# 11.10: Example of the Design against First Yield

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Safety of pressure vessels and piping systems is critical in design of offshore, chemical and nuclear installation. The simplest problem in this class of structures is a thick pipe loaded by an internal pressure \(p\). The tube is assumed to be infinitely long and the internal and external radii are denoted respectively by \(a\) and \(b\). In the cylindrical coordinate system \((r, \theta, z), \sigma_{zz} = 0\) for the open-ended short tube and \(\sigma_{rr} = \sigma_r\) and \(\sigma_{\theta\theta} = \sigma_{\theta}\) are the principal radial and circumferential stresses. The material is elastic up to the point of the first yield. The objective is to determine the location where the first yield occurs and the corresponding critical pressure \(p_y\).

The governing equation is derived by writing down three groups of equations:

Geometrical relation:

\[\epsilon_r = \frac{d}{dr} u, \quad \epsilon_{\theta} = \frac{u}{r} \]

where \(u\) is the radial component of the displacement vector, \(u = u_r\). The hoop component is zero because of axial symmetry.

Equilibrium:

\[\frac{d}{dr} \sigma_r + \frac{\sigma_r − \sigma_{\theta}}{r} = 0 \]

Elasticity law:

\[\sigma_r = \frac{E}{1 − \nu^2} (\epsilon_r + \nu\epsilon_{\theta}) \]

\[\sigma_{\theta} = \frac{E}{1 − \nu^2} (\epsilon_{\theta} + \nu\epsilon_r) \]

There are five equations for five unknowns, \(\sigma_r\), \(\sigma_{\theta}\), \(\epsilon_r\), \(\epsilon_{\theta}\) and \(u\). Solving the above system for \(u\), one gets

\[r^2 \frac{d^2}{dr^2} u + r \frac{d}{dr} u − u = 0 \]

The solution of this equation is

\[u(r) = C_1r + \frac{C_2}{r} \]

where \(C_1\) and \(C_2\) are integration constants to be determined from the boundary conditions. The stress and displacement boundary condition for this problem are

\[(T − \sigma_r) = 0 \quad \text{ or } \quad \delta u = 0 \]

In the case of pressure loading, the stress boundary condition applies:

\[\text{at } r = a \quad \sigma_r = −p_a \]

\[\text{at } r = b \quad \sigma_r = −p_b \]

The minus sign appears because the surface traction \(T\), which in our case is pressure loading, acts in the opposite direction to the unit normal vectors n, see Figure (12.15). In the present case of internal pressure, \(\sigma_r(r = a) = −p\) and \(\sigma_r(r = b) = 0\). The radial stress is calculated from Equations (**11.5.3**) and (**11.6.2-11.6.3**)

\[\sigma_r = \frac{E}{1 − \nu^2} \left[(1 + \nu)C_1 − (1 − \nu) \frac{C_2}{r^2} \right] \]

The integration constants can be easily calculated from two boundary conditions, and the final solution for the stresses is

\[\sigma_r(r) = \frac{a^2p}{b^2 − a^2} \left(1 − \frac{b^2}{r^2} \right) \]

\[\sigma_{\theta}(r) = \frac{a^2p}{b^2 − a^2} \left(1 + \frac{b^2}{r^2} \right) \]

Eliminating the term \((b/r)^2\) between the above two equations gives the straight line profile of the stresses, shown in Figure (\(\PageIndex{2}\)).

\[\sigma_r + \sigma_{\theta} = 2p \frac{1}{\left(\frac{b}{a}\right)^2 − 1} \]

It is seen that the stress profile is entirely in the second quadrant and the tube reaches yield at \(r = a\), for which the stresses are

\[\sigma_r = −p \]

\[\sigma_{\theta} = p \frac{b^2 + a^2}{b^2 − a^2} \]

In the case of the Tresca yield condition

\[|\sigma_{\theta} − \sigma_r| = \sigma_y \]

The dimensionless yield pressure is

\[\frac{p}{\sigma_y} = \frac{1}{2} \left[ 1 − \left(\frac{a}{b}\right)^2 \right] \]

The von Mises yield condition predicts

\[\frac{p}{\sigma_y} = \frac{1 − (a/b)^2}{\sqrt{3 + (a/b)^4}} \]

For example, if \(\frac{a}{b} = \frac{1}{2}\), the first yield pressure according to the von Mises yield condition is \(\frac{p_y}{\sigma_y} = \frac{3}{7}\) while the Tresca yield criterion predicts \(\frac{p_y}{\sigma_y} = \frac{3}{8}\). The difference between the above two cases is 14%.