김경종
28 KiB
Figure 6.12-1. Independent displacement modes of a bilinear element.
6.12-2 are called hourglass modes because of their physical shape. Each of these distortions, as well as each straining mode in Fig. 6.12-1, would be considered the same mode if nodal d.o.f. were reversed—that is, if \{d\} of that mode were replaced by -\{d\} .
Elements need not be rectangular in order to display a mechanism. Imagine, for example, that displacements are u = a_{1}\xi\eta and v = a_{2}\xi\eta , where a_{1} and a_{2} are constants. Then, at \xi = \eta = 0 , we have u_{, \xi} = u_{, \eta} = v_{, \xi} = v_{, \eta} = 0 ; hence, according to Eqs. 6.3-17 and 6.3-18, \epsilon_{x} = \epsilon_{y} = \gamma_{xy} = 0 at the Gauss point of an order 1 rule, regardless of the shape of the element.
Consider next the quadratic plane element, having either eight or nine nodes, and integrated with a 2 by 2 Gauss rule (Fig. 6.12-3). Displacements in the nine-node element of Fig. 6.12-3b are [6.11]
\begin{array}{l} u = 3 \xi^ {2} \eta^ {2} - \xi^ {2} - \eta^ {2} \\ v = 0 \end{array} \tag {6.12-2}
At the Gauss points of a 2 by 2 rule—that is, where \xi and \eta are \pm 1/\sqrt{3} —one finds u_{, \xi} = u_{, \eta} = v_{, \xi} = v_{, \eta} = 0 . Therefore, according to Eqs. 6.3-17 and 6.3-18, strains are zero at these points, for any geometric shape of the undeformed ele
Figure 6.12-2. (a) Mesh of four bilinear elements, showing Gauss points of an order 1 rule in each element (squares). (b,c,d) Possible mechanisms (“hourglass” modes).
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y,v 4 7 3 8 η ξ 9 6 1 5 2
Eight- or nine-node elements
(a)
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c,u
Nine-node element only
(b)
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Pure geometric diagram with curved and straight lines forming a grid-like structure (no text or symbols)
Nine-node element only
(c)
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Pure geometric diagram of a symmetrical curved shape with dashed and solid lines, no text or symbols present
Eight- and nine-node elements
(d)
Figure 6.12-3. Possible mechanisms (“hourglass” modes) in quadratic elements integrated by an order 2 rule. Gauss points are shown by squares.
ment. A similar v field is possible (Fig. 6.12-3c). Thus we have identified two mechanisms.
mechanisms.
The foregoing two mechanisms are not possible in the eight-node element because the \xi^{2}\eta^{2} term is not present (see Eqs. 6.6-1). However, yet another mechanism is possible in both eight-node and nine-node elements (Fig. 6.12-3d). Its displacement field is simple to state for a square element; it is
u = \xi (3 \eta^ {2} - 1) \quad \text { and } \quad v = \eta (1 - 3 \xi^ {2}) \tag {6.12-3}
Again, strains are zero at the Gauss points of a 2 by 2 rule. This mechanism is usually not of great concern because two adjacent elements cannot both have such a mode, as may be seen by trying to connect two deformed elements. Thus an instability present in individual elements is not present in the mesh.
Summing up, Fig. 6.12-3 identifies three element instabilities in quadrature elements arising from a 2 by 2 Gauss quadrature rule. The element stiffness matrix has rank 12 for both eight-node and nine-node elements (rank equals order less the number of rigid-body and instability modes). None of these instabilities exists if the Gauss rule is 3 by 3 or greater.
if the Gauss rule is 5 by 5 of greater. A mesh that has no mechanisms may yet behave badly because restraints on the mechanisms are weak. Consider Fig. 6.12-4a. Elements may be the four-node elements of Fig. 6.12-2 or the nine-node elements of Fig. 6.12-3, respectively integrated by one-point and four-point rules. Load P is concentrated and applied centrally rather than being distributed across the right end. Mechanisms are not possible because all nodes at the left support are fixed. However, this restraint
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P L_T
(a)
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E₁>>E₂ E₁ E₂ P
(b)
Figure 6.12-4. Problems that involve “near mechanisms,” if reduced integration is used. In (b), elements that were initially rectangular are here shown deformed.
becomes weaker with increasing distance from the support. Near the load, distortions of the type shown in Figs. 6.12-2b and 6.12-3b become pronounced. Indeed, for a 2 by 24 mesh, the computed displacement of load P may be over 500 times the displacement predicted by the elementary formula PL/AE [6.12].
