(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

Question

(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 [latex]P_{W}[/latex] of [latex]240 MN/m^{2}[/latex] 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 [latex]P_{A}[/latex] of [latex]580 MN/m^{2}[/latex]. Determine the residual stresses produced at the bore of the cylinder when the autofrettage pressure is removed and hence deterniine the new value of the safety factor at the bore when the working pressure [latex]P_{W} =240 MN/m^{2}[/latex] is applied. The yield stress of the cylinder material is [latex]σ_{y} = 850 MN/m^{2} [/latex]and axial stresses may be ignored.

Accepted Answer

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

From eqn[latex](10.5)^{†}[/latex]

[latex]p=p_{0}\sqrt{1-\frac{y^{2} }{b^{2} } }[/latex] (10.5)

[latex]\sigma _{rr}=-P\left[\frac{\left(R_{2}/r ight) ^{2} -1}{K^{2}-1 } ight]=\frac{-240}{8.24} \left[\frac{0.19^{2} }{r^{2} }-1 ight][/latex]
[latex]=-240 MN/m^{2} [/latex] at the bore surface (r = 0.0625 mm)

and from eqn . [latex]( 10.6)^{†}[/latex]

[latex]P=\frac{1}{2}\pi bLp_{0}[/latex] (10.6)

[latex]\sigma _{ heta heta }=-P\left[\frac{\left(R_{2}/r ight) ^{2} +1}{K^{2}-1 } ight]=\frac{240}{8.24} \left[\frac{0.19^{2} }{r^{2} }+1 ight][/latex]
[latex]=298.3 MN/m^{2} [/latex] at the bore surface

Thus, assuming axial stress will be the intermediate stress [latex](σ_{2})[/latex] value, the critical stress conditions for the cylinder at the internal bore are [latex]σ_{1} = 298.3 MN/m^{2}[/latex] and[latex]σ_{3} = -240 MN/m^{2}[/latex].

:. Applying the Tresca theory of failure ([latex]σ_{1}- σ_{3} = σ_{y}/n)[/latex])

Safety factor [latex] n=\frac{850}{298.3-(-240)} =1.58[/latex]

(b) Autofrettage conditions

From eqn 3.37 the radius [latex]R_{p}[/latex] of the elastic/plastic interface under autofrettage pressure of [latex]580 MN/m^{2} [/latex]will be given by:

[latex]P_{A}=\frac{\sigma _{y} }{2}\left[\frac{K^{2}-m^{2} }{K^{2} } ight] +\sigma _{y} \log _{e} m[/latex] (3.37)
[latex]580 imes 10^{6} =\frac{850 imes 10^{6} }{2}\left[\frac{3.04^{2}-m^{2} }{3.04^{2} } ight]+850 imes 10^{6} \log _{e} m[/latex]

By trial and error:

m	[latex]850\log _{e}m[/latex]	[latex]\frac{850}{2} \left[\frac{3.04^{2}-m^{2} }{3.04^{2} } ight] [/latex]	[latex]P_{A}[/latex]
1.6	399.5	307.3	706.8
1.4	286.0	334.8	620.8
1.3	223 .0	347.3	570.3
1.33	242.4	343.6	585.6
1.325	239.2	344.2	583.4

:. to a good approximation m = 1.325 = [latex]R_{p}/R_{1}[/latex]

:. [latex]R_{p} =1.325 imes 62.5=82.8 mm[/latex]

:. From eqns 3.38 and 3.39 stresses in the plastic zone are:

[latex]\sigma _{r} =\sigma _{y} \left[\log _{e} (\frac{1}{R_{p} } )-\frac{1}{2}(1-\frac{x^{2}_{P} }{x^{2}_{2} } ) ight][/latex] (3.38)
[latex]\sigma _{H} =\sigma _{y} \left[1+\log _{e} (\frac{r}{R_{p} } )-\frac{1}{2}(1-\frac{x^{2}_{P} }{x^{2}_{2} } ) ight][/latex] (3.39)

[latex]\sigma _{rr}=850 imes 10^{6} \left[\log _{e}\left(\frac{r}{82.8} ight)-\frac{1}{2 imes 190^{2} } \left(190^{2} - 82.8^{2} ight) ight][/latex]
[latex]=850 imes 10^{6} \left[\log _{e}\left(r/82.8 ight) -0.405 ight][/latex]
and [latex] \sigma _{ heta heta } =850 imes 10^{6}\left[\log _{e}\left(r/82.8 ight)+0.595 ight][/latex]

:. At the bore surface where r = 62.5 mm the stresses due to,autofrettage are:

[latex]\sigma _{rr}=-850 MN/m^{2} [/latex] and [latex] \sigma _{ heta heta } =266.7 MN/m^{2}[/latex]

Residual stresses are then obtained by elastic unloading of the autofrettage pressure, i.e. by applying [latex]\sigma _{rr}= +580 MN/m^{2}[/latex] at the bore in eqns (10.5) and [latex]( 10.6)^{†}[/latex]; i.e. by proportions:

[latex]p=p_{0}\sqrt{1-\frac{y^{2} }{b^{2} } } [/latex] (10.5)

[latex]P=\frac{1}{2}\pi bLp_{0} [/latex] (10.6)

[latex]\sigma _{rr}=850 MN/m^{2} [/latex] and [latex]\sigma _{ heta heta } =-298.3 imes \frac{580}{240} =-721 MN/m^{2}[/latex]

Giving residual stresses at the bore of

[latex]\sigma ^{′}_{rr} =580-580=0[/latex]
[latex]\sigma ^{′}_{ heta heta } =266.7-721=-453 MN/m^{2}[/latex]

Working stresses are then obtained by the addition of elastic loading stresses due to an internal working pressure of [latex]240 MN/m^{2}[/latex]

i.e. from part (a) [latex]\sigma _{rr}=-240 MN/m^{2} [/latex] and [latex] \sigma _{ heta heta } =298.3 MN/m^{2}[/latex]

:. final working stresses are:

[latex]\sigma _{rr_{w}} =0-240=-240 MN/m^{2}[/latex]
[latex]\sigma _{ heta heta_{w} } =298.3-454.3=-156 MN/m^{2}[/latex]

:. New safety factor according to Tresca theory

[latex]n=\frac{850}{-156-(-240)} =10.1[/latex]

N.B. It is unlikely that the Tresca theory will give such a high value in practice since the axial working stress (ignored in this calculation) may well become the major principal stress [latex]\sigma _{rr} [/latex]in the working condition and increase the magnitude of the denominator to reduce the resulting value of n.

[latex]^{†}[/latex] E.J. Hearn, Mechanics of Materials I, Butterworth-Heinemann, 1997

Question 3.7: (a) A thick cylinder, inside radius 62.5 mm and outside radi...

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