Post on 27-Nov-2014
Common design parameters:
0.018260911.596639
Required bending coefficient, 4.5063529Minimum flexural reinforcement ratio (NSCP Section 410.6.1)
0.004793 0.0050725
Slab Designation: 2S1-2S2/3S1-3S2/RS1-RS2Check if one-way or two-way slab
Short span 2S1/2S4: 3.15
Short span 2S2/2S3: 3.125
Long span 2S1-2S4: 7.25
0.4344828 < 0.50 therefore, one wayMinimum slab thickness ( NSCP Table 409-1 for one-end continuous)
97.630952 try 100 mmDead load
Slab weight = 2.95Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Live loadProduction area = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Check limitations (NSCP Section 408.4.3)a) LL/DL = 0.9677419 < 3 b) (L-S)/S = 0.008 < 0.2c) Load is uniformly distributed
Design moments (NSCP Section 408.4.3)-M1 = 6.24456+M2 = 10.70496-M3 = 14.868236+M4 = 9.21875-M5 = 13.4090909
Check effective depth to control deflection based on maximum factored moment = 14.868236
60.547451
74 > 57.80134 mm OKCheck effective depth for shear
Shear at supports, R = 27.35712
55.472586 > 26.40941 kN OKCompute steel reinforcements
Section s1 6.24456 1.26705625 0.00472 0.0062933 465.70648 242.72799 say 200 mm2 10.70496 2.17209642 0.0082662 0.0082662 611.70117 184.79612 say 150 mm3 14.868236 3.01684847 0.0117284 0.0117284 867.9043 130.24477 say 100 mm4 9.21875 1.87053608 0.007067 0.007067 522.95567 216.15599 say 200 mm5 13.4090909091 2.72077975 0.010497 0.010497 776.77651 145.52448 say 100 mm
Design of Floor Slabs (fc’ = 28 MPa, fy = 276 MPa)
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =
Ru =
r min = > 1.4/ fy =
La =
La =
Lb =
Ratio, m = La/Lb =
Minimum t =
Required effective depth, d req'd =
Actual effective depth, d form =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
6.24456
1.26705620.00472
Steel area, 465.70648
Area of 12 mm f bars: 113.04Required spacing, s = 242.72799Therefore, use 12 mm bars @ 0.2 m top bars @ discontinuous edge
Compute temperature bars
250
Using 10 mm f bars: 78.5Req'd spacing: s = 314Max spacing: s = 625Therefore, use 10 mm bars @ .25m o.c. temperature bars
Loads transmitted to supporting beams by 2S12G-1,3/1-2, 3G-1,3/1-2, RG-1,3/1-2:DL = 7.812LL = 7.562B-1,2/1-2, 3B-1,1/1-2, RB-1,2/1-2:DL = 8.9838LL = 8.694
Loads transmitted to supporting beams by 2S22B-1,2/1-2, 3B-1,2/1-2, RB-1,2/1-2, 2G-2/1-2, 3G-2/1-2, RG-2/1-2:DL = 7.812LL = 7.56
Slab Designation: 2S3Check if one-way or two-way slab
3.15
5.25Ratio, m = 0.6 > 0.50 (two-way)
Minimum slab thicknessMinimum t = 97.222222 try 100 mm
Dead loadSlab weight = 2.95Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Live loadProduction area = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Design momentscase 4 m = 0.6( - ) Ma = 13.33838( + ) Ma = 9.0766267( - ) Mb = 4.579344( + ) Mb = 3.363948At discontinuous edge: ( - ) Ma = 3.0255422At discontinuous edge: ( + ) Mb = 1.121316
Check effective depth for flexure
57.347922
74 > 56.5114 mm OK
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
As =
Ab =
For fy = 276 MPa, Ast = 0.002bt =
Ab =
Check span in short direction, La =
Clear span in long direction, Lb =
Required effective depth, d req'd =
Actual effective depth, d form =
Check effective depth for shear
21.172032
