Design of Rectangular Water Tank

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 L/B= 6/4= 1.5 <2. The top portion of side walls will be designed as a continuous frame. bottom 1 m or H / 4 whichever is more is designed as cantilever. H/4= 3.5 / 4 = 0.875 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

3

KN / m2

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

6m E

A

F

2.5 m

3.5 m

24.5 KN / m

2

1m

D 34.3 KN / m2

Elevation

Plan

Fixed end moments :2 MAB = w x l / 12 = 2 = 24.5 x 6 / 12 =

MAD =

73.5

KNm

-32.66

KNm

w x l2 / 12 = 2 = 24.5 x 4 / 12 =

Kani's Method :-

0 0

D

- 32.66 -12.25 -12.25 0 -57.16

Rotation factor at Joint A

-3/10 40.84 -2/10

A

73.5 -8.17 -8.17 0 57.16

0 0

B

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 2 = 24.5 x 6 / 8 - 57.16 53.09 KNm

B.M. at centre of short span =

w x l 2 / 8 - 57.16 2 = 24.5 x 4 / 8 - 57.16 -8.16 KNm

Direct tension in long wall = =

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 short wall =

Design of Long Walls :At support M= 57.16 KNm T= 49 KN Tension on liquid face. From Table 9-6 Q = 0.306 Assuming d / D = 0.9 D=

√M / Q x b 6 = √57.16 x 10 / 0.306 x 1000 =

Take D = 450 mm From Table 9-5

432.2

mm,

Assuming d / D = 0.9 d = 450 - 25 - 8 = 417 mm

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 = 146.15273 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 tension on remote face T= 49 KN 53.09 / 49 e= M/T= Line of action of forces lies outside the section = 1.08 m 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 6

43.51 x 10 / 190 x 0.89 x 417 mm2 617

Ast2 for direct tension = T / σst =

49 x 10 3 / 150

=

327 617 + 327

mm2

=

944

mm2

Total Ast1 + Ast2 =

Provide 16 mm O bar spacing of bar = Area of one bar x 1000 / required area in m 2 / m = 200.96 x 1000 / 944 = 212.88136 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 = 139.55556 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 = 218.05556 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= 57.16 T= 73.5 From Table 9-5 Ast1 for moment = M / σst x j x d = =

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 = 130.6632 mm Provide 16 mm O bar @ 130 mm C/C…marked(b) = 1546 mm2 / m. At centre M= T=

8.16 73.5

KNm KN

tension on liquid face

mm2

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

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


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 = 139.55556 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 = 218.05556 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 Cantilever moment : Yw x H / 6 , whichever is greater. M= Yw x H x h2 / 6 OR = 9.8 x 3.5 x 1 / 6 = 9.8 x 3.5 / 6 ,tension on liquid face. = 5.72 KNm = 5.72

mm2

From Table 9-5 Ast for moment = 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 ,172 mm2 bothway = 344 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 /172 = 292 mm Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. mm2 Ast = 346 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 = 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 = = ly / lx =

1.5 x 6.25 9.38 KN / m

6.5 / 4.5 = 1.4 AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.

Table 26 Mx =

αx x w x

αx =

0.085

αy =

0.056

lx2

αy x w x lx2

My = 2

2 = 0.056 x 9.38 x 4.5 = 10.64 KNm

= 0.085 x 9.38 x 4.5 = 16.15 KNm From Table 6-3 ,Q = 2.76

√M / Q x b 6 = √16.15 x 10 / 2.76 x 1000

drequired =

= 76.50 mm dshort = 150 - 15(cover) - 5 …………(O.K.) = 130 > 76.50 mm 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 2 Q = M / bD Q = M / bD2 Pt Pt 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

Grade of concrete

Grade of steel

σcbc

M20

Fe250 Fe415

7 7

σst

k j Q N / mm N / mm2 For members less than 225mm thickness and tension on liquid face M20 Fe250 7 115 0.445 0.851 1.33 Fe415 7 150 0.384 0.872 1.17 For members more than 225mm thickness and tension away from liquid face 2

125 190

0.427 0.329

0.858 0.89

of forces lies outside the section

D/2

e=M/T

TABLE 9-3 Minimum Reinforcement for Liquid Retaining Structures % of reinforcement Thickness, mm Mild Steel HYSD bars 100 0.3 0.24 150 0.286 0.229 200 250 300 350 400

0.271 0.257 0.243 0.229 0.214

0.217 0.206 0.194 0.183 0.171

1.28 1.03

450 or more

0.2

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

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

A

150 150 1 : 4 : 8 P.C.C. 6000

150 450

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)

(c) (d)

8 O @ 130 c/c (d) 12 O @ 130 c/c (e)

4000

1000 vv

(a)

(b)

( a ) 1000

(d)

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 250

415

500

550

15 20 25

2.22 2.96 3.70

2.07 2.76 3.45

2.00 2.66 3.33

1.94 2.58 3.23

30

4.44

4.14

3.99

3.87

ion )

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) Grade Concrete Material M20 (given) Grade HYSD reinforcement Fe 415 (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 L/B= 8 / 3.6 = 2.22 >2. The long walls are designed as vertical cantilevers from the base. The short walls are designed as supported on long walls. If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m. bottom 1 m or H / 4 whichever is more is designed as cantilever. H/4= 3.0 / 4 = 0.75 m bottom h = 1 m will be designed as cantilever. Moments and tensions : Maximum B.M. in long walls at the base = (1 / 6 ) x Yw x H3 3 = ( 1 / 6 ) x 9.8 x 3 = 44.1 KNm.

Maximum ( - ve ) B.M. in short walls at support = Yw x ( H - h ) x B2 / 12 2 = 9.8 x ( 3 - 1 ) x 4 / 12 = 26.13 KNm.

Maximum ( + ve ) B.M. in short walls at centre = Yw x ( H - h ) x B2 / 16 2 = 9.8 x ( 3 - 1 ) x 4 / 16 = 19.60 KNm.

For bottom portion M= Yw x H x h2 / 6 = 9.8 x 3.0 x 1 / 6 = 4.90 KNm Direct tension in long wall =

Direct tension in short 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

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

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

√M / Q x b = √44.1 x 10 6 / 0.306 x 1000

D=

= Take D =

379.6

mm, d = 400 - 25 - 8 = 367 mm

400 mm

From Table 9-5 ,

Ast for Moment

Ast = M / σst x j x d 6 = 44.1 x 10 / 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 = 218.75 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 3 = 35.28 x 10 / 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 = 146.9 mm Provide 8 mm O bar @ 140 mm c/c on each face = 357 mm2 on each face

Design of short walls :-

At support

M= T=

From Table 9-5 Ast1 for moment =

26.13 19.6

KNm KN

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

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

19.6 x 10 3 / 150 131 mm2 544 + 131

=

675

mm2


Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 675 = 167.47 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

3 3 2 = ( 1 / 3 ) x 1000 x ( 203.56 + 196.44 ) + ( 13.33 - 1 ) x 706 x 163.44 = 5.34E+09 + 2.33E+08 4 = 5.57E+09 mm

fct = T / AT 3 = 19.6 x 10 / 408705 = 0.048 N / mm2

fcbt = M x ( d - x ) / Ixx 6 9 = 26.13 x 10 x 163.44 / 5.57 x 10 = 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 ………………….. ( O. K. ) 0.4912 ≤ 1 At centre : M= T= From Table 9-5 Ast1 for moment =

19.6 19.6

KNm KN

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


19.6 x 10 3 / 150 131 mm2 408 + 131 539

mm2


Area of one bar x 1000 / required area in m 2 / m = 113.04 x 1000 / 539 = 209.72 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 3 = 19.6 x 10 / 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 = 146.9 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 6 = 4.9 x 10 / 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 ,172 mm2 bothway = 344 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 /172 = 292 mm Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. mm2 Ast = 346 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

√ 6 = √18.76 x 10 / 2.76 x 1000

drequired = M / Q x b =

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


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 .

+ 0.4 = 4.0 m.

ion to short walls.

ase to top.For economy, the educed as we have done for

x

x 163.442

Table 9-2 Permissible concrete stresses in calculations relating to resistance to cracking Permissible stresses in N / mm2 Grade of Direct Tension due to Shear stress concrete ‫ﺡ‬v = V / b j d tension σct bending σcbt M15 1.1 1.5 1.5 M20 1.2 1.7 1.7 M25 1.3 1.8 1.9 M30 M35 M40

1.5 1.6 1.7

2.0 2.2 2.4

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

2000

150 150

150

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

8000

400

Section A-A

B

2000

8 O @ 140 mm c/c

400 900

3600

16 O @ 200 c/c ( chipiya )

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

A

A

12 O @ 200 c/c 900

900 vv

vv

400 2000

2000

B 8000

400

Sectional plan

400

6 O @ 150 c/c

8 O @ 110 c/c 150

16 O @ 200 c/c ( chipiya ) 8 O @ 140 c/c 3000

900

900

Base details not shown for clarity

900

Base details not shown for clarity

900

150 150 150 400

3600 Section B- B

400 150

effective span = Live load = Floor finish = material

Design of simply supported one way slab 4 m supported on masonry wall of 230 mm thickness 2.5 KN / m2 ( given ) 1 KN / m2 ( given ) 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 6 2 Mu / b x d2 = 24 x 10 / 1000 x (160) = 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 ] = 0.289%

( given )

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 ) > 0.12 % ( minimum steel for Fe415) = 0.128 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

3 = 24 x 10 / 1000 x 160 N / mm2 < ( ‫ﺡ‬C ) N / mm2 ( too small ) = 0.150

For bars at support d = 160 As = 231

mm 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 2 for Pt = 0.144 ‫ﺡ‬c = 0.28 N / mm

25 difference -0.05 IS 456-2000 clause 40.2.1.1 for 180 mm slab depth 20 difference k = 1.24 ? 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 mm2 As = 231 Mu1 = 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

Vu = 24

-0.04

OR

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 small or 300 mm i.e. 300 mm = 3 x 160 = 480 mm spacing provided = 170 < 300 mm ……………….( O.K.) mm Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is s = 5 x 160 = 800 mm spacing provided = 230 < 450 mm ……………….( O.K.) mm 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 %

Pt = = = From equation Pt =

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.

100 x As / b x d 100 x 231 / 1000 x 160 0.14 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck

Mu1 / b x d2 = 0.49

we get ,

2 -6 Mu1 = 0.49x 1000 x 160 x 10

= 12.54

KNm

160 180

v

v

400

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

v

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

4000

v

400

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 two-way 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.

ng development length , l0 sumed as 8 O for HYSD ally end anchorage is not and 12 O for mild steel ( provided usually whose e length 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 ( given ) mild steel grade Fe250 ( given ) Live Load As per IS 875 Solution : For slab S2 live load = 2 KN /m2 For slab S1 live load = 3 KN /m2 Assume 120 mm thick slab For S2

( Balcony slab ) DL LL

self load = 0.12 x 25 = 3 floor finish = live load = Total 1.5 ( 4 + 2 ) =

For S1

1 0 4 Pu = (6+3) DL self load = 0.12 x 25 = 3

ly = 6m

0

KN / m2

0 2 2 KN / m2 LL 0

KN / m2 KN / m2 KN / m2

KN / m2

floor finish = 1 0 KN / m2 live load = 0 3 KN / m2 Total 4 3 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 Reaction at A = w x l / 2 - Moment @ B at distance 3 m shear = = 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 2 = 12.06 x 1.34 - 9 x 1.34 / 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 = 2.8 x 1.2 + 10.5 x 1.2 / 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 6 / 2.22 x 1000

drequired =

= dprovided = = 2 Mu / b x d ( + ) = = Pt =

70.14 mm, 120 - 15 - 6 ( assume 12 O bar ) ……………….( O.K.) 99 mm, 6 8.08 x 10 / 1000 x (99)2 0.82 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 = 195.76 mm Provide 10 mm O bar@170 mm c/c = 462 mm2. ( alternate bent up ) 6 2 Mu / b x d2 ( - ) = 10.92 x 10 / 1000 x (99) = 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 ) 2 2 Total 231 + 332 mm = 563 mm 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 say 600 mm = 598 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 = = = From equation Pt = At A ,

Half bars bent = 462 / 2 = 231 mm2 )

100 x As / b x d 100 x 231 / 1000 x 99 0.233

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

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

= 0.87 x 250 x 231 x 99 ( 1 - 250 x

fy / fck we get ,

2

Mu1 / b x d = 0.487

= 2

Mu1 = 0.487 x 1000 x 99 x 10 = 4.77 KNm Vu = 12.06 KN Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd

-6

4.974

= 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 514.18 + 12 O ≥ 54.4 O 42.4 O ≤ 514.18 which gives ……………….( O.K.) O ≤ 12.13 mm Half bars bent = 462 / 2 = 231 mm2 ) P = At B , 100 x As / b x d t = 100 x 231 / 1000 x 99 = 0.233 From equation OR 2 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d ) Mu1 = 0.87 x fy x Ast x d ( 1 - fy x Ast / b

= 0.87 x 250 x 231 x 99 ( 1 - 250 x

fy / fck we get ,

Mu1 / b x d2 = 0.487

= 2

-6

Mu1 = 0.487 x 1000 x 99 x 10 = 4.77 KNm Near point of contraflexure i.e. 0.15 x l from B Vu = 17.14 - ( 0.15 x 3 ) x 9 = 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

……………….( O.K.)

