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 & 5903C 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 )
7
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
12
Reference
Calculations
Output
13
Reference
Calculations
Output
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Reference
Calculations
Output
15
Reference
Calculations
Output
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