An phu apartment 15 floors

63 8.1.1 Calculated internal force combinations .... 74 8.2.1 Calculated internal force combinations .... Table1.1 Building elevation table + Area of the typical floor: 1917 m21.2 Applie

OVERVIEW OF CONSTRUCTION ARCHITECTURE

General information

- Name of the Project: An Phu Apartment Building

- Address: Duong Van An, An Phu Ward, District 2, Ho Chi Minh City

Figure 1.1 Location of the building on Google Map

+ Stories: total 18 floors:1 ground floor (8 apartments and parking), 15 typical floors

+ Area of the typical floor: 1917 m 2

Applied load

Static load include: Self-weight of the building; finish, brick wall, MEP, other devices

Standard live load operation depending on the function of each room (according to

Because the total height of the building is larger than 40m, the wind load acting on the structure includes both static and dynamic components of the wind load (HCMC)

Standard wind pressure Wo = 0.83 kNm 2 (wind pressure partition II-A).

Materials

− Concrete: Concrete grade B25 → 𝑅 𝑏 = 14.5 MPa 𝑅 𝑏𝑡 = 1.05 MPa, 𝐸 𝑏 = 30×10 3 MPa

− Steel ỉ ≤ 10: use steel grade CB240-T with 𝑅 𝑆 = 210 MPa, 𝑅 𝑠𝑤 = 170 MPa

− Steel ỉ > 10: use steel grade CB300-T with 𝑅 𝑆 = 260 MPa, 𝑅 𝑠𝑤 = 210MPa.

DESIGNING STAIRCASE

General information

The typical staircase (from 1 st floor to the roof floor) of this building consists of 2 clauses with 10 steps each.

Data for calculating

− Steel ỉ > 10: use steel grade CB300-T with 𝑅 𝑆 = 260 MPa, 𝑅 𝑠𝑤 = 210MPa

300 = 19.444 → So n = 20 steps The angle of the ladder: cos α = 0.857

Choose the thickness of stair slab: hs = 𝐿 𝑜

Choose beam dimension of landing: hb= 𝐿 𝑜

10÷13= (0.207 ÷ 0.27)𝑚 → So we choose hb =0.25 m bb = ℎ

In theory, we should choose hinge connection to design By theory, in case that ℎ 𝑏

For structures where the height exceeds three stories, a fixed connection is typically utilized Although the ideal scenario involves constructing both the columns and stairs simultaneously, in practice, the columns are erected first, followed by the installation of the stairs.

So we'll take the hinge connection to design the stairs.

Designing typical staircases

Figure 2.1 General information of the typical staircase

Figure 2.2 Calculated diagram of the typical staircase

Dead load δg (mm) g (kN/m 3 ) n g tc s (kN/m 2 ) g tt s

Table 2.1 Applied load on landing

Table 2.2 Applied load on ladders

Calculated live load: 𝑝 𝑡𝑡 = 𝑝 𝑡𝑐 𝛾 = 3 × 1.2 = 3.6 kN/m 2 δg (mm) g (kN/m 3

- Handrail’s standard self-weight: 𝑞 𝑙𝑐 = 3kN/m → 𝑔 𝑙𝑐 = 𝑞 𝑙𝑐

2.3.2 Landing and ladders reinforcement arrangement

Cut a strip with 1m width (b = 1m) to design reinforcement

Because the stairs have the same 2 clauses, we calculate for 1 side and then get the same result for the other side

Consider a random position far from A as a distance x, then calculate the moment at that position: 𝑀 𝑥 = 𝑅 𝐴 cos 𝛼 − 𝑞 2 𝑥 2

The maximum moment at the span could be determine by: ‘the derivative of monent is the shear force and the shear force there must be 0’

Take the derivative of Mx with respect to x and let that derivative equal 0, we have x:

Choose a = 15mm → h0 = hs – a = 120 – 15 = 105mm b = 1000mm

Table 2.3 Landing and ladders reinforcement arrangement

Self-weight of the beam: 𝑔 𝑏 = 𝑏 𝑏 (ℎ 𝑏 − ℎ 𝑠 )𝑛𝛾 𝑏𝑡 = 0.536 𝑘𝑁/𝑚

Self-weight of the wall brick: 𝑔 𝑡 = 𝑏 𝑡 ℎ 𝑡 𝑛𝛾 𝑡 = 6.93 𝑘𝑁/𝑚

Total load applied on beam: 𝑞 = 𝑔 𝑏 + 𝑔 𝑡 + 𝑅 𝐵 = 21.02 𝑘𝑁/𝑚

𝑏𝑅 𝑏 ℎ 0 2 ; 𝜉 = 1 − √1 − 2𝛼 Where: hb = 250 mm bb = 150 mm

Figure 2.4 Calculated diagram of the beam

(mm)   As (mm 2 ) Choosing rebar

Table 2.4 Landing beam reinforcement arrangement

Choose rebar for stirrup ỉ6, n = 2, st = 300mm

With Qdb > Qmax = 28.38 kN, so we don’t need to calculate shear force reinforcement.

Check the width of the stairs' cracks ( 2 nd Limited State)

According to TCVN 5574:2018, the limit crack width is [an] = 0.3 mm (CB240-V,

Tensile component: 1,2 φ1 = 1 Factor when the effect of short-term temporary loads and short-term effects of regular loads and long-term temporary loads η: The coefficient is as follows:

With reinforced bar with ledge: 1.0

With smooth round steel bar: 1.3

With fiber reinforcement with ledges or cables: 1.2

With plain reinforcement: 1.4 d: reinforced glass μ: Reinforced content σ a : The stress of the outermost S reinforcement bar is calculated according to the formula

M: Standard Moment works on the lake wall in 1 m width z: The distance from the center of the S reinforcement area to the set point of the synergy in the compression-resistant area of the concrete section above the crack is determined as follows: h1 = 2a: for rectangular components

The relative compression zone height of concrete is calculated as follows:

In the formula, the second rank number is assigned a "+" sign when pre-compression force is present, while it receives a "-" sign in the case of pre-traction This adjustment is made for the bendable component, resulting in the second rank number being equal to 0 The value of φf is calculated using a specific formula.

Where β: The coefficient is as follows:

For heavy concrete & light concrete: 1.8

For hollow concrete and honeycomb concrete: 1,4

𝑏ℎ 𝑜 2 𝑅 𝑏 es tot: The eccentricity of the vertical force N to the rebar parting center of gravity, corresponding to it is momen M (Due to the bendable structure, es tot = 0)

The pre-stretch reinforcement area is defined as zero, while the compression-resistant height of the wing section I and T is also zero The coefficient characteristics of the plastic state of the compressed concrete area depend on environmental humidity and the properties of long-term and short-term loads Specifically, the coefficient υ is 0.15 for long-term loads and 0.45 for short-term loads in environments with humidity levels exceeding 40%.

Table 2.5 The 2 nd Limited State checking results of staircase

Eb (MPa) 30000 30000 30000 α 7 7 7 b (mm) 1000 1000 150 h (mm) 120 120 250 a (mm) 15 15 20 a' (mm) 15 15 20 h1 (mm) 30 30 40 h'f(mm) 30 30 40 ho(mm) 105 105 230

Mtt (kNm) 20.73 11.84 19.16 à 0.00829 0.00538 0.01165 δ 1 1 1 φ1 1 1 1 η 1 1 1 d (mm) 10 10 16 β 1.8 1.8 1.8 δ ' 0.130 0.074 0.166 φf 0 0 0 λ 0 0 0 υ 0.15 0.15 0.15 ξ 0.215 0.184 0.247 z (mm) 93.69 95.35 201.58 σa (kN/m 2 ) 254303.0 219873.7 236402.6 acrc (mm) 0.139 0.134 0.132

Satisfy ([acrc]>acrc) OK OK OK

DESIGNING WATER TANK

Data for calculating

− Concrete: Concrete grade B15 → 𝑅 𝑏 = 8.5MPa 𝑅 𝑏𝑡 = 0.75 MPa,

Locate at the elevator core: a x b x h = 7 x 4 x 1.5 (m)

Height of the foot column: hfc = 1m l1 x l2 = 3.5 x 4 (m) l2/l1 = 4/3.5 = 1.14 < 2 → 2-ways slab (No.9 diagram)

Figure 3.1 Beam arrangement of water tank

Figure 3.2 Calucalted diagram of 2-ways slab (No.9 diagram)

Design tank lid

Choose the thickness of the tank lid hnb = 0.07m = 70mm

3.2.1 Applied loads δg (mm) g (kN/m 3 ) n g tc s (kN/m 2 ) g tt s (kN/m 2 )

Table 3.1 Self-weight apply on tank lid

Standard live load: p tc = 0.75 kN/m 2 (Safety factor n = 1.3)

Calculated live load: p tt = n p tc = 0.75 x 1.3 = 0.975 kN/m 2

Total load: q = g tt s + p tt = 3.728 kN/m 2

Moment at the span: M1 = mi1.P; M2 = mi2.P

Moment at the support : MI = ki1.P; MII = ki2.P

Cut a strip with the width b = 1000 mm to calculate

Table 3.2 Reinforcement arrangement of tank lid

Design tank wall

Choose the thickness of the tank lid htb = 0.07m = 70mm

Standard wind pressure Wo = 0.83 kNm 2 (wind pressure partition II-A)

Considering the most unfavorable field, the slab is under the influence of water pressure and suction wind, so the applied load has a trapezoidal shape

Figure 3.4 Tank wall calculated diagram

The standard wind pressure for partition II-A is calculated as W0 = 0.83 kNm², with a height-dependent coefficient k = 1.37 at an elevation of 58.5 meters, as referenced in Table 5 of TCVN 2737:1995 This coefficient varies based on the topography and altitude of the location being assessed Additionally, the aerodynamic coefficient for suction is determined to be c = 0.6.

Heaped live load of water:

p = 1000 kG/m 3 = 10 kN/m 3 : Specific gravity of water np = 1.1 h = 1.5m

Cut a strip follow the shorter direction with the width b = 1000 mm to calculate

Table 3.3 Reinforcement arrangement of tank wall

Design tank bottom

Choose the thickness of the tank bottom hdb = 0.1m 0mm

Table 3.4 Self-weight of the tank bottom Live load:

Heaped live load of water:

𝑝 𝑛 = 𝑛 𝑝 𝛾 𝑝 ℎ = 16.5 𝑘𝑁/𝑚 2 For the bottom plate, the repair load is not included because the tank does not contain water when repairing (according to TCVN 5574-2018)

Total load: q = g tt s + pn = 21.28kN/m 2

Moment at the span: M1 = mi1.P; M2 = mi2.P

Moment at the support : MI = ki1.P; MII = ki2.P

Cut a strip with the width b = 1000 mm to calculate

Table 3.5 Reinforcement arrangement of tank bottom

Design beam

Figure 3.7 Beam of tank bottom dimension 3.5.1 Applied loads

Self-weight of beams by SAP 2000 program

DN1: Triangle distributed load qDN1 = 4.71kN/m

DN2: Trapezium distributed load qDN2 = 5.21kN/m

DN3: Trapezium distributed load qDN3 = 10.08 kN/m

DD1: Triangle distributed load qDD1 = 31.35 kN/m

DD2: Trapezium distributed load qDD2 = 34.18 kN/m

DD3: Trapezium distributed load qDD3 = 56.33 kN/m

Figure 3.6 Beam of tank lid dimension

Figure 3.8 Distribution loads calculated diagram 3.5.2 Reinforcement arrangement

Span 4.04 100 300 280 0.0606 0.0626 57 2ỉ10 157 0.56 Support 7.21 100 300 280 0.1082 0.1148 105 2ỉ10 157 0.56 DN2 Span 10.41 100 300 280 0.1562 0.1708 156 2ỉ10 157 0.56 DN3 Span 21.59 200 300 280 0.1620 0.1778 326 2ỉ16 402 1.44

Table 3.7 Water tank’s beam reinforcement arrangement

Check the shear strength of the reinforcement and concrete:

Q db = √8R bt bh 0 2 R sw nA sc

Q max = 112.65 kN > Q db = 139.26 kN → Satisfy

Design column for water tank

For simplicity in calculation and approximate results, we consider the column as a compressive member at the center and ignore the moment due to wind load

Choose the column size 300 x 300 (mm) and arrange 4ỉ6 (Asc = 113 mm 2 )

Displacement

Figure 3.9 Tank bottom displacement by SAP2000

Determine the displacement of bottom slab by Sap2000: f = 0.0016m = 1.6 mm

Table 3.8 Water tank’s beam displacement checking

Check the width of the stairs' cracks ( the 2 nd Limited State)

According to TCVN 5574:2018, the limit crack width of water tank (level 3) is:

Tensile component: 1,2 φ1 = 1 Factor when the effect of short-term temporary loads and short-term effects of regular loads and long-term temporary loads η: The coefficient is as follows:

With reinforced bar with ledge: 1.0

With smooth round steel bar: 1.3

With fiber reinforcement with ledges or cables: 1.2

With plain reinforcement: 1.4 d: reinforced glass μ: Reinforced content σ a : The stress of the outermost S reinforcement bar is calculated according to the formula

M: Standard Moment works on the lake wall in 1 m width

The distance from the center of the reinforced area to the synergy set point in the compression-resistant region of the concrete section above the crack is defined as h1 = 2a for rectangular components.

The relative compression zone height of concrete is calculated as follows:

In the given formula, the second rank number is marked with a "+" sign when there is pre-compression force and a "-" sign when pre-traction is present For the bendable component, this second rank number equals 0 The value of φf is calculated using a specific formula.

Where β: The coefficient is as follows:

For heavy concrete & light concrete: 1.8

For hollow concrete and honeycomb concrete: 1,4

𝑏ℎ 𝑜 2 𝑅 𝑏 es tot: The eccentricity of the vertical force N to the rebar parting center of gravity, corresponding to it is momen M (Due to the bendable structure, es tot = 0)

The pre-stretch reinforcement area is defined as zero, with the compression-resistant height of the wing section also at zero The coefficient characteristics of the plastic state of the compressed concrete area depend on environmental humidity and the load's long-term and short-term properties For long-term loads, the coefficient is υ = 0.15, while for short-term loads in environments with humidity exceeding 40%, the coefficient is υ = 0.45.

SPECIALIZATION TANK BOTTOM WALL TANK

Eb (MPa) 30000 30000 30000 30000 30000 30000 α 7 7 7 7 7 7 b (mm) 1000 1000 1000 1000 1000 1000 h (mm) 100 100 100 100 70 70 a (mm) 15 15 15 15 10 10 a' (mm) 15 15 15 15 10 10 h1 (mm) 30 30 30 30 20 20 h'f(mm) 30 30 30 30 20 20 ho(mm) 85 85 85 85 60 60

Mtt (kNm) 5.93 4.52 13.69 10.49 2.73 0.96 à 0.00394 0.00295 0.00887 0.00887 0.00472 0.00235 δ 1 1 1 1 1 1 φ1 1 1 1 1 1 1 η 1 1 1 1 1 1 d (mm) 8 8 10 10 6 6 β 1.8 1.8 1.8 1.8 1.8 1.8 δ ' 0.097 0.074 0.223 0.171 0.089 0.032 φf 0 0 0 0 0 0 λ 0 0 0 0 0 0 υ 0.15 0.15 0.15 0.15 0.15 0.15 ξ 0.139 0.119 0.192 0.209 0.162 0.113 z (mm) 79.08 79.95 76.83 76.12 55.14 56.61 σa (kN/m 2 ) 265252 265857 250887 262938 217821 219031 acrc (mm) 0.157 0.162 0.135 0.141 0.114 0.124

Satisfy ([a]>acrc) OK OK OK OK OK OK

Table 3.9 The 2 nd Limited State checking results of water tank

CHAPTER 4: DESIGN SLABS FOR TYPICAL FLOORS

Calculated datas

Choose preliminary slab thickness: b = 300mm (Flat slab)

Height of typical floors H = 3500mm

Layer δ g (mm)  g (kN/m 3 ) n g tc s (kN/m 2 ) g tt s (kN/m 2 )

Table 4.1 Dead load of the slab

To effectively distribute wall loads, it is essential to convert them into an evenly distributed load across the floor and boundary beams In a typical floor scenario, the total length of partition walls measures 418 meters, while the floor area is 1,917 square meters.

- Apply distributed on slab (0.2m width, 3.2m height):

Live load: p tc (kN/m 2 ) n p tt (kN/m 2 )

Design flat slabs

Figure 4.1 3D Model of the flat slab by SAFE

The selected floor structure system is a beamless floor structure (also known as a mushroom floor or flat slabs) without a column cap

Two primary methods for calculating the reinforcement in two-way slab structures are recognized in the standards of the US, Australia, and several European countries: the direct design method and the equivalent frame method.

In modern design consultancy and engineering, computers play a crucial role in the construction industry Advanced structural computing software has simplified the resolution of complex 3-dimensional structural problems using finite element methods Consequently, traditional manual calculation methods have become less relevant This project utilizes Safe software to assess the internal strength of the slab effectively.

A flat slab structure is a two-way slab system that bends in two directions Unlike traditional two-way slabs with support beams on all four sides, a flat slab transmits the load to column strips aligned with the rows of columns These column strips function similarly to beams in conventional slab systems, effectively distributing the load across the structure.

In design practice, slabs are segmented into strips defined by the columns and the space between them The column strip, located along a row of columns, typically extends 1/4 of the span on either side of the column's centerline The mid-span strip is formed between the two column strips, functioning similarly to beams that support loads at the ends of the columns.

Due to the building's 2-way symmetrical plan on typical floors, we can calculate the floor area once and apply the same results to the other areas.

Diagram that divides calculated trips:

Figure 4.2 Slab trips layout in 2 directions 4.2.2 Load input

When subjected to horizontal loads, the internal forces in the floor are minimal, as these loads are primarily transferred to the core structure Consequently, the internal forces in the floor mainly arise from vertical loads Therefore, in floor calculations, it is sufficient to consider only vertical loads, disregarding the effects of horizontal forces.

Figure 4.3 Illustration of the case of DL

1 DL DEAD Dead load and self-weight

2 LL1 LIVE Load sorted in the form of a chessboard cell type 1

6 FLL LIVE Full applied live load

Figure 4.4 Illustration of the case of LL1

Figure 4.8 Illustration of the case of FLL

Internal forces determination

Utilize SAFE software to compute internal forces in floors based on the specified cases The floor should be segmented into strips, with the width defined as a shell element, ensuring this division is both reasonable and precise to enhance result accuracy The internal force results are then summarized by generating a moment envelope chart to highlight the maximum values.

Figure 4.9 Enveloped moment diagram by X-direction (MAX)

Figure 4.10 Enveloped moment diagram by X-direction (MIN)

Figure 4.11 Enveloped moment diagram by Y-direction (MAX)

Figure 4.12 Enveloped moment diagram by Y-direction (MIN)

Design reignforcement

Arrange reinforcement by following fomulars:

Table 4.5 Design upper layer reinforcement in the x-direction

Table 4.6 Design lower layer reinforcement in the x-direction

Table 4.7 Design upper layer reinforcement in the y-direction

Table 4.8 Design lower layer reinforcement in the y-direction

(mm 2 )   x span 4.3 650 300 280 0.006 0.006 59 ỉ20@100 3142 1.73 support 137.92 1300 300 280 0.093 0.098 1992 ỉ20@100 0.00 y span 13.08 1000 300 280 0.012 0.012 181 ỉ18@200 1272 0.45 support 16.79 2000 300 280 0.007 0.007 231 ỉ18@200 0.00

Table 4.9 Design staircase slab reinforcement

Deformation cheking

Check at the location with the largest deflection (according to the drawing)

Figure 4.13 Maximum displacement of the slab

According to the Figure 4.13, we have fmax = 0.0078m = 7.8mm

Penetration resistance

Inspect the floor position located on the wall pole between axis 2, as this area poses the highest risk of penetration throughout the entire floor space If this inspection meets safety standards, further checks at other locations are unnecessary.

Penetration resistance condition:𝑃 𝑥𝑡 ≤ 0.75𝑅 𝑏𝑡 𝑏 𝑡𝑏 ℎ 0 (according to TCVN 5574 -1991) With btb = 2bc + 2hc + 4ho = 5720mm

To enhance the structural integrity of the column and accommodate internal forces caused by concrete shrinkage and temperature changes, a layer of steel structure ỉ12@200 is placed on top of the column to ensure a stronger connection.

The length of the bars is c ≥ 0.35.l = 0.35.9 = 3.15 m → 𝐜𝐡𝐨𝐨𝐬𝐞 c = 3.2m

Shear strength

Maximum shear force: Qb = 364.81 kN

The smallest shear capacity of concrete: 𝑄 𝑏 ≤ 𝜑 𝑏3 (1 + 𝜑 𝑓 + 𝜑 𝑛 )𝑅 𝑏𝑡 𝑏ℎ 0

Coefficient 𝜑 𝑛 , considering the vertical force effect In this case 𝜑 𝑛 = 0

The coefficient 𝜑 𝑓 takes into account the effect of the compressive flange in the T- section, the I-section is determined according to the formula: 𝜑 𝑓 = 0.75 (𝑏′ 𝑓 −𝑏)ℎ′ 𝑓

The conclusion is that the concrete is not damaged on the inclined section (The slab structure is not damaged by the cutting force.)

Reinforced concrete components exposed to cutting forces must be carefully evaluated for durability, particularly in the inclined strip between oblique cracks This assessment should adhere to the condition: 𝑄 𝑏 ≤ 0.3𝜑 𝑤1 𝜑 𝑏1 𝑅 𝑏 𝑏ℎ 0, which ensures that the main compression stress resistance is maintained.

Coefficient 𝜑 𝑤1 , taking into account the effect of stirrup perpendicular to the component longitudinal axis, is determined according to the formula:

For the slab without stirrup, we take 𝜑 𝑤1 = 1

Coefficient 𝜑 𝑏1 is determined by the formula: 𝜑 𝑏1 = 1 − 𝛽𝑅 𝑏

(β = 0.02: for heavy concrete, small granular concrete, honeycomb concrete β = 0.01: for light concrete )

So the condition of limiting the width of the tilted fissure is completely satisfied.

WIND LOAD OF THE BUILDING

Characteristics of dynamics of the building

See the building as a cantilever bar with limited concentrated volume Considering the system consists of a cantilever bar with n mass concentration points with the corresponding mass M1, M2 , , Mn

The system's oscillating general differential equation when bar volume is ignored

[M], [C], [K]: mass matrix, resistance coefficient, rigidity of the system

Ü, U̇, U: vector acceleration, velocity, displacement of the coordinates determines the order of freedom of the system

W ′ (τ): vector force impact is located at the corresponding coordinates

The system's own frequency and oscillation form are determined from a homogeneous, unobstructed differential equation (excluding the C resistance factor):

The condition of existence fluctuates is that the equation (4) exists non-trivial solutions

Because of [y] ≠ 0 so [K] – ω 2 [M] = 0 (5) there must be an solution, the condition is:

CANTILEVER BAR CALCULATED DIAGRAM WITH LIMITED FOCUS MASS POINTS

ij: Displacement at point j caused by the unit force located at point i

i: Frequency of individual oscillations (Rad/s)

The equation (6) allows for the determination of the real, positive value of n distinguished by ω By knowing the frequency of individual oscillations, the corresponding oscillating cycle can be calculated using the formula: Ti = 2π ωi.

Works with n orders of freedom will have their own oscillating frequency, corresponding to each individual oscillating frequency i will have a separate form of oscillation

Replace all i values into fomular (4), separate forms of oscillation are determined by this fomular

For large high-rise buildings, calculating oscillations can be complex; therefore, utilizing software to identify distinct forms of oscillation is beneficial for the project.

Etabs 9.0.4 software to calculate the frequency and pattern of oscillations according to the theory presented above

5.1.2 Calculate individual types of oscillations

The building's bearing structures are represented in a 3-dimensional model, utilizing frame elements for columns and beams, while floors and hard walls are depicted using shell elements.

The concentrated volume declared when the analysis fluctuates to be 100% static load and 50% live load

Figure 5.1 Individual basic types of oscillation

Survey the first 12 separate oscillation modes

Table 5.1 Period and frequency values are exported from Etabs of 12 modes

Static wind load components

According to TCVN 2737:1995, the standard static component value of the Wj wind load at zj height compared to the standard mark determined by the formula:

The equation Wj = Wo.k(zj).c represents the wind pressure on a structure, where k is the coefficient that adjusts for changes in wind pressure based on the height zj in terrain type B, indicating that the building is situated on an open plot The aerodynamic coefficient c, calculated as the sum of cđón (0.8) and chút (0.6), totals 1.4.

Wo: standard wind pressure value Select Wo = 0.83 kN/m 2 (corresponding to the IIA area)

(kN/m 2 ) S xj (m 2 ) W xj (kN) S yj

Table 5.2 The static component value of the wind load is attributed to the

Dynamic wind load components

According to TCVN 229:1999, it is essential to consider the dynamic wind load when calculating multi-storey buildings exceeding 40 meters in height Therefore, for a building standing at 60 meters, the impact of the dynamic component of wind load must be factored into the structural calculations.

The sensitivity of a building to wind load dynamics determines whether to consider only the pulse component of wind speed or to also include the building's inertial force in the dynamic wind load composition.

The sensitivity of the building is evaluated by examining the correlation between its fundamental individual oscillation frequency values, particularly the first individual oscillation frequency, and the limit frequency of fL=1.3 Hz, as outlined in Table 2 of TCVN 229:1999.

Compare the first individual oscillation frequency f1 with the fL limit frequency: f1=0.625 (s) < fL = 1.3 (Hz)

Thus, the dynamic component of the wind load must take into account the impact of both the wind velocity pulse and the inertial force of the work

According to the separate oscillation frequency summary Table 4.1 above we have: f3 = 0.972< fL = 1,3 < f4 = 2.386 (Hz)

Therefore, it is necessary to calculate the dynamic composition of the wind corresponding to the first 3 separate oscillations

Among the three types of oscillations, the second type (MODE 2) exhibits minimal displacement in the X and Y directions, making it negligible for calculation purposes A summary of the two significant oscillation types is presented in the following table.

Table 5.3 Two types of oscillations to calculate

The standard dynamic component value of the wind load acting on the j part (with z height) corresponding to the first form of oscillation is determined according to the formula (4.3) TCXD 229:1999

W p(ij) : Force, unit is daN or kN depending on the unit of calculation WFj

The concentrated mass of the second part of the work, denoted as M j (T), is essential for understanding the dynamic behavior of structures The dynamic coefficient, ξ i, corresponds to the i-th mode of oscillation, while Ψ i represents a coefficient derived from dividing the work into n segments, assuming that the wind load remains constant within each segment Additionally, y ij indicates the ratio of horizontal displacement at the center of the j-th segment related to the i-th mode of oscillation.

Coefficient Ψ i can be determine by the fomular:

The standard dynamic component value of wind load on the j part of a building, which varies with different oscillation forms due to wind velocity pulses, is calculated using a specific formula.

