2 x 57938.1 / 2 ( 57938.1 + 426006 ) = 0.1197 as per IS 456-2000 fig. 27 lef / l = 1.06 < 1.2 ……………….consider 1.2 ley = 1.2 x unsupported length = 1.2 ( 7000 -420 ) = 7896 mm ley / b = 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. ℓex / D = 6000 / 400 = 15.0 > 12 ℓey / b = 5000 / 300 = 16.7 > 12 Therefore the column is slender about both the axes. Additional moments Max = ( Pu D / 2000 ) x ( ℓex / D )2 = ( 1500 x 400 / 2000 ) x (15)2 x 10-3 = 67.5 KNm May = ( Pu b / 2000 ) x ( ℓey / b )2 = ( 1500 x 300 / 2000 ) x (16.7)2 x 10-3 = 62.75 KNm The above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000 but 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 ). Ag = 400 x 300 Puz = 0.45 fck Ac + 0.75 f = 120000 mm2 From chart 63 , puz / Ag = 22.5 N/mm2 Puz = 22.5 x 120000 x 10-3 = 2700 KN Calculation of Pb :

Ac = 400 x 300 - 120000*3/100 OR

= 116400 mm2 Puz = 0.45 x 30 x 116400 x 10-3 + 0.75 x 4 = 1571.4 + = 2692

KN

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 Pb ( about xx-axis ) = ( k1 + k2 p / fck ) fck b D Pbx = ( 0.196 + 0.203 x 3 /30 ) 30 x 300 x 400 x10-3 = 779 KN Pb ( about yy-axis ) = ( k1 + k2 p / fck ) fck b D Pby = ( 0.184 + 0.028 x 3 /30 ) 30 x 300 x 400 x10-3 = 672 KN Kx = ( Puz - Pu ) / ( Puz - Pbx ) = ( 2700 - 1500 ) / ( 2700 - 779 ) = 1200 / 1921 = 0.625 Ky = ( Puz - Pu ) / ( Puz - Pby ) = ( 2700 - 1500 ) / ( 2700 - 672 ) = 1200 / 2028 = 0.592 The additional moments calculated earlier , will now be multiplied by the above values of k . Max = 67.5 x 0.625 = 42.2 KNm May = 62.75 x 0.592 = 37.15 KNm The 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 ) Mux = ( 0.6 x 40 - 0.4 x 22.5 ) = 15.0 KNm Muy = ( 0.6 x 30 - 0.4 x 20 ) = 10.0 KNm The 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. ex = ( ℓ / 500 ) + ( D / 30 ) =

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

ey = ( ℓ / 500 ) + ( b / 30 ) = ( 7000 / 500 ) + ( 300 / 30) = 24 Moments due to minimum eccentricity : 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 Total moments for which the column is to be designed are : Mux = 41.0 + 42.2 = 83.2 KNm Muy = 36.0 + 37.15 = 73.15 KNm The section is to be checked for biaxial bending Pu / fck b D = 1500 x 103 / ( 30 x 300 x 400 )

= 0.417 p / fck = 3 / 30 = 0.10 referring to chart 45 (d' / D = 0.15 ) , Mu / fck b D2 = 0.104 Mux1 = 0.104 x 30 x 300 x 4002 = 149.8

KNm

referring to chart 46 (d' / D = 0.2 ) , Mu / fck b D2 = 0.096 Muy1 = 0.096 x 30 x 400 x 3002 = 103.7 KNm Mux / Mux1 = 83.2 / 149.8 =

0.56

Muy / Muy1 = 73.15 / 103.7 = 0.71 Pu / Puz = 1500 / 2700 = 0.56 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. for Pu / Puz = 0.56 , αn = 1.602 From , IS 456-2000 , Clause 39.6 αn = 1.602 check : Mux

αn

+

Mux1 1.602 ( 0.56 ) 0.395 = 0.972

+ +

≤1

Muy Muy1

αn

0.2 difference

0.34

0.04 difference

?

≤1

1.602 ( 0.71 ) 0.577 ……………………..( O.K.)

As = p x b x D / 100 = 3.0 x 300 x 400 / 100 = 3600 mm2 Provide 25 mm diameter bar ( 3600 / 491 ) = 8 no. = 3928 mm 2 Ties : O min = 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 300

500

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

300

Note that the distance between corner bars in one face is more than 48 O tr ( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used . Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient. Design of slender ( Long ) columns ( with Uniaxial bending ) Size of column 230 x 450 mm Column of a braced frame Concrete grade M 20 Characteristic strength of reinforcement 415 N/mm 2 HYSD reinforcement Unsupported length in both the direction = 5.0 m Factored load Pu = 1000 kN Factored.moment in the direction of larger dimension M uxx = 80 kNm at top and 60 KNm at bottom. Factored.moment in the direction of shorter dimension M uyy = 40 kNm at top and 30 KNm at bottom. The column is bent in double curvature and is slender about both the axis. 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. Solution : Assume adjustment factor k = 0.8 for the first trial . Additional moments Max = ( Pu D / 2000 ) x ( ℓex / D )2 = ( 1000 x 450 / 2000 ) x (13.2 )2 x 10-3 = 39.2 KNm May = ( Pu b / 2000 ) x ( ℓey / b )2 = ( 1000 x 230 / 2000 ) x (15.6 )2 x 10-3 = 28 KNm About XX Pu = 1000 KN Muxx = Mi = = = =

