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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