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Published byArianna Blacksmith Modified over 3 years 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) a) 50 mm width x 200 mm deep X (i) Ix = bd3/12 = 50 x 2003 / 12 = x 106 mm4 (i) Zx = bd2/6 = 50 x 2002 / 6 = x 103 mm3 b) 100 mm width x 100 mm deep X (i) Ix = bd3/12 = 100 x 1003 / 12 = x 106 mm4 (i) Zx = bd2/6 = 100 x 1002 / 6 = x 103 mm3

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**Q2 Referring to 1 (a) and 1 (b): a) which is heavier? b)**

both have the same cross-sectional area so both have the same weight per unit length b) which is stiffer and by what proportion? I(a)x =33.3 x 106 mm I(b)x = 8.3 x 106 mm4 200 x 50 is 4 times stiffer (about the X-X axis) c) which is stronger and by what proportion? Z(a) x =333.3 x 103 mm Z(b)x = x 103 mm3 200 x 50 is 2 times stronger (in bending)

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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. How wide does the beam have to be? I x = 33.3 x 106 mm4 I x = bd3 / 12 33.3 x 106 = b x 1503 / 12 b = 33.3 x 106 / (1503 / 12) = 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. How wide does the beam have to be? Z x = x 103 mm3 Z x = bd2 / 6 333.3 x 103 = b x 1502 / 6 b = x 103 / (1502 / 6) = 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 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 Assume a live load of 1.5 kPa and a total dead load (incl self-weight of the joists) of 0.4 kPa

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**Q4 (cont.) tributary area 200 x 50 mm timber joists @ 600mm crs 3600**

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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 m2 Total loading = ( ) = 1.9 kPa (kN/ m2) Total load (distributed) on one joist (W = wL) = 1.9 x 2.16 = kN Maximum bending moment = WL/8 (where W =wL) = x 3.6 / 8 = 1.85 kNm Bending stress, f = M / Z = 1.85 / x 103 bring it all to N and mm - multiply top line x 1000 (kN to N) and 1000 (m to mm) = 5.6 N/mm2 = 5.6 MPa (ii) Is F8 strong enough? (Softwood grade F8 is capable of taking 8 MPa) 5.6 < 8. So this is within the capacity of F8 timber

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**Q4 (b) b) (i) What is the maximum deflection?**

the Modulus of Elasticity of F8 timber can be taken as 9000 MPa For simply supported beam with UDL Maximum deflection = 5 W L3 / 384 EI = 5 x 4104 x / 384 x 9000 x 33.3 x 106 (using N and mm everywhere) = 8.4 mm (ii) Is the calculated deflection within the allowed limit Allowed limit = Span / 300 = 3600 / 300 = 12 mm So this is within the allowed limit

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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 joists @ 600mm crs tributary area for beam = 7 x 3.6 = 25.2 m2

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**Q5 (cont1.) a) Find a Universal Beam section strong enough?**

Tributary area for one beam = 7 x 3.6 = 25.2 m2 Total load one beam = 1.9 x 25.2 = kN Max bending moment of beam = WL/8 = x 7/ 8 = 41.9 kNm 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) Bending stress f = M / Z (want f to be 200 or less) Z must be at least Z > = 41.9 x 106 / 200 >= x 103 mm3 Looking up the Table of Universal Beams we find that a 200UB25.4 has Z = 232 x 103. The depth of the beam d = 203 mm

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**Q5 (cont2.) b) Is this section stiff enough? If not, upsize it.**

span-to-depth ratio = 7000 / 203 = 34.5 this looks a bit optimistic. Let us check the deflection From the table, Ix of the section = 23.6 x 106 mm4 deflection = 5 WL3 / 384 EI = 5 x x / 384 x x 23.6 x 106 = 45.3 mm allowable deflection = span / = 7000 / 500 = 14.0 mm clearly the section is not stiff enough. Working with defl = 14, we get 14 = 5 WL3 / 384 EIxnew (alternatively) Ixnew / Ix = 45.3 / 14 Ixnew = 3.24 x Ix = 3.24 x 23.6 = x106mm4 From table find that 310UB40.4 has Ix = 85.2 x106 mm4

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**Q5 (cont3.) c) What span-to-depth ratio do you end up with**

for the steel beams? (310UB40.4) depth of beam d = 304 mm span-to-depth ratio = 7000 / 304 = 23

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