김경종
29 KiB
TABLE 5.10 Smallest six frequencies (in Hz) of the 16-element mesh in Fig. 5.41(b) using a consistent mass matrix ^{\dagger}
| Mode number | 16-element model | 16 × 64 element model, $^{\ddagger}$ 3 × 3 Gauss integration | |
| 3 × 3 Gauss integration | 2 × 2 Gauss integration | ||
| 1 | 112.4 | 110.5 | 110.6 |
| 2 | 634.5 | 617.8 | 606.4 |
| 3 | 906.9 | 905.5 | 905.2 |
| 4 | 1548 | 958.4 $^{\S}$ | 1441 |
| 5 | 2654 | 1528 | 2345 |
| 6 | 2691 | 2602 | 2664 |
^{\dagger} The element consistent mass matrices are always integrated using 3 \times 3 Gauss integration.
^{\dagger} We include the results using a fine mesh (with 64 elements replacing each nine-node element of the 16-element mesh) for comparison purposes.
^{§} Spurious i.e., phantom mode.
behavior see, for example, W. K. Liu, Y.K. Hu, and T. Belytschko [A] and W. J. T. Daniel and T. Belytschko [A].
In the above discussion we focused attention on the evaluation of the element stiffness matrices. Considering the element force vectors, it is usually good practice to employ the same integration scheme and the same order of integration as for the stiffness matrices. For the evaluation of an element mass matrix, it should be recognized that for a lumped mass matrix only the volume of the element needs to be evaluated correctly and for the consistent mass matrix the order given in Table 5.9 is usually appropriate. However, special cases exist in which for the sufficiently accurate evaluation of a consistent mass matrix a higher-order integration may be necessary than in the calculation of the stiffness matrix.
EXAMPLE 5.42: Evaluate the stiffness and mass matrices and the body force vector of element 2 in Example 4.5 using Gauss numerical integration.
The expressions to be integrated have been derived in Example 4.5,
\mathbf {K} = E \int_ {0} ^ {8 0} \left(1 + \frac {x}{4 0}\right) ^ {2} \left[ \begin{array}{c} - \frac {1}{8 0} \\ \frac {1}{8 0} \end{array} \right] \left[ \begin{array}{c c} - \frac {1}{8 0} & \frac {1}{8 0} \end{array} \right] d x \tag {a}
\mathbf {M} = \rho \int_ {0} ^ {8 0} \left(1 + \frac {x}{4 0}\right) ^ {2} \left[ \begin{array}{c} 1 - \frac {x}{8 0} \\ \frac {x}{8 0} \end{array} \right] \left[ \left(1 - \frac {x}{8 0}\right) \frac {x}{8 0} \right] d x \tag {b}
\mathbf {R} _ {B} = \frac {1}{1 0} \int_ {0} ^ {8 0} \left(1 + \frac {x}{4 0}\right) ^ {2} \left[ \begin{array}{c} 1 - \frac {x}{8 0} \\ \frac {x}{8 0} \end{array} \right] d x \tag {c}
The expressions in (a) and (c) are integrated exactly with two-point integration, whereas the evaluation of the integral in (b) requires three-point integration. A higher-order integration is required in the evaluation of the mass matrix because this matrix is obtained from the displacement interpolation functions, whereas the stiffness matrix is calculated using derivatives of the displacement functions.
Using one-, two-, and three-point Gauss integration to evaluate (a), (b), and (c) we obtain One-point integration:
\mathbf {K} = \frac {1 2 E}{2 4 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \frac {\rho}{6} \left[ \begin{array}{c c} 4 8 0 & 4 8 0 \\ 4 8 0 & 4 8 0 \end{array} \right]; \quad \mathbf {R} _ {B} = \frac {1}{6} \left[ \begin{array}{c} 9 6 \\ 9 6 \end{array} \right]
Two-point integration:
\mathbf {K} = \frac {1 3 E}{2 4 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \frac {\rho}{6} \left[ \begin{array}{c c} 3 7 3. 3 & 3 4 6. 7 \\ 3 4 6. 7 & 1 0 1 3. 3 \end{array} \right]; \quad \mathbf {R} _ {B} = \frac {1}{6} \left[ \begin{array}{c} 7 2 \\ 1 3 6 \end{array} \right]
Three-point integration:
\mathbf {K} = \frac {1 3 E}{2 4 0} \left[ \begin{array}{c c} 1 & - 1 \\ - 1 & 1 \end{array} \right]; \quad \mathbf {M} = \frac {\rho}{6} \left[ \begin{array}{c c} 3 8 4 & 3 3 6 \\ 3 3 6 & 1 0 2 4 \end{array} \right]; \quad \mathbf {R} _ {B} = \frac {1}{6} \left[ \begin{array}{c} 7 2 \\ 1 3 6 \end{array} \right]
It is interesting to note that with too low an order of integration the total mass of the element and the total load to which the element is subjected are not taken fully into account.
Table 5.9 summarizes the results of an analysis for the appropriate integration orders in the evaluation of the stiffness matrices of two-dimensional elements. Of course, the information given in the table is also valuable in deducing appropriate orders of integration for the calculation of the matrices of other elements.
EXAMPLE 5.43: Discuss the required integration order for the evaluation of the MITC9 plate and isoparametric three-dimensional elements shown in Fig. E5.43.
Consider the plate element first. The integration in the r, s plane corresponds in essence to the evaluation of the nine-node element in Table 5.9. In general, this integration order of 3 \times 3 will also be effective when the element is used in distorted form.
The required integration order for the evaluation of the stiffness matrix of the three-dimensional solid element can also be deduced from the information given in Table 5.9. The displacements vary linearly in the r direction; hence, two-point integration is sufficient. In the
text_image
z y x t s r w β α
(a) MITC9 plate bending element
text_image
w v u x y t s r z
(b) Three-dimensional solid element
Figure E5.43 Elements considered in Example 5.43
t, s planes, i.e., at r equal to constants, the element displacements correspond to those of the eight-node element in Table 5.9. Hence, 2 \times 3 \times 3 Gauss integration is required to evaluate the element stiffness matrix exactly.
5.5.6 Reduced and Selective Integration
Table 5.9 gives the recommended Gauss numerical integration orders for two-dimensional isoparametric displacement-based elements, and the recommended orders for other elements can be deduced (see Example 5.43). With these integration orders (referred to as "full" integration), the element matrices of geometrically undistorted elements are evaluated exactly, whereas for geometrically distorted elements a sufficiently accurate approximation is obtained (unless the geometric distortions are very large, in which case a higher integration order is recommended).
However, in view of our discussion in Section 4.3.4, we recall that the displacement formulation of finite element analysis yields a strain energy smaller than the exact strain energy of the mathematical/mechanical model being considered, and physically, a displacement formulation results in overestimating the system stiffness. Therefore, we may expect that by not evaluating the displacement-based element stiffness matrices accurately in the numerical integration, better overall solution results can be obtained. This should be the case if the error in the numerical integration compensates appropriately for the overestimation of structural stiffness due to the finite element discretization. In other words, a reduction in the order of the numerical integration from the order that is required to evaluate the element stiffness matrices exactly (for geometrically undistorted elements) may be expected to lead to improved results. When such a reduction in the order of numerical integration is used, we refer to the procedure as reduced integration. For example, the use of 2 \times 2 Gauss integration (although not recommended for use in practice; see Section 5.5.5) for the nine-node isoparametric element stiffness matrix corresponds to a reduced integration. In addition to merely using a reduced integration order, selective integration may also be considered, in which case different strain terms are integrated with different orders of integration. In these cases of reduced and selective numerical integration the specific integration scheme should be regarded as an integral part of the element formulation.
The key question as to whether a reduced and/or selectively integrated element can be recommended for practical use is: Has the element formulation (using the specific integration procedure) been sufficiently tested and analyzed for its stability and convergence? If tractable, a mathematical stability and convergence analysis is of course most desirable.
A natural first step in such an analysis is to view the reduced and/or selectively integrated element as a mixed element (see D. Chapelle and K. J. Bathe [E]). (An example is the two-node mixed interpolated beam element in Section 5.4.1 further mentioned below.) Once the exact equivalence between the reduced and/or selectively integrated element and a mixed formulation has been identified, the second step is to analyze the mixed formulation for stability and convergence and in this way obtain a deep understanding of the element based on the reduced/selective integration.
Since there are many possibilities for assumptions in mixed formulations, it is natural to assume that there exists a mixed formulation that is equivalent to the reduced/selectively integrated element and seek that formulation for analysis purposes. However, the mere fact
that in general reduced/selectively integrated elements can be regarded as mixed formulations does not justify the use of reduced integration because of course not every mixed formulation represents a reliable and efficient finite element scheme. Rather, this equivalence (with the details to be identified in each specific case) only points toward an approach for the analysis of reduced/selectively integrated elements.
It also follows that once a complete equivalence has been identified, we can consider the reduced/selective integration as merely an effective way to accurately calculate the finite element matrices of the mixed formulation, and we adopt here this view and interpretation of reduced and selective integration.
A relatively simple example is the isoparametric two-node beam element based on one-point (r-direction) Gauss integration. In Example 4.30 and Section 5.4.1 we showed that this element is completely equivalent to the beam element obtained using the Hu-Washizu variational principle with linearly varying transverse displacement w, section rotation \beta , and a constant shear strain \gamma within each element. The stability and convergence of the element were considered in Section 4.5.7 where we showed that the ellipticity and inf-sup conditions are satisfied.
Let us consider the following additional example to emphasize these observations.
EXAMPLE 5.44: A simple triangular plate bending element can be derived using the isoparametric displacement formulation in Section 5.4.2 but integrating the stiffness matrix terms with one-point integration. This integration evaluates the stiffness matrix terms corresponding to bending exactly, whereas the terms corresponding to the transverse shear are integrated approximately. Hence, the element stiffness matrix is based on reduced integration (or we may also say selective integration because only the shear terms are not integrated exactly).
Derive a variational formulation and the stiffness matrix for this element.
The element and its variational formulation have been presented by J.-L. Batoz, K. J. Bathe, and L. W. Ho [A]. We note that the element is a natural development when we are aware of the success of the one-point integrated isoparametric beam element (see Example 4.30 and Sections 4.5.7 and 5.4.1). This beam element has a strong variational basis, the mathematical analysis ensures good convergence properties, and computationally the element is simple and effective.
For the development of the variational basis of this plate element, we note that the one-point integration implicitly assumes a constant transverse shear strain (as in the isoparametric one-point integrated two-node beam element). Referring to Example 4.30, we can therefore directly establish the variational indicator for the plate element as
\tilde {\Pi} _ {\mathrm{HR}} ^ {*} = \int_ {A} \left(\frac {1}{2} \kappa^ {T} \mathbf {C} _ {b} \kappa + \gamma^ {T} \mathbf {C} _ {s} \gamma^ {\mathrm{AS}} - \frac {1}{2} \gamma^ {\mathrm{AS} T} \mathbf {C} _ {s} \gamma^ {\mathrm{AS}}\right) d A - \int_ {A} w p d A + \text { boundary terms } \tag {a}
where \kappa, \mathbf{C}_b, \gamma, \mathbf{C}_s have been defined in (5.95) to (5.97) and \gamma^{\mathrm{AS}} contains the assumed transverse shear strains
\boldsymbol {\gamma} ^ {\mathrm{AS}} = \left[ \begin{array}{l} \gamma_ {x z} ^ {\mathrm{AS}} \\ \gamma_ {y z} ^ {\mathrm{AS}} \end{array} \right] = \text { constant }
The relation in (a) is a modified Hellinger-Reissner functional. Substituting the interpolations for w , \beta_x , and \beta_y into \kappa and \gamma , integrating over the element midsurface area A , and invoking the stationarity of \tilde{\Pi}_{\mathrm{HR}}^* with respect to the nodal point variables \hat{\mathbf{u}} ,
\hat {\mathbf {u}} = \left[ \begin{array}{c} w _ {1} \\ \theta_ {x} ^ {1} \\ \theta_ {y} ^ {1} \\ \cdot \\ \cdot \\ \dot {\theta} _ {y} ^ {3} \end{array} \right]
and \gamma^{\mathrm{AS}} , we obtain
\left[ \begin{array}{c c} \mathbf {K} _ {b} & \mathbf {G} ^ {T} \\ \mathbf {G} & - \mathbf {D} \end{array} \right] \left[ \begin{array}{c} \hat {\mathbf {u}} \\ \boldsymbol {\gamma} ^ {A S} \end{array} \right] = \left[ \begin{array}{c} \mathbf {R} \\ \mathbf {0} \end{array} \right]
where \mathbf{K}_b = \int_A\mathbf{B}_b^T\mathbf{C}_b\mathbf{B}_b dA
\mathbf {D} = \int_ {A} \mathbf {C} _ {s} d A = A \mathbf {C} _ {s}
\mathbf {G} = \mathbf {C} _ {s} \int_ {A} \mathbf {B} _ {s} d A
and B_{b} and B_{s} are strain-displacement matrices,
\kappa = \mathbf {B} _ {b} \hat {\mathbf {u}}
\gamma = \mathbf {B} _ {s} \hat {\mathbf {u}}
Using static condensation, we obtain the stiffness matrix of the element with respect to the nodal point variables only,
\mathbf {K} = \mathbf {K} _ {b} + \mathbf {G} ^ {T} \mathbf {D} ^ {- 1} \mathbf {G}
As we discussed in Section 5.4.2, the pure displacement-based isoparametric plate element (i.e., using full numerical integration for the bending and transverse shear terms in the displacement-based stiffness matrix) is much too stiff (displays the shear locking phenomenon). The presentation in Example 5.44 shows that the one-point integrated element has a variational basis quite analogous to the basis of the one-point integrated isoparametric beam element. However, whereas the beam element is reliable and effective, the plate element stiffness matrix in Example 5.44 has a spurious zero eigenvalue and hence the element is unreliable and should not be used in practice (as was pointed out by J.-L. Batoz, K. J. Bathe, and L. W. Ho [A]).
The important point of this example is that a variational basis of an element might well exist, but whether the element is useful and effective can of course be determined only by a deeper analysis of the formulation.
The equivalence between a certain isoparametric reduced or selectively integrated displacement-based element and a mixed formulation may also hold only for specific geometric element shapes and may also no longer be valid when nonisotropic material laws (or geometric nonlinearities) are introduced. An analysis of the effects of each of these conditions should then be performed.
5.5.7 Exercises
5.53. Evaluate the Newton-Cotes constants when the interpolating polynomial is of order 3, i.e., \psi(r) is a cubic.
5.54. Derive the sampling points and weights for three-point Gauss integration.
5.55. Show that 3 \times 3 Gauss numerical integration is sufficient to calculate the stiffness and mass matrices of a nine-node geometrically undistorted displacement-based element for axisymmetric analysis.
5.56. Show that 2 \times 2 Gauss integration of the stiffness matrix of the eight-node plane stress displacement-based square element results in the spurious zero energy mode shown. (Hint: You need to show that B\hat{u} = 0 for the given displacements.)
text_image
u 2u 2u Symmetric deformations
5.57. Consider the 9/3 u/p element and show that 3 \times 3 Gauss integration of a geometrically undistorted element gives the exact stiffness matrix. Also, show that 2 \times 2 Gauss integration is not adequate.
5.58. Identify the required integration order for full integration of the stiffness matrix of the six-node displacement-based triangular element when using the Gauss integration in Table 5.8.
natural_image
Simple geometric triangle diagram with six vertices and connecting lines (no text or labels)
Plane stress element
5.59. Consider the nine-node plane stress element shown. All nodal point displacements are fixed except that u_{1} is free. Calculate the displacement u_{1} due to the load P.
(a) Use analytical integration to evaluate the stiffness coefficient.
(b) Use 1 \times 1 , 2 \times 2 , and 3 \times 3 Gauss numerical integration to evaluate the stiffness coefficient. Compare your results.
text_image
E = 200,000 v = 0.30 6 10 u₁, P Thickness t = 1.0
5.60. Consider the evaluation of lumped mass matrices for the elements shown in Table 5.9. Determine suitable Gauss integration orders for the evaluation of these matrices.
