김경종
26 KiB
4.23 Consider the uniform, three-node bar element of Problem 4.3. Let the element be fixed at the left end and loaded by a uniformly distributed axial load of intensity q . The respective rows of [k] are [7, -8, 1] , [-8, 16, -8] , and [1, -8, 7] , each times AE / 3L . Calculate u_2 and u_3 if nodal loads are (a) calculated in consistent fashion. (b) of magnitude qL / 3 at each of the three nodes.
4.24 Verify the correctness of the three nodal loads shown in Fig. 4.3-4.
4.25 (a) Compute loads allocated to nodes 4, 7, and 3 in Fig. 4.3-4 by an edge-normal traction that varies linearly with x , from intensity -\overline{q} (compression) at node 4 to intensity +\overline{q} (tension) at node 3.
(b) Combine the result of part (a) with the loads of Eq. 4.3-11 to determine \{\mathbf{r}_e\} for a traction that varies linearly from intensity q_{4} at node 4 to intensity q_{3} at node 3.
4.26 (a) Force F acts on one edge of the plane bilinear element at y = b / 2 , as shown. What 8 by 1 load vector \{\mathbf{r}_e\} results?
(b) What would be the three nonzero nodal loads if the left edge had three uniformly spaced nodes? (See Eq. 4.3-10.)
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y,v a a F 4 3 b x,u b 1 2
Problem 4.26
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y,v a a 4 3 a/2 b x,u 4 Q 3 10 1 2 b
Problem 4.27
4.27 A 10-unit force acts at point Q in the plane bilinear element shown. What 8 by 1 load vector \{\mathbf{r}_e\} results?
4.28 Let a concentrated force F act in the y direction on the top edge of the element in Fig. 4.3-4. Where on this edge—that is, at what value of x / L —must F act if nodal loads at nodes 4 and 7 are to be equal? What then is the load at node 3?
4.29 A uniform body force F_{x} acts in the positive x direction on a plane rectangular bilinear element (Fig. 4.2-4). Use Eq. 4.1-6 to compute values of the nodal loads in terms of F_{x} , a, b, and uniform element thickness t.
4.30 Verify the nodal loads shown in Fig. 4.3-5b by integration of [N]^{T}\{\Phi\} over a rectangular face. Shape functions are given in Eqs. 6.6-1. For simplicity you may use \xi = x and \eta = y as coordinates whose origin is at the center of a square face two units on a side.
4.31 Imagine that the temperature of a uniform beam element varies linearly through its depth, from -T along the lower surface to +T along the upper surface. Nodal moments, but not nodal forces, appear in \{r_{e}\} . Compute these moments by use of mechanics of materials arguments rather than by use of Eq. 4.1-6.
4.32 Use the virtual work concept to compute the moment M_{1} = qL^{2} / 12 at node 1 in Fig. 4.3-6a. That is, compute work done by load q as it moves through displacements created by virtual rotation \delta \theta_{1} (with other nodal d.o.f. kept at zero), and equate it to work done by M_{1} in moving through displacement \delta \theta_{1} .
4.33 A concentrated lateral force F is applied to a standard beam element at a distance x from the left end. For what value of x between 0 and L will \{\mathbf{r}_e\} contain the largest magnitude of moment at node 1? Suggestion: Note the interpretation made in Fig. 4.3-2b.
4.34 Let a uniformly distributed load of intensity q act over only the left half of a beam element. Compute the 4 by 1 load vector \{\mathbf{r}_e\} . Check that loads in \{\mathbf{r}_e\} are statically equivalent to the original load q .
4.35 The uniform cantilever beam shown carries Model the beam by a single beam element.
(a) Calculate w_{2} (the lateral deflection at the right end). Use the consistent load vector at node 2. Express your answer in terms of P, L; E, I, and x .
(b) Again calculate w_{2} , but use load lumping: let the lateral force be Px / L and ignore the moment load at node 2.
(c) Compute the exact w_{2} according to elementary beam analysis.
(d) Compute the ratio of the finite element w_{2} to the exact w_{2} . Do this for part (a) and for part (b). Plot these ratios versus x / L .
(e) Compute the bending moment at the left end, first as given by a consistent loads, then as given by lumped loads. Compute the ratio of each moment to the exact value, and plot the two ratios versus x for 0 < x < L .
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x P L
Problem 4.35
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z M_c x L/2 L/2
Problem 4.36
4.36 A uniform simply supported beam is loaded by moment M_c at midspan, as shown. If the entire beam is modeled by one element, what rotation at midspan is computed? (The exact answer is \theta = -ML / 12EI .)