Mi + k x Max 0.6 Mu2 + 0.4 Mu1 0.6 x 80 - 0.4 x 60 48 - 24 24 KNm < 0.4 x 80 = 32 KNm Mi = 32 KNm

Take , Note that Mu1 is considered negative as the column bends in double curvature . Mu,xx = 32 + 0.8 x 39.2 = 63.36 KNm.

< 80 KNm

Mu,xx = 80 KNm Take , About YY Pu = 1000 KN Muyy = Mi + k x May Mi = = = =

0.6 Mu2 + 0.4 Mu1 0.6 x 40 - 0.4 x 30 24 - 12 12 KNm < 0.4 x 40 = 16 KNm Mi = 16 KNm

Take , Mu,yy = 16 + 0.8 x 28 = 38.4 KNm. Mu,yy = 40 KNm Take , Finally design the column for Pu = 1000 KN

< 40 KNm

Mu,xx = 80 KNm Mu,yy = 40 KNm For the first trial , assume uniaxial bending about y axis for the following values . P'u = 1000 KN M'uy = ( 230 / 450 ) ( 80 + 40 ) = 61.33 KNm d' / D = 50 / 230 = 0.22 Say 0.2 P'u / fck 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 = 0.129 p / fck = from chart 46 , SP : 16 = 0.162 p = 0.162 x 20 = 3.24 Asc = ( 3.24 /100 ) x 230 x 450 = 3353 mm2 Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel are usually conservative . Let us try 4 - 25 O + 4 - 20 O = 3220 mm 2 Now check the assumed section as follows : Puz = 0.45 fck Ac + 0.75 fy Asc Ac = 230 x 450 - 3220 = 100280 mm2 Puz = 0.45 x 20 x 100280 x 10-3 + 0.75 x 415 x 3220 x 10-3 = 902.52 + 1002.225 = 1904.7 KN Pb = ( k1 + k2 p / fck ) fck b D Asc = ( p / 100 ) x b x D p / fck = ( 3220 x 100 ) / ( 230 x 450 x 20 )

= 0.156 For d' / D = 0.2 , k1 = 0.184 and k2 = -0.022 from table 60 ,SP : 16 Pb = ( 0.184 - 0.022 x 0.156 ) x 20 x 230 x 450 x 10-3 = 0.181 x 2070 = 373.8 KN k = ( Puz - Pu ) / ( Puz - Pb ) = ( 1904.7 - 1000 ) / ( 1904.7 - 373.8 ) = 904.7 / 1530.9 = 0.59 Design for Pu = 1000 KN Mux = 32 + 0.59 x 39.2 = 55.13 KNm < 80 Muy = 16 + 0.59 x 28 = 32.52 KNm < 40

Take , Mux = 80 KNm Take , Muy = 40 KNm

For p / fck = 0.162 and Pu / fck b D = 0.48 , the reinforcement being equally distributed , the moment capacities can be found out as follows : About XX d' / D = 50 / 450 = 0.11 ≈ 0.15 From chart 45 , SP : 16 Mux1 / fck b D2 = 0.145 Mux1 = 0.145 x 20 x 230 x 4502 x 10-6 = 135.07 KNm About YY d' / D = 50 / 230 = 0.22 ≈ 0.2 From chart 46 , SP : 16 Muy1 / fck b D2 = 0.13 Muy1 = 0.13 x 20 x 450 x 2302 x 10-6 = 61.89 KNm Check : Pu / Puz = 1000 / 1904.7 = 0.525 From , IS 456-2000 , Clause 39.6 αn = 1.542 check : α Mux n

+

Mux1

Muy Muy1

1.542

0.446 0.956

+ ≤1

≤1

1.542

+

80 135.07

αn

0.2 difference 0.075 difference

40 61.89 0.51 …………….( O.K.)

230

0.34 ?

230

Provide 4 - 25 O + 4 - 20 O equally distributed. Ties : Minimum diameter Otr = 25 / 4 = 6.25 mm Use 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 Note that the distance between corner bars in one face is less than 48 O tr ( 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 sufficient.

)

IS 456-2000 clause 25

Note :- A column may be considered as short when both slender lex / D and ley / b ≤ 12 OR lex / ixx < 40 where, For Circular column 10 . ( Given )

KNm at bottom.

KNm at bottom.

lex = effective length in respect of the major axis D = depth in respect of the major axis ley = effective length in respect of the minor axis b = width of the member ixx = radius of gyration in respect of the major axis. iyy = radius of gyration in respect of the minor axis.

IS 456 : 2000 Table 28 Effective Length of Compression Memb 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.

