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Loading combinations
Own weight in z-direction
g
k,z
: cos15° ∙ 0,9m ∙ 0,3 kN/m
2
= 0,26 kN/m
Own weight in y-direction
g
k,y
: sin15° ∙ 0,9m ∙ 0,4 kN/m
2
= 0,07 kN/m.
Snow load at roof level
q
k
=
μ
1
∙
C
e
∙
s
k
Form factor
μ
1
= 0,8, when roof angle is less than 30° and in normal
conditions
C
e
= 1,0.
q
k
= 0,8 ∙ 1,0 ∙ 2,5 N/m
2
= 2 kN/m
2
(horizontal projection).
q
k,z
= cos15° ∙ cos15° ∙ 2kN/m = 1,68 kN/m
q
k,y
= cos15 ∙ sin15° ∙ 2kN/m = 0,45 kN/m
The most critical ultimate limit state (ULS) load combination:
E
d,z,ULS
=
γ
G
∙
g
k,z
+
γ
Q
∙
q
k,z
E
d,z,ULS
= 1,15 ∙ 0,26 kN/m
2
+ 1,5 ∙ 1,68 kN/m
2
= 2,82 kN/m
E
d,y,ULS
=
γ
G
∙
g
k,y
+
γ
Q
∙
q
k,y
E
d,y,ULS
= 1,15 ∙ 0,07 kN/m
2
+ 1,5 ∙ 0,45 kN/m
2
= 0,76 kN/m
Axial compression
N
c,d
=
γ
Q
∙
N
c,k
= 1,5 ∙ 3 kN/m
2
= 4,5 kN
Note: Safety factors γ
G
and γ
Q
are according to Finnish national annex of Eurocode 0.
Most critical serviceability limit state (SLS) load combination:
E
d,z,SLS
=
γ
G
∙
g
k,z
+
γ
Q
∙
q
k,z
E
d,z,ULS
= 1,0 ∙ 0,26 kN/m
2
+ 1,0 ∙ 1,68 kN/m
2
= 1,94 kN/m
ULS design
Bending moment resistance in y-direction
M_(d,z) = E_(d,z,ULS)∙L2/8 = 2,82kN/m∙〖(4m)〗^2/8 = 5,64 kNm
σ_(m,y,d)=M_(d,z)/W_y =(5,64 kNm)/(4,32〖∙10〗^5 mm^3 )=13,1 N/mm^2
f_(m,0,edge,d)=k_mod/γ_M ∙k_h∙f_(m,0,edge,k)=0,8/1,2∙1,034∙44 N/mm^2 =30,3 N/mm^2
Bending moment resistance in z-direction at centre support of a 2-span beam
M_(d,y)=E_(d,y,ULS)∙〖(L/2)〗^2/8 = 0,76 kN/m∙〖(4 m/2)〗^2/8 = 0,38 kNm
σ_(m,z,d)=M_(y,d)/W_z =(0,38 kNm)/(8,10〖∙10〗^4 mm^3 )=4,7 N/mm^2
f_(m,0,flat,z,d)=k_mod/γ_M ∙f_(m,0,flat,z,k)=0,8/1,2∙48 N/mm^2 =32,0 N/mm^2
Lateral torsional buckling (LTB) is prevented at the middle of the span.
The purlin is loaded from the compression side and supported against torsion at the main supports and in
the middle of the span. According to Table 6.1 of EN1995-1-1, for uniformly distributed load, the effective
length is
L
ef
= 2000 mm+2∙240 mm = 2480 mm.
σ_(m,y,crit)=M_(z,crit)/W_y =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y )
(4.42)
σ_(m,y,crit)= (π√(10600 N/mm^2∙1,82∙〖10〗^6 〖 mm〗^4∙400 N/mm^2∙6,5
mm∙〖4,32∙10〗^5 mm^3 )
σ_(m,y,crit)=21,6N/mm^2
λ_rel=√((k_h∙f_(m,k))/σ_(m,y,crit) )=√((1,03∙44 N/mm^2)/(21,6 N/mm^2 ))=1,45
(4.41)
9. CALCULATION EXAMPLES OF LVL STRUCTURES
d,y
=
d,y,ULS
∙ (L/2)
2
/8 = 0,76 kN/m ∙ (4 m/2)
2
/8 = 0,38 kNm
m,z,d
=
y,d
= 0,38 kNm 8,10 ∙ 10
4
mm
3
= 4,7 N/mm
2
m,0,flat,z,d
=
mod M
∙
m,0,flat,z,k
= 0,8 1,2 ∙ 48 Nmm
2
= 32,0 N/mm
2
m,y,crit
=
z,crit y
=
�
0,05 0,05 tor
ef y
(4.42)
m,y,crit
= π�10600 N/mm
2
∙ 1,82 ∙ 10
6
mm
4
∙ 400 N/mm
2
∙ 6,56 ∙ 10
6
∙ mm
4
2480 mm ∙ 4,32 ∙ 10
5
mm
3
m,y,crit
= 21,6N/mm
2
rel
= �
h
∙
m,k m,y,crit
= �
1,03∙44 N/mm
2
21,6 N/mm
2
= 1,45
(4.41)
1
2
1
(4.40)
d,y d,y,ULS
∙ (L/2)
2
/8
0,76 k / ∙ (4 m/2)
2
/8
0,38 k
,z,d y,d z
0,38 kNm 8,10 ∙ 10
4 3
4,7 N/mm
2
,0,flat,z,d
od M
∙
,0,flat,z,k
0,8 1,2 ∙ 48 Nmm
2
32,0 N/m
2
,y,crit
z,crit y
�
, 5 0,05 tor
ef y
(4.42)
,y,crit
π 10600 N/
2
∙ 1,82 ∙ 10
6
mm
4
∙ 400 N/mm
2
∙ 6,56 ∙ 10
6
∙
4
2480 m ∙ 4,32 ∙ 10
5
mm
3
,y,crit
21,6 /
2
rel
h
∙
m,k m,y,crit
1,03∙44 N/ m
2
21,6 N/
2
1,45
(4.41)
1
2
1
(4.40)
d,z
=
d,z,ULS
∙ 2/8 = 2,82kN/m ∙ (4m)
2
/8 = 5,64 kNm
m,y,d
=
d,z y
= 5,64 kNm 4,32 ∙ 10
5
mm
3
= 13,1 N/mm
2
m,0,edge,d
=
mod M
∙
h
∙
m,0,edge,k
= 0,8 1,2 ∙ 1,034 ∙ 44 Nmm
2
= 30,3 N/mm
2
d,y
=
d,y,ULS
∙ (L/2)
2
/8 = 0,76 kN/m ∙ (4 m/2)
2
/8 = 0,38 kNm
m,z,d
=
y,d z
= 0,38 kNm 8,10 ∙ 10
4
mm
3
= 4,7 N/mm
2
m,0,flat,z,d
=
mod M
∙
m,0,flat,z,k
= 0,8 1,2 ∙ 48 Nmm
2
= 32,0 N/mm
2
m,y,crit
=
z,crit y
=
�
0,05 0,05 tor
ef y
(4.42)
m,y,crit
= π�10600 N/mm
2
∙ 1,82 ∙ 10
6
mm
4
∙ 400 N/mm
2
∙ 6,56 ∙ 10
6
∙ mm
4
2480 mm ∙ 4,32 ∙ 10
5
mm
3
m,y,crit
= 21,6N/mm
2
rel
= �
h
∙
m,k m,y,crit
= �
1,03∙44 N/mm
2
21,6 N/m
2
= 1,45
(4.41)
when 1,4 <
rel,m
,
crit
=
1
rel,m 2
∙=
1 1,45
2
= 0,48
(4.40)
192
LVL Handbook Europe