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Date : 2016-06-23 12:24:21

Introduction:

The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed constant across the wall thickness.

 

Description:

 

Consider a thin ring or cylinder as shown in Fig below subjected to a radial internal pressure p caused by the centrifugal effect of its own mass when rotating.

The centrifugal effect on a unit length of the circumference is

p = m ω2 r

Fig: Thin ring rotating with constant angular velocity ω

 

Here the radial pressure ‘p' is acting per unit length and is caused by the centrifugal effect if its own mass when rotating.

Thus considering the equilibrium of half the ring shown in the figure,

2F = p x 2r (assuming unit length), as 2r is the projected area

F = pr

Where F is the hoop tension set up owing to rotation.

The cylinder wall is assumed to be so thin that the centrifugal effect can be assumed constant across the wall thickness.

 

F = mass x acceleration = m ω2 r x r

 

This tension is transmitted through the complete circumference and therefore is resisted by the complete cross –sectional area.

hoop stress = F/A = m ω2 r2 / A

 

Where A is the cross – sectional area of the ring.

 

Now with unit length assumed m/A is the mass of the material per unit volume, i.e. the density ρ .

hoop stress = ρ ω2 r2

σH = ρ . ω2 . r2

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