A similar situation is depicted in Fig. 6.12-4b [4.6]. A 2 by 2 Gauss rule is used to integrate [k] of each element. The stiff element, shown shaded, is weakly restrained by soft elements connected to the fixed boundary, allowing the mode of Fig. 6.12-3d (with signs of {d} reversed) to become pronounced, although not unbounded.
Elements for solids, for plate bending, and for nonstructural problems can also suffer from instabilities. Methods for detecting and controlling these modes are similar to methods used for plane elements.
A conservative analyst will avoid using any element that contains a possible instability because its dangers may not be foreseen.
Control of Instabilities. Various control methods have been proposed. Their goal is to eliminate instability by providing restraint, but without simultaneously stiffening the element's response to "legitimate" modes that are already working well. In what follows we summarize an effective method, with particular reference to a rectangular bilinear element. The method adds "hourglass stiffness" to an element integrated by one-point quadrature. The resulting element is inexpensive to formulate and works very well.
For simplicity, consider only the x -direction nodal displacements \{\mathbf{d}_x\} of the eight modes shown in Fig. 6.12-1. For modes 1 and 8, \{\mathbf{d}_x\} = \{\mathbf{0}\} . An arbitrary combination of modes 2 through 6 is
\left\{\mathbf {d} _ {x} \right\} = a _ {2} \left\{ \begin{array}{l} 1 \\ 1 \\ 1 \\ 1 \end{array} \right\} + a _ {3} \left\{ \begin{array}{c} 1 \\ 1 \\ - 1 \\ - 1 \end{array} \right\} + a _ {4} \left\{ \begin{array}{c} - 1 \\ 1 \\ 1 \\ - 1 \end{array} \right\} + a _ {5} \left\{ \begin{array}{c} 1 \\ - 1 \\ - 1 \\ 1 \end{array} \right\} + a _ {6} \left\{ \begin{array}{c} - 1 \\ - 1 \\ 1 \\ 1 \end{array} \right\} \tag {6.12-4}
where the a_{i} are constants. Mode 7 is
\{\mathbf {d} _ {x} \} _ {7} = a _ {7} \left[ \begin{array}{l l l l} 1 & - 1 & 1 & - 1 \end{array} \right] ^ {T} \tag {6.12-5}
To provide mode 7 with the stiffness it lacks under one-point quadrature, we form the “stabilization matrix”
[ \mathbf {k} ] _ {7} = \{\mathbf {d} _ {x} \} _ {7} \{\mathbf {d} _ {x} \} _ {7} ^ {T} \tag {6.12-6}
A similar matrix [k]_{8} , containing a constant a_{8} , serves to restrain mode 8. To the stiffness matrix computed by one-point quadrature, we now add [k]_{7} and [k]_{8} . It is possible to choose values of a_{7} and a_{8} such that a rectangular element displays the exact strain energy in states of pure bending.
Note that mode 7 is orthogonal to all other modes—that is,
\{\mathbf {d} _ {x} \} _ {i} ^ {T} \{\mathbf {d} _ {x} \} _ {i} = \{\mathbf {0} \} \quad \text { for } \quad i = 1, 2, 3, 4, 5, 6, 8 \tag {6.12-7}
Orthogonality prevents [k]_{7} from stiffening modes other than mode 7. That this is so may be seen by computing nodal forces \{\bar{r}\}_{i} associated with matrix [k]_{7} ,
\{\overline {{{\mathbf {r}}}} \} _ {i} = [ \mathbf {k} ] _ {7} \{\mathbf {d} _ {x} \} _ {i} = \{\mathbf {d} _ {x} \} _ {7} \{\mathbf {d} _ {x} \} _ {7} ^ {T} \{\mathbf {d} _ {x} \} _ {i} = \{\mathbf {d} _ {x} \} _ {7} (0) = \{\mathbf {0} \} \tag {6.12-8}
for i = 1 through 6 and for i = 8 .
for r = 1 through 0 and for r = 0 . The foregoing control method can be generalized to elements having more than four nodes and to elements of arbitrary shape [6.11, 6.13, 13.49, 13.52-13.54].
