Lecture 22: Finite element method: Axial vibrations of bars

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Lecture 22: Finite element method: Axial vibrations of bars Reading materials: Section 9.1

1. Introduction Discretization Assembly and solution

2. Governing equations Axial vibrations of a long slender bar

initial conditions

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boundary conditions

u(x0, t)=ux0

3. Axial vibration element

Finite element approximation

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4. Energy kinetic energy due to distributed mass

kinetic energy due to concentrated mass

Total kinetic energy

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Strain energy

work done by the distributed applied force

work done by the concentrated loads

Total work done

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5. Equations of motion U=Us-W

6. Assembly and Solution The element equations are derived in the above with local coordinates. Assemble the element equations into the global equations based on the global coordinates. Apply the boundary conditions. Numerically solve the equations of motion.

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7. Example 1

General element equations

Element 1

⎛ 1.869 0.9345 ⎞⎛ u&&1 ⎞ ⎛ 490000 − 490000 ⎞⎛ u1 ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ & & u 0 . 9345 1 . 869 490000 490000 − ⎝ ⎠⎝ 2 ⎠ ⎝ ⎠⎝ u 2 ⎠ ⎝ 0 ⎠ Element 2

⎛ 1.335 0.6675 ⎞⎛ u&&2 ⎞ ⎛ 350000 − 350000 ⎞⎛ u2 ⎞ ⎛ 0 ⎞ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ + ⎜⎜ & & u − 0 . 6675 1 . 335 350000 350000 ⎠⎝ u3 ⎠ ⎝ 0 ⎠ ⎝ ⎠⎝ 3 ⎠ ⎝

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Element 3

⎛ 0.801 0.4005⎞⎛ u&&3 ⎞ ⎛ 210000 − 210000⎞⎛ u3 ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ & & u 0 . 4005 0 . 801 − 210000 210000 ⎝ ⎠⎝ 4 ⎠ ⎝ ⎠⎝ u 4 ⎠ ⎝ 0 ⎠ Considering concentrated mass and force on node 4, the element equations for Element 3 could be written as:

⎛ 0.801 0.4005 ⎞⎛ u&&3 ⎞ ⎛ 210000 − 210000 ⎞⎛ u 3 ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ & & ⎝ 0.4005 100.801⎠⎝ u 4 ⎠ ⎝ − 210000 210000 ⎠⎝ u 4 ⎠ ⎝ F (t )⎠ Form of global equations

Assemble of elements Element 1

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Element 2

Element 3

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Global equations of motion

Apply BCs

Note: for nonzero BCs

Solutions Newmark’s method

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