# Application of reduced integration in linear shell problems

Kamal Hassan

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## To cite this version:

Kamal Hassan. Application of reduced integration in linear shell problems. [Research Report] IRIA-RR-362, IRIA. 1979, pp.15. [hal-04716614]

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APPLICATION OF
REDUCED INTEGRATION
IN LINEAR SHELL PROBLEMS

Kamal HASSAN

15p.

Rapport de Recherche N° 362

Septembre 1979
APPLICATION OF REDUCED INTEGRATION IN LINEAR SHELL PROBLEMS

Kamal Hassan

IRIA/LABORIA

(presented by Roland Glowinski)

Résumé :

Le but de ce rapport est de montrer une méthode pratique de détermination de l'ordre minimum d'intégration numérique qui est nécessaire pour la solution du problème linéaire de coques. On va utiliser le modèle de KOITER mais en fait ce que l'on propose s'applique également à la plupart des autres modèles. Pour approcher la solution on a utilisé deux types d'éléments finis : l'élément $P_5$- d'ARGYRIS et l'élément $P_3$- complet de HSIEH CLOUGH et TOCHER.

Abstract :

The aim of this paper is to give a practical determination of the minimal order of numerical integration necessary for the solution of linear shell problems. We shall use the model of KOITER, but in fact our methodology can be applied to most shell models. To approximate the solution we have used two finite elements : the $P_5$- ARGYRIS element and the complete $P_3$- HSIEH CLOUGH and TOCHER element.
# INTRODUCTION

As it is well known, a considerable amount of time is required by numerical integration in most finite element problems ; in shell problem particularity this time aspect becomes very important (see Table 5). Therefore we need numerical integration formulae which use as few points as possible preserving at the same time the rate of convergence and the degree of accuracy resulting from exact integration. The results in this direction can be summarized in the following statement [3-4] :

> If the finite element approximates the solution $u$ by a piece wise polynomial of degree $p$, and, if $m$ is the maximum order of differentiation occuring in the bilinear from $a(u,v)$. Then, the minimum order of numerical integration needed to approximate $a(.,.)$ could not be less than $2(p-m)$.

In this note, we shall show the influence of the order of numerical integration on the number of rigid movements of the shell ; we shall then use it as a criterion for the determination of the necessary order. Our analysis will be based on the "simple" example of a right circular cylinder, but the results can be applied to any general shell. We shall use the model of KOITER (refs. [1,2,6,7]), but in fact our methodology can be applied to most shell models. To approximate the solution we have used the two finite elements : the $P_5$-ARGYRIS element, and the complete $P_3$-HSIEH-CLOUGH-TOCHER element (abbreviated as HCT-C). (Refs. [4,5,6]).

We begin by a brief description of the geometry of the cylinder in sec. 1, then we give the variational formulation of the problem in sec. 2 In sec.3 we describe how to determine the necessary order of numerical integration; obtained results of numerical experiments are presented.
# 1- THE GEOMETRY OF THE CYLINDER

<figure><img src="image_4.png" /><figcaption>Figure 1</figcaption></figure>

Let $\Omega = ]0, 2\pi[ \times ]-H[ \subset \mathbb{R}^2$, with boundary $\Gamma$, where $\mathbb{R}^2$ is the usual euclidean plane. The middle surface of the cylinder $S$ is, by definition, the image of $\bar{\Omega} = \Omega \cup \Gamma$, by the mapping :

$$ \vec{\phi}: \Omega \subset \mathbb{R}^2 \longrightarrow \mathbb{R}^2 $$

$$ (x, y) \longrightarrow \vec{\phi}(x, y) = \cos x \ \vec{e}_1 + \sin x \ \vec{e}_2 + y \ \vec{e}_3 $$

Where $\mathbb{R}^3$ is the usual euclidean space, and where $\vec{e}_i$, $1 \le i \le 3$ is an orthonormal basis of the space. The local reference system of coordinates is :

$$ \vec{a}_1 = \frac{\partial \vec{\phi}}{\partial x}, \quad \vec{a}_2 = \frac{\partial \vec{\phi}}{\partial y}, \quad \vec{a}_3 = \frac{\vec{a}_1 \times \vec{a}_2}{|\vec{a}_1 \times \vec{a}_2|}, $$
Where $|·|$ is the euclidean norm in $\mathbb{R}^3$, and where $\times$ denotes the usual exterior product. The contravariant basis $a^i$ coincide with the covariant basis $\dot{a}^i$ in the case of our cylinder

The first fundamental form $(a_{\alpha\beta})$ of the cylinder is :

$$ (a_{\alpha\beta}) = \begin{pmatrix} R^2 & 0 \\ 0 & 1 \end{pmatrix} $$

Where R is the radius of the cylinder.

