structural design of a four storey office building
TRANSCRIPT
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
2 Dongyezhan Liu 1421941
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______
✓
3 Dongyezhan Liu 1421941
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)
4 Dongyezhan Liu 1421941
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.
5 Dongyezhan Liu 1421941
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
6 Dongyezhan Liu 1421941
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
7 Dongyezhan Liu 1421941
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 26 Maximum negative bending moment ...................................................................... 35
Figure 27 Maximum positive bending moment ....................................................................... 36
Figure 28 Maximum bending moment diagram for level1-3 ................................................... 39
Figure 29 Maximum negative bending moment ...................................................................... 40
Figure 30 Maximum positive bending moment ....................................................................... 41
Figure 31 Maximum negative bending moment ...................................................................... 42
Figure 32 Maximum positive bending moment ....................................................................... 43
Figure 33 Maximum bending moment diagram for roof (transversal section) .................. 46
Figure 34 Maximum negative bending moment for transversal sction ................................... 47
8 Dongyezhan Liu 1421941
Figure 35 Maximum positive bending moment for transversal section ................................... 48
Figure 36 Maximum negative bending moment for transversal section ................................. 49
Figure 37 Maximum positive bending moment for transversal section ................................... 50
Figure 38 Maximum bending moment diagram for level 1-3 (transversal section) ................. 54
Figure 39 Maximum negative bending moment for transversal section ................................. 54
Figure 40 Maximum positive bending moment ....................................................................... 56
Figure 41 Maximum negative bending moment for transversal section ................................. 57
Figure 42 Maximum positive bending moment for transversal section ................................... 58
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
9 Dongyezhan Liu 1421941
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
10 Dongyezhan Liu 1421941
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)
11 Dongyezhan Liu 1421941
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)
12 Dongyezhan Liu 1421941
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.
13 Dongyezhan Liu 1421941
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 f𝑐′ =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
14 Dongyezhan Liu 1421941
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.
15 Dongyezhan Liu 1421941
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.
16 Dongyezhan Liu 1421941
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:
17 Dongyezhan Liu 1421941
𝐿𝑛
𝑏𝑤≤ 25
𝐿𝑛ℎ
𝑏𝑤2
≤ 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=500Mpaboth end continuous h=600. 700, 800mm
h ≥𝐿
25=
10000
25= 400𝑚𝑚
Design beams: 350×800 mm
mmh
bw 4002
mmbbb wce 5005050
mmbb
LLn ce 95005001000022
Check: 2575.23400
9500n
wb
L OK
Check: 10078.20nh2
wb
LOK
Check: 3350
800232
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
18 Dongyezhan Liu 1421941
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:
19 Dongyezhan Liu 1421941
𝜑𝑒 =1.8
𝐴= 0.12 = 0.122 ⟹ 𝜑𝑒 = 0.25𝑘𝑝𝑎 (𝜑𝑒 ≤ 0.25)
A=10×8=80m2
For level3, 2, & 1:
𝑄𝑢 = 1 × 3 = 3𝑘𝑝𝑎
𝜑𝑎 = 0.3 +3
√𝐴= 0.41 ⟹ 𝜑𝑎 = 0.635𝑘𝑝𝑎 (0.5 ≤ 𝜑𝑎 ≤ 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
20 Dongyezhan Liu 1421941
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:
𝜑𝑒 =1.8
𝐴= 0.12 = 0.122 ⟹ 𝜑𝑒 = 0.25𝑘𝑝𝑎 (𝜑𝑒 ≤ 0.25)
A=10×8=80m2
For level3, 2, & 1:
𝑄𝑢 = 1 × 3 = 3𝑘𝑝𝑎
𝜑𝑎 = 0.3 +3
√𝐴= 0.41 ⟹ 𝜑𝑎 = 0.635𝑘𝑝𝑎 (0.5 ≤ 𝜑𝑎 ≤ 1)
21 Dongyezhan Liu 1421941
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 ↓
22 Dongyezhan Liu 1421941
𝑊1: 8 × 0.5 × (3.5
2+
4.5
2) = 16𝑘𝑛 ↓
𝜑𝑒𝑄:
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 = 𝑮𝒃𝒆𝒂𝒎𝒔 + 𝑮𝒄𝒐𝒍𝒖𝒎𝒏𝒔 + 𝑮𝒔𝒍𝒂𝒃𝒔 + 𝑮𝒈𝒍𝒂𝒛𝒊𝒏𝒈 + 𝝋𝒆𝑸
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 = 𝐺𝑏𝑒𝑎𝑚𝑠 + 𝐺𝑐𝑜𝑙𝑢𝑚𝑛𝑠 + 𝐺𝑠𝑙𝑎𝑏𝑠 + 𝐺𝑔𝑙𝑎𝑧𝑖𝑛𝑔 + 𝜑𝑒𝑄
𝑾𝒕 = 𝑾𝒓𝒐𝒐𝒇 + 𝑾𝟑 + 𝑾𝟐 + 𝑾𝟏 = 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]
Level Wi[kn] Hi[m] Wihi[knm] Fi= 0.92v +𝑊𝑖ℎ𝑖
∑ 𝑊𝑖ℎ𝑖+ 0.08𝑣 (𝑅𝑜𝑜𝑓 𝑜𝑛𝑙𝑦)[𝑘𝑛]
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 Dongyezhan Liu 1421941
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] 𝑊𝑖𝑑𝑖2 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√∑(𝑊𝑖𝑑𝑖
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
𝐶𝑑(𝑇) =𝐶(𝑇1)𝑆𝑝
𝐾𝜇≥ (
𝑍
20+ 0.02) 𝑅𝑢 𝑏𝑢𝑡 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 0.03𝑅𝑢
𝑆𝑝 = 0.7 (𝜇 = 3) 𝐾𝜇 = 𝜇 = 3 (𝑆ℎ𝑎𝑙𝑙𝑜𝑤 𝑠𝑜𝑖𝑙, 𝑇 ≥ 0.7𝑠)
Check: 𝐶𝑑(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
Level Wi[kn] Hi[m] Wihi[knm] Fi= 0.92v +𝑊𝑖ℎ𝑖
∑ 𝑊𝑖ℎ𝑖+ 0.08𝑣 (𝑅𝑜𝑜𝑓 𝑜𝑛𝑙𝑦)[𝑘𝑛]
Roof 1500.716 15 22510.74 126
T Shallow soil
0.9 1.29
1 1.19
24 Dongyezhan Liu 1421941
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 Dongyezhan Liu 1421941
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 Dongyezhan Liu 1421941
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+𝜑𝑐𝑄 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 Dongyezhan Liu 1421941
Figure 18 Maximum shear force diagram for longitudinal section
28 Dongyezhan Liu 1421941
Figure 19 Maximum axial load diagram for longitudinal section
29 Dongyezhan Liu 1421941
5.2. Transversal section
Figure 20 Maximum bending moment diagram for transversal section
30 Dongyezhan Liu 1421941
Figure 21 Maximum shear force diagram for transversal section
31 Dongyezhan Liu 1421941
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:
𝑑𝑏
ℎ𝑐≤ 6𝛼𝑡 ×
√𝑓𝑐′
𝑓𝑦(1+𝑓𝑠𝑓𝑦
) where 𝛼𝑡 = 1 (𝑜𝑛𝑒 𝑤𝑎𝑦𝑓𝑟𝑎𝑚𝑒) 𝑓𝑠 = 0.5𝑓𝑦
32 Dongyezhan Liu 1421941
Therefore, 𝑑𝑏 = (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 Dongyezhan Liu 1421941
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.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=371488000
0.85 × 350 × 684 × 30= 60.85𝑚𝑚
jd=d −𝑎
2= 760 − 60.85 ÷ 2 = 729.575𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
371488000
500 × 729.575= 1018.36𝑚𝑚2
AD20=314.16mm2
As/AD20=3.24 says 4 bars
→4HD20 is required (Asreq=1256.64mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
34 Dongyezhan Liu 1421941
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 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
As the picture above shows, the maximum positive bending moment is 285.748knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=336744000
0.85 × 350 × 684 × 30= 55.068𝑚𝑚
jd=d −𝑎
2= 760 − 55.068 ÷ 2 = 732.446𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
336744000
500 × 732.446= 917.92𝑚𝑚2
AD20=314.16mm2
As/AD20=2.9 says 3 bars
35 Dongyezhan Liu 1421941
→3HD20 is required (Asreq=942.478mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 942.478 × 732.466 × 500 = 345.166𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.1.3. . Point ○c of maximum negative moment case 1.35G:
Figure 26 Maximum negative bending moment
As the picture above shows, the maximum positive bending moment is 530.063knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=623604000
0.85 × 350 × 684 × 30= 102.151𝑚𝑚
36 Dongyezhan Liu 1421941
jd=d −𝑎
2= 760 − 102.151 ÷ 2 = 708.924𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
623604000
500 × 708.924= 1759.3𝑚𝑚2
AD20=314.16mm2
As/AD20=5.6 says 6 bars
→6HD20 is required (Asreq=1884.956mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1884.956 × 708.924 × 500 = 668.145𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.1.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:
Figure 27 Maximum positive bending moment
As the picture above shows, the maximum positive bending moment is 228.885knm.
