(Luận văn) luxury apartment brg park residence

INTRODUCTION

DEMAND AND PURPOSE OF THE CONSTRUCTION

- In the face of rapid population growth, the demand for land to build houses is increasing, but many people cannot afford to buy land to build houses To solve this problem, the solution to building high-rise apartment buildings and developing residential planning is reasonable today In addition, the investment in the construction of high-rise housing projects to replace low-rise buildings and degraded residential areas also help to change the face of the urban landscape to match the stature and position of the city At the same time, it also helps create job opportunities for many people.

- Therefore, BRG PARK RESIDENCE commercial and service apartment complex was born to contribute to solving the above goals This is a modern high-rise building, fully equipped, beautiful landscape, and includes entertainment, commercial, and shopping suitable for living, entertaining, and working, a high-rise apartment building is designed and construction with high quality, fully equipped to serve the living needs of the people.

PROJECT INTRODUCTION

- Located in Tang Nhon Phu A Ward, District 9, Ho Chi Minh City, surrounded by full services, entertainment, transportation, appropriate education.

Figure 1: The geographical location of the project

- BRG PARK RESIDENCE has become a bright spot attracting many customers, families, and businesses to settle down and develop.

- The apartments here are reasonably designed with small and medium size, convenient for many customers and families, especially businesspeople, office workers or working from home.

- With a modern and luxurious design BRG PARK RESIDENCE promises to be a complex that provides a safe and comfortable living environment, fully meeting the needs of long-term settlement and investment.

ARCHITECTURE SOLUTIONS

- BRG PART RESIDENCE commercial and service apartment complex includes 2 basements and 18 floors.

+ Dimension of the floor basement: 86 (m) × 40.2 (m).

+ Dimension of the typical floor: 34 (m) × 40.2 (m).

- It includes 2 elevator cores, 2 exit staircases.

- Stories from 1 to 4 are the commercial center.

- Stories from 5 to 18 are apartments and technical rooms.

- The basement is located at the height of -5.80 (m) and is arranged with ramps from the ground to the main direction to facilitate the circulation up and down the basement We see that the function of the project is a high-class apartment building, so most of the basement area is used for parking, because the target customers of the project are high-income people, so the arrangement of basement space Parking for cars is necessary, besides arranging motorbike parking Arrange the gene boxes reasonably and create the coolest space possible for the basement.

- Stories from 1 to 4 are considered as common living areas for the whole block, beautifully decorated Arrange the dining, refreshment area and stage, common living space for the 1st floor of the block In general, it is easy to operate and manage when arranging rooms like the existing architecture.

- The technical floor is used to control and operate the machines in the building and only technical staff are allowed to enter.

- Stories from 5 to 17, including 3-star luxury apartments with full amenities and interior finishing, below is the floor plan that shows the function of the block, the apartments are arranged reasonable around the common path to help convenient traffic between the two blocks along with efficiency in the process of using the building.

- The 18th floor is the roof floor.

CAPSTONE PROJECT INSTRUCTOR: TRAN TUAN KIET Assoc Prof

- The building has the form of a vertical block, the height of the building is 64.6 (m).

- The facade of the building is in harmony with the surrounding landscape.

- The main materials used in the project are granite stone, water-based paint, stainless steel glass frame and soundproof safety glass to create a harmonious and elegant color.

- The building has a modern architectural shape suitable for the nature of a high-class apartment complex combined with a commercial center The use of new materials for the facade such as Granite, high-class tiles along with thick glass panels create a luxurious look for an architectural work, which is a construction trend today.

- Using and fully exploiting modern features with large glass doors, exterior walls are finished with water-based paint The reinforced concrete roof has a waterproof and heat- insulating layer Brick wall, plastering, water painting, wall color painting.

- The horizontal transportation system in the building is the corridor system.

- Standing traffic system includes elevator operating 24/24, 2 emergency stairs In which, 5 elevators on each side are arranged in the middle and run along the height of the building with the remaining 2 stairs arranged on the opposite side.

- The elevator system is designed to be comfortable, convenient, and suitable for the needs of the building.

- Lighting system: Requires standard illuminance for each area.

- Emergency and exit lighting system: Arranged in sensitive areas such as lobby, corridor, staircase, and crowded places.

- Socket power supply system, air conditioning system, water heater: The most reasonable and optimal arrangement in the working space and safely grounded.

- Electrical cabinets and power cables: With a separate power supply transformer for the office block, the backup generator system provides 100% power for the office block when the electrical cabinet power is evenly distributed on all floors, each floor will be installed 1 separate galvanometer.

- Daily activities: Water supply for buildings is connected from the City's water supply network on the residential route through the water meter to the underground water tank Then use the pressure pump to supply water to the entire toilet area on the upper floor and reserve water by the roof water tank.

- The water supplied after passing the water meter enters the domestic water tank and the reserve water tank for firefighting located in the basement, then uses a pressure pump to supply water to the wall firefighting system and the Sprinkler firefighting system in the basements floor.

- The building's domestic water is put into the wastewater treatment tank and then discharged to the City's drainage system Drainage gas holes designed with closed lids are drained on a separate route out of the building and connected to the area sewer.

- Rainwater drainage: Rainwater from the roof and balcony is collected into ỉ200ữ ỉ400 pipes and drained to the basement and then to the manhole to collect water from there through the drainpipe connected to the regional drainage system.

- Rainwater on the ground will drain into the manhole to collect surface water from which it will drain into the area network.

- Four sides of the work are installed windows to get light in the rooms There are also air conditioners in the rooms.

- The fire extinguishers are placed in a conspicuous position and can fight fires at all positions of the building, each floor has one location of the fire box Use CO2 & ABC chemical fire extinguishers.

- Based on regulations on fire protection for high-rise office buildings, it is necessary to have a sprinkler system with automatic sprinkler heads for working rooms, corridors, basements for parking and large halls where people gather for activities The sprinkler automatic fire fighting main pipeline connected to the water inlet throat located at the entrance gate of the work so that the fire truck can supply water to fight the fire.

- The lightning protection system used for this building is an active lightning collector

CIPROTEC ESE code CPT-1, protection radius Rp = 65 (m) The lightning rod is mounted on the 5 (m) high pole, and the entire base needle is placed on the highest reinforced concrete roof of the building.

- Bare copper lightning conductor, cross section 70 (mm²) is led from the top of the lightning rod to the ground by 2 lines in the ỉ42 pipe along the stairs & elevator walls.

- Before hitting the ground, the cables must be connected through a test box with stainless steel casing at a height of 1m above the finished surface.

- Particularly, the lightning escape route along the staircase wall will be installed with an additional lightning counter with a height of 2 (m) from the finished surface Test box & lightning counter installed in the wall.

- Use heat-welding method to connect the lightning conductors to the terminals of the piles.

- Before hitting the ground, the cables must be connected through a stainless-steel test box at a height of 2 (m) above the finished surface.

- All equipment, metal structural frames on the roof must be connected to the nearest lightning conductor by 50 (mm²) bare copper wire.

- Using heat-welding method to connect the grounding pile system to the lightning conductor.

- The down-conductor that will be connected to the grounding ground independently of the grounding ground resistance must have a value of less than 1 Ohm (Rrtx < 1 Ohm).

STRUCTURE SOLUTION OF PROJECT

OVERALL

- Structure system of building is frame structure and pier wall.

- The flat roof is reinforced concrete and is waterproofed.

- Stairs made of reinforced concrete block Water tank is made of reinforced concrete, used to store water, alternately supplying water for the use of all floors Covering walls and partition walls between apartments are 200 (mm) thick, room walls are 100 (mm) thick.

- To ensure the structural requirements, the full-block concrete rib structure is a reasonable choice for this work, with a low floor height, to create space, the structural plan is flat floor Typical floor calculations are as follows:

+ Floor plan and calculation diagram.

+ Calculation of reinforcement for the floor.

+ Check the deflection of the floor.

CHOOSE STRUCTURE SOLUTION

2.1)Main load-bearing structural system:

- Based on the working diagram, the structure of high-rise buildings can be classified as follows:

+ Basic structural systems: Frame structure, pier wall structure, rigid core structure and tube structure.

+ Mixed structural systems: Frame - brace structure, frame – pier wall structure, core pipe structure and combined pipe structure.

- In which the pier wall structure (also known as the rigid wall) is a wall system that is both responsible for bearing vertical loads and as a system for bearing horizontal loads This is the type of structure that, according to many foreign documents, is very suitable for high-rise apartments The outstanding advantage of this structural system is that it does not need to use the floor beam system, so it is optimally combined with the plan that is not obstructed by the beam system, so the height of the house is reduced The load- bearing wall structure system combined with the floor system forms a multi-compartment box system with large spatial rigidity, high monolith city, good horizontal rigidity, and large bearing capacity, especially horizontal loads.

- The pier wall structure has good earthquake resistance According to the results of research on damage caused by earthquakes, for example, “the February 1971

STUDENT: NGUYEN HOANG AN 17140001 earthquake in California”, “the December 1972 earthquake in Nicaragua”, “the 1977 earthquake in Romania” showed that the building structures with rigid wall structures are only slightly damaged while those with frame structures are severely damaged or completely collapsed.

→ Therefore, this is the chosen structural solution for the project.

- Structure of beam and slab floor system.

- Advantages: The calculation is simple, the slab thickness is small, so it saves concrete and reinforcement materials As a result, the entire frame slab is significantly reduced in load due to the floor's own load Currently, rib flooring has been widely used in our country as well as other countries with diverse construction technologies, skilled and professional workers, so it is convenient for technical selection and construction organization.

- Disadvantages: The beam height and the deflection of the floor slab are large when exceeding the large aperture, leading to a large building height, which is detrimental to the structure of the building when subjected to horizontal loads and does not save material costs But above the beams are mostly covering walls (beams are hidden in the wall) separating the spaces, so it still saves space.

- The structure consists of a beam system perpendicular to each other in two directions, dividing the slab into four-sided manifest boxes.

- Advantages: Avoiding the case of having too many columns inside, so it saves space and has beautiful architecture, suitable for works with high aesthetic requirements and large use space such as halls and bridges club.

- Cons: Complicated construction techniques On the other hand, when the floor plan is too wide, it is necessary to arrange additional main beams Therefore, it also cannot avoid limitations because the main girder height must be high to reduce deflection.

- The structure consists of floor slabs placed directly on columns (with column caps or with column caps).

- Advantages: The structure height is small, so the height of the floor clearance can be increased Space saving use Easily divide the use space Suitable for works with medium aperture (6-8m) Beautiful architecture, suitable for modern architectural works.

- Cons: Large floor thickness should cost materials, large self-load causing waste

Advanced technology and construction requirements Currently, the number of projects in Vietnam that can use this type of floor is limited, but soon, beamless floors combined with prestressed floors will be widely used and bring high efficiency in construction economic and technical for our country.

+ The intended use of the project.

+ Architectural and structural characteristics, load of the work.

+ Basis of preliminary analysis above.

→ Choose the full block frame structure for designing building.

- In the construction, the floor system has a great influence on the spatial working of the structure.

- Choosing the right floor plan is very important Therefore, it is necessary to have the right analysis to choose a suitable plan for the structure of the work.

- Because the project is a high-rise building, at the same time to ensure the aesthetic drawing of the apartments, the main structural solution of the project is selected as follows:

+ Foundation structure of bored piles.

+ The building structure is load-bearing wall structure, including a pier wall system.

PRINCIPLES OF STRUCTURAL CALCULATION

- When designing, it is necessary to create a structure diagram, section size and reinforcement arrangement to ensure durability, stability, and spatial stiffness in terms of overall as well as individual structural parts Ensuring sufficient bearing capacity must be in both construction and use phases.

- When calculating the design of reinforced concrete structures, it is necessary to satisfy the requirements for calculation according to 2 groups of limit states.

- To ensure the bearing capacity of the structure, specifically to ensure the structure:

- Not damaged by loads and impacts.

- No loss of shape or position.

- No damage when the structure is fatigued.

- Not damaged by the simultaneous impact of force factors and adverse environmental influences.

- To ensure the normal operation of the structure, it is necessary to limit:

- The crack does not expand beyond the allowable limit or does not appear cracks.

- There are no distortions beyond the allowable limit such as deflection, rotation angle,sliding angle, oscillation.

METHODS OF DETERMINATION OF INTERNAL FORCE

- Internal force is determined by manual calculation method with the following tasks:

- Separation of components in the building by linearity and locality.

- Select the appropriate calculation scheme.

- Solve internal forces according to the lookup table or mechanical formulas.

- However, the solution time is long and complicated, easy to make mistakes when calculating and the accuracy is not high, or it is too safe because the calculation scheme is often chosen as the mount, the ideal match is just an assumption of the boundary conditions are not so ideal In some cases, the load is only approximated And the solution formulas are only true for the condition when the material is still working in the elastic region.

- Therefore, students combine solving internal forces by manual and software methods (solved by FEM finite element method).

- The software results are reliable when several deformation criteria are met by the line of action of the load, the magnitude of the deformation is consistent with the position of the specific force, and the internal force will be different from the calculation The software model considers the influence of the components together, if the internal force is much different from the calculation, there will be reasonable evaluations and explanations for the choice.

- Within the scope of this project, students use the following software to analyze the internal forces of the model:

+ ETABS 20: Finite element software analyzes the working of the whole building. + SAFE 20: finite element software specializing in the analysis of plate members (floor slabs, foundations, ).

CAPSTONE PROJECT INSTRUCTOR: TRAN TUAN KIET Assoc Prof.

MATERIALS

Table 1: The values of concrete

1 R b = 17 (MPa) Slab, beam, pier wall, staircase, mat foundation, bored

2 Sand cement mortar Cement mortar for construction, plastering of house

Table 2: The values of steel

CHOOSE PRELIMINARY DIMENSION OF STRUCTURE

6.1)Choose preliminary dimension of beam:

- Dimensions of the component cross-section are as follows:

+ Preliminary section of beam according to empirical formula (preliminary according to 2 conditions of deflection and strength condition as follows:

+ The dimension of the beam cross-section is preliminarily determined through the girder span (experimental formula) so that the necessary clearance in the floor height is sufficient and the bearing capacity is sufficient.

Table 3: Dimension section of main beam

Length of Height h c Width b c Section

Table 4: Dimension section of secondary beam

Length of Height h c Width b c Section

- Let h s be the thickness of the slab, hs is selected according to the condition of bearing capacity and convenience for construction, in addition h s > h min TCVN 5574:2018

+ h min = 50mm for residential floors and public buildings.

+ h min = 60mm for factory floor.

+ h min = 70mm for slabs made from lightweight concrete.

- The slab thickness is preliminarily determined by the formula:

❖ L 1 : calculated span in the short side direction

- Consider the floor slab with the largest dimension: 12 (m) × 10 (m):

→ So the floor works in 2 directions, choose m = 50:

→ Choose h s = 180 (mm) (satisfy the condition h s > h min = 50 (mm) for residential floors and public buildings).

6.3) Choose preliminary dimension of pier wall:

- Preliminary dimensions of ladder walls and core: According to TCVN 198 - 1997:

- Holes (doors) in the bulkheads shall not significantly affect the load-carrying performance of the bulkheads and shall be reinforced with structural measures for the area around the openings.

- The wall thickness (b) is chosen not less than 150 mm and not less than 1/20 of the floor height.

- Choose thickness of pier wall:

6.4) Choose thickness of cover concrete layer:

- The large thickness of protective concrete is determined based on the following criteria:

- QCVN 06-2010/BXD – National technical regulation on fire safety for houses and constructions.

- The construction site is in Ho Chi Minh, far from the area with concrete penetration such as the coast, river region, …

Table 5: The thickness of concrete cover layer

2 Structure contact of soil, lining concrete layer 35 (mm)

- For the assembled structure, the minimum value of the thickness of the concrete layer protecting the load-bearing reinforcement shown in Table 2.8 is reduced by 5 (mm).

- For structural reinforcement, the minimum value of the thickness of the protective concrete layer is taken to be reduced by 5 (mm) compared to the required value for the load-bearing reinforcement.

- In all cases, the thickness of the protective concrete layer should also be taken not less than the diameter of the reinforcement bar and not less than 10 (mm).

- The thickness of the protective concrete layer at the top of the pre-stressed members over the length of the stress transmission zone to be taken is not less than 3d and not less than 40 (mm) for bar reinforcement and not less than 20 (mm) for cables It is allowed to take the thickness of the protective concrete layer of the section in the

STUDENT: NGUYEN HOANG AN 17140001 support for prestressed reinforcement with or without anchors like that of the section in the span for prestressed members with transmitted support internal forces. concentrated when there are steel details in the support and reinforcement limiting horizontal deformation (horizontal welded wire mesh or longitudinal reinforcement) arranged according to the instructions in 10.3.4.10 TCVN 5574-2018.

- In members with pre-stressed longitudinal reinforcement on concrete and located in cages, the distance from the surface of the member to the surface of the cage should be taken not less than 40 mm and not less than the width (diameter) telescopic tube, up to the side – not less than half the height (diameter) of the cage When the prestressed reinforcement is in the grooves or outside the member section, the thickness of the protective concrete layer created by the subsequent spray method or other method is taken to be not less than 20 (mm).

- The thickness of the protective concrete layer for the belt reinforcement, distributed reinforcement and structural reinforcement should be taken not less than the diameter of these reinforcements and not less than:

- When the height of the member section is less than 250 (mm): 10 (mm) (15 (mm)).

- When the height of the member section is 250 (mm) or more: 10 (mm) (20 (mm)).

DESIGN STAIRCASE

LOADING

Figure 8: The section of landing layers

Table 6: Static load on landing

Table 7: Static load on the ladder

No Layer γ δ đ1 g tc Overload g tt

- Live load is taken according to TCVN 2737-1995, for stairs is p tc = 3 kN/m2, overload coefficient is taken as 1.2.

CALCULATE REINFORCEMEN

- The ladder specification for uniformly distributed load Cut a strip of width b = 1 (m) for calculation.

- Choose the link between the ladder and the landing as a pinned -

Some concepts of calculating stairs:

❖ If h d /h s < 3 then the connection between the ladder and the landing is considered a pinned

❖ If h d /h s > 3 then the connection between the ladder and the landing is concsidered a restraint.

+ Above is the concept of calculation in some reference textbooks However, in fact, calculating stairs has some inadequacies in the calculation scheme as follows:

+ In the all-concrete structure, there is no link that is absolutely absolute and absolutely joint The connection between the ladder slab and the landing beam is a semi- intermediate connection between the clamp and the joint (rigid connection),

CAPSTONE PROJECT INSTRUCTOR: TRAN TUAN KIET Assoc Prof it depends on the relative stiffness between the ladder slab and the incident beam, if If h d /h s < 3 then close is the pinned and vice versa Therefore:

- In the case if the connection between the ladder slab and the beam is considered to be clamped, it will lead to lack of web steel and excess structural support steel damaged due to lack of steel at the belly of the ladder.

- In case if the connection between the ladder slab and the beam is considered a joint, it will lead to lack of bearing steel and excess belly steel → The structure is not damaged but only causes cracking at the knee (due to lack of bearing steel) and gradually returns to normal matching diagram However, in reality, if the stairs are cracked at the pillow, it will lead to the lining tiles to peel off, so it is not allowed to crack the stairs in the design.

- In a multi-storey building structure, columns and beams are constructed on each floor, and the ladder is an independent structure that is constructed later Therefore, it is very difficult to ensure the rigidity of the ladder and ladder beams and walls, as well as the steel anchoring according to the designed calculation scheme (this happens very often during construction at the construction site).

- Conclusion: Based on the above analysis, the calculation is in favor of safety, ensuring usability when the building is subjected to the most unfavorable loads, as well as ensuring the aesthetics of the stairs during the period of use Students choose the diagram of 2 pinned to calculate but still arrange structural steel on the pillow (ỉ10a200) to prevent cracking for the stairs.

Figure 9: Calculation diagram of staircase

3.1) Calculate internal force by SAP2000 software:

Figure 10: Static load on part 1 of staircase

Figure 11: Static load on part 2 of staircase

Figure 12: Live load on part 1 of staircase

Figure 13: Live load on part 2 of staircase

Figure 14: Momen on part 1 of staircase

Figure 15: Momen on part 2 of staircase

Figure 16: Shear force on part 1 of staircase

Figure 17: Shear force on part 2 of staircase

- Momen maximum on ladder is: M max = 17.61 (kNm).

- Choose moments at support and span:

Table 8: Result of calculation reinforcement for ladder

- Self-weight of landing beam:

- Loads transmitted by the ladder:

- Loads transmitted by the floor:

- Loads transmitted from the ladder to beam:

❖ q: is total load on the floor: 4.95 + 3.6 = 8.55 (kN/m 2 ).

- Total load on landing: q = 1.2375 + 21.93 + 4.51 = 27.68 (kN/m).

Figure 19: Shear force on landing Table 9: Result of calculate reinforcement for landing

- The beam has Q max = 31.61 (kN).

- Check the cutting ability of concrete (According to TCVN 5574-2018):

→ Therefore, concrete is not capable of withstanding shear Need arrange stirrup.

- Determine the number of reinforcement steps according to the structural conditions:

→ Choose S = 100 (mm) arragne at L/4 head beam.

→ Choose S = 200 (mm) arragne at L/4 middle beam.

→ Arrange ∅8a100 for support, ∅8a200 for span

Figure 20: The maximum deflection value of staircase

- The upper deflection is determined under ideal conditions, having a stiffness EJ If replacing with reinforced concrete material with hardness B, considering:

- Considering the presence of reinforcement

- Occurrence of cracks in the tensile zone of the section

- The maximum deflection of the stairs according to TCVN 5574 - 2018 is:

2500 max = 200 = 200 5 (mm) f = 1.29 (mm) < f max = 12.5 (mm).

Figure 21: The section of staircase

Figure 22: The typical structure staircase

DESIGN SLAB FOR THE TYPICAL FLOOR

MODELLING SLAB BY SAFE SOFTWARE

- Floor design plan using SAFE software to calculate internal forces.

- The method of calculating internal force by the finite element method is being widely used today due to taking advantage of the strong computing power of the computer With specialized software such as SAFE.

- Calculation sequence in SAFE software:

❖ Step 2: Assign loads to the model.

❖ Step 3: Run and export output result.

Figure 25: The typical floor in SAFE

Figure 26: The 3D of typical floor in SAFE

CALCULATE REINFORCEMENT

4.1)Check short-term deflection of slab:

Figure 31: The value of short-term deflection

- Check floor deflection with SAFE Floor deflection is mainly due to static load, wall load and live load with standard load according to Limit State 2.

- Floor deflection in SAFE: f = 13.403 (mm).

- According to TCVN 5574:2018, the allowed deflection of the floor satisfies the expression f < f gh , in which f gh l is the defined deflection:

4.2) Check long-term deflection of slab:

Figure 32: The value of long-term deflection

- The long-term deflection of the floor is determined by the formula: f = f1 – f2 + f3

❖ f1 is the short-term deflection of the entire load.

❖ f2 is the deflection due to short-term effects of long-term loads

❖ f3 is the deflection due to long-term effects of long-term loads

- According to Table 4, Section 4.2.11 Vietnamese Standard 5574-2018, Note 2: when under the effect of permanent load, short-term temporary load, and long-term temporary load The deflection of beams or slabs should not exceed 1/150 span.

- Limit deflection of floor slab:

- Based on the above calculation results, the long-term deflection of the floor: f = 66.433 (mm) → f < f gh , so the floor satisfies the condition of long-term deflection.

- Divide the Strip to get the floor force from SAFE:

+ To facilitate the calculation of reinforcement later, proceed to divide the floor into strips In a strip of strips, the internal force is integrated Therefore, if dividing a strip too wide, the calculation of reinforcement later will be inaccurate and dangerous, and if dividing a strip with a width that is too small, the calculation will be complicated and very time consuming.

+ Divide the strip with a strip width of L/2 m across the beam floor plan in the X and

Y directions (In principle, we should divide the strip so that the moment graph is nearly identical, do not divide the place with negative moments and positive is 1 range because the SAFE software accumulates the average of the moment values.)

