construction organization content design construction schedule and site logistics

Cast-in-place concrete is selected to be used in this project. Vertical transport equipment:Required height of tower crane:Hyc=hct + hat + hck + ht = 33+1+1.5+1.5=37mWhere:hct: height o

DEPARTMENT OF CONSTRUCTION TECHNOLOGY AND MANAGEMENT

MINI - PROJECTCONSTRUCTION ORGANIZATION

Content: Design Construction Schedule and Site Logistics

Supervisor: Pham Nguyen Van Phuong

Hanoi, 2021

1.Introduction to the building

- Structure of the building is monolithic reinforced concrete structure The building has 9storeys and 12 spans All the geometric data of the building is illustrated in the below table.

- Construction and site conditions of the building:

 Geological properties: soil grade II with high strength of soil There’s no need toreinforce foundation pit.

 The level of water table is by far greater than the level of foundation pit

Reinf.Ratio µ% 1

3.Foundation structure

a) Depth of foundation:

H = 3t = 120 + m*10 = 120 + 3*10 = 150 (cm) m  choose t = 50 (cm)b) Dimensions of foundation are chosen following to the formula below:

;

*) At axis A&E:+

Thus, choose:

*) At axis B&D:+

Thus, choose:

*) At axis C:++

Thus, choose:

4 Ground

Thickness of lining concrete:

Choose: Thickness of reinforced concrete

Choose:

- Heat proofing :12+n/3=12+4/3=13.33Choose : 15 cm

- 2 layers of traditional terra-cotta floortile ( ceramic tile)

6 Wall

- Plaster 40% of external wall area ; 50% of internal wall area.

- Paint 6% of external wall area ; 1% of internal wall area.

- Doors ( and windows) take 60% of external wall area ; 10% of internal wall area.

- Power, water factor 0,32 working hour per 1m of floor. : 2

7 Construction plan and location of building

a) Construction plan

Figure 10: Construction plan

Figure 11: Building location on construction plan

X1 =10+5n=10+5x4= 30 (m) X2 =15+n/2=15+4/2= 17 (m)

Table 226: Labour resource for formwork dismantlement of each partition

QuantityTotal 100(m2)

11

Use tower crane as vertical transport equipment.

Because the length of the building is quite great, hence, tower crane with large radius will bearranged

Cast-in-place concrete is selected to be used in this project. Vertical transport equipment:

Required height of tower crane:

Hyc=hct + hat + hck + ht = 33+1+1.5+1.5=37(m)Where:

hct: height of buildinghat: safety distance

hck: height of structure componentsht: height of hanging kitsWorking range of tower crane:

L=43.42m: the length of the building

: width of scaffold : safety distance

b=1m : distance from centre of tower crane to the edge of tower crane

 Select tower cranebase on the following requirements:- Maximum operating radius: R =31.43 (m)yc Hoisting height: H=37 (m)

(ZoomLion’s product catalouge) with following parameters:- Load capacity: Q=2.8(T)

- Maximum operating radius: R=35 (m)- Hoisting height: H=40.8 (m)- Speed:

o Hoisting speed: 48 m/mino Lowering speed: 48 m/mino Trolleying speed: 30-60 m/mino Slewing speed: 0-0.72 r/min

 Cycle of tower crane: 1

Where:

(3 4)

E is simultaneous gesture factor E=0.8

t1=10(s): time for hanging the bucket on the lifting hook.: hoisting time.

60 4 54( )0,6

: slewing time to pouring position.: time for trolley get to pouring position.

: time for lowering the bucket to constructing position.

t6 = 120s: pouring concrete time.

: time for lifting bucket to former position : time for trolley come back to the start point.



- Productivity of tower crane per working shift (for concrete transporting only)N=Q x n x k x k x Tck tttg

Legend: +) t =0.8 x 408.175=326.54(s) : time period of a cycleck +) n =3600/326.54= 11.02: number of cycle in an hourck +) k =0.8 : weighting factortt

+) k =0.9 : factor considering time-used of cranetg +) T=8 (h): duration of a working shift => N= 2.8 x 11.02 x 0.8 x 0.9 x 8 = 177.73(T/shift)*Volume to be transported in 1 shift:

+ Concrete volume: Q = 39.83 x 2.5 = 99.575 (T)bt+ Weight of reinforcement: Q = 3.57 (T)ct

+ Weight of formwork: Q = 762x0.03x0.65=14.859 (T)vk+ Maximum total volume to be transported in 1 shiftQ = 99.575 + 3.57+ 14.859= 118.004 T < Qtower-crane = 177.73 TSo the selected tower crane ensures productivity

6 Choose a concrete mixer

- Using commercial concrete.

C ROOF CONSTRUCTION1 List of tasks

1 Pouring waterproof concrete layer 50mm2 Pouring heat resistant concrete layer 150mm

4 Paving the ceramic tiles

Figure 25: Cross-section of roof layers

2 Calculation of work volume

Table 27: Work volume of roof construction

No Name of tasks

Unit VolumeLengt

3 Brick wall (110mm thick) 122.56 0.11 1.1 m³ 14.83

2 Heat resistant concrete m³ 116.32 AF.32310 1.66 193.093 Brick wall (110mm thick) m³ 14.83 AE.81310 1.72 25.51

3 Division of construction segment

Table 29: Work volume of each partition

Unit AmountArea

Waterproof concrete 202.79 0.05 m³ 10.14Heat resistant concrete 202.79 0.15 m³ 30.42Brick wall (110mm thick) 4.50 1.1 m³ 4.95

Waterproof concrete 173.81 0.05 m³ 8.69Heat resistant concrete 173.81 0.15 m³ 26.07Brick wall (110mm thick) 2.14 1.1 m³ 2.35

day

Man-1 Heat resistant concrete m³ 30.42 AF.32310 1.66 50.50Brick wall (110mm thick) m³ 4.95 AE.81310 1.72 8.51

Waterproof concrete m³ 10.37 AF.32310 1.66 17.21Heat resistant concrete m³ 31.10 AF.32310 1.66 51.63Brick wall (110mm thick) m³ 2.76 AE.81310 1.72 4.75

Waterproof concrete m³ 9.67 AF.32310 1.66 16.05Heat resistant concrete m³ 29.00 AF.32310 1.66 48.14Brick wall (110mm thick) m³ 2.62 AE.81310 1.72 4.51

Heat resistant concrete m³ 26.07 AF.32310 1.66 26.68Brick wall (110mm thick) m³ 2.35 AE.81310 1.72 4.04

D FINISHING WORK

1 Work breakdown structure

1 Constructing inside wall-110

3 Installing MEP system4 Plastering inside wall-1105 Painting inside wall-1106 Installing door and window7 Plastering outside wall8 Painting outside

2 Calculation of work volume

a) Brick wall and installation of doors and windows

- Brick wall run along all axes: 220mm for exterior wall and110mm for interior wall.

- Doors and windows take: 60% of external wall and 10% of internal wallTable 231: Volume of 110mm-thick internal wall and area of internal doors

Volumedoors brick

wallofbrick

110-Internal wall: 50% plastered and 220-external wall: 40% plastered

Table 253: Area of plastering for 110- internal wall

Dimensions (m)

Area of Area ofbrickwall plasterHeight Lengt

6 2-12(spanBC-CD) 3.2 3.25 205.92 102.96

wall plasterHeight Lengt

c) Inside painting and outside painting

- Inside building is painted 2 layers: first is base layer and second is cover layer 1% of internal wall is painted.

- Outside building is painted 2 layers: first is base layer and second is cover layer 6% of external wall is painted.

Table 275: Area of inside painting

Area of Area ofbrick painting

Height Length

paintingHeight Length (m²) (m²)

CD)

d) Mechanical, electricity and pipe installation

MEP task consumes 0.32 working hour per 1m of floor Base on the area of floor, labour2resource can be determined.

Table 297: Labour resource for MEP

Element Dimension (m) Quantity NumberArea

floor Norm hoursMan Man-dayLength Widtth per floor of floor (m²) (hour)

3 Calculation of labour resource

Table 308: Labour resource for finishing tasks ( 9 floors)

day1 Building 110-wall m³ 867.33 AE.81310 1.72 1491.81

5 Doors & windows m² 2905.21 AH.32111 0.25 726.30

4 Choose hoist

- Mass of wall built in 1 shift is: =(867.33+297.56)/(9x4)=32.36 m3

-According to basic construction norms, 1m3 of walls need 550 bricks, 0.3m3 of mortar.1m3 has a weight of 2T

-Total weight is Q = 32.36x2 = 64.72 T/shift1

- The volume of plastering in 1 shift is =3944.19/(9x4)=109.56 m2

-According to the basic construction norm, 1m2 of plaster requires 0.02m3 of mortar,1m3 has a weight of 2T.

