Structural design of a four storey office building

1
Dongyezhan Liu 1421941
Faculty of Technology and Built Environment
DEPARTMENT OF CIVIL ENGINEERING
ENGINEERING DEVELOPMENT PROJECT (ENGG
MG7001/MG7101)
Structural design of a four storey
office building
Dongyezhan Liu
Disclaimer: Dongyezhan Liu
This document is a report on << Proposal for Structural design of a four storey office
building>> that was carried out as part of a student learning exercise. It has been marked and
awarded a grade. However, regardless of the grade awarded, there is no guarantee that the
contents of this document will be free from errors, inconsistencies, or discrepancies. While this
document may contain findings and recommendations that could be of use to the client, or indeed
anyone else reading this report, neither Unitec, the author of this report, nor any of the persons
mentioned under the Acknowledgements section of this document, shall bear any responsibility or
liability; should the client or anyone else, upon implementing the design or utilising any of the
findings and recommendations contained within this document, incur any harm, damage, liability,
injury, or any other kind of loss whatsoever.
October 2016

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Permission Statement
I hereby, give ------- OR don’t give -------- permission for my research/design project report
entitled:
_____________Structural design of a four storey office building
to be held in the Unitec Library.
Course: ___________Bachelor of Engineering Technology (Civil) ________
Year of completion: _____2016___
Department/School of _________Engineering_________________
I agree to this research being consulted for research or study purposes only provided that due
acknowledgement of its use is made where appropriate and any copying is made in
accordance with the Copyright Act 1994.
I agree to this research being available for interlibrary loan.
I have made all efforts to ensure that the information contained in the report is accurate. I
will not be held responsible for any inaccuracies. Also, I agree to this work being copied for
archive, external moderation, monitoring, promotional and future learning purposes.
Name: ________Dongyezhan Liu______
Signed: _______Dongyezhan Liu_______ Date: _____10/10/2016______
✓

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Declaration for Ethical Approval of Research/Design Project
I ________Dongyezhan Liu_________ declare that this research or design project:
(Student’s full name)
Either
 Does not involve humans as participants?
Or
 Has ethical approval from UREC (as included in this project report)?
Signed: ________Dongyezhan Liu_______ Date: _____10/10/2016______
(Signature)

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Executive Summary
The design of this project is about a 24×30 mm2
four storey reinforced concrete building
located on Auckland centre region. This project is focusing on structural frame analyses and
critical beams and columns design based on calculating the permanent, imposed and
Earthquake actions through utilizing Multi-frame program. While this project is in process, to
get start with the design ideas, the New Zealand standards, design concept, structural factors
and possible procedures relating to this project have been studied and learned to get
methodology and literature view done. The basic theory of this project is to design the
actions based on concrete frame. Meanwhile, the maximum bending moment, shear force
and axial load diagram are analysed through the different situations from combinations of
actions. The critical columns and beams design are undertaken from the BM, SF and AL
diagram. The capacity of beams and columns will be counted. All of these are compliant with
NZS 1170 series -Structural design actions and NZS 3101:2006, Concrete Structures Standard.
The detailed drawings are included.
Acknowledgement
I would like to express my heartfelt thanks to my tutor Dr. Sherif Beskhyroun. This project
cannot be successfully completed without his great guidance and patience. In the process I
deal with this project, I have received many constructive suggestions and effective feedbacks
from him.
Finally, I would like to thank my parents and family for all the supports throughout this
project.

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Table of Contents
1. Introduction........................................................................................................................8
1.1. Description of the building..........................................................................................8
1.2. Literature views.........................................................................................................10
1.3. Objectives..................................................................................................................11
1.4. Reinforced concrete..................................................................................................11
1.5. Structural considerations and assumptions..............................................................12
1.5.1. Gravity load........................................................................................................12
1.5.2. Material properties............................................................................................13
1.6. Durability...................................................................................................................14
1.7. Fire resistance ...........................................................................................................14
2. Methodology.....................................................................................................................14
2.1. New Zealand standards.............................................................................................14
2.2. Multi-frame ...............................................................................................................15
3. Frame building ..................................................................................................................15
3.1. Frame building...........................................................................................................15
3.2. Preliminary member sizes.........................................................................................16
4. Actions ..............................................................................................................................17
4.1. Longitudinal actions ..................................................................................................17
4.1.1. Permanent actions G .........................................................................................17
4.1.2. Imposed action Q...............................................................................................18
4.2. Transversal action .....................................................................................................19
4.2.1. Permanent actions G .........................................................................................19
4.2.2. Imposed action Q...............................................................................................20
4.3. Earthquake action .....................................................................................................21
5. Structure analysis .............................................................................................................25
5.1. Longitudinal section ..................................................................................................26
5.2. Transversal section....................................................................................................29
6. Beams Design....................................................................................................................31
6.1. Design Beams for longitudinal section......................................................................31
6.1.1. Roof level (Reinforcement bars)........................................................................32

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6.1.2. Roof level (Stirrups) ...........................................................................................37
6.1.3. Level 1-3 (Reinforcement bars)..........................................................................39
6.1.4. Level 1-3 (Stirrups).............................................................................................44
6.2. Design Beams for transversal section .......................................................................46
6.2.1. Roof level (Reinforcement bars)........................................................................46
6.2.2. Roof level (Stirrups) ...........................................................................................52
6.2.3. Level 1-3 (Reinforcement bars)..........................................................................54
6.2.4. Level 1-3 (Stirrups).............................................................................................59
7. Column Design..................................................................................................................61
7.1. Marginal columns (Longitudinal section)..................................................................61
7.1.1. Roof level ...........................................................................................................61
7.1.2. Level 3 ................................................................................................................63
7.1.3. Level 2 ................................................................................................................65
7.1.4. Level 1 ................................................................................................................67
7.2. Marginal columns (Transversal section) ...................................................................69
7.2.1. Roof level ...........................................................................................................69
7.2.2. Level 3 ................................................................................................................71
7.2.3. Level 2 ................................................................................................................73
7.2.4. Level 1.....................................................................................................................76
7.3. Internal columns .......................................................................................................78
7.3.1. Roof level ...........................................................................................................78
7.3.2. Level 3 ................................................................................................................79
7.3.3. Level 2 ................................................................................................................81
7.3.4. Level 1 ................................................................................................................82
8. Conclusions.......................................................................................................................83
9. References........................................................................................................................84
10. Appendices....................................................................................................................84
10.1. Longitudinal section combinations........................................................................84
10.1.1. 1.35G ..............................................................................................................84
10.1.2. 1.2G+1.5Q1 .....................................................................................................86
10.1.3. 1.2G+1.5Q2 ....................................................................................................87
10.1.4. 1.2G+1.5Q1+1.5Q2.........................................................................................89
10.1.5. G+Eu+𝜑𝑐𝑄......................................................................................................90

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10.2. Transversal section................................................................................................92
10.2.1. 1.35G ..............................................................................................................92
10.2.2. 1.2G+1.5Q1 ....................................................................................................93
10.2.3. 1.2G+1.5Q2 ....................................................................................................95
10.2.4. 1.2G+1.5Q1+1.5Q2.........................................................................................96
10.2.5. G+Eu+𝜑𝑐𝑄......................................................................................................98
Figure 1 longitudinal section ......................................................................................................9
Figure 2 Transversal section.......................................................................................................9
Figure 3 Plan view ....................................................................................................................10
Figure 4 Table 3.2 from the NZS1170.1 structural design action.............................................12
Figure 5 Table 3.1 from NZS1170.0 general principles.............................................................13
Figure 6 Table3.6 from NZS3101:2006 Part1 ...........................................................................14
Figure 7 Floor plan....................................................................................................................16
Figure 8 Clause 9.4.1.2 Beams with rectangular cross sections （NZS 3101）.......................16
Figure 9 Permanent actions G of longitudinal section .............................................................18
Figure 10 Imposed actions Q of longitudinal section ...............................................................19
Figure 11 Permanent actions G of transversal section.............................................................20
Figure 12 Imposed actions Q of transversal section.................................................................21
Figure 13 Earthquake actions (Longitudinal section) ...............................................................24
. Figure 14 Deflection due to Earthquake actions (Longitudinal section).................................24
Figure 15 Earthquake actions (Transversal section).................................................................25
Figure 16 Deflection due to Earthquake actions (Transversal section)....................................25
Figure 17 Maximum bending moment diagram for longitudinal section.................................26
Figure 18 Maximum shear force diagram for longitudinal section ..........................................27
Figure 19 Maximum axial load diagram for longitudinal section .............................................28
Figure 20 Maximum bending moment diagram for transversal section ..................................29
Figure 21 Maximum shear force diagram for transversal section............................................30
Figure 22 Maximum axial load diagram for transversal section...............................................31
Figure 23 Maximum bending moment diagram for roof..........................................................32
Figure 24 Maximum negative bending moment......................................................................33
Figure 25 Maximum positive bending moment .......................................................................34
Figure 28 Maximum bending moment diagram for level1-3 ...................................................39
Figure 33 Maximum bending moment diagram for roof （transversal section） ..................46
Figure 34 Maximum negative bending moment for transversal sction ...................................47

