PROJECT OVERVIEW
Project introduction
Nowadays, in the process of integration of the country, the economy is growing, leading to the improvement of people’s living standards A large part of the people have the need to find a place to live with a clean environment, many convenient service to support their career, which requires the birth of many high-class apartments In that trend, many companies build high - class apartment buildings to meet the living needs of people.
With the demand for housing increasing while the land fund in the city center is getting less and less, the projects to build high - rise apartment buildings in the suburbs are reasonable and encouraged to invest The above projects, at the same time, contribute to the creation of an urban face if well organized and in harmony with the surrounding landscape.
Thus, the investment in the construction of Tan Tao 1 apartment complex is in line with the policy of encouraging investment of Ho Chi Minh City, meeting the urgent needs of people's housing and promoting economic development, completing the system, urban infrastructure
Address: 9/11 Ton Duc Thang street, District 1, Ho Chi Minh City.
Scale of construction works
According to Appendix II - Circular No 06/2021/TT - BXD dated June 30, 2021 of the Minister of Construction: Civil construction - grade 2 (10.000 m 2 ≤ slab ≤ 30.000 m 2 or 8 floors ≤ number of floors ≤ 24 floors).
The building has a height of 61.2 m (from the height of 0.00 m, not including the basement).
The construction area of the project is: 1440.25 2 1.2.4 Construction function: o Basement floor: Garage arrangement. o 1 st floor: Commercial center, service, office o 2 nd – 16 th floors: Apartment layout.
Architectural solutions
The plan is rectangular with the area of the land as above (1440.25 m 2 ).
The basement is located at the height of -3.60 m, arranged with 02 ramps from the ground to the basement (slope i = 18%) in 2 directions in and out from the main road, the entrance and exit are arranged appropriately to avoid causing damage, clutter is difficult to manage We see that because the main function of the building is to rent out apartments, the basement is mostly used for parking, arranging reasonable gain boxes and creating the most open space possible for the basement The system of stairs and elevators is arranged right at the basement entrance, users can see it immediately when entering to serve travel.
The first floor is considered as the common living area of the whole block, beautifully decorated with: stainless steel columns, a book display area and a living room creating a common living space for the ground floor of the block Especially, the building management room is located where guests can see if there is a need and the internal area of the building is arranged in an area with separate entrance In general, it is easy to operate and manage when arranging rooms like the existing architecture.
Floor (floors 2-16) this is the floor plan that shows us the most clearly the function of the block, along with the location adjacent to the road at both ends of the building, the function of the house is highly effective.
1.3.2 Construction traffic solutions: o Vertical walkways: there are 6 elevator rooms, 2 stairs o Horizontal walkways:
Corridor is the main walkway.
STRUCTURAL OVERVIEW
Analysis and selection of structural
Vertical load bearing system and horizontal bearing system
The vertical structural system has the main load-bearing role in the structure of high- rise buildings:
Receiving loads from slabs, beams to the foundations, to the ground.
Receiving horizontal loads acting on the construction (distributed between columns, wall and transmitted to the foundations).
Linked with slabs, beams to form a rigid frame system, keeping the overall stability of the building, limiting the vibration and displacement of the top of the building.
A structural system consisting of slabs, beams, combined with vertical bearing systems such as columns, wall which has the function of receiving used loads and transmitting them to beams and then transmitting them to column, diaphragm vertical structures Slabs are considered as rigid plates that are solidified horizontally to distribute loads to vertical structures.
The scale of the building is 2 basements and 17 floating floors, the total floating height is 61.2 m, choose wall system as the load-bearing structure for the work (the frame is subjected to vertical loads and the diaphragm can both withstand vertical loads and loads horizontal weight as well as other impacts while increasing the rigidity of the structure).
Choose the selected beam, slab plan to design for this project, both to meet the requirements of load-bearing, deflection due to the not too large column pitch, and to ensure the requirements of usability.
The vertical structure is the elevator rigid core system At the same time, a diaphragm is arranged around the stairs to increase the ability to withstand lateral loads in the short way of the building and the torsional resistance of the building.
Normally, the foundation of a high-rise building has to bear large compressive force, besides the earthquake load also creates a large horizontal force for the building, so the proposed solutions for the foundation include:
21 o Deep foundation: bored pile foundation, barret pile foundation, reinforced concrete pile foundation, prestressed centrifugal pile foundation. o Shallow foundation: 1-way strip foundation, 2-way strip foundation, raft foundation. o Choose an option: Compare solutions for pressed pile foundations or bored piles.
Materials used
Concrete used according to tables 7 section 6, TCVN 5574 - 2018
Concrete used for structure includes the following types:
Table 2.1 The compressive strength of concrete for the components
The work uses ribbed steel CB400-V (6 ≤ d ≤ 50 mm) and plain steel CB300-T 6 ≤ f
CB400-V ribbed steel (6 ≤ d ≤ 50 mm) has the following parameters:
Table 2.2 Specifications of CB400-V and CB300-T
Elastic modulus of steel 2×10 5 2×10 5 (MPa)
Preliminary section
Where: o = 0.8 ÷ 1.4 Depends on the load. o = 30 ÷ 35 For beam-type slabs where l is the slab span. o = 40 ÷ 45 For a four-sided slab where l is the span in the short-way direction. o ℎ = 50For flat roof. o The 2-way working slab with the largest short side 1 = 3750for calculation (Take the short edge of the largest slab).
We calculate preliminary the beam height according to the formula:
We calculate preliminary the beam width according to the formula:
Table 2.4 Beam cross-sectional dimentions of typical slab
Types of beams Beam span Preliminary dimension
Selecting the column section size will be do according to the following steps: o Step 1: Determine the vertical transmission area A s o Step 2: Preliminary calculation of total slab load q, including dead load and live load, q = 12.5 kN/m 2 to calculate the preliminary cross-section. o Step 3: Count the number of floors above the columns under consideration (n s ) o Step 4: The force acting on the column then. o Step 5: The effect of horizontal load equal to the coefficient k that varies depending on the column position, the preliminary column cross-section is calculated by:
Where: k is the influence coefficient of the moment, k = 1.1 ÷ 1.5, take k = 1.5 for the side column, k = 1.2 for the middle column, k = 1.1 for the corner column.
N: compressive force is approximated as follows:
In order to match the bearing capacity and create a better architectural space, the column cross section will be reduced about 3 floors 1 time from the foundation to the roof However, sudden hardness changes should be avoided The stiffness of the upper floor structure is not less than 70% of that of the adjacent lower layer If 3 floors decrease in hardness continuously, the total reduction will not exceed 50% (TCXD 198:1997, section 2.5.4)
Table 2.5 Column section preliminary of 2,5 – axis Column 2,5 – axis (B,C,D,E –
N (kN/m 2 ) A tt (m 2 ) A choose Dimension axis) (m 2 ) (mm 2 )
Table 2.6 Column section preliminary of remaining columns
Remaining columns N (kN/m 2 ) A tt (m 2 ) A choose Dimension
Based on architectural drawings and active loads, we preliminarily select the dimensions of the rigid wall According to Article 3.4.1 TCXD 198 – 1997; specified diaphragm thickness (b) choose not less than 150 (mm) and not less than 1/20 th of the floor height.
Where: o : Slab area of each floor o = 0.015
In addition, the selection of diaphragm in the corner and rigid diaphragm at core sections depends on the beam width: o Diaphragm in the corner: = 300 o Rigid diaphragm at core: = 300
LOAD AND IMPACT
Load classification
Loads are classified according to the duration of the load, divided into regular loads and temporary loads (long-term, short-term and special).
The self-weight of the building parts includes the supporting and covering structures, the mass and the pressure of the soil.
The long-term temporary loads include the weight and impact of equipment and machinery during use, impact due to temperature, humidity, etc.
Short-term loads include the weight of people, materials, accessories, tools and fixtures within the scope of equipment servicing and repair, and wind loads Loads generated during fabrication and erection of building structures.
