Formwork 3rd Edition Worked Examples

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Worked Examples to Formwork A guide to good practice (3rd Edition) These worked examples are a companion document to The Concrete Society publication Formwork – A guide to good practice (3rd Edition). They include worked examples of formwork design and back propping calculations. Formwork is the key to successful and economic concrete construction. It has a dominant influence on the appearance and accuracy of finished concrete which in turn affects the ease with which the following trades can complete their work. It generally accounts for a third or more of the value of the structure. The design and construction of formwork is an essentially practical subject and relies on the engineering judgement and expertise of those involved. Developments in materials and equipment for formwork are constantly extending the specifier’s and contractor’s options, in terms of concrete finish, speed and economy of construction, etc. The guide brings together the practical and engineering aspects of formwork in a way which will be of particular value to all concerned with the specification, design, manufacture, construction and use of formwork for buildings and civil engineering structures. These worked examples have been prepared with the approval of the Construction Standing Committee of The Concrete Society by a working party chaired by Eur Ing Peter Pallett, Temporary Works Consultant, and Lecturer. Assistance was invited from experts in the industry including contractors, architects, consulting engineers, specialist suppliers and trade associations.

Formwork A guide to good practice (3rd Edition) Worked Examples Prepared by a Working Party of The Concrete Society

CS 169 First published April 2012 Reprinted with minor amendments July 2012 © The Concrete Society

The Concrete Society Riverside House, 4 Meadows Business Park, Station Approach Blackwater, Camberley, Surrey, GU17 9AB Tel: +44 (0)1276 60 7140 Fax: +44 (0)1276 60 7141 Email: [email protected] Visit: www.concrete.org.uk

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Acknowledgements

The Concrete Society gratefully acknowledges the contribution made by members of the Working Party and in particular Eur Ing P.F. Pallett in the preparation of this companion document. The Concrete Society also wishes to express its gratitude to all those participating organisations and individuals who have assisted in the preparation of the worked examples.

CS 169 First published April 2012 Reprinted with minor amendments July 2012 ISBN 978-1-904482-69-7 © The Concrete Society

The Concrete Society

Riverside House, 4 Meadows Business Park, Station Approach Blackwater, Camberley, Surrey, GU17 9AB Tel: +44 (0)1276 60 7140 Fax: +44 (0)1276 60 7141 Email: [email protected] Visit: www.concrete.org.uk Further copies may be obtained from The Concrete Bookshop: www.concretebookshop.com

All rights reserved. Except as permitted under current legislation no part of this work may be photocopied, stored in a retrieval system, published, performed in public, adapted, broadcast, transmitted, recorded or reproduced in any form or by any means, without the prior permission of the copyright owner. Enquiries should be addressed to The Concrete Society. Although the Concrete Society (limited by guarantee) does its best to ensure that any advice, recommendations or information it may give either in this publication or elsewhere is accurate, no liability or responsibility of any kind (including liability for negligence) howsoever and from whatsoever cause arising, is accepted in this respect by the Society, its servants or agents. Note on photographs: The Committee recognise that some photographs may show breaches of current safety regulations but the photographs have been retained in the guide to illustrate particular items of interest. Readers should also note that all Concrete Society publications referenced in Section 9 are subject to revision from time to time and should therefore ensure that they are in possession of the latest version. Printed by Berforts - Information Press Ltd, Eynsham, UK.

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Worked Examples Worked Examples to Formwork – a guide to good practice The following worked examples of formwork design and backpropping calculations are based on the information from the Concrete Society’s book “Formwork - a guide to good practice” Edition Three. The seven worked examples of formwork design and backpropping are:

Page

2

1

Double-faced wall formwork, with wind stability

2

Square column formwork

15

3

Forces - Single-faced wall formwork

24

4

Bridge deck soffit formwork with void formers

30

5

Backpropping Calculations 205mm slab One set formwork Two sets backprops

6

Backpropping Calculations 175mm slab One set formwork Two sets backprops

7

Backpropping Calculations 175mm slab Two sets formwork

39 50 63

The Section and Figure numbers stated in the worked examples refer to the Formwork Guide and the Clause numbers refer to BS 5975. It is confirmed that the calculations in the worked examples have been checked to the relevant category, and that a Design Check Certificate has been issued. All rights reserved. Except as permitted under current legislation no part of this work may be photocopied, stored in a retrieval system, published, performed in public, adapted, broadcast, transmitted, recorded or reproduced in any form or by any means, without the prior permission of the copyright owner. Enquiries should be addressed to The Concrete Society. Although the Concrete Society (limited by guarantee) does its best to ensure that any designs, advice, recommendations or information it may give either in this publication or elsewhere are accurate, no liability or responsibility of any kind (including liability for negligence) howsoever and from whatsoever cause arising, is accepted in this respect by the Society, its servants or agents.

1

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Worked Example One

Worked Example One – Double faced Wall Formwork DESIGN BRIEF

Output

Design the formwork for an insitu reinforced concrete wall 5.15m high, 10m 5.15m long and 450mm wide. The base slab with high an integral 150mm high kicker has already been cast. There are no features 10m long required on the wall. The client’s specification states “Finish Class F4” with the deflection limited to 1/270th of the span of any formwork member. Through ties are allowed. The concrete will be CEM I with admixtures but no retarder. The work will take place in April. The site is on the outskirts of Lichfield in the West Midlands, and is about 120 km from the nearest sea, in a flat river flood plain with an altitude of 55m above sea level.

δ ≤ 1/270

The TWD has been advised that concreting will be by skip at an assumed volume rate of 9.0 cubic metres per hour. The concrete temperature for T = 10°C April is assumed to be 10°C. A proprietary vertical steel soldier system will be used with a tie rod system having a stated safe working load of Tie SWL 90kN per tie. The site crane has been rated at 2000kg at all radii. = 90 kN The TWD has selected the face contact material to be Canadian Douglas Fir plywood 7ply, 19mm, sanded, but unknown mill source.

19mm 7ply

The walings will be timber of Strength Class C16.

C16

The Design Check Category to Table 29 required is Check Two.

Cat 2

SIZE OF FORMWORK By inspection of the length of wall it would suit handling the form in two panels, each at least 5.0m long and 5.0m high. If the forms are designed slightly longer, then it will allow either overlap onto previous wall, or provision for fixing the stopend on to the face. (See Fig. 77) CONCRETE AND CONCRETE PRESSURE CALCULATION Using the concrete information in the Design Brief and the CIRIA method in Section 4.4 for ascertaining pressure gives:Concrete density 25 kN/m3 Concrete Group A (Basic Concrete) Volume rate of delivery is 9.0m3/h Plan area of pour = 10 x 0.45 = 4.5m2 (A)

2

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Worked Example One

Hence rate of rise ( R)

Volume rate 9.0   2.0m/h ( R) Plan Area 4.5

Thus from Table PAA (Section 4.4.3.1 ), for temperature 10°C and for H = 5.0m gives Pmax = Maximum concrete pressure = 60 kN/m2 The equivalent fluid head of concrete is The concrete pressure diagram is as opposite

PLYWOOD

Output Pmax 60 kN/m2

Pressure 60   2.4m Density 25.0

2400

5000

60 kN/m2

From Brief use Canadian Douglas Fir 7 ply 19mm plywood. As mill not known use structural properties for wall formwork Table 15 assuming the plywood spanning in its strong direction (See Fig. 39) with the sheets fitted vertically (i.e. parallel) on to horizontal walings gives:Bending Stiffness parallel (EI) = 2.20 kNm2/m Moment of Resistance parallel (fZ) = 0.483 kNm/m Shear Load parallel (qA) = 7.14 kN/m The plywood will span over several walings. The width of the bearing area of the plywood on to the waling will affect the safe shear span of the plywood. We will assume that each waling will be nominal 75mm wide. Thus B > (2 x t) i.e. 75 > (2 x 19) means that Appendix B Loading Case 58 can be used.

Appx B LC 58

Thus if the centre/centre ply span is L, the load per metre width of plywood is w = 60 x 1.0 = 60 kN/m. Maximum Moment in plywood is:Mmax = - 0.095 w L2 = 0.095 x 60 L2 = 5.70 x L2 Thus 0.483 = 5.70 x L2 Hence L = 0.291m = 291mm (Note the –ve sign indicates the hogging moment at the support. The plywood is considered symmetrical so the limit is ± 0.483 kNm/m.) Using plywood sheets 2440mm long x 1220mm wide (8’ x 4’) standard modules of the bearers/walings, assuming one is fitted at the plywood joints, gives 9 spans of 271mm or 8 spans of 305mm - Try 8 spans. As the value of 291mm is close to 305mm, try to use the larger, more economic, span by checking shear and deflection limits.

3

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Worked Example One The maximum shear force is thus Ss = 0.525 w ( L – B – t ) Ss = 0.525 x 60.0 x (0.305 – 0.075 – 0.019) = 6.65 kN cf allowable of 7.14 kN therefore OK in shear at 305 c/c.

Output LC 58

Check deflection in a span with limit set as 1 / 270th.

wL4 60.0  0.3054  0.0066  EI 2.20 305 = 1.13 mm = 0.00126m = 1.26mm cf allowable 270 thus   0.0066

FAIL

Now check if deflection is reduced if we had chosen 9 spans of 271mm.

60.0  0.2714 wL4  0.0066  2.20 EI 271 = 0.00097m = 0.97mm cf allowable = 1.0 mm hence OK 270 thus   0.0066

δ= 0.97mm ply span 271mm

Notes (1)

Actual checking the face finish with a straight edge would measure the range. (see Appendix B and Figure 158) This would increase the value of deflection measured.

(2)

The waling spacing was calculated for the maximum concrete pressure that occurs in the lower 2.6m of the form. In setting out the bearer/waling positions it is normal practice to increase the spacing towards the top, thus reducing the number of walings. It is important that the design stage considers the detail of the plywood on the walings. If the bottom of the form overlaps the kicker by 50mm the form height is 5050mm. If the bottom sheet of plywood has a waling flush to the edge, and there is a butt joint at the top of the sheet at the joint, then the actual plywood span is now:2440 - 75 2 = 267 mm 9 Thus assuming the height of 5050 gives a possible waling spacing of:- ( 10 x 267) + (3 x 305) + (4 x 357)

(

(3)

4

)

When using values from Appendix B, for Moment, Reactions, Shear etc the factors are ALL in decimals and the maximum value for a particular case will be the largest factor. For example at Loading Case 38 the maximum moment is 0.105wL2 as the factor 0.105 is larger than 0.079, 0.078, 0.033 and 0.046.

Form height 5.05m actual ply span is 267 mm

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Worked Example One

Output

WALINGS

Assume 75 x 150 timbers of strength class C16 will be used. The permissible structural values are taken from Table 8 in Section 3.3.1.4 which assumes wall formwork with load sharing as four or more walings will be in contact with the plywood. Allowable bending stiffness = 128.77 kNm2 Allowable moment of resistance = 1.906 kNm Allowable shear load = 10.37 kN Allowable bearing stress = 2980 (2,300)kN/m2 Applied distributed loading on one waling gives: Maximum distributed load on one waling = 60.0 x 0.267 w = 16.02 kN/m

w = 16.02 kN/m

Each waling will be at least 5.0m long. Select by inspection soldier centres of say 1.1m with walings spanning between soldiers. This has the advantage that standard 38mm scaffold boards can be used for access and working platforms as 1100mm is less than the maximum span of 1.2m for a grade 1.2m board (BS 2482). The form is split into two panels for handling and ideally will need to suit plywood modules. Assume one central join and the necessity to cut one plywood sheet gives an optimum panel length of 5.49m. (i.e. four + one cut in half = 4.5 x 1220 = 5490mm) From Section 5.2.2.1 and Figure 77 for wall thicknesses up to 450mm and panel lengths greater than the form height, the stopends may be secured to the panel ends. In waling design it is good practice to limit any cantilever to about one third of the adjacent span: i.e 1100/3 = 367mm approximate maximum cantilever Consider one panel with five soldiers (A to E) with horizontal bearers/walings. Any cantilevers are assumed to be approximately equal and no greater than about one third span. Although the panel will be 5.49m long, it is only loaded over the 5.0m length, gives by inspection a layout of:10.000m overall pour 5.000 m

centreline

w = 16.02 kN/m 240

A

1100

1100

B

1100

C

1100

D

Plan of waling

360

E

5

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Worked Example One

Analysis of the continuous beam could be by using a computer analysis program, or simply use Loading Case 37 from Appendix B2 of the Formwork Guide. This example will use the latter i.e. LC 37.

Output Use LC 37

This assumption gives results which, although approximate, are sufficiently accurate for the general design of members in formwork. Using LC 37 with MA = ME

L = 1.10m and w = 16.02 kN/m, gives:

= - 0.0556 wL2 = - 0.0556 x 16.02 x 1.12 = - 1.07 kNm

MB = MD = - 0.0911 wL2 = - 0.0911 x 16.02 x 1.12 = - 1.77 kNm MC

= - 0.079 wL2 = - 0.079 x 16.02 x 1.12 = - 1.53 kNm

MAB = MDE =

0.052 wL2 =

0.052 x 16.02 x 1.12 = 1.01 kNm

MBC = MCD =

0.040 wL2 =

0.040 x 16.02 x 1.12 = 0.78 kNm

Thus the maximum waling bending moment (see note (3) on page 4) occurs at support “B” and will be approximately -1.77 kNm. (< allowable moment of resistance 1.906 kNm (page 5). ) Therefore OK.

Max BM -1.77 kNm

(in practice it is not necessary to write down all the values of BM in the calculations. “By inspection the max. BM is . . . . “ ) CHECK SHEAR VALUES

SA = SE = 0.333 wL

=

0.333 x 16.02 x 1.1 = 5.87 kN

SAB = SED = 0.465 wL =

0.465 x 16.02 x 1.1 = 8.19 kN

SBA = SDE = 0.535 wL =

0.535 x 16.02 x 1.1 = 9.43 kN

SBC = SDC = 0.513 wL

=

0.513 x 16.02 x 1.1 = 9.04 kN

SCB = SCD = 0.487 wL =

0.487 x 16.02 x 1.1 = 8.58 kN

The maximum shear value is 9.43 kN and the permissible shear load is 10.37 kN (page 5). Therefore OK. (in practice it is not necessary to write down all the values of shear force in the calculations. “By inspection the max. shear force is . . . “ )

Max Shear 9.43 kN

CHECK THE REACTIONS AT SUPPORTS (i.e. Soldier positions)

RA = RE = 0.798 wL

= 0.798 x 16.02 x 1.1

RB = RD = 1.048 wL

= 1.048 x 16.02 x 1.1

RC = 0.976 wL

6

= 0.976 x 16.02 x 1.1 Total

= 14.06 kN 14.06 kN = 18.47 kN 18.47 kN = 17.20 kN 82.26 kN

Max Reaction 18.47 kN

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Worked Example One

As a check the total actual applied load is 5.00 x 16.02 = 80.10 kN. Thus the actual is 2.6% less than the assumed design load. This is because of the assumption that the cantilever was 1/3 rd of span, i.e. 367mm at each end cf 240mm and 360mm. Note that the shear values will also not sum exactly to the reactions for similar reasons.

Output

CHECK DEFLECTIONS

By inspection of LC 37 the maximum deflection occurs in span AB. wL4 Thus deflection in 1100mm is 0.00387 EI 4 16.02  1.1 = 0.00387     0.70mm 128.77 1 1  0.70    (Section 2.7 ) Therefore OK. L 1100 1571 270

δAB = 0.70mm

SOLDIERS AND TIES Assume a proprietary soldier is used (See Section 3.2.3) and that it comprises two parts each 2.7m long, placed on the panels vertically at the five positions A to E. The selection of the vertical spacing of the tie rods will be influenced by many factors and will often be determined by experience. The principal factors are:1. Proprietary supplier’s soldier data giving pre-determined positions of the ties 2. The concrete pressure diagram with reduction of pressure near to the top of the wall. 3. Size and load capacity of the ties. Note that different types of waler plate have different allowable loads which may limit the tie rod capacity. 4. Pre-determined positions as specified by the client to coincide with horizontal features on the wall. 5. Economics of material and labour costs for fixing the tie rod assemblies and the subsequent making good of tie hole positions. In certain cases fewer, higher load, but more expensive tie rods might be a more economic solution. 6. The position of the walings and any joints in the soldier. Maximum point load from walings at 267mm c/c on to the soldiers is their reactions at RB and RD of 18.47 kN on soldiers at B and D. By interpolation the maximum equivalent distributed load on the most 18.47 heavily loaded soldier is w = = 69.18 kN/m 0.267

Equiv. udl soldiers 69.18kN/m

7

SOLDIER LOADING DIAGRAM

Output

By inspection of the pressure diagram, and from experience consider installing ties at positions M to Q as shown in diagram.

