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Published byArianna Blacksmith Modified about 1 year ago

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WORKSHEET 5 PROPERTIES OF SECTIONS AND DEFLECTIONS

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Q1 Find the Moment of Inertia, I, and the Section Modulus, Z, of the following timber sections: (include units) (i) I x = bd 3 /12 = 50 x / 12= 33.3 x 10 6 mm 4 (i) Z x = bd 2 /6 = 50 x / 6= x 10 3 mm 3 (i) I x = bd 3 /12 = 100 x / 12= 8.3 x 10 6 mm 4 (i) Z x = bd 2 /6 = 100 x / 6= x 10 3 mm 3 a) 50 mm width x 200 mm deep X X b) 100 mm width x 100 mm deep X X

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Q2 a) which is heavier? b) which is stiffer and by what proportion? Referring to 1 (a) and 1 (b): both have the same cross-sectional area so both have the same weight per unit length 200 x 50 is 4 times stiffer (about the X-X axis) c) which is stronger and by what proportion? 200 x 50 is 2 times stronger (in bending) I(a) x =33.3 x 10 6 mm 4 I(b) x = 8.3 x 10 6 mm 4 Z(a) x =333.3 x 10 3 mm 3 Z(b) x = x 10 3 mm 3

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Q3 (a) a) You want a beam as stiff as a 200 x 50 mm beam, but you only have space for 150 mm depth. I x = 33.3 x 10 6 mm 4 I x = bd 3 / x 10 6 = b x / 12 b = 33.3 x 10 6 / (150 3 / 12) How wide does the beam have to be? = 118 mm

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Q3 (b) b) You want a beam as strong as a 200 x 50 mm beam, but you only have space for 150 mm depth. Z x = x 10 3 mm 3 Z x = bd 2 / x 10 3 = b x / 6 b = x 10 3 / (150 2 / 6) How wide does the beam have to be? = 89 mm (in both cases it requires a substantial increase in width)

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Q3 (c) c) why don’t we use an even deeper and narrower beam, e.g. a 300 x 20 mm beam? a) the beam may buckle b) the depth may be a problem

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Q4 You could try using Multiframe for the following questions Use Douglas Fir (the Modulus of Elasticity in the Wood Sections Library is 4500KSI = ~7000MPa) a span-to-depth ratio for timber floor joists - 18:1 Assume a live load of 1.5 kPa and a total dead load (incl self-weight of the joists) of 0.4 kPa Check this out for 200 x 50 mm softwood joists at 600 mm centres spanning 3.6 m in the upper floor of a house

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Q4 (cont.) tributary area 200 x 50 mm timber 600mm crs

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Q4 (a) a) (i) What is the maximum bending stress in the joists? Tributary area = 3.6 x 0.6 = 2.16 m 2 Total loading = ( ) = 1.9 kPa (kN/ m 2) Total load (distributed) on one joist (W = wL) = 1.9 x 2.16 = kN Bending stress, f = M / Z = 1.85 / x 10 3 bring it all to N and mm - multiply top line x 1000 (kN to N) and 1000 (m to mm) = 5.6 N/mm 2 = 5.6 MPa 5.6 < 8. (ii) Is F8 strong enough? ( Softwood grade F8 is capable of taking 8 MPa) So this is within the capacity of F8 timber Maximum bending moment = WL/8 (where W =wL) = x 3.6 / 8 = 1.85 kNm

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Q4 (b) b) (i) What is the maximum deflection? For simply supported beam with UDL Maximum deflection = 5 W L 3 / 384 EI = 8.4 mm Allowed limit = Span / 300 (ii) Is the calculated deflection within the allowed limit So this is within the allowed limit the Modulus of Elasticity of F8 timber can be taken as 9000 MPa = 5 x 4104 x / 384 x 9000 x 33.3 x 10 6 (using N and mm everywhere) = 3600 / 300= 12 mm

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Q5 In a two-storey house, you want to span across a double garage 7m wide. You decide to do this with steel beams at 3.6 m centres. (The timber joist system in Q1 spans between them) 3.6 m 7 m steel beams timber 600mm crs 3.6 m tributary area for beam = 7 x 3.6 = 25.2 m 2

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Q5 (cont1.) Tributary area for one beam = 7 x 3.6 = 25.2 m 2 Total load one beam = 1.9 x 25.2 = kN Max bending moment of beam = WL/8 = 41.9 kNm = x 7/ 8 a) Find a Universal Beam section strong enough? (assume a maximum allowable stress of 200MPa for Grade 300 steel. Use the Table given or a BHP catalogue) >= x 10 3 mm 3 Looking up the Table of Universal Beams we find that a 200UB25.4 has Z = 232 x The depth of the beam d = 203 mm Bending stress f = M / Z (want f to be 200 or less) Z must be at least Z > = 41.9 x 10 6 / 200

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Q5 (cont2.) span-to-depth ratio = 7000 / 203 b) Is this section stiff enough? If not, upsize it. deflection = 5 WL 3 / 384 EI = 34.5 clearly the section is not stiff enough. Working with defl = 14, we get this looks a bit optimistic. Let us check the deflection From the table, I x of the section = 23.6 x 10 6 mm 4 = 5 x x / 384 x x 23.6 x 10 6 = 45.3 mm allowable deflection = span / 500 = 7000 / = 5 WL 3 / 384 EI xnew (alternatively) I xnew / Ix = 45.3 / 14 = 76.4 x10 6 mm 4 I xnew = 3.24 x I x = 3.24 x 23.6 From table find that 310UB40.4 has I x = 85.2 x10 6 mm 4 = 14.0 mm

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Q5 (cont3.) span-to-depth ratio = 7000 / 304 c) What span-to-depth ratio do you end up with for the steel beams? (310UB40.4) = 23 depth of beam d = 304 mm

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