Biện pháp tổ chức thi công khối K0 cầu Hưng Hà (bảng tính + bản vẽ)

The socialist republic of viet nam
Céng hßa x· héi chđ nghĩa việt nam
Ministry of transport
Bộ giao thông vận tải
Project management unit no.1
Ban quản lý dự án 1

Hung ha bridge construction project
Dự án xây dựng cầu hưng hà

----------------------

Calcualtion sheet
Construction scaffolding of k0
Segment of the pier p25

Bảng tính
đà giáo thi công khối ®Ønh trô
Trô P25

Ha Noi, 01/2017


The socialist republic of viet nam
Céng hßa x· héi chđ nghĩa việt nam
Ministry of transport
Bộ giao thông vận tải
Project management unit no.1
Ban quản lý dự án 1

Hung ha bridge construction project

Dự án xây dựng cầu hưng hà

----------------------

Calcualtion sheet
Construction scaffolding of k0
Segment of the pier p25

Bảng tính
đà giáo thi công khối đỉnh trụ
Trụ P25

tư vấn giám sát-consultant

Ha Noi, 01/2017

Nhà thầu-contractor


hung ha bridge construction project
item : main bridge construction
calcultion sheet of scaffolding for k0 construction
A .GENERAL
I. general introduction
- Hung Ha bridge is located at KM28+335.105, a part of Project of Hung Ha bridge and approach road
construction connecting Ha Noi - Hai Phong expressway and Cau Gie - Ninh Binh expressway.

II. Design standard:
+ Design procedure of bridge construction and auxiliary equipment 22TCN - 200:1989.
+ Design standard of bridge and culvert 22TCN-18-79.

+ Design draflange of Hung Ha bridge
+ Other document (it will be cited when applying)
III. Material
- Shaped steel material
+ Yield strength

Fy =

+ Calculates strength

Ru =

2000 kG/cm2

+ Unit weight

s =

7850 kG/m3

+ Elastic modulus

Es =

235 Mpa

200000 Mpa

IV. Data
1. Load and load factor : (Table 1.13 - 22TCN - 200-89)

- Table 1.13 - Standard 22TCN 200 - 89 :
Load

Notation

Standard

Overload

Load of structure which being construction

nbt

1.0

1.1

Load of structure and auxiliary structure

npt

1.0

1.2

Load cause pouring and vibrating concrete

ndd

1.0

1.3

Load cause people and slight equipment

ntb

1.0

1.3

Unit

Quantity

T/m3

2.50

Length of flange

m

14.00

Concrete volume of left web

m3

19.28

Concrete weight of left web

T

48.21

Concrete volume of right web

m3

19.28

Concrete weight of right web

T

48.21

m

3.25

- Load applying to triangle scaffolding system of K0 for 1 cantilever arm:
Item

Notation

Flange of girder
Unit weight of Concrete

Cantilever part
Length of cantilever

Page 1


Volume of Bottom concrete
Weight of bottom concrete

BTD

m3

30.02

T

75.05

3

Concrete volume of left web

m

17.42

T

43.55

m3

17.42

T

43.55

m3

16.75

T

41.88

m3

17.48

T

43.70

Concrete volume of left web + cantilever

m3

8.95

Weight of left web

T

22.38

Concrete volume of right web + cantilever

m3

8.95

Weight of right web

T

22.38

m3

117.00

T

292.49

Concrete weight of left web

BTST

Concrete volume of right web
Concrete weight of right web

BTSP

Concrete volume of middle web
Concrete weight of middle web

BTSG

Volume of cap concrete
Weight of cap concrete

BTN

Concrete volume of 1 cantilever end
Concrete weight of 1 cantilever end
Inside scaffolding (cantilever)

DGT

T

3.49

Inside formwork (cantilever)

VKT

T

2.58

Scaffolding at 1 flange side

DGC

T

5.67

Formwork at 1 flange side

VKC

T

2.94

Scaffolding shoring bottom formwork

DGD

T

10.00

Bottom formwork

VKD

T

11.03

T

35.71

Total weight of scaffolding + formwork
2. Checking step.

