1. Hey everyone, i'm brand new to this site and am requiring some help and a push in the right direction. Here's my problem, I have a solid circular shaft of 200mm diameter, and it is to be replaced with a hollow shaft having a diameter ratio of 2. I need to determine the dimensions of the hollow shaft if the maximum stress is to be the same as for the solid shaft. I also need to know what the percentage saving in weight of the hollow shaft compared with the solid one. 2. I think the relevent equations are: Tau(max) = T/J x r(max) J = Pi(r^4)/2 J = Pi(ro^4 - ri^4)/2 3. I have used the above equations to establish tau of the solid shaft, but then do I use a the second J equation to find the inner radius given that I know the diameter ratio is 2 ( meaning 400mm diameter?? is this correct?). Do i also assume torque is the same for both the hollow and solid shaft? Can anyone give me a gentle shove in the right direction?
The problem states, for the hollow versus solid shaft, hold torque constant, and set outside diameter of the hollow shaft equal to two times inside diameter of the hollow shaft. Your relevant equations are correct. See if you can figure it out now.
Thanks for your help nvn, but I am still stuck... I have used the equations J(soild)/r(solid) = J(hollow)/r(hollow) (I have cancelled out the Torques and the stresses from the equations due to the fact they must remain constant to both shafts) I have then transposed the J (Pi(ro^4 - ri^4)/2) to make ro the subject - redifining ri^4 in terms of ro given tha diameter ratio of 2 in order to be able to do it. I end up with the answer of 159mm = ro I know this is wrong because my tutor has attached the answer to work to and he says the answer should read as ri=102mm; ro=204mm; 22% saving in weight. Any ideas what i'm doing wrong?
In mathematics, that is called an unknown, not a subject. You made ro the unknown. OK, so it sounds like you need to check your algebra and arithmetic. Keep trying.
So I know: J(solid) r(Solid) J/r (solid) = 3.14x10^-3 Then by transforming J(hollow) into Pi(ro^4 - ri^4)/2 and letting that = r(hollow) x 3.14x10^-3, I can then transpose the equation to find ro. Is this all correct? I find the transposition to be 0.01256 x 2r(hollow) = Pi x ro^4(hollow) So then.... ro^3 = 4x10^-3 and... ro = (4x10^-3) cube rooted (not sure of keyboard notation for that one!) All seem correct?
Thanks! I still cant get the answer... I feel so dumb! I have used the equation to get the new J/r(solid) figure of 1.57x10^-3 Then I get the full transposition for the outer radius of the hollow shaft (to get 'ro') to be: ro= (4x(1.57x10^-3)/Pi)^(0.33333) When I apply this formula, it still doesnt seem to work out. Am I doing something fundamentally wrong with the transpostion? Just to clarify, should I be using the a constant figure (eg 1) for the torque figure, or can I simply cancel it out from both sides of the equation as said above? likewise with the shear stress figure?
Yes, you can cancel out torque and shear stress; that's correct. Your equation in the second sentence of post 3 is correct. Therefore, just keep checking your algebra and arithmetic, and try it again.
I think i'm having trouble re-arranging the J(hollow) formula to find the unknown ro. Is the answer which I was provided with by my tutor actually correct? 102mm id 204mm od? the transposition I keep coming to is the post above is that correct?
Yes, the answer of your tutor is correct to three significant digits. Your current answer is incorrect. Starting from the equation in the second sentence of post 3, take a new sheet of scratch paper and very carefully rework the algebra, step by step, to solve for ro.
I've tried to derive, re-derive and re-arrange the formula ll week and i'm still struggling. Just to be clear, the R (hollow) which I use for the main equation is the same as Radius Outer (Ro) isn't it? I just cant seem to get it right, any more hints?
Drew112233: Yes, r (hollow) is an ambiguous name for ro. Show your work, so I can see your algebra and arithmetic, step by step, to find out where the mistake is occurring.
This problem offers an interesting note on "design optimization" or "light weight design" or whatever other term we might want to invoke. A simple solid shaft 200 mm in diameter is going to be replaces by a much more complex shaft almost 320 mm in diameter with a bore of 159 mm in order to save 22% on the weight. The cost of manufacturing this hollow shaft will be substantially more, as well as the fact that it will take a significantly larger space. It is important that we know how to make such designs, but it is also important that we know when they are worth making and when they are to be avoided! Only rarely and in special circumstances would this be a good engineering choice.
This is very similar to a problem I had in Sept 07. In this case, I needed the smallest hollow shaft required to transmit angular information down a shaft, from one end to the other, under torsional load, with a limit on angular displacement error. (And it scales as D^-4, yikes!) You're problem is a bit simpler. I have a polar moment of inertia formula for a hollow round shaft given as [tex]J = \left( \pi / 32 \right) \left( D^4 - d^4 \right) [/tex] D, and d are the outside and inside diameters, respectively. This differs from yours. Why?
There really is no difference. Drew stated his J expression in terms of radii, where as the second expression given by Phrak is in terms of diameters. The (1/2)^4 makes the 1/16 difference where they appear to disagree.
Thanks, DrD. Missed that. This is really become an algebra problem I see, and the factor scales-away anyway. What you say about optimization is a good one. Almost invariably one runs into competing constraints, or competing optimizations, in space, strength, cost, etc.
The miniscule weight savings in this textbook problem is misleading. A scaling of radius by 50% represents a strength increase by a factor of 5. Weight, by comparison scales by a factor of 2.25. The outside of a shaft takes most of the stress and transmits most of the torque. The inner 75% of a shaft is wasted material. Anyway, strength per unit weight scales as the square of the diameter, as does rigidity per unit weight.