Western Sydney University: Concrete Structures Major Design Project

This project focuses on the structural design of an eight-story office building located in Sydney's Central Business District, with a car park below street level. The assignment involves the detailed design of building elements including ground floor slabs, columns below the ground floor, and beams. The design process encompasses load calculations, considering dead, live, and wind loads, along with the application of relevant Australian standards (AS 3600, AS 1170). The analysis includes slab design for both corner and edge slabs, considering bending moments, shear forces, and reinforcement spacing. Beam design covers both continuous and simply supported beams, determining bending moments, shear forces, and deflection checks. Column design involves the analysis of axial compression, bending moments, and the creation of interaction diagrams to ensure structural integrity. Step-by-step calculations were performed to produce reliable values that are adoptable for the construction process.

CONCRETE STRUCTURES
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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 caseof 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 Mx=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
α2=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

ptmin =0.2 ( D
d )2 f 'cs
f sy
ptmin =0.2 ( 200
174 )
2 0.6 × √ 32
500 =1.793 ×10−3
pt < ptmin ∴ use ptmin
In the x-direction
Ast= ptmin ×bd 1.793 ×10−3 ×1000 ×174=312 mm2 /m
∴ provide N 12 @340 mm spacing340 m m2 /m
Checking spacing Cl.9.4.1b
Spacing ¿ Lesserof [ 2 D ; 300 ] mm
¿ Lesserof [400 ; 300]
Therefore use minimum spacing of 300 mm
∴ provide N 12 @300 mm spac ing 367 mm2 /m
∴ Therefore the spacingis okCheck the transverse steel for cracking control
In the y-direction
Ast= ptmin ×bd 1.793 ×10−3 ×1000 ×152=273 mm2 /m
∴ provid e N 12 @ 400mm spacing 275m m2 /m

Checking spacing Cl.9.4.1b
Spacing ¿ Lesserof [ 2 D ; 300 ] mm
¿ Lesserof [400 ; 300]
Therefore, use minimum spacing of 300 mm
∴ provide N 12 @300 mm spacing367 m m2 /m
∴ Therefore the spacingis okCheck the transverse steel for cracking control
Location direction pt Ast N12@ s Ast actual position
Supports X 1.793 ×10−3 312 300 367 top
Mid-
region
X 1.793 ×10−3 312 300 367 bottom
supports Y 1.793 ×10−3 273 300 367 top
Mid-
region
Y 1.793 ×10−3 273 300 367 bottom
Check shear
V* = LyFd
2 = 5 ( 11.532 )
2 =28.83 KN
Vuc=β 1 β 2 β 3 bvd 0 fcv ( Ast
bvdo )
1 /3
Cl.8.2.7
β 2=β 3=1

do=200−20−6=174 mm β 1=1.1 (1.6− do
1000 )≥ 1.1
∴ β 1=1.1 (1.6− 174
1000 )=1.67>1.1
bv=1000 mm
Ast = 367 mm2/m
f ' cv = ( f ' c )
1
3 = ( 32 )
1
3 =3.17 MPa< 4 MPa
∴ Vuc=1.59 ×1.0 × 1.0× 1000× 154 ×3.17 ( 367
1000 ×154 )1
3 =103.8 kN
0.5 ∅ Vuc=0.5 ×0.7 × 103.8=36.33 kN ∅ =0.7 Table 2.2.2
∴ Vuc=¿36.33 KN¿ V∗¿ 28.83 kN
Therefore, no shear reinforcement required.
Slab
Panel 2 (edged slab)
2short edge discontinuous
Ly
Lx = 5000
5000 =1<2 Two way slab

Effective depth, d= D−cover −0.5 main ¯diameter
d=200−20−0.5 ×12 where cover (A1) = 20mm, main bar=12mm
design force per meter width, 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 Table 2.3.2
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 (Office building) From AS 1170.0:2002 Cl.4.2.2 – Table 4.1
Effective design service Load, F¿= ( 1.0+kcs ) g+(ψs +k cs ψ1 ) q Cl.8.5.3.2