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

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

From eqn(10.5)^{†}

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

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

and from eqn . ( 10.6)^{†}

P=\frac{1}{2}\pi bLp_{0}               (10.6)

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

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

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

Safety  factor   n=\frac{850}{298.3-(-240)} =1.58

(b) Autofrettage conditions

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

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

By trial and error:

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

:.  R_{p} =1.325\times 62.5=82.8 mm

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

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

\sigma _{rr}=850\times 10^{6} \left[\log _{e}\left(\frac{r}{82.8} \right)-\frac{1}{2\times 190^{2} } \left(190^{2} – 82.8^{2} \right) \right]
=850\times 10^{6} \left[\log _{e}\left(r/82.8\right) -0.405\right]
and    \sigma _{\theta \theta } =850\times 10^{6}\left[\log _{e}\left(r/82.8\right)+0.595\right]

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

\sigma _{rr}=-850 MN/m^{2}  and \sigma _{\theta \theta } =266.7 MN/m^{2}

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

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

P=\frac{1}{2}\pi bLp_{0}                                      (10.6)

\sigma _{rr}=850 MN/m^{2}  and  \sigma _{\theta \theta } =-298.3\times \frac{580}{240} =-721 MN/m^{2}

Giving residual stresses at the bore of

\sigma ^{′}_{rr} =580-580=0
\sigma ^{′}_{\theta \theta } =266.7-721=-453 MN/m^{2}

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

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

:. final working stresses are:

\sigma _{rr_{w}} =0-240=-240 MN/m^{2}
\sigma _{\theta \theta_{w} } =298.3-454.3=-156 MN/m^{2}

:. New safety factor according to Tresca theory

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 \sigma _{rr} in the working condition and increase the magnitude of the denominator to reduce the resulting value of n.

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