LVL Handbook EUROPE

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4. STRUCTURAL DESIGN OF LVL STRUCTURES
4.3.9 Stability of LVL members
LVL cross sections are usually slender, as it is economical to
produce panel billets and cut thin and high or deep beams and
stud dimensions. Stability calculation is therefore particularly
important for LVL beams.
Column stability and lateral torsional stability shall be ver-
ified using the characteristic stiffness properties E0,05 and G0,05.
4.3.9.1 Members subjected to combined bending
and axial compression or tension
Roof rafters of pitched roofs are typical members which shall
be analysed for combined bending and compression. Accord-
ing to Eurocode 5, the expressions (4.17) & (4.18) or (4.19) &
(4.20) shall be fulfilled.
For combined bending and axial tension the following ex-
pressions shall be satisfied:
σ_(t,0,d)/f_(t,0,d) +σ_(m,y,d)/(k_(m,α (4.17) (EC5 6.17)
σ_(t,0,d)/f_(t,0,d) +k_m σ_(m,y,d) (4.18) (EC5 6.18)
For combined bending and axial compression the following
expressions shall be satisfied:
(σ_(c,0,d)/f_(c,0,d) )^2+σ_(m,y,d)/( (4.19) (EC5 6.19)
(σ_(c,0,d)/f_(c,0,d) )^2+k_m σ_(m, (4.20) (EC5 6.20)
where
km makes allowance for re-distribution of stresses and the
effect of inhomogeneities of the material in cross-section.
For rectangular LVL cross sections km = 0,7 and for other
cross sections km = 1,0; and
km,α is a factor for combined stresses in tapered beams, see
subsection 4.3.11. For straight beams km,α =1,0.
4.3.9.2 Columns subjected to either compression or
combined compression and bending
According to Eurocode 5, the expressions (4.20) & (4.21) shall
be fulfilled.
σ_(c,0,d)/〖k_(c,y)∙f〗_(c,0,d) +σ_(m (4.29) (EC5 6.23)
σ_(c,0,d)/〖k_(c,z)∙f〗_(c,0,d) +k_m∙σ (4.30) (EC5 6.24)
When both λrel,z ≤ 0,3 and λrel,y ≤ 0,3 the stresses should sat-
isfy the expressions (4.19) and (4.20) of combined bending and
axial compression. In all other cases the stresses, which will
be increased due to deflection, should satisfy the expressions
(4.29) and (4.30).
k_(c,y)=1/(k_y+√(k_y^2-λ_(rel,y)^2 )) (4.31)(EC5 6.25)
k_(c,z)=1/(k_z+√(k_z^2-λ_(rel,z)^2 (4.32) (EC5 6.26)
k_y=0,5(1+β_c (λ_(rel,y)-0,3)+λ_(rel,y)^2 (4.33) (EC5 6.27)
k_z=0,5(1+β_c (λ_(rel,z)-0,3)+λ_(rel,z)^2 (4.34) (EC5 6.28)
Factor βc is 0,10 for LVL members within the straightness
limit of L/500. The limit is defined in Eurocode 5, Section 10 as
the deviation from straightness measured midway between the
supports of frame members, columns and beams where lateral
instability can occur.
The relative slenderness ratio should be taken as:
λ_(rel,y)=λ_y/π √(f_(c,0,k)/E_0,05 ) (4.35) (EC5 6.21)
λ_(rel,z)=λ_z/π √(f_(c,0,k)/E_0,05 ) (4.36) (EC5 6.22)
Where
λy and λrel,y are the slenderness ratio corresponding to
bending about the y-axis (deflection in the
z-direction);
λz and λrel,z are the slenderness ratio corresponding to
bending about the z-axis (deflection in the
y-direction); and
E0,05 is the characteristic value of modulus of elasticity
parallel to the grain.
For rectangular cross section the slenderness of a member is
defined as:
λ=l_c/i=l_c/√((I/A) )=l_c/√((((bh^3)/12))/bh)=√12 (4.37)
where
lc is the buckling length; and
h is the height of the member in the direction of the buckling
analysis.
The buckling coefficient kc for different LVL classes and slen-
derness ratios are shown in Figure 4.18 and Table 4.8. The val-
ues are very similar for all shown LVL classes.
More advanced instructions on determining the effect of
an angle to the grain on LVL strength properties may be found
in the manufacturers’ technical documentation.