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Figure 35 Maximum positive bending moment for transversal section...................................48
Figure 36 Maximum negative bending moment for transversal section .................................49
Figure 38 Maximum bending moment diagram for level 1-3 (transversal section) .................54
Figure 43 Maximum shear force diagram for level 1-3 (transversal section)...........................59
Figure 44 Columns design for the whole building....................................................................61
1.Introduction
1.1. Description of the building
In recent years, Auckland city has been growing rapidly. And, more and more people decide
to work in the CBD. Therefore, there is a need to construct more office building regarding
market demand. My project is to design a four storey reinforced concrete building in central
Auckland.
In my design, the structural design will be divided mainly into two parts: manual structural
analyses and finite program (Multi-frame 2D). In other words, this design will focus on slabs,
columns, and beams. This building is 24*30m. And, the story height is 4.5m. The strand
height will be 3.5m for each level. The short edge will have 3 bays. Therefore, for each bay

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will be 8*10m in a rectangular shape. This building could be demonstrated in Multi-frame 2D.
Figure 1 longitudinal section
Figure 2 Transversal section

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Figure 3 Plan view
In the design part, the earthquake and wind actions will be considered in order to design the
resistance of reinforced concrete frame.
For more details, the capacity of the maximum load will be required to calculate to analyse
bending moment, shear force and axial force at critical sections under the ultimate strength
design. Due to safety consideration, there are two vital elements related to against failures.
This consideration includes serviceability limit state and ultimate limit state. The serviceability
limit state is to remain the elastic ensuring that the durability of the structures is on the
allowed range of normal working conditions. The ultimate limit state is providing ductility
prevented collapsed.
All the elements will be concerned with New Zealand concrete standards and structural
design action standards.
1.2. Literature views
The design of structures included the related elements are required to meet the standard for
stability, stiffness, strength, ductility, durability, robustness and fire resistance. (NZS3101:
2006)

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Reinforced concrete is one of the most widely used composite materials in modern building
constructions. It utilizes the concrete in resisting compression forces, and steel bars or wires,
to resist reasonable tension forces. (Noel, 1993)
For the earthquake actions, it is defined in the equivalent static forces. To analyse the
equivalent static forces, each level of the structure needs to be acted simultaneously. In this
part, the horizontal seismic shear will be considered for calculating the combination of
horizontal design action coefficient and total seismic weight of the building. (NZS170. 5:2002)
To satisfy the static equilibrium of the horizontal forces for each level of the building, the
shear force, V, is required to be equal in magnitude and opposite in direction to the full set of
equivalent static lateral forces acting at various heights above ground level (Lusa, 2016)
The destruction of the building of reinforced concrete elements in flexure could exist in 3
ways, tension, compression and balanced failure. They have been dictated by the volume of
longitudinal reinforcement in tension. (Lusa, 2016) (p=As/(b×d)) In this equation, the ratio of
reinforcement equals to the area of tension steel divided by effective section area of
concrete. (NZS3101: 2006)
The shear walls can be very efficient in resisting lateral loads originating from wind or
earthquakes. Well-designed shear walls in seismic areas can provide adequate structural
safety and give a great measure of protection against costly non-structural damage during
moderate seismic disturbances. (Park & Pauley, 1975)
1.3. Objectives
In my project, the design will be undertaken by the following approach:
Design the permanent and imposed actions
Design the earthquake and wind actions
Analyse the maximum axial force, shear force and bending moment using the multi-frame 2D
Design the specified columns
Design the specified beams
1.4. Reinforced concrete
It is most widely used materials in the world especially in construction (economic building
materials)
The maintenance cost of reinforced concrete is very low. And, in structure like footings,
dams, piers etc. reinforced concrete is the most economical construction material. Compared
to the use of steel in structure, reinforced concrete requires less skilled labour for the
erection of structure. Furthermore, Reinforced concrete, as a fluid material in the beginning,
can be economically moulded into a nearly limitless range of shapes.
It has great fire and weather resistance than other normal materials. (Such as: steel and
timber)

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Reinforced concrete has a high compressive strength and adequate tensile strength
compared to other building materials. Due to the provided reinforcement, reinforced
concrete can also withstand a good amount tensile stress.
1.5. Structural considerations and assumptions
1.5.1. Gravity load
 Permanent load
Double Tee Floors – 3.79 Kpa
Ceiling and services – 0.3 Kpa
Partition wall – 0.4 Kpa
 Imposed load
The purpose of this building is used for office building only. The heavy loading duty cannot be
applied on the level of building.
The roof level is assumed as other floors.
Figure 4 Table 3.2 from the NZS1170.1 structural design action
 Earthquake load
Important level =3 as the office building is high consequence for loss of human life.

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Figure 5 Table 3.1 from NZS1170.0 general principles
Z = 0.13 (Auckland region)
Site subsoil class C (shallow soil)
Design working life = 50 years, according to NZS3101:2006 Part1, The provisions of this
section shall apply to the detailing and specifying for durability of reinforced and pre-stressed
concrete structures and members with a specified intended life of 50 or 100 years.
Compliance with this section will ensure that the structure is sufficiently durable to satisfy the
requirements of the NZ Building Code throughout the life of the structure, with only normal
maintenance and without requiring reconstruction or major renovation. The 50 years
corresponds to the minimum structural performance life of a member to comply with that
code.
Limited ductile frame for the structure design
 Wind load
Due to the non-critical case compared with the earthquake load, the wind load is not took
into account in this project.
1.5.2. Material properties
-
Concrete density 𝜌𝑐 =24 kn/m3
- Concrete compressive strength f 𝑐
′
=30 Mpa (not less than 25MPa and not greater
than 100MPa)
- Modulus of elasticity for concrete Ec = (fc’)1/2 * 3200 + 6900 (25084MPa)
- Modulus of Elasticity for steel Es = 200000 MPa
- Main bending& shear reinforcement bar steel Grade 500E
- Main stirrups steel Grade 500

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1.6. Durability
According to NZS3101:2006 Part1, this project building is located at non-aggressive soil with
A2 exposure classification and is situated at protected environment.
Due to the standard, for fc’ = 30Mpa, the minimum required concrete cover is set up as
30mm for beams and columns.
Figure 6 Table3.6 from NZS3101:2006 Part1
1.7. Fire resistance
The fire resistance is assumed as 60 minutes throughout the office building.
2.Methodology
All the procedures of this project are analysed, calculated and considered through the New
Zealand Standards. For the details, auto CAD will be use to demonstrate specified drawings.
And, the analyses of the maximum of axial forces, shear forces and bending moments will be
utilized in Multi-frame 2D.
2.1. New Zealand standards
NZS1170: 2002 part 0: General principles
It provides general procedures and criteria for the structural design of a building or structure
in the limit states format. It covers limit states design, actions, and combinations of actions,
methods of analysis, robustness and confirmation of design. The objective of this standard is
to provide designers with general procedures and criteria for the structural design of
structures. It outlines a design methodology applied in accordance with established
engineering principles.

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NZS1170: 2002 part 1: Permanent, imposed and other actions
It specifies the varied situations in terms of the limit state design of structures used by
permanent, imposed, liquid pressure, ground water, rainwater pounding and earth pressure
actions.
NZS1170: 2002 part 2: Wind actions
It provides that the procedures to determine the winds speed resulting in any directions of
the building. Also, it shows that the criteria of wind actions undertaken in reasonable
structural design between different wind zone. (Except tornadoes)
NZS1170: 2002 part 5: Earthquake actions- New Zealand
It establishes the requirements of structural design for period of vibration, horizontal seismic
shear and equivalent static horizontal force.
NZS3101: 2006 Concrete structures standard (Part1- The design of concrete structures)
It outlines the verified methodology and compliant criteria of designing reinforced and pre-
stressed concrete structures with New Zealand Building Code. Additionally, it also has
summary tables to guide engineers design from aspects of beams, columns and connections.
2.2. Multi-frame
The Multi-frame program is the basic software which has been taught in the Unitec structure
class to design the structural frame. It includes the majority of materials about beams and
columns section. In this software, the permanent, imposed and earthquake action could be
inputted onto the different level of building. Also, there is possible to put all the basic static
actions into combination static actions to prepare the various situations. Also, the maximum
bending moment, shear force and axial load could be analysed for calculating the capacity for
critical beams and columns.
3. Frame building
3.1. Frame building
The frame is illustrated as below.

