Question

Rubber Steel 2r Cross-Section Problem 2. A rubber cylinder of height h and radius r fits perfectly inside of a steel cylinder (see figure), such that: (1) there is no gap between the steel and the rubber, (2) the rubber is initially stress free, and (3) the contact between the rubber and steel is frictionless. The steel cylinder is resting on the floor, and the rubber itself has material properties E and v. A compressive force F is applied to the top of the rubber cylinder. Derive a formula for the lateral pressure p between the rubber and the steel. Derive a formula for the shortening Δ of the rubber cylinder. a. b.

Answer 1

A The stress due to the lateral pressure sigma_x = -p stress due to the compressive force(F) sigma_y = -F/A = -F/(pi r^2) Now Hooks law for three dimension strain sigma x = 1/E(sigma_x - v sigma_y - v sigma_2) here in ques gives = > cylinder is rigid then sigma x = 0 0 = 1/E(sigma_x - v sigma_y - v sigma_z) 0 = -p - v(-F/pui r^2) - v(-p) p = VF/(1 - v)pir^2 b rubber strain in y direction then sigma y = 1/E(sigma_y - v sigma_x - v sigma_z) (Delta h/h) = 1/E(-F/pi r^3 - v(-p) - v(-p)) Delta or Delta h = h/E(z vp - F/pi r^2) (or) Delta or Delta h = h/E(2v^2 F/(1 - v) - pi r^2 - F/pi r^2) Delta = hf/pi r^2 E(2V^2/1 - v - 1)