Thesis civil eng [revised i]

1
St ID：2018400151
Shenyang Agriculture University
Graduation Thesis
(2017-2021 Batch)
Title：Design of Single Villa Residence in Community of Shenyang
School： Institute Of Water Conservancy
Major： Civil Engineering
St. Name： Chandio Naeem Ahmed
hang Yuqing
Z
：
Professor
Date： 10/06/2021

2
Title
Design of Single Villa Residence in
Community of Shenyang
Start
date
March 5, 2021
Student
Name
Chandio Naeem Ahmed
Supervisor
name
Zhang Yuqing
1. Graduation dissertation (design) main content
Constructing Design of Single Villa Residence in Community of Shenyang includes thorough
investigation and collection of all technical data that includes height of entire building, height of
eaves and foundation depth. It involves the elevation and selection of building design along
with calculated structural plan and creates a digital construction structure.
The architectural design includes:
1. Project overview
2. General plan design
3. Plan design
4. Elevation design
5. Section design
Structural design contains:
1. Structural layout and calculation diagram
2. Load calculation
3. Structural calculation
4. Stair structure design
5. Basic design
2. Graduation dissertation (design) basic requirement
a. Collection of information through research
b. Drawing architectural plans including elevation, section and other
details
c. Structure diagrams contains beam layouts, slabs, columns,
reinforcements and large scale node drawings
d. Detailed architectural design description
e. Structure design specification
f. Thesis following prescribed format of school and essay should be
minimum 15,000 words.

3
3. Graduation dissertation (design) schedule
Stage Stage name Date
1 Complete design drawings such as building
plan
2021. 3.19—
2021.4.29
2 Improve related design content
2021.3.19 —
2021.4.29
3 Study related Literature from publications
2021. 4.30—
2021. .5.2
4 Write initial draft of the paper
2021.5.3—
2021. 5.26
5 Revision and improvements in paper
2021. 5.26—
2021. 6.1
6 Paper completion 2021. 6.10
Message：
3. Information to be collected and main references Supervisor assignment）
 Encyclopedia of current building design codes (school library)
 Atlas of related building structures (school library)
 Architectural design materials collection (1-6) (school library)
 Housing Architecture
 Architecture Journal (School Library)
 World Architecture (School Library)
 Construction Economy (School Library)
 Encyclopedia of current structural design specifications (school library)
 Reinforced concrete structure
 10. Masonry structure

4
Shenyang agriculture university
Graduation dissertation (design) title Selected and approved form
Title
Design of Single Villa Residence
in Community of Shenyang Title source
ID for student 2018400151 Name
Chandio
Naeem
Ahmed
Major
Civil
Engineering
Supervisor name Zhang Yuqing Professional title Lecturer
Research contents
The main horizontal loads of this building are wind loads and horizontal
earthquakes. In the calculation process, the internal force analysis of the frame
under the horizontal load is first carried out, and then the vertical internal force
is calculated. The column axial force, the column end bending moment and the
column and shear force are mainly calculated. Beam end shear and bending
moment. At the same time, load combination calculations are carried out.
During the combination, the maximum negative bending moment and the
maximum positive bending moment at the beam end, the maximum positive
bending moment in the span and the maximum shear force at the beam end are
the main checking calculation steps. In the foundation design process, the
foundation bearing capacity must be checked first, and finally the foundation
bottom surface reinforcement can be carried out.
Research plan Ready to consult reference phase, to be completed before March 19, 2021.
Architectural design stage: preliminary formulation of the building's
dimensions, materials and elevation, etc., expected to be completed before May
20, 2021.
Structural design stage: calculation and check calculation basis and check
whether each component is qualified expected to be completed before June
2021.
Characteristic Use cast-in-place concrete frame structure, that comes with good seismic
performance, mature construction technology, abundant raw material resources,
relatively high cost and energy consumption, integrity, rigidity, fire resistance
and durability, and the frame structure has a room layout Flexible and other
features.
Supervisor
Comments
Teaching and
research office
comments
Department
comments

5
Name
Chandio Naeem
Ahmed Major Civil Engineering
Supervisor
name
Zhang Yuqing
Professional
title
Lecturer To guide the
number of
graduates this
year
11
Title
Design of Single Villa Residence in Community of
Shenyang
Guidelines
processing
Time Location Guidelines content
2021.3.10
2021.3.26
2021..4.6
2021.5.10
2021.5.25
2021.5.30
2021..6.2
2021.6.8
Decentralized guidance
WeChat
Collective Guidance
WeChat Group
Collective Guidance
WeChat Group
Collective coaching of
Tencent meetings
Decentralized guidance
WeChat
Collective Guidance
WeChat Group
Collective Guidance
WeChat Group
Collective coaching of
Tencent meetings
Explanation of the task book
Inspection and Suggestions on the First Draft
of Architectural Drawings
Review and suggestion of architectural
drawings
Review and suggestion of structural drawings
Checking the final draft of architectural
drawings
Drawing inspection and suggestion
Modification and format modification of
calculation sheet
Final review and revision suggestions for
calculations and drawings
Student signature：
Date:
Supervisor signature：
Date:
Director of teaching and research office signature：
Date:

6
Reviewer comments：
Reviewer (signature)：
Date:
Graduation dissertation (design) examination and comments form
Title: Design of Single Villa Residence in Community of Shenyang
Name: Chandio Naeem Ahmed ID for student：2018400151 Major: Civil
Engineering
Supervisor Comments：
Supervisor (signature)：
Date:

7
Mark：
The defense committee comments：
Chairman (signature）：
Date

8
Table of Contents
Abstract..........................................................................................................................................................................9
Preface..........................................................................................................................................................................10
CHAPTER 1: ARCHITECTURAL DESIGN .............................................................................................................12
1.1 Research content and design scheme.....................................................................................................12
1.2 Design Information................................................................................................................................12
1.3 Project Overview ...................................................................................................................................12
1.4 Architectural Design..............................................................................................................................12
1.5 Room Layout .........................................................................................................................................18
1.6 Engineering Structure............................................................................................................................19
1.7 Fire Protection and Safety .....................................................................................................................21
CHAPTER 2: STRUCTURAL DESIGN ....................................................................................................................22
2.1 Selection of structural scheme and material ..........................................................................................22
2.2 Structure layout scheme and structure selection....................................................................................22
2.3 Selection and Size Estimation of Main Components.............................................................................30
2.4 Internal force analysis of frame under load ...........................................................................................34
2.5 Calculation of internal force and top lateral displacement underearthquake load ...........................35
2.6 Calculation of internal force and lateral displacement of framestructure under wind
load ..............................................................................................................................................................37
2.6.3 Internal Force Calculation Under Wind Load ....................................................................................39
2.7 Internal force analysis of transverse frame under vertical load.............................................................39
2.8 Combination of internal forces of transverse frame ..............................................................................45
2.9 Section design........................................................................................................................................49
2.10 Foundation design................................................................................................................................53
2.11 Staircase Design...................................................................................................................................58
2.12 Design of cast in place concrete slab...................................................................................................62
References....................................................................................................................................................................67
Acknowledgement.......................................................................................................................................68

Residential Building Design (Study Area is Shenyang)
9
Abstract
With the rapid development of our country's economy the rapid development of market
economy has created a large number of villa houses, but in the design, form is still using the
western classicism, modernism and international style, whether it is architectural modeling,
architectural color, or indoor functional area planning generally tend to a model. And my
design style is modern style, modern style villa design is more aimed at modern aesthetic, the
pursuit of fashion characteristics at the same time to create a comfortable living environment,
different from the traditional Chinese villa style, modern style villa design more diversified,
diversified overall more avant-garde and modern society more in line with. So, with this villa
design, I More attention is paid to the pertinence of villa architecture, because it is not only
the place where people live, but also as a place for visitors to relax, so my design is to
integrate modern elements of the single-family villa design.
For the convenience of use, this single-family villa has two floors, adopts asymmetric
design, highlights its independence, emphasizes simplicity and generosity, and the large
terrace outside the second bedroom is the highlight of this design. The bright spot of this
design gives a refreshing visual feeling under the premise of taking into account the structural
safety. With a grid, at the same time pay attention to the independence of single-family villa,
beautiful, set up more personalized appearance design, with targeted supporting facilities.
The concrete parameters of the building are as follows: the height of the building is 7.7
meters, the basic wind pressure is 0.35 KPa, the basic snow pressure is 0.40 KPa the
calculation of the structure design is based on the manual calculation of one product frame to
check the internal force load and the use of PKPM computer software to assist, the two
combined to verify the rationality and feasibility of the design.
Design is mainly aimed at Shenyang area, the address I choose in the scenic Qipanshan
near Qipanshan, because of the characteristics of the daily wind and the characteristics of
Shenyang winter low temperature and snow, because the address is located in the northeast of
the climate cold and dry characteristics, relative structural insulation performance is more
important. I designed it specifically.
.
Key words: Villa; Frame structure; Design description; Modern style

10
Preface
The graduation design of the single-family villa I chose this time uses all the knowledge
I have learned in the past four years in university and the comprehensive application under the
guidance of teachers. This is not only the design of a villa, but also the integration of what I
have learned in the past four years. I mainly use the design based on three major studies,
including the principle of reinforced concrete and housing architecture I think the design
industry is very important this time.
With the development of society, the country has made great breakthroughs in various
fields, especially in architecture. In recent years, new structural forms have emerged, and new
software has been used to assist building construction. In order to promote the integration and
development of BIM Technology and architectural design course teaching, this paper starts
from the necessity and significance of teaching reform combining the two. In recent years, the
architectural form of villa has become more and more diverse. In the process of architectural
design, I think the purpose of the building is to provide accommodation for people, so we
should adhere to the people-oriented concept and realize the improvement of the overall
beauty in the case of meeting the needs of residents. It is particularly important for the villa I
designed this time, not only to meet the site selection of the building where the villa is located,
but also to improve the room quality The layout of the villa is particularly exquisite, and the
materials used in the building are also the top priority, so I chose the villa as my graduation
design in the process of topic selection, because the villa is different from the standard
symmetrical structure building, the design of each node in the structural part is more difficult
and more challenging, and the design of architectural appearance is also a test of my
understanding of architectural design in four years of University With imagination, this
design can let my imagination and creativity play out better. At the same time, it can also
make the flexible use of the knowledge I have learned in the aspect of structural calculation,
and make it run through the knowledge I have learned in the whole university for four years.
Through four years of study, although I have mastered the basic knowledge of civil
engineering, and have done a lot of relevant course design in my junior and senior year, I still
have to make sufficient preparation for graduation design. First, I investigated a lot of
knowledge about villas on the Internet, understood the design style of villas at home and
abroad, and also consulted my tutor Zhang Yuqing about villas In the end, combined with
these materials, we began to design the scheme.
The architectural design part mainly uses the BIM software Revit 2016 learned in the
University. In the process of design, I found many shortcomings. By consulting professional
teachers and looking for specifications, I can correct my mistakes a little bit. I think this
graduation design experience will lay a solid foundation for my future work and prepare for

11
my future.
In the design process, I encountered many technical and knowledge difficulties that I didn't
encounter in the past, but they were successfully overcome through the careful guidance of
teacher Zhang Yuqing. At the same time, I also refer to the information on the Internet to
pave the way for my villa graduation design. A lot of information also let me know more
about the hard work of people engaged in the civil engineering industry. Construction is
always carried out on the premise of safety first, so as a person who will be engaged in this
industry, I will cherish the graduation design experience more. Lay a solid foundation for my
future.

12
CHAPTER 1: ARCHITECTURAL DESIGN
1.1 Research content and design scheme
There are many differences between villa and ordinary residence, and there are many
similarities and differences between functional areas “[1]”. The residential building is located
in Shenyang City, Liaoning Province. In the design, the use of courtyard and homestead in
Northeast China should be fully considered to meet the requirements of its use function and
overall structure. The design of the villa is mainly divided into two parts, the overall
architectural design and structural design. And use the relevant software to draw its drawings.
The architectural drawings include: plan, elevation, section, structural node detail, roof
drainage, general plan, door and window list and architectural design description; the structural
drawings include foundation layout, foundation node detail, beam and slab layout, beam and
slab reinforcement, etc.
1.2 Design Information
According to the assignment book provided by Mr. Zhang Yuqing, we know the design
data and parameters of the overall climate conditions in Shenyang.
1.3 Project Overview
The title of this graduation project is the design of the single villas in the happy
community of Shenyang. The address is located near the beautiful Qipan mountain in
Shenyang. The overall number of floors of the building is 2, the height of each floor is 3.5m
3M, the total height of the building is 7.7m (calculated from the outdoor floor), the total
building area is 387.00m2
; the building structure is frame structure; the indoor design
elevation of the ground floor is + 0.500; the indoor and outdoor height difference is 500mm;
the design service life is 50 years.
1.4 Architectural Design
Architectural design of this building consists of Graduation design mission book which
selects the building designs such as <<Building design fire code>> GB50016-2014,
<<Building floor design specifications>> GB50037-2013, <<Technical specifications for
roofing engineering>> GB50345-2012.
The project name is “Shenyang Shenbei New Area Villa Design”, This project adopts the
framework structure system, the building life is 50 years, the building fire resistance grade
two.it building area is about 378 square meters. The project is designed in accordance with the
requirements of gb50763-2012 of the barrier-free design code in the following areas. a,
building entrance and door b, entrance platform c, waiting hall d, elevator car e, public walkway,

13
f, public toilet g, doors and windows. The passageways, floors and walkways of the buildings
are carried out in accordance with the barrier-free design specifications, and there are elevators
with barrier-free label. In the waiting hall, elevator car should be equipped with the
corresponding configuration. the width of passages and walkways in public areas > 1.5m.
passages, walkways and ground construction should meet the requirements for barrier-free use
and specifications. Wheelchair-accessible door fans in public facilities are equipped with
cross-handles and closed-door handles, observing the installation of 0.35m high under the glass
aluminum alloy or stainless-steel door guards. For doors and windows, aluminum alloy or
stainless-steel door guards are necessary.
1.4.1 General layout design
If we want to ensure the rationality of architectural graphic design, we need to do a
good job in the analysis of the conception elements of architectural graphic design “[2]”.
According to, according to the task, the design area is in Shenyang, Northeast China, so we
need to consider the heat preservation measures, pay attention to the cold characteristics of
winter in Northeast China, because it is a villa, so I choose to be located in Qipanshan,
Shenyang.
Figure 1.1: General Architectural Layout
Qipanshan is a 4A scenic spot. The air quality and scenery nearby are very suitable for the
construction of villas. In the general layout, I chose the open area instead of the mountain,
which is more conducive to the construction. The specific general layout is shown in Figure

14
1.1.
1.4.2 Building Design
The building design of this building is completely in strict accordance with the template on
the whole. At the same time, considering the Chinese traditional culture's stress on house
Feng Shui, the layout of rooms is carried out. Sufficient consideration is given to the space
configuration of each room. According to the functional division of different rooms, the
requirements of residents on the room function are satisfied as far as possible on the premise
of ensuring safety, and at the same time in this Architectural design, I also designed the
terrace part on the second floor, which I think is a highlight of the villa. The open space
makes people feel comfortable and adds more fun (hanging garden). Make the architectural
design more humanized and comfortable.
Figure 1.2: Ground Floor Plan

15
Figure 1.3: First Floor Plan

16
1.4.3 Facade Design
In order to ensure the overall beauty and independence of villa design, different from the
symmetry of traditional villa, this design adopts asymmetric design. Considering the characteristics
of cold and dry climate in winter in Northeast China, compared with the humid environment in South
China, thermal insulation performance should be considered in the overall structure design. Design
on the premise of meeting the structural safety. I designed several balconies on the second floor. In
order to facilitate the outdoor experience of the residents, the modern style is adopted to give people a
feeling different from the traditional villa.
Figure 1.4: North Façade
Figure 1.5: South Façade

17
z
Figure 1.6: West Façade
Figure 1.7: East Façade
1.4.4 Section design
Section design is also an important part of architectural design, which can directly reflect the

18
z
internal combination relationship of buildings.
1.5 Room Layout
Villa as a whole is a two-story building, facing south from north. The choice of rooms is
mainly composed of bedroom, living room, dining room, storage room, functional room,
kitchen and toilet. In order to consider the lighting, the bedroom part is mainly selected on the
sunny side of the house, and the functional room is placed on the north side of the building. In
order to give the residents a good life experience, I set up a fitness and multimedia room.
1.5.1 Bedroom
The main function of the living room is to provide a place for residents to sleep and rest. Four bedrooms
on the two floors are all located in the south. Considering heating and lighting, the opening of bedrooms
on the first floor is 4800 mm, the depth is 3600 mm and the opening is 4000 mm, the depth is 3600 mm.
In the second-floor bedroom, the bay is 3000mm, the depth is 3600mm and the bay is 3900mm, the depth
is 3600mm. The balcony is connected to the outside of the second-floor bedroom, so the design is
considering that the residents can have abetter outdoor experience. For the sake of safety, parapets and
railings are arranged aroundthe balcony.
1.5.2 Multimedia room
The opening and depth of the multimedia room are 4800 mm and 3600 mm respectively.
The overall space is large enough. In order to satisfy the spiritual life of modern people, I set
up a multimedia room on the back of the second floor. The room is equipped with projection
curtain wall and audio equipment. At the same time, it meets the needs of watching movies
and singing, so that residents can experience the fun of cinema and KTV without leaving
home.
1.5.3 Gymnasium
For modern people, the fast-paced life is that people have less and less time to exercise,
and health is the foundation of everything. So for this purpose, I built a fitness room to make
it convenient for people who want to exercise but have no time to go out.
1.5.4 Restaurant and kitchen
The kitchen and dining room are connected in one without partition in the middle,
which can not only increase the use area of the room, but also facilitate the use of residents.
For modern people, the use of the kitchen is not only a place for cooking food, but also a
space for communication with family members.
1.5.5 Wash Room
Toilet design in the building is particularly important, not only to consider the number
of layouts in the overall space, but also to do a good job in the ground waterproof design.
1.5.6 Store Room
Storage room is also very important in the whole building. It can not only make the

19
z
overall layout of the room tidier, but also better organize the goods.
1.6 Engineering Structure
1.6.1 Foundation
The foundation of this design adopts the independent foundation under the column.
Combined with the overall robustness of the structure, the multi-objective optimization design
of the independent foundation is carried out to determine the best solution for the design of
the independent foundation under the column “[2]”. On the choice of materials, I choose the
strength of C20 concrete.
1.6.2 Walls
Internal and external walls are made of 200 mm thick aerated concrete blocks, and the
external walls are provided with thermal insulation layer. The overall height of the wall is 900
mm, and the wall is made of 200 mm thick aerated concrete blocks. Internal and external 20
thick cement mortar painting.
Figure 1.8: Sectional View of the Wall
1.6.3 Floor Structure
According to the structure and fire rating requirements, each floor adopts 100 thick
cast-in-place reinforced concrete slab, paved with ceramic tiles.
1.6.4 Roof Structure
50 cm thick C20 fine aggregate concrete protective layer
1:1 cement sand compaction
60 cm thick 3:7 grey soil

20
Figure 1.9: Cross-section of Roof
Figure 1.10: Cross-section of Roof daughter

21
40 cm thick fine stone concrete protective layer
20 cm thick cement mortar for specified slope
80 cm thick slag cement insulation layer
150 cm thick water reinforced mixed house panels
15 cm thick mixed lime mortar plastering
1.6.5 Stairs Structure
The stairs structure for the building is shown below in figure 1.10.
Figure 1.11: Structure of Stair steps
1.7 Fire Protection and Safety
Fire safety is also the most important part to be considered in this design. There are
four fire extinguishers on each floor. At the same time, the kitchen on the first floor is a high-
risk place for fire, so it is particularly important.

22
CHAPTER 2: STRUCTURAL DESIGN
2.1 Selection of structural scheme and material
China is an earthquake prone country, so the reinforced concrete frame structure with
good seismic performance is widely used in buildings”[3]”. In this design, it is initially
determined to be a frame structure, and the section size of the beam is determined according to
the span of the beam. The concrete grade of the beam and column is C30, the reinforcement
HRB400, and the stirrup HRB335. The foundation and foundation cushion are constructed
separately with C20 concrete foundation, column bars are inserted and poured to the top of the
foundation.
2.2 Structure layout scheme and structure selection
2.2.1 Overall bearing scheme of structure
The overall structure adopts the frame bearing scheme, and the layout of frame beam and
frame column is shown in the structural layout
Figure 2.1: Ground Floor Structural Layout

23
Figure 2.2: First Floor Structural Layout
Figure 2.3: Structural Plan of the building

24
Figure 2.4: Structural Plan for the roof beams of building

25
Figure 2.5: Structural Plan for the Foundation of building

26
Figure 2.6: Structural drawing for the Column#1 with its Foundation

27

28

29

30
2.3 Selection and Size Estimation of Main Components
2.3.1 Selection of main components
(1) Beam, slab and column, the overall structure is cast-in-place reinforced concrete.
(2) Wall: aerated concrete block.
(3) Thickness of wall: outer wall: 200 mm, inner wall: 200 mm.
(4) Foundation: natural foundation shallow foundation (independent foundation under
column)
2.3.2 Estimation of section size of beam and column
(1) Main load bearing frame: select the frame with larger span for calculation
L=5700mm h= (1/8~1/12) L=712mm~475mm
h=500mm. b= (1/2~1/3) h=250mm~167mm
b=250mm
b>200mm Therefore, the section size of the initial part of the main frame beam is
selected as：b × h=250mm × 500mm
A~B Span beam l0=3600mm，take h=500mm，take 𝑏 =250mm
B~C Span beam l0=1800mm，take h=500mm，take 𝑏 =250mm
C~D Span beam l0=3600mm，take h=500mm，take 𝑏 =250mm
(2) Frame column:
h= (1/15~1/20) H=250mm~165mm take b=250mm
b= (1~2/3) h takeb=h
① According to this value, the seismic grade of the building is grade 0.85 ，C30
concrete is selected
Fc=14.3
According to the requirements of code for seismic design of buildings:
Considering that the building is located in the earthquake area, the size of column section
should not be less than 300 mm. Therefore, the section size of the primary column is as
follows: b*h=350mm*350mm。
2.3.3 Roof and floor loads
Dead load:

31
roofing:
Roof (not accessible)
40 thick C20 fine aggregate concrete protective layer 22×0.04=0.88 KN/m2
4-thick APP modified asphalt waterproof layer 0.40 KN/m2
10 thick cement mortar leveling layer 20×0.01=0.20 KN/m2
80 thick polystyrene insulation layers 0.5×0.08=0.04 KN/m2
10 thick 1:3 cement mortar leveling layer 20×0.01=0.20 KN/m2

32
100 thick cast-in-place concrete roof slab 25×0.10=2.50 KN/m2
20 thick mixed mortar plastering 20×0.02=0.40 KN/m2
Total 4.92 KN/m2
Floor:
Tile floor 0.55 KN/m2
20 thick 1:3 cement mortar leveling layer 20X0.02=0.40 KN/m2
100 thick reinforced concrete floor 25X0.10=2.50 KN/m2
20 thick mixed mortar plastering 20X0.02=0.40 KN/m2
2.3.4 Self weight of components
(1) Column weight
Upper column: side column：0.35*0.35*25*3=9.18KN/pieces
Center pillar：0.35*0.35*25*3=9.18KN/pieces
Bottom column: column: 0.35 * 0.35 * 25 * 3.5 = 10.72KN/piece
(2) Load unit of reinforced concrete beam（KN/m)
Beam: 250mm * 500mm beam dead weight 0.25*0.5*30=3.75
Beam side painting 2*（0.5-0.1）*0.02*20=0.32
Total 3.75+0.32=4.07
(3) Representative value of gravity load on walls, doors and windows
（1）External wall brick（KN/m2)
External wall: aerated concrete block with dead weight of 6KN/m2
200 thick brick wall 0.2*6=1.2
8 thick exterior wall facing brick 0.50
One layer of plain cement slurry 20*0.01=0.2
50 thick polystyrene board insulation 0.5*0.05=0.03
10 thick 1:2.5 cement mortar leveling 20*0.01=0.2
Total 2.13
Internal wall practice unit（KN/m2)
Interior wall: aerated concrete block with dead weight of
6KN/m2
Cement mortar wall
Slurry finish
200 thick brick wall
0.2*6=1.2
20 thick plastering on both sides
0.02*20*2=0.8
Total 2.0
The gravity load per unit area of wooden door is;

33
0.2KN/m2；
The gravity load per unit area of plastic steel window is taken as
0.4KN/m2；
The load area of steel door is unit of gravity
0.4KN/m2
1. Gravity load of bottom wall, doors and windows:
（1） Door and window opening area:
S Outside=1.2*1.5*5+1.5*2*2+0.9*2.1*1+1.6*2.4+2.4*2.1=25.77m2
S Within=0.9*2.1*4=7.56m2
（2） Wall area (including door and window openings)
S out=（12.37*3.5+15.37*3.5）*2=194.18m2
S in=（1.8+0.2）*3*3.5+3.6*3.5*3+4.8*3.5+4.5*3.5+3.1*3.5*2=113.05m2
（3） Gravity load of doors, windows and walls:
(194.18-25.17) *2.13+(113.05-7.56) *2+7.56*0.2+25.77*0.4=582.80KN
2. Gravity load of top wall, doors and windows:
（1） Door and window opening area:
S out=1.2*1.5*6+2*1.5*3+1.6*2.4+3.6*2.1=25.8m2
S in=0.9*2.1*6=11.34m2
（2） Wall area (including door and window openings)
S out=（3.6*2+1.8+0.2）*2*3+15.37*2*3=147.42m2
S in=4.6*2*3+3.6*3+2.8*3+3.6*3*3=79.2m2
（3） Gravity load of doors, windows and walls:
（147.42-25.8）*2.13+（79.2-11.34）*2.0+25.8*0.4+11.34*0.2=407.36KN
Live load：
Roof： 0.5kN/m2
Floor： 2.0kN/m2
Roof snow load： 0.5kN/m
The variable load value of the top roof is as follows: （15.37*11.12-1.6*3.7）*0.5=82.49KN
The value of uniform live load on the floor is: （（3.6*2+1.8*2）*15.37-0.9*3.65+1.3*4.8）
*2.0=337.9KN
2.3.5 Calculation of representative value of gravity load
Roof: representative value of gravity load = (standard of structure weight and standard of
component weight) + (standard value of 0.5 * snow load).
The self-weight of the structure and the self-weight of the components are half of the
self-weight of the top layer.
The calculation of load standard value (unit: KN) is shown in Table 1
2.3.6 Load stratification summary
Top floor: 735.87+0.5*82.49+217.35+0.5*250.80+0.5*407.36=1323.54KN

34
Bottom Floor: 677.49+0.5*337.9+170.325+113.40+0.5*(277.2+250.8）+(582.8+407.36）
*0.5=1864.25KN
G floor =1323.5+1864.25=3187.78KN
2.4 Internal force analysis of frame under load
The linear stiffness of frame beam and frame column is calculated as follows
𝐸𝐼
𝑖 =
𝑙
2.4.1 Calculation of linear stiffness of frame beam and column
Calculation formula of linear stiffness;
AB Span beam：𝑖 =
𝐸𝐼
=
2∗3.0∗107∗0.25∗0.53
= 2.44 ∗ 104𝐾𝑁/. 𝑚
𝑙 12∗3.6
BC Span beam：𝑖 =
𝐸𝐼
=
2∗3.0∗107∗0.25∗0.53
= 4.88 ∗ 104𝐾𝑁/. 𝑚
𝑙 12∗1.8
CD Span beam：𝑖 =
𝐸𝐼
=
2∗3.0∗107∗0.25∗0.53
= 2.44 ∗ 104𝐾𝑁/. 𝑚
𝑙 12∗3.6
Calculation of linear stiffness of columns
Bottom column：𝑖 =
𝐸𝐼
=
3.0∗107∗0.35∗0.353
= 3.75 ∗ 104𝐾𝑁/. 𝑚
𝑙 12∗3.5
Second floor column: 𝑖 =
𝐸𝐼
=
3.0∗107∗0.35∗0.353
= 4.4 ∗ 104𝐾𝑁/. 𝑚
𝑙 12∗3.0
2.4.2 Calculation of lateral stiffness D value
In the graduation project of civil engineering major, D-value method is a common manual
calculation method, which is used to calculate the moment diagram of frame structure under
horizontal load. [3]
The D-value method is used to calculate the Shear Force of each column
Formula used：𝐷 = 𝛼*12i𝑐/ℎ𝑖2（2.4.2）

35
Table 2.1: D values for shear force
Numb
er of
Layers
A B C D £ D
k a D k a D k a D k a D
1 0.6 0.4 1579 2.2 0.6 2240 2. 0.6 22 0.6 0.4 1579 7640
2
0.7 0.2
1589
1.6 0.4 2641 1.6 0.4 2641 0.7 0.2 1584 8452
4 7 6 5 5 6 5 5 4 7 9 8
2.5 Calculation of internal force and top lateral displacement underearthquake
load
2.5.1 Transverse natural period
The calculation results of the hypothetical lateral displacement distance of the structure vertex
are shown in Table 2.2
𝑛
𝑉𝑔𝑖 = ∑𝐺𝑘
𝑘=𝑖
Table 2.2: Vertex Displacement
Arra
nge
ment
Gi (kN) n
Gi (kN)
i
Di Relative
displacement
between
Gi
layers (m)
i (m)

36
i i1
D
1 1323.5
4
1323.54 7640
8
0.017 0.055
2 1864.2
5
3187.78 8452
8
0.038 0.038
2.5.2 Calculation of lateral seismic force
In the code for seismic design of buildings (GBJ11-89), the bottom shear method is used
to calculate the horizontal seismic action, and the seismic action is designed according to 7
degrees. The results of the total horizontal seismic standard value of the structure are shown
in Table 2.3.
Table 2.3: lateral seismic action and seismic Shear Force of each floor
Serial 𝒉𝒊 𝑯𝒊 𝑮𝒊 𝑮𝒊𝑯𝒊 𝑮𝒊𝑯𝒊
𝒏
/ ∑𝑮𝒊 𝑯𝒊
𝒊=𝟏
𝑭𝒊 𝑽𝒊
1 3 6.5 1323.54 8528 0.57 347.07 347.07
2 3.5 3.5 1864.25 6524.88 0.43 261.83 608.90
2.5.3 Checking calculation of frame seismic deformation
The elastic relative rotation angle between layers under seismic load shall meet the

37
requirements of the following formula.
𝟏
𝜽𝒆 ≤ [𝜽𝒆] =
𝟓𝟓𝟎
（2.5.3-1）
Table 2.4: Seismic Shear force of each floor frame under horizontal seismic action
Sr # 𝑽𝒊 𝑫𝒊 𝑽𝒊/𝑫𝒊 𝒉𝒊 𝜽𝒆
1 347.07 76408 0.00441 3 1/976
2 608.90 84528 0.00725 3.5 1/640
2.5.4 Analysis of Internal Force of Frame Under Horizontal Earthquake
Table 2.5: calculation of shear force and bending moment of frame column
Descriptions serial h 𝑽𝒊 𝑫𝒊
∑ 𝑫
D/∑𝑫 𝑽im k 𝒚𝟎 M M
A column 2 3.0 347.07 15849 84528 0.19 65.08 0.74 0.40 78.09 117.14
1 3.5 608.90 15796 76408 0.21 125.88 0.65 0.65 286.38 154.20
B column 2 3.0 347.07 26415 84528 0.31 108.46 1.66 0.42 136.66 188.72
1 3.5 608.90 22408 76408 0.29 178.57 2.21 0.55 343.75 281.25
C column 2 3.0 347.07 26415 84528 0.31 108.46 1.66 0.42 136.66 188.72
1 3.5 608.90 22408 76408 0.29 178.57 2.21 0.55 343.75 281.25
D column 2 3.0 347.07 15849 84528 0.19 65.08 0.74 0.40 78.09 117.14
1 3.5 608.90 15796 76408 0.21 125.88 0.65 0.65 286.38 154.20
Table 2.6: Calculation of bending moment, shear force and axial force at beam end
Serial AB Cross BC Cross CD Cross Axial force of
column
𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb 𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb 𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb Side
pillar
Central
column
1 78.0
9
65.07 3.6 39.8 216.9
2
216.
92
1.8 241 65.0
7
78.09 3.6 65.0
7
-39.8 -201.2
2 232.
29
156.8
6
3.6 163.
2
312.3 312.
3
1.8 462.9 156.
86
232.2
9
3.6 146.
86
-203 -500.9
2.6 Calculation of internal force and lateral displacement of framestructure
under wind load

38
𝑘=𝑖
2.6.1 Load calculation
For the horizontal wind load, because the wind load is perpendicular to the building
surface, and the size of the wind load is related to the wind area, wind shock coefficient 𝛽𝑍,
wind load shape coefficient 𝜇𝑠, wind pressure height variation coefficient 𝜇𝑧 and the local
wind pressure of the building. The calculation formula of wind load is: 𝐹𝑖 = (𝑕𝑖 + 𝑕𝑗)/2
× 𝛽𝑍𝜇𝑆𝜇𝑍𝜔0, where (𝑕𝑖 + 𝑕𝑗)/2 represents the wind area, 𝑕𝑖 ，ℎ𝑗 which is the height of the
upper and lower layers respectively. For the bottom layer, the indoor and outdoor height
difference should be added. For the top layer, it is twice the height of the parapet, and B is the
wind width.
In this design, the building height of the house is 7.7m < 30m, and the influence of
pulsation can be ignored, taking 𝛽𝑧=1.0. According to load code for building structures (GB
5009-2001), the standard value of wind load is calculated by the following formula.
𝜔𝑘 = 𝛽𝑧𝜇𝑠𝜇𝑧𝜔𝑜
Where，𝜔𝑜— Basic wind pressure（kN/m2
）；
𝜇𝑠— Wind load shape coefficient;
𝜇𝑧— Wind pressure height variation coefficient;
𝛽𝑧—z The wind-induced vibration coefficient at the height.
(1) Calculate the wind load 𝑞𝑧 = 15.37*0.35𝛽𝑧𝜇𝑠𝜇𝑧 at the elevation of each floor.
According to the height 𝐻𝑖 of the elevation of each floor, check 𝜇𝑧 and substitute it into the
above formula to get the elevation 𝑞1𝑧 of each floor. As shown in table 2.7. Among them,
𝑞1𝑧 is windward and 𝑞2𝑧is leeward.
Table 2.7: Wind load calculation
Serial 𝐻𝑖（m) 𝜇𝑧 𝛽𝑧 𝑞1𝑧KN/M 𝑞2𝑧KN/M
Top Floor 6.5 1.0 1.0 5.594 3.496
First floor 3.5 1.0 1.0 5.594 3.496
First floor：
5.594+5.594 3 5.594+5.594 3.496+3.496 3
𝐹 = 0.5 [( ) + 5.594] ∗ + ∗ 1 +
2 2 2 2
+3.496
2
* 0.5 ∗ +
2
3.496+3.496 ∗ 1 = 22.725𝐾𝑁 First floor:
2
5.594 + 5.594 5.594 + 5.594 3.496 + 3.496 3.496 + 3.496
𝐹2 = 0.5 [(
2
) + ] ∗ 3.5 + [ + ]
2 2 2
* 0.5 ∗ 3.5 = 31.815𝐾𝑁
2.6.2 Calculation of horizontal displacement of frame under wind load
According to the calculated concentrated horizontal load of nodes in each floor, the
formula𝑉𝑖 = ∑
𝑛 𝐹𝑘
The calculation process is shown in Table 2.8 below.
Table 2.8: Calculation of story shear force and lateral displacement
[ ]

39
Serial Fi（KN） Vi（KN）
∑ 𝐷(N/mm)
𝛥𝜇𝑖（mm） 𝜇𝑖（mm） 𝛥𝜇𝑖/ℎ𝑖
2 22.725 22.725 83889 1.6 5.3 0.533× 10−3
1 31.815 54.54 78661 3.7 3.7 1.05× 10−3
2.6.3 Internal Force Calculation Under Wind Load
The results are shown in table 2.9 and table 2.10
Table 2.9: Calculation of shear force and bending moment of frame column
Column Serial h 𝑽𝒊 𝑫𝒊
∑ 𝑫
D/∑𝑫 𝑽im k 𝒚𝟎 M M
A
column
2 3.0 22.725 15849 84528 0.19 4.26 0.74 0.40 5.11 7.67
1 3.5 54.540 15796 76408 0.21 11.28 0.65 0.65 25.65 13.81
B
column
2 3.0 22.725 26415 84528 0.31 7.10 1.66 0.42 8.95 12.36
1 3.5 54.540 22408 76408 0.29 15.99 2.21 0.55 30.79 25.19
C
column
2 3.0 22.725 26415 84528 0.31 7.10 1.66 0.42 8.95 12.36
1 3.5 54.540 22408 76408 0.29 15.99 2.21 0.55 30.79 25.19
D
column
2 3.0 22.725 15849 84528 0.19 4.26 0.74 0.4 5.11 7.67
1 3.5 54.540 15796 76408 0.21 11.28 0.65 0.65 25.65 13.81
Table 2.10: Calculation of bending moment and shear force at beam end and axial force
of column under wind load
Serial Left side beam Middle beam Right side beam Axial force of
column
𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb 𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb 𝑀𝑙
𝑏 𝑀𝑟
𝑏 L Vb Side
pillar
Central
column
1
5.1
1
4.2
6
3.6 2.6
0
14.2
0
14.2
0
1.8 15.8 4.2
6
5.1
1
3.6 2.6
0
-2.6 -13.2
2
39.
46
18.
66
3.6 16.
1
37.3
2
37.3
2
1.8 41.5 18.
66
39.
46
3.6 16.
1
-18.7 -38.6
2.7 Internal force analysis of transverse frame under vertical load
2.7.1 Load calculation

40
Figure 2-3 calculation diagram under dead load
(1) Dead load calculation
Figure 2.10: First Floor Structural Layout
The mid span bending moment is calculated according to the internal force balance
condition. The mid span bending moment in the simply supported state is calculated first, and
then the mid span bending moment is calculated according to the mid span bending moment
in the simply supported state and the bending moment at both ends of the support 𝑀中 =
𝑀0 − (𝑀𝑟 − 𝑀𝑙)/2。
The shear force calculation formula is：𝑉 =
𝑞𝑙
−
∑ 𝑀
𝑉 =
𝑞𝑙
+
∑ 𝑀 𝑉′
= 𝑉 − 𝑞 ∗ 𝑏/2
𝐿 2 𝑙
，
𝑅 2 𝑙
，
The calculation formula of axial force is ：N=V+PP- The calculation of vertical
concentrated load at the top of column is shown in the table below.
Table 2.11: Bending Moment with dead loads
Compon
ent
A B C
Lay
er
Column Top C Bottom
C
Top C Bottom
C
Top
C
Bottom
C
Top
C
Bottom
C
2 -5.19 -5.42 3.22 3.45 -3.0
6
-3.25 5.19 5.05
1 -3.37 -1.69 2.15 1.08 -2.0
0
-0.99 3.19 1.62
Beam AB Span beam BC Span beam CD Span beam
Left
End
MID Right
End
Left
End
MID Right
End
Left
End
MID Right
End
2 6.39 -10.3 -11.82 7.50 -4.2 -7.13 11.98 -10.4 -6.65

41
0
1 10.29 -13 -15.74 8.34 -5.3 -7.47 15.42 -12.7 -10.48
Table 2.12: Shear under dead load
Compo
nent
A B C
Lay
er
Column Top C Bottom
C
Top
C
Bottom
C
Top C Bottom C Top
C
Bottom
C
2 2.95 -2.95 -1.8
5
1.85 1.75 -1.75 -2.84 2.84 2.95
1 1.23 -1.23 -0.7
9
0.79 0.73 -0.73 -1.17 1.17 1.23
Left
End
MID Righ
t
End
Left
End
MID Right
End
2 17.06 20.16 7.59 7.20 20.28 17.24
1 25.37 28.49 6.89 5.98 27.77 24.95
Figure 2.11: Dead load bending moment diagram

42
(2) Live load calculation
Figure 2.12: Calculation diagram under live load
Table 2.13: Bending moment under live load
Compon
ent
A B C
Lay
er
Column Top C Bottom
C
Top C Bottom
C
Top
C
Bottom
C
Top
C
Bottom
C
2 -0.8 -1.2 0.5 0.6 -0.5 -0.6 0.9 1.30 -0.8
1 -0.5 -1.0 0.7 0.4 -0.7
0
-0.70 1.00 0.50 -0.5
Left
End
MID Right
End
Left
End
MID Right
End
Left
End
MID Right
End
2 0.9 -1 -1.5 1.70 -0.6
0
-1.5 1.4 -1.1 -1.0
1 2.3 -3.4 -1.5 3.3 -0.8 -3.0 4.0 -3.5 -2.5

43
Figure 2.13: Envelope diagram of live load bending moment
Table 2.14: Shear under live load
Compone
nt
A B C
Laye
r
Column Top C Bottom
C
Top C Bottom
C
Top
C
Bottom
C
Top
C
Bottom
C
2 0.5 -0.5 -0.4 0.4 0.5 -0.5 -0.6 0.60 0.
5
1 0.3 -0.3 -0.2 0.2 0.10 -0.10 -0.3 0.3 0.
3
Left
End
MID Right
End
Left
End
MID Right
End
2 1.5 0.3 1.7 1.30 1.7 0.2
1 5.1 1.2 4.00 2.2 6.0 1.4

44
Figure 2.14: Envelope diagram of live load shear
Table 2.15: Column axial force under constant load and live load
Compon
ent
A B C
Lay
er
Column Top C Bottom
C
Top C Bottom
C
Top
C
Bottom
C
Top
C
Bottom
C
2 41.26 52.28 50.95 61.98 64.3
9
75.41 49.14 60.16 41.26
1 107.85 120.41 133.85 146.41 163.
36
175.91 128.51 141.06 107.85
Left
End
MID Right
End
Left
End
MID Right
End
Left
End
MID Right
End
2 3.2 3.2 5.0 5.0 5.6 5.6 3.2 3.20 3.2
1 15.0 15.0 18.3 18.3 22.0
0
22.00 14.6 14.6 15.0

45
Figure 2.15: Axial Force Diagram
2.8 Combination of internal forces of transverse frame
The most unfavorable combination of internal forces for frame beams is as follows:
For beam end section: Mmax Nmax, Vmax
For cross section between beams: Mmax Vmax
The partial coefficients of basic combined loads according to load specification 3.2.5 are
as follows
1. partial coefficient of permanent load:
The combination coefficient of variable load effect control is 1.2
The combination coefficient of permanent load effect control is 1.35
The partial coefficient of variable load is generally taken as 1.4
The calculation is as follows: table 16, 17, 18, 19

