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’ ) =