김경종
26 KiB
14.25 (a) A three-node plane triangular element is constrained to move only in its plane. Write the formula for [\mathbf{k}_{\sigma}] in terms of a 2 by 2 matrix [s] and matrix [B] of Eq. 5.4-3. Let \{\mathbf{d}\} = \left[u_1 \quad u_2 \quad u_3 \quad v_1 \quad v_2 \quad v_3\right]^T .
(b) Let the same element be allowed only deflection w = w(x, y) normal to its plane. What then is the formula for [\mathbf{k}_{\sigma}] ? Let \{\mathbf{d}\} = \begin{bmatrix} w_1 & w_2 & w_3 \end{bmatrix}^T .
14.26 (a) Show that Eqs. 14.4-11 and 14.4-12 yield Eq. 14.4-13.
(b) Derive Eq. 14.4-14 from Eq. 14.4-13.
14.27 Show that Eq. 14.4-12 yields \epsilon_{x} = 0 for the following rigid-body rotations. In addition, make a sketch that shows the displaced position of the bar.
(a) u_{2} = -1, w_{2} = 3 , and L = 5 .
(b) u_{2} = -5, w_{2} = 5 , and L = 5 .
(c) u_{2} = -10, w_{2} = 0 , and L = 5 .
(d) u_{2} = -L(1 - \cos \theta), w_{2} = L \sin \theta.
14.28 For each of the rigid-body rotations stated in Problem 14.27, show that Eq. 14.4-13 yields R_{x} = R_{z} = 0 .
14.29 If \Delta represents an axial-displacement increment, the associated force in a uniform bar is F = AE\Delta/L . Show that Eq. 14.4-14 yields this result, whether the reference configuration of the bar in Fig. 14.4-1b is defined by \theta = 0 or by \theta = \pi/2 . Assume that strains are small.
Section 14.5
14.30 (a) Equation 14.5-8 has two roots, of which the lower, \lambda = \lambda_{\mathrm{cr}} , is given by Eq. 14.5-9. What is the other root and the corresponding mode shape?
(b) Verify that use of [k_{\sigma}] from Eq. 14.2-11 yields P_{cr} = -3EI/L^{2} for the problem of Fig. 14.5-2. Again use a single element.
14.31 Use the [k_{\sigma}] derived in Problem 14.8 to determine the buckling load in Fig. 14.5-2. Use one element. The only d.o.f. needed are w_{2} and \theta_{2} .
14.32 The two pin-connected bars shown may be considered rigid and weightless. The linear springs each have stiffness k .
(a) Determine the buckling load P_{\mathrm{cr}} .
(b) In the buckling mode, determine w_{3} if w_{2} = 1 . Sketch this mode, and show that "kickoff" forces [\mathbf{K}_{\sigma}][w_{2} - w_{3}]^{T} are equal in magnitude to forces in the deflected springs.
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z, w a a Rigid Rigid P 1 2 3 k k x
Problem 14.32
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z, w ← a → ← a → EI EI P 1 2 3 x F
Problem 14.33
14.33 The uniform beam shown is fixed at the left end, simply supported at the right end, and divided into two identical beam elements, each of length a .
(a) Set up a three-equation system [K + K_{\sigma}]\{D\} = \{R\} , then eliminate d.o.f. \theta_{2} and \theta_{3} by condensation, leaving w_{2} as the only d.o.f.
(b) Determine P_{\mathrm{cr}} (for the case F = 0 ) and compute its percentage error.
(c) Plot w_{2} / a versus P / P_{\mathrm{cr}} if F = 0.10EI / a^2 .
14.34 Determine buckling loads of the uniform columns shown. Express P_{\mathrm{cr}} in terms of E, I, and L . Use the [\mathbf{k}_{\sigma}] of Eq. 14.2-9.
(a) For Fig. (a), use \theta_{1} and \theta_{2} as d.o.f.
(a) For Fig. (a), use \theta_{1} and \theta_{2} as d. on.
(b) For Fig. (b), impose symmetry about the midpoint, so that \theta_{1} and w_{2} are the only d.o.f. needed. Node 2 is not a hinge.
(c) For Fig. (c), let the linear spring have stiffness k = 2EI / L^3 . The lower end is fully fixed. No axial d.o.f. is needed.
(d) For Fig. (d), let the rotational spring have stiffness k = EI / L . The lower
(d) For Fig. (d), let the rotational spring have stiffness k = EI / L . The lower end is fully fixed. No axial d.o.f. is needed.
