(Đồ án hcmute) capstone project grand mercure apartment

Construction introduction

In response to the country's ongoing integration, industrialization, and modernization, there is a growing need to replace low-rise buildings in degraded residential areas with high-rise constructions This trend reflects the development of contemporary living standards, leading to the establishment of the Grand Mercure apartment building.

Situated in the Bac Ha District, this project boasts a stunning and open location that enhances the overall residential planning with a blend of harmony, rationality, and modernity.

Urban infrastructure

- The work is located on the main road, convenient for the supply of materials and traffic outside the building

- The electricity and water supply system in the region has been completed, meeting the requirements for construction

The construction site is flat and free from any existing structures or underground works, making it highly suitable for the overall project layout and construction activities.

Architectural solution

Functional plan and subdivition

- The work plan is rectangular with gouges, length 53.4m, width 36m, and construction land area is 1922.4m 2

The building features 20 floors, including a semi-basement, with a height of 73.1 meters from the 0.00m level to the roof The 0.00m level is aligned with the natural ground level, which is 1.50 meters lower than the ground floor The basement is situated at -1.50m.

The basement features a centrally located elevator surrounded by parking spaces, while technical systems, including domestic water storage tanks, pumping stations, and wastewater treatment facilities, are strategically arranged to reduce pipeline length Additionally, the design incorporates essential components such as high voltage and low voltage stations, along with a dedicated fan room for optimal functionality.

- Ground floor: used as a supermarket to serve the needs of buying and selling, entertainment services for households as well as the general needs of the area

- Floor 2 - 17: arrange apartments to serve the needs of living

- Rooftop: layout of technical rooms, machines, air-conditioners, satellite equipment,

A straightforward ground solution that maximizes space for apartment arrangements involves using lightweight materials for partitions, allowing for flexible organization that aligns with current trends and preferences, and can be easily adapted in the future.

Appearance

The design features a striking shape that ascends directly from the traditional architecture below, embodying a modern yet powerful aesthetic This harmonious blend of strength and softness highlights the project's scale and significance, aligning with the country's overarching development strategy.

Front elevation

CAPSTONE PROJECT INSTRUCTOR: Assoc Prof TRAN TUAN KIET

STUDENT: HUYNH GIA HUY ID: 17149016 2

- Using and fully exploiting the modern features with large glass doors, outer walls are completed with water paint.

Transport system

- Horizontal traffic in each unit is a corridor system

The vertical transport system in the building includes a staircase and elevators, designed for efficiency and convenience The staircase features a main commuter ladder and an emergency exit, while the elevator system comprises two main lifts and one larger medical and service lift Centrally located, the elevators are surrounded by apartments arranged around a corridor, minimizing travel distances and ensuring a well-ventilated environment.

Technical solution

Power system

- The system receives electricity from the general electrical system of the town into the house through the electric machine room

- From here, electricity will be transmitted around the building through the internal grid

- In addition, when there is a power failure you can immediately use a backup generator located in the basement to generate.

Water supply and sewerage system

Water is sourced from the regional supply system and stored in a basement water tank An automatic pumping system then distributes the water to each room via the main plumbing network located near the service area.

- After being treated, the wastewater is fed into the area's general drainage system.

Fire prevention, emergency exit

- Reinforced concrete (reinforced concrete) works with hollow brick walls that are both sound and heat insulation

- Fireproof boxes are arranged along the corridor with CO2 cylinders

- All floors have two stairs to ensure escape when there is a fire incident

- In addition, there is a large fire protection lake on the top of the roof

Lighting protection

- Option for the Dynasphire ball active air-termination system set up in the rooftop and the copper grounding system is designed to minimize the risk of lightning strikes.

Garbage drainage system

Each floor's garbage is collected and disposed of in a discreet garbage chute located in the basement, which includes a dedicated waste removal section These garbage chutes are designed with care to prevent odors and minimize environmental pollution.

Climate characteristics of the construction area

Lao Cai features a tropical monsoon climate, yet its inland location and complex terrain lead to variable weather conditions that differ across time and space Notably, temperature fluctuations can be sudden, with Sa Pa experiencing extreme drops below 0°C, occasionally resulting in snowfall.

- Lao Cai climate is divided into two seasons: the rainy season starts from April to

October, the dry season starts from October to March next year The average temperature is located in the highlands from 150C - 200C (particularly in Sa Pa from

140C - 160C and in no month it goes over 200C), the average rainfall is from 1,800mm

-> 2,000mm Average temperature is located in the low area from 230C - 290C, the average rainfall is from 1,400mm - 1,700mm

Mist is prevalent across the province, often reaching significant thickness During severe cold spells, frosts can occur in the high mountains and sheltered valleys, typically lasting for 2 to 3 days.