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

KN

Shear stress ‫ﺡ‬v = Vu / b x d 3 = 13.09 x 10 / 1000 x 99 N / mm2 < ‫ﺡ‬c ……………….( O.K.) = 0.132

Pt = 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 2 for Pt = 0.233 ‫ﺡ‬c = 0.34 N / mm IS 456-2000 clause 40.2.1.1 for 120 mm slab depth k = 1.3 Design shear strength = 1.3 x 0.34 N / mm2 = 0.442 Span BC : Vu = 17.14 KN

0.1 difference 0.02 difference

0.07 ?

> ‫ﺡ‬v ……………….( O.K.)

Shear stress ‫ﺡ‬v = Vu / b x d 3 = 17.14 x 10 / 1000 x 99 N / mm2 < ‫ﺡ‬c ……………….( O.K.) = 0.173

Pt = 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 = 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 2 for Pt = 0.569 ‫ﺡ‬c = 0.48 N / mm

0.014

0.25 difference 0.08 IS 456-2000 clause 40.2.1.1 for 120 mm slab depth 0.181 difference k = 1.3 ? 0.0579 Design shear strength = 1.3 x 0.48 N / mm2 > ‫ﺡ‬v ……………….( O.K.) = 0.624 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 small or 300 mm i.e. 297 mm = 3 x 99 = 297 mm spacing provided = 170 < 297 mm ……………….( O.K.) mm (2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is s = 5 x 99 = 495 mm spacing provided = 150 < 450 mm ……………….( O.K.) mm

S3

Column 300 x 300

1.2m 0.15m

B3

B1

B2

ly = 6m

0.15m

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 ( 1 - 0.0389 ) KNm

-6

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

f slab or 300 mm whichever is small or 300 mm i.e. 297 mm ……………….( O.K.) e depth 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 i.e. two-way slab. =1 < 2 Middle strip : IS 456-2000 Table -26 Two adjacent edges Discontinuous 0.047 x 11.625 x 52 = Mu1 , Mu3 ( - ) = αx x w x lx2 = 2

lx2

0.035 x 11.625 x 5 = Mu2 ( + ) = αy x w x = From Table 6-3 For M15 mix and Fe415 steel, Q = 2.07

13.66

KNm

10.17

KNm

√M / Q x b = √13.66 x 10 6 / 2.07 x 1000

drequired =

= 81.23 mm, dprovided for positive moment reinforcement = 150 - 15 - 10 - 5 ( assume 10 O bar ) > 81.23 mm ……………….( O.K.) = 120 mm, dprovided for negative moment reinforcement = 150 - 15 - 5 > 81.23 mm ……………….( O.K.) = 130 mm, 6 2 2 Mu / b x d ( + ) = 10.17 x 10 / 1000 x (120) = 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. 6 2 Mu / b x d2 ( - ) = 13.66 x 10 / 1000 x (130) = 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 = 279.11 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 = 267.23 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 = 279.11 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 2 for Pt = 0.258 , ‫ﺡ‬c = 0.354 N / mm 0.25 difference 0.11 IS 456-2000 clause 40.2.1.1 0.242 difference ? -0.1065 for 150 mm slab depth k = 1.3 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 for 150 mm slab depth k = 1.3 Design shear strength = 1.3 x 0.356 = 0.463 N / mm2 ……………….( O.K.) Actual shear stress = Vu / b x d 3 = 31.795 x 10 / ( 1000 x 130 ) N / mm2 < ‫ﺡ‬c ……………….( O.K.) = 0.245

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 2 for Pt = 0.209 , ‫ﺡ‬c = 0.321 N / mm

0.07 0.041 difference ? 0.1 difference

IS 456-2000 clause 40.2.1.1 for 150 mm slab depth k = 1.3 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 for 150 mm slab depth k = 1.3 Design shear strength = 1.3 x 0.326 = 0.424 N / mm2 Actual shear stress = Vu / b x d

-0.0287

3 = 29.06 x 10 / ( 1000 x 120 ) N / mm2 < ‫ﺡ‬c ……………….( O.K.) = 0.242 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 Pt = = = From equation Pt =

(HYSD Fe415 steel )

100 x As / b x d 100 x 251 / 1000 x 120 0.209 OR 50 1-√1-(4.6 / fck) x (Mu / b x d ) 2


= 0.87 x 415 x 251 x 120 ( 1 - 415

fy / fck we get ,

2

Mu1 / b x d = 0.711

= 2

-6

10.875

Mu1 = 0.711 x 1000 x 120 x 10 = 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 458.36 + 8 O ≥ 54.3 O 46.3 O ≤ 458.36 which gives ……………….( O.K.) O ≤ 9.90 mm Vu = 11.25 x ( 4.5 / 2 ) = 25.31 KN Short span Assuming L0 = 8 O Pt = = = From equation Pt =

(HYSD Fe415 steel )



= 0.87 x 415 x 462 x 160 ( 1 - 415

fy / fck we get ,

Mu1 / b x d2 = 0.9595

= 2

Mu1 = 0.9595 x 1000 x 160 x 10 = 24.56 KNm Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd =56 O

-6

26.689

= 24.554 (From Table 7-6 )

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 1261.5 + 8 O ≥ 56 O 48 O ≤ 1261.5 which gives ……………….( O.K.) O ≤ 26.28 mm 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 ……………….( O.K.) = 40.65 < 42.9 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 s = 3 x 130 = 390 mm spacing provided = 200 < 300 mm ……………….( O.K.) mm (2) Distribu. bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is s = 5 x 120 = 600 mm

625 A

3750

S1

625

Edge strip 8 O @ 260 c/c

Middle Strip


8 O @ 200 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

625

Edge strip Middle Strip Edge strip

3750

8 O @ 260 c/c

B

625

8 O @ 260 c/c

spacing provided = 260 < 450 mm ……………….( O.K.) mm 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. 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 = 279.11 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

lab

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


-6


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

-6

625

1500

1500

500

3750

1500

S1

625


Middle Strip

8 O @ 240 c/c

625

1500

1000

3750

Edge strip Middle Strip Edge strip

1000 625

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

……………….( O.K.)


8 O @ 140 c/c

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 ) 2 = ( 6 + 4.5 ) KN / m Consider 1 m wide strip of the slab.

given

( 6 + 4.5 ) KN / m

1 A

3m

2 B

3m C

2

3m D

2

3m E

1

3m F

The factored moments at different points using the coefficients are as follows : 2 2 Mu1 ( + ) = ( 1 / 12 ) x w ( DL ) x l + ( 1 / 10 ) x w ( LL ) x l = = = Mu2 ( + ) =

( 1 / 12 ) x 6 x 32 + ( 1 / 10 ) x 4.5 x 32 4.5 + 4.05 8.55 KNm ( 1 / 16 ) x w ( DL ) x l2 + ( 1 / 12 ) x w ( LL ) x l2

= = = Mu3 ( - ) =


= = = Mu4 ( - ) =


2 2 = ( 1 / 12 ) x 6 x 3 + ( 1 / 9 ) x 4.5 x 3 = 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 6 / 2.07 x 1000

drequired =

= 69.2 mm, Try 110 mm overall depth dprovided = 110 - 15 - 5 ( assume 10 O bar ) ……………….( O.K.) = 90 mm, 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 moment Mu/(b x d2 ) Pt point Steel Provided KNm mm2 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

3 = 18.9 x 10 / 1000 x 90 N / mm2 < ( ‫ﺡ‬C ) N / mm2 ( too small ) = 0.210

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 N / mm2 for Pt = 0.397 ‫ﺡ‬c = 0.42 0.25 difference 0.11 IS 456-2000 clause 40.2.1.1 for 110 mm slab depth 0.103 difference k = 1.3 ? 0.0453 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 mm 2 ) Vu = 18.9 - 0.15 x 3 x 10.5 = 14.18 KN Actual Shear stress = Vu / b x d 3 = 14.18 x 10 / 1000 x 90 N / mm2 < ( ‫ﺡ‬C ) N / mm2 ( too small ) = 0.158

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 N / mm2 for Pt = 0.324 ‫ﺡ‬c = 0.38

0.25 difference 0.11 IS 456-2000 clause 40.2.1.1 for 110 mm slab depth 0.176 difference k = 1.3 ? 0.0774 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 2 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d ) Mu1 = 0.87 x fy x Ast x d ( 1 - fy x Ast / b = 0.87 x 415 x 292 x 90 ( 1 - 415

fy / fck we get ,

Mu1 / b x d2 = 1.064

= 2

Mu1 = 1.064 x 1000 x 90 x 10 = 8.62 KNm Assuming L0 = 8 O (HYSD Fe415 steel ) Development length of bars Ld = O σs / 4 x ‫ﺡ‬bd

-6

9.4884

= 8.64

= O x 0.67 x 415 / 4 x 1

=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 845.783 + 8 O ≥ 56.4 O 48.4 O ≤ 845.78 which gives ……………….( O.K.) O ≤ 17.47 mm 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 792.102 + 12 O ≥ 56.4 O 44.4 O ≤ 792.1 which gives O ≤ 17.84 mm Check for deflection : -

……………….( O.K.)

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 small or 300 mm i.e. 270 mm = 3 x 90 = 270 mm spacing provided = 220 < 270 mm ……………….( O.K.) mm (2 )Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is = 450 mm = 5 x 90 spacing provided = 160 < 450 mm ……………….( O.K.) mm

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

Near middle of end span

Dead load and imposed load ( fixed )

+ 1 / 12

+ 1 / 16

-1 / 10

imposed load not fixed )

+ 1 / 10

+ 1 / 12

-1 / 9

(

At middle of At support next to interior span the end support

NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span. Table 13 Shear Force coefficients At end support

Type of load Dead load and imposed load ( fixed ) imposed load not fixed )

(

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

t reinforcement ( 292 mm 2 )


-6

s critical in span AB

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

efficients Support moments At other interior supports

-1 / 12 -1 / 9 shall be multiplied by the total

icients 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. given material M15 grade concrete HYSD reinforcement of grade Fe415 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 2 0.104 x 11.25 x 4.52 = Mux = αx x w x lx = Muy = αy x w x lx2 = From Table 6-3

2

0.046 x 11.25 x 4.5 = Q = 2.07

23.7

KNm

10.48

KNm

√M / Q x b = √23.7 x 10 6 / 2.07 x 1000

drequired =

= 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. 6 2 Mu / b x d2 ( short ) = 23.7 x 10 / 1000 x (160) = 0.926 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck = 50 1-√1-(4.6 / 15) x (0.926)

……………….( O.K.)