Special loads are defined due to explosion, load due to serious violation of technological process, equipment malfunction, temporary damage Impact of ground deformation caused by changes in soil structure (landslide or wet settlement).
Dead load
3.2.1 Self-weight of the structure:
Standard specified gravity of reinforcement concrete: 25kN/m 3 , coefficient “ = 1.1” Weight of soil cover: 20 kN/m 3 , coefficient “
= 1.1” Weight of water: 10 kN/m 3 , overload coefficient “ = 1.0” The self-load of the building will depend on the geometrical dimensions of each member and the ETABS structural software automatically calculates the self-load.
3.2.2 Weight of retaining wall and fixed wall:
Where: o : The overall coefficient. o : Width of wall o “ℎ = ℎ − ℎ ”: Wall above the beam.
Total weight of the partition wall is converted into a uniformly distributed load on the slabs by taking the entire wall load placed on the slab consideration, multiplying it by a coefficients of 1.10 to consider the effect of concentrated load in the a wall array on the slab, and then dividing it by the area slab with walls to get a uniformly distributed load simply calculate.
Where: o : The overall coefficient. o = 18 / 2 : Specified gravity of wall 100 mm o : Thickness of wall. o : Length of wall o : Slab area.
Table 3.1 Load acting on slab Thickness Specified Height Length Area
No of wall gravity of wall of wall wall load wall load slab coefficient
200mm wall railing terrace 1.2m high placed on side beams:
Wall load acting on beams:
The load acting to construction:
Ceramic tiles 10 mm Morta 20 mm Reinforcement concrete 150 mm Morta 15 mm
Figure 3.1 Structural layers of ordinary slab Table 3.2 Dead loads on the slab bedroom, kitchen, dining room and living room
No Material gravity dead load dead load kN/m 3 m kN/m 2 kN/m 2
4 Total finish dead load (not including the
1.33 1.64 weight of the slab itself)
Table 3.3 Dead loads acting on the toilet slab
No Material gravity dead load dead load kN/m 3 m kN/m 2 kN/m 2
5 Total finish dead load (not including the
Table 3.4 Dead loads acting on the terrace slab
No Material gravity dead load dead load kN/m 3 m kN/m 2 kN/m 2
5 Total finish dead load (not including the
1,7 2.122 weight of the slab itself)
Table 3.5 Dead loads acting on the roof slab
No Material gravity dead load dead load kN/m 3 m kN/m 2 kN/m 2
4 Total finish dead load (not including the
1.2 1.522 weight of the slab itself)
Table 3.6 Dead loads acting on the basement slab
No Material gravity dead load dead load kN/m 3 m kN/m 2 kN/m 2
Live load
The live load acting on the building is based on TCVN 2737 - 1995 and the function of each work area, the live load value for each functional area is as follows:
Table 3.7 Summary table of active loads Functions of the n Long - term live Short - term live rooms (kN/m 2 ) load (kN/m 2 ) load (kN/m 2 ) (kN/m 2 )
Living room, dining room, 1.5 1.3 0.3 1.2 1.95 bathroom
Lobby, corridor, apartment floor 3 1.2 1 2 3.6 stairs
Analysis of building dynamics
For high-rise buildings, the active load including static load (static load, live floorload) and dynamic load (wind load, earthquake) is a type of load that changes with time, so it is necessary to consider the work load the process of being subjected to loads over time Dynamic problems are different from static problems: For dynamic problems, the state of stress and strain of the system also changes with time, the motion of the system with large acceleration and the inertia force depends on the motion acceleration, so it is necessary to considering the inertia force as well as the drag force.
The free oscillation patterns of the building are determined on the simplified structure diagram of the finite mass concentration point Each type of oscillation is characterized by the amplitude of the oscillations of the mass concentration points and the individual frequency of oscillations. o The work is considered as a cantilever bar embedded in the foundation o The mass concentration points are set corresponding to the floor level. o Mass concentrated at a point of mass concentration taken from half of the height of the lower floor to half of the height of the upper floor, including the mass of the structure itself and the floor structure layers, the volume of walls and walls floor fixed compartment and part live load.
We accept the following assumptions: o The floor beam is extremely rigid in its plane. o The entire volume of each floor is concentrated on the floor level. o The vertical displacement of the structure is small compared the horizontal displacement.
The difference between dynamic and static problems: o Load changes over time (may change in setting point, magnitude, direction and direction of impact) leading to changes in internal force Therefore, the result of texture analysis must be a function of time. o When the structure moves with acceleration, it will generate an inertial force (F ma), the static equilibrium equations are correct only when this component of inertial force is taken into account.
For diagrams with more than 3 mass concentration points, determining the specific vibration pattern of the building requires a large number of calculations.
Therefore, students use the help of ETABS software
3.4.2 The volume of participation fluctuates:
Standard volume value for each slab:
The mass of participating oscillations declared during oscillation analysis is 100% dead load and 50% live load for dynamic wind; 100% dead load and 24% live load when calculating earthquakes.
Figure 3.2 Declare Mas Source Data in Etabs
Buildings that declare both dynamic wind and earthquake loads should choose the number of vibrations to be considered in accordance with the conditions under consideration:
According to wind dynamics, the first number of cycles need considered is satisfied:
3.4.4 Space frame model: o Select the model, declare the grid unit o Material declaration o Declare section o Model building o Assign cross-section of floor beams, columns and walls o Structural analysis o Check model and run
Figure 3.3 Model of building in Etabs
Table 3.8 Period and frequency table of oscillation
Table 3.9 Frequency period and percentage of mass involved in fluctuating
Fluctuating modes: 1, 2, 3 có < = 1.3 o Mode 1 oscillate to Y-direction o Mode 2 oscillate to X-direction o Mode 3 oscillate to Z-direction
According to the experience formula, the period of Mode 1 is in the range (0.1 ÷ 0.14)
Where n is the number of floors (n = 17) We have: Mode 1 = 1.916 is in the range:
Figure 3.6 Mode 1 oscillate to Y-direction Figure 3.5 Mode 2 oscillate to X-direction
Figure 3.7 Mode 1 oscillate to Z-direction
Wind load
3.5.1 Static component of wind load:
0 − Standard wind pressure is determined from the treated wind velocity on the basis of wind speed monitoring data at a height of 10m above the benchmark (average speed is about 3 seconds, exceeded once in 20 years on average) corresponding to the terrain of wind pressure zoning.
The wind pressure value is determined for each wind pressure partition:
Table 3.10 Table of wind pressure according to the map of wind pressure zoning
With the location of the project located in District 1, Ho Chi Minh City, the wind pressure area is determined to be II-A:
Select the method of inputting the static component of the wind load as the force concentrated at the geometric center of the floor in two directions.
Where: o − Reliability coefficient of wind load (buildings with more than 50 years of existence, choose = 1.2). o − The aerodynamic coefficient is taken according to the windward side, taking the sum for both the intake and thrust winds = 0.8 + 0.6 = 1.4 o − Wind height of the j th floor o − Be the j th floor where the static wind component needs to be calculated, the approximation can be taken as H j equal to the height of the floor. o − Windward width of the j th floor.
36 o − Relative height of the upper floor compared to the lower floor. o − Coefficient taking into account the change of wind pressure with altitude apply formula A.23 in TXCD 229:1999.
Table 3.11 Wind load according Y – direction High level Height catch Width catch
STORY of slab k t the wind the wind m (m) (m) kN/m 2 kN kN
Table 3.12 Wind load according X – direction High level Height catch Width catch
STORY of slab k t the wind the wind m m m kN/m 2 kN kN
3.5.2 Dynamic component of wind load:
According to [TCVN 229-1999]: Instructions for calculating the dynamic component of wind loads according to the standard TCVN 2737 - 1995 The building has a height of
H = 60.6m > 40m, so the calculation must take into account the dynamic wind component for the building.
Calculation basis: According to [TCVN 229 -1999: Instructions for calculating the dynamic component of wind loads according to the standard TCVN 2737-1995].
The standard value of the dynamic component of the wind acting on element j of the ith vibration pattern is determined by the formula:
Where: o − Concentrated mass of the j th element. o − The dynamic coefficient corresponding to the ith vibration is determined based on the graph of the dynamic coefficient o Building by Reinforcement Concrete so we take = 0.3
38 o The parameter is determined by the formula:
Where: o − Reliability coefficient of load. o 0 ( 2 ) − Standard wind pressure value 0 = 83 ( 2 ) o − i th Specific frequency of oscillation. o − The factor is determined by dividing the building into several parts, to the extent that each part of the wind load can be considered as constant The coefficient is determined by the formula:
=1 o − The standard value of the dynamic component of the wind load acting on the j th part of the building, corresponding to different types of oscillations only taking into account the influence of the wind velocity pulse, whose dimension is force It determined by the formula:
= × × × o − The standard value of the static component of the wind pressure acting on the j th part of the building. o − It is dynamic pressure coefficient of wind load at the height corresponding to the j th part of the building. o − Area of the windward surface corresponding to the jth element of the building o The calculated value of the dynamic component of the wind load is determined by the formula:
Where: o = 1.2 − Reliability coefficient of load. o − Wind load adjustment coefficient according to the use time of the building, taking = 1.
Table 3.14 Calculated wind load dynamics in the Y-direction
No Story M j (kg) j 1X 1X W Fj (kN) y ji y ji W Fj y ji 2 M j W pjiX (kN) W pjiX tt (kN)
Table 3.16 Calculated wind load dynamics in the X-direction
No Story M j (kg) j 1y 1y W Fj (kN) y ji y ji W Fj y ji 2 M j W pjiY (kN)
Equivalent horizontal static force method:
This method of analysis is applicable to houses whose response is not significantly affected by vibrations of higher order than the fundamental in each principal direction.
The requirements of this clause are considered satisfied if the building structure meets both of the following conditions:
There are fundamental oscillation periods T1 in two main directions less than the following values:
Satisfy the criteria for vertical regularity given in 4.2.3.3 TCVN 9386:2012
This method of analysis should be applied to buildings that do not satisfy the conditions stated in 4.3.3.2.1 TCVN 9386:2012 when applying the equivalent lateral static force analysis method.
The response of all types of vibrations that contribute significantly to the overall response of the house must be considered.
The requirements given in 4.3.3.3.1 TCVN 9386:2012 may be satisfied if one of the following two conditions is met:
The sum of the effective masses of the considered vibration patterns accounts for at least 90 % of the total mass of the structure;
All oscillations with an effective mass greater than 5 % of the total mass are taken into account.
When using the spatial model, the above conditions need to be checked for each required direction.
If the requirements specified in 4.3.3.3.1 TCVN 9386:2012 cannot be satisfied (e.g. in buildings and buildings where torsional vibrations contribute significantly) then the minimum number of vibration patterns k is considered in the calculation when analyzing. space needs to satisfy both of the following conditions:
: the number of oscillations to be considered in the calculation
: the number of floors above the foundation or top of the underlying hardware
: period of oscillation of the k th form
→ Based on the construction characteristics, the method of vibration response spectrum is selected to calculate earthquakes.
, , , : has been defined in section 3.2.2.2 TCVN 9386:2012
= 0.2: coefficient corresponding to the lower bound of the horizontal design spectrum
In which: (according to TCVN 9386:2012)
Direction: horizontal (without considering the vertical)
In Y-direction: according to article 4.3.3.5.2 TCXDVN 375 - 2006 if the is greater than 0.25g, the vertical component of the seismic action should be taken into account.
Load combination
In TCVN: 2737-1995, two types of combinations are distinguished: basic combinations and special combinations The basic set includes: o Basic combination 1 (main combination) includes: permanent load (DL) and a temporary load, the combination factor is taken as 1. o Basic combination 2 (sub-combination): with 2 or more types of loads, the calculated values of temporary loads or their respective internal forces must be multiplied by the same combination factor: short-term transient load multiplied by a composite factor of 0.9.
For special combinations, including earthquake loads, the combination of internal forces should comply with TCVN: 9386 – 2012
Table 3.17 Load combination by ultimate limit state ULTIMATE LIMIT STATE (ULS): THGH I
SW Finish WALL LIVE 1 LIVE 2 WIND X WIND Y EQX EQY
Table 3.18 Load combination by service limit state SERVICE LIMIT STATE (SLS): THGH II
SW Finish WALL LIVE 1 LIVE 2 WIND X WIND Y EQX EQY
Where: o SW: Self-weight (permanent load) o Finish: Finish layers (temporary load) o WALL: wall load o LIVE 1, 2 : Live load o WLX: Wind load in direction X (temporary load) o WLY: Wind load in direction Y (temporary load) o EQX: Earthquake in direction X (special load) o EQY: Earthquake in direction Y (special load)
STAIRS
General concept
Stairs are the main means of vertical traffic of the building, formed from consecutive steps to form the body (side) of the ladder, the steps are connected by landings and projections to form the stairs Stairs are an important element in terms of architectural use and art, enhancing the aesthetics of the building.
Structure of stair: Ladder (structural depth), landing, beam, handrail
Stair structure
4.2.1 Floor plan and section of stairs:
4.2.2 Choose the preliminary size of the stairs:
The height of typical floor: ℎ = 3600
The stair has 18 steps, each skirtboard has 9 steps o Width of step: = 300 o Height of step: ℎ = 190
The landing is connected directly to the rigid wall, so there is no need to arrange the landing beam
According to “Kết cấu bê tông cốt thép Tập 3 – Võ Bá Tầm” book o Tilt angle of skirtboard:
Preliminary selection of the section
Preliminary selection of the skirtboard:
Preliminary selection of the beam:
Note: According to “Kết cấu bê tông cốt thép Tập 3 – Võ Bá Tầm” book, at case 2:
Skirtboard and beam are continuous.
Figure 4.4 Detail structure of stairs layers
Active load
4.3.1 Load acting on the landing:
Self-weight of layers of skirtboard:
Table 4.1 Load acting on landing
Load Material Thickness n Calculated load m kN/m 3 kN/m 2
Table 1 TCVN 2737-1995 Reliability factor for loads due to the mass of the building material and soil
Material and soil Reliability factor
Concrete with a volumetric weight greater than 1600 kg/m 3 , reinforced concrete, reinforced stone bricks and wood 1.1
Concrete with a volumetric mass of not more than
1600 kg/m3, separating materials, plasters and finishes (plates, shells, roll materials, coatings,
In work 1.3 primers, etc.) depending on production conditions
4.3.2 Load acting on the skirtboard:
Self-weight of layers of skirtboard:
For granite tiles with thickness :
For concrete cement with thickness :
For step floor have section ℎ × :
Table 4.2 Load active on skirtboard
Load Material Thickness n Calculated load m kN/m 3 kN/m 2
Load the transmission skirtboard onto the beam:
Note: Load the transmission skirtboard onto the beam by reaction at the support with the value 52.12 kN/m as the figure 4.3.3
Weight of rigid wall acting on beam:
Table 4.3 Dead load acting on the slab Structural layer Thickness (m) Self-weight n Calculated load
Live load acting on the slab: = × = 0.3 × 1.2 = 0.36
Load transmission slab onto the beam:
For simplicity and safety, we consider the load from the skirtboard to be evenly distributed throughout the beam:
Calculation diagram
With the support of Sap2000 software, solve internal forces for the two-headed linkage cases as follows:
Figure 4.6 Fixed connection moment diagram
Figure 4.7 Pin connection moment diagram
Figure 4.8 Pin and roller connection moment diagram
Figure 4.9 Fixed connection shear force diagram
Figure 4.10 Pin connection shear force diagram
Figure 4.11 Pin and roller connection shear force diagram
Figure 4.12 Beam of stair diagram
Figure 4.13 Moment diagram of beam
Figure 4.14 Shear force diagram of beam
Internal force of landing, skirtboard results:
Based on the above diagrams, as well as to satisfy the condition that the connection is not completely fixed or pin, redistribute the moment as follows:
Calculation of reinforcement
Calculate reinforcement according to TCVN 5574-2018
= 25 → ℎ 0 = ℎ − = 150 − 25 = 125 With the condition in work: = 0.9;
Table 4.4 Calculating reinforcement for skirtboard Location
4.5.2 Check the shear resistance of the skirtboard:
According to Article 8.1.3 TCVN 5574-2018, when calculating the strength of reinforced concrete members under shear force, we need to calculate according to the following 2 conditions: o Calculation of strength of flexural members along the concrete strip between inclined sections (to check the cross-section, if it is not satisfied, increase the strength level or section size):
→ So the skirtboard slab is able to withstand shear according to the condition of calculating the strength according to the concrete strip between inclined sections o Calculation of strength of flexural members according to inclined section subjected to shear According to Article 8.1.3.3.1 TCVN 5574-2018, the shear force resisted by concrete in the skirtboard slab without reinforcement must satisfy the following conditions:
Calculated according to flexural structure, rectangular cross-section
Calculate reinforcement according to TCVN 5574-2018:
Table 4.5 Calculating reinforcement for beam Moment
Extra reinforcement 1 16 at bottom beam
Shear force of concrete resistance:
Calculated according to flexural structure, rectangular cross-section
Distance between stirrups at support: = 150
Distance between stirrups at span: = 250
Check the conditions (according to TCVN 9386-2012 and TCVN 5574-2018):
TYPICAL FLOOR DESIGN
General concept
Floor is a structure that directly supports internal loads in a direction perpendicular to the floor surface In multi-storey buildings, the floor is responsible for the horizontal rigid wall to transmit horizontal loads (wind, earthquake, etc.) to the frame structure.
