Design of Columns and Beams for a Multi-Storey Building

This paper focuses on the design of columns and beams for a multi-storey building. It covers topics such as loading, slab design, moments, bending design, and shear reinforcement. Detailed calculations and recommendations are provided for each step of the design process.

CONCRETE STRUCTURES
By Name
Course
Instructor
Institution
Location
Date

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

( 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
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
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.
Beam
Transfer of Loads(Primary)

Load transfer figure from AS 3600 figure 7.3.4
Edge beams
Continuous Beam B1 (L Beam)
Given Details:
Beam Width , b=500 mm
Assuming beam depth, D=500 mm
Therefore:
d= D−cover −stirrup−halfthediameterofthebar
d=500−40−12− 16
2 =440 mm

Dead Load, G= 5 ×0.2 ×24
4 + ( 0.5× 0.3 ×24 )+ 1 ×1
4 =10 kPa
Live Load, Q=3.0 kPa
The load conditions that are considered:
w=1.2 G+ 1.5Q
W = ( 1.2× 10 ) + ( 1.5× 3 )
∴ W=16.5 kN /m
External Beam (continuous Beam)/Secondary

Continuous beam with 3 equal spans
Moment, M ¿=αF l2 where l is the clear length, 4500
Figure showing Factors of moment
Supports (A, B)
Mid-span (C, D)
Shear Force, V ¿=βFl
Figure showing Factors of shear force

α M ¿=αF l2
(kNm)
β V ¿=βFl
kN
A -1/16 -20.88 ½ 37.13
C +1/11 +30.38 1/7 10.61
B1 -1/10 -33.41 1.15/2 42.70
B2 -1/10 -33.41 ½ 37.13
D +1/16 20.88 1/8 9.28
Checking for the deflections:
By applying the deemed to comply ratio, according to AS3600, Cl.8.5.4
Lef
d ≤ [ K 1 ( Δ
L ef ) befEc
K 2 Fd , ef ] 1
3
Lef =LesserofLorLn + Ds
Where : ln=L−2 × ( 500
2 )=5000−500=4500 mm
Lef =Lesserof [ 5000 ; 4500+200 ]
¿ Lesserof [5000 ; 4700]

∴ Lef =4700 mm
∆
Lef = 1
250 Table 2.3.2
Ec=30100 MPa , when f ' c=32 MPa Table 3.1.2
Fd , ef =eff ective design service load :
Fd , ef = ( 1+Kcs ) G+ ( Ψs+ KcsΨL ) Q
Where:
Kcs=
[ 2−1.2 ( Asc
Ast ) ] ≥0.8 Cl.8.5.3.2
Kcs=2>0.8 Cl.4.2.2 – Table 4.1
Ψs=0.7and ΨL=0.4 (Floor for an office building)
∴ Fd , ef = (1+2 ) 10+¿
K1:
P=0.005¿ AS 3600
β= bef
bw
Where:
bw=500 mm
bef =bw+0.1 aCl.8.8b

a=0.7 L (For Continuous Beam)
∴ a=0.7 ×5000=3500 mm
So, bef =500+ ( 0.1 ×3500 )=850 mm
∴ β= 850
500 =1.7
So,
p=0.005>0.001 ( f ' c )
1
3
( β )
2
3
= 0.001× ( 32 )
1
3
( 1.7 )
2
3
=0.0022
Therefore: K 1= ( 5−0.04 f ' c ) p+0.0022 ≤ 0.1
( β )
2
3
K 1= ( 5−0.04 ×32 ) 0.005+0.0022 ≤ 0.1
( 1 ,.7 )
2
3
∴ K 1=0.0208<0.0702
K 2= 2.4
384
Lef
d ≤ [ K 1 ( Δ
L ef ) befEc
K 2 Fd , ef ] 1
3
4700
440 ≤
[ 0.0208 × ( 1
250 )× 850 ×30100
( 2.4
384 )× 34.5 ]1
3