-Total weight is Q = 109.56 x 0.02 x 2 = 4.38 T/ca2-Total weight to lift is Q = Q1+Q2= 64.72+4.38=69.1 T/shift

+ Lifting load: 0.5 T+ Design lifting speed: 7 m/s+ Maximum lifting height: 50 m+ Machine weight: 5.7 billion

+ Capacity of the hoist is calculated by the formula: N = q x n x k x k12Where:

+ k1 = 0.8 coefficient of using the hoist.+ k2 = 0.8 time utilization factor.+ q = 0.5 (T)

+ n = 3600 / Tck with Tck = t + t + t + t1234

+ t : loading and unloading time , t = 3 minutes = 180 s11+ t : lifting and lowering time, t = 2x24/7= 7 s22Tck = 180 +7= 187 s

Instead: n = 3600 / 187= 19 cycles/h.So: N = 0.5x19x0.8x0.8 = 6 (T/h)

Productivity in 1 shift: Nca = 8 x 6 = 48 (T) So we choose these 2 hoists to satisfy theworking requirements Arrange the hoist at the positions as shown on the constructionsite drawing, ensuring convenience for construction.

*) Hoist transporting people

In addition, to serve traffic to high floors, we also use a PGX (800-16) passenger hoist.

5.Choose Mortar Mixer

The volume of mortar for each segment of each work is calculated as the table below:

Table 319: Schedule calculation for one floor

4 Design warehouse for materials and components.5 Design temporary houses on the construction site.6 Design of electricity and water supply networks.

7 Design the system of occupational safety and environmental sanitation.

3 CONSTRUCTION ORGANIZATION FOR MAIN TASKS:3.1.Locate and plan layout on site:

-Based on given data, the construction site layout is presented below:

Y1 = 34 (m) Y = 70 (m)2

Use fixed-tower crane There are regulations while using tower crane such as:

- The standing of the crane must be the most beneficial in terms of working, convenient in hoisting, mounting or transmaterials, components, etc., with a large reach covering the entire project.

- The standing of the crane must ensure the safety of the crane, the construction site and the workers, and facilitate theand dismantling of the crane.

- Economical: make full use of the crane power, have a reasonable service radius, high productivity

3.3.Traffic road calculation

Construction site traffic system includes temporary road system, built for construction work The temporary road system inc- Off-site road: is the road connecting the construction site with the existing public road network.

- On-site road: is the network of roads within the construction site, also known as internal roads.

When designing the construction site road, it must comply with the current regulations of the Ministry of Transport and otheregulations of the state At the same time, when designing the construction road network plan, it is necessary to follow the gprinciples as follows:

- Maximize the use of existing local roads and combine the use of permanent roads to be built, under the project's plannpre-constructing a part of these roads, to serve the construction.

- Based on the diagrams and flow of goods transportation to rationally design the road network, ensure convenient transof materials, equipment, etc., and reduce the number of loading and unloading times to the maximum.

- To ensure safety and increase transportation productivity, it is advisable to design a one-way street when possible.- Under normal conditions, with a 1-lane road, the parameters of the road width are taken as follows:

- Road width: b = 3,75 (m)- Side road width: c = 2.1,25 = 2,5 (m)- Road-bed width: B = b + c = 6,25 (m)

- The curved radius of the line where the angle is taken is R = 15(m) - Road slope: i = 3%

- Road material.

- Thoroughly level the ground, then spread a layer of sand 15-20(cm) thick, carefully compact the rocks about 20-30(cmof the rocks and spread 4x6 rocks.

