MEM 423: Mechanics of VibrationsContinuous systems (pde)Dr. A. YousuffDept. MEMDrexel Universityoverview Longitudinal vibration of a bar/rod (8.3) wave equation Torsional vibration of a shaft (8.4) wave equation Lateral vibration of beams 4thorder pde Orthogonality of mode shapesYousuff MEM 423: Vibrations 2Longitudinal Vibration of a bar/rod Consider an elastic bar:Yousuff MEM 423: Vibrations 3( , ) : f x t external force per unit length( , ) : u x t axial displacement: ( ) Pforce longitudinal: mass density PdP dxxc=c( ) : - A x cross sectional area: axial stress o: ' E Young s modulus:uaxial strainxccuP A EAxoc= =c&note the directions of P P dP +Governing pdeYousuff MEM 423: Vibrations 4Newtons law in x-direction:As before, this can be expressed as22( ) ( ) ( )u ux A x EA x fdxt x xc c c (= + (c c c Assuming a uniform bar2 22 2( , ) ( , )( , )u x t u x tEA fx t Ax tc c+ =c cWave equationYousuff MEM 423: Vibrations 5f(x,t)=0 free vibration equation:The solution, as before, is( )( , ) ( ) ( ) cos sin cos sinx xu x t U x T t A B C t D tc ce ee e| |= = + + |\ . U(x) represents the normal modeInitial conditions:00( , 0) ( )( , 0) ( )u x u xux u xt=c =c Boundary conditionsend conditionsboundary conditionsfrequency equationmode shapenatural frequenciesYousuff MEM 423: Vibrations 6Fixed-freeFree-freeFixed-fixed(0, ) 0( , ) 0u tul tx=c=c(0, ) 0( , ) 0utxul txc=cc=c(0, ) 0( , ) 0u tu l t ==cos 0lce =sin 0lce =sin 0lce =(2 1)( ) sin2n nn xUxlot +=( ) cosn nn xUxlot=( ) cosn nn xUxlot=(2 1);20,1, 2,...nn clnte+==;2,1, 2 0 ,...nn clnte ==;21, 2, 3,...nn clnte ==BC masses at endYousuff MEM 423: Vibrations 71m1k1c2m2k2c(0, ) u t( , ) u l tLuP EAxc=cRuP EAxc=c1 1 1(0, )(0, ) (0, ) (0, )u tEA k u t c u t mu txc= + +c 2 2 2( , )( , ) ( , ) ( , )u l tEA k u l t c u l t m u l txc= c Fixed-free barYousuff MEM 423: Vibrations 8boundary conditions:(0, ) 0( , ) 0u tul tt=c =c (2 1)0,andcos 0; 0,1,...2nl n cA B nc c le e te+= = = = 0(2 1) (2 1) (2 1)( , ) sin cos sin2 2 2n nnn x n ct n ctu x t C Dl l lt t t=+ + + (= + ( 00002 (2 1)( )sin24 (2 1)( )sin(2 1) 2lnlnn xC u x dxl ln xD u x dxn c lttt+=+=+}}1 2( , ) ( ) ( ) w x t w x ct w x ct = + +Wave-equationYousuff MEM 423: Vibrations 9The solution to wave equation can be expressed as0 0 01 1( , ) [ ( ) ( )] ( )2 2x ctx ctw x t wx ct wx ct w dco o+= + + +}Lateral Vibration of a beam Consider a flexible beam:Yousuff MEM 423: Vibrations 10( , ) : f x t external force per unit length( , ) : w x t lateral displacement( , ) : V x t shear force: mass density , V MdV dx dM dxx xc c= =c c ( ) : - A x cross sectional area( ) : I x moment of inertia: ' E Young s modulus( , ) : M x t bending moment22( , ) ( ) ( , )wM x t EI x x txc=cpde of the beamYousuff MEM 423: Vibrations 112 2 22 2 2( , ) ( , )( ) ( ) ( , )w x t w x tEI x A x f x tx x t (c c c+ = (c c c The force and moment (about O) equations yield:22( ( ) ) ( )0 ( ) ( ) ( , )2wA x dx V dV V fdxtdxM dM V dV dx fx t dx Mc= + + +c= + + + With V=cM/ cx, and ignoring (dx)2, 2 22 2( ) ( , )w MA x fx tt xc c= +c cUsing thin beam theoryEquation of motionYousuff MEM 423: Vibrations 12For a uniform beam:The equation for free vibration (f(x,t)=0) isInitial conditions:00( , 0) ( )( , 0) ( )w x wxwx wxt=c = c Solution to free-vibration eqns. Well use the separation of variables. w(x,t) = W(x)T(t) Sub. in pde, and solve for W and T.Yousuff MEM 423: Vibrations 132 4 224 21

c d W d TaW dx T dte = = =442 24422220;0d WWAdxc EId TTdt|e e|e = = =+ =Solution, contd.Yousuff MEM 423: Vibrations 14( ) cos sin T t A t B t e e = +1 2 3 4( ) cos sin cosh sinh W x C x C x C x C x | | | | = + + +Determine | from boundary conditions, and 2 4 EIAe |=Common boundary conditions:Fixedpinnedfree0 w= 0wxc=c0 w=220wEIxc=c220wEIxc=c220wEIx x| | c c= |c c\ .Yousuff MEM 423: Vibrations 15Orthogonality of mode shapesYousuff MEM 423: Vibrations 16Let Yi(x) be the ithmode shape, i=1,2, Then, the mode shapes satisfy the orthogonality condition:0( ) ( ) 0;li jY Y d i j o o o = =}0( ) ( ) .li i iY Y d depends on C o o o}Note: