EVNGENCO3/VTTPMB Vinh Tan 4 Thermal Power Plant

DMPP Consortium

Calculation Sheet for Fire Fighting Pump House

A. APPLIED STANDARD
- TCVN 4453:1995 Monlithic concrete and reinforced concrete structures - Codes for construction,
check and acceptance

(Appendix A - Data to design scaffolding formwork for concrete and reinforced concrete
monolithic)

- TCVN 2737 :1995 Loads and operation. Standard for design

- TCXDVN 5574 : 2012 Concrete structure and Reinforced concrete structure. Standard for design.

- TCXDVN 5575 : 2012 Steel structure. Standard for design

B. FORMWORK DESIGN:

I. SLAB FORMWORK:

Using plywood class VI:

σ  135kG/cm  2

γ  450kG/m  3


E  1.1  10 5 kG/cm 2 
Wood plate is formed by small wood plates. The thickness of wood plate is 1.8 cm.
The space between girders is calculated to ensure flexure & deformed.

1. Loads:

No Load Formula n qtc(kG/m2) qtt(kG /m2)

1 Weight of formwork itself qtc1 =   b h 1.1 8.1 8.91

Weight of reinforcement &
q 2   btct  hs
tc
2 1.2 390 468
concrete
Dynamic load of worker &
3 q3tc 1.3 250 325
equipment
Load of the concrete
4 q4tc 1.3 200 260
compaction
5 Load of concreting method q5tc 1.3 400 520

Total load q  q1  q 2  0.9(q 3  q 4  q 5 ) 1,163.1 1,471.4

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The load acting on the beam power is evenly distributed static load qtt consists of concrete floor,
formwork and live loads during construction.
+ Static load:
Including load of concrete, re-bar and formwork.
- load of formwork:
qtt1 = n    h = 1.145010.018=8.91 (kG/m) .
qtc1 =   h = 4500.018=8.1 (kG/m2).
- load of concrete, re-bar: the thick of floor 150mm.
qtt2 = n  h  sàn = 1.20.152600=468(kG/m).
qtc2 = h  sàn b = 0.1526001=390(kG/m).
Final static load: qtt = qtt1+ qtt2 = 8.91+468=476.91 (kG/m).
Final standard load: qtc = qtc1+ qtc2 = 8.1+390=398.1 (kG/m).
+ Dynamic load:
Including load by worker, machine, curing & compacting concrete to formwork.
- Dynamic load by worker, machine :
p3 = n .ptc =1.3250=325(kG/m2).
In which dynamic standard load by worker, machine ptc=250kG/m2
- Dynamic load by pouring & compacting concrete
p4 = n .ptc = 1.3(200 + 400) = 780 (kG/m2).
In which dynamic standard load by compacting concrete 200kG/m2, & pouring concrete 400kG/m2
Final load capacity:
qtt = q1 +q2 +0.9(p3 +p4 ) = 8.91+468+0.9  (325+780)=1,471.4 (kG/m2).
Total standard load capacity
qtcs = 8.1+390+0.9*(250+600) =1,163.1(kG/m2).

At a 1m wide strip formwork perpendicular to the purlins → diagram calculation is constantly

bearing beams and purlins and the load evenly distributed.
qtts =1,471.4 kG/m
qtcs =1,163.1 kG/m

2. Check the thickness of Plywood sheathing and space of Joist

2.1. Verification for bending stress:

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 σ (*)
M max
σ
W
q tt  l 2
In which: M max  (kG/cm )
10
b  h2 100  1.82
W   54(cm 3 ) ;
6 6
10  σ  W 10  135  54
From (*) so: l    70.39(cm)
q tt 14.714
2.2. Verification for deflection:
f   f  (**)

Deformed of slab formwork:  f  

l
400
q tcl 4
Max. deformed of slab formwork: f 
128.E.J
b.h 3 100  1.83
J   48.6(cm 4 )
12 12
128.E.J 3 128  1.1  10 5  48.6
from (**) so: l  3   52.79(cm)
400.q tc 400  11.63
Finally, space between girder: l=30 cm

3. Check the Joists:

- Use a steel box 50x50x1.5mm as a joist.
t = 21000 x 0.9= 1890 kG/cm2
E =2.1*106 kG/cm2.
(r  c 3  (r  d  2)  (c  2  d 3 ) (5  53  (5  0.15  2)  (5  2  0.15) 3 )
Jx    11.42cm 4
W 12
2  J 2  11.42
W   4.57
c 5
- Load capacity:
+ Load capacity effects to girder by space between girder l = 30cm.
+ Calculating map

Formwork of floor
Hozizontal beam
Vertical beam

+ private load of girder:

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q6 = n *b *h* t
in which:
safety factor: n =1,1
steel nature weight t = 2235 kG/cm2
b, h width & height of girder. Choose bxh=5x5cm
qttxg = 1.1*0.05*0.05*2235 =6.146 kG/m
=>Total load capacity into girder:
qtt = qtts.l+ qttxg=1471.4x0.3+6.146=447.57 kG/m
qtc = qtcs.l+ qtcxg=1163.1x0.3+6.146/1.1=354.52 kG/m

3.1. Verification for bending stress:

qtt  l2 2
M max   R  W(kG / cm)
10
Sufficient condition:
q tt .l 2 4.4757  120 2
M max    6,445kG.cm
10 10
Interia moment of formwork

 1410kG/cm 2  σ  1890kG/cm 2

M max 6,445
σ tt  
W 4.57
3.2. Verification for deflection:
qtc * l 4
f=
128* E * J
3.54  120 4
f   0.24cm  f   l/400  120/400  0.3cm
128  2.1  10 6  11.42
So the section of girder bh = 55x0.15 cm and space l=120cm is OK

4. Check the Stringer:

- Use a steel box 50x100x2.2mm as a stringer.
t = 21000 x 0.9= 1890 kG/cm2
E =2.1*106 kG/cm2.
(r  c 3  (r  d  2)  (c  2  d 3 ) (5  10 3  (5  0.22  2)  (10  2  0.22) 3 )
Jx    84.65cm 4
W 12
2  J 2  84.65
W   16.93cm3
c 10
- Load to column:
+ Angles vertical load is distributed over a wide range of approximately 1 to 2 head teachers from
the list l = 120 cm
+ Map calculation purlins along the beam simply place them on top of the pillow is the teaching
load concentrated passed down from horizontal purlins (at purlin bearing most dangerous).
There diagram properties:

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P P P P P P P P P

BPAL = 1200 BPAL = 1200 BPAL = 1200 BPAL = 1200

Mmax Mmin Mmax Mmin Mmax Mmin Mmax

BPAL = 1200 BPAL = 1200 BPAL = 1200 BPAL = 1200

- load into column:

P  q dn  L dn  447.57  1.2  537.08kG
L = 1.2m.
Moment Mmax, Mmin:
Mmax1 = 0.19*P*L = 0.19*537.08*1.2 = 122.45 ( kG.m)
Mmax2 = 0.12*P*L = 0.12*537.08*1.2 = 77.34 ( kG.m)
Mmin = 0.13*P*L = 0.13*537.08*1.2 = 83.78 ( kG.m)
- Private load:
q6 = n *b *h* t
in which:
safety factor: n =1,1
steel nature weight t = 2235 kG/cm2
b, h là width & height of girder.
qbt = 1.1  0.05  0.1  2235 =12.29 kG/m
q bt .l 2 12.29  1.2 2
M bt    1.77kG.cm
10 10
- Max moment of column: Mmax = Mmax1+Mbt
 Mmax = 122.45+1.77 = 124.22 ( kG.m)

4.1. Verification for bending stress:

 8.532kG/cm 2  σ  1890kG/cm 2 OK

M max 124.22
σ tt  
W 16.93
=> choosing stringer with secttion (5  10) is OK.
4.2. Verification for deflection:
qtc * l 4
f=
128* E * J
3.54  120 4
f   0.048cm  f   l/400  120/400  0.3cm
128  2.1  10 6  84.65
with: f < [f] so column with section (5  10) is OK