5.61. Consider the plate bending element formulation in Example 5.29. Assume that the element stiffness matrix is evaluated with one-point Gauss integration. Show that this element has spurious zero energy modes. ^{12}
5.62. Consider the plate bending element formulation in Example 5.29 and assume that the bending strain energy is evaluated with 2 \times 2 Gauss integration and the shear strain energy is evaluated with one-point Gauss integration. Show that this element has spurious zero energy modes. ^{12}
5.6 COMPUTER PROGRAM IMPLEMENTATION OF ISOPARAMETRIC FINITE ELEMENTS
In Section 5.3 we discussed the isoparametric finite element formulation and gave the specific expressions needed in the calculation of four-node plane stress (or plane strain) elements (see Example 5.5). An important advantage of isoparametric element evaluations is the similarity between the calculations of different elements. For example, the calculation of three-dimensional elements is a relatively simple extension from the calculation of two-dimensional elements. Also, in one subroutine, elements with a variety of nodal point configurations can be calculated if an algorithm for selecting the appropriate interpolation functions is used (see Section 5.3).
The purpose of this section is to provide an actual computer program for the calculation of the stiffness matrix of four-node isoparametric elements. In essence, SUBROUTINE QUADS is the computer program implementation of the procedures presented in Example 5.5. In addition to plane stress and plane strain, axisymmetric conditions can be considered. It is believed that by showing the actual program implementation of the element, the relative ease of implementing isoparametric elements is best demonstrated. The input and output variables and the flow of the program are described by means of comment lines.
SUBROUTINE QUADS (NEL, ITYPE, NINT, THIC, YM, PR, XX, S, IOUT) QUA00001
C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
P R O G R A M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
T O CALCULATE ISOPARAMETRIC QUADRILATERAL ELEMENT STIFFNESS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . M A T R I X F O R A X I S Y M M E T R I C , P L A N E S T R E S S , A N D P L A N E S T R A I N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Condition S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - INPUT VARIABLES - - . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NEL = NUMBER OF ELEMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I TYPE = ELEMENT TYPE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E Q. 0 = AXISYMMETRIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . EQ. 1 = PLANE STRAIN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .NINT = GAUSS NUMERICAL INTEGRATION ORDER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .THIC = THICKNESS OF ELEMENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . YM = YOUNG'S MODULUS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . PR = POISSON'S RATIO . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .XX(2,4) = ELEMENT NODE COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S(8,8) = STORAGE FOR STIFFNESS MATRIX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .I OUT = UNIT NUMBER USED FOR OUTPUT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
C. -- OUTPUT -- QUA00024
C. S(8,8) = CALCULATED STIFFNESS MATRIX QUA00025
