1. Falsework By Paul Markham

2. Definition of falsework from BS5975:2008 Falsework : temporary structure used to support a permanent structure while it is not self supporting. The term is used particularly for the temporary support to concrete slabs. 2

3. Definition of falsework from BS5975:2008 Falsework : temporary structure used to support a permanent structure while it is not self supporting. The term is used particularly for the temporary support to concrete slabs. 3

4. Falsework 4

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10. Formwork 10

11. Limit state or permissible stress? BS5975:2008 written in terms of permissible stress but large parts refer to British Standards which were withdrawn on 31 March 2010. BSI has an obligation to remove anything which contradicts Eurocodes so it is likely that the next revision of BS5975 will remove the references to permissible stress codes. It would be possible to design the timber secondaries to BS EN 1995 (Eurocode 5). However, Ply Soldiers All have to be designed in accordance with permissible stress. Tie rods 11

12. Falsework Typically the falsework design makes an allowance of 1.5kPa for controlled heaping of concrete. 12

13. Falsework 13

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20. Falsework 20

21. Permanent falsework 21

22. Permanent falsework 22

23. Permanent Falsework 23

24. Permanent falsework 24

25. Permanent Falsework 25

26. Permanent Falsework 26

27. Falsework – design of primary member Slab 500 thick. Weight density of concrete = 25kN/m 3 Design to EC5: Unfavourable permanent action,  G = 1.35 Unfavourable variable action,  Q = 1.5 Design pressure, P d = weight of concrete + weight of falsework + allowance for live load. = 0.5 x 25 x 1.35 + 0.5 x 1.35 + 1.5 x 1.5 = 19.8kPa Say props on 1.5 x 1.5m grid. Therefore primary beams must span 1.5m and characteristic line load on primary, w = 19.8 x 1.25 (allowance for continuity of secondaries) x 1.5 = 37.1kN/m. Select suitable timber: 27

28. Timber design to EC5 Try twin 147 x 72 timbers – 28

29. Timber design to EC5 Need a picture of the falsework From Formwork a Guide to Good Practice, for a two span continuous beam: Deflection, d = 0.00541 W L 3 / EI Second moment of area, I = bd 3 / 12 = 2 x 72 x 147 3 / 12 = 38.1 x 10 6 mm 3 Therefore, d = 0.00541 (37.1) x 1500 3 / (7.4 x 10 3 x 38.1 x 10 6 ) = 2.4mm 1500/2.4 = Span/625 – OK 29

30. 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 30

31. Timber design to EC5 Clause 3.2 (3) Depth factor, k h,m = lower of (150/h) 0.2 and 1.3. Where h is the depth (150/h) 0.2 = (150/147) 0.2 = 1.004 31

32. Timber design to EC5 Modification factor for duration of load and moisture content, 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 32

33. Timber design to EC5 33

34. 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 = 1.1. 34

35. Timber design to EC5 k crit is a factor to account for lateral buckling see Clause 6.3.3 (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 = 1.0. 35

36. Timber design to EC5  M is the familiar partial factor for material properties (Table 2.3) So use  M = 1.3. 36

37. 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 /  M = 1.004 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 72 x 147 2 / 6 = 518 600mm 3. Therefore M d = 13.3 x 518600 = 6.9kNm. Applied moment, M = wl 2 /8 = 37.1 x 1.5 2 /8 = 10.43kNm therefore fails 37

38. Timber design to EC5 Check shear: F = 0.375 w l = 0.375 x 37.1 x 1.5 = 20.9kN Applied shear stress  d = 3 x F / 2 x A = 3 x 20.9 x 10 3 / (2 x 2 x 72 x 147) = 1.48 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 38

39. 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 39

40. Falsework 40

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43. Falsework Any questions? 43