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II. GIRDER FORMWORK :

Girder RG1 section: 300x500mm
1. Loads:

No Load Formula n qtc( kG/m2) qtt( kG/m2)

1 Weight of formwork itself qtc1 =   b h 1.1 2.43 2.673

Weight reinforcement & tc

2 q 2  γ btct  h d 1.2 390 468
concrete
Dynamic load of worker &
3 q3tc 1.3 250 325
equipment
Load of the concrete
4 q4tc 1.3 200 260
compaction

5 Load of concreting method q5tc 1.3 400 520

Total load q  q1  q 2  0.9(q 3  q 4  q 5 ) 1,157.3 1,465.17

The load acting on the beam power is evenly distributed static load qtt consists of concrete floor,
formwork and live loads during construction.
+ Static load:
Including load of concrete, re-bar and formwork.
- load of formwork:
qtt1 = n    h = 1.14500.30.018=2.673 (kG/m) .
qtc1 =   h = 4500.3x0.018=2.43 (kG/m2).
- load of reinforce concrete
qtt2 = n  h  sàn = 1.20.3x0.52600=468(kG/m).
qtc2 = h  sàn b = 0.3x0.526001=390(kG/m).
+ Dynamic load:
Including load by worker, machine, curing & compacting concrete to formwork.
- Dynamic load by worker, machine :
p3 = n .ptc =1.3250=325(kG/m2).
In which dynamic standard load by worker, machine ptc=250kG/m2
- Dynamic load by pouring & compacting concrete
p4 = n .ptc = 1.3(200 + 400) = 780 (kG/m2).
In which dynamic standard load by compacting concrete 200kG/m2, & pouring concrete 400kG/m2
Final load capacity:
qtt = q1 +q2 +0.9(p3 +p4 ) = 2.673+468+0.9  (325+780)=1,465.17(kG/m2).
Total standard load capacity
qtcs = 2.43+390+0.9*(250+600) =1,157.3(kG/m2).

2. Check the thickness of Plywood sheathing and space of Joists

2.1. Verification for bending stress:

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 σ (*)
M max
σ
W
q tt  l 2
In which: M max  (kG/cm)
10
b  h2 30  1.82
W   16.2(cm 3 ) ;
6 6
10  σ  W 10  135  16.2
From (*) so: l    38.64(cm)
q tt 14.65
2.2. Verification for deflection: f  f (**)

Deformed of slab formwork: f   l

400
q tc l 4
Max. deformed of girder formwork: f 
128.E.J
b.h 3 30  1.83
J   14.58(cm 4 )
12 12
128.E.J 128  1.1  10 5  14.58
from (**) so: l  3  3  35.4(cm)
400.q tc 400  11.57
Finally, space between girder: l=30 cm

3. Check the Joists:

- Cross-section steel girder: bh = 55cm,:
t = 21000 x 0.9= 1890 kG/cm2
E =2.1*106 kG/cm2.
(r  c 3  (r  d  2)  (c  2  d 3 ) (5  53  (5  0.15  2)  (5  2  0.15) 3 )
Jx    11.42cm 4
W 12
2  J 2  11.42
W   4.57
c 5
- Load capacity:
+ Load capacity effects to girder by space between girder l = 30cm.
+ Calculating map

Formwork of floor
Hozizontal beam
Vertical beam

Picture: Load map

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+ private load of girder:

q6 = n *b *h* t
in which:
safety factor: n =1,1
steel nature weight t = 2235 kG/cm2
b, h width & height of girder. Choose bxh=5x5cm
qttxg = 1.1*0.05*0.05*2235 =6.146 kG/m
=>Total load capacity into girder:
qtt = qtts.l+ qttxg=1465.17x0.3+6.146=445.7 kG/m
qtc = qtcs.l+ qtcxg=1157.3x0.3+6.146/1.1=352.69kG/m

3.1. Verification for bending stress:

qtt  l2 2
M max   R  W(kG / cm)
10
Sufficient condition:
q tt .l 2 4.457  120 2
M max    6,418kG.cm
10 10
Interia moment of formwork

 1404kG/cm 2  σ  1890kG/cm 2

M max 6418
σ tt  
W 4.57
3.2. Verification for deflection:
qtc * l 4
f=
128* E * J
in which: E - steel elastic module; E = 2.1 106 kG/cm2
J - Moment of inerita of wood plate width:
3.54  120 4
f   0.24cm  f   l/400  120/400  0.3cm
128  2.1  10 6  11.42
So the section of girder bh = 55x0.15 cm and space l=120cm is OK

4. Check the Stringer:

Choosing steel box: 50x100x2.2 mm:
t = 21000 x 0.9= 1890 kG/cm2
E =2.1*106 kG/cm2.
(r  c 3  (r  d  2)  (c  2  d 3 ) (5  103  (5  0.22  2)  (10  2  0.22) 3 )
Jx    84.65cm 4
W 12
2  J 2  84.65
W   16.93
c 10

- Load to column:
+ Angles vertical load is distributed over a wide range of approximately 1 to 2 head teachers
from the list l = 120 cm

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+ Map calculation purlins along the beam simply place them on top of the pillow is the
teaching load concentrated passed down from horizontal purlins (at purlin bearing most
dangerous). There diagram properties:

P P P P P P P P P

BPAL = 1200 BPAL = 1200 BPAL = 1200 BPAL = 1200

Mmax Mmin Mmax Mmin Mmax Mmin Mmax

BPAL = 1200 BPAL = 1200 BPAL = 1200 BPAL = 1200

Picture: Load presses into column

- load into column:

P  q dn  L dn  445.7  1.2  534.84kG
L = 1.2m.
Moment Mmax, Mmin:
Mmax1 = 0.19*P*L = 0.19*534.84*1.2 = 121.94 ( kG.m)
Mmax2 = 0.12*P*L = 0.12*534.84*1.2 = 77.02 ( kG.m)
Mmin = 0.13*P*L = 0.13*534.84*1.2 = 83.4( kG.m)
- Private load:
q6 = n *b *h* t
in which:
safety factor: n =1,1
steel nature weight t = 2235 kG/cm2
b, h là width & height of girder.
qbt = 1.1  0.05  0.1  2235 =12.29 kG/m
q bt .l 2 12.29  1.2 2
M bt    1.77kG.cm
10 10
- Max moment of column: Mmax = Mmax1+Mbt
 Mmax = 121.94+ 1.77 = 123.7 ( kG.m)

4.1. Verification for bending stress:

 7.3kG/cm 2  σ  1890kG/cm 2

M max 123.7
σ tt  
W 16.93
=> choosing column with secttion (5  10) is OK.
4.2. Verification for deflection:
qtc * l 4
f=
128* E * J
in which:
qtc = P/1.2 = 445.7 +12.29/1.1 = 456.87 kG/m

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E - steel elastic module; E = 2.1 106 kg/cm2

4.57  120 4
f   0.04cm  f   l/400  120/400  0.3cm
128  2.1  10 6  84.65
with: f < [f] so column with section (5  10) is OK

II.COLUMN FORMWORK:
Section 400x500mm
1. Loads:

No Load Formula n qtc( kG/m2) qtt( kG/m2)

1 Lateral pressures Ptt= . .(0.27V+0.78)k1.k2 1.3 2992.5 3890

Load of the concrete

2 q1tc 1.3 200 260
compaction
Load of concreting
3 q2tc 1.3 400 520
method
Wind load
4 W  0.5  W0  k  c 335.24 335.24
(TCVN 2737:1995)
Total load q  P  0.9(q1  q 2  W) 4893.7 3996.2

