131022 Beams, Columns And Footings
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DESIGN OF ROOF BEAM (RB-1) Assume Dimension
200 mm x 300 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 7.46 kn 1.68 kn/m Pu(from right) = 3.40 kn 0.58 kn/m
Wt. of Beam Wt. of Truss Total Live Load Wu = =
2.26 kn/m 3.23 kn/m 1.4DL + 1.7LL = 8.66 kn/m
Design Moment Mu= 14.33 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = = M1
0.90(20.7) 89.93 kn-m
( 200 mm ) ( 300 mm )^2 ( 0.334 )
>
(1-0.59*0.334)
Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 14.33*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 300 mm )^2 (1-0.59*w) w = 0.032 P = wfc'/Fy = 0.032 (20.7)/275 = 0.0024 Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 300 mm ) = 306.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.52 Bars Say 2 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.90 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF ROOF BEAM (RB-2) Assume Dimension
150 mm x 300 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of floor Wt. of Stair Total Live Load Wu = =
1.26 0.11 0.60 1.97
kn/m kn/m kn/m kn/m
3.00 kn/m 1.4DL + 1.7LL = 7.86 kn/m
1.4(1.97) + 1.7(3.00)
Design Moment Mu= WL2/8
= 3.55 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 150 mm ) ( 300 mm )^2 ( 0.334 ) (1-0.59*0.334) = 67.45 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 3.55*10^6 = 0.90(20.7)(w) ( 150 mm ) ( 300 mm )^2 (1-0.59*w) w = 0.01 P = wfc'/Fy = 0.01 (20.7)/275 = 0.0008 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 150 mm ) ( 300 mm ) = 229.50 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.14 Bars Say 2 Bars THEREFORE USE 150 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.90 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-1) Assume Dimension
200 mm x 400 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 33.64 kn 1.92 kn/m 3.96 kn/m
Wt. of Beam Wt. of Slab Total Live Load Wu = =
1.4DL + 1.7LL 13.57 kn/m
5.88 kn/m 3.14 kn/m =
Design Moment Mu= 34.67 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 400 mm )^2 ( 0.334 ) (1-0.59*0.334) = 159.88 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 34.67*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 400 mm )^2 (1-0.59*w) w = 0.06 P = wfc'/Fy = 0.06 (20.7)/275 = 0.0045 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 400 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 3 Bars THEREFORE USE 200 mm X 400 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
3.33 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-2) Assume Dimension
200 mm x 400 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Wall Wt. of Slab Total Live Load Wu = =
Point Load Pu(from left) = 16.35 kn 1.92 kn/m 8.25 kn/m 3.36 kn/m 13.53 kn/m
1.4DL + 1.7LL 23.46 kn/m
2.66 kn/m =
1.4(13.53) + 1.7(2.66)
Design Moment Mu= 31.31 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 400 mm )^2 ( 0.334 ) (1-0.59*0.334) = 159.88 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 31.31*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 400 mm )^2 (1-0.59*w) w = 0.054 P = wfc'/Fy = 0.054 (20.7)/275 = 0.0041 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 400 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 4 Bars THEREFORE USE 200 mm X 400 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
2.70 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-3) Assume Dimension
200 mm x 350 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Slab
1.68 kn/m 3.72 kn/m
Total Live Load Wu = =
5.40 kn/m
2.95 kn/m 1.4DL + 1.7LL = 12.58 kn/m
Design Moment Mu= WL2/8
= 10.63 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 122.41 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 10.63*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.024 P = wfc'/Fy = 0.024 (20.7)/275 = 0.0018 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 350 mm ) = 357.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.78 Bars Say 2 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 3-16mm Bars at Top & 2-16mm at Bot.
2.60 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-4) Assume Dimension
150 mm x 350 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Slab
1.26 kn/m 1.15 kn/m
Total Live Load Wu = =
2.41 kn/m 0.91 kn/m 1.4DL + 1.7LL = 4.92 kn/m
1.4(2.41) + 1.7(0.91)
Design Moment Mu= WL2/8
= 1.38 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 150 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 91.81 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 1.38*10^6 = 0.90(20.7)(w) ( 150 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.004 P = wfc'/Fy = 0.004 (20.7)/275 = 0.0003 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 150 mm ) ( 350 mm ) = 267.75 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.33 Bars Say 2 Bars THEREFORE USE 150 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.50 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF CONTINUOUS BEAM (CB-1) Assume Dimension
200 mm x 350 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 8.67 kn 1.92 kn/m
Wt. of Beam
Total Live Load Wu = =
1.92 kn/m
1.4DL + 1.7LL 2.69 kn/m
0.00 kn/m =
Design Moment Mu= 10.00 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 122.41 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 10.00*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.017 P = wfc'/Fy = 0.017 (20.7)/275 = 0.0013 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 350 mm ) = 357.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.78 Bars Say 4 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
1.00 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF COLUMN Assume Dimension Abd =
(C-1) 400 mm x 400 mm
Span (L)=
###
LOADINGS Dead Load PDEADLOAD = 170.00 kn Live Load PLIVELOAD = 30.00 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 72860.69 mm^2 Abd Ag <
Pu
= 289.00 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 1600.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 7.96
A16 = 201.06 mm^2 Say
8 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 400 mm Cocrete Column Provided With 8-16mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(C-2)
Assume Dimension Abd = 80000 mm^2 LOADINGS
200 mm x 400 mm
Dead Load PDEADLOAD = 63.65 kn Live Load PLIVELOAD = 19.84 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 30969.07 mm^2 Abd Ag <
Span (L)=
Pu
= 122.84 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 800.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 3.98
A16 = 201.06 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 400 mm Cocrete Column Provided With 6-16mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(C-3)
Assume Dimension Abd = 62500 mm^2 LOADINGS
250 mm x 250 mm
Dead Load PDEADLOAD = 12.00 kn Live Load PLIVELOAD = 0.00 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 4235.50 mm^2 Abd Ag <
Span (L)=
Pu
= 16.80 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 400.00 mm^2 N#16
.:Try =
OK!
12.00 mm Bars As/A12 = 3.54
A12 = 113.10 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 192.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 250 mm X 250 mm Cocrete Column Provided With 4-12mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(PC-1)
Assume Dimension Abd = 70000 mm^2 LOADINGS
200 mm x 350 mm
Dead Load PDEADLOAD = 10.84 kn Live Load PDEADLOAD = 1.82 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 4606.11 mm^2 Abd Ag <
Span (L)=
Pu
= 18.27 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 700.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 3.48
A16 = 201.06 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 350 mm Cocrete Column Provided With 4-12mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN FOOTING (F-1) LOADINGS: Max. Column Load : 37.39 kn Weight of Footing : 8.64 kn Af =( 1.2 m^2) Total loads (Pu) : 46.03 kn Allowable Soil Bearing Capacity (Assumed q b) : 150 kpa = Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= Pu/Qe Qu = Pu/Af
= 0.32 m^2 = 38.36 kpa
Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 46.03 kn - 38.36 kpa Vn= Vu/Φbd = 0.039 Mpa Design of Moment Mu= 38.36 kpa (1.20 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
<
Af
Ok!
<
Qe
Ok!
[(0.4 m)^2] = < Vc
39.89 kn Ok!
[(0.4 m)^2/2] = 3.68 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 3.68*10^6 =0.90(20.7)(w) ( 1200 mm ) ( 1200 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1200 mm) ( 225 mm ) = 1377.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 6.85 Bars Say 7 Bars THEREFORE USE 1200 mm^2 X 300 mm Thick Column Footing Provided With 7-16 mm Each Way
(1-0.59*w) 0.000316
1-0.59*w) <
Pmin!
DESIGN OF COLUMN FOOTING (F-2) LOADINGS: Max. Column Load Weight of Footing
: :
37.39 kn 8.64 kn
Qe =
Af =( 1.2 m^2)
Total loads (Pu) : 46.03 kn Allowable Soil Bearing Capacity (Assumed q b)
Af2= :
150 kpa
= Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= 1.02 m^2 SAY 1.10 m^2 Af < = qb Qu = Pu/Af 38.36 kpa < Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 46.03 kn - 38.36 kpa Vn= Vu/Φbd = 0.039 Mpa Design of Moment Mu= 38.36 kpa (1.20 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
[(0.4 m)^2] = < Vc
Ok! Ok!
39.89 kn Ok!
[(0.4 m)^2/2] = 3.68 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 3.68*10^6 =0.90(20.7)(w) ( 1200 mm ) ( 1200 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1200 mm) ( 225 mm ) = 1377.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 6.85 Bars Say 7 Bars THEREFORE USE 1200 mm^2 X 300 mm Thick Column Footing Provided With 7-16 mm Each Way
(1-0.59*w) 0.000316
P Af22
+
1.02 m^2
1-0.59*w) <
Pmin!
6Pe Af23
DESIGN OF COLUMN FOOTING (F-3) LOADINGS: Max. Column Load : 12.00 kn Weight of Footing : 4.80 kn Af =( 0.8 m^2) Total loads (Pu) : 16.80 kn A =( 0.64 m^2) Allowable Soil Bearing Capacity (Assumed q b) : 150 kpa = Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= Pu/Qe Qu = Pu/Af
= 0.12 m^2 = 21.00 kpa
Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 16.80 kn - 21.00 kpa Vn= Vu/Φbd = 0.020 Mpa Design of Moment Mu= 21.00 kpa (0.80 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
<
Af
Ok!
<
Qe
Ok!
[(0.4 m)^2] = < Vc
13.44 kn Ok!
[(0.4 m)^2/2] = 1.34 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 1.34*10^6 =0.90(20.7)(w) ( 800 mm ) ( 800 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 800 mm) ( 175 mm ) = 714.00 mm^2 No. of Bars Needed : Use 12 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 3.55 Bars Say 5 Bars THEREFORE USE 800 mm X 800 mm X 250 mm Thick Column Footing Provided With 5-12 mm Each Way
(1-0.59*w) 0.000316
1-0.59*w) <
Pmin!
DESIGN OF WALL FOOTING (WF-1) LOADINGS: 6" CHB Uniform Load : 1.29 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.69 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip P + Qe = B
:
150 kpa
6Pe B2
B=
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 10.33 kpa B (1 m) Mu = Qu*(B*0.5-0.15)2/2 = 0.052 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.05*10^6 = 0.90(20.7)(w) ( 500 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00014 P = wfc'/Fy = 0.00014 (20.7)/275 = 0.000011 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 500 mm ) ( 200 mm ) = 510.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.54 Bars Say 3 Bars THEREFORE USE 500 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
THEREFORE USE 1000 mm^2 X 300 mm Thick Column Footing Provided With 6-16 mm Each Way
0.05 m
Pmin!
SAY
0.50 m
DESIGN OF WALL FOOTING (WF-2) LOADINGS: 6" CHB Uniform Load : 1.29 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.69 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip 3.69 k/m B= = 0.03 m SAY 145.2 kpa
:
150 kpa
0.50 m
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 10.33 kpa B (1 m) Mu = Qu*(B*0.5-0.15)2/2 = 0.052 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.05*10^6 = 0.90(20.7)(w) ( 500 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00014 P = wfc'/Fy = 0.00014 (20.7)/275 = 0.000011 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 500 mm ) ( 200 mm ) = 510.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.54 Bars Say 3 Bars THEREFORE USE 500 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
Pmin!
DESIGN OF WALL FOOTING (WF-3) LOADINGS: 4" CHB Uniform Load : 0.83 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.23 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip 3.23 k/m B= = 0.02 m SAY 145.2 kpa
:
150 kpa
0.40 m
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 11.29 kpa B (1 m) Mu = Qu*(B*0.4-0.15)2/2 = 0.014 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.01*10^6 = 0.90(20.7)(w) ( 400 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00005 P = wfc'/Fy = 0.00005 (20.7)/275 = 0.000004 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 400 mm ) ( 200 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 3 Bars THEREFORE USE 400 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
THEREFORE USE 1000 mm^2 X 300 mm Thick Column Footing Provided With 6-16 mm Each Way
Pmin!
DESIGN OF SLAB (S1) Min. Thickness L1 = 1.35 m t = (L/20)(0.4+fy/700) LOADINGS:
L2 = 2.80 m 53.52 mm
Dead Load Wt. of Slab Wt. of Floor Finish
2.40 kpa 0.60 kpa
Total Dead Load
3.00 kpa
Live Load Residential
1.90 kpa
Wu = 1.4DL+1.7LL Wu = 7.43 kn-m
Say
100 mm
= 7.43 kpa (per 1 m strip)
Slab Aspect Ratio m = L2/L1 =
2.07 > 0.50 One-way Slab
Design Moment Mu= WL2/8 d = t-20-0.5(Φbar)
= 0.93 kn-m = 75 mm Mu= Φfc'bd^2(w)(1-0.59w)
0.93*10^6 w P Pmin As=pbd= S= Thickness 100 mm S1
= 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 (1-0.59*w) = 0.0089 = wfc'/Fy = 0.0089 (20.7)/275 = 0.0007 = 1.4/Fy = 0.0051
0.0051
( 1000 mm ) ( 75 mm ) = Use 10 mm Ø Bars; Ab = 78.5(1000)/ 382.50 = 205.23 mm At Short Span At Support 10 mm Φ at 200 mm o.c.
at Mid Span 10 mm Φ at 200 mm o.c.
382.5 mm^2 78.54 mm^2 SAY 200 mm
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
<
Pmin!
