Design of Rectangular Water Tank

Capacity 80000 Litres (given)

Material M20 (given)

Fe 415 Grade HYSD reinforcement (given)

80400 Litres L / B = 6 / 4 = 1.5 < 2 .

H / 4 = 3.5 / 4 = 0.875 m

2.5 x YW = 2.5 x 9.8

= 24.5 KN / m2 where Yw is unit weight of water = 9.8 KN / m

3 6 m 6 m A E F 3.5 m 24.5 KN / m2 D 34.3 KN / m2 Elevation Plan MAB = = 73.5 KNm MAD = = -32.66 KNm - 32.66 73.5 0 -12.25 -8.17 0 0 -12.25 -8.17 0 D 0 A 0 B -57.16 57.16

Fixed end moments

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

Volume = 6 x 4 x 3.35 x 10 3 =

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

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

The top portion of side walls will be designed as a continuous frame. bottom 1 m or H / 4 whichever is more is designed as cantilever. bottom 1 m will be designed as cantilever.

Rotation factor at Joint A

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

Kani's Method :-24.5 x 62 / 12 = 24.5 x 42 / 12 = w x l2 / 12 = w x l2 / 12 = -3/10 40.84 -2/10 2.5 m 1 m

Joint Member Relative Stiffness( K ) ∑ K Rotation Factor u =(-1/2) k / ∑ K AB I / 6 - 2 / 10 AD I / 4 - 3 / 10 MA F = 40.84 KNm MAB = = = 57.16 MAD = = (- 32.66 ) + 2 x (- 12.25 ) + 0 = -57.16 = 53.09 KNm = -8.16 KNm = 49 KN = 73.5 KN M = 57.16 KNm T = 49 KN Q = 0.306 D =

√

M / Q x b =

√

57.16 x 10 6 / 0.306 x 1000 = 432.2 mm, Take D = 450 mm d = 450 - 25 - 8 = 417 mm Direct tension in long wall =

A 5 * I / 12

Sum of FEM

73.5-32.66

B.M. at centre of long span =

B.M. at centre of short span = MAB F + 2 MAB' + MBA' 73.5 + 2 x (- 8.17 ) + 0 MAD F + 2 MAD' + MDA'

Direct tension in short wall = Yw ( H - h ) x L / 2 24.5 x 6 / 2 = Design of Long Walls

:-24.5 x 4 / 2 = w x l 2 / 8 - 57.16 w x l 2 / 8 - 57.16 Yw ( H - h ) x B / 2 24.5 x 42 / 8- 57.16 24.5 x 62 / 8- 57.16

Tension on liquid face.

Assuming d / D = 0.9 At support

From Table 9-6

From Table 9-5

M / σst x j x d =

= 1048 mm2

Ast2 for direct tension = T / σst =

= 327 mm2

= 1375 mm2

=

= 146.15273 mm

Provide 16 mm O bar @ 130 mm C/C…marked(a) = 1546 mm2 / m. At centre M = 53.09 KNm T = 49 KN e = M / T = = 1.08 m E = e + D / 2 - d b = = 888 mm D = 49 x 0.888 d = 43.51 KNm d' M / σst x j x d = = 617 mm2

Ast2 for direct tension = T / σst = = 327 mm2 = 944 mm2 = = 212.88136 mm Total Ast1 + Ast2 = 617 + 327 Provide 16 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 200.96 x 1000 / 944 E = e + D / 2 - d 1080 + 450 / 2 - 417 modified moment Ast1 for moment = 43.51 x 10 6 / 190 x 0.89 x 417 49 x 10 3 / 150

Larger steel area is provided to match with the steel of short walls. 49 x 10 3 / 150

Total Ast1 + Ast2 = 1048 + 327

i.e.tension is small

Line of action of forces lies outside the section spacing of bar = Area of one bar x 1000 / required area in m2 / m

57.16 x 10 6 / 150 x 0.872 x 417 Ast1 for moment =

53.09 / 49 200.96 x 1000 / 1375 Provide 16 mm O bar

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

= 720 mm2

360 mm2

=

= 139.55556 mm

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

=

= 218.05556 mm

Provide 10 mm O bar @ 200 mm C/C…marked(c) = 392 mm2 Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005 mm2

385 mm2 mm2 M = 57.16 KNm T = 73.5 KN M / σst x j x d = = 1048 mm2

Ast2 for direct tension = T / σst =

= 490 mm2

= 1538 mm2

=

= 130.6632 mm

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

M = 8.16 KNm

T = 73.5 KN

200.96 x 1000 / 1538

tension on liquid face 57.16 x 10 6 / 150 x 0.872 x 417

73.5 x 10 3 / 150 Total Ast1 + Ast2 = 1048 + 490

Provide 16 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392 Design of short walls

:-At support

tension on liquid face From Table 9-5

A_st1 for moment =

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 /360

Horizontal steel :-

Liquid face ( d) - Provide 8 mm O @ 130 mm C/C = Provide 10 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 78.50 x 1000 /360

Vertical Steel ( c) Distribution steel =

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

On each face = Provide 8 mm O bar

M / σst x j x d =

= 150 mm2

Ast2 for direct tension = T / σst =

= 490 mm2

= 640 mm2

=

= 176.625 mm

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

= 720 mm2

360 mm2

=

= 139.55556 mm

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

=

= 218.05556 mm

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

Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869 mm2 385 mm2 mm2 M = OR Y_w x H / 6 , whichever is greater. = = 9.8 x 3.5 / 6 = 5.72 KNm = 5.72 78.50 x 1000 /360 Horizontal steel :-

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

Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392 spacing of bar = Area of one bar x 1000 / required area in m2 / m

50.24 x 1000 /360

Vertical Steel ( c) Provide 10 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m From Table 9-3 minimum reinforcement 0.16 %

Distribution steel = 0.16 / 100 x 1000 x 450 On each face =

Provide 8 mm O bar

Total Ast1 + Ast2 = 150 + 490 Provide 12 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 113.04 x 1000 / 640

From Table 9-5 Ast1 for moment =

8.16 x 10 6 / 150 x 0.872 x 417

73.5 x 10 3 / 150

Bottom 1 m will be designed as cantilever

,tension on liquid face. Cantilever moment :

-Yw x H x h 2

/ 6 9.8 x 3.5 x 1 / 6

M / σst x j x d = = 105 mm2 = 720 mm2 360 mm2 = = 218 mm

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

0.229% = = 344 mm2 = = 292 mm Ast = 346 mm 2 l_x = 4 + 0.15 = 4.15 say 4.5 m ly = 6 + 0.15 = 6.15 say 6.5 m 3.75 KN / m2 1.0 KN / m2 1.5 KN / m2 6.25 KN / m2 PU = = 9.38 KN / m l_y / l_x = 6.5 / 4.5 = 1.4

From Table 9-3 minimum reinforcement 0.16 % From Table 9-5

Ast for moment =

5.72 x 10 6 / 150 x 0.872 x 417

spacing of bar = Area of one bar x 1000 / required area in m2 /m 78.50 x 1000 /360 each face Distribution steel = 0.16 / 100 x 1000 x 450 On each face = Provide 10 mm O bar Base slab

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

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

0.229 / 100 x 1000 x 150 Provide 8 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 /172

Dead Load : self 0.15 x 25 = floor finish =

1.5 x 6.25

,172 mm2 bothway

Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. Designed section,Elevation etc. are shown in fig.

Live load =

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

For 1 m wide strip

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

Table 26 αx = 0.085 α_y = 0.056 Mx = My = = = = 16.15 KNm = 10.64 KNm

√

M / Q x b =

√

16.15 x 10 6 / 2.76 x 1000 = 76.50 mm d_short = = 130 > 76.50 mm dlong = 130 - 10 = 120 mm = 0.96 Pt = fy / fck = 415 / 20 = = 0.29% = 377 mm2 = = 208 mm Provide 10 mm O bar @ 210 mm c/c = 374 mm2 . = 0.74 Pt = fy / fck = 415 / 20 …………(O.K.) Larger depth is provided due to deflection check.

