An duong vuong apartment

Comstruction method Slab method, Solid slab is concrete slab and beam combined Frame method, cast in place reinforced concrete wall Foundation method, pored pile 1.6.2.. Choose primary s

OVERVIEW OF ARCHITECTURE

Construction introduction

In recent years, the country's trend towards integration, industrialization, and modernization has led to a significant shift in urban development, emphasizing the need for high-rise buildings to replace deteriorating low-rise residential areas This evolution in construction is exemplified by the emergence of the An Duong Vuong apartment building, which addresses these pressing urban needs.

Situated in the Bac Ha District, this project boasts a stunning and open location, enhancing the overall residential planning with a harmonious blend of modernity and thoughtful design.

Urban infrastructure

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

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

The construction site features a flat terrain free of existing structures and underground utilities, making it highly suitable for efficient construction and effective overall planning.

Architectural solution

The work plan is rectangular with gouges, length 64.5m, width 34.4m, and construction land area is 2218.8m2

The building features 19 floors, including a semi-basement, with a total height of 64.5 meters from the designated 0.00m level to the roof The 0.00m level aligns with the natural ground level, which is 1.50 meters lower than the ground floor The basement is situated at a depth of -1.50m.

The basement features a centrally located elevator surrounded by parking spaces Technical systems, including domestic water storage tanks, pumping stations, and wastewater treatment stations, are strategically arranged to minimize pipeline length Additionally, high voltage and low voltage stations, along with a fan room, are incorporated into the design for efficient operation.

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

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

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

A straightforward ground solution that utilizes lightweight materials for partitions allows for the flexible organization of apartment spaces, aligning with current trends and preferences while enabling easy future modifications.

The design features a striking silhouette that ascends from traditional architecture, blending modernity with a powerful yet gentle aesthetic This harmonious balance reflects the project's scale and significance in alignment with the country's development strategy.

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

Horizontal traffic in each unit is a corridor system

The vertical transport system in the building includes stairs and elevators, featuring a primary commuter staircase and an emergency exit The elevator system comprises two main lifts and a larger medical and service lift, strategically located in the center of the structure This design ensures that apartments are arranged around the core, optimizing travel distance and providing convenience, efficiency, and excellent ventilation.

Technical solution

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

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

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

1.4.2 Water supply and sewerage system

Water is sourced from the regional supply system and directed to a basement water tank An automatic pumping system then distributes the water to each room via the main plumbing system adjacent to the service room.

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

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

Fireproof boxes are arranged along the corridor with CO2 cylinders

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

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

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

Each floor has designated garbage chutes that lead to a waste removal area in the basement, ensuring efficient disposal These chutes are designed with discretion to prevent odors and minimize environmental pollution.

Climate characteristics of the construction area

Lao Cai experiences a tropical monsoon climate, characterized by variable weather conditions due to its inland location and complex terrain This region often sees sudden temperature fluctuations, with daily highs and lows that can be extreme, such as in Sa Pa, where temperatures can drop below 0°C, occasionally resulting in snowfall.

Lao Cai experiences a distinct climate with two main seasons: the rainy season from April to October and the dry season from October to March In the highlands, temperatures range between 15°C to 20°C, with Sa Pa averaging between 14°C to 16°C, and never exceeding 20°C The region receives an average annual rainfall of 1,800mm to 2,000mm In contrast, the lowland areas have higher temperatures, averaging between 23°C to 29°C, with rainfall ranging from 1,400mm to 1,700mm.

Mist is prevalent across the province, often appearing quite dense in certain areas During severe cold spells, frost can form in the high mountains and sheltered valleys, typically persisting for 2-3 days.

Design solutions

Based on geological survey documents, architectural design documents, impact load on the works, the structure design plan is selected as follows:

Slab method, Solid slab is concrete slab and beam combined

Frame method, cast in place reinforced concrete wall

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

Software for use in analyzing and calculate

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

Calculate reinforcement and foundation: Excel

Reference Viet Nam standard

TCVN 2337-1995 Tải trọng và tác động

TCXD 229-1999 Tính toán thành phần động của tải gió

TCVN 5574-2018 Kết cấu bê tông cốt thép

TCXD 198-1997 Nhà cao tầng bê tông toàn khối

TCVN 9362-2012 Tiêu chuẩn thiết kế và nền nhà công trình

TCVN 10304-2014 Tiêu chuẩn thiết kế móng cọc

TCVN 9386-2012 Thiết kế công trình chịu tải động đất.

Structural solution

1.9.1 Choose primary section of slab

Choose floor thickness depending on span and applied load

The slab thickness is determined by the empirical formula:

Whereas: m = ( 40 ÷ 50 ) for four-sided manifest, Li = 8.2m short side length of typical floor

1.9.2 Choose primary section of beam

Table 1.1: Fomular of sectional selection

Section of Beam Type of beam

Width b Simple-span Mult-span

15 Select the span of the main beam to calculate: L = 8200mm

1.9.3 Choose primary section of column

Preliminary selection of column sizes is based on experience or approximate formula Determine the vertical transmission area

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

Count the number of floors on the section under consideration m

Take into account the impact of the horizontal load, the coefficient k, which varies depending on the column position c b

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

=   qi: distributed load on slab (hoạt tải + tĩnh tải) ni: number of floor

Si: is the transmission area of the floor to the column qi (10÷15 daN/m 2 ) Choose qi = 14 daN/m 2

Table 1.2: Primary section of middle column

Story Atr.tải Q N  A tt b × h Ac choose m 2 daN/m 2 daN cm 2 cm cm cm 2

Table 1.3: Primary section of edge column

Story Atr.tải Q N  A tt b × h Ac choose m 2 daN/m 2 daN cm 2 cm cm cm 2

1.9.4 Choose primary section of wall

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

DESIGN OF STAIRCASE

Geometry of staircase and calculation free-body diagram

Choose the staircase at axis D-E to calculate:

Table 2.1: General geometry of staircase

Height of one flight v 2 h = ht

Number of rises on each flight n

Length of flight stair (Along with diagonal axis): L= +L 1 L 2 =3.6 2.45= 6.05 (m)+

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

Choose the thickness of staircase slab: 170 mm

Figure 2.1: Layout of stair case

Choose preprimery geometry of beam: b 0 mm; h @0 mm b d

Loading on staircase

Illustrate the incline angle of stair Tan ( ) 1750 0.5 26 57 ' 0.891

 = = = →  Figure 2.2: Structure of landing Table 2.2: Structure of the flight

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

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

' tt 3.6 kN/m2 p 2.2.2 Loading of diagonal slab

Figure 2.3: Structure of riser Table 2.3: Equivalent thickness of each structure layer

Thickness of grantite layer Thickness of plater Thickness of brick layer b td h cos 2

Table 2.4: Dead load on staircase slab

Total gravity uniform load combined riser: 0.27 kN/m 7.664

Live load consist of 2 main elements:

Distributed load on 1m long along with diagonal slab: tcn tc m p =p 1 cos =  3 1 0.891=2.7 kN/m ttn tt m p =p 1 cos =3.6 1 0.891 3.24 kN/m  12

Seria l Structure Dead load g tt kN/m

Total load tt tt tt q =g +p kN/m

Analise the modeling with ETAB

The ladder works as a bending member

Each individual must develop their own calculation scheme, which will guide the construction process The nature of the connection between the ladder and the beam, or the ladder and the wall—whether it is joint (fixed or movable)—is a complex issue that relies heavily on the designer's concept.

Figure 2.7: Dead load Figure 2.6: Live load

Figure 2.5: Moment diagram Figure 2.4: Shear diagram

Calculation of staircase as same as the bending element

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

We modeling used spin connection so we distributed moment shown below: span max

Calculate reinforcement

2.4.1 Calculate reinforcement for landing and flight

With: b00 mm; h o 0 15 155 mm− The result of calculation was shown in table below:

As mm 2 μ % Choose Asbt mm 2 Support M(n) 13.808 405.05 0.26 ỉ10a150 523

2.4.2 Calculate reinforcement for the beam of the landing and flight

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

Choose preprimary section of beam b h 0 400 mm

Figure 2.9: Free body diagram of beam D1

= = Maximum negative moment of beam D1:

M = =  Figure 2.11: Moment diagram of beam D1

Reinforcement grade CB-400V → Rs = 350 Mpa

Figure 2.10: Shear diagram of beam D1

Table 2.7: Calculate of reinforcenment of beam D1

As mm 2 μ % Choose Asbt mm 2

The shear capaity of concrete:

 + + =      So must be calculate stirrup for beam

Coefficient  w 1 consider the effect of stirrup to axis

Q   R bh =       The beam is not damaged by primary compressive stress

Shear resistance of the belt: w 175 2 28.3

= =   Shear resistance of the belt and concrete:

Q =  + + R bh q There is no need to calculate the shear reinforcement

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

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

The article discusses key parameters in concrete stress analysis, including Rb and Rbt, which represent the axial compressive stress of concrete, and Rsw, indicating the tension stress of reinforcement It highlights the coefficient Φf that affects the compression wing in the cross-section T, along with the span of stirrup and the β coefficient influenced by concrete Additionally, it mentions the φn coefficient related to axial force, and specifies φb3 as 0.6 and φb2 as 2 for heavy concrete applications.

DESIGN OF ROOF WATER TANK

Architecture require

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

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

The preliminary calculation of water demand for the apartment building reveals that it consists of 16 floors, with the second floor and above designated for residential apartments Each floor accommodates 8 apartments, and with an average of 4 residents per apartment, the total number of occupants significantly influences the overall water demand.

Avarage water consumsion: q sh 0 1/person.day.night

3 max day night sh day /1000 150 416 1.35 / 1000 74.3 m /

The size of the water tank: V =LBH =9.6 4.1 2  x.7 m /day.night 3

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

Data of calculation

Water tanks are divided into three types:

With geometry of water tank a = 9.6 m; b = 4.1 m; h = 2 m → Low water tank

To minimize thickness and deflection in large roof water tanks (spanning over 7 meters), it is essential to implement a girder system for both the cover plate and the bottom plate.

To optimize the design of the cover slab, treat it as a floor plan divided into three tiles measuring 2.8 x 4.1 meters and 4 x 4.1 meters The initial selection of the tank cover thickness can be determined using the specified formula.

In which l1 is length of shorter side, l1 = 4.1 m, l2 is length of longer side l2 = 5 m bn h 1 (5000 4100) (81 101) mm

 Choose thickness of cover slab hbn = 100 mm

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

Choose thickness of water tank wall hbt = 150mm

Preliminary selection of thickness of bottom slab

The bottom plate must support the weight of the concrete and withstand significant water pressure of 20 kN/m² at a depth of 2 meters Therefore, it is crucial to ensure adequate anti-cracking and waterproofing measures Consequently, the initial thickness of the bottom plate is carefully determined to meet these structural requirements.

Choose thickness of bottom slab hbd = 150 mm

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

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

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

Calculation of cover slab

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

Including the self-weight of the structural layers

Table 3.1: Static loading of the slab

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

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

1.22 2 l = 4.1=  →The cover plate works in 2 directions

Prilimanary of top cover panel beam: Dbn : 200 400 mm

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

Figure 3.2: Free-body diagram 3.3.3 Internal forces

Maximum positive moment between span:

Maximum negative moment between support:

In which: αi1, αi2, β i1, β i2: are the coefficients looking up the table according to the diagram

P is total pressure on slab: P=qL L1 2 =(g+p L L) 1 2

Table 3.2: Internal forces of cover panel

 =  =  =  =  Reinforced the checking hole 600 x 600 by 4ỉ12 b s o b s

 m  Choose As As choose choose

Calculation of wall plate

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

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

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

Level of the top water tank: z = 64.5 m

Wind pressure constant throughout the height of the tank wall

Inflow wind load: W h =nW kc 0 =1.2 0.55 1.5235 0.6   =0.603306 kN/m 2

Outflow wind load: W d =nW kc 0 =1.2 0.55 1.5325 0.8   =0.804408 kN/m 2

Table 3.4: Static loading of the slab

Static loading of the slab with strip 1 m: N bt =glb=5.06x1.8x1=9.11 kN

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

Consider the ratio:L h 2 Wall slab with length 9.6 m work on 1 diretion

Wall slab with length 4.1 m work on 1 diretion

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

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

→ Since the sides are roughly the same size, just calculate for the 9.6 m side slab and the same layout for the 4.1 m side slab

Load combination: Full filled water tank + Inflow wind load

Figure 3.3: Free body diagram 3.4.3 Internal forces

Internal forces be calculated by fomular according to table 6 [KCBTCT tập 3 Võ Bá Tầm] linear calculation result was solved by super position method

Figure 3.4: Internal forces 3.4.4 Calculation of reinforcement

33.6 128 n nhip nhip gio nhip nuoc q h Wh

M =M +M = + Maximum negative moment between support:

15 8 n goi goi gio goi nuoc q h Wh

(kNm) mm mm mm cm2  a cm2

Calculation of bottom slab

The bottom slab is completely concreted with bottom beams, using the beam system qh²/8

Figure 3.5: Bottom slab of water tank Table 3.6: Static loading of the slab

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

1.2 2 l = 4.1=  bottom slab work on 2 directions

Figure 3.6: FBD of bottom slab

Maximum negative moment between support:

In which: mi1, mi2, k i1, k i2 are the coefficients looking up the table according to the diagram

P is total pressure on slab:P=qL L1 2 =(g+p L L) 1 2

Table 3.7: Internal forces of bottom panel

As Choose As choose choose

Calculation of water tank beam system

• Loading on top cover beam

DN2, DN3: Trapozoidal distributed load; DN1: Triangle distributed load

= =  • Wind load on beam system

Inflow wind load: W h =nW kcB 0 =1.2 0.55 1.5235 0.6 0.9    =0.5429754 kN/mOutflow wind load: W d =nW kc 0 =1.2 0.55 1.15235 0.8 0.9    =0.7239672 kN/m

Table 3.9: Total loading on top cover beam system

DL LL IWL OWL kN/m kN/m

DD2, DD3: Trapozoidal distributed load

Dead load: g xl bd 1 6.16 4.1 g 13.09 kN/m

= =  Live load: p xl bd 1 18 4.1 p 23.38kN/m

= =  • Wind load on beam system

• Seft weight of wall slab:q bt =glb=3.169 2 1  =5.7 kN/m

Table 3.10: Total loading on beam system

Shape DL LL IWL OWL kN/m

Students address the challenges of spatial working beam systems by utilizing Etabs 9.7.1 software to model the frame system effectively.

Figure 3.9: Modeling of water tank

Figure 3.13: Moment DN1 and DD1

Figure 3.16: Shear force DN1 and DD1

Figure 3.17: Shear force DN2and DD2

Figure 3.18: Shear force DN3 and DD3 3.6.4 Calculate reinforcement

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

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

• Determine constructive span of stirrup:

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

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

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

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

Axial stress condition (kN) (mm)

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

Check deflection and deformation of bottom slab

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

In which:  : Coefficient based on ratio of 2

L [Table appendix 17, Kết cấu bê tông cốt thép tập 3, Võ Bá Tầm]

L2/L1 = 1.22→  =0.0024728 q c : standard load evenly distributed on the bottom plate,q tc 54 kN/m 2

D: Cylindrical stiffness , determined by the formula

So the deflection of the bottom plate:

Because of slab with L Mmax

Conclusion: Pile is satisfied for constructing reasons

Constructing process (Lift the pile to construction site)

Conclusion: Pile is satisfied for constructing reasons

Calculate reinforcement for hanging purpose

One-side force of crane apply on hanging bar:

Cross section area of hanging bar:

Length of fixed segment of hanging bar: n n

Design of foundation F1

Table 6.7: Reaction forces column C26 foundation F1

Story Point Load FX FY FZ MX MY MZ

6.5.2 Verify number of pile for foundation F1

Total axial force apply on foundation F1: Ntt = 7911.69 kN

Preliminary selection for number of piles: tt pile c,d

Selection of geometry of pile cap and depth:

Geometry of pile cap: Bpcap × Lpcap × Hpcap = 3.2 m × 3.2 m × 2 m

Depth of foundation F1: Hmax tt = 228.08 KN tt o m min d h h 0.7tg(45 ) 2H

 = 3.09 m hm = 3.5 m > hmin = 3.09 m → Satisfied low depth pile foundation

Total axial loading lean on Bot of pile cap

Determine the values of pmax(j) and pmin(j): y max x max max,min 2 2 coc i i

Therefor Σxi 2 = 6.48 m 2 , Σyi 2 = 8.64 m 2 , xmax = 1.2 m, ymax = 1.2 m

Pmin = 1015.5 KN > 0 ➔ pile is not plucked

6.5.3 Verify strength condition and settlement

Average friction angle of each soil layers :