6.13 REMARKS ON STRESS
COMPUTATION
Element stresses follow from Eq. 4.7-1, with the substitution \{\epsilon\} = [\mathbf{B}]\{\mathbf{d}\} :
\{\pmb {\sigma} \} = [ \mathbf {E} ] ([ \mathbf {B} ] \{\mathbf {d} \} - \{\pmb {\epsilon} _ {0} \}) + \{\pmb {\sigma} _ {0} \} \tag {6.13-1}
Here, in isoparametric elements, [B] is a function of the natural coordinates and \{\sigma\} contains stresses referred to the global coordinate system xyz. Where in the element should stresses be calculated? For isoparametric elements, it often happens that stresses (especially shear stresses) are most accurate at Gauss points of a quadrature rule one order less than that required for full integration of the element stiffness matrix.
Consider Fig. 6.13-1. Sides of a bilinear element remain straight during deformation. A typical element, deformed by bending moment but with rigid-body motion removed, is shown in Fig. 6.13-1b. Displacements in the element are u = -a_1\xi \eta and v = 0 , where a_1 is a positive constant. Thus shear strain \gamma_{xy} is proportional to \xi . On the neutral surface of bending, \gamma_{xy} displays the sawtooth pattern seen in Fig. 6.13-1c. Only at \xi = 0 in each element is \gamma_{xy} correctly computed (as zero) under pure bending deformation. In a general problem of plane stress analysis, where bending can occur in both directions (modes 7 and 8 of Fig. 6.12-1 simultaneously), the best computation point for \gamma_{xy} in a bilinear element is at \xi = \eta = 0 . This is the Gauss point location of an order 1 rule, which is one order less than the order 2 rule of full integration.
A similar circumstance occurs with the eight-node and nine-node quadratic elements. In the beam of Fig. 6.13-2, the exact \gamma_{xy} is constant along the x axis. In the quadratic element, \gamma_{xy} along the x axis displays the parabolic distributions shown. However, one finds that the quadratic element displays the correct \gamma_{xy} at the Gauss points of a 2 by 2 quadrature rule. In other problems of stress analysis, normal strains can also display parabolic variations, and again the most accurate strains are to be found at the Gauss points of an order 2 rule.
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y x,u P L L P
(a).
(b)
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γxy Finite element Exact 0 L 2L x
(c)
Figure 6.13-1. (a) Beam loaded in bending. (b) Bending distortion of a typical bilinear element. (c) Shear strain along the x axis.
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y x L L V
(a)
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γxy Finite element Exact 0 L 2L x
(b)
Figure 6.13-2. (a) Beam loaded by transverse tip force V. (b) Shear strain along the x axis.
In elements based on displacement fields, one expects stresses to be less accurate than displacements, as explained in Section 3.5. However, in the foregoing examples, stresses are “superaccurate” or “superconvergent” at the Gauss points because there they have the same degree of accuracy as displacements. Indeed, in unusual situations it may happen that stresses are more accurate than displacements. For example, in Fig. 6.12-4a, stresses may be substantially correct at Gauss points (of an order 1 or order 2 rule, for four- and nine-node elements, respectively), although displacements are grossly in error. This is possible because the modes that permit excessive displacements produce zero strain at the Gauss points.
The theory of locating error-minimal points for stress computation is explained elsewhere [6.14,6.15]. One discovers that these points are Gauss points: at \xi = \eta = 0 in bilinear (plane) and trilinear (solid) elements, and where \xi, \eta , and \zeta are \pm 1/\sqrt{3} in eight- or nine-node quadratic (plane) and 20- or 21-node quadratic (solid) elements. These conclusions are rigorously true for rectangular elements. For distorted elements, Gauss points may not be optimal locations but they remain very good choices.
Stresses at Gauss points can be interpolated or extrapolated to other points in the element. The result obtained is usually more accurate than the result of evaluating Eq. 6.13-1 directly at the point of interest. The interpolation–extrapolation process is explained as follows.
Imagine that stresses have been computed at the four Gauss points of a plane element (points 1, 2, 3, and 4 in Fig. 6.13-3). We now wish to interpolate or extrapolate these stress values to other points in the element. In Fig. 6.13-3, coordinate r is proportional to \xi and s is proportional to \eta . At (say) point 3, r = s = 1 and \xi = \eta = 1/\sqrt{3} . Therefore, the factor of proportionality is \sqrt{3} ; that is,
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D η r = 1 ξ = 1 G C η = 1 4 s 3 s = 1 H r F ξ 1 *P 2 A E B
Figure 6.13-3. Natural coordinate systems used in extrapolation of stresses from Gauss points.
r = \xi \sqrt {3} \quad \text { and } \quad s = \eta \sqrt {3} \tag {6.13-2}
Stresses at any point P in the element are found by the usual shape functions,
\sigma_ {P} = \sum N _ {i} \sigma_ {i} \quad \text { for } \quad i = 1, 2, 3, 4 \tag {6.13-3}
where \sigma is \sigma_x, \sigma_y , or \tau_{xy} . The N_i are the bilinear shape functions given by Eq. 6.3-2, but now written in terms of r and s rather than \xi and \eta ; that is,
N _ {i} = \frac {1}{4} (1 \pm r) (1 \pm s) \tag {6.13-4}
In Eq. 6.13-3, the N_{i} are evaluated at the r and s coordinates of point P. For example, let point P coincide with corner A. To calculate stress \sigma_{xA} at corner A from \sigma_{x} values at the four Gauss points, we substitute r = s = -\sqrt{3} into the shape functions, and obtain
\sigma_ {x A} = 1. 8 6 6 \sigma_ {x 1} - 0. 5 0 0 \sigma_ {x 2} + 0. 1 3 4 \sigma_ {x 3} - 0. 5 0 0 \sigma_ {x 4} \tag {6.13-5}
For solids, an interpolation-extrapolation formula similar to Eq. 6.13-3 is based on stresses at eight Gauss points and the trilinear N_{i} of Eq. 6.7-6.
In Section 4.7 we advised that usually the temperature field used for thermal stress analysis should have the same competence as the element strain field. Accordingly, if element stresses are based on Gauss point values, thermal strains \{\epsilon_0\} in Eq. 6.13-1 should also be based on Gauss point values.
6.14 EXAMPLES. EFFECT OF ELEMENT GEOMETRY
Simple test problems show how accuracy is affected by element distortion, changes in Gauss quadrature rule, and changes in element aspect ratio. Our examples are two-dimensional, but the trends displayed pertain to three-dimensional elements as well.
Example Problems. Table 6.14-1 illustrates the behavior of the bilinear element when its [k] is formed by four-point Gauss quadrature. Results are expressed as
TABLE 6.14-1. STRESSES AND DEFLECTIONS IN CANTILEVER BEAMS OF CONSTANT THICKNESS UNDER TRANSVERSE TIP LOAD P. LENGTH = 10, DEPTH = 2, \nu = 0.25 . VALUES BY BEAM THEORY = 1.000, OF WHICH 3% OF v_{A} IS DUE TO TRANSVERSE SHEAR DEFORMATION.
 |  |  | |||
| $\sigma_{xB}$ | $v_A$ | $\sigma_{xC}$ | $v_A$ | $\sigma_{xC}$ | $v_A$ |
| 0.096 | 0.091 | 0.727 | 0.682 | 0.301 | 0.494 |
the ratio of computed value to the value given by beam theory. We see that square elements are better than elongated elements, and that geometric distortion stiffens the element and makes answers less accurate.
The principal failing of the bilinear element is that under pure bending loads, for which \gamma_{xy} should be zero, the element displays substantial values of \gamma_{xy} except at its center, as noted in connection with Fig. 6.13-1. This defect, known as parasitic shear, makes the element too stiff in bending. An improved form of the element discussed in Section 8.3.
Table 6.14-2 illustrates the behavior of eight-node and nine-node versions of the quadratic element [6.12]. All nodes at the left end are fixed. Load P on the right end is allotted to nodes in the proportion 1–8–1, which is consistent with a parabolic distribution of shear stress. Side nodes are midway along the sides. Point B is a Gauss point of a 2 by 2 quadrature rule. Results are expressed as the ratio of computed value to the value given by beam theory.
When elements are rectangular, we see that eight-node and nine-node elements have comparable accuracy. Both become stiffer if the quadrature rule used to generate [k] is changed from 2 by 2 to 3 by 3.
