9. CALCULATION EXAMPLES OF LVL STRUCTURES These structural calculation examples for LVL are based on Eurocodes (EN1990, EN1991 and EN1995) and the additional instructions given in Chapters 4-5. Where information from National annexes is required, the Finnish annex or the default values of Eurocodes have been used. The calculations make references to the equation numbers of the Chapters 4 - 6. The examples are chosen to demonstrate the calculations methods. Therefore some of the component sizes may not necessarily ideal for practice and those cases have comments of possibly better suitable sizes at the end of each example. 9.1 LVL 48P JOIST FLOOR Residential floor: span L = 4500 mm; width b = 5000 mm; 45x240 mm LVL 48P joists at spacing s = 400 mm; 22 mm chipboard decking. Support length 45 mm. Live load qk = 2,0kN/m2; partition load g 2,k = 0,3 kN/m2 and self-weight g1,k = 0,6 kN/m2. Service class SC1. Joist properties: Bending strength edgewise fm,0,edge,k = 48 N/mm2 Shear strength edgewise fv,0,edge,k = 4,2 N/mm2 Compression perpendicular to the grain edgewise fc,90,edge,k = 6 N/mm2 Modulus of elasticity E0,mean = 13800 N/mm2 Modulus of rigidity G0,edge,mean = 600 N/mm2 Area of cross section A = b∙h = 10800 mm2 Section modulus W = b∙h2/6 = 4,32∙105 mm3 Moment of inertia I = b∙h3/12 = 5,18∙107 mm4 Moment stiffness of the joist EI = 13800 N/mm2 ∙ 5,18∙107 mm4 = 7,15∙1011 Nmm2 Shear rigidity of the joist GA = 600 N/mm2 ∙ 10800 mm2 = 6,48∙106 N Modification factor kmod for medium-term, SC1 = 0,8 Modification factor kdef for SC1 = 0,6 Material safety factor γM (default value in EC5) = 1,2 Size effect factor kh = (300/240)0,15 = 1,034 Loading combinations The most critical ultimate limit state (ULS) load combination: E_(d,ULS) = γ_G∙(g_(1,k)+g_(2,k))+ γ_Q∙q_k (4.1) E_(d,ULS) =1,15∙(0,6 kN/m^2+0,3 kN/m^2 )+1,5∙2,0 kN/m^2 E_(d,ULS)= 4,03 kN/m^2 Note: Safety factors γG and γQ are according to Finnish National annex of Eurocode 0. The most critical serviceability limit state (SLS) load combination: E_(d,SLS) = γ_G∙(g_(1,k)+g_(2,k))+ γ_G∙q_k (4.1) E_(d,SLS) =1,0∙(0,6 kN/m^2+0,3 kN/m^2 )+1,0∙2,0 kN/m^2 E_(d,SLS)= 2,9 kN/m^2 d,ULS = G ∙ ( 1,k + 2,k)+ Q∙ k (4.1) d,ULS =1,15∙ (0,6 kN/m2 + 0,3 kN/m2 )+1,5∙ 2,0 kN/m2 d,ULS = 4,03 kN/m2 d,SLS = G ∙ ( 1,k + 2,k)+ G ∙ k (4.1) d,SLS =1,0∙ (0,6 kN/m2 + 0,3 kN/m2 )+1,0∙ 2,0 kN/m2 d,SLS = 2,9 kN/m2 d = d,ULS ∙ ∙ 2/8 = 4,03kN/m2 ∙ 0,4m∙ (4,5m)2/8 = 4,1 kNm 4,1 kNm d,ULS = G ∙ ( 1,k + 2,k)+ Q∙ k (4.1) d,ULS =1,15∙ (0,6 kN/m2 + 0,3 kN/m2 )+1,5∙ 2,0 kN/m2 d,ULS = 4,03 kN/m2 d,SLS = G ∙ ( 1,k + 2,k)+ G ∙ k (4.1) d,SLS =1,0∙ (0,6 kN/m2 + 0,3 kN/m2 )+1,0∙ 2,0 kN/m2 d,SLS = 2,9 kN/m2 d = d,ULS ∙ ∙ 2/8 = 4,03kN/m2 ∙ 0,4m∙ (4,5m)2/8 = 4,1 kNm = d = 4,1 kNm = 9,5 N/mm LVL Handbook Europe 179
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