37 Dongyezhan Liu 1421941
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=269272000
0.85 × 350 × 684 × 30= 44.11𝑚𝑚
jd=d −𝑎
2= 760 − 44.11 ÷ 2 = 708.924𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
269272000
500 × 708.924= 737.945𝑚𝑚2
AD20=314.16mm2
As/AD20=2.3 says 3 bars
→3HD20 is required (Asreq=942.478mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 942.478 × 737.945 × 500 = 347.748𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.2. Roof level (Stirrups)
38 Dongyezhan Liu 1421941
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
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
331729
350×760= 1.247𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
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 = 𝑣𝑐𝑏𝑤𝑑 = 0.105√30 × 350 × 760 = 153.608𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 331.729 − 153.608 = 178.122kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
178122= 335 𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 330𝑚𝑚
Check: s= 330mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 330
500= 79.07𝑚𝑚2 < 𝐴𝑉 = 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
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
397415
350×760= 1.494𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
1884.956
350×760= 0.0071
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.141√𝑓𝑐
′
39 Dongyezhan Liu 1421941
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.141√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.141√30 × 350 × 760 = 205.229𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 397.415 − 205.229 = 192.185kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
192185= 310𝑚𝑚
Check: s= 310mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 310
500= 74.28𝑚𝑚2 < 𝐴𝑉 = 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 Dongyezhan Liu 1421941
6.1.3.1. Point○a of maximum negative moment case 1.2G+1.5Q2:
Figure 29 Maximum negative bending moment
As the picture above shows, the maximum positive bending moment is 449.84knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=529224000
0.85 × 350 × 684 × 30= 86.69𝑚𝑚
jd=d −𝑎
2= 760 − 86.69 ÷ 2 = 716.654𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
529224000
500 × 716.654= 1476.93𝑚𝑚2
AD20=314.16mm2
As/AD20=4.7 says 5 bars
→5HD20 is required (Asreq=1570.796mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
41 Dongyezhan Liu 1421941
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1570.796 × 716.654 × 500 = 562.859𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.3.2. Point○b of maximum positive moment case 1.2G+1.5Q2:
Figure 30 Maximum positive bending moment
As the picture above shows, the maximum positive bending moment is 392.742knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=462049000
0.85 × 350 × 684 × 30= 75.687𝑚𝑚
jd=d −𝑎
2= 760 − 75.687 ÷ 2 = 722.156𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
462049000
500 × 722.156= 1279.64𝑚𝑚2
AD20=314.16mm2
As/AD20=4.1 says 5 bars
42 Dongyezhan Liu 1421941
→5HD20 is required (Asreq=1570.796mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1570.796 × 722.156 × 500 = 562.859𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.3.3. Point○c of maximum negative moment case 1.2G+1.5Q1+1.5Q2:
Figure 31 Maximum negative bending moment
As the picture above shows, the maximum positive bending moment is 674.36knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=793365000
0.85 × 350 × 684 × 30= 129.96𝑚𝑚
43 Dongyezhan Liu 1421941
jd=d −𝑎
2= 760 − 129.96 ÷ 2 = 695𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
793365000
500 × 695= 2283𝑚𝑚2
AD20=314.16mm2
As/AD20=7.3 says 8 bars
→8HD20 is required (Asreq=2513.274mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 2513.274 × 695 × 500 = 873.388𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.1.3.4. Point○d of maximum positive moment case 1.2G+1.5Q2:
Figure 32 Maximum positive bending moment
As the picture above shows, the maximum positive bending moment is 339.337knm.
𝑀𝑛 =𝑀∗
𝜑=
339.337
0.85= 399.22𝑘𝑛𝑚
44 Dongyezhan Liu 1421941
d = h − 𝐶𝑐 −𝐷
2= 800 − 30 −
20
2= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=399220000
0.85 × 350 × 684 × 30= 65.39𝑚𝑚
jd=d −𝑎
2= 760 − 65.39 ÷ 2 = 727.302𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
399220000
500 × 727.302= 1097.81𝑚𝑚2
AD20=314.16mm2
As/AD20=3.5 says 4 bars
→4HD20 is required (Asreq=1256.637mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1256.637 × 727.302 × 500 = 456.977𝑘𝑛𝑚 > 𝑀𝑛 (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 Dongyezhan Liu 1421941
Vn =𝑉∗
𝜑=
335.079
0.75= 446.772𝑘𝑛
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
446772
350×760= 1.68𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
1570.796
350×760= 0.0059
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.129√𝑓𝑐
′
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.129√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.129√30 × 350 × 760 = 188.022𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 446.772 − 188.022 = 258.75kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
258750= 230𝑚𝑚
Check: s= 230mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 230
500= 55.11𝑚𝑚2 < 𝐴𝑉 = 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
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
499589
350×760= 1.878𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
2513.274
350×760= 0.0094
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.164√𝑓𝑐
′
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.164√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.164√30 × 350 × 760 = 239.644𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 499.589 − 239.644 = 259.946kn
46 Dongyezhan Liu 1421941
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
259946= 220𝑚𝑚
Check: s= 220mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 220
500= 52.72𝑚𝑚2 < 𝐴𝑉 = 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 Dongyezhan Liu 1421941
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.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=257702000
0.85 × 350 × 684 × 30= 42.214𝑚𝑚
jd=d −𝑎
2= 760 − 42.214 ÷ 2 = 738.893𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
257702000
500 × 738.893= 697.54𝑚𝑚2
AD20=314.16mm2
As/AD20=2.2 says 3 bars
→3HD20 is required (Asreq=942.478 mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
48 Dongyezhan Liu 1421941
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 942.478 × 738.893 × 500 = 348.195𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.2.1.2. point ○b of maximum positive moment case 1.35G:
Figure 35 Maximum positive bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 252.614 knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=297193000
0.85 × 350 × 684 × 30= 48.683𝑚𝑚
jd=d −𝑎
2= 760 − 48.683 ÷ 2 = 735.659𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
297193000
500 × 735.659= 807.96𝑚𝑚2
AD20=314.16mm2
As/AD20=2.57 says 3 bars
49 Dongyezhan Liu 1421941
→3HD20 is required (Asreq=942.478mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 942.478 × 735.659 × 500 = 346.671𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.2.1.3. Point○c of maximum negative moment case 1.35G:
Figure 36 Maximum negative bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 402.042 knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=472991000
0.85 × 350 × 684 × 30= 77.48𝑚𝑚
50 Dongyezhan Liu 1421941
jd=d −𝑎
2= 760 − 77.48 ÷ 2 = 721.26𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
472991000
500 × 721.26= 1311.57𝑚𝑚2
AD20=314.16mm2
As/AD20=4.17 says 5 bars
→5HD20 is required (Asreq=1570.796 mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1570.796 × 721.26 × 500 = 566.476𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.2.1.4. Point ○d of maximum positive moment case 1.2G+1.5Q1+1.5Q2:
Figure 37 Maximum positive bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 180.401 knm.
𝑀𝑛 =𝑀∗
𝜑=
180.401
0.85= 212.236𝑘𝑛𝑚
51 Dongyezhan Liu 1421941
d = h − 𝐶𝑐 −𝐷
2= 800 − 30 −
20
2= 760𝑚𝑚
Assume jd= 0.9d= 760×0.9= 684mm
∑ 𝐹 = 0 C=T
a =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=212236000
0.85 × 350 × 684 × 30= 34.766𝑚𝑚
jd=d −𝑎
2= 760 − 34.766 ÷ 2 = 742.617𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
212236000
500 × 742.617= 571.59𝑚𝑚2
AD20=314.16mm2
As/AD20=1.82 says 2 bars
→2HD20 is required (Asreq=628.319mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 > 𝐴𝑠𝑟𝑒𝑞 ( 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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=212236000
0.85 × 350 × 682.2 × 30= 34.7858𝑚𝑚
jd=d −𝑎
2= 760 − 34.7858 ÷ 2 = 740.571𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
212236000
500 × 740.571= 573.17𝑚𝑚2
As/AD24=1.27 says 2 bars
→2HD24 is required (Asreq=904.779 mm2)
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
52 Dongyezhan Liu 1421941
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 904.779 × 740.571 × 500 = 335.026𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
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
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
317101
350×760= 1.192𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
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 = 𝑣𝑐𝑏𝑤𝑑 = 0.105√30 × 350 × 760 = 153.608𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 317.101 − 153.608 = 163.494kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
163494= 360𝑚𝑚
Check: s= 360mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 360
500= 86.266𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾
Thus, R10@ 360 c/c
53 Dongyezhan Liu 1421941
6.2.2.2. Point ○b of column face SF case 1.35G:
Asreq=1570.796 mm2
𝑉∗ = 284.524𝑘𝑛
Vn =𝑉∗
𝜑=
284.524
0.75= 379.365𝑘𝑛
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
379365
350×760= 1.426𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
1570.796
350×760= 0.0059
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.129√𝑓𝑐
′
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.129√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.129√30 × 350 × 760 = 188.022𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 379.365 − 188.022 = 191.343kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
191343= 310𝑚𝑚
Check: s= 310mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 310
500= 74.28𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾
Thus, R10@ 310 c/c
54 Dongyezhan Liu 1421941
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:
Figure 39 Maximum negative bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 337.041 knm.
