Formwork 3rd Edition Worked Examples

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Formwork

A guide to good practice (3rd Edition)

Worked Examples

Prepared by a Working Party of The Concrete Society 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.

Worked Examples to Formwork

A guide to good practice (3rd Edition)

The Concrete Society

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Email: [email protected] Visit: www.concrete.org.uk CS 169

First published April 2012

Reprinted with minor amendments July 2012 © The Concrete Society

9 781904 482697 ISBN 978-1-904482-69-7

The Concrete Society

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Tel: +44 (0)1276 60 7140 Fax: +44 (0)1276 60 7141

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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. CS 169

First published April 2012

Reprinted with minor amendments July 2012 ISBN 978-1-904482-69-7

© The Concrete Society

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.

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:

1

Double-faced wall formwork, with wind stability

2

Square column formwork

3

Forces - Single-faced wall formwork

4

Bridge deck soffit formwork with void formers

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

Page

2

15

24

30

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.

Worked Example One – Double faced Wall Formwork

DESIGN BRIEF Output

Design the formwork for an insitu

reinforced concrete wall 5.15m high, 10m long and 450mm wide. The base slab with an integral 150mm high kicker has

already been cast. There are no features 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.

5.15m high 10m long

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 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 90kN per tie. The site crane has been rated at 2000kg at all radii.

T = 10°C Tie SWL = 90 kN The TWD has selected the face contact material to be Canadian Douglas

Fir plywood 7ply, 19mm, sanded, but unknown mill source. The walings will be timber of Strength Class C16.

19mm 7ply 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

Hence rate of rise ( R)   5 . 4.0 9 Area Plan rate Volume _{2.0m/h ( R)} Output

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

Pmax

60 kN/m2

The equivalent fluid head of concrete is 2.4m 0 . 25 60 Density Pressure _{ }

The concrete pressure

diagram is as opposite 2400

5000

PLYWOOD 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.

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_.

thus 20 . 2 305 . 0 0 . 60 0066 . 0 EI wL 0066 . 0 4    4   = 0.00126m = 1.26mm cf allowable =113 270 305 _{. mm FAIL} Now check if deflection is reduced if we had chosen 9 spans of 271mm. thus 20 . 2 271 . 0 0 . 60 0066 . 0 EI wL 0066 . 0 4    4   = 0.00097m = 0.97mm cf allowable ₂₇₀271 =1.0mm hence OK 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:-

( )

₌₂₆₇ 9 2 75 -2440 mm

Thus assuming the height of 5050 gives a possible waling spacing of:- ( 10 x 267) + (3 x 305) + (4 x 357)

(3) 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

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

Output

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 of5.49m.

(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 1100 1100 1100 1100 360 A B C D E Plan of waling

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

MA = ME = - 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.

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

Max BM -1.77 kNm

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 = 14.06 kN 14.06 kN RB = RD = 1.048 wL = 1.048 x 16.02 x 1.1 = 18.47 kN 18.47 kN RC = 0.976 wL = 0.976 x 16.02 x 1.1 = 17.20 kN Total 82.26 kN Max Reaction 18.47 kN

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. Thus deflection in 1100mm is EI wL 00387 . 0 4 = 0.70mm 77 . 128 1 . 1 02 . 16 00387 . 0   4  270 1 1571 1 1100 70 . 0 L   

 _{(Section 2.7 ) Therefore OK.}

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 heavily loaded soldier is w = ₀18_.267.47 = 69.18 kN/m

Equiv. udl soldiers 69.18kN/m

SOLDIER LOADING DIAGRAM

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

Output 350 Q 1500 2400 P 1200 joint O 5000 1200 2600 N 1000 M 69.18 0 150

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) 69.18 (13.01) (6.85) (10.38) 41.79 50 (50.15) (83.02) (69.18) (30.29) 1500 1200 1200 1000 150 Q P O N 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

2700 _Soldier 2700 _Soldier

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 KPO = KON = 1_1.2= 0.83 KNM = 0.75 11 = 0.75

Distribution Factors DFPQ = _(0.50₊.5_0.83)= 0.38 hence DFPO = 0.62

DFOP = _(0.830.₊83_0.83)= 0.50 hence DFON = 0.50

DFNM = _(0.750.₊75_0.83)= 0.47 hence DFNO = 0.53

DF’s

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 FEM PO Approx. = -0.0833 wL2 _{- 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 FEM OP Approx. = 0.0833 wL2 _{+ 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 FEM ON, FEM NO Approx. = -0.0833 wL2