- Calculation of reinforcement is performed as for flexural members with dimensions:

- Calculation of floor steel is done as follows:

According to section 8.1.2.2.3 Vietnamese Standard 5574-2018, the value of is determined by the formula:

❖ , is the relative strain of the tensile reinforcement when the stress is equal to R s

❖ 2 = 0.0035 is the relative strain of concrete under compression when the stress is equal to Rb for concrete with compressive strength class B60 or less

- Reinforcement content: calculated reinforcement and layout content must satisfy the following conditions:

❖ à min : minimum reinforcement ratio, usually taken àmin = 0.05%

Figure 35: Momen on strip X-direction

Slab Possition Moment b h s h o As Steel As choose

(kNm) mm mm mm (cm 2 ) ỉ a (cm 2 )

Figure 37: Momen on strip Y-direction

Slab Possition Moment b h s h o As Steel As choose

(kNm) mm mm mm (cm 2 ) ỉ a (cm 2 )

4.4) Check the shear resistance of slab:

Figure 38: The result of shear force

- According to Section 8.1.3.3.1 TCVN 5574-2018, it is stipulated: When there is no transverse reinforcement, the calculation shall be carried out according to condition (89) of this section with Q SW taken as zero.

- Calculation of reinforced concrete members subjected to bending according to inclined section is carried out according to the following conditions (condition 89 of

❖ Q sw : shear force carried by transverse reinforcement in inclined section.

❖ Q b : shear force carried by the concrete in the inclined section

- The shear force Q b is determined by the following formula:

- But Q b is not more than 2.5 R bt bh o and not less than 0.5R bt bh o

- We have the largest sheer force of the floor Q max = 186.89 (kN)

Q = 186.89 (kN) < Qb = 0.5 R bt bh o = 440(kN)

→ So the floor has enough shear resistance.

4.5) Check crack width in slab:

- Crack resistance grade according to TCVN 5574-2018 in section 8.2.2.1.3 we have: acrc1 = 0.4 and acrc2 = 0.3.

Figure 39: Short-term crack width max = 0.2 (mm)

- Conclusion: after declaring the parameters including cracking and creeping for the floor, we get the largest crack 0.2 (mm) < [acr1] = 0.3 (mm).

Figure 40: Long-term crack width max = 0.1 (mm)

- Conclusion: after declaring the parameters including cracking and creeping for the floor, we get the largest crack 0.1 (mm) < [acr2] = 0.4 (mm).

→ Satisfy the cracking condition for slab.

Figure 41: Reinforcement X-direction of the typical floor

Figure 42: Reinforcement Y-direction of the typical floor

DESIGN FRAME STRUCTURE

COMBINATION LOAD

- The combination of horizontal components of seismic action can be done as follows:

- The maximum value of each effect acting on the structure due to two horizontal components of the seismic action, which can be determined by the square root of the sum of the squares of the values of the action effects due to each component lying horizontal caused.

- The rule above generally gives results in favor of safety.

- The seismic load combination is determined by the square root method of sum squares: ax = ±√ 2 + 2

❖ ax : Consequential values of the maximum impact are caused by the simultaneous action of horizontal seismic forces in both main directions.

❖ and : respectively, are the values of the effects caused by the seismic forces acting in the x-x and y-y directions.

❖ In practice, earthquake forces acting in two perpendicular horizontal directions are not always in phase with each other Therefore, TCVN 9386 - 2012 allows to

STUDENT: NGUYEN HOANG AN 17140001 use another combination plan in which 100% of seismic effects in one direction are combined with 30% of seismic effects in the perpendicular direction.

TT Load due to the weight of the floor itself (calculated by the software)

Loads due to the self-weight of floor finishes include (floor tiles, lining mortars, plasters, equipment piping loads, etc.)

TX Load due to the weight of the wall itself on the beam or floor.

HT1 Live loads acting on the floor are valid < 2 / 2

HT2 Live loads acting on the floor are valid ≥ 2 / 2

GTX Static component of wind load in the X direction

GTY Static component of wind load in the Y direction

GDX Live component of wind load in the X direction

GDY Live component of wind load in the Y direction

DDX Earthquake in the X direction

DDY Earthquake in the Y direction

Table 31: Combinations follow Limit State 1

No Name Combination Load case Overload coefficient form

3 Comb1 ADD TT; TTCT; TX; HT1; HT2 1.1; 1.2; 1.1; 1.3; 1.2

4 Comb2 ADD TT; TTCT; TX; GX 1.1; 1.2; 1.1; 1.2

5 Comb3 ADD TT; TTCT; TX; GX 1.1; 1.2; 1.1; -1.2

6 Comb4 ADD TT; TTCT; TX; GY 1.1; 1.2; 1.1; 1.2

7 Comb5 ADD TT; TTCT; TX; GY 1.1; 1.2; 1.1; -1.2

8 Comb6 ADD TT; TTCT; TX; HT1; HT2; 1.1; 1.2; 1.1; 1.17;

12 Comb10 ADD TT; TTCT; TX; DDX; DDY 1.1; 1.2; 1.1; 1; 0.3

13 Comb11 ADD TT; TTCT; TX; DDX; DDY 1.1; 1.2; 1.1; 1; -0.3

14 Comb12 ADD TT; TTCT; TX; DDX; DDY 1.1; 1.2; 1.1; -1; 0.3

3 ADD TT; TTCT; TX; DDX; DDY 1.1; 1.2; 1.1; -1; -0.3

4 ADD TT; TTCT; TX; DDY; DDX 1.1; 1.2; 1.1; 1; 0.3

5 ADD TT; TTCT; TX; DDY; DDX 1.1; 1.2; 1.1; 1; -0.3

6 ADD TT; TTCT; TX; DDY; DDX 1.1; 1.2; 1.1; -1; 0.3

7 ADD TT; TTCT; TX; DDY; DDX 1.1; 1.2; 1.1; -1; -0.3

TT; TTCT; TX; HT1; HT2; 1.1; 1.2; 1.1; 0.39;

28 Enve ENVE Comb1 to Combo25 (not

Table 32: Combinations follow Limit State 2

3 Comb1 ADD TT; TTCT; TX; HT1; HT2 1;1;1;1;1

4 Comb2 ADD TT; TTCT; TX; GX 1;1;1;1

5 Comb3 ADD TT; TTCT; TX; GX 1; 1; 1; -1

6 Comb4 ADD TT; TTCT; TX; GY 1;1;1;1

7 Comb5 ADD TT; TTCT; TX; GY 1;1;1;-1

8 Comb6 ADD TT; TTCT; TX; HT1; HT2; GX 1; 1; 1; 0.9; 0.9;

9 Comb7 ADD TT; TTCT; TX; HT1; HT2; GX 1; 1; 1; 0.9; 0.9; -

10 Comb8 ADD TT; TTCT; TX; HT1; HT2; GY 1; 1; 1; 0.9; 0.9;

11 Comb9 ADD TT; TTCT; TX; HT1; HT2; GY 1; 1; 1; 0.9; 0.9; -

12 Comb10 ADD TT; TTCT; TX; DDX; DDY 1; 1; 1; 1; 0.3

13 Comb11 ADD TT; TTCT; TX; DDX; DDY 1; 1; 1; 1; -0.3

14 Comb12 ADD TT; TTCT; TX; DDX; DDY 1; 1; 1; -1; 0.3

15 Comb13 ADD TT; TTCT; TX; DDX; DDY 1; 1; 1; -1; -0.3

16 Comb14 ADD TT; TTCT; TX; DDY; DDX 1; 1; 1; 1; 0.3

17 Comb15 ADD TT; TTCT; TX; DDY; DDX 1; 1; 1; 1; -0.3

18 Comb16 ADD TT; TTCT; TX; DDY; DDX 1; 1; 1; -1; 0.3

19 Comb17 ADD TT; TTCT; TX; DDY; DDX 1; 1; 1; -1; -0.3

20 Comb18 ADD TT; TTCT; TX; HT1; HT2; 1; 1; 1; 0.3; 0.3; 1;

22 Comb20 ADD TT; TTCT; TX; HT1; HT2; 1; 1; 1; 0.3; 0.3; -

28 Enve ENVE Comb1 to Combo25 (not

MODEL BUILDING BY ETABS

Figure 50: The 3D model in ETABS

4.1) Check the displacement of the top building:

- 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 the structure are carried out in accordance with current design standards In addition, high-rise building structures must also satisfy the following requirements:

- 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:

- According to the table of displacement results of the construction, the largest horizontal displacement at the top of the work in the x and y directions is F x = 41.422

- Height of building in ETABS: H = 70 (m).

Figure 51: The result of diaphragm center of mass displacements X-direction

Figure 52: The result of diaphragm center of mass displacements Y-direction

4.2) Anti-roll test for construction:

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

→ No need to check the anti-roll condition.

CALCULATE THE REINFORCEMENT FOR BEAM

Table 33: : Reinforcement for main beam

Table 34: : Reinforcement for secondary beam

- Based on data output from ETABS:

- Beam has the largest shear force: Q = 676.94 (kN).

- Shear strength of concrete: Q b = φ b3 (1 + φ f + φ n )R bt bh o (According to formula 76, section 6.2.3.3 of TCVN 5574 – 2018) (φ b3 = 0.6, φ n = 0, φ f = 0, see section 6.2.3.3 of TCVN 5574 – 2018).

→ Need to calculate belt reinforcement for beams.

- Belt steel is arranged to satisfy the belt pitch s = min(s tt ,s max ,s ct ,s dd )

❖ Calculating the largest step of the belt: smax ( b4 =1.5)

❖ The step of the belt is selected by structure:

→ Step belt reinforcement: s = min (183.3; 379; 250; 150) = 150 (mm).

5.2.2) Check the condition after selecting the belt reinforcement:

According to formula 72, Section 6.2.3.2 of TCVN 5574 – 2018:

- Arranged on 2 supports of beams 1 L is 8 and step 150 mm.

- Arranged on span of beams 1 2 L is 8 and step 250 mm because choose to follow the structure.

- Length of anchor or connecting reinforcement: = ( + ) and not less than =

- Anchor and connection of reinforcement in tensile zone: ≥ 30

- Anchor and connection of reinforcement in compression zone: ≥ 20

- The base reinforcement anchor length is calculated as follows: ×

❖ , : is the circumferential area of the cross section of the reinforcement, respectively.

❖ : The reinforcement adhesion strength is calculated as follows:

❖ 1 : is the coefficient considering the effect of the reinforcement surface type, taking 1=2.5 for non-prestressed, hot-rolled ribbed reinforcement

❖ 2 : is the factor that considers the influence of the diameter of the reinforcement, taking 2=1 for the diameter of blur more than 32mm

5.3.2) Calculation of reinforced anchor length:

- The length of the reinforcement anchorage calculated according to the requirements of the reinforcement, considering the solution for the structure of the anchorage area of the member, is determined by the formula:

L an = αLL o,an As,cf (mm)

❖ , , : are the calculated and actual cross-sectional areas of reinforcement, respectively

❖ : is the coefficient considering the stress state of concrete and reinforcement. Including the influence of the solution for the structure of the anchor area of the member to the length of the anchorage area.

❖ For reinforced bars without prestressing, when anchoring ribbed bars with straight anchoring or smooth reinforcement with hooks or U-bending without additional anchor details, = 1 for reinforcement bars tension and take = 0.75 for compressive reinforcement bars.

→ For convenience of calculation, we take the length of the reinforcement anchor as follows:

5.3.3) Calculating the length of the reinforcement connection:

- The joints of reinforcement bars in tension or in compression must have an overlap length not less than the value of length L lap , determined by the formula:

❖ , , : are the calculated and actual cross-sectional areas of reinforcement, respectively

❖ 2 is the coefficient that considers the stress state of the reinforcement bar, the structural solution of the member in the area connecting the bars, the number of connected steel bars in a section compared to the total number of steel bars in this section, approx distance between connected steel bars

❖ When connecting ribbed reinforcement with straight ends, as well as connecting plain steel bars with hooks or U-bends without additional anchoring details, the coefficient for tensile steel is equal to 2 = 1.2 for resistant reinforcement bars tensile and take 2 = 0.9 for compressive reinforcement bar

→ For convenience of calculation, we take the length of the reinforcement anchor as follows:

CALCULATION OF FRAME

❖ Calculation of pier wall reinforcement:

- Walls are one of the important bearing structures in high-rise buildings However, the calculation of reinforcement has not been specifically mentioned in the design standards of Vietnam.

- Therefore, within the scope of this project, the method "assumption of the moment boundary region" is used to calculate the reinforcement for the pier wall.

- Tensile stress due to reinforcement Compressive stress borne by concrete and reinforcement.

- Consider a rigid wall with loads N z , M y

Figure 54: Division of load-bearing zones on the horizontal and vertical sections of the wall

6.2) Calculation steps for longitudinal reinforcement for pier walls:

❖ Step 1: Assume the length B of the moment boundary region Considering the wall is subjected to axial force N and bending moment in the plane My, this moment is equivalent to a pair of torques placed at the two boundary regions of the wall.

❖ Step 2: Determine the tensile or compressive force in the boundary and middle region. l,r = × ±

❖ Ab: Area of the wall boundary area

❖ A: Area of the whole wall

❖ Step 3: Calculate the area of reinforcement in tension and compression.

- Calculation of reinforcement for boundary areas such as columns in tension - compression at the center The bearing capacity of the column in tension - compression at the center is determined by the formula:

❖ Rb, Rs: Calculated compressive strength of concrete and reinforcement.

❖ Ab, As: cross-sectional area of BT boundary area and longitudinal reinforcement

❖ ≤ 1: coefficient of reduction of bearing capacity due to longitudinal bending

(coefficient of longitudinal bending) Determining φ according to the empirical formula, can only be used when: 14 ≤ ≤ 104

❖ l o : calculated length of the column.

❖ i min : radius of inertia of the section in the thin direction → i min = 0.288b

❖ From the above formula, we deduce the area of compressive reinforcement:

- When N < 0 (tensile boundary area), due to the initial assumption: tensile stress due to reinforcement, the area of tensile reinforcement is calculated according to the following formula:

❖ Step 4: Check the reinforcement content If not satisfied, the size B of the boundary area must be increased and then recalculated from step 1 The length B of the boundary area has the maximum value of L/2, if this value is exceeded, the wall thickness should be increased.

- When calculating Fa < 0: put compressive reinforcement according to structure

According to TCVN 5574-2018 Structural steel for hard walls in moderate and strong earthquake zones.

- Horizontal reinforcement: content 0.4% but not less than 1/3 of the content of longitudinal reinforcement.

- In this wall internal force calculation, choose the structural longitudinal steel content of the regions:

❖ Step 5: Check the remaining wall as the center of compression In case the concrete has enough bearing capacity, the compressive reinforcement in this area is placed according to the structure.

- At any cross-section of the wall, reinforcement of belt steel at both ends of the wall must be reinforced Because the local stress (the shear stress and the normal stress in the plane) usually arises at the two ends of the Check for restriction conditions:

+ Concrete does not fail due to primary compressive stress:

Q max < Q1 = b3 (1 + + ) bt bh (with = 0.1 bt bh ≤ 0.5)

- If both conditions (1) and (2) are satisfied, just put the belt according to the structure.

❖ Step 7: Arrange reinforcement for rigid wall:

- The distance between longitudinal and transverse reinforcement bars must not be greater than the smallest of the following two values: { s ≤ 1.5b s ≤ 30 cm

- The arrangement of reinforcement needs to comply with “TCVN 5574:2018” as follows:

+ Two layers of wire mesh must be laid The diameter of reinforcement is chosen not less than 10mm and not more than 0.1b.

+ Selected vertical reinforcement content: 0.6(%) ≤ μ ≤ 3.0(%) (for moderate strong earthquakes).

+ Horizontal reinforcement chooses not less than 1/3 of longitudinal reinforcement with content ≤ 0.4% (for moderate and strong earthquakes) Use 10, n = 2.

- Measures should be taken to increase the cross-section in the boundary area of the pier walls.

- At the corners connecting the rigid walls together, connecting belts must be arranged.

- Because the moment can change direction, the reinforcement in the boundary area A s = max(A nen s , A keo s ); middle area reinforcement As’.

6.3)Calculation of horizontal reinforcement for pier wall: b3(1+ f + n ) b Rbt bho Qmax 0.3 wl b1 b R b bh o

❖ f = 0: factor considering the effect of the blade in compression

ℎ ≤ 0.5 : factor considering the longitudinal force effect

- The distance between the transverse reinforcements according to the calculation on the most dangerous inclined section:

- Maximum distance between transverse reinforcements in terms of shear concrete: s max 1.5(1 + n

- The design distance of the crossbar is: s = min(stt ,smax ,sct )

- Diameter of crossbar: choose ỉ = 12 mm and arrange all the reinforcement with about s = 200 (mm).