Next in Table 6.14-2, elements are made trapezoidal by moving nodes C and D horizontally to positions L/4 and 3L/4, where L is the length of the beam. The final mesh in Table 6.14-2 introduces one curved interelement boundary by moving node E left of center an amount L/20. We see that 2 by 2 is the preferred integration rule, and that the eight-node element is much more sensitive to geometric distortion than the nine-node element. In one case, stress \sigma_{xB} in the eight-node element is not even of the correct sign.
The obvious lesson is that an ideal element is compact, straight-sided, and has equal corner angles. Of course, elements must be distorted to some extent in modeling an actual structure, but gratuitous distortion is to be avoided. In particular, if an element side is curved to model the curved boundary of a structure, other element sides that form interelement boundaries should be straight.
Quadratic triangles (Sections 5.5 and 6.8) have approximately the same accuracy as the nine-node element. For example, in the second case in Table 6.14-2, let
TABLE 6.14-2. STRESSES AND DEFLECTIONS IN TWO-ELEMENT CANTILEVER BEAMS OF CONSTANT THICKNESS UNDER TRANSVERSE TIP LOAD P. LENGTH = 100, DEPTH = 10, \nu = 0.30 . VALUES BY BEAM THEORY = 1.000 (IN WHICH THE TRANSVERSE-SHEAR CONTRIBUTION TO v_{A} IS NEGLECTED). SKETCHES ARE NOT TO SCALE.
| Element Type | Gauss Rule | $\sigma_{xB}$ | $v_A$ | $\sigma_{xB}$ | $v_A$ | $\sigma_{xB}$ | $v_A$ |
| 8 node | 2 × 2 | 1.000 | 0.968 | 0.051 | 0.362 | -0.048 | 0.430 |
| 8 node | 3 × 3 | 1.129 | 0.930 | 0.048 | 0.161 | 0.050 | 0.221 |
| 9 node | 2 × 2 | 1.000 | 1.006 | 1.125 | 1.109 | 0.958 | 0.955 |
| 9 node | 3 × 3 | 1.141 | 0.954 | 0.687 | 0.791 | 0.705 | 0.737 |
each quadrilateral be divided along its shorter diagonal, to which a midside node is added. Thus we produce four straight-sided quadratic triangles, which yield v_{A} = 0.796 .
Geometric Distortion: Examples and Tests. Elements in Fig. 6.14-1 have very poor geometry. Such elements should not be used. But if used, and if surrounded by elements of acceptable geometry, stresses will be poor in and very near the distorted element, but reasonable at some distance away because of Saint-Venant's principle.
In Fig. 6.14-1, Gauss points of a 2 by 2 rule lie at centers of the small black quadrilaterals. Dashed lines are lines of constant \xi and constant \eta . Dashed lines would be parallel if the element were rectangular (or if the elements were sketched in \xi \eta space, where each element is square). The first element in Fig. 6.14-1 could be a bilinear element or a quadratic element with midside nodes. The remaining three elements are distortions of quadratic rectangles with midside nodes, respectively created by moving one corner node, one side node, and two side nodes.
In Fig. 6.14-1a there is a singularity at node 3, where the Jacobian determinant J is zero. Elsewhere in the element, J > 0 . If the \xi \eta system were made left-handed, by numbering nodes clockwise around the element but leaving shape
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4 η 3 ξ 1 2
(a)
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4 7 3 η 8 1 ξ 6 5 2
(b)
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4 8 1 5 2 η 3,7 ξ 6
(c)
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4 8 η ξ 6,7 1 5 2 3
(d)
Figure 6.14-1. Badly shaped elements, showing Gauss points bounded by lines of constant \xi and constant \eta . (a) Interior angle at node 3 is 180^{\circ} . (b) Node 1 moved to the center of the original rectangle. (c) Node 7 moved from top midside to corner 3. (d) Two side nodes moved to center of the original rectangle.
functions and the xy system unchanged, we would find J < 0 within the element, and all diagonal coefficients of the element [k] would be negative.
In the latter three elements of Fig. 6.14-1, part of each element falls outside the intended element boundaries and J < 0 at one of the Gauss points. These distortions do not prevent the elements from passing constant-strain patch tests, but drastically reduce the ability of the elements to represent more complicated states of deformation [6.16].
A user-oriented finite element program performs tests on the geometry of elements. Clearly it is easy to check that interior corner angles of quadrilaterals are not far from 90^{\circ} , that side nodes are not far from the midpoint of a straight line between adjacent corner nodes, and that J is positive at each Gauss point and not greatly different from the value of J at other Gauss points. It is extra trouble, but perhaps advisable, to check that J is also positive at vertex nodes [6.16].
