Don't know about the 2nd bit, I'm 1st year HND n haven't had strength of materials advanced yet. I think the wall thickness could be 12.2mm
Divide your UTS by 7.5 to get your max permissible. The force required to lift 9000kg is 88290N (m * g) From that the area of contact required can be calculated as 1.655x10^-3m^2
Cross sectional area is given by pi*(outside^2-inside^2)/4 transpose that and I got an outside diameter of 92mm. Subtract 80mm inside dia and you get your 12.2mm thickness.
Again, I'm just 1st year of HND so if I'm wrong I'm sorry but I think what I've done is good there.....
A thick walled hydraulic cylinder is to be made of internal diameter of 80 mm and its function will be to lift a platform of mass 9000 kg. Determine the required wall thickness, if its cylinder’s wall steel chosen has a U.T.S. of 400 MN/m2 and a factor of safety of 7∙5 to 1 is to be employed.
Also calculate the new internal diameter when pressurised, given that the steels Poisson’s ratio is quoted as n = 0∙35 and Young’s Modulus is E = 200 GN/m2