Two 18 x 21 x 25 mm rubber mounts are used to isolate the vibrational motion of a machine from its supports. A 420-N applied load P causes the rigid upper frame (point O) to deflect 1 mm downward. Determine: (a) average shear strain in the rubber mounts, and (b) average shear stress in the rubber mounts.

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

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

trig 15.3 cos