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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.

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Answer:

a=42.86 mm

b=160.7 mm

Explanation:

Using the attached figure

Shearing strain=[tex]\frac {\sigma}{a}=\frac {\tau}{G}[/tex]

[tex]a=\frac {G\sigma}{\tau}=\frac {12 Mpa\times 0.005 m}{1.4 Mpa}=0.042857143  m\approx 42.86 mm[/tex]

The shearing stress will be given by

[tex]\tau=\frac {0.5P}{A}=\frac {P}{2bc}[/tex]

[tex]b=\frac {P}{2c\tau}=\frac {45000 N}{2\times 0.1\times 1.4\times 10^{6}}=0.160714286  m\approx 160.7 mm[/tex]

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