Spørgsmål 1 (5%) Forklar med relevant argumentation, at den stationære temperaturfordeling i områdets indre er bestemt ved følgende randværdiproblem

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1 SKRIFTLIG EKSAMEN I NUMERISK DYNAMIK Bygge- og Anlægskonstruktion, 8. semester Fredag den 9. juni 006, kl Alle hjælpemidler er tilladt OPGAVE y u = 0 isoleret rand r u = u 0 θ 0 θ c c u = 0 x Figuren viser en sektor af et ringformet plant område med indre radius c og ydre radius c. Centervinklen af sektorområdet er θ 0. Der indlægges et polært (r, θ)-koordinatsystem som vist på figuren. Langs pollinierne θ = 0 og θ = θ 0 holdes temperaturen konstant på værdien u = 0. Langs den indre periferi r = c antages området perfekt isoleret. Langs den ydre periferi r = c holdes temperaturen konstant på værdien u = u 0. Spørgsmål (5%) Forklar med relevant argumentation, at den stationære temperaturfordeling i områdets indre er bestemt ved følgende randværdiproblem u r + u r r + u ] c [ r θ = 0, r, c ] c [ u = 0, r, c, θ ]0, θ 0 [, θ = 0 θ = θ 0 u = u 0, r = c, θ ]0, θ 0 [ u r = 0, r = c, θ ]0, θ 0 [ Spørgsmål (5%) Bestem den stationære temperaturfordeling i området ved hjælp af separationsmetoden.

2 OPGAVE Givet et begyndelsesværdiproblem bestemt ved følgende bevægelsesligning og begyndelsesbetingelser for et udæmpet system af frihedsgrad } ẍ + x = e t, t > 0 x(0) =, ẋ(0) = 0 hvor ẋ = d x betegner differentiation med hensyn til tiden t. dt Spørgsmål (5%) Bestem løsningen til begyndelsesværdiproblemet ved hjælp af Laplacetransformation. Hjælp: Følgende opløsning i stambrøker kan være nyttig ved opgavens løsning s + s + (s + )(s + ) = s + + s s + + s + OPGAVE 3 Begyndelsesværdiproblemet i opgave betragtes igen, idet der nu tilstræbes en numerisk løsning. Spørgsmål (5%) Bestem funktionsværdien x(0.) ved hjælp af Newmark tidsintegration med parameterværdierne β =, γ = og tidsskridtet t = 0.. Spørgsmål (5%) Følgende spørgsmål ønskes besvaret med relevant argumentation: : Er algoritmen numerisk stabil med det anførte tidsskridt? : Er algoritmen forbundet med numerisk dæmpning? 3: Er algoritmen forbundet med periodefejl, og bestem i givet fald periodetilvæksten pr. periode?

3 OPGAVE 4 Givet et generelt egenværdiproblem defineret ved følgende masse- og stivhedsmatricer 0 M = 3 0, K = Spørgsmål (0%) Beregn de to laveste egensvingningsformer og tilhørende egenværdier ved subspace iteration, idet der benyttes følgende startbasis Φ 0 = Φ () 0 Φ () 0 = 0 0 Opgaven betragtes som løst, når iteration er gennemført. Det understreges, at uagtet en analytisk løsning på problemet kan tilvejebringes, ønskes kun den numeriske løsning bestemt. OPGAVE 5 Det generelle egenværdiproblem defineret i opgave 4 betragtes igen. Spørgsmål (0%) Undersøg ved hjælp af Gauss faktorisering eller Sturmsekvens check, hvor mange egenværdier, der er mindre end.. Spørgsmål (5%) Hvad slutter man ud fra analysen i spørgsmål om egenværdierne bestemt ved subspaceiteration i opgave 4?

4 SOLUTIONS PROBLEM Question : The considered boundary value problem reads u r + u r r + u ] c [ r θ = 0, r, c ] c [ u = 0, r, c, θ ]0, θ 0 [, θ = 0 θ = θ 0 u = u 0, r = c, θ ]0, θ 0 [ u r = 0, r = c, θ ]0, θ 0 [ The stationary temperature distribution is determined by Laplacet s differential equation u = 0. With the Laplace operator specified in polar coordinates the differential equation in () is obtained, cf. Section 3., p. 56. The temperature is prescribed at the value u = 0 along the polar lines θ = 0 and θ = θ 0, and at the value u = u 0 along the outer periphery, which specify the Dirichlet boundary conditions of the problem. The isolated inner periphery implies that no heat transfer takes place through this boundary, corresponding to the gradient of the temperature distribution in the direction of the outward directed normal vanishes, i.e. u u = 0. Since n = r, the Neumann boundary condition = 0 is valid at n r the inner periphery. () Question : The separation method is used, i.e. we search for product solutions u(r, θ) = R(r)Θ(θ) to the partial differential equation in (). Insertion and separation of the variables provides the following ordinary differential equations R (r)θ(θ) + r R (r)θ(θ) + r R(r)Θ (θ) = 0 R (r) + r R (r) R(r) = Θ (θ) Θ(θ) = λ r R (r) + rr (r) λ R(r) = 0 () Θ (θ) + λ Θ(θ) = 0 (3)

5 The general solution of (3) reads Θ(θ) = c 3 cos(λ θ) + c 4 sin(λ θ) (4) The boundary condition u(r, θ) = u(r, 0) = 0 along the polar line θ = 0 implies that Θ(0) = c 3 = 0. The boundary condition u(r, θ 0 ) = 0 along the polar line θ = θ 0 then implies that Θ(θ 0 ) = c 4 sin(λ θ 0 ) = 0 λθ 0 = nπ λ = λ n = nπ θ 0, n =,,... (5) Hence, solutions to (3), which fulfills the boundary conditions along the polar lines θ = 0 and θ = θ 0, have the form Θ n (θ) = c 4 sin(λ n θ), n =,,... (6) () is a Cauchy-Euler differential equation, cf. p. 94. The auxiliary equation reads, see (4.7-) m(m ) + m λ = 0 m = { λ λ (7) The general solution becomes, cf. (4.7-) R n (r) = c r λn + c r λn (8) Then, product solutions which satisfy the partial differential equation in (), and the boundary conditions along the polar lines θ = 0 and θ = θ 0 has the form u n (r, θ) = λ n = nπ θ 0 (A n r λn + B n r λn ) sin(λ n θ), n =,,... (9) where A n = c c 4 and B n = c c 4. From the superposition principle follows that any linear combination of solutions of type (7) also fulfills the partial differential equation and the boundary conditions along the polar lines θ = 0 and θ = θ 0. Then, the general solution may be written as u(r, θ) = u n (r, θ) = n= n= (A n r λn + B n r λn ) sin(λ n θ) (0) What remains is to determine the expansion coefficients A n and B n, so the boundary conditions at the outer and inner periphery are fulfilled. These become

6 ) u(c, θ) = (A n c λn + B n c λn sin(λ n θ) = u 0 () n= u(r, θ) ( ( c ) λn ( c ) ) λn r = A n λ n Bn λ n sin(λ n θ) = 0 () r=c/ n= () must hold for arbitrary θ. This can only be fulfilled if the term within the bracket vanishes, leading to the condition ( c ) λn ( c ) λn A n λ n Bn λ n = 0 ( c B n = ) λn An (3) Then, () may be written u 0 = n= ) C n sin (nπ θθ0 C n = c λn( + λn) A n (4) (4) is a Fourier sine series with the period p = θ 0, cf. Definition.6. The coefficients C n become, cf. (.3.4), (.3.5) C n = θ 0 θ0 0 u 0 sin (nπ )dθ θθ0 cos(nπ) = u 0 sin(nπξ)dξ = u 0 0 nπ = u 0 ( ) n nπ Insertion of (3), (4) and (5) into (0) provides the following final result for the temperature distribution (5) ( ) n u(r, θ) = u 0 nπ n= ( r c ) λn ( + λ n r λn ) c) sin (nπ θθ0 + λn, λ n = nπ θ 0 (6) PROBLEM Question : The Laplace transformation is applied to the left and right hand sides of the ordinary differential equation leading to L{ẍ(t)} + L{x(t)} = L{e t } ( ) s X(s) sx(0) ẋ(0) + X(s) = s + ()

7 where Theorems 7., p.30 and 7.4, p. 37 have been used. Insertion of the initial conditions in () provides the following solution for the Laplace transform X(s) = L{x(t)} of the solution X(s) = s + s + (s + )(s + ) = s + + s s + + s + () Then, the solution of the initial value problem follows from Theorem 7.3, p. 34 x(t) = ( ) e t + cos t + sin t (3) PROBLEM 3 Question : The "matrices" of the equation of motion read M =, C = 0.0, K = [], f n = e tn x 0 =, ẋ 0 = 0 } () It follows that the undamped eigenfrequency and the damping ratio of the system becomes ω = =, ζ = 0.0 = 0.0 () Next, the algorithm in Box., p. 34 is followed. The initial acceleration is calculated from (-0) using f 0 = e 0 =.0000 ẍ 0 = M ( f 0 Cẋ 0 Kx 0 ) = 0 (3) The dynamic mass matrix becomes, cf. (-7) M = M + γ tc + β t K =.0008 (4) At first the predictors of the displacement and velocity are calculated, cf. (-5), (-6) x = x 0 + ẋ 0 t + ( ) β t ẍ 0 =.0000 (5) x = ẋ 0 + ( γ ) tẍ 0 = (6) Next, the acceleration at t = t = t = 0. follows from (-7) with f = e 0. = ẍ = M ( f C x K x ) = (7)

8 The updated displacement and velocities at the time t = 0. follow from (-8) and (-9) x = x + β ẍ t = (8) ẋ = x + γ ẍ t = (9) Next, the same loop is repeated with t = 0., which means that f = e 0. = The following results are obtained x = ẋ = (0) ẍ = 0.80 The corresponding exact results at t = 0. become, cf. Problem, Eq. 3 x = ẋ = ẍ = () Question : Since β = < γ =, the algorithm can at most be conditional stable, cf. (-78) or Box.7. The 6 upper bound on the time step causing instability follows from (-79) t ω γ β = =.909 () From (-74) follows that λ D = for γ =. This implies that the amplitudes during eigenvibrations neither increases nor decreases as long as () is fulfilled for the time step. Hence, the algorithm is free of numerical damping. The relative period error follows from (-98) T ( T = ( γ 36γ + ) + ) 96 β κ + O ( κ 4) = O ( κ 4) (3) where κ = ω t = 0.. Hence, the period error is very small for β = and γ =. The indicated parameter combination constitutes the so-called Fox-Gordon algorithm within the Newmark family.

9 PROBLEM 4 Question : The algorithm is performed as described in Box 7.3. At first the matrix A = K M is calculated, cf. (5-4) 0 A = = () At first the simultaneous inverse vector iteration is performed, cf. (7-4) Φ = AΦ 0 = = () The projected mass and stiffness matrices become, cf. (7.3) M = K = = T = T The solution of the corresponding generalized eigenvalue problem (7-3) becomes R = , Q = (3) (4) The estimate of the lowest eigenmode after the st iteration becomes, cf. (7-35) Φ = Φ Q = = (5) Correspondingly, after the nd, 3rd and 4th iteration steps the following matrices are calculated R =, Q = Φ = (6)

10 R 3 = , Q 3 = Φ 3 = (7) R 4 = , Q 4 = Φ 4 = (8) R 8 = , Q 8 = Φ 8 = (9) As seen the subspace iteration process determines the st eigenvalue after the nd iteration with the indicated number of figures, the st eigenvector and the nd eigenvalue after 4 iterations, and the nd eigenvalue after the 8th iterations. PROBLEM 5 Question : Initially, the matrix K µm is calculated for µ = K () = K. M =. 3 0 =..6.0 () At first, the problem is solved by Gauss factorization. The procedure in Box 3., eqs. (3-7), (3-9), (3-0) is followed.

11 L = 0 0 (.) 0 ( 0.4) (.) 0 ( 0.4) K () = L K() = L (). = L = () L = K (3) = L K() = S = L () = L L = L = (3) From this follows that L = , D = (4) Since, two components in the main diagonal of D are negative, it is concluded that two eigenvalues are smaller than µ =.. Alternatively, the analysis may be performed by a Sturm sequence check. The Sturm sequence becomes P (3) (.) =, sign ( P (3) (.) ) = + P () (.) = 0.4, sign ( P () (.) ) = P () (.) = 0.4 (.6) (.) = 4., sign ( P () (.) ) = P (0) (.) = det ( K. M ) = 9.84, sign ( P (0) (.) ) = + (5) Hence, the sign sequence of the Sturm sequence becomes + +. Since, two sign changes occur in this sequence, it is concluded that two eigenvalues are smaller than µ =.. Question : Since, the eigenvalues ρ = and ρ = obtained by the subspace iteration in Problem 4 are both smaller than µ =., it is concluded that ρ and ρ are identical to the two lowest eigenvalues λ and λ of the considered general eigenvalue problem, and that the 3rd eigenvalue λ 3 is larger than µ =..