So we have a compound cylinder, 100 mm internal diameter, 200 mm common diameter and an outer diameter of 300 mm. The pressure created by shrinking the outer cylinder on the inner cylinder is 30 MPa.
If the maximum hoop stress on the outer cylinder is 110 MPa, what is the maximum internal pressure the cylinder can widthstand? [79 MPa]
Also find the hoop stress at the outer diameter of the inner cylinder. [-18 MPa]
Obviously we have the lamé equation, where σr = A - B/r2 and σh = A + B/r2
Then you have a number of boundary conditions. So as far as I figure it we have
r = ri σr = -Pi
r = rc σr = -30 MPa
r = rc σr = -30 MPa, σh = 110 MPa
r = ro σr = 0
the problem I am having here is that with A and B being different for the 2 cylinders, it appears that there are too many boundary conditions for the outer cylinder and not enough for the inner one?
The Attempt at a Solution
My attempts so far have basically just centred around solving A and B for the outer cylinder using the various boundary conditions, but it is all a bit pointless because I don't know which ones you are supposed to use and I can't solve for the inner tube because I can't see how I have the relavent information. If anyone could help me with this I would be very appreciative.