αy x w x lx 2 0.085 x 9.38 x 4.52 0.056 x 9.38 x 4.52 αx x w x lx 2 From Table 6-3 ,Q = 2.76 Mu / b x d 2 (short) = 16.15 x 106 / 1000 x 130 x 130 50 1-

√

1-(4.6 / fck) x (Mu / b x d2) Mu / b x d 2 (long) = 10.64 x 106 / 1000 x 120 x 120 50 1-

√

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

√

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

√

1-(4.6 / 20) x (0.96) 50 [(1-0.88) x 20 / 415 ] 78.50 x 1000 /377 Ast (short) = 0.29 x 1000 x 130 / 100 Provide 10 mm O bar

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

= = 0.22% = 264 mm2 = = 190 mm = 264 mm2 . 50 [(1-0.91) x 20 / 415 ] Provide 8 mm O bar @ 190 mm c/c Provide 8 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 /264

B

4 m

Q = M / bD2 Pt Q = M / bD 2 Pt 0.75 0.3 0.4 0.295 0.289 0.8 0.305 0.37 0.299 0.272 0.85 0.31 0.355 0.302 0.258 0.9 0.314 0.335 0.306 0.246

Balanced Design Factors for members in bending For M20 Grade Concrete Mix

Mild steel HYSD bars d / D

TABLE 9-5 TABLE 9-6

Members in bending ( Cracked condition ) Coefficients for balanced design

Grade of concrete Grade of steel σ_cbc N / mm2 σ_st N / mm2 k j Q M20 Fe250 7 115 0.445 0.851 1.33 Fe415 7 150 0.384 0.872 1.17

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

M20 Fe250 7 125 0.427 0.858 1.28 Fe415 7 190 0.329 0.89 1.03 D / 2 e = M / T 0.214 0.24 0.229 0.217 0.206 0.194 0.183 0.171 300 350 400 0.3 0.286 0.271 0.257 0.243 0.229 Thickness, mm 100 150 200 250 TABLE 9-3

Minimum Reinforcement for Liquid Retaining Structures % of reinforcement

Mild Steel HYSD bars Line of action of forces lies outside the section

0.2 0.16 450 or more

3500 150 1 : 4 : 8 P.C.C. 150 450 6000 450 150 Elevation 1500 450 1000 4000 1000 450 1500 1500 6000 450 450 Section A-A Table 6-3 A ₁₅₀₀ 1500 _A 150 150

8 O @ 290 c/c both ways top and bottom 10 O @ 200 c/c 10 O @ 200 c/c - shape - shape 10 O @ 210 c/c 8 O @ 190 c/c 150 Free board v v v v 1000 16 O @ 130 c/c (a) 10 O @ 200 c/c both faces (c) 12 O @ 130 c/c (e) 16 O @ 200 c/c (b) 8 O @ 130 c/c (d) ( a ) _{( b ) ( d )} ( a ) ( c ) ( d )

250 415 500 550

15 2.22 2.07 2.00 1.94

20 2.96 2.76 2.66 2.58

25 3.70 3.45 3.33 3.23

30 4.44 4.14 3.99 3.87

Limiting Moment of resistance factor Q lim, N / mm 2

fy, N / mm 2

fck N / mm 2

TABLE 9-5

Members in bending ( Cracked condition ) Coefficients for balanced design

Pt,bal

1.36 0.98

1.2 0.61

(given) Material M20 (given) Fe 415 (given) 86400 Litres L / B = 8 / 3.6 = 2.22 > 2 . H / 4 = 3.0 / 4 = 0.75 m = = = 44.1 KNm. = = = 26.13 KNm. = = = 19.60 KNm. For bottom portion

M = OR = = = 4.90 KNm = 4.90 KNm = = 35.28 KN = = 19.6 KN Volume = 3.6 x 8 x 3.0 x 10 3 =

The short walls are designed as supported on long walls.

If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m. The long walls are designed as vertical cantilevers from the base.

bottom 1 m or H / 4 whichever is more is designed as cantilever. bottom h = 1 m will be designed as cantilever.

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

Solution

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

Grade HYSD reinforcement Size of tank : 3.6 m x 8.0 m x 3.0 m high

Maximum B.M. in long walls at the base (1 / 6 ) x Yw x H 3 ( 1 / 6 ) x 9.8 x 33 Moments and tensions :

Maximum ( - ve ) B.M. in short walls at support Yw x ( H - h ) x B

2 / 12 9.8 x ( 3 - 1 ) x 42 / 12

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

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

Direct tension in short wall= Yw ( H - h ) x 1 Yw x ( H - h ) x B 2 / 16 9.8 x ( 3 - 1 ) x 42 / 16 Yw x H x h 2 / 6 9.8 x 3.0 x 1 / 6 Yw x H / 6 , whichever is greater 9.8 x 3.0 / 6 9.8 x ( 3 - 1 ) x 1 9.8 x ( 3 - 1 ) x 3.6 / 2

M ( - ) = 44.1 KNm T = 35.28 KN D =

√

M / Q x b = = 379.6 mm, Take D = 400 mm d = = 367 mm Ast = = = 918.68 mm2 = = 218.75 mm = 1005 mm2 . As = = 684 mm2 . 342 mm2 . = T / σst = = 235 mm2 . = = 146.9 mm = 357 mm2 Provide 16 mm O bar @ 200 mm c/c

√

44.1 x 10 6 / 0.306 x 1000 400 - 25 - 8 M / σst x j x d Design of long walls :

-( water face )

( perpendicular to moment steel ) Assume d / D = 0.9 Q = 0.306 200.96 x 1000 / 918.68 ( 0.171 / 100 ) x 1000 x 400 From Table 9-6 From Table 9-5 , Provide 16 mm O bar spacing of bar = Ast for Moment Provide 8 mm O bar

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.

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

Steel required for direct tension

35.28 x 103 / 150

……… ( 2 ) 44.1 x 10 6 / 150 x 0.872 x 367

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

From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension. spacing of bar = Area of one bar x 1000 / required area in m2 / m

50.24 x 1000 / 342

on each face = ……… ( 1 )

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

on each face Design of short walls

:-M = 26.13 KNm T = 19.6 KN = = 544 mm2 T / σst = = 131 mm2 = 675 mm2 = = 167.47 mm = 706 mm2. 1000 203.56 400 367 163.44 = 13.33 = = 203.56 mm D - x = 196.44 mm d - x = 163.44 mm AT = = = 408705 mm2 Ixx = Provide 12 mm O bar@160 mm c/c checking : At support From Table 9-5 Ast1 for moment = 113.04 x 1000 / 675 19.6 x 10 3 / 150 M / σst x j x d 26.13 x 10 6 / 150 x 0.872 x 367

Ast2 for direct tension =

Area of one bar x 1000 / required area in m2 / m Total A_st1 + A_st2 = 544 + 131 Provide 12 mm O bar spacing of bar = ( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 ) ( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 ) b x D + ( m - 1 ) x A_st 1000 x 400 + (13.33 - 1 ) x 706 modular ratio m = 280 / 3 x σcbc x = b x D 2 / 2 + Ast ( m - 1 ) x d b x D + ( m - 1 ) x Ast ( 1 / 3 ) x b x ( x3 + ( D - x )3 ) + ( m - 1 ) x A_st x ( d - x )2

= = 5.34E+09 + 2.33E+08 = 5.57E+09 mm4 fct = T / AT = = 0.048 N / mm2 fcbt = = = 0.767 N / mm2 check : ≤ 1 0.4912 ≤ 1 M = 19.6 KNm T = 19.6 KN = = 408 mm2 T / σst = = 131 mm2 = 539 mm2 = = 209.72 mm = 565 mm2. A_s = = 684 mm2 . 342 mm2 . 26.13 x 106 x 163.44 / 5.57 x 10 9 ( fct / σct ) + ( fcbt / σcbt ) ( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) ≤ 1 ( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442 19.6 x 10 3 / 408705 M x ( d - x ) / Ixx From Table 9-2 At centre : From Table 9-5 Ast1 for moment = M / σst x j x d 0.04 + 0.4512 ≤ 1 ……….. ( O. K. ) Provide 12 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 113.04 x 1000 / 539

19.6 x 10 6 / 150 x 0.872 x 367

Ast2 for direct tension =

19.6 x 10 3 / 150 Total Ast1 + Ast2 = 408 + 131

on each face = ……… ( 1 )

Provide 12 mm O bar @ 200 mm c/c From Table 9-3

Distribution steel = 0.171 % for 400 mm depth ( 0.171 / 100 ) x 1000 x 400

= T / σst = = 131 mm2 . = = 146.9 mm = 357 mm2 M = 4.9 KNm Ast = = = 102 mm2 = 357 mm2 0.229% = = 344 mm2 = = 292 mm Ast = 346 mm 2 lx = 3.6 + 0.4 = 4 say 4 m ly = 8 + 0.15 = 8.15 say 8.5 m 3.75 KN / m2 1.0 KN / m2 1.5 KN / m2 6.25 KN / m2

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

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

Dead Load : self 0.15 x 25 = floor finish =

Live load = Provide 8 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 /172

Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom. Designed section,Elevation etc. are shown in fig.

