HW2. CEE 247. Structural Dynamics. UCI. Fall 2006

Nasser Abbasi

Problem 3.2


problem_3_2.png

Solution

The rotating mass generates a force of MATH where $\bar{\omega}$ is the angular speed of the rotating weight and $e$ is the distance of the mass from the center of the motor.

Hence the vertical load is MATH But MATH, hence the vertical force is MATH

The physical idealized model is


standard_model.png

Hence the equation of motion can now be written from the free body diagram as (where $M$ is the mass of the electric motor) and assuming the mass is moving to the right, and taking $u$ relative to the static equilibrium position.


free_body.png

MATH

The above ODE has the solution

MATH

Where MATH, where MATH and MATH and MATH and MATH and MATH

At steady state, the transient solution decays to zero thanks to the negative exponential term in it, and the solution becomes

MATH

Which has amplitude of $u_{0}$. Hence we need to evaluate $u_{0}$

MATH

But motor runs at $900$ rpm, hence MATH We are given that $W^{\prime}e=1$ lb.in, Hence MATH

Now we need to find $k$, the stiffness of the beam against bending. For this geometry, MATH

But $E=30,000$ ksi for steel, and for $W8\times31$ from tables we find $I_{xx}=110$ in$^{4}$, hence

MATH

Now to find $\omega_{n}.$Recall that that $W=1000$ lb, hence

MATH Hence MATH

Putting all these together we obtain

MATH

Hence MATH

(Note: The back of the book gives $u_{0}=0.0037$, I think the book used a slightly different steel table to obtain $I_{xx}$ which could have been slightly different than the one I used.)

Problem 3.3


problem_3_3.png

solution

From the free body diagram:


free_body.png

We see that the forces transmitted to the support are

MATH

Where $u$ here is taken as the steady state solution from problem 3.2, which is

MATH

Where MATH, and MATH Differentiate the above equation and substitute the results back into the $f_{tr}$ equation we obtain

MATH

Where MATH

Hence we see that the maximum force transmitted to the supports are given by

MATH

We now plug into the above equation the results we obtain from problem 3.2 to determine $f_{tr_{0}}$. All the variables in the above expression are known, which are repeated here

MATH

We just need to find the damping $c$. Since MATH and MATH, hence

MATH

Hence MATH

Now substitute all the above values into equation (1) we obtain

MATH

Problem 3.8


problem_3_8.png

solution

The physical idealized system is the following


standard_model_prob_3_8.png

Where in the above, $u\,$ is the absolute displacement of the tower, and $u_{g}$ is the absolute displacement of the ground. The free body diagram is


free_body_prob_3_8.png

Applying newton's second law we obtain

MATH

Expand and rearrange

MATH

Let the relative motion between the mass and the ground be $u_{r}$, hence $u_{r}=u-u_{g}$ or $u=u_{r}+u_{g}$, similarly

MATH

And

MATH

Using the above expressions for $u,\dot{u},\ddot{u}$, we can now rewrite (1) as

MATH

Expand (2) and cancel terms we obtain

MATH

Or

MATH

The above is the equation of motion of the tower using relative displacement. Hence we can view the term $m\ddot{u}_{g}$ as the effective force acting on the tower due to the acceleration of the ground.

Now using the fact that the ground motion is harmonic, we can write

MATH

Where $u_{g_{0}}$ is the maximum amplitude of the ground displacement, and $\bar{\omega}$ is the ground motion frequency. Hence from the above we obtain that MATH

Plug the above into (3) we obtain

MATH

The above is now in standard 2nd order linear system, the steady state solution for $u_{r}$ is

MATH

Where MATH and $u_{0}=u_{st}R_{d}$, where MATH, and MATH

MATH

But we are told that MATH, Hence the above becomes

MATH

We are given that MATH

and MATH

Then MATH

Then MATH

and since $\zeta=0.1$ we obtain

MATH

Problem 3.9

Determine the transmissibility of the above problem

Solution

We need to determine first the expression that represents the force that is transmitted to the ground. From the idealized system diagram


standard_model_prob_3_8.png

We see that the force transmitted to the support is MATH

But MATH

Where MATH where $F_{0}$ here is the effective force.

Hence

MATH

Where MATH

Hence Max force transmitted is $A_{tr}$

MATH

But MATH hence

MATH

But MATH hence

MATH

Now, since from the earlier problem we found that MATH, and given that $\zeta=0.1,$ then

MATH

Problem 3.12


problem_3_12.png

Solution

We start with the expression for the maximum amplitude steady state displacement given by

MATH

We have 2 cases, case 1 is when MATH (resonance), and the second case is when MATH. Hence we obtain the following equation

MATH

Hence MATH

Problem 3.17


problem_3_17.png

Solution

Considering maximum amplitude of steady state is given by

MATH

But MATH

We are given one case where $u_{r}$ (resonance) and another case where $u_{r\neq1}$.When $r=1$, we obtain

MATH

When $r\neq1$ we write (where we call $u_{0}$ when $r=r_{1}$ as $u_{1})$

MATH

Square equation (1) and (2) and divide by each others we obtain

MATH

Hence

MATH

Since MATH

Hence MATH

But MATH at resonance, hence

MATH

But from first part, we found expression for $\zeta$, which we plug into the above to obtain

MATH

Hence

MATH