9. CALCULATION EXAMPLES OF LVL STRUCTURES
Loading combinations
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 ∙ 2,75 N/m
2
= 2,2 kN/m
2
.
The most critical ultimate limit state (ULS) load combination:
E_(d,ULS )= γ_G∙g_k+ γ_Q∙q_k
(4.1)
E_(d,ULS )= 1,15∙(5m∙1,0 kN/m^2 )+1,5∙5m∙2,2 kN/m^2=22,3 kN/m
Note: Safety factors γ
G
and γ
Q
are according to Finnish National annex of Eurocode 0.
The most critical serviceability limit state (SLS) load combination:
E_(d,SLS) = γ_G∙g_k + γ_Q∙q_k
(4.1)
E_(d,SLS) = 1,0∙(5m∙1,0 kN/m^2+1,0∙5m∙2,2kN/m^2=16,0 kN/m
ULS design
Bending moment resistance
M_d= E_(d,ULS)∙s∙L^2/8 = 22,3kN/m∙〖(2,3m)〗^2/8 = 14,7 kNm
σ_(m,d)=M_d/W=(14,7 kNm)/(6,75〖∙10〗^6 mm^3 )=21,8 N/mm^2
f_(m,0,edge,d)=k_mod/γ_M ∙k_h∙f_(m,0,edge,k)=0,8/1,2∙1,00∙44 N/mm^2 =29,3 N/mm^2
σ_(m,d)≤f_(m,0,edge,d) →OK
Lateral torsional buckling
The lintel beam is laterally supported to wall studs in 600mm spacing and the load is applied via them.
Therefore the effective length is
L
ef
= 600mm (See table 4.9).
σ_(m,crit)=M_(y,crit)/W_y =(π√(E_0,05 I_z G_0,05 I_tor ))/(l_ef W_y )
(4.42)
σ_(m,crit)= (π√(10600 N/mm^2∙2,28∙〖10〗^6 mm^4∙400N/mm^2 ∙8,20∙〖10〗^6∙
〗^5 mm^3 )
σ_(m,crit)=72,2 N/〖mm〗^2
λ_rel=√(f_(m,k)/σ_(m,crit) )=√((44 N/mm^2)/(72,2N/mm^2 ))= 0,78
(4.41)
when 0,75<λ_(rel,m)≤1,4 ,k_crit=1,56-0,75∙λ_(rel,m)=1,56-0,75∙0,78=0,97
d,ULS
=
G
∙
k
+
Q
∙
k
(4.1)
d,ULS
= 1,15 ∙ �5m ∙ 1,0 kNm
2
� + 1,5 ∙ 5m ∙ 2,2 kN/m
2
= 22,3 kN/m
d,SLS
=
G
∙
k
+
Q
∙
k
(4.1)
d,SLS
= 1,0 ∙ (5m ∙ 1,0 kN/m
2
+ 1,0 ∙ 5m ∙ 2,2kN/m
2
= 16,0 kN/m
d
=
d,ULS
∙ ∙
2
/8 = 22,3kN/m ∙ (2,3m)
2
/8 = 14,7 kNm
m,d
= = 14,7 kNm 6,75 ∙ 10
6
mm
3
= 21,8 N/mm
2
m,0,edge,d
=
mod M
∙
h
∙
m,0,edge,k
= 0,8 1,2 ∙ 1,00 ∙ 44 Nmm
2
= 29,3 N/mm
2
m,d
≤
m,0,edge,d
→ OK
m,crit
=
y,crit y
=
�
0,05 z 0,05 tor
ef y
(4.42)
m,crit
= π�10600 N/mm
2
∙ 2,28 ∙ 10
6
mm
4
∙ 400Nmm
2
∙ 8,20 ∙ 10
6
∙ mm
4
600mm ∙ 6,75 ∙ 10
5
mm
3
m,crit
= 72,2 /
2
= �
m,k m,crit
= �
44 N/mm
2
72,2N/mm
2
= 0,78
(4.41)
when 0,75 <
rel,m
≤ 1,4 ,
crit
= 1,56 − 0,75 ∙
rel,m
= 1,56 − 0,75 ∙ 0,78 = 0,97
crit
∙
m,d
= 0,97 ∙ 29,3 N/mm
2
= 28,6 N/mm
2
m,d
≤
crit
∙
m,d
→
d
=
d,ULS
∙ ∙ /2 = 22,3kN/m ∙ 2,3m/2 = 25,6 kN