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 = 175.22 mm Provide 10 mm O bar@170 mm c/c = 462 ( short span ) mm2. 6 2 Mu / b x d2 ( long ) = 10.48 x 10 / 1000 x (150) = 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 = 232.59 mm Provide 8 mm O bar@ 230 mm c/c = 218 ( long span ) mm2. 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 ) = 25.31 KN Long span Assuming L0 = 8 O

(HYSD Fe415 steel )

Pt = = = From equation Pt =


Mu1 = 0.87 x fy x Ast x d ( 1 - fy x = 0.87 x 415 x 218 x 150 (

fy / fck we get ,

2

Mu1 / b x d = 0.5023

= 2


-6

11.806

= 11.331 (From Table 7-6 )

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 11.30 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 580.4 + 8 O ≥ 56 O 48 O ≤ 580.403 which gives ……………….( O.K.) O ≤ 12.09 mm Vu = 11.25 x ( 4.5 / 2 ) = 25.31 KN Short span Assuming L0 = 8 O Pt = = = From equation Pt =

(HYSD Fe415 steel )


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

fy / fck we get ,

2

Mu1 / b x d = 0.9595

= 2


-6

= 24.554 (From Table 7-6 )

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O ≥ 56 O 1261.5 + 8 O ≥ 56 O 48 O ≤ 1261.48 which gives ……………….( O.K.) O ≤ 26.28 mm 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 3 = 25.31 x 10 / 1000 x 150

26.689

= 0.169

N / mm

2

< ( ‫ﺡ‬C ) N / mm2

( too small )

100 x As / b x d = 100 x 218 / 1000 x 150 = 0.145 from table 7-1 2 for Pt = 0.145 ‫ﺡ‬c = 0.28 N / mm 25 difference -0.05 IS 456-2000 clause 40.2.1.1 20 difference ? -0.04 for 180 mm slab depth k = 1.24 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 = 12.011 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 25 difference -0.05 for 180 mm slab depth k = 1.24 20 difference ? -0.04 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 ……………….( O.K.) = 28.00 < 28.8 Check for cracking : IS 456-2000 , clause 26.3.3 (1) Main bars : maximum spacing permitted for short span steel = 3 x effective depth of slab or 300 m = 3 x 160 = 480 spacing provided = 170 < 300 mm ……………….( O.K.) mm (2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth of slab or 450 m = 5 x 150 = 750 spacing provided = 230 < 450 mm ……………….( O.K.) mm

Table - 26 Bending moment coefficients for rectangular panels supported on four sides with provi Case No. 1

2

3

4

5

6

Type of Panel and Moments considered

8

9

( Values of ly / lx ) 1.3

1.0

1.1

1.2

Interior Panels: Negative moment at continuous edge Positive moment at mid-span One Short Edge Discontinuous:

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

0.028

0.033

0.039

0.044

0.047 0.035

0.053 0.040

0.060 0.045

0.065 0.049

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 0.043

0.064 0.048

0.071 0.053

0.076 0.057

0.043

0.051

0.059

0.065

0.056

0.064

0.072

0.079

One long Edge Discontinuous:

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

7

short span coefficient αx

Three Edges Discontinuous (One Long Edge Continuous): Negative moment at continuous edge Positive moment at mid-span Three Edges Discrmntinuous (One Short Edge Continuous) : Negative moment at continuous edge Positive moment at mid-span Four-Edges Discontinuous: Positive moment at mid-span

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


-6

( 1 - 0.0402 ) KNm


-6

( 1 - 0.0799 ) KNm

A 350

6550

350

350 460

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

4300

690

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 c/c

v

v

8 O @ 230 c/c 350

4300

Section A-A

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

350

n four sides with provision for torsion at corners

ort span coefficient αx ( Values of ly / lx ) 1.4

Long span coefficient αy for all values of ly / lx

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 0.060

0.084 0.064

0.091 0.069

0.097 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

s 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 (Given ) Floor finish = 1 KN / m2 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 / m2 Floor finish = 1 + 0 KN /m2 Live load = 0 + 3 KN / m2 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

B

6m

6m

22 KN / m

36 KN / m

6m

A

B10

6m C

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

36 KN / m

B

A

B

C

162

162

A1 = ( 2 / 3 ) x Base x h1 = ( 2 / 3 ) x 6 x 162 = 648 a1 = 3 m

A2 = ( 2 / 3 ) x Base x h1 = ( 2 / 3 ) x 6 x 162 = 648 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 = ( 2 / 3 ) x 6 x 162 = 648 a2 = 3 m

A3 = ( 2 / 3 ) x Base x h3 = ( 2 / 3 ) x 6 x 99 = 396 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 ] …………………….( 2 )

MB + 4MC = - 522

As MD = 0

4MB + MC = - 648

…………………….( 1 )

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

(-)

(-)

(+) (-) 96

138

99 (+)

51

45 (+)

(+)

D

96 162

93

6m C

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

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

138

MB = - 138 KNm

115

85

82

VA x 6 -36 x 6 x 3 = -138 VA = 85 KN VA + VB = 36 x 6

2.36 m

50 101 131

MC = - 96 KNm

VB = 216 - 85 VB = 131 KN

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.) 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 mm2

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

bf / bw = 1650 / 230 = 7.17

For bf /bw = 7

Df / d = 120 / 535 = 0.224

0.01 diff.

2

Mu,lim / fck bw d = 0.671

( As per SP:16 ,Table 58 ) 0.006 diff. 2

Mu.lim = 0.671 x 15 x 230 x 535 x 10

-6

= 662.596564 KNm > 111.6 KNm

0.014 ? -0.0084

For 0.224

0.6566

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

If Mu < Mu,lim :

Ast = Mu / 0.87 x fy x lever arm where lever arm = d - Df / 2

= 535 - 120 / 2 = 535 - 60

6

Ast = 111.6 x 10 / 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 ) mm2 . = 301 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 2 Ast = ( 0.549 / 100 ) x 230 x 535 = 676 mm

20 % of steel should be carried through the span = 0.2 x 676 = 135 mm 2 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 . In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ℓ ) from A and at 900 mm ( 0.15 ℓ ) from B.

The moment of resistance of remaining bars = 0.87 fy Ast ( d - Df / 2 ) -6

= 0.87 x 415 x 339 ( 535 - 60 ) x 10 = 58.14 KNm Note that lever arm assumed here is d - Df / 2 Theoretical cut off point can be found out from case ( c ) as 58.14 = 89.6 x - 36 x2 / 2 i. e. x2 - 4.98 x + 3.23 = 0 x = ( -b + √ b2 - 4ac ) / 2 a = 4.98 + √ 4.982 - 4 x 1 x 3.23 / 2

or

= ( 4.98 - √ 4.982 - 4 x 1 x 3.23 ) / 2

= ( 4.98 + 3.45 ) / 2 = ( 4.98 - 3.45 ) / 2 x = 4.22 m x = 0.765 According to curtailment rules, the bars are cut off at 12 O or d eff whichever is greater i.e. 535 mm From A , curtail 3 - 12 O bars at 765 - 535 = 230 mm ( i.e. 230 - 115 = 115 mm from face of support say 300 m and at 4220 + 535 = 4755 mm ( i.e. 1245 mm from B say 300 mm ). The curtailment rules are to be satisfied . The third rule that the continuing bars provide double the area required for flexure at cut off point and the shear does not exceed 3 / 4 th that permitted. Moment at 0.230 m from A 2 = 89.6 x 0.23 - ( 0.23 / 2 ) x 36 KNm < ( 1 / 2 ) x 58.14 KNm ………….( O.K.) = 19.66 Moment at 4.755 m from A 2 = 89.6 x 4.755 - ( 4.755 / 2 ) x 36 KNm < ( 1 / 2 ) x 58.14 KNm ………….( O.K.) = 19.07 Minimum design shears at cut off points, referring to design shear diagram . Vu = 4 / 3 [ 81.32 ] = 108.43 KN at 0.23 m from A < 89.6 KN ,take 81.32 KN Vu = 4 / 3 [ 81.56 ] = 108.75 KN at 1.245 m from B. < 126.4 KN Negative moment reinforcement 20 % of steel should be carried through the span = 0.2 x 676 = 135 mm 2

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 . For negative moment top is tension and bottom is compression. The theoretical cut-off points and actual positions for curtailing 40 % bars i.e. 2 - 16 O may now be determine Remaining bars 2-10 O & 2-16 O provide 157 + 402 = 559 mm 2 area. The strength of section is controlled by top reinforcement only and is given approximately by M.R. = 0.87 fy Ast ( d - Df / 2 ) M.R. = 0.87 x 415 x 559 ( 535 - 60 ) x 10-6 = 0.20 x 475 = 95.00 KNm The theoretical cut off can be given by an equation 89.6 x - 36 x2 / 2 = -95 [ case ( a ) ]

x2 - 4.98 x - 5.28 = 0 x = ( -b + √ b2 - 4ac ) / 2 a 2 = 4.98 + √ 4.98 + 4 x 1 x 5.28 / 2 = ( 4.98 + 6.776 ) / 2 x = 5.88 m from A

i.e. 0.12 m from B Cut off 3 - 16 O bars at 120 + 535 ( deff ) = 655 mm say 900 mm from B . At 900 mm from B i.e. at 5.1 m from A , the moment is 89.6 x 5.1 - 36 x 5.12 / 2 = -11.22 KNm Moment capacity required at actual cut off point as per curtailment rules = 2 ( -11.22 ) = - 22.44 KNm Moment capacity of the section = - 95 KNm ………………( O.K. ) Note that shear capacity will be increased as per positive reinforcement curtailment check. Let us now determine the point of curtailment of 80 % of bars. Remaining bars are 2-10 O , i.e. 157 mm2 Moment capacity = 0.87 fy Ast ( d - Df / 2 ) = 0.87 x 415 x 157 ( 535 - 60 ) x 10-6 = 0.057 x 475 = 27.08 KNm Distance from A for this moment 89.6 x - 36 x2 / 2 = - 27.08 x2 - 4.98 x - 1.50 = 0 x = ( -b + √ b2 - 4ac ) / 2 a 2 = 4.98 + √ 4.98 + 4 x 1 x 1.50 / 2 = ( 4.98 + 5.55 ) / 2 m from A x = 5.27 i.e. at 0.73 m from B .

Cut off the bars at 730 + 535 ( deff ) = 1265 mm say 1500 mm from B. 20 % bars shall be continued throughout the span. Note that this is required for case ( c ) in span BC where point of contraflexure lies at 3 m from B. Continue 2 no. 10 mm O bars giving Ast = 157 mm2 area . These will be used as anchor bars. ( c ) design for shear reinforcement : At A , Vu = 89.6 KN also Vu = 81.32 KN upto 0.23 m from A 100 As / bd = 100 x 3 x 113 / 230 x 535 = 0.28 ( From IS 456-2000 , table 19 table 7-1 )

0.25 diff. 0.22 diff.

0.11 ?

Design ( permissible )shear strength of concrete , ‫ﺡ‬C, = 0.36 N / mm2 Actual shear strength ‫ﺡ‬v = Vu / bd 3 = 81.32 x 10 / ( 230 x 535 )

-0.10

= 0.661

N / mm2

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

OR

> ‫ﺡ‬c

Vus = ( 0.661 - 0.36 ) 230 x 535 x 10-3

= 81.32 - 0.36 x 230 x 535 x 10-3 = 37.038 KN = 81.32 44.298 = 37.02 KN Assuming 6 mm O two - legged stirrups , Asv = 57 mm2 , fy = 415 N / mm2 . Sv = 0.87 fy Asv d / Vus 3

= 0.87 x 250 x 57 x 535 / 37.02 x 10 = 179 mm From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 mm O stirrups = 0.87 Asv fy / 0.4 b = 0.87 x 57 x 250 / 0.4 x 230 = 134.8 mm spacing should not exceed ( i ) 450 mm ( ii ) 0.75 d = 0.75 x 535 = 401.25 mm ( iii ) ≤134.8 mm ( minimum ) ( iv ) 179 mm ( designed ) Minimum shear reinforcement of 6 mm O @ 130 mm c/c will be used . capacity of section with minimum shear reinforcement Vus = 0.87 fy Asv d / Sv -3 = ( 0.87 x 250 x 57 x 535 / 130 ) x 10 > 37.02 KN ………………….( O.K.) = 51 KN At B , Vu = 126.4 KN also Vu = 108.75 KN upto 1.245 m from B.

