箱涵支架设计计算

Formwork & Falsework Design for Single cell Box Culvert
Reference

Calculations Structures
5025C at 50+423, 5204C at 52+137, 5212C at 52+767 , 5812C at 58+994 & 59

03C at 59+092
Refer attched table No. 01

Output

Consider RCBC 5812C at 58+994 as a sigle cell typical structure for the calculation

Data

Top Slab Thickness Bottum Slab Thickness Wall Thickness Clear Span Clear Height Wall Height

= = = = = =

750 700 600 9930 5465 5200

mm mm mm mm mm mm

Codes
Generally the designs are undertaken to the following standards and modified wherer shown (a). BS 5975:1996 Code of practice for Falsework (b). BS 5950 : Part 1 , 1990 Structural Use of Steel Work in Building (c). BS EN 10210 Circular Hollow Hot Form Section (d). CIRIA Report 108 Concrete Pressure on Formwork

Design of Top Slab Introduction
Loads from wet concrete, formwork and working loads are directly rest on 15 mm plywood. The plywood is supported by 50mm diameter and 2 mm thick G.I. pipes, which are spaced at 150mm centers. Then the GI pipes are supported by 2/50 G.I. pipes spaced at 775 mm centers. The G.I. supports to 50 G.I. pipes at 775 mm and 785 mm centers. the total loads from concrete, working and formwork will be transformed to these scafoliding frams and then to bottom slab

Load Evaluation
BS 5975 1996 cl. 8.3.1 cl. 8.3.2

Concrete Density Weight of formwork Live load on formwork Dead Load

= = = = =

25 2 1.5

KN / m3 KN / m2 KN / m2
slab thickness

25* 0.750 + 2 20.75 KN / m2

1

Reference

Calculations
Formwork/falsework are designed for nominal loads ( unfactored ) Unfactored Load Intensity (n1) = = = 1.0gk +1.0 qk 1.0 * 20.75 + 1.0 * 1.5 22.25 KN / m2

Output

Design of 15 mm Plywood
Plywood is supported by 50 mm GI pipes 15 mm Flywood 150 mm Considering 1m wide strip UDL Check for Bending Maximum Bending Moment (M) = = = Maximum Bending Stress (s) = 0.063 (w) = 22.25 KN / m 50 mm G.I. pipes

WL2 8

KNm
KNm

M Z
0.037 ? 103 ?103 ?1000 ?152 ? ? ? 6 ? ?
1.67 N / mm
2

=

Nmm
Plywood

=

<

11.37

N / mm

2

15mm ok.

(Allow. Bending Stress) Therefore Bending Stress is Satisfactory. Check for Shear Maximum Shear on plywood = =

WL 2
12.98 ? 0.150 2

= 1.669 Shear Stress =

KN

0.974 ?103 1000 ?15

N / mm 2

= 0.111

N / mm2 ? 2.58 N / mm2
(Allow. Shear Stress) OK

2

Reference Check for Deflection

Calculations
Maximum Deflection =

Output W = 22.250 L = 150 E = 7000

1 WL4 384 EI

KN / m mm
N / mm 2

=

? ? 4 ? ? 13.82 ? 150 1 ? ? 384 ? ? 1000 ?153 ? ? ? 7000 ? ? ? ?? ? 12 ? ?? ?

?

?

= Span/Deflection =

0.0149 150

mm

0.0149 Span/Deflection = Hence 15 mm plywood is satisfactory 10067

>

200

OK

Design of GI Pipes
The plywood is supported by 50 mm diameter and 2 mm thick GI pipes, which are spaced at 150mm centers. Then the GI pipes are supported by 2/50 GI pipes spaced at 775 mm centers.

Check for Bending 0.150

2/50 G.I. Pipes G.I. Pipe

0.150 0.775 Maximum Bending Moment (M) =

WL2 8

=

12.98 ? 0.150 ? 0.7752 KNm 8
0.251

= Bending Stress (s) Check for G.I Pipes G. I. Pipe
Hand Book of structural steel work A.D. weller 1997

KNm

=

M Z

48.3 thickness 2 mm

I?

? ? d 41 ? d 4 2 ?
64
3 3.1428 48.34 ? 44.34

?

?

Reference

I?

? ? d 41 ? d 4 2 ?
Output

Calculations 64

I?
=
EN 10210

3.1428 48.34 ? 44.34 64 ?10
?4

?

?

7.81 cm4 3.238 cm3 .

Z= Therefore Bending Stress ( s) =

0.146 3.238 ?10?6

KN / m2

= Bending Stress (s) BS 5975 Ann-B 45.114 N/mm2

77385

KN / m2

<

Permissible Stress

< 125 N/mm2

Therefore Bending Stress is Satisfactory. Check for Shear Maximum Shear on GI Pipe Shear force = =

WL 2

12.98 ? 0.150 ? 0.775 2
KN

= 0.754 Shear Stress = 2.59

N / mm2 ? 93 N / mm2
(Allow. Shear Stress) OK

Check for Deflection Maximum Deflection =

1 WL4 384 EI

W = 3.338 L = 775

mm

KN / m
N / mm 2

EI = 2898 x 107 =
4 1 ? 1.947 ? 775 ? 384 ? 2898 ? 107 ?

?

?? ?
? ?

= Span/Deflection = 775 0.063 Span/Deflection =

0.063

mm

12279

>

200

OK

Hence GI pipe 150 c/c and supporting at 775 mm are satisfactory

4

Reference

Calculations Design of 2/50 G.I pipes just under the GI pipe

Output

150 2/50 GI Pipes 785 mm

2/50 GI Pipes 0.775 m 0.775 m

Load from GI pipe is a UDL W= W= Maximum Bending Moment (M) =

12.98 ? 0.775 KN / m
10.06

KN / m

WL2 8
10.06 ? 0.7852 8

=

KNm

= Bending Stress (s) Check for G.I Pipes G. I. Pipe
Hand Book of structural steel work A.D. weller 1997

0.775

KNm

=

M Z

48.3 thickness 2 mm

I?

? ? d 41 ? d 4 2 ?
64 3.1428 48.34 ? 44.34 64 ?10
?4

I?
=
EN 10210

?

?

7.81 cm4

Z=

3.238 cm3 .

5

Reference Therefore Bending Stress ( s)

Calculations
=

Output

0.775 2 2 ? (3.238 ?10?6 ) KN / m

= Bending Stress (s) BS 5975 Ann-B

119652

KN / m2

<

Permissible Stress

119.652 N / mm2 ? 125 N / mm2
Therefore Bending Stress is Satisfactory. Check for Shear Maximum Shear on GI Pipe Shear force = =

WL 2
10.06 ? 0.785 2

= 3.949 Shear Stress = 6.79

KN

N / mm2 ? 93 N / mm2
(Allow. Shear Stress) OK

Check for Deflection Maximum Deflection =

1 WL4 384 EI

W = 10.060 L = 785

mm

KN / m
N / mm 2

EI = 2898 x 107 =
4 1 ? 10.060 ? 785 ? ? ? 384 ? (2 ? 2898 ?107 ) ? ? ?

?

?

= Span/Deflection = 785 0.172 Span/Deflection =

0.172

mm

4574

>

200

OK

Hence 2/50 GI pipe 775 c/c and supporting at 785mm are satisfactory

6

Reference

Calculations Design of GI Pipe Supports.
0.785 m

Output

0.775 m 0.775 m

0.785 m Load on a Leg = = Assume - Effective length of GI pipe Radius of Gyration Slenderness ratio = = 12.98 x 0.775 x 0.785 7.90 1500

KN

KN
Scaffold leg

mm 1.62 cm

??
= =

Le r
(150 / 1.62) 93

BS 5950 CL:7.4

Capacity of GI pipe

Pc ? Ag ? pc
= = 290.9 x 124 36.07 KN 36

Capacity of a Leg( GI Pipe Scaffolding leg)

=

KN

capacity ok.

Therefore the condition is Satisfactory.

Design of Walls
Lateral pressure from wet concrete are directly applied to 15 mm plywood. The plywood is supported by 50mm diameter and 2 mm thick GI pipes, which are spaced at 150mm centers. Then the GI pipes are supported by 2(50) GI pipes, spaced at 600 mm centers.

Maximum concrete pressure on side formwork
CIRIA Report 108

P max ? D ?C1 R ? C2 K H ? C1 R ? ............. ?1? ? ? ? ?
Where :D= C1 = C2 = R= 24 kN/m3 ( Unit weight of concrete ) 1 ( as a wall ) 0.45 ( retarded concrete ) 1.5 m/hr. ( concrete pouring rate )

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Reference

Calculations
H= T= K= 1.730 m ( Wall height ) 32 0C ( concrete temparature at pouring ) 36 Tc +16 K= 36 32 +16 K= 0.5625 2 2

Output

P max ? 24 ?1 1.5 ? 0.45 ? 0.5625 1.730 ? 1 1.5 ? ? ? ? ?

Pmax. =

33.71

KN / m2

Design of Side Plywood
P max D
50mm G.I. pipes

? 1.41

0.320 150 150 150 15mm Flywood 50 mm GI Pipe Considering 1m wide strip Check for Bending UDL for 1m strip Maximum Bending Moment (M) (w) = = = 33.71 2(50) mm Gi pipe

33.71 KN / m2

KN / m

WL 8

2

=

42.67 ? 0.1502 KN .m 8 0.09 KNm

Maximum Bending Stress (s)

=

M Z
0.09 ?106 ? 1000 ?152 ? ? ? 6 ? ?

=

N / mm 2

8

Reference

Calculations
= Maximum Bending Stress (s) 2.40

Output

N / mm

2

<

Allow. Bending Stress

2.40 N / mm2

? 11.37 N / mm2

Therefore Bending Stress is Satisfactory. Calculation for Shear Maximum Shear Force = = = Shear Stress (s) = =

33.71? 0.150 2
2.53

WL 2

KN
N / mm 2

2.53 ?103 1000 ?15
0.17

N / mm 2

0.17 N / mm2

? 2.58 N / mm2
(Allow. Shear Stress) OK

Check for Deflection Maximum Deflection =

1 WL4 384 EI

W = 33.71 L = 150 E = 7000

=

? 4 1 ? 33.71?150 ? 1000 ?153 384 ? 7000 ? ? 12 ?

?

?

? ? ? ? ? ?

KN / m mm N / mm 2

= Span/Deflection =

0.023

mm

150 0.023
6645

Span/Deflection = Hence 15 mm plywood is satisfactory

>

200

OK

Design of GI Pipes
The plywood is supported by 50mm diameter and 2 mm thick GI pipes, which are spaced at 150mm centers.

9

Reference Check for Bending Maximum Bending Moment (M)

Calculations

Output

=

WL2 8
33.71? 0.150 ? 0.6002 8
0.228

= = Bending Stress (s) Check for G.I Pipes G. I. Pipe
Hand Book of structural steel work A.D. weller 1997

KNm

KNm

=

M Z

48.3 thickness 2 mm

I?

? ? d 41 ? d 4 2 ?
64 3.1428 48.34 ? 44.34
7.81 64 ?10
4 ?4 cm

I?
=
EN 10210

?

?

Z= Therefore Bending Stress ( s) = = Bending Stress (s)

3.238 cm3 .

0.228 3.238 ?10?6
70413.84

KN / m2 KN / m2

<

Permissible Stress

BS 5975 Ann-B

70.41 N / mm2 ? 125 N / mm2
Therefore Bending Stress is Satisfactory. Check for Shear Maximum Shear on GI Pipe Shear force = =

WL 2

33.71? 0.150 ? 0.600 2
KN

= 1.517 Shear Stress = 5.21

N / mm2 ? 93 N / mm2
(Allow. Shear Stress) OK

10

Reference Check for Deflection

Calculations

Output

Maximum Deflection =

1 WL4 384 EI

W = 5.057 L = 150

mm

KN / m
N / mm 2

EI = 2898 x 107 =
4 1 ? 5.057 ? 600 ? 384 ? 2898 ? 107 ?

?

?? ?
? ?

= Span/Deflection =

0.059

mm

600 0.059

Span/Deflection =

10188

>

200

OK

Hence GI pipe 150 c/c and supporting at 600 mm are satisfactory Check for M12 Tie bar

600 mm

460 mm Force on M12 Tie bar = = Direct Tensile Stress on M12 Thread bars = 33.71 x 0.460 x 0.600 9.30

KN

9.3 ? 103 113.098

N / mm 2

= Braking Load of Thread bars = Permissible Stress on Thread bars = Direct Tensile Stress on M12 tie bar F.O.S <

82 44.6 274

N / mm 2
KN

N / mm 2

Permissible Stress on Tie bar

=
=

274 82
3.3

Hence M12 tie bars are satisfied

11

Reference

Calculations Design of GI Pipes Support for walls

Output

BS 5975 Cl: 6.3.1.3

Consider 1m strip of the wall and assume all structure sway at one side 0.7 (100As)1/3 (fcu)1/3 Vertical force of the wall = = = Vertical force of the top slab = = Horizontal force of the wall = = Force on GI pipes/Adjutable jack = R Cos 45 = Reaction on GI Pipe Concrete weight + F/W weight + Live Load

{(1.730 ? 0.255 ?1? 24) ? 2 ? (1.730 ?1? 2) ? (1.730 ?1?1.5)}? 2
40.21

KN

(3.560 ? 0.395 ?1? 24) ? (3.560 ?1? 2) ? (3.560 ?1?1.5)
46.21

KN

86.42 ? 2.5%
2.16

KN

R Cos 45 2.16 3.05

R

=

KN

BS 5950 CL:7.4

Capacity of GI pipe

Pc ? Ag ? pc

( Effective length 1.8 m, Radious of Giration 1.62

?= 111)

Pc ? 290.9 ?133

Pc
Capacity of Ajestable jack

= =

38.69 38.69

KN KN

Reaction on GI pipes <

Capacity of GI pipes

Provide one GI pipe support as sway bracing @ 1000 c/c

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Reference

Calculations

Output

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Calculations

Output

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Calculations

Output

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Calculations

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