Minimum steel =

0.229 / 100 x 1000 x 150

,172 mm2 bothway Base slab

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

Steel required for direct tension 19.6 x 103 / 150

50.24 x 1000 / 342

on each face Bottom cantilever

Provide 8 mm O bar @ 140 mm c/c on each face ……… ( 2 )

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

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

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

on each face From Table 9-5

M / σst x j x d

4.9 x 10 6 / 150 x 0.872 x 367 Minimum steel = 342 mm2 on each face.

PU = = 9.38 KN / m = 18.76 KNm = 16.88 KN drequired =

√

M / Q x b =

√

18.76 x 10 6 / 2.76 x 1000 = 82.44 mm dprovided = = = 1.13 Pt = fy / fck = 415 / 20 = = 0.34% = 439 mm2 = = 114 mm = 457 mm2 . = 180 mm2 = Provide 8 mm O bar @ 110 mm c/c Provide 6 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m Distribution steel = ( 0.12 / 100 ) x 1000 x 150

28.26 x 1000 /180 Maximum moment = 9.38 x 42 / 8

Maximum shear = 9.38 x 3.6 / 2 From Table 6-3 ,Q = 2.76

Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcement Provide 8 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 /m 50.24 x 1000 /439

50 1-

√

1-(4.6 / 20) x (1.13)

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

Ast = 0.34 x 1000 x 129 / 100

…………(O.K.) Larger depth is provided due to deflection check.

Mu / b x d 2

= 18.76 x 106 / 1000 x 129 x 129

50 1-

√

1-(4.6 / fck) x (Mu / b x d2) 129 > 82.44

Design for flexure :

150 - 15(cover) - 6 1.5 x 6.25 For 1 m wide strip

= 157 mm

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

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

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.

M15 1.1 1.5 1.5 M20 1.2 1.7 1.7 M25 1.3 1.8 1.9 M30 1.5 2.0 2.2 M35 1.6 2.2 2.5 M40 1.7 2.4 2.7 Table 9-2 ( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442 Grade of concrete Permissible stresses in N / mm2 Direct tension σct Tension due to

bending σcbt Shear stress ﺡv = V / b j d

Permissible concrete stresses in calculations relating to resistance to cracking

150 3000 150 1 : 4 : 8 P.C.C. 150 150 400 8000 400 Section A-A 2000 400 900 3600 900 400 2000 2000 8000 400 Sectional plan 400 150 3000

Base details not shown for clarity

8 O @ 140 c/c (chipiya)

2000 2000

8 O @ 290 c/c both ways top and bottom 12 O @ 200 c/c - shape 8 O @ 110 c/c 6 O @ 150 c/c v v v v 900 12 O @ 160 c/c(chipiya) 16 O @ 200 c/c ( chipiya ) 150 8 O @ 140 mm c/c 8 O @ 140 c/c 12 O @ 200 c/c A A B B 8 O @ 140 c/c 8 O @ 140 c/c 8 O @ 110 c/c 6 O @ 150 c/c 16 O @ 200 c/c ( chipiya ) 900 900 8 O @ 140 c/c

150 150

150 400 3600 400 150 Section B- B

Base details not shown for clarity

( given ) 2.5 KN / m2 _{( given )} 1 KN / m2 _{( given )} ( given ) ( given ) = 25.2 = 158.7 mm D = = 178.7 mm 180 mm 4.5 KN / m2 1.0 KN / m2 2.5 KN / m2 Total 8.0 KN / m2 1.5 x 8 = 12 KN / m w x l2 / 8 = 12 x 42 / 8 = 24 KNm w x l / 2 = = 24 KN d = = 160 mm = 0.94 Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) f_y / f_ck = 415 / 15 = = 0.289%

Design of simply supported one way slab

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

solution : -

Assume 0.4 % steel , a trial depth by deflection criteria modification factor = 1.26

Floor finish =

material M15 grade concrete

HYSD reinforcement grade Fe415

drequired = 4000 / 25.2

158.7 + 15 ( cover ) + 5 ( assume 10 O bar ) Assume an overall depth =

( span / d ) ratio permissible = 1.26 x 20

IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement for simply supported, basic span / effective depth ratio = 20

Maximum moment =

Maximum shear =

12 x 4 / 2 Design for flexure :

-DL = 0.18 x 25 = Floor finish = Live load = Factored load =

Consider 1 m length of slab

50 [(1-0.84) x 15 / 415 ] 180 - 15 - 5 Mu / b x d 2 = 24 x 10 6 / 1000 x (160)2 50 1-

√

1-(4.6 / 15) x (0.94)

= 462 mm2 = = 170 mm = 462 mm2 . = 0.128 = 216 mm2 = = 233 mm = 218 mm2 . Vu = 24 KN = = 0.150 N / mm2 d = 160 mm As = 231 mm 2 . = 0.144 6 x β β = = 0.8 x 15 / 6.89 x 0.144 = 12.1 6 x 12.1 = 0.277 N / mm2 0.289 x 1000 x 160 / 100 Provide 10 mm O bar

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

78.50 x 1000 /462

Provide 10 mm O bar @ 170 mm c/c Ast =

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

Distribution steel = ( 0.12 /100 ) x 1000 x 180 Half the bars are bent at 0.1 l = 400 mm , and

remaining bars provide 231 mm2 area

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

spacing of bar =

50.24 x 1000 /216 Provide 8 mm O bar @ 230 mm c/c

Area of one bar x 1000 / required area in m2 /m Maximum spacing = 5 x 160 = 800 or 450 mm i.e. 450 mm Provide 8 mm O bar

100 x As / b x d = 100 x 231 / 1000 x 160 For bars at support

Check for shear :

-Actual Shear stress = V_u / b x d

24 x 103 / 1000 x 160

for Pt = 0.144 ﺡc = 0.28

IS 456-2000 Table 19 from table 7-1

<

(

ﺡ

C )N / mm

2

( too small )

Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 0.8 x fck / 6.89 Pt , but not less than 1.0

IS 456-2000 clause 40.2.1.1 -0.05 k = 1.24 ? -0.04 = 0.347 N / mm2 ……….( O.K.) 8 O As = 231 mm 2 Mu1 = OR = { 1 - (415 x 231 / 15 x 1000 x 160 ) } = 13.34 { 1- 0.0399 } = 12.812 KNm Vu = 24 KN 693.875 + 8 O ≥ 56 O 48 O ≤ 693.88 O ≤ 14.46 mm ……….( O.K.) 20 Pt = = 0.289

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

= 28.4

= 25.00 < 28.4 ……….( O.K.)

