Answer:
a) γ =0.055556
b) t = 0.4 MPa
Explanation:
Given:
- The dimensions of rubber block : 18 x 21 x 25
- A load was applied at upper frame P = 420 N
- The rubber deflects dx = 1 mm downwards
Find:
(a) average shear strain in the rubber mounts
(b) average shear stress in the rubber mounts.
Solution:
- For average shear strain we have the definition:
γ = dx / y
Where,
γ: The shear strain.
dx : Deflection along the shear force
y : The length perpendicular to deflection.
- From given data we have dx = 1mm, and the dimension of block perpendicular to deflection is the a dimension. Hence, dx = 0.001 and y =0.018 m:
γ = 0.001 / 0.018 = 0.055556
- The average shear stress along the mating flat surface. We have from definition:
t = F_shear / Area
- Where, F_shear: The shear force on each rubber block is P/2.
Hence,
t = (P/2) / b*c
Plug values in:
t = (420/2) / (0.021*0.025)
t = 0.4 MPa
