김경종
411 lines
32 KiB
Markdown
411 lines
32 KiB
Markdown
<!-- source-page: 301 -->
|
||
|
||
$$
|
||
[ \mathbf {E} ] = G \left[ \begin{array}{c c c c c c} 4 / 3 & - 2 / 3 & - 2 / 3 & 0 & 0 & 0 \\ - 2 / 3 & 4 / 3 & - 2 / 3 & 0 & 0 & 0 \\ - 2 / 3 & - 2 / 3 & 4 / 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array} \right] + B \left[ \begin{array}{c c c c c c} 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] \tag {9.4-7}
|
||
$$
|
||
|
||
or, abbreviated, $[E] = G[E_{G}] + B[E_{B}]$ . The element stiffness matrix (Eq. 4.1-5) becomes
|
||
|
||
$$
|
||
[ \mathbf {k} ] = G \int_ {V _ {e}} [ \mathbf {B} ] ^ {T} [ \mathbf {E} _ {G} ] [ \mathbf {B} ] d V + B \int_ {V _ {e}} [ \mathbf {B} ] ^ {T} [ \mathbf {E} _ {B} ] [ \mathbf {B} ] d V \tag {9.4-8}
|
||
$$
|
||
|
||
Therefore, structural equations have the form
|
||
|
||
$$
|
||
(G [ \mathbf {K} _ {G} ] + B [ \mathbf {K} _ {B} ]) \{\mathbf {D} \} = \{\mathbf {R} \} \tag {9.4-9}
|
||
$$
|
||
|
||
As $\nu$ approaches 0.5, bulk modulus $B$ approaches infinity. Therefore, $B[\mathbf{K}_B]$ acts as a penalty matrix that enforces the constraint of incompressibility. As $\nu$ approaches 0.5, numerical trouble becomes more likely, and finally the mesh "locks"—unless $[\mathbf{K}_B]$ is singular.
|
||
|
||
Remarks. In the notation of Eq. 9.3-3, Eqs. 9.4-4 and 9.4-9 correspond to homogeneous constraints, $\{Q\} = \{0\}$ . Therefore, Eqs. 9.4-4 and 9.4-9 arise naturally from minimization of a potential $\Pi_{p}$ that can be stated in the form
|
||
|
||
$$
|
||
\Pi_ {p} = \beta \int_ {V} (F + \alpha H) d V + P \tag {9.4-10}
|
||
$$
|
||
|
||
Here F and H are proportional to strain energy densities, $\beta$ is a common factor of F and H, $\alpha$ is the penalty number, and P represents work done by loads of all types (P need not be detailed here). There is no unique way to choose $\beta$ , but if possible it should be chosen in such a way that $\alpha$ is dimensionless. A dimensionless $\alpha$ makes it easier to select a numerical value of $\alpha$ such that ill-conditioning is avoided (see guidelines at the end of this section).
|
||
|
||
For the Mindlin beam, the correspondence between Eqs. 9.4-2 and 9.4-10 can be as follows. Replace dV by dx in Eq. 9.4-10, and let
|
||
|
||
$$
|
||
F = \frac {E}{2} \theta_ {, x} ^ {2} \quad H = \frac {E}{2 L _ {T} ^ {2}} (w _ {, x} - \theta) ^ {2} \tag {9.4-11}
|
||
$$
|
||
|
||
$$
|
||
\beta = \frac {b t ^ {3}}{1 2} \alpha = \frac {1 0 L _ {T} ^ {2} G}{t ^ {2} E} = \frac {5}{1 + \nu} \left(\frac {L _ {T}}{t}\right) ^ {2}
|
||
$$
|
||
|
||
where t is the beam depth (thickness) and $L_{T} = \Sigma L_{i}$ is the total length of the beam after elements of length $L_{i}$ have been assembled. $^{3}$ We know that beam deflections depend more strongly on $L_{T}$ and t than on the beam width b. Accordingly, b is excluded from F, H, and $\alpha$ , so that the effect of the $L_{T}/t$ ratio is apparent.
|
||
|
||
<!-- source-page: 302 -->
|
||
|
||
Since the factor $5 / (1 + \nu)$ varies little for valid choices of $\nu$ , we see that $\alpha$ depends strongly on $(L_T / t)^2$ . When $(L_T / t)^2$ is large, the penalty number $\alpha$ is large. (Note that when $L_T / t$ approaches infinity so does $L_i / t$ , so the foregoing arguments could be restated using element length rather than $L_T$ .)
|
||
|
||
The incompressible case, Eq. 9.4-9, can be obtained from Eq. 9.4-10 if
|
||
|
||
$$
|
||
F = \frac {E}{4 (1 + \nu)} \{\boldsymbol {\epsilon} \} ^ {T} \left[ \mathrm{E} _ {G} \right] \{\boldsymbol {\epsilon} \} \quad H = \frac {E}{2} \{\boldsymbol {\epsilon} \} ^ {T} \left[ \mathrm{E} _ {B} \right] \{\boldsymbol {\epsilon} \} \tag {9.4-12}
|
||
$$
|
||
|
||
$$
|
||
\beta = 1 \quad \alpha = \frac {1}{3 (1 - 2 \nu)}
|
||
$$
|
||
|
||
The penalty number $\alpha$ becomes large as $\nu$ approaches 0.5.
|
||
|
||
Constraints and Quadrature Points. We wish to show that the number of penalty function constraints is proportional to the number of sampling points used to integrate element matrices.
|
||
|
||
Numerical integration of stiffness matrices (such as $[K_{b}]$ , $[K_{s}]$ , $[K_{G}]$ , and $[K_{B}]$ in Eqs. 9.4-4 and 9.4-9) corresponds to evaluation of energy $\Pi_{p}$ by numerical integration. Thus Eq. 9.4-10 can be written as
|
||
|
||
$$
|
||
\Pi_ {p} \approx \beta \sum_ {I = 1} ^ {\text {numel}} \left[ \sum_ {i = 1} ^ {n} \left(F _ {I} J _ {I}\right) _ {i} W _ {i} + \alpha \sum_ {j = 1} ^ {m} \left(H _ {I} J _ {I}\right) _ {j} T _ {j} \right] + P \tag {9.4-13}
|
||
$$
|
||
|
||
where $(F_{I}J_{I})_{i}$ and $(H_{I}J_{I})_{j}$ are values of functions F and H times the Jacobian J, evaluated at the ith or jth sampling point, and $W_{i}$ and $T_{j}$ are positive weights (or weight products) appropriate to the integration rule. If n = m and sampling points i and j are the same, the integration scheme is uniform; otherwise it is selective. The integration scheme is called full if enough sampling points are used to provide exact integration of all stiffness coefficients of an undistorted element (e.g., a rectangular element). If fewer sampling points are used—that is, if either n or m is reduced—the integration scheme is called reduced and $\Pi_{p}$ is said to be under-integrated. In what follows we assume that kinematic modes of deformation are either impossible or are suppressed by boundary conditions (kinematic modes are discussed in Section 6.12).
|
||
|
||
Now consider the effect of letting $\alpha$ become very large in Eq. 9.4-10. If the correct $\Pi_{\rho}$ is to be closely approximated, a large $\alpha$ must be associated with a zero or near-zero value of the second summation in Eq. 9.4-13. Since $(J_{I})_{j} > 0$ and $T_{j} > 0$ , the desired condition is that $(H_{I})_{j} = 0$ . From equations such as Eqs. 9.4-11 and 9.4-12, we see that $H = 0$ implies satisfaction of the constraint (zero shear strain and zero volume change in the respective cases). Thus, in these examples, each integration point used to evaluate the penalty matrix imposes a constraint, and the total number of constraints in the structure is the number of elements times the number of penalty integration points per element.
|
||
|
||
The preceding discussion suggests that locking difficulties may be avoided by applying reduced integration to terms that yield a penalty matrix. Thus, for example, we would use a one-point rule to evaluate $U_{s}$ of Eq. 9.4-2b when the shape functions of Eq. 9.4-3 are used (for which two-point integration would be exact). Further discussion and other examples appear in Section 9.5.
|
||
|
||
In the foregoing examples, material property matrices used in generating the
|
||
|
||
<!-- source-page: 303 -->
|
||
|
||
penalty matrix have rank one (E in Eq. 9.4-11, $[E_{B}]$ in Eq. 9.4-12). In other problems, the analogous matrices may have rank greater than one. Then the number of penalty constraints per element may be greater than the number of integration points [9.8]. This happens in certain plate-bending formulations, where an integration point constrains two transverse shear strains, namely $\gamma_{yz}$ and $\gamma_{zx}$ (see Section 11.3). In some problems it is conceivable that the number of constraints will exceed the number of d.o.f. This circumstance implies that some of the constraints are redundant.
|
||
|
||
Guideline for Choice of $\alpha$ . If computer words carry approximately p decimal digits, experience has shown that $\alpha$ should not exceed $10^{p/2}$ if ill-conditioning and numerical difficulty are to be avoided. If this guideline is followed, coefficients of [K] in Eq. 9.3-3 influence the latter p/2 digits in computer words used to store the complete matrix $[K + C^{T}\alpha C]$ . Typically $10^{p/2}$ is $10^{3}$ to $10^{4}$ in single precision and $10^{6}$ to $10^{7}$ in double precision. If material properties yield a larger value of $\alpha$ , it is best to lower $L_{T}/t$ (Eq. 9.4-11) or $\nu$ (Eq. 9.4-12) artificially so that the guideline is satisfied.
|
||
|
||
The foregoing choice of $\alpha$ is made after one has chosen an integration rule that avoids locking of the mesh. Even without locking, the penalty matrix enforces constraints, and it is to avoid numerical difficulty associated with these remaining constraints that one takes care in the choice of $\alpha$ .
|
||
|
||
# 9.5 CONSTRAINT COUNTING
|
||
|
||
We seek a guideline for choosing a suitable numerical integration formula in problems where penalty constraints arise naturally. Specifically, if the number of constraints is proportional to the number of sampling points used to integrate the penalty matrix, how many points per element should be used?
|
||
|
||
In what follows we continue to assume that all weight factors in quadrature rules are positive. Otherwise, it is conceivable that terms of a summation will cancel one another. This would confuse the counting rule.
|
||
|
||
Mesh Locking. Let the beam in Fig. 9.5-1a be built of the shear-flexible beam elements of Fig. 9.4-1. Support conditions suppress w and $\theta$ d.o.f. at the fixed
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
L
|
||
L
|
||
L
|
||
t
|
||
P
|
||
①
|
||
②
|
||
Nel
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
Nes
|
||
N2
|
||
es
|
||
Nes
|
||
</details>
|
||
|
||
(b)
|
||
Figure 9.5-1. (a) Cantilever beam built of $N_{el}$ elements. (b) An $N_{es}$ by $N_{es}$ mesh of bilinear elements. The case $N_{es} = 4$ is shown.
|
||
|
||
<!-- source-page: 304 -->
|
||
|
||
end, leaving $2N_{e\ell}$ active d.o.f. in $\{D\}$ . Now imagine that $[K_{s}]$ of Eq. 9.4-4 is generated by use of two sampling points per element (thus, $[K_{s}]$ is integrated exactly). Then, if the beam is slender, there are $2N_{e\ell}$ penalty constraints. All d.o.f. $\{D\}$ are now used to satisfy the constraint $w_{,x} - \theta \approx 0$ , and the computed deflection of load P is nearly zero. This conclusion is unchanged by changing the supports. With no supports, rigid-body motion of the “locked” beam becomes possible.
|
||
|
||
The same situation prevails if the beam is built of bilinear plane elements (four nodes, eight d.o.f., as in Table 6.14-1) integrated with a 2 by 2 Gauss rule. Now the beam contains $4N_{e\ell}$ active d.o.f. However, owing to the “parasitic shear” discussed in Section 8.3, there are $4N_{e\ell}$ constraints if the beam is thin. Again the mesh locks as the length-to-thickness ratio becomes large.
|
||
|
||
In Fig. 9.5-1b there are two d.o.f. per node and therefore $2N_{es}^{2}$ active d.o.f. Let the material be nearly incompressible; that is,
|
||
|
||
$$
|
||
\epsilon_ {V} = u _ {, x} + v _ {, y} + w _ {, z} \approx 0 \tag {9.5-1}
|
||
$$
|
||
|
||
where $\epsilon_{V}$ is the volumetric strain. If the plane strain condition $w_{,z} = 0$ prevails, then the penalty constraint $u_{,x} + v_{,y} \approx 0$ is enforced at each integration point. With a 2 by 2 integration rule there are $4N_{es}^{2}$ constraints, and the mesh is locked.
|
||
|
||
In the foregoing examples, the ratio of number of d.o.f. to number of penalty constraints is 1/1 in Fig. 9.5-1a and 1/2 in Fig. 9.5-1b. Note that these same ratios can be determined from a single element. Consider the addition to the mesh of a single element, such as the element shaded in Fig. 9.5-1b. It brings two additional d.o.f. to the mesh, and under a 2 by 2 Gauss rule it also brings four additional constraints, for a d.o.f.-to-constraint ratio of 1/2, as previously determined. For the mesh as a whole, the 1/2 ratio would still be approximately correct if support conditions are changed, provided that $N_{es}$ is large. Accordingly, we will henceforth presume that meshes contain many elements, and do our “constraint counting” by examining the additional d.o.f. and constraints that are brought to a mesh by adding a single element (or perhaps a single “macroelement” built of subelements).