- Checking of rigidity of separate part of scaffolding system under the affect of concrete weight
and gravity load of itself component.
- Checking of deformation after checking of rigidity;
- Checking of stability for compression bar of scaffolding system
with load and load combination which shown in table 19 - TCVN 200-89.
3. Checking items:
- (i) Calculation of internal force and checking load capacity of separate elements of scaffolding system
- (ii) Calculation of internal force and checking load capacity of high strength bar PC32
- (iii) Calculation of internal force and checking load capacity of pile cap (pier column)
under effect of tensile force of bar PC32

Page 2


b . Checking of structure
I. general layout

+21.659
B11

+21.640

+21.619

I200x100x5.5/8

I200x100x5.5/8

L = 6.42m

B11

L = 6.42m

I200x100x5.5/8
B2

B3
L = 12 m
Lan can thép

I300x150x6.5/9
L = 12.0m

Ván khuôn đáy
Formwork

Parapet
2I500x200x10/16

2H350&H300
B1'

L = 20.0 m

B1

L = 20.0 m
Triangle flame

Khung tam gi¸c
Thanh N2

Bar N2
Triangle flame

Thanh PC 36

Khung tam gi¸c

Bar PC36

Thanh N3
Khung tam gi¸c

Bar N3

Triangle flame
Bar N1

Thanh N1
H350x350x12/19

H350x350x12/19

C1

C1
L=6.925m

L=6.925m

Xuyên tâm / Radial D16
L100x100x10
l2
L = 1.579 m

I150x75x5/7
T6

L100x100x10
L1
L thay đổi

b

b

I200x100x5.5
B8
L = 5.34 m

Shape scaffolding
KT 1.5x1.5x1.5m

Đà giáo khoang
KT 1.5x1.5x1.5m

H300x300x10/15
B4

L = 15.0 m
H300x300x10/15

2I500x200x10/16
B1

L = 4.00 m
H300x300x10/15

L = 20.0 m
H300
G1

L = 5.904 m

L = 10 m

H300x300x10/15
L = 5.610 m

L­íi an toµn
Safety mesh
C1

H350x350

H350x350

L=6.925m

L=6.925m

C1

Page 3


- Expanded triangle frame structure of pier is final bearing structure of scaffolding for K0 structure
- The bearings of triangle frame work relying on friction with pier column under affect of PC32 tensile force
- Total 6 triangle frames deviding equally for 2 cantilever flanges.
3. Construction sequence
- Setting triangle frame and scaffolding for bottom formwork.
- Setting scaffolding, formwork for 2 flanges
- Pouring concrete for K0 segment stage 1
- Pouring concrete for K0 segment stage 2
- Pouring concrete for K0 segment stage 3

II. Calculation of beam system t6 - I150x75x5/7
1. General layout
b9

b9

L5

Shape scaffolding
KT 1.5x1.5x1.5m

B8

B4

H300x300x10/15

H300x300x10/15

B7

B7

L = 4.00 m

L = 4.00 m
H300x300x10/15

H300x300x10/15

L = 5.904 m

L = 5.904 m

B5

B5
H300x300x10/15

H300x300x10/15

L = 5.610 m

L = 5.610 m

B6

B6

2. Applied load
- Load applying to beam T6 include:
+ Uniform load of flange concrete
+ Uniform load of flange formwork
- Split 1 band width L=1.5m:
- Load of flange concrete at each section matched to flange thickness:
+ Flange thickness h3=0.25 m

Pbt3 = nbt.(L.h3).bt =

1.03 (T/m)

nbt : overload factor of concrete load

nbt =

1.1

bt : Unit weight of concrete

bt =

2.50 (T/m3)

Page 4


Pbt2 = nbt.(L.h2) =

+ Flange thickness h2=0.33 m

1.36 (T/m)

na : overload factor of concrete load
Pbt1 = nbt.(L.h1) =

+ Flange thickness h1=0.55 m

nbt =

1.1

nbt =

1.1

npt =

1.2

ndd =

1.3

ndd =

1.3

2.27 (T/m)

na : overload factor of concrete load
- Load of flange formwork:
Pvk = npt.L.0,06 =

0.11 (T/m)

npt : overload factor of structure which be building
Choose formwork have weight: 0.06T/m2
- Vertical load by vibrating of concrete mixture:
qdd = 200 (kg/m2)
Pdd = ndd.L.qdd =

0.39 (T/m)

ndd : overload factor of structure which be building
- Vertical load by people and slight equipment:
qng = 250 (kg/m2)
Png = nng.L.qng =

0.49 (T/m)

ndd : overload factor of structure which be building
3. Calculation diagram

Pbt1

Pbt2

Pbt3

Pvk+Pdd+Png

1500

1500

Use Midas 6.30, we have:

Page 5

Moment diagram (T.m)

Shear diagram (T)

Page 6


Bearing reaction diagram (T)

Deformation diagram (m)

Page 7


4. Checking of bar T6 :
a.Bar I150x75x5/7 has geometric feature:
L=

3.00

(m)

F=

17.90

(cm2)

Wx =

88.80

(cm3)

Jx =

666.00

(cm4)

Sx =

50.00

(cm3)

b. Internal force:
- Maximum moment applying to bar T6:
M=

0.68

(T.m)

2.46

(T)

- Maximum shear force:
Q=

c. Checking of bending-tensile strength:
- Checking formula:

σ max =

M max
 Ru
W

In there:
s :

stress

Mmax:

Maximum moment

Wx :

bending section modulus

R

steel intension. R = 1900 (kG/cm2).

:

766kg/cm2

s =



W=

<

R=

88.80

2000kg/cm2

(cm3)

OK



d. Checking of shear strength:
- Checking formula:

τ=

Q max ×S
 0.6×R u
b×J x

=

1200

(kG/cm2)

In there:
Qmax: Maximum shear force
S : Static moment

S=

50.0

cm3

b: web width

b=

0.50

cm

Jx: Inertia moment

Jx=

666.0

cm4

R: steel intensity. R = 1900 (kG/cm2).


 =

369kg/cm2

<

0,6.[R] =

1200kg/cm2



OK

e. Checking of deflection:
- Checking condition:
fmax  [L/250]

Page 8


Calculation deflection follow Midas:

f max =

1.00

(mm)

L=

1.50

(m)

[L/250]=

6.00

(mm)

<

[L/250]

6.00mm

Calculation length:
Allowable deflection:


fmax

=

1.00mm

OK



f. Checking of rigidity of compression diagonal strut bar:
+ The diagonal strut bar is compressed:

1.64 T

a.Shaped steel L100x100x10 has geometric feature:
Area of cross section

F=

19.0

cm2

Inertial moment

Jx =

175.0

cm4

Jy =

175.0

cm4

+ According standard, for material work steadily:

smax=P/ F ≤ [s]«d =j.[Ru]
+ In there :

- smax : calculation stress of maximum compression bar
- Ru : allowable stress of steel ; Ru = 2000( kg/cm2)
- j : stress reduction factor - Checking table
j : depending on material and slenderness l

λ=

μ.l
=
i min

i min =

J min
=
F

With

L=

104.06

3.03

157.90

cm

Least radius of gyration of beam cross section

m : Factor depending on connection type; with connection at 1 lugs m = 2


smax =

86.34

Checking tablej =
Kg/cm2

<

0.38

j.[Ru] =

760.00

Kg/cm2

OK

Page 9


II. Calculation of beam system B4 - H300x300x10/15 L = 15m :
1. General layout
trô p24

trô p26

pier p24

pier p26

GhÕ kÝch 250

I200x100x5.5/8
B9

L = 16 m
Đà giáo khoang
Formwork

H300x300x10/15
L = 15.0 m

B4

Lưới an toàn

Safety mesh

Hệ đà giáo tam giác

Hệ đà giáo tam giác

Triangle scaffolding system

Triangle scaffolding system

Th©n trơ

a

Pier body

2. Applied load
- Applied load to beam B4 is centre-point load which be transferred from leg of scaffolding for flange.
- Load of shaped scaffolding to 1 side of glange has value:
Pdg = npt.DGC =

