A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rotates at 800 r/min. Calculate 1. the magnitude of the force and the stress in the rim if the density of the materials is 7800 kg/m3. 2. the total change in diameter. Young's modulus for the material is 200 GPa.

Question

Engineering

Valery Downs

2 August, 12:27

A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rotates at 800 r/min. Calculate 1. the magnitude of the force and the stress in the rim if the density of the materials is 7800 kg/m3. 2. the total change in diameter. Young's modulus for the material is 200 GPa.

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Amiah · Answer 1

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius. The mass is computing by the rim density and its volume:

m=p. V

m=p * (A*R)

Where A is the cross-sectional area in meters:

m = ((0.015m*0.200m) * 0.6m) * (7800 kg/m^3) = 28.08 kg

The angular speed in rad/s is:

ω = 800r/m. 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg) * (133.33 rad/s) ^2 * (0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π. R. b)

Where b is rim's with equal to 200mm:

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa = 10000000Pa+397887Pa = 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d = 2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt. d / (2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr = 0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr = 0.00498m = 4.98mm [/tex]