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Solution Manual for The Finite Element Method: Basic Concepts and Applications with MATLAB, MAPLE, and COMSOL,, 3rd Edition

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    2.1 Fill in the details leading to Eq. (2.17) and use it to find the weighted residuals formulation of Example 2.1 Solution: a) dx ck = 0 (Eq.l.U) Integration by parts: udv=uv '%Jw .Set " - ^ W and ^ ^ ^ _ ^ ^ , t h e n v=-K^ ^"'^ du =^dx dx dx d^T dx^ dx - (j>{x) -K V dJ^ dx dx dx -K dT L d^ dx dx dx Substitute into Eq. 2.13 and we get ° dx dx dx = 0 {Eq.lAl) b) 7(0) = and -K dT dx x=L KT-TJ Replacing in Eq. 2.17 K dx dx dx- ^Qdx-^(L)h(T(L)-TJ = 0

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    2.2 Use Eq. (2.17) to find the weighted residuals formulation of Eqs. (2.1) the heat flux q. Solution: - (2.3) in terms of cL ^d(l>dT K •''^ dx dx rL dx- ^Qdx + ^ -K dT_ dx = 0 The heat flux term at x = L must be zero, and dT -K dx Hence x=Q L ^dldT^^^^ f i ^ ^ dx + ^(0) q ^ dx dx •'^

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    2.3 Obtain expression (2.30). Solution: The weighted residuals form for the second term is >(X) dx = 0 After integration by parts on the first term K dx dx dx- ipQ dx + ^ •K dT_ dx = 0 L/2 Note: The boundary terms are set to zero i f no fluxes are prescribed. The functions for element 2 are J-/ Let T{x) = -{L-x)a,+j(x-LI2)a, {Eq.2.\9) Using E q . 2.28 1 L 2 2 " dx< . L/2 2 _ L _ L L_ a. L/2 -(x-L/2) ^ L dx- Or 4K rL " 1 - I dx- '^2 1 2Q rL < L/2 - 1 1 a. L ' L/2 (x-L/2) < > — < 0 y 0 V J 0 Integrating we have 2K ' 1 - f < « 2 1 > dx-< 0 0 L - 1 1 1 ^3, 4 h/2 1

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    Consider the equation + w + x = 0, 0 < j c < l with w(0) = u ( l ) = 0. Assume an approximation for u{x) = i(x) with ^ ( x ) = x ( l - The residual is R{u, x) - -2a^ + a , - x ) ] + X and the weight Wi{x) = (j) \{x) = JC (1 - x). Find the solution from the following integral .Solution: rl rl W{x) R(u, x)dx= x(l - x)(-2a^ + a^x(l -x) + x)dx = 0 Or rl 0 W(x) R(u, x)ck= ^ (-2a^x + (3a, + l)x^ - 2a^ + l)x^ + a^x"^)ck = 0 -a, 4 - - ( 3 a + l ) - - ( 2 a , + l ) + - a = — + — = 0 1 3V 1 ^ 4 V 1 ; ^ 1 10 ' 12 5 therefore 5 ' 1 8 18 Compare with the exact solution given by u (x) X sinl Evaluating at x=l/2 u{VI7) — 0.069444 ^^^^ "^^^^ error. w * ( l / 2 ) = 0.069747

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    2.5 Use the weak statement formulation to find solutions to the equation dx" w(0) = 0, = 1, 0 < j c < l du dx (a) Using linear interpolation functions, as explained in Section 2.3, use (i) one element and (ii) two elements. (b) Using simple polynomial functions that satisfy the boundary conditions at the left- hand side, i.e., (i) u(x) = a\x, hence ^ i ( x ) = W\{x) = x (ii) u{x) = (afix+a2X^, hence (f>\{x) = Wx{x) = x, (l >2 {x )=W2ix )-^x\ (c) Using circular functions that satisfy the boundary conditions at the left-hand side, i.e., (i) u{x) = a/ sin (n/2Lx) (ii) u{x) = «y sin (n/2Lx) +aa2 sin (3n/2Lx) (d) Find the analytical solution and compare with results from (a), (b) and (c). Solution a) The residual function is ]^(u^x) = ^ + 1 The weak form after integration by parts is dx^ f dW du ^ dx- du + W dx- -W — Jo ^ dx dx J dx = 0 i) One element solution u{x) = ( l - x ) ^ ! +X<32 j ; [ - 1 \\dx< dx-' 0 0 [ - 1 \\dx< dx-' . — < > 1 _ [ - 1 \\dx< J O 1 0 Integrating flf] -<22 = - 1 / 2 —a^^ + ^2 = 1 / 2 since a | = 0 then a 2 = l / 2 or u(x) = x/2

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    Two element solution. Element 1: H / 2 ~-2~ [ - 2 2]dx\ rl /2 ~l-2x~ 0] ol " 2 - 2 " - 1 / 4 [ - 2 2]dx\ > + dx-< > — < > or < > .0 _ 2 _ [^2 J 0 _ 2 x _ oj OJ _ - 2 2 _ - 1 / 4 Element 2 ^ 2 = 2 ( l - x ) ^ 3 - 2 ( x - l / 2 ) r - 9 i - 2 2]dx 1/2 - 2 2 a-, 1/2 2 ( 1 - x ) 2 ( x - l / 2 ) J Assemble dx- the > or 2 - 2 - 2 2 ^2 «3 - 1 / 4 3 / 4 elements " 2 - 2 0 " - 1 / 4 ^ - 2 4 - 2 < «2 - 1 / 2 > 0 - 2 2 _ 3 / 4 > since a^^Q weget a2=ll^ and a^=lf2 u(x) = 3 x - l x / 4 0 < x < l / 2 l / 2 < x < l b) Using polynomials i) ^ j = x u[x) = a^x (la^ + x)dx-l = 0 , a^=\l2 , i / ( x ) = x / 2 , same as using one element. ii) ^ i - x , (l>2=x^ ' u{x) = a^x + a2X^ F o r W = ^ | r [1 • (<3i + 2 « 2 ^ ) + - 1 = 0 r i r F o r W = ^ 2 Jo • ( a , + 2 a 2 x ) + x ^ k/x - 1 = 0 T h e system becomes ^1 +^2 = 1 / 2 4 a i + —^9 = 2 / 3 or c) a = 0 , ^2 = 1 / 2 <3«(i w ( x ) = X ^ / 2

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    Using circular functions i) ^ } = sin — X v2 J , u {x) = ai sin — X v2 y , d>\ — cos ^ 2 — X V 2 y ^ 2 — f l j cos ;7r — X 2 + sm — X v2 y ti6c-sin — X v 2 / - 0 0 Use the identities 2 1 cos axdx = — (si n ax cos ax - ax) and 2a • • sin V n — X 2 cos 8 ; T — X v 2 , 1 / + 0 ^ 2 cos n — X V2 )) - s m — X v 2 y du dx 2 7t sin — X cos — X U J ;r l 2 j .Thei = 0 or hence a, » 0.295 and M ( X ) = 0.295 s i n — X V 2 J ii) ^ 2 = s m F o r r = ^ : For r = ^2 : X d) ^ 2 f3^ sin — X + A T sin X sin l 2 y I 2 j cos 3 — c o s — X V 2 y flj cos V — X v2 y + < 3 r 2 cos X + sin 3 — X 2 ^1 cos V + ^2 cos 3;r — X v 2 y ( i r - s i n •X J J + sin 3 ;7- •X 2 (ir-sin / V dx - 0 0 3;r \du — X 1 J dx 2 — flfj + 0 • (32 H 1 = 0 (aj is unchanged) 8 n 97r^ 2 0 - a , + — + 1 = 0 . fl2~-0-109 and ^ 8 ^ 3;r ^ w ( x ) « 0.295 s i n - 0 . 1 0 9 s i n — X v2 y Analytical 3;r X solution -.2 = 1 , integrating w ( x ) = — + CjX + C2 and applying boundary conditions u ( x ) =

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    Notice that all the approximations are exact at x = 1/2, and the Galerkin approximations using x 2 and X both yield the exact solution. Also notice that the boundary condition at x = 1 is satisfied only approximately. I n case a), the first solution gives ^ dx _ i . and the second solution gives 2 du dx JC-l _ ^ which is getting better.This point is discussed in more detail in problem 2.6. 4