t,0,d t,0,d+ m,y,d m,αm,y,d + m m,z,d m,z,d≤ 1 (4.17) (EC5 6.17)
t,0,d t,0,d+ m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 (4.18) (EC5 6.18)
� c,0,d c,0,d� 2 + m,y,d m,αm,y,d + m m,z,d m,z,d≤ 1 (4.19) (EC5 6.19)
� c,0,d c,0,d� 2 + m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 (4.20) (EC5 6.20)
t,0,d t,0,d+ m,y,d m,αm,y,d + m m,z,d m,z,d≤ 1 (4.17) (EC5 6.17)
t,0,d t,0,d+ m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 4.18 (EC5 6.18)
� , , , ,� + m,y,d ,α,y,d + m , , , ,≤ 1 19 E 19
� c,0,d c,0,d�2 + m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 (4.20) (EC5 6.20)
t,0,d t,0,d+ m,y,d m,αm,y,d + m m,z,d m,z,d≤ 1 (4.17) (EC5 6.17)
t,0,d t,0,d+ m m,y,d m,αm,y, + m,z,d m,z, ≤ 1 (4.18) (EC5 6.18)
� c,0,d c,0,d�2 + m,y,d m,αm,y,d + m m,z,d m,z,d≤ 1 (4.19) (EC5 6.19)
� c,0,d c,0,d�2 + m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 (4.20) (EC5 6.20)
t,0,d t,0,d+ m,y,d m,αm,y,d + m ,z, m,z,d≤ 1 (4.17) (EC5 6.17)
t,0, t,0,d+ m m,y,d m,αm,y,d + ,z,d m,z,d≤ 1 (4.18) (EC5 6.18)
� c,0,d c,0,d�2 + m,y,d m,αm,y,d + m ,z, m,z,d≤ 1 (4.19) (EC5 6.19)
� c,0,d c,0,d�2 + m m,y,d m,αm,y,d + m,z,d m,z,d≤ 1 (4.20) (EC5 6.20)
c,0,d c,y∙ c,0,d + m,y,d m,α∙ m,y,d + m ∙ m,z,d m,z,d≤ 1 (4.29) (EC5 6.23)
c,0,d c,z
∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1 (4.30) (EC5 6.24)
c,y = 1
y+� y2− rel,y 2 (4.31) (EC5 6.25)
1 (4.32) (EC5 6.26)
c,0,d c,y∙ c,0,d + m,y,d m,α∙ m,y,d + m ∙ , , m,z,d≤ 1 (4.29) (EC5 6.23)
c,0,d c,z∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + , , m,z,d≤ 1 (4.30) (EC5 6.24)
c,y = 1
y+� y2− rel,y 2 (4.31) (EC5 6.25)
(4.32) (EC5 6.26)
c,0,d c,y∙ c,0,d + m,y,d m,α∙ m,y,d + m ∙ m,z,d m,z,d≤ 1
c,0,d c,z
∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1
c,y = 1
y+� y2− rel,y 2
c,z = 1
z+� z2− rel,z 2
y = 0,5�1 + c� rel,y − 0,3� + rel,y 2 �
z 0,5�1 + c� rel,z − , � rel,z 2 �
rel,y = y � c,0,k 0,05
rel,z = z � c,0,k 0,05
= c = c
�� � = c
�� ℎ 3 12 � ℎ = √12 � c ℎ �
c,0,d c,y∙ c,0,d + ,y,d ,α
∙ ,y,d + ∙ m,z,d m,z,d≤ 1
c,0,d c,z∙ c,0,d ∙ ,y,d ,α
∙ ,y,d ,z,d m,z,d
c,y 1
y y2 rel,y 2
c,z 1
z � z2 rel,z 2
y , �1 c� rel,y , � rel,y 2 �
z = 0,5�1 + c� rel,z − 0,3� + rel,z 2 �
rel,y y c,0,k 0,05
rel,z = z � c,0,k 0,05
c c
� � c � ℎ 3 12 � ℎ c
ℎ
c,0,d c,y∙ c,0,d + m,y,d m,α∙ m,y,d + m ∙ m,z,d m,z,d≤ 1
c,0,d c,z∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1
c,y = 1
y+� y2− rel,y 2
c,z = 1
z+� z2− rel,z 2
y = 0,5�1 + c� rel,y − 0,3� + rel,y 2 �
z = 0,5�1 + c� rel,z − 0,3� + rel,z 2 �
rel,y = y � c,0,k 0,05
rel,z = z � c,0,k 0,05
= c = c
�� � = c
�� ℎ 3 12 � ℎ = √12 � c
ℎ�
c,0,d c,y∙ c,0,d + m,y,d m,α∙ m,y,d + m ∙ ,z,d ,z,≤ 1
c,0,d c,z
∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1
c,y = 1
y+� y2− rel,y 2
c,z = 1
z+� z2− rel,z 2
y , c rel,y , rel,y 2 �
z = 0,5�1 + c� rel,z − 0,3� + rel,z 2 �
rel,y = y � c,0,k 0,05
rel,z = z � c,0,k 0,05
= c = c
�� � = c
�� ℎ 3 12 � ℎ = √12 � c
ℎ�
c,0,d c,y∙ c,0,d + m,y,d m,α m,y,d+ m ∙ m,z,d m,z,d 1
c,0,d c,z
∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1
,y y y rel,y
c,z = 1
z+� z2− rel,z 2
y = 0,5�1 + c� rel,y − 0,3� + rel,y 2 �
z = 0,5�1 + c� rel,z − 0,3� + rel,z 2 �
rel,y y c,0,k 0,05
rel,z = z � c,0,k 0,05
= c = c
�� � = c
�� ℎ 3 12 � ℎ = √12 � c
ℎ �
c,0,d c,y∙ c,0,d m,y,d m,α∙ m,y,d m ∙ m,z,d 1
c,0,d c,z
∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d m,z,d m,z,d ≤ 1
c,y 1
y y2 rel,y 2
c,z 1
z+� z2− rel,z 2
y 0,5�1 c� rel,y 0,3� rel,y 2 �
z 0,5�1 c� rel,z 0,3� rel,z 2 �
l,y y c,0,k 0
rel,z = z � c,0,k 0,05
= c = c
�� � c
�� ℎ 3 12 � ℎ √12 � c ℎ �
c,0,d c,y∙ , , + m,y,d m,α∙ m,y,d + m ∙ , , , ,
c,0,d c,z∙ c,0,d + m ∙ m,y,d m,α∙ m,y,d + m,z,d m,z,d≤ 1
c,y 1
y+� y2− rel,y 2
c,z = 1
z+� z2− rel,z 2
y = 0,5�1 + c� rel,y − 0,3� + rel,y 2 �
z = 0,5�1 + c� rel,z − 0,3� + rel,z 2 �
y y � c,0,k ,
rel,z = z � c,0,k 0,05
= c = c
�� � = c
�� ℎ 3 12 � ℎ = √12 � c
ℎ �
LVL Handbook Europe 127