Calculate Shear Stress for a Glued Solid Rod in .375 in. D Hole

[BigRedTruck]

Homework Statement

A Solid rod with .375 in. D sits glued in the hole with .38 in. D. There is glue only on the annulus of the rod. A force is applied to the back side of the rod until the glue bond is fully sheared and the rod falls out.

Find the shear stress

Homework Equations

τ=F/A psi
Annulus A = π((D2/2)^2-(D1/2)^2) in^2
Force is observed values where the glue sheared in lbf

The Attempt at a Solution

Rod D Hole D Annulus A Force Shear
0.375 0.38 0.002964878 219.9 74168.31

Some other observed forces and calculated shear values
F tau
130.3 43947.84
71.61 24152.76
31.12 10496.22
52.06 17558.90


I don't have anyone around to check my numbers and these values seem really large

 

[PhanthomJay]

The shear stress depends on the area of the glued surface around the rod, not the end area, You will need to be given the length of the glued section of the rod (the embedded depth or hole depth) to solve the problem by finding the surface area of the glue.

( If its super glue, it'll pull out pretty easily, because although it has pretty good tensile strength, it has rotten strength in shear)!

 

[BigRedTruck]

Ok so I need to change my stress formula to tau=Force/(width of the bond (D2-D1) * height of the bond)
in the first case it would be 219.9 lbf /(.005 in.*.27 in.)=162888.9 psi

I've been looking around most everything I've seen are lap joints

 

[PhanthomJay]

No, you are not taking the inner surface area at the glue/rod interface. Picture unwrapping the glue into a flat shape. The surface area is the rod circumference times the hole depth. Shear areas are parallel to the shear forces.