Statically indeterminant torsion question

[xJJx]

Hi, I'm struggling with this question on an exam pastpaper. I have attached the question and worked solutions below and i am stuck on Q10 c). I'm not sure how they knew that the torque is constant along the whole shaft. I thought a stepped shaft has different torques for its different sections since the length & cross sectional area of the sections are different. Can someone please explain this to me, thank you

 

[PhanthomJay]

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I'm not sure how they knew that the torque is constant along the whole shaft. I thought a stepped shaft has different torques for its different sections since . Can someone please explain this to me, thank you

For the first case when the shaft is loaded in tension, do you expect different tension forces in different sections since the cross sectional area of the sections are different? Don't confuse forces or torques with stresses or deformations.

 

[xJJx]

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For the first case when the shaft is loaded in tension, do you expect different tension forces in different sections since the cross sectional area of the sections are different? Don't confuse forces or torques with stresses or deformations.

Thank you for taking the time to help. One of the assumptions for applying the torsion equation to a section is that the cross-sectional area must be uniform throughout that section, but it's not uniform in a stepped shaft. Are the torques carried by each section of the shaft different to each other ONLY when the torque is applied at an intermediate position on the shaft (i.e not at one of the ends)?

 

[Chestermiller]

If you do a moment balance on a portion of the shaft that encompasses the change from one cross section to the other, what do you get?

 

[xJJx]

If you do a moment balance on a portion of the shaft that encompasses the change from one cross section to the other, what do you get?

T1=T2 in order for the shaft to be in eqm right?

 

[xJJx]

If you do a moment balance on a portion of the shaft that encompasses the change from one cross section to the other, what do you get?

I'm also confused at how this shaft can be in eqm if the torques are both rotating clockwise; sorry haha, torsion is my Achilles heel

 

[Chestermiller]

T1=T2 in order for the shaft to be in eqm right?

Correct.

 

[Chestermiller]

I'm also confused at how this shaft can be in eqm if the torques are both rotating clockwise; sorry haha, torsion is my Achilles heel

It's not drawn very clearly. The one on the left is supposed to indicate counter clockwise.

 

[xJJx]

I see, that makes more sense now! But when teaching us statical indeterminancy, my professor said when there's an applied torque at the change of section (image uploaded), one of the shaft sections twists one way and the other section twists the other way by the same amount, however looking at the direction of the arrows and eqm equation, this doesn't look like it is the case as they look like they are going in the same direction. Do you know why he did this?

 

[PhanthomJay]

I see, that makes more sense now! But when teaching us statical indeterminancy, my professor said when there's an applied torque at the change of section (image uploaded), one of the shaft sections twists one way and the other section twists the other way by the same amount, however looking at the direction of the arrows and eqm equation, this doesn't look like it is the case as they look like they are going in the same direction. Do you know why he did this?

oh now I think I know why you titled this as a statically indeterminate problem. Your first example appears to be a statically determinate shaft fixed at one end and free at the other end with a Torque applied at the free end. For a statically determinate problem, the variation in cross sections doesn't matter when computing torques at various sections of the shaft. You just use the equilibrium equations and free body diagrams to solve for the twisting moments at any section. If for this statically determinate shaft you were to apply a clockwise torque T at some point between the ends of the shaft, then the torques left of that point at any section toward the fixed end would be T counterclockwise , and there would be no torques in the shaft right of that point toward the free end . This is strictly from equilibrium considerations using sum of torques = 0.

Now in your last example with the shaft fixed at both ends, you have a statically INdeterminate problem, which is whole new ballgame, because internal and end reaction torques are very much dependent on the various section parameters (J, G, and L). So in general T1 and T2 are not equal, and you need to look at twist deformation angles to solve the problem. Here both end reaction moments have the same direction as you note, and the angle of twist at the change in section shapes is the same on each section at that point, one considered positive and the other negative, by convention.

 

[xJJx]

oh now I think I know why you titled this as a statically indeterminate problem. Your first example appears to be a statically determinate shaft fixed at one end and free at the other end with a Torque applied at the free end. For a statically determinate problem, the variation in cross sections doesn't matter when computing torques at various sections of the shaft. You just use the equilibrium equations and free body diagrams to solve for the twisting moments at any section. If for this statically determinate shaft you were to apply a clockwise torque T at some point between the ends of the shaft, then the torques left of that point at any section toward the fixed end would be T counterclockwise , and there would be no torques in the shaft right of that point toward the free end . This is strictly from equilibrium considerations using sum of torques = 0.

Now in your last example with the shaft fixed at both ends, you have a statically INdeterminate problem, which is whole new ballgame, because internal and end reaction torques are very much dependent on the various section parameters (J, G, and L). So in general T1 and T2 are not equal, and you need to look at twist deformation angles to solve the problem. Here both end reaction moments have the same direction as you note, and the angle of twist at the change in section shapes is the same on each section at that point, one considered positive and the other negative, by convention.

Thank you so much for replying! Your description really helped clear things up for me; initially I was confused because I thought all stepped shafts can't have uniform torque but I understand this isn't the case now. From what you're saying, if a clockwise torque is applied at an INTERMEDIATE position (i.e NOT at the ends) of a shaft fixed at one end and free at the other, the torque which gets transferred along the shaft to the fixed end would be = applied torque (but in opposite direction) and there would be no torque towards the free end? sorry I don't mean to repeat what you're saying, I just want to clarify that you can only really have a statically indeterminant shaft if the shaft doesn't have any free ends right?

 

[PhanthomJay]

Correct on all counts.

 

[xJJx]

Correct on all counts.

Again, can't thank you enough! I understand torsion much more now. Also, these scenarios are the same for a hollow shaft too right? I can't find anywhere online that explicitly states that but it seems to be the case. The only thing that would be different is J (polar second moment of area) for a hollow shaft, from what I've seen so far.

 

[PhanthomJay]

Again, can't thank you enough! I understand torsion much more now. Also, these scenarios are the same for a hollow shaft too right? I can't find anywhere online that explicitly states that but it seems to be the case. The only thing that would be different is J (polar second moment of area) for a hollow shaft, from what I've seen so far.

You are correct again. Keep on plugging!

 

[xJJx]

You are correct again. Keep on plugging!

Yayy, okay last thing I'm going to bother you with promise! haha (sorry): I've attached a question I'm stuck on, it's part (c) of the attachment. I've attached both my solutions (blue writing) and my professor's (black writing); instead of substituting P/w into T like how he did, I kept T as it is and used it to calculate the power at the end but my answer is wrong and I don't understand why

 

[PhanthomJay]

I didn't check the math, but it doesn't make a difference where you make the substitution for P = Tw. Probably a math error by someone.