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

 

[rb1957]

sure, one pin with two loads ... call that double shear if you like, clearly you do; i won't.

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

 

[desertfox]

Hi RB1957

I can't post a sketch at present however let me try and describe what I believe,
Assume a torque of 100Nmm acting on a shaft of say 10mm diameter which is sleeved and pinned on one side only,so the pin can only transmit the torque through one interface, take moments to calculate the load on the pin at the interface and we obtain

100Nmm = 5mm * unknown force

Force = 100/5. ...... 20N

Now consider the pin passes through the shaft and creates two interfaces the reaction at each intersection becomes

Force= 100/10 ......... 10N

So we could increase the torque to 200Nmm to achieve a reaction of 20N at each interface thus increasing the capacity of the pin to transmit torque.

Now if the pin in the latter example fails and produces three pieces what failure mode would you call it, if it isn't double shear?

 

[rb1957]

no, i do not call a pin reacting a couple "double shear". i can see how you'd define double shear by the number of pieces the pin would break into, but i don't.

can we just agree to disagree ?

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

 

[hydtools]

Pressure in the hub.
The force on the pin in the hub acts at the mid-point in the hub thickness at diameter (D1 + D)/2. That force x (D1+d)/2 = T, or that force = 2T/(D1+D)

The pressure p2 = that force divided by the pin projected area(d*(D1-D)/2) or
p2 = (2T/(D1+D))/(d*(D1-D)/2) = 4T/(D1+D)*d*(D1-D) = 4T/(d*(D1^2 - D^2))

remember (D1+D)*(D1-D) = D1^2 - D^2

Ted

 

[desertfox]

Hi rb1957

Yes no problem. We can agree to disagree!

Regards

Desertfox

 

[rb1957]

i think you can get the same result using bending stresses, like i did for p1.

more than one way to skin a cat (but cats like none of them)

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

 

[hydtools]

rb1957, the OP asked for compression/pin bearing stresses, not pin bending stress.

Ted

 

[rb1957]

i'm tired ... yes, dear ...

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