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Mechanics of Materials - Torsion

  1. Dec 17, 2011 #1

    Femme_physics

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    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Dec 17, 2011 #2

    NascentOxygen

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    FAB is the force acting at the radius of the shaft. You used its torque.

    0.15 is in mm, SI formulae require metres.

    Fix these and you'll get 56.6 MPa
     
  4. Dec 17, 2011 #3

    PhanthomJay

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    Your first relevant equation is not correct and your second seems to be an allowable shear stress which is not applicable in this problem.
    For a circular member subject to torsion from 'twisting' moments about the long axis, you should look up the formula for max shear torsional stress. The formula you noted applies for average transverse shear streses parallel to the cross sectional area.
     
    Last edited: Dec 18, 2011
  5. Dec 19, 2011 #4

    Femme_physics

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    Yes, and I believed I used the right value.
    I fixed it to "M" but I still get the same numbers just the decimal point shifting.

    Not correct by not being relevant or not correct by not correct? My teacher gave us those formulas.

    This formula?

    http://img42.imageshack.us/img42/2406/etaz.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Dec 19, 2011 #5

    PhanthomJay

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    not relevant Yes....where Io is the polar moment of inertia about the longitudinal axis of the shaft.
     
    Last edited by a moderator: May 5, 2017
  7. Dec 19, 2011 #6

    NascentOxygen

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    You used the right value of torque, but you didn't convert it to its tangential force equivalent at the radial distance from the shaft's axis. The formula you have been given requires the tangential force applied at the shaft surface. Force = torque / radius
    I just rechecked, and I get book's answer.
    I verified your teacher's formula before I replied the first time, to confirm it derives from the general formula. It's correct.
     
  8. Dec 19, 2011 #7

    I like Serena

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    Erm...
    [tex]{T/r \over \pi r^2} = {300 Nm / 0.015 m \over \pi (0.015 m)^2} = 28.3 MPa \ne 56.6 MPa[/tex]
     
  9. Dec 19, 2011 #8

    NascentOxygen

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    Ah, there's a 2 that has gone missing somewhere. The polar moment of a round shaft is Pi/2 times r^4. The teacher's formula seems to be out by a factor of two.

    I guess this means no one (except me) will have got the right answer?
     
  10. Dec 19, 2011 #9

    I like Serena

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    I believe PhantomJay had it right when he said Fp needed another formula that included a polar moment, which she just gave.

    The first formula in the OP is apparently for the shear tension, not the shear torsion.
     
  11. Dec 19, 2011 #10

    NascentOxygen

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    Determine maximum shear stress on shaft, [itex]\tau _{max}[/itex], from the general formula:

    [itex]\frac T J = \frac {\tau_{max}} {r} \; \;\; where\;\; for\;\; a\;\; solid\;\; cylindrical\; shaft\;\;J = \frac {\mathrm{\pi}} {2}\, r^{4}[/itex]
     
    Last edited: Dec 19, 2011
  12. Dec 23, 2011 #11

    Femme_physics

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    Is this the formula you need to use to solve it? Because I don't see it in my formula sheet.
     
  13. Dec 23, 2011 #12

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    No, this is not the formula you need to solve it.

    It was derived from your formula ##\tau_s={F_{max} \over A_s}## and I showed it was wrong.
    You cannot use that formula here, since it applies to linear shear (by a regular force) and not to torsional shear (by rotational torque).


    The formula you should use is the one given by NascentOxygen in his last post.
    This is effectively the same as the formula you showed in your answer to PhantomJay, except that you did not specify what ##I_0## was.
     
    Last edited: Dec 23, 2011
  14. Dec 24, 2011 #13

    Femme_physics

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    Moment of inertia of a non-hollow cylinder.

    http://img444.imageshack.us/img444/9633/mm2vh.jpg [Broken]


    But I don't have the mass in the question, so am not sure how am I supposed to use it.

    Still quite off

    http://img404.imageshack.us/img404/2576/69366748.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  15. Dec 24, 2011 #14

    I like Serena

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    Are you sure it is the moment of inertia?
    Isn't it the "polar" moment of inertia?



    Suppose you moved the decimal point 6 places to the left? :wink:
     
    Last edited by a moderator: May 5, 2017
  16. Dec 24, 2011 #15

    Femme_physics

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    Oh yes, it's polar. *smacks forehead*

    Ow.....been a while since I smacked my forehead.

    http://img443.imageshack.us/img443/734/offufs.jpg [Broken]

    Still off...darn it


    I didn't move anything. That's the result.
     
    Last edited by a moderator: May 5, 2017
  17. Dec 24, 2011 #16

    I like Serena

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    :smile:


    In your result you have the unit [MPa], but it should be [Pa].
    You still need to convert to [MPa], that is, move the decimal point 6 places over.
     
  18. Dec 24, 2011 #17

    Femme_physics

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    How do I make sure to always get the result in MPa? Do I use mm instead of m? BTW I edited my post above.
     
  19. Dec 24, 2011 #18

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    Yep. That should do the trick.
    In your post above, you should use mm.
     
  20. Dec 24, 2011 #19

    Femme_physics

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  21. Dec 24, 2011 #20

    I like Serena

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    Yep. You've got it down!

    http://paniagua.es/wp-content/uploads/2011/03/smileygafas-115x115.png
     
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