A 2 cm by 2 cm by 2 cm rubber engine mount (E = 30 MPa, Poisson ratio = 0.4999)

Question

A 2 cm by 2 cm by 2 cm rubber engine mount (E = 30 MPa, Poisson ratio = 0.4999) must support a 300 kg motor. What (maximum) deflection (in cm) will occur for a.) vertical and b.) horizontal mountings. Assume the weight is uniformly distributed across a face of of the material. Note, in the vertical mount the motor sits atop the mount, while in the horizontal geometry the motor is mounted on the side of the mount, which is supported on the opposite side. c) Is the engine mount properly designed for horizontal mounting? Explain briefly why.

Explanation / Answer

Since i am not understanding the set-up properly for second part , i am just solving the first part. You can give me half points for that , in case no one else gives a full answer.

P=load = 300 g = 300 * 10 = 3000N

A= area = 0.02*0.02 = 4 *10^-4 m^2

Stress= load/area = P/A = 3000/4*10^-4 = 7.5 * 10^6 Pa

Strain = Stress/E = 7.5 * 10^6 / 30 *10^6 = 0.25

Increase in lenght = strain * original lenght = 0.25* 2 = 0.5 cm(answer for a)

not sure for this part still giving a try

b)

transverse decrease in length = 0.499 * axial change in length = 0.25cm

So according to me design is proper in horizontal direction as decrease in length is very minimal.

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