C. IMPLICIT DOUBLE PRECISION (A-H,O-Z) QUA00026
C. THIS PROGRAM IS USED IN SINGLE PRECISION ARITHMETIC ON CRAY QUA00030
C. EQUIPMENT AND DOUBLE PRECISION ARITHMETIC ON IBM MACHINES, QUA00031
C. ENGINEERING WORKSTATIONS AND PCS. DEACTIVATE ABOVE LINE FOR QUA00032
C. SINGLE PRECISION ARITHMETIC. QUA00033
C. DIMENSION D(4,4),B(4,8),XX(2,4),S(8,8),XG(4,4),WGT(4,4),DB(4) QUA00034
C. MATRIX XG STORES GAUSS - LEGENDRE SAMPLING POINTS QUA00035
C. DATA XG/ 0.D0, 0.D0, 0.D0, 0.D0, -.5773502691896D0, QUA00036
1 .5773502691896D0, 0.D0, 0.D0, -.7745966692415D0, 0.D0, QUA00037
2 .7745966692415D0, 0.D0, -.8611363115941D0, QUA00040
3 -.3399810435849D0, .3399810435849D0, .8611363115941D0 / QUA00041
C. MATRIX WGT STORES GAUSS - LEGENDRE WEIGHTING FACTORS QUA00042
C. DATA WGT / 2.D0, 0.D0, 0.D0, 0.D0, 1.D0, 1.D0, QUA00043
1 0.D0, 0.D0, .5555555555556D0, .8888888888889D0, QUA00044
2 .5555555555556D0, 0.D0, .3478548451375D0, .6521451548625D0, QUA00045
3 .6521451548625D0, .3478548451375D0 / QUA00046
C. OBTAIN STRESS - STRAIN LAW QUA00047
C. F=YM/(1.+PR) QUA00048
G=F*PR/(1.-2.*PR) QUA00049
H=F + G QUA00050
C. PLANE STRAIN ANALYSIS QUA00051
C. D(1,1)=H QUA00052
D(1,2)=G QUA00053
D(1,3)=0. QUA00054
D(2,1)=G QUA00055
D(2,2)=H QUA00056
D(2,3)=0. QUA00057
D(3,1)=0. QUA00058
D(3,2)=0. QUA00059
D(3,3)=F/2. QUA00060
IF (ITYPE.EQ.1) THEN QUA00061
THIC=1. QUA00062
GO TO 20 QUA00063
ENDIF QUA00064
ENDIF QUA00065
D(1,4)=G QUA00066
D(2,4)=G QUA00067
D(3,4)=0. QUA00068
D(4,1)=G QUA00069
D(4,2)=G QUA00070
D(4,3)=0. QUA00071
D(4,4)=H QUA00072
IF (ITYPE.EQ.0) GO TO 20 QUA00073
C. AXISYMMETRIC ANALYSIS QUA00074
C. D(1,4)=G QUA00075
D(2,4)=G QUA00076
D(3,4)=0. QUA00077
D(4,1)=G QUA00078
D(4,2)=G QUA00079
D(4,3)=0. QUA00080
D(4,4)=H QUA00081
C. FOR PLANE STRESS ANALYSIS CONDENSE STRESS-STRAIN MATRIX QUA00082
C. DO 10 I=1,3 QUA00083
A=D(I,4)/D(4,4) QUA00084
DO 10 J=I,3 QUA00085
D(I,J)=D(I,J) - D(4,J)*A QUA00086
10 D(J,I)=D(I,J) QUA00087
C.
C CALCULATE ELEMENT STIFFNESS
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P(1,1) = 0.25* SP
P(1,2) = -P(1,1)
P(1,3) = -0.25* SM
P(1,4) = -P(1,3)
C
C 2. WITH RESPECT TO S
C
P(2,1) = 0.25* RP
P(2,2) = 0.25* RM
P(2,3) = -P(2,2)
P(2,4) = -P(2,1)
C
C EVALUATE THE JACOBIAN MATRIX AT POINT (R,S)
C
10 DO 30 I=1,2
DO 30 J=1,2
DUM = 0.0
DO 20 K=1,4
20 DUM=DUM + P(I,K)*XX(J,K)
30 XJ(I,J)=DUM
C
C COMPUTE THE DETERMINANT OF THE JACOBIAN MATRIX AT POINT (R,S)
C
DET = XJ(1,1)* XJ(2,2) - XJ(2,1)* XJ(1,2)
IF (DET.GT.0.00000001) GO TO 40
WRITE (IOUT,2000) NEL
GO TO 800
C
C COMPUTE INVERSE OF THE JACOBIAN MATRIX
C
40 DUM=1./DET
XJI(1,1) = XJ(2,2)* DUM
XJI(1,2) = -XJ(1,2)* DUM
XJI(2,1) = -XJ(2,1)* DUM
XJI(2,2) = XJ(1,1)* DUM
C
C EVALUATE GLOBAL DERIVATIVE OPERATOR B
C
K2=0
DO 60 K=1,4
K2=K2 + 2
B(1,K2-1) = 0.
B(1,K2) = 0.
B(2,K2-1) = 0.
B(2,K2) = 0.
DO 50 I=1,2
B(1,K2-1) = B(1,K2-1) + XJI(1,I) * P(I,K)
50 B(2,K2) = B(2,K2) + XJI(2,I) * P(I,K)
B(3,K2) = B(1,K2-1)
60 B(3,K2-1) = B(2,K2)
C
C IN CASE OF PLANE STRAIN OR PLANE STRESS ANALYSIS DO NOT INCLUDE
C THE NORMAL STRAIN COMPONENT
C
IF (ITYPE.GT.0) GO TO 900
C
COMPUTE THE RADIUS AT POINT (R,S)
C
XBAR=0.0
DO 70 K=1,4
70 XBAR=XBAR + H(K)*XX(1,K)
C
EVALUATE THE HOOP STRAIN-DISPLACEMENT RELATION
C
IF (XBAR.GT.0.00000001) GO TO 90
C
FOR THE CASE OF ZERO RADIUS EQUATE RADIAL TO HOOP STRAIN
C
DO 80 K=1,8
80 B(4,K)=B(1,K)
GO TO 900
C
NON-ZERO RADIUS
C
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