4.2 Direct Integration Techniques
137
The assumed value of ¨
x i + 1 is substituted into Eqs. (4.16) and (4.17), along with
the already computed values of x i , ˙
x i and ¨
x i , to obtain ˙
x i + 1 and x i + 1 . These values
of x i + 1 and ˙
x i + 1 are substituted into the equation of motion, to obtain ¨
x i + 1 . If this
value of ¨
x i + 1 does not tally with the previous value of ¨
x i + 1 , then the process is to
be repeated with the new value of ¨
x i + 1 . This is to be continued till two successive
values of ¨
x i + 1 agree within reasonable limits. This method is demonstrated in the
following example.
Example 4.3 Solve the problem of Example 4.1 (given in Fig. 4.2) by the linear
acceleration method.
The equation of motion for this problem is
¨
x =
1
2
F(t) − 1000x
(4.18)
x i = ˙
x i = 0
and
t = 0.02 s
Assume ¨
x 1 = ¨
x 2 = 25.
Then substituting these values into Eqs. (4.16) and (4.17), we get
˙
x 2 = 0 +
1
2
× 0.02 × 2 × 25 = 0.5
and x 2 = 0 + 0 +
1
6
× (0.02)
2 [2 × 25 + 25] = 0.005.
Substituting the above values x 2 and ˙
x 2 into Eq. (4.18), gives
¨
x 2 = 30 − 1000 × 0.005 = 25
The value ¨
x 2 is identical to the assumed value. Therefore, calculation may be
started for the next time step, i.e. t = 2 × 0.02 s.
Let ¨
x 2 = ¨
x 3 = 25.
Then from Eqs. (4.16) and (4.17)
˙
x 3 = ˙
x 2 +
1
2
( ¨
x 2 + ¨
x 3 ) t
= 0.5 +
1
2
(2 × 25) × 0.02 = 1.0
and
x 3 = x 2 + ˙
x 2 t +
1
6
(2 ¨
x 2 + ¨
x 3 ) ( t)
2
= 0.005 + 0.5 × 0.02 +
1
6
(2 × 25 + 25) (0.02)
2