The floor is also subject to internal forces due to uneven settlement, internal forces arising from temperature changes
The floor is absolutely rigid in its plane (horizontal plane) Excluding curvilinear strain (out of the floor plane) on the elements Ignore the impact of the stiffness of this floor on the floors next to it
Computational hypothesis
The floor is absolutely rigid in its plane (horizontal plane) Excluding curvilinear strain (out of the floor plane) on the elements Ignore the impact of the stiffness of this floor on the adjacent floors.
All load bearing system components on each floor have the same horizontal displacement.
When a horizontal load is applied, this load will be transmitted to the building in the form of forces distributed on floors and floors, transferring these forces to columns and walls.
The axial deformation of the floor, beam is considered negligible.
Dead loads acting on the floor
Dead loads acting on the floor are shown in Chapter 3, Section 3.2.
Live load acting on the floor
Live load acting on the floor are shown in Chapter 3, Section 3.3.
Set up calculation scheme
The 3rd floor model is exported from ETABS to SAFE including the force on the floor.
Figure 5.1 3D model of the 3rd floor floor in SAFE
Figure 5.2 Floor plan of the 3 rd floor in SAFE
Load cases are exported from ETABS:
Figure 5.3 Finish load acting on 3 rd floor
Figure 5.4 Live load (LL2) acting on floor
Figure 5.6 Wall load acting on floor
Figure 5.8 Draw trip sequences in the y direction with the width of 1m to get the internal force
Figure 5.9 Draw trip sequences in the x direction with the width of 1m to get the internal force
Floor test by limit state I
5.6.1 Sequence of calculation of reinforcement for floor:
The calculation is done as follows:
The selected floor reinforcement content must satisfy the following requirements: (according to section 10.3.3.1 TCVN 5574:2018)
5.6.2 Calculation and arregement rebar for floor:
From the long strips of the floor Take internal force and cross-section, calculate steel as a beam Calculate steel for floor 3 with moment value at span M 1 =7.139 kN/m
Table 5.1 Spreadsheet of floor steel in X direction
Slab Section M Steel kN/m Φ (mm) a As (mm ) 2 %
Table 5.2 Spreadsheet of floor steel in Y direction
Slab Section M Steel kN/m Φ (mm) a As (mm ) 2 %
FRAME DESIGN
Shape and size
Tan Tao apartment according to architectural drawings include:
The structural system used is the frame structure system - hard wall (hard core).
Section size
Dimensions of space frame members have been preliminarily selected in Chapter 2
Evaluation of frame internal force results
Figure 6.1 Name of frame element
Figure 6.2 Force chart along the column, the wall (kN)
Figure 6.3 Diagram of the moment of beams, columns and frames (kN/m)
Figure 6.4 Diagram of shearing force of beams and columns (kN)
Figure 6.5 Wall shear force diagram frame B
Figure 6.6 Diagram of wall moment axis frame (kN/m)
Figure 6.7 Diagram of wall axial force
6.3.2 Displacement of the top of the building:
According to TCVN 5574:2018 table M.4 Appendix M, the ratio between the horizontal displacement of the top of the building D under the effect of wind load and the height up to the top of the work from the foundation surface H needs to be satisfied.
H: is the height of the building from the top of the foundation.
: is the horizontal displacement at the top of the structure.
The reactions in the two modes of vibration are considered to be independent of each other.
Check 4 combination SLS2, SLS3, SLS4 and SLS5:
Story Diagram Combo Load UX UY mm mm mm mm
→ The displacement at thep top is sastified
6.3.3 Check the relative horizontal displacement between floors:
Relative horizontal displacement between floors due to wind:
According to Appendix M.4.4 TCVN5574:2018, table M4: “Limited displacement and deflection in the horizontal direction of the building, individual members and conveyor supports due to wind loads, foundation inclination and the effects of temperature and climate ”
Relative horizontal displacement of a floor in a multi-storey house with “walls, partitions made of bricks, gypsum concrete, reinforced concrete panels” with limited displacement:
Where: ℎ is the height of the floors in a multi-storey building: For lower floors – equal to the distance from the foundation surface to the axis of the roof floor beams: For the remaining floors, equal to the distance between the axes of the beams of each floor, multi-storey building taken as the distance from the top of the foundation to the axis of the roof support beams
Table 6.2 Horizontal limit displacement f u according to structural requirements
Houses, walls and partitions Connection between Limit wall and frame displacement
2 One floor of a multi-storey house: Soft ℎ /300 a Walls and partitions made of bricks, gypsum concrete, Hard ℎ /500 reinforced concrete panels b Walls (natural stone cladding)
Hard ℎ /700 made of bricks ceramic
(with self-loading walls) = 15 Soft ℎ /200 floor height, m ≥ 30 ℎ /300
Take SLS2 (x-axis), SLS4 (y-axis) Combo to check condition:
Table 6.3 Combination table for checking floor displacement due to wind X, Y
Story Height displacement displacement Check by GX by GY / mm mm mm mm
Relative horizontal displacement between floors due to earthquake
According to section 4.4.3.2 TCVN 9386:2012, Limit the relative horizontal displacement between floors
For buildings with non-structural members of brittle material attached to the structure:
For buildings with non-structural parts of flexible materials:
For buildings with non-structural parts fixed so as not to affect structural deformation or buildings without non-structural parts
Where: dr: relative design horizontal displacement between floors, determined as the difference of mean horizontal displacements “ds” at the ceiling and floor of the floors under consideration.
= = × ( 4.3.4.1 9386: 2012) ds: displacement of a point of the structural system caused by the design seismic action
= 3.9: displacement behavior coefficient, assumed to be “q” unless otherwise specified dc: displacement of the same point of the structural system is determined by linear analysis based on the design response spectrum.
: The reduction factor takes into account the lower periodicity of the seismic action relative to the damage limitation requirement, depending on the seismic hazards and the significance of the structure (Appendix E).
Note to section 4.4.3.2 TCVN 9386:2012 clearly states: Different values of v depending on seismic hazards and building importance are recommended as follows: 0.4 for importance levels I and II and = 0.5 for importance levels III and IV h: height of story
≤ × =0.0032×ℎ dc: is the dislocation displacement from the Etab model
Table 6.4 Combination table checking floor displacement due to earthquake X, Y
Story Height displacement displacement Check by DDX by DDY h mm mm mm mm
In order for the structure not to overturn when subjected to earthquakes, the following conditions must be satisfied: (according to section 2.6.3 TCVN 198:1997)
: is moment resist overturn : is moment overturn
Note: according to section 3.2 TCVN 198 - 1997: High-rise buildings - Design of reinforced concrete structures of the whole block: reinforced concrete high-rise buildings with a height-to-width ratio greater than 5 must be tested for anti-rollover under impact earthquake dynamics and wind loads When calculating the anti-rolling moment, the live load of the floors is taken into account by 50%, and the final load is taken at 90%.
GEOGRAPHY OF CONSTRUCTIONS
Introduction to engineering geology
The land area is expected to build works at the corner of Ton Duc Thang and Le Thanh Tong streets, District 1, Ho Chi Minh City with an area of about 900m².
The survey area is an area with relatively flat topography In general, the difference in terrain elevation in the proposed construction area is almost negligible There are solid construction works around the construction area.
Students selected borehole BH3 of the geological survey results of Tan Tao Apartment Building (District 1, Ho Chi Minh City) to conduct calculations for the foundation design for this project Because the HK3 borehole passes through many layers and has a length of sandy soil layer.
The top part is 4 layers of soil distributed from the current ground to about 35m depth, as follows: o Layer 1: Filled soil (fs) is distributed from the current ground to a depth of
1.4m/1.8m. o Layer 2: Very plastic clay mixed with organic fine sand (OH) in flowing plastic state, distributed from 1.4m/1.8m to a depth of 4.7m/5.3m. o Layer 3: Fine sand, poorly mixed powder (SP-SM), medium density distributed from depth 6.5m/7.5m to 11.2m/13.2m and thin lenses 3a (CL), 3b (SC) ). o Layer 4: Fine sand mixed with clay, powder (SC-SM), medium density, distributed from 9.0m/13.2m to a depth of 33.5/36.5m and thin lenses 4a (CH),4b (CL)
Comment:Except for layers 1 and 2 which have poor and unstable composition with thin thickness, the main components of this section are fine-grained sand mixed with clay and powder The fine sandy soil layer has a medium density with medium favorable geotechnical conditions for foundation design and construction.
Soil layer 2 is a soft soil that needs attention in the design and construction of the foundation In addition, it is necessary to pay attention to the non-cohesive quality of the fine-grained sandy soil layers distributed near the ground, during the excavation and construction of the basement.
The second part is three layers of hard clay, distributed from a depth of 35m to 50m, as follows: o Layer 5: Flexible clay mixed with fine sand (CL), semi-hard to hard state, distributed from depth 33.5m/36.5m to 39.0m/41.0m. o Layer 6: Very plastic clay (CH), hard state, distributed from depth to 45m.0m. o Layer 7: Plastic clay mixed with fine-grained sand (CL), hard state, distributed from 45m to 48.7m/51.4m depth.
Comment: The soil layers of this part have a stable distribution depth and thickness with clay composition, semi-hard to hard powder state The thickness of this section varies from 12m to 16m This section has good geological conditions for construction, which is convenient for the design and construction of deep pile foundations.
The last part is the dense to very dense sandy soil layers, distributed from a depth of about 50m to the end drilling depth (100m), including the following soil layers: o Layer 8: Fine sand mixed with fine clay (SC-SM) medium density, distributed from depth 49.3m/51.4m to 57.0m/61.3m and lens 8a (SC). o Layer 9: Fine-grained sand mixed with clay and clay (SC-SM) is very dense, distributed from a depth of 57.0m/61.3m to 89.0m. o Layer 10: Fine sand mixed with clay, powder (SC-SM) with very tight density, distributed from a depth of 89.0m to more than 100m.
Comment: The soil layers of this part have a stable distribution of thickness and depth, with the main component being fine-grained sand with very dense density This is the soil with very good engineering geological conditions, which is convenient for the design and construction of deep pile foundations.
Groundwater depth was recorded after the completion of drilling 24 h and 1 week thereafter The results are presented in the following table:
The existing underground water level is about 3-4m
Theoretical basis of geological data statistics
7.2.1 Processing and statistics to calculate the foundation:
Geological survey records for foundation design have a large number of boreholes and a large number of soil samples in a large soil layer The problem is that for these soil layers, we have to choose the criteria that represent the foundation Initially, when drilling, sampling was based on the observation of color change, the grain that we divided into each layer of soil.
According to TCVN 9362 - 2012, it is called a engineering geology class when the set of values with its mechanical - physical characteristics must have a sufficiently small coefficient of variation Therefore, we must exclude samples with data that differ from the average value for a geological unit.
Statistics is a very important job in the calculation of the foundation, requiring high accuracy
7.2.2 Determine the standard value, the calculated value of the physical and mechanical properties of the soil:
Division of geological units (engineering geological classes)
Where: σ - the mean square error of the feature
A i - eigenvalue a characteristic of sample i from a particular experiment n - the number of experiments is determined, where n is the number of samples in the same class
In the sample set of a soil layer, if there is a coefficient of variation ≤ [ ], the condition of a geological unit is satisfied, and if > [ ], we must exclude data with large errors Where [ ] is the allowable coefficient of variation, depending on each type of characteristic
Table 7.2 Coeficient of variation Characteristic of soil Coefficient of variation
Raw error exclusion: exclude the Ai error from the set when:
1 v - Statistical criteria determined depends on the number of experimental samples, look up from the following table:
If there is an error, remove the Ai error and recalculate the above values.
7.2.3 Determination of standard characteristics and calculated values:
Determine the standard values of and by the method of least squares of direct shear tests for all experimental values of τ in a engineering geology unit.
The relationship between τ and σ is determined by the Coulomb equation (soil strength condition):
Use the least squares method to find c and φ:
Taking the partial derivative of both sides with respect to c and tgφ, respectively, we get:
From here we deduce the equation of two unknowns c and tgφ:
Solving the equation we get:
Where: − is the accuracy index, calculated by the following formula:
Where: − coefficient depends on the confidence probability α
When calculating the ground by intensity (TTGH1), then α = 0.95
When calculating the ground according to deformation (TTGH2), then α = 0.85
(n – 1) with R, γ The coefficient corresponds to the confidence probability α (n – 2) with c, φ α = 0.85 α = 0.95
When calculating statistics, the number of samples n ≥ 6 is the limit state statistics If n < 6 then we conduct statistical test ν < [ν] and take the standard value equal to the mean value (density γ, humidity W ) With the adhesive force c and the internal friction angle υ from the fast undrained shear test, the number of test samples is 1 (corresponding to 3 pairs (σ, τ): n=3), then only the standard value is calculated The number of test samples is 1 (corresponding to 3 pairs of numbers (σ, τ): n=3), only standard values are calculated, the number of test samples is greater than or equal to 2 (corresponding to 6 pairs (σ, τ): n
≥ 6), then perform statistics according to the limit state.
When looking up the table t α note n-1, n-2 Using the LINEST function in EXCEL to support statistics of the stick force c and the friction angle in φ
When making statistics for the criteria c, φ, we must initially check the statistics for each pressure level to know if there is any type of sample or not.
BORED PILE FOUNDATION PLAN DESIGN
Introduce about bored piled foundation
Bored piles are piles constructed by the method of drilling holes first in the soil, then the holes are filled with concrete Hole making is done by drilling, closing or by other excavation methods The current common diameter of stuffed piles is 600, 800, 1000,
When designing and constructing, it is necessary to have a thorough understanding of the ground conditions as well as the characteristics of construction technology to ensure the quality regulations of the piles.
