MIT2_092F09_lec16 - 2.092/2.093 Finite Element Analysis of...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
2.092/2.093 Finite Element Analysis of Solids & Fluids I Fall ‘09 Lecture 16 - Solution of Dynamic Equilibrium Equations, cont’d Prof. K. J. Bathe MIT OpenCourseWare Reading assignment: Sections 9.1-9.3 Recall from our last lecture the general dynamic equilibrium equation and initial conditions: 0 MU ¨ + CU ˙ + KU = R ( t ) ; 0 U , U ˙ (1) This equation can be solved by: Mode superposition Direct integration Mode Superposition i = ω i 2 i (2) The ω i 2 are the eigenvalues and φ i are the eigenvectors for this system. Solve for ω i 2 , φ i : ω 2 ω 2 . . . ω 2 1 2 n 0 ���� ���� ���� for φ 1 for φ 2 for φ n where each φ i refers to a mode shape. Aside: Consider, picking “a” φ , = α ˜ (3) 1 where α is a nonzero scalar. Obviously, K α φ = ˜ = R . If φ ˜ is an eigenvector, then the load R 1 obtained using φ ˜ gives us back the vector φ ˜ (now scaled by α ). We also used orthonormality to establish that: φ T i j = δ ij φ T i j = ω i 2 δ ij The definition of the Rayleigh quotient is φ T ρ ( φ ) = φ T where φ can be any vector. So, we have ρ ( φ i ) = ω i 2 ω 1 2 ρ ( φ i ) ω n 2 Recall that the strain energy for any vector φ is 1 2 φ T . Thus, the strain energy corresponding
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern