## Info

2tr x 120 x 106

In order to determine the residual stresses after unloading, the unloading process is assumed completely elastic.

Thus, unloading from condition (b) is equivalent to applying a moment of 15 kNm of opposite sense to the loading moment on a complete elastic bar. The effective stress introduced at the outer surface by this process is thus given by the simple torsion theory

J 7T x (504 - 404)10-12 15 x 103 x 50 x 10"3 x 2

The unloading stress distribution is then linear, from zero at the centre of the bar to 129 MN/m2 at the outside. This can be subtracted from the partially plastic loading stress distribution as shown in Fig. 3.43 to produce the residual stress distribution shown.

Similarly, unloading from the fully plastic state is equivalent to applying an elastic torque of 15.33 kNm of opposite sense. By proportion, from the above calculation,

15.33 2

equivalent stress at outside of shaft on unloading = x 129 = 132 MN/m

Subtracting the resulting unloading distribution from the fully plastic loading one gives the residual stresses shown in Fig. 3.44.

120 129 9

120 129 9

Example 3.7

(a) A thick cylinder, inside radius 62.5 mm and outside radius 190 mm, forms the pressure vessel of an isostatic compacting press used in the manufacture of ceramic components. Determine, using the Tresca theory of elastic failure, the safety factor on initial yield of the cylinder when an internal working pressure Pw of 240 MN/m2 is applied.

(b) In view of the relatively low value of the safety factor which is achieved at this working pressure the cylinder is now subjected to an autofrettage pressure PA of 580 MN/m2. Determine the residual stresses produced at the bore of the cylinder when the autofrettage pressure is removed and hence determine the new value of the safety factor at the bore when the working pressure Pw = 240 MN/m2 is applied.

The yield stress of the cylinder material is ay = 850 MN/m2 and axial stresses may be ignored.

Solution

(a) Plain cylinder - working conditions K — 190/62.5 = 9.24

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