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Breaking solid shaft in torsion purposefully

  1. Aug 26, 2016 #1
    I am trying to design a drive shaft that breaks when i apply a specific torque to protect fro over torquing. I made a few samples, and now i am trying to compare the results to my equations. My problem is, none of the equations seem to match the actual results. Your help would be appreciated:

    I made a bar that was 0.078in. in diameter, necked it down to 0.067in. diameter (.0335in. radius)

    In testing, I found that the bar shears/breaks at around 9.6 in-lbs.

    τ = T*r/J = 9.6in-lbs*.0335in/((π/2)*.0335in^4) = 164ksi

    Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi, and i thought, "Great! 3.5% error on my yield strength", but then i remembered the 0.577*σ rule. (making shear stress only .577*170 = 98ksi). I'd expect this to break at 5.7in-lbs using that conversion. Can any explain why my test results don't match up to my theoretical values?

    (I considered using a stress concentration factor, but that would only make my answer more puzzling because it would increase the discrepancy where τ(max) = τ(nom)*K and K>1)


    Thank you
     
  2. jcsd
  3. Aug 26, 2016 #2

    berkeman

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    Welcome to the PF.

    Couldn't you just use a mechanism design similar to a click-type ratcheting torque wrench? It has the advantage of being non-destructive...
     
  4. Aug 26, 2016 #3
    Thank you for the welcome. I've used PF for years, but finally came across an issue that i didn't see a solution to already.

    I could use a ratchet, but that wouldn't work for this application. I'm really interested in just leaving the part behind if that torque is overloaded.
     
  5. Aug 26, 2016 #4

    SteamKing

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    In my experience, designing a structure (like a shaft) to fail at a precise value of shear is very difficult.

    Most standards organizations dealing with structural design tend to use relatively high factors of safety (compared with tensile loading) when assessing the shear strength of a material. AISC typically places the max shear stress for steel at 0.4 * sy, while tensile strength can be 0.6 * sy minimum.
     
  6. Aug 26, 2016 #5

    berkeman

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    So maybe the best engineering solution would be a click torque wrench design that permanently disengages after the first "click"...
     
  7. Aug 27, 2016 #6

    Nidum

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    Conventional practice is to use a shear pin . More predictable and reliable because of the simpler nature of the shearing action and in any case you can always find a suitable pin configuration by trial and error without damaging the parent components .

    There are many other possibilities .

    For example - two face to face wave cams held together by an axial spring . This is re-useable - it will slip above a certain torque but continue to drive when torque is reduced again .

    Another one is the friction drive - a shaft light press fit in a tube . Works well with plastic components .

    Tell us the actual problem ?
     
  8. Aug 27, 2016 #7

    jack action

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    Actually, this rule is for the yield strength (i.e. when it begins to permanently deform) of steel. For the ultimate strength of steel it is 0.75*σ. The constant can be higher for other materials. Note that these relationships are very approximate for use only as an rule of thumb estimate if no other source of information is available.

    Note also that there is a difference between the apparent stress (F/A0) and the actual stress (F/A).
     
  9. Aug 29, 2016 #8
    The design is already set for its application. I am trying to better understand how to make it work reliably. The actual problem is that both FEA and hand calculations don't line up with the actual part.

    Looks like this may actually make my problem worse. If the actual stress is higher than what I calculate, shouldn't the bar break even sooner?

    So are you saying that there is a factor of safety placed on my certificate of material properties?
     
  10. Aug 29, 2016 #9

    SteamKing

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    No, what I'm saying is that your material properties are based on a tensile test, not a shear test. To use those test results for calculating the max. allowable shear stress, the figure 0.4 sy is supposed to account for going from a tensile loading to a shear loading and any other unknown factors. The test figures on your material data sheets are the actual stress results from the tension test, but no one normally designs to the yield strength of the material.

    For design in bending, where there are tensile and compressive stresses, you can design to 0.6 sy, where sy is the min. yield strength of the material. That extra 0.2 sy on the allowable stress is used as a cushion to guard against the vagaries of shear loading.

    With your shaft, you want it to fail when a certain torque figure is encountered. Stressing a bar axially to yield doesn't necessarily produce failure; steel has a different figure called the rupture stress, which is the stress measured just before the test piece breaks.
     
  11. Aug 29, 2016 #10

    Nidum

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    What is the actual configuration of the shaft and surrounding parts - can you supply a drawing ?
     
  12. Aug 29, 2016 #11
    For my test I put the fixed end in a drill press and the rotating end in a torque gauge and twisted the shaft (by hand, with the drill press off) until it came apart. I measure at almost 9.6 in-lbs every time i repeated the test. Obviously the drive shaft shears off at the necked down area. Wrapping my head around why the shaft withstood a higher shear than expected is why i am reaching out to you all.

    Thank you for your help so far. I've been reading around in my old machine design books and doing other research, but haven;t come up with much yet. I appreciate the outside perspectives.
     

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  13. Aug 29, 2016 #12

    Nidum

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    I'll give you a clue :

    A simple beam does not fail when stress in outer fibres reaches yield stress . In fact you can usually load a beam a good bit more before actual failure occurs . Think about why this is and how you can apply similar thinking to torsional failure situations .
     
    Last edited: Aug 29, 2016
  14. Aug 29, 2016 #13
    Are you suggesting that the failure is not happening out the outside radius? or possible there is some necking occurring which diminishes the radius? I think this is a possibility, however, to get a number that is relatively close to what I'd expect, the radius would need to change from 0.0335in to 0.0200 in. (which isn't impossible).

    Is it possible that i am incurring some cold work into the material as well during my machining operation? it IS a 300 series steel I am working with.
     
  15. Aug 29, 2016 #14

    jack action

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    Yes, your number would be higher because of the smaller «true» area, but if you look at the diagram carefully, you'll notice that the «true» stress at point #3 (rupture) is also higher than the given ultimate strength. All in all, it may even out.
     
  16. Aug 29, 2016 #15

    Nidum

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    @thissong

    As load is increased stress in outer fibres increases until it reaches yield stress .

    What happens when load is increased further ?
     
  17. Aug 30, 2016 #16

    JBA

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    Apart from the above issue, you state "Now, my material states that I have an tensile strength of 170 ksi and yield strength of 140 ksi". What is the reference that is giving you those values? I am asking this because most standard material specs for purchased materials are statements of minimum material properties and generally the actual properties of the purchased materials exceed those quoted minimum values.
     
  18. Aug 31, 2016 #17

    Nidum

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    As well as everything already mentioned there is the problem of ensuring that the shaft is subject to a pure torque . Any significant bending forces could change the value of the breaking torque considerably . Made worse in this particular situation because the shaft is so small in diameter and has a built in stress raiser .

    There are really many problems in trying to get an exact figure for the breaking torque of this shaft . The best you will ever manage is to find upper and lower bounds for the value .
     
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