 = + + + 18.1 o Length of expanding span: tb x Lcoc tan

=   =2.4 m Length, width of assuming foundation

Checking pressure at the Bot of assuming foundation

Total standard loading lean on bottom of foundation tt tc N

Waf = Laf × Baf × Zi × γi @128 KN

Eccentricity along with X direction: tc y x tc af e M

Eccentricity along with Y direction: tc x y tc af e M

Standard pressure at the bottom of assuming foundation tc tc af x y max pcap pcap pcap pcap

 = + + +  735.8 kN/m 2 pcap tc tc af x y min pcap pcap pcap

 = + − −  733.6 kN/m 2 tc tc tc tb ( max min) / 2

Soil bearing capacity under pile tip

In which: tc 2 tc 2 max tc 2 min tc 2 tc 2 tb

Thus, the ground under the conventional foundation block satisfies the condition of stability

Determine settlement of assuming foundation

The self-soil pressure of the bottom of assuming foundation block: σo bt = 547.6 kN/m 2

The stress causing settlement at the bottom of assuming foundation block : σo gl = σtc tb - σo bt = 734.7- 547.6= 187.1 kN/m 2

To analyze the soil layer beneath a conventional foundation block, divide it into multiple layers, each measuring 1 meter in thickness Calculate the stress responsible for settlement until the condition σ n bt ≥ 5σ n gl is met, indicating the point at which settlement ceases Utilize the formula bt bt i i 1 − i ih for accurate calculations.

 =  :settlement stress at the bottom of the second layer i koi:Look up table accoring to ratio of qu qu

B Table 6.8: Settlement result of foundation F1

Vị trí Z (m) Z/B K0 σibt σigl E σibt/σigl Si kN/m 2 kN/m 2 kN/m 2

At a depth of 4 m from the foundation, then σn bt > 5σn gl

The settlement of the foundation is calculated according to the formula:

S = 2.29 cm < [Sgh] = 8 cm → Satisfied settlement condition

In order to ensure that the pile cap has only compressive stress, the height of the pile cap must satisfy the following conditions: L - 2bm < (bc + 2ho)

In structural engineering, the center of gravity for tensile reinforcement is calculated as 110 ho = 1.8 m, which is the distance from the outer edge of the compressive concrete area The distance from the outermost edge of the pile to the edge of the pile cap is measured at 0.2 m, while the width of the column is established at 0.6 m.

Examinate pile reaction on pile cap

Settlement of single pile following to the formula B.1 – Appendix B- TCVN

= 9 = The load acted on pile (kN)

EYoung modulus of the material pile (kN/m 2 ) pile 3

The pile is not plucked

Table 6.9: Reinforcement result for foundation F1

Section Moment b (mm) h 0 (mm) As (cm 2 ) Choose As choose

Figure 6.6: Moment on X-direction Figure 6.5: Moment on Y-direction

Geometry of pile cap: Bpcap × Lpcap × Hpcap = 4.8 m × 6 m × 2 m

Depth of foundation F2: Hmax tt = 69.57 KN tt o m min d h h 0.7tg(45 ) 2H

Total axial loading lean on bottom of pile cap

Determine the values of pmax(j), and pmin(j): y max x max max,min 2 2 coc i i

In wich Σxi 2 = 40.32 m 2 , Σyi 2 = 14.4 m 2 , xmax = 2.4 m, ymax = 1.2 m

Average friction angle of each soil layers :

Length, width of assuming foundation:

Checking pressure on bottom of assuming foundation

Total standard loading lean on bottom of assuming foundation tt tc N

Wqu = Lqu × Bqu × Zi × γi R166 KN

Eccentricity along with X direction: tc y x tc qu e M

Eccentricity along with Y direction: tc x y tc qu e M

Standard pressure at the bottom of assuming foundation tc qu y tc x max m m m m

 = + + +  794.4 kN/m 2 tc qu y tc x min m m m m

Bearing capacity of soil under pile tip

=     +    +  In which: tc 2 tc 2 max tc 2 min tc 2 tc 2 tb

 =   Thus, the ground under the conventional foundation block satisfies the condition of stability

The stress causing settlement at the bottom of assuming foundation block : σo gl = σtc tb - σo bt = 676.14- 547.6= 128.54 kN/m 2

To analyze the soil layer beneath a conventional foundation block, divide it into multiple layers, each with a thickness of 1 meter Calculate the stress responsible for settlement until the condition σ n bt ≥ 5σ n gl is met, indicating the point at which settlement ceases Utilize the koi lookup table based on the ratio of qu to qu for accurate results.

Table 6.11: Settlement result of foundation F2

Pos Z (m) Z/B K0 σibt σigl E σibt/σigl Si kN/m 2 kN/m 2 kN/m 2

In order to ensure that the pile cap has only compressive stress, the height of the pile cap must satisfy the following conditions: L - 2bm < (bc + 2ho)

L = 4.8 m (Length of pile cap) ho = 2 - 0.2 = 1.8 m bm = 0.6 m (distance from the outermost edge of the pile to the edge of the pile cap) bc = 0.85 m (width of column)

Settlement of single pile following to the formula B.1 – Appendix B- TCVN

The spring stiffness: Q a 1148 k 142 kN/mm

Table 6.12: Reinforcement result for foundation F2 Section Moment b (mm) h (mm) As (cm 2 ) Choose As choose

Preliminary selection for number of piles:

Geometry of pile cap: Bpcap × Lpcap × Hpcap = 3.2 m ×4.4 m × 2 m

Depth of foundation F3 with Hmax tt = 149.47 KN tt o m min d h h 0.7tg(45 ) 2H

W = Bpcap × Lpcap × Hpcap × γpcap = 3.2×4.4×2× 25 = 704 KN

Total axial loading lean on bottom of pile cap

Determine the values of pmax(j) and pmin(j): y max x max max,min 2 2 coc i i

Therefor Σxi 2 = 21.6 m 2 , Σyi 2 = 11.52 m 2 , xmax = 1.8 m, ymax = 1.2 m

Average friction angle of each soil layers:

 = + + + 18.1 o Length of expanding span: tb x Laf tan

Length, width of assuming foundation

Wqu = Lqu × Bqu × Zi × γi = 46147 KN

Eccentricity along with X direction: tc y x tc qu e M

Eccentricity along with Y direction: tc x y tc qu e M

Standard pressure at the bottom of assuming foundation tc qu y tc x max m m m m

 = + + +  835.4 kN/m 2 tc qu y tc x min m m m m

In which: tc 2 tc 2 max tc 2 min tc 2 tc 2 tb

835.4 kN/m 1.2R 3698.1 kN/m σ 5.3 kN/m >0 831.4 kN/m R 3081.8 kN/m

 =   Thus, the ground under the conventional foundation block satisfies the condition of stability

The stress causing settlement at the bottom of assuming foundation block : σo gl = σtc tb - σo bt = 676- 547.6= 128.4 kN/m 2

To optimize the foundation block's soil layer, divide it into multiple 1-meter thick layers Calculate the stress that leads to settlement until the condition σ n bt ≥ 5σ n gl is met, indicating the point at which settlement ceases.

 =  : settlement stress at the bottom of the second layer i koi:Look up table accoring to ratio of qu qu

Table 6.14: Settlement result of foundation F3

Positiion Z (m) Z/B K0 σibt σigl E σibt/σigl Si kN/m 2 kN/m 2 kN/m 2

In order to ensure that the pile cap has only compressive stress, the height of the pile cap must satisfy the following conditions:

The pile cap has a length of 4.4 meters, with the center of gravity of the tensile reinforcement positioned 1.8 meters from the outer edge of the compressive concrete area The distance from the outermost edge of the pile to the edge of the pile cap measures 0.2 meters, while the column width is 0.7 meters.

Table 6.15: Reinforcement result for foundation F3

Section Moment b (mm) h (mm) As

Design of pit foundation

Table 6.16: Reaction result of core wall

6.8.2 Verify number of pile for pit foundation and arrangement

Total axial force apply on foundation Ntt = 28583.7 kN

Preliminary selection for number of piles: tt pile tk

Figure 6.15: Layout of pit foundation

Geometry of pile cap: Bđ × Lđ × Hđ = 4.4 m × 8 m × 2 m

Depth of pit foundation with Hmax tt = 1069.2 kN tt o m min d h h 0.7tg(45 ) 2H

Average friction angle of each soil layers:

Length of expanding span: tb x Lcoc tan

Length, width of assuming foundation:

