9. CALCULATION EXAMPLES OF LVL STRUCTURES (4.39) ((13,1 N/mm^2 )/(0,48∙30,3 N/mm^2 ))^2+(0,42 N/mm^2 )/(0,16∙19,3 N/mm^2 ) Shear resistance V_(d,y) = E_(d,ULS)∙s∙L/2 = 2,92 kN/m∙4 m/2 = 6,2 kN τ_(v,d)=〖3∙V〗_(d,y)/(2∙A)=(3∙6,2 kN)/(2 ∙10 800〖 mm〗^2 )=0,9 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 perpendicular to grain F_(c,90,d) = V_(d,y)=6,2 kN σ_(c,90,d)=F_(c,90,d)/A_ef =(6,2 kN)/(45 mm∙(15 mm+45 mm))=1,2 N/mm^2 σ_(c,90,d)≤k_(c,90)∙f_(m,0,edge,d) →OK SLS design Instantaneous deflection w_inst = w_(inst,g) + w_(inst,q) w_(inst,g)=(5〖∙g〗_(d,z,SLS)∙L^4)/(〖384∙E〗_mean∙I)+mm+0,13 mm=1,75 mm (4.74) w_(inst,q)=(5〖∙q〗_(d,z,SLS)∙L^4)/(〖384∙E〗_m w_inst =1,75 mm+8,45 mm=10,2 mm Final deflection w_(net,fin) = (1+k_def)∙w_(inst,g) + (1+ψ_2∙k_def)∙w_(inst,q (4.73) Note: For the snow load in Finnish national annex: ψ2 = 0,2 w_(net,fin) = (1+0,6)∙1,75 mm + (1+0,2∙0,6)∙8,45 mm = 12,9 mm Requirement: w_(net,fin)≤L/250=4000/250=16 mm→OK 0,16 mm ∙ 19,3 m Nm2 +0,7∙ 30,3 m m Nm m2 + 32,0mmmNm2 = 0,14 + 0,30 + 0,15 = 0,59→OK c,0,d c,y∙ c,0,d + m,y,d m,y,d + m∙ m,z,d m,z,d ≤1 (4.30) 0,42 m Nm2 0,83∙ 19,3 m Nm2 + 13,1 m Nm2 30,3 m Nm2 +0,7∙ 4,7 m Nm2 32,0 m Nm2 = 0,03 + 0,43 + 0,10 = 0,56→OK � m,y,d crit∙ m,o,edge� 2 + c,0,d c,z∙ c,0,d ≤1 (4.39) � 13,1 m Nm2 0,48∙ 30,3 m Nm2� 2 + 0,42 m Nm2 0,16∙ 19,3 m Nm2 = 0,82 + 0,14 = 0,96 →OK d,y = d,ULS ∙ ∙ /2 = 2,92 kN/m∙ 4 m/2 = 6,2 kN v,d = 3∙ d,y 2∙ = 3∙ 2 6,2 kN ∙ 10 800 mm2 = 0,9 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,y = 6,2 kN c,90,d = c,90,d ef = 6,2 kN 45 mm∙ (15 mm+ 45 mm) = 1,2 N/mm2 0,106,42 mm ∙ 19,3 m Nm2 +0,7∙ 1 3 3 0 , , 1 3 m Nm mm2 + 342, 7, 0mmmNm2 = 0,14 + 0,30 + 0,15 = 0,59→OK c,0,d c,y∙ c,0,d + m,y,d m,y,d + m∙ m,z,d m,z,d ≤1 (4.30) 0,42 m Nm2 0,83∙ 19,3 m Nm2 + 13,1 m Nm2 30,3 m Nm2 +0,7∙ 4,7 m Nm2 32,0 m Nm2 = 0,03 + 0,43 + 0,10 = 0,56→OK � m,y,d crit∙ m,o,edge� 2 + c,0,d c,z∙ c,0,d ≤1 (4.39) � 13,1 m Nm2 0,48∙ 30,3 m Nm2� 2 + 0,42 m Nm2 0,16∙ 19,3 m Nm2 = 0,82 + 0,14 = 0,96 →OK d,y = d,ULS ∙ ∙ /2 = 2,92 kN/m∙ 4 m/2 = 6,2 kN v,d = 3∙ d,y 2∙ = 3∙ 2 6,2 kN ∙ 10 800 mm2 = 0,9 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,y = 6,2 kN c,90,d = c,90,d ef = 6,2 kN 45 mm∙ (15 mm+ 45 mm) = 1,2 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k =1,0∙ 0 1 , , 8 2∙ 6,0 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK inst = inst,g + inst,q inst,g =5∙ d,z,SLS∙ 4 384∙ mean∙ + 6 5∙ d,z,SLS∙ 2 8∙ mean = 1,62 mm + 0,13 mm = 1,75 mm (4.74) inst,q = 5∙ d,z,SLS ∙ 4 384∙ mean ∙ + 6/5∙ d,z,SLS ∙ 2 8∙ mean = 7,83 mm + 0,62 mm = 8,45 mm inst = 1,75 mm + 8,45 mm = 10,2 mm net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) For the snow load in Finnish national annex: ψ2 = 0,2 net,fin = (1+0,6)∙ 1,75 mm + (1 + 0,2∙ 0,6)∙ 8,45 mm = 12,9 mm Requirement: net,fin ≤2 L 50 =4 2 0 5 0 0 0 = 16 mm→OK , ∙ , , , , ∙ ∙ , , , 0 1 , , 8 2 , , ≤ , ∙ , , , , , , 5∙ , , ∙ 384∙ ∙ ∙ , , ∙ 8∙ , ∙ , , ∙ ∙ ∙ ∙ , , ∙ ∙ , ∙ , ∙ ∙ , , , 2 L 50 4 2 0 5 0 0 0 , ∙ , , , , ∙ ∙ , , , ∙ 0 1 , , 8 2∙ , , ≤ , ∙ , , , , , , 5∙ , , ∙ 384∙ ∙ ∙ , , ∙ 8∙ , ∙ , , ∙ ∙ ∙ ∙ , , ∙ ∙ , ∙ , ∙ ∙ , , ∙ ∙ ∙ , 2 L 50 4 2 0 5 0 0 0 229 (253) c,90,d = d,y = 6,2 kN c,90,d = c,90,d ef = 6,2 kN 45 mm ∙ (15 mm+ 45 mm) = 1,2 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k = 1,0 ∙ 0 1 , , 8 2 ∙ 6,0 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK SLS design Instantaneous deflection inst = inst,g + inst,q inst,g = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 1,62 mm+ 0,13 mm = 1,75 mm (4.74) inst,q = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 7,83 mm + 0,62 mm = 8,45 mm inst = 1,75 mm + 8,45 mm = 10,2 mm Final deflection net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) Note: For the snow load in Finnish national annex: ψ2 = 0,2 net,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm Requirement: net,fin ≤2 L 50 , 4 2 0 5 0 0 0 =16 mm→OK 9.5 Wall stud Load-bearing internal wall is a centre support of an intermediate floor of a 2 storey one family house. 45x120mm LVL 32 P wall stud L is 2700mm and they are at s = 600mm spacing. Each stud is loaded by the self-weight of the structure gk is 5kN and imposed load qk is 11kN. Eccentricity ez of the loading is assumed to be ¼ of the stud width = 120mm/4 = 30mm. Buckling is prevented by wall panelling in the weaker direction. 229 (253) c,90,d = d,y = 6,2 kN c,90,d = c,90,d ef = 6,2 kN 45 mm ∙ (15 mm+ 45 mm) = 1,2 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k = 1,0 ∙ 0 1 , , 8 2 ∙ 6,0 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK SLS design Instantaneous deflection inst = inst,g + inst,q inst,g = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 1,62 mm+ 0,13 mm = 1,75 mm (4.74) inst,q = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 7,83 mm + 0,62 mm = 8,45 mm inst = 1,75 mm + 8,45 mm = 10,2 mm Final deflection net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) Note: For the snow load in Finnish national annex: ψ2 = 0,2 = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm Requirement: net,fin ≤2 L 50 , 4 2 0 5 0 0 0 =16 mm→OK 9.5 Wall stud Load-bearing internal wall is a centre support of an intermediate floor of a 2 storey one family house. 45x120mm LVL 32 P wall stud L is 2700mm and they are at s = 600mm spacing. Each stud is loaded by the self-weight of the structure gk is 5kN and imposed load qk is 11kN. Eccentricity ez of the loading is assumed to be ¼ of the stud width = 120mm/4 = 30mm. Buckling is prevented by wall panelling in the weaker direction. 0,16 mm 30,3 m m Nm m + 32,0mmmNm , , , ∙ , , , , , , , , , , , , ∙ , , , , , ∙ , , , , ∙ ∙ , ∙ , ∙ , , , ∙ , , , = 0 1 , , 8 2 , ≤ , , , , , , , , , , , ∙ , , , , ∙ ∙ , , , , , , ∙ , , , , , , 5∙ , , ∙ ∙ ∙ , , ∙ 8∙ , ∙ , , ∙ ∙ ∙ ∙ , , ∙ ∙ , ∙ , ∙ ∙ , , , 4 2 0 5 0 0 0 230 (255) c,90,d = d,y = 6,2 kN c,90,d = c,90,d ef = 6,2 kN 45 mm ∙ (15 mm+ 45 mm) = 1,2 N/mm2 c,90 ∙ c,90,edge,d = c,90 ∙ mod M ∙ c,90,edge,k = 1,0 ∙ 0 1 , , 8 2 ∙ 6,0 N/mm2 = 4 N/mm2 c,90,d ≤ c,90 ∙ m,0,edge,d →OK S design antaneous deflection inst = inst,g + inst,q inst,g = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 1,62 mm+ 0,13 mm = 1,75 mm (4.74) inst,q = 5 ∙ d,z,SLS ∙ 4 384 ∙ mean ∙ + 6/5 ∙ d,z,SLS ∙ 2 8 ∙ mean = 7,83 mm + 0,62 mm = 8,45 mm inst = 1,75 mm + 8,45 mm = 10,2 mm al deflection net,fin = (1+ def) ∙ inst,g + (1+ 2 ∙ def) ∙ inst,q (4.73) Note: For the snow load in Finnish national annex: ψ2 = 0,2 net,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm Requirement: net,fin ≤2 L 50 , 4002050mm=16 mm→OK 5 Wall stud d-bearing internal wall is a centre support of an intermediate floor of a 2 storey one family se. 45x120mm LVL 32 P wall stud L is 2700mm and they are at s = 600mm spacing. h stud is loaded by the self-weight of the structure gk is 5kN and imposed load qk is 11kN. entricity ez of the loading is assumed to be ¼ of the stud width = 120mm/4 = 30mm. kling is prevented by wall panelling in the weaker direction. 194 LVL Handbook Europe
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