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Considered Section For Floor Beams

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Project Title: PROPOSED TWO-STOREY COMMERCIAL BUILDING

Owner: Heirs of Luis T. Yap


Location: Brgy. Zone II, Sogod, Southern Leyte
Civil Engineer:

CONSIDERED SECTION FOR FLOOR BEAMS:


COMPUTED MOMENTS:
MH1 = -125.55 KNm
MH2 = - 138.067 KNm
MH3 = -100.51 KNm
MH4 = -13.60 KNm

DESIGN OF BEAM(FB-1)
BEAM SPECIFICATIONS: Code: NSCP 2001
Effective Length = 4.6 m Width = 250 mm
Total Depth = 400 mm Min. Concrete Cover = 40 mm
Diameter of Stirrups = 10 mm Diameter of Bar (Assumed) = 16 mm
Effective Depth = 342 mm
fy = 275 Mpa
f'c = 20.7 Mpa

CONSIDERING BEAM H2-H3: Load, Shear & Moment Diagram


DESIGN FACTORED LOADS:
@ support Mu = 138.067 KN.m
@ midspan Mu = 75.748 KN.m
Vu = 152.207 KN
STEEL REINFORCEMENT @ SUPPORT:
Diameter(mm) Number of Deformed Bar
16 9
20 7

STEEL REINFORCEMENT @ MIDSPAN:


Diameter(mm) Number of Deformed Bar
16 5
20 4

MAXIMUM ALLOWABLE SPACING OF STIRRUPS = 85.5 mm

DESIGN OF BEAM(FB-2)
BEAM SPECIFICATIONS: Code: NSCP 2001
Effective Length = 2.65 m Width = 250 mm
Total Depth = 400 mm Min. Concrete Cover = 40 mm
Diameter Of Stirrups = 10 mm Diameter Of Bar (Assumed) = 16 mm
Effective Depth = 342 mm
fy = 275 Mpa
f'c = 20.7 Mpa
CONSIDERING BEAM H3-H4: Load, Shear & Moment Diagram

DESIGN FACTORED LOADS:


@ support Mu = 100.51 KN.m
@ midspan Mu = 14.192 KN.m
Vu = 107.94 KN
STEEL REINFORCEMENT @ SUPPORT:
Diameter(mm) Number of Deformed Bar
16 7
20 5

STEEL REINFORCEMENT @ MIDSPAN:


Diameter(mm) Number of Deformed Bar
16 3
20 2

MAXIMUM ALLOWABLE SPACING OF STIRRUPS = 171 mm

CONSIDERED SECTION FOR THE ANALYSIS OF ROOF BEAMS:


COMPUTED MOMENTS:
MJ1 = -96.51 KNm
MJ2 = -106.92 KNm
MJ3 = -75.68 KNm
MJ4 = -20.483 KNm

DESIGN OF BEAM(RB-1)
BEAM SPECIFICATIONS:
Effective Length = 4.6 m Width = 250 mm
Total Depth = 350 mm Min. Concrete Cover = 40 mm
Diameter of Stirrups = 10 mm Diameter of Bar (Assumed) = 16 mm
Effective Depth = 292 mm
fy = 275 Mpa
f'c = 20.7 Mpa
CONSIDERING BEAM J2-J3: Load, Shear & Moment Diagram

DESIGN FACTORED LOADS:


@ support Mu = 106.92 KN.m
@ midspan Mu = 59.092 KN.m
Vu = 123.48 KN

STEEL REINFORCEMENT @ SUPPORT:


Diameter(mm) Number of Deformed Bar
16 9
20 6

STEEL REINFORCEMENT @ SUPPORT:


Diameter(mm) Number of Deformed Bar
16 5
20 3

MAXIMUM ALLOWABLE SPACING OF STIRRUPS = 146 mm


DESIGN OF TWO-WAY SLAB
SLAB SPECIFICATIONS:
Effective Width(la) = 4.6 m Ratio la/lb = 1.08
Effective Length (lb) = 4.26 m Width = 1000 mm
Min. Concrete Cover = 20 mm Diameter of Bar (assumed) = 12 mm
fy = 230 Mpa
f'c = 20.7 Mpa

SLAB DIMENSION:
Design Thickness = 125.00 mm
Effective Depth = 87 mm
Tributary Width = 1000 mm

DESIGN LOADS:
Weight Of Slab = 2.94 KN/m Total Dead Load = 5.38 KN/m
Accessories = 2.44 KN/m Live Load = 1.90 KN/m
Wu = 10.77 KN/m

Location Mu Rn p use p As S (mm) use


S(mm)
@ continuous edge
short direction 11.85 1.74 0.00798 0.00798 693.93 162.98 162.98
long direction 8.40 1.23 0.00556 0.00609 529.57 213.57 213.00
@ discontinuous edge
short direction 1.56 0.23 0.00100 0.00609 529.57 213.57 213.00
long direction 1.82 0.27 0.00117 0.00609 529.57 213.57 213.00
@ mid-span
short direction 4.67 0.69 0.00304 0.00609 529.57 213.57 213.00
long direction 5.47 0.80 0.00357 0.00609 529.57 213.57 213.00