4.37 Verify the percentage errors listed at the end of section 16.5. Use beam theory rather than equations [\mathbf{K}]\{\mathbf{D}\} = \{\mathbf{R}\} .
4.38 Consider the bending moment M at the left end of the distance of the cantilever beam element in Fig. 4.3-7a. Compute M from the equation M = EI[\mathbf{B}]\{\mathbf{d}\} , where [\mathbf{B}] is given by Eq. 4.2-4 and nonzero d.o.f. are those used in Eq. 4.3-13. What percentage errors in M are given by the consistent loading and by the lumped loading in Fig. 4.3-7?
Section 4.4
4.39 Imagine that the right-hand element in Fig. 4.4-2 is not an incompatible element, but rather the bilinear element of Fig. 4.2-4. Let vertical displacements \bar{v} be imposed as shown at nodes 5 and 6. The remaining d.o.f. are zero.
(a) Show that edges 3–5 and 4–6 remain straight.
(b) Compute the stresses on both sides of interelement boundary 3-4. Express your answers in terms of E , \nu , \overline{\nu} , and the coordinates.
4.40 A uniform beam is modeled by bilinear plane elements, as shown. For each of the loadings (1), (2), and (3), answer the following questions about stresses displayed by the finite element solution. In parts (c) and (d), plot stresses qualitatively (without numerical calculation).
(a) Is \sigma_{x} continuous across the interelement boundary at A ? Explain.
(b) Is \sigma_y zero at B ? Explain.
(c) Plot \tau_{xy} along the line y = 0 .
(d) Plot \sigma_{x} along the line y = 0 .
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y,v B A C D x,u
{1}
(2)
{3}
Problem 4.40
4.41 Apply the displacement field of the bilinear element (Eq. 4.2-12) to the trapezoidal element shown. Replace the a_{i} by nodal d.o.f. u_{i} , then evaluate u along the line x = y . Hence, demonstrate that this element is incompatible. (The isoparametric formulation, Chapter 6, can produce a compatible element of this shape.)
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y,v 1 1 4 3 1 2 x,u
Problem 4.41
4.42 Assume that a plane body is isotropic and that body forces F_{x} and F_{y} are constant. In terms of F_{x}, F_{y}, E , and \nu , what must be the values of a_{4} and a_{8} in Eq. 4.2-12 if the differential equations of equilibrium are to be satisfied?
Section 4.5
4.43 The bar element shown is uniform and has nodes 1 and 2. Let the assumed axial displacement field have the form u = \lfloor 1 - x^2 \rfloor \lfloor a_1 - a_2 \rfloor^T . First, replace
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L x,u 1 2
Problem 4.43
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L/2 → L/2 → → x,u 1 2
Problem 4.44
d.o.f. a_1 and a_2 by nodal d.o.f. u_1 and u_2 . Then determine the strain-displacement matrix [B] and the element stiffness matrix [k], in terms of A, E, and L . What defects do you see in these results, and what is their source?
4.44 Repeat Problem 4.43 with reference to the element shown (nodes at x = \pm L / 2 ) and the assumed axial displacement field u = \lfloor x - x^2 \rfloor \lfloor a_1 - a_2 \rfloor^T .
4.45 (a) Compute the axial deflection at x = 2L and the axial stress at x = 0 in the uniform two-element model shown. Use the element stiffness matrix derived in Problem 4.43 and evaluate stress by the calculation \sigma_x = E[\mathbf{B}]\{\mathbf{d}\} . Are the results correct?
E[\mathbf{B}]\{\mathbf{d}\} . Are the results correct. (b) Repeat part (a), but use the element stiffness matrix derived in Problem 4.44.
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y L L P x,u 1 2 3
Problem 4.45
4.46 If each of the two coefficients xy in Eq. 4.2-12 is replaced by x^2 + y^2 , the element becomes incompatible. Why? Suggestion: Let two adjacent elements have two corner nodes in common. Along the boundary between elements, d.o.f. of these corner nodes must produce the same boundary displacement in each element if the elements are to be compatible. But how many d.o.f. are needed to define a quadratic curve? And what does this imply?
4.47 It is proposed that a beam element be based on a cubic polynomial but that d.o.f. are to be only lateral displacements w_{i} , where i = 1, 2, 3, 4 . Nodes are to be at either end and at the third points. What convergence criterion is violated by this element?
