4. STRUCTURAL DESIGN OF LVL STRUCTURES The calculations are made assuming that the floor is unloaded, i.e. only the mass of the floor and other permanent actions are accounted for. Note: in some National Annexes a part of the live load is also taken into consideration, e.g. in Finland 30 kg/ m2 31. For a rectangular floor with span l, the fundamental frequency f1 may be approximately calculated as follows: f_1=π/(2l^2 ) √((EI)_l/m) (4.78) (EC5 7.5) where m is mass per unit area [kg/m2]; l is the floor span [m]; and (EI)l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: v=(4 ∙ (0,4 + 0,6 n_40 ))/(m ∙ b ∙ l + 200) (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; n40 in the number of first-order modes with natural frequencies up to 40 Hz; b is floor width [m]; m is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: n_40={((4 (4.80) (EC5 7.7) where (EI)b is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: w=min{█((F ∙ l^2)/( (4.81) (EC5 NA, Finland)31 or w=(F ∙ l^2)/(43,6 ∙〖 k_(δ )∙ ( (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] k_δ=∜((EI)_b/(EI)_l ) (4.83) with the limitation kδ ≤ b/l For multiple span floor additional instructions can be found, e.g., from the National annex of Austria 33. For residential floors with a fundamental frequency less than 8Hz (f1 ≤ 8Hz) a special investigation should be made. Instructions for 4,5 Hz ≤ f1 ≤ 8 Hz cases are defined, e.g., in the National Annexes of Austria33 or Germany. In practice, their requirements can be fulfilled only when the own weight of the floor is >250 kg/m2, which is quite heavy for an LVL floor structure. 4.4 COMBINED CROSS SECTIONS 4.4.1 Basic principles Glued composite cross sections utilize the joints between members, significantly increasing the stiffness and resistance of the whole cross section compared to the members acting separately. This composite action can be calculated for a mechanically jointed cross section, but the influence of joint slip must then be taken into consideration and, therefore, the overall stiffness is much lower. The specific properties of the composite cross section that are essential to the structural analysis – effective stiffness EIeff, normal stresses from bending moment, and shear stress at the glued joints – can be defined according to equations (4.84) – (4.88). The effective stiffness EIeff of a glued composite cross section is calculated according to equation: 〖EI〗_eff= ∑_i▒〖 E_(i ) I_i+E_i A_i e_i^2 〗 (4.84) where EIeff is the effective stiffness of the composite cross section [Nmm2]; Ei is the modulus of elasticity of a part i [N/mm2]; Ii is the moment of inertia of a part i [mm4], for rectangular cross section Ii = bi∙hi 3/12, where b i is the width [mm] of the part and hi is the height [mm] of the part; Ai is the cross-sectional area of a part i [mm2]; and ei is the eccentricity of the part i = distance between the centre of gravity of part i and neutral axis of the entire composite cross section [mm]. The location of the neutral axis of a composite cross section related to the bottom of the section is: e_0=(∑_i▒〖E_i ∙ A_i ∙ a_i 〗)/(∑_i▒〖E_i ∙ A_i 〗) (4.85) where ai is the distance between the centre of gravity of part i and the bottom of the entire composite cross section [mm]. eff = ∑ i i + i i i 2 i Ii is the moment of inertia of a part i [mm4], for rectangular cros bi∙hi 3/12, where bi is the width [mm] of the part and hi is the h part; Ai is the cross-sectional area of a part i [mm2]; and ei is the eccentricity of the part i = distance between the centre and neutral axis of the entire composite cross section [mm]. The location of the neutral axis of a composite cross section related to the section is: 0 = ∑ i ∙ i ∙ i i∑ i ∙ i i (4.85) where ai is the distance between the centre of gravity of part i and the entire composite cross section [mm]. Normal stress from bending moment is calculated for composite cross se the equation: i,d(z) = i ∙ (z)i ∙ d eff (4.86) 163 (255) (4.78) (EC5 7.5) where m is mass per unit area [kg/m2]; l is the floor span [m]; and (EI)l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: = 4 ∙ (0,4 + 0,6 40) ∙ ∙ + 200 (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; n40 in the number of first-order modes with natural frequencies up to 40 Hz; b is floor width [m]; m is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: 40 = {((40 1) 2 − 1) ( )4 ( ) ( ) } 0,25 (4.80) (EC5 7.7) where (EI)b is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: =min { ∙ 2 42 ∙ ∙ ( )l ∙ 3 48 ∙ ∙ ( )l (4.81) (EC5 NA, Finland)31 or = ∙ 2 43,6 ∙ ∙ ( )l (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] = √( )b ( )l 4 (4.83) 163 (255) (4.78) (EC5 7.5) where m is mass per unit area [kg/m2]; l is the floor span [m]; and l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: = 4 ∙ (0,4 + 0,6 40) ∙ ∙ + 200 (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; n40 in the number of first-order modes with natural frequencies up to 40 Hz; b is floor width [m]; m is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: 40 = {((40 1) 2 − 1) ( )4 ( ) ( ) } 0,25 (4.80) (EC5 7.7) where (EI) is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: =min { ∙ 2 42 ∙ ∙ ( )l ∙ 3 48 ∙ ∙ ( )l (4.81) (EC5 NA, Finland)31 = ∙ 2 43,6 ∙ ∙ ( )l (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] = √( )b ( )l 4 (4.83) 163 (255) (4.78) (EC5 7.5) where is mass per unit area [kg/m2]; l is the floor span [m]; and (EI)l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: = 4 ∙ (0,4 + 0,6 40) ∙ ∙ + 200 (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; n40 in the number of first-order modes with natural frequencies up to 40 Hz; is floor width [m]; is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: 40 = {((40 1) 2 − 1) ( )4 ( ) ( ) } 0,25 (4.80) (EC5 7.7) where (EI)b is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: =min { ∙ 2 42 ∙ ∙ ( )l ∙ 3 48 ∙ ∙ ( )l (4.81) (EC5 NA, Finland)31 or = ∙ 2 43,6 ∙ ∙ ( )l (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] = √( )b ( )l 4 (4.83) 163 (255) (4.78) (EC5 7.5) where is mass per unit area [kg/m2]; l is the floor span [m]; and l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: = 4 ∙ (0,4 + 0,6 40) ∙ ∙ + 200 (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; 40 in the number of first-order modes with natural frequencies up to 40 Hz; is floor width [m]; is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: 40 = {((40 1) 2 − 1) ( )4 ( ) ( ) } 0,25 (4.80) (EC5 7.7) where (EI) is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: =min ∙ 2 42 ∙ ∙ ( )l ∙ 3 48 ∙ ∙ ( )l (4.81) (EC5 NA, Finland)31 = ∙ 2 43,6 ∙ ∙ ( )l (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] = √( )b ( )l 4 (4.83) 163 (255) (4.78) (EC5 7.5) where m is mass per unit area [kg/m2]; l is the floor span [m]; and (EI)l is the equivalent plate bending stiffness of the floor about an axis perpendicular to the beam direction calculated for 1 metre wide section [Nm2/m]. For a rectangular floor with overall dimensions b x l, simply supported along all four edges, the value may, as an approximation, be taken as: = 4 ∙ (0,4 + 0,6 40) ∙ ∙ + 200 (4.79) (EC5 7.6) where v is the unit impulse velocity response [m/Ns2]; n40 in the number of first-order modes with natural frequencies up to 40 Hz; b is floor width [m]; m is the mass [kg/m2]; and l is the floor span [m]. The value of n40 may be calculated from: 40 = {((40 1) 2 − 1) ( )4 ( ) ( ) } 0,25 (4.80) (EC5 7.7) where (EI)b is the equivalent plate bending stiffness of the floor about an axis parallel to the beam direction calculated for a 1 metre wide section [Nm2/m] and (EI) b<(EI)l. The deflection under F =1 kN point load can be calculated from equations: =min { ∙ 2 42 ∙ ∙ ( )l ∙ 3 48 ∙ ∙ ( )l (4.81) (EC5 NA, Finland)31 or = ∙ 2 43,6 ∙ ∙ ( )l (4.82) (EC5 NA, Austria)33 where s is the spacing of the floor beams [m] = √( )b ( )l 4 (4.83) Figure 4.28. Recommended range of and relationship between a and b. Performance improves in the arrow 1 direction and decreases in the arrow 2 direction (EC5 Figure 7.2). The calculations are made assuming that the floor is unloaded, i.e. only the mass of the floor and other permanent actions are accounted for. Note: in some National Annexes a part of the live load is also taken into consideration, e.g. in Finland 30 kg/m2 31. For a rectangular floor with span l, the fundamental frequency f1 may be approximately calculated as follows: 1 = 2 2√( )l 138 LVL Handbook Europe
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