= 157 + 603 = 760 mm2 2-10 O + 3-16 O 100 As / bd = 100 x 760 / 230 x 535 = 0.62 ( From IS 456-2000 , table 19 table 7-1 )

0.25 diff. 0.13 diff.

0.08 ?

-0.04

Design ( permissible )shear strength of concrete , ‫ﺡ‬C, = 0.50 N / mm2 Actual shear strength ‫ﺡ‬v = Vu / bd 3 = 108.75 x 10 / ( 230 x 535 ) N / mm2 > ‫ﺡ‬c = 0.884

shear design is necessary . At support , Vus = Vu - ‫ﺡ‬c b d -3 = 108.75 - 0.50 x 230 x 535 x 10 = 108.75 61.525 = 47.23 KN

OR

Vus = ( 0.884 - 0.50 ) 230 x 535 x 10-3 = 47.251

KN

Assuming 6 mm O two - legged stirrups , Asv = 57 mm2 , fy = 415 N / mm2 . Sv = 0.87 fy Asv d / Vus 3

= 0.87 x 250 x 57 x 535 / 47.23 x 10 = 140 mm From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 mm O stirrups = 0.87 Asv fy / 0.4 b = 0.87 x 57 x 250 / 0.4 x 230 = 134.8 mm spacing should not exceed ( i ) 450 mm ( ii ) 0.75 d = 0.75 x 535 = 401.25 mm ( iii ) ≤134.8 mm ( minimum ) ( iv ) 140 mm ( designed ) Minimum shear reinforcement of 6 mm O @ 130 mm c/c will be used . capacity of section with minimum shear reinforcement Vus = 0.87 fy Asv d / Sv -3

= ( 0.87 x 250 x 57 x 535 / 130 ) x 10 > 47.25 KN = 51 KN ( d ) Check for development length : At A, Mu1 = 58.14 KNm ( Calculated previously )

………………….( O.K.)

Vu = 89.6 KN L0 = 12 O Ld = 56.4 O check

1.3 Mu1 / Vu + L0 ≥ Ld

1.3 x 58.14 x 106 / 89.6 x 103 + 12 O ≥ 56.4 O 843.549107 ≥ 44.4 O O ≤ 19 mm ………………….( O.K.) At point of contraflexure i.e. 1.245 m from B Mu1 = 58.14 KNm Vu = 126.4 - 1.245 x 36 = 81.58 KN

115.6

L0 = 12 O ( L0 is much more than this . It can be taken equal to deff.However to permit the lapping of bars we shall consider 12 O . ) check 1.3 Mu1 / Vu + L0 ≥ Ld 1.3 x 58.14 x 106 / 81.58 x 103 + 12 O ≥ 56.4 O 926.477078 ≥ 44.4 O O ≤ 21 mm ………………….( O.K.) ( e ) Check for deflection : Basic span / d ratio = 26 100 Ast / b d = 100 x 678 / 1650 x 535 = 0.077 modification factor = 2 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement bw / bf = 230 / 1650 = 0.14 < 0.3 IS 456-2000 clause 23.2.1 fig-6 ,for flanged beam

Reduction factor = 0.8 Span / d permissible = 26 x 2 x 0.8 = 41.6 Actual span / d = 6000 / 535 = 11.21 ………………………( safe ) ( d ) Check for cracking (spacing of bars ) : Clear distance between bars = ( 230 - 50 - 3 x 12 ) / 2 = 72 mm Minimum clear distance permitted = hagg + 5 mm = 20 + 5 = 25 mm or 12 mm ( O of bar ) i.e. 25 mm . Maximum clear distance permitted ( cracking - table 8-1 , IS 456-200 , table 15 ) ………………………( safe ) = 180 mm B

A

2 - 10 O + 2 - 16 O

2 - 10 O

300

1500

1500 900

6- 12 O

900

2 - 10 O + 3 - 16 O ( Extra ) 1500 1500 900 900 2 - 10 O

3 - 12 O

300

6000

6- 12 O

300

6000 A

D N S

2 - 10 O

6000

B

6 mm O @ 130 mm c/c

D N S

6 mm O @ 130 mm c/c

D N S

6 mm O @ 130 mm c/c

2-10 O + 3

2 - 10 O 120

120

450

450 20 O pin 6 - 12 O

3-

230

230

Section A - A

Section B - B

3m B11

B12 3m

6m

6m

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

6m

A

) -6 A2a2 / ( I2L2)

22 KN / m

36 KN / m

B

6m

6m C

Using three moment equation for span ABC MA ( L1 / I1 ) + 2MB ( L1 / I1 + L2 / I2 ) + MC ( L2 / I2 ) = - 6 A1a 22 KN / m

B

A

B

99

A1 = ( 2 / 3 ) x Base x h1 = ( 2 / 3 ) x 6 x 99 = 396 a1 = 3 m ) ] - [ 6 x 648 x 3 / ( I x 6) ]

A2 = ( 2 / 3 ) x Base x h1 = = a2 = 3 m

MA ( 6 / I ) + 2MB ( 6 / I + 6 / I ) + MC ( 6 / I ) = - [ 6 x 396 x 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 A2a

) -6 A3a3 / ( I3L3)

36 KN / m

C

B

C

162

A2 = ( 2 / 3 ) x Base x h2 A3 = ( 2 / 3 ) x Base x h3 = ( 2 / 3 ) x 6 x 162 = = 648 = a2 = 3 m a3 = 3 m ) ] - [ 6 x 396 x 3 / ( I x 6) ]

MB ( 6 / I ) + 2MC ( 6 / I + 6 / I ) + MD ( 6 / I ) = - [ 6 x 648 x 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

69

( 162 - 69 = 93 KNm )

48

( 99 - 48 = 51 KNm )

21

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

22 KN / m

36 KN / m

47

58

(+)

(+)

6m C104 (-)

99 (+)

47 (+)

(-)

(-)

104

104

104

X dist. 6 - X dist.

85

104

108

48.67

83.33

131

131 X =85 ( 6 - X )

2.21 m

216 X = 510 108

X = 2.36 m

83.33

MC = - 104 KNm

MB = - 104 KNm

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

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

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 VB = 191.33 KN VB + 83.33 = 191.33 VB = 108 KN

ts. Case ( a ) gives maximum ecause of symmetry. Case ( b ) ) gives maximum positive

ative moment = 138 KNm. ases , redistribute ( increase or

( As per IS 456-2000 , Clause 37.1.1 )

ve been reduced.

-2000 ,Clause 23.1.2 , Note )

IS 456-2000 ,Clause 37.1.1 Redistribution of m Continuous Beams and Frames ( b ) The ultimate moment of resistance provided member is not less than 70 percent of the mome obtained from an elastic maximum moment diagr appropriate combinations of loads. ( c ) The elastic moment at any section in a m particular combination of loads shall not be re than 30 % of the numerically largest moment giv the elastic maximum moments diagram for member , covering all appropriate combination of IS 456-2000 ,Clause 23.1.2 Effective width of f In the absence of more accurate determination , width of flange may be taken as the following but greater than the breadth of the web plus half the distances to the adjacent beams on either side :

( As per IS456-2000 ,Clause 26.5( a ) ) For bf /bw = 8

In the absence of more accurate determination , width of flange may be taken as the following but greater than the breadth of the web plus half the 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 Df

0.016

Where ,

?

bf = effective width of flange ,

-0.0096

l0 = distance between points of zero moments in

0.743

bw = breadth of the web ,

0.0717 0.671 ced beam) as explained below.

Df = thickness of flange , and b = actual width of the flange. NOTE - For continuous beams and frames , ' l0 ' assumed as 0.7 times the effective span.

IS 456-2000 Clause 26.5 Requirements of Rein Structural Members

26.5.1 Beams 26.5.1.1 Tension Reinforcement a ) Minimum reinforcement - The minimum are

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 we d = effective depth , and fy = characteristic strength of reinforcement in N /

mm ( 0.15 ℓ ) from B.

b ) Maximum reinforcement - The maximum ar 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 2

3 - 16 mm O extra at top = 603 mm , pport B and remaining 2 - 16 mm O

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

mm ( 0.15 ℓ ) from B.

For singly reinforced rectangular sections

fy, N / mm2

fck N / mm2

250

415

15 20 25 30

2.22 2.96 3.70 4.44

2.07 2.76 3.45 4.14

check for shear

IS 456-2000 , Tab 2

( 4.98 - √ 4.98 - 4 x 1 x 3.23 ) / 2

Table 7-1

( 4.98 - 3.45 ) / 2 m hichever is greater i.e. 535 mm 5 = 115 mm from face of support say 300 mm)

Design shear strength of conc Concrete Pt = 100 x As bxd M15 ≤ 0.15 0.28 0.25 0.35 0.50 0.46 0.75 0.54 1.00 0.60 1.25 0.64 1.50 0.68 1.75 0.71 2.00 0.71 2.25 0.71 2.50 0.71 2.75 0.71

tinuing bars provide double the area 3 / 4 th that permitted. 0.745

110.4 KN

126.4 KN

89.6 KN 3.51 m

6 KN ,take 81.32 KN 2.49 m

126.4 KN

89.6 KN 110.4 KN

3.00 0.71 The above given table is based on the f Design shear strength ‫ﺡ‬c =

3 - 16 mm O extra at top = 603 mm2 , pport B and remaining 2 - 16 mm O p is tension and bottom is compression. % bars i.e. 2 - 16 O may now be determined .

β=

IS 456-2000 , Tab Table 7-2

s given approximately by

Maximum shear stress , Concrete grade

( ‫ﺡ‬c )max N/mm

2

M15 2.5

IS 456-2000 Clause 26.5.1.6 Minimum Minimum shear reinforcement in the for such that :

where, Asv = total cross-sectional area of st

Sv = stirrup spacing along the leng b = breadth of the beam or breadt fy = characteristic strength of the s

which shall not be taken great

rules = 2 ( -11.22 ) = - 22.44 KNm

ment curtailment check.

check for development length IS 456-2000 clause 26.2.1 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

Development length of bars Ld =

Concrete grade ( ‫ﺡ‬bd N / mm

2

M15 1.0

M20 1.2

M25 1.4

M30 1.5

M35 1.7

M40 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 %. σs = 0.87 x fy For mild steel Fe250

required for case ( c ) in span BC

σs = 0.67 x fy For Fe415 IS 456-2000 clause 26.2.3.3 Ld ≤ M1 / V + L0

ll be used as anchor bars.

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

Tension bars

Compression bars

M15

M20

M15

M20

250

55 O

26 O

44 O

37 O

415

56 O

47 O

45 O

38 O

0.36 ) 230 x 535 x 10

500

-3

0.50 ) 230 x 535 x 10-3

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 simply supported continuous

7 20 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 N / mm2 250 415 500

-30 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

/ mm2 .