DESIGN OF SLAB (S2) Min. Thickness
t = 100 mm d = 75 mm L1 = 1.70 m L2 = 2.60 m
LOADINGS: Dead Load Wt. of Slab Wt. of Floor Finish Elect. & Other Fixtures Total Dead Load
2.40 0.60 0.15 3.15
Live Load Residential
1.90 kpa
Wu = Wu = Slab Aspect m=
kpa kpa kpa kpa
1.4DL+1.7LL = 7.64 kpa 7.64 kn-m (per 1 m strip) Ratio L1/L2 = 0.65 > 0.50 Two-way Slab
At Continuous End Ms = 0.069 (7.64 Mb = 0.022 (7.64 At Midspan Ms = 0.076 (7.64 Mb = 0.024 (7.64 Design Moment Mu= Φfc'bd^2(w)(1-0.59w)
kpa) 1.70 m = kpa) 2.60 m =
1.52 kn-m 1.14 kn-m
kpa) 1.70 m = kpa) 2.60 m =
1.68 kn-m 1.24 kn-m
(at Continuous Edge short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 1.52*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.015 P = wfc'/Fy = 0.015 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY Design Moment Mu= Φfc'bd^2(w)(1-0.59w) (at Midspan short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 1.68*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.02 P = wfc'/Fy = 0.02 (20.7)/275 = Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY
Thickness 100 mm S2
At Short Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
long direction) (1-0.59*w) 0.001129
<
Pmin!
<
Pmin!
382.50mm^2 200 mm
(1-0.59*w) 0.001505
382.50mm^2 200 mm
At Discontinuous End At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
DESIGN OF SLAB (S3) Min. Thickness
t = 100 mm d = 75 mm L1 = 2.80 m L2 = 3.80 m
LOADINGS: Dead Load Wt. of Slab Wt. of Floor Finish Elect. & Other Fixtures Total Dead Load
2.40 0.60 0.15 3.15
Live Load Residential
1.90 kpa
Wu = Wu = Slab Aspect m=
kpa kpa kpa kpa
1.4DL+1.7LL = 7.64 kpa 7.64 kn-m (per 1 m strip) Ratio L1/L2 = 0.74 > 0.50 Two-way Slab
At Continuous End Ms = 0.069 (7.64 Mb = 0.022 (7.64 At Midspan Ms = 0.076 (7.64 Mb = 0.024 (7.64 Design Moment Mu= Φfc'bd^2(w)(1-0.59w)
kpa) 2.80 m = kpa) 3.80 m =
4.13 kn-m 2.43 kn-m
kpa) 2.80 m = kpa) 3.80 m =
4.55 kn-m 2.65 kn-m
(at Continuous Edge short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 4.13*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.04 P = wfc'/Fy = 0.04 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY Design Moment Mu= Φfc'bd^2(w)(1-0.59w) (at Midspan short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 4.55*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.045 P = wfc'/Fy = 0.045 (20.7)/275 = Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY
Thickness 100 mm S3
At Short Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 175 mm o.c. 175 mm o.c.
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
long direction) (1-0.59*w) 0.003011
<
Pmin!
<
Pmin!
382.50mm^2 200 mm
(1-0.59*w) 0.003387
382.50mm^2 200 mm
At Discontinuous End At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
200 mm x 300 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 7.46 kn 1.68 kn/m Pu(from right) = 3.40 kn 0.58 kn/m
Wt. of Beam Wt. of Truss Total Live Load Wu = =
2.26 kn/m 3.23 kn/m 1.4DL + 1.7LL = 8.66 kn/m
Design Moment Mu= 14.33 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = = M1
0.90(20.7) 89.93 kn-m
( 200 mm ) ( 300 mm )^2 ( 0.334 )
>
(1-0.59*0.334)
Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 14.33*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 300 mm )^2 (1-0.59*w) w = 0.032 P = wfc'/Fy = 0.032 (20.7)/275 = 0.0024 Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 300 mm ) = 306.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.52 Bars Say 2 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.90 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF ROOF BEAM (RB-2) Assume Dimension
150 mm x 300 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of floor Wt. of Stair Total Live Load Wu = =
1.26 0.11 0.60 1.97
kn/m kn/m kn/m kn/m
3.00 kn/m 1.4DL + 1.7LL = 7.86 kn/m
1.4(1.97) + 1.7(3.00)
Design Moment Mu= WL2/8
= 3.55 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 150 mm ) ( 300 mm )^2 ( 0.334 ) (1-0.59*0.334) = 67.45 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 3.55*10^6 = 0.90(20.7)(w) ( 150 mm ) ( 300 mm )^2 (1-0.59*w) w = 0.01 P = wfc'/Fy = 0.01 (20.7)/275 = 0.0008 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 150 mm ) ( 300 mm ) = 229.50 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.14 Bars Say 2 Bars THEREFORE USE 150 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.90 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-1) Assume Dimension
200 mm x 400 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 33.64 kn 1.92 kn/m 3.96 kn/m
Wt. of Beam Wt. of Slab Total Live Load Wu = =
1.4DL + 1.7LL 13.57 kn/m
5.88 kn/m 3.14 kn/m =
Design Moment Mu= 34.67 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 400 mm )^2 ( 0.334 ) (1-0.59*0.334) = 159.88 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 34.67*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 400 mm )^2 (1-0.59*w) w = 0.06 P = wfc'/Fy = 0.06 (20.7)/275 = 0.0045 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 400 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 3 Bars THEREFORE USE 200 mm X 400 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
