HW 3. CEE 247. Structural Dynamics. UCI. Fall 2006
Nasser Abbasi
Integration by parts is used in many problems below to solve . I derive it once.
Let and hence and
Hence
Hence the integral becomes
The above is the form to remember.
or
For example, when , , we obtain
Solution
We first assume that the initial absolute state of the girder is , and
This is the force load diagram
The intercept is and the slope is hence since the general line equation for is , we see that the equation for force loading is
First we draw the physical model diagram
Using Duhamel integral, the displacement is (using the assumption of no damping)
Substitute (1) into the above and carry the integration.
But hence the above becomes
Now to find the stiffness
Hence
Now substitute the above results for and in equation (2), and evaluate at we obtain
part(b)
To find maximum displacement we use the response spectrum shown on page 107 of the 5th edition of the text book. First we find the natural period
Hence
Hence from the spectrum on page 107, we see that approximately
But
Hence
This is a small program to plot u(t) itself. We see that became maximum before . maximum as at about
solution
fig P4.5 is
Hence we need to find
For and for an undamped simple oscillator, using Duhamel integral, the displacement is
hence the above becomes
Now we find
Hence
Now we do the case for
Hence
But hence the above becomes
Notice there is a sign difference with the answer on the back of the book. The back of the book gives
I think the answer in the back of the book is wrong. One way to obtain the book answer from my answer is to replace by .
Frame shown in problem 4.3 above is subjected to sudden acceleration of 0.5 g applied to the foundation. Determine the maximum shear force in the columns. Neglect damping.
solution
The equation for motion when the system is subjected to ground acceleration can be written as
Where is the relative motion of the girder to the ground, and is the ground acceleration (absolute). Hence is the effective force
Hence this is the same problem as
which has the solution
Hence
But from problem 4.3, we calculated to be lb/in hence
Now maximum shear is given by , hence for the left column we have (I will take absolute value of displacement, since we are only interested in maximum value, the sense of shear is not relevant).
and for the right column
solution
lb
lb
lb/in
Let
Hence for
For
For an undamped simple oscillator, using Duhamel integral, the displacement is
Hence for
Hence
Note that and
Now for
But the free vibration response is , using
and the second integral is
Simplify to
Hence for is by putting the above result back into (1) we obtain
But , hence
simplify
Hence
Now
Note that at we have
and at we have
Now for since no force is applied, we use the free vibration solution using the above and as initial conditions
Now that we have for each time segment, we can plot the solution. Here it is for up to sec
Here is the solution for up to