.However to permit

or tension reinforcement 2.1 fig-6 ,for flanged beam

ar ) i.e. 25 mm . ………………………( safe )

At least 20 % of steel placed over support B to be continued through beam Cut off not more than 40 % of steel placed over support B

0.25ℓ1

0.25ℓ2 0.15ℓ1 0.15ℓ2

10 O

12 O

0.1ℓ1

300

0.15ℓ1 0.15ℓ2 ℓ1

6 mm O @ 130 mm c/c

D N S

10 O + 3-16 O Extra at top

3 - 12 O

Cut off not more than 40 % of steel at B

ℓ2 Cut off not more than 70 % of steel required at mid span

m positive moment in span BC

Case ( c ) Maximum positive mom

22 KN / m

36 KN / m

6m

6m

A

D

t equation for span ABC L1 / I1 + L2 / I2 ) + MC ( L2 / I2 ) = - 6 A1a1 / ( I1 L1 ) -6 A2a2 / ( I2L2)

Using three moment equation for MA ( L1 / I1 ) + 2MB ( L1 / I1 + L2 / I

36 KN / m

C

A

162

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

= - [ 1188 ] - [ 1944 ] As MA = 0

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

…………………….( 1 )

4MB + MC = - 522

t equation for span BCD L2 / I2 + L3 / I3 ) + MD ( L3 / I3 ) = - 6 A2a2 / ( I2 L2 ) -6 A3a3 / ( I3L3)

Using three moment equation for MB ( L2 / I2 ) + 2MC ( L2 / I2 + L3 / I

22 KN / m

D

B

99

A3 = ( 2 / 3 ) x Base x h3 ( 2 / 3 ) x 6 x 99 396 a3 = 3 m / I + 6 / I ) + MD ( 6 / I ) = - [ 6 x 648 x 3 / ( I x 6 ) ] - [ 6 x 396 x 3 / ( I x 6) ]

= - [ 1944 ] - [ 1188 ]

A2 = ( 2 / 3 ) x Base x h2 = = a2 = 3 m MB ( 6 / I ) + 2MC ( 6 / I + 6 / I ) + 6MB + 24MC + 6MD = - [ 1188 ] -

…………………….( 2 )

As MD = 0

MB + 4MC = - 522

…………………….( 1 )

4MB + MC = - 522

…………………….( 2 )

MB + 4MC = - 522

MB from Equation ( 2 ) into Equation ( 1 )

By putting Value of MB from Equa 4(- 522 - 4MC ) + MC = - 522 - 2088 - 16 MC + MC = - 522 15 MC = - 1566 MC = - 104.4 KNm

MC into Equation ( 1 )

By putting Value of MC into Equa 4MB + ( - 104.4 ) = - 522 4 MB = - 417.6 MB = - 104.4 KNm

22 KN / m

36 KN / m

6m

A

D

99 (+)

47 (+)

6m

162 (+)

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

110 (+)

MB = - 104 KNm

48.67

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

MB = - 104 KNm

90.67

2.52 m

MC = - 104 KNm

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

VD x 6 -36 x 6 x 3 = -104

Vc = 191.33 KN

VD = 90.67 KN

VC + 83.33 = 191.33

VC + VD = 36 x 6

VC = 108 KN

VC = 216 - 90.67

MC = - 104 KNm 48.67 x 12 + VB x 6 - 36 x 6 x 3 - 22 x 6 x 9 = - 104 VB = 191.33 KN VB + 83.33 = 191.33 VB = 108 KN

00 ,Clause 37.1.1 Redistribution of moments in us Beams and Frames ltimate moment of resistance provided at any section of a not less than 70 percent of the moment at that section rom an elastic maximum moment diagram covering all e combinations of loads. elastic moment at any section in a member due to a combination of loads shall not be reduced by more of the numerically largest moment given anywhere by c maximum moments diagram for the particular covering all appropriate combination of loads. 00 ,Clause 23.1.2 Effective width of flange ence of more accurate determination , the effective ange may be taken as the following but in no case an the breadth of the web plus half the sum of the clear to the adjacent beams on either side :

VC = 125.33 KN

ence of more accurate determination , the effective ange may be taken as the following but in no case an the breadth of the web plus half the sum of the clear to the adjacent beams on either side : -beams , bf = ( l0 / 6 ) + bw + 6 Df

-beams , bf = ( l0 / 12 ) + bw + 3 Df

ve width of flange ,

ce between points of zero moments in the beam ,

ess of flange , and width of the flange. or continuous beams and frames , ' l0 ' may be as 0.7 times the effective span.

00 Clause 26.5 Requirements of Reinforcement for

l Members

ension Reinforcement um reinforcement - The minimum area of tension reinforcement

e less than that given by the following :

mum area of tension reinforcement , h of the beam or the breadth of the web of T- beam ,

cteristic strength of reinforcement in N / mm 2 .

um reinforcement - The maximum area of tension ment shall not exceed 0.04 b D.

d = 100 x 0.85 / 250 = 0.34 steel , Fe415 grade d = 100 x 0.85 / 415 = 0.205 steel , Fe500 grade

d = 100 x 0.85 / 500 = 0.17 Table 6-3

ment of resistance factor Q lim, N / mm2

nforced rectangular sections

fy, N / mm2 500

550

2.00 2.66 3.33 3.99

1.94 2.58 3.23 3.87

IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Concrete grade M20 M25 M30 M35 M40 0.28 0.29 0.29 0.29 0.30 0.36 0.36 0.37 0.37 0.38 0.48 0.49 0.50 0.50 0.51 0.56 0.57 0.59 0.59 0.60 0.62 0.64 0.66 0.67 0.68 0.67 0.70 0.71 0.73 0.74 0.72 0.74 0.76 0.78 0.79 0.75 0.78 0.80 0.82 0.84 0.79 0.82 0.84 0.86 0.88 0.81 0.85 0.88 0.90 0.92 0.82 0.88 0.91 0.93 0.95 0.82 0.90 0.94 0.96 0.98

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 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 M20 2.8

M25 3.1

M30 3.5

M35 3.7

M40 4.0

S 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

total cross-sectional area of stirrup legs effective in shear , stirrup spacing along the length of the member , breadth of the beam or breadth of the web of flanged beam , and characteristic strength of the stirrup reinforcement in N / mm 2 which shall not be taken greater than 415 N / mm 2 .

ever is greater

of steel placed over support B

Cut off not more than 40 % of steel at B

n 70 % of steel required at mid span

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

36 KN / m

22 KN / m

6m

B

6m

6m C

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

22 KN / m

B

162

C

99

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 99 648 = 396 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 396 x 3 / ( I x 6) ] 6MA + 24MB + 6MC = - [ 1944 ] - [ 1188 ] 4MB + MC = - 522

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) 22 KN / m

36 KN / m

C

C

D 162

99

A2 = ( 2 / 3 ) x Base x h2 ( 2 / 3 ) x 6 x 99 396 a2 = 3 m

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

MB ( 6 / I ) + 2MC ( 6 / I + 6 / I ) + MD ( 6 / I ) = - [ 6 x 396 x 3 / ( I x 6 ) ] - [ 6 x 648 x 3 / ( I x 6) ] 6MB + 24MC + 6MD = - [ 1188 ] - [ 1944 ] …………………….( 2 )

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 = - 104.4 KNm By putting Value of MC into Equation ( 1 ) 4MB + ( - 104.4 ) = - 522 4 MB = - 417.6 MB = - 104.4 KNm 36 KN / m

6m

162

36 KN / m

22 KN / m

B 104 (-)

6m

99

6m C 104

162 D (+)

(-)

110

110

(+)

(+) (-)

104

(-) 5.0

104

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

104

104 125.33

MB = - 104 KNm

66

66 90.67 125.33

MC = - 104 KNm

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

VD x 6 -36 x 6 x 3 = -104

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

VD = 90.67 KN

Vc = 191.33 KN

VC + VD = 36 x 6

VC + 125.33 = 191.33

VC = 216 - 90.67

VC = 66 KN

VC = 125.33 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

( 162 - 52 = 110 KNm )

125.33 X =90.67 ( 6 - X )

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 6 2 2 Mu / b d == 104.4 x 10 / ( 230 x 515 ) Mu,lim = 2.22 x 230 x 5152 x 10-6 < 2.22 ( Table 6-3 ) = 1.71 = 135.42 KNm The section is singly reinforced ( under-reinforced ) Mu < Mu,lim 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 2 Ast = ( 0.93 / 100 ) x 230 x 515 = 1102 mm 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 3 = 69.6 x 10 / ( 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 ) 0.25 difference = 1.06 2 N / mm > ‫ﺡ‬v 0.19 difference design ( permissible ) shear strength ‫ﺡ‬c = 0.61 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

v

v 230

Stirrups 6 O @ 130 c/c

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 2 2 Mu = w x ℓ / 2 = 20 x 3 / 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

2.07 x 230 x 5172 x 10-6

Mu,lim =

6 2 Mu / b d2 =90 x 10 / ( 230 x 517 )

= 127.26

KNm

= 1.46 < 2.07 ( Table 6-3 ) Mu < Mu,lim 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 2 Ast = ( 0.464 / 100 ) x 230 x 517 = 552 mm 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 = ( ‫ﺡ‬bd = 1.6 x 1 = 1.6 N / mm

330.5

N / mm2

2

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 + 16 = = 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 N / mm2 = 12.64 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 -3 Fbt = 160 x 201 x 10 = 32.16 KN

550

N / mm2 ( area of 16 O bar = 201 mm2 )

Bearing stress = Fbt / r O 3 = 32.16 x 10 / 180 x 16 2 2 = 11.16 N / mm < 12.64 N /mm The arrangement is thus satisfactory. ( c )Check for shear : Vu = 60 KN

……………………..( safe )

Actual shear strength ‫ﺡ‬v = Vu / bd 3 = 60 x 10 / ( 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 ) 0.25 difference = 0.507 2 N / mm < ‫ﺡ‬v 0.243 difference design ( permissible ) shear strength ‫ﺡ‬c = 0.46 shear design is necessary . Vus = ( 0.505 - 0.46 ) x 230 x 517 x10-3 At support , Vus = Vu - ‫ﺡ‬c b d OR -3 = 60 - 0.46 x 230 x 517 x 10 = 5.35 = 60 54.699 = 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 3

= 0.87 x 250 x 56 x 517 / 5.3 x 10 = 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 ………………………( safe ) < 8.4 ( 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. 50 279

295

r 108 r

3-16 O

internal radius r = 180 550

295 150 500 Stirrups

2-10 O

3000 mm 6 mm O @ 130 mm c/c

consider self-weight. 23.2 KN /m

3m

69.6 KN

104.4 KNm

-

2.22 x 230 x 5152 x 10-6

IS 456-2000 , Table 19 Table 7-1 Pt = Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Concrete grade 100 x M15 M20 M25 M30 M35 As

≤ 0.15

, table 19 table 7-1 ) 0.04 ?

nsion reinforcement

………………………( safe )

4-20 O v

v

2-12 O

-0.0304

0.28

0.28

0.29

0.29

0.29

0.25 0.35 0.36 0.36 0.37 0.37 0.50 0.46 0.48 0.49 0.50 0.50 0.75 0.54 0.56 0.57 0.59 0.59 1.00 0.60 0.62 0.64 0.66 0.67 1.25 0.64 0.67 0.70 0.71 0.73 1.50 0.68 0.72 0.74 0.76 0.78 1.75 0.71 0.75 0.78 0.80 0.82 2.00 0.71 0.79 0.82 0.84 0.86 2.25 0.71 0.81 0.85 0.88 0.90 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

( given )

2.07 x 230 x 5172 x 10-6

alent to an anchorage

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

16 25

IS 456-2000 Clause 26.2.2.5 Bearing stresses at bends The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall b calculated as given below :

16 25

The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall b 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 ck / [ ( 1+ 2O/a Where, fck is the characteristic strength of concrete and

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

, table 19 table 7-1 ) 0.08 ?

-0.0778

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

KN

nsion reinforcement

………………………( safe )

3-16 O

2-10 O

230

ble 19

crete , ‫ﺡ‬C, N / mm2 e grade 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 x fck ( √ 1 + 5 x β - 1 )

ut not less than 1.0

stresses at bends ds and hooks described in IS : 2502-1963 ess inside a bend in any other bend shall be

ds and hooks described in IS : 2502-1963 ess inside a bend in any other bend shall be

n a bar or group of bars,

ize of bar of equivalent area

stress shall not exceed 1.5 f ck / [ ( 1+ 2O/a ) ]

een bars or group of bars , for a bar or group mber a shall 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 2 Mu = w x ℓ / 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

OR Mu,lim =

6 2 Mu / b d2 = 203.69 x 10 / ( 230 x 545 )

= 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 = 141.41

2.07 x 230 x 5452 x 10-6

= 141.41

KNm

Mu > Mu,lim 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 consider d' / d = 0.1 . = 0.07 Stress in compression steel , fsc = 353 N / mm2 ( refer to table 6-6 ) Asc = Mu2 / ( fsc x ( d - d' ) ) 6 = 62.28 x 10 / 353 ( 545 - 40 ) mm2 = 349

Corresponding tension steel Ast2 = Asc fsc / 0.87 fy = 349 x 353 / ( 0.87 x 415 ) mm2 = 341.2 Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) ) 6 = 141.41 x 10 / ( 0.87 x 415 ( 545 - 0.42 x 0.48 x 545 ) ) 6 = 141.46 x 10 / 361.05 ( 435.13 ) = 900 mm2

Ast = Ast,lim + Ast2 . 2 = 900 + 341.2 mm = 1241.2 mm2 Provide mm2 Asc = 2-16 O = 402 mm2 Ast = 4-20 O = 1256

( 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 3 = 162.95 x 10 / ( 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 N / mm2 < ‫ﺡ‬v design ( permissible ) shear strength ‫ﺡ‬c = 0.6 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 -3 = 139.27 - 0.6 x 230 x 545 x 10 = 64.06 KN Assuming 8 mm O two - legged stirrups , Asv = 100 mm2 , fy = 415 N / mm2 .