8.1.1 Advantages : o During construction, there is no impact on the surrounding environment o The bearing capacity of the pile is very large. o The amount of steel in bored piles is small, mainly to support horizontal loads o
Capable of constructing piles when passing through interspersed hard soil layers
8.1.2 Disadvantages: o High cost due to complicated construction techniques, although the design of reinforcement in piles is very economical. o The method to check the quality of bored pile concrete is very complicated by ultrasonic method or pile static load test. o The side friction of the pile body can be significantly reduced compared to the driven pile and the pressed pile due to the hole drilling technology.
Concrete used for bored piles is ordinary concrete In addition to the strength condition, the concrete must have a large slump to ensure the continuity of the pile As a rule, the concrete grade used for bored piles is generally not lower than 20MPa and the slump is shown in Table 8.1.3
Table 8.1 Punching concrete bored piles
Pouring freely in water, the reinforcement has a large gap that allows the 7.5-12.5 concrete to move easily.
The reinforcement spacing is not large enough to allow the concrete to move easily, when the pile head is in the temporary bulkhead area 10-17.5
When the pile diameter is 15 Normally, the concrete of bored piles has a cement content of not less than 350kg/m3 To avoid segregation due to high slump concrete or dehydration in high temperature conditions, appropriate admixtures should be used.
The reinforcement of bored piles is determined according to the calculation, and at the same time, the structural requirements must be satisfied:
In the case of piles in tension, the longitudinal cotost should be arranged throughout the length of the pile When longitudinal reinforcement is said it is necessary to weld according to the bearing requirement When the pulling force is small, the longitudinal reinforcement is arranged to the necessary depth so that the pulling force is completely eliminated through pile friction.
For axially compressed piles, the reinforcement content is not less than 0.2-0.4%.
The diameter of reinforcement is not less than 10mm and is arranged evenly around the pile circumference.
For piles subjected to horizontal loads, the reinforcement content is not less than 0.4- 0.65% Bolts are usually Φ6-Φ10, spacing 200-300mm A single ring belt or a discontinuous spiral belt can be used If the length of the steel cage is greater than 4m, to increase the stiffness of the whole block, add steel belt Φ12 2m apart, and these belts are used to attach the rollers to create a layer of reinforcement protection.
The thickness of the longitudinal reinforcement protection layer of bored piles is not less than 50mm
Usually bored piles are made holes from the ground level, the soil in the piles is removed The phenomenon of soil stretching during construction will cause tensile stress to the pile and it persists until the pile is loaded enough to bear the above-mentioned tensile force until this force is suppressed due to the transmission of the work load.
For tall buildings, the horizontal force is small, so usually the steel cage is staggered or completely cut off the earthquake (based on the moment diagram and the shear force of the pile in the problem of solving piles subjected to horizontal loads).
8.1.5 Contents of foundation and pile foundation design:
The foundation and rest pile foundations are calculated according to the limit state. Limit state I (TTGH I) (intensity): o Bearing capacity of piles according to soil condition o Strength of materials o Stability of piles and foundations
Limit state II (TTGH II) (deformation): o Pile foundation settlement o Horizontal displacement of piles and pile foundations
Structure and size
Student choose height for all cap: 2.2m
Calculated tensile strength 14.5 × 10 3 / 2 Calculated compressive strength 1.05 × 10 3 / 2
Choose height of pile cap: H d = 2.2 m
Elevation of the bottom of the pile cap:= −7.2 − 2.2 = −9.4
The pile is attached to the cap: 0.1 m
Pile head smashing section: 0.7 m (the reinforcement section anchored to the cap approximately 40d)
The total section of the pile head and the pile attached to the cap: 0.7 + 0.1 = 0.8 m Pile length from the bottom of the cap: = 53 − 9.4 = 43.6
Depth of pile tip: -53m The tip of the pile is forbidden to enter the 8 th layer of soil, which is a layer of fine sand mixed with clay-powder, in a tight state, with an SPT value of 32~47m.
Choose protective concrete layer: 30mm
The content of reinforcement in the pile is taken according to the structure (0.4% ÷ 0.65%) but must not be less than the minimum content due to the pile when subjected to horizontal loads (0.2% ÷ 0.4%) Choose 14 20 ( = 4398 2 ), reinforcement content
Figure 8.1 Detail of bored pile
Figure 8.2 Geological cross-section and piling depth -53m
Calculating the bearing capacity of a single pile
8.3.1 Piles bearing capacity by material:
The bearing capacity of pile according to material is determines as follow:
= 0.85: is the working condition coefficient of concrete
′ = 0.7: is the effect coefficient of pouring concrete in the narrow space of the pit and the wall pipe when pouring concrete into the borehole cage under the drilling fluid.
= 0.0044 2 : is cross-sectional area of longitudinal reinforcement
= ×1 4 2 − 0.0044 = 0.781 2 : is cross-section area of concrete
: coefficient of reduction of bearing capacity due to longitudinal bending effect according to TCVN 5574:2012 for concrete and reinforced concrete structures according to design standards
Because the pile is plugged into the mud layer, it is necessary to consider the longitudinal bending coefficient calculated as follows:
= 3.25 × 10 7 : Elastic modulus of the pile material
Moment of inertia of horizontal piles:
: is the conventional width of the pile in (m), for piles with a minimum pile diameter of 1m take = + 1 = 1 + 1 = 2
Radius of inertia of pile cross section:
When calculating according to material strength, it is permissible to view the pile as a hard fix in the soil at a cross section located about l1 from the base of the station determined by the formula
: Deformation coefficient determined according to the instructions in Appendix A
0 = 0.2: the length of the pile from the bottom of the cap to the leveling height
( / 4 ): Proportional coefficient, taken depending on the type of soil surrounding the pile according to Table A.1, TCVN 10304: 2014
Figure 8.3 Coefficient K according to table A1 TCVN 10304:2014
Table 8.2 Coefficient K in each soil layer around the pile
4 Poorly graded fine sand with silt (SP-SM),
0.43 12000 27.5 330000 yellow-brown, brown-gray, medium dense.
5 Lean clay with fine sand (CL), yellow-gray,
0.27 14280 4.5 64260 green-gray, very stiff to hard.
6 Fat clay (CH), yellow-gray, red-brown, 0.31 14760 4 59040 green-gray, hard.
7 Fine sandy lean clay (CL), yellow-gray,
8A Very clayey fine sand (SC), yellow-gray,
-0.07 12000 2.3 27600 green-gray, medium dense to dense
8 Silty, clayey fine sand (SC-SM), yellow-
8.3.2 Piles bearing capacity according to ground criteria:
Pile bearing capacity according to soil and mechanical indicators:
For bored piles, the bearing capacity according to the physico-mechanical properties of the ground soil is determined according to Section 7.2.3 TCVN 10304 – 2014
= 1: is the working condition coefficient of the pile
: is the coefficient of working condition of the soil under the pile tip, = 0.9 for the case of using the method of pouring concrete under water, = 1 for other cases
: is the cross-sectional area of the pile tip
: is the circumference of the cross-section of the pile body
: is the coefficient of working conditions of the soil on the pile body, depending on the method of creating holes and concrete pouring conditions, look up table 5, page
: is the average resistance strength of the soil layer “i” on the pile body, look up table 3, page 25, TCVN 10304-2014
: is the length of the pile in the "i" soil layer.
: is the resistance strength of the soil under the pile tip, determined as follows:
1 , 2 , 3 , 4 : are dimensionless coefficients that depend on the value of friction angle in the calculation 1 of the ground and are taken according to Table 6 (page 30, TCVN 10304-2014)
′ 1 : is the calculated density of the soil under the pile tip (taking into account the buoyancy effect in the water-saturated soil).
1 : is the average calculated density (in layers) of the soil above the pile tip (taking into account buoyancy in water-saturated soils)
: is the diameter of closed or pressed piles, bored piles and pipe piles, the diameter of the extension (for piles with fore extension) or the borehole diameter for piles - pillars, bonded to the soil by cement-sand
ℎ: is the pile lowering depth, from the natural ground or the design ground (when there is an excavation design) to the tip of the pile or to the bottom of the fore extension.
