A pressure vessel with a solid wall has an ri of 8 in., ro of 17.6 in., and σy = 80,000 psi. Plot σθ and σr when P = P* and when P = 60,000 psi.

Question

A pressure vessel with a solid wall has an [latex]r_i[/latex] of 8 in., [latex]r_o[/latex] of 17.6 in., and [latex]σ_y[/latex] = 80,000 psi. Plot [latex]σ_θ[/latex] and [latex]σ_r[/latex] when P = P[latex]^*[/latex] and when P = 60,000 psi.

Accepted Answer

From Eq. 15.3,

[latex]P^{*}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(\frac{r_{o}^{2}-r_{i}^{2}}{r_{i}^{2}} ight)\left(\frac{r_{i}}{r_{o}} ight)^{2}[/latex] (15.3)

[latex]\begin{aligned}P^* &=\frac{80,000}{\sqrt{3}}\left(\frac{17.6^{2}-8^{2}}{8^{2}} ight)\left(\frac{8}{17.6} ight)^{2} \&=36,650 psi\end{aligned}[/latex]

and from Eq. 15.1,

[latex]\begin{aligned}&\sigma_{ heta}=P^{\prime}\left(1+\frac{r_{o}^{2}}{r^{2}} ight) \&\sigma_{r}=P^{\prime}\left(1-\frac{r_{o}^{2}}{r^{2}} ight) \&\sigma_{l}=P^{\prime}\end{aligned}[/latex] (15.1)

where

[latex]P^{\prime}=P\left(\frac{r_{i}^{2}}{r_{o}^{2}-r_{i}^{2}} ight)[/latex]

[latex]\sigma_{ heta}=9540\left(1+\frac{17.6^{2}}{r^{2}} ight)[/latex]

[latex]\sigma_{r}=9540\left(1-\frac{17.6^{2}}{r^{2}} ight)[/latex]

These two values are shown in Fig. 15.3.

From Eq. 15.7 with P = 60,000 psi and [latex]\sigma_{y}[/latex] = 80,000 psi, a trial-and-error calculation gives [latex] ho=11.50[/latex] in. Hence from Eq. 15.5,

[latex]P=\frac{\sigma_{y}}{\sqrt{3}}\left(1-\frac{ ho^{2}}{r_{o}^{2}}-2 \ln \frac{r_{i}}{ ho} ight)[/latex] (15.7)

where [latex]\sigma_{ heta}, \sigma_{r}, \sigma_{l}[/latex] = hoop, radial, and longitudinal stress, respectively (psi)

[latex]\sigma_{y}[/latex] = yield stress of material (psi)

[latex]r_{o}[/latex] = outside radius of shell (in.)

[latex]r_{i}[/latex] = inside radius of shell (in.)

r = radius at any point in the shell (in.)

[latex] ho[/latex] = elastic-plastic interface radius (in.)

[latex]\begin{aligned}&\sigma_{ heta}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(\frac{ ho^{2}}{r_{o}^{2}} ight)\left(1+\frac{r_{o}^{2}}{r^{2}} ight) \&\sigma_{r}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(\frac{ ho^{2}}{r_{o}^{2}} ight)\left(1-\frac{r_{o}^{2}}{r^{2}} ight) \&\sigma_{l}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(\frac{ ho^{2}}{r_{o}^{2}} ight)\end{aligned}[/latex] (15.5)

[latex]\sigma_{ heta}=19,720\left(1+\frac{17.6^{2}}{r^{2}} ight)[/latex]

[latex]\sigma_{r}=19,720\left(1-\frac{17.6^{2}}{r^{2}} ight)[/latex]

and from Eq. 15.6

[latex]\begin{aligned}&\sigma_{ heta}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(1+\frac{ ho^{2}}{r_{o}^{2}}+2 \ln \frac{r}{ ho} ight) \&\sigma_{r}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(-1+\frac{ ho^{2}}{r_{o}^{2}}+2 \ln \frac{r}{ ho} ight) \&\sigma_{l}=\left(\frac{\sigma_{y}}{\sqrt{3}} ight)\left(\frac{ ho^{2}}{r_{o}^{2}}+2 \ln \frac{r}{ ho} ight)\end{aligned}[/latex] (15.6)

[latex]\sigma_{ heta}=46,190\left(1.427+2 \ln \frac{r}{11.5} ight)[/latex]

[latex]\sigma_{r}=46,190\left(-0.573+2 \ln \frac{r}{11.5} ight)[/latex]

A plot of these equations is shown in Fig. 15.3.

The two plots in Fig. 15.3 are significant because they show the redistribution of the stress pattern as the inner region of the cylinder becomes plastic. Also note the reduction of the stress at the inner surface and the increase of the stress at the elastic-plastic boundary as the pressure is increased beyond P[latex]^*[/latex].

Question 15.1: A pressure vessel with a solid wall has an ri of 8 in., ro o...

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