4.48 What quadratic terms are omitted from the displacement field of the solid trilinear element (Eq. 4.2-16)? What cubic terms are omitted?
4.49 In three dimensions, which of the cubic terms would you add to a complete quadratic field (10 terms), if there are to be no preferred directions and the total number of terms in the expansion is to be (a) 11, (b) 13, (c) 14, (d) 16, (e) 17, (f) 19?
Section 4.6
4.50 Imagine that the mesh of Fig. 4.6-1 passes the patch test for constant values of \sigma_x , \sigma_y , and \tau_{xy} . If now all nine nodes are assigned displacements consistent with a field of constant strain (e.g., Eq. 4.2-9), what loads at node 5 should result from the calculation [K]{D}, and why?
4.51 For the four-element patch shown in Fig. 4.6-1, determine the nodal loads appropriate to the following patch tests. In each case show the loads on a sketch.
(a) Uniaxial stress \sigma_y = \sigma_c .
(b) Shear stress \tau_{xy} = \tau_c .
4.52 Sketch an assembly of hexahedral solid elements, suitable for use as a patch test mesh. Let the elements have corner nodes only and some corner angles other than 90^{\circ} . Show supports and nodal loads appropriate to a test for uniform tensile stress \sigma_{z} .
4.53 For the beam element of Fig. 4.2-2, consider the lateral displacement field
w = \frac {L - x}{L} w _ {1} + \frac {x}{2} \left(\frac {L - x}{L}\right) \theta_ {1} + \frac {x}{L} w _ {2} - \frac {x}{2} \left(\frac {L - x}{L}\right) \theta_ {2}
(a) Show that this field includes the required rigid-body motion capability.
(b) If nodal d.o.f. consistent with a state of constant curvature are prescribed, is the correct w_{xxx} obtained?
(c) Determine the element stiffness matrix based on the given field. What defects does it have?
4.54 Divide the element shown into four elements by connecting midpoints of opposite sides. Subdivide the new elements again in the same way. In the limit, what shape does each element approach?
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Simple geometric diagram of a quadrilateral with four vertices marked by dots (no text or labels)
Problem 4.54
4.55 Imagine that the four-element patch of Fig. 4.6-1 is to be tested to see how well it models the pure bending states given in parts (a) and (b). For each of these states, what should be the nodal loads? Rearrange support conditions as may be convenient and appropriate.
(a) \sigma_{x} varies linearly from -6 at y = 0 to +6 at y = 4 .
(b) \sigma_{y} varies linearly from -4 at x = 0 to +4 at x = 6 .
4.56 In Fig. 4.6-3, let u vary linearly with s along edge 1-2. Interpolate u from nodal d.o.f. u_{1} and u_{2} . Assume that all material points in the element move only radially, so that v = 0 and u = u(r) . Show that circumferential strain \epsilon_{\theta} becomes independent of s as a becomes much larger than h.
Section 4.7
4.57 Nodal d.o.f. of (say) plane elements need not be restricted to displacements u and v. One might also use all four first derivatives of u and v, for a total of six d.o.f. per node. Such an approach has both advantages and disadvantages. What do you think they are?
4.58 Model the bar of Fig. 4.7-1 by two elements, each of length L_{T}/2 . Solve for the nodal d.o.f. and the resulting stress in the upper element. Augment this stress by the appropriate form of Eq. 4.7-3. Is the exact stress field obtained?
4.59 Consider the bending moment at the center of the uniformly loaded cantilever beam of Fig. 4.3-7.
(a) What is the exact value?
(a) What is the exact value.
(b) What value is given by M = EI[\mathbf{B}]\{\mathbf{d}\} and the exact values of d.o.f. w_{2} and \theta_{2} ? Use a one-element model.
(c) What value is given by adding the result of part (b) to the bending moment at the center of a uniformly loaded clamped-clamped beam of length L ?
4.60 In Problem 4.35e, nodal d.o.f. calculated by use of consistent nodal loads produce a bending moment M = (2Lx^2 - x^3)P / L^2 at the fixed end of the cantilever beam. If this M is augmented by the bending moment at the left end of a clamped-clamped beam under load P at position x , is the exact moment produced?
4.61 Calculate the bending moment at x = L / 4 in the one-element simply supported beam of Problem 4.36. Compute this moment by adding (a) M caused by nodal d.o.f. \theta_{1} and \theta_{2} , and (b) M in a fixed-fixed beam of length L with M_{c} applied at midspan.