|
||
|
||
Desirable Constraint Ratios. We define the constraint ratio $r$ as the ratio of the number of active d.o.f. in $\{\mathbf{D}\}$ to the number of penalty constraints. Locking occurs if $r \leq 1$ . If $r$ is slightly greater than unity, the mesh does not lock, but poor results are likely because most d.o.f. are occupied in satisfying penalty constraints; few d.o.f. are left to model the elastic behavior of the system. Extensive numerical testing has shown that near-optimal constraint ratios are $r = 2/1$ for two-dimensional problems and $r = 3/1$ for three-dimensional problems. In each case these ratios correspond to the number of differential equations of equilibrium (two and three for plane and solid problems, respectively) divided by the number of constraint conditions on the system of governing differential equations (one constraint; Eq. 9.5-1 for incompressibility or $\gamma_{zx} = 0$ for Mindlin beams).
|
||
|
||
A favorable constraint ratio is usually achieved by underintegration or by selective reduced integration—that is, by use of m < n in Eq. 9.4-13. In the context of Eq. 9.4-9 this means using a lower-order Gauss quadrature rule for $[K_{B}]$ than for $[K_{G}]$ .
|
||
|
||
If quadrature points are unsymmetrically distributed, such an element may not be geometrically isotropic. This should be of no consequence, because lack of geometric isotropy is annoying only in a coarse mesh, and a rather fine mesh is needed to produce accurate results if the material is nearly incompressible.
|
||
|
||
<!-- source-page: 305 -->
|
||
|
||
Further Examples. With the Mindlin beam element of Fig. 9.4-1, it suffices to use uniform integration with $m = n = 1$ — that is, a single Gauss point for both $[\mathbf{k}_b]$ and $[\mathbf{k}_s]$ . This integration is exact for $[\mathbf{k}_b]$ because $\theta_{,x}$ is not a function of $x$ . However, $w_{,x} - \theta$ is linear in $x$ , so $[\mathbf{k}_s]$ is underintegrated by a single Gauss point. Thus two d.o.f. and one constraint are added to the mesh by each element, for an ideal constraint ratio of $r = 2/1$ . As an exercise, one can show that in place of Eq. 9.4-5, one now obtains a far more accurate result for a tip-loaded cantilever element. Additional discussion of transverse shear constraints can be found in [9.7,9.8] and in Section 11.3.
|
||
|
||
Figure 9.5-2 shows elements that might be used for nearly incompressible media (Eq. 9.4-9). Plane and solid elements have respectively two and three d.o.f. per node. For the bilinear and trilinear elements, it is appropriate to use one-point integration to obtain $[K_{B}]$ : thus, for Figs. 9.5-2a and 9.5-2b, we obtain r = 2/1 and r = 3/1, respectively, which are the optimal ratios. In Fig. 9.5-2c, depending on whether the element has eight or nine nodes, uniform 3 by 3 integration gives r = 6/9 or r = 8/9, respectively, and selective reduced integration with a 2 by 2 rule for the penalty matrix gives r = 6/4 or r = 8/4. The latter is optimal.
|
||
|
||
For convenience, triangles in Fig. 9.5-2d are considered as a two-element patch that brings eight d.o.f. to a mesh. No integration rule for the penalty matrix is entirely satisfactory: three points per triangle gives r = 8/6 (too low), and one point per triangle gives r = 8/2 (too high). Indeed, were the triangles to have vertex nodes only, even one point per triangle would be too high: thus r = 2/2, which means that a mesh would lock. These difficulties with triangles for plane problems suggest that tetrahedra for solid problems would not work well. However, the question of what approach is best for solid problems is not yet settled.
|
||
|
||
In distorted meshes, one may find that reduced integration is not adequate to represent element volume exactly. This difficulty disappears with mesh refinement if subdivision causes elements to become parallelograms or parallelepipeds. The difficulty can be avoided by use of the consistent penalty method, which has additional attributes to recommend it (see Section 9.6).
|
||
|
||
# 9.6 ADDITIONAL TECHNIQUES FOR INCOMPRESSIBLE MEDIA
|
||
|
||
Deviatoric–Dilatational Splitting. It is convenient to regard stresses $\{\sigma\}$ as being composed of a deviatoric state $\{\sigma_{D}\}$ (which produces no change of volume) and a dilatational state $\{\sigma_{V}\}$ (which produces no change of shape). Thus, from Eq. 9.4-7,
|
||
|
||
$$
|
||
\{\boldsymbol {\sigma} \} = \{\boldsymbol {\sigma} _ {D} \} + \{\boldsymbol {\sigma} _ {V} \} = G [ \mathrm{E} _ {G} ] \{\boldsymbol {\epsilon} \} + B [ \mathrm{E} _ {B} ] \{\boldsymbol {\epsilon} \} \tag {9.6-1}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram showing a square with a vertical hatched border (no text or symbols)
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple line drawing of a 3D cube with shaded edges and vertices (no text or symbols)
|
||
</details>
|
||
|
||
(b)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram of a square frame with hatched sides and a small circle at the center (no text or symbols)
|
||
</details>
|
||
|
||
(c)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram showing a square with diagonal line and marked points, no text or symbols present
|
||
</details>
|
||
|
||
(d)
|
||
Figure 9.5-2. (a) Plane bilinear element. (b) Solid trilinear element. (c) Plane quadratic element. (d) Two plane triangular elements.
|
||
|
||
<!-- source-page: 306 -->
|
||
|
||
We discover that $\{\sigma_{V}\}$ contains three equal normal stresses and no shear stresses. Specifically, in terms of the volumetric strain $\epsilon_{V}$ stated in Eq. 9.5-1, nonzero stresses in $\{\sigma_{V}\}$ are
|
||
|
||
$$
|
||
\sigma_ {x V} = \sigma_ {y V} = \sigma_ {z V} = \lambda , \quad \text { where } \quad \lambda = B \epsilon_ {V} \tag {9.6-2}
|
||
$$
|
||
|
||
in which $\lambda$ is called the hydrostatic pressure function and B is the bulk modulus, defined in Eq. 9.4-6. Equations 9.6-1 and 9.6-2 yield $\lambda = (\sigma_{x} + \sigma_{y} + \sigma_{z})/3$ . (In the theory of plasticity, where $\{\sigma_{D}\}$ and $\{\sigma_{V}\}$ are also used, $\lambda$ is called the mean stress.) With $\lambda$ from Eq. 9.6-2, Eq. 9.6-1 assumes the form
|
||
|
||
$$
|
||
\{\boldsymbol {\sigma} \} = \left\{\boldsymbol {\sigma} _ {D} \right\} + \lambda \left[ \begin{array}{l l l l l l} 1 & 1 & 1 & 0 & 0 & 0 \end{array} \right] ^ {T} \tag {9.6-3}
|
||
$$
|
||
|
||
where $\{\sigma_D\} = G[\mathbf{E}_G]\{\epsilon\}$ and $G[\mathbf{E}_G]$ is stated in Eq. 9.4-7. For a completely incompressible material $\lambda$ is a “volumetric stress,” which can be regarded as a system of pressures that keeps the body from dilatating.
|
||
|
||
One may argue that if Poisson's ratio approaches 0.5 while all other parameters of the problem are held fixed, the incompressible pressure field is, at each material point, the limit of the slightly compressible pressure $\lambda$ computed from Eq. 9.6-2 [9.9]. The penalty method provides a way to "perturb" the exactly incompressible solution slightly and thus obtain $\lambda$ as a good approximation of the exactly incompressible pressure.
|
||
|
||
Pressure Calculation in the Penalty Method. In the penalty method, one can calculate pressure $\lambda$ by evaluating $\epsilon_{V}$ from the displacement field and then using Eq. 9.6-2. The error in $\lambda$ associated with use of a penalty constraint rather than an exact constraint is of order $10^{-p/2}$ if $\alpha$ is chosen as $10^{p/2}$ , as suggested at the end of Section 9.4. $^{5}$ However, this error estimate is valid only if $\lambda$ is calculated at locations where the constraint is enforced—that is, at the Gauss points used to evaluate element matrices $[k_{B}]$ . For the penalty method, then, the (reduced) volumetric integration points of Eq. 9.4-13 play a three-part role: that of “pressure points,” that of constraint points, and that of volumetric integration points.
|
||
|
||
If pressures are desired at other points, techniques of Section 6.13 can be used to extrapolate pressures to displacement nodes or to any other points in an element. Expressions analogous to Eq. 6.13-5 can be developed for any of the elements discussed in Section 9.5. However, for many of those elements, pressures are susceptible to an instability akin to the hourglassing discussed in Section 6.12. Unfortunately, this is particularly true of elements such as the bilinear and quadratic elements of Fig. 9.5-2 when the standard one-point or four-point reduced formulas are selectively applied to the volumetric terms of Eq. 9.4-13. One way to avoid this is by postprocessing the computed pressures using an “averaging” or “smoothing” scheme to smooth out the penalty pressures. This can be rigorously justified in terms of error analysis [9.11-9.13], and is not difficult to implement [9.7,9.14].
|
||
|
||
<!-- source-page: 307 -->
|
||
|
||
Consistent Penalty Method. The consistent penalty method provides an alternative way to calculate $[K_{B}]$ . Reduced integration is not required, and incompressibility constraints are imposed at certain “pressure points” rather than at integration points. Thus, as compared with the preceding penalty method, constraint points are divorced from integration points, and the choice of integration rule for $[K_{B}]$ is not dictated by constraint counting. Constraint counting is still used, but now to achieve a balance between the number of pressure points and the number of displacement d.o.f. If pressure points are well chosen, the aforementioned pressure instabilities will not arise. A full description of the method is beyond the scope of this book: it is a special case of the “[B] method” of [9.7], and details appear in [9.10]. The following summary is offered.
|
||
|
||
Let $\lambda_{i}$ represent hydrostatic pressure at the pressure points, such as the three shown in Fig. 9.6-1. Pressure $\lambda$ over the element is interpolated from these $\lambda_{i}$ . Thus $\lambda$ is interpolated independently of d.o.f. at element nodes. The formulation procedure leads to equations like Eq. 9.2-2, except that (a) $\{\mathbf{Q}\} = \{\mathbf{0}\}$ , and (b) the lower-right submatrix [0] is replaced by a square nonsingular matrix that represents slight compressibility.
|
||
|
||
An important feature of pressure points is that they are not shared by adjacent elements, even if pressure points are placed on element boundaries. Thus the pressure points $\lambda_{i}$ are internal d.o.f. that can be eliminated by condensation before elements are assembled (see Eq. 8.1-1, and let $\{d_{c}\} = \{\lambda\}$ ). The result of condensation and assembly is a set of equations like Eq. 9.4-9, obtained with computational efficiency comparable to that of the reduced-integration penalty method, and with an improved $[K_{B}]$ .
|
||
|
||
The nine-node, three-pressure-point element of Fig. 9.6-1, in consistent penalty form and with $\alpha = 10^{p/2}$ , is the best element known for two-dimensional incompressible elasticity and fluid flow. That the constraint ratio is r = 8/3, rather than the optimal r = 2/1, evidently causes no ill effects. There are rigorous error bounds for the element, showing that the pressure is as accurate as strains in compressible elasticity [9.10]. There are no difficulties with spurious pressure modes.
|
||
|
||
In three-dimensional problems, the situation is not nearly as clear. There appears to be no three-dimensional analogue of the two-dimensional quadratic element with three pressure points. At present, we recommend the eight-node brick of Fig. 9.5-2b with one pressure point, using the consistent formulation or the selective/reduced formulation (if the elements are not severely distorted by isoparametric transformations). For undistorted elements these two formulations are identical [9.7,9.10]. Unfortunately, this element has spurious modes, and pressure smoothing is required. We recommend the pressure-smoothing scheme described in [9.14]. Development of good elements for incompressible media in three dimensions is an active area of current finite element research.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>natural_image</summary>
|
||
|
||
Simple geometric diagram with a rectangle and four marked points (no text or symbols)
|
||
</details>
|
||
|
||
Figure 9.6-1. A plane element with nine nodes and three pressure points, used in a consistent penalty method [9.10].
|
||
|
||
<!-- source-page: 308 -->
|
||
|
||
# PROBLEMS
|
||
|
||
# Section 9.1
|
||
|
||
9.1 In Fig. 9.1-1, let members CA, AB, and BD be identical. Also assume that conditions $v_{A} = v_{B} = 0$ have already been imposed, so that [K] operates on d.o.f. $u_{A}, \theta_{A}, u_{B}$ , and $\theta_{B}$ . Write [K], then condense it to a 3 by 3 matrix by imposing the constraint $u_{A} = u_{B}$ , so that $\{D\}$ becomes $\{D\} = \left[u_{A} \quad \theta_{A} \quad \theta_{B}\right]^{T}$ .
|
||
9.2 Let the quadratic element of Eqs. 6.6-1 have straight sides and midside nodes. Imagine that the displacement of each side node is to be the average of displacements of the two adjacent corner nodes. Write the appropriate form of Eq. 9.1-4. (You may wish to consider the $u_{i}$ separately from the $v_{i}$ .) What kind of element will be produced by applying these constraints? Verify your prediction.
|
||
9.3 Write Eq. 7.8-5 in the form of Eq. 9.1-2. How many equations of constraint are there?
|
||
9.4 Write the specific form of Eq. 9.1-2 appropriate to Fig. 8.1-2; that is, write the equation that joins the two frames at $A$ with a hinge connection. For simplicity, include in your equations only the three d.o.f. of each frame at $A$ (six d.o.f. altogether). Identify matrices $[\mathbf{C}_r]$ and $[\mathbf{C}_c]$ in your formulation.
|
||
9.5 The bar element shown has axial stiffness $k = AE / L$ and axially directed d.o.f. $u_{1}$ and $u_{2}$ . Using the procedure of Section 9.1, solve for $u_{1}$ if the constraint $u_{2} = 0$ is imposed.
|
||
9.6 (a) Let $\{\mathbf{Q}\}$ be nonzero in Eq. 9.1-1. Hence, derive the equation analogous to Eq. 9.1-6. (b) Let the constraint $u_{2} = \overline{u}$ be applied in Problem 9.5. Use the method of part (a) to determine $u_{1}$ .
|
||
9.7 Element 2 is to be connected to element 1, as shown. Explain in detail how to treat the stiffness matrix of element 2 before assembly so that node 3 is constrained to lie on linear edge 1–2. (D.o.f. of nodes 1 and 2, but not of node 3, are to appear in the structural equations.)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
P → 1 A,E 2 → x,u
|
||
</details>
|
||
|
||
Problem 9.5
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
1
|
||
L
|
||
a
|
||
①
|
||
②
|
||
3
|
||
2
|
||
</details>
|
||
|
||
Problem 9.7
|
||
|
||
9.8 Three nodes lie on an $x$ axis at coordinates $x_1, x_2,$ and $x_3$ . Write a relation in the form of Eq. 9.1-1 that constrains their $x$ -direction displacements to be directly proportional to $x$ .
|
||
9.9 If Fig. 9.1-2 were to represent three identical beam elements under lateral load, what constraint among d.o.f. would be enforced by the “same node number” device (described below Eq. 9.1-9) applied to nodes 2 and 3? Would [k] of the right element contribute to stiffness of the structure? If so, in what way?
|
||
|
||
<!-- source-page: 309 -->
|
||
|
||
9.10 The two rigid links AB and BC are connected by a hinge at B and are supported by identical springs at A, B, and C, as shown. Write the structural equations that use $v_{A}$ , $v_{B}$ , and $v_{C}$ as d.o.f. Then impose the constraint that the hinge does not allow relative rotation between the two links, and solve for $v_{A}$ , $v_{B}$ , and $v_{C}$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y,v
|
||
L
|
||
L
|
||
A
|
||
B
|
||
C
|
||
P
|
||
x
|
||
k
|
||
k
|
||
k
|
||
</details>
|
||
|
||
Problem 9.10
|
||
|
||
9.11 The two-element uniform cantilever beam shown is built of standard beam elements (Eq. 4.2-5). Imagine now that element 2-3 is to be made rigid. Impose this constraint, eliminate $w_{3}$ and $\theta_{3}$ , solve for $w_{2}$ and $\theta_{2}$ , and compare these results with the prediction of elementary beam theory.
|
||
|
||
9.12 Two identical standard beam elements (Eq. 4.2-5) are joined at node 2 and are simply supported at nodes 1, 2, and 3, as shown. Node 3 is loaded by moment $M_0$ .
|
||
|
||
(a) Impose the constraint $\theta_{2} = \theta_{3}$ and solve for all three rotational d.o.f. in terms of $M_0, L, E$ , and $I$ .
|
||
(b) Sketch a free-body diagram of the constrained two-element structure.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
1
|
||
2
|
||
3
|
||
P
|
||
a
|
||
a
|
||
</details>
|
||
|
||
Problem 9.11
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
1
|
||
2
|
||
3
|
||
M₀
|
||
L
|
||
L
|
||
</details>
|
||
|
||
Problem 9.12
|
||
|
||
# Section 9.2
|
||
|
||
9.13 The following question is strictly mathematical, and serves as a review of the Lagrange multiplier method. What is the area of the largest rectangle that can be inscribed in the ellipse $(x / a)^2 + (y / b)^2 - 1 = 0$ ?
|
||
9.14 Solve the problem of Fig. 9.1-3 (i.e., impose $v_{1} = v_{2}$ ) by use of a Lagrange multiplier.
|
||
9.15 A bar of axial stiffness $k = AE / L$ lies along the $x$ axis and is allowed only axial displacements $u$ . Its right end carries a force $P = 3$ in the $+x$ direction. Its left end (node 1) is to be displaced two units to the right. Impose displacement $u_{1} = 2$ and solve for $u_{2}$ by use of a Lagrange multiplier.