= 0.0833 x 69.18 x 1.22 _{= -8.30 kNm} FEM NM Approx. = 0.125 wL2 = 0.125 x 69.18 x 1.02 _{= 8.64 kNm} FEM MN Approx. = -0.50 wL2 = -0.50 x 69.18 x 0.152 _{= -0.78 kNm} Appdx B LC 12 LC 2 + 9 LC 2 + 9 LC 2 LC 4 LC 3

Q P O N M DF’s 0.38 0.62 0.50 0.50 0.53 0.47 FEM Distn +6.04 +0.22 -+7.406.61 +0.35 +0.45 -8.30 +8.30 +0.45 -0.03 +0.78 +0.39 -8.64 -0.02 -0.78 CO Distn -0.09 +0.23 +0.18 -0.14 -0.08 -0.02 +0.23 -0.08 -0.12 -0.11 CO Distn +0.02 -0.04 -0.07 +0.02 +0.06 -0.06 -0.04 +0.07 +0.02 +0.02 +6.19 -6.19 +7.94 -7.94 +8.36 -8.36 0.78 -0.78 Max BM 8.36 kNm ^4.13 ν 4.13 ^5.16 ^6.62 v6.62 v 5.16 ^ 6.62 ^ 6.97 v 6.97 v 6.62 ^ 8.36 ^0.78 v 0.78 v 8.36 Elastic Shear ^10 10 ^20.19 ^25.07 ^25.07 ^ 6.65 ^13.30 ^41.51 ^41.51 ^34.59 ^34.59 10.38 Static Shear 5.97 24.32 30.26 39.83 41.16 41.86 42.17 27.01 10.38 Summation Shear 5.97 54.58 80.99 84.03 37.39 Q P O N M Reactions

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.

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 5.9 54.6 82.1 81.3 39.0

(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 ).

Mom Distn

Computer Joint OK

Free span bending moments :-

(Values of T, w and L taken from Loading diagram on page 8)

MQP approximately 0.128 TL = 0.128 x 30.29 x 1.5 = 5.82 kNm MPO approximately 0.125 wL2 + 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 MON approximately 0.125 wL2 = 0.125 x 69.18 x 1.22 = 12.45 kNm MNM approximately 0.125 wL2 =0.125 x 69.18 x 1.02 = 8.65 kNm Output Appdx B LC 8 LC 1 + 8 LC 1 LC 1 - 8.36 -7.94 - 6.19 - 0.78 0 Q P O N M + 2.73 +3.52 + 4.30 + 4.08 B.M. Diagram showing net B.M’s

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)

WEIGHT OF WALL FORM

The weight of one form panel, 5.05m high and 5.49m long will be 5.05 x 5.49 x 60 = 1665 kg = 1665₁₀₀ = 16.65 kN (one face) (See Section 4.2.2 Table 20 ) Hence crane limit OK.

Output 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                     1000 55 1 8 . 21 0 . 1 1000 A 1 v T

Swind wind b,map = 23.0

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.)

cp,net=1.7

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. 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

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

Wind o/t per panel 47.41 kNm

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

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) 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.)

Min Stab o/t / panel 12.86 kNm

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

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 kNm Using two push-pull props connected to a point 3.0m up the form on one 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:-

Horizontal force  2 3 94 . 60

m

10.16 kN

Therefore load in prop is



10

.

16



2

₁



20.32 kN

Push/pull Prop swl 20.32 kN The anchorage for each prop will be required to resist the design

forces as follows:

Vertical =

20

₂

.

32



3

= ± 17.60 kN per anchorage

Horizontal = ± 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 2 3m 1 60° Anchorage 1730

Warning : If using kentledge blocks in lieu of anchorages, check the

Worked Example Two – Square Column Formwork

DESIGN BRIEF Output

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.

The concrete will contain a blend of

Portland cement with less than 40% fly ash and will have a pumping admixture, but no retarder. The work will be carried out during a short construction period in COLUMN (square) summer and the expected concrete temperature will be 15O_{C. The form will be filled in just over one}

hour.

The contract has already purchased stocks of 18mm Finnish WISA-Form Birch through plywood in 1220 x 2440 sheet sizes. It is 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.

(This example only designs the formwork.)

CONCRETE PRESSURE CALCULATIONS

Volume of concrete in the column is (4.60-0.10) x 0.9752 _{= 4.28m}3

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 O_C.

20 x 20 chamfers 100mm kicker Group B T = 15O_C Ply 18mm WISA-Form Birch 13ply C24 timber

Using Table PBB, in Section 4.4.3.1, for a column, and for a height of

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.

Output

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).

Ply fitted with 2440 vertically

Assume 75mm x 150mm backing timbers of strength class C24.

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 :- 75 x 150 20 x 20 Hardwood fillet 150 45 A B See detail 975 1375 95kN/m2 4500 3800

Side (A) 250mm Side (B) of 4 5 75 975  _{= 224mm}

(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:-

 = 63 . 3 250 . 0 95 0066 . 0 EI wL 0066 . 0  4    4 = 0.00067m = 0.67mm Allowable is 270 250 = 0.93mm



OK.

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:-

Moment = 0.125 wL2 _{= 0.125 × P} L × 0.502 = 0.968kNm/m Thus PL = ₂ 50 . 0 125 . 0 968 . 0  = 30.98kN /m 2

Thus the timbers could be reduced at a level 25

98 .

30 _{= 1240mm from} 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

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.

Output

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 MBC = 1.07 + 0.21 = 1.28 MCB = 1.07 + 0.31 = 1.38 MCD = 0.89 + 0.09 = 0.98 MDC = 0.89 + 0.13 = 1.02 MDE = 1.07 0.20m (4.75) (14.25) (1.76) (14.30) (3.13) (12.81) (5.69) (5.91) (1.53) 12.81kN/m 19.06kN/m 23.75kN/m 1.00m 4.38kN/m 1.35m 0.75m 0.60m 0.7m A B C D E

Stiffness kAB = ¾ × 35 . 1I = 0.556 kBC = 1I = 1.00 .0 KDE = ¾ × 60 . 0I = 1.250 kCD = 0.I = 1.333 75 Distribution factors BA/BC 0.36/0.64

CB/CD 0.43/0.57 DC/DE 0.52/0.48 Output A B C D E DF 1 0.36 0.64 0.43 0.57 0.52 0.48 +0.36 -0.36 +0.48 -0.48 Co -0.18 +0.24 FEM +2.02 -1.28 +1.38 -0.98 +1.02 -1.07 -0.20 -0.36 -0.17 -0.23 -0.10 -0.09 Co -0.09 -0.18 -0.05 -0.12 +0.03 +0.06 +0.10 +0.13 +0.06 +0.06 Co +0.05 +0.03 +0.03 +0.07 -0.02 -0.03 -0.03 -0.03 -0.04 -0.03 BM +0.36 -0.36 +1.65 -1.65 +1.13 -1.13 +0.89 -0.89 +0.48 -0.48 0.36kNm 1.65kNm 1.13kNm 0.89kNm 0.48kNm Shear 1.53 2.96 2.96 6.41 6.41 7.15 7.15 7.13 7.13 4.75 (static) 1.90 3.79 1.04 2.09 0.594 1.17 (elastic) 0.96 0.96 0.52 0.52 0.32 0.32 0.68 0.68 1.53 3.90 7.71 7.97 7.98 8.06 8.00 7.81 6.45 4.75 A B C D E Reactions 5.43kN 15.68kN 16.04kN 15.81kN 11.20kN Moments Shears 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.

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.

Max. B.M. 1.65kNm

Shear 8.06 kN

DESIGN OF STEEL YOKES – HORIZONTAL

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 Output

Consider Side A and Side B (see plan layout on page 16).

The loading patterns for Side (A) is:- See page 16

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:-

Thus the tie load for each side is 2

4

16.04  = 32.08kN.

From Section 3.5.3 the minimum factor of safety on mild steel ties is 2.5. 10.44kN 14.43kN 14.43kN 10.44kN 14.43kN 237.5 225 225 225 225 237.5 1375 H J (B) 8.02kN 16.04kN 16.04kN 8.02kN 16.04kN 187.5 250 250 250 250 187.5 1375 F G (A)

Thus the required minimum failure load is- 32.08 × 2.5 = 80kN

(Note: Typical proprietary M20 ties state failure loads of 120kN.) 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:- MFG = 32.08 ×

2

1.375 - (8.02 × 0.500) - (16.04 × 0.250) = 22.06 - 4.01 - 4.01 = 14.04 kNm

And for Side B (HJ) by taking moments about centre of span gives:- MHJ = 32.08 ×

2

1.375 - (10.44 × 0.450) - (14.43 × 0.225) = 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 Thus, y e rl = 1900 = 92 20.6

From BS 5975, Table A.1, the permissible bending stress, pbc = 155N/mm2

Actual bending stress, fbc =

Z M = 6₃ 10 77.3 2 10 14.11    _{= 91.27N/mm}2

_

_OK

(Note the “2 × “ in bottom line for the twin channels)

Output Tie load 32.08kN Max BM is 14.11kNm Bending stress 91.27N/mm2

Check bearing stress on backing timbers:-

fb = 106 2 65 72 16.04 _   = 1715kN/m 2

_

_OK.

(Allowable is 3250kN /m2 _{without wane and 2570kN/m}2 _{with wane.)}

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

Bearing stress 1715kN/m2

Output

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

See BS 5975 Annex J for web buckling and web crushing formulae. Web buckling R =



b n1



t_w p_c where t_w = 5.5mm



b n1



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

From BS 5975 Annex J, Table J.1, the effective length of the web divided by radius of gyration is l ≈ _ry

w

t D

7.0  _{= 159}

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 = 1000 37 5.5 167  = 33.98kN per channel Actual load per channel is

2

32.08 = 16.04kN

Thus, web stiffeners not required for web buckling.

Web stiffeners NOT required for buckling 2No. 125x65x15 PFC 75x150 timber 65 100x100x6 plate washer M20 All Thread rod

Web tw = 5.5mm b + n1 NA 30 b + n2 125 21.5

Web crushing R =



b n2



t_w p_b where tw = 5.5mm



b n2



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

1000 210 5.5

104  _{= 120.12kN per channel} 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.

Output Use Twin 125x65 PFC Web stiffeners not required

Worked Example Three – Forces in Single-faced Wall Formwork

DESIGN BRIEF

Alternative One Alternative Two

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

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_.

{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.

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

Output 2.85m high Pmax 60 kN/m2 100mm kicker Class F2

275 The TWD has been advised that concreting will be by skip. The

design will assume a concrete temperature for April of 10°C. T = 10°C Output Assume the self weight of the formwork in both Alternatives is

45 kg/m2 _{= 0.45 kN/m}2_{. See Table 20 Section 4.2.2.}

Ignore the self weight of the working platform.

SW

0.45kN/m2

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. 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.

Pmax 60 kN/m2 R = 2 m/hr Fill in 80 mins minimum The equivalent fluid head of concrete is 2.4m

0 . 25 . 60 Density Pressure _{ }

The concrete pressure

and the resulting 2400 2750

force diagram Ft

350 1250 Fr

60 kN/m2 _base

Concrete pressure develops a horizontal force of:- Ft = ×24×1= 2 60 _{. 72 kN acting +350+100=1250} 3 2400 _mm Fr = 60 x 0.35 x 1 = 21 kN acting +100=275 2 350 _mm

Total force from concrete pressure FT = 72 + 21 = 93 kN

Output Lateral

force 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 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

Alternative One 170 + 225 + 800₂ = 795mm Alternative Two 350 + 800₂ = 750mm

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 Tuplift = 95.₁78_.225-1.83=78.19 kN/m run

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.}

Tuplift

78.19 kN/m

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 = 95.₁78_.925+0.61 = 50.07 kN

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:- PropDE = 50.07 225 . 1 282 .

2 _{ = 93.26 kN per metre run} The vertical load downwards at position E is thus:-

Vertical LoadE = ₁1_..₂₂₅925×50.07= 78.68 kN per metre run

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 78.19 kN

Lateral Shear Force at C 42.93 kN Axial Load in prop DE 93.26 kN Vertical load at position E 78.68 kN Horizontal force at position E 50.07 kN

Fixings Output per metre

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 the tie assembly will be 93  2 = 131.50 kN per metre of wall. This load in the tie assembly will in turn generate a downward vertical 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.}

