Engineering Applications of Linear Algebra

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Terence Potter
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1 Engineering Applications of Linear Algebra -Continuu Mechanics: Stresses and Principal Aes - Vibrating Systes

2 Stresses and Principal Aes ɶ = ɶ : Stress Tensor ij = ji

3 = e + e + e = e + e + e = e + e + e e e e e e e e = e = ɶ e e

4 ɶ ɶ ɶ n = n n n = 0 n n = ɶ n = 0 ( ) 0 n n = 0

5 = = 0,, = Stress nvariants = + + = + + =

6 + + = 0 Let,, be the roots ( )( )( ) = 0 ( ) ( ) = 0 = + + = + + =

7 . :. = 0 0 = + + = = = = = = 8 = 4 = 6

8 + + = = = ( ɶ ) n 0 = 4 = = = 4 : n =,, = : n =,, = : n = 0,,

9 Vibrating Systes d d d α α α dt dt dt = a + a + + a n n d d d α α... α... dt dt dt = a + a + + an n... d d d α α α dt dt dt n n n +... n n... n + + = a + a + + a nn n

10 A a... an.. =.... a... a n nn n n. =.. n n d ( ) d ( ) d ( ) ( ) α + α α L α α... α dt dt dt L =A n engineering practice A is a syetric atri. a ij = a ji

11 A is a syetric atri and can be diagonalized y an orthogonal Modal atri Q:. D = Q AQ = Qz L = A L [ Q ] = A ( Q ) z z QL [ z] = AQz [ ] ( z ) L = Q AQ z [ z] L = Dz D λ 0 λ. = λβ.. 0 λ n Lz λ z Lz λ z =... Lz n λn z n Lz = λ z β β β

12 Free Undaped Vibration: L = A L d dt d = ɺɺ = A dt Lz d z β = λ z β β dt = λ z β = β β β OR zɺɺ = λ z β β β β =,,..., n z ( ) = c sin λ t + φ β β β β ω β = λ β z = c ( ω φ ) sin t + β β β β

13 ω, ω,..., ω,..., ωn β are called Natural Frequencies. Corresponding to each natural frequency there eist an eigenvector which is called a Mode. The iniu natural frequency (corresponding to iniu absolute eigenvalue) is called Fundaental Frequency and its corresponding eigenvector is called Fundaental Mode. z = Qz } = y sin ( ω t + φ ) = c sin ( ω t + φ ) β β β β β β d dt = A ( ) A ( ) β β y β β ω y sin ω t + φ sin ω t + φ = ω y A y λy = A y =

14 ( ) ( y y ) ( t ) ( t ), =, sin ω + φ sin ω + φ = 0 α β α β α β α α β β n = α = c α α at t = 0 : (0), ɺ (0) are given n n ( ω φ ) = c = c y sin t + α α α α α α α = α = n = (0) y sin ( ) α = ( φ ) c α α α 0, y = c sinφ β β β

15 n α = ( ) ( ω t φ ) = c y sin + α α α α n d ɺ = = c ω y cos ω + φ dt n = α = α = ɺ (0) ω y cos ɺ 0, y = c ω cos φ φ c β β β β β β ( t ) α α α α α ( φ ) c α α α α ( ) ( 0 ), ω 0, β < y β > = tan < ɺ y > β ( ) < 0, y β > = { < ( 0 ), y β > λ β

16 Eaple Find natural frequencies and vibration odes of the following syste: k k k k ɺɺ = k 0 + k ( ) ( ) ɺɺ = k + k ( ) ( ) ɺɺ = k + k 0 ( ) ( )

17 α = k ɺɺ = A ɺɺ = α + α ɺɺ = α α + α ɺɺ = α α α α 0 A= α α α 0 α α ( A λ ) = ( λ α ) α ( λ α ) det 0 λ = α λ = + λ = ( ) ( ) = 0 α α ω = λ k ω = ω = ω = + = ( ) ( ) k k

18 ( ) sin t ω φ = + ( ) 0 sin t ω φ = + 0 ( ) sin t ω φ = +

19 Eaple Find the subsequent otion of the following syste at tie t>0 = = k = k = k y (0) = y (0) = y ( t ) k y ɺ (0) = 6 y ɺ (0) = 6 y ( t )

20 yɺɺ = y + y y yɺɺ ( ) = y y ( ) yɺɺ = 5y + y yɺɺ = y y yɺɺ y ɺɺ 5 y = y y=ay ɺɺ 5 λ λ det ( A λ ) = 0 = 0 λ + 7λ + 6 = 0 λ λ = = 6 λ = ω = λ = 6 ω = 6 λ = λ =

21 y t = a sin ω t + φ + a sin ω t + φ ( ) ( ) ( ) y t a sin t a sin 6 t ( ) = ( ) ( ) + φ + + φ y ( 0 ) = a sin φ a + sin φ a sin φ a sin φ = + : 5 a sin 5 φ = 5 : a sin φ = 0 4 a sin φ = a sin φ = 0

22 y t = a sin ω t + φ + a sin ω t + φ ( ) ( ) ( ) yɺ t = a ω cos ω t + φ + a ω cos ω t + φ ( ) ( ) ( ) yɺ ( 0) = aω cosφ + aω cosφ yɺ ( 0) = aω cosφ aω + cosφ 6 = a cos 6 cos φ a + φ 6 : 5 a cos 0 φ = a cosφ = : a 6 cos φ = 6 4 a cosφ =

23 π φ =, φ = 0, a =, a = y t t π t ( ) = sin + + sin ( 6 ) y ( t ) sin t π = + sin 6 t y ( t ) sin t π = + + sin 6 t y ( t ) = cost sin 6t y ( t ) = cost + sin 6 t

Math Matrix Algebra

Math Matrix Algebra Math 44 - Matrix Algebra Review notes - (Alberto Bressan, Spring 7) sec: Orthogonal diagonalization of symmetric matrices When we seek to diagonalize a general n n matrix A, two difficulties may arise: