김경종
384 lines
27 KiB
Markdown
384 lines
27 KiB
Markdown
<!-- source-page: 141 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
q1
|
||
q2
|
||
L
|
||
s
|
||
y
|
||
x
|
||
=
|
||
L/6 (2q1 + q2)
|
||
L/6 (q1 + 2q2)
|
||
</details>
|
||
|
||
Figure 4.3-3. Linearly varying load on a linear edge, and the consistent nodal loads produced.
|
||
|
||
where
|
||
|
||
$$
|
||
u = \frac {L - x}{L} u _ {1} + \frac {x}{L} u _ {2} \quad \text { and } \quad q = \frac {L - x}{L} q _ {1} + \frac {x}{L} q _ {2} \tag {4.3-9}
|
||
$$
|
||
|
||
Substitution of Eqs. 4.3-7 and 4.3-9 into Eq. 4.3-8 shows that Eq. 4.3-8 is indeed satisfied.
|
||
|
||
Plane Elements. Figure 4.3-3 shows a distributed load of intensity q that acts normal to a linear edge of a plane element. The inclination of the edge with respect to global coordinates xy does not matter. So long as edge-normal displacement varies linearly with edge-tangent coordinate s, loads are allocated to nodes as shown. This is of course the same allocation as seen in Eq. 4.3-7. An edge-parallel traction that varies linearly with s would be allocated to nodes in the same proportions.
|
||
|
||
Figure 4.3-4 shows a uniformly distributed load of intensity q that acts normal to a quadratic edge. That is, the edge-normal displacement v is $v = \left[N\right]\left[v_{4} \quad v_{7} \quad v_{3}\right]^{T}$ , where
|
||
|
||
$$
|
||
\lfloor \mathrm{N} \rfloor = \left\lfloor \frac {2 x ^ {2}}{L ^ {2}} - \frac {x}{L} \quad 1 - \frac {4 x ^ {2}}{L ^ {2}} \quad \frac {2 x ^ {2}}{L ^ {2}} + \frac {x}{L} \right\rfloor \tag {4.3-10}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
L
|
||
2
|
||
L
|
||
2
|
||
q
|
||
b
|
||
4
|
||
7
|
||
y
|
||
x
|
||
3
|
||
8
|
||
6
|
||
a
|
||
a
|
||
1
|
||
5
|
||
2
|
||
=
|
||
qL
|
||
6
|
||
2qL
|
||
3
|
||
qL
|
||
6
|
||
4
|
||
7
|
||
3
|
||
8
|
||
6
|
||
1
|
||
5
|
||
2
|
||
</details>
|
||
|
||
Figure 4.3-4. Uniform traction of intensity q on a quadratic edge, and the consistent nodal loads produced.
|
||
|
||
<!-- source-page: 142 -->
|
||
|
||
The edge is straight and node 7 is at the midpoint. Work-equivalent loads at nodes 4, 7, and 3 are
|
||
|
||
$$
|
||
\left\{\mathbf {r} _ {e} \right\} = \int_ {- L / 2} ^ {L / 2} \left\lfloor \mathbf {N} \right] ^ {T} q d x = \left\{ \begin{array}{l} q L / 6 \\ 2 q L / 3 \\ q L / 6 \end{array} \right\} \tag {4.3-11}
|
||
$$
|
||
|
||
We see that a uniform traction does not produce the same force at each node on a quadratic edge.
|
||
|
||
Nodal loads produced by a concentrated force within a plane element are evaluated by means of Eq. 4.3-4. Shape functions for the rectangular element of Fig. 4.3-4 are given by Eqs. 6.6-1 if one substitutes $\xi = x / a$ and $\eta = y / b$ . If a concentrated force $F$ acts toward (say) the right at the center of this element (where $\xi = \eta = 0$ in Eqs. 6.6-1), Eq. 4.3-4 dictates that rightward forces $F / 2$ appear at nodes 5, 6, 7, and 8 and that leftward forces $F / 4$ appear at nodes 1, 2, 3, and 4. The nodal forces sum to $F$ , as they must, but the appearance of nodal forces whose sense is opposite to that of $F$ is unexpected from the standpoint of “common sense.” Note that in a nonstructural problem, the analogue of a concentrated force $F$ is a point source or a sink.
|
||
|
||
If a uniform body force acts on a plane four-node rectangular element, one-quarter of the total force appears at each node. If the body force is again uniform and the element again plane and rectangular, but now with eight nodes as in Fig. 4.3-4, the total force is allocated to nodes in the proportions shown on the upper face of the element in Fig. 4.3-5b.
|
||
|
||
Solid Elements. Again we use Eq. 4.1-6. If there is a body force $\{\mathbf{F}\}$ , integration spans the entire element volume and all shape functions of the element are involved. If there is a traction $\{\Phi\}$ on one face, integration spans only the loaded face, and only the shape functions associated with nodes on that face are involved.
|
||
|
||
Imagine, for example, that a uniform normal stress $\sigma_{c}$ acts on a rectangular face that has corner and midside nodes (Fig. 4.3-5). Shape functions for this face may be taken from Eqs. 6.6-1 with $\xi = x / a$ and $\eta = y / b$ . The last integral in Eq. 4.1-6 yields the nodal loads shown in Fig. 4.3-5. The total force is $4P + 4Q = \sigma_{c}A$ , as required, but loads at corner nodes have a direction opposite to that of $\sigma_{c}$ .
|
||
|
||
|
||
|
||
$$
|
||
P = \frac {\sigma_ {c} A}{3}
|
||
$$
|
||
|
||
$$
|
||
Q = \frac {\sigma_ {c} A}{1 2}
|
||
$$
|
||
|
||
Figure 4.3-5. (a) Uniform stress $\sigma_{c}$ on a rectangular face area of $A$ . Side nodes are at midsides. (b) Consistent nodal loads.
|
||
|
||
<!-- source-page: 143 -->
|
||
|
||
|
||
Figure 4.3-6. Consistent nodal loads associated with loads q (constant), P, and M on a standard four-d.o.f. beam element (Eq. 4.2-3).
|
||
|
||
Beam Elements. Figure 4.3-6 shows nodal loads produced by typical loading patterns on a beam element. Nodal loads are calculated by use of Eqs. 4.1-6, 4.3-4, and 4.3-5. Specifically,
|
||
|
||
$$
|
||
\left\{\mathbf {r} _ {e} \right\} _ {4 \times 1} = \int_ {0} ^ {L} \left\lfloor \mathbf {N} \right] ^ {T} q d x + \left\lfloor \mathbf {N} \right\rfloor_ {L / 2} ^ {T} P + \left\lfloor \frac {d \mathbf {N}}{d x} \right\rfloor_ {L / 2} ^ {T} M \tag {4.3-12}
|
||
$$
|
||
|
||
Loads $\{r_{e}\}$ are of course work-equivalent to the original loads q, P, or M, in the sense defined in connection with Eq. 4.3-3. They are also statically equivalent; that is, loads $\{r_{e}\}$ produce the same resultant force and the same resultant moment about an arbitrary point as do the original loads, as the reader can easily show.
|
||
|
||
Error Produced by Lumping. The following example shows the merit of using consistent nodal loads rather than a lumping. Consider the uniformly loaded cantilever beam of Fig. 4.3-7a. Consistent nodal loads for a one-element model are shown in Fig. 4.3-7b. Taking [k] from Eq. 4.2-5 and fixing the left end of the beam, we arrive at the following set of equations to be solved for $w_{2}$ and $\theta_{2}$ ,
|
||
|
||
$$
|
||
\frac {E I}{L _ {T} ^ {3}} \left[ \begin{array}{c c} 1 2 & - 6 L _ {T} \\ - 6 L _ {T} & 4 L _ {T} ^ {2} \end{array} \right] \left\{ \begin{array}{l} w _ {2} \\ \theta_ {2} \end{array} \right\} = \left\{ \begin{array}{l} q L _ {T} / 2 \\ - q L _ {T} ^ {2} / 1 2 \end{array} \right\} \tag {4.3-13}
|
||
$$
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
z,w
|
||
q
|
||
x
|
||
LT
|
||
</details>
|
||
|
||
(a)
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
qL_T
|
||
2
|
||
qL_T^2
|
||
12
|
||
2
|
||
L_T
|
||
</details>
|
||
|
||
(b)
|
||
|
||
|
||
(c)
|
||
Figure 4.3-7. (a) Uniformly loaded cantilever beam. (b) Consistent loads at node 2 of a one-element model. (c) Lumped (inconsistent) loads at node 2.