= 3 x 160 = 480 mm

170 mm

5 x effective depth of slab or 450 mm whichever is small = 5 x 160

230 mm

Check for development length : -Assuming L0 =

for 180 mm slab depth Design shear strength = 1.24 x 0.28

0.87 x 415 x 231 x 160

= 56 O ( from Table 7-6 )

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

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

modification factor = 1.42

For tying the bent bars at top , provide 8 mm O @ 230 mm c/c ( span / d ) ratio permissible = 1.42 x 20

Actual (span / d ) ratio = 4000 / 160

Main bars : maximum spacing permitted =

Distribution bars : maximum spacing permitted = spacing provided =

25 difference 20 difference

(HYSD Fe415 steel ) For continuing bars

0.87 x fy x Ast x d { 1 - ( fy x Ast / fck x b x d ) }

The depth could be reduced

1.3 x ( 12.81 x 106 / 24 x 103 ) + 8 O ≥ 56 O which gives ……….( O.K.) = 800 mm ……….( O.K.) spacing provided = < 300 mm

3 x effective depth of slab or 300 mm whichever is small or 300 mm i.e. 300 mm Check for cracking :

-< 450 mm IS 456-2000 , clause 26.3.3

Development length of bars Ld = O σs / 4 x ﺡbd 1.3 x ( M_u1 / V_u ) + L₀ ≥ L_d

NOTE : -

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

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

Pt = = = 0.14 P_t = 50 1-

√

1-(4.6 / fck) x (M_u / b x d2) fy / fck we get , 0.49 Mu1 = = 12.54 KNm 160 180

5 x effective depth of slab or 450 mm whichever is small

0.49x 1000 x 1602 x 10-6

400

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 From equation Mu1 / b x d 2 = 4000 400 ……….( O.K.) or 450 mm i.e. 450 mm ……….( O.K.)

3 x effective depth of slab or 300 mm whichever is small or 300 mm i.e. 300 mm v v v v v 8 O @ 230 c/c 10 O @ 170 c/c ( alternate bent )

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

centre to centre distance shall be used. 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, 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 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 ( a ) Simply Supported Beam or Slab

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

supports are wider than I/12 of the clear span or 600 mm whichever is less, the effective span shall be taken as under:

( b )Continuous Beam or Slab - In the case of continuous beam or slab, if the width of the support is less than l/12 of the clear span, the

effective span shall be as in (a). If the Effective Span

IS 456-2000 clause 22.2

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 l_y / l_x < 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.

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.

material ( given ) ( given ) ( Balcony slab ) DL LL For S2 3 0 KN / m 2 1 0 KN / m2 0 2 KN / m2 Total 4 2 KN / m2 Pu = ( 6 + 3 ) KN / m2 DL LL For S1 3 0 KN / m 2 1 0 KN / m2 0 3 KN / m2 Total 4 3 KN / m2 Pu = (6 + 4.5) KN / m2 1.875 KN / m Pu = 2.8 KN / m 9 KN/m 6 KN/m A 3m B 1.2m C 9 KN/m 2.8 KN A 3m B 1.2m C

considering fig (a)

wx l2 / 2 = 6 x 1.22 / 2 mild steel grade Fe250

1.5 ( 4 + 3 ) = Weight of parapet 0.075 x 25 x 1 =

1.5 x 1.875 = Consider 1 m long strip

(1) To get maximum positive moment in slab S2 only dead load on slab S1 and total load on slab S2 shall be considered

(b) Loads for maximum negative

moment,maximum shear for cantilever span and maximum reaction at support B

cantilever moment = Solution : -

For slab S2 live load = 2 KN /m 2 For slab S1 live load = 3 KN /m

2 Assume 120 mm thick slab

self load = 0.12 x 25 = floor finish = live load = 1.5 ( 4 + 2 ) = self load = 0.12 x 25 = Live Load As per IS 875

ly = 6m

(a) Loads for maximum positive moment Design of Cantilever one way slab

floor finish = live load =

10.5 KN/m

M15 grade concrete used for residential purpose

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

SS2 2

S2 S1

= 4.32 KNm shear = = = 12.06 KN = 1.34 m = = 8.08 KNm = = 10.92 KNm Vu,BA = = = 17.14 KN Vu,BC = = = 15.4 KN 10.92 KNm Q = 2.22 = = 70.14 mm, = 99 mm, = 0.82 Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) fy / fck = 250 / 15 = = 0.405% = 401 mm2 Moment steel : 50 1-

√

1-(4.6 / 15) x (0.82) 50 [(1-0.865) x 15 / 250 ] 10.5 x 1.2 + 2.8 Maximum moment = Mu / b x d 2 ( + ) = 120 - 15 - 6 ( assume 12 O bar ) Ast ( + ) = 0.405 x 1000 x 99 / 100 ……….( O.K.) 8.08 x 10 6 / 1000 x (99)2 From Table 6-3 dprovided =

√

10.92 x 10 6 / 2.22 x 1000 Maximum shear at B 9 x 3 / 2 + 10.92 / 3 w x l / 2 + Moment @ B at distance 3 m w x l + 2.8

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

Maximum positive moment = 12.06 x 1.34 - W x l2 / 2 12.06 x 1.34 - 9 x 1.342 / 2

√

M / Q x b 9 x 3 / 2 - 4.32 / 3 w x l / 2 - Moment @ B at distance 3 m Reaction at A = 2.8 x 1.2 + 10.5 x 1.22 / 2 Maximum negative moment = w x l + w x l2 / 2

(2) To get maximum negative moment and maximum shear at B, Point of zero shear from A = 12.06 ( KN ) / 9 ( KN / m )

= = 195.76 mm = 462 mm2. = 1.11 Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) fy / fck = 250 / 15 = = 0.564% = 558 mm2

= 231 mm2. ( bent bars extended ) = = 231 mm2 remaining area = 558 - 231 = 327 mm2 = = 346 mm = 332 mm2. ( Extra ) = 180 mm2. = = 157 mm = 188 mm2. For negative moment reinforcement

Total 231 + 332 mm2 = 563 mm2 steel provided

( from Table 7-6 ) 28.26 x 1000 / 180

Provide 6 mm O bar@150 mm c/c

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

Distribution steel = ( 0.15 /100 ) x 1000 x 120 Provide 6 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m For mild steel minimum reinforcement 0.15 %

Provide 12 mm O bar

Provide 12 mm O bar@340 mm c/c

spacing of bar = Area of one bar x 1000 / required area in m2 / m Area provided = Area of one bar x 1000 / spacing of bar in m Provide 10 mm O bar Mu / b x d 2 ( - ) = 10.92 x 10 6 / 1000 x (99)2 78.5 x 1000 / 401 113.04 x 1000 / 327

Note that at simple support , the bars are bent at 0.1 l whereas at continuity of slab it is bent at 0.2 l ( alternate bent up ) 78.5 x 1000 / 340 Provide 10 mm O bar@340 mm c/c 50 [(1-0.812) x 15 / 250 ] Ast ( - ) = 0.564 x 1000 x 99 / 100 Provide 10 mm O bar@170 mm c/c Provide 10 mm O bar

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

= = 54.4 O Ld = 54.4 x ( 10 + 12 ) / 2 = 598 mm. 12 O (mild steel ) At A , Pt = = = 0.233 OR Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) Mu1 = fy / fck = we get , 0.487 = 4.974 Mu1 = = 4.78 = 4.77 KNm Vu = 12.06 KN =54.4 O 514.18 + 12 O ≥ 54.4 O 42.4 O ≤ 514.18 O ≤ 12.13 mm At B , Pt = = = 0.233 OR Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) Mu1 = fy / fck = we get , 0.487 = 4.974 Mu1 = = 4.78 = 4.77 KNm Vu = 13.09 KN =54.4 O 473.72 + 12 O ≥ 54.4 O 42.4 O ≤ 473.72 O ≤ 11.2 mm which gives Mu1 / b x d 2 = ……….( O.K.) 0.487 x 1000 x 992 x 10-6

Development length of bars Ld = O σs / 4 x ﺡbd = O x 0.87 x 250 / 4 x 1 Near point of contraflexure i.e. 0.15 x l from B

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

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

100 x As / b x d Half bars bent = 462 / 2 = 231 mm 2 ) 100 x 231 / 1000 x 99 From equation 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

which gives ……….( O.K.)

