V_(d,(x=323mm))=0,4 m∙4,03 kN/m^2 ∙(1-(0,3 m+0,045 m/2)/(4,5 m))=3,13 kN M_(d,(x=323mm))=(0,4 m∙4,03 kN/m^2 ∙(0,3 m+0,045 m/2))/2∙(4,5 m-(0,3 m+(0,045 m)/2)) M_(d,(x=323mm))=1,1 kNm F_(t,90,d)=(3,13 kN∙0,06 m)/(4∙0,24 m)∙[3-((0,06 m)/0,24m)^2 ]+0,008∙(1,1 kNm)/(0,09 m)=0,57 kN+0,10 kN=0,67 kN σ_(t,90,d)=F_(t,90,d)/(0,5∙l_(t,90)∙b∙k_(t,90) )=(0,67 kN)/(0,5∙0,15 m∙0,045 m∙1,0)=0,20 N/mm^2 ≤f_ (t,90,d)→OK The verification of shear stress concentration at the hole edge shall fulfil the condition: τ_d=k_τ∙(1,5∙V_d)/(b∙(h-h_d ) )≤f_(v,d) (4.62) Where k_τ=1,85∙(1+a/h)∙(h_d/h)^0,2=1,85∙(1+(0,14 m)/(0,24 m))∙((0,06 m)/(0,24 m) (4.63) τ_d=k_τ∙(1,5∙V_d)/(b∙(h-h_d ) )=2,22∙(1,5∙3,13 kN)/(45 mm∙(240 mm-60 mm) )= f_(v,0,edge,d)=k_mod/γ_M ∙f_(v,0,edge,k)=0,8/1,2∙4,5 N/mm^2 =3,0 N/mm^2 τ_d≤f_(v,0,edge,d)→OK Bending stress at the location of a rectangular hole is verified by the equations: (M_d/W_n +M_(o,d)/W_o )/f_(m,d) ≤1 and (M_d/W_n +M_(u, (4.64 and 4.65) Where W_n=(b∙(h^2-h_d^2 ))/6=(45 mm∙((240 mm)^2 〖-(60 mm)〗^2 ))/6=4,0 (4.66) M_(o,d)=A_o/(A_u+A_o )∙V_d∙a/2=(b∙h_ro)/(b∙h_ro+b∙h_ru )∙V_d∙a/2 (4.67) M_(o,d)=(45 mm∙90 mm)/(45 mm∙90 mm+45 mm∙90 mm)∙3,13 kN∙(140 mm W_o=(b∙h_ro^2)/6=(45 mm∙(90 mm)^2)/6=60750 mm^3 (4.69) f_(m,0,d)=k_mod/γ_M ∙k_h∙f_(m,0,k)=0,8/1,2∙1,03∙32 N/mm^2 =22,1 N/mm^2 (M_d/W_n +M_(o,d)/W_o )/f_(m,d) =((1,1 kNm)/(4,05∙〖10〗^5 mm^3 )+(0, (22,1N/mm^2 )=(4,5N/mm^2)/(22,1N/mm^2 )=0,20≤1→OK Since the hole is at the centre line of the cross section, only one of the equations (4.64 and 4.65) needs to be verified. The 140x60 mm hole 300 mm from the support edge fulfils the requirements in LVL 36 C joist. 9. CALCULATION EXAMPLES OF LVL STRUCTURES d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ �1−0,3 m +4,05,0m45 m/2� = 3,13 kN d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ (0 2 ,3 m + 0,045 m/2)∙ �4,5 m−�0,3 m + 0,04 2 5 m�� d,(x=323mm) = 1,1 kNm t,90,d = 3,13 kN∙ 4 0,06 m ∙ 0,24 m ∙ �3−�00,,0264mm�2� +0,008∙ 10,1,0k9Nmm = 0,57 kN + 0,10 kN = 0,67 kN t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 = 0,67 kN 0,5∙0,15 m∙0,045 m∙1,0 =0,20m N m2 ≤ t,90,d →OK d = τ ∙ 1,5∙ d ∙(ℎ−ℎd) ≤ v,d (4.62) τ =1,85∙ �1+ ℎ� ∙ �ℎdℎ �0,2 =1,85∙ �1+0,14 0 m ,24 m� ∙ �0,06 0 m ,24 m�0,2 =2,22 (4.63) d = τ ∙ 1,5∙ d ∙ (ℎ−ℎd) =2,22∙ 1,5∙ 3,13 kN 45 mm∙ (240 mm−60 mm) = 1,3 m N m2 v,0,edge,d = mod M ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,5 m N m2 = 3,0 N/mm2 d ≤ v,0,edge,d →OK d n+ o,d o m,d ≤1 and d n+ u,d u m,d ≤1 (4.64 and 4.65) n = ∙�ℎ2−ℎd2 6 � =45 mm∙�(240 mm)2−(60 mm)2� 6 =4,05∙ 105 mm3 (4.66) o,d = o u+ o ∙ d ∙ 2 = ∙ℎro ∙ℎro+ ∙ℎru ∙ d ∙ 2 (4.67) o,d = 45 mm∙ 90 mm 45 mm∙ 90 mm+ 45 mm∙ 90 mm∙ 3,13 kN∙ 1402mm = 0,11 kNm o = ∙ℎr2 o 6 =45 mm∙(90 mm)2 6 = 60750 mm3 (4.69) m,0,d = mod M ∙ h ∙ m,0,k = 0 1 , , 8 2∙ 1,03∙ 32 m N m2 = 22,1 N/mm2 d n + o,d o m,d = 4,05 1,1 kNm ∙ 105 mm3 + 600,71510kmNmm3 22,1N/mm2 = 4,5N/mm2 22,1N/mm2 =0,20≤1→OK d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ �1−0,3 m +4,05,0m45 m/2� = 3,13 kN d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ (0 2 ,3 m + 0,045 m/2)∙ �4,5 m−�0,3 m + 0,04 2 5 m�� d,(x=323mm) = 1,1 kNm t,90,d = 3,13 kN∙ 4 0,06 m ∙ 0,24 m ∙ �3−�00,,0264mm�2� +0,008∙ 10,1,0k9Nmm = 0,57 kN + 0,10 kN = 0,67 kN t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 = 0,67 kN 0,5∙0,15 m∙0,045 m∙1,0 =0,20m N m2 ≤ t,90,d →OK d = τ ∙ 1,5∙ d ∙(ℎ−ℎd) ≤ v,d (4.62) τ =1,85∙ �1+ ℎ� ∙ �ℎdℎ �0,2 =1,85∙ �1+0,14 0 m ,24 m� ∙ �0,06 0 m ,24 m�0,2 =2,22 (4.63) d = τ ∙ 1,5∙ d ∙ (ℎ−ℎd) =2,22∙ 1,5∙ 3,13 kN 45 mm∙ (240 mm−60 mm) = 1,3 m N m2 v,0,edge,d = mod M ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,5 m N m2 = 3,0 N/mm2 d ≤ v,0,edge,d →OK d n+ o,d o m,d ≤1 and d n+ u,d u m,d ≤1 (4.64 and 4.65) n = ∙�ℎ2−ℎd2 6 � =45 mm∙�(240 mm)2−(60 mm)2� 6 =4,05∙ 105 mm3 (4.66) o,d = o u+ o ∙ d ∙ 2 = ∙ℎro ∙ℎro+ ∙ℎru ∙ d ∙ 2 (4.67) o,d = 45 mm∙ 90 mm 45 mm∙ 90 mm+ 45 mm∙ 90 mm∙ 3,13 kN∙ 1402mm = 0,11 kNm o = ∙ℎr2 o 6 =45 mm∙(90 mm)2 6 = 60750 mm3 (4.69) m,0,d = mod M ∙ h ∙ m,0,k = 0 1 , , 8 2∙ 1,03∙ 32 m N m2 = 22,1 N/mm2 d n + o,d o m,d = 4,05 1,1 kNm ∙ 105 mm3 + 600,71510kmNmm3 22,1N/mm2 = 4,5N/mm2 22,1N/mm2 =0,20≤1→OK ,( ) ∙ 4,03 m kN2 ∙ � 0,3 m +4,05,0m45 m/2 ,( ) ∙ 4,03 m kN2 ∙ (0 2 ,3 m + 0,045 m/2)∙ � 0,3 m + 0,04 2 5 m ,( ) t,90, ∙ ∙ ∙ � 00,,0264mm 2� ∙ 10,1,0k9Nmm = 0,57 kN + 0,10 kN = 0,67 kN t,90, t,90,d 0,5∙ t,90∙ ∙ t,90 0,67 0,5∙0,15 m∙0,045 m∙1,0 m N m2 ≤ t,90, τ ∙ 1,5∙ d ∙( d) ≤ v, (4.62) τ ∙ � ℎ� ∙ � d 0,2 ∙ � 014 ,24 � ∙ �006 ,24 0,2 (4.63) τ ∙ 1,5∙ ∙ ( ∙ 1,5∙ ∙ 60 mm) = 1,3 m N m2 v,0,edge, ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,5 m N m2 2 ≤ v,0,edge, d n o,d o ,d d n u,d u ,d (4.64 and 4.65) ∙�ℎ2 d2 45 (240 )2 (60 )2� 6 ∙ 5 3 (4.66) o, o u+ o ∙ ∙ 2 ∙ℎro ∙ℎro+ ∙ℎru ∙ ∙ 2 (4.67) o, ∙ ∙ ∙ ∙ ∙ o ∙ℎr2 o 6 45 (90 )2 6 3 (4.69) ,0, ∙ ∙ ,0,k = 0 1 , , 8 2∙ ∙ 32 m N m2 2 o, o , = 4,05 1,1 kNm ∙ 5 3 + 600,71510kmNmm3 2 22 d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ �1−0,3 m +4,05,0m45 m/2� = 3,13 kN d,(x=323mm) = 0,4 m∙ 4,03 m kN2 ∙ (0 2 ,3 m + 0,045 m/2)∙ �4,5 m−�0,3 m + 0,04 2 5 m�� d,(x=323mm) = 1,1 kNm t,90,d = 3,13 kN∙ 4 0,06 m ∙ 0,24 m ∙ �3−�00,,0264mm�2� +0,008∙ 10,1,0k9Nmm = 0,57 kN + 0,10 kN = 0,67 kN t,90,d = t,90,d 0,5∙ t,90∙ ∙ t,90 = 0,67 kN 0,5∙0,15 m∙0,045 m∙1,0 =0,20m N m2 ≤ t,90,d →OK d = τ ∙ 1,5∙ d ∙(ℎ−ℎd) ≤ v,d (4.62) τ =1,85∙ �1+ ℎ� ∙ �ℎdℎ �0,2 =1,85∙ �1+0,14 0 m ,24 m� ∙ �0,06 0 m ,24 m�0,2 =2,22 (4.63) d = τ ∙ 1,5∙ d ∙ (ℎ−ℎd) =2,22∙ 1,5∙ 3,13 kN 45 mm∙ (240 mm−60 mm) = 1,3 m N m2 v,0,edge,d = mod M ∙ v,0,edge,k = 0 1 , , 8 2∙ 4,5 m N m2 = 3,0 N/mm2 d ≤ v,0,edge,d →OK d n+ o,d o m,d ≤1 and d n+ u,d u m,d ≤1 (4.64 and 4.65) n = ∙�ℎ2−ℎd2 6 � =45 mm∙�(240 mm)2−(60 mm)2� 6 =4,05∙ 105 mm3 (4.66) o,d = o u+ o ∙ d ∙ 2 = ∙ℎro ∙ℎro+ ∙ℎru ∙ d ∙ 2 (4.67) o,d = 45 mm∙ 90 mm 45 mm∙ 90 mm+ 45 mm∙ 90 mm∙ 3,13 kN∙ 1402mm = 0,11 kNm o = ∙ℎr2 o 6 =45 mm∙(90 mm)2 6 = 60750 mm3 (4.69) m,0,d = mod M ∙ h ∙ m,0,k = 0 1 , , 8 2∙ 1,03∙ 32 m N m2 = 22,1 N/mm2 d n + o,d o m,d = 4,05 1,1 kNm ∙ 105 mm3 + 600,71510kmNmm3 22,1N/mm2 = 4,5N/mm2 22,1N/mm2 =0,20≤1→OK , , , , , , ,5∙ , ∙ ∙ , , , , , , , , ,5∙ ∙ , , ,, ,, , ∙ ∙ , , , ∙ , , , , , , d n o,d o , d n u,d u , ∙�ℎ , + ∙ ∙ℎ ∙ℎ + ∙ℎ ∙ , ∙ℎ , , ∙ ∙ , , , , 208 LVL Handbook Europe
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