1.design Of Slab 4j4n28
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Design of Floor Slabs (fc’ = 28 MPa, fy = 276 MPa) Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = 11.6 Strength ratio, m = fy / (0.18)(fc’ ) = Ru = Required bending coefficient, 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) ρ min = 0 > 1.4/ fy = 0.01
Slab Designation: 2S1-2S2/3S1-3S2/RS1-RS2 Check if one-way or two-way slab La = Short span 2S1/2S4: 3.15 L = Short span 2S2/2S3: 3.13 a Lb = Long span 2S1-2S4: 7.25 Ratio, m = La/Lb = 0.43 < 0.50 therefore, one way Minimum slab thickness ( NS Table 409-1 for one-end continuous) Minimum t = 97.63 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Check limitations (NS Section 408.4.3) a) LL/DL = 0.97 < 3 b) (L-S)/S = 0.01 < 0.2 c) Load is uniformly distributed Design moments (NS Section 408.4.3) -M1 = 6.24 +M2 = 10.7 -M3 = 14.87 +M4 = 9.22 -M5 = 13.41 Check effective depth to control deflection based on maximum factored moment = Required effective depth, d req'd = 60.55 Actual effective depth, d form = 74 > 57.80134 mm OK Check effective depth for shear Shear at s, R = 27.36 Shear capacity, Vc = 55.47 > 26.40941 kN OK Compute steel reinforcements Mu Ru As Req'd ρ Use ρ Section s 1 6.24 1.27 0 0.01 465.71 242.73 say 200 mm 2 10.7 2.17 0.01 0.01 611.7 184.8 say 150 mm 3 14.87 3.02 0.01 0.01 867.9 130.24 say 100 mm 4 9.22 1.87 0.01 0.01 522.96 216.16 say 200 mm 5 13.41 2.72 0.01 0.01 776.78 145.52 say 100 mm
14.87
Sample computations: For Mu = 6.24 Moment resistance coefficient, Ru = 1.27 Steel Ratio, ρ = 0 As = Steel area, 465.71 Ab = Area of 12 mm f bars: 113.04 Required spacing, s = 242.73 Therefore, use 12 mm bars @ 0.2 m top bars @ discontinuous edge Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = 250 A = Using 10 mm f bars: 78.5 b Req'd spacing: s = 314 Max spacing: s = 625 Therefore, use 10 mm bars @ .25m o.c. temperature bars Loads transmitted to ing beams by 2S1 2G-1,3/1-2, 3G-1,3/1-2, RG-1,3/1-2: DL = 7.81 LL = 7.56 2B-1,2/1-2, 3B-1,1/1-2, RB-1,2/1-2: DL = 8.98 LL = 8.69 Loads transmitted to ing beams by 2S2 2B-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 = 7.81 LL = 7.56
Slab Designation: 2S3 Check if one-way or two-way slab Check span in short direction, La = Clear span in long direction, Lb = Ratio, m = 0.6 > 0.50 (two-way) Minimum slab thickness Minimum t = 97.22 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Design moments case 4 m= 0.6 (-) Ma = 13.34 (+) Ma = 9.08 (-) Mb = 4.58 (+) Mb = 3.36 At discontinuous edge: ( - ) Ma = At discontinuous edge: ( + ) Mb = Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form =
3.15 5.25
3.03 1.12 57.35 74 > 56.5114 mm OK
Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc =
Compute steel reinforcements Mu Section ( - ) Ma,cont 13.34 ( + ) Ma,mid 9.08 ( - ) Mb,cont 4.58 ( + ) Mb,mid 3.36 ( - ) Ma,discont 3.03 ( - ) Mb,discont 1.12 *d = 74 - 12 =
21.17 2.62 55.47 > 21.58682 kN OK
Ru 2.71 1.84 0.93 0.97 0.61 0.23
Req'd ρ 0.01 0.01 0 0 0 0 62
Use ρ 0.01 0.01 0 0 0 0
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.1 m o.c top bars
As s 1029.86 109.76 say 100 mm 514.54 219.69 say 200 mm 200 565.2 say 300 mm 200 565.2 say 300 mm 200 565.2 say 300 mm 200 565.2 say 300 mm (long direction steel is placed on top of short direction steel at midspan) 250 13.34 2.71 0.01 1029.86 113.04 109.76
