Formwork By Paul Markham

Definition of formwork from BS5975:2008 Formwork (also forms, shutters or shuttering): structure, usually temporary, but in some cases wholly or partly permanent, used to contain poured concrete to mould it to the required dimensions and support it until it is able to support itself. 2

Formwork 3

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Timber formwork 8

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Formwork The loads that must be considered in formwork and false work fall into five categories; 1.Self weight of the formwork. 2.Operating loads/live loads. 3.Environmental loads. 4.Horizontal loads. 5.Loads due to concrete. 10

Formwork 11

Formwork 12

Limit state or permissible stress? BS5975 is written in terms of permissible stress but large parts refer to British Standards which were withdrawn on 31 March It would be possible to design the timber secondaries to BS EN 1995 (Eurocode 5). However, often aluminium secondaries used and Ply Soldiers All have to be designed in accordance with permissible stress. Tie rods Therefore probably easiest to design using permissible stress and using aluminium secondaries. However, sometimes it will be necessary to use timber secondaries in which case mixed permissible and partial factor design is required. 13

Timber Most timber is now S4S – sawn four sides, this reduces the section sizes slightly so that a section nominally 100 x 50mm is actually 95 x 47mm (20% reduction in second moment of area). Common timber grades for formwork use are C16 and C24 CConifer(D, deciduous) 16 or 24 Characteristic bending strength in MPa. Allowable strength is about a third of the characteristic strength but other factors apply for moisture content and duration of loading etc. Use charts to EC5 for rapid design. 14

Timber design charts 15

Single sided formwork H Raking prop Horizontal reaction Uplift 16

Single sided formwork 17

Hy-rib 18

Hy-rib Expanded metal Lower pressures 19

Formwork – supplier’s software 20

Formwork 21

Formwork 22

Formwork 23

Software 24

Software 25

Climbing Formwork 26

Climbing Formwork 27

Climbing Formwork 28

29 Climbing device ACS 100 upper climbing head hydraulic cylinder lower climbing head climbing rail climbing shoe PERI ACS – Automatic Climbing System.

Climbing Formwork 30

Formwork 31

Formwork 32

Formwork 33

Formwork 34

Formwork – RMD Alsec 35

Formwork – RMD Albeam 36

Formwork – RMD Alform Compare these figures with characteristic loads 37

RMDK Soldiers – allowable loads 38

Plywood 39

Timber design to EC no 145 x 47 timbers Design loads: Bending moment, M = 3.66kNm Shear Force, F = 15.2kN Deflection, d = W L 3 / EI (for a two span continuous beam from Formwork a Guide to Good Practice) Second moment of area, I = bd 3 / 12 = 2 x 47 x / 12 = 23.9 x 10 6 mm 3 Therefore, d = (15 x 1.2) x / (7.4 x 10 3 x 23.9 x 10 6 ) = 0.95mm 2 40

Timber design to EC5 C24 timber: Characteristic bending strength, f m,k = 24 MPa (BS EN 338) Various factors must be applied to this to get the design bending strength: k h,m Depth factor k mod Modification factor for duration of load and moisture content k sys System strength factor k crit Factor for lateral buckling  M Partial factor for material properties 41

Timber design to EC5 Clause 3.2 (3) k h,m = lower of (150/h) 0.2 and 1.3. Where h is the depth (150/h) 0.2 = (150/145) 0.2 =

Timber design to EC5 k mod : Therefore, medium term loading. The falsework will be erected outdoors and subject to rain. Therefore Service Class 3. Use Table 3.1 to find k mod 43

Timber design to EC5 44

Timber design to EC5 k sys System strength factor – Clause 6.6 The secondaries can be used to distribute the loads from one member to the adjacent members. They will be overloaded but this is an accidental situation and that overloading would be acceptable. Therefore use k sys =

Timber design to EC5 k crit is a factor to account for lateral buckling see Clause (5) The primaries in this falsework will be located in U-heads which will provide torsional restraint at the supports. There will be sufficient friction with the secondaries to prevent lateral displacement of the top of the timber. Therefore use k crit =

Timber design to EC5  M is the familiar partial factor for material properties (Table 2.3) So use  M =

Timber design to EC5 Design bending strength, f m,d = k h,m x k mod x k sys x k crit x f m,k / g M = x 0.65 x 1.1 x 1.0 x 24 / 1.3 = 13.3MPa Therefore the bending resistance, M d = f m,d x w y Where w y is the section modulus. w y = N x b x h 2 / 6 = 2 x 47 x / 6 = mm 3. Therefore M d = 13.3 x = 4.4kNm > applied moment, therefore bending is ok. 48

Timber design to EC5 Check shear: F = 15.2kN Applied shear stress  d = 3 x F / 2 x A = 3 x 15.2 x 10 3 / (2 x 2 x 47 x 145) = 1.65 MPa C24 timber: Characteristic shear strength, f v,k = 2.5MPa (BS EN 338) Various factors must be applied to this to get the design shear strength: k mod Modification factor for duration of load and moisture content k sys System strength factor  M Partial factor for material properties Factor of 1.5 allowed by BS 5975 for falsework 49

Timber design to EC5 Therefore, design shear strength, f m,d = k mod x k sys x f v,k x 1.5 /  M = 0.65 x 1.1 x 2.5 x 1.5 / 1.3 = 2.06MPa > Applied, therefore ok Deflection: Permissible deflection = smaller of 5mm or span/270 Span/270 = 1200 / 270 = 4.4mm. Therefore, permissible deflection = 4.4mm > calculated deflection (0.9mm) Therefore deflection is ok. 50

Formwork Any questions? 51