Section modulus of assuming foundation

Height of assuming foundation: Haf = 33 m

Cross section area of assuming foundation: A af = L B af af =117.76 m 2

Mass of soil in the conventional foundation block at the bottom of the platform and on the platform:

Weight of pile: W pile =n pile bt  A L pile pile 360 kN

Weight of pile cap: W pcap =  bt h pcap A pcap = 1760 kN

Weight of soil: W dc =  1 h A d d +n A c c h i  = ' i 3475.8 kN

Total weight of assuming foundation: Waf = Wd + Wpile + Wpcap - Wdc = 75709 kN

Loading on bottom of assuming foundation:N tc d =N tc +W qu 528.75 kN

Standard pressure at the bottom of assuming foundation tc tc tc qu y tc x max qu qu x y

 = + + +  919.1 kN/m 2 tc tc tc qu y tc x min qu qu x y

=     +    +  In which: tc 2 tc 2 max tc 2 min tc 2 tc 2 tb

Thus, the ground under the conventional foundation block satisfies the condition of stability

The self-soil pressure of the bottom of assuming foundation block:σo bt = 547.6 kN/m 2

The stress causing settlement at the bottom of assuming foundation block : σo gl = σtc tb - σo bt = 813.4 – 547.6 = 265.8 kN/m 2

To analyze the settlement of a conventional foundation block, divide the underlying soil layer into multiple segments, each with a thickness of 1 meter Calculate the stress leading to settlement until the condition σ n bt ≥ 5σ n gl is met, indicating the point at which settlement ceases The formula used for this calculation is bt bt i i 1 − ihi.

 =   : settlement stress at the bottom of the second layer i koi:Look up table accoring to ratio of qu qu

B Table 6.17: Settlement result of pit assumong foundation

Pos Z (m) Z/B K0 σibt σigl E ibt/σigl Si kN/m 2 kN/m 2 kN/m 2

Settlement of single pile following to the formula B.1 – Appendix B- TCVN 10304:2014:

EYoung modulus of the material pile (kN/m 2 ) pile

Figure 6.17: Moment on X,Y-direction Table 6.18: Reinforcement result for pit foundation

Section Moment b (mm) h (mm) As

[1] TCVN 2737-1995: Tiêu chuẩn thiết kế tải trọng và tác động

[2] TCVN 5574-2018: Kết cấu bê tông và bê tông cốt thép

[3] TCVN 198-1997: Nhà cao tầng – Thiết kế kết cấu bêtông cốt thép toàn khối

[4] TCVN 229:1999 Chỉ dẫn tính toán thành phần động của tải trọng gió theo TCVN

2737:1995 - NXB Xây Dựng - Hà Nội 1999

[5] TCVN 9386-2012: Thiết kế công trình chịu động đất

[6] TCVN 10304-2014: Móng cọc – Tiêu chuẩn thiết kế

[7] Tiêu chuẩn Anh BS 8110-1997 (Dùng thiết kế kết cấu khung với sự trợ giúp của phần mềm Etabs)

[8] GS.TS Nguyễn Đình Cống – Tính toán thực hành cấu kiện BTCT – Tập 1 Nhà xuất bản xây dựng

[9] GS.TS Nguyễn Đình Cống – Tính toán thực hành cấu kiện BTCT – Tập 2 –

Nhà xuất bản xây dựng

[10] GS.TS Nguyễn Đình Cống – Tính toán tiết diện cột BTCT – Nhà xuất bản xây dựng

[11] Võ Bá Tầm – Kết cấu bê tông cốt thép Tập 2 (Cấu kiện nhà cửa) – Nhà xuất Đại Học

[12] Võ Bá Tầm – Kết cấu bê tông cốt thép Tập 3 (Cấu kiện đặc biệt) – Nhà xuất Đại Học

[13] Hướng dẫn kết cấu nhà cao tầng BTCT chịu động đất theo TCXDVN 375-2006

- Nhà xuất bản xây dựng

[14] GS.TS NguyễnVăn Quảng - Nền và móng các công trình dân dụng và công nghiệp – Nhà xuất bản xây dựng

[15] Nguyễn Văn Hiệp - Vấn đề tổ hợp tải trọng cho nhà nhiều tầng, Tạp chí xây dựng số

[16] PGS.TS – Nguyễn Lê Nin – Động đất và thiết kế công trình chịu động đất –