The tip of the pile plugged into the 8 th layer of soil is tight sand:
Table 8.3 Resistance coefficient of loose soil under the pile tip (M1)
Layer 4 Layer 5 Layer 6 Layer 7 Layer 8a
Layer 4 Layer 5 Layer 6 Layer 7 Layer 8a
Table 8.4 Soil resistance according to physico-mechanical criteria
Type of soil Layer Depth Average depth % m m m / / kN/m
Poorly graded fine sand with silt 19 21 20 2 0.43 0.8 30 30.0 48
(SP-SM), yellow-brown, brown-
Lean clay with fine sand (CL), 36.5 38.5 37.5 2 0.08 1 103.5 103.5 207 yellow-gray, green-gray, very stiff 5 38.5 40.5 39.5 2 0.08 1 106.3 106.3 212.6 to hard 40.5 41 40.75 0.5 0.08 1 108.05 108.1 54.025
Fat clay (CH), yellow-gray, red-
Fine sandy lean clay (CL), yellow-
Very clayey fine sand (SC),
156.754 250.8064 yellow-gray, green-gray, medium 8A
Silty, clayey fine sand (SC-SM),
Table 8.5 Summary table of calculation parameters
Parameters in the formula Value Unit
→Load bearing capacity according to the physical-mechanical norms of the ground:
= 1 × (1 × 1309.474 × 0.781 + 3.14 × 2459.369) = 8745.119 Extreme bearing capacity of piles according to ground strength
According to Appendix G.2 TCVN 10304-2014, the ultimate bearing capacity of piles according to the criterion of ground strength is as follows:
: is the cross-sectional area of the pile tip
: is the perimeter of the pile cross-section
: is the length of the pile in the "i" soil layer
: is the shear strength (due to unit friction) of the “i” soil layer on the pile body
: is the resistance strength of the soil under the pile tip determined by the formula
′ , ′ : are the bearing capacity coefficients of the soil under the pile tip (looked up in Appendix B TCXD 205: 1998 "Figure B3").
According to figure 8.3.2.a (Figure B3, TCXD 205:1998), we have formula:
− 3 0 where 1 ′ is internal friction angle of the soil beforce lowering the pile.
′ , : is the effective mantle pressure at the pile tip level (with a value equal to the effective vertical soil stress at the pile tip level).
For adhesive soils the strength of non-drained shear resistance under the tip of the pile:
′ = 6: for large diameter bored piles
For loose soil (c = 0), the shear strength below the pile tip:
If the pile tip depth is less than , then ′ , is taken as the value of the mantle pressure at the pile tip depth If the pile tip depth is greater than , take the value of ′ , as the mantle pressure at depth Determine the values of and coefficients k and ′ , in Table G.1, TCVN 10304-2014:
Soil state / Closed Bored pile Closed Bored pile tightness pile and Barrette pile and Barrette
Shear resistance due to friction on pile body
For cohesive soils, the undrained shear strength on the pile body in the ith layer is determined by method
, : Undrained resistance strength of the “i” soil layer
: The coefficient depends on the characteristics of the soil layer above the cohesive soil layer, the type of pile and the method of pile lowering, the consolidation of the soil during construction and the method of determining c u (look-up graph G1 "Appendix G" ).
For loose soil (c = 0) the average shear strength on the pile body in the “i” th layer determined by method :
: The coefficient of horizontal pressure of the soil on the pile body
: Mean vertical effective stress in the “i” th layer
: is the friction angle between the soil and the pile, normally the concrete pile is taken equal to the internal friction angle of the soil , for steel piles taken as 2 ⁄3
According to the above formula, the deeper the depth, the shear strength on the pile body increases, however, it only increases to the limit depth Z L about 15~20 times the pile diameter d and then does not increase anymore Therefore, the shear resistance on the pile body in loose soil is calculated as follows:
On piles with depth less than Z L :
On piles with depth equal to and greater than Z L :
Strength of undrained shear strength below pile tip:
Table 8.7 Table of coefficients for calculating shear resistance under pile tip
Table 8.8 Table of C u parameters from the UU experiment in the summary table
Table 8.9 Calculation table of hip resistance of cohesive soil layer.
Table 8.10 Table of hip resistance of loose soil layers.
→ Total hip resistance around the pile:
→ Bearing capacity according to ground strength criteria:
Pile bearing capacity according to standard penetration test SPT
Formula of the Japanese Institute of Architecture (1988):
, : Length of pile in loose soil layer “i”
, : The length of the pile in the cohesive soil layer "i"
: The resistance strength of the soil under the pile tip is determined as follows:
When the pile tip is in loose soil, = 300 for driven piles and = 150 for bored piles.
When the pile tip is in the cohesive soil = 9 for driven piles and = 6 for bored piles.
: Average SPT is in the range of 1d below and 4d above the pile tip.
, : Average resistance strength on piles in loose soil layer “i”:
, : Resistance strength on the pile segment located in the “i”th cohesive soil layer:
, : The undrained shear strength of cohesive soil can be determined from the SPT index in cohesive soil , = 6.25 ,
, : Average SPT index in loose soil layer “i”.
: The correction factor for the driven pile depends on the ratio of undrained shear strength and the mean value of the effective normal vertical stress, determined based on the figure below:
= 1: Coefficient of adjustment according to thinness h/d of driven piles, bored piles Specific calculation
The tip of the pile is forbidden to enter the soil layer 7 so we have the resistance calculated as follows:
With is SPT index in the cohesive soil layer under the pile tip
TCVN 10304 - 2014 stipulates that when calculating the bearing capacity of piles according to the Japanese formula, is the undrained shear strength of cohesive soil that can be determined from the SPT index.
Table 8.11 Resistance strength on the pile segment in the cohesive soil layer “i”
Table 8.12 Resistance strength on the pile segment in the loose soil layer “i”
→ The ultimate bearing capacity of piles according to the Japanese formula:
Computational bearing capacity of compressive pile
, : calculated value of the compressive load acting on the pile (the value of the load transmitted to the pile when the pile is working)
0 : coefficient of working condition, taking into account the factor of increasing the homogeneity of the ground when using pile foundation, taken as 1 for single pile and taken as 1.15 in multi-pile foundation
: reliability coefficient on the importance of the works, taken as 1.2, 1.15, 1.1 corresponding to the importance of the works of grade I, II, III (Appendix E)
, : calculated value of the compressive bearing capacity of the pile
: The confidence coefficient is taken as follows:
In case of piles subjected to compressive load in the foundation of high or low calyx with the base of the base lying on a highly deformed soil layer, the value of is taken depending on the number of piles in the foundation.
= 1.4: Foundation with at least 21 piles
, : The standard value of the pile beaing capacity is determined from the eigenvalues of the ultimate compressive bearing capacity (minimum value)
Table 8.13 Summary table of extreme bearing capacity
Calculated bearing capacity Extreme bearing capacity
According to the intensity criterion 9747.448528
According to Japanese SPT experiment 13292.04554
Table 8.14 Summary table of design bearing capacity of piles in pile group
Figure 8.6 The layout of bored piles
Foundation design F1
The foundation is located at the boundary position, subject to the reaction from the foot at the intersection between the 2 axes 6 and B
Table 8.15 F1 foundation calculation load combination
− − − − kN kN/m kN/m kN kN
8.4.3 Calculate and arrange the number of piles in the work:
Calculate the number of piles in the cap
Design foundation with bored piles with diameter = 1 with pile length = 43.6
Maximum calculated compressive force value: = 10943.10
Number of pile in the cap:
Arrangement of piles in the cap
Distance between two piles from 3 ÷ 6
Distance from the center of the pile to the edge of the platform is 1
With the number of piles as above, students choose the distance of the piles, arrange the piles from which to calculate the size of the cap.
