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 Pchar = 9.0 kN (downward)
Serviceability load Pserv = 2.5 kN (downward)
| Load combination | Design Load (kN) | Design strength (kN) = Pchar x phi x K1 |
|---|---|---|
| 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.
