Toque vs Shear Stress 2

[LoBue]

I have an application where I am attaching a hex nut to a motor shaft with Loctite. The motor is rated to 2.3 IN-LBS and the loctite has a compressive shear strength of 3,495 PSI.

The equation for shafts, Shear Stress Max = (Torque x Radius of Shaft)/(Polar moment of inertia) only takes into account the cross-section. Does anybody know of an equation that will take into account the length of engagement as well?

Thank You

 

[zekeman]

If you are driving the load through the nut, the shear stress is approximately shearforce/shear area or
T/(2pir^2l) where
pi=3.1416...
T=design torque
r=pitch radius of nut
l= width of nut
I would use 5 times the rated torque for motor startup and another 2 times for safety considerations or 10 times the rated. In any event, 3495 psi should be ok in your case.

 

[rb1957]

wouldn't it be the area of the thread faces that was reacting the applied torque ?

actuately this would be pi*d*(w*n) ...
d is diameter of the nut (say mean thread diameter)
w is the width of a thread
n is the number of threads engaged

conservatively, and simply, use pi*d*l ...
d is the diameter of the nut
l is the length of engagement

 

[JStephen]

You're assuming that the compressive shear strength of the Loctite is also the capacity of the bond. Is that necessarily the case?

Also, does it make any difference that it's a dynamic loading? All sorts of stuff will come lose when being vibrated that wouldn't in a static loading situation.

 

[LoBue]

I may have mislead everyone by my use of the term "hex nut". The part that is going onto the motor shaft has an internal hex cut into it and a thru hole. This thru hole is a locational slip fit onto the motor shaft. I will be using the loctite to adhere this part to the shaft. Sorry for not being more clear.

Joe

 

[rb1957]

so the locite is on a surface ... pi*d*l
d = diameter of hole
l = length of engagement

 

[Cockroach]

Use the length of engagement as one diameter external thread. This is the default length used in industry given the lack of user input into programs computing shear stress on studs and nuts.

I would treat your application similar to torsion on a box/pin system. Apply a moment to the pin and get an axial load, this would directly lead to stress once you compute the pin shear area.

Note that the shear area of the box (nut) is always greater than the pin (screw). You need only worry of pin failure since the nut can take much more load.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada