Đồ án kĩ thuật thi công 2 tiếng anh đại học xây dựng construction

NATIONAL UNIVERSITY OF CIVIL ENGINEERING
DIVISION OF CONSTRUCTION TECHNOLOGY AND MANAGEMENT

CONSTRUCTION TECHNOLOGY II PROJECT
Requirement: Design assembly method for Industrial building using
prefabricated elements

Instructor
Student
Class
Student ID

VUONG DO TUAN CUONG
55XE

Hanoi, 11/2014
I.

ANALYZING TYPICAL FIGURES OF THE BUILDING


One-storey industrial building with 2 spans and 13 frames constructed by assembly method
from a range of different structure members including concrete columns, roof and wall panels,
crane runway beams, concrete roof frames, and opening windows. These components have
already been fabricated in factories, then delivered to construction site for installing.
Length of building: 12x6 = 72 (m), so there should be one settlement joint.
Width of building: L = 2 x 18 = 36 (m).
1.

Building cross-section:

BUILDING CROSS SECTION


BUIDLING PLAN
2.

Structure parameters:


General Information

Number of storeys

1

Number of spans

2

Number of column: A;B;C;D

13

Side-columns
Concrete columns
Middle columns

Concrete roof frame

Crane runway beam

H(m)

13.3

h(m)

10.0

W(T)

7.8

H(m)

13.3

h(m)

10.0

W(T)

8.7

L(m)

18

a(m)

2.45

W(T)

5.0

L(m)

6

h(m)

0.8

W(T)

3.3


Opening window

Roof and wall panel

l(m)

6

b(m)

2.6

W(T)

1.2

Dimension(m)

3x6

W(T)

2.3



II.

SELECTING EQUIPMENTS:

1. Selecting hanging and tying equipments:
1.1. Columns:
Columns is light and have brackets => use steel cable and friction belt to hang them
vertically.

a) Middle columns:
Tensile force is calculated by the formula:

Ptt
S =k×

m × n × cosβ

Qct
S

Legend:
k – Safety ratio, k = 6

Ptt = 1.1× 8.7 = 9.57 T
β - Inclined angle of cable and vertical direction, β=00
m - Ratio related to the difference in the value of tensile force
within the two branches of cable, m=1
n - Number of steel cable, n=2
=> S = 6 ×

9.57
= 28.71 T
1× 2 × Cos 0

Selecting cable 6x37x1 with diameter D=26mm, tensile strength (140kg/cm2), failure force
F= 29000 (kg)
 Cable weight: γ=2.38 kg/m


 Length of each steel cable: lbc= 1.5 + 3.3 = 4.8 (m)
 Frictional belt weight: qfb=30 kg
 Weight of hanging and tying equipment:
qe = 2 × γ × lbc + q fb = 2 × 2.38 × 4.8 + 30 = 52.85 kg

b) Side-columns (C1)

Tensile force is calculated by the formula:

S =k×

Ptt
m × n × cosβ

Qct
S

Legend:
k – Safety ratio, k = 6

Ptt = 1.1× 7.8 = 8.58 T
β - Inclined angle of cable and vertical direction, β=00
m - Ratio related to the difference in the value of tensile force
within the two branches of cable, m=1
n - Number of steel cable, n=2
=> S = 6 ×

8.58
= 25.74 T
1× 2 × Cos 0

Selecting cable 6x37x1 with diameter D=26mm, tensile strength (140kg/cm2), failure force
F= 29000 (kg)
 Cable weight: γ=2.38 kg/m
 Length of each branch of steel cable: lbc= 1.5 + 3.3 = 4.8 (m)
 Frictional belt weight: qfb=30 kg
 Weight of equipment for hanging and tying:

qe = 2 × γ × lbc + q fb = 2 × 2.38 × 4.8 + 30 = 52.85 kg


1.2. Crane runway beam
Crane runway beam (CRB) is a structure member working in horizontal direction, so we
choose an ordinary equipment for hanging and tying with semi-automatic lock. The way how
to hang and tie is shown in the following figure

where:
1234-

Steel cushion block
Steel cable
Semi-automatic lock
Pipe section for inserting cable

Crane runway beam (CRB) is a structure member working in horizontal direction, so we
choose an ordinary equipment for hanging and tying with semi-automatic lock.
Tensile force of cable is determined by the below formula:

S =k×

Ptt
m × n × cosβ

Legend:
k – Safety ratio, k = 6

Ptt = 1.1× 3.3 = 3.63 T
β - Inclined angle of cable and vertical direction, β=450

m - Ratio related to the difference in the value of tensile force within the
two branches of cable, m=1


n - Number of steel cable, n=2
=> S = 6 ×

3.63
= 15.4 T
1× 2 × Cos 45

Selecting cable 6x37x1 with diameter D=17,5 mm, tensile strength (140 kg/cm2), failure
force F= 12750 (kg)
Length of each branch of steel cable: l= 1.5 + 2.4/Cos(45) = 4.89 (m)
Cable weight: γ=1.06 kg/m
Frictional belt weight: qfb=30kg
 Weight of equipment for hanging and tying
qe = γ.l + qfb= 1.06 x 4.89 x 2 + 30 x 2= 70.84 kg = 0.071 T
1.3. Concrete roof panels:
Panels have 2 dimesions 3x6 (m) with the weigh of 2.3 T ,we use four steel cables with a
self-balacing ring.

Tensile force in each cable is calculated by:

S =k×

Ptt
m × n × cosβ

Ptt = 1.1× 2.3 = 2.53 T

m=0,75 (four cables-branches)
n=4
β= 450 (for safety purpose, calculate with β=450)
=> S = 6 ×

2.53
= 7.16 T
0, 75 × 4 × cos 45

Selecting cable 6x37x1 with diameter D=13mm, tensile strength (140kg/cm2), failure
force F= 7200 (kg)


 Length of each steel cable: l= 6 (m)
 Cable weight: γ=0.59 kg/m)
 Weight of equipment for lifting:
qe= 0.59 x 6 x 4= 14.2 (kg) = 0,01 (T)

1.4. Concrete wall panels:
Panels have 2 dimesions 3x6 (m) with the weight of 2.3T ,we use a cluster of two
cables with a self-balacing ring.

steel

Tensile force in each cable is calculated by:

S =k×

Ptt
m × n × cosβ

Ptt = 1.1× 2.3 = 2.53 T
m=1
n=2
β= 450 (acctually β do not equal to 45 0,but higher. However, for safety
reason, calculate with β=450)
=> S = 6 ×

2.53
= 10.73 T
1× 2 × cos 45

Selecting cable 6x37x1 with diameter D=17.5mm, tensile strength (140kg/cm2), failure
force F= 12750 (kg)
 Length of each steel cable: l= 4.5 (m)
 Cable weight: γ=1.06 (kg/m)


 Weight of equipment for lifting:
qe= 4.5 x 1.06 x 2= 9.54 (kg) = 0,01 (T)
1.5. Steel frame:
Because contructing these components at high place, steel roof frames need to be preerected on the ground in order to ensure safety and then be hanged and installed
simultaneously. Tools for hanging and tying are equipped semi-automatic lock and selfbalancing ring.
The steel roof frames have long span, so hanging bar is used to assist the process of
erecting. Supported positions are chosen at joints of frames to avoid moment and shear force
appearing within frames.

The steel roof frames have L=18m in length, we choose hanging bar with code 195946R11 ( [Q]= 10T, G=0,455 (T)
The tensile force in each branch of two cables is determined by the below formula:

S = k×

Ptt
m × n × cosβ

Legend:
Ptt= 1,1.P= 1.1 x (5+1.2)= 6.82 (T)
n= 2
m= 1
β=30o
 S = 6×

6.82
= 23.63 T
1× 2 × cos300


Selecting cable 6x37x1 with diameter D=24 mm, tensile strength (140 kg/cm2), failure
force F= 24300 (kg)
Length of each branch of steel cable l= 5.0 (m)
Frictional belt weight: qfb=30 kg
Cable weight: γ=1.99 kg/m
 Weight of lifting equipment
qe= γ.l + qfb+ G= 1.99 x 5 x 2 + 30 x 2 + 455= 534.9 (kg) = 0.53 (T)
2. Lifting and installing data calculation:
Selecting crane tower base on the following parameters:
Hrq – height of hook
Lrq - length of working jib
Qrq - lifting capacity
Rrq - working radius

2.1. Calculating lifting & installing data of column
Assembling column without obstacles
=> working jib with αmax = 750
a) Middle columns:
Lifting capacity :
Qrq = Ptt + qe = 9.57 + 0.05= 9.62 T
Using geometric method we have crane data diagram :
Requirement height:
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 0
a : distance from ground to height of lifting cable a = 0.5 m
hm : height of element hm = 13.3 m
he : length of lifting cable he = 1.5 m
hh : height of pulley, hook hh = 3.47 m
→ H rq = 0 + 0.5 + 13.3 + 1.5 + 3.47 = 18.77( m)


Jib length :

H rq − hc

18.77 − 1.5
= 17.80(m)
sin 75
sin 750
Where: hc – level of the slewing ring of the crane (from the ground surface).
Lrq =