55 Dongyezhan Liu 1421941
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=396519000
0.85 × 350 × 684 × 30= 64.953𝑚𝑚
jd=d −𝑎
2= 760 − 64.953 ÷ 2 = 727.523𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
396519000
500 × 727.523= 1090.05𝑚𝑚2
AD20=314.16mm2
As/AD20=3.47 says 4 bars
→4HD20 is required (Asreq=1256.637 mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1256.637 × 727.523 × 500 = 457.116𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
56 Dongyezhan Liu 1421941
6.2.3.2. Point ○b of maximum positive moment case 1.2G+1.5Q2:
Figure 40 Maximum positive bending moment
As the picture above shows, the maximum positive bending moment is 318.837 knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=375102000
0.85 × 350 × 684 × 30= 61.445𝑚𝑚
jd=d −𝑎
2= 760 − 61.445 ÷ 2 = 729.278𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
375102000
500 × 729.278= 1028.7𝑚𝑚2
AD20=314.16mm2
As/AD20=3.27says 4 bars
→4HD20 is required (Asreq=1256.637mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
57 Dongyezhan Liu 1421941
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1256.637 × 729.278 × 500 = 458.219𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.2.3.3. Point○c of maximum negative moment case 1.2G++1.5Q1+1.5Q2:
Figure 41 Maximum negative bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 523.994 knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=616464000
0.85 × 350 × 684 × 30= 100.982𝑚𝑚
jd=d −𝑎
2= 760 − 100.982 ÷ 2 = 709.509𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
616464000
500 × 709.509= 1737.72𝑚𝑚2
AD20=314.16mm2
As/AD20=5.5 says 6 bars
58 Dongyezhan Liu 1421941
→6HD20 is required (Asreq=1884.956 mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 1884.956 × 709.509 × 500 = 668.697𝑘𝑛𝑚 > 𝑀𝑛 (𝑂𝐾)
6.2.3.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:
Figure 42 Maximum positive bending moment for transversal section
As the picture above shows, the maximum positive bending moment is 269.456 knm.
𝑀𝑛 =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 =𝑀𝑛
0.85𝑏𝑓𝑐′𝑗𝑑
=317007000
0.85 × 350 × 684 × 30= 51.928𝑚𝑚
59 Dongyezhan Liu 1421941
jd=d −𝑎
2= 760 − 51.928 ÷ 2 = 734.063𝑚𝑚
𝐴𝑠 =𝑀𝑛
𝑓𝑦𝑗𝑑=
317007000
500 × 734.063= 863.74𝑚𝑚2
AD20=314.16mm2
As/AD20=2.75says 3 bars
→3HD20 is required (Asreq=942.478mm2)
Check: 𝐴𝑚𝑖𝑛 =√𝑓𝑐
′
4𝑓𝑦𝑏𝑤𝑑 =
√30
4×500× 350 × 800 = 728.471𝑚𝑚2
𝐴𝑚𝑎𝑥 =10+𝑓𝑐
′
6𝑓𝑦𝑏𝑤𝑑 =
40
6×500× 350 × 800 = 3546.667𝑚𝑚2
𝐴𝑚𝑖𝑛 ≤ 𝐴𝑠𝑟𝑒𝑞 ≤ 𝐴𝑚𝑎𝑥 (𝑂𝐾)
𝑃𝑚𝑎𝑥 =10 + 𝑓𝑐
′
6𝑓𝑦=
40
6 × 500= 0.013 < 0.025 (𝑂𝐾)
M = 𝐴𝑠𝑟𝑒𝑞𝑗𝑑𝑓𝑦 = 942.478 × 728.471 × 500 = 345.906𝑘𝑛𝑚 > 𝑀𝑛 (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
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
429575
350×760= 1.615𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
1256.437
350×760= 0.0047
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.117√𝑓𝑐
′
60 Dongyezhan Liu 1421941
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.117√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.117√30 × 350 × 760 = 170.815𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 429.575 − 1570.815 = 258.76kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
258760= 230𝑚𝑚
Check: s= 230mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 230
500= 55.115𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾
Thus, R10@ 230 c/c
6.2.4.2. Point ○b of column face SF case 1.35G:
Asreq=1884.956 mm2
𝑉∗ = 362.965𝑘𝑛
Vn =𝑉∗
𝜑=
362.965
0.75= 483.965𝑘𝑛
𝑣𝑛 =𝑉𝑛
𝑏𝑤𝑑=
483965
350×760= 1.819𝑀𝑝𝑎
ρ =𝐴𝑠
𝑏𝑤𝑑=
1884.956
350×760= 0.0071
𝑣𝑐 = (0.07 + 10𝜌)√𝑓𝑐′ = 0.141√𝑓𝑐
′
Check 0.08√𝑓𝑐′ < 𝑣𝑐 = 0.141√𝑓𝑐
′ < 0.2√𝑓𝑐′ 𝑂𝐾
𝑉𝑐 = 𝑣𝑐𝑏𝑤𝑑 = 0.141√30 × 350 × 760 = 205.229𝑘𝑛
𝑉𝑠 = 𝑉𝑛 − 𝑉𝑐 = 483.965 − 205.229 = 278.724kn
Use 2 legged R10 → 𝐴𝑉 = 157.08𝑚𝑚2
𝑉𝑠 = 𝐴𝑉𝑓𝑦𝑡𝑑
𝑠→ 𝑠 = 𝐴𝑉𝑓𝑦𝑡
𝑑
𝑉𝑠=
157.08×500×760
278724= 210𝑚𝑚
Check: s= 210mm<𝑑
2=
760
2= 380𝑚𝑚 𝑂𝐾
61 Dongyezhan Liu 1421941
Check:
𝐴𝑉,𝑚𝑖𝑛 =1
16√𝑓𝑐
′𝑏𝑤𝑠
𝑓𝑦𝑡=
1
16√30
350 × 210
500= 74.28𝑚𝑚2 < 𝐴𝑉 = 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 Dongyezhan Liu 1421941
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= 4.439.452
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*
V468.031kn where =0.75
Vn= bd
Vn2.044Mpa
458500
243
bd
DAP S 0.0059
Vb= ''
109.0p1007.0 cc ff )( Check: '''
2.0109.008.0 ccc fff OK
Kn=
2'
*
50030
27263831
31
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 Dongyezhan Liu 1421941
Steel shear stress:
Vs= Vn- Vc/2= 2.044-0.785/2= 1.651Mpa
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
399.2064500
500500651.1
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
288214
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 250mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 250 mm
500
120500651.1min
fyt
bsVsAvreq 198.143mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
7.1.2. Level 3 Reinforcement bars
(Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
Mdes= 246.286knm Ndes=655.828kn
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
(Cl 10.3.8.1) Minimum longitudinal reinforcement in nominally ductile design 0.008Ag
64 Dongyezhan Liu 1421941
From Column chart 500/30/0.8 & 0.9 → minimum pt= 0.008
(N* & Mcorres) case1.2G+1.5Q1+1.5Q2= (Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
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= 4.439.452
2000
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V425.74kn where =0.75
Vn= bd
Vn1.859Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
65582831
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
65 Dongyezhan Liu 1421941
Av=
532.1764500
500500124.1
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
221072
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 330mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 330 mm
500
120500124.1min
fyt
bsVsAvreq 198.143mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 169.471mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041mm< Avprov= 314.16mm2 OK
4legged R10 @ 120c/c
7.1.3. Level 2 Reinforcement bars
(Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
Mdes= 274.86knm Ndes=1052.32kn
𝑁∗
𝜙.𝑏.ℎ =
1052320
0.85× 500×500 = 4.95 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 274860000
0.85×500×5002 = 2.59 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= 0.014
(N* & Mcorres) case1.2G+1.5Q1+1.5Q2= (Ncorresponding & M*) case1.2G+1.5Q1+1.5Q2
Thus, the maximum pt= 0.014
66 Dongyezhan Liu 1421941
Asreq= pt×b×h= 0.014×500×500= 3500mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 7.739.452
3500
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V456.153kn where =0.75
Vn= bd
Vn1.992Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
105232031
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
114.1864500
500500894.1
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
770225
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 320mm
10db = 10 x 24 = 240 mm
67 Dongyezhan Liu 1421941
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 320 mm
500
120500.4891min
fyt
bsVsAvreq 198.143mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041 mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
7.1.4. Level 1 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
Mdes= 235.584knm Ndes=1101.533kn
𝑁∗
𝜙.𝑏.ℎ =
1101533
0.85× 500×500 = 5.18 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 235584000
0.85×500×5002 = 2.22 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= 0.012
(N* & Mcorres) case1.2G+1.5Q1+1.5Q2
Mdes= 168.908knm Ndes=1441.401kn
𝑁∗
𝜙.𝑏.ℎ =
1441401
0.85× 500×500 = 6.78 MPa
68 Dongyezhan Liu 1421941
𝑀∗
𝜙.𝑏.ℎ2 = 168908000
0.85×500×5002 = 1.59 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
Thus, the maximum pt= 0.012
Asreq= pt×b×h= 0.012×500×500= 3000mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 6.639.452
3000
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V314.663kn where =0.75
Vn= bd
Vn1.374Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
144140131
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
69 Dongyezhan Liu 1421941
Av=
995.1014500
5005000.816
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
59047
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 1220mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 1220 mm
500
120500816.0min
fyt
bsVsAvreq 97.915 mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041mm< Avprov= 314.16mm2 OK
4legged R10 @ 120c/c
7.2. Marginal columns (Transversal section)
7.2.1. Roof level Reinforcement bars
(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
(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= 0.014
70 Dongyezhan Liu 1421941
(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
Thus, the maximum pt= 0.014
Asreq= pt×b×h= 0.014×500×500= 3500mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 7.739.452
3500
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V362.186 kn where =0.75
Vn= bd
Vn1.582 Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
26721231
31
Agf
N
c
1.107
Vc=KaKnVbAcv=1×1.107×0.708×500×458=179.465 kn
71 Dongyezhan Liu 1421941
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
719.1484500
50050019.1
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
59047
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 390mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 390 mm