Loads transmitted to ing beams 2G-3/3,3G-3/3,RG-3/3,2B-2/3,3B-2/3,RB-2/3: DL = 7.34 LL = 7.11 2G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1: DL = 7.81 LL = 7.56
Slab Designation: 2S4/3S4/RS4 Check if one-way or two-way slab Check span in short direction, La = 3.1 Clear span in long direction, Lb = 5.25 Ratio, m = 0.59 > 0.50 (two-way) Minimum slab thickness Minimum t = 96.67 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Design moments case 9 m= 0.6 (-) Ma = 12.92
(+) Ma = (-) Mb = (+) Mb = At discontinuous edge: ( - ) At discontinuous edge: ( + )
7.03 2.5 2.34 Mb = Mb =
0.78 0.78
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form =
56.44 74 > 53.76681 mm OK
Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid ( - ) Ma,discont
22.01 1.17 55.47 > 19.87212 kN OK
Ru 12.92 7.03 2.5 2.34 0.78
*d = 74 - 12 =
2.62 1.43 0.51 0.68 0.16
Req'd ρ 0.01 0.01 0 0 0
Use ρ 0.01 0.01 0 0 0
62
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.1 m o.c. top bars
As 995.28 394.6 200 200 200
s 113.58 say 100 mm 286.47 say 250 mm 565.2 say 300 mm 565.2 say 300 mm 565.2 say 300 mm
(long direction steel is placed on top of short direction steel at midspan) 250 12.92 2.62 0.01 995.28 113.04 113.58
Loads transmitted to ing beams 2G-2/3,3B-2/3,3G-2/3,3B-2/3,RG-2/3,RB-2/3: DL = 7.23 LL = 6.99 2G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1: DL = 7.81 LL = 7.56
Slab Designation: 2S5 Check if one-way or two-way slab Check span in short direction, La = 2.45 Clear span in long direction, Lb = 2.5 Ratio, m = 0.98 > 0.50 (two-way) Minimum slab thickness Minimum t = 49.44 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96
Live load Production area =
4.8
Factored loads Factored deal load = Factored live load = Total factored load =
6.94 8.16 15.1
case 9 m= (-) Ma = (+) Ma = (-) Mb = (+) Mb = At discontinuous edge: ( - )
0.98 5.68 2.48 2.96 2.18
Design moments
Mb =
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid ( - ) Ma,discont
Ru 5.68 2.48 2.96 2.18 0.73
37.41 74 > 30.98426 mm OK 12.69 5.81 55.47 > 11.45173 kN OK
Req'd ρ 1.15 0.5 0.6 0.63 0.15
*d = 74 - 12 =
0.73
0 0 0 0 0 62
Use ρ 0.01 0 0 0 0
As 422.15 250 200 200 200
s 267.77 say 300 mm 452.16 say 300 mm 565.2 say 300 mm 565.2 say 300 mm 565.2 say 300 mm
(long direction steel is placed on top of short direction steel at midspan) **temperature controls, Ast = 0.002bt = 250 Sample computations: For Mu = 5.68 Moment resistance coefficient, Ru = 1.15 Steel Ratio, ρ = 0 A = Steel area, 422.15 s Ab = Area of 12 mm f bars: 113.04 Required spacing, s = 267.77 Therefore, use 12 mm bars @ 0.3 m o.c. bottom bars Loads transmitted to ing beams 2G-2/3,2B-3: DL = LL = 2G-7/2, 2B-4: DL = LL =
Slab Designation: Check if one-way or two-way slab
2S6/3S6
5.16 5 1.05 1.02
Clear span in short direction, La = Clear span in long direction, Lb =
2.45 6.55
Ratio, m = 0.37 < 0.50 (one-way) Minimum slab thickness Minimum t = 89.36 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Check limitations (NS Section 408.4.3) a) LL/DL = 0.97 < 3 b) (L-S)/S < 0.20 c) Load is uniformly distributed Compute moments (NS Section 408.4.3) For span ledd than 3 m: At s, - M1 = 7.56 At midspan, + M2 = 6.48 Check effective depth to control deflection based on maximum factored moment = Required effective depth, d req'd = 43.16 Actual effective depth, d form = 74 > 47.88771 mm OK Check effective depth for shear Shear at s, R = 18.5 Shear capacity, Vc = 55.47 > 20.66823 kN OK Compute steel reinforcements Mu Ru As Req'd ρ Use ρ Section s 1 7.56 1.53 0.01 0.01 425.19 265.86 say 200 mm 2 6.48 1.31 0 0 362.61 311.74 say 200 mm
Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.2 m o.c. top bars Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = Ab = Using 10 mm f bars: 78.5 Req'd spacing: s = 392.5 Max spacing: s = 625 Therefore, use 10 mm bars @ .25 mm o.c temp bars Loads transmitted to ing beams 2B-4, 3B-4: DL = 6.08 LL = 5.88 2G-6/2, 3G-6/2: DL = 6.99
7.56 1.53 0.01 425.19 113.04 265.86
200
7.56
LL =
6.76
Slab designation: CS1 Check if one-way or two-way slab Clear span in short direction, La = 1 Slab is one way since it is catilever slab Minimum slab thickness min t = 69.5 try 100 mm Dead load Slab weight = 2.36 Live load Canopy = 1.9 Factored loads Factored dead load = 3.3 Factored live load = 3.23 Total factored load = 6.53 Compute moments At : (-) M1 = 3.27 Check effective depth for flexure Required effective depth, d = 28.38 Actual effective depth, d = 74 > 29 OK Check effective depth for shear At : V1 = 6.53 Shear capacity: φVc = 55.47 > 6.534 OK Flexural steel at Bending coefficient, Ru = 0.66 Steel ratio, ρ = < ρ min 0 (4/3)r = 0 Steel area, As = 240.37 Area of 12-mm bars: Ab = 113.04 Spacing of 12-mm f bars, s = 470.27 Maximum spacing, s = 300 Therefore, use 12 mm bars @ .3 m o.c top bars Spacing of 10-mm f temperature bars For fy = 276 MPa, Ast = 0.002bt = 200 A = Using 10 mm f bars: 78.5 b Req'd spacing: s = 392.5 Max spacing: s = 500 Therefore, use 10 mm bars @ .25 mm o.c temp bars Loads transmitted to ing beams 2G-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 = 2.36 LL = 1.9
Slab Designation: RS6 Check if one-way or two-way slab Clear span in short direction, La = Clear span in long direction, Lb = Ratio, m = Minimum slab thickness Minimum t = Dead load Slab weight =
2.45 3.1 0.79 > 0.50 (two-way) 61.67 try 2.95
100 mm
Cement finish = Ceiling suspended loads = Total dead load =
1.53 0.48 4.96
Live load Same as lower floor =
4.8
Factored deal load = Factored live load = Total factored load =
6.94 8.16 15.1
Factored loads
Design moments case 2 m= (-) Ma = (+) Ma = (-) Mb = (+) Mb = Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid
Ru 5.89 3.83 3.92 2.54
0.79 use m = 0.90 5.89 3.83 3.92 2.54 38.12 74 > 38.83294 mm OK 13.14 7.4 18.93 > 12.35558 kN OK
Req'd ρ 1.2 0.78 0.8 0.73
*d = 74 - 12 =
0 0 0 0
Use ρ 0.01 0 0 0
62
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = 0 Steel area, As = 438.77 Area of 12 mm f bars: Ab = 113.04 Required spacing, s = 257.63 Maximum spacing, s = 300 Therefore, use 12 mm bars @ 0.2 m o.c top bars Loads transmitted to ing beams RG-6/2, RB-4: DL = 4.01 LL = 3.88 RB-1/3, RG-2/3: DL = 2.61 LL = 2.53
Slab Designation: US1 Check if one-way or two-way slab Clear span in short direction, La =
As 438.77 282.25 289.19 266.39
s 257.63 say 200 mm 400.5 say 400 mm 390.89 say 300 mm 424.34 say 400 mm
(long direction steel is placed on top of short direction steel at midspan) 250 5.89 1.2
3.15
Clear span in long direction, Lb = Ratio, m = Minimum slab thickness Minimum t = Dead load Slab weight = Ceiling suspended loads = Total dead load =
6.55 0.48 < 0.50 (one-way) 91.13 try
100 mm
2.36 0.48 2.84
Live load Upper deck =
1.9
Factored loads Factored deal load = Factored live load = Total factored load = Compute moments (NS Section 408.4.3) For span less than 3 m: At s, - M1 = At midspan, + M2 =
3.98 3.23 7.21
5.96 5.11