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    2.6 Subdivide the interval 0 < x < Z into 10 linear elements, construct the element equations for four consecutive elements starting with element four, and show that none of the parameters aj is related to more than two additional parameters in the equations. The only ones involved being ai-i and a,+] (as a consequence of this, the final system of linear equations will involve a tri-diagonal matrix which is easy to solve). Solution: For an element e^^ ={x/Xj^ <x<Xi^_^_^) the element stiffness matrix is obtained from r^k+'i ^ . All the stiffness matrices must be equal, so let's calculate the first dx dx one. A 1 1^ ^ 1 =1 X ^2 = 10 L and after assembling 1 0 ^ thus 1 - 1 0 0 r i / l O L - 1 2 - 1 0 K 0 1 2 - 1 10 ~ L 10 L 0 0 - 1 2 10 10 dx=- lOK 0 0 0 0 0 0 0 0 0 0 - 1 0 - 1 1 a , a-, a 10 a 11 • 1 - 1 - 1 1 Note that each equation / involves only three unknowns and a^^^, so the stiffness matrix is tri-diagonal.

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    2.7 The Rayleigh-Ritz formulation for the problem of solving Eq. (2.1) with boundary conditions (2.2) and (2.3) can be stated as: Minimize the fiinctional F{T) = £ 2 dx -QT\dx-Tq .t=0 over all functions T{x) with square integrable first derivatives that satisfy Eq. (2.3) at x = Z. Approximate T{x) using two linear elements as we did before and replace into Eq. (2.1) to obtain F{T) = F(«], ai, a^). Now minimize F{a\, a^, ai) as a function of three variables. Show that the final system of equations is identical to Eq. (2.31), thus, in this case, the Galerkin method and the Rayleigh-Ritz method are equivalent. Solution: Element I : (b^ =1 X ^ L , 2 > 2 = — X L rLIlK V — a, + — ^2 L ^ L ^ \ -Q Jo 2 (1) Element2: 2 2 r 2 ^ 1 X V L J 2 ^ K. .2 QL. . a^+-xa2 ]fh-a^g = —(a2-ai) ——(a^+a2)-a^q K L 'L K_ 111 2 2 2 fln H Qn L ^ L ^ \ -Q 2 X L - x - 1 L a^\bc-a^q = —{a^-a2) — ; ^ ( « 2 + « 3 ) (2) F(T) = (1) + (2) = ^{a2-a,f+j {a^ - «2 ) ' ( ^ i + 2^2 + «3 ) " ^ ^ / 2 2 2 \ F ( r ) = F ( a j , a 2 , ) = — (^fli - 2(2 ,<32 + 2 ^2 - 2fl2<^3 + ^3 j The conditions for a minimum are dF _dF _ dF da^ 5^2 ^^3 _ Q thus {a,-a2)-^-q = 0 L dF 2K da2 L dF IK L (-flj +2^2 "^^2) ( a 3 - a , ) - ^ = 0 QL In Matrix form 2K L "1 - 1 0"" a2 a - 1 2 - 1 < a2 2 > + < 0 > 0 - 1 1 4 0 - 1 1 1 0 which is identical to Eq. 2.31.

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    2.8 Let us consider once more Problem 2.3 above. Assume that w(x) = aix^ and Wi(x) = x^, but this time use E q . (2.11) and proceed as in Problem 2.2 in order to obtain a solution. Show that this solution is identical to that obtained in Problem 2.3(b)(ii), and hence, the process of integration by parts does not introduce any changes. Solution: - 1 dx = rl x\2a -\)dx = 0 '2 3 x'^ — x a, 2 1 = 0 , —a — = 0 , a, = 1 / 2 , w(; 3 3 '

