Design of Columns and Beams for a Multi-Storey Building

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Introduction
The project aimed at the design of a building which was to be eight storeys in height.The
building was to be used as an office at every level and one of its level being used as the car par
just below the level of the street(Khatib 2016).The construction of the building had been
identified to be at the Central Business District of Sydney.This particular paper however has
focussed on the design of the columns that were to be located below the ground floor including
the beams and the columns as had been shared below

Figure 1: The section to be considered for the design purposes
DESIGN
Layout plan

Loading
Super imposed = 1.0kpa
Live Load= 3.0kpa
ρw =24+0.6 v assume v=0.5%
¿ 24 × 0.6 ×05=24.3
gk = ( 24.3× 0.2 ) +1 =5.86kpa
qk=¿ 3.0 kpa

Load combination
ULS=1.2 gk +1.5 qk
1.2 ×5.86+1.5 ×3=11.532
Slab
Panel 1 (corner slab)
2 adjacent continuous
Ly
Lx = 5000
5000 =1<2
Force design for every meter of the width noted as Fd:
Fd=1.2G+1.5 Q
Fd=1.2× 5.86+1.5 ×3=11.532kN /m
Minimum effective Depth for the deflection check
Lef
d ≤ K 3 K 4 [ ( ∆
Lef )Ec
Fd . ef ]1/ 3
Allowable deflection, ∆
Lef
= 1
250
Lef =LesserofLorLn + Ds

¿ Lesserof 5000∨4700
∴ Lef =4700 mm
k cs=2−1.2 Asc
Ast
≥ 0.8 say 2.0 Cl .8 .5 .3.2
ψs =0.7 , ψ1=0.4 (Building type-Office) From AS 1170.0:2002 Cl.4.2.2 –
Effective design service Load, F¿= ( 1.0+ kcs ) g+(ψs +k cs ψ1 ) q Cl.8.5.3.2
( 1.0+2.0 ) ×5.86+ ( 0.7+2.0 ×0.4 ) ×3=22.08 kN /m2
k3 =1.0 For two way slab
k 4=2.95
Ec=30100 case of f ' c=32 MPa
d ≥ Lef
k3 k 4
3
√ Ec ( ∆
Lef )
F¿
= 4700
1.0× 2.95
3
√ 30100 × 1
250
22.08 × 10−3
=90.5 mm
Effective depth, d= D−cover −0.5 main ̄diameter
d=200−20−0.5 ×12 where cover (A1) = 20mm, main bar=12mm
∴ dx=200−20−6=174 mm
d y=174−12=152 mm
∴ d x , d y <90.5 mm Acceptable

Moments
Ly/Lx =1.0
M*x = β 1 Fd Lx2 Cl.6.10.3.2
M*y = β 1 Fd Lx2
Coefficient βx=0.035
β y=0.035
Positive moment
M x
¿=0.035× 11.532× 52=10.09 kNm
M y
¿ =0.035 ×11.532 ×52=10.09 kNm
Negative moment (supports ) Cl.6.10.3.2 (B & C)
Interior support M x=1.33 × 10.09=−13.42 kNm
Exterior support M x=0.5× 10.09=−5.045 kNm
Bending design
ξ= α2 f 'c
f sy
α 2=1.0−0.003 f '
c
1.0−0.003× 32=0.904 say 0.85

γ=1.05−0.007 f '
c
1.05−0.007 ×32=0.91 say 0.85
assum ing ∅=0.8
ξ= α2 f 'c
f sy
=0.85 × 32
500 =0.0544
pt =ξ− √ξ2− 2 ξM
φb d2 f sy
Bottom
In the x- direction d x=174 mm
pt =0.0544− √0.05442− 2 ×0.0544 × 10.09× 106
0.85 ×1000 ×1742 × 500 =0.00078
In the y-direction d=152mm
pt =0.0544− √ 0.05442− 2× 0.0544 ×10.09 ×106
0.85 ×1000 ×1522 ×500 =0.0001
Top
In the x-direction
pt =0.0544− √0.05442− 2 ×0.0544 × 13.42× 106
0.85 ×1000 ×1742 × 500 =0.001
In the y-direction
pt =0.0544− √0.05442− 2× 0.0544 ×5.045 ×106
0.85 ×1000 ×1522 ×500 =0.0005