Need Compressive Stress Equation Explained. Please Help!

[jus2sho]

Ok, so I have a shaft and a hub that is connected with a solid dowel pin in double shear. I need to calculate the compressive stress in the shaft and in the hub under a torque load. I've found an equation that I believe may be my answer, but can someone please break the equation down for me so I can understand what's actually going on here? Stress is normally calculated by σ = P / A and I understand that T = P*D, but where does the Area fit into the equation below? Please see the attachments for further details. Thanks in advance for any help!

Pressure in Shaft:
- This equation can be rearranged to (6*P)/(d*D). (d*D) is the Area, but how does the multiplier 6 come into play?

Pressure in Hub:
- Totally lost on this equation. Please explain.

Site For Reference

 

[desertfox]

Hi

Can you explain why you think you have compressive stress in the shaft?

 

[chicopee]

I think that he is referring to the bearing stress that the pin is causing to the holes of the hub and shaft.

 

[jus2sho]

Yes, I do mean the bearing stress. Just can't figure out how the equations above are derived.

 

[racookpe1978]

Where "in the hub" do you think this "pressure stress" from the formula is being calculated with the formula?

 

[jus2sho]

At the contact surface between the pin-hub or pin-shaft. The formula was actually found on mitcalc.com.

 

[desertfox]

Hi jus2sho

I've tried to find a proof for the formula's you posted with no luck, other than working through them myself which I haven't got time to do, however I believe the formula's to be correct if that helps.

 

[jus2sho]

I believe they're correct as well. I'd just like help if someone could explain how they're derived.

 

[rb1957]

"how does the multiplier 6 come into play?"
from bending of a rectangular section ... stress = My/I = M(d/2)/(b*d^3/12) = 6M/(b*d^2)
so p1 is the bending stress on the inner shaft as the shaft applies the torque to the pin.

the pin takes the torque from the inner shaft to the outer, as a couple. the pin is not in double shear; yes, there are two faces in shear, but this is not double shear.

see if you can take my description of p1 and figure out p2. hint, the outer shaft isn't a rectangle, but two rectangles.

another day in paradise, or is paradise one day closer ?

 

[desertfox]

Hi rb1957

Take a look at this link, I believe the pin is in double shear

https://books.google.co.uk/books?id...ge&q=taper pin in sleeved shaft joint&f=false

 

[rb1957]

no, double shear means two shear faces are reacting the same load, so a fastener can rect something like double the single shear allowable.

this fastener the shear faces are reacting a couple; not double shear.

another day in paradise, or is paradise one day closer ?

 

[hydtools]

Where did the 6 come from? Imagine force distribution on the pin rectangular area within the shaft diameter D. That force distribution is triangular with the largest value at the shaft diameter and zero at the shaft center. The resultant force of this triangle is applied at a distance of D*2/3 from the shaft center. That distance is D/2 * 2/3 = D/3. The resultant force is T/(D/3), torque divided by the moment arm of the resultant force. The shear produced by that force is tau*d*D/2 = T*3/D. tau = 6*T/(d*D^2).

Ted

 

[rb1957]

sure, there's more than one way to malipulate the math to get "6" ...
"triangular distribution of stress/pressure" = elastic bending distribution.

another day in paradise, or is paradise one day closer ?

 

[jus2sho]

rb1957 nailed it. Thanks! It makes perfect sense where the "6" comes from. Now I just have to figure out the second equation...

 

[desertfox]

Hi

Here is an example showing the formula the op originally posted

https://books.google.co.uk/books?id...stresses in tapered pin under torsion&f=false

Not though again the author states the pin is in double shear.

RB1957

The reason I believe the pin is in double shear is that when the pin passes through both sides of the shaft, it only takes half of the total load due to the total torque at each interface, now if the pin did not pass right through the shaft, ie a blind pin, then the force at the single interface due to the applied torque would be double that of the pin passing through both sides of the shaft.

 

[hydtools]

rb1957, I don't understand your using a rectangular cross section beam equation for a round pin. It seems that a better choice would be that for a round cross section beam.

The pin is in double shear.

Ted

 

[rb1957]

the rectangle is the section through the pin, reacting the torque with a linear stress field (aka bending stress), just like you proposed.

the pin is not in double shear ... the two active shear faces are reacting a couple; for me that is not double shear. for me double shear is when a pin has a load of 2P applied and reacts P on two faces.

another day in paradise, or is paradise one day closer ?

 

[desertfox]

Hi RB1957

Well effectively that's what happens, if the pin is blind and only through one side it see's twice the force that it would see than if it were through both sides.
Another way of defining single and double shear is the number of pieces the pin break into on failure, so in single shear it would break in two pieces and in double shear it would break in three, now if the pin is through both sides of the shaft then clearly it would break into three pieces, would you agree?

 

[rb1957]

what are you trying to express when you say "double shear" ?

to me it's saying that the pin has two faces resisting the same load (2P applied, P reacted on 2 faces) so that the pin can support twice(ish) the usual load (ie 2*the single shear allowable).

i would not call this application double shear.

"if the pin is blind and only through one side" ... IMHO, no ... now you have two studs (as near as i can understand your description) each supporting 1/2 of the couple ... in single shear. I'm understanding the loading as torque coming down the shaft, and the pin takes the torque from the shaft to the cylinder.

another day in paradise, or is paradise one day closer ?

 

[hydtools]

What is a couple? Two forces acting at some distance apart. Those two forces each act on the pin, lets say at the shaft/collar interface, creating shear loads on the pin where it crosses that interface twice. I see then double shear, one pin with two shear loads.

Ted