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 = 0.24 kNm
Z = b.d2 / 6 = 60750 mm^3
k1 = 1.0 for short term load
fb = 14 MPa 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 = 0.53 kNm
ΦNncx = Φ.k1.k8.fc.A = 23.8 kN
ΦNncy = Φ.k1.k8.fc.A = 45.4 kN
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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