Please negate any shear forces that arise from sliding The p

Solution

The maximum pressure, P_max = 1 GPa. Assuming the load to be under yield load and remains within the elastic limit of the stress strain curve.

Using Hook\'s law,

sigma = E x epsilon

Replacing the above eqn with max pressure, 1 GPa and find the strain induces.

epsilon (longitudinal) = sigma / E = 1 / 207 = 0.0048

Using the poisson ratio find the lateral strain of the material,

nu = lateral strain / longitudinal strain = lateral strain / 0.0048

lateral strain = nu x 0.0048 = 0.28 * 0.0048 = 0.00135

The lateral strain reflects the change in radius of the pin while also impacts the contact area of the pin with the disk. The infinite contact radius is replaced with the change in radius due to lateral strain.

lateral strain = del R / R

del R = lateral strain * R

del R = 0.00135 * 5 = 0.00676 mm

The surface area in contact is given by,

A = pi [ (r+del r)^2 - (r)^2 ]

A = pi * del r ( del r + 2r)

Replace r and del r with 150 mm and 0.00676 mm.

A = 3.14 * 0.00676 ( 0.00676 + 2*150) = 6.36 mm^2

Now the normal stress acting on the surface area A = 6.36 mm^2, rotating at an angular speed is the net force given by,

F = F applied - mu * Fapplied

The force F is given by,

sigma = F / A, where sigma is 1GPa

1 = F / 6.36

F = 6.63 * 10^9 N

The applied force, F applied = F / 1 - mu

F applied = 6.63 * 10^ 9 / (1- 0.28) = 8.844 x 10^9 N