Lecture12-Prof.Ju

Lecture12-Prof.Ju - CEE M237A MAE M269A Lecture 12...

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CEE M237A / MAE M269A Lecture 12: Vibrations of Structural Systems Professor J. Woody Ju
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Vibration of Structural Systems y Some structures are composed of various members, whose vibrational characteristics are governed by different equations y The natural vibration of such systems can be analyzed by assembling their individual components y In this section, some examples are given to illustrate the analysis of such systems 2
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Example 1–Two Distinct Beams with an Additional Massless Spring 3 44 12 11 1 1 1 2 2 2 2 2 The EOM for the two distinct beam 0 ; 0 s are: ww EI Aw xx ρρ ∂∂ += + = ±±
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) y Boundary Conditions at 4 () ( ) 2 22 12 2 2 2 23 2 11 2 2 , 0, 0 ; 0 0, , ; 0 wL t wt E I x w L t EI xx == ∂∂ = = 1 0 (no displ., no moment) x = and (no moment, no shear): xL =
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) y The four interface conditions (displ., slope, moment, shear-spring) are: 5 () ( ) 11 2 22 2 2 2 12 2 33 2 1 1 ,0 , , , , , wL t w t t w t EI xx t w t t w t kw L t = ∂∂ = = −+ =
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) y The homogeneous solutions are: 6 () ( ) ( ) ( ) () ( ) [ ] ()( ) [ 11 1 1 1 2 1 1 31 1 4 1 1 22 5 2 2 6 2 2 2 72 12 2 ,s i n c o s cosh sinh i n c ; os cosh wT t wx t T t C x C x Cx C x t T tC x C x Wx w T t W x ββ β ⇒= + + = + =+ = + ] 28 2 2 42 2 2 sinh where ; AA EI ρρ βω + ==
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) 7 () 1 12 3 2 23 1 11 2 3 2 1 22 2 2 2 3 3 2 1 1 1 4 1 1 The BCs at 0 are : 0 =0 0 0 0 0 ,s i n 0 The BCs at sinh are : 0 0 x WC C CC dW EI C C dx xL dW L dx dx Wx t C x C x ββ = →+ = = = =→ + = ⇒∴ = = =+ −=
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) 8 () ( ) 11 2 2 12 22 2 2 2 33 2 1 1 four interface cond The are: 0 0 0 0 itions WL W dW L dW dx dx dW L dW EI dx dx kW L dx dx = = = −+ =
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) 9 () 5 22 6 7 8 5 6 7 8 11 1 4 1 1 6 7 2 1 4 1 1 5 8 1 2 2 1 4 1 1 2 1 Therefore, we have: sin cos cosh sinh 0 cos sin sinh cosh 0 sin sinh 0 cos cosh 0 sin sinh CL C L C L C L C L C L C L C L C C C LC C EI C L E ββ β −− + + = −++ + = +− = −+ = + ( ) 3 1 67 3 2 1 1 1 4 1 1 1 1 5 8 3 1 0 cos sin cosh sinh 0 where CC I L C L L C C kEI βα α −= = =
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Example 1–Two Beams with an Additional Massless Spring (Cont’d) 10 These six eqns. may be cast in matrix form The determinant of the 6 by 6 coeff. matrix must vanish in order to admit non-trivial solutions The roots of this determinant can be extracted by the method regula falsi
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Example 2–Two Distinct Beams with an Axial Force Member 11 () 4 4 2 33 2 For the two beams and the st 0 1,2 rut 0 : i ii i i i i w EI Aw i x u EA Au z ρ += = = ±±
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Example 2–Two Beams with an Axial Force Member (Cont’d) y
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This note was uploaded on 05/18/2011 for the course MAE 269A taught by Professor Ju during the Spring '11 term at UCLA.

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Lecture12-Prof.Ju - CEE M237A MAE M269A Lecture 12...

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