500
12050019.1min
fyt
bsVsAvreq 142.77 mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
7.2.2. Level 3 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
Mdes= 199.312knm Ndes=506.22 kn
𝑁∗
𝜙.𝑏.ℎ =
506220
0.85× 500×500 = 2.38 MPa
72 Dongyezhan Liu 1421941
𝑀∗
𝜙.𝑏.ℎ2 = 199312000
0.85×500×5002 = 1.88 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= 0.014
(N* & Mcorres) case1.2G+1.5Q1+1.5Q2
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
From Column chart 500/30/0.8 & 0.9 → minimum pt=0.013
Thus, the maximum pt= 0.014
Asreq= pt×b×h= 0.014×500×500= 3500mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 7.739.452
3500
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V362.186kn where =0.75
Vn= bd
Vn1.582Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
73 Dongyezhan Liu 1421941
Kn=
2'
*
50030
64976231
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
.9481414500
500500136.1
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
157911
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 460 mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 460 mm
500
120500136.1min
fyt
bsVsAvreq 136.27 mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041 mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
7.2.3. Level 2 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
74 Dongyezhan Liu 1421941
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
(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= 0.012
(N* & Mcorres) case1.2G+1.5Q1+1.5Q2
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
From Column chart 500/30/0.8 & 0.9 → minimum pt=0.011
Thus, the maximum pt= 0.012
Asreq= pt×b×h= 0.012×500×500= 3000mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 6.639.452
3000
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V314.663 kn where =0.75
Vn= bd
Vn1.5374Mpa
75 Dongyezhan Liu 1421941
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
388.104331
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
04.1094500
500500872.0
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
84860
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 850 mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 850 mm
500
120500872.0min
fyt
bsVsAvreq 104.678mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041 mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
76 Dongyezhan Liu 1421941
7.2.4. Level 1 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
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
(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) case1.2G+1.5Q1+1.5Q2
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
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
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= 4.439.452
2000
24
D
sreq
A
A The minimum reinforce bars is 8.
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 Dongyezhan Liu 1421941
Vn= 75.0
214.376*
V285.834 kn where =0.75
Vn= bd
Vn1.248Mpa
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
142794931
31
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
Use 4 legged R10 stirrups Avprov= 314.16mm2
Av=
497.864500
500500692.0
4fy
bhs
Vmm < Avprov= 314.16mm2 OK
31091
45850016.314
Vs
dfAS
s
dfAV
ytvprov
req
req
ytvprovs 2310 mm
10db = 10 x 24 = 240 mm
Smin= mm1254
500
4
b Smin=120mm
mm1503
458
3
d
Sreq = 2310 mm
500
120500692.0min
fyt
bsVsAvreq 83.037 mm < Avprov= 314.16mm2 OK
For anti-buckling:
500
12050030
16
1
16
1 min'
vmin
yt
cc
f
sbfA 41.079mm< Avprov= 314.16mm2 OK
78 Dongyezhan Liu 1421941
24500135
120500248
135
A minb D
df
sfA
byt
y
te134.041 mm< Avprov= 314.16mm2 OK
4 legged R10 @ 120c/c
7.3. Internal columns
7.3.1. Roof level Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
Mdes= 105.789knm Ndes=454.292kn
𝑁∗
𝜙.𝑏.ℎ =
454292
0.85× 500×500 = 2.14 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 105789000
0.85×500×5002 = 1 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= 0.012
(N* & Mcorres) case1.35G
Mdes= 36.89knm Ndes=600.272kn
𝑁∗
𝜙.𝑏.ℎ =
600272
0.85× 500×500 = 2.82 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 36890000
0.85×500×5002 = 3.17 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
Thus, the maximum pt= 0.012
Asreq= pt×b×h= 0.012×500×500= 3000mm2
(Cl 10.3.8.2) Minimum number of longitudinal reinforcement bars
N= 6.639.452
3000
24
D
sreq
A
A The minimum reinforce bars is 8.
Thus, use 8D24
Stirrups
79 Dongyezhan Liu 1421941
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*
V160.127 kn where =0.75
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
60027231
31
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.