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Figure 7 Floor plan
3.2. Preliminary member sizes
In the AS/NZS 3101: part 1:2006 concrete standard, the table 2.1 & the clause 9.4.1. Design
of reinforced concrete beams will be used as main design principles for my project.
According to this:
Figure 8 Clause 9.4.1.2 Beams with rectangular cross sections （NZS 3101）
The depth, width and clear length between the faces of supports of members with
rectangular cross sections, to which moments are applied at both ends by adjacent beams,
columns or both, shall be such that:

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𝐿𝑛
𝑏𝑤
≤ 25
𝐿𝑛ℎ
𝑏 𝑤
2
≤ 100
Therefore, the beam sizes will be undertaken as:
From Table 2.1, fy= 500Mpa one end continuous
h ≥
𝐿
20
=
10000
20
= 500𝑚𝑚
fy=500Mpaboth end continuous h=600. 700, 800mm
h ≥
𝐿
25
=
10000
25
= 400𝑚𝑚
Design beams: 350×800 mm
mm
h
bw 400
2

mmbbb wce 5005050 
mm
bb
LLn ce
950050010000
22

Check: 2575.23
400
9500n

wb
L
OK
Check: 10078.20
nh
2

wb
L
OK
Check: 3
350
800
232 
b
h
OK
For the column sizes, from clause 10.4. Dimensions of columns and piers.
mmbbb wce 5005050 
Therefore, columns will be: 500×500mm
4.Actions
4.1. Longitudinal actions
4.1.1. Permanent actions G
Slabs

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Double Tee Floors = 3.79Kpa
Ceiling & services = 0.3Kpa
Partition wall= 0.4Kpa
mknGfloors /92.35879.34.03.0  ）（
Beams
6.72kn/m240.80.351beams G
Columns
mG /kn6245.05.01columns  Numbers of columns: 4×4=16
Weight:
Roof: 6 ×
3.5
2
= 10.5𝑘𝑛 total: 10.5×16= 168kn
Level3: 6 × （
3.5
2
+
3.5
2
） = 21𝑘𝑛 total: 21×16= 336kn
Level2: 6 × （
3.5
2
+
3.5
2
） = 21𝑘𝑛 total: 21×16= 336kn
Level1: 6 × （
3.5
2
+
4.5
2
= 24𝑘𝑛 total: 24×16= 384kn
Glazing& Certain wall
Gglazing=0.5Kpa
Figure 9 Permanent actions G of longitudinal section
4.1.2. Imposed action Q
For roof level:

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𝜑 𝑒 =
1.8
𝐴
= 0.12 = 0.122 ⟹ 𝜑 𝑒 = 0.25𝑘𝑝𝑎 (𝜑 𝑒 ≤ 0.25)
A=10×8=80m2
For level3, 2, & 1:
𝑄 𝑢 = 1 × 3 = 3𝑘𝑝𝑎
𝜑 𝑎 = 0.3 +
3
√𝐴
= 0.41 ⟹ 𝜑 𝑎 = 0.635𝑘𝑝𝑎 (0.5 ≤ 𝜑 𝑎 ≤ 1)
Figure 10 Imposed actions Q of longitudinal section
4.2. Transversal action
4.2.1. Permanent actions G
Slabs
Double Tee Floors = 3.79Kpa
Ceiling & services = 0.3Kpa
Partition wall= 0.4Kpa
mknGfloors /9.441079.34.03.0  ）（
Beams
6.72kn/m240.80.351beams G
Columns
mG /kn6245.05.01columns  Numbers of columns: 4×4=16
Weight:
Roof: 6 ×
3.5
2
= 10.5𝑘𝑛 total: 10.5×16= 168kn

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Level3: 6 × （
3.5
2
+
3.5
2
） = 21𝑘𝑛 total: 21×16= 336kn
Level2: 6 × （
3.5
2
+
3.5
2
） = 21𝑘𝑛 total: 21×16= 336kn
Level1: 6 × （
3.5
2
+
4.5
2
= 24𝑘𝑛 total: 24×16= 384kn
Glazing& Certain wall
Gglazing=0.5Kpa
Figure 11 Permanent actions G of transversal section
4.2.2. Imposed action Q
For roof level:
𝜑 𝑒 =
1.8
𝐴
= 0.12 = 0.122 ⟹ 𝜑 𝑒 = 0.25𝑘𝑝𝑎 (𝜑 𝑒 ≤ 0.25)
A=10×8=80m2
For level3, 2, & 1:
𝑄 𝑢 = 1 × 3 = 3𝑘𝑝𝑎
𝜑 𝑎 = 0.3 +
3
√𝐴
= 0.41 ⟹ 𝜑 𝑎 = 0.635𝑘𝑝𝑎 (0.5 ≤ 𝜑 𝑎 ≤ 1)

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Figure 12 Imposed actions Q of transversal section
4.3. Earthquake action
Seismic weight:
Beams:
0.8 × 0.35 × 24 × (10 − 0.5) × (8 − 0.5) × 9 = 4309.2kn
Columns:
𝑊𝑟𝑜𝑜𝑓: (0.5 × 0.5 × 24) ×
3.5
2
× (4 × 4) = 168kn
6 × （
3.5
2
+
3.5
2
） = 21kn 𝑊3: 21×16= 336kn
6 × （
3.5
2
+
3.5
2
） = 21𝑘𝑛 𝑊2: 21×16= 336kn
6 × （
3.5
2
+
4.5
2
= 24𝑘𝑛 𝑊1: 24×16= 384kn
Glazing& Certain wall:
𝑊𝑟𝑜𝑜𝑓: 8 × 0.5 ×
3.5
2
= 7𝑘𝑛 ↓
𝑊3: 8 × 0.5 × (
3.5
2
+
3.5
2
) = 14𝑘𝑛 ↓
𝑊2: 8 × 0.5 × (
3.5
2
+
3.5
2
) = 14𝑘𝑛 ↓

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𝑊1: 8 × 0.5 × (
3.5
2
+
4.5
2
) = 16𝑘𝑛 ↓
𝜑 𝑒 𝑄:
φe = 0.3 From (NZS1170.5)
Wroof=0.3 × (0.25 × 8) × 10 × 3 = 18kn
W3=0.3 × (0.635 × 8) × 10 × 3 = 45.72kn
W2=0.3 × (0.635 × 8) × 10 × 3 = 45.72kn
W1=0.3 × (0.635 × 8) × 10 × 3 = 45.72kn
𝑾𝒊 = 𝑮 𝒃𝒆𝒂𝒎𝒔 + 𝑮 𝒄𝒐𝒍𝒖𝒎𝒏𝒔 + 𝑮 𝒔𝒍𝒂𝒃𝒔 + 𝑮 𝒈𝒍𝒂𝒛𝒊𝒏𝒈 + 𝝋 𝒆 𝑸
𝑊𝑟𝑜𝑜𝑓 = 4309.2 + 2994.38 + 168 + 14 + 18 =7503.58kn
𝑊3 = 4309.2 + 2994.38 + 336 + 28 + 45.72 =7713.3kn
𝑊𝑟𝑜𝑜𝑓 = 4309.2 + 2994.38 + 336 + 28 + 45.72 =7713.3kn
𝑊𝑟𝑜𝑜𝑓 = 4309.2 + 2994.38 + 384 + 32 + 45.72 =7765.3kn
𝑊𝑖 = 𝐺 𝑏𝑒𝑎𝑚𝑠 + 𝐺𝑐𝑜𝑙𝑢𝑚𝑛𝑠 + 𝐺𝑠𝑙𝑎𝑏𝑠 + 𝐺 𝑔𝑙𝑎𝑧𝑖𝑛𝑔 + 𝜑 𝑒 𝑄
𝑾 𝒕 = 𝑾 𝒓𝒐𝒐𝒇 + 𝑾 𝟑 + 𝑾 𝟐 + 𝑾 𝟏 = 𝟑𝟎𝟔𝟗𝟓. 𝟒𝟖𝒌𝒏
V = 𝐶(𝑇) 𝑊𝑡 = 30695.08𝑘𝑛 (Assume𝐶(𝑇) = 1)
Level Wi[kn] Hi[m] Wihi[knm] Fi= 0.92v +
𝑊𝑖ℎ𝑖
∑ 𝑊𝑖ℎ𝑖
+ 0.08𝑣 (𝑅𝑜𝑜𝑓 𝑜𝑛𝑙𝑦)[𝑘𝑛]
Roof 7503.58 15 112553.7 13125.075
3 7713.3 11.5 88702.95 8408.524
2 7713.3 8 61706.4 5849.408
1 7765.3 4.5 34943.85 3312.474
∑ 297906.9 ∑ 30695.48 √ OK
For internal frame: [Fi/5]
𝑊𝑖ℎ𝑖
∑ 𝑊𝑖ℎ𝑖
Roof 1500.716 15 22510.74 2625.015
3 1542.66 11.5 17740.59 1681.7
2 1542.66 8 12341.28 1169.882