The area element $dS$ of the cylinder is given by :

$$ dS = \sqrt{a} dx dy, $$

where :

$$ a = \det(a_{\alpha\beta}) = a_{11} a_{22} - (a_{12})^2 $$

as a rule, greek letters $\alpha, \beta, \gamma, ...$, are used for indices 1,2 and latin letters $i,j,k,...$, are used for indices 1,2,3. For these indices we also use the convention of summation over repeated indices.

## 2 - VARIATIONAL FORMULATION OF THE PROBLEM

Let $\Omega = ]0, 2\pi[ \times ]0, 1[\Gamma, \vec{u}$ (the displacement vector of the point $\vec{\phi}(x,y)$ (see figure 1)) is given by :

$$ \vec{u} = u_i \cdot \vec{a}^i, \quad i = 1, 2, 3, $$

where $u_i$ is a mapping :

$$ u_i : (x,y) \in \Omega \longrightarrow u_i(x,y) \in \mathbb{R} $$

In addition let : $\vec{u} \in \vec{V} = [H^1(\Omega)]^2 \times H^2(\Omega)$, with the following boundary conditions :

$$ \vec{u}|_{\Gamma_0} = 0, \quad \partial_{\nu} u_3|_{\Gamma_0} = 0, $$
where :

e: is the thickness of the shell supposed to be constant,

E : is the Young modulus, and

ν : is the Poisson coefficient.

### 3-MATRIX SINGULARITY DUE TO NUMERICAL INTEGRATION

From the application of finite elements in linear problems we obtain a system of linear equations :

$$Kx = y.$$

In our case, we have a direct interpretation for the above symbols :

K : stiffness matrix,

x : general displacement vector which contains all nodal parameters,

y : general force vector.

Now we use numerical integration for the determination of the elements of K and y. It is known [8] that the real influence of numerical integration is felt in the evaluation of the matrix K, which, if we calculate its elements exactly, will be a regular (a well known property of conforming finite elements). Moreover, if we do not consider boundary conditions the matrix K has, at most six zero eigenvalues (see BATHE and WILSON for the case of plane stress analysis). This number corresponds to the six possible rigid movements of the shell [5,7] :

$$V = A + B \times \phi$$

→ →

where A and B are constant vectors in $\mathbb{R}^3$, x... denotes the exterior product in $\mathbb{R}^3$, φ the mapping defining the middle surface of the shell. Numerically, this means that if we have more than 6 "small" eigenvalues, our matrix is "almost" singular. In order to avoid this situation we suggest to calculate the eigenvalues of the elementary stiffness matrix using different schemes of integration, whose order is between 2p and 2(p-m) (see "Introduction"), and according to this "test" we can determine practically and "safely" a convenient numerical integration scheme. We give in table 2 the possible convenient schemes for ARGTRIS and HCT-C elements [10-11]. The schemes are symmetric in natural coordinates, so that if a point λ = (λ₁, λ₂, λ₃) occurs, so do all its permutations.
If the $\lambda_i$, $i = 1, 2, 3$ are distinct, there are six points; if two $\lambda_i$ coincide, there are three; the point $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$, if used in the scheme; is taken once. The name of the formulae $S-I-J$ are given in such a way that I gives the number of points used while I gives the maximum order of polynomial to be integrated exactly by the formula.

### Remark 3-1:

The reader can see that we have only given the "optimal" existing formulae, so that we have maximum precision with minimum number of integrating points using only positive weights.

<table><caption>**Table 2**

**NUMERICAL INTEGRATION FORMULAE**</caption><thead><tr><th rowspan="2">Formula</th><th colspan="2">Natural Coordinates (λ<sub>1</sub>, λ<sub>2</sub>, 1-λ<sub>1</sub>-λ<sub>2</sub>)</th><th rowspan="2">Weighting coefficients</th><th rowspan="2">Multi-plicity</th></tr><tr><th>λ<sub>1</sub></th><th>λ<sub>2</sub></th></tr></thead><tbody><tr><td>S-3-2</td><td>0,66666 66666 666667</td><td>0,16666 66666 666667</td><td>0,33333 33333 333333</td><td>3</td></tr><tr><td rowspan="2">S-6-4</td><td>0,81624 75729 804585</td><td>0,09157 62135 0977073</td><td>0,10995 17436 553218</td><td>3</td></tr><tr><td>0,10810 30181 680702</td><td>0,44594 84909 1596490</td><td>0,22338 15896 780115</td><td>3</td></tr><tr><td rowspan="3">S-7-5</td><td>0,33333 33333 333333</td><td>0,33333 33333 333333</td><td>0,22500 00000 00000</td><td>1</td></tr><tr><td>0,79742 69853 53087</td><td>0,10128 65073 23456</td><td>0,12593 91805 44927</td><td>3</td></tr><tr><td>0,05971 53717 89770</td><td>0,47014 20641 05115</td><td>0,13239 41527 89506</td><td>3</td></tr><tr><td rowspan="3">S-12-6</td><td>0,87382 19710 169965</td><td>0,06308 90144 9150177</td><td>0,05084 49063 702055</td><td>3</td></tr><tr><td>0,50142 65090 581342</td><td>0,24928 67451 709329</td><td>0,11678 62757 263407</td><td>3</td></tr><tr><td>0,63650 24991 213939</td><td>0,31035 24510 3377294</td><td>0,08285 10756 183929</td><td>6</td></tr><tr><td rowspan="5">S-16-8</td><td>0,33333 33333 333333</td><td>0,33333 33333 333333</td><td>0,14431 56076 777862</td><td>1</td></tr><tr><td>0,65886 13844 964797</td><td>0,17056 93077 517601</td><td>0,10321 73705 347184</td><td>3</td></tr><tr><td>0,89890 55433 659397</td><td>0,05054 72283 1703103</td><td>0,03245 84976 2319813</td><td>3</td></tr><tr><td>0,08141 48234 1455473</td><td>0,45929 25882 927229</td><td>0,09509 16342 67285</td><td>3</td></tr><tr><td>0,00839 47774 09957211</td><td>0,72849 23929 554041</td><td>0,02723-03141 744349</td><td>6</td></tr><tr><td rowspan="6">S-19-9</td><td>0,33333 33333 333333</td><td>0,33333 33333 333333</td><td>0,09713 57962 82799</td><td>1</td></tr><tr><td>0,02063 49616 02524</td><td>0,48968 25191 98738</td><td>0,03133 47002 27139</td><td>3</td></tr><tr><td>0,12582 08170 14126</td><td>0,43708 95914 92937</td><td>0,07782 75410 04774</td><td>3</td></tr><tr><td>0,62359 29237 61934</td><td>0,18820 35356 19033</td><td>0,07964 77389 27210</td><td>3</td></tr><tr><td>0,91054 09732 11094</td><td>0,04472 95133 94453</td><td>0,02557 76756 58698</td><td>3</td></tr><tr><td>0,03683 94120 54736</td><td>0,22196 29891 60766</td><td>0,04328 35393 77289</td><td>6</td></tr><tr><td rowspan="6">S-25-10</td><td>0,33333 33333 333333</td><td>0,33333 33333 333333</td><td>0,08174 33291 46286</td><td>1</td></tr><tr><td>0,71567 77978 36872</td><td>0,14216 11010 56564</td><td>0,04595 73636 04745</td><td>3</td></tr><tr><td>0,93588 92535 66112</td><td>0,03205 53732 16944</td><td>0,01335 29688 13150</td><td>3</td></tr><tr><td>0,14813 28857 83821</td><td>0,32181 29952 88835</td><td>0,06390 49063 96424</td><td>6</td></tr><tr><td>0,02961 98894 88730</td><td>0,36914 67818 27811</td><td>0,03418 46481 62959</td><td>6</td></tr><tr><td>0,02836 76653 39938</td><td>0,16370 17337 37182</td><td>0,02529 77577 07288</td><td>6</td></tr></tbody></table>
## 4. NUMERICAL RESULTS

In table 3, we give the number of "small" eigenvalues obtained by a QR algorithm for the elementary stiffness matrix. As it is clear from this table the scheme S-12-6 is not suitable for the ARGYRIS element; of course this does not imply that if one uses this scheme we will not obtain approximate and even sometimes "good" results (see example 2).

<table><thead><tr><th rowspan="2">Formula</th><th rowspan="2"></th><th colspan="2">N° of small eigenvalues</th></tr><tr><th>≤ λ<sub>max</sub> × 10<sup>-7</sup></th><th>≤ λ<sub>max</sub> × 10<sup>-8</sup></th></tr></thead><tbody><tr><td rowspan="4">HCT-C</td><td>S-3-2</td><td>6</td><td>5</td></tr><tr><td>S-6-4</td><td>6</td><td>4</td></tr><tr><td>S-7-5</td><td>6</td><td>4</td></tr><tr><td>S-12-6</td><td>6</td><td>4</td></tr><tr><td rowspan="3">ARGYRIS</td><td>S-12-6</td><td>15</td><td>9</td></tr><tr><td>S-16-8</td><td>10</td><td>6</td></tr><tr><td>S-25-10</td><td>10</td><td>6</td></tr></tbody></table>

Number of small eigenvalues corresponding to the following data :

$$E = 2 \times 10^{11} N/m^2$$

$$e = 0,01 m$$

$$R = 1 m$$

$$v = 0,3$$

The triangle used is the right isoscles triangle of side lengthes equal 1(λ<sub>max</sub> is the maximum obtained eigenvalue).

### Table 3

Before giving the examples we make two important remarks :

### Remark 4.1

We have used a general program (refs.[12-13-14] which does not take in consideration the particularity of the cylinder especially its simple geometry ; therefore the time calculation have a generale sense.
### Remark 4.2.

All calculations of the HCT-C element were done on the CII-IRIS 80, the unit of computational time of which is in minutes, while that of ARGYRIS were made by IBM-370-168 the unit of computational time being in seconds.

### Examples:

Two examples are given to illustrate the effect of the numerical integration order on the final solution. We have used two examples for which the exact solution is known.

#### Example 1 (see [7]) using the notation of sec. 2:

Given:

$$f^1 = 0,$$

$$f^2 = \frac{4Eev}{(1-v^2)R} (H^2 - y^2)y,$$

$$f^3 = \frac{Ee}{12R^2(1-v^2)} \left\{ \left( 12 + \frac{e^2}{R^2} \right) (H^2 - y^2)^2 + 8 v e^2 (H^2 - 3y^2) + 24 e^2 R^2 \right\}.$$

Where $H$ is constant (see figure 1).

THEN, the exact solution is:

$$u_1 = u_2 = 0,$$

$$u_3 = (H^2 - y^2)^2.$$

For the ARGYRIS element, we almost recover the exact solution using S-16-8 and S-25-10 formula (see table 2). But S-12-6 gives us (as it is expected from our eigenvalues analysis) poor results (in the middle of the domain $\Omega$ we obtain as relative errors $E_r(u_3) = 0,003$ and $E_r(u_3,yy) = 0,014$. Moreover, we obtain worse results in the boundary; for example at $x = \pi$, $y = H$ relative error $E_r(u_3) = 0,084$, these results corresponds to a number of elements equal to 8, $H = \frac{1}{2}$ (see figure 1)).

For the HCT-C element see table 4 and figure 2.
<table><thead><tr><th>N° of<br>elements</th><th>S-3-2</th><th>S-6-4</th><th>S-7-5</th><th>S-12-6</th></tr></thead><tbody><tr><td>8</td><td>2,34 × 10<sup>-2</sup></td><td>2,36 × 10<sup>-2</sup></td><td>2,40 × 10<sup>-2</sup></td><td>2,39 × 10<sup>-2</sup></td></tr><tr><td>16</td><td>5,02 × 10<sup>-3</sup></td><td>3,46 × 10<sup>-3</sup></td><td>3,46 × 10<sup>-3</sup></td><td>3,45 × 10<sup>-3</sup></td></tr><tr><td>32</td><td>1,05 × 10<sup>-3</sup></td><td>2,39 × 10<sup>-4</sup></td><td>2,53 × 10<sup>-4</sup></td><td>2,53 × 10<sup>-4</sup></td></tr><tr><td>72</td><td>4,45 × 10<sup>-4</sup></td><td>2,33 × 10<sup>-4</sup></td><td>2,33 × 10<sup>-4</sup></td><td>2,32 × 10<sup>-4</sup></td></tr></tbody></table>

(Example 1, HTC-C)

Relative error : $E_p(u) = \left| \frac{u_3 - u_{3h}}{u_3} \right|$ at the mid-point of Ω
Table 4.

<figure><img src="image_5.png" /><figcaption>Figure 2.</figcaption></figure>

Example 1 HTC-C

THE INFLUENCE OF DIFFERENT SCHEMES OF NUMERICAL
INTEGRATION.
<table><thead><tr><th rowspan="2">N° of<br>elements</th><th colspan="2">S-12-6</th><th colspan="2">S-16-8</th><th colspan="2">S-25-10</th></tr><tr><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th></tr></thead><tbody><tr><td>16</td><td>1,2 × 10<sup>-3</sup></td><td>28,07</td><td>6,9 × 10<sup>-5</sup></td><td>35,90</td><td>1,2 × 10<sup>-6</sup></td><td>47,62</td></tr><tr><td>32</td><td>7,3 × 10<sup>-5</sup></td><td>58,71</td><td>6,7 × 10<sup>-5</sup></td><td>67,91</td><td>6,7 × 10<sup>-5</sup></td><td>92,71</td></tr><tr><td>42</td><td>3,7 × 10<sup>-6</sup></td><td>95,02</td><td>2,47 × 10<sup>-6</sup></td><td>106,97</td><td>2,48 × 10<sup>-6</sup></td><td>134,14</td></tr></tbody></table>

(Example 2, $H = \frac{1}{2}$, ARGYRIS element)

$$ \text{Relative error } E_r(u)_- = \left| \frac{u_3 - u_{3h}}{u_3} \right| $$
at (0; 0,25)

<table><caption>Table 5</caption><thead><tr><th rowspan="2">N° of<br>elements</th><th colspan="2">S-12-6</th><th colspan="2">S-16-8</th><th colspan="2">S-25-10</th></tr><tr><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th><th>E<sub>r</sub>(u)</th><th>Time<br>(s)</th></tr></thead><tbody><tr><td>16</td><td>1,8 × 10<sup>-2</sup></td><td>28,07</td><td>9,1 × 10<sup>-3</sup></td><td>35,90</td><td>9,2 × 10<sup>-3</sup></td><td>47,62</td></tr><tr><td>32</td><td>2,4 × 10<sup>-3</sup></td><td>58,71</td><td>2,4 × 10<sup>-3</sup></td><td>67,91</td><td>2,4 × 10<sup>-3</sup></td><td>92,71</td></tr><tr><td>48</td><td>9,8 × 10<sup>-4</sup></td><td>95,02</td><td>9,8 × 10<sup>-4</sup></td><td>106,97</td><td>9,8 × 10<sup>-4</sup></td><td>134,14</td></tr></tbody></table>

(Example 2, $H = \frac{1}{2}$, ARGYRIS element)

$$ \text{Relative error } E_r(u) = \left| \frac{u_{3,yy} - u_{3,yyh}}{u_{3,yy}} \right| $$
at (0; 0,25)

<caption>Table 6</caption>
<table><thead><tr><th rowspan="2">N° of<br>elements</th><th colspan="2">S-3-2</th><th colspan="2">S-6-4</th><th colspan="2">S-7-5</th><th colspan="2">S-12-6</th></tr><tr><th>E<sub>r</sub>(u<sub>3</sub>)</th><th>Time<br>(s)</th><th>E<sub>r</sub>(u<sub>3</sub>)</th><th>Time<br>(M)</th><th>E<sub>r</sub>(u<sub>3</sub>)</th><th>Time<br>(M)</th><th>E<sub>r</sub>(u)</th><th>Time<br>(M)</th></tr></thead><tbody><tr><td>16</td><td>1,4 × 10<sup>-2</sup></td><td>3,93</td><td>4,2 × 10<sup>-3</sup></td><td>5,67</td><td>3,7 × 10<sup>-3</sup></td><td>6,23</td><td>4,8 × 10<sup>-3</sup></td><td>9,25</td></tr><tr><td>32</td><td>1,4 × 10<sup>-3</sup></td><td>6,80</td><td>6,2 × 10<sup>-4</sup></td><td>10,45</td><td>6,6 × 10<sup>-4</sup></td><td>11,59</td><td>5,2 × 10<sup>-4</sup></td><td>17,32</td></tr><tr><td>48</td><td>1,4 × 10<sup>-3</sup></td><td>8,48</td><td>1,2 × 10<sup>-3</sup></td><td>14,6</td><td>1,2 × 10<sup>-3</sup></td><td>16,11</td><td>1,2 × 10<sup>-3</sup></td><td>25,31</td></tr></tbody></table>

(Example 2, $H = \frac{1}{2}$, HCT-C element)

$$ \text{Relative error } E_r(u) = \left| \frac{u_3 - u_{3h}}{u_3} \right| \\ \text{at } (0; 0; 25) $$

<table><caption>Table 7</caption><thead><tr><th rowspan="2">N° of<br>elements</th><th colspan="2">S-3-2</th><th colspan="2">S-6-4</th><th colspan="2">S-7-5</th><th colspan="2">S-12-6</th></tr><tr><th>E<sub>r</sub>(u)</th><th>E<sub>r</sub>(u<sub>y</sub>)</th><th>E<sub>r</sub>(u)</th><th>E<sub>r</sub>(u<sub>y</sub>)</th><th>E<sub>r</sub>(u)</th><th>E<sub>r</sub>(u<sub>y</sub>)</th><th>E<sub>r</sub>(u)</th><th>E<sub>r</sub>(u<sub>y</sub>)</th></tr></thead><tbody><tr><td>48</td><td>4,7 × 10<sup>-3</sup></td><td>3,3 × 10<sup>-3</sup></td><td>3,7 × 10<sup>-3</sup></td><td>5,7 × 10<sup>-3</sup></td><td>3,7 × 10<sup>-3</sup></td><td>5,7 × 10<sup>-3</sup></td><td>3,7 × 10<sup>-3</sup></td><td>5,7 × 10<sup>-3</sup></td></tr></tbody></table>

(Example 2, $H = \frac{1}{2}$, HCT-C element)

$$ \text{Relative error } E_r(u) = \left| \frac{u_3 - u_{3h}}{u_3} \right| \\ E_r(u_y) = \left| \frac{u_{3,y} - u_{3,h}}{u_{3,y}} \right| \\ \text{at } (0; \frac{1}{24}) $$

<caption>Table 8</caption>
### Example -2

Given :

$$f = 0,$$

$$f = - \frac{\pi E e v}{HR(1-\nu^2)} \sin\left(\frac{2\pi}{H} y\right),$$

$$f = - \frac{Ee}{1-\nu^2} \left\{ \frac{1}{R^2} \left( i + \frac{e^2}{12R^2} \right) \sin^2\left(\frac{\pi}{H} y\right) - \frac{\pi^2 e^2}{3H^2} \left[ \frac{\nu}{R^2} + \frac{2\pi^2}{H^2} \right] \cos\left(\frac{2\pi}{H} y \right) \right\}.$$

Then the exact solution is :

$$\begin{aligned} u_1 &= u_2 = 0, \\ u_3 &= \sin^2\left(\frac{\pi}{H} y\right). \end{aligned}$$

See tables 5,6 for the results of ARGYRIS element, and tables 7,8 for HCT-C element.

## CONCLUSION.

For the choice of a convenient numerical integration formula, it would be sufficient to focus the study of the element stiffness matrix of a typical triangle. A safe decision as to the feasibility of the formula should be based on the number of "small" eigenvalues of the above matrix.

## ACKNOWLEDGEMENT

The support of this work by the Electricité de France- Chatou-, is gratefully acknowledgement. I also would like to think Professors P.G. CIARLET and R. GLOWINSKI and Doctor J.M. BOISSERIE for their helpful suggestions.
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