Half bars bent = 462 / 2 = 231 mm2 )

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

Development length of bars Ld = O σs / 4 x ﺡbd = O x 0.87 x 250 / 4 x 1 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 100 x 231 / 1000 x 99 From equation Mu1 / b x d 2 = 0.487 x 1000 x 992 x 10-6 1.3 x ( 4.77 x 106 / 12.06 x 103 ) + 12 O ≥ 54.4 O Check for development length :

-Assuming L0 =

100 x As / b x d

O x 0.87 x 250 / 4 x 1

say 600 mm

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

13.09 KN = = 0.132 N / mm2 < ﺡc Pt = = 0.233 6 x β β = = = 7.47 6 x 7.47 = 0.34 N / mm2 IS 456-2000 clause 40.2.1.1 0.07 k = 1.3 ? 0.014 = 0.442 N / mm2 > ﺡv Vu = = = 0.173 N / mm2 < ﺡc Pt = = 0.569 6 x β β = = = 3.06 6 x 3.06 = 0.487 N / mm2 Span AB :

Check for shear : -

Use Vu = 13.09 KN Shear stress ﺡv = Vu / b x d

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

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

Design shear strength ﺡc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 7.47 - 1 )

IS 456-2000 Table 19 from table 7-1 13.09 x 10 3 / 1000 x 99

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

Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 )

……….( O.K.)

0.8 x 15 / 6.89 x 0.233

……….( O.K.) for Pt = 0.233 ﺡc = 0.34

0.1 difference for 120 mm slab depth 0.02 difference

Span BC :

17.14 KN Shear stress ﺡv = Vu / b x d

Design shear strength = 1.3 x 0.34

Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 0.8 x fck / 6.89 Pt , but not less than 1.0 0.8 x 15 / 6.89 x 0.569

17.14 x 10 3 / 1000 x 99

……….( O.K.) 100 x As / b x d = 100 x 563 / 1000 x 99

Design shear strength ﺡc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 3.06 - 1 )

IS 456-2000 Table 19 from table 7-1 for Pt = 0.569 ﺡc = 0.48

IS 456-2000 clause 40.2.1.1 0.08 k = 1.3 ? 0.0579 = 0.624 N / mm2 > ﺡv 20 P_t = = 0.467

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

= 40

= 30.30 < 40 ……….( O.K.)

7 P_t =

= 0.569

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

= 12.6

= 12.12 < 12.6 ……….( O.K.)

= 3 x 99 = 297 mm

170 mm

5 x effective depth of slab or 450 mm whichever is small = 5 x 99

150 mm

0.25 difference for 120 mm slab depth 0.181 difference Design shear strength = 1.3 x 0.48

modification factor = 2

( span / d ) ratio permissible = 2 x 20 Actual (span / d ) ratio = 3000 / 99

……….( O.K.) Check for deflection :

-Basic ( span / d ) ratio =

100 x Ast / b x d = 100 x 462 / 1000 x 99 For span AB :

(1) Main bars : maximum spacing permitted = 3 x effective depth of slab or 300 mm whichever is small ( span / d ) ratio permissible = 1.8 x 7

Actual (span / d ) ratio = 1200 / 99 For span BC :

Basic ( span / d ) ratio =

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

modification factor = 1.8

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

or 300 mm i.e. 297 mm

spacing provided = < 297 mm ……….( O.K.)

(2 )Distribution bars : maximum spacing permitted =

= 495 mm Check for cracking :

1.2m 0.15m 0.15m 0.15m 0.15m 1.2m lx = 3m ly = 6m S2 S1 S3 B1 B2 B3 B4 1m high parapet Column 300 x 300

KNm KNm ( 1 - 0.0389 ) 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 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 )

1200 1200

150 3000 150 1200

5 x effective depth of slab or 450 mm whichever is small 3 x effective depth of slab or 300 mm whichever is small

……….( O.K.) or 300 mm i.e. 297 mm ……….( O.K.) 120 125 600 300 or 450 mm i.e. 450 mm 10 O @ 340 c/c (bent) + 12 O @ 340 c/c (extra) 10 O @ 170 c/c 6 O @ 150 c/c 6 O @ 150 c/c

Material 3.75 KN / m2 1.00 KN / m2 3.00 KN / m2 Total 7.75 KN / m2 Pu = = 11.625 KN / m2 ly / lx = 5 / 5 = 1 < 2 13.66 KNm 10.17 KNm = = 81.23 mm, = = 120 mm, = 130 mm, = 0.706 Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) fy / fck = 415 / 15 = = 0.208% 50 1-

√

1-(4.6 / 15) x (0.706) 50 [(1-0.885) x 15 / 415 ] > 81.23 mm ……….( O.K.) > 81.23 mm ……….( O.K.) Mu / b x d 2 ( + ) = 10.17 x 10 6 / 1000 x (120)2 From Table 6-3 drequired =

√

M / Q x b =

For M15 mix and Fe415 steel, Q = 2.07

√

13.66 x 10 6 / 2.07 x 1000 dprovided for positive moment reinforcement

dprovided for negative moment reinforcement 150 - 15 - 5 150 - 15 - 10 - 5 ( assume 10 O bar ) IS 456-2000 Table -26 αx x w x lx 2 = 0.047 x 11.625 x 52 = αy x w x lx 2 = 0.035 x 11.625 x 52 = Two adjacent edges Discontinuous

Mu2 ( + ) =

i.e. two-way slab.

Mu1 , Mu3 ( - ) =

floor finish = Live Load = 1.5 x 7.75

Middle strip :

HYSD reinforcement of grade Fe415 Solution : -

Assume 150 mm thick slab Self load 0.15 x 25 =

Design of Continuous Two-way slab Banking hall

slab is restrained with edge beams

= 250 mm2 = = 200.96 mm = 251 mm2. = 0.81 Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) fy / fck = 415 / 15 = = 0.240% = 312 mm2 = = 161 mm = 335 mm2. For HYSD Fe415

= 180 mm2 = = 126 mm2 = = 279.11 mm = 193 mm2. = 188 mm2. Provide 8 mm O bar@ 260 mm c/c

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 Provide 8 mm O bar

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

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

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 / 180

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 / 312 Provide 8 mm O bar@ 150 mm c/c Minimum steel = ( 0.12 / 100 ) x 1000 x 150 50.24 x 1000 / 250 Provide 8 mm O bar@ 200 mm c/c Mu / b x d 2 ( - ) = 13.66 x 10 6 / 1000 x (130)2 ( 1 / 2 ) x 251 50 1-

√

1-(4.6 / 15) x (0.81) 50 [(1-0.867) x 15 / 415 ] Ast = 0.24 x 1000 x 130 / 100 Provide 8 mm O bar Provide 8 mm O bar Ast = 0.208 x 1000 x 120 / 100

= = 267.23 mm = 193 mm2. = 94 mm2. = = 279.11 mm = 193 mm2. S.F. = = = 31.795 KN = 0.258 N / mm2 _{0.25 difference 0.11} IS 456-2000 clause 40.2.1.1 0.242 difference ? -0.1065 k = 1.3 = 0.460 N / mm2 OR 6 x β β = = 6.75 6 x 6.75 = 0.356 IS 456-2000 clause 40.2.1.1 k = 1.3 = 0.463 N / mm2 = = 0.245 N / mm2 < ﺡc ……….( O.K.) Design shear strength = 1.3 x 0.356

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

31.795 x 103 / ( 1000 x 130 ) Design shear strength ﺡ_c = 0.85 √ 0.8 x f_ck ( √ 1 + 5 x β - 1 )

0.8 x fck / 6.89 Pt , but not less than 1.0 Design shear strength ﺡc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 6.75 - 1 )

for 150 mm slab depth from table 7-1

for Pt = 0.258 , ﺡc = 0.354

for 150 mm slab depth Design shear strength = 1.3 x 0.354

……….( O.K.)

