Detailed Structural Design and Analysis of Slabs, Beams, and Columns

This document presents a comprehensive structural design analysis, covering the design of slab panels, beams, and columns. The slab design includes calculations for both corner and edged slabs, detailing load combinations, effective depths, bending moments, and reinforcement requirements for both top and bottom steel, as well as shear checks. Beam design focuses on continuous and simply supported beams, incorporating deflection checks, bending moment calculations, shear force analysis, and the determination of required reinforcement. The column design involves the analysis of axial compression, bending capacity, and the construction of an interaction diagram to assess the structural behavior under combined loading conditions. The analysis adheres to relevant Australian Standards (AS 3600) and includes detailed calculations for various load cases, ensuring structural integrity and compliance with design codes. The document provides a thorough examination of each structural element, offering insights into the design process and essential considerations for civil engineering applications.

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

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
( 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 condition 6 Of Table 9.3.4.2
Ec=30100 caseof f ' c=32 MPa Table 3.1.2

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 Table 6.10.3.2(A)
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
assuming ∅ =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 spacing367 m m2 /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
∴ provide N 12 @ 400 mm spacing 275 mm2 /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
( 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 condition 6 Of Table 9.3.4.2
Ec=30100 caseof f ' c=32 MPa Table 3.1.2
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 Table 6.10.3.2(A)
β y=0.035
Positive moment

M x=M y=0.034 ×11.532 ×52 =9.8 kNm
Negative moment
Interior support M x=1.33 × 9.8=−13.04 kNm
Exterior support M x=0.5× 9.8=−4.9 kNm
Bending design
pt =ξ− √ξ2− 2 ξM
ϕb d2 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
assuming ∅ =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 × 9.8× 106
0.85 ×1000 ×1742 × 500 =0.00078
In the y-direction d=152mm
pt =0.0544− √ 0.05442− 2× 0.0544 ×9.8 × 106
0.85 ×1000 ×1522 ×500 =0.0001
Top
In the x-direction
pt =0.0544− √0.05442− 2× 0.0544 ×13.04 × 106
0.85 ×1000 ×1742 × 500 =0.001
In the y-direction
pt =0.0544− √0.05442− 2× 0.0544 ×5.4 × 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