STRUCTURAL

DESIGN ANALYSIS

PROPOSED TWO(2)-STOREY
RESIDENTIAL BUILDING

Prepared by:

JULIE ANNE CLAIRE A. LORILLA

Civil Engineer
PRC No.: 162198 Validity: JULY 6,2021

PTR No.: 2185852 Date JANUARY 4, 2021

Issued:
TIN No.: 739-542-579 Place
Issued:
 DESIGN CRITERIA
I. DESIGN LOADING
1. DEAD LOADS
A. Concrete = 24.00 kN/m³
B. Soil = 18.30 kN/m³
C. Steel = 77.00 kN/m³
D. CHB 100mm thk. = 2.15 kPa
E. CHB 150mm thk. = 2.75 kPa
F. Wall Finish = 0.50 kPa
For each face plastered, 0.24 kPA
G. Floor / Floor Finish = 1.85 kPa
Ceramic/Quarry Tile, 0.45 kPa
Mortar, 0.35 kPa / 12.50 mm thk.
Thickness: 50.00 mm thk.
H. Ceiling = 0.20 kPa
Gypsum board (per mm thickness), 0.008 kPa
Thickness: 12.50 mm thk.
Suspended Steel Channel System, 0.10 kPa

2. LIVE LOADS
A. Residential
1) Basic Floor = 1.90 kPa
Area 2.90
= kPa
2) Exterior
Balconies = 1.90 kPa
3) Deck = 1.90 kPa
4) Storage = 1.00 kPa

II. DESIGN STRESSES

1. CONCRETE
A. For Footing, Columns, Beams, and fc' = 20.70 MPa
Slabs.
Compressive Strength @ 28 DAYS

2. REINFORCING STEEL BAR

A. For Footing, Columns and Beams
For bars 16mmØ and Greater Intermediate Grade fy = 275.00 MPa
Deformed Bar
B. For Slab on Fill, Suspended Slabs, Ties and Stirrups
For bars 12mmØ and Lower Intermediate Grade fy = 230.00 MPa
Deformed Bar
C. For Trusses, Purlins, Bracings and Struts fy = 248.00 MPa

III. DESIGN SOIL BEARING

1. ALLOWABLE SOIL BEARING PRESSURE qa = 96.00 kP
a
NOTE: No Footing Shall Rest on Fill
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F1
Location: B3
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 2.10 m Service LC, Po = 232.032 kN
Width, W= 2.10 m Ftng. Self Wt. Wfn 37.044 kN
=
Depth, d= 0.350 m Total, Po Po t 269.076 kN
=
Eff. Depth, d eff. 0.267 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 369.000 kN
Min. Fdn. Depth D min. = 1.16 m Ftng. Self Wt. Wfu 44.453 kN
=
Column, b= 0.350 m Total, Pu Pu t 413.453 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 2.01 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 2.01 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.555 kN/m² b. Along Width
a. Along Length,
USBP, qu = 93.754A = W x
kN/m² A= Ly
= 2.10 x 0.608 = 2.10 x 0.608
= 1.277 m² = 1.277 m²
Vu = qu A Vu = qu A
= 93.75 x 1.277 = 93.75 x 1.277
= 119.724 kN = 119.724 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 2.10 x 0.267 / 6 = √20.70 x 2.10 x 0.267 / 6
= 425.172 kN = 425.172 kN
x = L/2 - b/2 - d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 2.10/2 - 0.350/2 -
0.267 = 0.75 x 425.17 / 119.72 = 2.10/2 - 0.350/2 - 0.267 = 0.75 x 425.17 / 119.72
= 0.608 m = 2.66 >1 .'. Adequate! = 0.608 m = 2.66 >1 .'. Adequate!
Punching Shear:
x= b+ d A LW - xy Vc = √fc' bo d / 3
=
= 0.350 + 0.267 = (2.10 x 2.10) - (0.617 x 0.617) = √20.70 x 2.468 x 0.267 / 3
= 0.617 m = 4.029 m² = 999.356 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.267 = 93.75 x 4.029 = 0.75 x 999.36 / 377.74
= 0.617 m = 377.735 kN = 1.98 >1 .'. Adequate!
bo = 2x + 2y
= (2 x 0.617) + (2 x 0.617)
= 2.468 m
Reinforcing Steel Bars:
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 93.75 x 1.838 = 93.75 x 1.838
= 172.320 kN-m = 172.320 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 172.32x10^6 / = 172.32x10^6 /
(0.90 x 2.10 x 0.267²) (0.90 x 2.10 x 0.267²)
= 1.2789 MPa = 1.2789 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.2789) / (0.85 x [(2 x 1.2789) / (0.85 x
20.70)]} 20.70)]}
x = L/2 - b/2
= 0.004833 = 0.004833
= 2.10/2 - 0.350/2
use ρ = 0.005091 use ρ 0.005091
= 0.875 m =
A= Wx As = ρ W d y = W/2 - As = ρ L d
= 2.10 x 0.875 h/2
= 1.838 m² = 0.005091 x 2.10 x 0.267 = 2.10/2 - 0.350/2 = 0.005091 x 2.10 x 0.267
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F1
Location: B4
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 2.00 m Service LC, Po = 191.277 kN
Width, W= 2.00 m Ftng. Self Wt. Wfn 33.600 kN
=
Depth, d= 0.350 m Total, Po Po t 224.877 kN
=
Eff. Depth, d eff. 0.267 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 304.000 kN
Min. Fdn. Depth D min. = 1.07 m Ftng. Self Wt. Wfu 40.320 kN
=
Column, b= 0.350 m Total, Pu Pu t 344.320 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.84 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.84 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.555 kN/m² b. Along Width
a. Along Length,
USBP, qu = 86.080A = W x
kN/m² A= Ly
= 2.00 x 0.558 = 2.00 x 0.558
= 1.116 m² = 1.116 m²
Vu = qu A Vu = qu A
= 86.08 x 1.116 = 86.08 x 1.116
= 96.065 kN = 96.065 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 2.00 x 0.267 / 6 = √20.70 x 2.00 x 0.267 / 6
= 404.926 kN = 404.926 kN
x = L/2 - b/2 - d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 2.00/2 - 0.350/2 -
0.267 = 0.75 x 404.93 / 96.07 = 2.00/2 - 0.350/2 - 0.267 = 0.75 x 404.93 / 96.07
= 0.558 m = 3.16 >1 .'. Adequate! = 0.558 m = 3.16 >1 .'. Adequate!
Punching Shear:
x= b+ d A LW - xy Vc = √fc' bo d / 3
=
= 0.350 + 0.267 = (2.00 x 2.00) - (0.617 x 0.617) = √20.70 x 2.468 x 0.267 / 3
= 0.617 m = 3.619 m² = 999.356 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.267 = 86.08 x 3.619 = 0.75 x 999.36 / 311.52
= 0.617 m = 311.524 kN = 2.41 >1 .'. Adequate!
bo = 2x + 2y
= (2 x 0.617) + (2 x 0.617)
= 2.468 m
Reinforcing Steel Bars:
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 86.08 x 1.650 = 86.08 x 1.650
= 142.032 kN-m = 142.032 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 142.03x10^6 / = 142.03x10^6 /
(0.90 x 2.00 x 0.267²) (0.90 x 2.00 x 0.267²)
= 1.1069 MPa = 1.1069 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.1069) / (0.85 x [(2 x 1.1069) / (0.85 x
20.70)]} 20.70)]}
x = L/2 - b/2
= 0.004160 = 0.004160
= 2.00/2 - 0.350/2
use ρ = 0.005091 use ρ 0.005091
= 0.825 m =
A= Wx As = ρ W d y = W/2 - As = ρ L d
= 2.00 x 0.825 h/2
= 1.650 m² = 0.005091 x 2.00 x 0.267 = 2.00/2 - 0.350/2 = 0.005091 x 2.00 x 0.267
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F3
Location: A3
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.80 m Service LC, Po = 157.666 kN
Width, W= 1.80 m Ftng. Self Wt. Wfn 31.104 kN
=
Depth, d= 0.400 m Total, Po Po t 188.770 kN
=
Eff. Depth, d eff. 0.317 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 281.000 kN
Min. Fdn. Depth D min. = 1.10 m Ftng. Self Wt. Wfu 37.325 kN
=
Column, b= 0.250 m Total, Pu Pu t 318.325 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.69 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.69 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.270 kN/m² b. Along Width
a. Along Length,
USBP, qu = 98.248A = W x
kN/m² A= Ly
= 1.80 x 1.233 = 1.80 x 0.408
= 2.219 m² = 0.734 m²
Vu = qu A Vu = qu A
= 98.25 x 2.219 = 98.25 x 0.734
= 218.012 kN = 72.114 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.80 x 0.317 / 6 = √20.70 x 1.80 x 0.317 / 6
= 432.679 kN = 432.679 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.80 - 0.250 -
0.317 = 0.75 x 432.68 / 218.01 = 1.80/2 - 0.350/2 - 0.317 = 0.75 x 432.68 / 72.11
= 1.233 m = 1.49 >1 .'. Adequate! = 0.408 m = 4.50 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.80 x 1.80) - (0.409 x 0.667) = √20.70 x 1.485 x 0.317 / 3
0.317/2
= 0.409 m = 2.967 m² = 713.920 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.317 = 98.25 x 2.967 = 0.75 x 713.92 / 291.50
= 0.667 m = 291.502 kN = 1.84 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.409) + 0.667
Reinforcing Steel Bars: = 1.485 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 98.25 x 2.790 = 98.25 x 1.305
= 274.112 kN-m = 128.214 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 274.11x10^6 / = 128.21x10^6 /
(0.90 x 1.80 x 0.317²) (0.90 x 1.80 x 0.317²)
= 1.6838 MPa = 0.7876 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.6838) / (0.85 x [(2 x 0.7876) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.80 - 0.250 = 0.006448 = 0.002931
= 1.550 m use ρ = 0.006448 use ρ 0.005091
A= Wx =
= 1.80 x 1.550 As = ρ W d y = W/2 - As = ρ L d
h/2
= 2.790 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F3
Location: C3
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.80 m Service LC, Po = 148.402 kN
Width, W= 1.80 m Ftng. Self Wt. Wfn 31.104 kN
=
Depth, d= 0.400 m Total, Po Po t 179.506 kN
=
Eff. Depth, d eff. 0.317 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 255.000 kN
Min. Fdn. Depth D min. = 1.05 m Ftng. Self Wt. Wfu 37.325 kN
=
Column, b= 0.250 m Total, Pu Pu t 292.325 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.65 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.65 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.270 kN/m² b. Along Width
a. Along Length,
USBP, qu = 90.224A = W x
kN/m² A= Ly
= 1.80 x 1.233 = 1.80 x 0.408
= 2.219 m² = 0.734 m²
Vu = qu A Vu = qu A
= 90.22 x 2.219 = 90.22 x 0.734
= 200.207 kN = 66.224 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.80 x 0.317 / 6 = √20.70 x 1.80 x 0.317 / 6
= 432.679 kN = 432.679 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.80 - 0.250 -
0.317 = 0.75 x 432.68 / 200.21 = 1.80/2 - 0.350/2 - 0.317 = 0.75 x 432.68 / 66.22
= 1.233 m = 1.62 >1 .'. Adequate! = 0.408 m = 4.90 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.80 x 1.80) - (0.409 x 0.667) = √20.70 x 1.485 x 0.317 / 3
0.317/2
= 0.409 m = 2.967 m² = 713.920 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.317 = 90.22 x 2.967 = 0.75 x 713.92 / 267.70
= 0.667 m = 267.695 kN = 2.00 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.409) + 0.667
Reinforcing Steel Bars: = 1.485 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 90.22 x 2.790 = 90.22 x 1.305
= 251.725 kN-m = 117.742 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 251.73x10^6 / = 117.74x10^6 /
(0.90 x 1.80 x 0.317²) (0.90 x 1.80 x 0.317²)
= 1.5463 MPa = 0.7233 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.5463) / (0.85 x [(2 x 0.7233) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.80 - 0.250 = 0.005894 = 0.002687
= 1.550 m use ρ = 0.005894 use ρ 0.005091
A= Wx =
= 1.80 x 1.550 As = ρ W d y = W/2 - As = ρ L d
h/2
= 2.790 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F4
Location: B1
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.70 m Service LC, Po = 134.888 kN
Width, W= 1.70 m Ftng. Self Wt. Wfn 24.276 kN
=
Depth, d= 0.350 m Total, Po Po t 159.164 kN
=
Eff. Depth, d eff. 0.267 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 237.000 kN
Min. Fdn. Depth D min. = 1.04 m Ftng. Self Wt. Wfu 29.131 kN
=
Column, b= 0.250 m Total, Pu Pu t 266.131 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.55 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.55 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.555 kN/m² b. Along Width
a. Along Length,
USBP, qu = 92.087A = W x
kN/m² A= Ly
= 1.70 x 0.458 = 1.70 x 1.083
= 0.779 m² = 1.841 m²
Vu = qu A Vu = qu A
= 92.09 x 0.779 = 92.09 x 1.841
= 71.736 kN = 169.532 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.70 x 0.267 / 6 = √20.70 x 1.70 x 0.267 / 6
= 344.187 kN = 344.187 kN
x = L/2 - b/2 - d
FS = фVc / Vu y= W-h- FS = фVc / Vu
= 1.70/2 - 0.250/2 - d
0.267
= 0.75 x 344.19 / 71.74 = 1.70 - 0.350 - 0.267 = 0.75 x 344.19 / 169.53
= 0.458 m
= 3.60 >1 .'. Adequate! = 1.083 m = 1.52 >1 .'. Adequate!
Punching Shear:

x= b+ d A LW - xy Vc = √fc' bo d / 3
=
= 0.250 + 0.267 = (1.70 x 1.70) - (0.517 x 0.484) = √20.70 x 1.485 x 0.267 / 3
= 0.517 m = 2.640 m² = 601.314 kN
y= h + Vu = qu A FS = фVc / Vu
d/2
= 0.350 + = 92.09 x 2.640 = 0.75 x 601.31 / 243.11
0.267/2
= 0.484 m = 243.110 kN = 1.86 >1 .'. Adequate!
bo = x + 2y
Reinforcing Steel Bars: = 0.517 + (2 x 0.484)
a. Along Length, = 1.485 m

b. Along Width,
Mu = qu A Mu = qu A
= 92.09 x 1.233 = 92.09 x 2.295
= 113.543 kN-m = 211.340 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 113.54x10^6 / = 211.34x10^6 /
(0.90 x 1.70 x 0.267²) (0.90 x 1.70 x 0.267²)
= 1.0410 MPa = 1.9376 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
x = L/2 - b/2
[(2 x 1.0410) / (0.85 x [(2 x 1.9376) / (0.85 x
= 1.70/2 - 0.250/2 20.70)]} 20.70)]}
= 0.725 m = 0.003905 = 0.007483
A= Wx use ρ = 0.005091 use ρ = 0.007483
= 1.70 x 0.725 As = ρ W d y= W-h As = ρ L d
= 1.233 m² = 0.005091 x 1.70 x 0.267 = 1.70 - 0.350 = 0.007483 x 1.70 x 0.267
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F5
Location: C4
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.70 m Service LC, Po = 134.888 kN
Width, W= 1.70 m Ftng. Self Wt. Wfn 27.744 kN
=
Depth, d= 0.400 m Total, Po Po t 162.632 kN
=
Eff. Depth, d eff. 0.317 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 237.000 kN
Min. Fdn. Depth D min. = 1.07 m Ftng. Self Wt. Wfu 33.293 kN
=
Column, b= 0.250 m Total, Pu Pu t 270.293 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.57 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.57 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.270 kN/m² b. Along Width
a. Along Length,
USBP, qu = 93.527A = W x
kN/m² A= Ly
= 1.70 x 1.133 = 1.70 x 0.358
= 1.926 m² = 0.609 m²
Vu = qu A Vu = qu A
= 93.53 x 1.926 = 93.53 x 0.609
= 180.133 kN = 56.958 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.70 x 0.317 / 6 = √20.70 x 1.70 x 0.317 / 6
= 408.641 kN = 408.641 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.70 - 0.250 -
0.317 = 0.75 x 408.64 / 180.13 = 1.70/2 - 0.350/2 - 0.317 = 0.75 x 408.64 / 56.96
= 1.133 m = 1.70 >1 .'. Adequate! = 0.358 m = 5.38 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.70 x 1.70) - (0.409 x 0.667) = √20.70 x 1.485 x 0.317 / 3
0.317/2
= 0.409 m = 2.617 m² = 713.920 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.317 = 93.53 x 2.617 = 0.75 x 713.92 / 244.76
= 0.667 m = 244.760 kN = 2.19 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.409) + 0.667
Reinforcing Steel Bars: = 1.485 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 93.53 x 2.465 = 93.53 x 1.148
= 230.544 kN-m = 107.369 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 230.54x10^6 / = 107.37x10^6 /
(0.90 x 1.70 x 0.317²) (0.90 x 1.70 x 0.317²)
= 1.4995 MPa = 0.6983 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.4995) / (0.85 x [(2 x 0.6983) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.70 - 0.250 = 0.005707 = 0.002592
= 1.450 m use ρ = 0.005707 use ρ 0.005091
A= Wx =
= 1.70 x 1.450 As = ρ W d y = W/2 - As = ρ L d
h/2
= 2.465 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F6
Location: C1
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.50 m Service LC, Po = 102.598 kN
Width, W= 1.50 m Ftng. Self Wt. Wfn 21.600 kN
=
Depth, d= 0.400 m Total, Po Po t 124.198 kN
=
Eff. Depth, d eff. 0.317 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 183.000 kN
Min. Fdn. Depth D min. = 1.05 m Ftng. Self Wt. Wfu 25.920 kN
=
Column, b= 0.250 m Total, Pu Pu t 208.920 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.37 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.37 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.270 kN/m² b. Along Width
a. Along Length,
USBP, qu = 92.853A = W x
kN/m² A= Ly
= 1.50 x 0.933 = 1.50 x 0.833
= 1.400 m² = 1.250 m²
Vu = qu A Vu = qu A
= 92.85 x 1.400 = 92.85 x 1.250
= 129.994 kN = 116.066 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.50 x 0.317 / 6 = √20.70 x 1.50 x 0.317 / 6
= 360.566 kN = 360.566 kN
x= L-b-d
FS = фVc / Vu y= W-h- FS = фVc / Vu
= 1.50 - 0.250 - d
0.317
= 0.75 x 360.57 / 129.99 = 1.50 - 0.350 - 0.317 = 0.75 x 360.57 / 116.07
= 0.933 m
= 2.08 >1 .'. Adequate! = 0.833 m = 2.33 >1 .'. Adequate!
Punching Shear:

x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.50 x 1.50) - (0.409 x 0.509) = √20.70 x 0.918 x 0.317 / 3
0.317/2
= 0.409 m = 2.042 m² = 441.332 kN
y= h + Vu = qu A FS = фVc / Vu
d/2
= 0.350 + = 92.85 x 2.042 = 0.75 x 441.33 / 189.61
0.317/2
= 0.509 m = 189.606 kN = 1.75 >1 .'. Adequate!
Reinforcing Steel Bars: bo = x + y
a. Along Length, = 0.409 +
0.509)
= 0.918 m

