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: 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