“Computational Civil Engineering - "Intersections" International Journal

244 C. C. Comisu2[ tan [ k] −ω[ m] −ω[ c]]{ ψ } = { 0}θ (9)iIf equation (9) is premultiplied by the transpose { ψobtain:Tω { ψi} []{ c ψi}T{ ψ } [[] k −ω[ m]]{ ψ }tanθi2iii} Tiand rearranged, we= (10)From equarion (10) it can be seen that each of the roots tagθiis acontinuous function of ω. For low values of ω, tagθ is small i.e. θ i is a smallangle. As ω increases and approaches ω 1 the undamped natural frequency, one ofthe roots θ i, which can be named θ 1, approaches the value π/2. When ω tends to ∞ ,θθ i = 1, ...n can be1 (ω) tends to π. In a similar manner the remaining roots ( )ii2plotted as a function of frequency ω i . Thus θ k is that root which has the value π/2 atthe undamped natural frequency ω = ω k .. If every element in the matrix [c] is multiplied by a constant factor, thenequation (8) shows that the roots tagθ will all be increased by the same ratio.iThus equation (9) which determines the mode shapes, will be multipliedthroughout by the same factor and the mode shape{ ψ i} will be unchanged.Equation (9) can be re-written as⎡⎢⎣2[] k − [ m][ c] ⎤ { ψ } = { 0}ω ⋅ω − ⎥ i(11)tanθi ⎦When ω is equal to one of the undamped natural frequency, say ω = ω 1 ,then one of the roots θ 1 is 90 0 . Thus equation (11) which determines the modeshape for this root becomes:2[ k] − [ m]]{ ψ } { 0}ω (12)1 i=It can thus be seen, that when the frequency is equal to one of theundamped natural frequency, the mode shape for one of the roots (which is equalro π/2) is identical to the principal or normal mode shape.Attention can now be paid to the force that is required to excite any onemode { ψ i} for the corresponding root tagθiat any one frequency. The forceration required can be calculated from equation (5) namely:2[ k] −ω[ m]]{ ψ } + ω θ [ c]{ ψ } = { Γ }cosθ siniiii(13)