3.33 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-2) Assume Dimension
200 mm x 400 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Wall Wt. of Slab Total Live Load Wu = =
Point Load Pu(from left) = 16.35 kn 1.92 kn/m 8.25 kn/m 3.36 kn/m 13.53 kn/m
1.4DL + 1.7LL 23.46 kn/m
2.66 kn/m =
1.4(13.53) + 1.7(2.66)
Design Moment Mu= 31.31 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 400 mm )^2 ( 0.334 ) (1-0.59*0.334) = 159.88 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 31.31*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 400 mm )^2 (1-0.59*w) w = 0.054 P = wfc'/Fy = 0.054 (20.7)/275 = 0.0041 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 400 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 4 Bars THEREFORE USE 200 mm X 400 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
2.70 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-3) Assume Dimension
200 mm x 350 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Slab
1.68 kn/m 3.72 kn/m
Total Live Load Wu = =
5.40 kn/m
2.95 kn/m 1.4DL + 1.7LL = 12.58 kn/m
Design Moment Mu= WL2/8
= 10.63 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 122.41 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 10.63*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.024 P = wfc'/Fy = 0.024 (20.7)/275 = 0.0018 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 350 mm ) = 357.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.78 Bars Say 2 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 3-16mm Bars at Top & 2-16mm at Bot.
2.60 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF FLOOR BEAM (B-4) Assume Dimension
150 mm x 350 mm
Span (L)=
LOADINGS Dead Load Wt. of Beam Wt. of Slab
1.26 kn/m 1.15 kn/m
Total Live Load Wu = =
2.41 kn/m 0.91 kn/m 1.4DL + 1.7LL = 4.92 kn/m
1.4(2.41) + 1.7(0.91)
Design Moment Mu= WL2/8
= 1.38 kn-m
Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 150 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 91.81 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 1.38*10^6 = 0.90(20.7)(w) ( 150 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.004 P = wfc'/Fy = 0.004 (20.7)/275 = 0.0003 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 150 mm ) ( 350 mm ) = 267.75 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.33 Bars Say 2 Bars THEREFORE USE 150 mm X 350 mm Conc. Beam Provided w/ 2- 16mm Bars at Top & 2-16mm at Bot.
1.50 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF CONTINUOUS BEAM (CB-1) Assume Dimension
200 mm x 350 mm
Span (L)=
LOADINGS Dead Load
Point Load Pu(from left) = 8.67 kn 1.92 kn/m
Wt. of Beam
Total Live Load Wu = =
1.92 kn/m
1.4DL + 1.7LL 2.69 kn/m
0.00 kn/m =
Design Moment Mu= 10.00 kn-m Pmax = 0.75Pb;
Pb = (0.85fc'B600)/[fy(600+275)] Pb = 0.85(20.7)(0.85)(600)/[275(600+275)] = 0.037 Pmax = 0.75Pb = 0.028 W= pfy/fc' = 0.025(275)/20.7 = 0.334 Moment Capacity of Concrete M1= Φfc'bd^2(w)(1-0.59w) = 0.90(20.7) ( 200 mm ) ( 350 mm )^2 ( 0.334 ) (1-0.59*0.334) = 122.41 kn-m M1 > Mu Reinf. Not Required, Use Min. Reinf. Mu= Φfc'bd^2(w)(1-0.59w) 10.00*10^6 = 0.90(20.7)(w) ( 200 mm ) ( 350 mm )^2 (1-0.59*w) w = 0.017 P = wfc'/Fy = 0.017 (20.7)/275 = 0.0013 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 200 mm ) ( 350 mm ) = 357.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 1.78 Bars Say 4 Bars THEREFORE USE 200 mm X 350 mm Conc. Beam Provided w/ 4- 16mm Bars at Top & 2-16mm at Bot.
1.00 m
1-0.59*0.334)
1-0.59*w) <
Pmin!
DESIGN OF COLUMN Assume Dimension Abd =
(C-1) 400 mm x 400 mm
Span (L)=
###
LOADINGS Dead Load PDEADLOAD = 170.00 kn Live Load PLIVELOAD = 30.00 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 72860.69 mm^2 Abd Ag <
Pu
= 289.00 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 1600.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 7.96
A16 = 201.06 mm^2 Say
8 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 400 mm Cocrete Column Provided With 8-16mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(C-2)
Assume Dimension Abd = 80000 mm^2 LOADINGS
200 mm x 400 mm
Dead Load PDEADLOAD = 63.65 kn Live Load PLIVELOAD = 19.84 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 30969.07 mm^2 Abd Ag <
Span (L)=
Pu
= 122.84 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 800.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 3.98
A16 = 201.06 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 400 mm Cocrete Column Provided With 6-16mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(C-3)
Assume Dimension Abd = 62500 mm^2 LOADINGS
250 mm x 250 mm
Dead Load PDEADLOAD = 12.00 kn Live Load PLIVELOAD = 0.00 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 4235.50 mm^2 Abd Ag <
Span (L)=
Pu
= 16.80 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 400.00 mm^2 N#16
.:Try =
OK!
