Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsToday’s Objectives:Students will be able to:a) Find the steady state response of 2DOF forced systemsb) Use Simulink for MDOF systemsForced two degree of freedom systemsRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsFinish Example from last time⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡++−+−++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡000022221122112211210θθGGxLkLkkLkLkLkLkkxJm&&&&2112inLLxxs+−=≈θθ22111212G2211 xso xLLLxLLLLxxLxxGG+++=−=−k1k2m, J0L1L2Let’s use the displacement at both ends as our coordinatesx1x2Note:∑∑∑∑=⇒==⇒=0G0 0MJMdtdLFxmFdtdPsysGsysxθ&&&&SoRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsFinish Example from last time022122111212=+++++xkxkxLLmLxLLmL&&&&2221112112000 LxkLxkLLxxJJMdtdLGsys−=+−=⇒=∑&&&&&&θx1x2k1x1k2x2022211112102210=+−+−+LxkLxkxLLJxLLJ&&&&In matrix form we get02122112112210210211212=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡++−++xxLkLkkkxxLLJLLJLLmLLLmL&&&&Notes22122111212 xkxkxLLmLxLLmLxmFdtdPGsysx−−=+++=⇒=∑&&&&&&Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsFinish Example from last time - Matlabevec1 =-0.0315 0.00320.0059 0.0574eval1 =4981.3 00 6785.3freq =70.578582.3732evec2 =-1.0000 -0.5187-0.7780 1.0000eval2 =4981.3 00 6785.3freq2 =70.578582.3732Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsForced vibrations – 2 DOF systemstsinFxxkkkkxxmmω⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡000121221212112121&&&&{}{}{}pcxxx+=tsinXXxxω⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧2121Consider the 2-DOF system with the following EOM:We know there will be a particular solution (satisfies right hand side) and a homogeneous solution (satisfied RHS = 0)Let’s look at the particular solution. Since the RHS is a “sin” let’s assumeSubstituting into EOM we get⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−−−0121222212122111FXXmkkkmkωωRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsSteady-state response (cont.)()[]ωZ()[]()[]()[]⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧−0011121FZdetZintadjoFZXXωωωFrom previous page⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−−−0121222212122111FXXmkkkmkωωSo, we can solve for {X1X2}Tby premultiplying by [Z(ω)]-1Quick matrix review of the inverse of a 2x2 matrix:⎥⎥⎦⎤⎢⎢⎣⎡−−−=⎥⎥⎦⎤⎢⎢⎣⎡−acbdbcaddcba11Rose-Hulman Institute of TechnologyMechanical EngineeringVibrations()[]()()()21222222111 `kmkmkwZdet +−−=ωω()[]()()()()()2222212122222121ωωωωωω−Ω−Ω=−−=mmmmwZdetThe determinant of [Z(ω)] isThis can be factored and written in terms of the natural frequencies as()⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧01121221212222221FmkkkmkZXXωωωSo we getX1 =X2 =Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsSketch these()()()22222121122221ωωω−Ω−Ω−=mmFmkX()()222221211122ωω−Ω−Ω=mmFkXX1ωΩ1Ω2X2ωΩ1Ω2These are frequency response plotsRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsSummary1. Find equations of motion: 2. Assume simple harmonic motion3. This gives4. Find natural frequencies5. Substitute in natural frequencies and find natural modes6. Write homogeneous solution1. Find equations of motion:2. To find the particular solution assume: 3. This gives4. Solve for {X}[]{}[]{}{}0=+ xKxM&&[][](){}{}02=+− XKMω[]{}[]{}02=+− KxMω[][](){}{}02=+−iiXKMωHomogeneous Solution2DOF Steady State Solution (with harmonic forcing)[]{}[]{}{}tsinFxKxMω=+&&{}{}tsinXxω=[][](){}{}FXKM =+−2ω{}[][](){}FKMX12−+−=ω{} { }tieXxω=Known!Forcing frequencyNatural frequencyNatural modeMagnitude of sinωtRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsThere is a nice website in Australiahttp://www.mech.uwa.edu.au/bjs/Vibration/TwoDOF/default.htmlRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsWhat do we do if we have damping?[]{}[]{}[]{}{}FxKxCxM=++&&&(){}()[](){}()[](){}sFsHsFsBsX ==−1[][][]()()(){}(){}sFsXKCsMssB=++⎥⎦⎤⎢⎣⎡4443444212If we haveIf we take the Laplace TransformsoIf we let s = jω()[]matrixfunction responsefrequency =ωjHWe can determine this from experimental data and use it for system IDRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsUsing SimulinkSometimes you have to add small damping to get rid of transient solution0 50 100 150 200 250 300-0.500.50 50 100 150 200 250 300-1010 50 100 150 200 250 300-0.500.50 50 100 150 200 250