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(Solved): B2. A timber soffit form was designed for casting a concrete slab of \( 6 \ma ...



B2. A timber soffit form was designed for casting a concrete slab of \( 6 \mathrm{~m} \) by \( 8 \mathrm{~m} \) on plane. The???????

B2. A timber soffit form was designed for casting a concrete slab of \( 6 \mathrm{~m} \) by \( 8 \mathrm{~m} \) on plane. The plywood sheeting is supported by rows of secondary timber bearers spaced at \( 450 \mathrm{~mm} \mathrm{c} / \mathrm{c} \), which rest on primary bearers spaced at \( 600 \mathrm{~mm} \mathrm{c} / \mathrm{c} \). The primary bearers are in turn supported by vertical props spaced at \( 1500 \mathrm{~mm} \mathrm{c} / \mathrm{c} \). (i) Draw labelled sectional sketches to show the soffit form components. (5 marks) (ii) Given the following design and formwork material information, determine the maximum thickness (T) of concrete slab, which could be supported by the given formwork structure. (20 marks) Hints: Assume all formwork components are continuous beams of multiple equal spans and the following formulas to be used. Max. Bending Moment for a continuous beam \( \cong \frac{1}{10} \omega L^{2} \) Max. Shear Force \( \cong 0.6 \omega L \) Max. Deflection \( \delta=0.007 \frac{\omega L^{4}}{E I} \) for plywood; and \( 0.006 \frac{\omega L^{4}}{E I} \) for secondary and primary bearers, where \( \omega \) = Uniformly Distributed Load (UDL), \( \mathrm{L}= \) Span


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Lond Calculations Dead fort of Stab 25 xx = 25x KN/m² Construction Load. 1.5 KN/m = Dead lead of form work = 0.5 kN/m² (Assure) Total load = ?? = 25x+
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