Symbol

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. ( i + 1 ) th floor

slab

Beam

ℓ

( i ) th floor

ly on four sides.

1 of the IS 456-2000

Ac + 0.75 fy Asc

00 - 120000*3/100

116400 x 10-3 + 0.75 x 415 x 3600 x 10-3 1120.5

Effective length of column ( ℓef ) : It is the distance between the points of zero column height .

SP : 16 ,Table 60 Slender compression members- Values of P Rectangular sections : Pb / fck b D = k1 + k2 . p / f Circular sections : Pb / fck D2 = k1 + k2 . p / fck Values of k1 Section 0.05

d' / D 0.10 0.15

0.219

0.207

0.196

0.172

0.160

0.149

250 415

0.05 -0.045 0.096

d' / D 0.10 -0.045 0.082

Rectangular ; equal reinforcement on four sides

500 250 415 500 250

0.213 0.215 0.424 0.545 0.193

0.173 0.146 0.328 0.425 0.148

Circular

415

0.410

0.323

500

0.543

0.443

Rectangular Circular Values of k2 Section

fy N/ mm

Rectangular ; equal reinforcement on two opposite sides

values of k .

al moments

2

IS 456-2000 clause 39.7.1 The additional moments Max and May shall be calculated by the following f

ode IS 456-2000 )

Max = ( Pu D / 2000 ) x ( ℓex / D )2 May = ( Pu b / 2000 ) x ( ℓey / b )2 Where , Pu = axial load on the member,

nimum eccentricity tial moment for mm

> 20 mm

mm

> 20 mm

ℓex = effective length in respect of the major axis , ℓey = effective length in respect of the minor axis , D = depth of the cross - section at right angles to the major axis , and 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 in that plane. It should otherwise be considered unbraced. 2 ) In the case of a braced column without any transverse loads

occurring in its height, the additional moment shall be added 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 end moment (assumed negative if the column is bent in double curvature). In no case shall the initial moment be less than 0.4 Mu2 nor the total moment including the initial moment be less than Mu2. For unbraced columns, the additional moment 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 39.7.1.1 The values given by equation 39.7.1 may be multiplied by the following fa K = ( Puz - Pu ) / ( Puz - Pb ) ≤1 where , Pu = axial load on compression member, Puz = as defined in 39.6, Puz = 0.45 fck Ac + 0.75 f -0.068

Pb = axial load corresponding to the condition of maximum compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outer most layer of tension steel.

8-25 O

8O@ 300 c/c (two sets )

used . l bars would be

g)

60 KNm at bottom. ( Given )

d 30 KNm at

re less than

al steel are

-0.1275

4-25 O + 4 -20 O 450

l diameter of bar )

e used . rnal bars would be

8 O @ 230 mm c/c ( three sets )

as short when both slenderness ratios where, For Circular column lex / D ≤

major axis

minor axis

e major axis.

e minor axis.

th of Compression Members ( Clause E-3 ) Symbol

Theoretical Value of effective length

Recommended Value of effective length

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

2.0 ℓ

2.0 ℓ

f compression member. slab

Beam

ance between the points of zero moment (contraflexure ) along the

s- Values of Pb = k1 + k2 . p / fck

k1 + k2 . p / fck /D 0.20 0.184 0.138 d' / D 0.15 -0.045 0.046

0.20 -0.045 -0.022

0.104 0.061 0.203 0.256 0.077

-0.001 -0.011 0.028 0.040 -0.020

0.201

0.036

0.291

0.056

alculated by the following formulae :

f the major axis ,

f the minor axis , at right angles to

en plane if lateral

by walls or

eral forces in nbraced.

transverse loads

hall be added nd 0.6 Mu2, the smaller

is bent in double

be less than al moment be

ional moment

ven level or storey, o deflect

ch column may

in the same

ultiplied by the following factor :

5 fck Ac + 0.75 fy Asc

Design of short eccentrically loaded square columns - Biaxial bending. Size 500 mm x 500 mm Axial factored load 1500 KN Factored moment Mux = 90 KNm ( Given ) Muy = 120 Knm moment due to minimum eccentricity is less than the applied moment. Material M15 grade concrete HYSD reinforcement of grade Fe415 Solution :Assume an axial load P'u of 1500 KN and a uniaxial moment M'ux = 90 + 120 = 210 KNm . P'u / ( fck x b x D ) = 1500 x 103 / ( 15 x 500 x 500 ) = 0.4 M'ux / ( fck x b x D2 ) = 210 x 106 / ( 15 x 500 x 5002 ) = 0.112 d' = 40 + 10 = 50 mm d' / D = 50 / 500 = 0.1 From chart 32 , SP-16 p / fck = 0.078 p = 0.078 x 15 = 1.17 As = 1.17 x b x D = 1.17 x 500 x 500 / 100 = 2925 mm2 Provide 4-25 mm O + 4-20 mm O = 3220 mm 2 , equally distributed p = 3220 x 100 / ( 500 x 500 ) p = 1.288 p / fck = 1.288 / 15 = 0.086 The assumed section is now checked. For p / fck = 0.086 and Pu / ( fck x b x D ) = 0.4 , The reinforcement being equally distributed , the moment capacities from Chart - 44 , SP-16 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 = 202.5 KNm Puz = 0.45 fck Ac + 0.75 fy Asc Ac = 500 x 500 - 3220 = 246780 mm2 Puz = 0.45 x 15 x 246780 x 10-3 + 0.75 x 415 x 3220 x 10-3 = 1665.77 = 2667.99 KN