(e) Repeat part (c) letting k \to \infty (top free to rotate but cannot translate).
(f) Repeat part (d) letting k \to \infty (top free to translate but cannot rotate).
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L P 1 2 P (a)
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P 2 k L 1
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P 2 k L 1
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L/2 L/2 P 1 2 3 P (b)
(c)
{d}
Problem 14.34
14.35 Repeat Problem 14.34, but use the [\mathbf{k}_{\sigma}] derived in Problem 14.8.
14.36 Repeat Problem 14.34, to the extent possible, if [\mathbf{k}_{\sigma}] is taken from Eq. 14.2-11.
14.37 (a) Consider use of the condensation technique (described by example in Eqs. 14.5-11 and 14.5-12). How will you decide which d.o.f. to eliminate?
(b) Using condensation, solve Problem 14.15(b).
(c) Using condensation, solve Problem 14.15(c).
(d) Using condensation to eliminate \theta_{1} , solve Problem 14.34(b).
(e) Using condensation to eliminate \theta_{1} , solve Problem 14.35(b).
14.38 The two slender bars shown, which are of lengths L and \alpha L but otherwise identical, are fixed at A and C and welded together at B , where they are simply supported and loaded by axial force P . The possibility of buckling is to be analyzed. Use the two-stage procedure suggested below Eq. 14.1-9: determine axial displacement at B in terms of P , hence compute axial bar forces and [\mathbf{k}_{\sigma}] for the bars, and finally seek P_{\mathrm{cr}} using \theta_2 as the only d.o.f.
14.39 The two-element frame shown is fixed at A and at C , and has the same EI throughout. The connection at B is rigid. Assume that d.o.f. that define axial strain are unnecessary because AE >> EI . Determine angle \beta that minimizes the buckling load P_{\mathrm{cr}} .
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A P B C L αL
Problem 14.38
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1.3a A B P β a C
Problem 14.39
14.40 Imagine that a diagonal stress stiffness matrix is proposed for the standard beam element, for which \{\mathbf{d}\} = \left[w_1 \quad \theta_1 \quad w_2 \quad \theta_2\right]^T . The form of [\mathbf{k}_{\sigma}] is [\mathbf{k}_{\sigma}] = [0 \quad c \quad 0 \quad c] , where c is a constant.
(a) Determine c by requiring that Eq. 14.2-7 be satisfied when w_{,x} is constant over the element.
(b) Use this [k_{\sigma}] to determine P_{cr} for a uniform pin-ended column. Consider a one-element model.
(c) Repeat part (b), but consider a two-element model: impose symmetry about the center of the column, so that nonzero d.o.f. of one-half the column are \theta_{1} and w_{2} .
Section 14.6
14.41 The weightless string shown is horizontal, under constant tension T , and carries two particles, each of mass m .
(a) Determine the static deflections of the particles caused by gravity.
(b) Determine the natural frequencies of vibration and the mode shapes.
14.42 Repeat Problem 14.41, but double the mass of the left-hand particle (to mass 2m).
14.43 A massless and flexible string of length 2L hangs from the ceiling. It carries two particles, each of mass m, one at the middle and the other at the lower end.
(a) What is the horizontal deflection of a small horizontal force Q applied to the lower end?
(b) What are the natural frequencies of vibration and the mode shapes?
14.44 The string shown is under tension T and has mass \rho per unit length. Use [M] and [K_{\sigma}] matrices associated with a cubic lateral-displacement field. Omit the conventional stiffness matrix [K]. Solve for the natural frequencies and mode shapes of small-displacement lateral vibrations. (The exact fundamental frequency is \omega_{1}^{2} = \pi^{2}T/4\rho a^{2} .)
(a) Use one element. Nonzero d.o.f. are then \theta_{1} and \theta_{2} .
(b) Use two elements and impose symmetry about the center. Nonzero d.o.f. to be used are then \theta_{end} and w_{center} .
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T m m T L L L
Problem 14.41
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T 2a
Problem 14.44
14.45 Solve Problem 14.44 using [k_{\sigma}] from Eq. 14.2-11, and (a) [M] based on a cubic lateral-displacement field, and (b) a lumped [M]. (That is, apply each of these mass matrices to each part of Problem 14.44.)
14.46 Model a simply supported beam by a single element. Let L = 1.0 \, \text{m} , A = 0.0002 \, \text{m}^2 , EI = 300.0 \, \text{N} \cdot \text{m}^2 , and \rho = 2100.0 \, \text{kg/m}^3 . Impose symmetry (and reduce the problem to a single d.o.f.) by setting \theta_2 = -\theta_1 .