2.616768
55.472586 > 21.58682 kN OK
Compute steel reinforcements
Section s( - ) Ma,cont 13.33838016 2.70643214 0.0104378 0.0139171 1029.8641 109.76206 say 100 mm( + ) Ma,mid 9.07662672 1.84169847 0.0069532 0.0069532 514.53946 219.6916 say 200 mm( - ) Mb,cont 4.579344 0.92917458 0.003435 0.00458 200 565.2 say 300 mm( + ) Mb,mid 3.363948 0.97235172 0.0035981 0.0047975 200 565.2 say 300 mm
( - ) Ma,discont 3.02554224 0.61389949 0.0022537 0.003005 200 565.2 say 300 mm( - ) Mb,discont 1.121316 0.22752131 0.0008283 0.0011044 200 565.2 say 300 mm
*d = 74 - 12 = 62 (long direction steel is placed on top of short direction steel at midspan)
250
13.33838
2.70643210.0104378
Steel area, 1029.8641
Area of 12 mm f bars: 113.04Required spacing, s = 109.76206Therefore, use 12 mm bars @ 0.1 m o.c top bars
Loads transmitted to supporting beams2G-3/3,3G-3/3,RG-3/3,2B-2/3,3B-2/3,RB-2/3:DL = 7.34328LL = 7.10642G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1:DL = 7.812LL = 7.56
Slab Designation: 2S4/3S4/RS4Check if one-way or two-way slab
3.1
5.25Ratio, m = 0.5904762 > 0.50 (two-way)
Minimum slab thicknessMinimum t = 96.666667 try 100 mm
Dead loadSlab weight = 2.95Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Live loadProduction area = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Design momentscase 9 m = 0.6( - ) Ma = 12.9183
In short direction: Va =
In long direction: Vb =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
**temperature controls, Ast = 0.002bt =
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
As =
Ab =
Check span in short direction, La =
Clear span in long direction, Lb =
( + ) Ma = 7.0289846( - ) Mb = 2.497824( + ) Mb = 2.339946At discontinuous edge: ( - ) Mb = 0.779982At discontinuous edge: ( + ) Mb = 0.779982
Check effective depth for flexure
56.437638
74 > 53.76681 mm OKCheck effective depth for shear
22.006528
1.17056
55.472586 > 19.87212 kN OKCompute steel reinforcements
Section s( - ) Ma,cont 12.91830016 2.62119555 0.0100873 0.0134497 995.27539 113.57661 say 100 mm( + ) Ma,mid 7.02898464 1.4262204 0.0053324 0.0053324 394.59644 286.46989 say 250 mm( - ) Mb,cont 2.497824 0.5068225 0.0018563 0.0024751 200 565.2 say 300 mm( + ) Mb,mid 2.339946 0.67636316 0.0024864 0.0033153 200 565.2 say 300 mm
( - ) Ma,discont 0.779982 0.15826272 0.0005753 0.0007671 200 565.2 say 300 mm
*d = 74 - 12 = 62 (long direction steel is placed on top of short direction steel at midspan)
250
12.9183
2.62119560.0100873
Steel area, 995.27539
Area of 12 mm f bars: 113.04Required spacing, s = 113.57661Therefore, use 12 mm bars @ 0.1 m o.c. top bars
Loads transmitted to supporting beams2G-2/3,3B-2/3,3G-2/3,3B-2/3,RG-2/3,RB-2/3:DL = 7.22672LL = 6.99362G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1:DL = 7.812LL = 7.56
Slab Designation: 2S5Check if one-way or two-way slab
2.45
2.5Ratio, m = 0.98 > 0.50 (two-way)
Minimum slab thicknessMinimum t = 49.444444 try 100 mm
Dead loadSlab weight = 2.95Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Required effective depth, d req'd =
Actual effective depth, d form =
In short direction: Va =
In long direction: Vb =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
**temperature controls, Ast = 0.002bt =
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
As =
Ab =
Check span in short direction, La =
Clear span in long direction, Lb =
Live loadProduction area = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Design momentscase 9 m = 0.98( - ) Ma = 5.6754262( + ) Ma = 2.4839401( - ) Mb = 2.96416( + ) Mb = 2.18272At discontinuous edge: ( - ) Mb = 0.7275733
Check effective depth for flexure
37.408077
74 > 30.98426 mm OKCheck effective depth for shear
12.692646
5.8097536
55.472586 > 11.45173 kN OKCompute steel reinforcements
Section s( - ) Ma,cont 5.675426176 1.1515758 0.0042786 0.0057047 422.15035 267.77189 say 300 mm( + ) Ma,mid 2.483940144 0.50400539 0.0018459 0.0018459 250 452.16 say 300 mm( - ) Mb,cont 2.96416 0.60144469 0.0022074 0.0029432 200 565.2 say 300 mm( + ) Mb,mid 2.18272 0.63091687 0.0023171 0.0030894 200 565.2 say 300 mm
( - ) Ma,discont 0.72757333333 0.14762871 0.0005366 0.0007154 200 565.2 say 300 mm
*d = 74 - 12 = 62 (long direction steel is placed on top of short direction steel at midspan)
250
5.6754262
1.15157580.0042786
Steel area, 422.15035
Area of 12 mm f bars: 113.04Required spacing, s = 267.77189Therefore, use 12 mm bars @ 0.3 m o.c. bottom bars
Loads transmitted to supporting beams2G-2/3,2B-3:DL = 5.1646LL = 4.9982G-7/2, 2B-4:DL = 1.054LL = 1.02
Slab Designation: 2S6/3S6Check if one-way or two-way slab
Required effective depth, d req'd =
Actual effective depth, d form =
In short direction: Va =
In long direction: Vb =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
**temperature controls, Ast = 0.002bt =
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
As =
Ab =
2.45
6.55Ratio, m = 0.3740458 < 0.50 (one-way)
Minimum slab thicknessMinimum t = 89.357143 try 100 mm
Dead loadSlab weight = 2.95Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Live loadProduction area = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Check limitations (NSCP Section 408.4.3)a) LL/DL = 0.9677419 < 3 b) (L-S)/S < 0.20c) Load is uniformly distributed
Compute moments (NSCP Section 408.4.3)For span ledd than 3 m:At supports, - M1 = 7.5551467At midspan, + M2 = 6.47584
Check effective depth to control deflection based on maximum factored moment = 7.5551467
43.160612
74 > 47.88771 mm OKCheck effective depth for shear
Shear at supports, R = 18.5024
55.472586 > 20.66823 kN OKCompute steel reinforcements
Section s1 7.55514666667 1.53298163 0.0057458 0.0057458 425.18632 265.85991 say 200 mm2 6.47584 1.31398425 0.0049001 0.0049001 362.60553 311.74373 say 200 mm
7.5551467
1.53298160.0057458
Steel area, 425.18632
Area of 12 mm f bars: 113.04Required spacing, s = 265.85991Therefore, use 12 mm bars @ 0.2 m o.c. top bars
Compute temperature bars
200
Using 10 mm f bars: 78.5Req'd spacing: s = 392.5Max spacing: s = 625Therefore, use 10 mm bars @ .25 mm o.c temp bars
Loads transmitted to supporting beams2B-4, 3B-4:DL = 6.076LL = 5.882G-6/2, 3G-6/2:DL = 6.9874
Clear span in short direction, La =
Clear span in long direction, Lb =
Required effective depth, d req'd =
Actual effective depth, d form =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
As =
Ab =
For fy = 276 MPa, Ast = 0.002bt =
Ab =
LL = 6.762
Slab designation: CS1Check if one-way or two-way slab
1Slab is one way since it is catilever slab
Minimum slab thicknessmin t = 69.5 try 100 mm
Dead loadSlab weight = 2.36
Live load Canopy = 1.9
Factored loads Factored dead load = 3.304Factored live load = 3.23Total factored load = 6.534
Compute momentsAt support: ( - ) M1 = 3.267
Check effective depth for flexureRequired effective depth, d = 28.381851Actual effective depth, d = 74 > 29 OK
Check effective depth for shearAt support: V1 = 6.534
55.472586 > 6.534 OKFlexural steel at support
0.66289260.0024362
(4/3)r = 0.00324826
240.37156
113.04Spacing of 12-mm f bars, s = 470.27194Maximum spacing, s = 300Therefore, use 12 mm bars @ .3 m o.c top bars
Spacing of 10-mm f temperature bars
200
Using 10 mm f bars: 78.5Req'd spacing: s = 392.5Max spacing: s = 500Therefore, use 10 mm bars @ .25 mm o.c temp bars
Loads transmitted to supporting beams2G-1,3/1-3, 3G-1,3/1-3, RG-1,3/1-3, 2G-7/1-2, 3G-7/1-2, RG-7/1-2:DL = 2.36LL = 1.9
Slab Designation: RS6Check if one-way or two-way slab
2.45
3.1Ratio, m = 0.7903226 > 0.50 (two-way)
Minimum slab thicknessMinimum t = 61.666667 try 100 mm
Dead loadSlab weight = 2.95
Clear span in short direction, La =
Shear capacity: fVc =
Bending coefficient, Ru =Steel ratio, r = < r min
Steel area, As =
Area of 12-mm bars: Ab =
For fy = 276 MPa, Ast = 0.002bt =
Ab =
Clear span in short direction, La =
Clear span in long direction, Lb =
Cement finish = 1.53Ceiling suspended loads = 0.48Total dead load = 4.96
Live loadSame as lower floor = 4.8
Factored loadsFactored deal load = 6.944Factored live load = 8.16Total factored load = 15.104
Design momentscase 2 m = 0.7903226 use m = 0.90( - ) Ma = 5.8930144( + ) Ma = 3.8266178( - ) Mb = 3.9190349( + ) Mb = 2.537655
Check effective depth for flexure
38.11842
74 > 38.83294 mm OKCheck effective depth for shear
13.136704
7.40096
18.925941 > 12.35558 kN OKCompute steel reinforcements
Section s( - ) Ma,cont 5.8930144 1.19572567 0.004447 0.0059293 438.77127 257.62853 say 200 mm( + ) Ma,mid 3.82661776 0.7764422 0.0028606 0.0038142 282.25038 400.49547 say 400 mm( - ) Mb,cont 3.91903488 0.79519416 0.0029309 0.0039079 289.18692 390.88905 say 300 mm( + ) Mb,mid 2.53765504 0.73351111 0.0026999 0.0035999 266.39172 424.33751 say 400 mm
*d = 74 - 12 = 62 (long direction steel is placed on top of short direction steel at midspan)
250
5.8930144
1.19572570.004447
438.77127
113.04Required spacing, s = 257.62853Maximum spacing, s = 300Therefore, use 12 mm bars @ 0.2 m o.c top bars
Loads transmitted to supporting beamsRG-6/2, RB-4:DL = 4.01016LL = 3.8808RB-1/3, RG-2/3:DL = 2.61392LL = 2.5296
Slab Designation: US1Check if one-way or two-way slab
3.15
Required effective depth, d req'd =
Actual effective depth, d form =
In short direction: Va =
In long direction: Vb =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
**temperature controls, Ast = 0.002bt =
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
Steel area, As =
Area of 12 mm f bars: Ab =
Clear span in short direction, La =
6.55Ratio, m = 0.480916 < 0.50 (one-way)
Minimum slab thicknessMinimum t = 91.125 try 100 mm
Dead loadSlab weight = 2.36
Ceiling suspended loads = 0.48Total dead load = 2.84
Live loadUpper deck = 1.9
Factored loadsFactored deal load = 3.976Factored live load = 3.23Total factored load = 7.206
Compute moments (NSCP Section 408.4.3)For span less than 3 m:At supports, - M1 = 5.9584613At midspan, + M2 = 5.1072525
Check effective depth for flexure
38.329504
74 > 33.85009 mm OKCheck effective depth for shear
Shear at supports, R = 11.34945
55.472586 > 10.13925 OKCompute steel reinforcements
Section s1 5.95846125 1.2090052 0.0044978 0.005997 443.77823 254.72182 say 300 mm2 5.1072525 1.03629018 0.0038402 0.0051202 378.89799 298.33887 say 300 mm
5.9584613
1.20900520.0044978
443.77823
Area of 12 mm f bars: 113.04Required spacing, s = 254.72182Therefore, use 12 mm bars @ 0.3 m o.c. top bars
Compute temperature bars
200
Using 10 mm f bars: 78.5Req'd spacing: s = 392.5Max spacing: s = 500Therefore, use 10 mm bars @ .3 mm o.c temp bars
Loads transmitted to supporting beamsUB-4, UB-5:DL = 4.473LL = 2.9925
Clear span in long direction, Lb =
Required effective depth, d req'd =
Actual effective depth, d form =
Shear capacity, Vc =
Mu Ru Req'd r Use r As
Sample computations: For Mu =
Moment resistance coefficient, Ru =Steel Ratio, r =
Steel area, As =
Ab =
For fy = 276 MPa, Ast = 0.002bt =
A+D31 =
Section Mu Ru Req'r p Use p As s1 5.8424 1.185455726 0.0044078 0.0058771 434.90459 259.91908 say 2502 10.0156 2.0322214106 0.0077077 0.0077077 570.36908 198.18746 say1503 14.0218 2.8451018586 0.0110116 0.0110116 814.86018 138.72319 say 100
4 8.7636 1.778183589 0.0067033 0.0067033 496.04522 227.88245 say 200
5 12.7471 2.5864580797 0.0099448 0.0099448 735.9185 153.60397 say 150
m = 11.6 Ab = 113.04
fy = 276 s req'd = 259.91908fc' = 28 s max = 300
Common design parameters:
0.018260911.596639
Required bending coefficient, Ru = 4.5063529
Minimum flexural reinforcement ratio (NSCP Section 410.6.1)p min = 0.004793 > 1.4/ fy = 0.0050725
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =
Common design parameters:
0.018260911.596639
Required bending coefficient, Ru = 4.5063529Minimum flexural reinforcement ratio (NSCP Section 410.6.1)
p min = 0.004793 > 1.4/ fy = 0.0050725
0.9 0.40.875
0.85 0.34
-0.37
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =
Common design parameters:
0.018260911.596639
Required bending coefficient, Ru = 4.5063529Minimum flexural reinforcement ratio (NSCP Section 410.6.1)
p min = 0.004793 > 1.4/ fy = 0.0050725
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =
Common design parameters:
0.018260911.596639
Required bending coefficient, Ru = 4.5063529Minimum flexural reinforcement ratio (NSCP Section 410.6.1)
r min = 0.004793 > 1.4/ fy = 0.0050725
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =
Common design parameters:
0.0182609
11.596639
Required bending coefficient, Ru = 4.5063529
Minimum flexural reinforcement ratio (NSCP Section 410.6.1)r min = 0.004793 > 1.4/ fy = 0.0050725
To control deflection, r ≤ 0.18fc’/ fy =
Strength ratio, m = fy / (0.18)(fc’ ) =
Common design parameters:
0.018260911.596639
Required bending coefficient, Ru = 4.5063529
Minimum flexural reinforcement ratio (NSCP Section 410.6.1)
r min = 0.004793 > 1.4/ fy = 0.0050725
To control deflection, r ≤ 0.18fc’/ fy =Strength ratio, m = fy / (0.18)(fc’ ) =