Load in anchor 132 kN/m

Diagram of Forces acting on system

c.o.l. c.o.l. Concrete Concrete Force Force Anchor Force F G H G H F

(a) Actual Force location (b) Resolved Forces for calculation

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)} + {1.2 x (2.140 + 0.180 – 1.885 – 350 - 2 800 )} - (93 x 0.255) = 95.78 - 0.45 + 0.38 - 23.72 = 71.99 kNm Hence the vertical reaction at H is VH =

885 . 1 99 . 71 _{= 38.19 kN.} Maximum VH 38 kN

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

Hence the vertical reaction at G is:- VG =

885 . 1 11 . 106 _{= 56.29 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 NOTE: To reduce tip deflection at B caused by elongation of the tie rod in tension under load, the anchor/tie asembly may be

pre-tensioned. 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.

Output per metre

Worked Example Four – Bridge Deck Soffit Formwork

DESIGN BRIEF Insitu deck Parapet added 1000 9400 6400 Foundations Output

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) 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.

4) The site is about 2 km from the 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.

< 2 yrs

25.10 AOD

2 km sea

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. ) 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.

Output Class F2 δ = 1/270 Timbers C16 Plywood Ultraform LOADING Parapet cast 1000 1000 after main deck slab Fixing 1000 Secondary 1200 1200 Primary 4600

{Comment: The fixing in face is to reduce differential movement

during future casting of parapet.}

Loading

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_.

Less void volume _π ×_r2 ×1_.0 _{  }         1.0 25 2 65 . 0 1416 . 3 2 8.30 kN/m2 Output

Hence equivalent solid slab load is 25.00 – 8.30 = 16.70 kN/m2

Plus self weight of soffit forms (Section 4.2.3) ) = 0.50 kN/m2

(Note this includes the plywood at 0.11 kN/m2 ₎

General 17.95 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_.

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.) Maximum

19.62 kN/m2

Hence maximum load for formwork on the primary bearers

(i.e. on forkheads) is 17.95 + 1.67 = 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.}

Secondy

27.92 kN/m2

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 fZ = 0.600 kNm/m Shear Load 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:- =406mm

6

2440 _{or =488mm}

5

2440 _{Try 406mm} 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_-

thus 21 . 3 406 . 0 53 . 27 0066 . 0 EI wL 0066 . 0 4    4   δ = 0.00153m = 1.53mm cf allowable 1.51 270 406 _{ mm ACCEPT} {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.}

δply

1.53mm Ply span

SECONDARY BEARERS

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.

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.}

Output Secondary

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 EI = 128.77 kNm2

Moment of Resistance fZ = 1.778 kNm Shear Load qA = 9.67 kN

Bearing stress 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. Second

Y

Max BM 1.63 kNm

Max Shear

8.16 kN 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.

Max

Reaction 10.51 kN Check bearing on twin 75 x 225 timbers. Using finished sizes from

Table 10 gives bearing stress = _{( )} = 072 0 × 072 0 × 2 51 10 . . . 1,014 kN/m2.

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

Second’y

With wane OK Check deflection in a span with limit set as 1 / 270th_:-

thus 77 . 128 20 . 1 34 . 11 00688 . 0 EI wL 00688 . 0 4    4   = 0.00126m = 1.26mm cf allowable 1200₂₇₀ =4.44mm OK. δsecond’y 1.26mm

PRIMARY BEARERS

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.

Output

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.

Primary Max BM 5.297 Nm 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. Max Shear 21.19 kN Check deflection in a span with limit set as 1 / 270th_-

thus 2 44 . 296 50 . 1 54 . 23 00688 . 0 EI wL 00688 . 0 4 4       = 0.00138m = 1.38mm cf allowable 1500₂₇₀ =5.55mm 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

{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                   1000 25 1 5 . 24 0 . 1 1000 A 1 v T

Swind wind b,map = 25.11

Swind

25.11 The reference height (z) is the height to top of side forms to deck

formwork of (6.4m + 1.0m) = 7.4m. z = 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

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

_q =0_.7×0_.613×_c ×_S2 =0_.7×0_.613×2_.41×25_.112 = wind EF p 652 N/m2 Max pres 0.652 kN/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.

(A) Wind on Soffit Bearers

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:-

cf = 2.0 and A ref = d1 x length of soffit considered

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

Output

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:-

( 5 5 -4 9. _{x (0.65 – 0.3)) + 0.3 = (0.88 x 0.35) + 0.3 = 0.60} 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.}