Structural Dynamics The spring force is given by and F(t) is the driving force. Start by applying Newton’s second law (F=ma). We will now look at free vibrations. Considering the free vibration of the mass—that is, when F(t) = 0. Spring mass system.
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Structural Dynamics
The spring force is given by and F(t) is the driving force. Start by applying Newton’s second law (F=ma).
We will now look at free vibrations.
Considering the free vibration of the mass—that is, when F(t) = 0.
Spring mass system.
Setting m = 0.
We obtain.
Free body diagrams of spring mass system.
The free vibration of the system will take the form of simple harmonic motionbelow.
Time/displacement curve.
The motion defined above is called simple harmonic motion. The displacementand acceleration are proportional but of opposite directions..
xm is the maximum displacement or amplitude of the vibration.
The period is the time necessary to make a full cycle.
Time/displacement curve.
All vibrations are damped to some degree by friction forces. These forces may be caused by dry or Coulomb friction between rigid bodies, by internalfriction between molecules within a deformable body, or by fluid frictionwhen a body moves in a fluid. These result in natural circular vibration frequencies that are less than those calculated using free vibration.
Lets start with a 1D bar to look at vibrations.
We will assume a displacement function taking the form of.
So the shape functions will be.
The strain/displacement relationship is given by.
1
2
uu
The stress/strain relationship is given by.
D B U
Time dependence: The bar is not in equilibrium under a time‐dependent forceso f1x f2x.
At each node ‘the external (applied) force minus the internal force is equal to the nodal mass times acceleration.’
We add this internal force F = ma to each nodal force to obtain.
The masses m1 and m2 are lumped to each of the nodes to obtain.
exf
The external nodal force becomes.
The element stiffness matrix is.
The lumped mass matrix is given by.
The acceleration term is.
Another way to handle the mass is through the developmentof the consistent mass matrix where use the shape function to model the mass along the bar.
We will use D’Alembert’s principle and introduce an effective body force Xe as.
the minus sign indicates that the acceleration produces D’Alembert’sbody forces opposite in the direction as the acceleration.
We will use and substitute
for {X} gives us.
Using and completing the derivatives.
The element mass matrix is given by.
Substituting for interpolation functions:
For a 1D bar one gets:
Complete the matrix multiplications.
Upon integration
Assembly
Numerical Integration in Time
Several different methods are available to do the numerical integration.
We will cover central‐difference and the Newmark‐Beta.
The central difference method is based on finite difference expressions in time for velocity and acceleration at time t given by the following:
With a Taylor expansion acceleration can be defined in terms of displacements.
Rearranging gives us.
This will be used to determine the nodal displacements in the next timestep i+1 knowing the displacements at time steps i and i – 1 and the acceleration attime i.
From
we obtain
To obtain an expression for , we first multiply above by the mass matrix M and then substitute
for into the equation at the top of the page to obtain:
and rearranging.
We start the computations by determining the displacement atEqn 1.
at t = 0 then is
Eqn. 1
==
Determine the displacement, velocity, and acceleration at 0.05‐s time intervals up to 0.2 s for the one‐dimensional spring‐mass oscillator subjected to the time‐dependent forcing function shown in the figure below. This forcing function is a typical one assumed for blast loads. The restoring spring force versus displacement curve is also provided. [Note that the bar in the figure represents a one‐element bar with its left endfixed and right node subjected to F(t) when a lumped mass is used.
Step 1
At time = 0, the displacement and velocity are
m = 31.83 lb-s/in2
Step 4
Use
Newmark‐Beta Method
Newmark‐Beta equations
where and are parameters chosen by the user. The parameter is generally chosen between 0 and 1/4, and is often taken to be 1/2.
= 1/2 and = 1/4 , gives stable analysis results.
Q = kdi is the restoring spring force.
Given we wish to solve this equation given
into gives
Rearranging and combining terms gives us.
we get.
This equation has a nontrivial solution if and only if the determinant of the coefficient matrix of is zero, so we must have.′
This equation has a nontrivial solution if and only if the determinant of the coefficient matrix of is zero, so we must solve.′
below has a
below. Eqn. 1 and either the lumped mass matrix or the consistent‐mass
matrix.
The lumped‐mass matrix will be used in this analysis with 2 elements.
Assemble the global stiffness matrix.
The mass matrices are.
Assembled
Assemble all the components and apply the boundary conditions(which is the same as:
Dividing above by AL and letting = E/ ( L2), we obtain:
Take the determinant of this
gives.so
The first and second natural frequencies are given by:
The natural frequencies are given by:
Hz.
We can obtain the displacements from here.
It is customary to specify the value of one of the natural modes for a given i or i . With 3
1 1 so from above equation 21 0.7
2/2
23
00
Substituting 2topequation.
Again assuming 31 1
2/2
23
00
Using the previous results we shall now solve a time dependentproblem using the same problem as the natural frequency workabove.
It has been shown that the time step must be less than or equal to 2 divided by the highest natural frequency when the central difference method is used.
In our case
Due to other practical consideration it is better to use.
An alternative guide (used only for a bar) for choosing the approximatetime step is
Cx is called the longitudinal wave velocity and is given by:
For convenience we shall use a time step value of 0.25 second.
Thus eliminating the first row and columns of above and rearranging.
Solve for d1 using.
Beam Element DynamicsThe lumped mass matrix for a beam element is given as follows for the beam element below.
The mass moment of inertia of half of the beam segment about each end node using.
The consistent mass matrix is determined by using the shape functions:
The mass matrix becomes.
Consider the two beams below each has mass density , modulus ofelasticity E, cross‐sectional area A, area moment of inertia I, and length 2L. he beam is discretized into (a) two beam elements of length L.
(a) Two‐Element Solution
Using boundary conditions d1y = 0, 1 = 0, d3y = 0, and 2 = 0 to reduce the matrices) as:
Dividing above by AL we get.
The exact solution is
The global K matrix becomes.
Evaluating the determinant.
‐ ‐ =0
‐ ‐ =0
‐ ‐ =0
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