# Fixing Selection Worked Example

Select a joist hanger to fix a 140×45 joist spanning 2700mm @ 450 crs to a timber beam. 1.5kPa domestic floor. Assume dry in service condition.

Floor joist reaction:

G = 0.5 kPa x 0.45 x 2.7 / 2 = 0.30kN

Qu = 1.5 kPa x 0.45 x 2.7 / 2 = 0.91kN

Qc = 1.8 kN / 2 = 0.9kN

From the LUMBERLOK Timber Characteristic Loadings Brochure, for a 47 x 90 joist hanger:

Characteristic strength P_{char} = 9.0 kN (downward)

Serviceability load P_{serv} = 2.5 kN (downward)

Load combination | Design Load (kN) | Design strength (kN) = P_{char} x phi x K_{1} |
---|---|---|

1.4G | 1.2G+1.6Qu | 1.2G+1.6Qc |

1.4 x 0.30 = 0.42 | 1.2 x 0.3 + 1.6 x 0.91 = 1.82 | 1.2 x 0.3 + 1.6 x 0.9 = 1.8 |

9.0 x 0.8 x 0.6 = 5.8 | 9.0 x 0.8 x 0.8 = 7.7 |

The servicability load published by LUMBERLOK gives the designer some idea of how much the joint will slip because at this load there is a straight line relationship between the load and deflection. The servicability load published is the load at a joint slip of 1mm. From this the actual slippage of the joint can be calculated.

At a load of 2.5kN there will be 1 mm slip, so at a load of G+0.7Q = 0.94kN there will be:

joint deflection = 0.94 x 1 / 2.5 = 0.4mm slip at each joist hanger.