6.80 (T)

npt : overload factor of formwork scaffolding
npt =

1.2

- Named bars B4 form inside of pier column to outside alternately B4-1 and B4-2
- Load from bearing reaction force of bar T6 transfering to bar B4 have alternately value:

PB4-1 =

1.77 (T)

PB4-2 = 4.61+1.16 =

5.77 (T)

So load of 1 leg of scaffolding transfering to B4:
Bar B4-1

Pn = Pdg /20 + PB4-1 =

2.11 (T)

Bar B4-2

Pn = Pdg /20 + PB4-2 =

6.11 (T)

3. Calculation diagram

Pn

Pn

Pn

Pn

Pn

Pn

Pn

Pn

Pn

Pn

Page 10


Use Midas 6.30, we have:
Maximum moment diagram (T.m)

Maximum shear diagram(T)

Page 11


Bearing reaction diagram (T)
With bar B4 - 1

With bar B4 - 2

Page 12

Deformation diagram (m)

4. Checking of bar B4 :
a.Bar H300x300x10/15 chas geometric feature:
L=

15.00

(m)

F=

119.80

(cm2)

Wx =

1360.00

(cm3)

Jx =

20400.00

(cm4)

Sx =

747.40

(cm3)

M=

3.42

(T.m)

Q=

6.26

(T)

b. Internal force :
- Maximum moment:

- Maximum shear :

c. Checking of bent member:
-Checking formula:

σ max =

M max
R

W

In there:
s :

stress

Mmax:

Maximum moment

Wx :

bending section modulus

R

steel intension. R = 1900 (kG/cm2).

:


s =

251kg/cm2

<

R=

W=

1360.00

(cm3)

2000kg/cm2



Ok

Page 13


d. Checking of shear member:
- Checking formula:

Q max ×S
 0.6×R =
b×J x

τ=

1200

(kG/cm2)

In there:
Qmax: maximum shear force

S : static moment

S=

747.4

cm3

b: web width

b=

1.00

cm

Jx: inertial moment

Jx=

20400.0

cm4



OK



OK

R: steel intension. R = 1900 (kG/cm2).
229kg/cm2

 =



<

0,6.[R] =

1200kg/cm2

f max =

1.00

(mm)

L=

3.20

(m)

[L/250]=

12.80

(mm)

<

[L/250]

12.80mm

e. Checking of deformation:
- Checking formula:
fmax  [L/250]
Midas result:
Calculation length:
Allowable deflection:
fmax



=

0.00mm

III. Calculation of frame system from 3 beams B5 , b6 , b7 - H300x300x10/15
1. General layout:
b9

b9

L5

Shape scaffolding
KT 1.5x1.5x1.5m

B8

B4

H300x300x10/15

H300x300x10/15
B7

B7

L = 4.00 m

L = 4.00 m
H300x300x10/15

H300x300x10/15

L = 5.904 m

L = 5.904 m

B5

B5

H300x300x10/15

H300x300x10/15

L = 5.610 m

L = 5.610 m

B6

B6

Page 14


2. Applied load
- Applied load to beam B4 is centre-point load which be transferred from leg of scaffolding for flange.
- Load from bearing reaction force of bar B4 have alternately value:
PB4-1 =

4.45 (T)

PB4-2 =

12.36 (T)

3. Calculation diagram
2331

1500

320

PB4-2

350

405

3955

700

200

PB4-1

Use Midas 6.30, we have:
Moment diagram (T.m)

Page 15


Longitudinal force diagram (T)

Shear diagram (T)

Page 16

Deformation diagram (m)

Bearing reaction diagram (T)

Page 17


4. Checking of bar B5, B6, B7 :
a.Bar H300x300x10/15 chas geometric feature:
L=

4.00

(m)

