TWO WAY _SLAB S1

7/30/2019 TWO WAY _SLAB S1

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SULAYMANYA PROJECT BHEL

DESIGN OF TWO WAY SLAB

DATA :

Concrete C30 fcu = 30 N/mm

Steel Fy 460 fy = 415 N/mm

Cover 30 mmClear Shorter Span Lx = 4.50 m Depth of Slab = 180 mm

Clear Longer Span Ly = 7.50 m Dia of reinf. = 12 mm

Eff. Depth of Slab dx= 144 mm

Eff. Depth of Slab dy= 132 mm

Eff.Shorter Span lx = 4.64 m

Eff. Longer Span ly = 7.63 m

LOAD CALCULATIONS:

Self wt. = 4.5 kN/m

Finishing = 0 kN/m

Dead load = 3.02 kN/m

Total DL 7.520 kN/m

Live load = 1.5 kN/m

Equipment = 0.00 kN/m

Total LL 1.500 kN/m

Load Factor DL LL

1.4 1.6

Factored Load 12.928 kN/m

Load considered for Design = 12.928 kN/m

Span Ratio ly/lx = 7.63/4.64

= 1.643

Support Condition 4 Two Adjacent edges Discontinuous

aNx = 0.083

aPx = 0.062

aNy = 0.045

aPy = 0.034

Shorter Span:

Negative Moment (Support) MNx = aNx aNx x wu x lx

= 0.083 x 12.928 x 4.64

= 23.187 kN.m

Positive Moment (Mid span) MPx = aPx x wu x lx

= 0.062 x 12.928 x 4.64

= 17.410 kN.m

120974050.xls.ms_office Two way project 12/17/2012


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Longer Span :

Negative Moment (Support) MNy = aNx aNy x wu x lx

= 0.045 x 12.928 x 4.64

= 12.547 kN.m

Positive Moment (Mid span) MPy = aPy x wu x lx

= 0.034 x 12.928 x 4.64

= 9.480 kN.m

Assume k' = 0.156 (Assuming redistribution does not exced 10%)

CASE : 1If k < k' , compression reinforcement is not required and:

z = d { 0.5 + sqrt (0.25 - ( k / 0.9)) } but not greater than (0.95 * d)

x = (d - z) / 0.45

Ast = Mu / ( 0.87 * fy * z )

CASE : 2

If k > k' , compression reinforcement is required and:

z = d { 0.5 + sqrt (0.25 - ( k' / 0.9)) }

x = (d - z) / 0.45

Lever arm Lv = d - ( 0.9 x/2)

Ast' = ((k - k' ) * fcu * b * d2

/ ( 0.87 * fy *( d - d' )))

Ast = (k' * fcu * b * d2

/ ( 0.87 * fy * z )) + Ast'

Shorter Span:

Negative Moment (Support) = 23.187 kN-mEff. Depth of Slab dx = 144.000 mm

k= Mu / fcu.bd

= 0.037

Our Case CASE :1

Z = 137.77 mm

Zmax = 136.80 mm

Ast req for design = 469.45 mm

% of Reinforcement = 0.326 %

Min. % of steel required = 0.13 %

Ast min. = 234 mm

Ast req. = 469.5 mm

Positive Moment (Mid span) = 17.410 kN-m

Eff. Depth of Slab dx = 144.000 mm

k= Mu / fcu.bd

= 0.028

Our Case CASE :1

Z = 139.374 mm

Zmax = 136.80 mm




Ast min. = 234 mm

Ast req. = 352.5 mm

Longer Span:

Negative Moment (Support) = 12.547 kN-m


Ast min. = 234 mmAst req. = 277.1 mm


Eff. Depth of Slab dy = 132.00 mm

k= Mu / fcu.bd

= 0.018



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Our Case CASE :1

Z = 129.28 mm

Zmax = 125.40 mm




Ast min. = 234 mm

Ast req. = 234.0 mm

Spacing of Reinforcements:-

Shorter Span (Support)

Maximum Spacing 3 x dx = 3 x 144 432 mm

or 750 mm

Required spacing for the ultimate moment 241 mm

Hence, required spacing = 241 mm

Provide 12 mm dia bars @ 200 mm

Clear spacing fs mm

Estimated design service stress in the tension reinforcement

fs = (5 x fy x As,req)/(8 x As,prov) x (1/bb)

As,req = 469.45 mmAs,pro = 565.49 mm

bb = (Moment at the section after redistribution)/(Moment at the section before redistribution)

Assuming bb = 1.00

fs = (5 x 415 x 469.45)/(8 x 565.49) x (1/1)

= 215.3263158

Clear spacing = 47000 /215.326 = 218 mm

Provided Spacing 200mm c/c < (218 + 12)mm c/c

Ok

So Provide 12 mm dia bars for Shorter Span (Support) @ 200 mm c/c

Shorter Span (Mid Span)

Maximum Spacing 3 x dx = 3 x 144 432 mmor 750 mm


Hence required spacing = 321 mm


Clear spacing fs mm



As,req = 352.48 mm

As,pro = 565.49 mm


Assuming bb = 1.00

fs = (5 x 415 x 352.48)/(8 x 565.49) x (1/1)

= 161.6753797

Clear spacing = 47000 /161.675 = 291 mm

Provided Spacing 200mm c/c < (291 + 12)mm c/c Ok

So Provide 12 mm dia bars for Shorter Span (Mid Span) @ 200 mm c/c

Longer Span (Support)

Maximum Spacing 3 x dy = 3 x 132 396 mm

or 750 mm



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Clear spacing fs mm



As,req = 277.12 mmAs,pro = 565.49 mm


Assuming bb = 1.00

fs = (5 x 415 x 277.12)/(8 x 565.49) x (1/1)

= 127.1069008

Clear spacing = 47000 /127.107 = 370 mm


So Provide 12 mm dia bars for Longer Span (Support) @ 200 mm c/c

Longer Span (Mid Span)


or 750 mm




Clear spacing fs mm



As,req = 234.00 mm

As,pro = 565.49 mm


Assuming bb = 1.00

fs = (5 x 415 x 234.00)/(8 x 565.49) x (1/1)= 107.3301147

Clear spacing = 47000 /107.330 = 438 mm


So Provide 12 mm dia bars for Longer Span (Mid Span) @ 200 mm c/c

Result

Shorter Span

Support 12 @ 200mm c/c

Mid span 12 @ 200mm c/c

Longer Span





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CHECK FOR DEFLECTION

Allowable span / depth Ratio l/d = 26

% of Tension reinforcement pt = 0.39

Corresponding Compression reinf. Bar dia = 12 @ 200mm c/c

% of Compression reinforcement pc = 0.393

Modification factor for Tension reinf. F1 = 1.45Modification factor for Comp. reinf. F2 = 1.12

Modified span/depth ratio = l/d x F1 x F2

= 26 x 1.45 x 1.12

= 42.02

Actual span/depth ratio = 4644/144

= 32.25

Actual span/depth ratio < Modified span/depth ratio

32.25 < 42.02 SAFE

12 @ 200mm c/c 12 @ 200mm c/c

12 @ 200mm c/c 12 @ 200mm c/cSHORTER SPAN



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Dialog3

Support Conditions

OK

Interior Panel

One Short Edge Discountinuous

One Long Edge Discountinuous

Two Adjacent edges Discountinuous

Four edges Discountinuous

Three edges Discountinuous (One S. Edge Conti.)

Two Long edges Discountinuous

Three edges Discountinuous (One L. Edge Conti.)

Two Short edges Discountinuous

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fy 500 415 250 pc F2 pc F2

fs 312.5 259.375 156.25 0.5 1.0 0 1 1.00

pt F1 0.25 1.08 1.08

fe500 fe415 fe250 0.5 1.14 1.14

0.025 6.202993 544.2161 -3.25164 0.75 1.2 1.20

0.2 1.378087 1.765946 3.89266 1 1.25 1.25

0.4 1.094346 1.325535 2.247002 1.25 1.29 1.29

0.6 0.976711 1.156779 1.801495 1.5 1.33 1.33

0.8 0.907498 1.060945 1.579327 1.75 1.36 1.36

1 0.860215 0.996885 1.441441 2 1.39 1.39

1.2 0.82509 0.950017 1.345464 2.25 1.42 1.42

1.4 0.797556 0.913697 1.273756 2.5 1.45 1.45

1.6 0.775149 0.884408 1.217545 2.75 1.47 1.47

1.8 0.756404 0.860089 1.171927 3 1.5 1.50

2 0.740388 0.839441 1.133923

2.2 0.726473 0.821599 1.101608

2.4 0.714218 0.80596 1.073673

2.6 0.703305 0.79209 1.049199

2.8 0.693494 0.779667 1.0275133 0.684603 0.768447 1.008115

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3


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pt f1

0.0889 2 1.99

0.2 1.605 1.63

0.4 1.2667 1.26

0.6 1.1063 1.10

0.8 1.025 1.02

1 0.9647 0.97

1.2 0.9235 0.93

1.4 0.8941 0.89

1.6 0.8706 0.86

1.8 0.8471 0.84

2 0.8294 0.83

2.2 0.8176 0.82

2.4 0.8059 0.802.6 0.7882 0.78

2.8 0.7706 0.74

3 0.7588 0.73

3.2

y = 9E-05x6 - 0.0007x5 + 0.0007x4 + 0.0167x3 - 0.0997x2 + 0.3308x + 1.0006

1

1.1

1.2

1.3

1.4

1.5

0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 2.25 2.5 2.75 3

ModificationFactor

% of Pc


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y = 0.0773x6 - 0.8184x5 + 3.4419x4 - 7.3509x3 + 8.4739x2 - 5.2412x + 2.3914

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3

ModificationFacto

r

% of Pt

%of Pt Vs Modification Factor for Fe415 Steel


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TUAS - BLOCK 3&4 CB - 70900 R-0 56

DESIGN OF TWO WAY SLAB

DATA :

Concrete C30 fcu = 30 N/mm

Steel 415 fy = 415 N/mm

Cover 30 mm

Clear Shorter Span Lx = 2.50 m Depth of Slab = 150 mmClear Longer Span Ly = 3.00 m Dia of reinf. = 10 mm

Eff. Depth of Slab dx= 115 mm

Eff. Depth of Slab dy= 105 mm

Eff.Shorter Span lx = 2.62 m

Eff. Longer Span ly = 3.11 m

LOAD CALCULATIONS:

Thickness of RCC slab = 0.15 m

Self wt of RCC slab = 0.15x25 = 3.75 kN/m

Total dead weight on floor slab = 3.75 kN/m

Self wt. = 3.75 kN/m

Dead load = kN/m

Total DL 3.750 kN/m

Live load = 1.5 kN/m

Total LL 1.500 kN/m

Load Factor DL LL

1.4 1.6

Factored Load 7.65 kN/m

Load considered for Design = 7.650 kN/m

Span Ratio ly/lx = 3.11/2.62

= 1.187

Support Condition 9 Four Edges Discontinuous

aNx = 0.000

aPx = 11.300

aNy = 0.000

aPy = 0.056

Shorter Span:

Negative Moment (Support) MNx = aNx aNx x wu x lx

= 0.000 x 7.650 x 2.62= 0.000 kN.m

Positive Moment (Mid span) MPx = aPx x wu x lx

= 11.300 x 7.650 x 2.62

= 591.130 kN.m

(Considering 15% Margin)

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TUAS - BLOCK 3&4 CB - 70900 R-0 57

Longer Span :

Negative Moment (Support) MNy = aNx aNy x wu x lx

= 0.000 x 7.650 x 2.62

= 0.000 kN.m

Positive Moment (Mid span) MPy = aPy x wu x lx

= 0.056 x 7.650 x 2.62

= 2.929 kN.m

Assume k' = 0.156 (Assuming redistribution does not exced 10%)

CASE : 1If k < k' , compression reinforcement is not required and:

z = d { 0.5 + sqrt (0.25 - ( k / 0.9)) } but not greater than (0.95 * d)

x = (d - z) / 0.45

Ast = Mu / ( 0.87 * fy * z )

CASE : 2

If k > k' , compression reinforcement is required and:

z = d { 0.5 + sqrt (0.25 - ( k' / 0.9)) }

x = (d - z) / 0.45

Lever arm Lv = d - ( 0.9 x/2)

Ast' = ((k - k' ) * fcu * b * d2

/ ( 0.87 * fy *( d - d' )))

Ast = (k' * fcu * b * d2

/ ( 0.87 * fy * z )) + Ast'

Shorter Span:

Negative Moment (Support) = 0.000 kN-mEff. Depth of Slab dx = 115.000 mm

k= Mu / fcu.bd

= 0.000

Our Case CASE :1

Z = 115.00 mm

Zmax = 109.25 mm




Ast min. = 195 mm

Ast req. = 195.0 mm


Eff. Depth of Slab dx = 115.000 mm

k= Mu / fcu.bd

= 1.490

Our Case CASE :2

Z = #NUM! mm

Zmax = 109.25 mm

Ast req for design = #NUM! mm

% of Reinforcement = #NUM! %


Ast min. = 195 mm

Ast req. = #NUM! mm

Longer Span:

Negative Moment (Support) = 0.000 kN-m


k= Mu / fcu.bd2

= 0.000Our Case CASE :1

Z = 105.000 mm

Zmax = 99.750 mm




Ast min. = 195 mm

Ast req. = 195.0 mm


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TUAS - BLOCK 3&4 CB - 70900 R-0 58


k= Mu / fcu.bd

= 0.009

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TUAS - BLOCK 3&4 CB - 70900 R-0 59

Our Case CASE :1

Z = 103.96 mm

Zmax = 99.75 mm




Ast min. = 195 mm

Ast req. = 195.0 mm

Spacing of Reinforcements:-

Shorter Span (Support)

Maximum Spacing 3 x dx = 3 x 115 345 mm

or 750 mm


345 mm


Clear spacing fs mm



As,req = 195.00 mmAs,pro = 523.60 mm


Assuming bb = 1.00

fs = (5 x 415 x 195.00)/(8 x 523.60) x (1/1)

= 96.60

Clear spacing = 47000 /96.597 = 487 mm

Provided Spacing 150mm c/c < (487 + 10)mm c/c

Ok

So Provide 10 mm dia bars for Shorter Span (Support) @ 150 mm c/c

Shorter Span (Mid Span)

Maximum Spacing 3 x dx = 3 x 115 345 mmor 750 mm

Required spacing for the ultimate moment #NUM! mm

#NUM! mm


Clear spacing fs mm



As,req = #NUM! mm

As,pro = 523.60 mm


Assuming bb = 1.00

fs = #NUM!

= #NUM!

Clear spacing = #NUM! #NUM! mm

#NUM! #NUM!

So Provide 10 mm dia bars for Shorter Span (Mid Span) @ #NUM! mm c/c

Longer Span (Support)


or 750 mm

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TUAS - BLOCK 3&4 CB - 70900 R-0 60


315 mm


Clear spacing fs mm



As,req = 195.00 mmAs,pro = 523.60 mm


Assuming bb = 1.00

fs = (5 x 415 x 195.00)/(8 x 523.60) x (1/1)

= 96.59710327

Clear spacing = 47000 /96.597 = 487 mm


So Provide 10 mm dia bars for Longer Span (Support) @ 150 mm c/c

Longer Span (Mid Span)


or 750 mm


315 mm


Clear spacing fs mm



As,req = 195.00 mm

As,pro = 523.60 mm


Assuming bb = 1.00

fs = (5 x 415 x 195.00)/(8 x 523.60) x (1/1)= 96.59710327

Clear spacing = 47000 /96.597 = 487 mm


So Provide 10 mm dia bars for Longer Span (Mid Span) @ 150 mm c/c

Result

Shorter Span


Mid span #NUM!

Longer Span



120974050.xls.ms_office 12/17/2012


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TUAS - BLOCK 3&4 CB - 70900 R-0 61

CHECK FOR DEFLECTION

Allowable span / depth Ratio l/d = 26

% of Tension reinforcement pt = 0.46

Corresponding Compression reinf. Bar dia = 10 @ 150mm c/c

% of Compression reinforcement pc = 0.455

Modification factor for Tension reinf. F1 = 2.00Modification factor for Comp. reinf. F2 = 1.13

Modified span/depth ratio = l/d x F1 x F2

= 26 x 2.00 x 1.13

= 58.85

Actual span/depth ratio = 2615/115

= 22.74

Actual span/depth ratio < Modified span/depth ratio

22.74 < 58.85 SAFE

10 @ 150mm c/c 10 @ 150mm c/c

10 @ 150mm c/c #NUM!SHORTER SPAN

120974050.xls.ms_office 12/17/2012


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TUAS - BLOCK 3&4 CB - 70900 R-0 62

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TUAS - BLOCK 3&4 CB - 70900 R-0 64

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TUAS - BLOCK 3&4 CB - 70900 R-0 65

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TUAS - BLOCK 3&4 CB - 70900 R-0 66

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BS 8110 : Part 1 : 1985 Table 3.15

Bending Moment Coefficients for Rectangular panels supported on four sides with provisions

for Torsion at corners.

ax ay

1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

1 N 0.031 0.037 0.042 0.046 0.050 0.053 0.059 0.063 0.032

P 0.024 0.028 0.032 0.035 0.037 0.040 0.044 0.048 0.024

2 N 0.039 0.044 0.048 0.052 0.055 0.058 0.063 0.067 0.037

P 0.029 0.033 0.036 0.039 0.041 0.043 0.047 0.050 0.028

3 N 0.039 0.049 0.056 0.062 0.068 0.073 0.082 0.089 0.037

P 0.030 0.036 0.042 0.047 0.051 0.055 0.062 0.067 0.028

4 N 0.047 0.056 0.063 0.069 0.074 0.078 0.087 0.093 0.045

P 0.036 0.042 0.047 0.051 0.055 0.059 0.065 0.070 0.034

5 N 0.046 0.050 0.054 0.057 0.060 0.062 0.067 0.070 0.000

P 0.034 0.038 0.040 0.043 0.045 0.047 0.050 0.053 0.034

6 N 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.045

P 0.034 0.046 0.056 0.065 0.072 0.078 0.091 0.100 0.034

7 N 0.057 0.065 0.071 0.076 0.081 0.084 0.092 0.098 0.000

P 0.043 0.048 0.053 0.057 0.060 0.063 0.069 0.074 0.044

8 N 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.058

P 0.042 0.054 0.063 0.071 0.078 0.084 0.096 0.105 0.044

9 N 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

P 0.055 0.065 0.074 0.081 0.087 0.092 0.103 0.111 0.056

As per BS 8110 : Part 1 : 1985 TABLE 3.16 Shear Coefficients for uniformly loaded Two - way Rectangular slabReference "Limit State Design of Reinforced Concrete by P.C. Varghese" Table 12.3 Pg.228

gx gy

1.0 1.1 1.2 1.3 1.4 1.5 1.75 2.0

1 C 0.330 0.360 0.390 0.410 0.430 0.450 0.480 0.500 0.330

D 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

2 C 0.360 0.390 0.420 0.440 0.450 0.470 0.500 0.520 0.360

D 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.240

3 C 0.360 0.400 0.440 0.470 0.490 0.510 0.550 0.590 0.360

D 0.240 0.270 0.290 0.310 0.320 0.340 0.360 0.380 0.000

4 C 0.400 0.440 0.470 0.500 0.520 0.540 0.570 0.600 0.400

D 0.260 0.290 0.310 0.330 0.340 0.350 0.380 0.400 0.260

5 C 0.400 0.430 0.450 0.470 0.480 0.490 0.520 0.540 0.000

D 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.260

6 C 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.400

D 0.260 0.300 0.330 0.360 0.380 0.400 0.440 0.470 0.000

7 C 0.450 0.480 0.510 0.530 0.550 0.570 0.600 0.630 0.000

D 0.300 0.320 0.340 0.350 0.360 0.370 0.390 0.410 0.29

8 C 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.45

D 0.290 0.330 0.360 0.380 0.400 0.420 0.450 0.480 0.3

9 C 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000D 0.330 0.360 0.390 0.410 0.430 0.450 0.480 0.500 0.330

MATERIAL PROPERTIES Edge conditions

concrete fck steel fy 1 Interior Panel

C15 15 Fy 250 250 2 One Short Edge Discontinuous

C20 20 Fy 460 460 3 One Long Edge Discontinuous

C25 25 4 Two Adjacent edges Discontinuous

C30 30 5 Two Short edges Discontinuous

C35 35 6 Two Long edges Discontinuous

C40 40 7 Three edges Discontinuous (One long edge continuous)

8 Three edges Discontinuous (One short edge continuous)

9 Four Edges Discontinuous

Documents

TWO WAY _SLAB S1