4 TEMPORARY HOUSE

4.1.Number of labours and staffs

- There are 5 main groups of labour on site:

2 Mixed material layer

- Based on the schedule diagram, the average number of type A labour is: A= N =93 (people)avr

- Assume the building is constructed in urban area (k%=25%), the number of type B labour is: B= k% x A =25%24( people)

- The number of type C labour is:

C= 6% x ( A+B)=6% x (93+24) = 7(people)- The number of type D labour is:

D= 5 % x ( A+B+C )= 5% x (93+24+7) = 7 (people)- The number of type E labour is:

E= 5 % x ( A+B+C+D) =5% x (93+24+7+7) =7 (people)

The total number of labour (including 2% of sick and 4% of temporary leave labour) is:G=1.06x(A+B+C+D+E)=1.06 x ( 93+24+7+7+7) =147 (people)

4.2.Temporary house area

- Accommodation for labours (minimum 4m2/labour): S =(A+B)x4=(93+24)x4 =468 (m )1 2

- Temporary house for technical staffs/engineers and office staffs (minimum 4m /staff):2 S =(C+D)x4=(7+7)x4 =52

- Temporary bathroom for labours (minimum 2.5m room/25 staff):2n=147/25 =6( room)

S3 =6 x2.5 = 15 (m )2

- Temporary canteen for labours (50m /100 labours):S4= 147/100 x50=73.5 (m )2

- Temporary WC for labours (minimum 2.5m room/25 labours):2n= 147/25 = 6(room)

S 

- Temporary head office: 28 16 (m )

S 

5 MATERIAL STORAGE AREA

The plan of storage is storing construction material enough for constructing one storey The mortar used for constructingand plastering is mark 75 cement-sand mortar The selected cement and sand are PC30 cement and fine black sand.

a) Determine the amount of materials to be stocked (Qdtr): depends on the following factors:- Time to receive materials and transport to the construction site is t = 1 day.1

- Time of loading and unloading is t = 1 day.2- Time for testing and classifying materials is t = 1 day.3- The time between receipts was t = 4 day.4- The reserve period is t = 5 days.5T== 12 days.

The amount of reserve material is determined by the formula: Q = qi.TWhere:

- Q : is the amount of material stored.idtr

- q : is the maximum amount of materials consumed daily.i- T : number of days of reserve, we take T = 6 (days).Sand, brick, stone take T = t1 + t5 = 6 daysdt- Reinforcement work: 3.57 T (segment 2 floors 1)-Concrete work: 39.81 m (segment 2 floors 1)3- Formwork work: 762 m (segment 2 floors 1)2

We have: the norm of 1m of concrete grade 250 has a grade of3 + 0.415 T of cement

+ 0.455 m3 of coarse sand+ 0.887 m3 of stone

Derive the amount of material for concrete in 1 shift:+ 16.52 T of cement

+18.11 m3 of coarse sand+ 35.31 m3 of stone- Wall construction work+ Wall 220 : 8.27 m3+ Wall 110 : 24.09 m3

Wall- 200 550 0,29

 Volume:

+Brick: 8.27x550+24.9x643=20560+Mortar: 8.27x0.29+24.09x0.23=7.939 (m )3Use mortar grade 75, norm 1 m3:

 Volume:

+Cement: 7.939x0.32=2.54 (T)+Black sand:7.939x1.09=8.65 (m )3-Plastering work

Plastering workArea (m )2Thickness (cm)Volume (m )3

- Use mortar grade 75, norm 1 m3:

Fine sand: 1.7 * 1,05 = 1.785 (m )3

-Formwork task:

+Formwork: 762 m => 22.86 m23 +Including shores: 1.4x22.86=32.004

-Volume of material used in a shift:.

Stored volume of materials:

No Material Unit Volume /1 day T ( day )dt Volume

- Storage area F is the area that directly used to store material:(m )2Where:

d: Quantity of material can be stored per m2

Qmax: Maximum estimated quatity of material at site storage

- Storage area S, including acess for loading, unloading, fire safety, etc.:S=αFWhere: α: Utilization factor

α =1.5-1.7: for general storehouses α =1.4-1.6: for warehouse

α =1.2-1.3: for storage yards storing structure members, component, etc α =1.1-1.2: for storage yards storing piles of material like sand, brick, etc.

No Material Unit Stored volume

Type of storage

Volume material/m² α

SelectS (m²)

6.2 Power of electrical appliligting equipment

Total(kW)