Page 269 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)
P. 269
256 Computational Methods Chap. 8
Solution: We first zero the largest off-diagonal term, whieh is a2^ = -0.7071.
1.0 -0.3536 0
A = -0.3536 r 1.0 -0.7071
0 i -0.7071 1.0
1 0 0
0 cos 6 —sin 6
0 sin 6 cos 6
2i?23 2 (- 0.7071)
tan 26 =
a. - 1
*22
26 = 90°
6 = 45°
sin 45° = cos 45° ==0.7071
1 0 0 '1 0 0
R. = 0 0.7071 -0.7071 « r = 0 0.7071 0.7071
0 0.7071 0.7071 0 -0.7071 0.7071
A, = R AR M
1 0 0 1.0 -0.3536 0 ■1 0 0
0 0.7071 0.7071 -0.3536 1.0 -0.7071 0 0.7071 -0.7071
0 -0.7071 0.7071 0 -0.7071 1.0 0 0.7071 0.7071
1.0 -0.250 I 0.250
-0.250 0.2929 1 0
0.250 0 1.7071
We thus find that in zeroing the term a22, = «32. we have introdueed a new nonzero
term ¿7,
2a. 2 (-0.250)
tan 26 = = -0.7071
«11 - «22 1 - 0.2929
26 = -35.26°
6 = -17.63°
sin 6 = -0.3029
cos 6 = 0.9530
0.9530 .3029 O'
R. = -0.3029 0.9530 0
0 0 1
1.097 0 0.2383
RiA.R^ = 0 0.2134 0.0757 = A,
0.2383 0.0757 1.7071