|
||
|
||
<!-- source-page: 144 -->
|
||
|
||
from which
|
||
|
||
$$
|
||
w _ {2} = \frac {q L _ {T} ^ {4}}{8 E I} \quad \text { and } \quad \theta_ {2} = \frac {q L _ {T} ^ {3}}{6 E I} \tag {4.3-14}
|
||
$$
|
||
|
||
These are the exact values of $w_{2}$ and $\theta_{2}$ . (Values of $w$ for $0 < x < L_{T}$ are not exact. The approximating field $w = \sum N_{i}d_{i}$ is cubic in $x$ , but the exact field for a uniformly distributed load is quartic in $x$ .)
|
||
|
||
If the beam is divided into two or more elements of equal length L, moment loads cancel at all interior nodes. Accordingly, lumped loading differs from consistent loading only in that lumped loading omits the clockwise moment $qL^{2}/12$ at the beam tip, thus causing tip deflection and tip rotation to be overestimated. With n the number of equal-length elements, lumped loading yields the following percentage errors in deflection and rotation at the right end.
|
||
<table><tr><td></td><td>n = 1</td><td>n = 2</td><td>n = 3</td><td>n = 4</td></tr><tr><td>Deflection error</td><td>33.3%</td><td>8.3%</td><td>3.7%</td><td>2.1%</td></tr><tr><td>Rotation error</td><td>50.0%</td><td>12.5%</td><td>5.6%</td><td>3.1%</td></tr></table>
|
||
|
||
Consistent loading produces exact values of end deflection and end rotation for all values of n.
|
||
|
||
# 4.4 EQUILIBRIUM AND COMPATIBILITY IN THE SOLUTION
|
||
|
||
In an exact solution, according to the theory of elasticity, every differential element of a continuum is in static equilibrium, and compatibility prevails everywhere. An approximate finite element solution does not fulfill these requirements in every sense. In the present section we note the extent to which equilibrium and compatibility conditions may be satisfied at nodes, across interelement boundaries, and within individual elements.
|
||
|
||
1. Equilibrium of nodal forces and moments is satisfied. The structural equations $\{R\} - [K]\{D\} = \{0\}$ are nodal equilibrium equations. Therefore, the solution vector $\{D\}$ is such that nodal forces and moments have a zero resultant at every node.
|
||
2. Compatibility prevails at nodes. Loosely speaking, elements connected to one another have the same displacements at the connection point. More precisely, elements are compatible at nodes to the extent of nodal d.o.f. they share. The latter statement allows the modeling of a physical hinge or roller between adjacent nodes that would otherwise be fully connected; one then connects some but not all nodal d.o.f. For example, if adjacent beam elements meet at a node where they share only translational d.o.f., a hinge connection is created.
|
||
|
||
<!-- source-page: 145 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>flowchart</summary>
|
||
|
||
```mermaid
|
||
graph TD
|
||
1 --> 2
|
||
2 --> 3
|
||
3 --> 4
|
||
1 -->|1| 2
|
||
2 -->|2| 3
|
||
3 -->|σx2| 4
|
||
4 -->|u4| 3
|
||
```
|
||
</details>
|
||
|
||
Figure 4.4-1. Differential element (shaded) that spans an interelement boundary.
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
2
|
||
4
|
||
6
|
||
y,v
|
||
x,u
|
||
1
|
||
3
|
||
5
|
||
v̄
|
||
v̄
|
||
</details>
|
||
|
||
Figure 4.4-2. Adjacent incompatible plane elements. All nodes but 5 and 6 have zero displacement.
|
||
|
||
3. Equilibrium is usually not satisfied across interelement boundaries. Figure 4.4-1 provides a simple example. Imagine that the elements are constant-strain triangles (Eq. 4.2-9) and that node 4 is the only node displaced, as shown. Then $\sigma_{x2}$ is the only nonzero stress, and the shaded differential element is not in equilibrium. Other types of finite elements behave similarly. (Interelement continuity of stresses may be displayed by elements that include strains among their nodal d.o.f., but such elements are not commonly used.)
|
||
|
||
When examining finite element stress output, one should not expect that stresses in adjacent elements will be the same along the common edge, that stresses at nodes will be the same in all elements that share the node, or that stress boundary conditions will be exactly satisfied (e.g., in Fig. 1.1-2b we will not compute precisely $\sigma_{x} = \tau_{xy} = 0$ along the right edge). For a properly constructed mesh these discrepancies will be small and will become smaller with mesh refinement.
|
||
|
||
4. Compatibility may or may not be satisfied across interlement boundaries. Interelement compatibility is satisfied by all elements we have discussed thus far. For example, with both the constant-strain triangle and the plane bilinear element (both discussed in Section 4.2), compatibility is guaranteed because element sides remain straight even after the element is deformed. (That sides remain straight can be shown by noting that edges are initially straight and displacements along a side are linear functions of the coordinates.)
|
||
|
||
Other elements, not yet discussed, may be incompatible. Several successful plate elements are incompatible in rotation about an interlement boundary. Figure 4.4-2 is another case in point. Stretching of the right edge causes vertical edges of the right element to bend, and the interelement gap (shaded) appears. (This element, called either incompatible or nonconforming, is discussed in Section 8.3.)
|
||
|
||
Incompatibilities between elements should tend toward zero as more and more elements are used to model a structure. Indeed, this must happen if an incompatible element is to be considered reliable enough for general use.
|
||
|
||
5. Equilibrium is usually not satisfied within elements. In the absence of body forces, the differential equations of equilibrium (Eqs. 1.6-2) are exactly satisfied by the constant-strain triangle (Eq. 4.2-9) but not by the bilinear rec-
|
||
|
||
<!-- source-page: 146 -->
|
||
|
||
tangle (Eq. 4.2-12) unless $a_4 = a_8 = 0$ . In general, satisfaction of the differential equations of equilibrium at every point in an element demands a relation among element d.o.f. that usually does not result from solution of the global finite element equations $[\mathbf{K}]\{\mathbf{D}\} = \{\mathbf{R}\}$ .
|
||
|
||
If exact results are to be approached as a mesh is refined, then satisfaction of the equilibrium equations must be approached throughout each element. For elements having few d.o.f., this means that a constant-strain condition must be approached within each element. Higher-order elements, such as the quadratic element of Fig. 4.3-4, can satisfy the differential equations of equilibrium while displaying either constant or linear-strain fields. However, an important property of any reliable element is that its displacement field be capable of representing all possible states of constant strain. $^{1}$
|
||
|
||
6. Compatibility is satisfied within elements. We require only that the element displacement field be continuous and single-valued. These properties are automatically provided by polynomial fields.
|
||
|
||
# 4.5 CONVERGENCE REQUIREMENTS
|
||
|
||
If a particular problem is repeatedly analyzed, each time using a finer mesh of elements, we generate a sequence of approximate solutions. How can we be assured that the sequence converges to the theoretically exact result? In the following, requirements for convergence are stated in general terms, then interpreted in terms appropriate to structural mechanics and elements based on displacement fields.
|
||
|
||
General. Let the field variable be $\phi = \phi(x,y,z)$ , and let there be a functional $\Pi = \Pi(\phi)$ that yields the governing differential equation of the physical problem from the stationary condition $d\Pi = 0$ . Assume that $\Pi$ contains derivatives of $\phi$ through order $m$ . If the exact $\phi$ is to be approached as the mesh is refined, then:
|
||
|
||
1. Within each element, the assumed field for $\phi$ must contain a complete polynomial of degree $m$ . (Completeness is discussed in Section 3.6.)
|
||
2. Across boundaries between elements, there must be continuity of $\phi$ and its derivatives through order $m - 1$ .
|
||
3. Let elements be used in a mesh (rather than tested individually), and let boundary conditions on the mesh be appropriate to a constant value of any of the $m$ th derivatives of $\phi$ . Then, as the mesh is refined, each element must come to display that constant value.
|
||
|
||
For example, if $\phi = \phi(x, y)$ and $\Pi$ contains first derivatives of $\phi$ , then the lowest-order acceptable field has the form $\phi = a_1 + a_2x + a_3y$ in each element, only $\phi$ itself need be continuous across interelement boundaries, and each element of an appropriately loaded mesh must display a constant value of $\phi_x$ (or of $\phi_y$ for other appropriate loading), at least as the mesh is refined.
|
||
|
||
<!-- source-page: 147 -->
|
||
|
||
Requirement 1 ensures that $\phi$ will be continuous within elements and is necessary (but not always sufficient) in order for Requirement 3 to be satisfied.
|
||
|
||
Requirement 2 is met at all stages of mesh refinement by compatible elements. Incompatible elements must become compatible as the mesh is refined ad infinitum.
|
||
|
||
Requirement 3, which must be met in the limit of mesh refinement, is also met even in a coarse mesh by most commonly used elements.
|
||
|
||
The order of differentiation $m$ can be determined by examination of either the governing differential equation or its associated functional $\Pi$ . The correspondence between differential equation and $\Pi$ is as follows: if derivatives of $\phi$ of order $2m$ appear in the differential equation, then derivatives of $\phi$ of order $m$ appear in $\Pi$ . If $2m$ is odd, so that no variational principle exists, one can yet obtain a finite element formulation by weighted residual methods such as the Galerkin method. Then one regards $m$ as the highest-order derivative of $\phi$ to be found in integral expressions used to generate the finite element matrices. Usually these expressions result from integrations by parts that reduce $m$ as much as possible.
|
||
|
||
Satisfaction of Requirements 1, 2, and 3 guarantees convergence to correct results, but says nothing about accuracy in a coarse mesh or the rate of convergence with mesh refinement. However, if the requirements are met at all stages of mesh refinement, and if each refinement is achieved by dividing the elements of the previous mesh into two or more elements, then convergence is monotonic [4.3]. This manner of subdivision means that each new mesh contains the old mesh, especially in the mathematical sense of having the old trial space embedded in the new.
|
||
|
||
Structural Mechanics. When elements are based on displacement fields, there is often more than one field required (e.g., fields for both u and v are needed in a plane problem). The order of differentiation, m, is determined from the strain energy term in $\Pi_{p}$ . Sometimes m has two values for one element. A case in point is a thin flat element that must both stretch and bend. Here m = 1 for displacements u and v tangent to the element midsurface and m = 2 for displacement w normal to the element midsurface. Interelement compatibility of u, v, w, $w_{,x}$ and $w_{,y}$ is required, at least as the mesh is refined ad infinitum.
|
||
|
||
Together, Requirements 1 and 3 say that a mesh of elements must, when given appropriate boundary conditions, display rigid-body motion or a state of constant strain. For the flat element mentioned in the preceding paragraph, we must find that strains $\{\epsilon\} = [B]\{d\}$ are zero when $\{d\}$ represents translation along (or small rotation about) any coordinate axis, and that the mesh gives constant values of strains $\epsilon_{x} = u_{,x}, \epsilon_{y} = v_{,y}$ , and $\gamma_{xy} = u_{,y} + v_{,x}$ when appropriate stretching loads are applied to boundaries of the mesh. When bending or twisting loads are applied, a constant-strain state must be observed in a layer parallel to the element mid-surface, which means that the mesh must be able to display constant curvatures $w_{,xx}$ and $w_{,yy}$ and the constant twist $w_{,xy}$ .
|
||
|
||
As an example of rigid-body motion, consider the constant-strain triangle, Eq. 4.2-9. The motion $u = a_{1}$ is a translation in the x direction, in which all points have displacement $a_{1}$ . Similarly, $v = a_{4}$ is a translation in the y direction. With $a_{6} = -a_{3}$ , the motion $u = a_{3}y$ and $v = -a_{3}x$ is a small clockwise rigid-body rotation through an angle $a_{3}$ about the point x = y = 0 (Fig. 4.5-1). One easily checks that $\epsilon_{x} = \epsilon_{y} = \gamma_{xy} = 0$ for this rotation.
|
||
|
||
All elements discussed in Section 4.2 can display rigid-body translation and
|
||
|
||
<!-- source-page: 148 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
u = a₃y
|
||
y
|
||
a₃
|
||
a₃
|
||
v = -a₃x
|
||
x
|
||
</details>
|
||
|
||
Figure 4.5-1. Rigid-body rotation through a small angle $a_{3}$ .
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
εₓ
|
||
Lₜ
|
||
a₂
|
||
L
|
||
x,u
|
||
s
|
||
q = q(x)
|
||
</details>
|
||
|
||
Figure 4.5-2. Axial strain $\epsilon_{x}$ in a bar under axial load q. A typical element has length L.
|
||
|
||
small rigid-body rotation without strain. An example of an element that cannot, at least in a coarse mesh, is an element used to model a shell of revolution—for example, a spherical shell [4.4]. The element displacement field is written in curvilinear coordinates rather than in Cartesian coordinates. When element nodal d.o.f. {d} represent a rigid-body translation along the axis of revolution, this shell element yields zero strains only if the arc subtended by the element meridian approaches zero. As the mesh is refined, the arc subtended by each element is defined by the convergence to correct results is obtained.
|
||
|
||
A physical explanation of Requirement 3 can be given with reference to Fig. 4.5-2. The structure is modeled by standard bar elements whose displacement field is of the form $u = a_1 + a_2s$ and whose strain is therefore $\epsilon_x = a_2$ . If $L_T >> L$ , the actual strain distribution over length $L$ departs very little from the constant value $\epsilon_x = a_2$ . Clearly, as $L$ shrinks, the actual curve can be matched arbitrarily closely in stairstep fashion. One could not converge to an arbitrarily close match by using an element based on the field $u = a_1 + a_2s^2$ . Here $\epsilon_x$ is not constant; it is $\epsilon_x = 2a_2s$ , for which $\epsilon_x = 0$ at the left end of every element. Such a defect is not correctible by mesh refinement.
|
||
|
||
Requirement 3 is stated in terms of a mesh of elements rather than in terms of a single element for the following reason. It is possible for an element to be based on a polynomial field that contains constant-strain terms, yet fail to display constant strain when a mesh of arbitrarily shaped elements is appropriately supported and loaded. None of the elements discussed in Section 4.2 exhibits this difficulty.
|
||
|
||
Most elements meet Requirement 3 when used as a coarse mesh, but some elements require mesh refinement. This matter is discussed in Section 4.6, in connection with the “weak” patch test.
|
||
|
||
Geometric Isotropy. Computed results for a given structure should not depend on how the mesh is oriented in global coordinates. For example, the computed displacement of load P in Fig. 4.5-3 should be the same whether the element has orientation (a) or orientation (b). Elements that are well behaved in this regard are called geometrically isotropic (also called geometrically invariant and spatially isotropic). Although geometric invariance is not required for convergence with mesh refinement, it is very desirable that elements have no “preferred directions” so that the user will not encounter results that seem puzzling or even alarming.
|
||
|
||
If a plane element is to be geometrically isotropic, it is not a displacement expansions for $u$ and $v$ have the same form and include terms sym-
|
||
|
||
<!-- source-page: 149 -->
|
||
|
||
|
||
Figure 4.5-3. A rectangular plane element, loaded by force P at one corner.
|
||
|
||
metric about the axis of a Pascal triangle (Fig. 3.6-1). The constant-strain triangle (Eq. 4.2-9) and the bilinear element (Eq. 4.2-12) both satisfy these requirements. In particular, to produce a bilinear element one should not supplement the complete linear fields of Eq. 4.2-9 with (say) $u = a_4x^2$ and $v = a_8x^2$ , as this would destroy geometric isotropy. One selects instead the modes $u = a_4xy$ and $v = a_8xy$ , which favor neither $x$ nor $y$ . If used in the problem of Fig. 4.5-3, the bilinear element yields the same displacement of load $P$ for orientation (a) and for orientation (b). (We have not yet explained how to deal with arbitrary orientations of this element; see Chapter 6.)
|
||
|
||
Similar remarks apply to solid elements. The trilinear element, Eq. 4.2-16, uses all linear terms, a balanced selection of quadratic terms, and a single cubic term that favors none of the coordinate directions over another.
|
||
|
||
# 4.6 THE PATCH TEST
|
||
|
||
The patch test was originated by Irons [4.5,4.6]. It is a simple test that can be performed numerically, so as to check the validity of an element formulation and its programmed implementation. We assume that the element is stable in the sense described below. Then, if the element passes the patch test; we have assurance that all convergence criteria noted in Section 4.5 are met. Therefore, when this type of element is used to model any other structure, mesh refinement will produce a sequence of approximate solutions that converges to the exact solution. In other words, the patch test serves as a necessary and sufficient condition for correct convergence of a finite element formulation. The test can be described in a general way, but we will describe it in terms appropriate to structural mechanics.
|
||
|
||
Procedure. One assembles a small number of elements into a “patch,” taking care to place at least one node within the patch, so that the node is shared by two or more elements, and so that one or more interelement boundaries exist. Figure 4.6-1 shows an acceptable two-dimensional patch, built of four-node elements of a type we have not yet discussed. Boundary nodes of the patch are loaded by consistently derived nodal loads appropriate to a state of constant stress. Internal nodes are neither loaded nor restrained. The patch is provided with just
|
||
|
||
<!-- source-page: 150 -->
|
||
|
||
|
||
|
||
<details>
|
||
<summary>text_image</summary>
|
||
|
||
y,v
|
||
1.5F
|
||
3
|
||
2
|
||
4
|
||
2
|
||
F
|
||
6
|
||
9
|
||
2
|
||
3
|
||
2F
|
||
5
|
||
8
|
||
1
|
||
2
|
||
1
|
||
1
|
||
4
|
||
7
|
||
F
|
||
x,u
|
||
Thickness = 1
|
||
F = 1/4 (4)(1)σc
|
||
F = σc
|
||
σc
|
||
</details>
|
||
|
||
Figure 4.6-1. Patch of four-node elements, loaded by forces $F$ consistent with the uniform stress state $\sigma_x = \sigma_c$ , $\sigma_y = \tau_{xy} = 0$ .
|
||
|
||
enough supports to prevent rigid-body motion. One next executes a standard solution and examines the computed stresses. If, throughout the element, computed stresses agree with exact stresses for the physical problem modeled, then the patch test is passed. By “agree” we mean exact agreement, allowing only for physical noise associated with the finite length of computer words.
|
||
|
||
The patch test must be repeated for all other constant-stress states demanded of the element being tested. In Fig. 4.6-1, a test for constant $\sigma_{x}$ is depicted; we must test the patch again for constant $\sigma_{y}$ , and again for constant $\tau_{xy}$ (each time using appropriate nodal loads). Solid elements must be patch-tested for constant states of $\sigma_{x}$ , $\sigma_{y}$ , $\sigma_{z}$ , $\tau_{xy}$ , $\tau_{yz}$ , and $\tau_{zx}$ . A patch of plate-bending elements must display constant bending moments $M_{x}$ and $M_{y}$ and constant twisting moment $M_{xy}$ .
|
||
|
||
It is only for convenience that we examine stress states rather than strains. It is typically, a computer program outputs stresses rather than strains. Computed nodal d.o.f. can also be examined; if they are incorrect, strains and stresses will also be incorrect. If displacements are correct but stresses are incorrect, one strain is that the stress calculation subroutine may be in error.
|
||
|
||
Support conditions must not prevent the constant state from occurring. In Fig. 4.6-1, complete fixity of nodes 1, 2, and 3 would be fatal to the patch test, as Poisson contraction in the y direction would be prevented along the left edge. It would be acceptable to impose the boundary condition $u_{3} = 0$ rather than a load at node 3—that is, to replace the force 1.5F by a roller support.
|
||
|
||
Stability. At the outset we assumed that the element to be patch-tested is stable. A stable element is one that admits no zero-energy deformation states when adequately supported against rigid-body motion. This matter is discussed in detail in Section 6.12. Unstable elements should be used with caution. They can produce an unstable mesh, whose displacements are excessive and quite unrepresentative of the actual structure.
|
||
|
||
of the actual structure.
|
||
|
||
Instabilities can be detected by an eigenvalue test (see Section 18.8). They can also be detected by a “perturbed” patch test, as follows [4.6]. Let the patch be just adequately supported, and add a small amount to one of the consistently derived nodal loads (e.g., apply an additional load equal to an existing load times
|