# On the Optimal Design of Steel Shells with Technological Constraints
## 5 Numerical Examples  
A full cylindrical shell composed of rings of different thicknesses is simply supported and subjected to the uniform internal pressure \(p_{0}\), as in Figure 2. The following initial data as dimensionless values are given as ring length \(X_{1}\), then \(X_{2}\), and \(P_{s}=p_{0}/\sigma_{0}\). The values of optimal ring thicknesses \(E_{1}\) and \(E_{2}\) were determined according to the procedure given in Section 4.  
**Example 1**.: _In this example, the dimensioning of the shell is calculated with the given data \(X_{1}\) = 0.08, \(X_{2}\) = 0.16, and \(P_{s}\) = 0.035. In the applied numerical procedure, the process is interrupted with the criterion of achieved accuracy \(\varepsilon=0.0001\). Table 1 shows the last iteration in which \(\xi\) was varied in the interval (0.743; 0.742) with \(n=10\)._  
The step in that iteration was \(h_{3}=10^{-4}\) and the numerical procedure was completed when \(\varepsilon\) = 0.0001 was reached. For \(\zeta\) = 0.7422, the ring thicknesses \(E_{1}\) = 0.02646 and \(E_{2}\) = 0.0307 were calculated. The minimum volume is \(\overline{V}=0.009145\) where it is \(\overline{V}=\frac{V}{2\pi R^{3}}\). The displacement rates \(\dot{w}_{i}\) (\(i\) = 1, 2) and its slopes \(P_{i}\) (\(i\) = 1, 2) corresponding to the circular plastic hinges of the shell collapse mechanism were calculated (see Table 2).  
In this way, the kinematic part of the solution of this example is satisfied.  
Bearing in mind that dimensionless quantities were taken into account in the calculation, we assume that it is \(\sigma_{0}\) = 235 Mpa and the radius of the shell R = 2 m. The thickness of the ring near the support is \(e_{1}\) = 52.92 mm, the thickness of the central ring is \(e_{2}\) = 61.4 mm, and imposed limit load \(p_{0}\) = 8.225 Mpa.  
**Example 2**.: _The following examples show the results of the optimal design of two shells with rings of different lengths, subjected by the same limit load \(p_{0}=0.0345\sigma_{0}\). The data for the first shell is \(X_{1}\) = 0.085 and \(X_{2}\) = 0.17 and for the second \(X_{1}\) = 0.09 and \(X_{2}\) = 0.18. In the upper part of Table 3, we represent the calculated thicknesses of these shells \(E_{1}\) and \(E_{2}\) and case when \(X_{1}\) = 0 for the shells of constant thicknesses \(E\) = \(E_{2}\)._  
\begin{table}
\begin{tabular}{l l l l l l} \hline \(\xi\) & \(E_{2}\) & \(E_{1}\) & \(\xi\cdot E_{2}\cdot E_{2}\) & \(E_{1}\cdot E_{1}\) & **VOLUME** \\ \hline
0.7429 & 0.03073 & 0.02635 & 0.000701410 & 0.000694219 & 0.004566007 \\
0.7428 & 0.03072 & 0.02636 & 0.000701118 & 0.000695064 & 0.004566942 \\
0.7427 & 0.03071 & 0.02638 & 0.000700826 & 0.000695907 & 0.004567874 \\
0.7426 & 0.03071 & 0.02640 & 0.000700534 & 0.000696753 & 0.004568809 \\
0.7425 & 0.03071 & 0.02641 & 0.000700242 & 0.000697590 & 0.004569744 \\
0.7424 & 0.03071 & 0.02643 & 0.000699947 & 0.000698453 & 0.004570687 \\
0.7423 & 0.03070 & 0.02644 & 0.000699653 & 0.000699302 & 0.004571624 \\
0.7422 & 0.03070 & 0.02646 & 0.000699363 & 0.000700150 & 0.004572559 \\
0.7421 & 0.03069 & 0.02648 & 0.000699072 & 0.000700999 & 0.004573497 \\ \hline \end{tabular}
\end{table}
Table 1: Last iteration for \(\xi\) of the numerical procedure.  
**Example 3**.: _The examples of the shells with the same limit load \(P_{\rm s}=0.035\sigma_{0}\) and the same length \(\overline{L}=2X_{2}\), but different ring lengths, were calculated (See Table 3, lower part). The following data for \(X_{1}\): 0 (the shell of constant thickness); 0.7; 0.095 and 0.11 have been taken. Comparing these shells, we conclude that the shell with a constant thickness has a maximum volume, \(\overline{V}=0.012616\)._