2700 Soldier

350

2400

Q

1500 P 1200 joint

5000 2600

O 1200

2700 Soldier

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Worked Example One

N 1000

69.18 0

M

150

Consider as a beam loaded with a concrete pressure and supported at the tie positions. The spans are not equal and there is a variation of load shape such that it is not feasible to analyse the soldier using the loading cases in the Formwork Guide. A computer beam analysis program can be used, or a manual calculation method such as by Moment Distribution. Although there is great dependence on computers such manual methods are still relevant, and will be used in this worked example. 1450

950

250 (0.09)

(13.01)

50

41.79 (30.29)

1500 Q

P

69.18

(6.85)

(50.15)

(83.02)

1200

1200 O

(10.38) (69.18)

1000 N

150 M

The Loading Diagram In the span OP the concrete pressure diagram changes direction 250mm from O. If the pressure diagram slope were extended to the tie position O, the increase in loading is only 0.09 kN. This simplified assumption will be used in calculations. The total load per soldier is:30.29 + 50.15 + 13.01 + 6.85 + 83.02 + 69.18 + 10.38 = 262.88 kN

8

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Worked Example One The anlysis of the soldier by moment distribution will determine the tie loads and required soldier characteristics.

Output

Moment Distribution Consider the end spans PQ and NM as propped cantilevers. These spans therefore have a stiffness reduction factor of 0.75 KQP = 0.75 1 1.5 = 0.5 Distribution Factors

KPO = KON = 1 1.2 = 0.83 0.5 = 0.38 (0.5 + 0.83) 0.83 DFOP = = 0.50 (0.83 + 0.83) 0.75 DFNM = = 0.47 (0.75 + 0.83)

DFPQ =

KNM = 0.75 1 1 = 0.75 hence DFPO = 0.62

DF’s

hence DFON = 0.50 hence DFNO = 0.53

Calculate fixed end moments prior to distribution using LC’s (Values of T, w and L taken from Loading Diagram on page 8 ) FEM PQ Approx. = 0.133 TL = 0.133 x 30.29 x 1.5 = 6.04 kNm 2 FEM PO Approx. = -0.0833 wL - 0.0667 TL = - (0.0833 x 41.79 x 1.22 ) – (0.0667 x (13.01 + 6.85 + 0.09) x 1.2) = -6.61 kNm 2 FEM OP Approx. = 0.0833 wL + 0.10 TL = (0.0833 x 41.79 x 1.22 ) + (0.10 x (13.01 + 6.85 + 0.09) x 1.2) = 7.40 kNm 2 FEM ON, FEM NO Approx. = -0.0833 wL = 0.0833 x 69.18 x 1.22 = -8.30 kNm 2 FEM NM Approx. = 0.125 wL = 0.125 x 69.18 x 1.02 = 8.64 kNm 2 FEM MN Approx. = -0.50 wL = -0.78 kNm = -0.50 x 69.18 x 0.152

Appdx B LC 12 LC 2 + 9

LC 2 + 9

LC 2

LC 4 LC 3

Moment distribution follows on next page

9

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Worked Example One Q

P

DF’s

0.38

O 0.62

0.50

N 0.50

0.53

M 0.47 +0.78

-6.61

+6.04 +0.22

FEM Distn

+7.40 +0.35

+0.45

-8.30 +0.45

+8.30 -0.03

+0.39 -8.64 -0.02

CO Distn

-0.09

+0.23 -0.14

+0.18 -0.08

-0.02 -0.08

+0.23 -0.12

-0.11

CO Distn

+0.02

-0.04 +0.02

-0.07 +0.06

-0.06 +0.07

-0.04 +0.02

+0.02

+6.19

ν

4.13

^4.13

^10 10 ^20.19 5.97

5.97

Q

24.32

-0.78

-6.19 +7.94

-7.94 +8.36

-8.36 0.78

^5.16 v6.62

^6.62 v 5.16

^ 6.62 ^ 6.97 v 6.97 v 6.62

^ 8.36 ^0.78 v 0.78 v 8.36

^25.07 ^25.07 ^ 6.65 ^13.30

^41.51 ^41.51

^34.59 ^34.59

30.26

41.16

42.17

54.58

39.83

80.99

P

O

41.86

84.03

N

27.01

-0.78

Max BM 8.36 kNm Elastic Shear

10.38

Static Shear

10.38

Summation Shear

37.39

M

Reactions and tie capacity Hence Maximum Reaction is at Soldier N of 84.03 kN. i.e. The Maximum tie load is 84.03 kN.

Reactions

Max tie load is 84.03kN

Check summation of reactions = 261.87 kN compared to loading diagram value on page 8 of ΣW = 262.88 kN. Hence OK. From the Design Brief tie SWL is 90 kN Hence OK. See Section 3.5 on Form Ties.

Tie SWL 90kN OK

Comment The reactions produced by moment distribution compare well with the results from a computer analysis of the more precise loading pattern.

Q

P

O

N

M

5.97

54.58

80.99

84.03

37.39

Mom Distn

5.9

54.6

82.1

81.3

39.0

Computer

(Computer analysis software used was BeamPal ) The computer also indicated that the bending moment at the soldier joint was 1.93 kNm. (Hence joint OK ).

10

Joint OK

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Worked Example One

Free span bending moments :(Values of T, w and L taken from Loading diagram on page 8)

Output Appdx B

MQP approximately 0.128 TL = 0.128 x 30.29 x 1.5 = 5.82 kNm 2 MPO approximately 0.125 wL + 0.128 TL = (0.125 x 41.79 x 1.22 ) + (0.128 x (13.01 + 6.85 + 0.09) x 1.2) = 10.58 kNm 2 2 MON approximately 0.125 wL = 0.125 x 69.18 x 1.2 = 12.45 kNm 2 2 MNM approximately 0.125 wL =0.125 x 69.18 x 1.0 = 8.65 kNm

LC 8 LC 1 + 8

- 6.19

-7.94

- 8.36

- 0.78

0 Q

LC 1 LC 1

+ 2.73

P

+3.52

B.M. Diagram showing net B.M’s

O

N + 4.30

M + 4.08

The maximum bending moment is 8.36 kNm occurring at support N. The net B.M. at the joint in the soldier is established by inspection from the diagram and is shown to be approximately 2.0 kNm. (Confirmed by computer analysis as 1.93 kNm. )

Joint BM 2.0 kNm

The selection of the soldier may be made at this stage, see also Section 3.2.3, or if predetermined in the design brief, the required properties are checked against the suppliers stated safe working values. As this design is using the actual applied loads it is checked against the stated swl, and NOT against the characteristic values of the items. Factors to be considered would be:A. Maximum positive moment in a span. (4.30 kNm approx.) B. Maximum negative (support) moment (8.36 kNm) which occurs in combination with a tie load of 84.03 kN. C. Moment at joint in soldier. (2.0 kNm approx.) D. Maximum tie rod load on soldier (84.03kN). Note:-Because of the nature of load spread from the waler plate, the precise evaluation of shear in the soldier is complex. Users should be aware that “shear in a soldier” is NOT the same as “tie rod capacity.” (See Section 3.2.3.1)

11

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Worked Example One WEIGHT OF WALL FORM

Output

The weight of one form panel, 5.05m high and 5.49m long will be 1665 = 16.65 kN (one face) 5.05 x 5.49 x 60 = 1665 kg = 100 (See Section 4.2.2 Table 20 ) Hence crane limit OK.

Weight of one panel 16.65 kN

WALL FORM STABILITY From the Design Brief the site is on the outskirts of Lichfield in the West Midlands, and is about 120 km from the nearest sea, in a flat river flood plain with an altitude of 55m above sea level. Consider the overturning moment and/or stability moment required for one panel of double faced formwork say 5.5m long. This assumes the worst case of one form panel erected on its own before the adjacent panel is fixed. Assume that one working platform is fitted, but in this calculation the self weight of the working platform is ignored. Using the simplified wind calculations for wall overturning given in Section 4.5.1.16.2 gives the following:

Use Section

4.5.1.16.2

Wind Factor Swind Using Figure 66, the fundamental basic wind velocity for Lichfield is:vb,map = 21.8 m/s (Note: A more conservative value would be to use the next highest isopleth as vb,map = 22 m/s) From Figure 67 the topographical factor for a flat river flood plain is:Twind = 1.00 (Fig 67 (a)) The Site altitude from brief was 55 m, hence A = 55. From Section 4.5.1.4

55  A    Swind  Twind  vb ,map   1     1.0  21.8   1  1000  1000   

Swind = 23.0

The reference height (z) is the height of formwork of 5.05m and length of panel exposed when one panel erected is 5.49m. Hence ratio of effective length / height is 1. Hence from Table 26 the net pressure coefficient is cp,net = 1.7.(Note if both panels were erected to full 10m long, the ratio becomes 2, giving cp,net = 1.6, hence using the single panel is worst case.) From the Brief the site is on the very outskirts of a town, and is considered in the country and more than 100 km from the closest shoreline. Hence from Table 25 combined exposure factor CEF = 1.90. The squared wind factor multiplied by combined exposure factor gives Swind x Swind x CEF = 23.0 x 23.0 x 1.90 = 1005

12

23.0

cp,net=1.7

CEF = 1.90

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Worked Example One Hence from Table 28 for h = 5m gives the wind overturning moment as 8.59 kNm for 1000 and 12.87 kNm for 1500. Hence interpolating for 1005 the maximum wind moment is 8.623 = 8.62 kNm per m.

Output

Maximum wind overturning moment per 5.5m long panel is:8.62 x 5.5m = 47.41 kNm / panel

Wind o/t per panel 47.41 kNm

From Table 28 the working wind moment per metre is 4.00 kNm/m hence working o/t moment per panel is:4.0 x 5.5m = 22kNm/panel

Wking o/t/ pnl 22 kNm

Minimum Stability Force The minimum stability force (Section 4.6.2) for wall formwork is 10% of formwork self weight. Allowing for two forms erected gives:Min Stability force = 16.65 x 2 x 10% = 3.33 kN (acting ¾ up form)

Min Stab o/t / panel 12.86 kNm

Hence minimum stability overturning is 5.15 x ¾ x 3.33 = 12.86 kNm (Note: the overturning is about the base so full height 5.15m used.)

Working Platform Loading Consider one working platform 800mm wide fitted to the top of the forms on the outside of the soldiers. See sketch at page 14. The self weight is ignored. The imposed loads will cause overturning and the approximate lever arm about the centre of the wall (say) will be:(800/2) + 225 + 150 + 20 + (450/2) = 1020mm = 1.02m From Section 4.3.2.2 the minimum working area load on any platform should be 0.75 kN/m2. Hence working area load per platform/panel is:0.8 x 5.5m x 0.75 = 3.30 kN / panel giving an overturning moment of 3.30 x 1.02m = 3.37 kNm. From Section 4.3.2.3 the imposed construction load during concrete placing in wall form should be 1.50 kN/m2. Hence imposed load per platform/panel is 0.8 x 5.5m x 1.5 = 6.60 kN giving an overturning moment of 6.60 x 1.02m = 6.73 kNm.

Nominal working o/turning 3.37 kNm

Full constn o/turning 6.73 kNm

The Three Stability Checks From Section 5.1.5 the three design checks for overturning are: ONE - Maximum wind plus nominal access load on platform 47.41 + 3.37 = 50.78 kNm /panel TWO - Working wind plus full construction load on platform 22.00 + 6.73 = 28.73 kNm /panel THREE - Minimum stabilty plus full construction load on platform 12.86 + 6.73 = 19.59 kNm /panel

13

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Worked Example One From Section 5.1.5 the minimum factor of safety on overturning is 1.2, hence for each 5.5m long panel, support is required to resist an overturning moment of 50.78 x 1.2 = 60.94 kNm / panel.

Output Design panel o/t 60.94 Using two push-pull props connected to a point 3.0m up the form on one kNm side only, at an angle of 60° to the horizontal. Suitable anchorage into a base slab is available. By Pythagoras Theorem the 60° propping forms a 2, 1, √3 triangle. Hence by similar triangles per prop position gives:60.94  10.16 kN Horizontal force 3m  2

2 Therefore load in prop is  10.16   20.32 kN 1

Push/pull Prop swl 20.32 kN

The anchorage for each prop will be required to resist the design forces as follows: Vertical = Horizontal

20.32  3 = ± 17.60 kN per anchorage 2

= ± 10.16 kN per anchorage

Anchor forces

Note: The factor of safety of 1.2 has already been included, so these values are the required restraint forces at the anchorage.

2.15m

√3 3m

2 1

60° 1730

Anchorage

Warning : If using kentledge blocks in lieu of anchorages, check the implications of the raking prop in tension. See Section 5.1.5

14

Warning

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Worked Example Two

Worked Example Two – Square Column Formwork DESIGN BRIEF

COLUMN (square)

Design the formwork for an in situ reinforced concrete column of square plan 975mm x 975mm and 4.60m high. There are six columns to be constructed. The Client’s specification requires an F3 high class finish without ties and no features on the faces. The corners are to be cast with 20mm x 20mm chamfers. The base to each column will have a 100mm high kicker for alignment of the formwork.

Output

20 x 20 chamfers 100mm kicker

The concrete will contain a blend of Portland cement with less than 40% fly ash Group B and will have a pumping admixture, but no retarder. The work will be carried out during a short construction period in summer and the expected concrete

temperature will be 15OC. The form will be filled in just over one hour.

T = 15OC

The contract has already purchased stocks of 18mm Finnish WISAForm Birch through plywood in 1220 x 2440 sheet sizes. It is 13ply.

Ply 18mm WISA-Form Birch 13ply

The TWD has decided to use Strength Class C24 backing timbers placed vertically and with structural steel yokes at varying centres horizontally to suit the concrete pressure. Stability and alignment will be from an erected scaffold around the column form, previously used for fixing the reinforcement.

C24 timber

(This example only designs the formwork.)

CONCRETE PRESSURE CALCULATIONS Volume of concrete in the column is (4.60-0.10) x 0.9752 = 4.28m3 and it is placed quickly in just over one hour. Thus the vertical rate of rise is about 4 metres per hour. The concrete is Group B at a temperature of 15 OC. Using Table PBB, in Section 4.4.3.1, for a column, and for a height of 4.5m and Rate of Rise of 4m/hr gives a design pressure of 95kN/m2.

Pmax = 95kN/m2

15

Output Design pressure, Pmax = 95kN/m2 The equivalent fluid head is, thus 95 = 3.80m 25 The shape of the pressure diagram is shown on the left.