F=

119.80

(cm2)

Wx =

1360.00

(cm3)

Jx =

20400.00

(cm4)

Sx =

747.40

(cm3)

(Calculation length)

b. Internal force :
- Maximum moment:
M=

5.56

(T.m)

N=

21.37

(T)

Q=

13.67

(T)

- Maximum longitudinal force :

- Maximum shear :

c. Checking of bent member:
-Checking formula:

s :
F

σ=

M max N max

 Ru
W
F

stress
area of bar cross section

Mmax

Maximum moment

Nmax

Maximum longitudinal force

Wx

bending section modulus

Ru

steel intension. R = 1900 (kG/cm2).

Checking with load combination in case of most unfavorable loading
Bar Element
name

Material

Bend section
modulus
Wx

Area
F

Moment
My

cm3

cm2

T.m

Longitudinal
force
P
T

stress
s

Checking

kg/cm2

B5

H 300

1360.0

119.8

4.86

21.37

535.73

ok

B6

H 300

1360.0

119.8

5.56

16.03

542.63

ok

B7

H 300

1360.0

119.8

3.65

12.43

372.14

ok

1200

(kG/cm2)

d. Checking of shear strength:
- Checking formula:

τ=

Q max ×S
 0.6×R =
b×J x

In there:
Qmax: Maximum shear force
S : Static moment

S=

747.4

cm3

b: web width

b=

1.00

cm

Page 18


Jx: Inertia moment

Jx=

cm4

20400.0

R: steel intensity. R = 2000 (kG/cm2).
501kg/cm2

 =



<

0,6.[R] =

1200kg/cm2

f max =

0.10

(mm)

L=

3.60

(m)

[L/250]=

14.40

(mm)

<

[L/250]

14.40mm



OK



OK

e. Checking of deformation:
- Checking condition:
fmax  [L/250]

Calculation deflection follow Midas:
Calculation length:
Allowable deflection:


fmax

=

4.00mm

f. Checking of rigidity of compression bar:
- B5 and B6 are compression bars, however B5 have larger compression and its bearing diagram is more
unfavorable, so we check the slenderness and compression-bending rigidity for B5
+ Bar B5 - H300x300x10/15, L=5.904 m has compression force:

21.37 T

+ Shaped steel H300xH300x10/15 has geometric feature:
Area of cross section

F=

119.8

cm2

Inertial moment

Jx =

20400.0

cm4

Jy =

6750.0

cm4

+ According standard, for material work steadily:

smax=P/ F ≤ [s]«d =j.[Ru]
+ In there :

- smax : calculation stress of maximum compression bar
- Ru : allowable stress of steel ; Ru = 2000( kg/cm2)
- j : stress reduction factor - Checking table
j : depending on material and slenderness l
L=

μ.l
λ=
=
i min

With

i min =

J min
=
F

590.4

cm

78.65

7.51

Least radius of gyration of beam cross section

m : Factor depending on connection type; with connection at 2 lugs m = 1
Relative eccentricity :
i = e0 / r =



smax =

3.06

Checking tablej =
178kg/cm2

<

0.22

j.[Ru] =

440kg/cm2



OK

Page 19


IV. Calculation of bolt connection and located pressure to concrte at bearings of b5,b6,b7

1. General layout

chi tiÕt / detail "b"
b

Bold D32
L = 800mm

b
mặt cắt /section b - b

Bu lông D32
L = 800mm

2. Applied load

- Each bearing used 04 bolts M32 length L=0.8 m
- Applied load to bolt is bearing reaction from internal force diagram of above frame
- Bearing reaction has value alternate:
+ Upper bearing : G1-1 =

13.67 (T)

- Tension bolt

G1-2 =

2.24 (T)

- Shear bolt

+ Below bearing : G2-1 =

-13.67 (T)

G2-2 =

16.06 (T)

- Shear bolt

3. Checking :
a. Checking of rigidity of bolt D32:
- Bolt D32 has geometric feature:
l=

0.80

(m)