|
||
9.16 To model the uniform cantilever beam shown, use a single standard beam element (Eq. 4.2-5). Use the Lagrange multiplier method to determine the deflection of load P under the following constraint conditions.
|
||
|
||
(a) The right end is to remain tangent to a straight line between nodes 1 and 2.
|
||
|
||
(b) The right end is to rotate half as much as the midpoint of the beam, but in the opposite direction.
|
||
|
||
<!-- source-page: 310 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
1
|
||
2
|
||
P
|
||
L
|
||
</details>
|
||
|
||
Problem 9.16
|
||
|
||
# Section 9.3
|
||
|
||
9.17 (a) Derive Eq. 9.3-3 from Eqs. 9.3-1 and 9.3-2. (b) Revise the argument associated with Eq. 9.3-7: do not make the simplifying assumptions $\lceil \alpha \rfloor = \lceil \mathbf{I} \rceil \alpha$ and $\{\mathbf{Q}\} = \{\mathbf{0}\}$ .
|
||
|
||
9.18 Solve the problem of Fig. 9.1-3 (i.e., impose $v_{1} = v_{2}$ ) by use of a penalty number.
|
||
|
||
9.19 Repeat Problem 9.15, but use a penalty number $\alpha$ instead of a Lagrange multiplier. For $k = 1$ , tabulate $u_{1}$ and $u_{2}$ for the values $\alpha = 1, 4, 10$ , and 100.
|
||
|
||
9.20 Use the penalty method to solve
|
||
|
||
(a) Problem 9.16a.
|
||
(b) Problem 9.16b.
|
||
|
||
# Section 9.4
|
||
|
||
9.21 Derive Eq. 9.4-5.
|
||
|
||
9.22 Use one-point quadrature to evaluate $U_{s}$ in Eq. 9.4-2. Hence, use the equations $\partial\Pi_{p}/\partial w_{2}=0$ and $\partial\Pi_{p}/\partial\theta_{2}=0$ to derive expressions for $w_{2}$ and $\theta_{2}$ for the problem posed in connection with Eq. 9.4-5. Compare these results with exact values.
|
||
|
||
9.23 (a) Let a Mindlin beam element (Fig. 9.4-1a) be simply supported at nodes 1 and 2 and loaded in pure bending. Use Eqs. 9.4-2 to show that exact integration gives an element strain energy U consistent with an effective moment of inertia $I_{e} = I(1 + GL^{2}/1.2Et^{2})$ , where $I = bt^{3}/12$ for the rectangular cross section.
|
||
|
||
(b) Show that if $U_{s}$ is integrated by one-point quadrature, the resulting $U$ is consistent with the exact moment of inertia.
|
||
|
||
9.24 Use Eqs. 9.4-2 and 9.4-3 to evaluate the stiffnesses $[k_{b}]$ and $[k_{s}]$ of the Mindlin beam element in the following ways, and determine the rank of each matrix.
|
||
|
||
(a) Evaluate $[\mathbf{k}_b]$ by one-point Gauss quadrature.
|
||
(b) Evaluate $[\mathbf{k}_s]$ by one-point Gauss quadrature. Call the result $[\mathbf{k}_s]_1$ .
|
||
(c) Evaluate $[\mathbf{k}_s]$ by two-point Gauss quadrature. Call the result $[\mathbf{k}_s]_2$ .
|
||
|
||
9.25 (a) Use $[\mathbf{k}_b]$ and $[\mathbf{k}_s]_1$ from Problem 9.24 to model a one-element cantilever beam fixed at node 1 (the left end). Load node 2 by moment $M$ only. Solve for $w_2$ and $\theta_2$ . Investigate what happens as $L$ becomes much larger than $t$ .
|
||
|
||
(b) Repeat part (a) but use $[\mathbf{k}_s]_2$ rather than $[\mathbf{k}_s]_1$ .
|
||
|
||
9.26 In Problem 9.25, compute the ratio of $w_{2}$ obtained from use of $[\mathbf{k}_s]_1$ to $w_{2}$ obtained from use of $[\mathbf{k}_s]_2$ . If $\nu = 0$ , for what value of $L / t$ is $w_{2}$ only $10\%$ in error?
|
||
|
||
9.27 A rectangular bilinear element (four nodes) is subjected to the constraint $\iint (\epsilon_x + \epsilon_y) dx dy = 0$ . Show that the same constraint is produced by setting $\epsilon_x + \epsilon_y = 0$ in a one-point Gauss quadrature rule.
|