4.2 Direct Integration Techniques
133
or
¨
x =
1
2
F (t) − 1000x
(4.11)
Time period of the system is
T = 2π
m
k
= 2π
2
2000
= 0.1987 s
Time interval t is chosen as
t =
1
10
× 0.1987 ∼ = 0.02 s
Another criterion is very important for selecting the time interval, i.e. time interval
should be such that it should represent the variation of the load with time and should
be capable of taking into account the sudden change in the loading function. In the
present problem with the time interval of 0.02 s, the time stations occur at sudden
breaks of the load function at t = 0.10 and 0.14 s.
The initial acceleration can be obtained from Eq. (4.11) above, given in Fig. 4.2b.
Noting that x 1 = ˙
x 1 = 0
¨
x 1 =
1
2
× 50 − 1000 × 0 = 25
From Eq. (4.4), x 2 =
1
2
× 25 × (0.02)
2
= 0.005 and
from Eq. (4.6), ˙
x 2 = 25 × 0.02 = 0.5.
Therefore at t = 0.02
¨
x 2 =
1
2
× 60 − 1000 × 0.005 = 25.
At t = 2 × 0.02, x 3 and ˙
x 3 be obtained from Eqs. (4.4) and (4.5)
x 3 = 0.005 + 0.5 × 0.02 +
1
2
× 25 × (0.02)
2
= 0.02
˙
x 3 = 2 × 25 × 0.02 + 0 = 1
Similarly, from Eq. (4.11), at t = 2 × 0.02
¨
x 3 =
1
2
× 70 − 1000 × 0.02 = 15
At t = 3 × 0.02