Question

Two blocks of rubber with a modulus of rigidity G 5 12 MPa are bonded to rigid supports and to a plate AB. Knowing that c 5 100 mm and P 5 45 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 5 mm.

Answer 1

a=42.86 mm

b=160.7 mm

Step-by-step explanation:

Using the attached figure

Shearing strain=
`\frac {\sigma}{a}=\frac {\tau}{G}`

`a=\frac {G\sigma}{\tau}=\frac {12 Mpa* 0.005 m}{1.4 Mpa}=0.042857143  m\approx 42.86 mm`

The shearing stress will be given by

`\tau=\frac {0.5P}{A}=\frac {P}{2bc}`

`b=\frac {P}{2c\tau}=\frac {45000 N}{2* 0.1* 1.4* 10^(6)}=0.160714286  m\approx 160.7 mm`