(20 marks) Question 7 is to be designed from unseasoned F14 hardwood member with
ID: 1765605 • Letter: #
Question
(20 marks) Question 7 is to be designed from unseasoned F14 hardwood member with the thickness of e beam has a clear span of 3.5 m and subjected to following unfactored loads. 100 mm. Th Permanent loads (G) -2.3 kN/m (50+ years duration) Imposed load (0) 5.2 kN/m (5 months duration) It was identified the critical load combination is 1.2G+1.50. This beam is a primary member in a non-residential structure without a post-disaster function. Take k12 1 (a) Find a suitable depth for the beam that complies strength limit state as per the (12 marks) (b) Determine maximum long-term deflection for the load G+0.40 (08 marks) Hint: Maximum deflection for a simply supported beam under uniformly distributed load is AS1720.1 wL4 given by EIExplanation / Answer
a) From AS1720.1 we can say that Ru = KmodRk where Rk is the basic characteristic strength of structural member,
Kmod =1.12 as given in the question.
Ru is the nominal capacity of the structural member
Given G = 2.3 KN/m and Q = 5.2 KN/m
Hence permanent load =2.3*3.5(span) =8.05 KN and imposed load = 5.2*3.5 =18.2 KN
Critical load combination Rk =1.2 G+1.5Q =1.2*8.05+1.5*18.2 =36.96 KN.
Hence Ru= 1.12*36.96 =41.39 KN.
Ru = strength capacity of the beam as per AS 1720.1*b*d where b=100mm as given in the question since placing the beam with the width of beam in the direction of thickness of hardwood member used gives more strength.
From this equation d can be determined.
b) Given load combination G+0.4Q = 8.05+0.4*18.2 =15.33 KN.
Maximum long term deflection = wL4/EI where w = uniformly distributed load
w = 15.33/3.5 = 4.38 KN/m, L =3.5m, E value for unseasoned F14 hardwood timber can be used and I =bd3/12
d determined from previous part will be used.
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