Sv = 0.87 fy Asv d / Vus 3

= 0.87 x 415 x 100 x 545 / 64.06 x 10 = 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 -3 -3 = (0.87 x 415 x 100 x 545 x 10 / 390 ) + 0.6 x 230 x 545 x 10 = 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 = hagg + 5 mm = 20 + 5 = 25 mm or 20 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.

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

2.07 x 230 x 5452 x 10-6

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 0.2 250 217 217 217 217 415 355 353 342 329 500 424 412 395 370 550 458 441 419 380

If Mu > Mu,lim : design the section either increasing the dimensions of section o Over reinforced section ( doubly-reinforced beam ).The additional moment of res needed is obtained by providing compression( top ) reinforcement and additional 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 19 table 7-1 )

nce d will be used for

face of the support

nsion reinforcement

mpression reinforcement

………………………( safe )

her increasing the dimensions of section or deign as rced beam ).The additional moment of resistance M u2 ession( top ) reinforcement and additional tensile ained below .

rced ( singly reinforced )

ced ( doubly 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. 2 Mu = w x ℓ / 8 2 = 30 x 6 / 8

= 135

KNm

Vu = w x ℓ / 2 = 30 x 6 / 2 = 90

KN

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

OR

6 2 Mu / b d2 = 135 x 10 / ( 230 x 565 )

= 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

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

Pt = 50

1 - √ 1 - ( 4.6 / 15 ) x ( 1.84 ) 415 / 15

= 0.614 2 Ast = ( 0.614 / 100 ) x 230 x 565 = 798 mm

As per IS 456-2000 clause 26.5.1.1 ( a ) Minimum steel , As = ( 0.205 / 100 ) x 230 x 565 = 266 mm2 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 90 KN , L0 = 12 O ( assume )

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 3 = 73.05 x 10 / ( 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 ) 0.25 difference = 0.309 2 N / mm design ( permissible ) shear strength ‫ﺡ‬c = 0.376 < ‫ﺡ‬v 0.191 difference shear design is necessary . Vus = ( 0.562 - 0.376 ) x 230 x 565 x10-3 At support , Vus = Vu - ‫ﺡ‬c b d 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 -3 = 2 x 201 x 0.87 x 415 x sin 45º x 10 ( 0.87 fy Asv sin α ) = 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 3

= 0.87 x 250 x 56 x 565 / 12.09 x 10 = 569.2 mm From IS 456-2000 clause 26.5.1.6 Spacing of minimum shear reinforcement using 6 O stirrups

= 24.17

= 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 -3

= 67.5 - 0.376 x 230 x 565 x 10 = 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

6 mm O @ 130 mm 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 t,lim , SP : 16 ,2.3

( 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 explained

Pt = 50 If Mu = Mu,lim :

2.07 x 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 deign a Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance M needed is obtained by providing compression( top ) reinforcement and additional tensile 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 Mu,lim / fck b d

2

Pt,lim fy / fck

250

415

500

550

0.148

0.138

0.133

0.129

21.93

19.86

19.03

18.2

Table 6-3 2

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

fy, N / mm2

fck N / mm2

250

415

500

550

15 20 25 30

2.22 2.96 3.70 4.44

2.07 2.76 3.45 4.14

2.00 2.66 3.33 3.99

1.94 2.58 3.23 3.87

Table 6-4

n , shear at distance d

Limiting Percentage of Reinforcement Pt,lim For singly reinforced rectangular sections

fy, N / mm2

fck

, table 19 table 7-1 ) 0.11 ?

250

415

500

550

15 20 25 30

1.32 1.75 2.19 2.63

0.72 0.96 1.20 1.44

0.57 0.76 0.95 1.14

0.50 0.66 0.83 0.99

-0.084 -3

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

KN

N / mm2

IS 456-2000 Clause 26.5 Requirements of Reinforcement for 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. Minimum steel % For mild steel

bars is not available

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

on reinforcement

………………………( safe )

Design bond stress (‫ﺡ‬bd ) for plain bars in tension Concrete grade ( ‫ﺡ‬bd

M15 2 N / mm 1.0

M20 1.2

M25 1.4

M30 1.5

M35 1.7

M40 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 %. σs = 0.87 x fy For mild steel Fe250 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 250 415 500

Tension bars M15 M20 55 O 26 O 56 O 47 O 69 O 58 O

check for shear

Compression bars

M15 44 O 45 O 54 O

M20 37 O 38 O 46 O

IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Pt = 100 x As Concrete grade bxd 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 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 such that : Asv / b sv ≥ 0.4 / 0.87 fy 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 2 fy = characteristic strength of the stirrup reinforcement in N / mm 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

he depth may be assumed

s factored moment )

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

tion (singly reinforced beam) as explained below.

explained in ( 1 ).

asing the dimensions of section or deign as m ).The additional moment of resistance M u2 op ) reinforcement and additional tensile ow .

a of tension reinforcement

ea of tension

hichever is greater

ups shall be provided

s effective in shear ,

web of flanged beam , and nforcement 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 = l0 / 6 + bw + 6 Df = 0.7 x 5000 / 6 + 230 + 6 x 120 = 1533.3333 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 = 175238.1 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 3 = 1.5 x ( ( 1 / 12 ) x 230 x 420 ) / 5000 = 426006 mm3 Column stiffness Kc = Ic / l = 1/12 x 400 x 2303 / 7000

= 57938.095 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 2 -3 = ( 1500 x 400 / 2000 ) x (15) x 10 = 67.5 KNm May = ( Pu b / 2000 ) x ( ℓey / b )2 2 -3 = ( 1500 x 300 / 2000 ) x (16.7) x 10 = 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 fy Asc 2 mm Ac = 400 x 300 - 120000*3/100 = 120000

From chart 63 , puz / Ag = 22.5 N/mm2 -3

OR

2 = 116400 mm

Puz = 22.5 x 120000 x 10 Puz = 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 36 = 2700 KN = 1571.4 + Calculation of Pb : = 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) = 27.333

ey = ( ℓ / 500 ) + ( b / 30 ) = ( 7000 / 500 ) + ( 300 / 30) = 24 Moments due to minimum eccentricity : -3 Mux = 1500 x 27.33 x 10 = 41.0 KNm > 15.0 KNm -3 Muy = 1500 x 24 x 10 = 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 3 Pu / fck b D = 1500 x 10 / ( 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 2 Mux1 = 0.104 x 30 x 300 x 400

= 149.8

KNm

referring to chart 46 (d' / D = 0.2 ) , Mu / fck b D2 = 0.096 2 Muy1 = 0.096 x 30 x 400 x 300 = 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 ( 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. 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 Muxx = 80 kNm at top and 60 KNm at bottom. Factored.moment in the direction of shorter dimension Muyy = 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 2 -3 = ( 1000 x 450 / 2000 ) x (13.2 ) x 10 = 39.2 KNm May = ( Pu b / 2000 ) x ( ℓey / b )2 2 -3 = ( 1000 x 230 / 2000 ) x (15.6 ) x 10 = 28 KNm About XX Pu = 1000 KN

Muxx =

Mi + k x Max

Mi = = = = Take ,

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

Note that Mu1 is considered negative as the column bends in double curvature . Mu,xx = 32 + 0.8 x 39.2 = 63.36 KNm. Mu,xx = 80 KNm Take , About YY Pu = 1000 KN Muyy = Mi + k x May

< 80 KNm

Mi = = = = Take ,

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

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 3 1000 x 10 / 20 x 230 x 450 = 0.48 P'u / fck b D = 6 2 M'uy / fck b D2 = 61.33 x 10 / 20 x 450 x 230 = 0.129

p / fck = p= = Asc =

from chart 46 , SP : 16 = 0.162 x 20 3.24 ( 3.24 /100 ) x 230 x 450

0.162

= 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 + 1002.23 = 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

Take , Mux = 80 KNm

Muy = 16 + 0.59 x 28 = 32.52 KNm < 40

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 2 -6 Mux1 = 0.145 x 20 x 230 x 450 x 10 = 135.07 KNm About YY d' / D = 50 / 230 = 0.22 ≈ 0.2 From chart 46 , SP : 16 Muy1 / fck b D2 = 0.13 2

Muy1 = 0.13 x 20 x 450 x 230 x 10 = 61.89 KNm Check : Pu / Puz = 1000 / 1904.7 = 0.525 From , IS 456-2000 , Clause 39.6 αn = 1.542 check : αn Mux

+

Mux1

-6

Muy Muy1

1.542 80 135.07 0.446

αn


0.34 ?

≤1

1.542

+ +

40 61.89 0.51

…………….( O.K.) 0.956 ≤1 230 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.

bending )

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, For Circular column / D ≤ 10 .

( Given )

and 22.5 KNm at bottom. and 20 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 Members ( C Degree of End Restraint of Symbol 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.

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

ed equally on four sides.

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

39.7.1.1 of the IS 456-2000

0.45 fck Ac + 0.75 fy Asc

400 x 300 - 120000*3/100 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3 1120.5 SP : 16 ,Table 60 Slender compression members- Values of P b

Rectangular sections : Pb / fck b D = k1 + k2 . p / fck Circular sections : Pb / fck D2 = k1 + k2 . p / fck Values of k1 Section Rectangular

0.05 0.219

d' / D 0.10 0.15 0.207 0.196

Circular

0.172

0.160

0.149

250 415 500 250 415 500 250 415

0.05 -0.045 0.096 0.213 0.215 0.424 0.545 0.193 0.410

d' / D 0.10 -0.045 0.082 0.173 0.146 0.328 0.425 0.148 0.323

500

0.543

0.443

Values of k2 Section fy N/ mm

Rectangular ; equal reinforcement on two opposite sides Rectangular ; equal reinforcement on four sides

e above values of k . Circular the initial moments of the code IS 456-2000 )

2

IS 456-2000 clause 39.7.1 The additional moments Max and May shall be calculated by the following formula Max = ( Pu D / 2000 ) x ( ℓex / D )2 May = ( Pu b / 2000 ) x ( ℓey / b )2 Where , Pu = axial load on the member,

from minimum eccentricity as the initial moment for

ℓex = effective length in respect of the major axis , mm

> 20 mm

mm

> 20 mm

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 factor : K = ( Puz - Pu ) / ( Puz - Pb ) ≤1 where , Pu = axial load on compression member, Puz = as defined in 39.6, -0.068

Puz = 0.45 fck Ac + 0.75 fy Asc

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

O tr

8O@ 300 c/c (two sets )

(two sets )

shall be used . r internal bars would be

bending )

top and 60 KNm at

at top and 30 KNm at

e axes are less than

( Given )

ut first trial steel are

-0.1275

4-25 O + 4 -20 O 450

gitudinal diameter of bar )

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

O tr s shall be used . for internal bars would be

sidered as short when both slenderness OR lex / ixx < 40 where, For Circular column lex

ct of the major axis

ct of the minor axis

ect of the major axis.

ect of the minor axis.

ve Length 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

ength of compression member. slab

Beam

s the distance between the points of zero moment (contraflexure ) along the

members- Values of P b

/ fck b D = k1 + k2 . p / fck

ck

D2 = k1 + k2 . p / fck d' / D 0.20 0.184 0.138 d' / D 0.15 -0.045 0.046 0.104 0.061 0.203 0.256 0.077 0.201

0.20 -0.045 -0.022 -0.001 -0.011 0.028 0.040 -0.020 0.036

0.291

0.056

hall be calculated by the following formulae :

espect of the major axis ,

espect of the minor axis , section at right angles to

d in a given plane if lateral rovided by walls or

st all lateral forces in dered unbraced. out any transverse loads oment shall be added 4 Mu1, and 0.6 Mu2,

nd Mu1 is the smaller

column is bent in double

moment be less than

the initial moment be

he additional moment

at any given level or storey, rained to deflect o for each column may

s acting in the same

ay be multiplied by the following factor :

Puz = 0.45 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 . 3 P'u / ( fck x b x D ) = 1500 x 10 / ( 15 x 500 x 500 ) = 0.4 6 2 2 M'ux / ( fck x b x D ) = 210 x 10 / ( 15 x 500 x 500 ) = 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 mm2 , 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 2 = 246780 mm Puz = 0.45 x 15 x 246780 x 10-3 + 0.75 x 415 x 3220 x 10-3 = 1665.8 = 2668

+

1002.2

KN

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


0.34 ?