46
Table 2.16: Internal Forces with different load combinations
Sr# secti
on
internal
force
SG
K
S
Q
K
SW
K
SE
K
1.2SGK+1.26(S
QK+SWK)
Γre[1.2(SGK+0.5S
QK)+1.3SEK]
1.35
SGK
+SQ
K
1.2S
GK
+1.4
SQK
1st A M 10. 2. 39. 232. 64.97 -34.47 236.78 -216.19 16.1 15.5
floor 29 30 46 29 9 7
V 25. 5. -16 -12. 16.58 57.16 12.81 37.44 39.3 37.5
37 10 .10 63 5 8
B 左 M -15 -1. -18 -15 -44.29 2.73 -167.78 138.10 -22.7 -20.9
.74 50 .66 6.86 5 9
V 28. 1. 16. 12.6 55.99 15.41 38.50 13.87 39.6 35.8
49 20 10 3 6 7
B 右 M 8.3 3. 37. 312. 61.19 -32.86 313.48 -295.50 14.5 14.6
4 30 32 30 6 3
V 6.8 4. -41 -19. -38.98 65.60 -10.95 26.96 13.3 13.8
9 00 .50 44 0 7
Stra MAB -13 -3. -19.88 -19.88 -13.23 -13.23 -20.9 -20.3
ddle .00 40 5 6
MBC -5. -0. -7.37 -7.37 -5.13 -5.13 -7.96 -7.48
30 80
2nd A M 6.3 0. 5.1 78.0 8.35 15.24 2.36 82.29 -69.9 9.53
floor 9 90 1 9 8
V 17. 1. -2. -4.5 21.98 19.09 25.64 11.62 20.4 24.5
06 50 60 2 4 3
B 左 M -11 -1. -4. -65. -16.93 -21.44 -10.71 -74.76 52.1 -17.4
.82 50 26 07 3 6
V 20. 0. 2.6 4.52 24.95 27.85 21.29 22.69 13.8 27.5
16 30 0 7 2
B 右 M 7.5 1. 14. 216. 9.88 29.03 -6.75 219.01 -203. 11.8
0 70 20 92 98 3
V 7.5 1. -15 -6.6 9.96 -8.66 31.16 1.13 14.0 11.9
9 70 .80 3 6 5
Stra MAB -10 -1. -13.62 -13.62 -13.62 -9.72 -9.72 -14.9
ddle .30 00 1
MBC -4. -0. -5.80 -5.80 -5.80 -4.05 -4.05 -6.27
20 60

47
Table 2.17: Moment and Axial Force Combination of A-Column in Transverse Frame
Arrang
ement
Sect
ion
Internal
force
SG
K
SQ
K
SW
K
SE
K
1.2SGK+1.26(S
QK+SWK)
[1.2(SGK+0.5SQ
K)+1.3SEK]
1.35
SGK
+SQ
K
1.2S
GK
+1.4
SQK
1 Top
Lay
er
M -5.1
9
-0.
80
-7.
67
-11
7.14
-16.90 2.43 -119.24 109.18 -7.81 -7.35
N 41.
26
3.2
0
2.6
0
39.8
0
56.82 50.27 77.38 -0.23 58.9
0
53.9
9
Bott
om
Lay
er
M -5.4
2
-1.
20
5.1
1
78.0
9
-1.58 -14.45 70.72 -81.56 -8.52 -8.18
N 52.
28
3.2
0
-2.
60
-39.
80
63.49 70.04 9.69 87.30 73.7
8
67.2
2
2 Top
Lay
er
M -3.3
7
-0.
50
13.
81
154.
20
12.73 -22.07 147.09 -153.60 -5.05 -4.74
N 107
.85
15.
00
18.
70
203.
00
171.88 124.76 301.74 -94.11 160.
60
150.
42
Bott
om
Lay
er
M -1.6
9
-1.
00
-25
.65
286.
38
-35.61 29.03 277.25 -281.19 -3.28 -3.43
N 120
.41
15.
00
-18
.70
-20
3.00
139.83 186.95 -82.81 313.04 177.
55
165.
49
Table 2.18: Moment and Axial Force Combination of Transverse Frame B-Column
Arrang
ements
Sec
tion
Internal
Force
SG
K
S
Q
K
S
W
K
SE
K
1.2SGK+1.26(S
QK+SWK)
Γre[1.2(SGK+0.5S
QK)+1.3SEK]
1.35
SGK
+SQ
K
1.2S
GK
+1.4
SQK
1 Top M 3.2 0.5 -12 -18 -11.08 20.07 -180.88 187.13 4.85 4.56
2 0 .36 8.7
2
N 50. 5.0 13. 69. 84.07 50.81 116.06 -19.85 73.7 68.1
95 0 20 70 8 4

48
Bot
tom
M 3.4
5
0.6
0
8.9
5
136
.66
16.17 -6.38 136.62 -129.87 5.26 4.98
N 61.
98
5.0
0
-13
.20
-69.
70
64.04 97.31 -9.93 125.99 88.6
7
81.3
8
2 Top M 2.1
5
0.7
0
25.
19
281
.20
35.20 -28.28 276.42 -271.92 3.60 3.56
N 133
.85
18.
30
38.
60
201
.20
232.31 135.04 324.87 -67.47 199.
00
186.
24
Bot
tom
M 1.0
8
0.4
0
30.
79
347
.75
40.60 -37.00 340.21 -337.90 1.86 1.86
N 146
.41
18.
30
-38
.60
-20
1.2
0
150.11 247.39 -56.17 336.17 215.
95
201.
31
Table 2.19: Shear Combination of A-Column in Transverse Frame
Arrangem
ents
SG
K
SQ
K
SW
K
SE
K
1.2SGK+1.26(SQK+
SWK)
[1.2(SGK+0.5SQK)+1.
3SEK]
1.35S
GK
+SQK
1.2SG
K
+1.4S
QK
2 -2.9
5
-0.
5
2.4
1
4.1
8
-1.13 -7.21 1.20 -6.96 -4.48 -4.24
1 -1.2
3
-0.
5
4.2
4
10.
56
3.24 -7.45 8.96 -11.63 -2.16 -2.18
Shear combination of B-column in transverse frame
2 1.8
5
0.4 0.9
0
7.1
2
3.86 1.59 8.79 -5.10 2.90 2.78
1 0.7
9
0.2 2.4
6
12.
36
4.30 -1.90 12.85 -11.25 1.27 1.23

2.9 Section design
2.9.1 Section design of frame beam
Take a floor AB span
The strength grade of concrete is C30, the longitudinal reinforcement is HRB400, and
the stirrup is HRB335.
2.6.1.1Calculation of flexural capacity of normal section of beam
Bending moment at support: a node MA=0.75X236.78=177.585KN·m
B node MB=0.75X167.78=125.84KN·m
𝑉𝐴𝐵=1.3X86.61-(65.63+0.7X12.80) =38.01KN
Bearing edge bending moment Mmax=165.08.57-38.01X0.25=155.56KN·m
Mmax =165.08.57-38.01X0.25=155.56KN·m
When the lower part of the beam is in tension, the section is designed as T-shaped
section.
𝑏′
=
𝑙
=
3.6
= 1.2𝑚
𝑓 3 3
ℎ0 = ℎ − 𝑎𝑠 = 500 − 30 = 470𝑚𝑚
ℎ′
/ℎ = 100/470 = 0.21 > 0.1 b ’
=1200mm
𝑓 0
𝛼 𝑓 ℎ′
（ℎ
f
− ℎ ′
/2) =1.0×14.3×1200×(470-100/2) =720.7KN·m>116.67 KN·m
1 𝑓 0 𝑓
The T-section belongs to the first kind.
𝛼𝑠 =
𝑀
𝛼 1𝑓𝑐𝑏𝑓′ℎ0
2
=
116.67×106
1.0×14.3×1200×4702 = 0.030
𝜉 = 1 − √1 − 2𝛼𝑠 = 1 − √1 − 2 × 0.030 = 0.0301
𝐴𝑠 =
𝛼 1𝑓𝑐𝑏𝑓′𝜉ℎ0
=
1.0 × 14.3 × 1200 × 0.030 × 470
= 672. 0𝑚𝑚2
𝐴𝑆𝑚𝑖𝑛
𝑓𝑦
=0.002bh=0.002×250×500=250mm2
360
𝐴𝑠𝑚𝑖𝑛=(0.518ft/fy)bh=（0.55×1.43/360）*250*500=257mm2
The diameter of three HRB400 steel bars is selected 18mm As=763mm2
𝐴𝑠′ =
𝑀
(ℎ0 − 𝑎𝑠′)
177.585 × 106
=
360 × (470 − 30)
= 1121𝑚𝑚2
The diameter of three HRB400 steel bars is selected 22mm As=1140mm2
Upper part of support BL
𝐴𝑠′ =
𝑀
=
(ℎ0 − 𝑎𝑠′)
125.84 × 106
360 × (470 − 30)
= 794𝑚𝑚2
Three HRB400 steel bars with a diameter of 20 mm are selected As=945mm2

𝜌 =
1140
250×470
= 0.97% > 0.3% Meet the structural requirements.
（2）AB Calculation of shear capacity of inclined section of span beam
V=148.71KN<0.25βcfcbh0=0.25*1.0*14.3*250*470=420KN
Therefore, the section size meets the requirements.
The stirrup in the densified area at the end of the beam is selected as double stirrup, and the
strength is hpb335.0.7ftbh0=0.7*1.43*250*665=166KN>Γ RE V=138.71KN Therefore, the
stirrups are only configured according to the structure. The reinforcement grade is HRB335
with the diameter of 8mm and the spacing of 100mm in the densified area, and HRB335 with
the diameter of 8mm and the spacing of 100mm in the non-densified area.
2.9.2 Section Design Of Frame Column
Table 2.20: Checking Calculation of Shear Span Ratio and Axial Compression Ratio
of Columns
Colum
N
Positio
n
B（mm） h（
0 mm） Fc(N/mm2
)
Mc
（KN*m）
Vc
/K
N
N（KN） Mc
/VC
h
0
N/fc
h
A Col Top F 350 320 14.3 137.33 67.50 424.50 3.76>2 0.24
<0.8
Bottom
F
350 320 14.3 347.33 81.04 2864.94 6.54>2 0.35
<0.8
B Col Top F 350 320 14.3 104.24 51.25 370.49 4.21>2 0.12
<0.8
Bottom
F
350 320 14.3 290.00 110.0 2784.96 4.27>2 0.66
<0.8
(1) Calculation of bearing capacity of frame column B-pillar normal section
According to the data in the calculation sheet, the most unfavorable internal force
combination of column B is selected for calculation and reinforcement checking
calculation.
Bending moment of joints at left and right beam ends ：-167.78+121.7X0.25=--137.35 KN·m
295.50-103.64X0.25=269.59 KN·m
B Bending moment of the joint at the ends of upper and lower columns ：
340.21/0.8-147.47/0.85X0.12=331.18 KN·m
-271.92+88.71X(0.65-0.1)=-223.12 KN·m
∑MB 柱=331.18+223.12=554.3 KN·m
∑MB 梁=167.78+313.48=481.26 KN·m
1.2∑MB 梁=1.2*481.26=577.51 KN·m
ΔMB =481.26.32-167.78.63=313.48 KN·m
At the b-joint, the bending moment is distributed to the ends of the upper and lower columns
according to the elastic calculation method
56

