For a given material and temperature, the uniaxial creep behavior follows Eq. 15.39(b).

$\varepsilon=\frac{\sigma}{E}+D_3 \sigma^\delta t^\phi$ (15.39b)

A thin-walled tubular pressure vessel of radius r and wall thickness b has closed ends and is made of this material. Develop an equation for the relative change in radius, Δr/r , as a function of time t and a constant pressure p in the vessel.

Question

For a given material and temperature, the uniaxial creep behavior follows Eq. 15.39(b).

[latex]\varepsilon=\frac{\sigma}{E}+D_3 \sigma^\delta t^\phi[/latex] (15.39b)

A thin-walled tubular pressure vessel of radius r and wall thickness b has closed ends and is made of this material. Develop an equation for the relative change in radius, Δr/r , as a function of time t and a constant pressure p in the vessel.

Accepted Answer

we first generalize the uniaxial stress–strain curve to an effective stress–strain curve:

[latex]\bar{\varepsilon}=\frac{\bar{\sigma}}{E}+D_3 \bar{\sigma}^\delta t^\phi, \quad \overline{\dot{\varepsilon}}=D_3 \phi \bar{\sigma}^\delta t^{\phi-1}[/latex]

The viscosity is thus both stress and time dependent:

[latex]\frac{1}{\eta}=\frac{\overline{\dot{\varepsilon}}}{\bar{\sigma}}=D_3 \phi \bar{\sigma}^{\delta-1} t^{\phi-1}[/latex]

This relationship can now be used with Eq. 15.64 to solve our particular problem, which involves stresses and strains as follows:

[latex]\begin{aligned}&\dot{\varepsilon}_x=\frac{1}{\eta}\left[\sigma_x-0.5\left(\sigma_y+\sigma_z ight) ight] \quad \quad (a) \&\dot{\varepsilon}_y=\frac{1}{\eta}\left[\sigma_y-0.5\left(\sigma_x+\sigma_z ight) ight] \quad \quad (b)\ &\dot{\varepsilon}_z=\frac{1}{\eta}\left[\sigma_z-0.5\left(\sigma_x+\sigma_y ight) ight]\quad \quad (c)\end{aligned}[/latex] (15.64)

[latex]\sigma_1=\frac{p r}{b}, \quad \sigma_2=\frac{p r}{2 b}, \quad \sigma_3 \approx 0, \quad \varepsilon_1=\frac{\Delta(2 \pi r)}{2 \pi r}=\frac{\Delta r}{r}[/latex]

Here, [latex]σ_{1} ext{ and } ε_{1}[/latex] are in the hoop direction, [latex]σ_{2}[/latex] is in the longitudinal direction, and these stresses are noted to be principal stresses.

Letting the (x, y, z) directions be the principal (1, 2, 3) directions, Eq. 15.64(a) gives

[latex]\dot{\varepsilon}_1=\frac{1}{\eta}\left[\sigma_1-0.5\left(\sigma_2+\sigma_3 ight) ight]=D_3 \phi \bar{\sigma}^{\delta-1} t^{\phi-1}\left[\frac{p r}{b}-0.5\left(\frac{p r}{2 b} ight) ight][/latex]

Also, [latex]\bar{\sigma}[/latex] is given by Eq. 15.67(a):

[latex]\begin{aligned}& \bar{\sigma}=\frac{1}{\sqrt{2}} \sqrt{\left(\sigma_1-\sigma_2 ight)^2+\left(\sigma_2-\sigma_3 ight)^2+\left(\sigma_3-\sigma_1 ight)^2} \quad \quad (a)\& \overline{\dot{\varepsilon}}=\frac{\sqrt{2}}{3} \sqrt{\left(\dot{\varepsilon}_1-\dot{\varepsilon}_2 ight)^2+\left(\dot{\varepsilon}_2-\dot{\varepsilon}_3 ight)^2+\left(\dot{\varepsilon}_3-\dot{\varepsilon}_1 ight)^2}\quad \quad (b)\end{aligned}[/latex] (15.67)

[latex]\bar{\sigma}=\frac{1}{\sqrt{2}} \sqrt{\left(\frac{p r}{b}-\frac{p r}{2 b} ight)^2+\left(\frac{p r}{2 b} ight)^2+\left(-\frac{p r}{b} ight)^2}=\frac{\sqrt{3} p r}{2 b}[/latex]

Substituting this [latex]\bar{\sigma}[/latex] into the expression for [latex]\dot{\varepsilon}_1[/latex] and simplifying gives

[latex]\dot{\varepsilon}_1=\frac{\sqrt{3}}{2} D_3 \phi t^{\phi-1}\left(\frac{\sqrt{3} p r}{2 b} ight)^\delta[/latex]

Since the pressure p is constant, we can integrate with respect to time to obtain the creep strain:

[latex]\varepsilon_{c 1}=\int_0^t \dot{\varepsilon}_1 d t=\frac{\sqrt{3}}{2} D_3 t^\phi\left(\frac{\sqrt{3} p r}{2 b} ight)^\delta[/latex]

We now need the elastic strain from Eq. 5.26. This is

[latex]\begin{aligned}\varepsilon_x & =\frac{1}{E}\left[\sigma_x-v\left(\sigma_y+\sigma_z ight) ight] \quad \quad (a)\ \varepsilon_y & =\frac{1}{E}\left[\sigma_y-v\left(\sigma_x+\sigma_z ight) ight] \quad \quad (b) \ \varepsilon_z & =\frac{1}{E}\left[\sigma_z -v\left(\sigma_x+\sigma_y ight) ight]\quad \quad (c)\end{aligned}[/latex] (5.26)

[latex]\varepsilon_{e 1}=\frac{1}{E}\left[\sigma_1-v\left(\sigma_2+\sigma_3 ight) ight]=\left(1-\frac{v}{2} ight)\left(\frac{p r}{b E} ight)[/latex]

Adding the elastic and creep strains finally gives the desired result:

[latex]\frac{\Delta r}{r}=\varepsilon_{e 1}+\varepsilon_{c 1}, \quad \frac{\Delta r}{r}=\left(1-\frac{ u}{2} ight)\left(\frac{p r}{b E} ight)+\frac{\sqrt{3}}{2} D_3 t^\phi\left(\frac{\sqrt{3} p r}{2 b} ight)^\delta[/latex]

Question 15.6: For a given material and temperature, the uniaxial creep beh...

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