Design solutions

Construction method

- Slab method, Solid slab is concrete slab and beam combined

- Frame method, cast in place reinforced concrete wall

Material for use

- Concrete used in the building is concrete with durability classes B25 and B20 with the following calculated parameters:

Software for use in analyzing and calculate

- Modeling of frame and slab of building: ETABS, SAFE, SAP

- Calculate reinforcement and foundation: Excel

Structural solution

Choose preliminary section of slab

- Choose floor thickness depending on span and applied load

- The slab thickness is determined by the empirical formula:

- Whereas: m = (40 ÷ 45) for four-sided manifest, Li = 8m short side length of typical floor

Choose preliminary section of beam

Table 1.1: Preliminary section of beam

Type of beam Height h Width b

- Select the span of the main beam to calculate: L = 8000mm

Choose preliminary section of column

- Preliminary selection of column sizes is based on experience or approximate formula

- Determine the vertical transmission area

- Preliminary total floor load, choose q = 1400 daN/m2

- Count the number of floors on the section under consideration m

- Considering the impact of the horizontal load, the coefficient k, which varies depending on the column position c b

- The column computation and section reselection will be performed again a again until the bearing capacity and architectural requirements are satisfied

 qi: distributed load on slab (live load + dead load)

 Si: is the transmission area of the floor to the column

 qi (10÷15 daN/m 2 ) Choose qi = 14 daN/m 2

Table 1.2: Preliminary section of central column

Story Atr.tải Q Number of floors

N  A tt b h Ac choose m 2 daN/m 2 m daN cm 2 cm cm cm 2

Table 1.3: Preliminary section of edge column

Choose preliminary section of core wall

- According to Article 3.4.1 of TCVN 198-1995, the thickness of the hard wall is not less than 150mm and not less than 1/20 of the floor height

Story Atr.tải Q Number of floors

N  A tt b h Ac choose m 2 daN/m 2 m daN cm 2 cm cm cm 2

DESIGN OF STAIRCASE

Geometry of staircase and calculation free-body diagram

- Choose the staircase at axis D-E to calculate:

Table 2.1: General geometry of staircase

Height of one flight v 2 h  ht

Number of rises on each flight n

- Length of flight stair (Along with diagonal axis): LL 1 L 2 3.62.4= 6 (m)

- Thickness of riser: Consider the plate of stair working on one-way,

- Choose the thickness of staircase slab: 180 mm

Figure 2.1: Layout of stair case

STUDENT: HUYNH GIA HUY ID: 17149016 8 h 500 b= (166.6

- Illustrate the incline angle of stair Tan   1900 27 49 ' 0.891

Loading on staircase

Table 2.2: Structure component of the landing flight

- According to TCVN 2737:1995, Table 3, we have: p =3 kN/m tc 2

- Loading factor 1.2 for the standard loading bigger than 2 kN/m 2

Thickness of granite layer Thickness of plaster Thickness of brick layer b td h cos 2

Table 2.3: Equivalent thickness of each structure layer

Total gravity uniform load combined riser: 0.27 kN/m 7.792

Table 2.4: Dead load on staircase slab

- Live load included of 2 main elements:

- Distributed load on 1m long along with diagonal slab:

- p tcn p 1 cos tc m    3 1 0.8912.7 kN/m ttn tt m p p 1 cos 3.6 1 0.891 3.24 kN/m  

Serial Structure Dead load g tt kN/m

Total load tt tt tt q g p kN/m

Analyze the modeling with ETAB

- The ladder works as a bending member

Each individual must develop their own calculation scheme, which serves as the foundation for constructing the structure The complexity of determining whether the connection between the ladder and the beam or the ladder and the wall is a joint (fixed or roller) or a mount relies heavily on the designer's conceptual approach.

- Calculation of staircase as same as the bending element

2,81 3 160 d s h h    the connection between slab and beam is pinned connection

- Connection at the supports of staircase not the fixed, not the spin only the middle of two these kinds so

- We modeling used pinned and roller connection so we distributed moment shown below:

Figure 2.4: Dead load Figure 2.5: Live load

STUDENT: HUYNH GIA HUY ID: 17149016 11 span max

Calculate reinforcement

2.4.1 Calculate reinforcement for landing and flight

Figure 2.6: Moment diagram Figure 2.7: Shear diagram

STUDENT: HUYNH GIA HUY ID: 17149016 12 b b o m 2 m s m R R b b o s ξγ R bh α = M , ξ = 1- 1-2α , A = , , γ R bh R      

- Moment in span: Mnh = 24.49 kNm

- Checking reinforcement content: min max

- The result of calculation was shown in table below:

As mm 2 μ % Choose Asbt mm 2

Table 2.6: Calculation reinforcement of flight

2.4.2 Calculate reinforcement for the beam of the landing and flight

- Note: For the landing beam is under reaction of flight so distributed load apply on main beam are reaction of flight and also the self-weight of beam

- Choose preprimary section of beam b h   200 400 mm 

Figure 2.9: Free body diagram of beam D1

- Weigh of wall build on beam:

- Loading the ladder plate transmitted to the ladder beam in the form of the support reaction in each 1 wide meter strip, will be reduced to a uniform distribution:

+ Maximum shearing force of beam D1: max 24.29 5.4

Figure 2.11: Shear force diagram of beam D1

 Reinforcement grade CB-400V → Rs = 350 Mpa

Figure 2.10: Moment diagram of beam D1

→ Choose 418 have As ch 18cm 2

As cm 2 μ % Choose Asbt cm 2

Table 2.7: Calculation reinforcement of beam D1

- The shear capacity of concrete:

- So must be calculate stirrup for beam

- Coefficient  w 1 consider the effect of stirrup to axis

- The beam is not damaged by primary compressive stress

- Shear resistance of the belt: w 175 2 28.3

- Shear resistance of the belt and concrete:

- There is no need to calculate the shear reinforcement

- So the layout of the reinforcement ỉ6a100 for the L/4, ỉ6a200 in middle with L/2

 Eb, Es yuong modulus of reinforcement and concrete; asw area of section reinforcement

 Rb, Rbt Axial compressive stress of concrete; Rsw tension stress of reinforcement

 Φf is the coefficient of effect on the compression wing in the cross-section T; span of stirrup;

 φn coefficient affected by axial force; φb3 = 0.6 with heavy concrete; φb2 = 2 with heavy concrete

DESIGN OF ROOF WATER TANK

Architecture require

- The roof water tank provides water for the daily needs of the building

- Roof water tank consists of 1 tank placed on the floor column system, at the position limited by the axis C'-B' and 3'-5'

The preliminary calculation of water demand for the apartment building, which consists of 17 floors with residential apartments starting from the second floor, indicates that there are 8 apartments on each floor With an average of 4 occupants per apartment, the total number of residents significantly contributes to the overall water demand.

- Average water consumption: q sh 150 1/person.day.night

3 max day night sh day /1000 150 544 1.35 / 1000 110.16 m /

- The size of the water tank: V  LBH    6 11 2 132 m /day.night 3

The water tank is fully constructed from concrete and features a secure lid Access to the tank is provided through a 600x600 mm inspection hole located at the corner An automatic pumping system efficiently pumps water from the tank twice daily.

Data of calculation

- Water tanks are divided into three types:

→ With geometry of water tank a = 10 m; b = 6 m; h = 2 m  Low water tank

To accommodate the large span of a roof water tank exceeding 7 meters, it is essential to utilize a girder system for both the cover plate and bottom plate This approach effectively minimizes the thickness and deflection of the structure, ensuring enhanced stability and performance.

To effectively design the cover slab, treat it as a floor plan by dividing it into three tiles measuring 2x6 meters and 6x6 meters For the initial selection of the tank cover thickness, utilize the specified formula to ensure structural integrity.

- In which l1 is length of shorter side, l1 = 5.4 m, l2 is length of longer side l2 = 6 m bn h 1 (5400 6000) (110 137) mm

→ Choose thickness of cover slab h bn = 120 mm

- Preliminary selection of the wall thickness of the tank wall according to the following formula: bt bt min

→ Choose thickness of water tank wall h bt = 150mm

- Preliminary selection of thickness of bottom slab

The bottom plate must support the weight of the concrete and withstand a significant water pressure of 20 kN/m² at a depth of 2 meters To ensure durability and prevent cracking, as well as to provide waterproofing, an appropriate thickness for the bottom plate is crucial.

→ Choose thickness of bottom slab h bd = 150 mm

 Concrete B25 : Rb = 14.5 MPa, Rbt = 1.05 MPa, Eb = 30x10 3 MPa

 Steel CB400-V : Rs = Rsc = 350 MPa; Rsw = 210 MPa; Es = 20x10 4 MPa

 Steel CB240-T : Rs = Rsc = 240 MPa; Rsw = 170 MPa; Es = 21x10 4 MPa.

Calculation of cover slab

- The cover panel with the tank wall and has the following dimensions:

Figure 3.1: Loading transfer to cover slab

- Including the self-weight of the structural layers

Table 3.1: Static loading of the slab

- The tank cover only has repair activities, no live load, we take the distributed live load as 0.75 kN/m 2 (TCVN 2737-1995)

- Effective fixed live load:p1.3 0.75 0.975 kN/m 2

1.1 2 l 5.4   The cover plate works in 2 directions

- Preliminary dimension of top cover panel beam: Dbn : 200 400 mm 

- Consider hd/hb >3  the connection between slab and beam is fixed connection calculation of the cover plate in the form of a 4-sided mounting list (diagram 9)

Figure 3.2: Free-body diagram internal forces

- Maximum positive moment between span:

- Maximum negative moment between support:

- In which: αi1, αi2, β i1, β i2: are the coefficients looking up the table according to the diagram 9 and the ratio L2/L1

- P is total pressure on slab: PqL L1 2 gp L L 1 2

Table 3.2: Internal forces of cover panel

- Reinforced the checking hole 600 x 600 by 4ỉ12 b s b o s

 m  Choose As As choose choose

Calculation of wall plate

- The pressure chart has the shape of a triangle that increases with depth

- In the bottom of water tank (z = 2m): p n     n h n p    10 2 1.1 22 kN/m  2

- Construction site placed in wind zone IA, so wind pressure: W o 0.55 kN/m 2

- Level of the top water tank: z = 73.1 m

- Wind pressure constant throughout the height of the tank wall

- Inflow wind load: W h nW kc 0 1.2 0.55 1.549 0.6   0.613 kN/m 2

- Outflow wind load: W d nW kc 0 1.2 0.55 1.549 0.8   0.817 kN/m 2

Table 3.4: Static loading of the wall plate

- Static loading of the slab with strip 1 m: N bt glb5.06 1.8 1  9.11 kN

- The wall is a structure subjected to compression and bending Compression force includes only the wall TLBT For simplicity in calculation, the wall is calculated as pure flexural member

- Wall slab with length 10 m work on 1 direction

- Wall slab with length 6 m work on 1 direction

- The connection between the wall plate and the cap beam is joint connection

- The connection between the wall and the bottom beam is the fixed connection

 Since the sides are roughly the same size, just calculate for the 10 m side slab and the same layout for the 6 m side slab

- Load combination: Full filled water tank + Inflow wind load

- Internal forces be calculated by formula according to table 6 [KCBTCT tập 3 Võ Bá

Tầm] linear calculation result was solved by super position method

- Maximum positive moment between span:

2.77 kNm 33.6 128 33.6 128 n nhip nhip gio nhip nuoc p h Wh

- Maximum negative moment between support:

15 8 15 8 n goi goi gio goi nuoc p h Wh

 tt %  sc % (kNm) mm mm mm (mm 2 )  a (mm 2 )

Table 3.5: Reinforcement result of cover slab

Calculation of bottom slab

- The bottom slab is completely concreted with bottom beams, using the beam system

Figure 3.5: Bottom slab of water tank

Table 3.4: Static loading of the slab

- Full filled water loading (h=2 m): p n     n h 1.1 10 2  22 kN/m 2

1.2 2 l  5  bottom slab work on 2 directions

Figure 3.6: FBD of bottom slab

 Maximum positive moment between span:

 Maximum negative moment between support:

- In which: mi1, mi2, k i1, k i2 are the coefficients looking up the table according to the diagram 9 and the ratio L2/L1

- P is total pressure on slab:PqL L1 2 gp L L 1 2

Table 3.6: Internal forces of bottom panel

% kNm mm mm mm 2  a mm 2

Table 3.7: Reinforcement result of bottom slab

Calculation of water tank beam system

 Loading of top cover beam:

Figure 3.7: Load transferring diagram of top cover

- Choose dimension of top cover beam is:

- Self-weight of the beam

- Loading on top cover beam

- DN2, DN3: Trapezoidal distributed load; DN1: Triangle distributed load

- Inflow wind load: W h nW kcB 0 x 1.2 0.55 1.546 0.6 6    3.67 kN/m

- Outflow wind load: W d nW kcB 0 x 1.2 0.55 1.546 0.8 6    4.89 kN/m

- Inflow wind load: W h nW kcB 0 y 1.2 0.55 1.546 0.6 11    6.73 kN/m

- Outflow wind load: W d nW kcB 0 y 1.2 0.55 1.546 0.8 11    8.97 kN/m

DL LL IWL OWL kN/m kN/m

Table 3.8: Total loading on top cover beam system

 Loading of bottom slab beam:

- DD2, DD3: Trapezoidal distributed load; DD1: Triangle distributed load

 Self-weight of wall slab:g t g bt  h 5.06 2 10.12 kN/m 

Table 3.9: Static loading of the wall plate

Shape DL LL IWL OWL kN/m

Table 3.10: Total loading on bottom cover beam system

- In fact, these beam systems work at the same time, so students solve the problem of spatial working beam systems by modeling the frame system with Etabs 9.7.1 software

Figure 3.9: Modeling of water tank

Figure 3.13: Moment DN1 and DD1

Figure 3.16: Shear force DN1 and DD1

- Calculated of main reinforcement and stirrups all according to TCVN 5574:2018

- Verify shear resistance of concrete:Q  b3 (1  n )R bh bt o arrange constructive of design

- Determine constructive span of stirrup:

In span ẳ L: h450,s ct min (h / 2;150); h450,s ct min(h / 3;300)

In span ẵ L: h300,s ct min(3h / 4;500);h  200 no needed to calculate

 Choose designed span of stirrup umin(s ,s ,s ct tt max )

- Verify axial stress condition:Q0.3  w1 b1 R bh b o

Table 3.11: Reinforcement result of beam roof water tank

Axial stress condition (kN) (mm)