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear Shear at s, R = Shear capacity, Vc = Compute steel reinforcements Mu Ru Req'd ρ Section 1 5.96 1.21 0 2 5.11 1.04 0
38.33 74 > 33.85009 mm OK 11.35 55.47 > 10.13925 OK Use ρ 0.01 0.01
Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = 0 Steel area, As = 443.78 Ab = Area of 12 mm f bars: Required spacing, s = 254.72 Therefore, use 12 mm bars @ 0.3 m o.c. top bars Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = A+D31 = Using 10 mm f bars: 78.5 Req'd spacing: s = 392.5 Max spacing: s = 500 Therefore, use 10 mm bars @ .3 mm o.c temp bars Loads transmitted to ing beams UB-4, UB-5: DL = 4.47 LL = 2.99
As 443.78 378.9 5.96 1.21
113.04
200
s 254.72 say 300 mm 298.34 say 300 mm
Section
Mu 1 2 3
Ru 5.84 10.02 14.02
Req'r p 1.19 2.03 2.85
Use p 0 0.01 0.01
As 0.01 0.01 0.01
434.9 570.37 814.86
s 259.92 say 250 198.19 say150 138.72 say 100
4 5 m= fy = fc' =
8.76 12.75 11.6 276 28
1.78 2.59
0.01 0.01 Ab = s req'd = s max =
0.01 0.01 113.04 259.92 300
496.05 735.92
227.88 say 200 153.6 say 150
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = 11.6 Strength ratio, m = fy / (0.18)(fc’ ) = Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
0.9 0.88 0.85
-0.37
0.4 0.34
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = 11.6 Strength ratio, m = fy / (0.18)(fc’ ) = Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
Slab Designation: 2S1-2S2/3S1-3S2/RS1-RS2 Check if one-way or two-way slab La = Short span 2S1/2S4: 3.15 L = Short span 2S2/2S3: 3.13 a Lb = Long span 2S1-2S4: 7.25 Ratio, m = La/Lb = 0.43 < 0.50 therefore, one way Minimum slab thickness ( NS Table 409-1 for one-end continuous) Minimum t = 97.63 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Check limitations (NS Section 408.4.3) a) LL/DL = 0.97 < 3 b) (L-S)/S = 0.01 < 0.2 c) Load is uniformly distributed Design moments (NS Section 408.4.3) -M1 = 6.24 +M2 = 10.7 -M3 = 14.87 +M4 = 9.22 -M5 = 13.41 Check effective depth to control deflection based on maximum factored moment = Required effective depth, d req'd = 60.55 Actual effective depth, d form = 74 > 57.80134 mm OK Check effective depth for shear Shear at s, R = 27.36 Shear capacity, Vc = 55.47 > 26.40941 kN OK Compute steel reinforcements Mu Ru As Req'd ρ Use ρ Section s 1 6.24 1.27 0 0.01 465.71 242.73 say 200 mm 2 10.7 2.17 0.01 0.01 611.7 184.8 say 150 mm 3 14.87 3.02 0.01 0.01 867.9 130.24 say 100 mm 4 9.22 1.87 0.01 0.01 522.96 216.16 say 200 mm 5 13.41 2.72 0.01 0.01 776.78 145.52 say 100 mm
14.87
Sample computations: For Mu = 6.24 Moment resistance coefficient, Ru = 1.27 Steel Ratio, ρ = 0 As = Steel area, 465.71 Ab = Area of 12 mm f bars: 113.04 Required spacing, s = 242.73 Therefore, use 12 mm bars @ 0.2 m top bars @ discontinuous edge Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = 250 A = Using 10 mm f bars: 78.5 b Req'd spacing: s = 314 Max spacing: s = 625 Therefore, use 10 mm bars @ .25m o.c. temperature bars Loads transmitted to ing beams by 2S1 2G-1,3/1-2, 3G-1,3/1-2, RG-1,3/1-2: DL = 7.81 LL = 7.56 2B-1,2/1-2, 3B-1,1/1-2, RB-1,2/1-2: DL = 8.98 LL = 8.69 Loads transmitted to ing beams by 2S2 2B-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 = 7.81 LL = 7.56
Slab Designation: 2S3 Check if one-way or two-way slab Check span in short direction, La = Clear span in long direction, Lb = Ratio, m = 0.6 > 0.50 (two-way) Minimum slab thickness Minimum t = 97.22 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Design moments case 4 m= 0.6 (-) Ma = 13.34 (+) Ma = 9.08 (-) Mb = 4.58 (+) Mb = 3.36 At discontinuous edge: ( - ) Ma = At discontinuous edge: ( + ) Mb = Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form =
