Sunday, 26 October 2014

Getting stressed on Southport Pier



I was looking at Southport Pier and trying to estimate the stress in one of the support pillars. I'd say that the sets of pillars were about 15 metres apart. Let's say the iron work had a width of 1cm and if the lattice spaces were taken into account, had a depth of 30cm. The volume of iron would be 0.045 cubic metres. Let's estimate the density of iron as 8000 kg per cubic metre. That gives a mass of iron held up by one pillar as 400kg to 1sf. On each set, the pillars are about 2 metres apart. Let's estimate the depth of the wooden deck as 5cm. The volume of wood would be 1.5 cubic metres. If the wood were able to float I could estimate it's density as 800 kg per cubic metre, giving a mass of 1200 kg. So let's say 1600kg is held up by one pillar. Pillars have a diameter of about 15cm. I'm going to assume that they are hollow and the iron is 2cm thick. This gives an area of cross-section as 0.008 square metres. Stress = force x cross-sectional area = 1600x10/0.008 = 2 million Newtons per square metre. The iron is in compression - it will take a much larger stress to break it than in tension, And then there's the tram that goes over periodically. Say that is 10 tonnes. The load will be spread over at least 4 pillars, giving at most an extra 2500kg per pillar but will more than double the stress.