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For the volume of Figure 5.8b, during a time interval dt, Equation 5.53 is expressed as∂q xq x A dt + Q A dx dt = U + q x +dx A dt(5.54)∂xwhereqx = heat flux across boundary (W/m2, Btu/hr-ft2);Q = internal heat generation rate (W/m3, Btu/hr-ft3);U = internal energy (W, Btu).The last term on the right side of Equation 5.54 is a two-term Taylor seriesexpansion of qx(x, t) evaluated at x + dx .

Note the use of partial differentiation,since for now, we assume that the dependent variables vary with time as well asspatial position.The heat flux is expressed in terms of the temperature gradient via Fourier’slaw of heat conduction:q x = −k x∂T∂x(5.55)153Hutton: Fundamentals ofFinite Element Analysis1545. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted Residualswhere k x = material thermal conductivity in the x direction (W/m-°C,Btu/hr-ft-°F) and T = T (x , t ) is temperature.

The increase in internal energy isU = c A dx dT(5.56)wherec = material specific heat (J/kg-°C, Btu/slug-°F); = material density (kg/m 3 , slug/ft 3 ).Substituting Equations 5.55 and 5.56 into 5.54 gives∂∂TQ A dx dt = c A dx dT +−k xA dx dt∂x∂x(5.57)Assuming that the thermal conductivity is constant, Equation 5.57 becomesQ = c∂T∂T 2− kx 2∂t∂x(5.58)For now we are interested only in steady-state heat conduction and for the steadystate ∂ T /∂t = 0, so the governing equation for steady-state, one-dimensionalconduction is obtained askxd2 T+Q=0dx 2(5.59)Next, the Galerkin finite element method is applied to Equation 5.59 toobtain the element equations. A two-node element with linear interpolation functions is used and the temperature distribution in an element expressed asT (x ) = N 1 (x )T1 + N 2 (x )T2(5.60)where T1 and T2 are the temperatures at nodes 1 and 2, which define the element,and the interpolation functions N1 and N2 are given by Equation 5.20.

As in previous examples, substitution of the discretized solution (5.60) into the governingdifferential Equation 5.55 results in the residual integrals:x2 d2 Tk x 2 + Q Ni (x) A dx = 0dxi = 1, 2(5.61)x1where we note that the integration is over the volume of the element, that is, thedomain of the problem, with dV = A dx.Integrating the first term by parts (for reasons already discussed) yieldsxx2x2dN i dTdT 1dx + Ak x A N i (x )− kx AQ N i (x ) dx = 0i = 1, 2dx x1dx dxx1x1(5.62)Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.5 One-Dimensional Heat ConductionEvaluating the first term at the limits as indicated, substituting Equation 5.60into the second term, and rearranging, Equation 5.58 results in the twoequationsx2x2dN 1 dN 1dN 2dT kx AT1 +T2 dx = AQ N 1 dx − k x A(5.63)dxdxdxd x x1x1x2kx Ax1x2dN 2 dN 1dN 2dTT1 +T2 dx = AQ N 2 dx + k x Adxdxdxdxx1x1(5.64)x2Equations 5.63 and 5.64 are of the form[k]{T } = { f Q } + { f g }(5.65)where [k] is the element conductance (“stiffness”) matrix having terms definedbyx2dN l dN mklm = k x Adxl, m = 1, 2(5.66)dx dxx1The first term on the right-hand side of Equation 5.65 is the nodal “force” vectorarising from internal heat generation with values defined byx2f Q1 = AQ N 1 dxx1(5.67)x2f Q2 = AQ N 2 dxx1and vector {fg} represents the gradient boundary conditions at the elementnodes.

Performing the integrations indicated in Equation 5.66 gives the conductance matrix ask x A 1 −1[k] =(5.68)L −1 1while for constant internal heat generation Q, Equation 5.67 results in the nodalvectorQ AL 2{ fQ} =(5.69)Q AL 2155Hutton: Fundamentals ofFinite Element Analysis1565. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted ResidualsThe element gradient boundary conditions, using Equation 5.55, described by −dT  dx q|x1x1=A{ f g } = kx A(5.70)−q|x2dT dx x2are such that, at internal nodes where elements are joined, the values for theadjacent elements are equal and opposite, cancelling mathematically. At externalnodes, that is, at the ends of the body being analyzed, the gradient values may bespecified as known heat flux input and output or computed if the specified boundary condition is a temperature.

In the latter case, the gradient computation isanalogous to computing reaction forces in a structural model. Also note that thearea is a common term in the preceding equations and, since it is assumed tobe constant over the element length, could be ignored in each term. However, aswill be seen in later chapters when we account for other heat transfer conditions,the area should remain in the equations as defined.

These concepts are illustratedin the following example.EXAMPLE 5.6The circular rod depicted in Figure 5.9 has an outside diameter of 60 mm, length of 1 m,and is perfectly insulated on its circumference. The left half of the cylinder is aluminum,for which kx = 200 W/m-°C and the right half is copper having kx = 389 W/m-°C. Theextreme right end of the cylinder is maintained at a temperature of 80°C, while the leftend is subjected to a heat input rate 4000 W/m2.

