Shear Stress on sleeve with grooves

AI Thread Summary
The discussion centers on the shear stress experienced by a sleeve with grooves in a butterfly valve assembly. The valve features a butterfly plate with a shaft that opens and closes it, and a sleeve is inserted with a slip fit, coated with adhesive for bonding. Calculations indicate that the adhesive bond strength of 2500 psi provides a safety margin, but concerns arise about the sleeve potentially spinning if the adhesive deteriorates due to torque from the shaft. Participants emphasize the need to relate the torque created by the shaft to the adhesive's bond strength and suggest calculating the net torque from the forces applied by the inlet pressure. The conversation highlights the complexity of visualizing the valve's construction and the importance of understanding the torque dynamics involved.
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1. There is a valve body in which there is a butterfly plate. A shaft goes through the butterfly plate allowing it to open and close. A sleeve is inserted into the I.D. of the butterfly plate that the shaft goes through. The I.D. of the hole (butterfly plate) that the sleeve is inserted is is 0.500 in. and the effective length the sleeve touches is 3.50 in. There will be a slip fit between the sleeve and butterfly plate. A adhesive that has a bond strength 2500 psi will be coated on the O.D. of the sleeve and inserted in the butterfly plate. The O.D of the sleeve will be 0.415 and has 5 grooves width of the grooves is 0.05". The inlet air pressure of 45 psi will be applied to the valve. The O.D. of the butterfly plate is 3.90 in.

2. σ = Load(lb)/Area3. What I have done so far is:
1) Load experienced on the whole plate, P = σ*Area = 45 psi * (3.90in)^2 * π/4) = 537 lb
2) Effect Area of adhesive on sleeve = 0.415in *π * 0.05in = 0.65 in^2
3) Calculate the pressure created on the sleeve
σ = 537lb/ 0.65 in^2 = 826 psi
4) Calculate the margin of safety from bond strength of adhesive
2500 psi / 826 psi = 3.02
I am not sure if this is correct though.

We know the sleeve cannot pull out or fallout. However, since there will be a torque created by the shaft opening or closing the butterfly plate, the sleeve is prone to spin freely if the adhesive deteriorates. The shaft is what rotates the butterfly plate. If there is torque created by the shaft and if the adhesive wore off then the sleeve MAY just spin around in the butterfly plate. I am not sure how to relate the torque to the bond strength of the adhesive.
 
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It's hard to visualize the construction of this valve from a verbal description. Is there a diagram or figure which shows the construction that you can post?
 
Here is a image! I hope this helps because I am stuck :(

309sbpy.jpg
 
There will be two sleeves inserted. One on each end.
 
wzzp144 said:
Here is a image! I hope this helps because I am stuck :(

309sbpy.jpg
The valve in the OP has an OD of 3.90 inches whereas the figure shows OD = 1.90 inches. Typo perhaps?
 
Yes sorry that is type. it will be 3.90" Outer diameter of butterfly plate
 
What am I missing here? The whole point of a butterfly valve in the first place is to avoid any torque on the actuating shaft. The torque on one wing of the plate is canceled out by the torque on the other wing.
 
Well I am trying to relate the bond strength of the adhesive to the force or torque? Let's say by chance the adhesive were to wear or deteriorate, it may cause the sleeve to spin inside the butterfly plate then. There will be torque created by the shaft to open and close the plate with the addition of an inlet pressure of 45 psi.
 
wzzp144 said:
There will be torque created by the shaft to open and close the plate with the addition of an inlet pressure of 45 psi.
That seems to be where the problem description and the solution attempt diverge.

Consider a view looking parallel to the shaft, as if you are looking down at the North pole. Due to the 45PSI you have some force applying a torque to the 'East' side of the butterfly. Also due to the 45PSI you have some force applying a torque to the 'West' side of the butterfly. Calculate the two torques and sum them, what due you get for the net torque?
 

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