Table 8.16 Parameter table of pile layout
Distance from center of piles 3 m
Distance from the center of the pile to the edge of the platform 1 m
The elevation of the bottom of the tower 9.4 m
Thickness of lining concrete layer 0.1 m
Arrange piles and cap foundations as shown:
Figure 8.7 Pile and foundation layout plan M1
Length of foundation cap in X direction: = 5
Length of foundation cap in Y direction: = 5
Use ETABS 2018 software to calculate the force acting on the pile head The base station is declared as a Slab element with a thickness of 2.2 m The piles under the foundation are declared as Column Supports with the form of Spring Constants at the pile ends with the spring stiffness calculated as follows:
: the single pile settlement corresponding to the design bearing capacity (which is the elastic settlement including the axial deformation when subjected to the longitudinal force of the pile, the elastic settlement of the ground around the pile and the elastic settlement of the pile tip).
According to Appendix B - TCVN 10304:2014: experience shows that the settlement of single pile depends on the size of the load and the diameter of the pile When the foundation has been safely designed according to the bearing capacity, the settlement of piles in sandy soil is usually small In this case the settlement of single pile can be calculated empirically according to the expression of Vesic (1977):
= = 4997.211 : load acting on the pile
: Cross-sectional area of the pile
= 3.25 × 10 7 : Elastic modulus of the pile material
Figure 8.8 Max pile head reaction (ULS8)
Figure 8.9 Min pile head reaction (ULS13)
Comment: The values of the pile head reaction are always smaller than the calculated allowable bearing capacity, and the pile does not bear small, the pile ensures the bearing conditions.
Check the bearing capacity of pile group
Bearing capacity of pile group:
Self-weight of foundation cap:
→ The bearing capacity of pile group is sastified
8.4.4 Stress test under conventional foundation block:
Define conventional block foundation according to Section 7.4.4 and Appendix C TCVN 10304-2014.
Length of pile in soil: = 43.6
Table 8.17 Table for determination of average friction angle
Conventional block foundation width in X direction:
Conventional block foundation length in Y direction:
Conventional foundation block bending moment:
Table 8.18 Calculation of soil volume in conventional foundation blocks
3b-cap Lean clay with fine sand (CL), red-brown, green-gray,
4 Poorly graded fine sand with silt (SP-SM), yellow-
27.5 10 275 brown, brown-gray, medium dense.
5 Lean clay with fine sand (CL), yellow-gray, green-
4.5 10.4 46.8 gray, very stiff to hard.
Fat clay (CH), yellow-gray, red-brown, green-gray,
Fine sandy lean clay (CL), yellow-gray, green-gray,
Very clayey fine sand (SC), yellow-gray, green-gray,
8 Silty, clayey fine sand (SC-SM), yellow-gray, grayish,
Total mass in the conventional foundation block:
Table 8.19 Table of loads to the bottom of the conventional foundation
Stress at the bottom of conventional block foundation:
Table 8.20 Stress table at the bottom of the conventional foundation block
Bearing capacity of the ground:
1 = 1.3: Coefficient of working conditions of the ground
2 = 1.3: Coefficient of working conditions of the work interacting with the ground
: Reliability coefficient, = 1 when the computed features are taken directly from the experiments
Table 8.21 Average stress of pile foundation, ′
Total from layer 4 to layer 8 488.72
At the tip of the pile in class 8 we have: (Check table 14 TCVN 9362-2012 for parameters A, B, D)
Table 8.22 Coefficients calculating the bearing capacity of the foundation (ULS II)
Bearing capacity of the ground:
8.4.5 Checking for settlement of conventional foundation blocks:
The average stress at the base of the conventional foundation:
Stress due to self-weight at the foundation bottom:
Void ration e, according to applied pressure kG/cm 2
8.4.6 Checking for shear condition for cap:
Take the maximum shear force in the strips with a width of 1 m in the x direction and in the y direction to check the shear strength of the foundation concrete.
According to Section 8.1.3.3 TCVN 5574:2018, for reinforced concrete members without reinforcement of the shear force, in order to ensure the durability on oblique cracks, it is necessary to calculate for the most dangerous oblique cracks according to the following conditions:
, 1 , 2 : are the existing shear resistance, the shear resistance, respectively maximum and maximum of the non-reinforced cross-section;
C is the vertical projection length of the most dangerous oblique crack;
2 = 1.5: is the coefficient taking into account the influence of longitudinal reinforcement, cohesion force and stress state characteristics of the compressive concrete lying on the oblique crack.
= 0.04: Thickness of protective concrete layer.
Shear strength of concrete on a strip of width b = 1m
Figure 8.11 MAX shear force diagram in the X direction
Satisfying shear conditions, the foundation does not need to arrange the reinforcement
Figure 8.12 MAX shear force diagram in the Y direction
Satisfying shear conditions, the foundation does not need to arrange the reinforcement 153
Based on the book "Kết cấu bê tông cốt thép 2" (Võ Bá Tầm), we have the following formulas and conditions:
For square foundation, we have:
Average perforation circumference of an obelisk:
= 0.85: coefficient of working condition of concrete.
= 9685.49 : Maximum axial force of column C22
Penetration test due to the reaction from the pile to the cap:
Step 1: Determine the penetration resistance zone.
Figure 8.13 An obelisk penetration of foundation F1 The penetration resistance zone: 0.5ℎ 0 = 0.5 × 2.16 = 1.08
Step 2: Determination of penetration resistance force:
Any group of piles located in the penetration resistance zone will not cause penetration, only those piles located outside the penetration zone will be counted.
There are some other cases like this F1 foundation, the penetration resistance zone is located in the center of the 4 piles, so the penetration force is still different from 0.
8.4.8 Calculation of reinforcement for the foundation cap:
The internal force calculation for reinforcement is done with the help of ETABS 2018 software Calculation of reinforcement for the envelope combination.
Figure 8.14 MAX moment diagram of foundation base F1 in X direction
Figure 8.15 MIN moment diagram of foundation base F1 in X direction
Table 8.25 Steel calculation in the X direction of foundation base F1
X- M b Choose direction kN/m mm mm 2 steel 2 (%)
Figure 8.16 MAX moment diagram of foundation base F1 in Y direction
Figure 8.17 MIN moment diagram of foundation base F1 in Y direction
Table 8.26 Steel calculation in the Y direction of foundation base F1
(%) direction kN/m mm mm 2 steel 2
8.4.9 Calculation of piles with horizontal load:
We check for the case of the maximum total shear force at the base of the cap Due to the small internal force, take the assigned computational load instead of the standard load.
Table 8.27 Table of the maximum shear force at the base of F1
Case − − − − kN/m kN/m kN kN
Model pile into bar element, surrounding ground is converted into springs with stiffness
The horizontal ground coefficient of a soil layer according to TCVN 10304:2014 is:
: The scale factor is taken depending on the type of soil surrounding the piles according to table A1 (TCVN 10304:2014).
: Depth of the pile cross-section in the soil, from the ground in the case of high pile foundations, or from the bottom of the cassock in the case of low pile foundations.
Table 8.28 Table for determining spring stiffness (F1) Point Depth
Figure 8.18 Horizontal displacement of pile head (F1)
Figure 8.19 Horizontal pressure chart from Sap2000 (F1)
Figure 8.20 Shear force chart from Sap2000 (F1)
Figure 8.21 Moment chart from Sap2000 (F1)
Horizontal displacement of Swivel angle Stress Moment Max shear pile head (m) (Rad) / 2 Max (kN/m) force (kN)
8.4.10 Checking pile horizontal displacement and ground stability around the pile:
Using the calculation results of piles bearing horizontal load from SAP2000 model with the foundation coefficient changing according to the depth of each soil layer for testing.
Because the deeper you go, the spring stiffness increases, so the moment, shear force, and spring reaction are very small Therefore, I only took 1/3 the upper of the pile to put in the explanation
Maximum stress in the ground at depth Z = -1m from the bottom of the cap(maximum spring reaction)
: cylindrical area (According to TCVN 205-1998, The calculated pressure is on the soil on the side of the pile, so the value is halved)
Check the horizontal displacement and rotation angle of the pile head:
Horizontal displacement, according to TCVN 10304:2014, section 11.12: Δ = 0 =8.01×10 −5