Falsework

By

Paul Markham

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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.

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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.

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Formwork

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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stress?BS5975:2008 written in terms of permissible stress but large parts refer to British Standards which were withdrawn on 31 March 2010.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, PlySoldiers All have to be designed in accordance with permissible stress.Tie rods

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FalseworkTypically the falsework design makes an allowance of 1.5kPa for controlled heaping of concrete.

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Falsework

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Permanent falsework

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Permanent falsework

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Permanent Falsework

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Permanent falsework

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Permanent Falsework

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Permanent Falsework

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Falsework – design of primary member

Slab 500 thick. Weight density of concrete = 25kN/m3

Design to EC5:Unfavourable permanent action, gG = 1.35

Unfavourable variable action, gQ = 1.5

Design pressure, Pd = 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:

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Timber design to EC5

Try twin 147 x 72 timbers –

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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 L3 / EI Second moment of area, I = bd3 / 12 = 2 x 72 x 1473 / 12 = 38.1 x 106mm3

Therefore, d = 0.00541 (37.1) x 15003 / (7.4 x 103 x 38.1 x 106) = 2.4mm1500/2.4 = Span/625 – OK

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Timber design to EC5

C24 timber: Characteristic bending strength, fm,k = 24 MPa (BS EN 338)

Various factors must be applied to this to get the design bending strength:

kh,m Depth factorkmod Modification factor for duration of load and moisture contentksys System strength factorkcrit Factor for lateral bucklinggM Partial factor for material properties

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Timber design to EC5

Clause 3.2 (3)Depth factor, kh,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

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Timber design to EC5

Modification factor for duration of load and moisture content, kmod :

Therefore, medium term loading .The falsework will be erected outdoors and subject to rain. Therefore Service Class 3.

Use Table 3.1 to find kmod

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Timber design to EC5

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Timber design to EC5

ksys 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 ksys = 1.1.

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Timber design to EC5

kcrit 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 kcrit = 1.0.

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Timber design to EC5

gM is the familiar partial factor for material properties (Table 2.3)

So use gM= 1.3.

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Timber design to EC5

Design bending strength, fm,d = kh,m x kmod x ksys x kcrit x fm,k / gM

= 1.004 x 0.65 x 1.1 x 1.0 x 24 / 1.3 = 13.3MPa

Therefore the bending resistance, Md = fm,d x Wy Where wy is the section modulus. wy = N x b x h2 / 6 = 2 x 72 x 1472 / 6 = 518 600mm3. Therefore Md = 13.3 x 518600 = 6.9kNm.

Applied moment, M = wl2/8 = 37.1 x 1.52/8 = 10.43kNm therefore fails

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Timber design to EC5

Check shear: F = 0.375 w l = 0.375 x 37.1 x 1.5 = 20.9kN

Applied shear stress td = 3 x F / 2 x A = 3 x 20.9 x 103 / (2 x 2 x 72 x 147) = 1.48 MPa

C24 timber: Characteristic shear strength, fv,k = 2.5MPa (BS EN 338)

Various factors must be applied to this to get the design shear strength:kmod Modification factor for duration of load and moisture contentksys System strength factorgM Partial factor for material propertiesFactor of 1.5 allowed by BS 5975 for falsework

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Timber design to EC5

Therefore, design shear strength,fm,d = kmod x ksys x fv,k x 1.5 / gM

= 0.65 x 1.1 x 2.5 x 1.5 / 1.3 = 2.06MPa > Applied, therefore ok

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Falsework

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Falsework

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Falsework

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Falsework

Any questions?