The standard value of the static component of wind load, denoted as Wj, is calculated as shown in the table above The dynamic pressure factor of the wind load, represented by ζj, varies with the height z and is specific to the j part of the building, influenced by the topographic form and height, as detailed in Table 3 of TCXD 229 – 1999.

The wind catchment area for part j of the building is denoted as S j (m²) The dynamic pressure space correlation coefficient (ν) for wind load varies based on the building's oscillation forms, which are determined by parameters such as ρ and χ, as outlined in Table 4 and Table 5 of TCXD 229 – 1999.

Table 5.5 𝚿 𝐢 values results Determine coefficient 𝛏 𝐢 :

The dynamic coefficient ξi determines depending on the εi parameter and loga reduction of the  oscillation

The building is reinforced concrete building, so we have δ = 0.3

The εi parameter is determined according to the formula:

: Wind load reliability coefficient  = 1.2 fi : Frequency of fluctuations separately i (Hz)

W0 : Standard wind load value Wo = 0.83 kN/m 2

The calculated value of the dynamic composition of the wind is determined:

: Wind load reliability coefficient  = 1.2 β: Wind load adjustment coefficient over time Take β = 1 (Assumed 50-year use period)

The calculation is summed up in the following the table below:

Table 5.7 The dynamic composition of the wind

From the left side From the right side WDX

Table 5.8 The dynamic values of the wind that imported into Etabs

FRAME DESIGNING

Frame model

Figure 6.1 Frame 3D model by Etabs

Space frame model built in ETABS software

Stories: total 18 floors:1 ground, 15 typical floors

Ord Name of load Types of load Meaning

1 DL DEAD Dead load and self-weight

2 LLEVEN LIVE Live load applied on even floors

3 LLODD LIVE Live load applied on odd floors

4 LL1 LIVE Live load applied on spans Type 1

5 LL2 LIVE Live load applied on spans Type 2

6 WLX WIND Wind load following X-direction

7 WLY WIND Wind load following Y-direction

8 WLXX WIND Wind load following opposite X-direction

9 WLYY WIND Wind load following opposite Y-direction

10 WDX WIND Dynamic wind load following X-direction

11 WDY WIND Dynamic wind load following Y-direction

12 WDXX WIND Dynamic wind load following opposite X-direction

13 WDYY WIND Dynamic wind load following opposite Y-direction

Table 6.1 Load cases to calculate by Etabs

COMB25 DL + 0,9 ( LLEVEN + LLODD + WLX )

COMB26 DL + 0,9 ( LLEVEN + LLODD + WLXX )

COMB27 DL + 0,9 ( LLEVEN + LLODD + WLY )

COMB28 DL + 0,9 ( LLEVEN + LLODD + WLYY )

Table 6.2 Load combinations to calculate by Etabs

Figure 6.2 Consider slabs to be absolutely rigid slabs type for calculation

Design boundary beams for typicals floors

Choose preliminary beam dimension: b x h = (mm)

Arrange longitudinal reinforcement by following fomulars:

𝑏ℎ 0 × 100 (%) Internal force used in the calculation of boundary beams taken from the Enveloped combination

Concrete is not damaged due to the main compression stress: Qmax< Qo = ko.Rk.b.ho (1) Shear capacity of concrete: Qmax < Q1 = k1.Rk.b.ho (2)

If both conditions (1) and (2) are satisfy, simply place the stirrup according to the structure

If one of 2 conditions is satisfy, the stirrup have to be calculated

Choose the stirrup diameter 8 (Asc = 50.3mm 2 ), n = 2 branches

When h  450mm : uct = min(h/2,150mm)

When h > 450mm : uct= min(h/3, 300 mm) = mm

At the middle of the spans:

When h > 300mm : uct = min(3h/4,500) = mm

The building features a 2-way symmetrical design on its typical floors, allowing us to calculate the floor area for just one section and then apply those results uniformly to the other areas.

Design rigid concrete walls for typicals floors

Consider rigid wall bears Nz, Mx Stress diagram at points on cross section of the wall

Divide the rigid wall into 5 areas to calculate and arrange the reinforcement by numbers in the figure above Each area’s section is b x 0.2h

The average stress of each section (b x 0.2h): σ = N

F: area of the cross section

Mx: Applied moment yi: distance between the center of the section and the center of the i area

The average stress of area 1,2: σ 𝑖 = N

J x y i (> 0 or < 0) (with i = 1,2) The average stress of area 3: σ 𝑖 = N

F > 0 (with i = 3) The average stress of area 4,5: σ 𝑖 = N

J x y i (> 0 or < 0) (with i = 4,5) Compressive/Tensive force of area 1, 2, 3, 4, 5: N i = 0.25h b σ i

Calculate reinforcement for areas 1, 2, 3, 4, 5, calculation as the center compression- bearing component:

Where: mb = 0.85 – Working condition of concrete

For area 1, 5: As = max(As1; As5)

For area 2,3,4: As’ = 2max(As2; As4) + As3

Total reinforced area of the section: 2As + As’

To enhance wall stability, it is essential to reinforce the stirrups at both ends of the wall This reinforcement is crucial due to the occurrence of local buckling, which typically manifests at the wall's ends where normal and shear stresses are most concentrated The pressure transmission is highest at these points, leading to potential structural weaknesses that need to be addressed.

- Concrete is not damaged due to the main compression stress: Qmax 5 have to check the anti-flipping condition because of the earthquake and wind load

In this project (An Phu Department), we have the ratio 𝐻

45= 1.33 < 5 → So we no need to check the anti-flipping condition

Conclusion: The overall stability of the building is guaranteed.

CONSTRUCTION METHOD

Construction method choice

Characteristic of the construction of all-block concrete is the process of installing formwork, mixing concrete, transportation, pouring and vibration

- Option 1: Manual construction: This option only applies to small buildings with little concrete volume, for cheap prices, but the quality is not high, does not reflect the

Zindustrial aspect This brings low economic efficiency

Mechanized construction with a semi-manual approach is ideal for large concrete structures, significantly reducing construction time while ensuring high-quality results This method not only enhances economic efficiency but also showcases advanced specialization within the industry, ultimately boosting labor productivity for construction workers.

From the advantages and disadvantages of the two options above, we choose option 2 is the most reasonable.

Pouring concrete phases

Phase 2 Pouring concrete - Slab Basement

Phase 3 Pouring concrete - Wall Basement

Phase 4 Pouring concrete - Slab Ground floor

Phase 5 Pouring concrete - Wall Ground floor

Phase 6 Pouring concrete - Slab Half floor

Phase 7 Pouring concrete - Wall Half floor

Phase 8 Pouring concrete - Slab floor 1

Phase 9 Pouring concrete - Wall floor 1

Phase 38 Pouring concrete - Slab floor Roof

Phase 39 Pouring concrete - Beam Water tank

Phase 40 Pouring concrete - Bottom slab Water tank

Phase 41 Pouring concrete - Side wall Water tank

Phase 42 Pouring concrete - Lid ofWater tank

Concrete volumn of elements

Rigid wall at the corner V 1 :

- Volumn of solid slab: Vsolid = Sslabhsan = 1917 x 0.3 = 575.1 m 3

- Calculated coefficient of flat slab: n = 0.36

- Volumn of typical flat slab: Vslab = nVsolid = 368.06 m 3

concrete pouring sequence

- Before pouring concrete must re-examine the shape, size and opening of the grooves of the formwork

- Do not go directly up to the steel and concrete pouring area especially with dirty slabs (it is necessary to bridge)

- Do not scrape when pouring concrete in areas with dense reinforced structures, narrow surfaces

- Pouring concrete must keep the correct order, the thickness of the pouring layer ensures that the concrete beams are good, do not pour the concrete is sized, stratified

Concrete should be poured continuously without interruption, especially in summer when temperatures reach 30°C, as stopping can compromise the integrity of the mix In cooler autumn or winter conditions, pauses should be limited to no more than one hour if temperatures drop below 25°C Proper procedures must be followed for stopping the concrete pour, which includes brushing off the surface mortar film, sanding the old concrete layer, rinsing, watering the cement, and then promptly pouring the new concrete.

Effective concrete compaction is essential for achieving optimal strength and durability, and it should be performed thoroughly without any missed areas It is crucial to ensure that the compaction process is conducted for the appropriate duration Additionally, when pouring concrete for structures such as blocks, water tanks, and floor plates in sanitary rooms, it is important to maintain a continuous pour without interruptions.

To ensure proper concrete curing, it is essential to water the surface regularly for the first seven days to maintain adequate humidity During the following week, watering should occur every three hours at night Always use clean water for this process Additionally, it is crucial to avoid applying heavy pressure to the mold during the initial curing days to prevent damage.

Ensure that the roof floor and toilet area floor, along with the water tank, are treated with waterproof cement according to standards, allowing the water to soak until it is fully drained.

Formwork structure for each components

Formwork have to be ensured requirements below:

- Must be the right size of the structural parts of the building

- Must be durable, hard, non-deformed, warped and must be stable

- Must be compact, light, handy and easy to disassemble

- The core slits must be tightly sealed so that the cement water does not flow out

- Can be reused many times To satisfy this requirement, the trunk after use must be shaved, cleaned and stored in the appropriate place

- When designing phase trunks, calculate for typical building parts and layout for other parts

Before installing formwork, it is essential to thoroughly assess the scaffolding's load-bearing capacity, durability, and overall stability Additionally, inspect all joining components, including latches, laces, and welds, to ensure they meet safety standards Never use any parts that do not comply with these requirements.

To ensure proper installation and inspection of formwork, it is essential to determine the necessary heights, including the foundation bottom, floor level, beam bottom, and floor bottom Marking the shaft and building heights in an optimal position facilitates accurate alignment and enhances the overall construction process.

- The surface of the formwork that comes into contact with the concrete should be non- stick resistant

The installation of side-mounted formwork for wall structures, floors, beams, and columns must be executed with early dismantling in mind, ensuring that the integrity of the remaining formwork and scaffolding, particularly at the base of beams, floors, and support columns, is not compromised.

- The support of the scaffolding must be firmly placed on a hard foundation, not slip and not deformed when subjected to the effects of the load and impact during construction

When installing formwork, it is essential to create appropriate drainage holes at the bottom to allow water and debris to escape during cleaning Prior to pouring concrete, these holes should be sealed, and care must be taken to leave openings for the steel reinforcements as specified in the design.

- While pouring concrete must be arranged to arrange people to regularly monitor the formwork of the anti-tree, when necessary to take timely and thorough remedial measures

- The formwork and scaffolding when completed must be tested according to TCVN 4453-95 before carrying out further work

12.4.2 Typical rigid wall formwork structure

Use steel formwork have section: 4001500 (mm), 3001500 (mm)

Steel grade CCT34: f = 210 MPa, E = 1.210 8 (kN/m 2 )

Specific weight of concrete:  = 25 kN/m 2

Load q tc (kN/m 2 ) n q tt (kN/m 2 )

Concussion when pouring concrete Pđ 4 1.3 5.2

Table 12.2 Applied load Design formwork plank:

Choose the distance between horizontal ribs l = 0.65m

Check the deflection according to the formula: f max = 5

Consider that the distance between vertical ribs is l = 0.6m

Check the strength ability: σ =M max

Consider that the distance between cables l = 1.2m

Choose section b x h x t = 50 x 100 x 5 (mm) (-shaped steel) and check

Assuming that the building that we are designing is the one in Ho Chi Minh City is the

Concentrated force of wind load:

W = W0S = 0.832.9= 4.81 (kN) Concentrated force of concrete when pouring:

Q = qSbt = 30.52/2 = 87.87 (kN) Cable design force is calculated by the formula: R = KS

M = 1: Coefficient of non-conditioning in the branches of the wire

N = 1: Number of branches Cable layout angle is 45 0 so we have:

→ Design force for cables: R = KS = 3.5  130.54 = 456.89 (kN)

We choose the diameter of the cable C30 (mm), the strength of the steel 165 (kG/mm 2 ) has the force to break the cable Rs = 510 kN

Use steel formwork have section 3001500 (mm)

Consider that the distance between braces l = 0.65m

The distance between posts is l = 0.8m

Load focus effect on the strutted bar according to horizontal direction:

Vertical force in the brace (arrange brace bar at an angle of 45 0 ):

Choose circle-section bar with diameter d = 0.04m = 4cm

𝜇 = 0.65 (consider that 2 heads of the bar are fixed connection)

Table 12.11  coefficient table for wood

According to Table above , we have  = 0.782 (with  = 52)

Section area of brace bar:

Use steel formwork have section 3001500 (mm), 4001500 (mm), 300500 (mm),

Table 12.12 Applied loads Design formwork plank:

Consider that the distance between longitudinal ribs is l = 0.9m

Consider that the distance between posts l = 1.1m

Select construction machines

The selected tower crane needs to meet the following requirements:

- Height: can bring the material to the highest position of the building, ensuring a safe distance

- Reach: can cover the entire scope of the construction site

- Crane strength: It is possible to lift the component with the largest weight with the greatest reach

When placing a tower crane, it's essential to ensure optimal construction conditions by avoiding entanglement with other construction vehicles and minimizing the rotation angle during transport Additionally, the crane's reach must extend to material warehouses and training grounds for components Therefore, positioning the tower crane at the building's façade allows for easy access to materials from outside transport and facilitates convenient crane removal.

Determine the parameters of the tower crane:

To determine the required height of a tower crane, the formula is H = hct + hck + hat + htb In this equation, hct represents the building's height at 60 meters, hck accounts for the maximum scaffolding height of 2 meters, hat denotes a safety height ranging from 1 to 1.5 meters, and htb is the height for hanging and tying, set at 1 meter.

