Civil Engineering Reference
In-Depth Information
1000
Y
K
2Y
=
C
2
2
X
(
1000
Y
)
C
C
Y
K
3Y
=
C
2
2
(
4000
X
)
Y
C
C
The external applied loads are:
F
CX
(t) = 0,0
F
CY
(t) = -2.0 kN
Using a mass of 0.1 without viscous damping, the first 10 iterations are:
M
X
=
0.1
M
Y
=
0.1
C
X
=
0.0
C
Y
=
0.0
X
t
X
F
x
(t) F
1
K
1X
F
2
K
2X
F
3
K
3X
∑FK
X
0.0 2000.0
0.00 8.944
8.944
-10.000
7.889 -78.885
1.0 1960.6
0.00 8.890
8.926
-10.000
7.816 -78.163
2.0 1843.0
0.00 8.711
8.866
-9.998
7.580 -75.796
3.0 1649.5
0.00 8.360
8.740
-9.994
7.106 -71.058
4.0 1385.1
0.00 7.729
8.488
-9.986
6.230 -62.304
5.0 1058.3
0.00 6.628
7.962
-9.978
4.612 -46.116
6.0
685.4
0.00
4.799
6.762
-9.971
1.590
-15.897
7.0
296.7
0.00
2.210
3.945
-9.965
-3.810
38.100
8.0
-54.0
0.00
-0.394
-0.855
-9.958
-11.207
112.074
9.0
-292.6
0.00
-1.987
-4.653
-9.947
-16.587
165.867
10.0
-365.4
0.00
-2.343
-6.025
-9.931
-18.298
182.984
t
Y
F
Y
(t)
F
1
K
1Y
F
2
K
2Y
F
3
K
3Y
∑FK
Y
Ÿ
0.0
0.0
-2.00
-4.472
4.472
0.000
0.000
-20.000
1.0
-10.0
-2.00
-4.580
4.508
-0.049
-0.121
-18.789
2.0
-38.8
-2.00
-4.910
4.624
-0.180
-0.466
-15.344
3.0
-82.9
-2.00
-5.488
4.859
-0.353
-0.981
-10.185
4.0
-137.2
-2.00
-6.346
5.287
-0.524
-1.583
-4.173
5.0
-195.7
-2.00
-7.488
6.051
-0.664
-2.102
1.015
6.0
-253.2
-2.00
-8.773
7.367
-0.762
-2.168
1.679
7.0
-309.0
-2.00
-9.753
9.189
-0.831
-1.395
-6.048
8.0
-370.8
-2.00
-9.992
9.963
-0.911
-0.940
-10.601
9.0
-443.3
-2.00
-9.801
8.852
-1.027
-1.976
-0.239
10.0
-516.0
-2.00
-9.722
7.982
-1.174
-2.914
9.140
Search WWH ::
Custom Search