-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 . 3 P'u / ( fck x b x D ) = 1500 x 10 / ( 15 x 300 x 500 ) = 0.67 6 2 2 M'u / ( fck x b x D ) = 72 x 10 / ( 15 x 500 x 300 ) = 0.107 d' = 40 + 10 = 50 mm d' / D = 50 / 300 = 0.167 ………………..use 0.2 From chart 34 , SP-16 p / fck = 0.167 p = 0.167 x 15 = 2.5 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 / f ck = 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 = 985.99 + 1222.6 = 2208.6 KN Pu / Puz = 1500 / 2208.6 = 0.68 From , IS 456-2000 , Clause 39.6 αn = 1.802 0.2 difference 0.12 difference check : αn αn Mux Muy 1 Mux1 Muy1

≤

+

1.802 60 146.25 0.2

+ +

0.33 ?

1.802 60 69.53 0.77

0.97 ≤1 Ties : Minimum diameter Otr = 25 / 4 = 6.25 mm 500

-0.198

Use 8 mm diameter M.S. ties. Spacing should not exceed lesser of ( i ) 300 mm ( ii ) 16 x 25 = 400 mm ( iii ) 48 x 8 = 384 mm Provide 8 mm O M.S. ties @ 300 mm c/c

500

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.

90 + 120 = 210 KNm .

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

αn Mux Mux1

αn

+

Muy Muy1

Mux , Muy =

≤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 αn vary linearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, α n is 1 .O; for values greater than 0.8, αn is 2.0.

25 O + 20 O

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

8O@ 350 c/c

300 / 500 ( 60 + 60 ) = 72 KNm .

8-25 O

ies shall be used .

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

, 0.4 , 0.6 , 0.8 )the values of αn vary

67 , 2.0 ) For values less than 0.2, α n 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 3 Pu / ( fck x b x D ) = 600 x 10 / ( 15 x 300 x 600 ) = 0.222 6 2 2 Mu / ( fck x b x D ) = 300 x 10 / ( 15 x 300 x 600 ) = 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.

ay be tied by open ties.

ies shall be used .

8-25 O + 2-12 O

8 mm O @ 300 mm c/c ( two sets )

( 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 = Ag + 7.936 2.2244 Ag 1800 x 103 = 1800 x 103 =

Ag 2 Ag = 177165 mm 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 10.1604

= 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 + 2.3356 Ag 1800 x 103 = 8.3328 1800 x 103 = 10.6684

Ag 2 Ag = 168729 mm If D is the diameter of the column ( ¶ / 4 ) x D2 = 168729 D = √ 214941 D = 463 mm Use 450 mm diameter column. 2 Ag = ( ¶ / 4 ) x 450 Then , = 158963 mm2 1800 x 10 = 1.05 [0.4 x 20 x (158963 - Asc ) + 0.67 x 415 x Asc ] Asc = 1335289 8.4 + 291.95 3

464711

=

283.55

Asc mm2

Asc = 1638.9 provide 20 mm diameter bars ( 1638.9 / 314 =) 6 No. Asc = 6 x 314 = 1884 mm2 Assume 8 mm O M.S. bars for helix at 40 mm clear cover . Dc = 450 - 40 - 40 = 370 mm 2 asp = ( ¶ / 4 ) x 8 = 50 mm2 Minimum ρs = 0.36 ( (Ag / Acr) - 1 ) fck / fy 2 2 = 0.36 ( ( 450 / 370 ) - 1 ) x 20 / 250 = 0.36 x 0.4792 x 20 / 250 = 0.0138 ρs = 4 x asp / p x Dc Now 0.0138 = 4 x 50 / p x 370 …………………….. ( 1 ) p = 39.17 mm 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.

Asc

gitudinal 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 Material M15 grade concrete mild steel Fe250 Solution : -

Given

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 = 3 7.292 Ag 1500 x 10 = 2 Ag = 205705 mm 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 2 mm Asc = 1765 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 2 mm = 2060 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

400 mm x 400 mm

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 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 the columns below have to be lapped with those in the

es shall be used .

20 O 16 O

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

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.

hen both slenderness re,

ntricity equal to the dimension / 30 , subject to ccentricity exceeds the

ed by the following

ement , shall be not less -sectional area of the

Design of Single flight staircase Riser = 200 mm Tread = 230 mm Nosing is provided to comply with minimum 250 mm tread size. Height of the floor = 3.2 m. ( given ) 2 Live load = 3 KN / m . Grade concrete Material M15 Fe415 HYSD reinforcement. Solution : Assume 180 mm thick waist slab. Inclined length of waist slab for one step 2 2 = √ 230 + 200 = 304.8 mm. Waist slab self weight in plan = 0.18 x 25 x 304.8 / 230 = 5.96 + 0 KN / m2. Weight of steps = (( 0 + 0.2 ) / 2 ) x 25 = 2.5 + 0 KN / m2. Floor finish length for each step ( 230 tread + 20 nosing ) + = 250 200 ( riser ) = 450 mm. Floor finish load 2 = ( 450 / 230 ) x 1 KN / m = 1.96 KN / m2. Parapet load is assumed to be uniformly distributed on the slab. Parapet load is 0.15 x 0.75 x 20 / ( 1 x 1 ) = 2.25 KN / m2. The total distributed loads on the stair are as follows : Loads on flight : waist slab 5.96 + 0 KN / m2. steps 2.5 + 0 KN / m2. floor finish 1.96 + 0 KN / m2. Parapet load 2.25 + 0 KN / m2. live load 0 + 3 KN / m2. Total 12.67 3 KN / m2. Factored load 1.5 ( 12.67 + 3 ) = 19 + 4.5 KN / m2. Loads on landing : waist slab ( 150 mm ) 0.15 x 25 = 3.75 + 0 KN / m2. floor finish 1.0 + 0 KN / m2. parapet 2.25 + 0 KN / m2. live load 0 + 3 KN / m2. 7 + 3 KN / m2. Factored load 1.5 ( 7 + 3 ) = 10.5 + 4.5 KN / m2.

Building inside

1

230

Also the end parapet point load 2.25 x 1.5 ( factored Value ) = 3.38 CD = KN / m. The loading diagram is shown in fig., for 1 m wide stair. Maximum ( + ) ve moment When cantilever has only dead load 2 2 MBC = w x ℓ / 2 + P x ℓ = 10.5 x 1.115 / 2 + 3.75 x 1.115 = 6.53 + 4.18 = 10.71 RA = w x ℓ / 2 - Moment @ B = ( 23.5 x 3.68 / 2 ) - ( 10.71 / 3.68 )

A

KNm.

= 43.24 2.91 = 40.33 KN. ( RA + Rx ) = 23.5 x ( x ) ) Point of zero shear at x = 40.33 / 23.5 = ( 40.33 + 0 ) = 23.5 x ( x ) ) 1.72 m. 2 Mu ( + ) = 40.33 x 1.72 - 23.5 x 1.72 / 2 ( moment is maximum when shear is zero ) = 69.37 34.76 = 34.61 KNm. Vu( AB ) = 40.33 KN. Maximum ( - ) ve moment at point B 2 w x ℓ / 2 + P x ℓ = 15 x 1.1152 / 2 + 3.75 x 1.115 Mu ( - ) = = 13.51 KNm. Vu( BA ) = w x ℓ / 2 + Moment @ B = ( 23.5 x 3.68 / 2 ) + ( 13.51 / 3.68 ) = 43.24 + 3.67 = 46.91 KN. Vu( BC ) = w x ℓ + P = 15 x 1.115 + 3.75 = 20.48 KN. Positive moment reinforcement : Mu = 34.61 KNm. d = 180 - 15 ( cover ) - 6 = 159 mm. 6 2 2 Mu / b x d = 34.61 x 10 / 1000 x 159 = 1.37 < 2.07 For M15 Fe415 , Q = 2.07 2 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d ) fy / fck = 50 1-√1-(4.6 / 15) x (1.37) 415 / 15 = 0.431 Ast = 0.431 x 1000 x 159 / 100 = 685 mm2. Provide 12 mm O bar Area of one bar = 113 spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 113 x 1000 /685 = 165 mm Provide 12 mm O bar @ 160 mm c/c = 706 mm2 . Check for deflection : Basic ( span / d ) ratio = 20 Pt = 100 x Ast / b x d = 100 x 706 / 1000 x 159 = 0.444 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement

modification factor = 1.25 ( span / d ) ratio permissible = 1.25 x 20 = 25 Actual (span / d ) ratio = 3680 / 159 = 23.14 < Negative moment reinforcement : Mu ( - ) = 13.51 KNm. d = 150 - 15 ( cover ) - 5 = 130 6 2 2 Mu / b x d = 13.51 x 10 / 1000 x 130 = 0.8 < 2.07 Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2)

25

……………….( O.K.)

mm. For M15 Fe415 , Q = 2.07

fy / fck 50 1-√1-(4.6 / 15) x (0.8) 415 / 15 = 0.237 Ast = 0.237 x 1000 x 130 / 100 =

= 308 mm2. Provide 10 mm O bar Area of one bar = 78.5 spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 78.5 x 1000 /308 = 255 mm Provide 10 mm O bar @ 200 mm c/c = 392 mm2 . Check for deflection : Basic ( span / d ) ratio = 7 Pt = 100 x Ast / b x d = 100 x 392 / 1000 x 130 = 0.302 IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement modification factor = 1.42 ( span / d ) ratio permissible = 1.42 x 7 = 9.94 Actual (span / d ) ratio = 1150 / 130 ……………….( O.K.) = 8.85 < 9.94 Check for shear : Vu = 40.33 At A , KN Shear stress ‫ﺡ‬v = Vu / b x d 3 = 40.33 x 10 / 1000 x 159

= 0.254

N / mm2

< ( ‫ﺡ‬C ) N / mm2

100 x As / b x d = 100 x 706 / 1000 x 159 = 0.444 from table 7-1 N / mm2 for Pt = 0.444 ‫ﺡ‬c = 0.435 IS 456-2000 clause 40.2.1.1

( as calculated below )

0.25 diff. 0.056 diff.

0.11 ?

for 180 mm slab depth k = 1.24 Design shear strength = 1.24 x 0.435 N / mm2 = 0.539 At B ,

Vu = 20.48

25 diff. -0.05 20 diff. ? > ‫ﺡ‬v ……………….( O.K.)

KN

Shear stress ‫ﺡ‬v = Vu / b x d 3 = 20.48 x 10 / 1000 x 130 N / mm2 < ( ‫ﺡ‬C ) N / mm2 = 0.158

100 x As / b x d = 100 x 392 / 1000 x 130 = 0.302 from table 7-1 N / mm2 for Pt = 0.302 ‫ﺡ‬c = 0.373 IS 456-2000 clause 40.2.1.1 k = 1.3 for 150 mm slab depth Design shear strength = 1.3 x 0.373 N / mm2 = 0.485 At point of contraflexure ( 0.15 ℓ from B ) Vu = 46.91 - ( 0.15 x 3.68 x 23.5 ) = 33.94 KN. Shear stress ‫ﺡ‬v = Vu / b x d

0.25 diff. 0.198 diff.

Pt = 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck 0.444= 50 1-√1-(4.6 / 15) x (Mu / b x d2 ) 415 / 15 ( Mu / b x d2 ) =

1.405

-6 Mu1 = 1.405 x 1000 x 159 x 159 x 10

= 35.52

KNm.

0.11 ?

> ‫ﺡ‬v ……………….( O.K.)

