Page 127 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

114 Transient Vibration Chap. 4

Solution: The natural period of the system is
2 tt _ 2 it
= 0.50
CO 477
The time increment is chosen as /i = 0.05, and the differential equation is reorga­
nized as
X = (x,t) = 2F(t) —IGtt^x

f
This equation is to be solved together with the recurrence equation, Eq. (4.7-7),
''/+1 = 2xi - I + h~f{x, t)
Because the force and the aceeleration are zero at / = 0, it is necessary to start
the computational process with Eqs. (4.7-9) and (4.7-10) and the differential equation:
•*2 = = 0.000417i'2

¿2 = 2F(0.05) - I61T-X2 = 50 - 158^2
Their simultaneous solution leads to
_ (0.05)V(0.05)
^ = 0.0195
" 3 + 8tt"(0.05)-
x, = 46.91
The flow diagram for the computation is shown in Fig. 4.7-6. With h = 0.05, the
time duration for the force must be divided into regions / = 1 to 5, 7 = 6 to 9, and
7 > 9. The index 7 controls the computation path on the diagram.
Shown in Figs. 4.7-7 and 4.7-8 are the computer solution and results printed on
a line printer. A smaller A/ would have resulted in a smoother plot.
Damped system. When damping is present, the differential equation
contains an additional term x- and Eq. (4.7-7) is replaced by
+ , = 2x, - x, _, + h^f{x,,Xi,t,) i > 2 (4.7-T)
We now need to calculate the velocity at each step as well as the displacement.
Considering again the first three terms of the Taylor series, Eq. (4.7-4), we
see that ^2 is available from the expansion of with i = 1:
ri . X
X2 = Xi -h Xj/Z + ^ / ( x , , Xj, C)
The quantity X2 is found from the second equation for Xy_j with i = 2:

X , = X 2 - X 2 h + • y / ( x , , i 2 ’ U)
With these results, X3 can be calculated from Eq. (4.7-7'). The procedure is thus
repeated for other values of x- and x^ using the Taylor series.