3.15 5.25
3.03 1.12 57.35 74 > 56.5114 mm OK
Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc =
Compute steel reinforcements Mu Section ( - ) Ma,cont 13.34 ( + ) Ma,mid 9.08 ( - ) Mb,cont 4.58 ( + ) Mb,mid 3.36 ( - ) Ma,discont 3.03 ( - ) Mb,discont 1.12 *d = 74 - 12 =
21.17 2.62 55.47 > 21.58682 kN OK
Ru 2.71 1.84 0.93 0.97 0.61 0.23
Req'd ρ 0.01 0.01 0 0 0 0 62
Use ρ 0.01 0.01 0 0 0 0
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.1 m o.c top bars
As s 1029.86 109.76 say 100 mm 514.54 219.69 say 200 mm 200 565.2 say 300 mm 200 565.2 say 300 mm 200 565.2 say 300 mm 200 565.2 say 300 mm (long direction steel is placed on top of short direction steel at midspan) 250 13.34 2.71 0.01 1029.86 113.04 109.76
Loads transmitted to ing beams 2G-3/3,3G-3/3,RG-3/3,2B-2/3,3B-2/3,RB-2/3: DL = 7.34 LL = 7.11 2G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1: DL = 7.81 LL = 7.56
Slab Designation: 2S4/3S4/RS4 Check if one-way or two-way slab Check span in short direction, La = 3.1 Clear span in long direction, Lb = 5.25 Ratio, m = 0.59 > 0.50 (two-way) Minimum slab thickness Minimum t = 96.67 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Design moments case 9 m= 0.6 (-) Ma = 12.92
(+) Ma = (-) Mb = (+) Mb = At discontinuous edge: ( - ) At discontinuous edge: ( + )
7.03 2.5 2.34 Mb = Mb =
0.78 0.78
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form =
56.44 74 > 53.76681 mm OK
Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid ( - ) Ma,discont
22.01 1.17 55.47 > 19.87212 kN OK
Ru 12.92 7.03 2.5 2.34 0.78
*d = 74 - 12 =
2.62 1.43 0.51 0.68 0.16
Req'd ρ 0.01 0.01 0 0 0
Use ρ 0.01 0.01 0 0 0
62
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.1 m o.c. top bars
As 995.28 394.6 200 200 200
s 113.58 say 100 mm 286.47 say 250 mm 565.2 say 300 mm 565.2 say 300 mm 565.2 say 300 mm
(long direction steel is placed on top of short direction steel at midspan) 250 12.92 2.62 0.01 995.28 113.04 113.58
Loads transmitted to ing beams 2G-2/3,3B-2/3,3G-2/3,3B-2/3,RG-2/3,RB-2/3: DL = 7.23 LL = 6.99 2G-6/1,3G-6/1,RG-6/1,2G-7/1,3G-7/1,RG-7/1: DL = 7.81 LL = 7.56
Slab Designation: 2S5 Check if one-way or two-way slab Check span in short direction, La = 2.45 Clear span in long direction, Lb = 2.5 Ratio, m = 0.98 > 0.50 (two-way) Minimum slab thickness Minimum t = 49.44 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96
Live load Production area =
4.8
Factored loads Factored deal load = Factored live load = Total factored load =
6.94 8.16 15.1
case 9 m= (-) Ma = (+) Ma = (-) Mb = (+) Mb = At discontinuous edge: ( - )
0.98 5.68 2.48 2.96 2.18
Design moments
Mb =
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid ( - ) Ma,discont
Ru 5.68 2.48 2.96 2.18 0.73
37.41 74 > 30.98426 mm OK 12.69 5.81 55.47 > 11.45173 kN OK
Req'd ρ 1.15 0.5 0.6 0.63 0.15
*d = 74 - 12 =
0.73
0 0 0 0 0 62
Use ρ 0.01 0 0 0 0
As 422.15 250 200 200 200
s 267.77 say 300 mm 452.16 say 300 mm 565.2 say 300 mm 565.2 say 300 mm 565.2 say 300 mm
(long direction steel is placed on top of short direction steel at midspan) **temperature controls, Ast = 0.002bt = 250 Sample computations: For Mu = 5.68 Moment resistance coefficient, Ru = 1.15 Steel Ratio, ρ = 0 A = Steel area, 422.15 s Ab = Area of 12 mm f bars: 113.04 Required spacing, s = 267.77 Therefore, use 12 mm bars @ 0.3 m o.c. bottom bars Loads transmitted to ing beams 2G-2/3,2B-3: DL = LL = 2G-7/2, 2B-4: DL = LL =
Slab Designation: Check if one-way or two-way slab
2S6/3S6
5.16 5 1.05 1.02
Clear span in short direction, La = Clear span in long direction, Lb =
2.45 6.55
Ratio, m = 0.37 < 0.50 (one-way) Minimum slab thickness Minimum t = 89.36 try 100 mm Dead load Slab weight = 2.95 Cement finish = 1.53 Ceiling suspended loads = 0.48 Total dead load = 4.96 Live load Production area = 4.8 Factored loads Factored deal load = 6.94 Factored live load = 8.16 Total factored load = 15.1 Check limitations (NS Section 408.4.3) a) LL/DL = 0.97 < 3 b) (L-S)/S < 0.20 c) Load is uniformly distributed Compute moments (NS Section 408.4.3) For span ledd than 3 m: At s, - M1 = 7.56 At midspan, + M2 = 6.48 Check effective depth to control deflection based on maximum factored moment = Required effective depth, d req'd = 43.16 Actual effective depth, d form = 74 > 47.88771 mm OK Check effective depth for shear Shear at s, R = 18.5 Shear capacity, Vc = 55.47 > 20.66823 kN OK Compute steel reinforcements Mu Ru As Req'd ρ Use ρ Section s 1 7.56 1.53 0.01 0.01 425.19 265.86 say 200 mm 2 6.48 1.31 0 0 362.61 311.74 say 200 mm
Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = As = Steel area, Ab = Area of 12 mm f bars: Required spacing, s = Therefore, use 12 mm bars @ 0.2 m o.c. top bars Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = Ab = Using 10 mm f bars: 78.5 Req'd spacing: s = 392.5 Max spacing: s = 625 Therefore, use 10 mm bars @ .25 mm o.c temp bars Loads transmitted to ing beams 2B-4, 3B-4: DL = 6.08 LL = 5.88 2G-6/2, 3G-6/2: DL = 6.99
7.56 1.53 0.01 425.19 113.04 265.86
200
7.56
LL =
6.76
Slab designation: CS1 Check if one-way or two-way slab Clear span in short direction, La = 1 Slab is one way since it is catilever slab Minimum slab thickness min t = 69.5 try 100 mm Dead load Slab weight = 2.36 Live load Canopy = 1.9 Factored loads Factored dead load = 3.3 Factored live load = 3.23 Total factored load = 6.53 Compute moments At : (-) M1 = 3.27 Check effective depth for flexure Required effective depth, d = 28.38 Actual effective depth, d = 74 > 29 OK Check effective depth for shear At : V1 = 6.53 Shear capacity: φVc = 55.47 > 6.534 OK Flexural steel at Bending coefficient, Ru = 0.66 Steel ratio, ρ = < ρ min 0 (4/3)r = 0 Steel area, As = 240.37 Area of 12-mm bars: Ab = 113.04 Spacing of 12-mm f bars, s = 470.27 Maximum spacing, s = 300 Therefore, use 12 mm bars @ .3 m o.c top bars Spacing of 10-mm f temperature bars For fy = 276 MPa, Ast = 0.002bt = 200 A = Using 10 mm f bars: 78.5 b Req'd spacing: s = 392.5 Max spacing: s = 500 Therefore, use 10 mm bars @ .25 mm o.c temp bars Loads transmitted to ing beams 2G-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 = 2.36 LL = 1.9
Slab Designation: RS6 Check if one-way or two-way slab Clear span in short direction, La = Clear span in long direction, Lb = Ratio, m = Minimum slab thickness Minimum t = Dead load Slab weight =
2.45 3.1 0.79 > 0.50 (two-way) 61.67 try 2.95
100 mm
Cement finish = Ceiling suspended loads = Total dead load =
1.53 0.48 4.96
Live load Same as lower floor =
4.8
Factored deal load = Factored live load = Total factored load =
6.94 8.16 15.1
Factored loads