b. Along Width,
Mu = qu A Mu = qu A
= 92.85 x 1.875 = 92.85 x 1.725
= 174.099 kN-m = 160.171 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 174.10x10^6 / = 160.17x10^6 /
(0.90 x 1.50 x 0.317²) (0.90 x 1.50 x 0.317²)
= 1.2833 MPa = 1.1807 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
x= L-b = [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
= 1.50 - 0.250 - √1 -
= 1.250 m [(2 x 1.2833) / (0.85 x [(2 x 1.1807) / (0.85 x
20.70)]} 20.70)]}
A= Wx
= 0.004850 = 0.004448
= 1.50 x 1.250
use ρ = 0.005091 use ρ = 0.005091
= 1.875 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F7
Location: A2
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.50 m Service LC, Po = 95.397 kN
Width, W= 1.50 m Ftng. Self Wt. Wfn 18.900 kN
=
Depth, d= 0.350 m Total, Po Po t 114.297 kN
=
Eff. Depth, d eff. 0.267 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 179.000 kN
Min. Fdn. Depth D min. = 0.96 m Ftng. Self Wt. Wfu 22.680 kN
=
Column, b= 0.250 m Total, Pu Pu t 201.680 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.31 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.31 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.555 kN/m² b. Along Width
a. Along Length,
USBP, qu = 89.636A = W x
kN/m² A= Ly
= 1.50 x 0.983 = 1.50 x 0.308
= 1.475 m² = 0.462 m²
Vu = qu A Vu = qu A
= 89.64 x 1.475 = 89.64 x 0.462
= 132.213 kN = 41.412 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.50 x 0.267 / 6 = √20.70 x 1.50 x 0.267 / 6
= 303.694 kN = 303.694 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.50 - 0.250 -
0.267 = 0.75 x 303.69 / 132.21 = 1.50/2 - 0.350/2 - 0.267 = 0.75 x 303.69 / 41.41
= 0.983 m = 1.72 >1 .'. Adequate! = 0.308 m = 5.50 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.50 x 1.50) - (0.384 x 0.617) = √20.70 x 1.385 x 0.267 / 3
0.267/2
= 0.384 m = 2.013 m² = 560.822 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.267 = 89.64 x 2.013 = 0.75 x 560.82 / 180.44
= 0.617 m = 180.437 kN = 2.33 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.384) + 0.617
Reinforcing Steel Bars: = 1.385 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 89.64 x 1.875 = 89.64 x 0.863
= 168.068 kN-m = 77.356 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 168.07x10^6 / = 77.36x10^6 /
(0.90 x 1.50 x 0.267²) (0.90 x 1.50 x 0.267²)
= 1.7463 MPa = 0.8038 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.7463) / (0.85 x [(2 x 0.8038) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.50 - 0.250 = 0.006701 = 0.002993
= 1.250 m use ρ = 0.006701 use ρ 0.005091
A= Wx =
= 1.50 x 1.250 As = ρ W d y = W/2 - As = ρ L d
h/2
= 1.875 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F7
Location: A4
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.50 m Service LC, Po = 104.453 kN
Width, W= 1.50 m Ftng. Self Wt. Wfn 18.900 kN
=
Depth, d= 0.350 m Total, Po Po t 123.353 kN
=
Eff. Depth, d eff. 0.267 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 194.000 kN
Min. Fdn. Depth D min. = 1.04 m Ftng. Self Wt. Wfu 22.680 kN
=
Column, b= 0.250 m Total, Pu Pu t 216.680 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 1.36 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 1.36 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.555 kN/m² b. Along Width
a. Along Length,
USBP, qu = 96.302A = W x
kN/m² A= Ly
= 1.50 x 0.983 = 1.50 x 0.308
= 1.475 m² = 0.462 m²
Vu = qu A Vu = qu A
= 96.30 x 1.475 = 96.30 x 0.462
= 142.045 kN = 44.492 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.50 x 0.267 / 6 = √20.70 x 1.50 x 0.267 / 6
= 303.694 kN = 303.694 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.50 - 0.250 -
0.267 = 0.75 x 303.69 / 142.05 = 1.50/2 - 0.350/2 - 0.267 = 0.75 x 303.69 / 44.49
= 0.983 m = 1.60 >1 .'. Adequate! = 0.308 m = 5.12 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.50 x 1.50) - (0.384 x 0.617) = √20.70 x 1.385 x 0.267 / 3
0.267/2
= 0.384 m = 2.013 m² = 560.822 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.267 = 96.30 x 2.013 = 0.75 x 560.82 / 193.86
= 0.617 m = 193.856 kN = 2.17 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.384) + 0.617
Reinforcing Steel Bars: = 1.385 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 96.30 x 1.875 = 96.30 x 0.863
= 180.566 kN-m = 83.109 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 180.57x10^6 / = 83.11x10^6 /
(0.90 x 1.50 x 0.267²) (0.90 x 1.50 x 0.267²)
= 1.8762 MPa = 0.8636 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.8762) / (0.85 x [(2 x 0.8636) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.50 - 0.250 = 0.007231 = 0.003221
= 1.250 m use ρ = 0.007231 use ρ 0.005091
A= Wx =
= 1.50 x 1.250 As = ρ W d y = W/2 - As = ρ L d
h/2
= 1.875 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F8
Location: C5
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.20 m Service LC, Po = 47.571 kN
Width, W= 1.20 m Ftng. Self Wt. Wfn 10.368 kN
=
Depth, d= 0.300 m Total, Po Po t 57.939 kN
=
Eff. Depth, d eff. 0.217 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 86.400 kN
Min. Fdn. Depth D min. = 0.76 m Ftng. Self Wt. Wfu 12.442 kN
=
Column, b= 0.250 m Total, Pu Pu t 98.842 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 0.93 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 0.93 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.840 kN/m² b. Along Width
a. Along Length,
USBP, qu = 68.640A = W x
kN/m² A= Ly
= 1.20 x 0.733 = 1.20 x 0.208
= 0.880 m² = 0.250 m²
Vu = qu A Vu = qu A
= 68.64 x 0.880 = 68.64 x 0.250
= 60.403 kN = 17.160 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.20 x 0.217 / 6 = √20.70 x 1.20 x 0.217 / 6
= 197.458 kN = 197.458 kN
x= L-b-d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.20 - 0.250 -
0.217 = 0.75 x 197.46 / 60.40 = 1.20/2 - 0.350/2 - 0.217 = 0.75 x 197.46 / 17.16
= 0.733 m = 2.45 >1 .'. Adequate! = 0.208 m = 8.63 >1 .'. Adequate!
Punching Shear:
x= b + A LW - xy Vc = √fc' bo d / 3
d/2 =
= 0.250 + = (1.20 x 1.20) - (0.359 x 0.567) = √20.70 x 1.285 x 0.217 / 3
0.217/2
= 0.359 m = 1.236 m² = 422.889 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.217 = 68.64 x 1.236 = 0.75 x 422.89 / 84.84
= 0.567 m = 84.839 kN = 3.74 >1 .'. Adequate!
bo = 2x + y
= (2 x 0.359) + 0.567
Reinforcing Steel Bars: = 1.285 m
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 68.64 x 1.140 = 68.64 x 0.510
= 78.250 kN-m = 35.006 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 78.25x10^6 / = 35.01x10^6 /
(0.90 x 1.20 x 0.217²) (0.90 x 1.20 x 0.217²)
= 1.5387 MPa = 0.6883 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 1.5387) / (0.85 x [(2 x 0.6883) / (0.85 x
x= L-b 20.70)]} 20.70)]}
= 1.20 - 0.250 = 0.005864 = 0.002554
= 0.950 m use ρ = 0.005864 use ρ 0.005091
A= Wx =
= 1.20 x 0.950 As = ρ W d y = W/2 - As = ρ L d
h/2
= 1.140 m²
 Project PROPOSED TWO(2) - STOREY RESIDENTIAL Owner JERSON MONTANO
Name: BUILDING :
SINDANGAN, ZAMBOANGA DEL NORTE Civil Engineer: JULIE ANNE LAIRE A. LORILLA
Location:

DESIGN ANALYSIS OF REINFORCED CONCRETE FOOTING

Footing: F9
Location: C5
I. FOOTING PROPERTIES
Design Footing: Unfactored Loads:
Length, L= 1.20 m Service LC, Po = 28.486 kN
Width, W= 1.20 m Ftng. Self Wt. Wfn 10.368 kN
=
Depth, d= 0.300 m Total, Po Po t 38.854 kN
=
Eff. Depth, d eff. 0.217 m Factored Loads:
=
Fdn. Depth, D= 1.50 m Service LC, Pu = 55.300 kN
Min. Fdn. Depth D min. = 0.51 m Ftng. Self Wt. Wfu 12.442 kN
=
Column, b= 0.250 m Total, Pu Pu t 67.742 kN
=
h= 0.350 m Minimum Footing
Dimension:
Material Length, L= 0.76 m
Properties:
Concrete, fc' = 20.70 Mpa Width, W= 0.76 m
Steel, fy = 275.00 Mpa Steel
Requirements:
Reduction Factor: Reinf Steel Bar Dia Ø= 16 mm
Bending, ф= 0.90 ρ bal = 0.85 β fc' 600 / fy(600+fy)
Shear, ф= 0.75 = 0.037292
bot steel cover, sc = 0.075 m ρ min = 1.40 / fy = 0.005091
II. DETAILED
Bearing Pressure:
CALCULATIONS ρ max = 75% ρ bal 0.027969
Wide Beam Shear: =
ESBP, qe = 66.840 kN/m² b. Along Width
a. Along Length,
USBP, qu = 47.043A = W x
kN/m² A= Ly
= 1.20 x 0.258 = 1.20 x 0.208
= 0.310 m² = 0.250 m²
Vu = qu A Vu = qu A
= 47.04 x 0.310 = 47.04 x 0.250
= 14.583 kN = 11.761 kN
Vc = √fc' W d / 6 Vc = √fc' L d /
6
= √20.70 x 1.20 x 0.217 / 6 = √20.70 x 1.20 x 0.217 / 6
= 197.458 kN = 197.458 kN
x = L/2 - b/2 - d
FS = фVc / Vu y = W/2 - h/2 - d FS = фVc / Vu
= 1.20/2 - 0.250/2 -
0.217 = 0.75 x 197.46 / 14.58 = 1.20/2 - 0.350/2 - 0.217 = 0.75 x 197.46 / 11.76
= 0.258 m = 10.16 >1 .'. Adequate! = 0.208 m = 12.59 >1 .'. Adequate!
Punching Shear:
x= b+ d A LW - xy Vc = √fc' bo d / 3
=
= 0.250 + 0.217 = (1.20 x 1.20) - (0.467 x 0.567) = √20.70 x 2.068 x 0.217 / 3
= 0.467 m = 1.175 m² = 680.572 kN
y= h + d Vu = qu A FS = фVc / Vu
= 0.350 + 0.217 = 47.04 x 1.175 = 0.75 x 680.57 / 55.28
= 0.567 m = 55.276 kN = 9.23 >1 .'. Adequate!
bo = 2x + 2y
= (2 x 0.467) + (2 x 0.567)
= 2.068 m
Reinforcing Steel Bars:
a. Along Length,
b. Along Width,
Mu = qu A Mu = qu A
= 47.04 x 0.570 = 47.04 x 0.510
= 26.815 kN-m = 23.992 kN-m
Ru = Mu / ф W d² Ru = Mu / ф L d²
= 26.82x10^6 / = 23.99x10^6 /
(0.90 x 1.20 x 0.217²) (0.90 x 1.20 x 0.217²)
= 0.5273 MPa = 0.4718 MPa
ρ act = 0.85 fc'/fy [1- √1 - ρ act = 0.85 fc'/fy [1- √1 -
(2Ru/0.85fc')] (2Ru/0.85fc')]
= [0.85(20.70/275.00)] {1 - √1 = [0.85(20.70/275.00)] {1 -
- √1 -
[(2 x 0.5273) / (0.85 x [(2 x 0.4718) / (0.85 x
20.70)]} 20.70)]}
x = L/2 - b/2
= 0.001947 = 0.001739
= 1.20/2 - 0.250/2
use ρ = 0.005091 use ρ 0.005091
= 0.475 m =
A= Wx As = ρ W d y = W/2 - As = ρ L d
= 1.20 x 0.475 h/2
= 0.570 m² = 0.005091 x 1.20 x 0.217 = 1.20/2 - 0.350/2 = 0.005091 x 1.20 x 0.217
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE COLUMN

Column: C2
Location: A3
I. COLUMN PROPERTIES
Material Dimensions: Steel Ratio: 1% -
Properties: 8%
Concrete, fc' = 20.70 MPa Base, b = 0.25 m SR = Ag / As
0
Main 1, fy = 275 MPa Height, h = 0.35 m = 1.84%
0
Main 2, fy = 275 MPa Length, L = 3.00 m
Shear, fy = 230 MPa
Reduction Reinforcing Steel Result Stresses:
Factors: Bar:
β= 0.85 MB1 RSBØ = 16 mm From Frame Analysis
Axial, ф = 0.70 Qty = 4 pcs Stresses Design F.S.
Flexure, ф = 0.90 MB2 RSBØ = 16 mm Pu kN 75.700 14.45
Shear, ф = 0.85 Qty = 4 pcs Mux kN-m 17.800 3.21
steel cover = 0.040 m Shear RSBØ = 10 mm Muy kN-m 18.200 6.26

Mz(kN-m)
My(kN-m) Fx(kN)
2 2 100 75.7
2 0 11.5
20 13.8 0 12.1 0 84.1 100
16 10 10
1
0
20 06 00 30
0 3 06 20 3 06 26 -7.92 1 2 -7.92 0 6
6 1 2 3 0 30 3
1 -7.57 50
10 0 10
10 20 - 2 10 10
2 -18.2 20 0 0 0
17.8
0
II. DETAILED CALCULATION
A. Compression B. Tension
Controlled Contolled
Pn = 0.80 [0.85 fc' (Ag-As) + Moment about X- Moment about Y-
Asfy] axis, axis,
= 1562.88 kN Layer 1: 3 d1' = 0.058 d1 = 0.067 Layer 3 d1' = 0.058 d1 0.117
1: =
Pu = Φ Pn Layer 2: 0 d2' = 0.000 d2 = 0.000 Layer 0 d2' = 0.000 d2 0.000
2: =
= 1094.02 kN Layer 3: 0 d3' = 0.000 d3 = 0.000 Layer 0 d3' = 0.000 d3 0.000
3: =
Stirrups Summary: d' = 0.058 c= 0.087 fs' = 200.0 d' = 0.058 c= 0.132 fs' = 336.3
: 0 6
16Db 256 mm 1 at 50 d= 0.192 a= 0.074 fs = 724.1 d = 0.292 a= 0.112 fs = 727.2
= 4 7
48db 480 mm 8 at 100 Cs 1 = 120.6 Ts 1 165.8 Cc = 455.7 Cs 1 = 165.88 Ts 1 165.88 Cc 492.6
= 4 = 8 1 = = 6
least = 250 mm rest at 250 Cs 2 = 0.00 Ts 2 0.00 Mnx = 63.40 Cs 2 = 0.00 Ts 2 0.00 Mny = 126.5
= = 1
Smax = 250 mm Cs 3 = 0.00 Ts 3 0.00 Mux = 57.06 Cs 3 = 0.00 Ts 3 0.00 Muy = 113.8
= = 6

Column: C-1
Location: Ground Floor Column @ Grid A2
I. COLUMN PROPERTIES
Material Dimensions: Steel Ratio: 1% -
Properties: 8%
Concrete, fc' = 20.70 MPa Base, b = 0.25 m SR = Ag / As
0
Main 1, fy = 275 MPa Height, h = 0.35 m = 2.76%
0
Main 2, fy = 275 MPa Length, L = 3.20 m
Shear, fy = 230 MPa
Reduction Reinforcing Steel Result Stresses:
Factors: Bar:
β= 0.85 Main 1, RSBØ 16 mm From Frame Analysis
=
Axial, ф = 0.70 Qty = 4 pcs Stresses Design F.S.
Flexure, ф = 0.90 Main 2, RSBØ 16 mm Pu kN 265.000 4.57
=
Shear, ф = 0.85 Qty = 8 pcs Mux kN-m 34.300 1.66
steel cover = 0.040 m Shear RSBØ = 10 mm Muy kN-m 29.900 3.81

Mz(kN-m)
My(kN-m) Fx(kN)
4 34.3 300 265 300
29.9 30 0 24 276
30 40
200
20 17.4 20 2 2 200
0 0 100
10 10 100
10 0 2 20
06 2 06 106 1 2 33.2 0 06 1006 1 2 0 6
1 2 0
10 2 2 100 -31 3-3.12 100
33.2 10 0 -20.3 0
20 -28.8 200 200
20 4 4
30 - -21.3 0 0 300

II. DETAILED CALCULATION

Project
A. Compression B. Tension
Name:
Controlled Contolled
Location:
Pn = 0.80 [0.85 fc' (Ag-As) + Moment about X- Moment about Y-
Asfy] axis, axis,
= 1728.49 kN Layer 1: 3 d1' = 0.058 d1 = 0.067 Layer 3 d1' = 0.058 d1 0.117
1: =
Pu = Φ Pn Layer 2: 0 d2' = 0.000 d2 = 0.000 Layer 0 d2' = 0.000 d2 0.000
2: =
= 1209.94 kN Layer 3: 0 d3' = 0.000 d3 = 0.000 Layer 0 d3' = 0.000 d3 0.000
3: =
Stirrups Summary: d' = 0.058 c= 0.087 fs' = 200.0 d' = 0.058 c= 0.132 fs' = 336.3
: 0 6
16Db 256 mm 1 at 50 d= 0.192 a= 0.074 fs = 724.1 d = 0.292 a= 0.112 fs = 727.2
= 4 7
48db 480 mm 8 at 100 Cs 1 = 120.6 Ts 1 165.8 Cc = 455.7 Cs 1 = 165.88 Ts 1 165.88 Cc 492.6
= 4 = 8 1 = = 6
least = 250 mm rest at 250 Cs 2 = 0.00 Ts 2 0.00 Mnx = 63.40 Cs 2 = 0.00 Ts 2 0.00 Mny = 126.5
= = 1
Smax = 250 mm Cs 3 = 0.00 Ts 3 0.00 Mux = 57.06 Cs 3 = 0.00 Ts 3 0.00 Muy = 113.8
= = 6
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE COLUMN

Column: C1
Location: B3
I. COLUMN PROPERTIES
Material Dimensions: Steel Ratio: 1% -
Properties: 8%
Concrete, fc' = 20.70 MPa Base, b = 0.30 m SR = Ag / As
0
Main 1, fy = 275 MPa Height, h = 0.30 m = 1.79%
0
Main 2, fy = 275 MPa Length, L = 3.00 m
Shear, fy = 230 MPa
Reduction Reinforcing Steel Result Stresses:
Factors: Bar:
β= 0.85 MB1 RSBØ = 16 mm From Frame Analysis
Axial, ф = 0.70 Qty = 4 pcs Stresses Design F.S.
Flexure, ф = 0.90 MB2 RSBØ = 16 mm Pu kN 55.300 20.23
Shear, ф = 0.85 Qty = 4 pcs Mux kN-m 36.300 2.43
steel cover = 0.040 m Shear RSBØ = 10 mm Muy kN-m 14.200 6.21

Mz(kN-m)
My(kN-m) Fx(kN)
2 2 80 66.2 8
40 36.3 32.1 14.2 0 55.3 0
0 13.1
40
10 10 4 4
20 0 0
20 08 0 3
0 3 07 20 07 208 -1.65 1 2 0 07
8 1 2 3 0 -
30 40 1.653
20 1 2 3
10 -15 10
2 2 8 8
40 - -33.4 40
0 0 0 0
35.5
II. DETAILED CALCULATION
A. Compression B. Tension
Controlled Contolled
Pn = 0.80 [0.85 fc' (Ag-As) + Moment about X- Moment about Y-
Asfy] axis, axis,
= 1598.07 kN Layer 1: 3 d1' = 0.058 d1 = 0.09 Layer 3 d1' = 0.058 d1 0.092
2 1: =
Pu = Φ Pn Layer 2: 0 d2' = 0.000 d2 = 0.00 Layer 0 d2' = 0.000 d2 0.000
0 2: =
= 1118.65 kN Layer 3: 0 d3' = 0.000 d3 = 0.00 Layer 0 d3' = 0.000 d3 0.000
0 3: =
Stirrups Summary: d' = 0.058 c= 0.109 fs' = 280.7 d' = 0.058 c= 0.109 fs' = 280.7
: 3 3
16Db 256 mm 1 at 50 d= 0.242 a= 0.093 fs = 732.1 d = 0.242 a= 0.093 fs = 732.1
= 1 1
48db 480 mm 8 at 100 Cs 1 = 165.8 Ts 1 165.8 Cc = 490.9 Cs 1 = 165.88 Ts 1 165.88 Cc 490.9
= 8 = 8 0 = = 0
least = 300 mm rest at 256 Cs 2 = 0.00 Ts 2 0.00 Mnx = 98.02 Cs 2 = 0.00 Ts 2 0.00 Mny = 98.02
= =
Smax = 256 mm Cs 3 = 0.00 Ts 3 0.00 Mux = 88.22 Cs 3 = 0.00 Ts 3 0.00 Muy = 88.22
= =

Column: C-1
Location: Ground Floor Column @ Grid A2
I. COLUMN PROPERTIES
Material Dimensions: Steel Ratio: 1% -
Properties: 8%
Concrete, fc' = 20.70 MPa Base, b = 0.35 m SR = Ag / As
0
Main 1, fy = 275 MPa Height, h = 0.35 m = 1.97%
0
Main 2, fy = 275 MPa Length, L = 3.20 m
Shear, fy = 230 MPa
Reduction Reinforcing Steel Result Stresses:
Factors: Bar:
β= 0.85 Main 1, RSBØ 16 mm From Frame Analysis
=
Axial, ф = 0.70 Qty = 4 pcs Stresses Design F.S.
Flexure, ф = 0.90 Main 2, RSBØ 16 mm Pu kN 354.000 4.39
=
Shear, ф = 0.85 Qty = 8 pcs Mux kN-m 41.200 4.09
steel cover = 0.040 m Shear RSBØ = 10 mm Muy kN-m 68.200 2.47

Mz(kN-m)
My(kN-m) Fx(kN)
60 60 400 354
80 68.2 8 400
40 38.2 34.4 40 370
47.7 0 20 20
4 4 20 20 0 0
0 0
10 07 20 00 20
0 2 08 10 08 17 -9.7 0 8
7 1 2 33.2 0 20 1 2
33.2 20
1 2 3-
20 93.. 20
40 40 -27.9 40 0 0
-49 -41.2
60 60 40 40
80 - 8 0 0
61.7 0
II. DETAILED CALCULATION
Project
A. Compression B. Tension
Name:
Controlled Contolled
Location:
Pn = 0.80 [0.85 fc' (Ag-As) + Moment about X- Moment about Y-
Asfy] axis, axis,
= 2221.15 kN Layer 1: 3 d1' = 0.058 d1 = 0.11 Layer 3 d1' = 0.058 d1 0.117
7 1: =
Pu = Φ Pn Layer 2: 2 d2' = 0.058 d2 = 0.11 Layer 2 d2' = 0.058 d2 0.117
7 2: =
= 1554.81 kN Layer 3: 0 d3' = 0.000 d3 = 0.00 Layer 0 d3' = 0.000 d3 0.000
0 3: =
Stirrups Summary: d' = 0.058 c= 0.132 fs' = 336.3 d' = 0.058 c= 0.132 fs' = 336.3
: 6 6
16Db 256 mm 1 at 50 d= 0.292 a= 0.112 fs = 727.2 d = 0.292 a= 0.112 fs = 727.2
= 7 7
48db 480 mm 8 at 100 Cs 1 = 165.8 Ts 1 165.8 Cc = 689.7 Cs 1 = 165.88 Ts 1 165.88 Cc 689.7
= 8 = 8 2 = = 2
least = 350 mm rest at 256 Cs 2 = 110.5 Ts 2 110.5 Mnx = 187.4 Cs 2 = 110.58 Ts 2 110.58 Mny = 187.4
8 = 8 6 = 6
Smax = 256 mm Cs 3 = 0.00 Ts 3 0.00 Mux = 168.7 Cs 3 = 0.00 Ts 3 0.00 Muy = 168.7
= 1 = 1
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: 2FB-1
Location: 3-AB
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa

Shear, fy = 230.00 Mz(kN-m) Mz(kN-m)

Reduction
MPa
Factor:
60 54.1 60 10 93.4
Flexure, ф = 0.90 β = 0.85 0 100
Torsion, ф = 0.85 steel cover = 0.040 40 40
5
m 20 8.07 20 0
50
Shear, ф = 0.85 max. agg. size = 0.025 m 25
06 20 20
Type of 007
Continuous 2 20 0. 07 20.219 1 2 2.63 0 8
Beam - 0.975 20
Beam: 0 5 5 -29
11.8
Dimensions: 40 0 -4 -50.3 50
Width, b = 0.250 m Length, L = 2.85 m 40 9.8
-54.4 60 10 10
Height, h = 0.400 m h min. = 0.333 0 0
m Reinf. Steel Bar Diameter:
Flexure = 16
mm Fy(kN) Fy(kN)
Shear = 10 150 10 10
mm 150 115 106 0 0
100 100
Effective Depth (m): 46.9 5
50 50 20.6 0
d1' = 0.058 m d1 = 0.342 50
m
2006 20 7 00 208
d2' = 0.099 m d2 = 0.301 0. 0.97 0 7 1 2 2.63 0
m 50 - -20.6
5 5 2 5 5
d3' = 0.140 m d3 = 0.260 20.4 -20.4 50
0 0
m 100
100
Result Stresses (from Frame Analysis): 10 -99.8 100
0
Stresses Left Midspan Right
- Mu 54.10 25.00 93.40
+ Mu 11.80 54.40
29.00
Vu 115.00 0.00 99.80

III. DETAILED CALCULATION

A. Design for Flexure
If pmin < pact < pmax, Singly Reinforced Beams (SRB) LEFT SUPPORT
If pact > pmax, Doubly Reinforced Beams (DRB) TOP BARS: BOTTOM BARS:
• Calculate ρ bal 0.85fc'β600 / fy(600+fy) = 0.037292 5 Layer 1 3 Layer
= 1
• Calculate ρ min 1.4 / fy = 0.005091 2 Layer 2 0 Layer
= 2
• Calculate ρ max = 0.75 p bal = 0.027969 0 Layer 3 0 Layer
Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right FACTOR OF SAFETY:
Support FS = 1.84 FS = 4.09
Area of Steel 1 As1 1005.310 603.186 1005.310
MIDSPAN
Area of Steel 2 As2 402.124 0.000 402.124
TOP BARS: BOTTOM BARS:
Area of Steel 3 As3 0.000 0.000 0.000
3 Layer 1 5 Layer
Steel Ratio ρ 0.017102 0.007055 0.017102 1
act 0 Layer 2 2 Layer
Reinf. Beam Type Singly R.B. Singly R.B. Singly R.B. 2
Stress Block a 87.990 37.710 87.990 0 Layer 3 0 Layer
Depth
Extreme Comp. Fiber to Neutral Axis c 103.518 44.365 103.518
FACTOR OF SAFETY:
Nominal Moment Mn 110.807 53.602 110.807
FS = 1.93 FS = 1.83
Design Moment Mu 99.730 48.250 99.730
RIGHT SUPPORT
Bottom Bar Reinforcements
TOP BARS: BOTTOM BARS:
Positive Moment (+Mu) Left Support Midspan Right
5 Layer 1 3 Layer
Support
1
Area of Steel 1 As1 603.186 1005.310 603.186
2 Layer 2 0 Layer
Area of Steel 2 As2 0.000 402.124 0.000 2
Area of Steel 3 As3 0.000 0.000 0.000 0 Layer 3 0 Layer
Steel Ratio ρ 0.007055 0.017102 0.007055
act FACTOR OF SAFETY:
Reinf. Beam Type Singly R.B. Singly R.B. Singly R.B. FS = 1.07 FS = 1.66
Stress Block a 37.710 87.990 37.710

B. Design for
Stirrups
• Calculate S = Av Fy d / Vs -153 mm
• Applied Force Vu max = 115.00 kN =
• Calculate Vc = 1/6 √fc'bd = 216.11 kN • Check Vs < 1/3 √fc'bd -80.82 < .'. Smax =
= 129.67 d/2
• Check Vu < 1/2 ΦVc = 115.00 > .'.
91.85 Required • Calculate Smax d/2 171 mm
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: 2FB-2
Location: 3-BC
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa

Shear, fy = 230.00 Mz(kN-m)

Reduction
MPa
Factor:
8 66.6 8
Flexure, ф = 0.90 β = 0.85 0 51.3 0
Torsion, ф = 0.85 steel cover = 0.040
4 4
m 0 0
Shear, ф = 0.85 max. agg. size = 0.025 m 1.3
0 9 2
Type of Continuous 208 1 2 3 3.33 0 09
Beam: Beam
4 -22.7 -27.9 -22.9 40
Dimensions: 0
Width, b = 0.250 m Length, L = 4.00 m
8 8
Height, h = 0.400 m h min. = 0.333 0 0
m Reinf. Steel Bar Diameter:
Flexure = 16
mm Fy(kN)
Shear = 10 10
100
mm 0
Effective Depth (m): 5 5
d1' = 0.058 m d1 = 0.342 0 13.7 0
m
200 2 09
d2' = 0.099 m d2 = 0.301 8 -13.7 1 2 3 3.33 0
m
50 50
d3' = 0.140 m d3 = 0.260 -55.7
m
Result Stresses (from Frame Analysis): 100
Stresses Left Midspan Right
- Mu 66.00 0.00 51.30
+ Mu 22.70 27.90 22.90
Vu 73.40 0.00 59.90

III. DETAILED CALCULATION

A. Design for Flexure
If pmin < pact < pmax, Singly Reinforced Beams (SRB) LEFT
SUPPORT If pact > pmax, Doubly Reinforced Beams (DRB) TOP BARS: BOTTOM BARS:
• Calculate ρ bal = 0.85fc'β600 / fy(600+fy) 0.037292 4 Layer 1 3 Layer 1
= 0.005091 2 Layer 2 0 Layer 2
• Calculate ρ min = 1.4 / fy 0.027969 0 Layer 3 0 Layer 3
=
6 Total 3 Total
• Calculate ρ
max = 0.75 p bal
= Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right Support
Area of Steel 1 FACTOR OF SAFETY:
804.248 804.248
FS = 1.31 FS = 2.13
As1
603.186
Area of Steel 2 As2 402.124 0.000
Area of Steel 3 As3 402.124
MIDSPAN
Steel Ratio ρ 0.000 0.000 0.000
TOP BARS: BOTTOM BARS:
act 0.014750 0.007055
3 Layer 1 4 Layer 1
Reinf. Beam Type 0.014750
0 Layer 2 2 Layer 2
Stress Block Depth a Singly R.B. - Singly
R.B. 0 Layer 3 0 Layer 3
Extreme Comp. Fiber to Neutral Axis c
75.420 37.710 75.420 3 Total 6 Total
Nominal Moment
88.729 44.365 88.729
96.415 53.602 96.415
Mn
86.780 48.250 86.780
Design Moment Mu
Bottom Bar Reinforcements
FACTOR OF SAFETY:
Positive Moment (+Mu) Left Support Midspan Right Support
FS = - FS = 3.11
Area of Steel 1 As1 Nominal Moment Mn
603.186 804.248 603.186
Area of Steel 2 As2 Design Moment Mu TOP BARS: BOTTOM BARS:
0.000 402.124 0.000
Area of Steel 3 As3 4 Layer 1 3 Layer 1
0.000 0.000 0.000
Steel Ratio ρ 2 Layer 2 0 Layer 2
0.007055 0.014750 0.007055
act 0 Layer 3 0 Layer 3
Singly R.B. Singly R.B. Singly R.B.
Reinf. Beam Type 6 Total 3 Total
37.710 75.420 37.710
Stress Block Depth a
44.365 88.729 44.365
Extreme Comp. Fiber to Neutral Axis c
FACTOR OF SAFETY:
FS = 1.69 FS = 2.11
Project
53.610 96.420 48.250 86.780 48.250
Name: RIGHT SUPPORT
53.610
Location:

B. Design for • Calculate S = Av Fy d / Vs -96 mm

Stirrups =
• Applied Force Vu max = 73.40 kN • Check Vs < 1/3 √fc'bd -129.76 < .'. Smax =
= 129.67 d/2
• Calculate Vc = 1/6 √fc'bd = 216.11 kN
• Calculate Smax d/2 171 mm
• Check Vu < 1/2 ΦVc = 73.40 < .'. Not = =
91.85 Required
• Summar
• Calculate Vn = Vu/ф = 86.35 kN y
• Calculate Vs = Vn - Vc = -129.76 kN
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: 2FB-3
Location: 1-BC
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa

Shear, fy = 230.00 Mz(kN-m)

Reduction
MPa
Factor:
8 8
Flexure, ф = 0.90 β = 0.85 61.5 0
0
Torsion, ф = 0.85 steel cover = 0.040 54.9
4
m 0 4
Shear, ф = 0.85 max. agg. size = 0.025 m 1.
0 6 2
Type of Continuous 202 1 2 3 3.85 0 03
Beam: Beam
4 -21.7 -25.5 -22.4
Dimensions: 0 40
Width, b = 0.250 m Length, L = 4.00 m
8 8
Height, h = 0.400 m h min. = 0.333 0 0
m Reinf. Steel Bar Diameter:
Flexure = 16
mm Fy(kN)
Shear = 10
mm
Effective Depth (m):
d1' = 0.058 m d1 = 0.342
m
2
d2' = 0.099 m d2 = 0.301 0
m
d3' = 0.140 m d3 = 0.260
m
Result Stresses (from Frame Analysis):
Stresses Left Midspan Right
- Mu 61.50 0.00 54.90
+ Mu 21.70 25.50 22.40
Vu 73.40 0.00 59.90
 8 73.4 8
0 0
4 4
0 11.4 0
Project 03 0 2
Name: 2 1 2 3 3.8 0
-11.4
Location: 40 5
III. DETAILED CALCULATION
A. Design for Flexure
8
If pmin < pact < pmax, Singly Reinforced Beams (SRB) LEFT -59.9
0
SUPPORT If pact > pmax, Doubly Reinforced Beams (DRB) TOP BARS: 80
BOTTOM BARS:
• Calculate ρ bal = 0.85fc'β600 / fy(600+fy) 0.037292 3 Layer 1 3 Layer 1
= 0.005091 2 Layer 2 0 Layer 2
• Calculate ρ min = 1.4 / fy 0.027969 0 Layer 3 0 Layer 3
=
5 Total 3 Total
• Calculate ρ
max = 0.75 p bal
= Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right Support
Area of Steel 1 FACTOR OF SAFETY:
603.186 603.186
FS = 1.19 FS = 2.22
As1
603.186
Area of Steel 2 As2 402.124 0.000
Area of Steel 3 As3 402.124
MIDSPAN
Steel Ratio ρ 0.000 0.000 0.000
TOP BARS: BOTTOM BARS:
act 0.012399 0.007055
3 Layer 1 3 Layer 1
Reinf. Beam Type 0.012399
0 Layer 2 2 Layer 2
Stress Block Depth a Singly R.B. - Singly
R.B. 0 Layer 3 0 Layer 3
Extreme Comp. Fiber to Neutral Axis c
62.850 37.710 62.850 3 Total 5 Total
Nominal Moment
73.941 44.365 73.941
81.328 53.602 81.328
Mn
73.200 48.250 73.200
Design Moment Mu
Bottom Bar Reinforcements
FACTOR OF SAFETY:
Positive Moment (+Mu) Left Support Midspan Right Support
FS = - FS = 2.87
Area of Steel 1 As1
603.186 603.186 RIGHT SUPPORT
Area of Steel 2 As2 603.186 TOP BARS: BOTTOM BARS:
Area of Steel 3 As3 0.000 402.124 0.000 3 Layer 1 3 Layer 1
Steel Ratio ρ 0.000 0.000 0.000 2 Layer 2 0 Layer 2
act 0.007055 0.012399 0 Layer 3 0 Layer 3
Reinf. Beam Type 0.007055 5 Total 3 Total
Stress Block Depth a Singly R.B. Singly R.B. Singly
Extreme Comp. Fiber to Neutral Axis c R.B. 37.710 62.850 37.710
FACTOR OF SAFETY:
Nominal Moment 44.365 73.941 44.365 FS = 1.33 FS = 2.15
53.610 81.330 53.610
Mn 48.250 73.200 48.250
Design Moment Mu

B. Design for • Calculate S = Av Fy d / Vs -96 mm

Stirrups =
• Applied Force Vu max = 73.40 kN • Check Vs < 1/3 √fc'bd -129.76 < .'. Smax =
= 129.67 d/2
• Calculate Vc = 1/6 √fc'bd = 216.11 kN
• Calculate Smax d/2 171 mm
• Check Vu < 1/2 ΦVc = 73.40 < .'. Not = =
91.85 Required
• Summar
• Calculate Vn = Vu/ф = 86.35 kN y
• Calculate Vs = Vn - Vc = -129.76 kN
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: 2FB-4
Location: C-13
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa

Shear, fy = 230.00 Mz(kN-m)

Reduction
MPa
Factor:
6 51.5 60
Flexure, ф = 0.90 β = 0.85 0 38.1
Torsion, ф = 0.85 steel cover = 0.040 40
4
m 20
0
Shear, ф = 0.85 max. agg. size = 0.025 m
2 20
Type of Continuous 20 0 1 2 09
Beam: Beam -20.6 3.17
-19.4
03 3
20 4
Dimensions:
Width, b = 0.250 m Length, L = 4.00 m
2 0
Height, h = 0.400 m h min. = 0.333 0 6
m Reinf. Steel Bar Diameter:
Flexure = 16
mm Fy(kN)
Shear = 10 8 8
mm 0 54 0
Effective Depth (m): 4 4
d1' = 0.058 m d1 = 0.342 0 12.6 0
m
2 0 2 09
d2' = 0.099 m d2 = 0.301 0 3 1 2 3
3.1 0
m -12.6
40 7
d3' = 0.140 m d3 = 0.260
m
Result Stresses (from Frame Analysis): 8 -61.5
0
Stresses Left Midspan Right 80
- Mu 38.40 0.00 51.50
+ Mu 20.60 20.60 19.40
Vu 54.00 0.00 61.50

III. DETAILED CALCULATION

A. Design for Flexure
If pmin < pact < pmax, Singly Reinforced Beams (SRB) LEFT SUPPORT
If pact > pmax, Doubly Reinforced Beams (DRB) TOP BARS: BOTTOM BARS:
• Calculate ρ bal 0.85fc'β600 / fy(600+fy) = 0.037292 4 Layer 1 3 Layer 1
= 0 Layer 2 0 Layer 2
• Calculate ρ min 1.4 / fy = 0.005091
= 0 Layer 3 0 Layer 3
• Calculate ρ max = 0.75 p bal = 0.027969 4 Total 3 Total
Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right Support FACTOR OF SAFETY:
Area of Steel 1 As1 804.248 603.186 804.248 FS = 1.64 FS = 2.34
Area of Steel 2 As2 0.000 0.000 0.000 MIDSPAN
Area of Steel 3 As3 0.000 0.000 0.000 TOP BARS: BOTTOM BARS:
Steel Ratio ρ 0.009406 0.007055 0.009406 3 Layer 1 4 Layer 1
act 0 Layer 2 0 Layer 2
Reinf. Beam Type Singly R.B. - Singly R.B.
0 Layer 3 0 Layer 3
Stress Block a 50.28 37.710 50.280
3 Total 4 Total
Depth 0
Extreme Comp. Fiber to Neutral Axis c 59.15 44.365 59.153
3 FACTOR OF SAFETY:
Nominal Moment Mn 70.07 53.602 70.079
FS = - FS = 3.06
9
Design Moment Mu 63.08 48.250 63.080 RIGHT SUPPORT
0 TOP BARS: BOTTOM BARS:
Bottom Bar Reinforcements
4 Layer 1 3 Layer 1
Positive Moment (+Mu) Left Support Midspan Right Support
0 Layer 2 0 Layer 2
Area of Steel 1 As1 603.186 804.248 603.186
0 Layer 3 1 Layer 3
Area of Steel 2 As2 0.000 0.000 0.000
4 Total 4 Total
Area of Steel 3 As3 0.000 0.000 201.062
Steel Ratio ρ 0.007055 0.009406 0.010148
FACTOR OF SAFETY:
act
Reinf. Beam Type Singly R.B. Singly R.B. Singly R.B. FS = 1.22 FS = 3.04

B. Design for
Stirrups
• Calculate S = Av Fy d / -86 mm
• Applied Force Vu max = 61.50 kN Vs =
• Calculate Vc = 1/6 √fc'bd = 216.11 kN • Check Vs < 1/3 √fc'bd -143.76 < .'. Smax =
= 129.67 d/2
• Check Vu < 1/2 ΦVc = 61.50 < .'. Not
91.85 Required • Calculate Smax d/2 171 mm
= =
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: 2FB-5
Location: A'-34
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa

Shear, fy = 230.00 Mz(kN-m)

Reduction
MPa
Factor:
4 32.2
Flexure, ф = 0.90 β = 0.85 0 30.3
40
Torsion, ф = 0.85 steel cover = 0.040
2
m 0
Shear, ф = 0.85 max. agg. size = 0.025 m 1.7
0 8 2
Type of Continuous 207 -6.03 1 2 3 - 0 11
Beam: Beam 5 2
2 -17.8 6.45
3.
Dimensions: 0 0
Width, b = 0.250 m Length, L = 4.00 m 5
4 4
Height, h = 0.400 m h min. = 0.333 0 0
m Reinf. Steel Bar Diameter:
Flexure = 16
mm Fy(kN)
Shear = 10 6 6
47.1
mm 0 0
Effective Depth (m):
4 4
d1' = 0.058 m d1 = 0.342 2 20
m 0 3.58
2 2 11
d2' = 0.099 m d2 = 0.301 0 07 -3.44 1 2 3 0
m 3.55
2
d3' = 0.140 m d3 = 0.260 20
m 0
40
Result Stresses (from Frame Analysis): 4
Stresse Left Midspan Right -46.1
s
- Mu
III. DETAILED 30.30
CALCULATION 0.00 32.20
+ Mu
A. Design for Flexure 6.03 17.80 6.45
LEFT SUPPORT
If pmin Vu
< pact < pmax,47.10 0.00 (SRB)
Singly Reinforced Beams 46.10
TOP BARS: BOTTOM BARS:
If pact > pmax, Doubly Reinforced Beams (DRB)
4 Layer 1 3 Layer 1
• Calculate ρ bal 0.85fc'β600 / fy(600+fy) = 0.037292
= 0 Layer 2 0 Layer 2
• Calculate ρ min 1.4 / fy = 0.005091 0 Layer 3 0 Layer 3
= 4 Total 3 Total
• Calculate ρ max = 0.75 p bal = 0.027969
Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right Support FACTOR OF SAFETY:
Area of Steel 1 As1 804.248 603.186 804.248 FS = 2.08 FS = 8.00
Area of Steel 2 As2 0.000 0.000 0.000 MIDSPAN
Area of Steel 3 As3 0.000 0.000 0.000 TOP BARS: BOTTOM BARS:
Steel Ratio ρ 0.009406 0.007055 0.009406 3 Layer 1 4 Layer 1
act 0 Layer 2 0 Layer 2
Reinf. Beam Type Singly R.B. - Singly R.B.
0 Layer 3 0 Layer 3
Stress Block a 50.28 37.710 50.280
3 Total 4 Total
Depth 0
Extreme Comp. Fiber to Neutral Axis c 59.15 44.365 59.153
3 FACTOR OF SAFETY:
Nominal Moment Mn 70.07 53.602 70.079
FS = - FS = 3.54
9
Design Moment Mu 63.08 48.250 63.080 RIGHT SUPPORT
0 TOP BARS: BOTTOM BARS:
Bottom Bar Reinforcements
4 Layer 1 3 Layer 1
Positive Moment (+Mu) Left Support Midspan Right Support
0 Layer 2 0 Layer 2
Area of Steel 1 As1 603.186 804.248 603.186
0 Layer 3 0 Layer 3
Area of Steel 2 As2 0.000 0.000 0.000
4 Total 3 Total
Area of Steel 3 As3 0.000 0.000 0.000
Steel Ratio ρ 0.007055 0.009406 0.007055
FACTOR OF SAFETY:
act
Reinf. Beam Type Singly R.B. Singly R.B. Singly R.B. FS = 1.96 FS = 7.48

B. Design for
Stirrups
• Calculate S = Av Fy d / -77 mm
• Applied Force Vu max = 47.10 kN Vs =
• Calculate Vc = 1/6 √fc'bd = 216.11 kN • Check Vs < 1/3 √fc'bd -160.70 < .'. Smax =
= 129.67 d/2
• Check Vu < 1/2 ΦVc = 47.10 < .'. Not
91.85 Required • Calculate Smax d/2 171 mm
= =
• Calculate Vn = Vu/ф = 55.41 kN
• Summar
• Calculate Vs = Vn - Vc = -160.70 kN y
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE BEAM

Beam: RB-1
Location: A-34
I. BEAM PROPERTIES
Material Properties: Shear and Moment Diagram:
Concrete, fc' = 20.70 MPa Flexure, fy = 275.00 MPa
Mz(kN-m)
Shear, fy = 230.00 MPa
Reduction Factor:
3 30
0 22.1 20.8
Flexure, ф = 0.90 β = 0.85 20
Torsion, ф = 0.85 steel cover = 0.040 2
0 10
m
30 2.06
Shear, ф = 0.85 max. agg. size = 0.025 1 3011
m 1 2 3 3.54
0 -6.9
Type of Continuous Beam -9.12 -7.46
Beam: 08 2
Dimensions: 1 0
Width, b = 0.200 m Length, L = 4.00 m
0 3
Height, h = 0.300 m h min. = 0.333
m Reinf. Steel Bar Diameter:
Flexure = 16 Fy(kN)
mm 29.5
3 30
Shear = 10 0
mm 20
2
Effective Depth (m): 4.06 10
0
d1' = 0.058 m d1 = 0.242
m 3 1 3011
d2' = 0.099 m d2 = 0.201
0 0 -4.06 1 2 3.54 10
3
m
d3' = 0.140 m d3 = 0.160
08 20
m 1 -28.3 30
Result Stresses (from Frame Analysis): 0
Stresses Left Midspan Right
- Mu 22.10 0.00 20.80
+ Mu 6.90 9.12 7.46
Vu 29.50 0.00 28.30