3800

4500

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Worked Example Two

95kN/m2

PLYWOOD The structural properties of the Finnish 18mm WISA-Form Birch through 13 ply face contact material are given in Appendix D, Table D-W. The plywood sheet size of 1220 x 2440 can be used most economically in this example by allowing sheets on two faces to be left uncut, thus only two faces need the plywood trimmed on one side to 975mm. This has the advantage that, because of the ‘lay-up’ of WISA-form plywood (with the strong direction in the 1220 direction), the face grain of the plywood will be parallel to the span. (See Section 3.3.2.4, Fig.39). Assume 75mm x 150mm backing timbers of strength class C24. See detail

75 x 150

150

A

20 x 20 Hardwood fillet

B

975 1375

From Appendix D, Table D-W:Bending stiffness parallel EI = 3.63kNm2/m Moment of resistance parallel fZ = 0.968kNm/m Shear force parallel qA = 15.96kN per m width Inspection of the proposed vertical timbers layout gives plywood spans of :-

16

45

Ply fitted with 2440 vertically

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Worked Example Two Side (A) 250mm 975  75  5 Side (B) of = 224mm 4 (Often, as in this case, the spacing will be different from side to side with the timbers designed for the worse case.) Using Loading Case 58 (Appendix B) for the plywood across 4 supports gives :Maximum moment = 0.095 wL2 Where w = 95kN/m Thus limiting span of the plywood is 0.095 × 95 × L2 = 0.968  Lmax = 0.328m = 328mm OK Checking the shear in the plywood for side (A) gives Maximum shear = 0.525 w (L-B-t) = 0.525 × 95 × (0.250-0.075-0.018) = 7.83kN c.f. allowable 15.96kN  OK. Checking deflection of the plywood gives:wL4 95  0.250 4  = 0.0066   0.0066  EI 3.63 = 0.00067m = 0.67mm 250  OK. Allowable is = 0.93mm 270 Check to see whether two timbers can be omitted near the top. The plywood would span three supports at 500mm. Thus, from Loading Case 21 (Appendix B) the relationship between the moment of resistance of the plywood and the concrete pressure at the limiting point (PL) is given by:= 0.125 × PL × 0.502 Moment = 0.125 wL2 = 0.968kNm/m 0.968 Thus PL = = 30.98kN /m2 2 0.125  0.50 30.98 Thus the timbers could be reduced at a level = 1240mm from 25 the top, by considering the moment of resistance only. In practice, on such a column of 4.6m height, all the timbers would be taken full height and not stopped short. The reaction on to the backing timbers is thus a maximum at the bottom of the form. UDL on timbers = 95 × 0.250 × 1.0 = 23.75kN/m. (The factor 1.0 is from Loading Case 58.)

Output Plywood span 250mm & 224mm

Ply  = 0.67mm

17

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Worked Example Two VERTICAL BACKING TIMBERS The Design Brief assumes 75 x 150 timbers of strength Class C24. There are more than 4 timbers and load sharing can be assumed. The structural properties of the timber are given in Section 3.3.1.4 Table 8 Bending stiffness EI = 158.04kN/m2 Moment of resistance fZ = 2.696kNm Shear force qA = 10.99kN Bearing stress = 3250kN/m2 (no wane) 2570kN/m2 (with wane) The selection of the yoke centres vertically is by experience, and will vary to suit the height of the column and the design concrete pressure. They will be closer together near the bottom of the column. The design procedure involves selecting an arrangement and carrying out a moment distribution to establish the design criteria for the vertical timber members. Note that the design will ignore the 100mm kicker and assume 4.6m loaded length. 23.75kN/m 19.06kN/m

(1.76)

(4.75)

(3.13)

12.81kN/m

(14.25) (14.30)

(5.69)

4.38kN/m

(12.81)

(5.91)

(1.53)

0.20m 0.7m

1.00m

1.35m

A

B

0.60m

0.75m

C

D

E

LOADING DIAGRAM

W = 1.53+5.91+5.69+12.81+3.132+14.30+1.76+14.25+4.75 = 64.13kN Assume 5No. yokes at A, B, C, D, and E. The fixed end moments are calculated using Appendix B, Loading Cases 2, 3, 4, 9, 10, and 12, assuming spans BA and DE are propped cantilevers. Moment MA = 0.36 and ME = 0.48 FEMs MBA = 1.00 + 1.02 = 2.02 = 1.28 MCB = 1.07 + 0.31 = 1.38 MBC = 1.07 + 0.21 MCD = 0.89 + 0.09 = 0.98 MDC = 0.89 + 0.13 = 1.02 MDE = 1.07

18

Output

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Worked Example Two I I = 0.556 kBC = = 1.00 1.35 1 .0 I I KDE = ¾ × = 1.250 kCD = = 1.333 0.60 0.75 Distribution factors BA/BC 0.36/0.64 CB/CD 0.43/0.57 DC/DE 0.52/0.48

Stiffness

A DF

1

B

C

D

0.36 0.64

0.43 0.57

0.52 0.48

+0.36 -0.36 Co FEM Co Co BM

Output

kAB = ¾ ×

+0.36 -0.36

E +0.48 -0.48

-0.18 +2.02 -1.28 -0.20 -0.36 -0.09 +0.03 +0.06 +0.05 -0.02 -0.03 +1.65 -1.65

+1.38 -0.17 -0.18 +0.10 +0.03 -0.03 +1.13

-0.98 -0.23 -0.05 +0.13 +0.03 -0.03 -1.13

+0.24 +1.02 -1.07 -0.10 -0.09 -0.12 +0.06 +0.06 +0.07 -0.04 -0.03 +0.89 -0.89

+0.48 -0.48

0.36kNm

1.65kNm

1.13kNm

0.89kNm

0.48kNm

Shear (static)

1.53 2.96 1.90

2.96 6.41 3.79 1.04

6.41 7.15 2.09 0.594

7.15 7.13 1.17

7.13 4.75

(elastic)

0.96 1.53 3.90

0.96 0.52 7.71 7.97

0.52 0.32 7.98 8.06

0.32 0.68 8.00 7.81

0.68 6.45 4.75

A

B

C

D

E

5.43kN

15.68kN

16.04kN

15.81kN

11.20kN

Reactions

The maximum Negative Bending Moment is 1.65kNm. (Allowable is 2.696kNm  OK) By inspection, the maximum ‘free’ bending moments in each of the spans between the supports A, B, C, D, and E have values which result in ‘nett’ bending moments in the spans that are less than the bending moments at the supports. If this is not apparent, by inspection, those values must be calculated and the worst case bending moment derived. The maximum Shear Force is 8.06kN. (Allowable is 10.99kN  OK) Check on loading gives: reactions = 5.43 + 15.68 + 16.04 + 15.81 + 11.20 = 64.16 Equates well with Loading diagram value, of 64.13.

Moments

Shears Reactions Max. B.M. 1.65kNm

Shear 8.06 kN

The maximum load on a yoke will be at yoke C. The load is 16.04kN per timber (at 250mm c/c). Check design of yoke and include a check on the bearing stress of the timber on to the yoke flanges.

19

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Worked Example Two DESIGN OF STEEL YOKES – HORIZONTAL

Output

Consider the use of twin 125mm x 65mm parallel flange channel(PFC) sections in Grade S275 mild steel for the horizontal yokes. End tie rods of 20mm dia. mild steel ‘all thread’ rod will connect opposite pairs of channels. (Note: There will actually be two horizontal yokes, one above the other, to restrain each pair of opposing formwork faces – see plan on page 16.) Properties of 125 x 65 PFC from Property Tables are :Weight = 14.8kg/m Depth of section (h) = 125mm Width of section (b) = 65mm Web thickness (s) = 5.5mm Flange thickness (t) = 9.5mm Radius of Gyration (ry) = 20.6mm Moment of Inertia (Ixx) = 483cm4 Elastic Modulus (Zxx) = 77.3cm3 Thus h/t = 125 = 13.16 9.5 Consider Side A and Side B (see plan layout on page 16). The loading patterns for Side (A) is:-

187.5

16.04kN

16.04kN

8.02kN

250

250

16.04kN 250

See page 16

8.02kN 187.5

250

1375

F

(A) G

The loading on Side (B) is determined by proportioning the loads in the ratio of the plywood spans (225/250). The loading patterns for Side (B) is:10.44kN 237.5

H

225

14.43kN

14.43kN 225

14.43kN 225

1375

10.44kN 225

237.5

(B) J

16.04  4 Thus the tie load for each side is = 32.08kN. 2 From Section 3.5.3 the minimum factor of safety on mild steel ties is 2.5.

20

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Worked Example Two Thus the required minimum failure load is32.08 × 2.5 = 80kN (Note: Typical proprietary M20 ties state failure loads of 120kN.)

Output Tie load 32.08kN

Check the actual bending moment for both Side A and Side B. For Side A (FG) by taking moments about centre of the span gives:1.375 MFG = 32.08 × - (8.02 × 0.500) - (16.04 × 0.250) 2 = 22.06 - 4.01 - 4.01 = 14.04 kNm And for Side B (HJ) by taking moments about centre of span gives:1.375 MHJ = 32.08 × - (10.44 × 0.450) - (14.43 × 0.225) 2 = 22.06 - 4.70 - 3.25 = 14.11 kNm Thus, the maximum bending moment is 14.11 kNm. From BS 5975, Annex K, the effective length of the twin channels, le, is le = 1.2 × L + 2 × D = 1.2 × 1375 + 2 × 125 = 1900mm l 1900 = 92 Thus, e = ry 20.6 From BS 5975, Table A.1, the permissible bending stress, pbc = 155N/mm2 14.11  106 M Actual bending stress, fbc = = = 91.27N/mm2OK 3 Z 2  77.3  10 (Note the “2 × “ in bottom line for the twin channels)

Check bearing stress on backing timbers:16.04 fb =  106 = 1715kN/m2  OK. 72  65  2 (Allowable is 3250kN /m2 without wane and 2570kN/m2 with wane.)

Max BM is 14.11kNm

Bending stress 91.27N/mm2

Bearing stress 1715kN/m2

(Note use of finished timber size (72mm) from 3.3.1.4, Table 10.) See sketch on next page.

21

Output

65

2No. 125x65x15 PFC

75x150 timber

Check the tie rod connection to the twin channel members. Assume a 100x100x6 plate washer.

b + n2

M20 All Thread rod 100x100x6 plate washer

21.5

30

125

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Worked Example Two

Web tw = 5.5mm

NA b + n1

See BS 5975 Annex J for web buckling and web crushing formulae. Web buckling R = b  n1  tw pc

where tw = 5.5mm

b  n1 

 125   6 = 30 + 2×   2  = 167mm

From BS 5975 Annex J, Table J.1, the effective length of the web 7.0  D divided by radius of gyration is l ≈ = 159 ry tw Thus, from BS 5975 Annex A, Table A.2, the allowable compressive stress is 37N/mm2 and the limiting value of web load is:R =

167  5.5  37 1000

Actual load per channel is

= 33.98kN per channel 32.08 = 16.04kN 2

Thus, web stiffeners not required for web buckling.

22

Web stiffeners NOT required for buckling

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Worked Example Two

Output

Web crushing R = b  n2  tw pb

where tw

= 5.5mm

b  n2

 21.5  = 30 + 2×   O  tan30  = 104mm

From BS 5975 Annex A, Section A.1 f), the bearing stress should not exceed 210N /mm2. Thus, the limiting value of web load is:R =

104  5.5  210 = 120.12kN per channel 1000

Actual load per channel is 16.04kN. Thus, web stiffeners not required for web crushing. Twin 125x65x15 Parallel Flange Channel (PFC), Grade S275 mild steel acceptable without web stiffeners.

Use Twin 125x65 PFC Web stiffeners not required

23

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Worked Example Three

Worked Example Three – Forces in Single-faced Wall Formwork DESIGN BRIEF

Alternative One

Alternative Two Output

Design the formwork for an insitu reinforced concrete wall 2.85m high, cast against an existing rock face without any fixings into the rock.

2.85m high

Two alternative support schemes are considered, either as conventional formwork with soldiers (Alternative One) or using proprietary formwork panels and a proprietary A frame system (Alternative Two). {Note: Always refer to the supplier for technical data on system.} The proprietary panels are stated in the supplier’s data sheets to have a maximum permissible design concrete pressure of 60 kN/m2.

Pmax 60 kN/m2

{Hence for this example the panel limitation will be applied to both alternatives.} The base slab with an integral 100mm high kicker has already been cast. Any anchors or ties necessary can be cast into the base as required.

100mm kicker

The client’s specification states “Finish Class F2” with the deflection limited to 1/270th of the span of any formwork member. There are no features required on the wall. Through ties are NOT allowed.

Class F2

As the exact thickness of the wall is not determined, the client intends to use a retarded Group B concrete. 24

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Worked Example Three

The TWD has been advised that concreting will be by skip. The design will assume a concrete temperature for April of 10°C.

Output T = 10°C

Assume the self weight of the formwork in both Alternatives is 45 kg/m2 = 0.45 kN/m2. See Table 20 Section 4.2.2.

SW 0.45kN/m2

Ignore the self weight of the working platform. During placing of concrete assume an imposed load on working platorms of 1.50 kN/m2 . See Section 4.3.2.3

Col 1.50 kN/m2

Consider design per metre run of wall. CONCRETE DESIGN PRESSURE and RATE of RISE CALCULATION Using the concrete information in the Design Brief and the method in Section 4.4, the stated maximum design pressure is 60 kN/m2.

Concrete density 25 kN/m3 Concrete Group B (Retarded Concrete) Concrete Temperature 10°C The full fluid head of concrete would be 2.75 x 25 = 68.75 kN/m2. Hence the limiting pressure is the proprietary formwork panels.

Pmax 60 kN/m2

Thus from Table RWB (Section 4.4.3.2 ), for temperature 10°C and for Design Pressure 60 kN/m2 gives R = HD for 2.5m height. and R = 1.6m/hr for 3m height. The notation HD implies that rate of rise is not relevant as pressure is nearly height x density (i.e. 2.5 x 25 = 62.5 ) and, as the wall is between 2.5m and 3.0m, by inspection a reasonable maximum rate of rise would be 2.0m/hr i.e. fillling the wall in about 1hr 20mins. The equivalent fluid head of concrete is

R = 2 m/hr Fill in 80 mins minimum

Pressure 60.   2.4m Density 25.0

The concrete pressure and the resulting 2400 force diagram

2750 Ft

350 60 kN/m2 Pressure diagram

1250

275

Fr

base Force diagram 25

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Worked Example Three

Concrete pressure develops a horizontal force of:60 2400 Ft = × 2.4 × 1 = 72 kN acting + 350 + 100 = 1250 mm 2 3 Fr = 60 x 0.35 x 1 = 21 kN acting

350 + 100 = 275 mm 2

Total force from concrete pressure FT = 72 + 21 = 93 kN Overturning moment from concrete pressure force about base is:OTM = (72 x 1.25 ) + ( 21 x 0.275) = 90.00 + 5.78 = 95.78 kNm/m

Output

Lateral force 93 kN OTM base

Platform Loading Construction operation load on platform is 1.50 x 0.8 = 1.2 kNm/m. Lever arm of loading about face of wall is 800 Alternative One 170 + 225 + = 795mm 2 800 Alternative Two 350 + = 750mm 2 Weight of Formwork Weight of formwork is 0.45 x 2.85 (say) x 1.0 = 1.28 kN/m run. Assume it acts approximately 80mm from face of wall. FORCES - ALTERNATIVE ONE

Take moments about E to find uplift force Tuplift at position C. The effect of the imposed load on the working platform cannot be considered in this case as it is acts as a restoring moment for considering uplift. The c.o.l. may actually not happen if there are only few people on the platform! Overturning moment about base is 95.78 kNm The restoring moment about position E is = ( Tuplift x 1.225 ) + {1.28 x (1.225 + 0.285 – 0.080)} = ( Tuplift x 1.225 ) + 1.83 Restoring moment must be ≥ overturning moment hence 95.78 - 1.83 Tuplift = = 78.19 kN/m run 1.225 Refer to Section 3.5.7 for relevant factors of safety on anchors. {Note: the minimum factor of safety of 1.2 (Section 5.1.5) refers to overturning of freestanding double faced formwork, and is not applicable when calculating the required uplift fixing force.}

26

Tuplift 78.19 kN/m

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Worked Example Three

Forces – Alternative One continued

Output

To establish maximum load in prop DE consider worst case of overturning about position C to find horizontal force Hd required at position D to stabilise the formwork. The c.o.l. is now relevant and is included in the calculation. Hence: 95.78 + {1.2 x (0.795 – 0.285) ≤ Hd x 1.925 Hence

