Shear resistance V_d = E_(d,ULS,fi)∙L/2= 14,6 kN/m∙4,0m/2 = 29,2 kN τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙29,2 kN)/(2 ∙26488 mm^2 )=1,7 N/mm^2 f_(v,d,fi)=(k_(mod,fi)∙k_fi)/γ_(M,fi) ∙f_(v,0,edge,k)=(1,0∙1,1)/1,0∙4,2 N/mm^2 =4,6 N/mm^2 τ_(m,d)≤f_(v,d,fi) →OK Compression perpendicular to grain When the main beam is supported on a wooden column which has the notional charring rate βn = 0,70mm/min, the support length becomes l_(support,fi)=100mm-0,70 mm/min∙30min+1,0∙7mm=72mm F_(c,90,d) = V_d = 29,2 kN σ_(c,90,d)=F_(c,90,d)/A_ef =F_(c,90,d)/(b∙(l_(support,fi)+15 mm) ) (4.14) σ_(c,90,d)=29,2kN/(77mm∙(72mm+15mm))=4,4 N/mm^2 k_(c,90)∙f_(c,90,d,fi)=(k_(c,90)∙k_(mod,fi)∙k_fi)/γ_(M,fi) ∙f_ k_(c,90)∙f_(c,90,d,fi)=(1,0∙1,0∙1,1)/1,0∙6 N/mm^2=6,6 N/mm^2 σ_(c,90,d)≤k_(c,90)∙f_(m,0,d,fi) →OK (4.13) Discussion According to EN1995-1-2:2004, clause 4.3.1 it is not necessary to verify compression perpendicular to the grain and shear resistance of a beam in the structural fire design. In this example they didn’t become critical, but in the detailing it shall be verified that the beam is securely supported also when the support length becomes smaller due to charring of the supports. 9. CALCULATION EXAMPLES OF LVL STRUCTURES rel =� m,k m,crit =�44 N/mm2 25,8 N/mm2 = 1,36 (4.41) when 0,75 < rel,m≤1,4 , crit =1,56−0,75∙ rel,m=1,56−0,75∙ 1,36=0,58 crit ∙ m,d,fi =0,58 ∙ 47,4 N/mm2 = 27,5 N/mm2 m,d = 19,2 N/mm2 ≤ crit ∙ m,d →OK (4.38) d = d,ULS,fi ∙ /2 = 14,6 kN/m∙ 4,0m/2 = 29,2 kN v,d = 3∙ d 2∙ = 3∙ 2 29,2 kN ∙ 26488 mm2 = 1,7 N/mm2 v,d,fi = mod,fi ∙ fi M,fi ∙ v,0,edge,k = 1,0∙1,0 1,1∙ 4,2 m N m2 = 4,6 N/mm2 m,d ≤ v,d,fi →OK support,fi =100mm−0,70 mm m in∙ 30min+1,0∙ 7mm=72mm c,90,d = d = 29,2 c,90,d = c,90,d ef = c,90,d ∙� support,fi+15 mm� (4.14) c,90,d =77mm 29,2kN ∙ (72mm + 15mm) = 4,4 N/mm2 c,90 ∙ c,90,d,fi = c,90 ∙ mod,fi ∙ fi M,fi ∙ c,90,edge,k c,90 ∙ c,90,d,fi = 1,0∙ 1,0∙ 1,0 1,1∙ 6 N/mm2 = 6,6 N/mm2 c,90,d ≤ c,90 ∙ m,0,d,fi →OK (4.13) rel =� m,k m,crit =�44 N/mm2 25,8 N/mm2 = 1,36 (4.41) when 0,75 < rel,m≤1,4 , crit =1,56−0,75∙ rel,m=1,56−0,75∙ 1,36=0,58 crit ∙ m,d,fi =0,58 ∙ 47,4 N/mm2 = 27,5 N/mm2 m,d = 19,2 N/mm2 ≤ crit ∙ m,d →OK (4.38) d = d,ULS,fi ∙ /2 = 14,6 kN/m∙ 4,0m/2 = 29,2 kN v,d = 3∙ d 2∙ = 3∙ 2 29,2 kN ∙ 26488 mm2 = 1,7 N/mm2 v,d,fi = mod,fi ∙ fi M,fi ∙ v,0,edge,k = 1,0∙1,0 1,1∙ 4,2 m N m2 = 4,6 N/mm2 m,d ≤ v,d,fi →OK support,fi =100mm−0,70 mm m in∙ 30min+1,0∙ 7mm=72mm c,90,d = d = 29,2 c,90,d = c,90,d ef = c,90,d ∙� support,fi+15 mm� (4.14) c,90,d =77mm 29,2kN ∙ (72mm + 15mm) = 4,4 N/mm2 c,90 ∙ c,90,d,fi = c,90 ∙ mod,fi ∙ fi M,fi ∙ c,90,edge,k c,90 ∙ c,90,d,fi = 1,0∙ 1,0∙ 1,0 1,1∙ 6 N/mm2 = 6,6 N/mm2 c,90,d ≤ c,90 ∙ m,0,d,fi →OK (4.13) crit ∙ m,d,fi =0,58 ∙ 47,4 N/mm = 27,5 N/mm m,d = 19,2 N/mm2 ≤ crit ∙ m,d →OK (4.38) d = d,ULS,fi ∙ /2 = 14,6 kN/m∙ 4,0m/2 = 29,2 kN v = 3∙ d 2∙ = 3∙ 2 29,2 kN ∙ 26488 mm2 = 1,7 N/mm2 v,d,fi = mod,fi ∙ fi M,fi ∙ v,0,edge,k = 1,0∙1,0 1,1∙ 4,2 m N m2 = 4,6 N/mm2 m,d ≤ v,d,fi →OK support,fi =100mm−0,70 mm m in∙ 30min+1,0∙ 7mm=72mm c,90,d = d = 29,2 c,90,d = c,90,d ef = c,90,d ∙� support,fi+15 mm� (4.14) c,90,d =77mm 29,2kN ∙ (72mm + 15mm) = 4,4 N/mm2 c,90 ∙ c,90,d,fi = c,90 ∙ mod,fi ∙ fi M,fi ∙ c,90,edge,k c,90 ∙ c,90,d,fi = 1,0∙ 1,0∙ 1,0 1,1∙ 6 N/mm2 = 6,6 N/mm2 c,90,d ≤ c,90 ∙ m,0,d,fi →OK (4.13) LVL Handbook Europe 213
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