Stt (mm) Smax (mm) Sct (mm) Schọn (mm) 1/4L 1/2L

Check deflection and deformation of bottom slab

- The deflection of the four-sided clamp panel is calculated according to the following formula:

  : Coefficient based on ratio of 2

L [Table appendix 17, Kết cấu bê tông cốt thép tập 3,

 q c : standard load evenly distributed on the bottom plate,q tc 20.54 kN/m 2

 D: Cylindrical stiffness , determined by the formula

- So the deflection of the bottom plate:

- Because of slab with L 40 m so there are two types of wind-load must be considering to building which are static wind-load and dynamic wind-load

- Wind load consist of two elements

- Static wind-load be calculated according to TCVN 2737:1995

- Static wind pressure calculated at altitude z is calculated by the formula: j o j J j

The wind pressure value (W o) for construction in Lao Cai province is determined based on the zoning map in appendix D and section 6.4 of TCVN 2737: 1995 In the IA area, the impact of storm winds is evaluated to be weak, with a calculated wind pressure of W o = 0.55 kN/m².

 k z : is the coefficient considering the change of wind pressure according to altitude, taken according to Table 5, TCVN 2737: 1995

 c: wind pressure coefficient, inflow coefficient c   0.8, outflow coefficient c   0.6

 The reliability coefficient of the wind load is  1.2

The wind-receiving area for each floor is determined by converting the static wind load into a concentrated force at the respective floor elevations, positioned at the rigid center of each floor This involves calculating the standard wind forces, denoted as Wtcx for the X direction and Wtcy for the Y direction.

Figure 4.2: Diagram of calculation of wind load impact on the project

 hj, hj-1, B is height of stories respectively j, j-1, and inflow width

Height Floor Height of floor

Z (m) …F H(m) kj (kG/m²) (kG/m²) (m) (kG/m²) GTX j

Height Floor Height of floor

Static wind load wind load X- directio n wind load Y- directio n

 Determine the oscillation frequency of model:

Figure 4.3: Result of frequency of model

Case Mode Period Frequency Comment sec Hz

→ Because this frame is reinforcement  =0.3; construction in wind area I-A Look up table 2 (TCVN 229:1999)  Limit value of oscillation frequency fL = 1.1

→ Construction calculated with 4 modes (mode 2, 6 for X-direction; mode 1, 4 for Y- direction)

- The height of building greater than 40m following to TCXD 229-1999, considering the effect of dynamic wind load, following to instruction the step for calculation as follow:

- “Note 10 at page 10: When the building is reinforcement concrete and brick stone, also steel building have the enclosure structure”

The dynamic component of wind load focuses solely on the impact of wind speed impulses The calculated dynamic wind pressure, denoted as Wpj, acting on the j section of the structure is determined using a specific formula.

 Wpj : pressure, the calculation unit is daN/m 2 or kN/m 2 depending on the calculation unit of Wj

The calculation unit for wind pressure acting on the j part of the building is expressed in daN/m² or kN/m², as specified by Article 4.10 of TCXD 229-1999, which outlines the standard values for the static component of wind pressure.

The coefficient of dynamic pressure of wind load, denoted as j, is dimensionless and corresponds to a specific height of the building Values for j are derived from TCVN 2737: 1995 and are provided in Table 3 of TCXD 229-1999.

 different types of building fluctuations, dimensionless In the formula, is obtained by

When the wind-receiving surface of a building is oriented rectangularly and aligned with the fundamental axes, the values for wind pressure are derived from Table 4 of TCXD 229-1999 Parameters ρ and γ should be determined as outlined in Table 5, with the values for the second and third vibrations set at ν2 = ν3 = 1.

Table 5 are taken according to TCVN 2737-1995

- The dynamic composition of the wind load must include the effect of the wind velocity impulse and the inertial force of the building

- When the fundamental natural frequency s, satisfies the inequality f s f L f s  1 , it is necessary to calculate the dynamic composition of the wind load with s first form of oscillation: pji j i i ji

 Wpj: Force, calculation unit is daN or kN depending on the calculation unit of WFj in the coefficient formula  i

 Mj: concentrating mass of part of the building jth; (t)

  i :dynamic coefficient corresponding to the ith oscillation, dimensionless depending on the parameters  i and logarithmiC REIduction of the oscillation i o i 940f

 γ =1.2 Safety factor of wind load

Table 4.3: Limited value of frequency f l

- Specific vibration frequency of the building:

- Calculating dynamic wind load with X direction on mode 2 and mode 6

- Calculating dynamic wind load with Y direction on mode 1 and mode 4

- ψ coefficient determine by formula: n ji Fj j=1 i n

WFj represents the standard value of the dynamic wind load component acting on the jth part of a building This value corresponds to various types of oscillations while considering the impact of wind velocity pulses It is measured in force units and is determined using a specific formula.