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

Check for shear : -At point 1 or 3

w x l / 2 + Moment @ point 1 or 3 in that span 11.625 x 5 / 2 + 13.66 / 5

100 x As / b x d = 100 x 335 / ( 1000 x 130 ) Provide 8 mm O bar

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 / 180

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

spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 / 188

Provide 8 mm O bar@ 260 mm c/c

This will be provided by minimum steel of edge strip, At corner B , steel required = ( 1/2 ) x 188

S.F. = w x l / 2 = = 29.06 KN = 0.209 N / mm2 _{0.1 difference 0.07} IS 456-2000 clause 40.2.1.1 0.041 difference ? -0.0287 k = 1.3 = 0.417 N / mm2 OR 6 x β β = = 8.33 6 x 8.33 = 0.326 IS 456-2000 clause 40.2.1.1 k = 1.3 = 0.424 N / mm2 = = 0.242 N / mm2 < ﺡc Vu = 29.06 KN

8 O (HYSD Fe415 steel ) Pt = = = 0.209 OR Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) Mu1 = fy / fck = we get , 0.711 = 10.875 Mu1 = = 10.246 = 10.24 KNm =54.3 O This is critical at point 4 or 5.

No bar is curtailed or bent up.

Mu1 / b x d 2

=

0.711 x 1000 x 1202 x 10-6

Development length of bars Ld = O σs / 4 x ﺡbd (From Table 7-6 ) Assuming L0 =

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) Actual shear stress = Vu / b x d

29.06 x 103 / ( 1000 x 120 )

……….( O.K.)

0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6 0.85 √ 0.8 x 15 ( √ 1 + 5 x 8.33 - 1 )

for 150 mm slab depth Design shear strength = 1.3 x 0.326

100 x As / b x d

100 x 251 / 1000 x 120 From equation

Check for development length : -At point 4 or 5 ,

1.3 x 0.321

Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 0.8 x f_ck / 6.89 P_t , but not less than 1.0 Design shear strength ﺡc =

At point 4 or 5

11.625 x 5 / 2

100 x As / b x d = 100 x 251 / ( 1000 x 120 ) from table 7-1

for Pt = 0.209 , ﺡc = 0.321

for 150 mm slab depth Design shear strength =

458.36 + 8 O ≥ 54.3 O 46.3 O ≤ 458.36

O ≤ 9.90 mm

Vu = = 25.31 KN

8 O (HYSD Fe415 steel ) Pt = = = 0.289 OR P_t = 50 1-

√

1-(4.6 / fck) x (M_u / b x d2) M_u1 = fy / fck = we get , 0.9595 = 26.689 M_u1 = = 24.554 = 24.56 KNm =56 O 1261.5 + 8 O ≥ 56 O 48 O ≤ 1261.5 O ≤ 26.28 mm 26 251 mm2. = 123 mm. P_t = 240.7 = 0.204

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

= 42.9 = 40.65 < 42.9 = 3 x 130 = 390 mm 200 mm = 5 x 120 = 600 mm < 300 mm ……….( O.K.)

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

IS 456-2000 , clause 26.3.3 Check for cracking :

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

spacing provided = 1.3 x ( M_u1 / V_u ) + L₀ ≥ L_d

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

which gives

Note that the bond is usually critical along long direction.

Actual (span / d ) ratio = 5000 / 123

……….( O.K.) 0.9595 x 1000 x 1602 x 10-6

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

……….( O.K.) Short span 11.25 x ( 4.5 / 2 ) 0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld 1.3 x ( 10.246 x 106 / 29.06 x 103 ) + 8 O ≥ 54.3 O which gives O σs / 4 x ﺡbd

( span / d ) ratio permissible = 1.65 x 26 Basic ( span / d ) ratio =

Mu1 / b x d 2 = Assuming L0 = 100 x As / b x d 100 x 462 / 1000 x 160 From equation

Check for deflection :

-100 x Ast / b x d = 100 x 251 / 1000 x 123 positive moment steel =

actual d = 150 - 15 -8 - 4

260 mm This is 180 mm2 = = 279.11 mm = 193 mm2. 625 3750 625 150 625 3750 625 Section A-A 50.24 x 1000 / 180

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

3750 625

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

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

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

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

625 Mi ddle Str ip S1 E dg e s trip E dg e s trip

Edge strip Middle Strip Edge strip

8 O @ 260 c /c 8 O @ 2 6 0 c /c 8 O @ 200 c /c 8 O @ 260 c/c 8 O @ 260 c/c 8 O @ 200 c/c A A B B 500 1500 1500 8 O @ 260 c/c 8 O @ 150 c/c 8 O @ 260 c/c 8 O @ 200 c/c

5 m

625 3750 625 5 m

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

5 m 5 m

Design of Continuous Two-way slab

Middle Strip S1 S1 E dg e s tr ip E dg e s tri p A B B 1 2 3 4 5 0.03 5 -0 .0 4 7 -0.047 0.035

KNm

( 1 - 0.0579 )

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck )

KNm

or 300 mm i.e. 300 mm

or 450 mm i.e. 450 mm ……….( O.K.)

5 x effective depth of slab or 450 mm whichever is small 3 x effective depth of slab or 300 mm whichever is small

( 1 - 0.0799 )

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck )

1000 625 3750 625 150 625 3750 625 1000 ……….( O.K.) 625 3750 625 Mi ddle Str ip S1 E dg e s trip E dg e s trip

Edge strip Middle Strip Edge strip

8 O @ 240 c /c 8 O @ 2 4 0 c /c 8 O @ 140 c /c 8 O @ 240 c/c 8 O @ 240 c/c 8 O @ 140 c/c 500 1500 1500 8 O @ 240 c/c 8 O @ 140 c/c 8 O @ 240 c/c 8 O @ 240 c/c 500 1500 1500 1500 1500 8 O @ 180 c/c

3 -0.047

given Material DL LL 3 0 KN / m2 1 0 KN / m2 0 3 KN / m2 Total 4 3 KN / m2 = 3m 3m 3m 3m 3m = = 4.5 + 4.05 = 8.55 KNm = = 3.38 + 3.38 = 6.75 KNm = = 5.4 + 4.5 = 9.9 KNm = = 4.50 + 4.50 = 9 KNm = = 18.9 KN KNm Q = 2.07 0.6 x 6 x 3 + 0.6 x 4.5 x 3 Maximum moment is Mu3 ( - ) = 9.9 From Table 6-3 ( 1 / 10 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32 Mu4 ( - ) = ( 1 / 12 ) x w ( DL ) x l 2 + ( 1 / 9 ) x w ( LL ) x l2 ( 1 / 12 ) x 6 x 32 + ( 1 / 9 ) x 4.5 x 32

Maximum shear is Vu(BA) = 0.6 x w x l + 0.6 x w x l Mu3 ( - ) = ( 1 / 10 ) x w ( DL ) x l 2 + ( 1 / 9 ) x w ( LL ) x l2 ( 1 / 12 ) x 6 x 32 + ( 1 / 10 ) x 4.5 x 32 Mu2 ( + ) = ( 1 / 16 ) x w ( DL ) x l 2 + ( 1 / 12 ) x w ( LL ) x l2 ( 1 / 16 ) x 6 x 32 + ( 1 / 12 ) x 4.5 x 32

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

The factored moments at different points using the coefficients are as follows : Mu1 ( + ) = ( 1 / 12 ) x w ( DL ) x l 2 + ( 1 / 10 ) x w ( LL ) x l2 Dead load 0.12 x 25 = floor finish = live load = factored load =

HYSD reinforcement of grade Fe415 M15 grade concrete

Solution :

-Try 120 mm thick slab

1.5 ( 4 + 3 ) ( 6 + 4.5 ) KN / m2 Consider 1 m wide strip of the slab.

