Question

Two blocks of rubber, each of width w = 60 mm, are bonded to rigid supports and to the movable plate AB. Knowing that a force of magnitude P = 19 kN causes a deflection δ = 3 mm of plate AB, determine the modulus of rigidity of the rubber used.

Answer 1

Answer:  modulus of rigidity of the rubber used = 12.6*10⁶ Pa or 12.6 MPa

               The dimension of the rubber for the cross sectional area were not provided therefore I have considered them to be as 150 mm (0.15 ) & 100 mm (0.1 mm)

Explanation:  Force exerted by the rubber blocks = (1/2)*P

                                                                                      = 0.5 * 19 * 1000 N

                                                                                      = 9.5*10³ N

                      Shear Stress τ = (9.5*10³ N)/(0.15 m * 0.10 m)

                                                = 6.33*10⁵ Pa

The deflection given is δ=3 mm and w= 60 mm

                                 γ = δ/w = (3*10⁻³ m)/(0.06 m)

                                    = 0.05

modulus of rigidity of the rubber = τ/γ = (6.33*10⁵ Pa)/(0.05)

                                                                  = 12.6*10⁶ Pa or 12.6 MPa