15.2<21.4
∴ Hence ,the deflection ˇis ok
Bending Check
Ast=4 × ( π
4 ) × ( 16 ) 2=804 mm2(Hegger et al 2013)
Asc=4 × ( π
4 )× ( 16 )2 =804 m m2
f ' c=32 MPa
fsy=500 MPa
α 2=1−0.003 f ' c AS3600 Cl.8.1.3
¿ 1− ( 0.003× 32 )
¿ 0.904 say 0.85limits 0.67 ≤ α ≤ 0.85

γ=1.05−0.007 f ' c
¿ 1.05− ( 0.007 ×32 )
¿ 0.826 within the limits0.67 ≤ γ ≤ 0.85
So, α 2=0.85∧γ =0.826
Taking assumption on Tensile Steel yields:
C= ( α 2 f ' c )( γ Kud)bef
C=0.85 ×32 ×0.826 × 850× Kud=19097.12 Kud
T = Astfsy
T =804 × 500=402× 103 N
For equilibrium, C=T
19097.12 Kud=402× 103
∴ dn= Kud=21.1 mm
0.003
dn = ε st
d−dn
0.003
21.1 = ε st
440−21.1
ε st=0.06> ε sy=0.0025 Where: ε sy= ( fsy
Ec )= 500
200 ×103 =0.0025
∴ Therefore the steel yields .

Check Ductility:
Ku= dn
d = 21.1
440 =0.047<0.4
Ductility is ok
Lever Arm, z=d−( γdn
2 )=440− ( 0.826 ×21.1
2 ) =431.3 mm
Capacity, Mu=TZ
Mu= ( 402×103 ) × 431.3
¿ 173.38 kNm
ΦMu=0.8 ×173.38=139 kNm
∴ ΦMu=139 kNm> M ¿=30.38 kNm
Design for shear
W =1.2G+1.5 Q
W = ( 1.2× 10 ) + ( 1.5× 3 )=16.5 kN /m
V ∗max= WL
2 = 16.5 ×5
2 =41.3 kN
do=d=440 mm
∴ V ∗¿
3000−440 = V ∗max
3000 ¿
V ∗¿ 41.3 × 2560
3000 =35.24 kN

Crushing check Cl.8.2.6
Φ Vumax=Φ 0.2 f ' c bv do
ΦVumax=0.7× 0.2 ×32× 500 ×440=985.6 kN
∴ ΦVumax=985.6 kN >V∗¿ 57.3 kN
Adequate shear Dimensions
The crushing of Web is not that much critical
Checking of Shear Strength.
Concrete shear capacity, V uc=β1 β2 β3 bd 3
√f ' c
3
√ Ast
bd × 10−3
β1=1.1 (1.6− 440
1000 )=1.14 use 1.1
β2=1
β3=2 × do
av
=2 × 0.56
2.5 =0.45 say 1
3
√f 'c=3
√32=3.17< 4 MPa acceptable
V uc=1.1× 1× 1× 250× 440 ×3.17 × 3
√ 804
500 × 440 ×10−3=137.5 kN
Φ 0.5 Vuc=0.5 ×0.7 ×137.5=48.125 kN
Since V ∗¿ 57.3 kN >Φ 0.5 Vuc=48.125 kN

∴ shear reinforcement required
Minimum strength with minimum reinforcement for a Beam:
Cl.8.2.9
Vumin=Vuc+0.1 √f ' cbvdo ≥ Vuc+0.6 bvdo
Since 0.1 √ 32=0.566
∴ Vumin=Vuc+0.6 bvdo
Vumin=137500+ ( 0.6 ×500 × 440 ) =269500 N =269.5 kN
ΦVumin=0.7 × 269.5=188.35 kN
Since V ∗¿ 57.3 kN <ΦVumin=188.65 kN
No More Shear Reinforcement is required.
Minimum shear reinforcement: Cl.8.2.8
Asv . min= 0.06 √ f ' cbvs
fsy . f ≥ 0.35 bvs
fsy . f
0.35 bvs
fsy . f = 0.35 ×500 × s
500 =0.35 s
Therefore, Asv . min=0.35 s
 Using 2-legged fitments N12:
Asv=2 × ( π
4 )× ( 12 )2=226 mm2
∴ S= Asvmin
0.35 = 226
0.35 =645.7 mm

S . max=min [0.75 D; 500]
¿ min [(0.75 ×500);500 ]
¿ min [ 375; 500 ]
¿ 375 mm
S=645.7 mm>S . max=375 mm
∴ useN 12 fittments @300 mmspacing .
Hogging Moment check
Ast=4 × ( π
4 ) × ( 16 ) 2=804 mm2
Asc=4 × ( π
4 )× ( 16 )2 =804 m m2

f ' c=32 MPa
fsy=500 MPa
α 2=1−0.003 f ' c AS3600 Cl.8.1.3
¿ 1− ( 0.003× 32 )
¿ 0.904 say 0.85limits 0.67 ≤ α ≤ 0.85
γ=1.05−0.007 f ' c
¿ 1.05− ( 0.007 ×32 )
¿ 0.826 withinthelimits 0.67 ≤ γ ≤ 0.85
So, α 2=0.85 andγ =0.826
Assuming Compression steel does NOT yield:
dsc =cover +stirrup+halfthediameterofthebar
dsc =40+12+ 16
2 =60
Cc+Cs=T
( α 2 f ' c ) ( γdn ) b+ Ascσsc= Astfsy
 Where: σ sc=Es ε u ( dn−dsc
dn )
( 0.85 ×32 ×0.826 × 500 dn ) +
( 804 ×200 ×103 ×0.003 × ( dn−60
dn ) )=402×103
11234 dn+ 482400− ( 28944000
dn )=402× 103

11234 d n2 +482400 dn−402 ×103 dn−28944000=0
11234 d n2 +80400 dn−28944000=0
d n2 +7.2 dn−2576.46=0
d n=−(7.2)± √ ( 7.2 )2−4 ( 1 ) (−2576.46 ¿)
2(1) ¿
∴ dn=47.3 mm∨−54.5 mm
ε sc=0.003 ( 47.3−50
47.3 )=−0.000171< ε sy=0.0025
Assumptions are correct.
ε st=0.003 ( d−dn
dn )=0.003 ( 440−47.3
47.3 )
¿ 0.025>ε sy=0.0025
∴ ThereforeTensile steel yields .
Capacity, Mu=Cc ( d− γdn
2 ) +Cs(d −dsc)
Mu=11234 × ( 440− 0.826 ×47.3
2 ) +482400 × ( 47.3−60
47.3 ) ×( 440−60)
¿−44.5 kNm

Simply Supported Beam/ internal beam B2 (T-Beam)c
Given Details:
Beam Width, b=500 mm
Assuming beam depth, D=400 mm
Therefore:
d= D−cover−stirrup−halfthediameterofthebar
d=500−40−12− 16
2 =440 mm
Dead Load, G= 7 ×0.2 ×24
2 + ( 0.5× 0.3 ×24 ) + 1 ×1
2 =20.9 kPa
Live Load, Q=3.0 kPa
The load conditions that are considered:
w=1.2 G+ 1.5Q

W = ( 1.2× 20.9 ) + ( 1.5× 3 )
∴ W =29.58 kN /m
Deflection check: AS3600, Cl.8.5.4
Lef
d ≤ [ K 1 ( Δ
L ef ) befEc
K 2 Fd , ef ] 1
3
Lef =LesserofLorLn + Ds
Where : ln=L−2 × ( 500
2 )=5000−500=4500 mm
Lef =Lesserof [ 5000 ; 4500+200 ]
¿ Lesserof [5000 ;4700]
∴ Lef =4700 mm
∆
Lef = 1
250 Table 2.3.2
Ec=30100 MPa , when f ' c=32 MPa Table 3.1.2
Fd , ef =effectivedesignserviceload :
Fd , ef = ( 1+Kcs ) G+ ( Ψs+ KcsΨL ) Q
Where:
Kcs=
[2−1.2 ( Asc
Ast ) ]≥0.8 Cl.8.5.3.2

