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)/Area 3. 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.