23
1 1553.06 4.5 6988.77 662.495
∑ 59581.38 ∑ 6139.092 √ OK
d4= 334.031 mm d3= 290.139 mm d2= 219.529 mm d1= 127.898 mm
Level Wi[kn] Fi[kn] di[m] 𝑊𝑖 𝑑𝑖
2 Fidi
Roof 1500.716 2625.015 0.334 167.414 876.755
3 1542.66 1681.7 0.29 129.738 487.693
2 1542.66 1169.882 0.22 74.665 257.374
1 1553.06 662.495 0.128 25.445 84.8
∑ 6139.092 √
OK
∑ 397.262 ∑ 1706.6
T = 2π√
∑(𝑊 𝑖 𝑑 𝑖
2)
𝑔 ∑(𝐹𝑖𝑑𝑖)
= 0.968𝑠
From NZS1170.5 Earthquake standard,
Thus, 𝐶ℎ(𝑇) = 1.222 when T=0.968s
Z= 0.13(Auckland)
Important level= 3
Design working life 50 years
Annual probability of exceedance 1/1000 →RS=Ru= 1.3
N(T, D)= 1
Annual probability of exceedance 1/250
Thus, 𝐶(𝑇) = 𝐶ℎ(𝑇) 𝑍𝑅𝑁(𝑇, 𝐷) → 𝐶(𝑇) =1.222×0.13×1=0.207
𝐶 𝑑(𝑇) =
𝐶(𝑇1) 𝑆 𝑝
𝐾 𝜇
≥ (
𝑍
20
+ 0.02) 𝑅 𝑢 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 0.03𝑅 𝑢
𝑆 𝑝 = 0.7 (𝜇 = 3) 𝐾𝜇 = 𝜇 = 3 (𝑆ℎ𝑎𝑙𝑙𝑜𝑤 𝑠𝑜𝑖𝑙, 𝑇 ≥ 0.7𝑠)
Check: 𝐶 𝑑(𝑇) =
0.207×0.7
3
= 0.048 ≥ 0.034 （ 𝑂𝐾）(≥ 0.039 𝑂𝐾)
Therefore, V = Cd(T)Wt = 0.048 × 6139.096 = 294.677kn
𝑊𝑖ℎ𝑖
∑ 𝑊𝑖ℎ𝑖
Roof 1500.716 15 22510.74 126
T Shallow soil
0.9 1.29
1 1.19

24
3 1542.66 11.5 17740.59 80.722
2 1542.66 8 12341.28 56.154
1 1553.06 4.5 6988.77 31.8
∑ 59581.38 ∑ 294.677 √ OK
d4= 16.033 mm d3= 13.927 mm d2= 10.537 mm d1= 6.139 mm
Figure 13 Earthquake actions (Longitudinal section)
. Figure 14 Deflection due to Earthquake actions (Longitudinal section)

25
Figure 15 Earthquake actions (Transversal section)
Figure 16 Deflection due to Earthquake actions (Transversal section)
5.Structure analysis
According to section 4 combinations of static actions from NZS1170.0 General Principles, to
use the combinations of actions for ultimate limit states in checking strength:

26
Ed= 1.35G permanent action only (does not apply to pre-stressing forces)
Ed= 1.2G+1.5Q permanent and imposed action
Ed= G+Eu+𝜑𝑐 𝑄 where 𝜑𝑐 = 0.4 (for office building, from Table 4.1 in NZS1170.0 General
Principles)
All the above combinations of static actions are analysed in Multi-frame.
5.1. Longitudinal section
Figure 17 Maximum bending moment diagram for longitudinal section

27
Figure 18 Maximum shear force diagram for longitudinal section

28
Figure 19 Maximum axial load diagram for longitudinal section

29
5.2. Transversal section
Figure 20 Maximum bending moment diagram for transversal section

30
Figure 21 Maximum shear force diagram for transversal section

31
Figure 22 Maximum axial load diagram for transversal section
6.Beams Design
6.1. Design Beams for longitudinal section
From NZS 3101 Part1: clause 9.3.8.4 Maximum diameter of longitudinal beam bar in internal
beam column joint zones. It says:
For nominally ductile structures the maximum diameter of longitudinal beam bars passing
through beam column joint zones shall not exceed the appropriate requirement given below
for internal beam column joints:
For the earthquake does not govern:
𝑑 𝑏
ℎ𝑐
≤ 6𝛼 𝑡 ×
√𝑓𝑐
′
𝑓𝑦(1+
𝑓 𝑠
𝑓 𝑦
)
where 𝛼 𝑡 = 1 (𝑜𝑛𝑒 𝑤𝑎𝑦𝑓𝑟𝑎𝑚𝑒) 𝑓𝑠 = 0.5𝑓𝑦

32
Therefore, 𝑑 𝑏 = (6 +
√30
500×1.5
) × 500 = 21.9𝑚𝑚 → Choose HD20
6.1.1. Roof level (Reinforcement bars)
Figure 23 Maximum bending moment diagram for roof

33
6.1.1.1. Point○a of maximum negative moment case 1.35G:
Figure 24 Maximum negative bending moment
As the picture above shows, the maximum positive bending moment is 315.765knm.
𝑀 𝑛 =
𝑀∗
𝜑
=
315.765
0.85
= 371.488𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
371488000
0.85 × 350 × 684 × 30
= 60.85𝑚𝑚
jd=d −
𝑎
2
= 760 − 60.85 ÷ 2 = 729.575𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
371488000
500 × 729.575
= 1018.36𝑚𝑚2
AD20=314.16mm2
As/AD20=3.24 says 4 bars
→4HD20 is required (Asreq=1256.64mm2
)
Check: 𝐴 𝑚𝑖𝑛 =
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2

34
𝐴 𝑚𝑖𝑛 ≤ 𝐴 𝑠𝑟𝑒𝑞 ≤ 𝐴 𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴 𝑠𝑟𝑒𝑞 𝑗 𝑑 𝑓𝑦 = 1256 × 729.575 × 500 = 458.171𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.1.2. Point of ○b maximum positive moment case 1.35G:
Figure 25 Maximum positive bending moment
𝑀 𝑛 =
𝑀∗
𝜑
=
285.748
0.85
= 336.744𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
336744000
0.85 × 350 × 684 × 30
= 55.068𝑚𝑚
jd=d −
𝑎
2
= 760 − 55.068 ÷ 2 = 732.446𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
336744000
500 × 732.446
= 917.92𝑚𝑚2
AD20=314.16mm2

35
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴 𝑠𝑟𝑒𝑞 𝑗 𝑑 𝑓𝑦 = 942.478 × 732.466 × 500 = 345.166𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.1.3. . Point ○c of maximum negative moment case 1.35G:
𝑀 𝑛 =
𝑀∗
𝜑
=
530.063
0.85
= 623.604𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
623604000
0.85 × 350 × 684 × 30
= 102.151𝑚𝑚

36
jd=d −
𝑎
2
= 760 − 102.151 ÷ 2 = 708.924𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
623604000
500 × 708.924
= 1759.3𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.1.1.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:

37
𝑀 𝑛 =
𝑀∗
𝜑
=
228.885
0.85
= 269.272𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
269272000
0.85 × 350 × 684 × 30
= 44.11𝑚𝑚
jd=d −
𝑎
2
= 760 − 44.11 ÷ 2 = 708.924𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
269272000
500 × 708.924
= 737.945𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.1.2. Roof level (Stirrups)

38
6.1.2.1. Point ○a of column face SF case 1.35G:
Asreq=942.478mm2
𝑉∗
= 248.797𝑘𝑛
Vn =
𝑉∗
𝜑
=
248.797
0.75
= 331.729𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
331729
350×760
= 1.247𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
942.478
350×760
= 0.0035
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.105√𝑓𝑐
′
Check 0.08√𝑓𝑐
′ < 𝑣𝑐 = 0.105√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.105√30 × 350 × 760 = 153.608𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 331.729 − 153.608 = 178.122kn
Use 2 legged R10 → 𝐴 𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴 𝑉 𝑓𝑦𝑡
𝑑
𝑠
→ 𝑠 = 𝐴 𝑉 𝑓𝑦𝑡
𝑑
𝑉𝑠
=
157.08×500×760
178122
= 335 𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 330𝑚𝑚
Check: s= 330mm<
𝑑
2
=
760
2
= 380𝑚𝑚 𝑂𝐾
Check:
𝐴 𝑉,𝑚𝑖𝑛 =
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 330
500
= 79.07𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 330 c/c
6.1.2.2. Point ○b of column face SF case 1.35G:
Asreq=1884.956mm2
𝑉∗
= 298.061𝑘𝑛
Vn =
𝑉∗
𝜑
=
298.061
0.75
= 397.415𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
397415
350×760
= 1.494𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
1884.956
350×760
= 0.0071
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.141√𝑓𝑐
′

39
′ < 𝑣𝑐 = 0.141√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.141√30 × 350 × 760 = 205.229𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 397.415 − 205.229 = 192.185kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
192185
= 310𝑚𝑚
Check: s= 310mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 310
500
= 74.28𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 310 c/c
6.1.3. Level 1-3 (Reinforcement bars)
Figure 28 Maximum bending moment diagram for level1-3

40
6.1.3.1. Point○a of maximum negative moment case 1.2G+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
449.84
0.85
= 529.224𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
529224000
0.85 × 350 × 684 × 30
= 86.69𝑚𝑚
jd=d −
𝑎
2
= 760 − 86.69 ÷ 2 = 716.654𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
529224000
500 × 716.654
= 1476.93𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2

41
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.1.3.2. Point○b of maximum positive moment case 1.2G+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
392.742
0.85
= 462.049𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
462049000
0.85 × 350 × 684 × 30
= 75.687𝑚𝑚
jd=d −
𝑎
2
= 760 − 75.687 ÷ 2 = 722.156𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
462049000
500 × 722.156
= 1279.64𝑚𝑚2
AD20=314.16mm2

42
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.1.3.3. Point○c of maximum negative moment case 1.2G+1.5Q1+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
674.36
0.85
= 793.365𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
793365000
0.85 × 350 × 684 × 30
= 129.96𝑚𝑚

43
jd=d −
𝑎
2
= 760 − 129.96 ÷ 2 = 695𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
793365000
500 × 695
= 2283𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴 𝑠𝑟𝑒𝑞 𝑗 𝑑 𝑓𝑦 = 2513.274 × 695 × 500 = 873.388𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.3.4. Point○d of maximum positive moment case 1.2G+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
339.337
0.85
= 399.22𝑘𝑛𝑚

44
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
399220000
0.85 × 350 × 684 × 30
= 65.39𝑚𝑚
jd=d −
𝑎
2
= 760 − 65.39 ÷ 2 = 727.302𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
399220000
500 × 727.302
= 1097.81𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.1.4. Level 1-3 (Stirrups)
6.1.4.1. Point ○a of column face SF case 1.2G+1.5Q2:
Asreq=1570.796mm2
𝑉∗
= 335.079𝑘𝑛

45
Vn =
𝑉∗
𝜑
=
335.079
0.75
= 446.772𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
446772
350×760
= 1.68𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
1570.796
350×760
= 0.0059
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.129√𝑓𝑐
′
′ < 𝑣𝑐 = 0.129√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.129√30 × 350 × 760 = 188.022𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 446.772 − 188.022 = 258.75kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
258750
= 230𝑚𝑚
Check: s= 230mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 230
500
= 55.11𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 230 c/c
6.1.4.2. Point ○b of column face SF case 1.2G+1.5Q1+1.5Q2:
Asreq=2513.274mm2
𝑉∗
= 374.692𝑘𝑛
Vn =
𝑉∗
𝜑
=
374.692
0.75
= 499.589𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
499589
350×760
= 1.878𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
2513.274
350×760
= 0.0094
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.164√𝑓𝑐
′
′ < 𝑣𝑐 = 0.164√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.164√30 × 350 × 760 = 239.644𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 499.589 − 239.644 = 259.946kn

46
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
259946
= 220𝑚𝑚
Check: s= 220mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 220
500
= 52.72𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 220 c/c
6.2. Design Beams for transversal section
6.2.1. Roof level (Reinforcement bars)
Figure 33 Maximum bending moment diagram for roof （transversal section）

47
6.2.1.1. Point○d of maximum negative moment case 1.2G+1.5Q1+1.5Q2:
Figure 34 Maximum negative bending moment for transversal sction
As the picture above shows, the maximum positive bending moment is 219.047 knm.
𝑀 𝑛 =
𝑀∗
𝜑
=
219.047
0.85
= 257.702𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
257702000
0.85 × 350 × 684 × 30
= 42.214𝑚𝑚
jd=d −
𝑎
2
= 760 − 42.214 ÷ 2 = 738.893𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
257702000
500 × 738.893
= 697.54𝑚𝑚2
AD20=314.16mm2
→3HD20 is required (Asreq=942.478 mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2

48
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.1.2. point ○b of maximum positive moment case 1.35G:
Figure 35 Maximum positive bending moment for transversal section
𝑀 𝑛 =
𝑀∗
𝜑
=
252.614
0.85
= 297.193𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
297193000
0.85 × 350 × 684 × 30
= 48.683𝑚𝑚
jd=d −
𝑎
2
= 760 − 48.683 ÷ 2 = 735.659𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
297193000
500 × 735.659
= 807.96𝑚𝑚2
AD20=314.16mm2

49
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.1.3. Point○c of maximum negative moment case 1.35G:
Figure 36 Maximum negative bending moment for transversal section
𝑀 𝑛 =
𝑀∗
𝜑
=
402.042
0.85
= 472.991𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
472991000
0.85 × 350 × 684 × 30
= 77.48𝑚𝑚

50
jd=d −
𝑎
2
= 760 − 77.48 ÷ 2 = 721.26𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
472991000
500 × 721.26
= 1311.57𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.1.4. Point ○d of maximum positive moment case 1.2G+1.5Q1+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
180.401
0.85
= 212.236𝑘𝑛𝑚

51
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
212236000
0.85 × 350 × 684 × 30
= 34.766𝑚𝑚
jd=d −
𝑎
2
= 760 − 34.766 ÷ 2 = 742.617𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
212236000
500 × 742.617
= 571.59𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝐴 𝑚𝑖𝑛 > 𝐴 𝑠𝑟𝑒𝑞 ( 𝑁𝑂𝑇 𝑂𝐾)
Thus, increase the dimensions.
Try HD24 AD24 = 452.389 mm2
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
24
2
= 758𝑚𝑚
Assume jd= 0.9d= 758×0.9= 682.2mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
212236000
0.85 × 350 × 682.2 × 30
= 34.7858𝑚𝑚
jd=d −
𝑎
2
= 760 − 34.7858 ÷ 2 = 740.571𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
212236000
500 × 740.571
= 573.17𝑚𝑚2
)
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)

52
6.2.2. Roof level (Stirrups)
6.2.2.1. Point ○a of column face SF case 1.35G:
Asreq=942.478mm2
𝑉∗
= 237.826𝑘𝑛
Vn =
𝑉∗
𝜑
=
237.826
0.75
= 317.101𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
317101
350×760
= 1.192𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
942.478
350×760
= 0.0035
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.105√𝑓𝑐
′
′ < 𝑣𝑐 = 0.105√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.105√30 × 350 × 760 = 153.608𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 317.101 − 153.608 = 163.494kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
163494
= 360𝑚𝑚
Check: s= 360mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 360
500
= 86.266𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 360 c/c

53
Asreq=1570.796 mm2
𝑉∗
= 284.524𝑘𝑛
Vn =
𝑉∗
𝜑
=
284.524
0.75
= 379.365𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
379365
350×760
= 1.426𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
1570.796
350×760
= 0.0059
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.129√𝑓𝑐
′
′ < 𝑣𝑐 = 0.129√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.129√30 × 350 × 760 = 188.022𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 379.365 − 188.022 = 191.343kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
191343
= 310𝑚𝑚
Check: s= 310mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 310
500
= 74.28𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 310 c/c

54
6.2.3. Level 1-3 (Reinforcement bars)
Figure 38 Maximum bending moment diagram for level 1-3 (transversal section)
6.2.3.1. Point○a of maximum negative moment case 1.2G+1.5Q2:

55
𝑀 𝑛 =
𝑀∗
𝜑
=
337.041
0.85
= 396.519𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
396519000
0.85 × 350 × 684 × 30
= 64.953𝑚𝑚
jd=d −
𝑎
2
= 760 − 64.953 ÷ 2 = 727.523𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
396519000
500 × 727.523
= 1090.05𝑚𝑚2
AD20=314.16mm2
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)

56
6.2.3.2. Point ○b of maximum positive moment case 1.2G+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
318.837
0.85
= 375.102𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
375102000
0.85 × 350 × 684 × 30
= 61.445𝑚𝑚
jd=d −
𝑎
2
= 760 − 61.445 ÷ 2 = 729.278𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
375102000
500 × 729.278
= 1028.7𝑚𝑚2
AD20=314.16mm2
As/AD20=3.27says 4 bars
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2

57
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.3.3. Point○c of maximum negative moment case 1.2G++1.5Q1+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
523.994
0.85
= 616.464𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
616464000
0.85 × 350 × 684 × 30
= 100.982𝑚𝑚
jd=d −
𝑎
2
= 760 − 100.982 ÷ 2 = 709.509𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
616464000
500 × 709.509
= 1737.72𝑚𝑚2
AD20=314.16mm2

58
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.3.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:
𝑀 𝑛 =
𝑀∗
𝜑
=
269.456
0.85
= 317.007𝑘𝑛𝑚
d = h − 𝐶𝑐 −
𝐷
2
= 800 − 30 −
20
2
= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =
𝑀 𝑛
0.85𝑏𝑓𝑐
′ 𝑗 𝑑
=
317007000
0.85 × 350 × 684 × 30
= 51.928𝑚𝑚

59
jd=d −
𝑎
2
= 760 − 51.928 ÷ 2 = 734.063𝑚𝑚
𝐴 𝑠 =
𝑀 𝑛
𝑓𝑦 𝑗 𝑑
=
317007000
500 × 734.063
= 863.74𝑚𝑚2
AD20=314.16mm2
As/AD20=2.75says 3 bars
)
√𝑓𝑐
′
4𝑓𝑦
𝑏 𝑤 𝑑 =
√30
4×500
× 350 × 800 = 728.471𝑚𝑚2
𝐴 𝑚𝑎𝑥 =
10+𝑓𝑐
′
6𝑓𝑦
𝑏 𝑤 𝑑 =
40
6×500
× 350 × 800 = 3546.667𝑚𝑚2
𝑃𝑚𝑎𝑥 =
10 + 𝑓𝑐
′
6𝑓𝑦
=
40
6 × 500
= 0.013 < 0.025 (𝑂𝐾)
6.2.4. Level 1-3 (Stirrups)
Figure 43 Maximum shear force diagram for level 1-3 (transversal section)
6.2.4.1. Point ○a of column face SF case 1.2G+1.5Q2:
Asreq=1256.437mm2
𝑉∗
= 322.181𝑘𝑛
Vn =
𝑉∗
𝜑
=
322.181
0.75
= 429.575𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
429575
350×760
= 1.615𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
1256.437
350×760
= 0.0047
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.117√𝑓𝑐
′

60
′ < 𝑣𝑐 = 0.117√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.117√30 × 350 × 760 = 170.815𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 429.575 − 1570.815 = 258.76kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
258760
= 230𝑚𝑚
Check: s= 230mm<
𝑑
2
=
760
2
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 230
500
= 55.115𝑚𝑚2
< 𝐴 𝑉 = 157.08𝑚𝑚2
𝑂𝐾
Thus, R10@ 230 c/c
Asreq=1884.956 mm2
𝑉∗
= 362.965𝑘𝑛
Vn =
𝑉∗
𝜑
=
362.965
0.75
= 483.965𝑘𝑛
𝑣 𝑛 =
𝑉𝑛
𝑏 𝑤 𝑑
=
483965
350×760
= 1.819𝑀𝑝𝑎
ρ =
𝐴 𝑠
𝑏 𝑤 𝑑
=
1884.956
350×760
= 0.0071
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐
′ = 0.141√𝑓𝑐
′
′ < 𝑣𝑐 = 0.141√𝑓𝑐
′ < 0.2√𝑓𝑐
′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐 𝑏 𝑤 𝑑 = 0.141√30 × 350 × 760 = 205.229𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 483.965 − 205.229 = 278.724kn
𝑑
𝑠
𝑑
𝑉𝑠
=
157.08×500×760
278724
= 210𝑚𝑚
Check: s= 210mm<
𝑑
2
=
760
2

61
Check:
1
16
√𝑓𝑐
′
𝑏 𝑤 𝑠
𝑓𝑦𝑡
=
1
16
√30
350 × 210
500
= 74.28𝑚𝑚2
< 𝐴 𝑉 = 50.32𝑚𝑚2
𝑂𝐾
Thus, R10@ 210 c/c
7.Column Design
7.1. Marginal columns (Longitudinal section)
b= 500mm h= 500mm Cc= 30mm
Assume D24 will be used (AD24=452.39 mm2
)
gh = h – 2cc – D = 500 – 2 x 30 - 24 = 416
g= gh/h= 416/500= 0.83
Figure 44 Columns design for the whole building
7.1.1. Roof level
 Reinforcement bars
(Ncorresponding & M*) case1.35G
Mdes= 256.865knm Ndes=272.638kn

62
𝑁∗
𝜙.𝑏.ℎ
=
272638
0.85× 500×500
= 1.28 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
256865000
0.85×500×5002 = 2.42 MPa
(Cl 10.3.8.1) Minimum longitudinal reinforcement in nominally ductile design 0.008Ag
From Column chart 500/30/0.8 & 0.9 → minimum pt= 0.008
(N*
& Mcorres) case1.35G= (Ncorresponding & M*) case1.35G
Thus, the maximum pt= 0.008
Asreq= pt×b×h= 0.008×500×500= 2000mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 4.4
39.452
2000
24

D
sreq
A
A
The minimum reinforce bars is 8.
Thus, use 8D24
 Stirrups
VE= 154.296kn N*
= 272.638kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×154.296= 351.203kn where Фo’=1.75
Vn= 
75.0
203.351*

V
468.031kn where  =0.75
Vn= 
bd
Vn
2.044Mpa



458500
243
bd
DA
P S
0.0059
Vb=
''
109.0p1007.0 cc ff  ）（ Check:
'''
2.0109.008.0 ccc fff  OK
Kn= 


 2'
*
50030
2726383
1
3
1
Agf
N
c
1.109
Vc=KaKnVbAcv=1×1.109×0.109 30 ×500×458=179.817kn
Vs=Vn- Vc= 468.031- 179.817= 288.214kn
Vc=Vb.Kn=  109.130111.0 0.785Mpa

63
Steel shear stress:
Vs= Vn- Vc/2= 2.044-0.785/2= 1.651Mpa
Use 4 legged R10 stirrups  Avprov= 314.16mm2
Av=
  399.206
4500
500500651.1
4fy
bhs





V
mm < Avprov= 314.16mm2
OK



288214
45850016.314
Vs
dfA
S
s
d
fAV ytvprov
req
req
ytvprovs 250mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 250 mm
  


500
120500651.1min
fyt
bsVs
Avreq 198.143mm < Avprov= 314.16mm2
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
fA 41.079mm< Avprov= 314.16mm2
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041mm< Avprov= 314.16mm2
OK
4 legged R10 @ 120c/c
7.1.2. Level 3
(Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
𝑁∗
𝜙.𝑏.ℎ
=
655.828
0.85× 500×500
= 3.09 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
246286000
0.85×500×5002 = 2.32 MPa

64
(N*
& Mcorres) case1.2G+1.5Q1+1.5Q2= (Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
Asreq= pt×b×h= 0.008×500×500= 2000mm2
N= 4.4
39.452
2000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 140.354kn N*
= 655.828kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×140.354= 319.305kn where Фo’=1.75
Vn= 
75.0
319.305*