III. DETAILED CALCULATION

A. Design for Flexure
If pmin < pact < pmax, Singly Reinforced Beams (SRB) LEFT
SUPPORT If pact > pmax, Doubly Reinforced Beams (DRB) TOP BARS: BOTTOM BARS:
• Calculate ρ bal = 0.85fc'β600 / fy(600+fy) 0.037292 2 Layer 1 2 Layer 1
= 0.005091 2 Layer 2 0 Layer 2
• Calculate ρ min = 1.4 / fy 0.027969 0 Layer 3 0 Layer 3
=
4 Total 2 Total
• Calculate ρ
max = 0.75 p bal
= Top Bar Reinforcements
Negative Moment (-Mu) Left Support Midspan Right Support
Area of Steel 1 FACTOR OF SAFETY:
402.124 402.124
FS = 1.71 FS = 3.27
As1
402.124
Area of Steel 2 As2 402.124 0.000
Area of Steel 3 As3 402.124
MIDSPAN
Steel Ratio ρ 0.000 0.000 0.000
TOP BARS: BOTTOM BARS:
act 0.018311 0.008308
2 Layer 1 2 Layer 1
Reinf. Beam Type 0.018311
0 Layer 2 2 Layer 2
Stress Block Depth a Singly R.B. - Singly
R.B. 0 Layer 3 0 Layer 3
Extreme Comp. Fiber to Neutral Axis c
62.850 31.425 62.850 2 Total 4 Total
Nominal Moment
73.941 36.971 73.941
42.039 25.024 42.039
Mn
37.840 22.530 37.840
Design Moment Mu
Bottom Bar Reinforcements
FACTOR OF SAFETY:
Positive Moment (+Mu) Left Support Midspan Right Support
FS = - FS = 4.15
Area of Steel 1 As1 Nominal Moment Mn
402.124 402.124 402.124
Area of Steel 2 As2 Design Moment Mu TOP BARS: BOTTOM BARS:
0.000 402.124 0.000
Area of Steel 3 As3 2 Layer 1 2 Layer 1
0.000 0.000 0.000
Steel Ratio ρ 2 Layer 2 0 Layer 2
0.008308 0.018311 0.008308
act 0 Layer 3 0 Layer 3
Singly R.B. Singly R.B. Singly R.B.
Reinf. Beam Type 4 Total 2 Total
31.425 62.850 31.425
Stress Block Depth a
36.971 73.941 36.971
Extreme Comp. Fiber to Neutral Axis c FACTOR OF SAFETY:
25.030 42.040 25.030
FS = 1.82 FS = 3.02
Project
22.530 37.840 22.530
Name: RIGHT SUPPORT
Location:
B. Design for • Calculate S = Av Fy d / Vs -100 mm
Stirrups =
• Applied Force Vu max = 29.50 kN • Check Vs < 1/3 √fc'bd -87.63 < .'. Smax =
= 73.40 d/2
• Calculate Vc = 1/6 √fc'bd = 122.34 kN
• Calculate Smax = d / 2 121 mm
• Check Vu < 1/2 ΦVc = 29.50 < .'. Not =
51.99 Required
• Summar
• Calculate Vn = Vu/ф = 34.71 kN y
• Calculate Vs = Vn - Vc = -87.63 kN
Project
PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name:
SINDANGAN,
Location: ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Engineer:

DESIGN ANALYSIS OF STEEL

DESIGN ANALYSIS OF PURLIN
I. MATERIAL PROPERTIES
Section: LC 100 x 50 x 15 x 2.0mm thk. Cee Purlins
Weight, Wt 3.51 kg/m Bay Length = 4.00 m
=
Section Modulus, Sx 16.20 x 10³ Spacing = 0.70 m
= mm³
Sy 5.20 x 10³ Roof Slope = 5.47 degree
= mm³
Flexural Stress, Fy 248.00 MPa Roof Dead Load = 0.10 kPa
=
Allowable Fbx = 0.66 Fy 163.68 MPa Roof Live Load = 1.00 kPa
Stresses: =
Fby = 0.75 Fy 186.00 MPa Wind Pressure = 1.60 kPa
=
II. DESIGN LOAD COMBINATIONS
A. Weight of Purlin
DL 3.51 x = 34.43 N/m Wpn = 34.43 cos 5.47° = 34.27 N/m
= 9.81
Wpt = 34.43 sin 5.47° = 3.28 N/m
B. Roof Loads
DL 0.70 x = 70.00 N/m Wrn = DL + LLR + = 1,330.00 N/m
= 0.10 0.5WL
LLR 0.70 x = 700.00 N/m = 1,330.00 cos = 1,323.94 N/m
= 1.00 5.47°
WL 0.70 x = 1,120.00 N/m Wrt = DL + LLR = 770.00 N/m
= 1.60
= 770.00 sin 5.47° = 73.40 N/m
III. DESIGN MOMENT
Mpx Wpn L² / 8 = 68.54 N-m Mrx = Wrn L² / 8 = 2,647.88 N-m
=
Mpy Wpt L² / 32 = 1.64 N-m Mry = Wrt L² / 32 = 36.70 N-m
=
IV. RESULT STRESSES
fbpx Mpx / Sx = 4.23 MPa fbrx = Mrx / Sx = 163.45 MPa
=
fbpy Mpy / Sy = 0.32 MPa fbry = Mry / = 14.12 MPa
= 0.5Sy
V. INTERACTION CRITERION
Note: 1/3 increase in allowable stresses is allowed.
fbpx + fbrx fbpy + fbry
+ ≤ 1.00
Fbx + 1/3 Fbx Fby + 1/3
Fby
4.23 + 163.45 0.32 +
+ 14.12 ≤ 1.00
163.68 + 54.56 186.00 +
62.00
0.83 < 1.00 .'. Adequate

DESIGN ANALYSIS OF TRUSS/RAFTER

I. MATERIAL PROPERTIES
Section: 2 L - LC 150 x 65 x 20 x 2.0mm thk. Cee Purlins
Weight, Wt = 4.76 kg/m Length = 3.60 m
Section Modulus, Sx = 28.60 x 10³ mm³ Tributary Width = 2.00 m
Sy = 8.10 x 10³ mm³ Roof Slope = 6.55 degree
Flexural Stress, Fy = 248.00 MPa Roof Dead Load = 0.10 kPa
Allowable Stresses: Fbx = 0.66 Fy = 163.68 MPa Roof Live Load = 1.00 kPa
Fby = 0.75 Fy = 186.00 MPa Wind Pressure = 1.60 kPa
II. DESIGN LOAD COMBINATIONS
A. Weight of Purlin
Wt = 34.43 x 2.00 x = 482.02 N Wpn = 482.02 cos 6.55° = 478.87 N/m
7
DLp = 482.02 / 3.60 = 133.89 N/m
B. Weight of Rafter/Truss
DLr/t = 4.76 x 9.81 x 2 = 93.39 N/m Wpn = 93.39 cos 6.55° = 92.78 N/m
C. Roof Loads
DL = 0.10 x 2.00 = 200.00 N/m Wrn = DL + LLR + 0.5WL = 3,800.00 N/m
LLR = 1.00 x 2.00 = 2,000.00 N/m = 3,800.00 cos 6.55° = 3,775.20 N/m
WL = 1.60 x 2.00 = 3,200.00 N/m
III. DESIGN MOMENT allowe =
Mpx = Wpn L² / 8 = d.
926.07 N-m
IV. RESULT STRESSES MPa fbrx
Mrx = = Mpx / 2Sx =
fbpx = Mpx / 2Sx = 16.19 Wrn L² /
V. INTERACTION CRITERION
Note: 1/3 increase in allowable stresses is 8
Project
Name:
6,115.82 106.
Location: 92 N-m MPa

fbx + frx
Fbx + 1/3 ≤ 1.00

Fbx
16.19 +
106.92 ≤ 1.00
163.68 +
54.56

0.56 < 1.00 .'. Adequate

Project PROPOSED TWO(2) - STOREY RESIDENTIAL BUILDING Owner: JERSON MONTANO
Name: SINDANGAN, ZAMBOANGA DEL NORTE Civil JULIE ANNE LAIRE A. LORILLA
Location: Engineer:

DESIGN ANALYSIS OF REINFORCED CONCRETE SLAB

TWO-WAY SLAB DESIGN
Slab: S-1
Location: AB-23
Slab Slab Properties: Span 0.75 ≈ 0.75
Type: Ratio:
Short Span = 3.00 m Ca dl 0.036 Ca ll = 0.04 Ca = 0.061
+ Mu = + Mu 9 - Mu
3 One Long Edge Long Span = 4.00 m Cb dl 0.013 Cb ll = 0.01 Cb = 0.036
Discontiuous = 6
Slab thk. = 0.125 m Steel Reinforcements: Temperature Bars:
DSBØ = 10 mm d eff. @ short 0.100 ρmin = 0.00609 Assumptio ρact As Smax
= n,
Material Unfactored Loads: d eff. @ long 0.090 ρmax = 0.03525 1/2 of 0.0030 305.0 258
Properties: = ρmin 5 0
Concrete, fc' 20.70 MPa Dead Load = 11.07 kN/m Along Short Along Long Span:
= Span:
Steel, fy = 230.0 MPa Live Load = 1.90 kN/m Location Mu ρact As Smax Location Mu ρact As Smax
0
Reduction Factor: Factored Loads: Mid 5.64 0.0027 609.0 129 Mid 2.27 0.0013 548.1 143
7 0 7 0
Bending, Φ = 0.90 Dead Load = 13.28 kN/m Cont. 8.96 0.0044 609.0 129 Cont. 5.29 0.0032 548.1 143
6 0 2 0
steel cover, sc 0.020 m Live Load = 3.04 kN/m Discont. 1.88 0.0009 609.0 129 Discont. 0.76 0.0004 548.1 143
= 1 0 5 0

Slab: S-1
Location: BC-23
Slab Slab Properties: Span 0.88 ≈ 0.85
Type: Short Span = 3.70 m Ratio: Ca dl = 0.036 Ca ll = 0.043 Ca = 0.066
Two Adjacent Edges + + Mu - Mu
Mu
4 Long Span = 4.20 m Cb dl = 0.019 Cb ll = 0.023 Cb = 0.034
Discontinuous Slab thk. = 0.125 m Steel Reinforcements: Temperature Bars:
DSBØ = 10 mm d eff. @ short = 0.100 ρmin = 0.00609 Assumption, ρact As Smax
Material Properties: Unfactored Loads: d eff. @ long = 0.090 ρmax = 0.03525 1/2 of ρmin 0.00305 258
305.00
Concrete, fc' = 20.70 MPa Dead Load = 11.07 kN/m Along Short Span: Along Long Span:
Steel, fy = 230.00 MPa Live Load = 1.90 kN/m Location Mu ρact As Smax Location Mu ρact As Smax
Reduction Factor: Factored Loads: Mid 8.33 0.00414 609.00 129 Mid 4.97 0.00302 143
548.10
Bending, Φ = 0.90 Dead Load = 13.28 kN/m Cont. 14.75 0.00749 749.00 105 Cont. 7.60 0.00468 143
548.10
steel cover, sc = 0.020 m Live Load = 3.04 kN/m Discont. 2.78 0.00135 609.00 129 Discont. 1.66 0.00100 143
548.10

