Design Aids

Column Design Example – Axial and bending load

Consider previous example with a short term transverse applied load causing a bending moment about
the X-X axis of 240Nm. Weak axis restraint 800mm.

M*x =0.24kNm
Z = b.d2 / 6 =60750mm^3
k1 =1.0for short term load
fb =14MPa for MSG8 timber

using k8 and Φ factors from example 1 we are considering the short term loading condition here, so the k1 factor used throughout the calculation will be k1 = 1.0

ΦMnx = Φ.k1.k8.fb.Z =0.53kNm
ΦNncx = Φ.k1.k8.fc.A =23.8kN
ΦNncy = Φ.k1.k8.fc.A =45.4kN
N*c / ΦNncx + M*x / ΦMnx =0.96< 1.0 so OK
N*c / ΦNncy + (M*x / ΦMnx)2 =0.45< 1.0 so OK

Example prepared by David Reid, Structural Engineer.

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