PROBLEMS
Section 6.2
6.1 (a) Determine [N] of Eq. 6.2-3 by following the formal procedure suggested below Eq. 6.2-1.
(b) Determine [N] of Eq. 6.2-3 by use of Lagrange's formula, as suggested below Eq. 6.2-1.
6.2 (a) Show that if x_{3} is at the middle of a bar of length L (Fig. 6.2-1a), then J = L / 2 in Eq. 6.2-5. Let node 1 have the arbitrary value x_{1} .
(b) How far from the center of the bar can node 3 be placed if, according to Eq. 6.2-4, strain \epsilon_{x} is to remain positive at the ends of the bar for arbitrary values of u_{1}, u_{2} , and u_{3} ?
6.3 Determine the element stiffness matrix [k] if x_{1} = 0 , x_{2} = L , and x_{3} = L / 2 in Fig. 6.2-1a. Let A and E be constant and do integrations explicitly.
6.4 Omit node 3 in Fig. 6.2-1a, so that the bar becomes a linear element with end nodes only. Derive the 2 by 2 stiffness matrix [k] by using the natural coordinate \xi .
6.5 The bar shown is fixed at both ends. It is modeled by one three-node element, whose shape functions are given by Eq. 6.2-3. Show that if the bar is uniform and loaded axially by its own weight, the exact stress distribution is obtained.
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L/2 L/2 1 3 2
Problem 6.5
Section 6.3
6.6 With reference to Fig. 6.3-1a, let x = \lfloor 1 \xi \eta \xi \eta \rfloor \lfloor a_1 a_2 a_3 a_4 \rfloor^T .
(a) Hence, write [A] in the relation \left\lfloor x_1 \quad x_2 \quad x_3 \quad x_4 \right\rfloor^T = [A] \left\lfloor a_1 \quad a_2 \quad a_3 \quad a_4 \right\rfloor^T .
(b) By inspection of Eqs. 6.3-2, write [\mathbf{A}]^{-1} in the relation x = \left\lfloor 1 \quad \xi \quad \eta \quad \xi \eta \right\rfloor [\mathbf{A}]^{-1} \left| x_1 \quad x_2 \quad x_3 \quad x_4 \right|^T .
(c) Check your answers by seeing if [\mathbf{A}][\mathbf{A}]^{-1} = [\mathbf{I}] .
6.7 Sketch a quadrilateral, with corners properly lettered and \xi \eta axes properly oriented, if shape functions are written as
N _ {A} = \frac {1}{4} (1 - \xi) (1 + \eta) \quad N _ {C} = \frac {1}{4} (1 - \xi) (1 - \eta)
N _ {B} = \frac {1}{4} (1 + \xi) (1 + \eta) \quad N _ {D} = \frac {1}{4} (1 + \xi) (1 - \eta)
6.8 The choice of natural coordinates made in Fig. 6.3-1 is not unique. As an alternative one could adopt natural coordinates r and s, as shown in the sketch for this problem. Write shape functions of the bilinear element in terms of r and s.
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s s = 1 4 3 r = 1 1 2 r y x
Problem 6.8
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B A D C
Problem 6.9
6.9 For the element shown, sketch the lines \xi = -0.5, 0.0 , and 0.5 and the lines \eta = -0.5, 0.0 , and 0.5. The N_{i} are given by Eq. 6.3-2. Let \xi = \eta = -1 at (a) point A , (b) point B , (c) point C , and (d) point D .
6.10 Show that y, \eta = J\xi_{,x} . Also write the remaining three similar relationships among the \xi and \eta derivatives of x and y and the x and y derivatives of \xi and \eta .
6.11 Sketch a bilinear element for which J is a function of \xi but not of \eta .
6.12 Both elements shown are square and two units on a side. Both are improperly numbered. For each, determine [J] and J , using the N_{i} of Eq. 6.3-2. What do the given numberings imply about the \xi \eta axes of the first element and the actual shape of the second element?
(a)
(b)
Problem 6.12
6.13 Evaluate [J] and J for each of the four elements shown. Also compute the ratio of element area to the area of a square two units on a side. How is this ratio related to J , and why?