𝑜
1 2
223.12
𝑀𝐵 𝑈𝑝𝑝𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛 = 577.51 ×
223.12 + 331.18
= 232.46𝐾𝑀 ⋅ 𝑚
MB下柱
 577.51
331.18
331.18  223.12
 345.04KM m 𝛾𝑅𝐸𝑀𝐵上柱 = 0.8 × 345.04 =
276.03𝐾𝑀 ⋅ 𝑚
𝑀 276.12 × 106
𝑒0 =
𝑁
=
1147.04 × 103 = 240.7𝑚𝑚
E 20mmAnd 1 / 30 of the eccentric dimension of the section, the larger one is chosen, that
is350/30=12mm, So take ea=20mm, The calculated length of the column is determined by
formula, in which the calculated length of the column is determined 𝜓𝑛 =
4.734×2
2.857+8.942
= 0.80
𝜓𝑐 = 0.80
𝑙0 = [1 + 0.15(𝜓𝑛 + 𝜓𝑐)] = [1 + 0.15 × (0.80 + 0.80)] × 3.3 = 4.09𝑚
𝑒𝑖 = 𝑒0 + 𝑒𝑎 = 240.7 + 20 = 360.7𝑚𝑚
So L0/h=3600/350=10.2>5, Therefore, the increase factor of eccentricity should be considered
𝜂。
=
0.5𝑓𝑐𝐴
=
0.5×14.3×350×350
= 1.42 > 1 ， take 1.0
1 𝑁 1147.04×103
lo/h=10.3<15
𝜂 =1 +
1
1400𝑒𝑖/ℎ𝑜
𝑙
( ℎ
) 𝜁 𝜁 = 1 +
1
1400×187.1/320
× 10.22 = 1.12
𝑒 =εei+h/2-as=1.12 × 187.1 + 350/2 − 30 = 354. 55𝑚𝑚
Symmetrical reinforcement ：𝜉 =
𝑁
𝑓𝑐𝑏ℎ0
=
1147.04×103
14.3×350×320
= 0.712 < 𝜉𝑏
𝐴𝑠 = 𝐴𝑠′ =
𝑁𝑒 − 𝛼1𝑓𝑐𝑏ℎ𝑜2(1 − 0.5𝜉)
𝑓𝑦
′
(ℎ𝑜 − 𝑎𝑠′)
1147.04 × 103 × 354.55 − 1.0 × 14.3 × 350 × 3202 × 0.712 × (1 − 0.5 × 0.712)
=
360 × (320 − 30)
= 1230.21mm2
Nmax M，N=1421.12KN
42.17+23.82X0.1=39.24 KN·m
32.8-25.64X0.55=16.07 KN·m
𝑙0 = 1.25 × 3.2 = 4𝑚
𝑀 38.58 × 106
𝑒0 =
𝑁
=
1421.12 × 103 = 27.09𝑚𝑚
𝑒𝑖 = 𝑒0 + 𝑒𝑎 = 27.09 + 20 = 47.09𝑚𝑚
As 𝑙0/ℎ = 4.125 × 103/350 = 8.25 > 5 Therefore, the eccentricity moment increase
coefficient is considered ε
𝜁 =
0.5𝑓𝑐𝐴
=
0.5×14.3×350×350
= 1.26 > 1 sp taking δ1 as 1.0
1 𝑁 1420.11×103
，
2

𝑜
1 2
S
C
lo/h=8.25<15 so taking δ2 as 1.0
ε=1 +
1
1400𝑒𝑖/ℎ𝑜
𝑙
( ℎ
) 𝜁 𝜁 = 1 +
1
1400×47.24/320
× 8.252 = 1.47
εei =1.47 × 47.24 = 69.44𝑚𝑚<0.3 ho=0.3X320=138mm
So it is small eccentric compression。
e=εei+h/2-as=69.44 + 320/2 − 40 = 279.44𝑚𝑚
𝑁 = 1420.11 < 𝜉𝑏𝑓𝑐𝑏ℎ0 = 0.550 × 14.3 × 350 × 320 = 1704𝐾𝑁
Therefore, the reinforcement is arranged according to the structure and meets the
requirements 𝜌𝑚𝑖𝑛，Unilateral reinforcement ratio 𝜌𝑚𝑖𝑛
choose 4@16 （AS=A ’
=908mm2
）
𝜌𝑠 =
1017+1608
= 1.05%
350×350
(1) Calculation of shear capacity of inclined section of B-column
Design value of upper column end bending moment: M t
=126.46 KN·m
Design value of column bottom bending moment: MC
b
=134.83 KN·m
126.46 + 134.83
𝑉 = 1.1 ×
2.65
= 110.50𝐾𝑁
𝛾𝑅𝐸𝑉
=
0.9×110.59×103
= 0.027 < 0.2 (meet the requirements)
𝛽𝑐𝑓𝑐𝑏ℎ0 1.0×14.3×350×350
𝑀𝐶 134.93 × 103
𝜆 =
𝑉𝐶ℎ
= = 2.6
110.94 × 320
𝑁 = 1420.00𝐾𝑁 > 0.3𝑓𝑐𝑏ℎ = 0.3 × 14.3 × 350*320 = 1072.5𝐾𝑁
𝐴 0.8×110.59×102−
1.05
×1.43×350×320−0.056×1420.11×103
4×50.3
= 2.65+1
< 0 𝜌𝑠𝑣 = = 0.2% >
𝑆 210×320
0.24
𝑓𝑡
= 0.24 ×
1.43
= 0.16%
350×200
𝑓𝑦 210
The stirrups are arranged according to the structure, and both the densified area and the
non-densified area are HRB335 bars with a diameter of 8mm and a spacing of 100 "mm.
Table 2.21: Column reinforcement calculation
Floo
rs
Colum
n
Section
N M e
a
δ
c
C
m
εn
s
e Bias type δ γr
e
As Reinforc
ement
Unilate
ral-ρ
ρ
2nd Top 197. 38. 2 1 0. 1. 65 Large 0.0 0. 65 4 16 0.2 1.
Floo Col 62 68 0 67 02 4.2 eccentricity 54 75 4.2 05
r
Botto 210. 11. 2 1 0. 1. 60 Large 0.0 0. 80 4 16 0.2 1.
m-Col 6 07 0 97 0 5.3 eccentricity 54 75 3.2 05
58
0
2

1st Top 1124 232 2 1 0. 1. 34 Large 0.2 91 4 16 0.2 1.
Floo Col .53 .46 0 73 09 5.2 eccentricit 14 1.3 05
r y
Botto 1420 345 2 1 0. 1. 36 Large 0.3 0. 92 4 16 0.2 1.
m-Col .11 .04 0 8 23 0.7 eccentricit 23 8 0.4 05
y
2.10 Foundation design
With the rapid development of China's social economy, people's living standard is on the rise
as a whole, and its requirements for the quality of life are also gradually increasing, especially
for the quality of housing construction. [8] the independent foundation under the column is
adopted for the foundation, the buried depth elevation of the foundation is - 1.5m, and the
height elevation of the bearing platform is - 0.9m
2.10.1 Calculation of foundation bearing capacity and determination of foundationbottom area
The site is flat, the surface of the area is 0.6 m thick miscellaneous fill, heavy 16KN/m2
then 3.4m Heavy miscellaneous fill18KN/m2，
the bearing capacity is 180kpa, It meets the
design requirements of shallow foundation of natural foundation of such buildings, and the
groundwater level is 5m below the surface.
A Column foundation calculation.
(1) Load:
M=29.03KN*M N=186.95KN V=11.65KN
（2 Estimation of foundation ground size A=L*B=1.5*1.5=2.25m2
The elastic resistance moment of the foundation bottom is
W=L*B2
*1/6=1.5*1.5*1.5/6=0.563
𝑓𝑎 = 𝑓𝑎𝑘 + (𝑑 − 0.5)
𝛾𝑚 = (16 × 0.6 + 18 × 0.9)/4 = 17.44𝐾𝑁/𝑚3
𝐺𝑘 = 𝛾𝑚𝑙𝑏ℎ = 17.44 × 1.5 × 1.5 × 1.5 = 58.86𝐾𝑁
2.10.2 Checking calculation of bearing capacity of foundation
𝑓𝑎 = 180 + 1.0 ∗ 17.44 ∗ （1.5-0.5） = 197.44𝑘𝑝𝑎
𝑀𝑘 = 29.03/1.35 + 11.65 × 1.4/1.35 = 33.59𝐾𝑁 ⋅ 𝑚
𝑁𝑘 = 186.95/1.35 + 58.86 = 197.34𝐾𝑁
Pkmax
Pkmin
Nk
l b

Mk
w

197.34
1.51.5

33.59
0.563
147.36KPA 1.2197.44236.92KPA
28.04KPA
𝑃𝑘 = (147.36 + 28.04)/2 = 87.7𝐾𝑃𝑎 < 𝑓𝑎 = 197.44𝐾𝑃𝑎
Meet the requirements
 


2.10.3 Checking calculation of foundation punching resistance
Fl = 0.7hp ftmh0
am=(at+ab)/2
F1=PJ*A1
So we only need to check the variable section.
The net reaction force of foundation ground is:
Pkmax
Pkmin
Nk
l b

Mk
w

197.34
1.51.5

33.59
0.563
147.36KPA 1.2197.44236.92KPA
28.04KPA
197.34 6 × 33.59
𝑃
1.52
1.52 172.67𝑎
𝑗𝑚𝑎𝑥
𝐴 = 1.5 × 0.1 = 0.15𝑚2
Calculation value of punching load：𝐹𝑙 = 172.67 × 0.15 = 25.9𝐾𝑁
Punching capacity：
1
𝑎𝑚 = 2
× (0.5 + 1.5) = 1𝑚
0.7𝛽𝑚𝑓𝑡𝑎𝑚ℎ0 = 0.7 × 1.0 × 1.1 × 1.0 × 103 × 465 = 358.05𝐾𝑁 > 𝐹𝐿 = 25.9𝐾𝑁
Meet the requirements.
2.10.4 Foundation reinforcement calculation
The calculation formula of reinforcement checking is as follows:
M =
1
2
2l  a'
(P  P )  (P  P )l
1
12 1 jmax j jmax j
Bending moment calculation:
xaxis,h0 = 565 mm
yaxis,h0 = 5 65mm
xaxis,h0 = 565mm
yaxis,h0 = 56 5 mm
Reinforcement calculation:
Reinforcement in X axis and reinforcement in Y axis
Load = 563
M1(kNm) = 16.353
AGx(mm*mm) = 160.232
Load = 562
M1(kNm) = 17.860
AGx (mm*mm) = 174.999
X Actual match:φ12@150(0.16%) y Actual match:φ12@150(0.16%)
60
 


