Question:
The mode shape reported in Table 10.4 for a pinned-free beam of \(\sqrt{3} x\) is a rigid-body mode.
Indicate whether the statement presented is true or false. If true, state why. If false, rewrite the statement to make it true.
Transcribed Image Text:
TABLE 10.4 End Conditions X=0 X=1 Fixed-fixed Natural frequencies and mode shapes for beams Five Lowest Natural Characteristic Frequencies Equation Wk Mode Shape cos A4 cosh A/4=1 =22.37 Pinned-pinned sin A1/4=0 Kinetic Energy Scalar Product = 61.66 (x,(x), x*(x)) [cosh Ax-cos Axa (sinhx - sin x)] S*x(x)x(x)dx = = 120.9 cosh A/4 sinh A - cos A4 sin A4 - 199.9 = 298.6 9.870 C, sin A/4x S*x(x)x(x)dx w=39.48 = 88.83 -157.9 Fixed-free cos 1/4 cosh 1/4 = -1 w=3.51 w = 22.03 = 61.70 120.9 - 199.9 =246.7 [cosh Ax cos xa(sinhx - sin x)] S*x(x)x(x)dx cos A+ cosh A - sin A/4 + sinhX}/* Free-free cosh 1/4 cos A/4=1 = 0 =22.37 1,3x (k = 1) [cosh Ax + cos Ax + a(sinh A*x + sin ^^x)] S*x(x)x(x)dx = 61.66 cosh A a = - cos A/4 Fixed-linear spring 3/4 (cosh A4 cos 1/4 + 1) = 0 120.9 w=199.9 For = 0.25 B = -B(cos A1/4 sin A1/4 - cosh 1/4 sin 11/4) w, 3.65 = 22.08 w=61.70 w = 120.9 w-199.9 sinh A sin A/4 [cos A*x-cosh Axa (sin x - sinh Ax)] x(x)x(x)dx cos A/4 + cosh A/4 = sinA}/4 + sinh)}/*