+

1002.23

Pu / Puz = 1500 / 2668 = 0.56 From , IS 456-2000 , Clause 39.6 αn = 1.602 check : αn Mux Muy Mux1 90 202.5 0.273

+

1.602

+ +

Muy1 120 202.5

αn

0.2 difference 0.04 difference

0.34 ?

-0.068

≤1

1.602

0.433

0.706 ≤1 Ties : Minimum diameter Otr = 25 / 4 = 6.25 mm Use 8 mm diameter M.S. ties. Spacing should not exceed lesser of ( i ) 500 mm ( ii ) 16 x 25 = 400 mm 500 ( iii ) 48 x 8 = 384 mm Provide 8 mm O M.S. ties @ 350 mm c/c

500 4-25 O + 4-20 O 8O@ 350 c/c

Design of short eccentrically loaded rectangle columns - Biaxial bending. Size 300 mm x 500 mm Axial factored load 1500 KN Factored moment Mux = 60 KNm ( Given ) Muy = 60 KNm moment due to minimum eccentricity is less than the applied moment. Material M15 grade concrete HYSD reinforcement of grade Fe415 Solution :Assume an axial load P'u of 1500 KN and a uniaxial moment M'uy = 300 / 500 ( 60 + 60 ) = 72 KNm . P'u / ( fck x b x D ) = = 2 M'u / ( fck x b x D ) = = d' = 40 + 10 = 50 mm d' / D = 50 / 300 = 0.167 From chart 34 , SP-16 p / fck = 0.167 p = 0.167 x 15 = 2.5

1500 x 103 / ( 15 x 300 x 500 ) 0.67 72 x 106 / ( 15 x 500 x 3002 ) 0.107 ………………..use 0.2

As = 2.5 x b x D = 2.5 x 300 x 500 / 100 = 3750 mm2 Provide 8-25 mm O = 3928 mm2 , equally distributed p = 3928 x 100 / ( 300 x 500 ) p = 2.62 p / fck = 2.62 / 15 = 0.175 The assumed section is now checked. About X d' / D = 50 / 500 = 0.1 , For p / fck = 0.175 and Pu / ( fck x b x D ) = 0.67 , Chart - 44 , SP-16 Mux1 / ( fck x b x D2 ) = 0.13 Mux1 = 0.13 x 15 x 300 x 5002 x 10-6 = 146.25 KNm About Y d' / D = 50 / 300 = 0.167 say 0.2 , For p / fck = 0.175 and Pu / ( fck x b x D ) = 0.67 , Chart - 46 , SP-16 Muy1 / ( fck x b x D2 ) = 0.103 Muy1 = 0.103 x 15 x 500 x 3002 x 10-6 = 69.53 KNm Pure axial load capacity Puz = 0.45 fck Ac + 0.75 fy Asc Ac = 300 x 500 - 3928 = 146072 mm2 Puz = 0.45 x 15 x 146072 x 10-3 + 0.75 x 415 x 3928 x 10-3 = 985.986 + 1222.59 = 2208.58 KN Pu / Puz = 1500 / 2208.6 = 0.68 From , IS 456-2000 , Clause 39.6 αn = 1.802 0.2 difference check : 0.12 difference αn αn Mux Muy 1 M M ux1

+

uy1

1.802 60 146.25 0.2

+ +

≤

1.802 60 69.53 0.77

0.33 ?

-0.198

0.97 ≤1 Ties : Minimum diameter Otr = 25 / 4 = 6.25 mm Use 8 mm diameter M.S. ties. Spacing should not exceed lesser of ( i ) 300 mm ( ii ) 16 x 25 = 400 mm 300 ( iii ) 48 x 8 = 384 mm Provide 8 mm O M.S. ties @ 300 mm c/c

500

Note that the distance between corner bars in one face is more than 48 O tr ( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used . Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.

0 = 210 KNm .

39.6 Members Subjected to Combined Axial Load and Biaxial Bending where , Mux

αn

Mux1

+

Muy Muy1

M

αn

≤1

Mux1, Muy1 =

αn is related to Pu/Puz where Puz = 0.45 fck Ac + 0.75 fy Asc For values of Pu / Puz = 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

1 .O; for values greater than 0.8, αn is 2.0.

4-25 O + 4-20 O

( least lateral dimension ) ( 16 times least longitudinal diameter of bar ) ( 48 times diameter of tie )

8O@ 350 c/c

0 ( 60 + 60 ) = 72 KNm .

8-25 O

( least lateral dimension ) ( 16 times least longitudinal diameter of bar ) ( 48 times diameter of tie )

be used .