- Lateral pressure:
Ptt= n . .(0.27V+0.78)k1.k2 ( V>=0.5, H=<4)
V= 1m/h
H= 3m
K1=1.2
K2=0.95
Ptt=1.3x2500x(0.27x1+0.78)x1.2x0.95=3890 kG/m2
Ptc=2992.5 kG/m2
- Dynamic load by compacting concrete
qtt1 = 1.3200 = 260 (kG/m2).
- Dynamic load by pouring concrete
qtt2 = 1.3400 = 520 (kG/m2).
- Wind load: (refer TCVN 2737:1990)
W  0.5  W0  k  c

W0  830Kg/m 2

k  1.07

C  0.8

W  0.5  830  1.07  0.8  335.24Kg/m 2

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Final load:
qtt = ptt +0.9(qtt1 +qtt2 +W) = 3890+0.9  (260+520+335.24)=4893.7 (kG/m2).
qtc = ptc +0.9(qtc1 +qtc2 +W) = 2992.5+0.9  (200+400+335.24)=3996.2 (kG/m2)

At a 1m wide strip formwork perpendicular to the purlins → diagram calculation is constantly

bearing beams and purlins and the load evenly distributed.
qtt =4893.7 kG/m
qtc =3996.2 kG/m

2. Check the thickness of Plywood sheathing and space of Studs

2.1. Verification for bending stress:

 σ (*)
M max
σ
W
q tt  l 2
In which: M max  (kG/cm)
10
b  h 2 50  1.82
W   27(cm 3 );
6 6
10  σ  W 10  135  27
From (*) so: l    27.29(cm)
q tt 48.94
2.2. Verification for deflection: f  f (**)

Deformed of slab formwork: f   l

400
q tc l 4
Max. deformed of slab formwork: f 
128.E.J
b.h 3 50  1.83
J   24.3(cm 4 )
12 12
128.E.J 128  1.1  10 5  24.3
from (**) so: l  3  3  27.76(cm)
400.q tc 400  40
Finally, space between girder: l=25 cm

3. Check the Studs:

- Use steel box 50x50x1.5mm as a stud
t = 21000 x 0.9= 1890 kG/cm2
E =2.1*106 kG/cm2
(r  c 3  (r  d  2)  (c  2  d 3 ) (5  53  (5  0.15  2)  (5  2  0.15) 3 )
Jx    11.42cm 4
W 12
2  J 2  11.42
W   4.57
c 5
- Load capacity:

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+ Load capacity effects to stud by space between studs l = 25cm.

+ Calculating map

Formwork of floor
Hozizontal beam
Vertical beam

+ Private load of girder:

q6 = n *b *h* t
in which:
safety factor: n =1,1
steel nature weight t = 2235 kG/cm2
b, h width & height of girder. Choose bxh=5x5cm
qttxg = 1.1*0.05*0.05*2235 =6.146 kG/m
=>Total load capacity into girder:
qtt = qtt.l+ qttxg=4893.7 x0.25+6.146=1229.57 kG/m
qtc = qtc.l+ qtcxg=3996.2 x0.25+6.146/1.1=1004.6 kG/m

3.1. Verification for bending stress:

Necessary condition:
Moment in girder
qtt  l2 2
M max   R  W(kG / cm)
10
Sufficient condition:
q tt .l 2 12.29  70 2
M max    6022kG.cm
10 10

 1317kG/cm 2  σ  1890kG/cm 2

M max 6022
σ tt  
W 4.57
3.2. Verification for deflection:
qtc * l 4
f=
128* E * J
in which: E - steel elastic module; E = 2.1 106 kg/cm2
10.05  70 4
f   0.08cm  f   l/400  70/400  0.175cm
128  2.1  10 6  11.42
So the section of girder bh = 55x0.15 cm and space l=70cm is OK

4. Check the Wales:

Use double steel box: 50x100x2.2 mm as double wales.
t = 21000 x 0.9= 1890 kG/cm2 và E =2.1*106 kG/cm2.