V
425.74kn where  =0.75
Vn= 
bd
Vn
1.859Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
 cf）（ Check:
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
6558283
1
3
1
Agf
N
c
1.262
Vc=KaKnVbAcv=1×1.262×0.708×500×458=204.688kn
Vs=Vn- Vc= 425.74-204.668= 221.072kn
Vc=Vb.Kn=  622.1087.0 0.894Mpa
Steel shear stress:
Vs= Vn- Vc/2= 1.859-0.894/2= 1.412Mpa

65
Av=
  532.176
4500
500500124.1
4fy
bhs





V
OK



221072
45850016.314
Vs
dfA
S
s
d
fAV
ytvprov
req
req
ytvprovs 330mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 330 mm
  


500
120500124.1min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
fA 169.471mm< Avprov= 314.16mm2
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041mm< Avprov= 314.16mm2
OK
4legged R10 @ 120c/c
7.1.3. Level 2
(Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
𝑁∗
𝜙.𝑏.ℎ
=
1052320
0.85× 500×500
= 4.95 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
274860000
0.85×500×5002 = 2.59 MPa
(N*
& Mcorres) case1.2G+1.5Q1+1.5Q2= (Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2

66
Asreq= pt×b×h= 0.014×500×500= 3500mm2
N= 7.7
39.452
3500
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 150.38kn N*
= 1052.32kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×150.38= 342.115kn where Фo’=1.75
Vn= 
75.0
342.115*

V
456.153kn where =0.75
Vn= 
bd
Vn
1.992Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
10523203
1
3
1
Agf
N
c
1.421
Vc=KaKnVbAcv=1×1.421×0.708×500×458=230.382kn
Vs=Vn- Vc= 456.153-230.382= 225.77kn
Vc=Vb.Kn=  .4211087.0 1Mpa
Steel shear stress:
Vs= Vn- Vc/2= 1.992-1 /2= 1.489Mpa
Av=
  114.186
4500
500500894.1
4fy
bhs





V
OK



770225
45850016.314
Vs
dfA
S
s
d
fAV ytvprov
req
req
ytvprovs 320mm
10db = 10 x 24 = 240 mm

67
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 320 mm
  


500
120500.4891min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041 mm< Avprov= 314.16mm2
OK
7.1.4. Level 1
(Ncorresponding & M*) caseG+Eu+Q
𝑁∗
𝜙.𝑏.ℎ
=
1101533
0.85× 500×500
= 5.18 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
235584000
0.85×500×5002 = 2.22 MPa
(N*
& Mcorres) case1.2G+1.5Q1+1.5Q2
𝑁∗
𝜙.𝑏.ℎ
=
1441401
0.85× 500×500
= 6.78 MPa

68
𝑀∗
𝜙.𝑏.ℎ2 =
168908000
0.85×500×5002 = 1.59 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
Asreq= pt×b×h= 0.012×500×500= 3000mm2
N= 6.6
39.452
3000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 103.735kn N*
= 1441.401 kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×103.735= 235.997kn where Фo’=1.75
Vn= 
75.0
235.997*

V
314.663kn where  =0.75
Vn= 
bd
Vn
1.374Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
14414013
1
3
1
Agf
N
c
1.577
Vc=KaKnVbAcv=1×1.577×0.708×500×458=255.616kn
Vs=Vn- Vc= 314.663-255.616= 59.047kn
Vc=Vb.Kn=  .5771087.0 1.116Mpa
Steel shear stress:
Vs= Vn- Vc/2= 0.816Mpa

69
Av=
  995.101
4500
5005000.816
4fy
bhs





V
OK



59047
45850016.314
Vs
dfA
S
s
d
fAV
ytvprov
req
req
ytvprovs 1220mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 1220 mm
  


500
120500816.0min
fyt
bsVs
Avreq 97.915 mm < Avprov= 314.16mm2
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041mm< Avprov= 314.16mm2
OK
4legged R10 @ 120c/c
7.2. Marginal columns (Transversal section)
7.2.1. Roof level
(Ncorresponding & M*) case 1.2G+1.5Q1+1.5Q2
Mdes= 219.047 knm Ndes=254.401kn
𝑁∗
𝜙.𝑏.ℎ
=
254401
0.85× 500×500
= 1.2 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
219047000
0.85×500×5002 = 2.06 MPa

70
(N*
& Mcorres) case1.35G
Mdes= 215.25 knm Ndes=267.212 kn
𝑁∗
𝜙.𝑏.ℎ
=
267212
0.85× 500×500
= 1.26 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
215250000
0.85×500×5002 = 2.03MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt=0.013
Asreq= pt×b×h= 0.014×500×500= 3500mm2
N= 7.7
39.452
3500
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 119.402 kn N*
= 267.212 kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×119.402= 271.64 kn where Фo’=1.75
Vn= 
75.0
271.64*

V
362.186 kn where  =0.75
Vn= 
bd
Vn
1.582 Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
2672123
1
3
1
Agf
N
c
1.107
Vc=KaKnVbAcv=1×1.107×0.708×500×458=179.465 kn

71
Vs=Vn- Vc= 362.186-179.465= 182.721 kn
Vc=Vb.Kn=  .1071087.0 0.784 Mpa
Steel shear stress:
Vs= Vn- Vc/2= 1.19 Mpa
Av=
  719.148
4500
50050019.1
4fy
bhs





V
OK



59047
45850016.314
Vs
dfA
S
s
d
fAV
ytvprov
req
req
ytvprovs 390mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 390 mm
  


500
12050019.1min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041mm< Avprov= 314.16mm2
OK
7.2.2. Level 3
Mdes= 199.312knm Ndes=506.22 kn
𝑁∗
𝜙.𝑏.ℎ
=
506220
0.85× 500×500
= 2.38 MPa

72
𝑀∗
𝜙.𝑏.ℎ2 =
199312000
0.85×500×5002 = 1.88 MPa
(N*
Mdes= 191.908 knm Ndes=649.762 kn
𝑁∗
𝜙.𝑏.ℎ
=
649762
0.85× 500×500
= 3.06MPa
𝑀∗
𝜙.𝑏.ℎ2 =
191908000
0.85×500×5002 = 1.81 MPa
Asreq= pt×b×h= 0.014×500×500= 3500mm2
N= 7.7
39.452
3500
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE=110.304 kn N*
= 649.762 kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×110.304= 271.64 kn where Фo’=1.75
Vn= 
75.0
271.64*

V
362.186kn where  =0.75
Vn= 
bd
Vn
1.582Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK

73
Kn= 


 2'
*
50030
6497623
1
3
1
Agf
N
c
1.26
Vc=KaKnVbAcv=1×1.26×0.708×500×458=204.275 kn
Vs=Vn- Vc= 362.186 -204.275= 157.911 kn
Vc=Vb.Kn=  .261087.0 0.892 Mpa
Steel shear stress:
Vs= Vn- Vc/2= 1.136 Mpa
Av=
  .948141
4500
500500136.1
4fy
bhs





V
OK



157911
45850016.314
Vs
dfA
S
s
d
fAV ytvprov
req
req
ytvprovs 460 mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 460 mm
  


500
120500136.1min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041 mm< Avprov= 314.16mm2
OK
7.2.3. Level 2

74
Mdes= 219.094 knm Ndes=808.283 kn
𝑁∗
𝜙.𝑏.ℎ
=
808283
0.85× 500×500
= 3.8 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
219094000
0.85×500×5002 = 2.0 6MPa
(N*
Mdes= 209.113 knm Ndes=1043.388 kn
𝑁∗
𝜙.𝑏.ℎ
=
1043388
0.85× 500×500
= 4.91 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
209113000
0.85×500×5002 = 1.97 MPa
Asreq= pt×b×h= 0.012×500×500= 3000mm2
N= 6.6
39.452
3000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE=123.506 kn N*
= 1043.388kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×123.506= 235.997 kn where Фo’=1.75
Vn= 
75.0
235.997*

V
314.663 kn where  =0.75
Vn= 
bd
Vn
1.5374Mpa

75



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
388.10433
1
3
1
Agf
N
c
1.417
Vc=KaKnVbAcv=1×1.417×0.708×500×458=229.803 kn
Vs=Vn- Vc= 314.663-229.803= 84.86 kn
Vc=Vb.Kn=  .4171087.0 1 Mpa
Steel shear stress:
Vs= Vn- Vc/2= 0.872Mpa
Av=
  04.109
4500
500500872.0
4fy
bhs





V
OK



84860
45850016.314
Vs
dfA
S
s
d
fAV
ytvprov
req
req
ytvprovs 850 mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 850 mm
  


500
120500872.0min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041 mm< Avprov= 314.16mm2
OK

76
7.2.4. Level 1
Mdes= 216.8 knm Ndes= 1114.559kn
𝑁∗
𝜙.𝑏.ℎ
=
1114559
0.85× 500×500
= 5.24 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
216800000
0.85×500×5002 = 2.04 MPa
(N*
Mdes= 125.161 knm Ndes=1427.949 kn
𝑁∗
𝜙.𝑏.ℎ
=
1427949
0.85× 500×500
= 6.72 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
125161000
0.85×500×5002 = 1.18 MPa
Asreq= pt×b×h= 0.008×500×500= 2000mm2
N= 4.4
39.452
2000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE=94.231 kn N*
= 1427.949 kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×94.231= 214.376 kn where Фo’=1.75

77
Vn= 
75.0
214.376*

V
285.834 kn where  =0.75
Vn= 
bd
Vn
1.248Mpa



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
14279493
1
3
1
Agf
N
c
1.57
Vc=KaKnVbAcv=1×1.57×0.708×500×458=254.743kn
Vs=Vn- Vc= 285.834 -254.743= 31.091 kn
Vc=Vb.Kn=  .571087.0 1.112 Mpa
Steel shear stress:
Vs= Vn- Vc/2= 0.692 Mpa
Av=
  497.86
4500
500500692.0
4fy
bhs





V
OK



31091
45850016.314
Vs
dfA
S
s
d
fAV
ytvprov
req
req
ytvprovs 2310 mm
10db = 10 x 24 = 240 mm
Smin= mm125
4
500
4
b
 Smin=120mm
mm150
3
458
3
d

Sreq = 2310 mm
  


500
120500692.0min
fyt
bsVs
OK
For anti-buckling:



500
120500
30
16
1
16
1 min'
vmin
yt
c
c
f
sb
OK

78





24500135
120500248
135
A minb D
df
sf
A
byt
y
te 134.041 mm< Avprov= 314.16mm2
OK
7.3. Internal columns
7.3.1. Roof level
𝑁∗
𝜙.𝑏.ℎ
=
454292
0.85× 500×500
= 2.14 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
105789000
0.85×500×5002 = 1 MPa
(N*
& Mcorres) case1.35G
𝑁∗
𝜙.𝑏.ℎ
=
600272
0.85× 500×500
= 2.82 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
36890000
0.85×500×5002 = 3.17 MPa
Asreq= pt×b×h= 0.012×500×500= 3000mm2
N= 6.6
39.452
3000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups

79
VE= 52.789 kn N*
= 600.272kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×52.789= 120.095 kn where Фo’=1.75
Vn= 
75.0
120.095*

V
160.127 kn where  =0.75



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
6002723
1
3
1
Agf
N
c
1.24
Vc=KaKnVbAcv=1×1.24×0.708×500×458=201.065kn
Vs=Vn- Vc= 160.127-201.065= -40.939kn
Thus, no need stirrups but constructive.
7.3.2. Level 3
𝑁∗
𝜙.𝑏.ℎ
=
1974727
0.85× 500×500
= 4.59 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
120900000
0.85×500×5002 = 1.14 MPa
(N*

80
𝑁∗
𝜙.𝑏.ℎ
=
1441401
0.85× 500×500
= 6.32 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
168908000
0.85×500×5002 = 0.09 MPa
Asreq= pt×b×h= 0.008×500×500= 2000mm2
N= 4.4
39.452
2000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 66.38kn N*
= 1342.712kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×66.38= 151.015kn where Фo’=1.75
Vn= 
75.0
151.015*

V
201.353kn where  =0.75



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
13427123
1
3
1
Agf
N
c
1.537
Vc=KaKnVbAcv=1×1.3537×0.708×500×458=249.215kn
Vs=Vn- Vc= 201.353-249.215= -47.863kn

81
7.3.3. Level 2
Mdes= 166.117knm Ndes= 1497.766kn
𝑁∗
𝜙.𝑏.ℎ
=
1497766
0.85× 500×500
= 7.05 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
166117000
0.85×500×5002 = 1.56 MPa
(N*
𝑁∗
𝜙.𝑏.ℎ
=
2123840
0.85× 500×500
= 9.99 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
30435000
0.85×500×5002 = 0.29 MPa
Asreq= pt×b×h= 0.008×500×500= 2000mm2
N= 4.4
39.452
2000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 94.426kn N*
= 2123.84kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE=1.3×1.75×94.426= 214.819 kn where Фo’=1.75

82
Vn= 
75.0
214.819*

V
286.428kn where  =0.75



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 


 2'
*
50030
21238403
1
3
1
Agf
N
c
1.85
Vc=KaKnVbAcv=1×1.85×0.708×500×458=299.875kn
Vs=Vn- Vc= 286.426-299.875= -13.449kn
7.3.4. Level 1
𝑁∗
𝜙.𝑏.ℎ
=
2029268
0.85× 500×500
= 9.55 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
204836000
0.85×500×5002 = 1.93 MPa
(N*
𝑁∗
𝜙.𝑏.ℎ
=
2912379
0.85× 500×500
= 13.71 MPa
𝑀∗
𝜙.𝑏.ℎ2 =
15532000
0.85×500×5002 = 0.15 MPa

83
Asreq= pt×b×h= 0.008×500×500= 2000mm2
N= 4.4
39.452
2000
24

D
sreq
A
A
Thus, use 8D24
 Stirrups
VE= 84.235kn N*
= 2912.379kn
d=h-Cc-D/2=500- 30- 24= 458mm
V*=1.3.Фo’.VE= 191.635kn where Фo’=1.75
Vn= 

*
V
255.513 kn where  =0.75



458500
243
bd
DA
P S
0.0059
Vb= 708.0p1007.0
'
''
2.0708.008.0 cc ff  OK
Kn= 
Agf
N
c
'
*
3
1 2.165
Vc=KaKnVbAcv=1×2.165×0.708×500×458=351kn
Vs=Vn- Vc= 255.513-351= -95kn
8.Conclusions

84
This project is being designed for the commercial office building located in Auckland centre
region. The dimensions of beams and columns are considered as 350×800 mm and 500×500
mm. The permanent, imposed and earthquake actions are calculated for creating the
structure frame. The structure frame is taken into account by two different aspects:
longitudinal and transversal section, which has the unique factors influenced by the various
length of bays for both sides. All the above components are relevant with the NZS 3101:2006
part 1 Concrete design and NZS 1170 series -Structural design actions.
In the result of load distribution applying on each level, for the permanent load distribution,
they are all equal to each other. The only different is the point load, which is calculated by
self-weight of glazing and columns. For the imposed load distribution, the main difference
could be observed by eyes. As roof level treated as other levels, the value of φe is remarkably
lower than other levels.
The internal columns design is done from longitudinal section due to the critical value
existing in this section. As the pt number is tiny caused by columns size w, all the
reinforcement bars are used as 8HD24.
9.References
Structural Concrete, Dr. Lusa Tuleasca handouts
Park, R., & Pauley, T. (1975). Reinforced concrete structures. Christchurch, New Zealand: John
Wiley & Sons Company.
Noel, J. (1993). Reinforced concrete design. Texas, USA: McGraw-Hill Company.
NZS3101: 2006 Concrete structures standard
NZS4203: 1992 Code of practice for general structural design and design loading for building
NZS1170: 2002 part 0: General principles
NZS1170: 2002 part 1: Permanent, imposed and other actions
NZS1170: 2002 part 5: Earthquake actions- New Zealand
10. Appendices
10.1. Longitudinal section combinations
10.1.1. 1.35G
 Bending moment

85
 Shear force
 Axial load

86
10.1.2. 1.2G+1.5Q1
 Bending moment
 Shear force

87
 Axial load
10.1.3. 1.2G+1.5Q2
 Bending moment

88
 Shear force
 Axial load

89
10.1.4. 1.2G+1.5Q1+1.5Q2
 Bending moment
 Shear force

90
 Axial load
10.1.5. G+Eu+𝜑𝑐 𝑄
 Bending moment

91
 Shear force
 Axial load

92
10.2. Transversal section
10.2.1. 1.35G
 Bending moment
 Shear force

93
 Axial load
10.2.2. 1.2G+1.5Q1
 Bending moment

94
 Shear force
 Axial load

95
10.2.3. 1.2G+1.5Q2
 Bending moment
 Shear force

96
 Axial load
10.2.4. 1.2G+1.5Q1+1.5Q2
 Bending moment

97
 Shear force
 Axial load

98
10.2.5. G+Eu+𝜑𝑐 𝑄
 Bending moment
 Shear force

99
 Axial load

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