 ξ j : is the dynamic pressure coefficient of the wind load, at the height z corresponding to the jth part of the building,

 υ: spatial correlation coefficient of dynamic pressure of wind load corresponding to other types of vibration of the building, determined according to table 4 TCXD 229: 1999

 S j : windward area of part j of the building, m 2 ; S =D ×Η j j j

- Dynamic component of wind loading due to pulse:

 Standard load acting on the i.th floor:W =W S pj pj j

- Wind pressure on ZOY with X direction

- Wind pressure on ZOX with Y direction

Height Floor Mass of floor

36.9 Story10 1961.979 -4.629 4.398 0.277 3.8 130.402 3.079 40.7 Story11 1961.979 -5.157 4.956 0.275 3.8 132.186 3.099 44.5 Story12 1933.014 -5.654 5.493 0.273 3.8 133.831 3.118 48.3 Story13 1906.183 -6.115 6.004 0.271 3.8 135.360 3.136 52.1 Story14 1906.183 -6.533 6.482 0.270 3.8 136.788 3.152 55.9 Story15 1834.400 -6.908 6.927 0.269 3.8 138.129 3.168

Table 4.4: Wind load result of mode 1

Height Floor Mass of floor

55.9 Story15 1834.400 -1.608 -0.985 0.269 3.8 138.129 5.076 59.7 Story16 1769.018 -3.572 -3.035 0.267 3.8 139.394 5.099 63.5 Story17 1769.018 -5.42 -5.066 0.266 3.8 140.592 5.120 67.3 Terrace 1769.018 -7.092 -7.013 0.265 3.8 141.729 5.141 71.1 Roofing 1700.930 -8.557 -8.823 0.264 2.9 142.812 3.938

Table 4.5: Wind load result of mode 2

Table 4.6: Calculation result for determine wind load

 Determine the dynamic component of the wind load:

- The calculated value of the dynamic component of the wind load is determined by the following formula: tt j i ji

Value of wind load dynamic component in X-

Value of wind load dynamic component in Y-Direction

Z(m) …F Mj (T) (kN) (kN) (kN) (kN)

Table 4.7: Dynamic component of windload

 Combination of static wind loaf and dynamic wind load:

- Internal force combination, displacement cause static and dynamic wind load

 X: Internal force cause by static and dynamic wind load

 X t : Internal force cause by static wind load acting on building

 X d : Internal force cause by dynamic wind load acting on building

 s: The number of the calculated oscillation

Static wind load Dynamic wind load

Z(m) …F Mj (T) (kN) (kN) (kN) (kN)

Table 4.8: Result of static and dynamic components of wind

 Internal force and displacement due to wind load:

To effectively analyze the combination of wind forces, utilizing ETABS software is essential due to the complexity and scale of the computational processes involved.

- The load internal force combination process is performed by the following steps:

- The static wind load in direction X: WTX

- The static wind load in direction Y: WTY

- The dynamic wind load in direction X: WDX

- The dynamic wind load in direction X: WDY

- Internal force combination of static and dynamic components of the throughput of the wind load: 2 COMBO

- Wind load on X-direction: WX = WTX “+” WDX

- Wind load on Y direction: WY = WTY “+” WDY

In the ETABS model, we will assign the static wind load at the geometric center of the building, ensuring that the wind is directed towards the center of mass for accurate structural analysis.

When designing high-rise buildings in earthquake-prone areas, accounting for seismic load is essential Earthquakes play a critical role in the structural requirements of such constructions, making it imperative to incorporate earthquake-resistant features.

- Calculation of earthquake force according to TCVN 9386: 2012 (Design of works subject to earthquakes)

- According to TCVN 9386: 2012, there are 2 methods of calculating earthquake load: equivalent horizontal static method and vibration spectrum analysis method

- With cycle T1(x) = 2.869s, T2(y) = 2.841s.The equivalent horizontal static method requirement is not fix: T 1 4T C 2.4s

  (4.3.3.2 TCVN 9386 : 2012) So in this project, the earthquake load will be calculated according to the vibration response spectrum analysis method (Article 4.3.3.3 of TCVN 9386: 2012)

- The calculation of earthquake load is done according to TCVN 9386: 2012 and the help of software ETABS

Table 4.9: Building characteristics and derivative parameters

According to TCVN 9386: 2012, the horizontal design spectrum Sd (T) for high-rise buildings can be determined using Excel functions, focusing solely on the horizontal components of earthquake effects.