The maximum distance from the crane's rotating center to the edge of the building or obstacles is 4 meters Additionally, the furthest distance from the building's edge to the placement point of the component must be considered in the intended direction of reach.

With determined parameters above, we select the crane tower which is satisfied all the requirements: QTZ – 6021

The maximum reached distance Rmax 60 m

The minimum reached distance Rmin 5 m

The lifting velocity Vlifted 20-80 m/min

The lowering velocity Vlowered 3 m/min

The crane (car) velocity V 27.5 m/min

Calculate the productivity and check the working capability of tower crane:

Q = 1.5T ktt = 0.5: utilization coefficient (by lifting capacity) ktg = 0.8: utilization coefficient (over period of time)

(T1: working time of the crane

T2: manual working time: removing, fitting, adjusting and placing the component in the right place) nck = 60/T = 7.5: number of cycles in 1 hour

→ Productivity of the crane in 1 hour: Ns hour = 1.5 x 7.5 x 0.5 x 0.8 = 4.5 T/h

→ Productivity of the crane in 1 shift: Ns shift = 8 Ns hour = 36 T/shift

Suppose the case of using only tower cranes to pour concrete beams and slabs: Each

Cycle for 1 time lifting up, turning and lowering down is T = 8mins

To transport 34.99 + 368.064 = 403.056 m 3 of concrete (beam and slab), the amount of bucket need to be transported is 404 buckets

So the implementation time: (404 x 8) / 60 = 54 hours → 7 shifts (8h/shift)

To ensure proper construction of slabs and beams, a stop circuit is essential; however, if this option is impractical, we utilize concrete pumps in conjunction with tower cranes as needed.

So that the time to pour concrete in 1 to 2 shifts is completely possible

Transporting concrete for pouring beams at elevated heights can be achieved through various methods, including tower cranes, concrete pumps, and hoists Each method presents unique advantages and disadvantages that should be considered for optimal efficiency and effectiveness in construction projects.

At the construction site, two hoists are utilized for transporting materials such as bricks, stones, and sand, as well as for moving personnel However, the hoist method requires the use of wheelbarrows, which necessitates considerable labor to transfer concrete from the hoist to the designated floor for pouring, making this option challenging to implement effectively.

Using a concrete pump for transporting concrete is an effective choice, especially when combined with tower cranes at the construction site for moving scaffolding, formwork, and reinforcement This coordinated approach enhances productivity and reduces the time required for concrete pouring.

Select the concrete pump B5RZ44-40 with parameters below:

To achieve a concrete output of 50 to 60 m³/h, the pouring time for slabs and beams is calculated to be approximately 6.7 hours based on a total volume of 403.56 m³ Utilizing a tower crane for support reduces the effective working time to around 6 hours.

Concrete pouring for the slab and boundary beams is scheduled from 6 AM to 12 PM, allowing for a one-hour lunch break and additional time to address any construction issues that may arise, ensuring timely project completion.

Due to the use of fresh concrete ordered at the factory, it is necessary to transport by specialized vehicle

Concrete volumn of typical slab and boundary beams is 404.056 m 3

Productivity of a truck is determined by the fomular: N = nqkt

Where: q: transported product weight (each truck carries 10m 3 of concrete) q = 10 x 2.5 = 25T kt = 0.7: utilization coefficient (over period of time) n: Amount of trips in 1 shift

Tch = tchat + tdo + tvandong + L/vdi + L/vve tchat: 10 mins (waiting for receiving mortar) tdo: 10 mins (waiting for pumping concrete) tvandong: 4 mins

L = 6km: Distance between the factory to the construction site vdi = vve = 30km/h (the average driving velocity in the city)

Amount of needed trucks in 1 shift: m = 404.056/70 = 5.75 trucks

We choose transported truck HOWO D10.34-40

- Rotation speed of mixing barrel: 0-16 rounds/min

Construction method

The building is a high-rise building, reinforced concrete frame, so the construction is very complicated and takes a lot of time, and resources, requiring close supervision of construction officiers

Use 2 meridians placed in 2 perpendicular directions to locate the core center of the wall, the molding marks, paint and mark these positions so that the teams and construction teams can easily accurately determine the required landmarks and locations

- Reinforcement must be used in the right data, type, diameter, size, quantity and location

- The reinforcement must be clean, not rusty, not dirty, especially grease

- When machining: Cutting, bending, pulling reinforced welding avoids changing the mechanical properties of reinforcement

Reinforcement is prepared at the base, shaped and sized according to design specifications, categorized by type, and bundled for efficient transport and installation using cranes.

For effective wall construction, it is essential to tie reinforcement before assembling the formwork Use 1mm soft steel wires for tying, ensuring that all joints meet technical specifications Additionally, concrete millet should be utilized to maintain the correct position and thickness of the protective layer around the reinforcement.

When reinforcing joints according to design standards, ensure that no more than 25% of the total area of the bearing reinforcement is joined with smooth round steel and no more than 50% with steel featuring a ledge The length of the tied connection must adhere to TCVN 4453-95, requiring a minimum of 250mm for tensile bearing steel and 200mm for compression-resistant steel.

The installation of reinforcements must ensure:

- The pre-installed parts must not affect, obstructing the rear installation parts

- Take measures to stabilize the reinforcement position, ensuring no deformation during construction

- After inserting and tying the stirrup, fix it temporarily then install the wall formwork

- Make sure the right shape and size are according to design requirements

- Ensure stable sustainability during construction

- Ensure tightness and easy dismantling

Method: Due to the installation of formwork after placing reinforcement, before grafting the formwork, it is necessary to clean the base of the column

- We pour in front of a wall with a height of 10-15 cm to make the rack, the formwork is accurate

- Wall formwork is machined in pieces according to wall size

- Pair the 3-sided box, thread the formwork box into the reinforced wall then install the other side

- Use the knob to fix the plank box, the distance of the plates according to calculations

- Reconfigure the wall heart position and stabilize the wall with oblique struts with adjustable lace and anchor wires

Commercial concrete is supplied by specialized companies and delivered to construction sites using dedicated vehicles To ensure a smooth and timely pouring process, it's crucial to pre-plan the optimal routes for concrete trucks Additionally, considering urban construction activities, scheduling the concrete pouring in advance is essential to prevent interruptions caused by transportation delays, particularly during rush hours that could lead to traffic congestion.

The transportation and pouring of concrete using tower cranes often face challenges such as slow speed and low productivity To enhance efficiency, it is crucial to organize the concrete pouring process meticulously and ensure that all preparations are complete, preventing any delays for the crane.

Concrete mortar is efficiently transported by truck and poured into a 1.5m³ mortar container, utilizing at least two containers to ensure continuous operation While one container is being loaded, the other is positioned for immediate use, allowing the crane to operate without delays It's crucial to have workers ready to adjust the lowering of the barrel accurately and remove the crane hook swiftly The crane instructor oversees the process, determining how to pour the concrete mortar—whether in one spot or multiple locations, and adjusting the thickness as needed for optimal results.

