SKRIFTLIG EKSAMEN I NUMERISK DYNAMIK Bygge- og Anlægskonstruktion, 7. semester Torsdag den 9. juni 23 kl. 9.-3. Alle hjælpemidler er tilladt OPGAVE f(x) x Givet funktionen f(x) x, x [, ] Spørgsmål (%) Foretag en udvikling af funktionen f(x) i en Fourier-Bessel række, idet der benyttes Bessel funktioner af orden n. Egenværdier udtages af randbetingelsen J (λ). Spørgsmål 2 (%) Foretag en udvikling af funktionen f(x) i en Fourier-Legendre række. Hjælp: Ved løsning af spørgsmål udnyttes, at funktionen er symmetrisk om x. Foretag indledningsvis på basis heraf en Fourier-Bessel udvikling af delfunktionen defineret på intervallet x [, ].
2 OPGAVE 2 Givet begyndelsesværdiproblemet y (x) 2y(x) + sin(πx), x ], [ y() hvor y (x) d dx y(x). Spørgsmål (5%) Løs det anførte begyndelsesværdiproblem ved hjælp af Laplace transformation. OPGAVE 3 Givet følgende begyndelsesværdiproblem for et system af frihedsgrad q(t)+.5 q(t)+ q(t) sint, t ], [ q() q() hvor q(t) d q(t) betegner differentiation mht. t. dt Spørgsmål (%) Bestem ved numerisk integration med den centrale differensmetode funktionsværdien q(.6) med anvendelse af tidsskridtet t.3. Spørgsmål 2 (5%) Undersøg om algoritmen er numerisk stabil med det specificerede tidsskridt.
3 OPGAVE 4 Givet et generelt egenværdiproblem defineret ved følgende masse- og stivhedsmatricer M 2 4, K 4 Spørgsmål (%) Bestem den laveste egenvektor og tilhørende egenværdi ved invers vektoriteration. Spørgsmål 2 (%) Bestem den største egenvektor og tilhørende egenværdi ved fremad (forward) vektoriteration. Spørgsmål 3 (%) Bestem den laveste egenvektor og tilhørende egenværdi ved Sturmsekvens iteration (teleskopmetoden). Det understreges, at uagtet en analytisk løsning på problemet let kan tilvejebringes, ønskes numeriske løsninger på alle 3 spørgsmål. Alle 3 spørgsmål betragtes som besvaret, når 2 iterationer er gennemført. Startvektorer for iterationerne vælges frit. OPGAVE 5 Givet et generelt egenværdiproblem defineret ved følgende masse- og stivhedsmatricer 2 4 M 4, K 2 4 Spørgsmål (2%) Bestem de 2 laveste egenvektorer og tilhørende egenværdier ved hjælp af subspace iteration. Spørgsmålet betragtes som besvaret, når blot en enkelt subspace iteration er gennemført. Start vektorbasis vælges frit.
4 SOLUTIONS PROBLEM Question : f(x) x Fig. : Restriction of function f(x) to the interval [,]. At first the Fourier-Bessel expansion is performed for the part of the function defined on the interval [,] as shown in Fig.. Since λ the Fourier-Bessel series follows from (.5.9-2) where f(x) c + c i J (λ i x), x [, ] () i2 c 2 2 c i 2 2 J 2 (λ i) x ( x ) dx 3 xj (λ i x) ( x ) dx, i 2, 3,... and the eigenvalues are defined from the roots of, cf. (6-4-6) and Table 6.2 (2) J (λ i) J (λ i ) λ 2 3.837 λ 3 7.56 λ 4.74 λ 5 3.324. (3)
5 The expansion () may be extended to the entire interval [, ] by replacing x with x, i.e. f(x) c + ( c i J λi x ), x [, ] (4) i2 Question 2: f(x) x Fig. 2: Function f(x) defined on entire interval [-,]. The Fourier-Legendre series follows from (.5.2-22) f(x) where c n P n (x), x [, ] (5) n c n 2n + 2, n odd f(x)p n (x)dx ( ) 2n + f(x)p n (x)dx, n even (6) In (6) it has been used that f(x) f( x) is an even function of x, and the Legendre polynomials are even or odd functions of x, depending on n is even or odd, cf. (6.4.9). The low-order expansion coefficients for even order of n may be evaluated as follows, cf. (6.4.9)