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    2.9 In example 2.2, first find the exact analytical solution, then calculate two more Galerkin approximations using Solution: The analytical solution is ^dWdu ii(^x) = —{^x-x ^y The Galerkin approximation is dx dx -W dx = 0' u(x) = (3] sin ;rx + sin 27tx , = sin TTX , sin ITTX a) rl oL 0 _ cos TTxf^a^ cos TTX + laj cos 2;rx) - sin nx dx = 0 ITV^ cos Inx cos nx + 2a2 cos Inx) - sin 2-;rx tir = 0 or — fli+0-a2 = 0 , a^= — u{x) - — sin ;rx n , <3«flf 0 • ^1 + 2:7r ^2 +0 = 0 , b) M (x) = aj sin:;rx + a2sin2 ;7rx + a3sin3 :7rx , f P j = s i n ; r x , W^^^VCLITIX , H^=sin3;Tx We only need to calculate , and remain the same due to the orthogonality of the sine functions. : 3;r^ cos 3;rx(fl} cos ;7rx +2^2 cos 2;rx4-3(33 cos 3 ; r x ) - s i n 3:^x r 2 ' 3;r ' Tin" , W3 (x) = — 3;r'* 1 . sin ;rx + — sin 3;rx 27

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    3.1 Repeat the development of expressions (3.7) to (3.12) using two linear elements to discretize the interval 0<x<L. Then combine the element shape functions to construct the global shape functions and compare your result to that obtained using expressions (3.2) to (3.4) directly, with n = 2. Solution: Let T(x) = a + j3x with Tix^) = T- and 7 ( ^ , ^ 1 ) = T-^^ • Hence a + fix^ = T- and a + y^jc^^j = T^^j. Therefore a = and p — ^+1 and X Rearranging T{x) = x-x^ ^. This expression is valid for any element, in r / N,(x) = \ N2 (X) = { particular if we have two elements N.(x) = \ x-^-x yXj -x^ J ^ X ^ X2 0 otherwise X X] ^ X2 - Xi y Xo - X y X3 - X2 y X | < X < X2 X2 < X < X3 0 otherwise X-X2 X2 < X < X3 0 otherwise N,{x) = If Xi = 0, X2 = 1 / 2 and X3 = L Then N,(x) = - ^ ( Z - 2 x ) 0<x<L/2 0 otherwise 2 — X 0<x<L/2 L -(L-x) L/2<x<L L 0 otherwise -{2x-L) L/2<x<L L 0 otherwise These expressions are exactly the same as those obtained from Eqs, 3.2 - 3.4.

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    3.2 Show that if in Eq. (3.6) we replace Tn+i = T L , the boundary condition (2.3) is automatically satisfied. Solution: n+l n J From Eq. 3.6 j ^ ^ ^ ) ^ ^ (^)7^. ^ £ j^j, ^^)T. + N^^^ (x)T^ • Evaluate at X = X„^l = L . Then ^ii^n+l) = ^ M i = h2,...,n and iV„^.i(x„^i) = 1 • Hence T {x^^^) = Tj^ .

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    3.3 Derive expression (3.25) from (3.24) for cubic elements. Then find a relation between global and local reference systems such as the ones given in Figure 3.3 and 3.6 for linear and quadratic elements. Solution: Let h be the size of the element. There are several ways to obtain the shape functions, the easiest is to recognize that AT. (jc^.) = ^-y, N-^ (x^ ) = 1 andN- (xj ) = 0 if / ^ J . So for /• = 1, i\^i (0) = l , N^(h /3) = N ^(2h/3) = N^(h) = 0 and therefore N^ix) must have the form TVj (x) = a(x-h/ 3Xx -2hl 3)(x- h) and imposing (0) = 1 we get ^ ^ 11? AT, (x) = — ^ (x - / 2 / 3)(x - Ih 1 3)(x - /z) = (1 - 3x / - 3x / 2/z)(l -xlh) 2h ah Similarly we get N2 (x) = -ax{x - 2h / 3)(x - /z), / 3) = 1 = y (-/? / 3)(-2h 1 3) so ^ ^ ^ 7 and iV2(x) = ^ x ( x - 2 / z / 3 ) ( x - ; 7 ) = — ( l - 3 x / 2 / z ) ( l - x / i ? 7 ) - ^^2(^) ^ 3 ( ^ ) 2h^ 2h^ h are similarly obtained and given by M^i^x) = -—{\-3xlh){\-xlh) and N^{x) = --{\-Zxlh){\-3xl2h) 2h h The relation between the local and global system in element is given by Local Global Node Node numbers numbers 1 3i+l 2 3i+2 3 3i+3 4 3i+4

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