Slab: S-1
Location: BC-23
Slab Slab Properties: Span 0.79 ≈ 0.75
Type: Ratio:
Short Span = 3.00 m Ca dl 0.036 Ca ll = 0.04 Ca = 0.061
+ Mu = + Mu 9 - Mu
3 One Long Edge Long Span = 3.80 m Cb dl 0.013 Cb ll = 0.01 Cb = 0.036
Discontiuous = 6
Slab thk. = 0.125 m Steel Reinforcements: Temperature Bars:
DSBØ = 10 mm d eff. @ short 0.100 ρmin = 0.00609 Assumptio ρact As Smax
= n,
Material Unfactored Loads: d eff. @ long 0.090 ρmax = 0.03525 1/2 of 0.0030 305.0 258
Properties: = ρmin 5 0
Concrete, fc' 20.70 MPa Dead Load = 11.07 kN/m Along Short Along Long Span:
= Span:
Steel, fy = 230.0 MPa Live Load = 1.90 kN/m Location Mu ρact As Smax Location Mu ρact As Smax
0
Reduction Factor: Factored Loads: Mid 5.64 0.0027 609.0 129 Mid 2.27 0.0013 548.1 143
7 0 7 0
Bending, Φ = 0.90 Dead Load = 13.28 kN/m Cont. 8.96 0.0044 609.0 129 Cont. 5.29 0.0032 548.1 143
6 0 2 0
steel cover, sc 0.020 m Live Load = 3.04 kN/m Discont. 1.88 0.0009 609.0 129 Discont. 0.76 0.0004 548.1 143
= 1 0 5 0

Slab: S-1
Location: BC-23
Slab Slab Properties: Span 0.95 ≈ 0.95
Type: Ratio:
Short Span 3.60 m Ca dl = 0.030 Ca ll 0.035 Ca 0.05
Two Adjacent Edges + + - Mu 0
4 = Long 3.80 m Mu Cb dl = 0.024 Mu = Cb 0.029 =
Discontinuous Slab thk. = 0.125 m Steel Reinforcements: Temperature Bars:
0.04
DSBØ = 10 mm d eff. @ short = 0.100 ρmin = 0.00609 Assumption, ρact As Smax
Material Properties: Unfactored Loads: d eff. @ long = 0.090 ρmax = 0.03525 1/2 of ρmin 0.00305 258
305.00
Concrete, fc' = 20.70 MPa Dead Load = 11.07 kN/m Along Short Span: Along Long Span:
Steel, fy = 230.00 MPa Live Load = 1.90 kN/m Location Mu ρact As Smax Location Mu ρact As Smax
Reduction Factor: Factored Loads: Mid 6.54 0.00323 609.00 129 Mid 5.94 0.00363 143
548.10
Bending, Φ = 0.90 Dead Load = 13.28 kN/m Cont. 10.58 0.00529 609.00 129 Cont. 9.52 0.00591 143
548.10
steel cover, sc = 0.020 m Live Load = 3.04 kN/m Discont. 2.18 0.00106 609.00 129 Discont. 1.98 0.00119 143
548.10
 ONE-WAY SLAB DESIGN
Slab: -
Location: -
Material Unfactored Loads: Slab Properties: Main Bar: Temp.
Properties: Bar:
Concrete, fc' = 0.00 MPa Dead Load = 0.00 kN/m Short Span = 0.00 m d eff. = 0.000 d eff. = 0.000
Steel, fy = 0.00 MPa Live Load = 0.00 kN/m Slab thk. = 0.000 m Mu = 0.00 Mu = 0.00
DSBØ = 0 mm ρact = 0.0000 ρact = 0.0000
0 0
Reduction Factor: Factored Loads: Steel Reinforcements: As = 0.00 Ast = 0.00
Bending, Φ = 0.00 Dead Load = 0.00 kN/m ρmin = 0.0000
Smax = 0 Smax = 0
0
steel cover, sc = 0.000 m Live Load = 0.00 kN/m ρmax = 0.0000
0
Project Owner:
Name:
Location: Civil
Engineer:
DESIGN ANALYSIS OF REINFORCED CONCRETE STAIR
REINFORCED CONCRETE SLAB
Location: Landing to Up Location: Landing to Down Stair
Stair
Slab Properties: Slab Properties:
Considering 1m Considering 1m strip,
strip,
Length = 3.80 m Rise = 0.200 m Length 2.200 m Rise = 0.200 m
0 =
Slab thk = 0.15 m Tread 0.250 m Slab thk 0.150 m Tread 0.250 m
0 = = =
Steps 16 set Steps 4 set
= =
Material Properties: Reduction Factor: Material Reduction Factor:
Properties:
Concrete, fc' = 20.7 MPa Bending, Φ 0.900 Concrete, fc' 20.70 MPa Bending, Φ 0.900
0 = = =
Main Steel, fy 75.00 MPa steel cover, sc 0.020 m Main Steel, fy 275.00 MPa steel cover, sc 0.020 m
= 2 = = =
Sec. Steel, fy =30.00 MPa Sec. Steel, fy 230.00 MPa
2 =
Unfactored Loads: Factored Loads: Unfactored Factored Loads:
Loads:
Dead Load = 7.23 kN/m Dead Load 8.68 kN/m Dead Load 5.79 kN/m Dead Load 6.95 kN/m
= = =
Live Load = 1.90 kN/m Live Load 3.04 kN/m Live Load 1.90 kN/m Live Load 3.04 kN/m
= = =
Steel Steel Reinforcements:
Reinforcements:
ρ min = 0.005091 ρ max = 0.027969 ρ min = 0.005091 ρ max 0.027969
=
Main Main
Bars, Bars,
Main RSBØ 12 mm d eff. = 0.124 m Main RSBØ 16 mm d eff. 0.122 m
= = =
Negative Moment, Positive Moment, Negative Positive Moment,
Moment,
- Mu 14.10 kN-m + Mu 21.15 kN-m - Mu 4.03 kN-m + Mu 6.04 kN-m
= = = =
ρ= 0.003819 ρ= 0.005823 ρ= 0.001103 ρ= 0.001661
Ast 631.28 mm² As = 722.05 mm² Ast 621.10 mm² As = 621.10 mm²
= =
Smax 179 mm Smax 157 mm Smax 324 mm Smax 324 mm
= = = =
Temperature Bars, Temperature Bars,
Sec. RSBØ = 10 mm Sec. RSBØ 10 mm
=
ρ= 1/2 of ρ As = 315.70 mm² ρ = 1/2 of ρ min As = 310.61 mm²
min
ρ= 0.002546 Smax = 249 mm ρ= 0.002546 Smax 253 mm
=

REINFORCED CONCRETE BEAM

Type of Beam: Simply Supported Beam Steel Reinforcements:
Dimensions: Design for
Flexure,
Width, b = 0.20 m Length, L 1.20 m Main RSBØ 16 mm ρ min 0.005091
0 = = =
Height, h = 0.30 m h min. = 0.100 m d eff. = 0.252 m ρ max 0.027969
0 =
Negative Positive Moment,
Moment,
Material Properties: Reduction Factor: - Mu 1.41 kN-m + Mu 2.11 kN-m
= =
Concrete, fc' = 20.7 MPa Bending, Φ 0.90 ρ act = 0.000090 ρ act = 0.000134
0 =
Main Steel, fy 75.00 MPa Shear, ф = 0.85 As 256.59 mm² As = 256.59 mm²
= 2 =
Sec. Steel, fy =30.00 MPa steel cover, sc 0.040 m Qty 2 L - 16mm RSBØ Qty = 2 L - 16mm RSBØ
2 = =
Design for Shear,
Uniform Distributed Load - 1 (UDL- Sec. RSBØ 10 mm S = AvFyd/Vs -246 mm
1): = =
Unfactored Loads, Factored Loads, Vu = 1.00 kN Vs < √fc'bd/3 -37.04 <
= 76.44
Dead Load = 7.23 kN/m Dead Load 8.68 kN/m Vc = √fc'bd/6 38.22 kN .'. Smax = d/2
= =
Live Load = 1.90 kN/m Live Load 3.04 kN/m Vu < 1/2 ΦVc 1.00 < Smax = d / 2 126 mm
= = 16.24 =
.'. Not Required Summar
y:
Uniform Distributed Load - 2 (UDL- Vn = Vu/ф 1.18 kN 1 @ 50 mm
2): =
Unfactored Loads, Factored Loads, Vs = Vn - Vc -37.04 kN 6 @ 100 mm
=
Dead Load = 5.79 kN/m Dead Load 6.95 kN/m Vs>2/3 √fc'bd -37.04 < 0 @ 150 mm
= = 152.87
Live Load = 1.90 kN/m Live Load 3.04 kN/m .'. Adequate rest @ 120 mm
=
 REINFORCED CONCRETE FOOTING
Design Footing: Design Loads: Steel
Requirements:
Length, = 1.20 m Unfactored RSBØ 12 mm ρ min = 0.005091
L Loads, =
Width, W = 0.80 m SLC, Pn = 20.82 kN d eff. 0.169 m ρ max 0.027969
= =
Depth, d = 0.250 m Ftng. Self Wt. = 5.76 kN Wide Beam
Shear,
Fdn. Depth, D = 0.80 m Total, Pn = 26.58 kN Vu = qu A = 6.47 kN
Min. Fdn. Depth = 0.52 m Factored Loads, Vc = 1/6 √fc' W d = 102.52 kN
Column, = 1.20 m SLC, Pu = 26.72 kN FS = ΦVc / Vu = 13.47 >1 .'.
b Adequate!
Column, = 0.15 m Ftng. Self Wt. = 6.91 kN Reinforcing Bars along
h Width,
Total, Pu = 33.63 kN Mu = qu A = 11.38 kN-m
Material Properties: Ru = Mu/Φ L = 0.369 MPa
d²
Concrete, Fc' = 20.70 Mpa Bearing Pressure: ρ act = 0.85fc'/fy [1- = 0.001356
√1-(2Ru/0.85fc')]
Steel, Fy = 275.00 Mpa ESBP, = 79.94 kN/m² As = ρ L d = 688.30 mm²
qe
USBP, = 35.03 kN/m² Qty = As / Ab = 7 L - 12mm RSBØ
qu
Reduction Factor: Reinforcing Bars along Length / Temperature
Bar,
Bending, ф = 0.90 Minimum Footing ρ = 1/2 of ρ min = 0.002546
Dimension:
Shear, ф = 0.85 Length, = 1.20 m Ast = ρ W d = 516.33 mm²
L
steel cover, sc = 0.075 m Width, W = 0.28 m Qty = As / Ab = 5 L - 12mm RSBØ