Using four equal-length elements, determine the steady-state temperature distribution in the cylinder.■ SolutionThe elements and nodes are chosen as shown in the bottom of Figure 5.9. For aluminumelements 1 and 2, the conductance matrices are[k al ] =kx A 1L −1−11=200(/4)(0.06) 2 1−10.25−11Alqin1120.25 m= 2.26Cu230.25 m340.25 m1−1qout450.25 mFigure 5.9 Circular rod of Example 5.6.−1W/ ◦ C1Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 20045.5 One-Dimensional Heat Conductionwhile, for copper elements 3 and 4,kx A 1[k cu ] =L −1−11389(/4)(0.06) 2 1=−10.25−111= 4.40−1−1W/ ◦ C1Applying the end conditions T5 = 80°C and q1 = 4000 W/m2, the assembled systemequations are  2.26 −2.260004000 T1  −2.26 4.52 −2.26  00   T2   0  (0.06) 2 0−2.26 6.66 −4.400  T3 =0 4 0T0−4.40 8.80 −4.40  4   0 80000−4.40 4.40−q511.31 0=00−0.0028q5Accounting for the known temperature at node 5, the first four equations can be written as  T1   11.31 2.26 −2.2600   0  −2.26 4.52 −2.260  T2 = 0T  −2.26 6.66 −4.40 0  3 T400−4.40 8.80352.0The system of equations is triangularized (used here simply to illustrate another solutionmethod) by the following steps.

Replace the second equation by the sum of the first andsecond to obtain  2.26 −2.2600T1   11.31  02.26 −2.260  T2 = 11.31 0−2.26 6.66 −4.40  T  0  3 00−4.40 8.80T4352.0Next, replace the third equation by the sum of the second and third  2.26 −2.2600T1   11.31    02.26 −2.260  T2 = 11.31 004.40 −4.40 T  11.31  3 T400−4.40 8.80352.0Finally, replace the fourth with the sum of the third and fourth to obtain  2.26 −2.2600T1   11.31  02.26 −2.260  T2 = 11.31 004.40 −4.40  T  11.31  3 0004.40T4363.31157Hutton: Fundamentals ofFinite Element Analysis1585. Method of WeightedResidualsCHAPTER 5Text© The McGraw−HillCompanies, 2004Method of Weighted ResidualsThe triangularized system then gives the nodal temperatures in succession asT4 = 82.57 ◦ CT3 = 85.15 ◦ CT2 = 90.14 ◦ CT1 = 95.15 ◦ CThe fifth equation of the system is−4.40T4 + 4.40(80) = −0.0028q 5which, on substitution of the computed value of T4, results inq 5 = 4038 .6 W/m 2As this is assumed to be a steady-state situation, the heat flow from the right-hand end ofthe cylinder, node 5, should be exactly equal to the inflow at the left end.

The discrepancyin this case is due simply to round-off error in the computations, which were accomplished via a hand calculator for this example. If the values are computed to “machineaccuracy” and no intermediate rounding is used, the value of the heat flow at node 5 isfound to be exactly 4000 W/m2. In fact, it can be shown that, for this example, the finiteelement solution is exact.5.6 CLOSING REMARKSThe method of weighted residuals, especially the embodiment of the Galerkinfinite element method, is a powerful mathematical tool that provides a techniquefor formulating a finite element solution approach to practically any problem forwhich the governing differential equation and boundary conditions can be written.

For situations in which a principle such as the first theorem of Castiglianoor the principle of minimum potential energy is applicable, the Galerkin methodproduces exactly the same formulation. In subsequent chapters, the Galerkinmethod is extended to two- and three-dimensional cases of structural analysis,heat transfer, and fluid flow. Prior to examining specific applications, we examine, in the next chapter, the general requirements of interpolation functions forthe formulation of a finite element approach to any type of problem.REFERENCES1.2.Stasa, F. L.

Applied Element Analysis for Engineers. New York: Holt, Rinehart, andWinston, 1985.Burnett, D. S. Finite Element Analysis. Reading, MA: Addison-Wesley, 1987.Hutton: Fundamentals ofFinite Element Analysis5. Method of WeightedResidualsText© The McGraw−HillCompanies, 2004ProblemsPROBLEMS5.1 Verify the integration and subsequent determination of c1 in Example 5.1.5.2 Using the procedure discussed in Example 5.4, determine three trial functions forthe problem of Example 5.1.5.3 It has been stated that the trial functions used in the method of weighted residualsgenerally satisfy the physics of the problem described by the differential equationto be solved. Why does the trial function assumed in Example 5.3 not satisfy thephysics of the problem?5.4 For each of the following differential equations and stated boundary conditions,obtain a one-term solution using Galerkin’s method of weighted residuals and thespecified trial function.

In each case, compare the one-term solution to the exactsolution.a.d2 y+ y = 2xdx 2y(0) = 00≤x ≤1y(1) = 0N 1 (x ) = x (1 − x 2 )b.d2 y+ y = 2 sin xdx 2y(0) = 00≤x ≤1y(1) = 0N 1 (x ) = sin xc.dy+ y 2 = 4xdxy(0) = 00≤x ≤1y(1) = 0N 1 (x ) = x 2 (1 − x )d.dy−y=2dxy(0) = 00 ≤ x ≤ 10y(10) = 0N 1 (x ) = x 2 (10 − x ) 2e.dyd2 y−3+y=xdx 2dxy(0) = 00≤x ≤1y(1) = 0N 1 (x ) = x (x − 1) 2159Hutton: Fundamentals ofFinite Element Analysis1605.

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