Use 4 legged R10 stirrups Avprov= 314.16mm2
4 legged R10 @ 240c/c
7.3.2. Level 3 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
Mdes= 120.9knm Ndes=974.727kn
𝑁∗
𝜙.𝑏.ℎ =
1974727
0.85× 500×500 = 4.59 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 120900000
0.85×500×5002 = 1.14 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) case1.2G+1.5Q1+1.5Q2
Mdes= 9.5knm Ndes=1342.712 kn
80 Dongyezhan Liu 1421941
𝑁∗
𝜙.𝑏.ℎ =
1441401
0.85× 500×500 = 6.32 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 168908000
0.85×500×5002 = 0.09 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
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= 4.439.452
2000
24
D
sreq
A
A The minimum reinforce bars is 8.
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*
V201.353kn where =0.75
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
134271231
31
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
Thus, no need stirrups but constructive.
Use 4 legged R10 stirrups Avprov= 314.16mm2
4 legged R10 @ 240c/c
81 Dongyezhan Liu 1421941
7.3.3. Level 2 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
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
(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) case1.2G+1.5Q1+1.5Q2
Mdes= 30.435knm Ndes=2123.84 kn
𝑁∗
𝜙.𝑏.ℎ =
2123840
0.85× 500×500 = 9.99 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 30435000
0.85×500×5002 = 0.29 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
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= 4.439.452
2000
24
D
sreq
A
A The minimum reinforce bars is 8.
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 Dongyezhan Liu 1421941
Vn= 75.0
214.819*
V286.428kn where =0.75
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn=
2'
*
50030
212384031
31
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
Thus, no need stirrups but constructive.
Use 4 legged R10 stirrups Avprov= 314.16mm2
4 legged R10 @ 240c/c
7.3.4. Level 1 Reinforcement bars
(Ncorresponding & M*) caseG+Eu+Q
Mdes= 204.836knm Ndes=2029.268kn
𝑁∗
𝜙.𝑏.ℎ =
2029268
0.85× 500×500 = 9.55 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 204836000
0.85×500×5002 = 1.93 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) case1.2G+1.5Q1+1.5Q2
Mdes= 15.532knm Ndes=2912.379 kn
𝑁∗
𝜙.𝑏.ℎ =
2912379
0.85× 500×500 = 13.71 MPa
𝑀∗
𝜙.𝑏.ℎ2 = 15532000
0.85×500×5002 = 0.15 MPa
From Column chart 500/30/0.8 & 0.9 → minimum pt<0
83 Dongyezhan Liu 1421941
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= 4.439.452
2000
24
D
sreq
A
A The minimum reinforce bars is 8.
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=
*V255.513 kn where =0.75
458500
243
bd
DAP S 0.0059
Vb= 708.0p1007.0' cf)( Check:
''2.0708.008.0 cc ff OK
Kn= Agf
N
c
'
*31 2.165
Vc=KaKnVbAcv=1×2.165×0.708×500×458=351kn
Vs=Vn- Vc= 255.513-351= -95kn
Thus, no need stirrups but constructive.
Use 4 legged R10 stirrups Avprov= 314.16mm2
4 legged R10 @ 240c/c
8. Conclusions
84 Dongyezhan Liu 1421941
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 Dongyezhan Liu 1421941
Shear force
Axial load
86 Dongyezhan Liu 1421941
10.1.2. 1.2G+1.5Q1
Bending moment
Shear force
87 Dongyezhan Liu 1421941
Axial load
10.1.3. 1.2G+1.5Q2 Bending moment
88 Dongyezhan Liu 1421941
Shear force
Axial load
89 Dongyezhan Liu 1421941
10.1.4. 1.2G+1.5Q1+1.5Q2 Bending moment
Shear force
90 Dongyezhan Liu 1421941
Axial load
10.1.5. G+Eu+𝜑𝑐𝑄 Bending moment
91 Dongyezhan Liu 1421941
Shear force
Axial load
92 Dongyezhan Liu 1421941
10.2. Transversal section
10.2.1. 1.35G Bending moment
Shear force
93 Dongyezhan Liu 1421941
Axial load
10.2.2. 1.2G+1.5Q1 Bending moment
94 Dongyezhan Liu 1421941
Shear force
Axial load
95 Dongyezhan Liu 1421941
10.2.3. 1.2G+1.5Q2 Bending moment
Shear force
96 Dongyezhan Liu 1421941
Axial load
10.2.4. 1.2G+1.5Q1+1.5Q2 Bending moment
97 Dongyezhan Liu 1421941
Shear force
Axial load
98 Dongyezhan Liu 1421941
10.2.5. G+Eu+𝜑𝑐𝑄 Bending moment
Shear force
99 Dongyezhan Liu 1421941
Axial load