- Earthquake load will be declared in the define section

Load combin ation type Case name Scale factor

8 COMB6 ADD TT;HT;WX 1;0.9;0.9

9 COMB7 ADD TT;HT;WX 1;0.9;-0.9

10 COMB8 ADD TT;HT;WY 1;0.9;0.9

11 COMB9 ADD TT;HT;WY 1;0.9;-0.9

12 COMB10 ADD TT;HT;WX;WY 1;0.9;0.63;0.63

13 COMB11 ADD TT;HT;WX;WY 1;0.9;0.63;-0.63

14 COMB12 ADD TT;HT;WX;WY 1;0.9;-0.63;0.63

15 COMB13 ADD TT;HT;WX;WY 1;0.9;-0.63;-0.63

18 COMB16 ADD TT;HT;QX 1;0.3;1

19 COMB17 ADD TT;HT;QY 1;0.3;1

20 COMB18 ADD TT;HT;QX;QY 1;0.3;1;0.3

21 COMB19 ADD TT;HT;QY;QX 1;0.3;1;0.3

20 COMBAO ENVE (COMB1;COMB2….COMB19) (1;1….;1)

Q Load cases Name Type Self weight Auto lateral load

1 Dead load TT Live 1.1 User defined

2 Live load HT Live 0 User defined

3 Dynamic load X WDX Wind 0 User defined

4 Dynamic load Y WDY Wind 0 User defined

5 Static load X WTX Wind 0 User defined

6 Static load Y WTY Wind 0 User defined

4.2.1 Checking top displacement of frame:

- Section 2.6.3 Structural inspection criteria, TCVN 198-1997 High-rise buildings – Design of reinforced concrete structures of the whole block stipulates:

Strength, deformation, overall stability, and local stability tests of high-rise structures are conducted in accordance with current design standards, ensuring compliance with essential requirements for safety and performance.

- Regarding the hardness test: The horizontal displacement at the top of the structure of the high-rise building calculated by the elastic method must satisfy the following conditions:

- According to section 2.6.3 TCVN 198-1997, the horizontal permissible displacement at the top of the work for the wall-frame structure is: 1

Figure 4.4: Top displacement of frame according to X-direction

Figure 4.5: Top displacement of frame according to Y-direction

4.2.2 Checking top acceleration of frame:

- According to TCXD 198:1997, the maximum acceleration of the movement at the top of the building under the influence of the wind meets the following conditions:

- In which: the allowable peak acceleration according to the standard is taken as 150 mm/s2

- Refer to the book DESIGN & CONSTRUCTION OF HIGH-RISE BUILDING -

Construction publishing house 1996, the work acceleration is calculated as follows:

 f: is the oscillation frequency of the form

 Aw: is the displacement of the dynamic wind

 Checking top acceleration according to X – direction:

 Checking top acceleration according to Y – direction:

4.2.3 Checking anti-roll of frame:

- According to TCXD 198 - 1997 high-rise buildings have a ratio of height to width:

→ No need to check the anti-roll condition

9386:2012 is the difference of the average horizontal displacements of story drifts at the ceiling and floor of the floor under consideration

4.2.4.2 Checking according to wind-load:

- According to the instructions in Table M.4, Appendix M - TCVN 5574:2018, the horizontal displacement of a floor of a multi-storey building is limited as follows:

- Walls and partitions made of bricks, gypsum concrete, reinforced concrete panels:

- Walls (natural stone tiles) made of ceramic tiles:

 d r : Floor displacement (relative horizontal displacement between floors) caused by wind load case It is the value of dr UX and dr UY from ETABS software

Table 4.13: Checking displacement of story according to wind load

4.2.4.3 Checking according to earthquake load:

- According to the instructions in 4.4.3.2 - TCVN 9386:2012, limit the relative horizontal displacement between floors (displacement of floor is as follows):

- For buildings with non-structural parts of brittle materials: d r  v 0.005h

- For buildings with non-structural parts made of flexible materials: d r  v 0.0075h

 d r : Displacement of floor (relative horizontal displacement between floors), is the difference of average horizontal displacements ds at ceiling and floor of the floor under consideration (d s q d d c )

 d s : Displacement of a point of the structural system caused by the design seismic action

 d q : Displacement behavior coefficient, assumed to be q, unless otherwise stated

 d c : Displacement of the same point of the structural system is determined by linear analysis based on response spectrum

The reduction factor, denoted as ν, is crucial for assessing the lower repetition period of seismic actions in relation to damage limitation requirements Its value varies based on seismic hazards and the importance of the building For important buildings classified as grade I and II, it is advisable to use ν = 0.4, while for significant structures of grade I and II, as well as levels III and IV, a value of ν = 0.5 is recommended.

