7.9 Direct Integration for Determining Response of MDF Systems
295
= 100
−0.00845
0.10234
− 20
0
0
− 1.0
0
0
=
−0.845
10.234
{ ˙
x} 1 = { ˙
x} 0 + 0.1{x} 0 + 0.1{ ¨
x} 1 = 0.1
−0.845
10.234
=
−0.0845
1.0234
Let us pass on to the next time interval and repeat the above steps
F
2
=
0
192
+
8 0
0 8
(100{x} 1 + 20{ ˙
x} 1 + { ¨
x} 1 )
=
0
192
+
8 0
0 8
100
−0.00845
0.10234
+ 20
−0.0845
1.0234
+
−0.845
10.234
=
−27.040
519.488
{x} 2 =
1
8 × 12781
118 −9
−9 109
−27.04
519.488
=
−0.0765
0.5562
{ ¨
x} 2 = 100
−0.0765
0.5562
−
−0.00845
0.10234
− 20
−0.0845
1.0234
− 1.0
−0.845
10.234
=
−4.274
14.681
{ ˙
x} 2 =
−0.0845
1.0234
+ 0.1
−0.00845
0.10234
+ 0.1
−4.274
14.681
=
−0.5127
2.5018
The above procedure can thus be repeated for other time steps.
Instead of working directly on the coupled equations of the MDF system, the
equations may be decoupled by using normal coordinates. The equation in each mode
can then be solved by Newmark’s method. The final result of the displacements for
such cases will be obtained by superimposing the effects of all the modes.