3 = 33.94 x 10 / 1000 x 159 N / mm2 < ( ‫ﺡ‬C ) N / mm2 = 0.213

100 x As / b x d = 100 x 706 / 1000 x 159 = 0.444 from table 7-1 N / mm2 for Pt = 0.444 ‫ﺡ‬c = 0.435 IS 456-2000 clause 40.2.1.1 for 150 mm slab depth k = 1.24 Design shear strength = 1.24 x 0.435 N / mm2 = 0.539 Check for development length : At A , Pt = 0.444 ,



0.25 diff. 0.198 diff.

> ‫ﺡ‬v ……………….( O.K.)

0.11 ?

Vu = 40.33

KN.

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 35.52 x 106 / 40.33 x 103 ) + 8 O ≥ 56.4 O ≥ 56.4 O 1144.95 + 8 O which gives

48.4 O ≤ 1144.95 O ≤ 23.66 mm

……………….( O.K.)

At point of contraflexure ( 0.15 ℓ from B ) Mu1 = Vu =

35.52 KNm. 33.94 KN.

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 35.52 x 106 / 33.94 x 103 ) + 8 O ≥ 56.4 O ≥ 56.4 O 1360.52 + 8 O 48.4 O ≤ 1360.52 O ≤ 28.11 which gives mm ……………….( O.K.) Development length for negative moment reinforcement …………………………………..( 1 ) = 56.4 x 10 = 564 mm Anchorage beyond point of contraflexure = 12 diameter i.e. 0.15 x 3680 + 12 x 10 mm from B …………………………………..( 2 ) = 672 mm. From ( 1 ) and ( 2 ) anchorage required = 672 mm. As a thumb rule provide anchorage of 1000 mm i.e. equal to the cantilever. The negative moment reinforcement is placed in the form of alternate chipiya as shown in fig. Distribution steel : According to IS 456-2000 clause 26.5.2.1 Minimum steel 0.15 % for Fe250 and 0.12 % for Fe415 In AB portion Minimum steel = ( 0.12 / 100 ) x 1000 x 180 = 216 mm2 Provide 8 mm O bar Area of one bar = 50.2 spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 50.2 x 1000 /216 = 232 mm Provide 8 mm O bar @ 230 mm c/c = 218 mm2 . In BC portion Minimum steel = ( 0.12 / 100 ) x 1000 x 150 = 180 mm2 Provide 8 mm O bar Area of one bar = 50.2 spacing of bar = Area of one bar x 1000 / required area in m 2 /m = 50.2 x 1000 /180 = 279 mm 10 mm O @ 200 c/c ( alternate Chipiya ) Provide 8 mm O bar @ 250 mm c/c = 200 mm2 . 1000 Sketch : 1000 15

16

1000 15 14 13

12 10 200

9 8

7

Chipiya Beam at slab 16level

11

230

16

8 mm O @ 230 mm c/c 12 mm O @ 160 mm c/c 180 mm thick waist slab

6 5 4 3 Plinth level 1

2

v Beam with minimum reinforcements

Wall from foundation

Note : At plinth level a small beam over the masonry is usually provided with minimum reinforcement and with minimum 230 mm x 230 mm size to give proper seating to the stair

Building inside

230 mm thick wall

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 150 mm thick masonry parapet 0.75 m high 3450

Beam at slab level

230

16

Landing at slab level 1000

( 3.75 + 0 ) KN

( 3.75 + 0 ) KN ( 19 + 4.5 ) KN / m (10.5 + 4.5 )

A

C

B

1.115 m

3.68 m

RA + Rx ) = 23.5 x ( x ) ) 40.33 + 0 ) = 23.5 x ( x ) ) um when shear is zero )

Table 6-3 2

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

fy, N / mm2

fck N / mm2

250

415

500

550

15 20 25 30

2.22 2.96 3.70 4.44

2.07 2.76 3.45 4.14

2.00 2.66 3.33 3.99

1.94 2.58 3.23 3.87

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

0.0246

≤ 0.15 0.25 0.50 0.75

Concrete grade M15 0.28 0.35 0.46 0.54

M20 0.28 0.36 0.48 0.56

M25 0.29 0.36 0.49 0.57

M30 0.29 0.37 0.50 0.59

M35 0.29 0.37 0.50 0.59

M40 0.30 0.38 0.51 0.60

-0.04

0.087

1.00 1.25 1.50

0.60 0.64 0.68

0.62 0.67 0.72

0.64 0.70 0.74

0.66 0.71 0.76

0.67 0.73 0.78

0.68 0.74 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 3.00

0.71 0.71

0.82 0.82

0.90 0.92

0.94 0.96

0.96 0.99

0.98 1.01

IS 456-2000 Clause 40.2.1.1 Overall depth of 300 or slab , mm 275 250 more k 1.00 1.05 1.10

225 1.15

200 1.20

175 1.25

0.087

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 Concrete grade

M15

( ‫ﺡ‬bd N / mm 1.0 2

M20

M25

M30

M35

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

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 / Tension bars Compression bars

mm2 250 415 500

chipiya as shown in fig.

10 mm O @ 200 c/c ( alternate Chipiya ) 1000 150

M15 55 O 56 O 69 O

M20 26 O 47 O 58 O

M15 44 O 45 O 54 O

M20 37 O 38 O 46 O

150 Chipiya

8 O @ 250 mm c/c

Beam at slab 16level

8 mm O @ 230 mm c/c

provided with minimum er seating to the stair

spans upto 10 m :

150 or less 1.30

26.2.1.1

plain bars in tension M40 1.9 for bars in compression

irming to IS : 1786-1985 , 0 %.

12 O , whichever is greater

ed by a compressive 3 M1 / V )

Design of dog-legged staircase 230 1

5

9 10

UP Floor

900

A 20

B 19

v

vv 300 900

15

Rise of step = 160 mm

900 Rise

1950

900

11

2250

150

230 230

> 200 mm < 230 mm

175 mm to 200 mm

Tread = 250 mm Tread 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 2 PU = 1.5 x 7.75 = 11.63 KN / m Span = 1950 + 150 = 2100 i.e. 2.1 m Consider 1 m length of slab M = w x l2 / 8 2 = 11.63 x 2.1 / 8 = 6.41 KNm Reinforcement will be in second layer, Assuming 12 mm O bars d = 150 - 15 (cover ) -12 - 6 = 117 mm From table 6-3 , Q = 2.22 for M15 Fe250 6 2 Mu / b x d2 = 6.41 x 10 / 1000 x 117 = 0.468 < 2.22 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 456-2000 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 mm2 . Provide 10 mm O bar @ 280 mm c/c = 280 Maximum spacing = 3 x d = 3 x 117 = 351

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 3 = 12.21 x 10 / 1000 x 117 N / mm2 < ( ‫ﺡ‬C ) N / mm2 ( too small ) = 0.104

< 0.28

for Pt = 0.224 ‫ﺡ‬c = 0.28

( from table 7-1 ) ……………….( O.K.)

Check for development length : Assuming L0 = 12 O (mild steel ) Pt = = = From equation Pt =

100 x As / b x d 100 x 280 / 1000 x 117 0.239 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck 2

Mu1 / b x d = 0.499 2 -6 Mu1 = 0.499 x 1000 x 117 x 10 = 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 43 O ≤ 727.19 which gives O ≤ 16.91 Check for deflection : Basic ( span / d ) ratio = 20

mm

……………….( O.K.)

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

<

……………….( O.K.)

40

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 KN / m2 self-load in plan = ( 296.8 / 250 ) x 0.15 x 25 = 4.45 Floor finish length for one step = 160 + 250 = 410 mm floor finish = ( 410 / 250 ) x 1 = 1.64 KN / m2 Weight of step ( ( 0 + 160 ) / 2000 ) x 25 = 2.00 KN / m2 Live load = Total = = Pu = 1.5 x 11.09 =

250 160

296.8

2

3.00 KN / m 4.45 + 1.64 + 2.00 + 3.00 KN / m2 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 3300

RA = RB = 0.525 x 11.63 + 16.64 x 2.25 / 2

525

525 = 900 / 2 +75

= 6.11 + 18.72 = 24.83 KN 2 2 Mu = 24.83 x 3.3/2 - 11.63 x (3.3/2) / 2 - ( ( 16.64 - 11.63 ) x ( 2.25/2 ) ) / 2 = 40.97 - 15.83 - 3.17 = 21.97 KNm From Table 6-3 Q = 2.22 drequired = = = dprovided = = Mu / b x d2 = = Pt =

√M / Q x b √21.97 x 10 6 / 2.22 x 1000 99.48 mm, 150 - 15 - 6 ……………….( O.K.) 129 mm, 6 21.97 x 10 / 1000 x (129)2 1.32 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 mm2 = 225 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 3 = 24.83 x 10 / 1000 x 129

N / mm2

= 0.192

< ( ‫ﺡ‬C ) N / mm2

( too small )

100 x As / b x d = 100 x 904 / 1000 x 129 = 0.70 from table 7-1 2 for Pt = 0.7 ‫ﺡ‬c = 0.524 N / mm IS 456-2000 clause 40.2.1.1 for 150 mm slab depth k = 1.3 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 = = = From equation Pt =

100 x As / b x d 100 x 904 / 1000 x 129 0.70 50 1-√1-(4.6 / fck) x (Mu / b x d2) fy / fck

we get ,

2

Mu1 / b x d = 1.344 2 -6 Mu1 = 1.344 x 1000 x 129 x 10 = 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.K.) O ≤ 27.62 mm 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 ……………….( O.K.) = 25.58 < 32.4 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 small = 3 x 129 = 387 mm or 300 mm spacing provided = 125 < 300 mm ……………….( O.K.) mm Distribution bars : maximum spacing permitted = 5 x effective depth of slab or 450 mm whichever is = 5 x 129 = 645 mm spacing provided = 220 < 450 mm ……………….( O.K.) mm

Rise Tread

> 200 mm < 230 mm

175 mm to 200 mm 250 mm to 280 mm

175 mm to 200 mm

250 mm to 280 mm

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

fck

N / mm

15 20 25

fy, N / mm2

2

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

check for shear IS 456-2000 , Table 19 Table 7-1 Design shear strength of concrete , ‫ﺡ‬C, N / mm2 Concrete grade Pt = 100 x As M15 M20 M25 M30 M35 bxd ≤ 0.15 0.28 0.28 0.29 0.29 0.29 0.25 0.35 0.36 0.36 0.37 0.37 0.50 0.46 0.48 0.49 0.50 0.50 0.75

0.54

0.56

0.57

0.59

0.59

1.00

0.60

0.62

0.64

0.66

0.67

1.25 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 2.50 0.71 0.82 0.88 0.91 2.75 0.71 0.82 0.90 0.94 3.00 0.71 0.82 0.92 0.96 The above given table is based on the following formula

0.90 0.93 0.96 0.99

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

M15 M20 M25 M30 ( ‫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

Concrete grade

2

M35 3.7

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

2

M15 1.0

M20 1.2

M25 1.4

M30 1.5

M35 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 %. σs = 0.87 x fy For mild steel Fe250 For Fe415 IS 456-2000 clause 26.2.3.3 Ld ≤ M1 / V + L0

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 250 415 500

55 O 56 O 69 O

26 O 47 O 58 O

44 O 45 O 54 O

37 O 38 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 three 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 275 more k 1.00 1.05

250 1.10

225 1.15

200 1.20

sketch : The reinforcements for both the flights are shown in fig. 700 700 10 O @ 280 c/c ( Landing ) 12 O @ 125 c/c 150 mm thick waist slab

v v v v v v

8 O @ 220 c/c 12 O @ 125 c/c

v

v

v v

v

v

12 O @ 125 c/c

Flight-B

vv v

v v v v

8 O @ 220 c/c

v 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

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

19

e , ‫ﺡ‬C, N / mm2 e grade 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 x fck ( √ 1 + 5 x β - 1 )

ut not less than 1.0

20

, N / mm2

M40 4.0

2.1.1

n bars in tension M40 1.9

rs in compression o IS : 1786-1985 , the

whichever is greater

compressive reaction M1

or spans upto 10 m :

be multiplied by hich case 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

150 v

vv

v v v v v v v

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

Design of Rectangular Water Tank

Recommend Documents