PaRtZ Posted April 27, 2010 Report Share Posted April 27, 2010 Hi guys Ive got to design a crane for my engineering degree (don't ask) Ive got to do stress calculations for areas I think are important, and since our crane exerts 53kNm on the cable drum to reel it in, I've concluded that I should analyse the effects of this torque on the shaft. So this is where Ive hit a stop, I cant find much information anywhere. Ive done stress analysis modules a couple of times, but it all involved bending and loading, not torsion. Or if it did, it was because of an applied load effecting it, not solely due to torsion (as typically they involve calculations for two or more stresses) so heres my work: Ive got a solid steel shaft of diameter 0.04m 0.60m long. Im trying to assume the worse case scenario that the cable gets stuck and the full 53kNm is applied to the shaft. Any help would be appreciated thanks Quote Link to comment Share on other sites More sharing options...
King C Posted April 27, 2010 Report Share Posted April 27, 2010 (edited) Are you sure that wouldnt just be a compressive force acting on it, or have you already done that bit. Im only in my first year so my knowledge of this subject is still somewhat limited. EDIT: ie there wouldnt be any torsion as there isnt any twisting? Edited April 27, 2010 by King C Quote Link to comment Share on other sites More sharing options...
PaRtZ Posted April 27, 2010 Author Report Share Posted April 27, 2010 I can see where your coming from and youre right, but ive possibly worded it wrong in the reasoning to why I want to calculate it. When the winch is activated, it is only connected to the drive at one end of the shaft (which is connected to the spur gear that creates the torque. At the other end is nothing. I wanted to work out if the drum didnt turn, what would the effect be on the length of shaft between the drum and the gear ... The more I think about it, the more complex the problem seems... I thought it was gna be easy to do...but its not so never mind. Thanks anyway Quote Link to comment Share on other sites More sharing options...
Tony Harrison Posted April 27, 2010 Report Share Posted April 27, 2010 It'll definitely be torsional stress, which (as my vague memory informs me) isn't overly complicated. Unfortunately my old textbooks are stored away somewhere, so I can't help (I didn't learn it well enough to remember 8 years later!!). You might also like to compare a solid shaft to a hollow one, as you may find a hollow one stronger. But you're bound to find something online. Quote Link to comment Share on other sites More sharing options...
King C Posted April 27, 2010 Report Share Posted April 27, 2010 I demand you draw me a picture! I'm interested now. For some reason I find it hard to picture scenarios but I find them easiest to work from. Its strange that Im willing to help with this even though I have Engineering assignments to do. Quote Link to comment Share on other sites More sharing options...
Trials Punk Posted April 27, 2010 Report Share Posted April 27, 2010 (edited) I could be watching porn right now...... Angle of twist=D x L .......................GxJ J=pie x D(to the power of 4) .................32 D = Diameter L = Length J = Second polar motion of area m4 for circular sections G=Modulous of rigidity (steel roughly 80GN/m2 I think thats right......... Excuse the full stops, couldn't get things under the line init! Edited April 27, 2010 by Trials Punk Quote Link to comment Share on other sites More sharing options...
Shaun H Posted April 27, 2010 Report Share Posted April 27, 2010 If you're talking about the shaft the drum spins on then it won't be experiencing a torque. If we're talking a driven shaft then the torque is pretty easy-peasy anyway! First, just need to calculate the moment exerted by the cable. If you want to impress your lecturer you may want to do calculations for a fully reeled in cable and completely reeled out. Whoops read your post properly and you already have a moment... Other than that, there's plenty on the web, (I've just had a quick look on google, "torque stress equation" brought up pretty much most of what's in my notes in the first result) to help you answer it, not to mention I'm sure you have your own notes. I'd be chucking in stuff about wind, earthquakes and anything else that will make your analysis stand out from the shitmunchers for max credit too. Quote Link to comment Share on other sites More sharing options...
Matt Vandart Posted April 28, 2010 Report Share Posted April 28, 2010 If you're talking about the shaft the drum spins on then it won't be experiencing a torque. If we're talking a driven shaft then the torque is pretty easy-peasy anyway! First, just need to calculate the moment exerted by the cable. If you want to impress your lecturer you may want to do calculations for a fully reeled in cable and completely reeled out. Whoops read your post properly and you already have a moment... Other than that, there's plenty on the web, (I've just had a quick look on google, "torque stress equation" brought up pretty much most of what's in my notes in the first result) to help you answer it, not to mention I'm sure you have your own notes. I'd be chucking in stuff about wind, earthquakes and anything else that will make your analysis stand out from the shitmunchers for max credit too. Quote Link to comment Share on other sites More sharing options...
shamus Posted April 29, 2010 Report Share Posted April 29, 2010 You're considering a driven shaft? With a maximum output of 53Kn? Does the drive come from either end of the shaft or is one end free? One Drive Point - Worst case would be the drum jamming on the shafts free end and the rotational force still being exerted at the driven end. Two Drive Points - (haven't seen many cranes using this system, too much to go wrong and expensive) worst case is the cable snagging in the middle and both drive points working against it. Again Google punched up most of the answers quite easily, just make sure you're analysing the correct scenario As Shaun H mentioned, overload calculations (shock loads and vibrations) and emergency situations (loss of a outrigger (subsiding/collision)) will be a massive bonus mark up! Quote Link to comment Share on other sites More sharing options...
PaRtZ Posted April 29, 2010 Author Report Share Posted April 29, 2010 Yeah I was going for the one point drive and it gets jammed. Im now gonna look at calculation of friction on interference fits and drum fail safes, so that if it were to jam something could catch the drum from turning, and just let the shaft turn inside th drum Shit idea but il get marks for thinking about it Quote Link to comment Share on other sites More sharing options...
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