Hi all,
I have two questions at uni which I am really struggling with, can anyone help at all with working out as I've had a go but I think my figures are all wrong!!
part a)
A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift. The total length of cable supporting the lift when it is at the ground floor is 14 m. The mass of the lift when full of passengers is 800 kg.The designer has decided to incorporate a safety factor of 10 into the lift cable, which means the cable must be able to withstand 10 times the load it will actually be exposed to in service, before it fails.
The steel selected for the cable has a failure stress of 1100 MN m– 2.
Using this information, calculate the required diameter of the cable. Show all your working. Assume that the cable is a single piece of steel, with a circular cross-section. Ignore any effect of the weight of the cable in your calculation.
The downward force F on the cable is calculated by multiplying the total mass m (expressed in kg) by g, the acceleration due to gravity:
F = m x g
Take the value of g to be 10 ms–2. (In reality there is extra force needed to accelerate the lift upwards, but as this is relatively small, there is no need to consider it here.)
Hint. The safety factor means that the cable will fail – the stress will reach its failure stress – when it is loaded to 10 times the intended design load. Use this to calculate the cross-sectional area of the cable, and so its diameter. I found the diameter to be just under 10 mm.
(10 marks)
part b)
By how much will the cable have extended owing to the weight of a full lift at the ground floor? Again, you should show all your working.
The Young’s modulus of steel is 210 GN m–2.
You need to calculate the strain in the cable arising from the load on it, and then the extension. Use a suitable value for the diameter if you have been unable to complete part (a).