STUDENT: HUYNH GIA HUY ID: 17149016 55 d rUX (mm) d rUY (mm) rUx.v r Ux.v

Attic 0.020838 0.034368 3.00E-02 2.12E-02 73.1 18.5 Satisfied Roofing 0.00024 0.000599 1.25E-03 1.25E-03 71.1 20 Satisfied Terrace 0.000275 0.000648 1.25E-03 1.25E-03 67.3 17.5 Satisfied

Table 4.14: Checking displacement of story according to earthquake load

4.2.5 Checking P-DELTA condition (second order effect analysis)

- The sensitivity of the relative horizontal displacement between floors is calculated according to the formula 4.28 TCVN 9386-2012: 0.1

  : is the sensitivity coefficient of the relative horizontal displacement between floors;

 P tot : is the total wall load at the floor under consideration and the floors above it when designed to withstand earthquakes;

 V tot : is the total shear force caused by the earthquake;

 h: is the height of the floor

 d r : is the difference of the average horizontal displacement at the center of mass at the ceiling and floor of the floor under consideration

 d rif : displacement according to the results of elastic analysis using design spectrum

+  0.1: no need to consider the P-Δ effect

+ 0.1  0.2: approximately multiply consequences effect by 1/(1-θ)

+  0.3: re-adjust the structural system and re-calculate and check

 Standard load combination to determine Ptot:

Ptot = Dead-load (self-weight + completion weight+ weight of wall…)+ n*Live-load

Figure 4.7: P tot result from ETABS programe

Determine Vtot with earthquake load PDDX, PDDY

With COMB PDDX = ADD(QX+0.3QY), COMB PDDYD(0.3QX+QY)

Attic 2 4578.6435 0.9612 1.4358 0.08495 0.0046 0.032091 0.001163 OK Roofing 3.8 26079.223 2.8199 4.2646 0.07395 0.0256 0.035919 0.008222 OK Terrace 3.8 50137.8525 4.7332 7.2492 0.07395 0.0046 0.04105 0.001667 OK

- Tensile plate should be calculated according to rectangular cross-section b×h

- The calculated reinforcement content and the layout content must meet the following condition: min max

 àmin: Minimum reinforcement content ratio: àmin = 0.05%

 àmax: Maximum reinforcement content ratio

- In case of positive moment in span: The flange participates in compression, so we consider steel according to the cross-section T

- Structural conditions to be included in the wing width calculation: b f = b d + 2S f

Mf = Rb × bf × hf × (h0–0,5.hf)

 If M ≤ Mf → neutral axis through the wing, then calculate the beam according to the rectangular section with dimensions (bf × h)

 If M  Mf → the neutral axis passes through the stiffness

- Tensile plate should be calculated according to rectangular cross-section b×h

- The calculated reinforcement content and the layout content must meet the following condition: min max

 àmin: Minimum reinforcement content ratio: àmin = 0.05%

 àmax: Maximum reinforcement content ratio

4.3.1.1 Effectuate the calculation of typical beam:

No Story Beam Span Location M

Table 4.16: Internal force of beam B1

 Calculate rebar at left support:

- Calculation of reinforcement along the problem of bending members of rectangular section b×h0×60(cm)

 Calculate rebar at right support:

- Calculation of reinforcement along the problem of bending members of rectangular section b×h0×60(cm)

- Calculate the reinforcement at the bearing section with the largest absolute value in the beam with the same span, then arrange for the remaining beams of the remaining floor

- Calculate stirrups of beam element

- Beam B14 (300×600) with maximum shear force: Qmax = 355.11 kN

- Checking condition of concrete strip between inclined sections (main compressive stress):

STUDENT: HUYNH GIA HUY ID: 17149016 61 b b1 b 0

- Checking condition of shear force on inclined section: b sw

- Considering the most dangerous case when Qb+Qsw reaches the minimum value

- Preliminary selection of belt steel diameter, number of branches, distance of belts are as follows:

Support h mm h mm s Span h mm h mm

- Replace to formula, we have:

- 0.5.R b.h bt 0 85.05 kNQ b 229.63kN2.5.R b.h bt 0 425.25 kN

- Checking condition of inclined section subjected to moment: s sw

- We have: M = 107.82 kNm < Ms + Msw = 115.74+20.49 = 136.23 kNm

→ Choosing this stirrup section is satisfied

Table 4.17: Reinforcement result of beam

The internal forces acting on columns consist of three key values: axial load (P), moment about the x-axis (Mx = M2), and moment about the y-axis (My = M3) Consequently, the design of the column is based on the principles of oblique eccentric compression and the arrangement of circumferential steel reinforcement.

In the analysis of the section with edges Cx and Cy, it is essential to consider the effects of compressive force N and bending moments Mx and My, along with the random eccentricities eax and eay By evaluating the bending in both directions, the coefficients ηx and ηy are determined, leading to an increase in the moments to Mx1 and My1.

DESIGN OF TYPICAL FLOOR

DESIGN OF FOUNDATION SYSTEM

CONSTRUCTION METHODS

Tiêu đề	Capstone Project Grand Mercure Apartment
Tác giả	Huynh Gia Huy
Người hướng dẫn	Assoc. Prof. Tran Tuan Kiet
Trường học	Ho Chi Minh City University of Technology and Education
Chuyên ngành	Automotive Engineering
Thể loại	Graduation project
Năm xuất bản	2023
Thành phố	Ho Chi Minh City

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Số trang	185
Dung lượng	11,2 MB