Design moments case 2 m= (-) Ma = (+) Ma = (-) Mb = (+) Mb = Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear In short direction: Va = In long direction: Vb = Shear capacity, Vc = Compute steel reinforcements Mu Section ( - ) Ma,cont ( + ) Ma,mid ( - ) Mb,cont ( + ) Mb,mid
Ru 5.89 3.83 3.92 2.54
0.79 use m = 0.90 5.89 3.83 3.92 2.54 38.12 74 > 38.83294 mm OK 13.14 7.4 18.93 > 12.35558 kN OK
Req'd ρ 1.2 0.78 0.8 0.73
*d = 74 - 12 =
0 0 0 0
Use ρ 0.01 0 0 0
62
**temperature controls, Ast = 0.002bt = Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = 0 Steel area, As = 438.77 Area of 12 mm f bars: Ab = 113.04 Required spacing, s = 257.63 Maximum spacing, s = 300 Therefore, use 12 mm bars @ 0.2 m o.c top bars Loads transmitted to ing beams RG-6/2, RB-4: DL = 4.01 LL = 3.88 RB-1/3, RG-2/3: DL = 2.61 LL = 2.53
Slab Designation: US1 Check if one-way or two-way slab Clear span in short direction, La =
As 438.77 282.25 289.19 266.39
s 257.63 say 200 mm 400.5 say 400 mm 390.89 say 300 mm 424.34 say 400 mm
(long direction steel is placed on top of short direction steel at midspan) 250 5.89 1.2
3.15
Clear span in long direction, Lb = Ratio, m = Minimum slab thickness Minimum t = Dead load Slab weight = Ceiling suspended loads = Total dead load =
6.55 0.48 < 0.50 (one-way) 91.13 try
100 mm
2.36 0.48 2.84
Live load Upper deck =
1.9
Factored loads Factored deal load = Factored live load = Total factored load = Compute moments (NS Section 408.4.3) For span less than 3 m: At s, - M1 = At midspan, + M2 =
3.98 3.23 7.21
5.96 5.11
Check effective depth for flexure Required effective depth, d req'd = Actual effective depth, d form = Check effective depth for shear Shear at s, R = Shear capacity, Vc = Compute steel reinforcements Mu Ru Req'd ρ Section 1 5.96 1.21 0 2 5.11 1.04 0
38.33 74 > 33.85009 mm OK 11.35 55.47 > 10.13925 OK Use ρ 0.01 0.01
Sample computations: For Mu = Moment resistance coefficient, Ru = Steel Ratio, ρ = 0 Steel area, As = 443.78 Ab = Area of 12 mm f bars: Required spacing, s = 254.72 Therefore, use 12 mm bars @ 0.3 m o.c. top bars Compute temperature bars For fy = 276 MPa, Ast = 0.002bt = A+D31 = Using 10 mm f bars: 78.5 Req'd spacing: s = 392.5 Max spacing: s = 500 Therefore, use 10 mm bars @ .3 mm o.c temp bars Loads transmitted to ing beams UB-4, UB-5: DL = 4.47 LL = 2.99
As 443.78 378.9 5.96 1.21
113.04
200
s 254.72 say 300 mm 298.34 say 300 mm
Section
Mu 1 2 3
Ru 5.84 10.02 14.02
Req'r p 1.19 2.03 2.85
Use p 0 0.01 0.01
As 0.01 0.01 0.01
434.9 570.37 814.86
s 259.92 say 250 198.19 say150 138.72 say 100
4 5 m= fy = fc' =
8.76 12.75 11.6 276 28
1.78 2.59
0.01 0.01 Ab = s req'd = s max =
0.01 0.01 113.04 259.92 300
496.05 735.92
227.88 say 200 153.6 say 150
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = 11.6 Strength ratio, m = fy / (0.18)(fc’ ) = Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
0.9 0.88 0.85
-0.37
0.4 0.34
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = 11.6 Strength ratio, m = fy / (0.18)(fc’ ) = Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) p min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
Common design parameters: 0.02 To control deflection, r ≤ 0.18fc’/ fy = Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
Common design parameters: To control deflection, r ≤ 0.18fc’/ fy = 0.02 Strength ratio, m = fy / (0.18)(fc’ ) = 11.6 Required bending coefficient, Ru = 4.51 Minimum flexural reinforcement ratio (NS Section 410.6.1) r min = 0 > 1.4/ fy = 0.01
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