(a) Determine the fundamental frequency \omega_{1} if there is no axial force.
(b) Determine the axial force that makes the frequency 347 rad/sec.
(c) Determine the frequency if the axial force is 1200 N in compression.
WEIGHTED RESIDUAL METHODS
The construction of approximate solutions of differential equations by means of weighted residual methods is summarized. The Galerkin method, which is the most popular weighted residual method, is used to produce finite element formulations.
15.1 INTRODUCTION
Thus far we have presented the finite element method as a Rayleigh–Ritz method—that is, as an approximation technique that is applied to a variational principle. A variational principle uses an integral expression, called a functional, that yields the governing differential equations and nonessential boundary conditions of a problem when operated upon by standard procedures of the calculus of variations. The principle of stationary potential energy is only one of many variational principles.
In an area of physical science other than structural mechanics, a variational principle may be unobtainable. This happens if the differential equation of the problem contains derivatives of odd order. A case in point is fluid mechanics, where, for some types of flow, all that is available are differential equations and boundary conditions. Yet the finite element method can still be applied by means of a weighted residual method. Like the Rayleigh–Ritz method, a weighted residual method uses integral expressions that contain the differential equations of a physical problem. Functional and residual formulations are both known as “weak” forms of stating the governing equations of a problem. The differential equations themselves comprise the “strong” form. (The weak form enforces conditions in an average or integral sense, whereas the strong form enforces them at every point.)
The following introductory treatment uses both structural and nonstructural problems to illustrate procedures.
15.2 SOME WEIGHTED RESIDUAL METHODS
This section presents an overview and uses the following notation:
u = dependent variable(s), for example, displacements of a point
x = independent variable(s), for example, coordinates of a point
f,g = functions of x , or constants, or zero
D, B = differential operators
Thus the governing differential equations and nonessential boundary conditions of an arbitrary physical problem are symbolized as
D u - f = 0 \quad \text { in domain } V \tag {15.2-1a}
B u - g = 0 \quad \text { on boundary } S \text { of } V \tag {15.2-1b}
For example, in beam bending Eq. 15.2-1a becomes EIw_{,xxx} = q , where w is lateral deflection and q is distributed lateral load. Thus D = EId^{4}/dx^{4} , u = w, and f = q. Equation 15.2-1b symbolizes two equations, namely, EIw_{,xx} - M_{B} = 0 and EIw_{,xxx} - V_{B} = 0 , where M_{B} and V_{B} are prescribed values of bending moment and transverse shear force at ends of the beam.
In general, the exact solution u = u(x) of Eq. 15.2-1a is unknown and is often difficult to determine. We seek instead an approximate solution, \bar{u} . Typically \bar{u} is a polynomial that satisfies essential boundary conditions and contains undetermined coefficients a_{1}, a_{2}, \ldots, a_{n} . Thus \bar{u} = \bar{u}(a, x) , and \bar{u} is “admissible” as defined in Section 3.2. To obtain an approximate solution we must determine values of the a_{i} such that u and \bar{u} are “close” in some sense.
If \bar{u} is substituted into Eqs. 15.2-1, equality does not prevail because \bar{u} is not exact. The discrepancy can be expressed as residuals R_{D} and R_{B} , which are functions of x and the a_{i} :
R _ {D} = R _ {D} (a, x) = D \bar {u} - f \quad (\text { interior residual }) \tag {15.2-2a}
R _ {B} = R _ {B} (a, x) = B \bar {u} - g \quad (\text {boundary residual}) \tag {15.2-2b}
In some physical problems it may happen that all boundary conditions are of the essential class. Then R_{B} need not enter; only R_{D} is used in determining the a_{i} of an approximation \bar{u} = \bar{u}(a, x) whose form satisfies essential boundary conditions a priori.
Residuals may vanish for some values of x , but they are not zero for all x unless \tilde{u} is the exact solution, \tilde{u} \equiv u . We presume that \tilde{u} is a good approximation of u if residuals are small. Small residuals can be achieved by various schemes, each of which is designed to produce algebraic equations that can be solved for the n coefficients a_i . Some popular schemes are summarized as follows. Their use is illustrated in Section 15.3.
Collocation. For n different values of x, the residuals are set to zero. The method is also called point collocation.
R _ {D} (a, x _ {i}) = 0 \quad \text { for } \quad i = 1, 2, \dots , j - 1 \tag {15.2-3a}
R _ {B} (a, x _ {i}) = 0 \quad \text { for } \quad i = j, j + 1, \dots , n \tag {15.2-3b}
Subdomain. Over n different regions of V and S, the integral of the residual is set to zero. The method is also called subdomain collocation.
\int_ {V _ {i}} R _ {D} (a, x) d V = 0 \quad \text { for } \quad i = 1, 2, \dots , j - 1 \tag {15.2-4a}
\int_ {S _ {i}} R _ {B} (a, x) d S = 0 \quad \text { for } \quad i = j, j + 1, \dots , n \tag {15.2-4b}
Least Squares. The a_{i} are chosen to minimize a function I:
\frac {\partial I}{\partial a _ {i}} = 0 \quad \text { for } \quad i = 1, 2, \dots , n \tag {15.2-5}
Function I is formed by integrating squares of the residuals,
I = \int_ {V} \left[ R _ {D} (a, x) \right] ^ {2} d V + \alpha \int_ {S} \left[ R _ {B} (a, x) \right] ^ {2} d S \tag {15.2-6}
where \alpha is an arbitrary scalar multiplier that may be used to achieve dimensional homogeneity and also serves as a penalty number. Larger values of \alpha increase the importance of R_{B} relative to R_{D} . The method is also called continuous least squares.
Least Squares Collocation. Equation 15.2-5 is still used, but I is redefined. It is now defined in terms of squared residuals at several points i, where i runs from 1 to m and m \geq n :
I = \sum_ {i = 1} ^ {j - 1} \left[ R _ {D} \left(a, x _ {i}\right) \right] ^ {2} + \alpha \sum_ {i = j} ^ {m} \left[ R _ {B} \left(a, x _ {i}\right) \right] ^ {2} \tag {15.2-7}
Equation 15.2-5 now yields n equations for the a_{i} , even when m > n. The method is also called point least squares and overdetermined collocation. If m = n, the method becomes simple collocation.
Galerkin. We select “weight functions” W_{i} = W_{i}(x) and set the weighted averages of residual R_{D} to zero. Or, in mathematical terms, we say that R_{D} is made orthogonal to the weight functions:
R _ {i} = \int_ {V} W _ {i} (x) R _ {D} (a, x) d V = 0 \quad \text { for } \quad i = 1, 2, \dots , n \tag {15.2-8}
In the Bubnov–Galerkin method, usually called simply the Galerkin method, weight functions W_{i} are coefficients of the generalized coordinates a_{i} . Thus W_{i} = \partial \bar{u} / \partial a_{i} . In the Petrov–Galerkin method, other forms of W_{i} are used.
In Galerkin methods, boundary residual R_{B} is used in combination with integration by parts, so as to introduce nonessential boundary conditions. The procedure is illustrated in subsequent examples.
Remarks. The commonality shared by the foregoing methods is that they all can be loosely symbolized as
\int_ {\Gamma} W _ {i} R d \Gamma = 0 \tag {15.2-9}
where R represents R_{D} and/or R_{B} and \Gamma represents V and/or S. In words, Eq. 15.2-9 says that over the region of interest, the weighted residual has an average value of zero (i.e., W_{i}R has zero average error). The various weighted residual methods differ in how W_{i} is defined [15.1]. In the collocation and subdomain
methods, the W_{i} are unit delta or step functions that are nonzero at certain points or over certain regions. In least squares methods, W_{i} = \partial R/\partial a_{i} . In the Galerkin method, W_{i} = \partial \bar{u}/\partial a_{i} .
In solving a problem, whether by the Rayleigh–Ritz method or a weighted residual method, we begin by establishing a trial family of solutions, \bar{u} = \bar{u}(a, x) . The a_{i} that define the best form of \bar{u} are chosen by the stationary–functional conditions \partial\Pi/\partial a_{i} = 0 for the Rayleigh–Ritz method or by Eq. 15.2-9 for a weighted residual method.
Galerkin's method yields a symmetric coefficient matrix if the system of differential equations and boundary conditions is self-adjoint [15.1]. If differential equations and a variational principle are both available, then the Galerkin method and the Rayleigh-Ritz method yield identical solutions when both use the same approximating function \bar{u} .
Least squares methods always produce a symmetric coefficient matrix. Other advantages of least squares include the avoidance of integration in least squares collocation and the “tuning” permitted by adjustment of \alpha in Eqs. 15.2-6 and 15.2-7. However, there are several disadvantages. Despite \alpha , the continuous least squares method may be too strongly influenced by less important residuals. (Separate \alpha_{i} may be applied to the separate residuals in least squares collocation.) The coefficient matrix tends to be ill conditioned. Because the weights are W_{i} = \partial R/\partial a_{i} , both R and the W_{i} contain derivatives of the same order. This means that in continuous least squares, integration by parts cannot reduce the highest-order derivatives of \bar{u} , and elements of higher order are therefore needed so as to achieve the necessary degree of interelement continuity. For example, a two-node, axially loaded bar element would require two d.o.f. at each node—namely, u and \varepsilon u_{,x} . This awkwardness could be avoided by reformulating the problem in terms of differential equations of lower order, but again more d.o.f. are required than in a simple displacement formulation, and element d.o.f. are a mixture of force and displacement quantities. Mixed formulations have not been popular in structural mechanics. The least squares collocation method does not require an initial reduction to first-order differential equations [15.4].
15.3 EXAMPLE SOLUTIONS
We illustrate the methods summarized in Section 15.2 by means of the following problem. Let the governing differential equation and nonessential boundary condition be ^{1}
u _ {, x x} + c x = 0 \quad \text { for } \quad 0 < x < L _ {T} \tag {15.3-1a}
u _ {, x} - b = 0 \quad \text { at } \quad x = L _ {T} \tag {15.3-1b}
where u has units of length, c is a constant having units (length) ^{-2} , and b is a dimensionless constant. The essential boundary condition is u = 0 at x = 0. A physical interpretation is given in Fig. 15.3-1. Equation 15.3-1a describes this
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y L_T σ_0 A, E q = q(x) x, u ∂²u/∂x² + q/AE = 0 for 0 < x < L_T E∂u/∂x = σ_0 at x = L_T
Figure 15.3-1. A physical interpretation of Eq. 15.3-1: axial displacement u of a uniform bar under linearly varying axial load q and end load \sigma_{0}A , where A is the cross-sectional area. E is the elastic modulus.
problem if q is the linear function q = q_0x and c = q_0 / AE . Numerical comparison of exact and approximate solutions appears in Table 15.3-1.
The exact solution of the problem is u = (3L_{T}^{2}cx - cx^{3} + 6bx)/6 , but we pretend that we do not know it. Instead, we seek two-parameter approximate solutions. Let the trial function be
\bar {u} = a _ {1} x + a _ {2} x ^ {2} \tag {15.3-2}
in which “best” values of a_{1} and a_{2} are required. Note that \tilde{u} satisfies the essential boundary condition u = 0 at x = 0. Substitution of Eq. 15.3-2 into Eqs. 15.3-1 yields the interior and boundary residuals
R _ {D} = 2 a _ {2} + c x \quad \text { and } \quad R _ {B} = (a _ {1} + 2 a _ {2} L _ {T}) - b \tag {15.3-3}
The nonessential boundary condition appears only at x = L_{T} , so R_{B} is evaluated at x = L_{T} .
As an alternative to Eq. 15.3-2, one can select a trial function \bar{u} that satisfies Eq. 15.3-1b a priori. Then R_{B} need not be used in subsequent manipulations. However, the present approach is more in accord with finite element applications of weighted residual methods.
Collocation. We arbitrarily elect to evaluate R_{D} at x = L_{T}/3 . Equations 15.3-3, with R_{D} = 0 and R_{B} = 0 , yield
a _ {1} = b + c L _ {T} ^ {2} / 3 \quad \text { and } \quad a _ {2} = - c L _ {T} / 6 \tag {15.3-4}
TABLE 15.3-1 RESULTS FOR THE PROBLEM OF EQ. 15.3-1 AND FIG. 15.3-1, FOR THE SPECIAL CASE b = 0, c = L_T = 1 .
| Quantity and Location | Exact | Collo- cation | Subdomain and Least Squares | Least Squares Collocation | Galerkin |
| u at $x = {L}_{T}/2$ | 0.2292 | 0.1250 | 0.1875 | 0.2500 | 0.2292 |
| u at $x = {L}_{T}$ | 0.3333 | 0.1667 | 0.2500 | 0.3333 | 0.3333 |
| ${u}_{,x}$ at $x = 0$ | 0.5000 | 0.3333 | 0.5000 | 0.6667 | 0.5833 |
| ${u}_{,x}$ at $x = {L}_{T}/2$ | 0.3750 | 0.1667 | 0.2500 | 0.3333 | 0.3333 |
| ${u}_{,x}$ at $x = {L}_{T}$ | 0 | 0 | 0 | 0 | 0.0833 |
Subdomain. We elect to evaluate Eq. 15.2-4a by integrating over the entire span, x = 0 to x = L_{T} . Thus Eq. 15.2-4a and R_{B} = 0 yield
a _ {1} = b + c L _ {T} ^ {2} / 2 \quad \text { and } \quad a _ {2} = - c L _ {T} / 4 \tag {15.3-5}
Least Squares. If \alpha = 1 / L_T the two terms in Eq. 15.2-6 have the same units, and
I = \int_ {0} ^ {L _ {T}} (2 a _ {2} + c x) ^ {2} d x + \frac {1}{L _ {T}} \left[ \left(a _ {1} + 2 a _ {2} L _ {T}\right) - b \right] ^ {2} \tag {15.3-6}
in which the latter term is already “integrated” over the point x = L_{T} . The equations \partial I/\partial a_{1} = 0 and \partial I/\partial a_{2} = 0 are found to yield Eqs. 15.3-5.
Least Squares Collocation. We arbitrarily elect to evaluate R_{D} at two points, and arbitrarily choose them to be x = L_{T}/3 and x = L_{T} . Residual R_{B} is as stated in Eq. 15.3-3. Again let \alpha = 1/L_{T} . The three residuals can be written in the form
\left\{ \begin{array}{l} R _ {D 1} \\ R _ {D 2} \\ R _ {B} \end{array} \right\} = \left[ \begin{array}{l l} 0 & 2 \\ 0 & 2 \\ 1 / L _ {T} & 2 \end{array} \right] \left\{ \begin{array}{l} a _ {1} \\ a _ {2} \end{array} \right\} - \left\{ \begin{array}{c} - c L _ {T} / 3 \\ - c L _ {T} \\ b / L _ {T} \end{array} \right\} \tag {15.3-7}
If all three residuals are set to zero, Eqs. 15.3-7 form an overdetermined system (three equations in two unknowns). A least squares solution requires that we form I = R_{D1}^{2} + R_{D2}^{2} + R_{B}^{2} and apply Eq. 15.2-5 for i = 1 and i = 2. We introduce symbols [Q], {a}, and {c} for arrays on the right-hand side of Eq. 15.3-7 and obtain the least squares solution for a_{1} and a_{2} as follows:
I = \{\mathbf {R} \} ^ {T} \{\mathbf {R} \} \quad \text { where } \quad \{\mathbf {R} \} = [ \mathbf {Q} ] \{\mathbf {a} \} - \{\mathbf {c} \} \tag {15.3-8a}
I = \{\mathbf {a} \} ^ {T} [ \mathbf {Q} ] ^ {T} [ \mathbf {Q} ] \{\mathbf {a} \} - 2 \{\mathbf {a} \} ^ {T} [ \mathbf {Q} ] ^ {T} \{\mathbf {c} \} + \{\mathbf {c} \} ^ {T} \{\mathbf {c} \} \tag {15.3-8b}
\left\{\frac {\partial I}{\partial \mathbf {a}} \right\} = \{\mathbf {0} \} \quad \text { yields } \quad [ \mathbf {Q} ] ^ {T} [ \mathbf {Q} ] \{\mathbf {a} \} = [ \mathbf {Q} ] ^ {T} \{\mathbf {c} \} \tag {15.3-8c}
In writing Eq. 15.3-8b we have used the relation \{\mathbf{c}\}^T [\mathbf{Q}]\{\mathbf{a}\} = \{\mathbf{a}\}^T [\mathbf{Q}]^T \{\mathbf{c}\} , which is true because each of the two matrix triple products is a scalar. In Eq. 15.3-8c, which is to be solved for \{\mathbf{a}\} , the coefficient matrix [\mathbf{Q}]^T [\mathbf{Q}] is symmetric and of the same order as \{\mathbf{a}\} . Applying Eq. 15.3-8c to Eq. 15.3-7, we obtain
a _ {1} = b + 2 c L _ {T} ^ {2} / 3 \quad \text { and } \quad a _ {2} = - c L _ {T} / 3 \tag {15.3-9}
Galerkin. From Eqs. 15.2-8 and 15.3-1a, the ith residual is
R _ {i} = \int_ {0} ^ {L _ {T}} W _ {i} (\bar {u}, _ {x x} + c x) d x \tag {15.3-10}
where i = 1,2 in the present example. It is standard practice in the Galerkin method to begin with integration by parts. A motivation is to reduce the order of differentiation in the integral. If derivatives of order 2m appear, the integral is defined if the integrand has continuous derivatives through order 2m - 1. In the