6 c ( 2 + ) c 2 ( 2 2+ ) c 4 ( 2 4+ ) ( x ) dx 2 ( x ) 2( 3x 2 ) dx 5 8 ( ) ( x 8 35x 4 3x 2 +3 ) dx 3 6 (7) PROBLEM 2 Question : Using Theorems 7.2 and 7.8 the Laplace transform of the differential equation becomes L{y (x)} 2L{y(x)} + L{sin(πx)} sy (s) y() 2Y (s)+ π s 2 + π 2 Y (s) ( ) π + s +2 s 2 + π 2 π 2 + π +4 π 2 +4 s +2 π π 2 +4 s s 2 + π 2 + 2π π 2 +4 s 2 + π 2 () where Y (s) L{y(x)} e sx y(x)dx (2) Then the solution follows from the inverse Laplace transform of (). Use of Theorem 7.3 provides y(x) π2 + π +4 π 2 +4 e 2x π π +4 cos(πx)+ 2π sin(πx) (3) 2 π +4 2
7 PROBLEM 3 The "matrix" formulation of the equation of motion reads, see Note, eq. () MÜ(t)+C U(t)+KU(t) } R(t) U() U, U() U () where } M [], C [.5], K [], R(t) [sin t] U [], U [] (2) The algorithm reads, cf. Note, eq. (5) U j+ A U j + A 2 U j + B R j, j,,... (3) where, cf. Note, eq. (6) ( ) ( ) A M + C K 2 M [.89578] t 2 2 t t 2 ( ) ( ) A 2 M + C M C [.985] t 2 2 t 2 2 t ( ) B M + C t 2 2 t [.8933] (4) At the evaluation of the "matrices" A, A 2, B the time step t.3 has been used. The startvalue U is calculated by Note, eq. (7) U U U t + ( 2 2Ü t Ü M C U ) KU + R() (5) Since, R() [sin()] [], it follows that U []. Then, based on (3) the following calculations are performed
8 } q(.3) U.89578.985 +.8933 sin(.). q(.6) U 2.89578..985 +.8933 sin(.3).264 (6) The analytical solution of the initial value problem is given as ( ( ) q(t) e ζt ζ ( ) ) cos ζ2 t + 2ζ sin ζ2 t cos t.3446 (7) ζ 2 2ζ where t.6 and ζ.25 have been inserted. As seen the numerical solution is not very accurate due to the rather large time step. Question 2: The period of the undamped eigenvibrations is T 2π 2π (8) Since, cf. Note, eq. (29) t.3 T 2π.5 π < π (9) it follows that the central difference method is numerical stable with the selected time step. PROBLEM 4 The following start vector is used both in Question and Question 2: Φ ()
9 Question : The matrix A becomes, cf. (8-4) A 4 2 4.66667 2 5.66667 (2) At the st and 2nd iteration step the following calculations are performed, see Box 8..66667 Φ 2 5.66667 2.66667 Φ 29.22 7.66667 2.66667 7.66667.66667.5653 Φ 2 2 5.66667.453.9658 Φ 2 39.628 2.86323 Φ T M Φ 29.22 (3).5653.453.9658 2.86323 Φ T 2 M Φ 2 39.628 (4).44.45484 Since Φ 2 has been normalized to unit modal mass, the Rayleigh quotient based on Φ 2 provides the following estimate for λ, cf. (7-25) T.44 4.44 ρ(φ 2 ).5883 (5).45484.45484 The exact solutions with the indicated accuracies were obtained after 4 iterations λ.5883, Φ ().4323.4553 (6) Question 2: The matrix B becomes, cf. (8-35) B 2 4 4 2.42857.7429.8574.42857 (7)
At the st and 2nd iteration step the following calculations are performed, see Box 8.3 2.42857.7429 Φ.8574.42857.7429 Φ 5.4286.42857 2.42857.7429.75593 Φ 2.8574.42857.8898.978 Φ 2 7.24.72893.7429.42857 Φ T M Φ 5.4286 (8).75593.8898.978.72893 Φ T 2 M Φ 2 7.24 (9).7438.275 Since Φ 2 has been normalized to unit modal mass, the Rayleigh quotient based on Φ 2 provides the following estimate for λ 2, cf. (7-25) ρ(φ 2 ) T.7438 4.7438.275.275 2.69799 () The exact solutions with the indicated accuracies were obtained after 5 iterations λ 2 2.6983, Φ (2).74224.283 () Question 3: At first a calculation with µ. is performed, which produces the following results 3.8. K.M..6 P (2) (.), sign(p (2) (.)) + P () (.) 3.8, sign(p () (.)) + P () (.) 3.8.6 (.) 2.7, sign(p () (.)) + (2)
Hence, the sign sequence of the Sturm sequence becomes +++, corresponding to the number of sign changes n sign in the sequence. From this is concluded that both eigenvalues are larger than µ.. Similar calculations are performed for µ.,.2,...,.6 µ. : Sign sequence + ++ n sign µ.2 : Sign sequence + ++ n sign µ.3 : Sign sequence + ++ n sign µ.4 : Sign sequence + ++ n sign µ.5 : Sign sequence + ++ n sign µ.6 : Sign sequence + + n sign (3) From this is concluded that the st eigenvalue is placed somewhere in the interval.5 <λ <.6. Next, similar calculations are performed for µ.5,.52,...,.59 µ.5 : Sign sequence + ++ n sign µ.52 : Sign sequence + ++ n sign µ.53 : Sign sequence + ++ n sign µ.54 : Sign sequence + ++ n sign µ.55 : Sign sequence + ++ n sign µ.56 : Sign sequence + ++ n sign µ.57 : Sign sequence + ++ n sign µ.58 : Sign sequence + ++ n sign µ.59 : Sign sequence + + n sign (4) From this is concluded that the st eigenvalue is confined to the interval.58 <λ <.59. Proceeding in this manner after 27 iterations the st eigenvalue is confined to the interval.5882 < λ <.5883. Setting λ.5883, the linear equation (-63) attains the form ( K.5883M) Φ() [ 3.68234.5883.5883.36468 ] [ Φ() Φ () 2 ] [ ] (5)
2 Setting Φ () the st equation provides Φ () 2 3.68234 (.5883) 3.7764 Φ () 3.7764 (6) Normalization to unit modal mass provides, cf. (6-54) Φ ().4323.4553 (7) which agrees with (6). PROBLEM 5 Question : The following start vector basis is used Φ [ Φ () Φ (2) ] () The matrix A becomes, cf. (6-44) 4 A K M 4 Then, the st iterated vector basis becomes, cf. (-4) 2.25 2.5.75 4 3. 9. 3. (2) 2.75 2.5.25 Φ [ Φ() Φ (2) ] AΦ.25 2.5.75 3. 9. 3..75 2.5.25 4.5 3 5. 9 (3) 4.5 2 The Rayleigh-Ritz analysis based on the calculated basis Φ provides the following projected mass and stiffness matrices, cf. (6-44), (-2), (-3)
3 4.5 3 M Φ T M Φ 5. 9 4.5 2 4.5 3 K Φ T K Φ 5. 9 4.5 2 2 4.5 3 4 5. 9 2 4.5 2 T 4 4.5 3 5. 9 4 4.5 2 The corresponding eigenvalue problem (-3) becomes K Q M Q R 25 74 [q () q (2) ] 7 69 [q () q (2) ] ρ, 74 44 69 43 ρ 2, R.9349 2.255 T, Q [.2598 ].5684.387.95943 [ 25 ] 74 74 44 [ 7 ] 69 69 43 (4) (5) The estimate of the lowest eigenvectors after the st iteration becomes, cf. (-34) 4.5 3 Φ Φ Q 5. 9 4.5 2.2598.5684.387.95943.2852.3245.42454.873 (6).2465.63899 Correspondingly, after the 2nd and 5th iteration steps the following matrices are calculated R 2.9328 2.443.2267.48 Φ 2.42727.6298.226.58493.9328 R 9 5 2..2252.49995 Φ 5.42735.3.2252.55, Q 2, Q 5 [.9328 ].345.25 2.5692 [.9328 ].. 2. (7) (8)
4 The subspace iteration process converged much faster to the st eigensolution than to th 2nd eigensolution. This is because the convergence rate to the st eigenvalue r, λ /λ 3.9328/3.57338 is much smaller than the the convergence rate to the 2nd eigenvalue r,2 λ 2 /λ 3 2./3.57338, cf. (-36). Finally, it should be checked that the calculated eigenvalues are indeed the lowest two by a Sturm sequence or Gauss factorization check. The 2nd calculated eigenvalue becomes ρ 2,5 2., so let µ 2.. Both K and M are on a three-diagonal form, so the Sturm sequence algorithm (-6) may be used on the matrix K 2.M, i.e..2 3. K 2.M 3. 7.4 3. 3..2 P (3) (2.), sign(p (3) (2.)) + P (2) (2.).2, sign(p (2) (2.)) P () (2.) (.2) ( 7.4) ( 3.) 2 8.3, sign(p () (2.)) P () (2.) (.2) ( 8.3) ( 3.) 2 (.2) 3.548, sign(p () (2.)) + (9) Hence, the sign sequence of the Sturm sequence becomes + +, corresponding to a number of sign changes n sign 2. From this is concluded that two eigenvalues are smaller than µ 2., and that the calculated eigenvalues are indeed the lowest two.