Kcs=2>0.8 AS 1170.0:2002 Cl.4.2.2 – Table 4.1
Ψs=0.7 Floor for an office building
ΨL=0.4
∴ Fd , ef = (1+2 ) 20.9+¿
In order to get the value of K1:
P=0.005 fromAS 3600
β= bef
bw
Where:
bw=500 mm
bef =bw+0.2 a Cl.8.8a
a=L ( forsimplysupportedbeam ) =5000 mm
So, bef =500+ ( 0.2 ×5000 )=1500 mm
∴ β= 1500
500 =3
So,
p=0.005>0.001 ( f ' c )
1
3
( β )
2
3
= 0.001× ( 32 )
1
3
( 3 )
2
3
=0.0015
Therefore:K 1= ( 5−0.04 f ' c ) p+0.002 ≤ 0.1
( β )
2
3

K 1= ( 5−0.04 ×32 ) 0.005+0.002 ≤ 0.1
( 3 )
2
3
∴ K 1=0.0206<0.0481
K 2= 2.4
384
Lef
d ≤ [ K 1 ( Δ
L ef ) befEc
K 2 Fd , ef ] 1
3
4700
440 ≤
[ 0.0206 × ( 1
250 )× 1500× 30100
( 2.4
384 )× 67.2 ]1
3
10.7<20.69
∴ This implies thaty the value of the deflection is ok.
Moment check

Ast=4 × ( π
4 ) × ( 16 ) 2=804 mm2
Asc=4 × ( π
4 )× ( 16 )2 =804 m m2
f ' c=32 MPa
fsy=500 MPa
α 2=1−0.003 f ' cAS3600 Cl.8.1.3
¿ 1− ( 0.003× 32 )
¿ 0.904 withinthelimits 0.67 ≤α ≤0.85
γ=1.05−0.007 f ' c
¿ 1.05− ( 0.007 ×32 )
¿ 0.826 withinthelimits 0.67 ≤ γ ≤ 0.85

So, α 2=0.85 andγ =0.826
Assuming Tensile Steel yields:
C= ( α 2 f ' c )( γKud)bef
C=0.85 ×32 ×0.826 × 1500× Kud=33700.8 Kud
T = Astfsy
T =804 × 500=402× 103 N
 For equilibrium, C=T
33700.2 Kud=402× 103
∴ dn= Kud=11.93 mm
0.003
dn = ε st
d−dn
0.003
11.93 = ε st
440−11.93
ε st=0.108> ε sy=0.0025 Where: ε sy= ( fsy
Ec )= 500
200 ×103 =0.0025
∴ Thereforethesteelyields.
Check Ductility:
Ku= dn
d = 11.93
440 =0.027< 0.4
Ductility is ok
Lever Arm:

z=d−( γdn
2 )=440− ( 0.826 ×11.93
2 ) =435 mm
Capacity, Mu=TZ
Mu= ( 402×103 )
¿ 174.9 kNm
ΦMu=0.8 ×174.9=139.9 kNm
∴ ΦMu=139.9 kNm> M∗¿ 84.13 kNm
Shear Check:
W =1.2G+1.5 Q
W = ( 1.2× 20.9 ) + ( 1.5× 3 ) =29.58 kN /m
V ∗max= WL
2 = 29.58 ×5
2 =73.95 kN
do=d=440 mm
∴ V ∗¿
3000−440 = V ∗max
300 ¿
V ∗¿ 73.95× 2560
3000 =63.104 kN
Check Web Crushing: Cl.8.2.6