A B C D E F

1 2 2 2 1

= = 69.2 mm, = 90 mm, P_t = 50 1-

√

1-(4.6 / fck) x (M_u / b x d2) fy / fck = 132 mm2 = 165 mm2 = = 171 = = 344 306 Table 8 mm O @ 220 c/c = 228 mm2 10 mm O @ 220 c/c = 357 mm2 10 mm O @ 220 c/c = 357 mm2 4 ( - ) 9.0 1.11 0.34 223 3 ( - ) 9.9 1.22 0.38 342 2 ( + ) 6.75 10 mm O @ 440 c/c + 8 mm O @ 440 c/c = 178 +114 = 292 mm2 (Half 10 O+half 8 O) 290 0.83 0.248 1 ( + ) 8.55 1.06 0.322 Factored moment KNm point Mu/(b x d2 ) P_t Ast mm2 Steel Provided dprovided = 110 - 15 - 5 ( assume 10 O bar )

……….( O.K.) Try 110 mm overall depth

spacing of bar = Area of one bar x 1000 / required area in m2 / m Ast = Pt x b x d / 100

drequired =

√

M / Q x b

√

9.9 x 10 6 / 2.07 x 1000

For Main steel ,HYSD Fe415 reinforcement

minimum steel area = ( 0.12 / 100 ) x 1000 x 110 For Distribution steel , mild steel Fe250 reinforcement

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

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

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

Provide 50 % Ast at end support top bars i.e. 292 / 2 = 146 mm 2

. Use 8 mm O

Use 6 mm O

spacing of bar = Area of one bar x 1000 / required area in m2 / m 28.26 x 1000 /165

Check for shear :

-spacing of bar = Area of one bar x 1000 / required area in m2 / m 50.24 x 1000 /146

Use 8 mm O @ 340 mm c/c = 148 mm2. Maximum shear = 18.9 KN

= = 0.210 N / mm2 d = 90 mm As = 357 mm 2 . = 0.397 6 x β β = = = 4.4 6 x 4.4 = 0.42 N / mm2 IS 456-2000 clause 40.2.1.1 0.11 k = 1.3 ? 0.0453 = 0.546 N / mm2 At point of contraflexure i.e. 0.15 x l from B

Vu = = 14.18 KN = = 0.158 N / mm2 d = 90 mm As = 292 mm 2 . = 0.324 6 x β β = = = 5.4 Actual Shear stress = V_u / b x d

18.9 x 103 / 1000 x 90

for 110 mm slab depth 0.103 difference Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 )

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

Design shear strength ﺡc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 4.4 - 1 ) 0.8 x 15 / 6.89 x 0.397

IS 456-2000 Table 19 from table 7-1 for Pt = 0.397 ﺡc = 0.42

<

(

ﺡ

C )N / mm

2

0.25 difference ( too small ) For bars at support

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

……….( O.K.) 18.9 - 0.15 x 3 x 10.5

Actual Shear stress = V_u / b x d

Design shear strength = 1.3 x 0.42

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

Design shear strength ﺡc = 0.85 √ 0.8 x fck ( √ 1 + 5 x β - 1 ) 14.18 x 103 / 1000 x 90

<

(

ﺡ

C )N / mm

2

( too small ) For bars at support

with positive moment reinforcement ( 292 mm2 )

0.8 x f_ck / 6.89 P_t , but not less than 1.0 0.8 x 15 / 6.89 x 0.324

6 x 5.4 OR = 0.39 N / mm2 IS 456-2000 clause 40.2.1.1 0.11 k = 1.3 ? 0.0774 = 0.494 N / mm2 Vu = = 13.28 KN At A , Pt = = = 0.324 OR Pt = 50 1-

√

1-(4.6 / fck) x (Mu / b x d 2 ) Mu1 = fy / fck = we get , 1.064 = 9.4884 Mu1 = = 8.64 = 8.62 KNm

8 O (HYSD Fe415 steel )

=56.4 O

845.783 + 8 O ≥ 56.4 O 48.4 O ≤ 845.78

O ≤ 17.47 mm

At support B, point of contraflexure is assumed at 0.15 x l from B Vu = = 14.18 KN Mu1 = 8.64 KNm L0 = 12 O 792.102 + 12 O ≥ 56.4 O 44.4 O ≤ 792.1 O ≤ 17.84 mm IS 456-2000 Table 19 from table 7-1

for Pt = 0.324 ﺡc = 0.38

0.25 difference for 110 mm slab depth 0.176 difference Design shear strength ﺡc = 0.85 √ 0.8 x 15 ( √ 1 + 5 x 5.4 - 1 )

0.4 x 6 x 3 + 0.45 x 4.5 x 3 = 100 x 292 / 1000 x 90 From equation

1.064 x 1000 x 902 x 10-6 Design shear strength = 1.3 x 0.38

……….( O.K.) 0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck ) 0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6 Mu1 / b x d 2 = Check for development length :

-Assuming L0 =

100 x As / b x d

Span AB is critical for checking this requirement At support A

0.4 x w x l + 0.45 x w x l

as before

( actual anchorage is more than 12 O but L0 is limited to 12 O or d , i.e. 90 mm whichever is greater )

1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

Development length of bars Ld = O σs / 4 x ﺡbd = O x 0.67 x 415 / 4 x 1 1.3 x ( Mu1 / Vu ) + L0 ≥ Ld

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

which gives ……….( O.K.)

1.3 x ( 8.64 x 106 / 14.18 x 103 ) + 12 O ≥ 56.4 O 18.9 - 0.15 x 3 x 10.5

which gives ……….( O.K.)

-26 Pt =

= 0.324

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

= 34.84

= 33.33 < 34.84 ……….( O.K.) 26

Pt =

= 0.253

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

= 41.6

= 33.33 < 41.6 ……….( O.K.)

= 3 x 90 = 270 mm

220 mm

5 x effective depth of slab or 450 mm whichever is small = 5 x 90

160 mm

……….( O.K.) Basic ( span / d ) ratio =

modification factor = 1.34

( span / d ) ratio permissible = 1.34 x 26

or 300 mm i.e. 270 mm 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 100 x Ast / b x d = 100 x 292 / 1000 x 90

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

Check for cracking :

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

modification factor = 1.6

( span / d ) ratio permissible = 1.6 x 26 Actual (span / d ) ratio = 3000 / 90 For span BC :

Basic ( span / d ) ratio =

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

Actual (span / d ) ratio = 3000 / 90

(2 )Distribution bars : maximum spacing permitted =

= 450 mm

Outer side Inner side IS 456-2000 Clause -22.5

( 22.5.1 ) Unless more exact estimates are made, for beams of uniform cross-section which support substantially uniformly distributed loads over three or more spans which do not differ by more than 15 percent of the longest, the bending moments and shear forces used in design may be obtained using the coefficients

Where coefficients given in Table 12 are used for calculation of bending moments, redistribution referred to in 22.7 shall not be permitted.

(22.5.2 ) Beams and Slabs Over Free End Supports given in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans the average of the two values for the negative moment at the support may be taken for design.

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

and I is the effective span, or such other restraining moment as may be shown to be applicable. For such a condition shear coefficient given in Table 13 at the end support may be increased by 0.05.

Where a member is built into a masonry wall 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

+ 1 / 12 + 1 / 10

Table 12 Bending Moment coefficients

Near middle of end span At middle of interior span Support moments At support next to

the end support Type of load

Span moments

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

Table 13 Shear Force coefficients + 1 / 16

+ 1 / 12

-1 / 10 -1 / 9 Dead load and imposed

load ( fixed ) imposed load (

not fixed )

Type of load

Dead load and imposed load ( fixed )

imposed load ( not fixed )

At support next to the end support

At end support

0.45 0.6 0.6

90 110 450 900 900 900 3000 3000 10 O @ 220 c/c 8 O @ 340 c/c 6 O @ 160 c/c 8 O @ 220 c/c 8 O @ 440 c/c + 10 O @ 440 c/c ( 0.3 l1 ) ( 0.3 l1 ) ( 0.3 l1 ) ( 0.15 l1 )

KNm

0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck )

0.87 x 415 x 292 x 90 ( 1 - 415 x 292 / 1000 x 90 x 15 ) x 10 -6 ( 1 - 0.0898 )

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

or 300 mm i.e. 270 mm 3 x effective depth of slab or 300 mm whichever is small Maximum positive moment occurs in span AB. Therefore, this check is critical in span AB

……….( O.K.)

( 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

Where coefficients given in Table 12 are used for calculation of bending moments, redistribution referred to in 22.7 shall not be permitted.

