LVL Handbook EUROPE

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9. CALCULATION EXAMPLES OF LVL STRUCTURES
Properties of a LVL 36C joist:
Bending strength edgewise fm,0,edge,k = 32 N/mm2
Tension in perpendicular to grain strength edgewise ft,90,edge,k = 5 N/mm2
Shear strength edgewise fv,0,edge,k = 4,5 N/mm2
Modification factor kmod for medium-term, SC1 = 0,8
Material safety factor γM (default value in EC5) =1,2
Size effect factor kh = (300/240)0,15 = 1,03
Tension stress perpendicular to the grain is verified by the equation
σt,90,d =F_(t,90,d)/(0,5∙l_(t,90)∙b∙k_(t,90) )≤f_(t,90,d)
(4.57)
where
kt,90 =min{█(1@(450/h)^0,5 )=1,0┤
(4.58)
σt,90,d = design value of tension stress perpendicular to the grain [N/mm2]
lt,90 = 0,5 ∙ (hd + h) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
Ft,90,d = design value of tension force perpendicular to the grain [N]
ft,90,d =k_mod/γ_M ∙f_(t,90,edge,k)=0,8/1,2∙5 N/mm^2 =3,3 N/mm^2
t,90,d = t,90,d
0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57)
t,90 = min � 1 �450 ℎ� 0,5 = 1,0 (4.58)
t,90 = 0,5 ∙ (ℎd + ℎ) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
t,90,d = mod M∙ t,90,edge,k = 0,8 1,2 ∙ 5 Nmm2 = 3,3 N/mm2
t,90,d = d∙ℎd
4∙ℎ ∙ �3 − �ℎd
ℎ �2� + 0,008 ∙ d
ℎr (4.59)
ℎr = 90 (4.60)
t,90,d = t,90,d
0,5∙ t,90 ∙ ∙ t,90 ≤ t,90,d (4.57)
t,90 = min � 1 �450 ℎ�0,5 = 1,0 (4.58)
t,90 = 0,5 ∙ (ℎd + ℎ) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
t,90,d = mod M∙ t,90,edge,k = 0,8 1,2 ∙ 5 Nmm2 = 3,3 N/ m2
t,90,d = d∙ℎd 4∙ℎ ∙ �3 − �ℎd ℎ �2� + 0,008 ∙ d ℎr (4.59)
ℎr = 90 (4.60)
t,90,d t,90,d
0,5∙ t,90 ∙ ∙ t,90 ≤ t,90,d (4.57)
t,90 = i � 1 �450 ℎ�0,5 = 1,0 (4.58)
t,90 = 0,5 ∙ (ℎd + ℎ) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
t,90,d = mod M∙ t,90,edge,k = 0,8 1,2 ∙ 5 Nmm2 = 3,3 N/mm2
t,90,d = d∙ℎ d
4∙ℎ ∙ �3 − �ℎd
ℎ �2� + 0,008 ∙ d
ℎr (4.59)
ℎr = 90 (4.60)
Tension in perpendicular to grain force Ft,90,d depends on the shear force Vd and bending moment M d at the
edge of the hole:
F_(t,90,d)=(V_d∙h_d)/(4∙h)∙[3-(h_d/h)^2 ]+0,008∙M_d/h_r
(4.59)
where
hr =90 mm (4.60)
t,90,d = t,90,d
0,5∙ t,90∙ ∙ t,90 ≤ t,90,d (4.57)
t,90 = min � 1 �450 ℎ�0,5 = 1,0 (4.58)
t,90 = 0,5 ∙ (ℎd + ℎ) = 0,5 ∙ (60 mm + 240 mm) = 150 mm (4.61)
t,90,d = mod M∙ t,90,edge,k = 0,8 1,2 ∙ 5 Nmm2 = 3,3 N/mm2
t,90,d = d∙ℎd
4∙ℎ ∙ �3 − �ℎd
ℎ �2� + 0,008 ∙ d
ℎr (4.59)
ℎr = 90 (4.60)
LVL Handbook Europe 207