9. EXEMPLES DE CALCUL DE STRUCTURES EN LAMIBOIS Vérification des ELU Résistance au moment de flexion M_d = E_(d,ULS)∙s∙L2/8 = 25,1 kN/m∙〖(4m)〗^2/8 = 50,3 kNm σ_(m,d)=M_d/W=(50,3 kNm)/(2,72〖∙10〗^6 〖 mm〗^3 )=18,5 N/mm^2 f_(m,0,edge,d)=k_mod/γ_M ∙k_h∙f_(m,0,edge,k)=0,8/1,2∙0,96∙44 N/mm^2 =28,1 N/mm^2 σ_(m,d)≤f_(m,0,edge,d) →OK Déversement La poutre faîtière est chargée par les chevrons du toit reliés aux côtés de la poutre à un espacement de 1 200 mm et ceux-ci agissent comme des supports contre le déversement, de sorte que la longueur efficace est Lef = 1 200 mm. σ_(m,crit)=M_(y,crit)/W_y =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y ) (4.42) σ_(m,crit)= (π√(10600 N/mm^2∙8,84∙〖10〗^6 〖 mm〗^4∙400N/m 〖2,72∙10〗^6 〖 mm〗^3 ) σ_(m,crit)=34,8 N/mm^2 λ_rel=√(f_(m,k)/σ_(m,crit) )=√((44 N/mm^2)/(34,8 N/mm^2 ))= 1,12 (4.41) quand 0,75<λ_(rel,m)≤1,4 ,k_crit=1,56-0,75∙λ_(rel,m)=1,56-0,75∙1,12=0,72 k_crit∙ f_(m,d)=0,72 ∙28,1 N/mm^2=20,1 N/mm^2 σ_(m,d)≤k_crit∙ f_(m,d)→OK (4.38) Résistance au cisaillement V_d = E_(d,ULS)∙s∙L/2 = 25,1kN/m∙4,0m/2 = 50,3 kN τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙50,3 kN)/(2 ∙40 800 mm^2 )=1,9 N/mm^2 f_(v,0,edge,d)=k_mod/γ_M ∙f_(v,0,edge,k)=0,8/1,2∙4,2 N/mm^2 =2,8 N/mm^2 τ_(m,d)≤f_(v,0,edge,d) →OK Compression, perpendiculaire au fil σ_(c,90,d)=F_(c,90,d)/A_ef =F_(c,90,d)/(b∙(l_support+15 mm) ) (4.14) σ_(c,90,d)=50,3kN/(2∙51mm∙(120mm+15mm))=3,7 N/mm^2 k_(c,90)∙f_(c,90,edge,d)=k_(c,90)∙k_mod/γ_M ∙f_(c,90,edge,k)=1,0∙0,8/1,2∙6 N/mm^2=4 N/mm^2 σ_(c,90,d)≤k_(c,90)∙f_(m,0,edge,d) →OK (4.13) = 25,1 kN/m d,SLS = G ∙ ( 1,k + 2,k)+ Q∙ k (4.1) d,SLS =1,0∙ (6m ∙ 1,0 kN/m2 + 0,2 kN/m )+ 6,0∙ 1,0∙ 2,0 kN/m2 d,SLS = 18,2 kN/m d = d,ULS ∙ ∙ 2/8 = 25,1 kN/m∙ (4m)2/8 = 50,3 kNm m,d = d = 2,7 5 2 0,3 kNm ∙ 106 mm3 = 18,5 N/mm2 m,0,edge,d = mod M ∙ h ∙ m,0,edge,k = 0 1 , , 8 2∙ 0,96∙ 44 m N m2 = 28,1 N/mm2 m,d ≤ m,0,edge,d →OK m,crit = y,crit y = � 0,05 0,05 tor ef y (4.42) m,crit = π�10600 N/mm2 ∙ 8,84∙ 106 mm4 ∙ 400N/mm2 ∙ 3,18∙ 107 ∙ mm4 1200 mm∙ 2,72∙ 106 mm3 m,crit = 34,8 N/mm2 rel =� m,k m,crit =�44 N/mm2 34,8 N/mm2 = 1,12 (4.41) when 0,75 < rel,m≤1,4 , crit =1,56−0,75∙ rel,m=1,56−0,75∙ 1,12=0,72 crit ∙ m,d =0,72 ∙ 28,1 N/mm2 = 20,1 N/mm2 m,d ≤ crit ∙ m,d →OK (4.38) d = d,ULS ∙ ∙ /2 = 25,1kN/m∙ 4,0m/2 = 50,3 kN v,d = 3∙ d 2∙ = 3∙ 2 50,3 kN ∙ 40 800 mm2 = 1,9 N/mm2 v,0,edge,d = mod M ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,2 m N m2 = 2,8 N/mm2 m,d ≤ v,0,edge,d →OK c,90,d = d = 50,3 c,90,d = c,90,d ef = c,90,d ∙� support+15 mm� (4.14) c,90,d = 50,3kN 2∙ 51mm∙ (120mm + 15mm) = 3,7 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k =1,0∙ 0 1 , , 8 2∙ 6 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK (4.13) , , , , ∙ , ∙ ∙ , ∙ , ≤ ∙ , , ∙ ∙ ∙ , ∙ ∙ ∙ ∙ , , , ∙ , , , = 0 1 , , 8 2∙ , ≤ , , , , , , , , , , , ∙� , , ∙ ∙ , ∙ , , , , ∙ ∙ , , , ∙ 0 1 , , 8 2∙ , , ≤ , ∙ , , , Ed,ELU d,SLS = G ∙ ( 1,k + 2,k)+ Q∙ k (4.1) d,SLS =1,0∙ (6m ∙ 1,0 kN/m2 + 0,2 kN/m )+ 6,0∙ 1,0∙ 2,0 kN/m2 d,SLS = 18,2 kN/m d = d,ULS ∙ ∙ 2/8 = 25,1 kN/m∙ (4m)2/8 = 50,3 kNm m,d = d = 2,7 5 2 0,3 kNm ∙ 106 mm3 = 18,5 N/mm2 m,0,edge,d = mod M ∙ h ∙ m,0,edge,k = 0 1 , , 8 2∙ 0,96∙ 44 m N m2 = 28,1 N/mm2 m,d ≤ m,0,edge,d →OK m,crit = y,crit y = � 0,05 0,05 tor ef y (4.42) m,crit = π�10600 N/mm2 ∙ 8,84∙ 106 mm4 ∙ 400N/mm2 ∙ 3,18∙ 107 ∙ mm4 1200 mm∙ 2,72∙ 106 mm3 m,crit = 34,8 N/mm2 Ed,ELU rel =� m,k m,crit =�44 N/ 2 34,8 N/mm2 = 1,12 (4.41) when 0,75 < rel,m≤1,4 , crit =1,56−0,75∙ rel,m=1,56−0,75∙ 1,12=0,72 crit ∙ m,d =0,72 ∙ 28,1 N/mm2 = 20,1 N/mm2 m,d ≤ crit ∙ m,d →OK (4.38) d = d,ULS ∙ ∙ /2 = 25,1kN/m∙ 4,0m/2 = 50,3 kN v,d = 3∙ d 2∙ = 3∙ 2 50,3 kN ∙ 40 800 mm2 = 1,9 N/mm2 v,0,edge,d = mod M ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,2 m N m2 = 2,8 N/mm2 m,d ≤ v,0,edge,d →OK c,90,d = d = 50,3 c,90,d = c,90,d ef = c,90,d ∙� support+15 mm� (4.14) c,90,d = 50,3kN 2∙ 51mm∙ (120mm + 15mm) = 3,7 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k =1,0∙ 0 1 , , 8 2∙ 6 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK (4.13) quand Manuel sur le Lamibois (LVL) – Europe 189
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