## Info

Now at r = R, i.e. at the outside of the disc, or = 0.

R3CO2

i.e. the collapse speed a>p is given by

For a disc with a central hole, (1) still applies, but in this case the value of the constant A is determined from the condition or = 0 at r = /?i the inside radius „ „

Again, or = 0 at the outside surface where r = R. Substituting in (1),

R3CO2 R3,Q)2

i.e. the collapse speed cop is given by

If a rotating disc is stopped after only partial penetration, residual stresses are set up similar to those discussed in the case of thick cylinders under internal pressure (auto-frettage). Their values may be determined in precisely the same manner as that described in earlier sections, namely, by calculating the elastic stress distribution at an appropriate higher speed and subtracting this from the partially plastic stress distribution. Once again, favourable compressive residual stresses are set up on the surface of the central hole which increases the stress range - and hence the speed limit - available on subsequent cycles. This process is sometimes referred to as overspeeding.

Mechanics of Materials 2 Disc edge

Mechanics of Materials 2 Disc edge Fig. 3.36. Residual stresses produced after plastic yielding ("overspeeding") of rotating disc with a central hole.

A typical residual stress distribution is shown in Fig. 3.36.

Examples

### Example 3.1

(a) A rectangular-section steel beam, 50 mm wide by 20 mm deep, is used as a simply supported beam over a span of 2 m with the 20 mm dimension vertical. Determine the value of the central concentrated load which will produce initiation of yield at the outer fibres of the beam.

(b) If the central load is then increased by 10% find the depth to which yielding will take place at the centre of the beam span.

(c) Over what length of beam will yielding then have taken place?

(d) What are the maximum deflections for each load case?

For steel crv in simple tension and compression = 225 MN/m2 and E — 206.8 GN/m2. Solution

(a) From eqn. (3.1) the B.M. required to initiate yielding is

But the maximum B.M. on a beam with a central point load is WL/4, at the centre.

The load required to initiate yielding is 1500 N. (b) If the load is increased by 10% the new load is

W' = 1500+ 150 = 1650 N The maximum B.M. is therefore increased to

= 825 Nm and this is sufficient to produce yielding to a depth d, and from eqn. (3.2), Mpp = ^ [3D2 - d2] = 825 Nm

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