Hd =

95.78 + 0.61 = 50.07 kN 1.925

At position C, the lateral force to be resisted is:FT – Hd = 93 – 50.07 = 42.93 kN This example uses a separate fixing in the slab at position C to resist the lateral shear force, hence the anchor for uplift is only designed to take tension. By Pythagoras Theorem the length DE is

1.2252 + 1.9252 = 2.282m

Hence by similar triangles the maximum load in prop DE is:2.282 PropDE =  50.07 = 93.26 kN per metre run 1.225 The vertical load downwards at position E is thus:1.925 × 50.07 = 78.68 kN per metre run Vertical LoadE = 1.225

Prop DE 93.26 kN/m run

Design of Wall Formwork The formwork would be designed in a similar way to Worked Example One using the appropriate concrete pressure and material properties. Output from Design The output from Alternative One gives following information for TWD to design the fixings and propping necessary per metre of wall as:Uplift at position C Lateral Shear Force at C Axial Load in prop DE Vertical load at position E Horizontal force at position E

78.19 kN 42.93 kN 93.26 kN 78.68 kN 50.07 kN

Fixings Output per metre

27

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Worked Example Three FORCES - ALTERNATIVE TWO

Output

Design philosophy The method of calculating the forces in Alternative Two is different from that in the previous calculations in Alternative One. The A frame in use is a proprietary welded structure with internal bracing. This means that it acts as a rigid body under load so that lateral forces, such as concrete pressure forces, are transmitted through the structure to the base, and allows the angled tie and anchor to resist the lateral forces with no residual overturning moment. {NOTE: The calculation in this example is intended to illustrate the magnitude of the forces in principle on a typical example. Always refer to the proprietary equipment supplier for the geometry and technical data for the A frame used. } Forces acting The maximum lateral force from the concrete pressure is 93 kN. The tie and anchor assembly at F acts at 45° to the base, so the axial Load in load in the tie assembly will be 93  2 = 131.50 kN per metre of wall. anchor This load in the tie assembly will in turn generate a downward vertical 132 kN/m load of a similar amount, i.e. 93 kN. Refer to Section 3.5.7 for relevant factors of safety on anchors and the tie rods used to connect to such anchors. {Note: the minimum factor of safety of 1.2 (Section 5.1.5) refers to overturning of freestanding double faced formwork, and is not applicable when calculating the required tie force.} Diagram of Forces acting on system c.o.l. self weight

Concrete Force Anchor Force

F

G

(a) Actual Force location

28

c.o.l. Concrete Force

H

self weight

F

G

H

(b) Resolved Forces for calculation

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Worked Example Three Forces Alternative Two continued

Output

To establish the maximum vertical load VH at position H take moments about G, allowing for the self weight of the formwork and including the c.o.l. as it is to the right of support G and increases the moment. (The self weight of the A frame is ignored) See (b) page 28. Moment about position G is 95.78 - {1.28 x (2.140 + 0.180 – 0.080 – 1.885)} 800 )} + {1.2 x (2.140 + 0.180 – 1.885 – 350 2 - (93 x 0.255) = 95.78 - 0.45 + 0.38 - 23.72 = 71.99 kNm 71.99 Hence the vertical reaction at H is VH = = 38.19 kN. 1.885 To establish the maximum vertical load VG at position G take moments about H, allowing for the self weight, but excluding the c.o.l. as it is to the right of support G and would decrease the moment. See (b) page 28. Moment about position H is:95.78 - {1.28 x (2.140 + 0.180 – 0.080)} - (93 x 2.140) = 95.78 - 2.87 – 199.02 = -106.11 kNm 106.11 Hence the vertical reaction at G is:- VG = = 56.29 kN. 1.885

Maximum VH 38 kN

Maximum VG 56 kN

There is no requirement to check the design of the proprietary suppliers A frame as it is a suppliers item, but the TWD MUST check that the loads applied in the Alternative Two design are acceptable for the type of A frame envisaged. Output from Design The output from Alternative Two gives the following forces per metre run of wall as information for the TWD to design the fixings and supports:Force in 45° angled tie/anchor assembly at position F 132 kN Vertical load at position G 56 kN Vertical load at position H 38 kN

Output per metre

NOTE: To reduce tip deflection at B caused by elongation of the tie rod in tension under load, the anchor/tie asembly may be pretensioned. A kicker must be used. Refer to the suppliers data. WARNING: The use of proprietary panel systems and A Frames will usually have predetermined positions for the A Frames.

29

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Worked Example Four

Worked Example Four – Bridge Deck Soffit Formwork Output

DESIGN BRIEF Insitu deck Parapet added

1000 9400 6400

Foundations Design the soffit formwork for an insitu reinforced concrete road bridge with voided cross-section spanning over a dual carriageway in a cutting. See isometric above for basic layout. 1) The falsework supports will be designed separately to BS 5975, and will be assumed erected for less than two years. {Comment: The plan layout of the falsework affects the location of the supports, and therefore the span of the bearers . In this example one of the support plan module dimensions will be assumed 1200mm. }

< 2 yrs

2) The structure is located at Park Bottom, Redruth, Cornwall in an area of open countryside with no wind breaks. 3) The altitude of the foundations is shown on the drawings as 25.10 AOD.

25.10 AOD

4) The site is about 2 km from the sea.

2 km sea

5) The parapet edge upstand will be cast separately after the bridge has taken up its self weight deflection. Hence the parapet is not considered in these soffit calculations. 6) The underside of the concrete deck is level and 6.4m above the falsework foundations.

30

H = 6.4m

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Worked Example Four 7) The client’s specification states soffit class “Finish Class F2” with the deflection limited to 1/270th of the span of any formwork member. 8) The proprietary polystyrene void formers are 650mm diameter and are assumed tied through the soffit to the secondary bearers. (See Section 3.11.1 Figure 56 Example 2. )

Output Class F2 δ = 1/270

9) The self weight of the void formers is ignored. 10) Concrete density assumed 25 kN/m3. 11) Timber used for the primary and secondary bearers are assumed constructional softwood timber to strength class C16. 12) The orientation of the timber bearers affects the design. In consideration of the physical positioning of the strapping to hold down the voids, it is preferable to place the secondary bearers at right-angles to the voids. This permits some lateral tolerance along the voids when fixing the straps. Hence primary bearers are parallel to the voids. 13) The plywood is assumed as Canply COFI-FORM SP Plus 17.5mm 7 ply plywood (e.g. Ultraform). See Appendix D Table D-S.

LOADING

Timbers C16

Plywood Ultraform Loading

1000

Parapet cast after main deck slab

1000

Fixing

1000

Secondary 4600

1200

1200

Primary

{Comment: The fixing in face is to reduce differential movement during future casting of parapet.} As the voids are at 1m centres, consider the load per square metre on the structure using the density of concrete from brief at 25 kN/m3. Solid concrete slab 1.0m thick

1.0 x 25 x 1 x 1 = 25.0 kN/m2

31

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Worked Example Four 2

 0.65  2 Less void volume π × r 2 × 1.0  3.1416     1.0  25  8.30 kN/m  2 

=

16.70 kN/m2

(Section 4.2.3) ) =

0.50 kN/m2

Hence equivalent solid slab load is 25.00 – 8.30 Plus self weight of soffit forms

(Note this includes the plywood at 0.11 kN/m2 ) Plus construction working area load (Section 4.3.2.2)

= 0.75 kN/m2

Hence load on general area of formwork is 17.95 kN/m2.

Output

General 17.95 kN/m2

Additional transient concrete load on 3m x 3m area based on 10% of weight 10% x 16.70 kN = 1.67 kN/m2 (Limits are 0.75 min to 1.75 max. ) (Refer to Section 4.3.2.4 and Table 21.) Hence maximum load for formwork on the primary bearers (i.e. on forkheads) is 17.95 + 1.67 = 19.62 kN/m2.

Maximum 19.62 kN/m2

{Comment: The 19.62 kN/m2 is the load that the falsework supports

at forkhead level. In addition the falsework will have to transmit its self weight, plus any platform loading from a soffit platform (See 4.3.2.2), to the foundations.} Design load on plywood When concreting the deck, there will effectively be a pressure head on the soffit plywood equivalent to the full pressure head of the 1.0m thick concrete slab. Hence the plywood requires to be designed not for the general area load, but for the full depth, plus self-weight and any relevant construction operations load. At worst case, the additional transient concrete load on a 3m x 3m area could be applied, hence the design load on plywood is:(25.0 x 1.0) + 0.75 + 1.67 + 0.11 = 27.53 kN/m2.

Ply 27.53 kN/m2

Design Load on Secondary Bearers In this example, the voids are strapped to the underside of the secondary bearers, hence the equivalent design load on the secondary bearers is (25.0 x 1.0) + 0.75 + 1.67 + 0.50 = 27.92 kN/m2. {Comment - Assumption to use the full weight of soffit forms.}

32

Secondy 27.92 kN/m2

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Worked Example Four PLYWOOD

Output

The plywood is 17.5mm COFI-FORM SP Plus plywood. It is assumed that the width of the bearers acting as secondaries supporting the plywood will be 75mm. It is also assumed that the plywood will span in its strong direction, i.e. with face grain parallel to the span. (See Figure 39) Structural safe working properties of the plywood are stated in Appendix D, Table D-S for the plywood as: Bending Stiffness

EI = 3.21 kNm2/m

Moment of Resistance Shear Load

fZ = 0.600 kNm/m qA = 8.62 kN/m

The plywood will span over more than 4 supports and B > 2t ( 75 > 38), hence Appendix B2 Loading Case 58 is applicable. Maximum Moment in plywood is Mmax = - 0.095 w L2 = 0.095 x 27.53 L2 = 2.615 x L2 Thus 0.600 = 2.615 x L2 Hence L = 0.479m = 479mm (Note the –ve sign indicates the hogging moment at the support. The plywood is considered symmetrical so the limit is ± 0.600 kNm/m.) Using 2440mm sheets of plywood a suitable module is thus:2440 2440 or Try 406mm = 406mm = 488mm 6 5 Check the shear on the 406mm span gives:Ss = 0.525 w ( L – B – t ) Ss = 0.525 x 27.53 x (0.406 – 0.075 – 0.019) = 4.51 kN cf allowable of 8.62 kN therefore OK in shear at 406 c/c Check deflection in a span with limit set as δ ≤ 1 / 270th-

wL4 27.53  0.4064  0.0066  EI 3.21 406  1.51 mm ACCEPT δ = 0.00153m = 1.53mm cf allowable 270 {Note – Very slightly over the deflection limit is reasonable as 0.02mm is hardly discernable in a span. The bending and shear are well within limits.} thus   0.0066

δply 1.53mm Ply span 406mm

33

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Worked Example Four SECONDARY BEARERS

Output

Use 75mm x 150mm strength class C16 constructional timber bearers at 406mm c/c. Try primaries at 1200mm apart to suit the falsework. Hence the secondaries span 1.20m.

Secondary

This also suits the use of standard 38mm scaffold boards to BS 2482 for erection as maximum span of Grade 1.2m board is 1.2m. {Comment: In this example the dimension was preset by the choice of modular support falsework. In practice this dimension would vary from contract to contract.} Hence from Section 3.3.1.4 Table 7, the safe structural properties for C16 load sharing 75 x 150 timbers in soffit formwork is:Bending Stiffness Moment of Resistance Shear Load Bearing stress

EI = 128.77 kNm2 fZ = 1.778 kNm qA = 9.67 kN 2,780 kN/m2 (without wane) 2,150 kN/m2 (wane permitted)

The design load on the secondary bearers is 27.92 kN/m2. Hence distributed load per bearer is 27.92 x 0.406 = 11.34 kN/m. Assuming the secondaries are supplied in minimum 3.95m lengths, and positioned over four primaries, Appendix B2 Loading Case 26 applies. Maximum bending moment is -0.10 w L2 = 0.10 x 11.34 x 1.202 = 1.633 kNm

< 1.778 hence OK.

Max Shear is 0.60 w L = 0.60 x 11.34 x 1.2 = 8.16 kN < 9.67 OK. Maximum Reaction from a secondary on to a primary is reduced because the actual value of w transmitted is reduced by the void uplift hence maximum reaction of bearer is 1.10 w L gives = 1.10 x 19.62 x 0.406 x 1.2 = 10.51 kN. Check bearing on twin 75 x 225 timbers. Using finished sizes from 10.51 = 1,014 kN/m2. Table 10 gives bearing stress = 2 × (0.072 × 0.072)

< 2,150 kN/m2 hence secondary timbers with wane permitted OK.

SecondY Max BM 1.63 kNm Max Shear

8.16 kN Max

Reaction 10.51 kN Second’y

With wane OK

Check deflection in a span with limit set as 1 / 270th:thus   0.00688

wL4 11.34  1.20 4  0.00688  EI 128.77

= 0.00126m = 1.26mm

34

1200 cf allowable = 4.44 mm 270

δsecond’y 1.26mm OK.

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Worked Example Four PRIMARY BEARERS

Output

Use twin 75mm x 225mm strength class C16 constructional timber bearers at 1.20m c/c spanning, say 1.5m. This dimension may be preselected by the choice of modular support falsework and vary by contract. Although there are twin timbers, load sharing is not considered for less than four members, hence from Section 3.3.1.4 Table 6, the safe structural properties for class C16 75 x 225 primary timbers in soffit formwork is:Bending Stiffness EI = 296.44 kNm2 Moment of Resistance fZ = 3.561 kNm Shear Load qA = 13.41 kN Bearing stress 2,530 kN/m2 (without wane) 1,950 kN/m2 (wane permitted) The design load on the primary bearers is 19.62 kN/m2. Hence the distributed load per bearer is 19.62 x 1.20 = 23.54 kN/m. Assuming the primaries are supplied in minimum 4.5m lengths, and positioned over four supports, Appendix B2 Loading Case 26 applies. Maximum bending moment is 0.10 w L2 = 0.10 x 23.54 x 1.502 = 5.297 kNm < (2 x 3.561) hence OK. Maximum Shear is 0.60 w L = 0.60 x 23.54 x 1.5 = 21.19 kN < ( 2 x 13.41) hence OK.

Primary Max BM 5.297 Nm Max Shear

21.19 kN

Check deflection in a span with limit set as 1 / 270ththus   0.00688

wL4 23.54  1.50 4  0.00688  EI 296.44  2

= 0.00138m = 1.38mm

cf allowable

1500 = 5.55 mm 270

OK

δprimary 1.38mm

Thus the twin primaries will safely span the 1.5m FALSEWORK

The falsework support grid is 1.20m x 1.50m.

1.2 x 1.5m grid

Allowing for random nature of secondary and primary bearers, the 10% continuity rule (Section 5.3.1 and Clause 19.3.3.2 in BS 5975) applies. Hence the likely design load in the forkheads to be supported by each of the falsework standards will be:1.20 x 1.50 x 19.62 x 1.1 = 38.85 kN

Forkhead Max 38.85 kN

35

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Worked Example Four

{Comment; The additional transient concrete load is considered in the maximum leg load as it acts on any 3m x 3m area, and would at one stage cover any 1.2 x 1.5m grid of supports.}

Output

Wind Forces on Soffit Formwork and Parapet Using the simplified method of wind force calculations to the latest BS EN 12811-1-4 given in Section 4.5, will give the lateral wind force on the formwork. The calculation of the additional wind force on the falsework structure would be to BS 5975, and is outside the scope of this brief. Wind Factor Swind Using Figure 66 the fundamental basic wind velocity for Park Bottom, Redruth, Cornwall gives Vb,map = 24.5 m/s. From Figure 67 the topographial factor for an area of open countryside with few hills is Twind = 1.0 (Fig 67 (a)) The site altitude from brief was 25.10m, hence A = 25.10 From Section 4.5.1.4 Swind  Twind  vb,map

A  25     1   1.0  24.5   1   = 25.11  1000   1000  

The reference height (z) is the height to top of side forms to deck formwork of (6.4m + 1.0m) = 7.4m. From the Brief the site is in open countryside with no wind breaks and about 2 km from the sea. Hence from Table 25 combined exposure factor for z = 5 is CEF = 2.18 and for z = 10 is CEF = 2.65 hence for z = 7.4 is CEF = 2.41 Peak velocity pressure Section 4.5.1.3 states the peak velocity pressure for structures erected for less than two years is given by 2 q p = 0.7 × 0.613 × cEF × Swind = 0.7 × 0.613 × 2.41 × 25.112 = 652 N/m2 Wind Forces The wind force is derived from FW,max = cscd x cf x qp x Aref x η (See Clause 17.5.1.10 in BS 5975.) The Structural factor cscd is usually unity. The total wind force on the deck soffit formwork will be the summation of the wind on the timber bearers, plus the wind on the side forms. The wind on the parapet upstand edges is not considered as it is cast seperately.

36

Swind 25.11

z = 7.4m

CEF = 2.41

Max pres 0.652 kN/m2

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Worked Example Four

(A) Wind on Soffit Bearers

Output

The wind will be acting parallel to the secondary bearers, hence from Clause 17.5.1.15.2 (a) Figure 9 in BS 5975 gives:and A ref = d1 x length of soffit considered cf = 2.0 hence A ref = (0.019 + 0.150 + 0.225) x 1.0m = 0.394 m2 / m run Maximum wind on bearers with η = 1 gives = cf x qp x Aref x η = 2.0 x 0.652 x 0.394 x 1.0 = 0.514 kN /m run

(B) Wind on side forms

Consider the two side forms projecting upwards 1.0m . The worst case for the 1.0m high side forms will be when both are erected without any voids or reinforcement fitted, and no inside upstand forms erected. Clause 17.5.1.14.3 Figure 10 in BS 5975 gives:windward cf = 1.8 and leeward cf = 1.8 x (shelter factor) The spacing/ height ratio is 9.4/1.0 = 9.4 gives by inspection of Figure 11 in BS 5975 a shelter factor:9 .4 - 5 ( x (0.65 – 0.3)) + 0.3 = (0.88 x 0.35) + 0.3 = 0.60 5 Hence leeward cf = 1.8 x 0.60 = 1.08 Maximum wind on side forms with η = 1 gives = cf x qp x Aref x η = (1.8 x 0.652 x 1.0 x 1.0 x 1.0 ) + (1.08 x 0.652 x 1.0 x 1.0 x 1.0 ) = 1.174 + 0.704 = 1.878 kN /m run Hence total wind load on deck formwork is 0.514 + 1.878 kN / m run = 2.39 kN / m run. As the falsework standards are at 1.5m centres the design lateral load for maximum wind from deck formwork only will be:2.39 x 1.5 = 3.58 kN per row of standards.

Maximum wind load 3.58 kN Per row

{Comment: This calculation has only considered wind across the structure. The TWD would also have to calculate the wind in the lengthwise direction of the bridge.}

37

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Worked Example Four

Working Wind Forces During working operations on site the working wind velocity pressure is limited to qw = 200 N/m2 = 0.20 kN/m2 (Clause 17.5.1.9 in BS 5975.) Hence proportioning the wind forces gives working wind load on deck 0.20 formwork as (0.514 + 1.878) × = 0.734 kN / m run 0.652 As the falsework standards are at 1.5m centres the design lateral load for working wind from deck formwork only will be 0.734 x 1.5 = 1.10 kN per row of standards

Output

Working wind load 1.10 kN Per row

Hence the supporting falsework designed to BS 5975 will need to resist the above lateral forces plus the wind loading on the falsework and any wind on the stripping-out working platform at all phases of construction. See Clause 19 of BS 5975. FALSEWORK DESIGN NOT COMPLETED IN FORMWORK GUIDE.

Void Flotation Assuming the voids are of circular cross section and that the flotation is taken as the full displacement weight of concrete (Archimedes), and ignoring the self weight of the voids gives:

Void flotation

2

 0.65  Flotation force per void is      1  25  8.31 kN/m  2  The brief stated that voids were restrained by strapping to the secondary bearers. These bearers are at 406mm c/c, hence assume strapped to alternate bearers, i.e. at 812mm centres. The factor of safety (Section 5.3.1) on holding down voids is 2.5. Hence, the flotation force to be resisted at each fixing position (See Figure 56 in Section 5.3.1) is 8.31 x 0.812 x 2.5 = 16.87kN {Note: The continuity factor 1.1 (Section 5.3.1) for continuous lengths of bearers / voids is taken account of within the 2.5 factor.} Note: (1) The fixings to the secondary bearers should be designed for this load as the design upwards load. (2) There will be an inherent factor of safety in any bolts, banding etc., so it is not necessary to apply a further factor on the flotation force calculated (i.e. 16.87kN) when selecting the fixing assembly.

38

Strap at 812 c/c Max. flotation load in fixing 16.87 kN

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Worked Example Five

Worked Example Five – Backpropping 205mm slab Example using Method One and Three of backpropping load calculations 205mm thick RC Slab - Four floors - One set of Formwork One / Two sets of backprops

DESIGN BRIEF To carry out the backpropping calculations only, on a FOUR storey insitu reinforced concrete flat slab building with columns on a regular grid. The permanent works drawing details a solid 205mm thick RC slab. Each slab will be cast and allowed to take up its deflected shape and become self-supporting prior to further construction of subsequent floors. The effects of not doing this are highlighted in the calculations. Assumptions 1) Assume a concrete density of 24 kN/m3. (Section 4.3.1) 2) Ignore the self weight of the formwork. 3) Any screeds to be added to the concrete floor are for future work and not included in these calculations. 4) Use only ONE set of formwork/falsework for slab construction. 5) A maximum of TWO sets of backprops could be made available. 6) Where a backprop is preloaded in position, it is assumed to be pre-loaded so that it imposes an upwards and downwards load of 0.50 kN/m2 to the slab above and below, respectively. 7) The slab that carries the falsework is known as “supporting slab”. 8) The completed slab behaves elastically under applied load. 9) An agreed procedure for striking and backpropping will be established following the outcome of these calculations. Method The calculation method used is the Method One tabular method of calculating backpropping loads given Table 34 at Section 5.4.2.3. The output of the calculations is expressed in load per unit area (kN/m2). Certain parts of the calculation give additional information using the equations from Method Three, reproduced at page 48. This example is in two parts so that the effects of applying construction operation loading on floors and preloading backprops are fully appreciated. The two parts are:PART ONE: Assumes that there will be a construction operation load on every floor at all stages of construction, and that the backprops are inserted hand tight. PART TWO: Assumes that there are no construction operation loads on general floors, but allowance is made for loads during placing of concrete. Backprops are inserted preloaded.

Output

39

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Worked Example Five LOADINGS

kN/m2 = 5.00 = 3.00 = 8.00

Output Design Service load 8.00 kN/m2

Weight of formwork / falsework = 0.00 Working area load (w.a.l.) (Section 4.3.2.2) = 0.75 {NOTE: the additional transient concrete load (Section 4.3.2.4) of 0.75kN/m2 is not considered in the backpropping calcs.} Construction operation load (c.o.l) on any floor (Service Class 1 ) = 0.75

w.a.l. 0.75 kN/m2

A. Permanent Works

Weight of slab 24kN/m3 x 0.205m Imposed load and finishes Total design service load

B. Temporary Works

Hence although the c.o.l. design load for the falsework will be the c.o.l 2 working area load plus transient concrete load (1.50 kN/m ),for 0.75 2 backpropping calculations it can be reduced to either 0.75 kN/m or be kN/m2 regarded as nil. See research and CS 140 Flat Slab Guide. Concrete strength The permanent works designer has completed the design based on an applied service load of 8.00 kN/m2 and a specified concrete strength. Striking Criteria See If a distributed load on a slab is less than the specified design load, then the required concrete strength for striking the slab at that stage Section may be reduced from the specified strength by a corresponding amount. 5.3.6.1

PART ONE - Calculations including c.o.l. and hand tight backprops. Calculations and Notation Storey heights are diagrammatical. (M1) is Method One. (M3) is Method Three. 6.50 Load in a prop or in falsework. concreting a slab 10.08 Load in a slab above the design load.

Stage

1

Operation

5.1

Erect falsework and pour slab ONE.

5.2

Strike formwork at 72%1 of 28d strength.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

1

0

Loads (kN) per square metre Load in floor slab(s)

Load in props 5.00 0.75 5.75

100%

Existing

Added

TOTAL

0

0

0

0

+5.75 +0.75 +5.00

5.75

-5.75

0

1

0

0

5.75

The 72% is ratio of loads as 5.75/8.00 = 72% of specified 28 day design strength.

40

5.75

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Worked Example Five Part One (c.o.l. and hand tight backprops) continued. Stage

5.3

5.4 (a)

Operation

Slab Level

Sketch of arrangement

Assumed % of wp transfer

Move formwork and insert hand tight backprops from 0 to 1 and pour slab TWO. Slab 1 is loaded

2

(M3)

1

66.7%

to 20% above design load!

0

33.3%

What happens if the backprops are accidentally removed first. Slab 1 is loaded

2

(-M1) 30%

1

70%

to 36% above its design load.

0

Loads (kN) per square metre Load in floor slab(s)

Load in props 5.00 0.75 5.75

Existing

Added

TOTAL

0

0

0

5.75

+3.83

9.58

1.92 0 +1.92 1.92 Slab 1 loaded 20% ABOVE design load 5.75 -0.58 5.17

0

+0.58

0.58

9.58

+1.342

10.92

1.92

-1.92

0

Slab 1 loaded 36% ABOVE design load

Hence try 5.4 by removing the falsework before disturbing the backprops.

5.4 (b)

5.5 (a)

Strike forms to slab 2 at 72% of 28d strength and move up. Leave b/props 0-1. Try using ONE set backprops and move up and pour slab THREE. Slab 2 is loaded to 22% above its design load.

0.00 2 1

0

+5.00 +0.75

5.75

9.58

-3.83

5.75

1.92 0

-1.92 0

0 0

5.75

+4.02

9.77

5.75

+1.73

7.48

0

0

0

0.00 0 3

(M1)

2

70%

5.00 0.75 5.75 1.73

1

30%

0

Slab 2 loaded 22% ABOVE design load

Hence try 5.4(b) with TWO sets of backprops and pour Slab 3.

5.5 (b)

USE TWO sets of backprops from 0 to 2 and pour slab THREE Slab 2 is loaded to 24% above its design load

3

(M3)

2

73%

5.00 +0.75 5.75

0

0

0

5.75

+4.20

9.95

5.75

+1.03

6.78

0.0

+0.52

0.52

1.55 1

18% 0.52

0

9%

Slab 2 loaded 24% ABOVE design load

Removing the backprops reverts the total load in ground slab to 0.00kN/m2, i.e. a reduction of 1.92kN/m2. Hence the load removed is 1.92kN/m2 distributed between the two remaining floors. 2

41

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Worked Example Five Part One (c.o.l. and hand tight backprops) continued. Stage

5.6 (a)

Operation

If site accidentally removed backprops from 0 to 1 first, slab 2 is loaded to 26% above its design load.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

3

(-M1) 12%

2

23%

1

65%

Loads (kN) per square metre Load in Load in floor slab(s) props 5.75 -0.063 5.69 1.55 -0.183 1.37

0

0

+0.063

0.06

9.95

+0.183 -0.06

10.07

6.78

+0.343

7.12

0.52

-0.52

0

Slab 2 loaded 26% ABOVE design load

Hence try 5.5(b) by removing the falsework before any backprops. 5.6 (b)

5.7

Allow slab 3 to gain 72% of 28d strength. Strike and move up. (Allow c.ol. on slab3 for working).

Move backprops from 0-1 up to 2-3 and Pour slab FOUR. The slab 3 is loaded to 19% above design load.

(M3) 3 2

73%

0

+5.75

5.75

9.95

-4.20

5.75

6.78

-1.03

5.75

0.52

-0.52

0

0

0

0

5.75

+3.74

9.49

5.75

+1.32

7.07

5.75

+0.69

6.44

0

0

0

0.00 1

18% 0.00

0

9%

4

(M1)

3

65%

5.00 +0.75 5.75 2.01

2

23% 0.69

1

12%

0

0

Slab 3 loaded

5.8

5.9

Allow slab 4 to gain 72% of 28d strength and strike first before removing backprops.

Remove backprops. Completed.

4

(M1)

3

65%

19% ABOVE design load

0

+5.75

5.75

9.49

-3.74

5.75

7.07

-1.32

5.75

6.44

-0.69

5.75

0 2

23% 0

1

12%

4

5.75

5.75

3

5.75

5.75

2

5.75

5.75

1

5.75

5.75

The assembly now has to carry the removed 0.52kN/m² between the three floors. As all the three floors elastically move, distribute as Method One. 3

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Worked Example Five Part One (c.o.l. and hand tight backprops) continued and Part Two. Output

Notes to Part One calculations: 1) Several “What If?“ scenarios have been included at stages 5.4(a), 5.5(a) and 5.6(a) as explanation of the complexity of backpropping and to highlight the use of correct procedures.

Falsework design 6.50 kN/m2

2) The likely maximum load on the falsework supports is 5.75kN/m2. For design, this load should be increased by the application of the transient insitu concrete load of 0.75 kN/m2.

Maximum Floor load 9.95 kN/m2

3) The maximum total construction load on a slab occurs at stage 5.5(b) of 9.95 kN/m². 4) The lowest supporting slab is often overloaded to above its design service load, and this fact should be made known to the Permanent Works Designer.

PART TWO - Calculations with preloaded backprops, including working area

loads (w.a.l.) but ignoring construction operation loads on floors.

Calculations and Notation (Storey heights are diagrammatical) (M1) is Method One (M3) is Method Three 6.50 Load in a prop or in falsework 10.08 Load in a slab above the design load concreting a slab

Stage

Operation

Slab Level

Sketch of arrangement

Assumed % of wp transfer

Existing

Added

TOTAL

0

0

0

0

+5.75

5.75

1

0

+5.00

5.00

0

5.75

-5.75

0

5.00

-0.50

4.50

0

+0.50

0.50

1

5.10

Erect props and pour slab ONE.

5.11

Slab hardens, w.a.l. removed and strike formwork at 62%4 of 28d strength.

0

5.12

4

Insert preloaded backprops and move falsework up onto slab 1.

Loads (kN) per square metre Load in floor slab(s)

Load in props 5.00 0.75 5.75

100%

2 0.00 1 +0.50 0

The 62% is ratio of loads as 5.00/8.00 = 62% of specified 28 day design strength. 43

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Worked Example Five Part Two (no c.o.l. and preloaded backprops) continued. Stage

Operation Pour Slab TWO.

5.13

5.14

5.15 (a)

Slab 1 is loaded slightly (4%) above its design load (Check with PWD). After pouring the slab the working area load becomes nil.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

2

(M3)

1

66.7%

0

33.3%

2

(M3)

1

66.7%

0

33.3% (-M1) 30%6

What happens if the backprops were accidentally removed first. Slab 1 is loaded

2

to 17% above its design load.

0

1

Loads (kN) per square metre Load in floor slab(s)

Load in props 5.00 +0.755 5.75 1.92 +0.50 2.42

Existing

Added

0

4.50

TOTAL 0

+3.83

8.33

0.50 +1.92 2.42 Slab 1 loaded ABOVE design load 0 0 5.00

1.67 +0.50 2.17

5.00 -0.65 4.35

70%4

4.50

+3.33

7.83

0.50

+1.67

2.17

0

+0.65

0.65

7.83

+1.52

9.35

2.17

-2.17

0

Slab 1 loaded 17% ABOVE design load

Hence try 5.14 by removing falsework before disturbing the backprops. 5.15 (b)

5.16

Strike forms to slab 2 at 62% of 28d strength and move up. Leave b/props 0-1. Insert preloaded b/props 1 to 2 and pour slab THREE. Slab 2 is loaded to 9% above its design load. (Check with PWD).

(M3)

0.00

1

66.7%

0 3

33.3% (M3)

2.17 -1.67 0.50

2

73%

1

18%

0

9%

2

5.00 +0.755 5.75 1.55 +0.50 2.05 0.52 +0.50 1.02

0

+5.00

5.00

7.83

-3.33

4.50

2.17 0

-1.67 0

0.50 0

5.00

+4.20 -0.50

8.70

4.50

+0.50 +1.03

6.03

0.50

+0.52

1.02

Slab 2 loaded 9% ABOVE design load

When c.o.l. is ignored on the completed floors, it is reasonable to allow at least the working area load of 0.75 kN/m2 on the formwork/falsework as the transient insitu concrete load is not present. 6 Removing the lower backprops, then two floors elastically move. 5

44

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Worked Example Five Part Two (no c.o.l. and preloaded backprops) continued. Stage

Operation

Slab Level 3

5.17

After pouring the slab the working area load becomes nil. Slab 2 is loaded to 2% slightly above its design load.

5.18 (a)

If you accidentally removed the backprops from 0 to 1 first, slab 2 is loaded to 5% above its design load (props 1 to 2 remain preloaded).

Sketch of arrangement

Assumed % of wp transfer

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL

(M3)

0

0

0

5.00

+3.65 -0.50

8.15

4.50

+0.50 +0.90

5.90

5.00 2

73%

1

18%

0

9%

3

(-M1) 12%

2

23%

1

65%

1.35 +0.50 1.85 0.45 +0.50 0.95

0.50 +0.45 0.95 Slab 2 loaded 2% ABOVE design load 0

+0.117

0.11

8.15

+0.227

8.37

5.90

+0.627

6.52

0.95

-0.95

0

5.00 -0.117 4.89 1.85 -0.337 1.52

0

Slab 2 loaded 5% ABOVE design load

Hence try 5.17 by removing the falsework before any backprops. 5.18 (b)

Allow slab 3 to gain 62% of 28d strength. strike and move up.

3

0

+5.00

5.00

8.15

-3.65

4.50

5.90

-0.90

5.00

0.95 0

-0.45 0

0.50 0

5.00

-0.50

4.50

4.50

+0.50

5.00

1

5.00

+0.50

5.50

0

0.50

-0.50

0

2 0.50 1 0.50 0 4 0.00 3

5.19

Move backprops up from O-1 to 2-3 and preload.

0.50 2 0.50

The assembly now has to carry the 0.50 + 0.45 = 0.95 kN/m² between the three floors. As all the three floors elastically move, distribute as Method One. See Note 4 to calculations.

7

45

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Worked Example Five Part Two (no c.o.l. and preloaded backprops) continued. Stage

Operation

Pour slab FOUR.

5.20

Slab 3 is loaded to only 3% slightly above design load. (Check with PWD.)

Slab Level

Sketch of arrangement

Assumed % of wp transfer

4

(M1) 100%

3

65%

2

23%

1

12%

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 5.00 +0.755 5.75 2.01 +0.50 2.51 0.69 +0.50 1.19

0

0

0

0

4.50

+3.74

8.24

5.00

+1.32

6.32

5.50

+0.69

6.19

0

0

0

Slab 3 loaded 3% ABOVE design load 4

5.21

After pouring the slab the working area load becomes nil. (The backprops remain preloaded.)

5.22

5.23

Allow slab 4 to gain 62% of 28d strength and strike first before removing backprops.

Remove backprops. Completed.

46

(M1)

0

0

0

4.50

+3.25

7.75

5.00

+1.15

6.15

5.50

+0.60

6.10

0

+5.00

5.00

7.75

-3.25

4.50

6.15

-1.15

5.00

1

6.10

-0.60

5.50

4

5.00

3

4.50

2

5.00

1

5.50

5.00 3

65%

2

23%

1

12%

1.75 +0.50 2.25 0.60 +0.50 1.10

4 3 0.50 2 0.50

5.00 +0.50

5.00 5.00

-0.50

5.00

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Worked Example Five Notes to Part Two calculations: 5) Several “What If?“ scenarios have been included at stages 5.15 (a) and 5.18(a) as explanation of the complexity of backpropping and to highlight the use of correct procedures. 6) The likely maximum load on the falsework supports is 5.75kN/m2. For design, this load should be increased by the application of the transient insitu concrete load of 0.75 kN/m2. 7) The maximum total construction load on a slab occurs at stage 5.16 of 8.70 kN/m2. 8) The justification to ignore the construction operations load (c.o.l.) is based on the research on an actual building (European Concrete Building Project). This demonstrated that when calculating backpropping loads, the construction operation loads on the existing slabs need not be included. They were not measured in practice. (See Section 7 and Figure 27 in CS 140 “Guide to Flat Slab Formwork and Falsework”.)

Output Falsework design 6.50 kN/m2

Maximum Floor load 8.70 kN/m2

9) In Part Two, the supporting slab is often overloaded to above its design service load, and this fact should be made known to the Permanent Works Designer. General Notes Applicable to all calculations: 10) What happens to the transfer of loads in slabs if a lower prop in compression is removed? As the floors are elastic, the lowest floor reverts to its unpropped state, and the reduction of load on that floor is distributed to the remaining floors. It is realistic to assume that Method One also works in reverse, so that the effect of the reduction on the lowest floor is distributed to the remaining floors in order of application. Hence if one level of propping remained, the ratio of 30/70 would apply, split with the lower floor carrying 70% of the removed load.

Output

11) Where a slab is loaded to above its design service load, the approval of the Permanent Works Designer should be sought. Note that loading a slab to above its design service load may be acceptable – See Annex E in CS 140 “Guide to Flat Slab Formwork and Falsework”.) continues

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Worked Example Five

General Notes – Method

Output

Three

12) Method Three at Section 5.4.2.3 uses simple equations, reproduced below, to generate loads in floors and backprops based on relative stiffness of floors. In this example the ground / foundations at level 0 are assumed rigid such that Slevel 0 = ∞. One level of backprops Load in backprops

wb1 

wp  S   3  s1   Ss2  



wp S    3  s1    



wp 3

 33.3%wp

where Ss2 = ∞ Hence the load on upper slab is 66.7% wp and load in props is 33.3% wp . Two Levels of backprops Load in upper backprops, where Ss3 = ∞ and Ss1 = Ss2 is given by

wb1 



wp      Ss1  Ss2    S   3  s1      Ss2    S     3  s2   Ss3     wp 4 1



3

wp 3.667



wp     3  1   1  Ss2    3        

 27%wp

Load in lower backprops, where Ss3 = ∞ and wb1 

wb2

 wp    3.667  wb1   9% w p   Ss2   Ss2     3  3       S s3  

wp 3.667

is given by

Hence load on upper slab is 73% wp load in upper props is 27% wp load on middle slab is 18% wp, load on lower props is 9% wp and load on lower slab is 9% wp . 13) Stages 5.3(a), 5.5(b), 5.13, 5.14, 5.16 and 5.17 use Method Three equations, see 12) above, assuming that the foundation / ground slab is infinitely stiff and Sground = ∞. See Section 5.4.2.5.

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Worked Example Five

Comments

Output

(a) Each site will be different - this is one example only. (b) Always liaise with the Permanent Works Designer (PWD) and the

Temporary Works Designer(TWD) for backpropping calculations. {The UK CDM Regulations place a duty on CDM co-ordinators to ensure this liaison takes place.} (c) Is the additional transient concrete load relevant in

backpropping calculations? In this example it represents a total load of 3m x 3m x 0.75 = 6.75 kN applied to the falsework. Assuming the columns are on, say, a 7m grid, this load effectively only applies 0.14 kN/m2 to the structure. As it is transient, it is reasonable to assume that it could be omitted in backpropping calculations for the slabs, but of course is relevant in the falsework design.

(d) The first, second and third floor slabs are loaded to above their

design capacity when backprops are inserted hand-tight. When the same arrangement is loaded with pretensioned backprops and the c.o.l. on completed floors ignored, the “above design capacity” is reduced, but not eliminated. (e) This example shows that during construction several floors will

be overloaded to above the service design load. Again, See both (d) and Notes 3 and 7.

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Worked Example Six

Worked Example Six - Backpropping Example using Method One and Three of backpropping load calculations 175mm thick RC Slab - Five floors - One set of Formwork - Two sets of backprops DESIGN BRIEF To carry out the backpropping calculations on a FIVE storey insitu reinforced concrete thin flat slab building with a regular column grid. The permanent works drawing details a solid 175mm thick RC slab. Each slab will be cast and allowed to take up its deflected shape and become self-supporting prior to further construction of subsequent floors. Assumptions 1) Assume a concrete density of 24 kN/m3. (Section 4.3.1) 2) The self weight of the formwork/falsework is 0.50 kN/m2. 3) Ignore the self weight of any backpropping. 4) When concreting a slab, the working area load (w.a.l.) of 0.75 kN/m2 is applied on the falsework. 5) Construction Operation Loads (c.o.l.) on floor slabs are ignored in Part Two of the backpropping calculations. 6) Ignore the additional transient insitu concrete load in the backpropping calculations. {Comment: It will be included in the falsework design.} 7) Any screeds to be added to the concrete floor are for future work and not included in these calculations. 8) Use only ONE set of formwork/falsework for slab construction. 9) A maximum of TWO sets of backprops could be made available. 10) Where a backprop is pretensioned in position in Part Two, it is assumed to be pre-loaded to 0.50 kN/m2. 11) The slab that carries the falsework is known as “supporting slab”. 12) The completed slab behaves elastically under applied load. 13) An agreed procedure for striking and backpropping will be established following the outcome of these calculations. Method The calculation method used is the Method One (M1) tabular method of calculating backpropping loads given Table 34 at Section 5.4.2.3. The output of the calculations is expressed in load per unit area (kN/m2). Certain parts of the calculation give additional information using the equations from Method Three, reproduced in Example 5 on page 48.

50

Output

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Worked Example Six This example is in two parts so that the effects of applying construction operation loading on floors and preloading backprops are fully appreciated. The two parts are:PART ONE:

Output

Assumes that there will be a construction operation load on every floor at all stages of construction, and that the backprops are inserted hand tight.

PART TWO: Assumes that there are no construction operation loads on general floors, but allowance is made for loads during placing of concrete. Backprops are inserted preloaded. kN/m2

Loadings A. Permanent Works

Weight of slab 24 x 0.175 Weight of finishes Super imposed floor load Total design service load

B. Temporary Works

= = = =

4.20 1.00 2.00 7.20

Weight of formwork / falsework = 0.50 Working area load (Section 4.3.2.2) = 0.75 Transient insitu concrete load = 0.00 (Would be 0.75 for 175mm slab thickness) Minimum imposed construction load on completed slab = 0.75 Concrete strength The permanent works designer has completed the design based on an applied service load of 7.20 kN/m2 and a specified concrete strength. Striking Criteria If a distributed load on a slab is less than the specified, then the required concrete strength for striking the slab at that stage may be reduced prorate to the specified strength. (See Section 5.3.6.1)

Design Service load 7.20 kN/m2

w.a.l 0.75 kN/m2

Backpropping Calculations and Notation Storey heights are diagrammatical. (M1) is Method One. (M3) is Method Three. concreting a slab 6.50 Load in a prop or in falsework 10.08 Load in a slab above the design service load

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Worked Example Six

PART ONE - Calculations with hand tight backprops and including the construction operation load on floors. Stage

6.1

Operation

Erect forms & falsework and pour slab ONE.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

1 100%

Loads (kN) per square metre Load in floor slab(s)

Load in props 0.75 4.20 0.50 5.45

0

Existing

Added

TOTAL

0

0

0

0

+5.45

5.45

0

+0.50 +4.20 +0.75

5.45

5.45

-5.45

0

0

0

0

5.45

+3.30

8.75

0

+1.65

1.65

0.50

6.2

Strike and move up formwork at 76%1 of 28d strength.

1 n/a 0

6.3

6.4

6.5 (a)

Insert b/props from 0 to 1. Pour slab TWO Slab 1 is loaded to 22% above design load!

2

(M3)

1

66.7% 1.65

CHECK with PWD.

0

33.3%

After pouring the slab the working area load becomes nil.

2

(M3)

1

66.7%

What happens if the backprops are

accidentally

0.75 4.20 4.95

Slab 1 loaded 22% ABOVE design load 0 0 4.20 5.45

+2.80

8.25

0

+1.40

1.40

0

+0.42

0.42

8.25

+0.98

9.23

1.40

-1.40

0

1.40 0 2

removed first. Slab 1 is loaded

1

to 28% above its design load.

0

33.3% (-M1) 30%2

4.20 -0.42 3.78

70%2

Slab 1 loaded 28% ABOVE design load

Hence start from 6.4 by removing falsework before disturbing the backprops. 0.50

6.5 (b)

Strike forms to slab 2 at 76%1 of 28d strength and move up. Leave the b/props 0-1.

2

1

+4.20 +0.50 +0.75

5.45

8.25

-3.30

4.95

1.40

-1.40

0

0 0

1

0

The 76% is ratio of loads as 5.45/7.20 = 76% of specified 28 day design strength. The assembly now has to carry the removed 1.40kN/m² between the two floors. As all the floors elastically move, distribute as Method One. 2

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Worked Example Six Part One (c.o.l. and hand tight backprops) continued. Stage

6.6 (a)

Operation

Slab Level

Sketch of arrangement

Assumed % of wp transfer

Try using ONE set of hand tight backprops and move to 1-2 Pour slab THREE. Slab 2 is loaded

3

(M1)

2

70%

to 24% above its design load.

0

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 4.20 0.75 4.95

0

0

0

5.45

+3.473

8.92

4.95

+1.48

6.43

0

0

0

1.48 1

30%

Slab 2 loaded 24% ABOVE design load

Hence try 6.5(b) with TWO sets backprops and Pour Slab 3.

6.6 (b)

Insert b/props 0 to 2 and pour slab THREE. Slab 2 is loaded to 26% above its design load.

CHECK with PWD.

6.7

After pouring the slab the working area load becomes nil. Slab 2 is loaded to 18% above its design load.

3

(M3)

2

73%

0.75 4.20 4.95

0

0

0

5.45

+3.61

9.06

4.95

+0.89

5.84

0

+0.45

0.45

1.34 1

18%

0

9%

0.45

3

(M3)

2

73%

Slab 2 loaded 26% ABOVE design load 0 0 0 4.20 5.45

+3.06

8.51

4.95

+0.76

5.71

0

+0.38

0.38

1.14 1

18% 0.38

0

9%

CHECK with PWD.

Slab 2 loaded

18% ABOVE design load

What happens if lower backprops 0 to 1 were accidentally removed first? 3

(-M1) 12%

2

23%

1

65%

If you

accidentally

6.8 (a)

remove backprops from 0 to 1 first, slab 2 is loaded to 19% above its design load.

0

4.20 -0.044 4.16 1.14 -0.134 1.01

0

+0.044

0.04

8.51

+0.094

8.60

5.71

+0.254

5.96

0.38

-0.38

0

Slab 2 loaded 19% ABOVE design load

Hence use 6.7 by removing the falsework before any backprops. The self weight of falsework is already on slab 2 so additional is 70% 4.95 on Slab 2 The assembly now has to carry the 0.38 = 0.38 kN/m² between the three floors. As all the three floors elastically move, distribute as Method One.

3 4

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Worked Example Six Part One (c.o.l. and hand tight backprops) continued. Stage

Operation

Slab Level

Sketch of arrangement

Assumed % of wp transfer (-M3)

6.8 (b)

Allow slab 3 to gain 76% of 28d strength. Strike and move up.

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 0.50

3

2

73%

1

18%

0

+0.75 +0.50 +4.20

5.45

8.51

-3.56

4.95

5.71

-0.76

4.95

0.38

-0.38

0

0

0

0

4.95 0.50

+3.22

8.67

4.95

+1.14

6.09

4.95

+0.59

5.54

0

0

0

0 0 0

6.9

Move up back props from 0 -1 to 2-3 and pour slab FOUR. The slab 3 is

4

9% (M1) 100%

3

65%

loaded to 21% above design

2

23%

load.

1

CHECK with PWD.

0

+0.75 4.20 4.955

1.73 0.59 12%

Slab 3 loaded

6.10

After pouring the slab the working area load becomes nil. The slab 3 is loaded to 14% above design load.

4

(M1)

3

65%

0

5.45

+2.73

8.18

4.95

+0.97

5.92

4.95

+0.50

5.45

1.47 2

23%

1

12%

0.50

Slab 3 loaded

3

14% ABOVE design load

0.50

4

6.11

0

4.20

(-M1)

Allow slab 4 to gain 76% of 28d strength. Strike and move up.

21% ABOVE design load

0

65%

0

+0.75 +4.20 +0.50

5.45

8.18

-3.23

4.95

5.92

-0.97

4.95

5.45

-0.50

4.95

0 2

23% 0.

1

12%

The falsework weight was taken on the slab 3 before it was backpropped, and is included in slab 3 load. 5

54

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Worked Example Six Part One (c.o.l. and hand tight backprops) continued. Stage

6.12

Operation

Move up back props from 1 -2 to 3 – 4 and pour slab FIVE. Slab 4 is loaded to 21% above design load. (Check with PWD).

6.13

After pouring the slab the working area load becomes nil. Slab 4 is loaded to 14% above design load.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

5

(M1) 100%

4

65%

3

23%

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 0.75 4.20 4.956

0

0

0

4.95 0.50

+3.22

8.67

4.95

+1.14

6.09

4.95

+0.59

5.54

4.95

0

4.95

1.73 0.59 2

12%

1

Slab 4 loaded 21% ABOVE design load 5

(M1)

0

0

0

4.95 0.50

+2.73

8.18

4.95

+0.97

5.92

4.95

+0.50

5.45

4.20 4

65% 1.47

3

23% 0.50

2

12%

Slab 3 loaded 14% ABOVE design load

Remove falsework to slab 5 before removing any backprops.

6.14

Remove falsework to Slab 5 at 68% of 28d strength, then remove backprops. Completed.

5

0

+4.95

4.95

4

8.18

-3.23

4.95

3

5.92

-0.97

4.95

2

5.45

-0.50

4.95

1

4.95

0

4.95

Notes to Part One calculations: 1) Several “What If?“ scenarios have been included as explanation of the complexity of backpropping and the use of correct procedures. 2) Stages 6.3, 6.4, 6.6(b) and 6.7 use Method Three equations which are reproduced in Example 5 page 48. 3) The calculations show that during construction ALL the floors will be stressed to above their design service load : Slab ONE 22% (6.3) Slab TWO 26% (6.6(b) ) Slab THREE 21% (6.9) and Slab FOUR 21% (6.12)

Output

Maximum loads Falsework

6.20

kN/m2 Backprops

1.73

kN/m2 6

The falsework weight was taken on slab 3 before it was backpropped, and is included in slab 4. 55

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Worked Example Six

PART TWO - Calculations with preloaded backprops, including working area loads (w.a.l.) but ignoring construction operation loads on floors. Calculations and Notation

Storey heights are diagrammatical

(M1) is Method One. (M3) is Method Three. 6.50 Load in a prop or in falsework. concreting a slab. 10.08 Load in a slab above the design load Stage

Operation

6.15

Erect forms & falsework and pour slab ONE.

6.16

Strike and move up formwork at 65%1 of 28d strength.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

1 100%

Loads (kN) per square metre Load in floor slab(s) Load in props 0.75 4.20 0.50 5.45

0 0.50

1

Existing

Added

TOTAL

0

0

0

0

+5.45 +0.50 +4.20

5.45

5.45

-5.45

0

4.20 0.50

-0.50

4.20

0

+0.50

0.50

0

4.70

n/a 0 0.50

6.17

Insert preloaded backprops 0 – 1.

1 0.50 0

6.18

6.19

6.20 (a)

1 2

Pour Slab TWO. Slab 1 is loaded slightly (4%) above its design load.

After pouring the slab the working area load becomes nil. What happens if the backprops were

accidentally

2

(M3)

1

66.7%

0

33.3%

2

(M3)

1

66.7%

0

33.3% (-M1) 30%2

2

removed! Slab 1 is loaded

1

to 16% above its design load.

0

70%2

+0.75 +4.20 4.95 +1.65 +0.50 2.15

0

4.20

0

+3.30

0.50 +1.65 2.15 Slab 1 loaded ABOVE design load 0 0 4.20 1.40 +0.50 1.90

4.20 -0.57 3.63

4.20

+2.80

7.00

0.50

+1.40

1.90

0

+0.57

0.57

7.00

+1.33

8.33

1.90

-1.90

0

Slab 1 loaded 16% ABOVE design load

The 65% is ratio of loads as (0.50+4.20)/7.20 = 65% of specified 28 day design strength. Removing the lower backprops, then two floors elastically move.

56

7.50

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Worked Example Six Part Two (no c.o.l. and preloaded backprops) continued. Hence from 6.19 remove falsework before disturbing the backprops . Stage

Operation

6.20 (b)

Strike forms to slab 2 at 65%3 of 28d strength and move up. Leave the preloaded b/props 0-1.

Slab Level

Sketch of arrangement

Assumed % of wp transfer (-M3)

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 0.50 0

+0.50 +4.20

4.70

7.00

-0.50 -2.80

3.70

1.40 0.50

-1.40

0.50

4.70

-0.50

4.20

1

3.70

+0.50 +0.50

4.70

0

0.50

-0.50

0

0

0

0

4.20

+3.474

7.67

4.75

+1.48

6.18

0

0

0

2

1

66.7% 0.50

0

33.3% 0.50

6.21

Move up backprops and install backprops to 1 to 2.

2 n/a

0.50

Hence pouring Slab 3 using only one set of backprops.

6.22

Pour Slab THREE. Slab 2 is loaded

3

(M1)

2

70%

to 7% above its design load.

1

Not preferred.

0

4.20 0.75 4.95 1.48

30%

Slab 2 loaded 7% ABOVE design load

Hence use 6.20(b) with TWO sets preloaded backprops. 0.50

6.23

Insert preloaded backprops 0 – 1.

2

4.70

-0.50

4.20

1

3.70

+0.50

4.20

0

0.50

0

0.50

n/a

0.50

Now pour Slab 3

3 4

The 65% is ratio of loads as 4.70/7.2 = 65% of 28 day strength ignoring the w.a.l. The self weight of falsework is already on slab 2 so additional is 70% 4.95 on Slab 2 57

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Worked Example Six Part Two (no c.o.l. and preloaded backprops) continued. Stage

6.24

Operation

Pour slab THREE (include w.a.l). Slab 2 is loaded to 8% above its design load.

CHECK with PWD. After pouring the slab the working area load becomes nil.

6.25

Slab 2 is loaded slightly (1%) above its design load.

CHECK with PWD.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

3

(M3)

2

73%

1

18%

0

9%

3

(M3)

2

73%

1

18%

0

9%

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 0.75 4.20 4.95 1.34 +0.50 1.84 0.45 +0.50 0.95

0

0

0

4.20

+3.61

7.81

4.20

+0.89

5.09

0.50

+0.45

0.95

Slab 2 loaded 8% ABOVE design load 0 0 0 4.20 1.14 +0.50 1.64 0.38 +0.50 0.88

4.20

+3.06

7.26

4.20

+0.76

4.96

0.50

+0.38

0.88

Slab 2 loaded slightly ABOVE design load

What happens if lower backprops 0 to 1 were removed first ? If you remove backprops

3

(-M1) 12%

2

23%

1

65%

accidentally

6.26 (a)

from 0 to 1 first, slab 2 is loaded to 4% above its design load (props 1 to 2 remain preloaded).

0

4.20 -0.115 4.09 1.64 -0.31 5 1.33

0

+0.115

0.11

7.26

+0.205

7.46

4.96

+0.575

5.53

0.88

-0.88

0

Slab 2 loaded 4% ABOVE design load

Hence correct procedure from 6.25 is to strike the falsework before removing any backprops.

The assembly now has to carry the 0.50 + 0.38 = 0.88 kN/m² between the three floors. As all the three floors elastically move, distribute as Method One. See Note 5 to calculations. 5

58

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Worked Example Six Part Two (no c.o.l. and preloaded backprops) continued. Stage

Operation

Slab Level

Sketch of arrangement

Assumed % of wp transfer (-M3)

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL 0.50

3

6.26 (b)

Allow slab 3 to gain 65% of 28d strength. Strike forms and move up.

2

73%

1

18%

0

9%

1.63 -1.13 0.50 0.88 -0.38 0.50

0

+0.50 +4.20

4.70

7.26

-3.06 -0.50

3.70

4.96

-0.76

4.20

0.88

-0.38

0.50

4.70

-0.50

4.20

3.70

+0.50

4.20

4.20

+0.50

4.70

0.50

-0.50

0

0

0

0

4.206

+3.22

7.42

4.20

+1.14

5.34

4.70

+0.59

5.29

0

0

0

0.50

6.27

Move up backprops from 0 – 1 and insert preloaded in 2 – 3.

3 0.50 2 0.50 1 0

6.28

Pour slab FOUR (include working area load). Slab 3 is loaded to only 3% slightly above design load.

(Check with PWD).

4

(M1) 100%

3

65%

2

23%

1

12%

0

0.75 4.20 4.956 1.73 +0.50 2.23 0.59 +0.50 1.09

Slab 3 loaded 3% ABOVE design load

Continues

6

The falsework weight was taken on slab 3 before it was backpropped, and is included in slab 3. 59

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Worked Example Six Part Two (no c.o.l. and preloaded backprops) continued. Stage

Operation

Slab Level 4

Sketch of arrangement

Assumed % of wp transfer

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL

(M1)

0

0

0

4.20

+2.73

6.93

4.20

+0.97

5.17

4.70

+0.50

5.20

0

+4.20 +0.50

4.70

6.93

-3.23

3.70

5.17

-0.97

4.20

5.20

-0.50

4.70

4.70

-0.50

4.20

3.70

+0.50

4.20

2

4.20

+0.50

4.70

1

4.70

-0.50

4.20

0

0

0

4.20

+3.22

7.42

4.20

+1.14

5.34

4.70

+0.59

5.29

4.20

6.29

After pouring the slab the working area load becomes nil. (The backprops remain preloaded).

3

65%

2

23%

1

12% (-M1)

1.47 +0.50 1.97 0.50 +0.50 1.00

0.50

4

6.30

Allow slab 4 to gain 65% of 28d strength. Strike and move up.

3

65% 0.50

2

23% 0.50

1

12% 0.50

6.31

6.32

Move up backprops from 1 – 2 and insert preloaded in 3 – 4.

Pour slab FIVE (include working area load). Slab 4 is loaded to only 3% slightly above design load.

(Check with PWD).

7

4 0.50 3 0.50

5

(M1) 100%

4

65%

3

23%

2

12%

0.75 4.20 4.957 1.73 +0.50 2.23 0.59 +0.50 1.09

Slab 3 loaded ABOVE its design load

The falsework weight was taken on slab 4 before it was backpropped, and is included in slab 4.

60

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Worked Example Six Part Two (no c.o.l. and preloaded backprops) continued. Hence slab 4 is slightly loaded to above its service design load – caused by the inclusion of the working area load. See 6.33 after pouring slab 5 is finished. Stage

Operation

Slab Level

Sketch of arrangement

5

Assumed % of wp transfer

Loads (kN) per square metre Load in floor slab(s) Load in props Existing Added TOTAL

(M1)

0

0

0

4.20

+2.73

6.93

4.20

+0.97

5.17

4.70

+0.50

5.20

4.20

6.33

After pouring the slab the working area load becomes nil. (The backprops remain preloaded.)

4

65%

3

23%

2

12%

1.47 +0.50 1.97 0.50 +0.50 1.00

Now allow the new slab to take up its deflected shape first by removing the falsework before removing the backprops. 5

6.34

Remove backprops. Completed.

0

+4.20

4.20

4

6.93

-0.50 -2.73

4.20

3

5.17

-0.97

4.20

2

5.20

-0.50 -0.50

4.20

1

4.20

0

4.20

Notes to Part Two - calculations:

Output

4) Several “What If?“ scenarios have been included as explanation of

Maximum loads

5) Stages 6.18, 6.19, 6.24 and 6.25 use Method Three equations

Falsework

the complexity of backpropping and the use of correct procedures.

reproduced at Worked Example 5 page 48.

6) The calculations show that during construction ALL the floors will

become slightly stressed to above their stated design service load during the placing of concrete:Slab ONE 4% (6.18) Slab TWO 8% (6.24) Slab THREE 3% (6.28) and Slab FOUR 3% (6.32)

6.20

kN/m2 Backprops

2.23

kN/m2

General notes applicable to Part One and Part Two on next page.

61

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Worked Example Six

General Notes Applicable to Parts One and Two 7) Not all the calculation stages shown are required. 8) What happens to the transfer of loads in slabs if a lower prop in

compression is removed? See Stages 6.5(a), 6.8(a), 6.20(a) and 6.26(a). It is equivalent to applying a tension load downwards on the lowest slab. As the slabs and props are all considered to be elastic, it is realistic to assume that Method One also works in reverse, so that the effect of a downwards pull on the lower floor is distributed in reverse in order of application. For one level of backprops 70/30 with lowest floor now carrying 70% of the removed load. For two levels of backprops 65/23/12 with lowest floor now carrying 65% of the removed load.

9) Where a slab is overloaded to above its design service load, this

fact should be made known to the Permanent Works Designer. Note that loading a slab to above its design service load may be acceptable – See Annex E in CS 140 “Guide to Flat Slab Formwork and Falsework”.)

Comments (a) Each site will be different - this is one example only. (b) Always liaise with the Permanent Works designer and the

Temporary Works designer for backpropping calculations. (c) The newly cast slab will nearly always have to be designed to support the weight of the formwork and falsework very shortly after being struck. This avoids having to lower the formwork / falsework to the ground, and allows it to be moved directly onto the new slab. (d) The backpropping calculations confirm that for the PWD design loading given, each slab may become loaded to above design load if working area load and/or construction operations loads are considered. Engineering judgement is required. The management of this risk is outside the scope of the Formwork Guide. (e) The structure can only be safely constructed using two sets of backprops.

62

Output

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Worked Example Seven

Worked Example Seven - Backpropping Example using Method One of backpropping load calculations 175mm thick RC Slab - Five or more floors - Two sets of Formwork/falsework This example, based on Example Six, is included to illustrate the effect on insitu floor slabs when using a construction technique, often favoured in developing countries, of using two sets of formwork/falsework without backpropping. The example is in three parts. In Part Two cast slabs are NOT allowed to take up their deflected shape prior to further loading from subsequent slabs. Where they are allowed to deflect prior to further loading the effects are shown in Part Three. ONE: Why such a technique was considered acceptable assuming rigid props. TWO: Calculations for the actual loading on the floor slabs without repropping. THREE: Calculations for the actual loading on the floor slabs allowing repropping.

WARNING - The Concrete Society does not recommend the procedures shown in this example. They are included to illustrate how loads greater than designed can be applied to the structure during construction. The example highlights the significant role of the Permanent Works Designer in considering the construction technique and establishing correct procedures for multi-storey construction. DESIGN BRIEF To carry out the backpropping calculations on a multi-storey insitu reinforced concrete thin flat slab building with more than five similar floors and a regular column grid. The permanent works drawing details a solid 175mm thick RC slab. Assumptions 1) Assume a concrete density of 24 kN/m3. (Section 4.3.1) 2) The self weight of the formwork/ falsework is 1.00 kN/m2. 3) When concreting the working area load (w.a.l.) of 0.75 kN/m2 is applied on the falsework. When not concreting no construction operation loads are considered. 4) Ignore the additional transient insitu concrete load in the backpropping calculations. {Comment: Will be included in the falsework design calculations.} 5) Any screeds to be added to the concrete floor are for future work and not included in these calculations. 6) Use only TWO sets of formwork/falsework for slab construction. 7) The completed slab behaves elastically under applied load. 8) An agreed procedure for striking will be established.

Output

63

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Worked Example Seven

PART ONE Theory assuming rigid backprops

Output

The previous theories of loads transferring equally between back propped slabs assumed rigid back props, and used a formwork self weight of 0.50kN/m2. This gave a simple load transmission methodology as described below:-

3 2 1

kN/m2

During Construction: Construction load on 3 Self weight of slab 3 Self weight of forms 2-3 Self weight of slab 2 Self weight of forms 1-2

= = = = =

0.75 4.20 0.50 4.20 0.50

Self weight of slab 1

=

4.20

Total is

14.35

Design Capacity Design Service Load slab 2 = 7.20 Design Service Load slab 1 = 7.20 Total is 14.40 The above analysis would indicate that using two sets of equipment would be acceptable as 14.40 > 14.35, and the floor slabs are OK. {Comment: The backpropping calculations shown in Part Two of this example use a high value of self weight of formwork at 1.00 kN/m2,

kN/m2

Applied 14.35 Design 14.40 Falsework Design 5.45 kN/m2

as formwork (0.50kN/m2) plus falsework (0.14kN/m3 x 3.5m = 0.49kN/m2) = 0.99 kN/m2.}

PART TWO : Backpropping Calculations - Without Repropping Method The calculation method used is the Method One tabular method of calculating backpropping loads given Table 34 at Section 5.4.2.3. The output of the calculations is expressed in load per unit area (kN/m2). Loadings kN/m2 A. Permanent Works Weight of slab 24 x 0.175 = 4.20 Weight of finishes = 1.00 Super imposed floor load = 2.00 Total design service load = 7.20 B. Temporary Works Weight of formwork / falsework = 1.00 Working area load (Section 4.3.2.2) = 0.75 Transient insitu concrete load = 0.00 Minimum imposed construction load on completed slab = 0.75 64

Service load 7.20 kN/m2 w.a.l. 0.75 kN/m2

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Worked Example Seven Part Two Continued Concrete strength The permanent works designer has completed the design based on an applied service load of 7.20 kN/m2 and a specified concrete strength.

Output

Striking Criteria If a distributed load on a slab is less than the specified, then the required concrete strength for striking the slab at that stage may be reduced pro rata to the specified strength. (See Section 5.3.6.1)

Backpropping Calculations and Notation Storey heights are diagrammatical. (M1) is Method One. (M3) is Method Three. 6.50 Load in formwork / falsework. concreting a slab. 10.08 Load in a slab above the design load.

Stage

7.1

7.2

7.3

7.4

Operation

Erect forms & falsework and pour slab ONE. Slab one hardens. (No w.a.l.)

Erect 2nd set of forms and pour slab TWO.

Slab two hardens. (No w.a.l.)

Slab Level

Sketch of arrangement

Assumed % of wp transfer

1 100% 0 1 n/a

Loads (kN) per square metre Load in floor slab(s)

Load in props 0.75 4.20 1.00 5.95 4.20 1.00 5.20

0 2

(M3)

1

66.7%

0.75 4.20 1.00 5.95

Existing

Added

TOTAL

0

0

0

0 0

+5.95 0

5.95 0

5.95

-0.75

5.20

0

0

0

0

+3.97

3.97

5.20

+1.98

7.18

0

0

0

0

+3.47

3.47

5.20

+1.73

6.93

5.20 +1.98 7.18 0

33.3%

2

(M3)

1

66.7%

4.20 1.00 5.20 5.20 +1.73 6.93

0

33.3%

As you only have two sets of formwork / falsework and no backprops, next stage must be to strike the lower set and put onto slab 2. 65

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Worked Example Seven Part Two Continued Stage

7.5

Operation

When you strike lower forms to slab 1 then Slab 1 is loaded to 6% above its design load.

Slab Level

2

Sketch of arrangement

Load in props

Existing

Added

TOTAL

(-M1) 30%

5.20

0

+1.781

1.78

3.47

+4.151

7.62

6.93

-5.93 -1.00

0

-1.781 3.42 1

70%

0 (M1)

7.6

7.7

When you place the lower set of forms up onto slab 2 the slab 1 is further loaded.

When you pour Slab 3 the slab 1 is now loaded to 30% above its design load.

Loads (kN) per square metre Load in floor slab(s)

Assumed % of wp transfer

2

Slab 1 loaded 6% ABOVE design load 1.00

70%

1.78

+0.70

2.48

7.62

+0.30

7.92

0

0

0

3.42 +0.30 3.72 1

30%

0 3

(M1)

2

70%

Slab 1 loaded 10% ABOVE design load 0.75 0 0 4.20 4.95 2.48

+3.47

5.95

7.92

+1.48

9.40

0

0

0

3.72 +1.48 5.20 1

30%

0

Slab 2 loaded 30% ABOVE design load 3

7.8

Slab three hardens. (No w.a.l.) Slab 1 is now loaded to 27% above its design load.

(M1)

0

0

0

2.48

+2.94

5.42

7.92

+1.26

9.18

0

0

0

4.20 2

70% 3.72 +1.26 4.98

1 0

30%

Slab 2 loaded 27% ABOVE design load

Again, as you only have two sets of formwork / falsework and no backprops, next stage must be to strike the lower set and put onto slab 3.

The assembly now has to carry the 6.93kN/m² capacity removed, less its own self weight, i.e. 5.93kN/m2 between the two remaining floors. As all the floors elastically move, using Method One. 1

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Worked Example Seven Part Two Continued Stage

7.9

Operation

When you strike the lower set of forms to slab 2 then Slab 2 is loaded to 14% above its design load.

Slab Level

3

Sketch of arrangement

Load in props

Existing

Added

TOTAL

(-M1) 30%

4.20

0

+1.192

1.19

5.42

+2.792

8.21

9.18

-3.98 -1.00

4.20

-1.192 3.01 2

70%

1 (M1) 3

7.10

When you place the forms onto slab 3 the slab 2 is further loaded.

Loads (kN) per square metre Load in floor slab(s)

Assumed % of wp transfer

Slab 1 loaded 14% ABOVE design load 1.00

70%

7.11

loaded to 39% above its design load.

+0.70

1.89

8.21

+0.30

8.51

4.20

0

4.20

3.01 +0.30 3.31 2

30%

1

When you pour Slab 4 the slab 2 is now

1.19

4

(M1)

3

70%

Slab 1 loaded 18% ABOVE design load 0.75 0 0 4.20 4.95 1.89

+3.47

5.36

8.51

+1.48

9.99

3.31 +1.48 4.79 2

30%

4

(M1)

3

70%

Slab 2 loaded 39% ABOVE design load

7.12

Slab four hardens. (No w.a.l.) Slab 2 is now loaded to 36% above its design load.

0

0

0

1.89

+2.94

4.83

8.51

+1.26

9.77

4.20 3.31 +1.26 4.57 2

30%

Slab 2 loaded 36% ABOVE design load

Again, as you only have two sets of formwork / falsework and no backprops, next stage must be to strike the lower set and put onto slab 4.

Removing the falsework reverts the total load in Slab 1 to its self weight of 4.20kN/m2, i.e. a reduction of 4.98kN/m2. This includes the falsework weight. Hence the effective load removed by removing the falsework is 4.98 – 1.00 = 3.98 kN/m2 distributed between the two remaining floors. 2

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Worked Example Seven Part Two Continued Stage

7.13

Operation

When you strike lower forms to slab 3 then Slab 3 is loaded to 12% above its design load.

Slab Level

4

Sketch of arrangement

Load in props

Existing

Added

TOTAL

(-M1) 30%

4.20

0

+1.373

1.37

4.83

+3.203

8.03

9.77

-4.57 -1.00

4.20

-1.373 2.83. 3

70%

2 (M1) 4

7.14

When you place the forms onto slab 4 the slab 3 is further loaded.

Loads (kN) per square metre Load in floor slab(s)

Assumed % of wp transfer

Slab 1 loaded 12% ABOVE design load 1.00

70%

7.15

loaded to 36% above its design load.

+0.70

2.07

8.03

+0.30

8.33

4.20

0

4.20

2.83 +0.30 3.13 3

30%

2

When you pour Slab 5 the slab 3 is now

1.37

5

(M1)

4

70%

Slab 1 loaded 16% ABOVE design load 0.75 0 0 4.20 4.95 2.07

+3.47

5.50

8.33

+1.48

9.81

3.13 +1.48 4.61 3

30%

Slab 2 loaded 36% ABOVE design load

7.16

Building continues using repeat stages as above

Notes to Part Two calculations: 1) The maximum load on the falsework supports occurs at stage 7.4 of 6.93kN/m2. For design, this load should be increased by the application of the transient insitu concrete load of 0.75 kN/m2. 2) The maximum total construction load on a slab is 9.99 kN/m². 3) The lowest supporting slab is often overloaded to above its design service load, and this fact should be made known to the Permanent Works Designer. See Annex E in CS 140 “Guide to Flat Slab Formwork and Falsework”.

Output Falsework design 7.68 kN/m2 Maximum Floor load 9.99 kN/m2

Removing the falsework reverts the total load in Slab 2 to its self weight of 4.20kN/m2, i.e. a reduction of 5.57kN/m2. This includes the falsework weight. Hence the effective load removed by removing the falsework is 5.57 – 1.00 = 4.57kN/m2 distributed between the two remaining floors.

3

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Worked Example Seven

PART THREE : Backpropping Calculations - With Repropping

Output

Method The calculation method used is the Method One tabular method of calculating backpropping loads given Table 34 at Section 5.4.2.3. The output of the calculations is expressed in load per unit area (kN/m2). Part Two highlighted the increase in design loads on floor slabs where the formwork/falsework was not destressed between uses. This part shows the difference when a completed floor is allowed to take up its deflected shape before repropping with the same falsework. In practice head jacks are all released and then all re-tightened to the deflected soffit to transmit further construction loads.

Stage

7.20

Operation

After a slab is cast and achieved 72%4 of strength, the falsework is de-stressed.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

Existing

Added

TOTAL

0

4.20

4.20

x+1

4.20

+1.00

5.20

x+0

4.20

0

4.20

x+2 1.00

(M1)

7.21

7.22

The falsework 1 to 2 is now reset to slab 2. When you place the 2nd set of forms onto slab 2 the slab 1 is further loaded.

When you pour Slab 3 the supporting slab is now loaded to 16% above its design load.

Loads (kN) per square metre Load in floor slab(s)

Load in props

x+2

Slab 1 loaded 6% ABOVE design load 1.00

70%

4.20

+0.70

4.90

5.20

+0.30

5.50

0

0

0

1.00 +0.30 1.30 x+1

30%

0 x+3

(M1)

x+2

70%

Slab 1 loaded 10% ABOVE design load 0.75 0 0 4.20 4.95 4.90

+3.47

8.37

5.50

+1.48

6.98

1.30 +1.48 2.78 x+1

30%

Slab 2 loaded 16% ABOVE design load

4

Allowing for the next set of falsework, strength needed is 5.20/7.20 = 72% of 28 day strength. 69

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Worked Example Seven Part Three Continued Stage

7.23

Operation

Slab three hardens. (No w.a.l.) Slab 2 is now loaded to 9% above its design load.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

x+3

(M1)

x+2

70%

Loads (kN) per square metre Load in floor slab(s)

Load in props

Existing

Added

TOTAL

0

0

0

4.90

+2.94

7.84

5.50

+1.26

6.76

4.20 1.30 +1.26 2.56 x+1

30%

Slab 2 loaded 9% ABOVE design load

7.24

7.25

After slab 3 hardens and achieved 72%5 of strength, the falsework 2 to 3 is destressed.

The falsework 2 to 3 is now reset to slab 3. Falsework 1 to 2 is removed.

7.26

The falsework removed is now placed on slab 3. Slab 2 is further loaded.

x+3

0

4.20

4.20

7.84

-2.94

4.90

6.76

-1.26

5.50

4.20

+0.096

4.29

4.90

+0.216

5.11

5.50

-0.30 -1.00

4.20

4.29

+0.70

4.99

5.11

+0.30

5.41

4.20

0

4.20

1.00 x+2 1.30 x+1 (-M1)

1.00

x+3

30%

1.00 -0.09 0.91

x+2

70%

x+1 (M1)

1.00

x+3

70%

0.91 +0.30 1.21

x+2

30%

x+1

Allowing for the next set of falsework, strength needed is 5.20/7.20 = 72% of 28 day strength. Removing the falsework reverts the total load in Slab 1 to its self weight of 4.20kN/m2, i.e. a reduction of 1.30kN/m2. This includes the falsework weight. Hence the effective load removed by removing the falsework is 1.30 – 1.00 = 0.30kN/m2 distributed between the two remaining floors.

5 6

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Worked Example Seven Part Three Continued Stage

7.27

Operation

When you pour Slab 4 the supporting slab 3 is now loaded to 18% above its design load.

Slab Level

Sketch of arrangement

Assumed % of wp transfer

x+4

(M1)

x+3

70%

Loads (kN) per square metre 0.75 4.20 4.95

0

0

4.99

+3.47

8.46

5.41

+1.48

6.89

1.21 +1.48 2.69 x+2

30%

x+4

(M1)

x+3

70%

Slab 2 loaded 18% ABOVE design load

7.28

Slab four hardens. (No w.a.l.) Slab 3 is now loaded to 10% above its design load.

0

0

0

4.99

+2.94

7.93

5.41

+1.26

6.67

4.20 1.21 +1.26 2.47 x+2

30%

Slab 2 loaded 10% ABOVE design load

7.29

Building continues using repeat stages as above

Notes to Part Three calculations: 1) The maximum load on the falsework supports occurs when you pour a slab, i.e. at stage 7.22 and 7.26 of 4.95 kN/m2. For design of falsework this should be increased by the self weight (1.00 kN/m2) and the application of the transient insitu concrete load of 0.75 kN/m2. 2) The maximum total construction load on a slab at stage 7.27 is 8.46 kN/m². 3) The supporting slab is often overloaded to above its design service load, and this fact should be made known to the Permanent Works Designer. See Annex E in CS 140 “Guide to Flat Slab Formwork and Falsework”.

Output Falsework design 6.70 kN/m2

Maximum Floor load 8.46 kN/m2

General Note – Method Three The equations used to generate load distribution when propping to a rigid ground/ foundation at level 0 are reproduced in Worked Example FIVE (see page 48).

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Worked Examples to Formwork A guide to good practice (3rd Edition) These worked examples are a companion document to The Concrete Society publication Formwork – A guide to good practice (3rd Edition). They include worked examples of formwork design and back propping calculations. Formwork is the key to successful and economic concrete construction. It has a dominant influence on the appearance and accuracy of finished concrete which in turn affects the ease with which the following trades can complete their work. It generally accounts for a third or more of the value of the structure. The design and construction of formwork is an essentially practical subject and relies on the engineering judgement and expertise of those involved. Developments in materials and equipment for formwork are constantly extending the specifier’s and contractor’s options, in terms of concrete finish, speed and economy of construction, etc. The guide brings together the practical and engineering aspects of formwork in a way which will be of particular value to all concerned with the specification, design, manufacture, construction and use of formwork for buildings and civil engineering structures. These worked examples have been prepared with the approval of the Construction Standing Committee of The Concrete Society by a working party chaired by Eur Ing Peter Pallett, Temporary Works Consultant, and Lecturer. Assistance was invited from experts in the industry including contractors, architects, consulting engineers, specialist suppliers and trade associations.

Formwork A guide to good practice (3rd Edition) Worked Examples Prepared by a Working Party of The Concrete Society

CS 169 First published April 2012 Reprinted with minor amendments July 2012 © The Concrete Society

The Concrete Society Riverside House, 4 Meadows Business Park, Station Approach Blackwater, Camberley, Surrey, GU17 9AB Tel: +44 (0)1276 60 7140 Fax: +44 (0)1276 60 7141 Email: [email protected] Visit: www.concrete.org.uk

FormworkWorked examplesCOVER.indd 1

ISBN 978-1-904482-69-7

9 781904 482697

18/07/2012 11:58:22