ΦVumax=Φ 0.2 f ' cbvdo
ΦVumax=0.7× 0.2 ×32× 500 ×440=985.6 kN
∴ ΦVumax=985.6 kN >V∗¿ 63.104 kN
Section Dimension adequate for the shear
Web Crushing is NOT Critical
Shear Strength of Concrete: Cl.8.2.7.1
Concrete shear capacity, V uc=β1 β2 β3 bd 3
√f ' c
3
√ Ast
bd × 10−3
β1=1.1 (1.6− 440
1000 )=1.276>1.1 use 1.1
β2=1
β3=2 × do
av
=2 × 0.56
2.5 =0.45 say 1
3
√f 'c=3
√32=3.17< 4 MPa acceptable
V uc=1.1× 1× 1× 250× 560 ×3.17 × 3
√ 804
500× 440 × 10−3=137.5 kN
Φ 0.5 Vuc=0.5 ×0.7 ×137.5=48.125 kN
Since V ∗¿ 88.3 kN >Φ 0.5 Vuc=48.125 kN
Shear Reinforcement is required.
Minimum Strength of a beam with minimum reinforcement: Cl.8.2.9

Vumin=Vuc+0.1 √ f ' c bvdo ≥Vuc+ 0.6 bvdo
Since 0.1 √ 32=0.566
∴ Vumin=Vuc+0.6 bvdo
Vumin=137500+ ( 0.6 ×500 × 440 ) =269500 N =269.5 kN
ΦVumin=0.7 × 269.5=188.35 kN
Since V ∗¿ 88.3 kN <ΦVumin=188.65 kN
 No More Shear Reinforcement is required.
Minimum shear reinforcement: Cl.8.2.8
Asv . min= 0.06 √ f ' cbvs
fsy . f ≥ 0.35 bvs
fsy . f
0.35 bvs
fsy . f = 0.35 ×500 × s
500 =0.35 s
Therefore, Asv . min=0.35 s
 Using 2-legged fitments N12:
Asv=2 × ( π
4 )× ( 12 )2=226 mm2
∴ S= Asvmin
0.35 = 226
0.35 =645.7 mm
S . max=min [0.75 D; 500]
¿ min [(0.75 × 400); 500]

¿ min [ 300; 500 ]
¿ 300 mm
S=645.7 mm>S . max=300 mm
∴ Hence ,useN 12 fittments@ 300 mmspacing .
Column
Column 1 on the corner Column:
Slab=5× 1
4 ×0.2 ×24=6 kPa
Beam=0.5× 0.4 × 24=4.8 kPa
Column=0.5 × 0.5× 24=6 kPa
DeadLoad , G=16.8 kPa
LiveLoad , Q=3.0 kPa
Slab=5× 5 × 1
4 ×0.2 ×24=30 kN 30
Beam=0.5× 0.4 × 24 ×( 5−0.25−0.25)=21.6 kN
Column=0.5 × 0.5× 24 × 4=24 kN
LiveLoad , Q=3.0 ×5 ×1.0=15 kN 15
N¿=2(30+21.6+ 24+15)=181.2 kN

W =1.2G+1.5 Q
W = ( 1.2× 16.8 ) + ( 1.5× 3.0 )=24.66 kN /m
Nuo= N∗¿
0.8 =181.2
0.8 =226.5 kN ¿
M uo= M∗¿
0.8 = 40.6
0.8 =50.75 kNm ¿
Column 2:
Slab=5× 1
2 ×0.2 ×24=12 kPa
Beam=0.5× 0.4 × 24=4.8 kPa
Column=0.5 × 0.5× 24=6 kPa
DeadLoad , G=22.8 kPa
LiveLoad , Q=3.0 kPa
Slab=5× 5 × 1
2 ×0.2 ×24=60 kN
Beam=0.5× 0.4 × 24 ×( 5−0.25−0.25−0.25)=20.4 kN
Column=0.5 × 0.5× 24 × 4=24 kN

LiveLoad , Q=3.0 ×5 ×1.0=15 kN
N¿=2(60+20.4 +24+15)=241.2kN
W =1.2G+1.5 Q
W = ( 1.2× 22.8 ) + ( 1.5 ×3.0 ) =31.86 kN /m
Nuo= N∗¿
0.8 =1074.6
0.8 =302kN ¿
M uo= M∗¿
0.8 = 40.9
0.8 =51.125 kNm ¿
Given Details:
 Beam concrete compressive strength, f ' c=32 MPa
 fsy=500 MPa
 4N16 been used
 Ast=Asc=804 mm2
 500 ×500 mm, Tied columns is symmetrically reinforced
Column on 7th level

In situ concrete is taken as 90% of cylinder strength (f’c) and is included in the coefficient α 1 :
α 1=1−0.003 f ' cCl.10.6.2.2
¿ 1− ( 0.003× 32 ) =0.904 Within the limits 0.72 ≤ α 1≤ 0.85
∴ α 1=0.85

Ultimate strength in compression without bending, Nuo:

Nuo= ( α 1 f ' c ) bD + Ascfsy+ Astfsy
Nu o= ( 0.85× 32× 500× 500 ) +2 ( 804 ×500 )
∴ Nuo=7600 kN
Strength reduction factor (Φ ¿in compression:
Φ=0.6From Table 2.2.2(AS3600:2002)
ΦNuo=0.6 ×7600=4560 kN
∴POINT A: (7600 kN, 0 kNm)
In terms of Strength reduction factor:(Fernández et al 2012)
∴POINT A: (4560 kN, 0 kNm)
Zero Stress in Tension Bars

The parameters in stress block Cl.10.6.2.5
α 1=1−0.003 f ' cCl.10.6.2.5.1
¿ 1− ( 0.003× 32 ) =0.904 limits 0.67 ≤ α 1 ≤0.85
∴ α 1=0.85
γ=1.05−0.007 f ' cCl.10.6.2.5.2
¿ 1.05− ( 0.007 ×32 )=0.826 Within the limits 0.67 ≤ α 1 ≤0.85
∴ γ =0.826

 The zero value for tension implies that the axis which is neutral actually passes via the bars
with the tension.
dn=Kud
∴ Ku=1
ε st=fst =0
dn=D−dst=d=500−60=440 mm
From Strain Diagram:
ε sc
d−dsc = εu
dn
ε sc = εu
dn ( d−dsc )= 0.003
440 ( 440−60 ) =0.0026
The yield strain for N bars: ε sy = fsy
E = 500
200 ×103 =0.0025
∴ ε sc=0.0026>ε sy=0.0025 fsc=fsy=500 MPa
Ultimate Strength of eccentrically loaded Column, Nu:
Nu=Cc+Cs
Nu= ( α 2 f ' c ) ( γKudn ) b + Ascfsy
¿ ( 0.85 ×32 ×0.826 × 1.0× 440 ×500 ) + ( 804 ×500 )
∴ Nu=5345 kN
∴ ΦNu=0.6 × 5345=3207 kN

 To determine eccentricity (e), taking moments about the level of tensile steel:
Nu h=Cc (d− γ ku d
2 )+Cs ( d−d sc )
5345 h=4943 (440− 0.826 ×1.0 × 440
2 )+ 402 ( 440−60 )
→ h=267.43 mm
∴ e=h−( D
2 −dst )
¿ 267.43−( 500
2 −60)
∴ e=77.43 mm
M u=Nu e=5345 ×0.07743=413.86 kNm
Strength reduction factor (Φ ¿in compression:
When Nu ≥ Nub :Φ=0.6 From Table 2.2.2(AS3600:2002)
ΦMu=0.6 × 413.86=248.32kN
∴POINT B: (5345 kN, 413.86 kNm)
In terms of Strength reduction factor:
POINT B: (3207 kN, 248.32 kNm)
Balanced Failure

Balance Failure occurs when reinforcing steel reaches the specified yield strain at the same time
as the concrete strain reaches the limiting value of 0.003.
ε u=ε cu εst =ε sy → f st ¿ f sy
kub = 0.003
0.003+εsy
; ε sy =0.0025 forgradeNsteel
kub =0.6
kub d=0.6× 440=264 mm
ε sc
kub d−dsc = εu
kub d
→ εsc = ε u
kub d ( k ub d−dsc )
¿ 0.003
264 ( 264−60 ) =0.00231
The yield strain for N bars:
ε sc =0.00231<εsy=0.0025
σ sc =Es× ε sc
σ sc =200,000× 0.00231=462 MPa

From Equilibrium:
Nub=Cc+ Cs−T s
Nub= ( α2 f c
' ) ( γ k u d ) + Asc σ sc −Ast f sy
¿ ( 0.85 ×32 ×× 0.826 ×1.0 ×264 × 500 ) + ( 804 × 462 ) − ( 804 × 500 )
∴ Nub=2935 kN
∴ Φ Nub=0.6 × 2935=1761 kN
Determining e:
Nu h=Cc (d− γ ku d
2 )+Cs ( d−d sc )
2935 h=2966 (440− 0.826 ×264
2 )+371 ( 440−60 )
→ h=382.5 mm
∴ e=h−( D
2 −dst )
¿ 382.5−( 500
2 −60)
∴ e=192.5 mm
Mu=Nu e=2935 ×0.1925=565 kNm

Strength reduction factor (Φ ¿in compression:
When Nu ≥ Nub :Φ=0.6From Table 2.2.2(AS3600:2002)
ΦMu=0.6 ×565=339 kN
∴POINT C: (2935 kN, 565 kNm)
 In terms of Strength reduction factor:
∴POINT C: (1761 kN, 339 kNm)
Pure Moment Capacity, M uo ( Nu=0)

Approximate Method is used for this section. The assumption made by this particular method is
that it is possible to completely ignore the compressive contribution of the steel hence, Cs=0
From Equilibrium:
Cc=T s
α 2 f c
' (γdn) b= Ast f sy
dn= 804 ×500
0.85× 32× 0.826 ×500 =35.8 mm
M uo= Ast f sy (d − γdn
2 )=804 ×500 (440− 0.826× 35.8
2 )
∴ Muo=171 kNm
Strength reduction factor ( Φ ¿ related with the bending action in the absence of
compression or axial tension
Φ=0.8 From Table 2.2.2(AS3600:2002)
Φ Muo=0.6 × 171=102.56 kN
∴POINT D: (0 kN, 171 kNm)
 When considering the reduction factor of the strength:
∴POINT D: (0 kN, 102.56 kNm)
Interaction Diagram
The interaction diagram is based on the following:
 Pure Axial compression, Nuo when M =0 (Squash Load Point)
 Zero Stress in Tension Bars

 Balanced Failure State (Balanced Point)( Doran and Cather 2013)
 Pure Moment Capacity Muo when N=0(Pure Bending Point)
Construction of Interaction Diagram
POINT Mu (kNm) Nu(kN ) Comments
A 0 7600 The Axial compression is pure,
Mu=0
B 413.86 5345 There is Zero Stress in Tension
Bars
C 565 2935 This is Balanced Failure
Condition
D 171 0 There is Pure Moment Capacity,
Nu=0
Strength Interaction Diagram

0 100 200 300 400 500 600
0
1000
2000
3000
4000
5000
6000
7000
8000
7600
5345
2935
0
STRENGTH INTERACTION DIAGRAM
Moment capacity ( )𝑴𝒖 𝒌𝑵𝒎
Axial Force Capacity ( )𝑵𝒖 𝒌𝑵
Conclusion
The design took into consideration all the required standards. Although there was slight variation
of the actual figures from the expected digits, it is important to note that such differences were
very insignificant. Step by step calculation produced reliable values that were adoptable for the
construction process.

REFERENCES
Doran, D. and Cather, B. eds., 2013. Construction materials reference book. Routledge.
Fernández Carrasco, L., Torrens Martín, D., Morales, L.M. and Martínez Ramírez, S.,
2012. Infrared spectroscopy in the analysis of building and construction materials (pp. 357-372).
InTech.
Hegger, M., Auch-Schwelk, V., Fuchs, M. and Rosenkranz, T., 2013. Construction materials
manual. Walter de Gruyter.
Khatib, J. ed., 2016. Sustainability of construction materials. Woodhead Publishing.