9. CALCULATION EXAMPLES OF LVL STRUCTURES Note: My,k value should be checked from the DoP of the nail supplier. The influence of rope effect based on the axial withdrawal capacity Fax,k of round nails is negligible. With these properties FV,nail,Rk is as the minimum of EN 1995-1-1, equation 8.6 failure modes (a)-(f) F_(V,nail,Rk)=min{█(4,43 (a)@2,89 (b)@1,55 (c)@1,58 (d)@1,13 (e)@0,85 (f))┤=0,85 kN Design resistance of the connection: R_d = k_mod/γ_M ∙n_ef∙F_(V,nail.Rk) n_ef=n^(k_ef ) kef = 1, when a nail row staggered perpendicular to grain by at least 1d. Without staggering in LVL edge face k_ef=min{█(1@1-0,03(20-a_1/d) )┤=1-0,03(20-50mm/3,1mm)=0,88 Nailing without staggering: R_d = k_mod/γ_M ∙n_ef∙F_(V,nail,Rk)=0,8/1,3∙6^0,88∙0,85 kN=0,62∙4,84∙0,85 kN=2,55 kN E_(d,ULS)>R_d→Not OK Nailing with staggering: R_d = k_mod/γ_M ∙n_ef∙F_(V,nail,Rk)=0,8/1,3∙6∙0,85 kN=3,1 kN Ed,ULS ≤ Rd → OK, staggering is required The canopy can be supported on a 51x300 mm LVL 48P ledger beam which is connected to 45 mm LVL 32P wall studs with 6pcs 3,1x90 mm round nails when the nail row is staggered perpendicular to grain by 1d. Discussion: The inclined screws connection has 16% higher capacity than the laterally loaded nailed connection in the example 9.6 and the leger beam depth is 100 mm smaller. However, the LVL 32P stud needs to be thicker due to the edge distance requirement a2,CG ≥4d of the screws. A laterally loaded screws connection would not be possible for the combination of screw size and LVL stud thickness, since the edge distance a4,c ≥ 7d at the LVL edge would not be fulfilled. V,nail,Rk =min⎪⎨ ⎩ ⎪⎧4,43 ( ) 2,89 ( ) 1,55 ( ) 1,58 ( ) 1,13 ( ) 0,85 ( ) = 0,85 kN d = mod M ∙ ef ∙ V,nail.Rk ef = ef ef =min� 1 1−0,03(20− 1/ ) =1−0,03�20−3 5 , 0 1 m m m m� =0,88 V,nail,Rk =min⎪⎨ ⎩ ⎪⎧4,43 ( ) 2,89 ( ) 1,55 ( ) 1,58 ( ) 1,13 ( ) 0,85 ( ) = 0,85 kN d = mod M ∙ ef ∙ V,nail.Rk ef = ef ef =min� 1 1−0,03(20− 1/ ) =1−0,03�20−3 5 , 0 1 m m m m� =0,88 V,nail,Rk =min⎪⎨ ⎩ ⎪⎧4,43 ( ) 2,89 ( 1,55 ( 1,58 ( 1,13 ( 0,85 ( = 0,85 kN d = mod M ∙ ef ∙ V,nail.Rk ef = ef ef =min� 1 1−0,03(20− 1/ ) =1−0,03�20−3 5 , 0 1 m m m m� =0,88 d = mod M ∙ ef ∙ V,nail,Rk=0,1 8,3∙ 60,88 ∙ 0,85 kN = 0,62∙ 4,84∙ 0,85 kN = 2,55 kN d,ULS > d →Not OK d = mod M ∙ ef ∙ V,nail,Rk=0,1 8,3∙ 6∙ 0,85 kN = 3,1 kN d = mod M ∙ ef ∙ V,nail,Rk=0,1 8,3∙ 60,88 ∙ 0,85 kN = 0,62∙ 4,84∙ 0,85 kN = 2,55 kN d,ULS > d →Not OK d = mod M ∙ ef ∙ V,nail,Rk=0,1 8,3∙ 6∙ 0,85 kN = 3,1 kN LVL Handbook Europe 205
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