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DMPP Consortium

(r  c 3  (r  d  2)  (c  2  d 3 ) (5  10 3  (5  0.22  2)  (10  2  0.22) 3 )

Jx    84.65cm 4
W 12
2  J 2  84.65
W   16.93cm3
c 10
- Load to column:
+ Angles vertical load is distributed over a wide range of approximately 1 to 2 head teachers
from the list l = 70 cm
+ Map calculation purlins along the beam simply place them on top of the pillow is the teaching
load concentrated passed down from horizontal purlins (at purlin bearing most dangerous).
There diagram properties:

- load into column:

P  q dn  L dn  4893.7  0.7  3425.6kG
L = 0.7m.
Moment Mmax, Mmin:
Mmax1 = 0.19*P*L = 0.19*3425.6*0.7= 455.6 ( kG.m)
Mmax2 = 0.12*P*L = 0.12*3425.6*0.7= 287.75 ( kG.m)

q
Mmin = 0.13*P*L = 0.13*3425.6*0.7 = 311.67 ( kG.m)
- Private load:

l
l
q6 = n *b *h* t
in which:
safety factor: n =1,1

l
steel nature weight t = 2235 kG/cm2 l
b, h là width & height of girder.
qbt = 1.1  0.05  0.1  2235 =12.29 kG/m

l
l

q bt .l 2 12.29  0.7 2
M bt    0.6kG.cm
10 10
- Max moment of column: Mmax = Mmax1+Mbt
 Mmax = 455.6+ 0.6 = 456.2 ( kG.m)

4.1. Testing flecure:

 26.9kG/cm 2  σ  1890kG/cm 2

M max 456.2
σ tt  
W 16.93
=> choosing column with secttion (5  10) is OK.
M=ql /10

4.2. Verification for deflection:

2

qtc * l 4
f=
128* E * J
(4.55  0.12/1.1)  70 4
f   0.005cm  f   l/400  70/400  0.175cm
128  2.1  10 6  84.65
with: f < [f] so column with section (5  10) is OK

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DMPP Consortium

C. CHECK STABLE CONDITION OF SCAFFOLDING SYSTEM

I. Technical Parameter of scaffolding

Outside diameter of main vertical support D 4.90 cm
Thickness of main vertical support t 0.20 cm
Inside diameter of main vertical support d 4.50 cm
Highest of biggest support H 300.00 cm
Width of frame W 120.00 cm
Diameter of horizontal bracing Dg 4.20 cm
Calculation strength of steel CT3: Rk 2,100 (kg/cm2)
Module Elasticity of steel CT3: E 2,100,000
Coefficient of working condition  0.9
Coefficient of calculation length  0.7
Coefficient of security (According to the standard design) n 1.4

II .II. Calculation:
1. Load capacity of system
* Original area (cm2) :
Ang = pi x D2/4 = 18.85
* Actual area (cm2) :
Ath =pi x (D2- d2 )/4 = 2.95
* Moment of inertia (cm4) :
J = pi x (D4 - d4 )/64 = 8.16
* Radius of inertia (cm) :
r= = 0.66
* Stable working height of the bearing system (cm):
lo = l x  = 112
* Slenderness of the bearing frame :
= lo/r = 169.7
* Slenderness conventional bearing frame :

o= = 5.37

* Buckling coefficient:
3100/o2 = 0.107
* Load capacity of system (kG)
[P] = Ath x R x /n = 3,985
* Stable condition (kG)
Pth =  x R x Ang x /n = 2,723

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DMPP Consortium

2. Check shoring system of Slab:

P=qttsx1.2x1.2=(1471.4+12.29)x1.2x1.2=2136.5 KG
* Verification for bearing capacity:
P<[P]: OK
* Verification for stable:
P<Pth: OK
3. Check shoring system of Girder:
P=qttdx0.3x1.2=(1465.2+12.29)x0.3x1.2=532 KG
* Verification for bearing capacity:
P<[P]: OK
* Verification for stable:
P<Pth: OK
 Scaffolding system is safety working

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