SKRIFTLIG EKSAMEN I NUMERISK DYNAMIK Bygge- og Anlægskonstruktion, 8. semester Fredag den 30. juni 2005, kl Alle hjælpemidler er tilladt

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1 SKRIFTLIG EKSAMEN I NUMERISK DYNAMIK Bygge- og Anlægskonstruktion, 8. semester Fredag den 3. juni 5, kl Alle hjælpemidler er tilladt OPGAVE u = y B u = u C A x c u = D u = Figuren viser en homogen cirkulær plade med radius c. Langs periferisegmentet AB holdes temperaturen konstant på værdien u. Langs den øvrige del af periferien BCD er temperaturen konstant lig. Spørgsmål (% Bestem den stationære temperaturfordeling i pladen.

2 OPGAVE f(t t T T T T T Spørgsmål (% Figuren viser grafen for en funktion f(t, der er periodisk med perioden T på intervallet [, [. Bestem den Laplacetransformerede af f(t. Spørgsmål (% Bestem dernæst den Laplacetransformerede af funktionen g(t = f(t + U(t τ t T e t/t hvor f(t er den i spørgsmål definerede funktion, τ er en positiv konstant, og U( er enhedsstepfunktionen defineret ved U(t τ = {, t < τ, τ t <

3 3 OPGAVE 3 Givet begyndelsesværdiproblemet ÿ(t +.ẏ(t + 4y(t = sin(.t, t ], [ y( =, ẏ( = hvor ẏ(t = d y(t betegner differentiation mht. t. dt Spørgsmål (5% Bestem funktionsværdien y(.6 ved numerisk integration med den generaliserede α-algoritme med λ =.8, og tidsskridtet t =.3. Spørgsmål (5% Følgende spørgsmål ønskes besvaret med relevant argumentation: : Er algoritmen numerisk stabil med det anførte tidsstep? : Er algoritmen forbundet med numerisk dæmpning, og i givet fald vil dette være af betydning?

4 4 OPGAVE 4 Givet et generelt egenværdiproblem defineret ved følgende masse- og stivhedsmatricer 4 M =, K = 4 Spørgsmål (% Bestem egenværdierne og de tilhørende egenvektorer normeret til modalmasse ved anvendelse af generaliseret Jacobiiteration. Opgaven betragtes som løst, når et sweep er gennemført. Det understreges, at uagtet en analytisk løsning på problemet let kan tilvejebringes, ønskes kun den numeriske løsning bestemt.

5 5 OPGAVE 5 Det generelle egenværdiproblem defineret i opgave 4 betragtes igen. Spørgsmål (% Undersøg ved hjælp af Gauss faktorisering hvor mange egenværdier, der er mindre end.3. Spørgsmål (% Besvar samme spørgsmål ved hjælp af et Sturmsekvens check.

6 6 SOLUTIONS PROBLEM Question : The boundary value problem for the stationary temperature in a circular plate of radius c is given as, cf. p. 56 u r + u r r + u =, r ], c[, θ [, π[ r θ { u, θ [, π/] u(c, θ = f(θ =, θ [π/, π[ ( The separation method is used, i.e. we search for product solutions u(r, θ = R(rΘ(θ to the partial differential equation in (. Insertion and separation of the variables provides the following ordinary differential equations R (rθ(θ + r R (rθ(θ + r R(rΘ (θ = R (r + r R (r R(r = Θ (θ Θ(θ = λ r R (r + rr (r λ R(r = ( Θ (θ + λ Θ(θ = (3 ( is a Cauchy-Euler differential equation, cf. p. 94. The auxiliary equation reads, see (4.7- m(m + m λ = m = { λ λ (4

7 7 Then, the general solution becomes, cf. (4.7- R(r = c r λ + c r λ (5 Since R(r as r, and λ >, the geometrical boundary condition that finite solutions should exist at r = compel us to impose c = in (5, so R(r = c r λ. The general solution of (3 reads Θ(θ = c 3 cos(λ θ + c 4 sin(λ θ (6 The temperature field must be continous at an arbitrary point. This means that the temperature at the positions (r, θ + and (r, θ + π must be the same, i.e. u(r, θ + = u(r, θ + π (7 Hence, the solutions searched for must be periodic in θ with the period π. In (6 this requires that λ = λ n = n, n =,,,... Then, product solutions which satisfy the partial differential equation in (, the geometrical boundary condition at r =, and the periodicity condition with respect to θ (7 has the form u n (r, θ = r (A λn n cos(λ n θ + B n sin(λ n θ λ n = n, n =,,,... (8 where A n = c c 3 and B n = c c 4. Notice that u (r, θ A. From the superposition principle follows that the following linear combination of solutions of type (8 equally fulfills the partial differential equation, the geometrical boundary condition at r = and the periodicity condition u(r, θ = u n (r, θ = A + n= n= r (A n n cos(n θ + B n sin(n θ (9 What remains is to chose the expansion coefficients A n and B n, so the boundary condition u(c, θ = f(θ in ( is fulfilled. This implies that

8 8 f(θ = A + n= c (A n n cos(n θ + B n sin(n θ ( ( represents a Fourier series of f(θ. Comparison with Definition.5, p.49, Eqs. (.-8 - (.-, provides with p = π A = a = π f(θdθ = π π π A n = a n c n = c n π B n = b n c n = c n π π π π π f(θ cos(n θdθ = c n π f(θ sin(n θdθ = c n π π/ π/ π/ u dθ = u 4 ( u cos(n θdθ = u nπ nπc sin n u sin(n θdθ = u nπc n ( ( nπ cos ( Hence, the final solution becomes u(r, θ = 4 u + u n= ( r n ( nπ sin nπ( c ( ( nπ cos(n θ + cos sin(n θ ( PROBLEM Question : Since the function f(t is periodic with the period T, Theorem 7., p. 345 may be used. Then, F(s = L { f(t } = [ e st] T/ = e st s s ( + e st/ e st T e st f(tdt = e st T/ ( + e st/ ( e st/ s( e st/ = e st dt = (

9 9 Question : At first the following identity is formulated t T e t/t = t τ + τ T ( t τ e (t τ+τ/t = e τ/t T e (t τ/t + τ T e (t τ/t ( Due to the linearity property of the Laplace transform, and the use of Theorem 7.7, p. 39, it then follows that L {U(t τ t } T e t/t = { } t τ e τ/t L T e (t τ/t U(t τ + τ { } T e τ/t L e (t τ/t U(t τ { } t e τ/t e τs L + τ } T e τ/t e τs L {e t/t T e t/t = (3 The following results may be derived { t L T e t/t } = { } t L T e t/t = T e st t T e t/t dt = T e u ue s u du, s = st ( s + = T ( (4 Ts + where Theorem 7.6, p. 34 has been used with a = and f(u = u. Similarly, using Theorem 7., p. 3 with a = } L {e t/t = T e u e s u du = T s + = T Ts + (5 Then, L {U(t τ t } T e t/t ( = e τ(s+/t T ( + τ Ts + Ts + (6

10 Finally, from ( and (6 L { g(t } = ( s ( T + e st/ + e τ(s+/t ( + τ Ts + Ts + (7 PROBLEM 3 Question : The "matrices" of the equation of motion read } M =, C =., K = [4], f n = sin(. t n x =, ẋ = ( It follows that the undamped eigenfrequency and the damping ratio of the system becomes ω = 4 =, ζ =. =.5 ( The integration parameters of the generalized α-method follows from Box.5, p.6 α f = λ D λ D + α m = λ D λ D + = = = 4 9 = 3 γ = + α f α m = = 3 8 β = ( γ + = ( = 8 (3 Next, the algorithm in Box.4, p. 57 is followed. The initial acceleration is calculated at first

11 ẍ = M ( f Cẋ Kx = (4 The dynamic mass matrix becomes, cf. (-7 M = ( ( ( α m M + αf γ tc + β t K =.7386 (5 Next, the acceleration at t = t = t =.3 follows from (-6 with f = and f = sin(..3 =.589 ẍ = ẍ + M [α f f + ( α f f Mẍ Cẋ Kx ( αf ( tcẍ + K( tẋ + t ẍ ] =.443 (6 The updated displacement and velocities at the time t =.3 follow from (-3 and (-4 ( ( x = x + ẋ t + β ẍ + β ẍ t =.3 (7 ( ( ẋ = ẋ + γ ẍ + γ ẍ t =.8 (8 Next, the same loop is repeated with t =.6, which means that f = sin(..6 =.95. The following results are obtained x =.636 ẋ =.6 (9 ẍ =.698 The corresponding exact results at t =.6 become x(.6 =.638 ẋ(.6 =.837 ẍ(.6 =.6637 (

12 Question : The criteria for unconditional stability of the generalized α-algorithm follow from (-3 γ = 8 α m = 3 β = 5 8 ( α f = 4 9 γ = 8 + α m α f = 8 The structural damping enhance the numerical stability of the algorithm. Hence, it can be concluded that the algorithm is numerical stable for an arbitrary time steps. Consequently, it will also be stable at the time step t =.3. The numerical damping ratio is of the order of magnitude ζ num = O ( κ, where κ = ω t. As a consequence ζ num is assumed to be very small for all practical time steps. Actually, with κ =.3 =.6 it is seen from Fig. -6 that this results in a numerical damping ratio, which is indeed ignorable compared to the structural damping ratio of ζ =.5.

13 3 PROBLEM 4 Question : The algorithm is performed as described in Box 6.4. The initial matrices are 4 M = M =, K = K = 4, Φ = ( In the st sweep the following calculations are performed for (i, j = (,, as specified by (6-, (6-, (6-3, (6-5: 4.5 ( a = = { α =.6667 β = ( b = = P =.6667, Φ = Φ P = M = P T M P =...5, K = P T K P = ( Next, the calculations are performed for (i, j = (, 3 :.3333 (.6667 a = = { α = (.6667 β =.44 b = = P =, Φ = Φ P = M = P T M P = , K = P T K P = (3

14 4 Finally, to end the st sweep the calculations are performed for (i, j = (, 3 : (. a = = { α = (. β =.4749 b = = P =.4749, Φ 3 = Φ P = M 3 = P T M P = , K 3 = P T K P = (4 At the end of the nd and 3rd sweep the following estimates are obtained for the modal matrix and the transformed mass and stiffness matrices Φ 6 = M 6 = , K 6 = Φ 9 = M 9 =..873., K 9 = (5 Presuming that the process has converged after the 3rd sweep the eigenvalues and normalized eigenmodes are next retrieved by the following calculations, cf. Box. 6.4

15 m = M 9 =..873., m = λ Λ = λ 3 = M 9 K 9 = λ... Φ = [ Φ ( Φ (3 Φ (] = Φ 9 m = (6 As a proof of the normalization of the eigenmodes to unit modal mass it can be verified that Φ T MΦ = I and Φ T KΦ = Λ.

16 6 PROBLEM 5 Question : Following the procedure in Box 3., eqs. (3-7, (3-9, (3-, the following calculations are performed.6.5. K ( = K.3M = K ( = L K ( = L = ( (.6 L ( = L =.5.6 ( L = ( From this follows that.6.5. K (3 = L K ( = S = L ( = L L = L = (....6 L = , D = 9.44 ( Since, two components in the main diagonal of D are negative, it is concluded that two eigenvalues are smaller than µ =.3.

17 7 Question : The Sturm sequence is calculated by the recursive algorithm ( K.3M = P (3 (.3 =, sign ( P (3 (.3 = + P ( (.3 =.6, sign ( P ( (.3 = P ( (.3 =.6.7 (.5 = 5.645, sign ( P ( (.3 = P ( (.3 =.3 ( (.5 (.6 = 4.466, sign ( P ( (.3 = + (4 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 =.3.