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= [(3 + v)R\ + (1 - v)fl2] at the centre and = (3 + v)^- [R2 - /?,]2 at r = J{R\R2)

For thick cylinders or solid shafts the results can be obtained from those of the corresponding disc by replacing v by v/(l — v), e.g. hoop stress at the centre of a rotating solid shaft is oH —

Rotating thin disc of uniform strength

For uniform strength, i.e. oH — ar — a (constant over plane of disc), the disc thickness must vary according to the following equation:

4.1. Thin rotating ring or cylinder

Consider a thin ring or cylinder as shown in Fig. 4.1 subjected to a radial 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 = ma>2r

Fig. 4.1. Thin ring rotating with constant angular velocity co.

Fig. 4.1. Thin ring rotating with constant angular velocity co.

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

2F — p x 2r (assuming unit length) 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.

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

F mco2r2

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 p.

hoop stress = peo2r2

42. Rotating solid disc

(a) General equations

Fig. 4.2. Forces acting on a general element in a rotating solid disc.

Fig. 4.2. Forces acting on a general element in a rotating solid disc.

Consider an element of a disc at radius r as shown in Fig. 4.2. Assuming unit thickness: volume of element = r 86 x Sr x 1 = r S6Sr mass of element = pr SdSr Therefore centrifugal force acting on the element

= mco2r

= prS0Sr(o2r = pr2co2808r

Now for equilibrium of the element radially se , ,

2oH6r sin — + orr89 — (o> + 8ar)(r + 8r)89 = pr to SO 8r

If 89 is small,

Therefore in the limit, as 8r 0 (and therefore 8or -> 0) the above equation reduces to dar 2 2

If there is a radial movement or "shift" of the element by an amount s as the disc rotates, the radial strain is given by ds 1

dr E

Now it has been shown in §9.1.3(a)+ that the diametral strain is equal to the circumferential strain.

0 0

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