Figure 2.16: Foundation Details
Figure 2.17: Foundation Reinforcement Diagram
B column for foundation calculation:

(1) Load:

62
M=37.04KN*M N=247.75KN V=12.85KN
（2）Estimation of foundation ground size A=L*B=1.5*1.5=2.25m2
The elastic resistance moment of the foundation bottom is
W=L*B2
*1/6=1.5*1.5*1.5/6=0.563
𝑓𝑎 = 𝑓𝑎𝑘 + (𝑑 − 0.5)
𝛾𝑚 = (16 × 0.6 + 18 × 0.9)/4 = 17.44𝐾𝑁/𝑚3
𝐺𝑘 = 𝛾𝑚𝑙𝑏ℎ = 17.44 × 1.5 × 1.5 × 1.5 = 58.86𝐾𝑁
2.10.5 Checking calculation of foundation bearing capacity
𝑓𝑎 = 180 + 1.0 ∗ 17.44 ∗ （1.5-0.5） = 197.44𝑘𝑝𝑎
𝑀𝑘 = 37.04/1.35 + 11.65 × 1.4/1.35 = 40.37𝐾𝑁 ⋅ 𝑚
𝑁𝑘 = 247.75/1.35 + 58.86 = 220.36𝐾𝑁
Pkmax
Pkmin
Nk
l b

Mk
w

247.75
1.51.5

40.37
0.563
170.32KPA 1.2197.44236.92KPA
35.63KPA
𝑃𝑘 = (170.32 + 35.63)/2 = 102.975𝐾𝑃𝑎 < 𝑓𝑎 = 197.44𝐾𝑃𝑎
Meet the requirements
2.10.6 Checking calculation of foundation punching resistance
Check the calculation according to the following formula in article 8.2.7 of basic code
Fl= 0.7hp ftmh0
am=(at+ab)/2
F1=PJ*A1
So we only need to check the variable section.
The net reaction force of foundation ground is:
Pkmax
Pkmin
Nk
l b

Mk
w

247.75
1.51.5

40.37
0.563
170.32KPA 1.2197.44236.92KPA
35.63KPA
247.75 6 × 40.37
𝑃 =
1.52 1.52 205.34𝑎
𝑗𝑚𝑎𝑥
𝐴 = 1.5 × 0.1 = 0.15𝑚2
Calculation value of punching load：𝐹𝑙 = 205.34 × 0.15 = 30.8𝐾𝑁
Punching capacity：
1
𝑎𝑚 = 2
× (0.5 + 1.5) = 1𝑚
0.7𝛽𝑚𝑓𝑡𝑎𝑚ℎ0 = 0.7 × 1.0 × 1.1 × 1.0 × 103 × 465 = 358.05𝐾𝑁 > 𝐹𝐿 = 25.9𝐾𝑁
Meet the requirements.
2.10.7 Foundation reinforcement calculation
The calculation formula of reinforcement checking is as follows
 

 


M =
1
2
2l  a'
(P  P )  (P  P )l
1
12 1 jmax j jmax j
2.11 Staircase Design

64
Figure 2.18: Stair Layout (Structutal Drawing)
Figure 2.19: Stair Side view with crossection (Structutal Drawing)
Figure 2.20: Stair Elevation (Structutal Drawing)
2.11.1 Design information
The overall stair adopts cast-in-place slab stairs, and the standard value of live load of
staircases is 3.5kN/m2

The concrete of stairs is C30，fc=14.3N/mm2
，ft=1.43N/mm2
Step reinforcement: HRB400 level，fy=360N/mm2
Reinforcement of platform slab：HRB400 level，fy=360N/mm2
Reinforcement of platform beam：
HRB400 level，
fy=360N/mm2
，
stirrup：
HRB400 level，
fy=360N/mm2
Method of stair surface course: 15mm thick granite surface course, the standard value of
self-weight is 28kN/m3
, the standard value of concrete self-weight is 25kN/m3
, and the
standard value of cement mortar at the bottom of slab is 20KN/m3
.
2.11.2 Load And Stress Calculation
The calculation diagram of stairs is as follows:
Figure 2.21: Calculation diagram of stairs
The calculation formula is as follows:
𝑕𝑕 = 𝑑𝑑 +
𝐴
𝑇𝐿
𝑞 =
1.2(𝑞 + 25 𝑋 𝑕𝑕)
𝑐𝑜𝑠𝜕 + 1.4𝑞
First standard level first run
Qb=10.579 Qbt=7.600；
Qp=7.600 Qw=7.000；

3300
3300
10.5
Figure 2.22: Loading on first Stairs
The first standard level, the second run
Qb=10.587 Qbt=7.600；
Qp=7.600 Qw=7.000；
10.5
14.41
Figure 2.23: Loading on second Stairs
2.10.2 Calculation of reinforcement area
Take the thickness of the plate h=100mm, and according to the design of one-way plate,
66

take the plate with width of b=1000mm for calculation. The calculated span l0=2600mm, the
thickness of protective layer is c=15mm, h0= h-c-d/2=100-15-8/2=81mm (d is the diameter of
reinforcement). Load calculation:
Standard No. of
floor Runs Asbd Asbf Aspd Aspf
number
1 1 4.15 3.29 0.00 0.00
1 2 5.07 4.01 0.00 0.00
Reinforcement results of terrace and platform:
Standard layer number running number: the bottom reinforcement of the platform with
the bottom reinforcement of the horizontal platform
1 1 #8@100 #8@200 #8@150 #8@180 #8@200
1 2 #10@150 #8@200 #8@100 nothing nothing
The diameter of stirrup is b×h=200mm×350mm, Calculation span l0=3600mm, Degree of
protective layer c=20mm and the diameter of stirrup is -d/2=350-20-8-18/2=313mm. (d is the
diameter of reinforcement).
Results of ladder beam reinforcement:
Standard floor number running number ladder beam 1 top longitudinal reinforcement
ladder beam 1 bottom longitudinal reinforcement ladder beam 1 stirrup ladder beam 2 bottom
longitudinal
stirrup
reinforcement ladder beam 2 top longitudinal reinforcement ladder beam 2
1 1 2#14 2#14 #8@200
1 2 2#14 2#14 #8@200
2.12 Design of cast in place concrete slab

Figure 2.24: Concrete slabs
2.12.1 Concrete Roof Design
The concrete roof slab is designed as two-way slab, and the specific parameters are as
follows
1. Basic information and load calculation
The concrete strength grade is C30, the reinforcement grade is HRB400, the slab
thickness is h = 100 "mm, the thickness of protective layer is ε = 15mm, the Poisson's ratio of
concrete is 0.2, and the design values of roof slab load in a, B and C zones are as follows:
Standard value of dead load: 4.92kN/m2, standard value of live load: 0.5kN/m2
g=1.3*4.92=6.396kN/m2
q=0.7*0.5*1.5=0.53KN/m2
𝑞=0.53/2=0.265KN/m2
2
g+q=6.926KN/m2
g+𝑞=6.661KN/m2
2
2. Calculation of Bending Moment
Grid plate in area A as an example, l01/ l02=3.6/4.8=0.75, the plate support condition is
fixed on four sides, and the bending moment coefficients under this support condition are
𝑙01mid span 0.04148, 𝑙02 mid span 0.00305, 𝑙01 bearing -0.08313, - 0.08313, 𝑙02bearing -
0.05587, - 0.05587, the bending moment coefficients under the condition of simple support
around are 𝑙01mid span 0.10474,01365.
𝑚1 = (0.04148+0.2×0.00305)(𝑔 + 𝑞/2)𝑙012+(0.10474+0.2×0.01365)𝑞/2𝑙012
=0.04205×6.661×3.6×3.6+0.10744×0.27×3.6×3.6=3.87kN.m
𝑚2 = (0.00305+0.2×0.04148)(𝑔 + 𝑞/2)𝑙012+(0.01365+0.2×0.10474)𝑞/2𝑙012
=0.01134×6.661×3.6×3.6+0.03461×0.265×3.6×3.6=1.10kN.m
𝑚′ =-0.08313× (𝑔 + 𝑞)012=-7.46kN.m； 𝑚′ =-0.05587× (𝑔 + 𝑞)𝑙012=5.01kN.m
1 2
C Grid l01/ l02=1.8/4.8=0.375，Support condition: four sides fixed support
𝑚1 = (0.04173+0.2×0.00262)(𝑔 + 𝑞/2)𝑙012+(0.10734+0.2×
𝑞/2𝑙012=3.96kN.m
𝑚2 = (0.00262+0.2×0.04173)(𝑔 + 𝑞/2)𝑙012+(0.01222+0.2×
0.10731)𝑞/2𝑙012=1.15kN.m
𝑚′ =-0.08312× (𝑔 + 𝑞)012=-6.27kN.m； 𝑚′ =-0.05599× (𝑔 + 𝑞)𝑙012=-3.54kN.m
1 2
B Grid d l01/ l02=3.6/5.7=0.63，Support condition: four sides fixed support
𝑚1 = (0.04107+0.2×0.00336)(𝑔 + 𝑞/2)𝑙012+(0.10217+0.2×
0.01488)𝑞/2𝑙012=5.76kN.m
𝑚2 = (0.00337+0.2×0.04108)(𝑔 + 𝑞/2)𝑙012+(0.01497+0.2×
68