d,ULS
=
G
∙
k
+
Q
∙
k
(4.1)
d,ULS
= 1,15 ∙ �5m ∙ 1,0 kNm
2
� + 1,5 ∙ 5m ∙ 2,2 kN/m
2
= 22,3 kN/m
d,SLS
=
G
∙
k
+
Q
∙
k
(4.1)
d,SLS
= 1,0 ∙ (5m ∙ 1,0 kN/m
2
+ 1,0 ∙ 5m ∙ 2,2kN/m
2
= 16,0 kN/m
d
=
d,ULS
∙ ∙
2
/8 = 22,3kN/m ∙ (2,3m)
2
/8 = 14,7 kNm
m,d
= = 14,7 kN 6,75 ∙ 10
6
mm
3
= 21,8 N/mm
2
m,0,edge,d
=
mod M
∙
h
∙
m,0,edge,k
= 0,8 1,2 ∙ 1,00 ∙ 44 Nmm
2
= 29,3 N/mm
2
m,d
≤
m,0,edge,d
→ OK
m,crit
=
y,crit y
=
�
0,05 z 0,05 tor
ef y
(4.42)
m,crit
= π�10600 N/mm
2
∙ 2,28 ∙ 10
6
mm
4
∙ 400Nmm
2
∙ 8,20 ∙ 10
6
∙ mm
4
600mm ∙ 6,75 ∙ 10
5
mm
3
m,crit
= 72,2 /
2
= �
m,k m,crit
= �
44 N/mm
2
72,2N/mm
2
= 0,78
(4.41)
when 0,75 <
rel,m
≤ 1,4 ,
crit
= 1,56 − 0,75 ∙
rel,m
= 1,56 − 0,75 ∙ 0,78 = 0,97
crit
∙
m,d
= 0,97 ∙ 29,3 N/mm
2
= 28,6 N/mm
2
m,d
≤
crit
∙
m,d
→
d
=
d,ULS
∙ ∙ /2 = 22,3kN/m ∙ 2,3m/2 = 25,6 kN
d,UL
=
G
∙
k
+
Q
∙
k
(4.1)
d,ULS
= 1,15 ∙ �5m ∙ 1,0 kNm
2
� + 1,5 ∙ 5m ∙ 2,2 kN/m
2
= 22,3 kN/m
d,SLS
=
G
∙
k
+
Q
∙
k
(4.1)
d,SLS
= 1,0 ∙ (5m ∙ 1,0 kN/m
2
+ 1,0 ∙ 5m ∙ 2,2kN/m
2
= 16,0 kN/m
d
=
d,ULS
∙ ∙
2
/8 = 22,3kN/m ∙ (2,3m)
2
/8 = 14,7 k
m,d
= = 14,7 kNm 6,75 ∙ 10
6
mm
3
= 21,8 N/mm
2
m,0,edge,d
=
mod M
∙
h
∙
m,0,edge,k
= 0,8 1,2 ∙ 1,00 ∙ 44 Nmm
2
= 29,3 N/mm
2
m,d
≤
m,0,edge,d
→ OK
m,crit
=
y,crit y
=
�
0,05 z 0,05 tor
ef y
(4.42)
m,crit
= π�10600 N/mm
2
∙ 2,28 ∙ 10
6
mm
4
∙ 400Nmm
2
∙ 8,20 ∙ 10
6
∙ mm
4
600mm ∙ 6,75 ∙ 10
5
mm
3
m,crit
= 72,2 /
2
= �
m,k m,crit
= �
44 N/mm
2
72,2N/mm
2
= 0,78
(4.41)
when 0,75 <
rel,m
≤ 1,4 ,
crit
= 1,56 − 0,75 ∙
rel,m
= 1,56 − 0,75 ∙ 0,78 = 0,97
crit
∙
m,d
= 0,97 ∙ 29,3 N/mm
2
= 28,6 N/mm
2
m,d
≤
crit
∙
m,d
→
d
=
d,ULS
∙ ∙ /2 = 22,3kN/m ∙ 2,3m/2 = 25,6 kN
d,ULS
=
G
∙
+
∙
( . )
,
S
= , 5 ∙ 5 ∙ , m
2
, ∙
∙ ,
2
,
,S S
∙
+
∙
( . )
,S S
, ∙ ( ∙ ,
2
, ∙
∙ ,
2
,
d,U S
∙ ∙
2
8
2,
(2,
)
2
/
4,
m,
= 1 , kNm ,75 ∙ 10
6 3
, /
2
m,0,e ge,d
∙
∙
,0,e ge,
0,8 1,2 ∙ ,
∙
Nmm
2
, /
2
,
,0,e ge,
,crit
=
y,crit y
=
�
0,05 z 0,05 tor
ef y
(4.42)
,crit
6
m
2
∙ ,2 ∙ 0
6 4
∙ 4 0m
2
∙ ,
∙
6
∙
4
∙ ,
∙
5
m
3
,crit
2,2
2
m,k ,crit
�
44 /
2
72,2 /
2
,
( . )
, 5
rel,
≤ , ,
crit
= ,
,
∙
rel,
,
,
∙ ,
,
crit
∙
,
= , 7 ∙
,
2
,
2
,
crit
∙
,
,
S
∙ ∙
,
m ∙ ,
,
LVL Handbook Europe
185




