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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
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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 req for design = 352.48 mm
% of Reinforcement = 0.245 %
Min. % of steel required = 0.13 %
Ast min. = 234 mm
Ast req. = 352.5 mm
Longer Span:
Negative Moment (Support) = 12.547 kN-m
Min. % of steel required = 0.13 %
Ast min. = 234 mmAst req. = 277.1 mm
Positive Moment (Mid span) = 9.480 kN-m
Eff. Depth of Slab dy = 132.00 mm
k= Mu / fcu.bd
= 0.018
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SULAYMANYA PROJECT BHEL
Our Case CASE :1
Z = 129.28 mm
Zmax = 125.40 mm
Ast req for design = 209.38 mm
% of Reinforcement = 0.159 %
Min. % of steel required = 0.13 %
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
Required spacing for the ultimate moment 321 mm
Hence required spacing = 321 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 = 352.48 mm
As,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 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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Required spacing for the ultimate moment 408 mm
Hence, required spacing = 396 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 = 277.12 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 277.12)/(8 x 565.49) x (1/1)
= 127.1069008
Clear spacing = 47000 /127.107 = 370 mm
Provided Spacing 200mm c/c < (370 + 12)mm c/c Ok
So Provide 12 mm dia bars for Longer Span (Support) @ 200 mm c/c
Longer Span (Mid Span)
Maximum Spacing 3 x dy = 3 x 132 396 mm
or 750 mm
Required spacing for the ultimate moment 483 mm
Hence, required spacing = 396 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 = 234.00 mm
As,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 234.00)/(8 x 565.49) x (1/1)= 107.3301147
Clear spacing = 47000 /107.330 = 438 mm
Provided Spacing 200mm c/c < (438 + 12)mm c/c Ok
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
Support 12 @ 200mm c/c
Mid span 12 @ 200mm c/c
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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
Page 61
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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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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 req for design = 0.00 mm
% of Reinforcement = 0.000 %
Min. % of steel required = 0.13 %
Ast min. = 195 mm
Ast req. = 195.0 mm
Positive Moment (Mid span) = 591.130 kN-m
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! %
Min. % of steel required = 0.13 %
Ast min. = 195 mm
Ast req. = #NUM! mm
Longer Span:
Negative Moment (Support) = 0.000 kN-m
Eff. Depth of Slab dy = 105.000 mm
k= Mu / fcu.bd2
= 0.000Our Case CASE :1
Z = 105.000 mm
Zmax = 99.750 mm
Ast req for design = 0.000 mm
% of Reinforcement = 0.000 %
Min. % of steel required = 0.13 %
Ast min. = 195 mm
Ast req. = 195.0 mm
Positive Moment (Mid span) = 2.929 kN-m
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Eff. Depth of Slab dy = 105.00 mm
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 req for design = 81.34 mm
% of Reinforcement = 0.077 %
Min. % of steel required = 0.13 %
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
Required spacing for the ultimate moment 403 mm
345 mm
Provide 10 mm dia bars @ 150 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 = 195.00 mmAs,pro = 523.60 mm
bb = (Moment at the section after redistribution)/(Moment at the section before redistribution)
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
Provide 10 mm dia bars @ 150 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 = #NUM! mm
As,pro = 523.60 mm
bb = (Moment at the section after redistribution)/(Moment at the section before redistribution)
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)
Maximum Spacing 3 x dy = 3 x 105 315 mm
or 750 mm
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TUAS - BLOCK 3&4 CB - 70900 R-0 60
Required spacing for the ultimate moment 403 mm
315 mm
Provide 10 mm dia bars @ 150 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 = 195.00 mmAs,pro = 523.60 mm
bb = (Moment at the section after redistribution)/(Moment at the section before redistribution)
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
Provided Spacing 150mm c/c < (487 + 10)mm c/c Ok
So Provide 10 mm dia bars for Longer Span (Support) @ 150 mm c/c
Longer Span (Mid Span)
Maximum Spacing 3 x dy = 3 x 105 315 mm
or 750 mm
Required spacing for the ultimate moment 403 mm
315 mm
Provide 10 mm dia bars @ 150 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 = 195.00 mm
As,pro = 523.60 mm
bb = (Moment at the section after redistribution)/(Moment at the section before redistribution)
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
Provided Spacing 150mm c/c < (487 + 10)mm c/c Ok
So Provide 10 mm dia bars for Longer Span (Mid Span) @ 150 mm c/c
Result
Shorter Span
Support 10 @ 150mm c/c
Mid span #NUM!
Longer Span
Support 10 @ 150mm c/c
Mid span 10 @ 150mm c/c
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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
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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
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