4.62 The uniform bar element shown is given the temperature variation T = cx / L , where c is a constant.
(a) Solve for u_{2} , then for axial stress \sigma_{x} = E(\epsilon_{x} - \epsilon_{x0}) .
(a) Solve for u_{2} , then for axial stress c_{x} and c_{y} for the constant temperature. (b) Repeat part (a), but for stress calculation use an appropriate constant temperature T_{0} rather than T = cx / L .
(c) Why do the results of parts (a) and (b) differ? Is either correct?
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y L 1 2 x,u
Problem 4.62
4.63 The sketch shows a uniform bar of length 2L, loaded by a uniformly distributed axial load of intensity q. Also shown is a two-element model and the consistently derived loads at nodes 2 and 3. Carry out two cycles of iterative improvement (see Eq. 4.7-7). To what result does the solution appear to be converging, and what then are the nodal average stresses?
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y q x,u 2L
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① qL ② qL/2 1 2 3 L L
Problem 4.63
STRAIGHT-SIDED TRIANGLES AND TETRAHEDRA
Natural coordinates are introduced. Triangles and tetrahedra are discussed, with the emphasis on triangles. Shape functions are general, but integration formulas and element matrices presented are restricted to elements having straight sides and evenly spaced side nodes. These restrictions are removed in Chapter 6.
5.1 NATURAL COORDINATES (LINEAR)
Natural coordinates are dimensionless. They are defined with reference to the element rather than with reference to the global coordinate system in which the element resides. They are used in preference to Cartesian coordinates because they simplify the process of formulating element matrices. The simplest instance, that of natural coordinates for a straight line, is considered in the present section. Similarly defined natural coordinates for triangles and tetrahedra are considered in Section 5.2. Differently defined natural coordinates for quadrilaterals and hexahedra are used in Chapter 6.
Natural coordinates \xi_{1} and \xi_{2} in Fig. 5.1-1 are defined as ratios of lengths:
\xi_ {1} = \frac {L _ {1}}{L} \quad \text { and } \quad \xi_ {2} = \frac {L _ {2}}{L} \tag {5.1-1}
Since L_{1} + L_{2} = L , coordinates \xi_{1} and \xi_{2} are not independent. They satisfy the constraint relation
\xi_ {1} + \xi_ {2} = 1 \tag {5.1-2}
Note that \xi_{1} and \xi_{2} are each either zero or unity at end points 1 and 2. Definitions
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0 • 1 P 2 x₁ L₂ L₁ x = ξ₁x₁ + ξ₂x₂ x₂ = x₁ + L
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L 1 2 ξ₁ = L₁/L 1 ξ₂ = L₂/L 1
Figure 5.1-1. Natural coordinates \xi_{1} and \xi_{2} along a straight line. Point P is arbitrarily located; that is, x can have any value.
5.1-1 are independent of global coordinate x . Even so, \xi_1 and \xi_2 can be used to state the position of the arbitrary point P in terms of x_1 and x_2 :
x = \xi_ {1} x _ {1} + \xi_ {2} x _ {2} \tag {5.1-3}
For example, the centroid of line 1-2 is at \xi_{1} = \xi_{2} = \frac{1}{2} , where x = (x_{1} + x_{2}) / 2 . Equations 5.1-2 and 5.1-3, and the inverse relations that state \xi_{1} and \xi_{2} in terms of x , are
\left\{ \begin{array}{l} 1 \\ x \end{array} \right\} = \left[ \begin{array}{l l} 1 & 1 \\ x _ {1} & x _ {2} \end{array} \right] \left\{ \begin{array}{l} \xi_ {1} \\ \xi_ {2} \end{array} \right\} \quad \text { and } \quad \left\{ \begin{array}{l} \xi_ {1} \\ \xi_ {2} \end{array} \right\} = \frac {1}{L} \left[ \begin{array}{l l} x _ {2} & - 1 \\ - x _ {1} & 1 \end{array} \right] \left\{ \begin{array}{l} 1 \\ x \end{array} \right\} \tag {5.1-4}
Equations 5.1-4 provide a linear mapping between the x and \xi coordinate systems. Interpolation along line 1-2 can be done in natural coordinates. Linear interpolation of a function \phi , from nodal values \phi_1 and \phi_2 , is
\phi = \lfloor \mathbf {N} \rfloor \left\{ \begin{array}{l} \phi_ {1} \\ \phi_ {2} \end{array} \right\}, \quad \text { where } \quad \lfloor \mathbf {N} \rfloor = \left\lfloor \xi_ {1} \quad \xi_ {2} \right\rfloor \tag {5.1-5}
In this instance the individual shape functions are N_{1} = \xi_{1} and N_{2} = \xi_{2} . Quadratic interpolation from nodal values \phi_{1} at \xi_{1} = 1 , \phi_{2} at \xi_{2} = 1 , and \phi_{3} at \xi_{1} = \xi_{2} = \frac{1}{2} , is
\phi = \xi_ {1} (2 \xi_ {1} - 1) \phi_ {1} + \xi_ {2} (2 \xi_ {2} - 1) \phi_ {2} + 4 \xi_ {1} \xi_ {2} \phi_ {3} \tag {5.1-6}
The three quadratic shape functions are shown in Fig. 5.1-2. Equation 5.1-6 can be derived by starting with \phi = a_{1}\xi_{1}^{2} + a_{2}\xi_{2}^{2} + a_{3}\xi_{1}\xi_{2} (which is quadratic in x ), evaluating the a_{i} by substitutions such as \phi = \phi_{1} at \xi_{1} = 1 , and using Eq. 5.1-2. Alternatively, one can use insight: by comparing curves in Figs. 5.1-1 and 5.1-2, we see that N_{1} = \xi_{1} - N_{3}/2 and N_{2} = \xi_{2} - N_{3}/2 .
Stiffness matrices can be formulated by use of natural cross-sections to imagine that line 1–2 in Fig. 5.1-1 is a two-node bar element of cross-sectional area A and elastic modulus E. From Eq. 5.1-5 with \phi = u , axial displacement is u = \xi_{1}u_{1} + \xi_{2}u_{2} . Therefore, the axial strain is
\epsilon_ {x} = \frac {d u}{d x} = \frac {\partial u}{\partial \xi_ {1}} \frac {\partial \xi_ {1}}{\partial x} + \frac {\partial u}{\partial \xi_ {2}} \frac {\partial \xi_ {2}}{\partial x} = u _ {1} \left(- \frac {1}{L}\right) + u _ {2} \left(\frac {1}{L}\right) \tag {5.1-7}
in which the factors \pm 1 / L are obtained from the second of Eqs. 5.1-4; that is,
\frac {\partial \xi_ {1}}{\partial x} = \frac {\partial}{\partial x} \left(\frac {x _ {2} - x}{L}\right) = - \frac {1}{L} \quad \text { and } \quad \frac {\partial \xi_ {2}}{\partial x} = \frac {\partial}{\partial x} \left(\frac {x - x _ {1}}{L}\right) = \frac {1}{L} \tag {5.1-8}
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1 3 2 ← L/2 ← L/2 → 1 N₁ = ξ₁(2ξ₁ - 1) N₂ = ξ₂(2ξ₂ - 1) 1 N₃ = 4ξ₁ξ₂
Figure 5.1-2. Quadratic shape functions over a span L in natural coordinates \xi_1 and \xi_2 , with 3 a central node.
The element stiffness matrix that operates on u_{1} and u_{2} is
[ \mathbf {k} ] = \int_ {L} A E \left\lfloor \mathbf {B} \right\rfloor^ {T} \left\lfloor \mathbf {B} \right\rfloor d L, \quad \text { where } \quad \left\lfloor \mathbf {B} \right\rfloor = \left\lfloor - \frac {1}{L} \quad \frac {1}{L} \right\rfloor \tag {5.1-9}
The familiar result, Eq. 2.4-5, follows immediately.
For elements of quadratic or higher order, the integrand of [k] contains functions of \xi_{1} and \xi_{2} . Integration of polynomials in \xi_{1} and \xi_{2} can be done by the formula
\int_ {L} \xi_ {1} ^ {k} \xi_ {2} ^ {\ell} d L = L \frac {k ! \ell !}{(1 + k + \ell) !} \tag {5.1-10}
where k and \ell are nonnegative integers and L is the distance between the end points \xi_{1}=1 and \xi_{2}=1 . When it appears, the factorial 0! is defined as unity. Integration in Eq. 5.1-9 involves integration of only dL. Therefore, k=\ell=0 in Eq. 5.1-10, and integration yields L. As additional examples of the application of Eq. 5.1-10,
\int_ {L} x d L = \int_ {L} \left(\xi_ {1} x _ {1} + \xi_ {2} x _ {2}\right) d L = \frac {L}{2} \left(x _ {1} + x _ {2}\right) \tag {5.1-11}
\int_ {L} \xi_ {1} \xi_ {2} ^ {2} d L = L \frac {2}{4 !} = \frac {L}{1 2} \tag {5.1-12}
5.2 NATURAL COORDINATES (AREA AND VOLUME)
As defined in Section 5.1, natural coordinates for a line are ratios of lengths. Analogous natural coordinates for triangles and tetrahedra are respectively defined as ratios of areas and ratios of volumes. In the present section we assume that sides of triangles and edges of tetrahedra are straight.
Area Coordinates. In Fig. 5.2-1, an arbitrarily located point P divides a triangle 1–2–3 into three subareas A_{1} , A_{2} , and A_{3} . Area coordinates are defined as ratios of areas ^{1} :
\xi_ {1} = \frac {A _ {1}}{A} \quad \xi_ {2} = \frac {A _ {2}}{A} \quad \xi_ {3} = \frac {A _ {3}}{A} \tag {5.2-1}
where A is the area of triangle 1–2–3. Since A = A_{1} + A_{2} + A_{3} , the \xi_{i} are not independent. They satisfy the constraint equation
\xi_ {1} + \xi_ {2} + \xi_ {3} = 1 \tag {5.2-2}
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3 Side 1 Side 2 A₂ P A₁ A₃ 1 2 Side 3 x y
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ξ₃=1 ξ₂ = 0 ξ₃ = 1/2 ξ₁ = 0 ξ₁ = 1 ξ₂ = 1/2 ξ₃ = 0 ξ₁ = 1/2
Figure 5.2-1. Natural (area) coordinates for a triangle.
The centroid of a straight-sided triangle is at \xi_1 = \xi_2 = \xi_3 = \frac{1}{3} .
The constraint equation and the linear relation between Cartesian and area coordinates are expressed by equations analogous to Eqs. 5.1-4,
\left\{ \begin{array}{l} 1 \\ x \\ y \end{array} \right\} = [ \mathbf {A} ] \left\{ \begin{array}{l} \xi_ {1} \\ \xi_ {2} \\ \xi_ {3} \end{array} \right\} \quad \text { and } \quad \left\{ \begin{array}{l} \xi_ {1} \\ \xi_ {2} \\ \xi_ {3} \end{array} \right\} = [ \mathbf {A} ] ^ {- 1} \left\{ \begin{array}{l} 1 \\ x \\ y \end{array} \right\} \tag {5.2-3}
where, with x_{ij} = x_i - x_j and y_{ij} = y_i - y_j ,
[ \mathbf {A} ] = \left[ \begin{array}{l l l} 1 & 1 & 1 \\ x _ {1} & x _ {2} & x _ {3} \\ y _ {1} & y _ {2} & y _ {3} \end{array} \right] \quad \text { and } \quad [ \mathbf {A} ] ^ {- 1} = \frac {1}{2 A} \left[ \begin{array}{l l l} x _ {2} \dot {y} _ {3} - x _ {3} y _ {2} & y _ {2 3} & x _ {3 2} \\ x _ {3} y _ {1} - x _ {1} y _ {3} & y _ {3 1} & x _ {1 3} \\ x _ {1} y _ {2} - x _ {2} y _ {1} & y _ {1 2} & x _ {2 1} \end{array} \right] \tag {5.2-4}
That the first of Eqs. 5.2-3 is correct may be checked by noting that it yields the correct x and y values at the vertices (and at other points, such as midsides), and that x and y vary linearly elsewhere.
Twice the area of triangle 1-2-3 is
2 A = \det [ \mathbf {A} ] = x _ {2 1} y _ {3 1} - x _ {3 1} y _ {2 1} \tag {5.2-5}
where x_{21} = x_2 - x_1 , and so on. The latter expression for 2A is neither unique nor obvious but can be obtained by manipulation of the determinant. If labels 1-2-3 are reversed, so that the order 1-2-3 is clockwise around the triangle, Eq. 5.2-5 gives a negative number for 2A .
The first of Eqs. 5.2-3 allows one to find the Cartesian coordinates of a point when its area coordinates are given. A point in the triangle is uniquely located by specifying any two of its area coordinates. If all three are specified they must satisfy the constraint equation, Eq. 5.2-2. Further discussion of interpolation in area coordinates appears in subsequent sections, where properties of specific elements are formulated.
Formulation of element matrices requires that a function \phi , expressed in terms of area coordinates, be differentiated with respect to Cartesian coordinates. By the chain rule, with \phi = \phi(\xi_1, \xi_2, \xi_3) ,