Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsToday’s Objectives:Students will be able to:a) Find the equations of motionb) Find the natural frequencies and mode shapesc) Find the time responseTwo degree of freedom systemsTorsional damper (vibration absorber)Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExamplesekktGModely1L1L2k1c1Gy2k2c2Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExampleFind the equation of motion for the system shown and put in 2ndorder matrix form∑→= FdtdPxsysk2m1m2k1x1 x2k3Free body diagramsx1x2Mass 1: x-direction:Mass 2: x-direction:∑→= FdtdPxsysRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample (continued)()()0012232222212111=−++=−++xkxkkxmxkxkkxm&&&&So the equations of motion are:We can write this in 2ndorder matrix form:⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡+−−++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡0000213222212121xxkkkkkkxxmm&&&&Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample (cont.)To make the algebra a bit easier let’s assume k1= k2= k3= k and m1= m2= m. Therefore our EOM is⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡0022002121xxkkkkxxmm&&&&or our individual equations are:02211=−+kxkxxm&&02122=−+kxkxxm&&Assume simple harmonic motiontitititieXxeXxeXxeXxωωωωωω2222212111 −=⇒=−=⇒=&&&&Or in a vector form this is {}{}{}{}titieXxeXxωωω2 −=⇒=&&(1)(2)Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample (finding natural frequencies)Substituting into Eq. 1 and 2 gives:⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡−−+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧−−⎥⎥⎦⎤⎢⎢⎣⎡002200212212XXkkkkXXmmωωor[][](){} {} 0or 0022002212=+−⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎟⎟⎠⎞⎜⎜⎝⎛⎥⎥⎦⎤⎢⎢⎣⎡−−+⎥⎥⎦⎤⎢⎢⎣⎡− XKMXXkkkkmmωωThis can be written as⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎥⎥⎦⎤⎢⎢⎣⎡+−−−+−00222122XXkmkkkmωωThus, we have two homogeneous equations and two unknowns. For a non-trivial solution we require …Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample (finding natural frequencies)For a non-trivial solution we require the determinant to be zero [][](){} 0detor 022222=+−=+−−−+−KMkmkkkmωωωThis gives()()022222=−+−+− kkmkmωω()034224=+− kkmmωωCharacteristic equation – roots determine the character of the response. We can solve this using the quadratic equations ()()mkmkmkmkmkmkmkm3or 2234442222=±=−±−−=ωorRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample (natural modes)So our two natural frequencies are: mkmk 3 and 21==ωωWe need to determine our natural modes. To do this we substitute the natural frequencies into the equations shown below. ()02212=−+− kXXkmω()02221=+−+− XkmkXωwe substitute ω = ω1, we choose X1= 1 and then we solve for X2(use either equation) ()()11221212=+−=+−=kkmmkXkkmXω(We would get the same answer if we used the second equation) Note that these equations are not independent so we only need one of them.So, if X1= 1 then X2= 1 so our first eigenvector or “natural mode” is:{}⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=111XRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsExample – natural modes (cont.)Similarly if we let ω = ω2, we choose X1= 1 and then we solve for X2we get()()112321222−=+−=+−=kkmmkXkkmXωSo, if X1= 1 then X2= 01. Another way of writing this is:{}⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧−=112XPhysical interpretation:kmmkx1=1x2=1kkmmkx1=1x2=-1Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsMode shapesMode 1:Mode 2:Rose-Hulman Institute of TechnologyMechanical EngineeringVibrationsMaple worksheetRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsFinal solution (as shown in the book) ()()()( )()()()()222111222112112111111111φωφωωωωω+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧−++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=+⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧−++⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧tcosCtcosCtsinBtcosBtsinAtcosAtxtxWhere A1, A2, B1and B2or C1,C2, φ1, and φ2are found using the initial conditions. NOTE: Any general motion can be considered to be a superposition of its normal modes. Therefore, once you find the eigenvectors {X}1and {X}2you can always write()(){}( ){}( ){} ( )(){}()()22221111222121211121φωφωωωωω+++=+++=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧tcosCXtcosCXtsinBtcosBXtsinAtcosAXtxtxRose-Hulman Institute of TechnologyMechanical EngineeringVibrationsSummary1. Find equations of motion: 2. Assume simple harmonic motion to get3. Find natural frequencies4. Substitute in natural frequencies and find natural modes5. Write final time response[]{}[]{}{}0=+xKxM&&[][](){}{}02=+− XKMω[]{}[]{}02=+− KxMω[][](){}{}02=+−iiXKMω()(){}( ){}( ){} (