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Max Max Max

This document summarizes the design of a concrete beam for a proposed two-storey, three-unit residence located in Paranaque City, Philippines. The design calculates the required steel reinforcement and stirrups based on the material strengths, shear and moment diagrams, and code requirements. At the supports, five 16mm diameter tension bars and three 16mm diameter compression bars are required. At midspan, three 16mm diameter tension bars and five 16mm diameter compression bars are needed, with 10mm diameter stirrups spaced at 270mm.

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Bryan Magnaye
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62 views2 pages

Max Max Max

This document summarizes the design of a concrete beam for a proposed two-storey, three-unit residence located in Paranaque City, Philippines. The design calculates the required steel reinforcement and stirrups based on the material strengths, shear and moment diagrams, and code requirements. At the supports, five 16mm diameter tension bars and three 16mm diameter compression bars are required. At midspan, three 16mm diameter tension bars and five 16mm diameter compression bars are needed, with 10mm diameter stirrups spaced at 270mm.

Uploaded by

Bryan Magnaye
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLS, PDF, TXT or read online on Scribd
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PROPOSED PROJECT: TWO-STOREY, THREE-UNIT RESIDENCE

LOCATION: YAKAL ST., PHASE 4, MARCELO GREEN VILLAGE, PARAÑAQUE CITY


OWNER: MS. TIA ROSE C. PERFECTO / MR. DON SANTIAGO C. PERFECTO / MR. JUNO C. PERFECTO

DESIGN OF BEAM
CONCRETE BEAM (2B1)
Material Strengths: Shear & Moment Diagram:
fc' = 21 Mpa
fy = 276 Mpa
Constant: *inclusive of wind and/or seismic load
β = 0.85 31.29 KN/m
Φ = 0.9
d = 350 mm 3.50 m
b = 150 mm
Assume 2 layer of steel bars: (d'=90)
d = 350 - 90 54.76
= 260 mm

Compute Pb 54.76
0.85 fc' β 600
Pb =
fy ( 600 - fy ) 22.36

0.85(21)(0.85)(600)
Pb =
(276)(600+276)

Pb = 0.0377

Compute Pmax 25.55 25.55


Pmax = 0.75 Pb
Pmax = 0.75 (0.0377)
Pmax = 0.0282

At the support: At midspan:


Mu = 25.55 KN,m Mu = 22.36 KN,m

Compute ω: Compute ω:
Mu = Φ fc' b d2 ω ( 1 - 0.59 ω) Mu = Φ fc' b d2 ω ( 1 - 0.59 ω)
25.554(1000000) = 0.9(21)(150)(260)(260)ω(1 - 0.59ω) 22.359(1000000) = 0.9(21)(150)(260)(260)ω(1 - 0.59ω)
0.59ω2 - ω = 0.1333 0.59ω2 - ω = 0.1167
0.59 0.59
ω2 - 1.69ω = 0.226 ω2 - 1.69ω = 0.198
Using pythagorean theorem Using pythagorean theorem
ω= 0.321 ω= 0.187

Compute P: Compute P:
ωfc' ωfc'
P= P=
fy fy
0.321 (21) 0.187 (21)
P= P=
276 276
P= 0.0244 < 0.0282 P= 0.0142 < 0.0282
Don't need compression bar Don't need compression bar

Compute area of steel (As): Compute area of steel (As):


As = Pbd As = Pbd
As = 0.0244(150)(260) As = 0.0142(150)(260)
As = 951.60 mm As = 554.90

Compute number of bar (N): Compute number of bar (N):


Try : 16 mm Ø bar Try : 16 mm Ø bar
As As
N= N=
(π/4)(d2) (π/4)(d2)
951.6 554.9
N= N=
(π/4)(16)(16) (π/4)(16)(16)
N= 4.733 N= 2.760

= say 5 bars = say 3 bars

Check for shear:


Vu = R - Wd
Vu = 54.7575 - (31.29)(260/1000) Compute spacing of stirrup:
Vu = 46.6221 KN Av fy d
S=
Vs
/fc' bd 157.08(276)(260)
Vc = S=
6 42190.16
/21 (150)(260) S= 267.17 mm = 270.00
Vc =
6
Vc = 29,786.74 KN d
Smax =
2 use 130 mm
ØVc 260
< Vu Smax =
2 2
0.85(29786.74) Smax = 130 mm = 130.00
< 46.62(1000)
2
12659.36 < 46622.10 KN DETAILS
It need stirrups
150 mm 150 mm
Vu
Vs = - Vc
Ø
46622.1
Vs = - 12659.3645
0.85 350 mm
Vs = 42190.16 KN

Use : 10 mm Ø bar: At support At midspan


2π d2
Av =
4 Tension Bar: 5 16mm dia. Bar 3
2π(10)(10) Comp. Bar: 3 16mm dia. Bar 5
Av =
4
Av = 157.08 mm2

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