12.00 mm Bars As/A12 = 3.54
A12 = 113.10 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 192.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 250 mm X 250 mm Cocrete Column Provided With 4-12mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN
(PC-1)
Assume Dimension Abd = 70000 mm^2 LOADINGS
200 mm x 350 mm
Dead Load PDEADLOAD = 10.84 kn Live Load PDEADLOAD = 1.82 kn PU = 1.4DL + 1.7LL Ag = Pu/{Φ0.8[0.85Fc'(1-Pg)+FyPg]} Ag = 4606.11 mm^2 Abd Ag <
Span (L)=
Pu
= 18.27 kn
Fc' Fy
= 20.70 Mpa = 275.00 Mpa
Φ (Tied Col)
= 0.70
Pg
= 0.01
OK!
NSCP PROVISION Abd 60000 < REINFORCEMENT As = PgAg = 700.00 mm^2 N#16
.:Try =
OK!
16.00 mm Bars As/A16 = 3.48
A16 = 201.06 mm^2 Say
4 Bars
FOR TIES Spacing of Lateral ties (Try 10 mm Ties) S = 16db = 256.00 mm S = 48 dt = 480.00 mm S = Least Dimension
=
250.00 mm
THEREFORE USE 200 mm X 350 mm Cocrete Column Provided With 4-12mm Vert. Bars and 10mm Ties at 250 mm
3.00 m
DESIGN OF COLUMN FOOTING (F-1) LOADINGS: Max. Column Load : 37.39 kn Weight of Footing : 8.64 kn Af =( 1.2 m^2) Total loads (Pu) : 46.03 kn Allowable Soil Bearing Capacity (Assumed q b) : 150 kpa = Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= Pu/Qe Qu = Pu/Af
= 0.32 m^2 = 38.36 kpa
Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 46.03 kn - 38.36 kpa Vn= Vu/Φbd = 0.039 Mpa Design of Moment Mu= 38.36 kpa (1.20 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
<
Af
Ok!
<
Qe
Ok!
[(0.4 m)^2] = < Vc
39.89 kn Ok!
[(0.4 m)^2/2] = 3.68 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 3.68*10^6 =0.90(20.7)(w) ( 1200 mm ) ( 1200 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1200 mm) ( 225 mm ) = 1377.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 6.85 Bars Say 7 Bars THEREFORE USE 1200 mm^2 X 300 mm Thick Column Footing Provided With 7-16 mm Each Way
(1-0.59*w) 0.000316
1-0.59*w) <
Pmin!
DESIGN OF COLUMN FOOTING (F-2) LOADINGS: Max. Column Load Weight of Footing
: :
37.39 kn 8.64 kn
Qe =
Af =( 1.2 m^2)
Total loads (Pu) : 46.03 kn Allowable Soil Bearing Capacity (Assumed q b)
Af2= :
150 kpa
= Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= 1.02 m^2 SAY 1.10 m^2 Af < = qb Qu = Pu/Af 38.36 kpa < Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 46.03 kn - 38.36 kpa Vn= Vu/Φbd = 0.039 Mpa Design of Moment Mu= 38.36 kpa (1.20 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
[(0.4 m)^2] = < Vc
Ok! Ok!
39.89 kn Ok!
[(0.4 m)^2/2] = 3.68 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 3.68*10^6 =0.90(20.7)(w) ( 1200 mm ) ( 1200 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1200 mm) ( 225 mm ) = 1377.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 6.85 Bars Say 7 Bars THEREFORE USE 1200 mm^2 X 300 mm Thick Column Footing Provided With 7-16 mm Each Way
(1-0.59*w) 0.000316
P Af22
+
1.02 m^2
1-0.59*w) <
Pmin!
6Pe Af23
DESIGN OF COLUMN FOOTING (F-3) LOADINGS: Max. Column Load : 12.00 kn Weight of Footing : 4.80 kn Af =( 0.8 m^2) Total loads (Pu) : 16.80 kn A =( 0.64 m^2) Allowable Soil Bearing Capacity (Assumed q b) : 150 kpa = Qe = 150 - 0.20 (24) = 145.2 kpa (Effective Soil Bearing) Area Required: Af2= Pu/Qe Qu = Pu/Af
= 0.12 m^2 = 21.00 kpa
Footing Thickness Allowable Punching Shear: Vc= √Fc'/3 = 1.52 Mpa Actual Punching Shear: Vu= 16.80 kn - 21.00 kpa Vn= Vu/Φbd = 0.020 Mpa Design of Moment Mu= 21.00 kpa (0.80 m) Moment Capacity of Concrete Mu= Φfc'bd^2(w)(1-0.59w)
<
Af
Ok!
<
Qe
Ok!
[(0.4 m)^2] = < Vc
13.44 kn Ok!
[(0.4 m)^2/2] = 1.34 kn-m
Mu= Φfc'bd^2(w)(1-0.59w) 1.34*10^6 =0.90(20.7)(w) ( 800 mm ) ( 800 mm )^2 w = 0.0042 P = wfc'/Fy = 0.0042 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 800 mm) ( 175 mm ) = 714.00 mm^2 No. of Bars Needed : Use 12 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 3.55 Bars Say 5 Bars THEREFORE USE 800 mm X 800 mm X 250 mm Thick Column Footing Provided With 5-12 mm Each Way
(1-0.59*w) 0.000316
1-0.59*w) <
Pmin!
DESIGN OF WALL FOOTING (WF-1) LOADINGS: 6" CHB Uniform Load : 1.29 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.69 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip P + Qe = B
:
150 kpa
6Pe B2
B=
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 10.33 kpa B (1 m) Mu = Qu*(B*0.5-0.15)2/2 = 0.052 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.05*10^6 = 0.90(20.7)(w) ( 500 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00014 P = wfc'/Fy = 0.00014 (20.7)/275 = 0.000011 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 500 mm ) ( 200 mm ) = 510.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.54 Bars Say 3 Bars THEREFORE USE 500 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
THEREFORE USE 1000 mm^2 X 300 mm Thick Column Footing Provided With 6-16 mm Each Way
0.05 m
Pmin!
SAY
0.50 m
DESIGN OF WALL FOOTING (WF-2) LOADINGS: 6" CHB Uniform Load : 1.29 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.69 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip 3.69 k/m B= = 0.03 m SAY 145.2 kpa
:
150 kpa
0.50 m
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 10.33 kpa B (1 m) Mu = Qu*(B*0.5-0.15)2/2 = 0.052 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.05*10^6 = 0.90(20.7)(w) ( 500 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00014 P = wfc'/Fy = 0.00014 (20.7)/275 = 0.000011 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 500 mm ) ( 200 mm ) = 510.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.54 Bars Say 3 Bars THEREFORE USE 500 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
Pmin!
DESIGN OF WALL FOOTING (WF-3) LOADINGS: 4" CHB Uniform Load : 0.83 kn/m Weight of Footing : 2.40 kn/m Total loads (Wu) : 3.23 kn/m Allowable Soil Bearing Capacity (Assumed q b) Effective Soil Bearing: = Qe = 150 - 0.20 (24) = 145.2 kpa Width of Footing Considering 1 m Strip 3.23 k/m B= = 0.02 m SAY 145.2 kpa
:
150 kpa
0.40 m
Ultimate Bearing Pressure 1.4DL + 1.7LL Qu = = 11.29 kpa B (1 m) Mu = Qu*(B*0.4-0.15)2/2 = 0.014 kn-m Mu= Φfc'bd^2(w)(1-0.59w) 0.01*10^6 = 0.90(20.7)(w) ( 400 mm ) ( 200 mm )^2 (1-0.59*w) w = 0.00005 P = wfc'/Fy = 0.00005 (20.7)/275 = 0.000004 Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 400 mm ) ( 200 mm ) = 408.00 mm^2 No. of Bars Needed : Use 16 mm Ø Bars; Ab = 201.00 mm^2 NBars= As/Ab = 2.03 Bars Say 3 Bars THEREFORE USE 400 mm WIDE X 200 mm Thick Wall Footing Provided With 3-16 mm Cont Bars w/ 10 mm Ø Tie Bars at 450 mm o.c.
<
THEREFORE USE 1000 mm^2 X 300 mm Thick Column Footing Provided With 6-16 mm Each Way
Pmin!
DESIGN OF SLAB (S1) Min. Thickness L1 = 1.35 m t = (L/20)(0.4+fy/700) LOADINGS:
L2 = 2.80 m 53.52 mm
Dead Load Wt. of Slab Wt. of Floor Finish
2.40 kpa 0.60 kpa
Total Dead Load
3.00 kpa
Live Load Residential
1.90 kpa
Wu = 1.4DL+1.7LL Wu = 7.43 kn-m
Say
100 mm
= 7.43 kpa (per 1 m strip)
Slab Aspect Ratio m = L2/L1 =
2.07 > 0.50 One-way Slab
Design Moment Mu= WL2/8 d = t-20-0.5(Φbar)
= 0.93 kn-m = 75 mm Mu= Φfc'bd^2(w)(1-0.59w)
0.93*10^6 w P Pmin As=pbd= S= Thickness 100 mm S1
= 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 (1-0.59*w) = 0.0089 = wfc'/Fy = 0.0089 (20.7)/275 = 0.0007 = 1.4/Fy = 0.0051
0.0051
( 1000 mm ) ( 75 mm ) = Use 10 mm Ø Bars; Ab = 78.5(1000)/ 382.50 = 205.23 mm At Short Span At Support 10 mm Φ at 200 mm o.c.
at Mid Span 10 mm Φ at 200 mm o.c.
382.5 mm^2 78.54 mm^2 SAY 200 mm
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
<
Pmin!
DESIGN OF SLAB (S2) Min. Thickness
t = 100 mm d = 75 mm L1 = 1.70 m L2 = 2.60 m
LOADINGS: Dead Load Wt. of Slab Wt. of Floor Finish Elect. & Other Fixtures Total Dead Load
2.40 0.60 0.15 3.15
Live Load Residential
1.90 kpa
Wu = Wu = Slab Aspect m=
kpa kpa kpa kpa
1.4DL+1.7LL = 7.64 kpa 7.64 kn-m (per 1 m strip) Ratio L1/L2 = 0.65 > 0.50 Two-way Slab
At Continuous End Ms = 0.069 (7.64 Mb = 0.022 (7.64 At Midspan Ms = 0.076 (7.64 Mb = 0.024 (7.64 Design Moment Mu= Φfc'bd^2(w)(1-0.59w)
kpa) 1.70 m = kpa) 2.60 m =
1.52 kn-m 1.14 kn-m
kpa) 1.70 m = kpa) 2.60 m =
1.68 kn-m 1.24 kn-m
(at Continuous Edge short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 1.52*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.015 P = wfc'/Fy = 0.015 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY Design Moment Mu= Φfc'bd^2(w)(1-0.59w) (at Midspan short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 1.68*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.02 P = wfc'/Fy = 0.02 (20.7)/275 = Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY
Thickness 100 mm S2
At Short Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
long direction) (1-0.59*w) 0.001129
<
Pmin!
<
Pmin!
382.50mm^2 200 mm
(1-0.59*w) 0.001505
382.50mm^2 200 mm
At Discontinuous End At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
DESIGN OF SLAB (S3) Min. Thickness
t = 100 mm d = 75 mm L1 = 2.80 m L2 = 3.80 m
LOADINGS: Dead Load Wt. of Slab Wt. of Floor Finish Elect. & Other Fixtures Total Dead Load
2.40 0.60 0.15 3.15
Live Load Residential
1.90 kpa
Wu = Wu = Slab Aspect m=
kpa kpa kpa kpa
1.4DL+1.7LL = 7.64 kpa 7.64 kn-m (per 1 m strip) Ratio L1/L2 = 0.74 > 0.50 Two-way Slab
At Continuous End Ms = 0.069 (7.64 Mb = 0.022 (7.64 At Midspan Ms = 0.076 (7.64 Mb = 0.024 (7.64 Design Moment Mu= Φfc'bd^2(w)(1-0.59w)
kpa) 2.80 m = kpa) 3.80 m =
4.13 kn-m 2.43 kn-m
kpa) 2.80 m = kpa) 3.80 m =
4.55 kn-m 2.65 kn-m
(at Continuous Edge short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 4.13*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.04 P = wfc'/Fy = 0.04 (20.7)/275 = Pmin = 1.4/Fy = 0.0051 Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY Design Moment Mu= Φfc'bd^2(w)(1-0.59w) (at Midspan short/long direction) Mu= Φfc'bd^2(w)(1-0.59w) 4.55*10^6 = 0.90(20.7)(w) ( 1000 mm ) ( 75 mm )^2 w = 0.045 P = wfc'/Fy = 0.045 (20.7)/275 = Pmin = 1.4/Fy = 0.0051
Reinforcement As=pbd= 0.0051 ( 1000 mm ) ( 75 mm ) = Spacing of Bars using 10 mm Φ As = 78.5 mm^2 S= 78.5(1000)/ 382.50 = 205.23 mm SAY
Thickness 100 mm S3
At Short Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 175 mm o.c. 175 mm o.c.
At Long Span At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
long direction) (1-0.59*w) 0.003011
<
Pmin!
<
Pmin!
382.50mm^2 200 mm
(1-0.59*w) 0.003387
382.50mm^2 200 mm
At Discontinuous End At Support at Mid Span 10 mm Φ at 10 mm Φ at 200 mm o.c. 200 mm o.c.
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