The concrete mortar container features a user-friendly mechanism that allows for separate loading and pouring of concrete, ensuring easy control Workers positioned on the working floors efficiently handle the concrete pouring process.

To enhance the efficient handling of concrete near dumping sites and prevent stratification during free falls from heights exceeding 3.5 meters, it is advisable to install supplementary equipment, including spill hoppers, elephant hoses, canvas pipes, and rubber tubes.

The concrete is poured into layers, the thickness of each layer pours 30-40cm, compacted thoroughly with a baton lagoon then pours the next layer of concrete

When pouring as well as when the concrete lagoon should pay attention not to cause a collision that misleads the reinforcement position

When pouring concrete, it is necessary to clean the concrete container to prepare for the next pour

Note : The quality and the slump of the concrete must be checked before using

Wall formwork is a type of non-bearing formwork, so after pouring concrete for 1 day we proceed to remove the formwork of walls

To prepare for the installation of beam and slab formwork, first remove the wall formwork, ensuring to leave a section above the top of the wall as specified in the design for connecting to the beam formwork.

The formwork is removed according to the principle: "Whichever is installed first, removed later, the latter is removed first"

The separation and counting of the formwork from the concrete must be done carefully to avoid damaging the formwork and chipping the concrete

To remove the formwork easily, people use nail plucks, pliers, crowbars, and other equipment

Note : It is necessary to carefully study the transmission in the installed formwork system to safely dismantle

12.6.2 Slabs and boundary beams construction method

Set up the scaffolding in sequence:

- Place the size foot scaffold (including the base and the size foot scaffold) linking the size foot scaffolds together with horizontal bracing and cross-bracing

- Insert the spear frame into each size foot scaffold

- Install horizontal and cross bracing rods

- The cage is articulated and tightened with the latch between the coupling, the frames are superimposed to the design position

- Adjust the height of the scaffolding system by size

Then proceed to place the bottom planks, planks, slab planks

Check the flatness and tightness of the mold

Before pouring concrete, ensure that the reinforcement is adequate in quantity, correct in type, and properly positioned Additionally, clean the reinforcement and moisten the surface of the mold.

Pour concrete with tower cranes similar to when constructing concrete columns.Concrete floor lagoons with table lagoons and concrete lagoons with baton lagoons

Before pouring the next subdivision of concrete, it is essential to properly prepare the stopping circuit This involves cleaning the area, roughening the surface, and wetting the cement to enhance adhesion before proceeding with the concrete pour.

Concrete maintenance and removal of formwork:

Concrete after pouring must have a reasonable maintenance process, must be moisturized for at least 1-2 or 7-10 days and nights depending on the type of component

In the initial two days after pouring concrete, it should be watered every two hours, starting with the first watering 4 to 7 hours post-application On subsequent days, watering should occur every 3 to 10 hours, adjusted based on air temperature, with less frequent watering during winter It is essential to wait until the concrete attains a strength of 24 kg/cm², which typically takes three days in winter, before allowing foot traffic on the surface.

The removal of bearing formwork is carried out when the concrete reaches 100% of the design intensity (about 24 days with a temperature of 200C) (78m span beams)

Remove the formwork according to the principles as stated in the section of removing the formwork of the wall.

Progress schedule of typical floor

Phase 8 and Phase 9 are picked to calculate the progress for typical floor with pouring concrete work of boundary beam, slab and rigid wall

Based on the volume of concrete in stages and the capacity of pouring commercial concrete Nca = 60 (m 3 / shift) of 1 machine-shift, we proceed to the concrete segment as follows: 𝑚 𝑖 = 𝑉 𝑖

Where: mi: amount of segment of each stage

Vi: volume of concrete of each stage

Weight of reinforcement (tons) Segments Concrete volume of each segment (m 3 )

Weight of reinforcement of each segment (tons)

Table 12.17 Concrete pouring work segment of phase 8 and 9

12.3.2 Scheduling and human resource determination

Slabs and beams should be removed after 7-10 days; however, it is not advisable to dismantle the entire structure at the point of the strutted column This is because the adjacent slabs and beams bear the load and are also subject to the operational loads of construction.

To ensure proper bonding, concrete must adequately cure when columns are placed directly on a solid foundation Depending on weather conditions, formwork can typically be removed after 1-2 days.

The norm of specific tasks based on Norms 1776/2007-Ministry of Construction

PHASE SEGMENT TASK Quantity Unit Norm Resource Duration

Table 12.18 Scheduling and human resource

[1] TCVN 2737-1995 – Tải trọng và tác động Tiêu chuẩn thiết kế

[2] TCXDVN 195-1997 – Kĩ thuật thiết kế và thi công nhà cao tầng

[3] TCXDVN 356-2005 – Kết cấu bê tông và bê tông cốt thép

[4] TCXDVN 229-1999 – Chỉ dẫn tính toán thành phần động của tải trọng gió

[5] TCXDVN 205-1998 – Móng cọc – Tiêu chuẩn thiết kế

[6] TCXDVN 326-2004 – Cọc khoan nhồi, tiêu chuẩn thi công và nghiệm thu

[7] TCXDVN 305-2004 – Bê tông khối lớn – qui phạm thi công và nghiệm thu

[8] Nguyễn Đình Cống (2007) Tính toán tiết diện cột bê tông cốt thép NXB Xây dựng

[9] Phan Quang Minh, Ngô Thế Phong, Nguyễn Đình Cống (2006) Kết cấu bê tông cốt thép, phần cấu kiện cơ bản NXB Khoa học và kỹ thuật

[10] Hoàng Nam Tập bài giảng bê tông cốt thép 1 và 2

[11] Võ Bá Tầm (2008) Kết cấu bê tông cốt thép Tập 2: cấu kiện nhà cửa NXB Đại học quốc gia Tp Hồ Chí Minh

[12] Nguyễn Tuấn Trung, Võ Mạnh Tùng Một số phương pháp tính cốt thép cho vách phẳng bê tông cốt thép

[13] Châu Ngọc Ẩn (2004) Cơ học đất NXB Đại học quốc gia Tp Hồ Chí Minh

[14] Châu Ngọc Ẩn (2005) Nền móng NXB Đại học quốc gia Tp Hồ Chí Minh

[15] Lê Văn Kiểm (2005) Thiết kế thi công NXB Đại học quốc gia Tp Hồ Chí Minh

Tiêu đề	An Phu Apartment 15 Floors
Tác giả	Duong Gia Han
Người hướng dẫn	PhD. Tran Tuan Kiet
Trường học	Ho Chi Minh City University of Technology and Education
Chuyên ngành	Civil Engineering
Thể loại	graduation project
Năm xuất bản	2022
Thành phố	Ho Chi Minh City

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Số trang	142
Dung lượng	17,06 MB