0.10208)𝑞/2𝑙012=1.8kN.m
𝑚′ =-0.08316× (𝑔 + 𝑞)012=-8.93kN.m； 𝑚′ =-0.0562× (𝑔 + 𝑞)𝑙012=-5.64kN.m
1 2
Grid section M/kN.m h0/mm Αs γs As/mm2
Reinforcement Actually As/mm2
A Plate
bottom
l01 3.87 80 0.057 0.96 187.25 #8@200 251.5
l02 1.10 72 0.019 0.98 59.83 #8@200 251.5
Board
surface
l01 -6.27 80 0.084 0.94 294.30 #8@160 314.4
l02 -3.54 80 0.053 0.96 190.21 #8@200 251.5
B Plate
bottom
l01 5.76 80 0.039 0.97 130.31 #8@200 251.5
l02 0.89 72 0.012 0.98 39.63 #8@200 251.5
Board
surface
l01 -8.93 80 0.059 0.96 199.81 #8@200 251.5
l02 -5.64 80 0.037 0.97 129.64 #8@200 251.5
C Plate
bottom
l01 3.96 80 0.055 0.96 177.52 #8@200 251.5
l02 1.15 72 0.026 0.98 58.32 #8@200 251.5
Board
surface
l01 -6.27 80 0.077 0.94 287.12 #8@160 314.4
l02 -3.54 80 0.051 0.96 165.34 #8@200 251.5
Table 2.22: Bending Moment Calculations
𝐴smin = 𝜌min𝑏𝑕 = max(0.2,45𝑓𝑡/𝑓𝑦)%𝑏𝑕 =0.2%×100×1000=200mm2
𝛼𝑠
=
𝑀
；
𝛼1𝑓𝑐𝑏𝑕0
2 S
 0.5(1 ) ；
𝐴
𝑠 =
𝑀
𝛾𝑆𝑓𝑦ℎ0
Effective height of concrete section 𝑕0 = 𝑕 − 𝜀 − 𝑑/2, where ε is the thickness of
protective layer and D is the diameter of reinforcement𝑙02 direction reinforcement placed in
𝑙01 direction, so 𝑙02 direction plate bottom reinforcement calculation 𝑕0 also needs to
subtract. The diameter of reinforcement in direction 𝑙01 is shown in table 2.22.
2.12.2 Design of concrete floor slab
The floor slab is designed as two-way slab
1. Basic information and load calculation
The concrete strength grade is C30, the reinforcement grade is HRB400, the slab thickness is
h = 100 mm, the thickness of protective layer is ε = 15 mm, the Poisson's ratio of concrete is
0.2, and the design values of roof slab load in A, B and C zones are as follows:
Standard value of dead load: 3.53kN/m2, standard value of live load: 2KN / m2
g=1.3×3.85=5.0kN/m2
q=1.5×2=3kN/m2
q/2=1.5kN/m2
g+q/2=6.5kN/m2
1- 2S

70
g+q=8.0kN/m2
Figure 2.25: Partition layout of roof slab
2. Bending moment calculation
Taking the grid plate in area A as an example, l01/ l02=3.6/4.8=0.75, the plate support
condition is fixed on four sides, and the bending moment coefficients under this support
condition are 𝑙01 mid span 0.04147, 𝑙02 mid span 0.00305, 𝑙01 bearing -0.08313, - 0.08313,
𝑙02 bearing - 0.05587, - 0.05587, the bending moment coefficients under the condition of
simple support around are 𝑙01 mid span 0.10474, 𝑙01365.
𝑚1 = (0.04148+0.2×0.00305)(𝑔 + 𝑞/2)𝑙012+(0.10474+0.2×0.01365)𝑞/2𝑙012
=0.04205×6.661×3.6×3.6+0.10744×0.27×3.6×3.6=6.20kN.m
𝑚2 = (0.00305+0.2×0.04148)(𝑔 + 𝑞/2)𝑙012+(0.01365+0.2×0.10474)𝑞/2𝑙012
=0.01134×6.661×3.6×3.6+0.03461×0.265×3.6×3.6=1.76kN.m
𝑚′ =-0.08313× (𝑔 + 𝑞)012=-8.9kN.m； 𝑚′ =-0.05587× (𝑔 + 𝑞)𝑙012=5.9kN.m
1 2
B Grid l01/ l02=3.6/5.7=0.63，Support condition: four sides fixed support
𝑚1 = (0.04107+0.2×0.00336)(𝑔 + 𝑞/2)𝑙012+(0.10217+0.2×
0.01488)𝑞/2𝑙012=5.75kN.m
𝑚2 = (0.00337+0.2×0.04108)(𝑔 + 𝑞/2)𝑙012+(0.01497+0.2×
0.10208)𝑞/2𝑙012=1.9kN.m
𝑚′ =-0.08316× (𝑔 + 𝑞)012=-8.65kN.m； 𝑚′ =-0.0562× (𝑔 + 𝑞)𝑙012=-5.83kN.m
1 2
C Grid l01/ l02=1.8/4.8=0.375，Support condition: four sides fixed support
𝑚1 = (0.04173+0.2×0.00262)(𝑔 + 𝑞/2)𝑙012+(0.10734+0.2×𝑞/2𝑙012=4.01kN.m
𝑚2 = (0.00262+0.2×0.04173) (𝑔 + 𝑞/2)012 +(0.01222+0.2×0.10731) 𝑞/
2𝑙012=1.27kN.m
𝑚′ =-0.08312× (𝑔 + 𝑞)012=-6.10kN.m； 𝑚′ =-0.05599× (𝑔 + 𝑞)𝑙012=-4.03kN.m
1 2

3. Reinforcement calculation
𝐴smin = 𝜌min𝑏𝑕 = max(0.2,45𝑓𝑡/𝑓𝑦)%𝑏𝑕 =0.2%×100×1000=200mm2
𝛼𝑠
=
𝑀
；
𝛼1𝑓𝑐𝑏𝑕0
2 S
 0.5(1 ) ；
𝐴
𝑠 =
𝑀
𝛾𝑆𝑓𝑦ℎ0
1.6.3 Daughter Wall Effective height of concrete section 𝑕0 = 𝑕 − 𝜀 − 𝑑/2, where ε is
the thickness of protective layer and d is the diameter of reinforcement 𝑙02 direction
reinforcement placed in 𝑙01 direction, so 𝑙02 direction plate bottom reinforcement
calculation 𝑕0 also needs to subtract. The diameter of reinforcement in direction 𝑙01 is
shown in table 2.23.
Table 2.23: Reinforcement Table of Plate
Grid section M/kN.m h0/mm αs γs As/mm2
Reinforcement Actually As/mm2
A Plate
bottom
l01 6.20 80 0.062 0.96 201.34 8@200 251.5
l02 1.76 72 0.024 0.97 65.23 8@200 251.5
Board
surface
l01 -8.9 80 0.093 0.94 300.1 8@160 314.4
l02 -5.9 80 0.061 0.96 201.21 8@200 251.5
B Plate
bottom
l01 5.75 80 0.044 0.97 140.23 8@200 251.5
l02 1.9 72 0.016 0.94 50.74 8@200 251.5
Board
surface
l01 -8.65 80 0.062 0.96 212.32 8@200 251.5
l02 -5.83 80 0.047 0.97 123.54 8@200 251.5
C Plate
bottom
l01 4.01 80 0.061 0.96 199.21 8@200 251.5
l02 1.27 72 0.026 0.98 70.3 8@200 251.5
Board
surface
l01 -6.10 80 0.091 0.94 296.4 8@160 314.4
l02 -4.03 80 0.059 0.96 201.5 8@200 251.5
1- 2S

72
References
[1] Z. Wei ,T. Liuquan, “ Traditional forms in the design of modern villa living space,”
Science and Technology Association Forum , pp. 185-186, 2007.
[2] L. Yanping, “Combination of functional areas in villas,” Journal of Design, vol 5, pp.
78-79, 2017.
[3] L. lang, “Analysis on elements of architectural plane design conception,” Journal of Henan
building materials, vol 5, p. 265, 2019.
[4] Z. MI, Z. Shaohua, Z. Zilan, H. B., D. Xiuli, “Robust design and analysis of independent
foundation under column,” Journal of Geotechnical mechanics, vol 40, 11, pp. 4506-4514,
2019.
[5] W. Bin, Z. Yu, “ Seismic performance analysis of reinforced concrete frame structure,”
Journal of Science and technology and Enterprise, vol 6, p. 120, 2015.
[6] S. Xufeng, “Analysis on the application of D-value method in graduation design of civil
engineering specialty,” Journal of Education and Teaching Forum, pp. 56-57, 2019 .
[7] Z. Jianfeng, G. Yinghao, J. yechao, “Research and Practice on teaching reform of deep
integration of Bim and villa architectural design,” Journal of Education Modernization ,
vol 6(65), pp. 93-94, 2019.
[8] Z. Xiang, “ Discussion on foundation design in building structure design,” Journal of
Building Materials and Decoration, , vol 5, pp. 111-112, 2021 .
[9] “Code for seismic design of buildings (GB 50011-2010),” Beijing: China Construction
Industry Press, 2010.
[10] “Ministry of construction of the people's Republic of China code for design of building
foundation (GB 5007-2011).,” China Construction Industry Press, Beijing, 2011.

Acknowledgement
Time flies like an arrow. The days of being a student are coming to an end. The only regret is
that due to the epidemic situation, we will be the first college students to graduate from home.
We can't express our gratitude to our classmates and all our teachers face to face, especially
our tutor. This will be the biggest legacy in my life I'm sorry.
Once again, I would like to thank for my cultivation. At the same time, I would also like to
thank the owners of all kinds of relevant literature and materials on the Internet for their
selfless sharing of their achievements. I have learned a lot from them, which is very helpful to
my graduation project. In the early stage of the design, due to the impact of the epidemic, a lot
of valuable time was wasted. I was only able to explore slowly at home, and the teacher could
only guide our graduation design through the network. Due to his limited understanding
ability, he took the trouble to explain it again and again for us. Although it was not face-to-
face, I could still feel it he is warm hearted, the network text is cold, but in this is not the
same, it is particularly warm and kind, whenever said some guidance voice in ourguidance
group, I will listen to his voice over and over again at home, listen to it again and again until I
fully understand the teacher's meaning, so that the teacher will not be responsible for us Heart.
In the middle and later stage of the design process, the problems gradually appeared one by
one. He will help me solve all the problems and give me constructive opinions, so that I can
carry on the graduation design smoothly. Thank you very much for your understanding,
tolerance and guidance in the past 14 weeks. Thank you for your hard work. Finally, I would
also like to thank the students who helped me. Thank you for teaching me that unity is
strength. Finally, I would like to thank myself for your perseverance in so long time. Come
on, it will be better in the future.
All the knowledge gained by the University in the past four years is shown in this design. The
days after that are still very long. What I have learned is far more than these dozens of pages.
I will go on with the harvest of the University and let it enrich my life in the future. Thanks
again to all the teachers, classmates and relatives who have helped me.

Graduation dissertation (design) commitment book
Title
Name Major
Supervisor name
Professional
title
Graduation dissertation (design) innovation：
Commitment content：
This graduation dissertation was independently completed under the
guidance of my supervisor. I did not copy others' achievements. The
individuals and groups that have made important contributions to the
research of this dissertation have been clearly indicated in this paper.
Student signature： Supervisor signature：
Date: Date:

Thesis civil eng [revised i]

Recommended

Recommended

More Related Content

Similar to Thesis civil eng [revised i]

Similar to Thesis civil eng [revised i] (20)

Recently uploaded

Recently uploaded (20)

Thesis civil eng [revised i]