Applications of Eigenvalues & Eigenvectors

Transcription

1 Applications of Eigenvalues & Eigenvectors Louie L. Yaw Walla Walla University Engineering Department For Linear Algebra Class November 17, 214

2 Outline 1 The eigenvalue/eigenvector problem 2 Principal stresses and directions in stress analysis 3 Fundamental frequencies and mode shapes in vibrations 4 Critical loads and buckled shapes in buckling analysis 5 Applications in electrical area/ feedback and control 6 Principal mass moment of inertia in 3D

3 1. The eigenvalue/eigenvector problem Find vector v and scalar λ that satisfies the following: where, K = nxn matrix v = nx1 vector, an eigenvector λ= a scalar, an eigenvalue Kv = λv (1) n vector and scalar pairs will satisfy equation (1). How to find such pairs is a linear algebra problem. Equation (1) arises on occasion during the solution of various types of engineering problems

4 1. The eigenvalue/eigenvector problem An example (3x3 matrix 3 eigenvalues and 3 eigenvectors): τ xx τ xy τ xz v 1 v 1 τ xy τ yy τ yz v 2 = τ v 2 (2) τ xz τ yz τ zz v 3 v v 1 v v 2 = τ v 2 (3) 5 9 v 3 v 3 MATLAB solution to eigenvalue problem: τ 1 = 2.17, v =.62, τ 2 = 9.29, v =.5, (4) τ 3 = 21.54, v =.78.31

5 2. Principal stresses and directions in stress analysis Consider a two dimension problem. A cantilever beam loaded at its free end. 4 Deflected Shape A Y X P y distance (in) Deflection Magnification Factor = x distance (in)

6 2. Principal stresses and directions in stress analysis z Stress block from point A y τ yy = τ yz = τ yx = 3.5 τ xz = τ zy = τ xx = 15 x τ xy = 3.5 τ zx = τ zz = Units of ksi τ xx τ xy τ xz τ = τ xy τ yy τ yz τ xz τ yz τ zz τ = A Stress Field τ xx (ksi) A Stress Field τ yy (ksi) A Stress Field τ xy (ksi)

7 2. Principal stresses and directions in stress analysis z MATLAB solution to eigen problem:.2166 τ 1 =.78, v 1 =.9763, y v 1. τ 1 =.78. τ 2 =., v 2 =., x 1. τ 3 = 15.8 τ 2 = v 3 v Units of ksi τ 3 = 15.8, v 3 =

8 3. Fundamental frequencies and mode shapes in vibrations Numerical Eigen problem solution 2D FEA of cantilever x & y dofs at 189 nodes (2 x 189 = 378 dofs) 378 eigenvalues & eigenvectors 378 frequencies & mode shapes Eigenvectors are 378th dimensional vectors! Kv = λv, K = 378x378 in this example! Finer Grid=More accurate First Five Mode Shapes of Cantilever Mode =1, Frequency = Hz Mode =2, Frequency = Hz 5 Mode =3, Frequency = Hz 1 Mode =4, Frequency = Hz 15 Mode =5, Frequency = Hz

9 4. Critical loads & buckled shapes in buckling analysis P P L 3 θ 3 k 3 L 3 θ 2 L 3 k 2 k 1 θ 1 undeformed deformed small angles assumed rotation spring A deformed shape equilibrium analysis results in the following: (k 1 + k 2 ) k 2 θ 1 k 2 (k 2 + k 3 ) k3 θ 2 = PL θ 1 θ 2 3 k 3 k3 θ 3 θ 3 Kθ = λθ

10 4. Critical loads & buckled shapes in buckling analysis 12 For the case of L = 1 inches k 1 = 3 kip-in/rad k 2 = 3 kip-in/rad k 3 = 3 kip-in/rad Eigenvalue = Pcr L 3 Eigenvector shape 2 Buckling Mode =1 Buckling Mode =2 Buckling Mode =3 4 P cr = P cr = P cr =

11 5. Applications in electrical engineering - feedback and control Outline of conceptual feedback and control Model dynamic system such as airplane, car, rocket M φ+c φ+kφ = F(t) The mathematical model of the system has inherent eigenvalues and eigenvectors Eigenvalues describe resonant frequencies where the system will have its largest, often excessive, response. We can choose F(t) to reduce the system response at the resonant frequencies.

12 5. Applications in electrical engineering - feedback and control Perhaps let F(t) = A φ+bφ and insert into dynamic system model The new system is M φ+ C φ+ Kφ = where C = C A = velocity dependent damping and K = K B = displacement dependent forcing We can adjust A and B so that the eigenvalues of the new barred system are different from the original system By doing so we cause (control) the system to avoid excessive vibration or instability R Q S

13 6. Principal mass moment of inertia in 3D 3D kinetics of a rigid body Inertia tensor (components dependent on XȲ Z coordinate axes orientation) I xx I xy I xz I = I yx I yy I yz I zx I zy I zz It is possible to orient the axes such that I x I = I y I z

14 6. Principal mass moment of inertia in 3D Ȳ Z Z v z Y v y v x X X In XȲ Z coordinate system I xx I xy I xz I = I yx I yy I yz I zx I zy I zz I x, I y, I z are eigenvalues v x, v y, v z are eigenvectors wrt XȲ Z In XYZ coordinates I x I = I y I z

15 Conclusion A few comments: Many applications of eigenvalues and eigenvectors in engineering K is not always symmetric eigenvalues are not always positive or real eigenvectors are orthogonal eigenvalues are invariant wrt to choice of coordinate axes

16 Questions. Kθ =?θ