(22.5.2 ) Beams and Slabs Over Free End Supports given in Table 12 and Table 13 respectively.

For moments at supports where two unequal spans the average of the two values for the negative moment at the support may be taken for design.

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

and I is the effective span, or such other restraining moment as may be shown to be applicable. For such a condition shear coefficient given in Table 13 at the end support may be increased by 0.05.

Where a member is built into a masonry wall 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

Table 12 Bending Moment coefficients

Support moments

At other interior supports

-1 / 12 -1 / 9 NOTE -For obtaining the bending moment, the coefficient shall be multiplied by the total design load and effective span.

Table 13 Shear Force coefficients

0.6 At all other

interior supports

900

3000 10 O @ 220 c/c ( 0.3 l1 )

material

l

x

=

say 4.5

m

l

y

=

say 6.75 m

4.5

KN / m

2

1.0

KN / m

2

2.0

KN / m

2

Total

7.5

KN / m

2

P

u

= 1.5 x 7.5

= 11.25

KN / m

l

y

/ l

x

=

= 1.5

M

ux

=

23.7

KNm

M

uy

=

10.48

KNm

Q = 2.07

=

= 107

mm,

= 160

mm,

160 - 10

= 150

mm,

= 0.926

P

t

= 50 1-

√

1-(4.6 / fck) x (M

u

/ b x d

2

)

f

y

/ f

ck

=

Solution :

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

Design of Simply supported two way slab

residential building drawing room 4.3 m x 6.55 m

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

M15 grade concrete

6.75 / 4.5

4.3 + 0.18 = 4.48

6.55+0.18 = 6.73

Live load ( residence ) =

For 1 m wide strip

Dead load : self 0.18 x 25 =

floor finish =

HYSD reinforcement of grade Fe415

given

IS 456-2000 Table -27

d

shortprovided

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

……….( O.K.)

> 107 mm

d

longprovided

=

> 107 mm

α

x

x w x l

x 2

=

α

y

x w x l

x 2

=

0.046 x 11.25 x 4.5

2

=

0.104 x 11.25 x 4.5

2

=

From Table 6-3

d

required

=

√

M / Q x b

√

23.7 x 10

6

/ 2.07 x 1000

Larger depth is provided due to deflection check.

Mu / b x d 2

( short ) =

23.7 x 10

6

/ 1000 x (160)

2

415 / 15

=

= 0.28%

= 448

mm

2

=

= 175.22

mm

= 462

mm

2

.

= 0.466

P

t

= 50 1-

√

1-(4.6 / fck) x (M

u

/ b x d

2

)

f

y

/ f

ck

=

415 / 15

=

= 0.134%

= 201

mm

2

= 216

mm

2

.

=

= 232.59

mm

= 218

mm

2

.

V

u

=

= 25.31 KN

8 O

(HYSD Fe415 steel )

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

A

st

( short ) = 0.28 x 1000 x 160 / 100

Provide 10 mm O bar

Mu / b x d 2 ( long ) =

10.48 x 10

6

/ 1000 x (150)

2

50 1-

√

1-(4.6 / 15) x (0.466

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

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

2

/ m

78.5 x 1000 / 448

Provide 10 mm O bar@170 mm c/c

( short span )

A

st

( long ) = 0.134 x 1000 x 150 / 100

For HYSD Fe415 minimum reinforcement 0.12 %

Provide 8 mm O bar

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

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

2

/ m

Provide 8 mm O bar@ 230 mm c/c

( long span )

50.24 x 1000 / 216

The bars cannot be bent or curtailed because if 50 % of long span bars are curtailed ,

the remaining bars will be less than minimum

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

negative moment due to slab nature.

Check for development length :

-Assuming L

0

=

P

t

=

=

= 0.145

OR

P

t

= 50 1-

√

1-(4.6 / fck) x (M

u

/ b x d

2

)

M

u1

=

f

y

/ f

ck

=

we get ,

0.5023

=

11.806

M

u1

=

= 11.331

= 11.30

KNm

=56 O

580.4 + 8 O

≥ 56 O

48 O ≤ 580.403

O ≤ 12.09

mm

V

u

=

= 25.31 KN

8 O

(HYSD Fe415 steel )

P

t

=

=

= 0.289

OR

P

t

= 50 1-

√

1-(4.6 / fck) x (M

u

/ b x d

2

)

M

u1

=

f

y

/ f

ck

=

we get ,

0.9595

=

26.689

M

u1

=

= 24.554

= 24.56

KNm

=56 O

1261.5 + 8 O

≥ 56 O

48 O ≤ 1261.48

O ≤ 26.28

mm

V

u

= 25.31

KN

=

From equation

0.87 x f

y

x A

st

x d ( 1 - f

y

x A

st

/ b x d x f

ck

)

1.3 x ( M

u1

/ V

u

) + L

0

≥ L

d

0.9595 x 1000 x 160

2

x 10

-6

Check for shear :

-Shear stress = V

u

/ b x d

Note that the bond is usually critical along long direction.

Development length of bars Ld =

O σ

s

/ 4 x ﺡ

bd

(From Table 7-6 )

M

u1

/ b x d

2

=

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

-6

M

u1

/ b x d

2

=

0.5023 x 1000 x 150

2

x 10

-6

1.3 x ( 24.56 x 10

6

/ 25.31 x 10

3

) + 8 O ≥ 56 O

which gives

……….( O.K.)

From equation

0.87 x f

y

x A

st

x d ( 1 - f

y

x A

st

/ b x d x f

ck

)

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

-6

100 x 462 / 1000 x 160

Short span

11.25 x ( 4.5 / 2 )

(From Table 7-6 )

100 x A

s

/ b x d

100 x 218 / 1000 x 150

Development length of bars Ld =

O σ

s

/ 4 x ﺡ

bd

1.3 x ( M

u1

/ V

u

) + L

0

≥ L

d

1.3 x ( 11.30 x 10

6

/ 25.31 x 10

3

) + 8 O ≥ 56 O

which gives

……….( O.K.)

Assuming L

0

=

100 x A

s

/ b x d

25.31 x 10

3

/ 1000 x 150

This is critical along long span

= 0.169

N / mm

2

= 0.145

N / mm

2 25 difference

-0.05

IS 456-2000 clause 40.2.1.1

20 difference

?

-0.04

k = 1.24

= 0.347

N / mm

2

OR

6 x β

β =

=

12.011

6 x 12.011

=

0.278

IS 456-2000 clause 40.2.1.1

25 difference

-0.05

k = 1.24

20 difference

?

-0.04

= 0.345

N / mm

2

20

P

t

=

= 0.289

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

= 28.8

= 28.00

<

28.8

= 3 x 160

= 480

170

mm

= 5 x 150

= 750

230

mm

for 180 mm slab depth

Design shear strength = 1.24 x 0.28

<

(

ﺡ

C

)

N / mm

2

( too small )

100 x A

s

/ b x d = 100 x 218 / 1000 x 150

0.8 x f

ck

/ 6.89 P

t

, but not less than 1.0

Design shear strength ﺡ

c

= 0.85 √ 0.8 x 15 ( √ 1 + 5 x 12.01 - 1 )

for 180 mm slab depth

……….( O.K.)

Design shear strength ﺡ

c

= 0.85 √ 0.8 x f

ck

( √ 1 + 5 x β - 1 )

from table 7-1

for P

t

= 0.145 ﺡ

c

= 0.28

IS 456-2000 , clause 26.3.3

Basic ( span / d ) ratio =

100 x A

st

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

modification factor = 1.44

Design shear strength = 1.24 x 0.278

……….( O.K.)

Check for deflection :

-This check shall be done along short span

( span / d ) ratio permissible = 1.44 x 20

Check for cracking :

-Actual (span / d ) ratio = 4480 / 160

……….( O.K.)

……….( O.K.)

3 x effective depth of slab or 300 mm whichever is small

spacing provided =

< 300 mm

……….( O.K.)

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

(2) Distribu. bars : maximum spacing permitted for long span steel = 5 x effective depth of slab or 450 mm whichever is small

spacing provided =

< 450 mm