9. CALCULATION EXAMPLES OF LVL STRUCTURES Properties of a LVL 36C joist: Bending strength edgewise fm,0,edge,k = 32 N/mm2 Tension in perpendicular to grain strength edgewise ft,90,edge,k = 5 N/mm2 Shear strength edgewise fv,0,edge,k = 4,5 N/mm2 Modification factor kmod for medium-term, SC1 = 0,8 Material safety factor γM (default value in EC5) =1,2 Size effect factor kh = (300/240)0,15 = 1,03 Tension stress perpendicular to the grain is verified by the equation σt,90,d =F_(t,90,d)/(0,5∙l_(t,90)∙b∙k_(t,90) )≤f_(t,90,d) (4.57) where kt,90 =min{█(1@(450/h)^0,5 )=1,0┤ (4.58) σt,90,d = design value of tension stress perpendicular to the grain [N/mm2] lt,90 = 0,5 ∙ (hd + h) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61) Ft,90,d = design value of tension force perpendicular to the grain [N] ft,90,d =k_mod/γ_M ∙f_(t,90,edge,k)=0,8/1,2∙5 N/mm^2 =3,3 N/mm^2 t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57) t,90 =min� 1 �4 ℎ 50�0,5 =1,0 (4.58) t,90 =0,5∙ (ℎd +ℎ)= 0,5∙ (60 mm + 240 mm) = 150 mm (4.61) t,90,d = mod M ∙ t,90,edge,k = 0 1 , , 8 2∙ 5 m N m2 = 3,3 N/mm2 t,90,d = d∙ℎd 4∙ℎ ∙ �3−�ℎdℎ �2� +0,008∙ dℎr (4.59) ℎr =90 (4.60) t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57) t,90 =min� 1 �4 ℎ 50�0,5 =1,0 (4.58) t,90 =0,5∙ (ℎd +ℎ)= 0,5∙ (60 mm + 240 mm) = 150 mm (4.61) t,90,d = mod M ∙ t,90,edge,k = 0 1 , , 8 2∙ 5 m N m2 = 3,3 N/mm2 t,90,d = d∙ℎd 4∙ℎ ∙ �3−�ℎdℎ �2� +0,008∙ dℎr (4.59) ℎr =90 (4.60) t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57) t,90 =min� 1 �4 ℎ 50�0,5 =1,0 (4.58) t,90 =0,5∙ (ℎd +ℎ)= 0,5∙ (60 mm + 240 mm) = 150 mm (4.61) t,90,d = mod M ∙ t,90,edge,k = 0 1 , , 8 2∙ 5 m N m2 = 3,3 N/mm2 t,90,d = d∙ℎd 4∙ℎ ∙ �3−�ℎdℎ �2� +0,008∙ dℎr (4.59) ℎr =90 (4.60) Tension in perpendicular to grain force Ft,90,d depends on the shear force Vd and bending moment Md at the edge of the hole: F_(t,90,d)=(V_d∙h_d)/(4∙h)∙[3-(h_d/h)^2 ]+0,008∙M_d/h_r (4.59) where hr=90 mm (4.60) t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57) t,90 =min� 1 �4 ℎ 50�0,5 =1,0 (4.58) t,90 =0,5∙ (ℎd +ℎ)= 0,5∙ (60 mm + 240 mm) = 150 mm (4.61) t,90,d = mod M ∙ t,90,edge,k = 0 1 , , 8 2∙ 5 m N m2 = 3,3 N/mm2 t,90,d = d∙ℎd 4∙ℎ ∙ �3−�ℎdℎ �2� +0,008∙ dℎr (4.59) ℎr =90 (4.60) LVL Handbook Europe 207
RkJQdWJsaXNoZXIy MjU0MzgwNw==