# 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.d^{2} / 6 = | 60750 | mm^3 |

k_{1} = | 1.0 | for short term load |

f_{b} = | 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 | ||

ΦM_{nx} = Φ.k_{1}.k_{8}.f_{b}.Z = | 0.53 | kNm |

ΦN_{ncx} = Φ.k_{1}.k_{8}.f_{c}.A = | 23.8 | kN |

ΦN_{ncy} = Φ.k_{1}.k_{8}.f_{c}.A = | 45.4 | kN |

N*c / ΦN_{ncx} + M*_{x} / ΦM_{nx} = | 0.96 | < 1.0 so OK |

N*c / ΦN_{ncy} + (M*_{x} / ΦM_{nx})^{2} = | 0.45 | < 1.0 so OK |

Example prepared by David Reid, Structural Engineer.

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