the resistance of LVL structures in demanding cases using, e.g., performance-based design methods. Note: For special cases where more advanced design methods are used, the report VTT-S-04746-16 also has information on the charring rate β0 in a test based on to a more stringent hydrocarbon (HC) time-temperature exposure curve (EN 1363-2:1999). 6.4.3 Design of unprotected beams and panels Since LVL beams are typically slender structures with largest available beam thicknesses up to 75 mm without multiple gluing, unprotected LVL beams cannot be designed for higher than 15 min fire resistance time requirements. The zero strength layer (k0 ∙ d0) reduces the thickness of an effective cross section significantly, making the beam even more slender in lateral torsional buckling analysis. Example: 63x300 mm LVL-P beam and 33 mm LVL-C panel in 15 minute fire exposure: Beam: d_(ef,beam)=β_n∙t+k_0∙d_0=0,70 mm/ min∙15min+15min/20min∙7mm=15,75mm Size of effective cross section of the beam in 3-side fire exposure: Width b: 63 mm - 2∙15,75 mm =31,5 mm Height h: 300 mm - 15,75 mm = 284 mm Design value of bending strength for LVL 48 P: f_(m,d,fi)=k_(mod,fi)∙(k_fi∙〖k_h∙f〗_(m,k))/γ_(M,fi) =1,0∙(1,1∙(300mm/284mm)^0,15∙44 N/mm^2 )/1, 6. PERFORMANCE OF LVL IN FIRE Figure 6.7. One-dimensional charring of LVL-C is linear in a 120 minutes fire exposure test according to the standardized time-temperature curve. Blue and red curves: exposure on the wide face of the specimens. Green and purple curves: exposure on the edge face of the specimens 39. 0 10 20 30 40 50 60 70 80 90 0 20 40 60 80 100 120 CHARRING DEPTH [mm] TIME [min] Wide face 2 Wide face 1 Edge face 1 Edge face 1 Figure 6.8. Effective cross section after 15 min fire exposure. m,d,fi = mod,fi ∙ fi ∙ h ∙ m,k M,fi =1,0∙ 1,1∙ � 32 08 04 mm mm�0,15 ∙ 44 m Nm2 1,0 =44,3 m N m2 ef,panel = 0 ∙ + 0 ∙ 0 = 0,65 mmmin∙ 15min + 1250mmiinn∙ 7mm = 15 mm m,d,fi = mod,fi ∙ fi ∙ m,k M,fi =1,0∙ 1,1∙ 36 m Nm2 1,0 = 39,6 mNm2 ef = n ∙ + 0 ∙ 0 = 0,70 mmmin∙ 30min+1,0∙ 7mm=28mm Figure 6.7. One-dimensional charring of LVL-C is linear in a 120 minute fire exposure test according to the standardized time-temperature curve. Blue and red curves: exposure on the wide face of the specimens. Green and grey curves: exposure on the edge face of the specimens 39. Note: For special cases where more advanced design methods are used, the report VTT-S04746-16 also has information on the charring rate β0 in a test based on to a more stringent hydrocarbon (HC) time-temperature exposure curve (EN 1363-2:1999). 6.4.3 Design of unprotected beams and panels Since LVL beams are typically slender structures with largest available beam thicknesses up to 75 mm without multiple gluing, unprotected LVL beams cannot be designed for higher than 15 min fire resistance time requirements. The zero strength layer (k0 ∙ d0) reduces the thickness of an effective cross section significantly, making the beam even more slender in lateral torsional buckling analysis. Example: 63x300 mm LVL-P beam and 33 mm LVL-C panel in 15 minute fire exposure: Beam: ef,beam= n ∙ + 0 ∙ 0 = 0,70 mmmin ∙ 15min + 1250mmiinn ∙ 7mm = 15,75mm Size of effective cross section of the beam in 3-side fire exposure: Width b: 63 mm - 2∙15,75 mm = 31,5 mm Height h: 300 mm - 15,75 mm = 284 mm Design value of bending strength for LVL 48 P: m,d,fi = mod,fi ∙ fi ∙ h ∙ m,k M,fi = 1,0 ∙ 1,1 ∙ (320804mmmm) 0,15 ∙ 44 m Nm2 1,0 =44,3 m N m2 Panel: Panel: d_(ef,panel)=β_0∙t+k_0∙d_0=0,65 mm/ min∙15min+15min/20min∙7mm=15 mm Effective thickness of the panel tpanel : 33 mm - 15 mm = 18 mm Design value of bending strength for LVL 36 C: f_(m,d,fi)=k_(mod,fi)∙(k_fi∙f_(m,k))/γ_(M,fi) =1,0∙(1,1∙36 N/ mm^2 )/1,0=39,6 N/mm^2 m,d,fi = mod,fi ∙ fi ∙ h ∙ m,k M,fi =1,0∙ 1,1∙ � 32 08 04 mm mm�0,15 ∙ 44 m Nm2 1,0 =44,3 m N m2 ef,panel = 0 ∙ + 0 ∙ 0 = 0,65 mmmin∙ 15min + 1250mmiinn∙ 7mm = 15 mm m,d,fi = mod,fi ∙ fi ∙ m,k M,fi =1,0∙ 1,1∙ 36 m Nm2 1,0 = 39,6 mNm2 ef = n ∙ + 0 ∙ 0 = 0,70 mmmin∙ 30min+1,0∙ 7mm=28mm m,d,fi = mod,fi ∙ fi ∙ h ∙ m,k M,fi =1,0∙ 1,1∙ � 32 08 04 mm mm�0,15 ∙ 44 m Nm2 1,0 =44,3 m N m2 ef,panel = 0 ∙ + 0 ∙ 0 = 0,65 mmmin∙ 15min + 1250mmiinn∙ 7mm = 15 mm m,d,fi = mod,fi ∙ fi ∙ m,k M,fi =1,0∙ 1,1∙ 36 m Nm2 1,0 = 39,6 mNm2 mm LVL Handbook Europe 163
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