8O@ 300 c/c (two sets )

Mux , Muy = moments about x and y axes due to design loads, Mux1, Muy1 = maximum uniaxial moment capacity for an axial load of Pu, bending about x and y axes respectively, and

.6 , 0.8 )the values of αn vary

) For values less than 0.2, αn is

Design of short eccentrically loaded columns - uniaxial bending. Size 300 mm x 600 mm Axial factored load 600 KN Factored moment 300 KNm ( Given ) moment due to minimum eccentricity is less than the applied moment. Material M15 grade concrete HYSD reinforcement of grade Fe415 Solution :Using 25 mm diameter bars with 40 mm clear cover d' = 40 + 12.5 = 52.5 mm d' / D = 52.5 / 600 = 0.088 ,say 0.1 Pu / ( fck x b x D ) = 600 x 103 / ( 15 x 300 x 600 ) = 0.222 2 Mu / ( fck x b x D ) = 300 x 106 / ( 15 x 300 x 6002 ) = 0.185 From chart 32 , SP : 16 p / fck = 0.1 p = 0.1 x 15 = 1.5 As = 1.5 x b x D = 1.5 x 300 x 600 / 100 = 2700 mm2 Provide 22 mm O ( 2700 / 380 = 8 no.) = 3041 mm 2 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. Provide 8- 25 O + 2 - 12 O = 4151 mm2 600 Ties : Minimum diameter Otr = 25 / 4 = 6.25 mm Use 8 mm diameter M.S. ties. Spacing should not exceed lesser of ( i ) 300 mm 300 ( ii ) 16 x 25 = 400 mm ( iii ) 48 x 8 = 384 mm Provide 8 mm O M.S. ties @ 300 mm c/c ( two sets ) Note that the distance between corner bars in one face is more than 48 O tr ( 600 - 80 -25 = 495 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used . Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.

d by open ties.

8-25 O + 2-12 O 8 mm O @ 300 mm c/c ( two sets )

be used .

( least lateral dimension ) ( 16 times least longitudinal diameter of bar ) ( 48 times diameter of tie )

Design of short circular column Working load = 1200 KN Assume emin < 0.05 D ( a ) lateral ties & ( b ) helical reinforcement Material M20 grade concrete HYSD steel Fe415 For Lateral reinforcement mild steel Fe250 Solution : Factored load = 1.5 x 1200 = 1800 KN ( a ) lateral ties : Pu = 0.4 fck Ac + 0.67 fy Asc Assume 0.8 % minimum steel. Asc = 0.008 Ag Then ,

Given

Ac = Ag - Asc = 0.992 Ag Substituting , we have 0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag 1800 x 103 = 1800 x 103 =

7.936

Ag +

1800 x 103 =

10.1604

Ag

2.2244

Ag

Ag = 177165 mm2 If D is the diameter of the column ( ¶ / 4 ) x D2 = 177165 D = √ 225688 D = 475 mm Use 475 mm diameter column. Asc = 0.008 x 177165 = 1417 mm2 Minimum 6 bars shall be used. Provide 16 mm diameter bars ( 1417 / 201 ) = 8 no. Asc = 8 x 201 = 1608 mm2 Use 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 similar member with lateral ties. Therefore Pu = 1.05 [ 0.4 fck Ac + 0.67 fy Asc ]

1800 x 103 = 1.05 [0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag ] Ag + 1800 x 103 = 8.3328 2.33562 Ag 1800 x 103 = 10.66842 Ag Ag = 168729 mm2 If D is the diameter of the column ( ¶ / 4 ) x D2 = 168729 D = √ 214941 D = 463 mm Use 450 mm diameter column. Ag = ( ¶ / 4 ) x 4502 Then , = 158962.5 mm2 1800 x 103 = 1.05 [0.4 x 20 x (158963 - Asc ) + 0.67 x 415 x Asc ] = 1335289

-

8.4

Asc

+

Asc = 283.55 Asc = 1638.903 mm2 provide 20 mm diameter bars ( 1638.9 / 314 =) 6 No. Asc = 6 x 314 464711

= 1884 mm2 Assume 8 mm O M.S. bars for helix at 40 mm clear cover . Dc = 450 - 40 - 40 = 370 mm asp = ( ¶ / 4 ) x 82 = 50 mm2 Minimum ρs = 0.36 ( (Ag / Acr) - 1 ) fck / fy = 0.36 ( ( 4502 / 3702 ) - 1 ) x 20 / 250 = 0.36 x 0.4792 x 20 / 250 = 0.013801 ρs = 4 x asp / p x Dc Now 0.0138 = 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 < Dc / 6 ( = 370 / 6 = 61.67 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.

291.953 Asc

al diameter of bar )

6 - 20 O

450

6 - 20 O

35 8 mm O @ 35 mm c/c

Design of short column Factored load = 1500 KN Assume emin < 0.05 D

Given

Material M15 grade concrete mild steel Fe250 Solution : Here , emin < 0.05 D , But emin = 20 mm

Therefore , size of column shall be minimum ( 20 / 0.05 = 400 )

400 mm x 400 mm

Assume 0.8 % minimum steel. Asc = 0.008 Ag Then , Ac = Ag - Asc = 0.992 Ag Pu = 0.4 fck Ac + 0.67 fy Asc = 0.4 x 15 x 0.992 Ag + 0.67 x 250 x 0.008 Ag Ag 5.952 Ag + 1.34 1500 x 103 = 1500 x 103 = 7.292 Ag Ag =

205705 mm2

If the column is to be a square , the side of column = √205705 = 453 mm Adopt 450 mm x 450 mm size column . Then , 1500 x 103 = 0.4 x 15 x ( 450 x 450 - Asc )+ 0.67 x 250 x Asc Asc + 167.5 1500 x 103 = 1215000 - 6 Asc Asc 285000 = 161.5 Asc = 1765 mm2 Provide 20 mm diameter bars = ( 1765 / 314 ) = 6 No. giving , Asc = 6 x 314 = 1884 mm2 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 Asc = 4 x 314 + 4 x 201 = 1256

+

804

= 2060 mm2 Lateral 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 .

Note that the distance between corner bars in one face is more than 48 O tr ( 450 - 80 -20 = 350 > 48 x 6 = 288 ). Therefore two sets of closed ties shall be used . Note that if this distance would be less than ( 48 x 6 ) mm , open ties for internal bars would be sufficient. 450

450

6 - 20 O

4 - 20 O + 4 - 16 O 450

450

6 mm O @ 250 mm c/c (double ties )

wrong arrangement

correct arrangement

IS 456-2000 clause 25 Note :- A column may be considered as short when both slenderness ratios lex / D and ley / b ≤ 12 OR lex / ixx < 40 where, lex = effective length in respect of the major axis D = depth in respect of the major axis ley = effective length in respect of the minor axis x 400 mm

b = width of the member ixx = radius of gyration in respect of the major axis. 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. IS 456-2000 clause 39.3 When emin ≤ 0.05 D , the column shall be designed by the following equation: Pu = 0.4 fck Ac + 0.67 fy Asc Ac = Area of concrete , Asc = Area of longitudinal reinforcement for columns. If emin > 0.05 D , the column shall be designed for moment also. IS 456-2000 clause 26.5.3.1 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. NOTE - The use of 6 percent reinforcement may involve practical difficulties in placing and compacting of concrete; hence lower percentage is recommended. Where bars from the columns below have to be lapped with those in the column under consideration, the percentage of steel shall usually not exceed 4 percent 26.5.3 columns 26.5.3.1 Longitudinal reinforcement a) The cross-sectional area of longitudinal reinforcement, shall be not less than 0.8 percent nor more than 6 percent of the gross crosssectional area of the column. NOTE - The use of 6 percent reinforcement may involve practical difficulties in placing and compacting of concrete; hence lower percentage is recommended. Where bars from

be used .

4 - 20 O + 4 - 16 O

6 mm O @ 250 mm c/c (double ties )

the columns below have to be lapped with those in the column under consideration, the percentage of steel shall usually not exceed 4 percent. b) In any column that has a larger cross-sectional area than that required to support the load, the minimum percentage of steel shall be based upon the area of concrete required to resist the direct stress and not upon the actual area. c)The minimum number of longitudinal bars provided in a column shall be four in rectangular columns and six in circular columns. d )The bars shall not be less than 12 mm in diameter. e) A reinforced concrete column having helical reinforcement shall have at least six bars of longitudinal reinforcement within the helical reinforcement. f )In a helically reinforced column, the longitudinal bars shall be in contact with the helical reinforcement and equidistant around its inner circumference. g )Spacing of longitudinal bars measured along the periphery of the column shall not exceed 300 mm. h )In case of pedestals in which the longitudinal reinforcement is not taken in account in strength calculations, nominal longitudinal reinforcement not less than 0.15 percent of the cross-sectional area shall be provided.

slenderness ratios

equal to the on / 30 , subject to ity exceeds the

e following

shall be not less nal area of the

Design of dog-legged staircase 230 UP Floor

1

9 10

5

A 20 v vv 300 vv

900 B

19

900

15

900

230 230

> 200 mm Tread < 230 mm

Rise of step = 160 mm

1950

900

11

2250

150

Rise

175 mm to 200 mm

Tread = 250 mm 250 mm to 280 mm Nosing is not provided ( given ) Material M15 Grade concrete mild steel reinforcement Fe250 Solution : 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 KN / m2 Floor finish = 1.00 KN / m2 Live load ( residence ) = 3.00 KN / m2 Total 7.75 KN / m2 PU = 1.5 x 7.75 = 11.63 KN / m2 Span = 1950 + 150 = 2100 i.e. 2.1 m Consider 1 m length of slab M = w x l2 / 8 = 11.63 x 2.12 / 8 = 6.41 KNm Reinforcement will be in second layer, Assuming 12 mm O bars d = 150 - 15 (cover ) -12 - 6 = 117 mm Mu / b x d2 = 6.41 x 106 / 1000 x 1172 = 0.468 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.468)

250 / 15 = 50 [(1-0.93) x 15 / 250 ] = 0.224% Ast = 0.224 x 1000 x 117 / 100 = 262 mm2 According to IS 456-200 clause 26.5.2.1 Minimum steel 0.15 % for Fe250 and 0.12 % for Fe415 Minimum steel = ( 0.15 / 100 ) x 1000 x 150 = 225 mm2 Provide 10 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 78.50 x 1000 /262 = 300 mm Provide 10 mm O bar @ 280 mm c/c = 280 Maximum spacing = 3 x d = 3 x 117 = 351

mm2 . mm

……………….( O.K.)

Check for shear : Vu = w x l / 2 = 11.63 x 2.1 / 2 = 12.21 KN Shear stress = Vu / b x d = 12.21 x 103 / 1000 x 117 = 0.104

N / mm2

for Pt = 0.224 ‫ﺡ‬c = 0.28

< ( ‫ ) ﺡ‬N / mm ( too small ) < 0.28 ( from table 7-1 ) 2

C

……………….( O.K.) Check for development length : Assuming L0 = 12 O (mild steel ) Pt = 100 x As / b x d = 100 x 280 / 1000 x 117 = 0.239 From equation Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck Mu1 / b x d2 = 0.499 Mu1 = 0.499 x 1000 x 1172 x 10-6 = 6.83 KNm Vu = 12.21 KN 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

1.3 x ( 6.83 x 106 / 12.21 x 103 ) + 12 O ≥ 55 O 727.191 + 12 O ≥ 55 O which gives Check for deflection : -

43 O ≤ 727.19 O ≤ 16.91

mm

……………….( O.K.)

Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 280 / 1000 x 117 = 0.239 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 2 ( span / d ) ratio permissible = 2 x 20 = 40 Actual (span / d ) ratio = 2100 / 117 = 17.95

<

40

……………….( O.K.)

Design of flight : Span and loading on both the flights are the same. Therefore same design will be adopted. Loads : Inclined length of waist slab for one step = √ 2502 + 1602 = 296.8 mm Assuming 150 mm thick waist slab self-load in plan = ( 296.8 / 250 ) x 0.15 x 25 = 4.45 Floor finish length for one step = 160 + 250 = 410 mm

KN / m2

floor finish = ( 410 / 250 ) x 1 = 1.64 Weight of step ( ( 0 + 160 ) / 2000 ) x 25 =

KN / m2 250 2.00 KN / m2 160 296.8 2 3.00 KN / m 4.45 + 1.64 + 2.00 + 3.00 KN / m2 11.09

Live load = Total = = Pu = 1.5 x 11.09 = 16.64

KN / m2

IS 456-2000 , clause 33.2 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. 16.64 KN / m 11.63 KN / m

11.63 KN / m

RA = 24.83 KN

RB = 24.83 KN 525

2250

525

525 = 900 / 2 +75

3300

RA = RB = 0.525 x 11.63 + 16.64 x 2.25 / 2 = 6.11 + 18.72 = 24.83 KN 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 = 40.97 - 15.83 - 3.17 = 21.97 KNm From Table 6-3 Q = 2.22 drequired = = = dprovided = = Mu / b x d2 = =

√M / Q x b √21.97 x 10 99.48

6

/ 2.22 x 1000

mm,

150 - 15 - 6 129 mm,

……………….( O.K.)

21.97 x 10 6 / 1000 x (129)2 1.32

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (1.32) 250 / 15 = 50 [(1-0.77) x 15 / 250 ] = 0.687% Ast = 0.687 x 1000 x 129 / 100 = 886 mm2 Provide 12 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 886 = 127.585 mm Provide 12 mm O bar@125 mm c/c = 904 mm2. Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement Distribution steel = ( 0.15 / 100 ) x 1000 x 150 = 225 mm2 Provide 8 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 50.24 x 1000 /225 = 223 mm Provide 8 mm O bar @ 220 mm c/c = 228 mm2 .

Check for shear : Vu = 24.83

KN

Shear stress = Vu / b x d = 24.83 x 103 / 1000 x 129 = 0.192

N / mm2

< (‫ﺡ‬

C

) N / mm2

( too small )

100 x As / b x d = 100 x 904 / 1000 x 129 = 0.70 from table 7-1 for Pt = 0.7 ‫ﺡ‬c = 0.524 N / mm2 IS 456-2000 clause 40.2.1.1 k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.524 = 0.681 N / mm2 ……………….( O.K.) Check for development length : Assuming L0 = 12 O (mild steel ) Pt = 100 x As / b x d = 100 x 904 / 1000 x 129 = 0.70 From equation Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck we get ,

Mu1 / b x d2 = 1.344 Mu1 = 1.344 x 1000 x 1292 x 10-6 = 22.37 KNm Vu = 24.83 KN

Development length of bars Ld =

O σs / 4 x ‫ﺡ‬bd

= O x 0.87 x 250 / 4 x 1

=54.4 O

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 22.37 x 106 / 24.83 x 103 ) + 12 O ≥ 54.4 O 1171.2 + 12 O ≥ 54.4 O 42.4 O ≤ 1171.2 which gives O ≤ 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 : Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 904 / 1000 x 129 = 0.70 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.62 ( span / d ) ratio permissible = 1.62 x 20

= 32.4 Actual (span / d ) ratio = 3300 / 129 = 25.58 < 32.4 ……………….( O.K.) The depth could be reduced Check 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 sm = 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 which = 5 x 129 = 645 mm spacing provided = 220 mm < 450 mm ……………….( O.K.)

to 200 mm to 280 mm

check for shear IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Pt = 100 x As bxd ≤ 0.15 0.25

Concrete grade M15 0.28 0.35

M20 0.28 0.36

M25 0.29 0.36

M30 0.29 0.37

M35 0.29 0.37

0.50 0.75

0.46 0.54

0.48 0.56

0.49 0.57

0.50 0.59

0.50 0.59

1.00

0.60

0.62

0.64

0.66

0.67

1.25 1.50 1.75

0.64 0.68 0.71

0.67 0.72 0.75

0.70 0.74 0.78

0.71 0.76 0.80

0.73 0.78 0.82

2.00

0.71

0.79

0.82

0.84

0.86

2.25 2.50 2.75 3.00

0.71 0.71 0.71 0.71

0.81 0.82 0.82 0.82

0.85 0.88 0.90 0.92

0.88 0.91 0.94 0.96

0.90 0.93 0.96 0.99

The above given table is based on the following formula Design shear strength ‫ﺡ‬c = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 6xβ β = 0.8 x fck / 6.89 Pt , but not less than 1.0 IS 456-2000 , Table 20 Table 7-2

Maximum shear stress , ‫ﺡ‬C, N / mm2 M30

M35

‫ﺡ‬c )max N / mm 2.5 2.8 3.1 3.5 check for development length IS 456-2000 clause 26.2.1 Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd

3.7

Concrete grade (

M15

M20

M25

2

IS 456-200 clause 26.2.1.1 Table 7-5 Design bond stress (‫ﺡ‬bd ) for plain bars in tension Concrete grade (

M15

M20

M25

M30

M35

‫ﺡ‬bd N / mm2

1.0 1.2 1.4 1.5 1.7 Note-1 : ‫ﺡ‬bd shall be increased by 25 % for bars in compression Note-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of ‫ﺡ‬bd shall be increased by 60 %. For mild steel Fe250 For Fe415 IS 456-2000 clause 26.2.3.3 Ld ≤ M1 / V + L0

σs = 0.87 x fy σs = 0.67 x fy

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 ) Table 7-6 Development length for single mild steel bars Tension bars Compression bars fy N / mm2 M15 M20 M15 M20 250 55 O 26 O 44 O 37 O 415 56 O 47 O 45 O 38 O 500 69 O 58 O 54 O 46 O check for deflection Basic values of span to effective depth ratios for spans upto 10 m : cantilever 7 simply supported 20 continuous 26 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. check for cracking (a)The horizontal distance between parallel main

reinforcement bars ( spacing )shall not be more than t times the effective depth of solid slab or 300 mm whichever is smaller. (b)The horizontal distance between parallel reinforcement bars ( spacing ) provided against shrinkage and temperature (distribution bar) shall not be more than five times the effective depth of a solid slab or 450 mm whichever is smaller. IS 456-2000 Clause 40.2.1.1 300 or Overall depth of slab , mm more 275 k 1.00 1.05

250 1.10

225 1.15

sketch : The reinforcements for both the flights are shown in fig. 700 700 10 O @ 280 c/c ( Landing )

v v v v v

200 1.20

10 O @ 280 c/c ( Landing )

vv v v v

12 O @ 125 c/c 150 mm thick waist slab

v

8 O @ 220 c/c

v

v v v

12 O @ 125 c/c

v

Flight-B

vv v v v v v v

8 O @ 220 c/c

v v

v

v

v

v

v

v

12 O @ 125 c/c 150 mm thick waist slab

10 O @ 280 c/c ( Landing )

Flight-A

or 300 mm whichever is small or 300 mm i.e. 300 mm ……….( O.K.) h of slab or 450 mm whichever is small or 450 mm i.e. 450 mm ……….( O.K.)

N / mm2 M40 0.30 0.38 0.51 0.60 0.68 0.74 0.79 0.84 0.88 0.92 0.95 0.98 1.01 1+5xβ-1)

ss than 1.0

m2 M40 4.0

n tension M40 1.9

mpression 786-1985 , the

er is greater

ssive reaction M1 /

s upto 10 m :

plied by se deflection

&

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.

175 1.25

150 or less 1.30

v

v

v

v

v

vv

150

v

vv

v v v v v v v

vv vv 10 O @ 280 c/c ( Landing ) 12 O @ 125 c/c 8 O @ 220 c/c