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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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___

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Name: ________Dongyezhan Liu______

Signed: _______Dongyezhan Liu_______ Date: _____10/10/2016______

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I ________Dongyezhan Liu_________ declare that this research or design project:

(Student’s full name)

Either

Does not involve humans as participants?

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Signed: ________Dongyezhan Liu_______ Date: _____10/10/2016______

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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.


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


6.1.2. 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


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


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


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


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)


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)


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.


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


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.


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.


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:


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


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:


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


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)


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 ↓


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]



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


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



Roof 1500.716 15 22510.74 126

T Shallow soil

0.9 1.29

1 1.19


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)


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:


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


Figure 18 Maximum shear force diagram for longitudinal section


Figure 19 Maximum axial load diagram for longitudinal section


5.2. Transversal section

Figure 20 Maximum bending moment diagram for transversal section


Figure 21 Maximum shear force diagram for transversal section


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


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


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

′


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 × 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


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


=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





′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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:



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


=623604000

0.85 × 350 × 684 × 30= 102.151𝑚𝑚


jd=d −𝑎

2= 760 − 102.151 ÷ 2 = 708.924𝑚𝑚

𝐴𝑠 =𝑀𝑛

𝑓𝑦𝑗𝑑=

623604000

500 × 708.924= 1759.3𝑚𝑚2

AD20=314.16mm2




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (𝑂𝐾)


6.1.1.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:




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


=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




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (𝑂𝐾)


6.1.2. Roof level (Stirrups)


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

′


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




𝑑

𝑉𝑠=

157.08×500×760

192185= 310𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


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


6.1.3.1. Point○a of maximum negative moment case 1.2G+1.5Q2:



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


=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




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2




′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (𝑂𝐾)


6.1.3.2. Point○b of maximum positive moment case 1.2G+1.5Q2:



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


=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





′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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:



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


=793365000

0.85 × 350 × 684 × 30= 129.96𝑚𝑚


jd=d −𝑎

2= 760 − 129.96 ÷ 2 = 695𝑚𝑚

𝐴𝑠 =𝑀𝑛

𝑓𝑦𝑗𝑑=

793365000

500 × 695= 2283𝑚𝑚2

AD20=314.16mm2




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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:



𝑀𝑛 =𝑀∗

𝜑=

339.337

0.85= 399.22𝑘𝑛𝑚


d = h − 𝐶𝑐 −𝐷

2= 800 − 30 −

20

2= 760𝑚𝑚

Assume jd= 0.9d= 760×0.9= 684mm

∑ 𝐹 = 0 C=T

a =𝑀𝑛


=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




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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


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




𝑑

𝑉𝑠=

157.08×500×760

258750= 230𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


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





𝑑

𝑉𝑠=

157.08×500×760

259946= 220𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


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）


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


=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


→3HD20 is required (Asreq=942.478 mm2)


′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2




′

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


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


=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





′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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


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


=472991000

0.85 × 350 × 684 × 30= 77.48𝑚𝑚


jd=d −𝑎

2= 760 − 77.48 ÷ 2 = 721.26𝑚𝑚

𝐴𝑠 =𝑀𝑛

𝑓𝑦𝑗𝑑=

472991000

500 × 721.26= 1311.57𝑚𝑚2

AD20=314.16mm2




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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:



𝑀𝑛 =𝑀∗

𝜑=

180.401

0.85= 212.236𝑘𝑛𝑚


d = h − 𝐶𝑐 −𝐷

2= 800 − 30 −

20

2= 760𝑚𝑚

Assume jd= 0.9d= 760×0.9= 684mm

∑ 𝐹 = 0 C=T

a =𝑀𝑛


=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




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


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


=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





′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (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




𝑑

𝑉𝑠=

157.08×500×760

163494= 360𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


16√𝑓𝑐

′𝑏𝑤𝑠

𝑓𝑦𝑡=

1

16√30

350 × 360

500= 86.266𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾

Thus, R10@ 360 c/c



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




𝑑

𝑉𝑠=

157.08×500×760

191343= 310𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


16√𝑓𝑐

′𝑏𝑤𝑠

𝑓𝑦𝑡=

1

16√30

350 × 310

500= 74.28𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾

Thus, R10@ 310 c/c


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:




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


=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




′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (𝑂𝐾)



6.2.3.2. Point ○b of maximum positive moment case 1.2G+1.5Q2:



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


=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



′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2




′

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:



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


=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





′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

6𝑓𝑦=

40

6 × 500= 0.013 < 0.025 (𝑂𝐾)


