Oar shaft strength

1 An oar is a lever, and structurally it is a beam stressed in bending. The front (towards the bow of the boat) face is in compression stress and the aft face is in tension.

2 This situation is dealt with using what engineers call simple beam theory. It allows calculation of the stress (force per square mm) in each part of the beam.

3 Testing samples of timber gives an idea of the strength of different timber species, within a certain range as samples vary. Most timbers are about twice as strong in tension as they are in compression.

4 When timber fails in tension there is a loud crack and a jagged gap appears. When it fails in compression there is no crack (maybe a creak) and no gap. Close inspection may reveal a thin line across the grain where the hollow wood fibres have buckled and collapsed. This is none the less a failure and must be avoided in oars.

5 When an oar breaks in bending the obvious failure is a tension crack on the rear face, but in fact it's likely that it has failed in compression on the other side, and then the undamaged wood is narrower and so the compression failure spreads until the reduced section fails on the tension side.

6 What this means is that to design an oar not to fail while rowing, it has to be made so the compression stress does not rise above what is allowable in that timber.

7 For at least a thousand years oars have been made without calculations, but if we want to be confident our design will not fail with fair use, we need the calculations.

8 To do this we need to know how hard the strongest rower can pull, so we can design for them. I have used 1000N which is about 102kg (force in the metric system is in Newtons, and 1kg = 9.81N). Atkinsopht uses 750N for an Olympic rower so 1000N is a lot.

9 Worst case is a 4.5m oar at a lowest gearing of 2.7, so an inboard of 4.5/(2.7+1)=1.216m. At 200mm in from the end of the oar the force is acting at a lever arm of 1.016m. The bending moment we need to resist at the thickest part of the oar is 1000N x 1.016m = 1016Nm.

10 Any oar shaft we design has to be able to take this bending moment (BM) at the oarlock and as the BM reduces along the shaft the section will also taper, but the stress needs to remain within the wood's limits.

11 If you don't like numbers you can skip this, but the conclusions will be the same:

For any beam cross section there is a second moment of area I. It depends on the shape of the section and its dimensions. it can be looked up on the internet, see [You must be registered and logged in to see this link.]
For any beam cross section there is also a number y which is just the distance from the neutral axis to the furthest out fibre, which is the most stressed and therefore the place which will fail first. Stress = Moment/(I x y).

12 Since we need to design the lightest possible outboard, we need to use light strong timber in the most efficient shape. It may surprise you to learn that a solid circle, one of the most common oar shaft sections, is one of the worst possible shapes. (fact not opinion) Inefficient beam sections have a lot of material not doing much, near the centre where it is nether being stretched nor squashed. Efficient beam sections like I-beams have a lot of their material a long way from the centre.

13 Square section beams are stronger than solid circular ones for the same weight, and rectangular ones which are deep in the direction of bending are more efficient still, which is why house beams are deeper than they are wide. We can make oars with wide rectangular sections but if too thin they become floppy. In my opinion they need to be no more than twice as wide (fore and aft) as they are deep (top to bottom).

14 Hollow sections are very efficient which is why modern carbon oars are thin wall hollow tubes. Most of their material is a long way from the centre unlike solid circles. Hollow oars can be made but it is controversial whether they are easy enough to make for community boatbuilders. We may need to come to an opinion on this.