Figure 55: Pier wall plan layout

Figure 56: Pier wall at axis 3-D Table 35: Result of detailed calculation of 3-D axis frame reinforcement

Rigid wall Boundary area Middle area Crossbar Belt core

DESIGN FOUNDATION

GEOLOGICAL DATA STATISTICS

- According to the results of field survey and laboratory test results Stratigraphy at the construction site is classified into the following classes:

+ Layer 1: Clay, gray brown; plastic.

+ Layer 2: Clay, white gray, red brown yellow; plastic.

+ Layer 3: Fine to coarse sand mixed with powder, a little clay with a few small pebbles, brown, yellow, brown, red; a bit stiff.

+ Layer 4: Clay, sub-clay, yellow brown, gray green; hard.

+ Layer 5: Fine sand mixed with powder, little red brown clay; very hard.

Figure 57: Engineering geological cross-section Table 36: Statistical design geological data.

CALCULATION OF LOAD RESISTANCE OF PILES

E: is the elastic modulus of the pile material (MPa) 32500

Calculated strength of transverse reinforcement (MPa) 175

- Based on geological parameters we have:

- Soil layer 5 (Fine sand mixed with powder, little red-brown clay, tight and hard), has an average SPT index of 29, this soil layer is quite hard, and it will be difficult to drive piles, so the construction cost will be high when constructing pressed piles.

- The construction area in District 9, Ho Chi Minh City has many residential areas around, dense housing density, the construction of pressed piles will affect the surrounding existing works.

- The project has 18 floors, 2 basements, so the load is quite large.

→ Using the bored pile method will be suitable for the above geology.

3.2.1) Choose the size, material, and depth of the pile:

Length of broken pile head (m) 1

Length of pile head attached to the pile grating (m) 0.5

Area of pile section Ab (cm 2 ) 0.785

- Laying depth, for the foundation under the pier wall:

- The height of the foundation: h d = 2 (m).

→ Choose a foundation depth of h m = 5.8 + 2 = 7.8 (m) (basement 2 has a height of -

- The top of the pile is on the elevation -7.8 (m) (compared with the natural ground).

- The bottom of the pile is on the elevation -40 (m) (compared with the natural ground).

- For ladder core foundation: ladder core foundation is at a height of -2 (m) lower than other foundations for pit construction.

- The height of the foundation: h d = 2.5 (m).

→ Choose a foundation depth of h m = 5.8 + 2.5 + 2 = 10.3 (m) (basement 2 has a height of -5.8 m above the natural ground).

- The top of the pile is on the elevation -10.3 (m) (compared with the natural ground).

- The bottom of the pile is on the elevation -42.5 (m) (compared with the natural ground).

3.2.2) Select pier wall for calculation:

- Students choose to calculate the foundation for typical walls in the building to arrange for the remaining walls.

- Students choose pier P1, P12 to calculate the foundation.

- Elevator core pier wall: because the staircase wall and the ladder core wall are placed quite close together, students put together one station for general calculation.

3.2.3) Types of loads used for calculation:

- Building foundations are calculated according to the most dangerous internal force value transmitted to the foot of the column, including:

- Students calculate with the above combination N max and then check with the remaining combination.

- The calculated load is used to calculate the foundation according to the first limit state From the table of internal force combinations, students choose the most dangerous combinations for calculation.

- The calculated load is used to calculate the foundation under the second limit state. From the table of internal force combinations, students choose the most dangerous combinations for calculation.

3.2.4) Load capacity according to the pile material:

- Formula for calculating load capacity:

❖ : Longitudinal bending coefficient of piles

❖ : 0.85, working condition coefficient for bored piles (section 7.1.9 TCVN

❖ ′ : 0.8, condition coefficient for pile construction in the foundations, drilling and concrete pouring into the borehole under water using wall pipes (Section 7.1.9 TCVN 10304 - 2014)

❖ A s : Area of pile cross-section, = 62.8 ( 2 )(20 20)

❖ R b : Calculated compressive strength of pile concrete.

❖ R s : Calculated compressive strength of reinforcement in pile

- Determining the working length of the pile (Article 7.1.8 TCVN 10304:2014), for all types of piles, when calculating according to the strength of the material, it is allowed to consider the pile as a rigid bar in the soil at the cross section located at a distance the bottom of the tower an interval l 1 determined by the formula:

❖ l o : is the length of the pile from the bottom of the tower to the leveling height l o = 0

❖ Distortion factor: = √ (according to Appendix A TCVN 10304:2014).

❖ E b = 32.5×10 6 (kN/m 2 ), material module for piles.

❖ I = 0.05×1 4 = 0.05 (m 4 ), moment of inertia of the pile cross-section.

- Determine the coefficient k, which is averaged over the soil layers (table A.1 TCVN 10304:2014) The main soil surrounding the pile body is mixed clay, choose k = 12000 kN/m 4

- Determine the thinness of the pile:

3.2.5) Load bearing capacity according to the physical and mechanical criteria of the ground:

- The load capacity for bored piles is calculated according to formula (12), section 7.2.3 TCVN 10304:2014.

❖ c : Coefficient of working condition of pile in soil, take = 1.

❖ γ cq : coefficient of working condition of the soil under the cape taking into account the influence of the pile lowering method on the soil resistance = 1 (Table 4 TCVN

❖ γ cf : The coefficient of working condition of the soil on the pile body taking into account the influence of the pile lowering method on the soil resistance = 1 (Table 4 TCVN 10304:2014).

❖ U: Perimeter of the cross section of the pile, = 3.1416 (m)

❖ l i : The length of the pile in the soil layer i

❖ f i : Average resistance strength of the soil layer i

❖ q b : The strength of soil resistance under the pile tip, taken according to table 2 TCVN 10304:2014 and interpolated Here the pile is plugged into the sand layer at a depth of over 40 m, so we can look up the table q b = 4500 kN/m 2

- Looking up Table 3 TCVN 10304: 2014, we have the following table: pile lowering depth -40m, pile diameter d = 1m Pile length L = 32.2 m

- Determined cf f i li (The ground is divided into homogeneous layers not exceeding 2m):

Table 40: Table of wall resistance according to physico-mechanical criteria

3.2.6) According to criteria of ground strength (Appendix G TCVN

- The formula for determining the ultimate load capacity R c, u according to TCVN

❖ q b : is the load-bearing strength of the soil under the pile tip, for pressed piles:

❖ c = 4.9 (kN/m 2 ): soil cohesion under pile tip.

❖ A b = 0.785 (m 2 ): Cross-sectional area of pile tip.

❖ u = 3.1416 (m): Perimeter of pile cross-section

❖ vp ′ : effective vertical soil stress at pile tip elevation due to self-weight

❖ , : load-bearing coefficient of the soil under the pile tip.

Table 41: The load-bearing coefficient of the soil under the pile tip according to

- Maximum bearing capacity of pile due to lateral friction: = ∑

- : the average resistance on the pile body on the soil layer i: + For loose soil:

❖ : coefficient of horizontal pressure of the soil layer i (Table 8 TCVN: 10304- 2014).

❖ a,i : friction angle between the soil and the pile in the loose soil layer i, with the pile being a reinforced concrete pile chosen: a, i = φ

❖ v,zi : vertical mean effective normal stress of soil layer i

❖ : is a coefficient that depends on the characteristics of the soil layer above the adhesive layer, the type of pile and the method of lowering the pile For bored piles, 0.3- 0.45 for hard plastic clay and 0.6-0.8 for soft plastic clay.

❖ c u, i : undrained cohesive force of soil layer i.

Table 42: Friction force of pile wall in sandy soil layer

Table 43: Friction force of pile wall in clay layer

- The bearing capacity of the pile according to the criterion of ground strength:

3.2.7) Load capacity of pile according to SPT test:

- According to the formula of the Japanese Institute of Architecture, the ultimate bearing capacity of piles , :

❖ : is the resistance strength of the soil under the pile tip determined as follows:

❖ When the tip of the pile is in the loose soil = 300 for pressure pile, = 150 for bored piles

❖ When the tip of the pile is in the cohesive soil = 9 for pressure pile, = 6 for bored piles

❖ A p = 0.785 (m 2 ): is the cross-sectional area of the pile tip.

- Nose resistance: (We use bored piles according to: (Section G.3.2 TCVN 10304- 2014)).

- The bearing capacity of the pile is created by the soil at the tip of the pile:

❖ f ci : is the resistance strength of the soil sticking on the pile body × ×

❖ α p and f L determined according to charts G.2a and G.2b in TCVN 10304-2014

❖ c ui : Undrained adhesion force of soil layer i (=6.25N ci kPa)

❖ f si is the resistance strength of loose soil on the pile body = 10 3 Table 44: Friction force of pile wall in sandy soil layer

Layer L i Depth N si f si f si × L i

Table 45: Friction force of pile wall in clay layer

- Load capacity according to SPT test:

3.2.8) Determination of design load capacity:

Table 46: Summary table of load capacity under the hard wall

- For the bearing capacity of the core foundation of the elevator is taken equal to the load capacity of the normal foundation.

3.2.9) Determine the allowable bearing capacity of the pile:

- The BRG PARK RESIDENCE construction are designed considering the effects of earthquakes due to the design load, as well as the effects of earthquakes determined according to Section 12.7 of TCVN 10304 -2014:

+for piles in the foundation at the base of the hard wall:

+ for piles in the foundation at least at the elevator core:

❖ : the coefficient of working conditions, considering the factor of increasing the homogeneity of the ground when using pile foundations, is taken as 1.15 in multi-pile foundations.

❖ : reliability coefficient on the importance of the work, taken as 1.25

❖ Preliminary selection of pile groups: ≤ 5, so= 1.75

❖ Preliminary selection of pile groups: 6 ≤ ≤ 10, so = 1.65 (Foundation M1, M2)

❖ Preliminary selection for the core foundation of the ladder is at least 21 piles therefore = 1.4 (Foundation MLT)

Figure 58: Foundation plan layout 3.2.10) Determine the stiffness of the spring:

- Theoretical basis for calculating the coefficient of ground:

❖ To determine the stiffness of the earth spring, we first need to determine the ground coefficient Currently, there are many different methods to calculate the background coefficient such as static compression method in the field, using empirical formulas of some authors (Vesic, Terzaghi, Bowles), table lookup method Each method has a certain level of reliability, the difference in calculation results is large, so choosing the

STUDENT: NGUYEN HOANG AN 17140001 right method is important In their work, students will use Bowles' empirical formula method to calculate the background coefficient.

- Base coefficient according to Bowles formula is calculated according to the formula:

❖ C: Unit conversion factor, for SI units take C = 40.

❖ c: The cohesive force of the soil layer above the point in which the base coefficient is being calculated

❖ , : Unitless coefficient, depending on the shape of the foundation:

❖ For square foundation, square pile, round foundation, round pile takes = 1.3

❖ For square foundations, square piles take = 0.8

❖ For round foundations, round piles take = 0.6

❖ The coefficient depends on depth: = × ( × )

- From the foundation coefficient, the pile spring stiffness can be calculated as follows: + Stiffness of pile spring according to pile body (horizontal):

+ Stiffness of pile spring according to pile body (vertical direction):

+ Spring stiffness at the pile tip (vertical):

❖ L i : Length of pile i is subdivided to assign ground springs.

Unit conversion coefficient C 40 Correction coefficients n 0.5

Unitless coefficient S c 1.3 Square pile edge (pile diameter) B 1 (m)

Unitless coefficient S γ 0.6 Foundation base elevation h -7.8 (m)

Table 47: Calculation results of the ground spring stiffness around the pile body along the vertical and horizontal

→ Stiffness of the pile kiln at the pile tip (vertical): = × = × ×

DESIGN FOUNDATION M1

4.1)Determine the number of piles and the arrangement of piles:

- Choose the internal force to calculate the foundation: (N max , M x tu , M y tu , Q x tu , Q y tu )

Table 48: Hard wall foot reaction P1

Wall Combo N max M x tu M y tu Q x tu Q y tu

- Piles load capacity used: R cd = N cd = 5087.18 (kN).

- Select the pile size and layout as follows:

- Distance between the two pile centers s = 3d = 3 x 1 = 3 (m) Distance from the center of the pile to the edge of the platform s = d = 1 (m).

Figure 60: Reaction force at pier wall

- We have: P max = 3477.19 kN < P tk = 5087.18 (kN) → Satisfy the condition that the pile is not damaged

- P min = 3407.64 kN > 0 → Satisfy the condition that the stake is not uprooted.

4.2)Check ground pressure under pile tip:

- Determine the conventional foundation block size:

- The concept of the pile and the soil between the piles working simultaneously as a homogeneous foundation block placed on the soil layer below the pile tip The transmission surface of the conventional foundation block is wider than the area of the base of the platform with the opening angle (according to section 7.4.4, TCVN 10304:2014).

❖ i : Friction angle in the calculation of each layer of soil with thickness li through which the pile penetrates;

❖ h i : Length of pile in the "i" soil layer

- The base area of the foundation block is estimated by the formula:

- According to 4.6.9, TCVN 9362 – 2012, calculated pressure acting on the ground:

❖ k tc : Reliability coefficient, ktc = 1.1 because the calculated features are taken directly from the statistical tables (Look up according to Article 4.6.11 TCVN 9362–2012)

❖ m 1 : Coefficient of working condition of foundation soil (laying foundation at soil layer 5), take m 1 = 1.2 (Look up Table 15 according to Article 4.6.10 TCVN 9362–2012)

❖ m 2 : Coefficient of working conditions of the work interacting with the ground, depending on the ratio of the size of the work, m 2 = 1 (See Table 15 according to Article 4.6.10 TCVN 9362–2012)

❖ Conventional foundation depth -40 m corresponding to the 5th soil layer with friction angle 27 o 39':

❖ A,B,D: The coefficient depending on the friction angle in the ground is taken according to table 14 TCVN 9362-2012: A = 0.929, B = 4.832, D = 7.31

❖ b: Small side of the base of conventional foundation b = 8.92 m

❖ II : The density of the soil layer from the bottom of the conventional foundation block downwards (layer 5), because the soil layer is below the groundwater level, so = 18.72 − 10 = 8.72( / 2 )

❖ ' II : Density of soil layers from the bottom of the conventional foundation block upwards because the soil layer is below the groundwater level, so:

- Mass of soil in conventional foundation block:

- Weight of soil occupied by piles:

- Weight of land occupied by the station: dd = × ℎ × = 8 × 5 × 2 × 8.72 = 697.6( )

- Load to the bottom of the platform:

- Anti-bending moment of conventional foundation block: × 2 8.92 × 11.92 2

- Stress at the bottom of the conventional foundation block:

- Stress at the bottom of foundation M1: ❖ Stable background conditions:

→ So, the base of the conventional foundation block satisfies the stability condition. 4.3) Settlement test for foundation:

- Divide the soil layer under the pile tip into many layers with thickness h i =0.5m Calculate

= 0 × ( −1) The stress causing settlement at the bottom of the “i” layer

❖ look up the table depending on the ratio and (Table C.1, TCVN 9362-2012)

→ Need to calculate settlement for the foundation.

- Determine the settlement of each element layer and calculate the total settlement, according to the following formula (Appendix C, TCVN 9362-2012);

Table 49: Relationship coefficient e - p of soil layer No 3

Table 50: Results of settlement calculation for foundation M1

- Considering at Z = 2.5 (m) from the bottom of the foundation there are:

→ Stop settlement at 5 th element layer.

4.4)Check the condition of penetration resistance:

→ The tower height satified the penetration condition.

4.5) Test the resistance to shearing of the foundation M1:

(According to formula 76, section 6.2.3.3 of TCVN 5574 – 2018 standard)

→ Satisfied No need to calculate belt.

4.6) Calculate reinforcement for mat foundation:

- The internal force to calculate the reinforcement for the foundation is taken from the Strips that evenly divide the foundation in the model.

Figure 62: Result of strip momen X-direction of M1

Figure 63: Result of strip momen Y-direction of M1

- Formula for calculating steel for foundation:

+ Choose a gt bottom layer a gt.d = a ngàm + 20 = 150 + 20 = 170 (mm)

+ Choose a gt top layer a gt.t = 50 (mm)

- The arrangement of structural steel in the upper layer for both the X and Y directions is: ỉ14a200.

Table 51: Calculate the reinforcement for M1

DESIGN FOUNDATION M2

5.1) Determine the number of piles and the arrangement of piles:

Table 52: Hard wall foot reaction P3

- The concept of the pile and the soil between the piles working simultaneously as a homogeneous foundation block placed on the soil layer below the pile tip The transmission surface of the conventional foundation block is wider than the area of the base of the platform with the opening angle (according to section 7.4.4, TCVN 10304:2014).

❖ m 2 : Coefficient of working conditions of the work interacting with the ground, depending on the ratio of the size of the work, m 2 = 1 (See Table 15 according to Article 4.6.10 TCVN 9362–2012)

- Weight of land occupied by the station: dd = × ℎ × = 8 × 8 × 2 × 8.72 = 1116.16( )

- Stress at the bottom of foundation M2: ❖ Stable background conditions:

→ So, the base of the conventional foundation block satisfies the stability condition. 5.3) Settlement test for foundation:

- Divide the soil layer under the pile tip into many layers with thickness h i =0.5m Calculate the stress causing settlement until the condition is satisfied n bt 5 n gl (location where settlement stops) with:

Table 53: Relationship coefficient e - p of soil layer No 3

Table 54: Results of settlement calculation for foundation M1

- Considering at z = 4(m) from the bottom of the foundation there are:

→ Stop settlement at 5 th element layer.

5.5) Test the resistance to shearing of the foundation M2:

Figure 67: Result of strip momen X-direction of M

Figure 68: Result of strip momen Y-direction of M1

Table 55: Calculate the reinforcement for M2

DESIGN FOUNDATION MLTM

6.1)Determine the number of piles and the arrangement of piles:

Table 56: Hard wall foot reaction PLTM

Figure 69: Foundation MTLM plan layout

- The concept of the pile and the soil between the piles working simultaneously as a homogeneous foundation block placed on the soil layer below the pile tip The transmission

STUDENT: NGUYEN HOANG AN 17140001 surface of the conventional foundation block is wider than the area of the base of the platform with the opening angle (according to section 7.4.4, TCVN 10304:2014).

❖ m 2 : Coefficient of working conditions of the work interacting with the ground, depending on the ratio of the size of the work, m 2 = 1 (See Table 15 according to Article 4.6.10

- Weight of land occupied by the station: dd = × ℎ × = 14 × 29 × 2.5 × 8.72 = 8850.8( )

- Stress at the bottom of foundation MLTM: ❖ Stable background conditions:

→ So, the base of the conventional foundation block satisfies the stability condition.

- Divide the soil layer under the pile tip into many layers with thickness h i =0.5m Calculate the stress causing settlement until the condition is satisfied n bt 5 n gl (location where settlement stops) with:

Table 57: Results of settlement calculation for foundation MLTM

- Considering at z = 3(m) from the bottom of the foundation there are:

→ Stop settlement at 2 nd element layer.

6.5) Test the resistance to shearing of the foundation MLTM:

Figure 71: Shear force at MLTM

- Foundation MLTM has Q max = 37060.396 kN

Figure 72: Result of strip momen X-direction of MLTM

Figure 73: Result of strip momen Y-direction of MLTM

Table 58: Calculate the reinforcement for MLTM

CONSTRUCTION RETAINING WALL PROCESS

CONSTRUCTION OF THE LEAD WALL

2.1) Mission of the lead wall:

- Helps to locate the diaphragm wall drilling plate, guide the digging bucket to go in the right direction, and also serve to support the CWS waterproof gasket assembly; as a support, positioning the steel cage during the connection of the steel cages of each excavator, supporting the concreting pipe during the concreting process, preventing the soil from slipping into the excavation pit due to travel it rains,…

- Conductor walls must satisfy the following requirements:

+ The guide wall is designed with a continuous wall using reinforced concrete and is constructed according to the drawings The ducting wall system is poured at the site to resist and hold the surrounding soil firmly Therefore, the conductive wall must be able to withstand soil and water pressure and construction loads.

+ The design of the ducting wall must ensure that the top of the ducting wall is at least

1.5m higher than the groundwater level At the same time, the leading wall must ensure that the construction is dry The calculation of the lead wall is based on the actual conditions outside the building and the ground conditions and construction equipment used for the construction site Conduit walls must be fully constructed, in terms of height and shape The conductive wall is high enough to sustain the bentonite solution and deep enough to resist soil erosion caused by the disturbance of the bentonite solution.

+ The lead wall also has the effect of keeping bentonite from being lost Bentonite in the trench is always about 20cm lower than the wall and at least 1m above the groundwater level.

+ The guide wall must be at least 10cm above the natural ground to prevent surface water as well as mud from falling in

+ The lead wall also acts as a standard ledge for inspection and digging.

2.2) The process of constructing the lead wall:

- Leveling the ground along the diaphragm wall route so that it is enough to build the lead wall on both sides, construction vehicles and equipment travel.

- Carrying out surveying work along diaphragm wall and lead wall (line installation, elevation, position, ).

- Use a bucket excavator to dig 1.1m deep above the natural ground After digging, proceed to repair the excavation pit and pour concrete lining.

- Simultaneously with that process, conduct reinforcement processing at the processing yard The reinforcement of the conductive wall is usually arranged and must be connected longitudinally to each other, ensuring the conductive wall to form a whole.

- After the lining concrete has set, proceed to turn on the ink on the lining concrete to locate the leading wall and formwork Then, use a drill to drill into the concrete lining to plug the 10mm diameter, 150mm long steel pieces as gauges.

- Lining thin plastic sheets into the ground, erecting reinforcement and formwork in the inner wall, using millets of diameter 60mm and thickness of 25mm.

- Then proceed to pour concrete leading wall, after the concrete has set for 24 hours, proceed to dismantle the formwork.

- Install the ground and support the leading wall with melaleuca slats 70 to 100mm in diameter, the distance between the wooden bars is about 2m, support 2 layers.

- The leading wall will be demolished to carry out the construction of the hat beam as well as the excavation work for the basement construction later.

- The correct and correct use of bentonite is decisive for the success or failure of the digging process for wall trenches The use of bentonite for retaining walls will account for a significant proportion of construction costs Therefore, when choosing bentonite to keep the trench wall, it is necessary to pay attention to the following factors:

+ The effect of keeping the wall from collapsing: the density of bentonite must be appropriate, the elevation of the bentonite in the trench must be at least 1.5m higher than the groundwater level to be able to create a waterproofing pressure on the vertical surface of the trench wall .

+ When there is bentonite in the trench, on the vertical surface of the trench wall will form a layer of mud slurry that is less permeable to water Bentonite penetrates into the trench wall, firmly adheres to the soil wall, and the mortar consolidates with the soil to increase the waterproofing ability and reduce the possibility of landslides.

+ The effect of removing the residue of the mud mortar: the mud mortar has stickiness, when the digging bucket digs the soil, the soil residue is mixed into the mud mortar and removed according to the process of exchanging mortar Because the residue is removed, it is not deposited, increasing the efficiency of the excavator.

+ The effect of mud mortar also cools the equipment, smoothes the cutting process of excavated soil, reduces the frictional force in the process of cutting the soil in the trench.

- Basically, the properties and process of using bentonite for diaphragm wall construction are similar to those for bored piles However, since the guide trench has only a limited depth, special attention should be paid to ensuring the bentonite level in the excavation pit.

- Bentonite, when put into the construction site before mixing, we must conduct inspection according to Vietnamese construction standards such as: manufacturer's certificate, production date, weight of bentonite bags, preparation method and preserve…

- Bentonite is prepared with clean water The proportion of bentonite pythons is prepared in such a way that it is capable of stabilizing the excavation wall.

- Carry out tests with appropriate instruments to determine if its parameters are within acceptable limits.

- Density of clean bentonite mortar must be checked regularly, must be checked daily to control the quality of bentonite mortar.

- The average value of the bentonite mortar parameters provided for the excavation is within the limit range.

- Test to determine density, viscosity, pH value is carried out from the beginning and test until the bentonite sample matches the table above If the requirements are not met, it must be repeated, and at the same time, the dirt in the solution must be removed by a dirt separator.

❖ Bentonite solution supplied to trenches:

- When the test results on density and viscosity, pH value meet the requirements, then this bentonite is allowed to be supplied to the trench.

- During the construction process of digging trenches for walls in the soil, due to contact with groundwater, sand, soil and concrete, the amount of expanded clay and additives (additives) in the bentonite is partly lost, or there is also some loss It is possible that due to mixing with the bottom sediment of the trench, contamination or loss of required features.

- Among the factors that strongly affect the properties of bentonite, the trenching method has the main and decisive influence.

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