216
6 Free Vibration of Multiple Degrees of Freedom System
V 2 = V 1 − m 1 y 1 p
2
= 0 − 2 × 0 × 8.525 = 0
Calculation for other quantities can be carried out as follows:
M 2 = M 1 + V 2 L 1 = 1 + 0 × 1 = 1
θ 2 = θ 1 + (M 1 + M 2 )
L 1
2E I
= 0 +
1 × 2
2 × 50
= 0.02
y 2 = y 1 + θ 1 L 1 +
M 1
3
+
M 2
6
L
2
1
E I
= 0 + 0 +
1
3
+
1
6
1
50
= 0.01
Similarly, different quantities at the free end will be
V 3 = V 2 − m 2 y 2 p
2
= 0 − 2 × 8.525 × 0.01 = − 0.1705
M 3 = M 2 + V 3 L 2 = 1 − 0.1705 × 1 = 0.8295
θ 3 = θ 2 + ( M 2 + M 3 )
L
2E I
= 0.02 + (1 + 0.8295)
1
2 × 50
= 0.0383
y 3 = y 2 + θ 2 L 2 +
M 2
3
+
M 3
6
L
2
2
E I
= 0.01 + 0.02 × 1 +
1
3
+
0.8295
6
1
50
= 0.0394
Next, assuming V 1 = 1 and M 1 = 0 and following steps similar to those presented
above
V 3 = 0.943
M 3 = 1.943
Therefore, the determinant given by Eq. (6.89b) will be
=
− 0.1705 0.943
0.8295 1.943
= − 1.1135
By assuming different frequencies, the above steps may be repeated.
6 Free Vibration of Multiple Degrees of Freedom System
V 2 = V 1 − m 1 y 1 p
2
= 0 − 2 × 0 × 8.525 = 0
Calculation for other quantities can be carried out as follows:
M 2 = M 1 + V 2 L 1 = 1 + 0 × 1 = 1
θ 2 = θ 1 + (M 1 + M 2 )
L 1
2E I
= 0 +
1 × 2
2 × 50
= 0.02
y 2 = y 1 + θ 1 L 1 +
M 1
3
+
M 2
6
L
2
1
E I
= 0 + 0 +
1
3
+
1
6
1
50
= 0.01
Similarly, different quantities at the free end will be
V 3 = V 2 − m 2 y 2 p
2
= 0 − 2 × 8.525 × 0.01 = − 0.1705
M 3 = M 2 + V 3 L 2 = 1 − 0.1705 × 1 = 0.8295
θ 3 = θ 2 + ( M 2 + M 3 )
L
2E I
= 0.02 + (1 + 0.8295)
1
2 × 50
= 0.0383
y 3 = y 2 + θ 2 L 2 +
M 2
3
+
M 3
6
L
2
2
E I
= 0.01 + 0.02 × 1 +
1
3
+
0.8295
6
1
50
= 0.0394
Next, assuming V 1 = 1 and M 1 = 0 and following steps similar to those presented
above
V 3 = 0.943
M 3 = 1.943
Therefore, the determinant given by Eq. (6.89b) will be
=
− 0.1705 0.943
0.8295 1.943
= − 1.1135
By assuming different frequencies, the above steps may be repeated.