F=

8.04

(cm2)

Wx =

3.28

(cm3)

Jx =

5.15

(cm4)

Jy =

5.15

(cm4)

Page 20

+ Calculation of upper bolt D32 follow tensile condition:
When external force has parallel direction with bolt body, applying and spliting the element of connection,
cause tension bolt.
- Tensile resitance of bolt is calculated as follow:

 N  tb = Abn  ftb
Abn - actual area of bolt body cross section (deduction by thread)
Standard TCVN 1916-1995 - Abn =

5.6

cm2

(Bolt D32)

ftb - calculate strength of material when tension work
Standard TCVN 1916-1995 - Abn =
=> :

[ N ] tb =

13.44

2400

Kg/cm2

(Bolt grade 4.6)

(T)

- External force applying to 1 in 4 bolts of bearing has value:



N max=G1-1/4

= 3.42

(T)

Nmax =

3.42(T)

<

[ N ] tb =

13.44(T)

OK



So bolt ensuring tensile resistance.
+ Calculation of upper bolt D32 follow shear condition (sliding):
When external force has perpendicular direction with bolt body, the body press closely hole torus,
bolt and connection plate sliding relatively, cause shear bolt

- Shear capacity of bolt is calculated as follow:

 N  vb = A  fvb   b  nv
fvb - calculate shear strength of bolt material
fvb =

Table 1.10 appendix I - Steel structure

1500kg/cm2

b - work condition factor of bolt connection
b=

Table 2.8 - Steel structure

0.9

(Bu l«ng th«)

A - area of cross section of bolt body (no thread part)

=>

Table 2.9-Steel structure

A=

7.06(cm2)

(Bu l«ng D32)

d - diameter of bolt body

d=

3.20(cm)

(Bu l«ng D32)

nv - number of calculate section of bolt

nv =

1

[ N ] vb =

9.53

(T)

- External force applying to 1 in 4 bolts of bearing has value:



N max=G2-2/4

= 4.02

(T)

Nmax =

4.02(T)

<

[ N ] tb =

9.53(T)



OK

So bolt ensuring shear capacity.
b. Checking of located pressure concrete at connection position:
- Checking condition:

smax = Nmax/ F< Rem
- With

Nmax concrete surface pressure cause tensioning bolt D32

Page 21


Nmax =

13.67

(T)

F : area of surface pressure = 300x300
F=

900

(cm2)

Rem : located pressure strength of concrete, with concrete 35Mpa (pile cap)
Rem=


smax =

350
15(Kg/cm2)

(Kg/cm2)
<

Rem=

350(Kg/cm2)



OK

(Note : factors and table in calculation sheet are chosen accuracy following "Steel structure elementary structure" - Science and engineering Publisher. Dr. phạm văn hội (Ed) - Dr. nguyễn quang viên - Master phạm văn tư - Engineer lưu văn tr­êng )

Page 22


V. calculation of inside formwork scaffolding - bar b11
1. General layout
Ván khuôn bản nắp

Cao độ bê tông đợt 3 : 21.644

Formwork
I200x100x5.5/8
b11

Cao độ bê tông đợt 2 : +20.194

L= 6.420 m
Kích ren/Jack
I200
b10
L=3.03

Đà giáo định hình
Shape scaffolding

Lan can thi công
Parapet

Lưới an toàn

Safety mesh

Lưới an toàn
Safety mesh

Hệ đà giáo
tam giác
Thang thi công
Ladder
Thân trụ
Pier body

I200x100x5.5
L100x100x10
Lan can thép

l4

L = 6.420m

b11 b11

I200x100x5.5
L = 6.420m

L=0.949m

L100x100x10
l5
L=2.990m

Steel parapet
Đệm kích/Wedge
H=0.25m
Đà giáo định hình
Shape scaffolding

Nêm thép

Nêm thép

Wedge

Wedge

H300x300x10/15
L = 15.0 m

Thang thi c«ng
Ladder

Page 23

B4