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STRUCTURAL TIMBER DESIGN TO EUROCODE 5: KEY FORMULAE
1.0 SECTIONS IN BENDING (ULS):
Design bending strength (no LTB): fm,y,d = [fm,y,k(kmod)(ksys)(kh)]/γM Design shear strength = fv,d = (fvk(kmod)(ksys))/γM Shear stress = τv,d = 1.5Vd/(befh) where bef = 0.67b Design bearing strength = fc,90,d = (fc,90,k(kmod)(ksys))/γM FOR LATERALLY UNRESTRAINED BEAMS: Global bending strength (with LTB) = k crit (f m,y,d ) Leff = L + 0.2h (for a simply supported beam)
σ m ,crit =
0.78 b
2
h Leff )^
E0.05 (for rectangular sections)
λrel, m=
f m,k
σm, crit
if λrel, m ≤ 0.75 then kcrit=1.
if 0.75< λrel ,m ≤1.4 thenkcrit =1.56−0.7 5 λrel ,m
2.0 SECTIONS IN BENDING (SLS):
Instantaneous deflection = flexural deflection + shear deflection final permanent load deflection = (instantaneous permanent load deflection) x (1+ kdef) final variable load deflection = (instantaneous variable load deflection) x (1 + 2 kdef)
3.0 AXIALLY LOADED SECTIONS (ULS):
TENSION:
Design axial tensile strength = ft,0,d = (ft,0,kkmodkhksys)/γM Design axial tensile stress = σt,0,d = Ft,0,d / Anet COMPRESSION: Design compression strength perpendicular to the grain (local) = fc,0,d = (fc,0,kkmodksys)/γM Where buckling is considered:
σ c , 0 ,d
kc ,, y f c , 0 , d
σm , y ,d
f m , y ,d
+km
σm , z , d
f m , z , d
*note, this is for major axis buckling, it may also be necessary to consider minor axis, i.e. kc,zfc,0,d The second and third terms reduce to zero if no bending is present i.e. if the load is purely axial and concentric Buckling ‘k’ factors:
k c, y=
k (^) y +√k (^) y 2
−λrel , y
2 k (^) y=0.5 (^) ( 1 +βe ( λrel , y−0.3) + λrel, y 2 )
λrel, y=
λy
f c , 0 ,k
E0.
βe = factor to allow for deviation in target dimensions βe = 0.2 for solid timber βe =0.1 for Glulam and LVL Slenderness λ= Le/i where Le = effective buckling length i = radius of gyration iy = major axis radius of gyration =
h
2 √ 3 for a rectangular section