6.2.3.4. Point ○d of maximum positive moment case 1.2G+1.5Q2:



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


=317007000

0.85 × 350 × 684 × 30= 51.928𝑚𝑚


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



′


√30

4×500× 350 × 800 = 728.471𝑚𝑚2


′


40

6×500× 350 × 800 = 3546.667𝑚𝑚2



′

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

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

′


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




𝑑

𝑉𝑠=

157.08×500×760

258760= 230𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾

Check:


16√𝑓𝑐

′𝑏𝑤𝑠

𝑓𝑦𝑡=

1

16√30

350 × 230

500= 55.115𝑚𝑚2 < 𝐴𝑉 = 157.08𝑚𝑚2 𝑂𝐾

Thus, R10@ 230 c/c


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




𝑑

𝑉𝑠=

157.08×500×760

278724= 210𝑚𝑚


2=

760

2= 380𝑚𝑚 𝑂𝐾


Check:


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


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


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


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




(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= 4.439.452

2000

24

D

sreq

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*

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



Av=

532.1764500

500500124.1

4fy

bhs


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


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f


24500135

120500248

135

A minb D

df

sfA

byt

y


4legged R10 @ 120c/c


(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



Asreq= pt×b×h= 0.014×500×500= 3500mm2


N= 7.739.452

3500

24

D

sreq

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*

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


Av=

114.1864500

500500894.1

4fy

bhs


770225

45850016.314

Vs

dfAS

s

dfAV

ytvprov

req

req

ytvprovs 320mm

10db = 10 x 24 = 240 mm


Smin= mm1254

500

4

b Smin=120mm

mm1503

458

3

d

Sreq = 320 mm

500

120500.4891min

fyt


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f


24500135

120500248

135

A minb D

df

sfA

byt

y

te134.041 mm< Avprov= 314.16mm2 OK



(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


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= 6.639.452

3000

24

D

sreq

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*

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



Av=

995.1014500

5005000.816

4fy

bhs


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


24500135

120500248

135

A minb D

df

sfA

byt

y


4legged R10 @ 120c/c

7.2. Marginal columns (Transversal section)


(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




(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= 7.739.452

3500

24

D

sreq

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*

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


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=

719.1484500

50050019.1

4fy

bhs


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


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f


24500135

120500248

135

A minb D

df

sfA

byt

y





Mdes= 199.312knm Ndes=506.22 kn

𝑁∗

𝜙.𝑏.ℎ =

506220

0.85× 500×500 = 2.38 MPa


𝑀∗

𝜙.𝑏.ℎ2 = 199312000

0.85×500×5002 = 1.88 MPa




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= 7.739.452

3500

24

D

sreq

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*

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


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


Av=

.9481414500

500500136.1

4fy

bhs


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


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f


24500135

120500248

135

A minb D

df

sfA

byt

y






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




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= 6.639.452

3000

24

D

sreq

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*

V314.663 kn where =0.75

Vn= bd

Vn1.5374Mpa


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


Av=

04.1094500

500500872.0

4fy

bhs


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


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f


24500135

120500248

135

A minb D

df

sfA

byt

y






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




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= 4.439.452

2000

24

D

sreq

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


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


Av=

497.864500

500500692.0

4fy

bhs


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


For anti-buckling:

500

12050030

16

1

16

1 min'

vmin

yt

cc

f



24500135

120500248

135

A minb D

df

sfA

byt

y



7.3. Internal columns




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= 6.639.452

3000

24

D

sreq

A


Thus, use 8D24

Stirrups


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.






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∗

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= 4.439.452

2000

24

D

sreq

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*

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







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∗

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= 4.439.452

2000

24

D

sreq

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


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







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∗

𝜙.𝑏.ℎ =

2912379

0.85× 500×500 = 13.71 MPa

𝑀∗

𝜙.𝑏.ℎ2 = 15532000

0.85×500×5002 = 0.15 MPa




Asreq= pt×b×h= 0.008×500×500= 2000mm2


N= 4.439.452

2000

24

D

sreq

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=

*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




8. Conclusions


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


Shear force

Axial load


10.1.2. 1.2G+1.5Q1

Bending moment

Shear force


Axial load

10.1.3. 1.2G+1.5Q2 Bending moment


Shear force

Axial load


10.1.4. 1.2G+1.5Q1+1.5Q2 Bending moment

Shear force


Axial load

10.1.5. G+Eu+𝜑𝑐𝑄 Bending moment


Shear force

Axial load


10.2. Transversal section

10.2.1. 1.35G Bending moment

Shear force


Axial load



Shear force

Axial load



Shear force


Axial load

10.2.4. 1.2G+1.5Q1+1.5Q2 Bending moment


Shear force

Axial load


10.2.5. G+Eu+𝜑𝑐𝑄 Bending moment

Shear force


Axial load

Structural design of a four storey office building

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Transcript of Structural design of a four storey office building