LVL Handbook EUROPE

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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
c,y∙ c,0,d m,y,d m,z,d (4.30)
0,42 Nmm2
0,83 ∙ 19,3 Nmm2 + 13,1 Nmm2
30,3 Nmm2 + 0,7 ∙ 4,7 Nmm2
32,0 Nmm2 = 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 Nmm2
0,48 ∙ 30,3 Nmm2�2 + 0,42 Nmm2
0,16 ∙ 19,3 Nmm 2 = 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 ∙ 6,2 kN2 ∙ 10 800 mm2 = 0,9 N/mm2
v,0,edge,d = mod M∙ v,0,edge,k = 0,8 1,2 ∙ 4,2 Nmm2 = 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,0,d c,y∙ c,0,d m,y,d m,y,d m m,z,d m,z,d (4.30)
0,42 Nmm2
0,83 ∙ 19,3 Nmm2 + 13,1 Nmm2
30,3 Nmm2 + 0,7 ∙ 4,7 Nmm2
32,0 Nmm2 = 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 Nmm2
0,48 ∙ 30,3 Nmm2�2 + 0,42 Nmm2
0,16 ∙ 19,3 Nmm 2 = 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 ∙ 6,2 kN2 ∙ 10 800 mm2 = 0,9 N/mm2
v,0,edge,d = mod M∙ v,0,edge,k = 0,8 1,2 ∙ 4,2 Nmm2 = 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 ∙ 0,8 1,2 ∙ 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 m + 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 sn l in Fi nish national a nex: ψ2 = 0,2
n t,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm
Requirement : net,fin ≤ L 250= 4000 250 = 16 mm → OK
, , , , , , , , , 0,8 1,2 ,
, , , , , ,
, ,
, ∙ , , ∙
∙ ∙ 6∙ , , ∙
∙ , , ,
, , , 4 5 , , 2 , , , 5
st , , ,
,fin , ,
ow load i i i i l n
e , = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0, , ,
i net,fin
c,90 ∙ c,90,e ge, = c,90 ∙ mo ∙ c,90,e ge,k , ∙ 0,8 1,2 ∙ , 2 2
c,90, c,90 ∙ ,0,e ge,
i st inst,g i st,
i st,g 5∙ d,z,SLS∙ 4
384∙ ean∙ 5∙ ,z,SLS∙ 2
8∙ ean , , , (4.74)
i st, = ∙ d,z,SLS ∙
3 ∙ ea ∙ + ∙ ,z,SLS ∙
∙ ea , , ,
i = 1,75 mm + 8, = 1 ,
net,fin = ( def) ∙ inst,g + (1 2 ∙ def ) ∙ inst,q ( . )
r th s l i Fi i natio l : 2 ,
n t,fi , ∙ , , ∙ , ) ∙ , ,
Req ir t: n ,fin L 250 = 4000 250 mm →
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 ∙ 0,8 1,2 ∙ 6,0 N/mm2 = 4 N/mm2
c,90,d ≤ c,90 ∙ m,0,edge,d → OK
LS design
stantaneous 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)in ,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
inal deflection
net,fin = (1 + def ) ∙ inst,g + (1 + 2 ∙ def) ∙ inst,q (4.73)
Note: For the snow load in Finnish national annex: ψ 2 = 0,2
net,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm
Requirement: net,fin ≤ L 250, 4000 250 = 16 mm → OK
.5 Wall stud
oad-bearing internal wall is a centre support of an intermediate floor of a 2 storey one family
ouse. 45x120mm LVL 32 P wall stud L is 2700mm and they are at s = 600mm spacing.
ach stud is loaded by the self-weight of the structure gk is 5kN and imposed load qk is 11kN.
ccentricity e z of the loading is assumed to be ¼ of the stud width = 120mm/4 = 30mm.
uckling 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 ∙ 0,8 1,2 ∙ 6,0 N/mm2 = 4 N/mm2
c,90,d ≤ c,90 ∙ m,0,edge,d → OK
LS design
stantaneous deflection
inst = inst,g + inst,q
,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
inal deflection
net,fin = (1 + def ) ∙ inst,g + (1 + 2 ∙ def) ∙ inst,q (4.73)
Note: For the snow load in Finnish national annex: ψ 2 = 0,2
net, in 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm
R quirement: net,fin ≤ L 250, 4000 250 = 16 mm → OK
.5 Wall stud
oad-bearing internal wall is a centre support of an intermediate floor of a 2 storey one family
ouse. 45x120mm LVL 32 P wall stud L is 700mm and they are at s = 600mm spacing.
ach stud is loaded by the self-weight of the structure gk is 5kN and imposed load qk is 11kN.
ccentricity e z of the loading is assumed to be ¼ of the stud width = 120mm/4 = 30mm.
uckling is prevented by wall panelling in the we ker direction.
c,0,d ,∙ , , ,y, , , m,z,
,
, , , N , , , ,N , , , ,
, ,
i ∙ , , , ,
, ∙ , ,
� 13,1 Nm , , , N ,
, , , ,
, , ∙ , ,
v,d , , , / 2
, , , , , , , , , ,
, , , ,
, , ,y ,
, , , , , ,
c,90 ∙ c,90,edge,d c,90 ∙ ∙ c,90,e ge,k , ∙ 0,8 1,2 ∙ , 2 m 2
c,90, c,90 ∙ ,0,e ge,
i st i st,g i st,q
i st,g 5∙ , , ∙ 4
384∙ ean∙ ∙ d,z,SLS∙ 2
8∙ ean , , ,
i st,q ∙ ,z,SLS ∙ 4
∙ ea ∙ ∙ ,z,SLS ∙
∙ ea , , ,
i t = , , ,
et,f ( e ) ∙ i , ( ∙ ) ∙ n t, ( . )
For t l i i i ti l : 2 ,
t,fi ( 0 ) 1 + ( , ) ∙ , ,
i , L 50 4000 250
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/mm 2
c,90 ∙ c,90,edge,d = c,90 ∙ mod M∙ c,90,edge,k = 1,0 ∙ 0,8 1,2 ∙ 6,0 N/mm2 = 4 N/mm2
c,90,d ≤ c,90 ∙ m,0,edge,d → OK
esign
taneous 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
eflection
net,fin = (1 + def ) ∙ inst,g + (1 + 2 ∙ def) ∙ inst,q (4.73)
Note: For the snow load in F nish national annex: ψ2 = 0,2
net,fin = (1 + 0,6) ∙ 1,75 mm + (1 + 0,2 ∙ 0,6) ∙ 8,45 mm = 12,9 mm
Req ireme : net,fin ≤ L 250, 4000 mm 250 = 16 mm → OK
Wall stud
bearing internal wall is a centre support of an interme iat floor of a 2 storey one family
. 45x120mm LVL 32 P wall stud L is 2700mm and they are at s = 600mm spacing.
stud is load d by the self-weight of th structure gk is 5kN and imposed load qk is 11kN.
tricity ez of the loading is ssum d to be ¼ of the stud width = 120mm/4 = 30mm.
ng is prevented by wall panelling in the weaker direction.
194 LVL Handbook Europe