0

=

Working radius of jib :
S = Lmin × cos 75 = 17.8 × cos 75 = 4.6(m)
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 4.6 + 1.5 = 6.1 m
b) Side-columns:
Lifting capacity: Qrq = Ptt + qe = 8.58 + 0.05 = 8.63 T
Requirement height:
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 0
a : distance from ground to height of lifting cable a = 0.5 m
hm : height of element hm = 13.3 m
he : length of lifting cable he = 3.47 m
hh : height of pulley, hook hh = 3.47 m
→ H rq = 0 + 0.5 + 13.3 + 1.5 + 3.47 = 18.77( m)
Jib length :


H rq − hc

18.77 − 1.5
= 17.80(m)
sin 75
sin 750
Where: hc – level of the slewing ring of the crane (from the ground surface).
Lrq =

0

=

Working radius of jib :
S = Lmin × cos 75 = 17.8 × cos 75 = 4.6(m)
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 4.6 + 1.5 = 6.1 m
2.2. Calculating lifting & installing data of crane runway beam

Assembling of crane runway beam without obstacles
=> Select working jib follow : αmax = 750
Lifting capacity : Qrq = Ptt + qe = 3.63+ 0.07 = 3.7 T
Requirement height:
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 10 – 0.7 = 9.3 m
a : height of lifting element a = 0.5 m


hm : height of element hm = 0.8 m
he : length of lifting cable he = 2.4 m
hh : height of pulley, hook hh = 1.5 m
→ H rq = 9.3 + 0.5 + 0.8 + 2.4 + 1.5 = 14.5( m)
Working radius :
Lrq =

H rq − hc

sin 750
With hc – level of the slewing ring of the crane (from the ground surface).
14.5 − 1.5
→ Lmin =
= 13.45(m)
sin 750
Working radius of jib :
S = Lmin x cos750 = 13.45 x cos750 = 3.48 m
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 3.4 + 1.5 = 4.98(m)

2.3. Calculating lifting & installing data of roof frame and opening window:
Assembling of crane runway beam without obstacles so we choose working jib as follow :
αmax = 750


Lifting capacity :
Qrq = Wm + qe = 1.1( Prf + Pow ) + qe = 6.82 + 0.53 = 7.35 T
Requirement height of hook :
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 13.3 – 0.7 = 12.6 m
a : height of lifting element a = 0.5 m
hm : height of element hm = 2.45 + 2.6 = 5.05 m
he : length of lifting cable he = 1 m
hh : height of pulley, hook hh = 1,5 m
→ H rq = 12.6 + 0.5 + 5.05 + 1 + 1.5 = 20.65( m)
Working radius :

Lrq =

H rq − hc

sin 750
With hc – level of the slewing ring of the crane (from the ground surface).
20.65 − 1.5
→ Lmin =
= 19.93( m)
sin 750
Working radius of jib :
S = Lmin x cos750 = 19.93 x cos750 = 5.16 m
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 5.16 + 1.5 = 6.66(m)










2.4. Calculating lifting & installing data of roof panel
Without fly jib HL = 17.75 m

















-0.3






Lifting capacity :
Qrq = Ptt + qe = 2.53 + 0.01 = 2.54 T
Requirement height of hook :
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 13,3 – 0.7 + 2.45 + 2.6 = 17.75 m
a : height of lifting element a = 0.5 m
hm : height of element hm = 0.4 m
he : length of lifting cable he = 6cos(45)=4.24 m

hh : height of pulley, hook hh = 1,5 m
→ H rq = 17.75 + 0.5 + 0.4 + 4.24 + 1.5 = 24.39(m)
Hch = HL + a + hm = 17.75 + 0.5 + 0.4 = 18.65 m
Best jib angle :
H −h
18.65 − 1,5
α tw = arctg 3 ch c = arctg 3
= 58.390
e+b
1+ 3
Working radius :
H ch − hc
e+b
+
0
sin 58.39 cos 58.390
With hc – level of the slewing ring of the crane (from the ground surface).
Lrq =


→ Lmin =

18.65 − 1.5 1 + 3
+
= 27.79(m)
0,851
0,524

Working radius of jib :
18.65 − 1.5

cos 58.390 = 10.55(m)
sin 58.390
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 10.55 + 155 = 12.05(m)
S=

Using fly jib αmax = 750






























-0.3








 

α tw = arctg 3

H ch − hc
= 750
'
e+b−l

→ l ' = 3.67 m
l = lmcos30 ( inclination angle between fly jib and horizontal direction)
Working radius :
’

0

H ch − hc e + b − l '
+
sin 750
cos 750
With hc – level of the slewing ring of the crane (from the ground surface)..
18.65 − 1.5 1 + 3 − 3.67
→ Lmin =
+
= 19.03(m)
0.966
0.259
Lrq =


Working radius of jib :
19.03 − 1.5
cos 750 + 3 + 1 = 8.70(m)
0
sin 75
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 8.70 + 1.5 = 10.2(m)
S=

2.4. Calculating lifting & installing data of wall panel

Assembling of crane runway beam without obstacles

=> Select working jib follow : αmax = 750
Lifting capacity : Qrq = Ptt + qe = 2.53+ 0.01 = 2.54 T
Requirement height:
Hrq = HL + a + hm + he + hh
With :
HL: height to install element HL = 13.3– 0.7 =12.6 m
a : height of lifting element a = 0.5 m
hm : height of element hm = 3 m
he : length of lifting cable he = 4.5cos(45)=3.18 m


hh : height of pulley, hook hh = 1.5 m
→ H rq = 12.6 + 0.5 + 3 + 3.18 + 1.5 = 20.78( m)
Working radius :
Lrq =

H rq − hc

sin 750
With hc – level of the slewing ring of the crane (from the ground surface).
20.78 − 1.5
→ Lmin =
= 19.96(m)
sin 750
Working radius of jib :
S = Lmin.cos750 = 19.96 x cos750 = 5.17 m
Minimum working radius of crane:
Rrq = S + r
→ Rrq = 5.17 + 1.5 = 6.67( m)
3. Crane selection:

CRANE SELECTION
Required parameters

ID

Component
Middle
Side

Qrq

Rrq

Hrq

Lmin

(T)
9.62
8.63

(m)
6.1
6.1

(m)
18.77
18.77

(m)
17.8
17.8
13.1
5
19.9
3
19.0
3
19.6
5

1

Column

2

Crane ruway beam

3.7

5

14.2

3

Roof frame

7.35

6.66

20.75

2.54

10.2

24.39

2.54

6.67

20.48

4

Roof panel

5

Fly jib

Wall panel

III.

Crane parameters

Q

R

H

(T)
9.7
8.7

(m)
7.5
8

(m)
21
20.5

MKG-10/18m

3.7

6

18

MKG-25BR/28.5m

7.5

9

27

MKG25BR/28.5m/5m

2.6

16

28

E1001D/25m

2.6

14.
5

21

Type of crane
MKG-25BR/18.5m

CONSTRUCTION METHOD:

Determine crane position based on lifting parameters of crane and construction site.
•

From crane operating diagram => Rmin (minimum radius for lifting elements) and
Rmax.

•

Determine working range of crane.

•

Determine moving diagram of crane.

1. Columns installation:
Crane MKG-25BR (L=18.5m) has Rmin=4.5m; Rmax=7.5m and 8m for middle and side
columns respectively.(self-weight of side and middle columns are 8.7 and 9.7 respectively.
1.1. Crane position:


For each axis A, B, or C, number of postions for crane to install columns is:
n=

14
= 5 positions
3

=> There are 5 x 3=15 postions in total to install all columns for the building.

1.2. Construction method:
Preparation:
Deliver columns to construction site by trucks. Use crane to lay them on site as shown

in the figure .
Mark column centers on foundation surface. Prepare supporting equipment.
Check columns’ dimensions, connection between columns and crane runway beams.
Prepare aggregates for concrete mixing batches.
Installation method :
Install lifiting system onto the columns’bodies by using friction belt, pour a thin
concrete layer into the foundation socket.
Hang lifting system onto the crane’s hook to lift the column 0.5m above the ground.
Use rope system to pull columns into their positions on the foundation. Slowly lower
the columns onto the foundation socket.
Use 5 wedges and 4 cables to temporarily fix each column. Adjust column’s level and
position.
Clean the columns’ bottoms and fix them to the foundation with fast setting mortar.
Fixing process includes 2 phases:
- Phase 1 : Pour mortar right beneath the wooden wedges.


-

Phase 2 : After concrete mortar reaches 80% strength, remove wooden wedges and
fill the foundation socket with mortar.

2. Crane runway beam installation:
2.1. Crane position:


Crane MKG10 (L=18m);with Q=3.7T => Rmin=4m; Rmax=6m.
Each position can be used to install 2 Crane runway beams.
=> There are 6 x 3=18 postions in total to install all crane runway beam for the building.

2.2. Construction method:
Preparation :
Transport CRB to construction site by trucks.


Check CRBs’ dimensions, connecting bolts.
Check the hanging system.
Prepare connecting components such as bolts, welder, welding machine.
Installation:
Hang and lift CRB up. Use rope system to adjust CRB positions on column’s console.
Use steel plate to adjust level of CRB (if necessary).
Use bolting and welding connections to fix the CRBs to the columns.

3. Roof frame and opening window installation:
3.1. Crane positions: