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Shear resistance
V_d = E_(d,ULS,fi)∙L/2= 14,6 kN/m∙4,0m/2 = 29,2 kN
τ_(v,d)=〖3∙V〗_d/(2∙A)=(3∙29,2 kN)/(2 ∙26488 mm^2 )=1,7 N/mm^2
f_(v,d,fi)=(k_(mod,fi)∙k_fi)/γ_(M,fi) ∙f_(v,0,edge,k)=(1,0∙1,1)/1,0∙4,2 N/mm^2 =4,6 N/mm^2
τ_(m,d)≤f_(v,d,fi) →OK
Compression perpendicular to grain
When the main beam is supported on a wooden column which has the notional charring rate
βn = 0,70mm/min, the support length becomes
l_(support,fi)=100mm-0,70 mm/min∙30min+1,0∙7mm=72mm
F_(c,90,d) = V_d = 29,2 kN
σ_(c,90,d)=F_(c,90,d)/A_ef =F_(c,90,d)/(b∙(l_(support,fi)+15 mm) )
(4.14)
σ_(c,90,d)=29,2kN/(77mm∙(72mm+15mm))=4,4 N/mm^2
k_(c,90)∙f_(c,90,d,fi)=(k_(c,90)∙k_(mod,fi)∙k_fi)/γ_(M,fi) ∙f_
k_(c,90)∙f_(c,90,d,fi)=(1,0∙1,0∙1,1)/1,0∙6 N/mm^2=6,6 N/mm^2
σ_(c,90,d)≤k_(c,90)∙f_(m,0,d,fi) →OK
(4.13)
Discussion
According to EN1995-1-2:2004, clause 4.3.1 it is not necessary to verify compression perpendicular to the grain
and shear resistance of a beam in the structural fire design. In this example they didn’t become critical, but
in the detailing it shall be verified that the beam is securely supported also when the support length becomes
smaller due to charring of the supports.
9. CALCULATION EXAMPLES OF LVL STRUCTURES when 0,75 < rel,m ≤ 1,4 , crit = 1,56 − 0,75 ∙ rel,m = 1,56 − 0,75 ∙ 1,36 = 0,58
crit ∙ m,d,fi = 0,58 ∙ 47,4 N/mm2 = 27,5 N/mm2
m,d = 19,2 N/mm2 ≤ crit ∙ m,d → OK (4.38)
d = d,ULS,fi ∙ /2 = 14,6 kN/m ∙ 4,0m/2 = 29,2 kN
v,d = 3 ∙ d 2 ∙ = 3 ∙ 29,2 kN2 ∙ 26488 mm2 = 1,7 N/ m2
v,d,fi = mod,fi ∙ fi
M,fi ∙ v,0,edge,k = 1,0 ∙ 1,1 1,0 ∙ 4,2 Nmm 2 = 4,6 N/mm2
m,d ≤ v,d,fi → OK
support,fi = 100mm − 0,70 mmmin ∙ 30min + 1,0 ∙ 7mm = 72mm
c,90,d = d = 29,2
c,90,d = c,90,d ef = c,90,d
∙� support,fi+15 mm� (4.14)
c,90,d = 29,2kN
77mm ∙ (72mm + 15mm) = 4,4 N/mm2
c,90 ∙ c,90,d,fi = c,90 ∙ mod,fi ∙ fi
M,fi ∙ c,90,edge,k
c,90 ∙ c,90,d,fi = 1,0 ∙ 1,0 ∙ 1,1 1,0 ∙ 6 N/mm 2 = 6,6 N/mm2
c,90,d ≤ c,90 ∙ m,0,d,fi → OK (4.13)
rel � m,k m,crit= � 44 N/mm2
25,8 N/mm2 = 1,36 (4.41)
when 0,75 < rel,m ≤ 1,4 , crit = 1,56 − 0,75 ∙ rel,m = 1,56 − 0,75 ∙ 1,36 = 0,58
crit ∙ m,d,fi = 0,58 ∙ 47,4 N/mm 2 = 27,5 N/mm2
m,d = 19,2 N/mm2 ≤ crit ∙ m,d → OK (4.38)
d = d,ULS,fi ∙ /2 = 14,6 kN/m ∙ 4,0m/2 = 29,2 kN
v,d = 3 ∙ d 2 ∙ = 3 ∙ 29,2 kN2 ∙ 26488 mm2 = 1,7 N/mm2
v,d,fi = mod,fi ∙ fi
M,fi ∙ v,0,edge,k = 1,0 ∙ 1,1 1,0 ∙ 4,2 Nmm2 = 4,6 N/mm2
m,d ≤ v,d,fi → OK
support,fi = 100mm − ,70 mmmin ∙ 30min + 1,0 ∙ 7mm = 72mm
c,90,d = d = 29,2
c,90,d = c,90,d ef = c,90,d ∙� support,fi+15 mm� (4.14)
c,90,d = 29,2kN
77mm ∙ (72mm + 15mm) = 4,4 N/mm2
c,90 ∙ c,90,d,fi = c,90 ∙ mo ,fi ∙ fi
M,fi ∙ c,90,edge,k
c,90 ∙ c,90,d,fi = 1,0 ∙ 1,0 ∙ 1,1 1,0 ∙ 6 N/mm2 = 6,6 N/mm2
c,90,d ≤ c,90 ∙ m,0,d,fi → OK (4.13)
d = d,ULS,fi ∙ /2 = 14,6 kN/m ∙ 4,0m/2 = 29,2 kN
v 3 ∙ d 2 ∙ = 3 ∙ 29,2 kN2 ∙ 26488 mm2 = 1,7 N/mm2
v,d,fi = mod,fi ∙ fi
M,fi ∙ v,0,edge,k = 1,0 ∙ 1,1 1,0 ∙ 4,2 Nmm2 = 4,6 N/mm2
m,d ≤ v,d,fi → OK
support,fi = 100mm − 0,70 mmmin ∙ 30min + 1,0 ∙ 7mm = 72mm
c,90,d = d = 29,2
c,90,d = c,90,d ef = c,90,d
∙� support,fi+15 m � (4.14)
c,90,d = 29,2kN
77mm ∙ (72mm + 15mm) = 4,4 N/mm2
c,90 ∙ c,90,d,fi = c,90 ∙ mod,fi ∙ fi
M,fi ∙ c,90,edge,k
c,90 ∙ c,90,d,fi = 1,0 ∙ 1,0 ∙ 1,1 1,0 ∙ 6 N/mm2 = 6,6 N/mm2
c,90,d ≤ c,90 ∙ m,0,d,fi → OK (4.13)
LVL Handbook Europe 213