 Use 125mm thick slab with reinforce 12 mm dia. spaced @ 150mm O.C. BW
DESIGN OF FOOTING(F1)

FOOTING SPECIFICATIONS:
PDL: 210 Kn column size: 400 mm X 400 mm
PLL: 55 Kn
SBP: 100 Kpa
Bar Diameter: 16 mm
Concrete Cover: 75 mm
Embedment Depth: 1.5 m
f'c: 20.7 Mpa
fy: 275 Mpa

SOLUTION:
I. Assume wt. of footing to be 10% of column load

Weight of footing = 21 Kn
Total Axial Load = 286 Kn

II. Compute for the required Area

Required Area = Total load/SBP = 1.95 m2

L2 = 1.95 m2
L= 1.39 m say 1.50

Try 1.5mX 1.5m footing


A = 2.25 mm2
Pu = 493.90 Kn

III. Compute for net upward soil pressure

qu = 219.51 Kpa

check with the allowable ultimate soil pressure

qunet = 223.65 Kpa < qu ok!!

IV. Compute "d" for punching

Allowable punching shear,Vc = (1 + 2/Bc)f'c0.5/6 > 1/3f'c.05


Bc = long side of coumn / short side of column = 1
Use Vc = 1.517 Mpa

Actual Punching Shear Vn = Vu /Øbod, Ø = .85 for shear

Assume d = 300 mm
Vu = 414612.37 N
bod = 483204 mm2
Vn = 0.86 Mpa Ok! Vc > Vn
Use depth = 300.00 mm
deff = 201.00 mm
V. Check fo wide-beam shear

Allowable wide beam shear,Vc =1/6f'c0.05


allowable Vc = 0.76 Mpa
Vu = 136.97 Kn
Vn = 0.45 Mpa Ok! Vc > Vn

VI. Compute for steel reinforcement

Design Moment Mu = qu(B)(X2/2)


X = 550 mm
Mu = 49.80 Knm
A. Solve for Ru; Mu=.90Rubd2

Ru= 0.91 Mpa


m = fy/.85fc' = 15.63
B. Solve for p= (1/m)(1-(1-(2Rum)/fy)1/2)
fy = 275 mpa
f'c = 20.7 mpa
p = 0.00341
C. Check if the beam needs compression steel by computing pmax;
pmax=.75pb
pb= .85f'cB1600
fy(600+fy)

B1 = .85 for f'c=<30Mpa


B1 = .85-.008(f'c-30) for f'c>30Mpa
B1 = 0.85
pb = 0.03729226
pmax = 0.027969195
pmin=1.4/fy

pmin = 0.00509 If p=<pmax SRRB


If p>pmax DRRB
D. p < pmax (Singly Reinforced)
Check, pmin = 0.00509
p= 0.00341
If p>pmin use p
If p<pmin use p = pmin

Use p = 0.00509
Solve for the required As = pbd

A= 1534.91 mm2
Diameter Number of Deformed bar
16 8
20 5

 Use 1.5 X 1.5 X .30m square footing reinforcement 8 - 16mm dia. Bothways
DESIGN OF STAIRCASE
STAIR SPECIFICATIONS
Effective Length = 3.4 m Width = 1000 mm
Total Depth = 150 mm Min. Concrete Cover = 20 mm
Diameter Of Bar (Assumed) = 16 mm
Effective Depth = 106 mm

SLAB DIMENSION
Design Thickness = 150 mm
Tributary Width = 1000 mm

DESIGN LOADS
Weight Of Slab = 3.53 KN/m Total Dead Load = 4.97 KN/m
Accessories = 1.44 KN/m Live Load = 4.80 KN/m
Wu = 15.12 KN/m
Mu = 21.85 KN.m
SOLUTION:
I. Compute for longitudinal steel reinforcement
A. Solve for Ru; Mu=.90Rubd2
Ru= 2.16 Mpa
m = fy/.85fc' = 13.07
B. Solve for p= (1/m)(1-(1-(2Rum)/fy)1/2)
p = 0.01005
C. Check if the beam needs compression steel by computing pmax;

pmax=.75pb
pb = 0.85f'cB1600
fy(600+fy)

B1 = .85 for f'c=<30Mpa


B1 = .85-.008(f'c-30) for f'c>30Mpa
B1 = 0.85
pb = 0.04701
pmax = 0.03525
pmin=1.4/fy
pmin = 0.00609 If p=<pmax SRRB
If p>pmax DRRB
D. p < pmax (Singly Reinforced)
Check, pmin = 0.00609
p= 0.01005
If p>pmin use p
If p<pmin use p = pmin
Use p = 0.01005
Solve for the required As = pbd
A = 1065.73 mm2 As = 300 mm2
Spacing of reinforcement S = Ao(1000)/As
S = 188.66 mm
Say 200mm
Temperature Bars As = 0.002bt
As = 300 mm2
S = 670.21 mm
Say 700 mm

 Use 150mm thk/ Slab with reinf. 16mm dia. Spaced @ 200mm O.C

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