9. CALCULATION EXAMPLES OF LVL STRUCTURES w_(inst,q)=(5〖∙q〗_(d,SLS)∙s∙L^4)/(〖384∙E〗_mean∙I)+〖〖6/5 ∙q〗_(d,SLS)∙s∙L〗^2/(〖8∙G〗_mean A)=5,97 mm+0,38 mm=6,35 mm w_inst = 2,86 mm+6,34 mm = 9,2 mm Requirement: w_inst≤L/400=4500/400=11,3 mm→OK Final deflection w_(net,fin) = (1+k_def)∙w_(inst,g) + (1+ψ_2∙k_def)∙w_(inst,q) (4.73) For the load category A: ψ2 = 0,3 w_(net,fin) = (1+0,6)∙2,86 mm + (1+0,3∙0,6)∙6,35 mm = 12,1 mm Requirement: w_(net,fin)≤L/300=4500/300=15 mm→OK Vibration design Lowest natural frequency f1 f_1=π/(2l^2 ) √((EI)_l/m) (4.79) m = g_1+g_2+30 kg/m^2 = 60 kg/m^2+30 kg/m^2+30 kg/m^2 = 120 kg/m^2 Note: In Finnish NA the share of live load qk in the frequency calculation is 30kg/m2 〖(EI)〗_l = EI ∙(1000/s) = 7,15∙1011 Nmm^2 ∙(1000/400 mm) 〖(EI)〗_l = 1,79∙〖10〗^6 Nm^2/m f_1=π/(2∙(4,5m)^2 ) √((1,79∙〖10〗^6 Nm^2/m)/(120 kg/m^2 ))=9,5Hz>8Hz→OK → The floor can be analyzed as a high frequency floor. Floor stiffness perpendicular to the span direction based on 22 mm chipboard decking: (EI)/m = 3500 N/mm^2∙1000 mm∙〖(22 mm)〗^3/12 = 3,11∙〖10〗^3 Nm^2/m For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the impulse velocity response v [m/Ns2] value may, as an approximation, be taken as: v=4(0,4+0,6n_40 )/(mbl+200) (4.80) n_40={((40/f_1 )^2-1) (b/l)^4 (EI)_l/(EI)_b }^0,25 (4.81) n_40={((40/9,5Hz)^2-1) 〖∙(5m/4,5m)〗^4∙(1,79∙〖10〗^6 Nm^2/m v=(4 (0,4+0,6 n_40 ))/(mbl+200)=(4 (0,4+0,6∙11))/(120∙5∙4,5+200 When a high value b = 150 is chosen from the Figure 4.28 and a conservative damping value ξ = 0,01 is used, the requirement for v is v≤〖150〗^((f_1 ξ-1) )=0,011→OK (4.78) inst = inst,g + inst,q inst,g =5∙ d,SLS∙ ∙ 4 384∙ mean∙ + 6 5∙ d,SLS∙ ∙ 2 8∙ mean (4.74) inst,g = 2,69 mm + 0,17 mm = 2,86 mm inst,q = 5∙ d,SLS ∙ ∙ 4 384∙ mean ∙ + 6/5 ∙ d,SLS ∙ ∙ 2 8∙ mean = 5,97 mm+ 0,38 mm = 6,35 mm inst = 2,86 mm + 6,34 mm = 9,2 mm Requirement: inst ≤ 400 =4 4 5 0 0 0 0 = 11,3 mm→OK net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) For the load category A: ψ2 = 0,3 net,fin = (1+0,6)∙ 2,86 mm + (1 + 0,3∙ 0,6)∙ 6,35 mm = 12,1 mm Requirement: net,fin ≤ 300 =4 3 5 0 0 0 0 = 15 mm→OK inst = inst,g + inst,q inst,g =5∙ d,SLS∙ ∙ 4 384∙ mean∙ + 6 5∙ d,SLS∙ ∙ 2 8∙ mean (4.74) inst,g = 2,69 mm + 0,17 mm = 2,86 mm inst,q = 5∙ d,SLS ∙ ∙ 4 384∙ mean ∙ + 6/5 ∙ d,SLS ∙ ∙ 2 8∙ mean = 5,97 mm+ 0,38 mm = 6,35 mm inst = 2,86 mm + 6,34 mm = 9,2 mm Requirement: inst ≤ 400 =4 4 5 0 0 0 0 = 11,3 mm→OK net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) For the load category A: ψ2 = 0,3 net,fin = (1+0,6)∙ 2,86 mm + (1 + 0,3∙ 0,6)∙ 6,35 mm = 12,1 mm Requirement: net,fin ≤ 300 =4 3 5 0 0 0 0 = 15 mm→OK , , , 5∙ , ∙ ∙ 384∙ ∙ ∙ , ∙ ∙ 8∙ , , ∙ , ∙ ∙ ∙ ∙ ∙ , ∙ ∙ ∙ : 400 4 4 5 0 0 0 0 , ∙ , ∙ ∙ , , ∙ ∙ ∙ : , 300 4 3 5 0 0 0 0 1 = 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 =���40 1�2 −1�� �4 ( )l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 48∙ ∙( )l (4.82) =� =� =0,2 1 = 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 =���40 1�2 −1�� �4 ( )l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 48∙ ∙( )l (4.82) =�( ) = 4 �3,11∙103 4 =0,2 1 = 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 =���40 1�2 −1�� �4 l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 48∙ ∙( )l (4.82) 1 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 1�2 −1�� �4 ( )l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 1 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 =���40 1�2 −1�� �4 ( )l b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 6 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 48∙ ∙( ) (4.82) 1 = 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 ( )l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m =4(0,4+0,6 40) +200 (4.80) 40 =���40 1�2 −1�� �4 ( )l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 ≤150( 1 −1) =0,011→ (4.78) =min� ∙ 2 42∙ ∙( )l ∙ 3 48∙ ∙( ) (4.82) 11 217 (255) 400, 4500 400 = 11,3 mm→OK Final deflection , ∙ , ∙ ∙ , , , 300, 4500 300 mm= 15 mm→OK Vibration design Lowest natural frequency f1 1 = 2 2 �( )l (4.79) = 1 + 2 + 30 kg/m2 = 60 kg/m2 + 30 kg/m2 + 30 kg/m2 = 120 kg/m2 Note: In Finnish NA the share of live load qk in the frequency calculation is 30kg/m2 ( ) l = ∙ (1000/ ) = 7,15∙ 1011 Nmm2 ∙ (1000/400 mm) ( )l = 1,79∙ 106 Nm2/m 1 = π 2∙ (4,5m)2�1,79∙ 106Nm2/m 120 kg/m2 = 9,5Hz > 8Hz →OK → The floor can be analyzed as a high frequency floor. Floor stiffness perpendicular to the span direction based on 22 mm chipboard decking: ( )/ = 3500 N/mm2 ∙ 1000 mm∙ (22 mm)3/12 = 3,11∙ 103Nm2/m For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the impulse velocity response v [m/Ns2] value may, as an approximation, be taken as: = 4(0,4+0,6 40) +200 (4.80) 40 40 1�2 −1�� �4 ( )l ( )b� 0,25 (4.81) 40 =���9, 4 5 0 Hz�2 −1� ∙ �4 5 ,5 m m�4 ∙ 1,79∙ 106Nm2/m 3,11∙ 103Nm2/m� 0,25 =11 = 4 (0,4+ 0,6 40) + 200 = 4 (0,4+ 0,6∙ 11) 120∙ 5∙ 4,5 + 200 = 0,010 When a high value b = 150 is chosen from the Figure 4.28 and a conservative damping value ξ = 0,01 is used, the requirement for is , , 2 l , ,6 +200 40 l , 9, 4 5 0 Hz 4 5 ,5 m m , LVL Handbook Europe 181
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