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MECH 375- MECHANICAL VIBRATIONS FINAL CRASH PART 4- Multi DOF Systems Determination of Natural Frequencies and Mode Shapes Concordia University CA$15.96   Add to cart
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MECH 375- MECHANICAL VIBRATIONS FINAL CRASH PART 4- Multi DOF Systems Determination of Natural Frequencies and Mode Shapes Concordia University

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MECH 375- MECHANICAL VIBRATIONS FINAL CRASH PART 4- Multi DOF Systems Determination of Natural Frequencies and Mode Shapes Concordia University

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  • November 21, 2023
  • 27
  • 2023/2024
  • Exam (elaborations)
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  • mech 375
  • mech 375 mechanical vibrations final crash
  • mech 375 mechanical vibrations

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MECH 375- MECHANICAL VIBRATIONS FINAL CRASH PART 4-
Multi DOF Systems Determination of Natural Frequencies and
Mode Shapes Concordia University

,Multi-Degree-of-Freedom Systems
For multi-degree-of-freedom systems, the equations of motion can be found by:

1. Newton’s method of force analysis
2. Lagrange’s equation for energy analysis.


Using Lagrange’s Equation to Derive Equations of Motion
Recall that kinetic energy is denoted by T and potential energy is denoted by U.

𝑑 ∂𝑇 ∂𝑇 ∂𝑈 ∂W
( )− + = 𝑄j =
𝑑𝑡 ∂𝑞j ∂𝑞j ∂𝑞j ∂𝑞j

Note that qj are the state-variable, where j = 1, 2, 3, …, n.

Method of Solving:

1. Find the kinetic energy (T) and potential energy (U) of the system. Solve for the work by an
external force (W) if there is any.
2. Apply Lagrange’s equation for each state variable qj.
3. May need to do a kinematic analysis.

Recall that all calculations must be done at the center of mass if the system contains a rigid body.


Influence Coefficients




The FBDs of the system:




2

, The Stiffness Coefficient Matrix [K]:

In general, F = kx, but x in this case will be set to 1. The stiffness coefficient matrix is defined as:

𝑘11 ⋯ 𝑘1𝑛
[𝐾] = [ ⋮ ⋱ ⋮ ]
𝑘𝑛1 ⋯ 𝑘𝑛𝑛

When setting x1 = 1 and x2 = x3 = 0, the coefficients being solved for through force balance are those in
the first row. Note that kij = kji.

The Flexibility Coefficient Matrix [a]:

In general, F = kx, but F in this case will be set to 1. The stiffness coefficient matrix is defined as:
𝑎11 ⋯ 𝑎1𝑛
[𝐾] = [ ⋮ ⋱ ⋮ ]
𝑎𝑛1 ⋯ 𝑎𝑛𝑛

When setting F1 = 1 and F2 = F3 = 0, the coefficients being solved for through force balance are those in
the first row. The coefficients aij are the displacements experienced by each block based on the force.
Note that aij = aji. Also note that:

[𝑎] = [𝐾]−1

Method of Solving:

For the Stiffness Coefficient Matrix:

1. Draw the FBDs with the including the force F. Do not include the inertial force.
2. Set the first block’s displacement to 1 and the others to zero, then do a force balance to solve
for kn1.
3. Repeat step 2 by setting the next block to 1 and the others to zero. Repeat until all blocks are
done.

For the Flexibility Coefficient Matrix:

1. Draw the FBDs with the including the force F. Do not include the inertial force.
2. Set the first block’s force F to 1 and the others to zero. Replace xn to ani, where i is the block’s
reference number, ie, first block has i = 1.
3. Solve the n number of equations simultaneously to find the coefficients.
4. Repeat steps 2 and 3 for different blocks.




3

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