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# Vibration of Continuous Structures

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What is a continuous structure?
How to analyse the vibration of string, bars and shafts?
How to analyse the vibration of beams?

#WikiCourses
https://wikicourses.wikispaces.com/Topic+Vibration+of+Continuous+Structures
https://eau-esa.wikispaces.com/Vibration+of+structures

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### Vibration of Continuous Structures

1. 1. Aero631 Dr. Eng. Mohammad Tawfik Vibration of Continuous Structures
2. 2. Aero631 Dr. Eng. Mohammad Tawfik Objectives • Derive the equation of motion for simple structures • Understand the concept of mode shapes • Apply BC’s and IC’s to obtain structure response
3. 3. Aero631 Dr. Eng. Mohammad Tawfik String and Cables
4. 4. Aero631 Dr. Eng. Mohammad Tawfik Strings and Cables • This type of structures does not bare any bending or compression loads • It resists deformations only by inducing tension stress • Examples are the strings of musical instruments, cables of bridges, and elevator suspension cables
5. 5. Aero631 Dr. Eng. Mohammad Tawfik The string/cable equation • Start by considering a uniform string stretched between two fixed boundaries • Assume constant, axial tension  in string • Let a distributed force f(x,t) act along the string f(x,t)  x y
6. 6. Aero631 Dr. Eng. Mohammad Tawfik Examine a small element of string xtxf t txw xFy   ),(sinsin ),( 2211 2 2     • Force balance on an infinitesimal element • Now linearize the sine with the small angle approximate sinx=tanx=slope of the string 1 2 2 1 x1 x2 = x1 +x w(x,t) f (x,t)
7. 7. Aero631 Dr. Eng. Mohammad Tawfik                   )( :about/ofseriesTaylortheRecall 2 1 112 xO x w x x x w x w xxw xxx             x t txw xtxf x txw x txw xx             2 2 ),( ),( ),(),( 12          2 2 ),( ),( ),( t txw txf x txw x               x t txw xtxfx x txw x x       2 2 ),( ),( ),( 1        
8. 8. Aero631 Dr. Eng. Mohammad Tawfik 0,0),(),0( 0at)()0,(),()0,( , ),(),( 00 2 2 22 2    ttwtw txwxwxwxw c x txw tc txw t         Since  is constant, and for no external force the equation of motion becomes: Second order in time and second order in space, therefore 4 constants of integration. Two from initial conditions: And two from boundary conditions: , wave speed
9. 9. Aero631 Dr. Eng. Mohammad Tawfik Physical quantities • Deflection is w(x,t) in the y-direction • The slope of the string is wx(x,t) • The restoring force is wxx(x,t) • The velocity is wt(x,t) • The acceleration is wtt(x,t) at any point x along the string at time t Note that the above applies to cables as well as strings
10. 10. Aero631 Dr. Eng. Mohammad Tawfik Modes and Natural Frequencies 2 2 2 2 2 2 2 2 2 )( )( )( )( 0 )( )( , )( )( )( )( =and=where)()()()( )()(),(                  tTc tT xX xX xX xX dx d tTc tT xX xX dt d dx d tTxXtTxXc tTxXtxw    Solve by the method of separation of variables: Substitute into the equation of motion to get: Results in two second order equations coupled only by a constant:
11. 11. Aero631 Dr. Eng. Mohammad Tawfik Solving the spatial equation:            n aX aX XX tTXtTX aaxaxaxX xXxX n           equationsticcharacteri 1 2 2121 2 0sin 0sin)( 0)0( ,0)(,0)0( 0)()(,0)()0( nintegratioofconstantsareand,cossin)( 0)()( Since T(t) is not zero an infinite number of values of  A second order equation with solution of the form: Next apply the boundary conditions:
12. 12. Aero631 Dr. Eng. Mohammad Tawfik Also an eigenvalue problem 0)()0(,0)(,)( ofvalueindexedtheofbecauseresultsindextheHere sin)(,1,2,3=For 2 2            nnnnnn nn XXxXXX x n x n axXn      The spatial solution becomes: The spatial problem also can be written as: Which is also an eigenvalue, eigenfunction problem where  =2 is the eigenvalue and Xn is the eigenfunction.
13. 13. Aero631 Dr. Eng. Mohammad Tawfik Analogy to Matrix Eigenvalue Problem happenalsowillexpansionmodal sfrequencieseigenvalueshapes,modebecomerseigenvecto rolesametheplays gnormalizinofconditiontheandresultsalsoityorthoganal reigenvectoandalso)( ioneigenfunctreigenvecto),( operatormatrix,conditionsboundaryplus,2 2 xX xX x A n ni       u
14. 14. Aero631 Dr. Eng. Mohammad Tawfik The temporal solution         1 22 )sin()cos()sin()sin(),( )sin()cos()sin()sin( sincossinsin),( )conditionsinitialfrom(getnintegratioofconstantsare, cossin)( 3,2,1,0)()( n nn nn nnnnnnn nn nnnnn nnn x n ct n dx n ct n ctxw x n ct n dx n ct n c xctdxctctxw BA ctBctAtT ntTctT         Again a second order ode with solution of the form: Substitution back into the separated form X(x)T(t) yields: The total solution becomes:
15. 15. Aero631 Dr. Eng. Mohammad Tawfik Using orthogonality to evaluate the remaining constants from the initial conditions                       010 0 1 0 2 0 )sin()sin()sin()( )0cos()sin()()0,( :conditionsinitialtheFrom 2,0 , )sin()sin( dxx m x n ddxx m xw x n dxwxw mn mn dxx m x n n n n n nm    
16. 16. Aero631 Dr. Eng. Mohammad Tawfik             3,2,1,)sin()( 2 )0cos()sin(c)( 3,2,1,)sin()( 2 3,2,1,)sin()( 2 0 0 1 0 0 0 0 0            ndxx n xw cn c x n cxw ndxx n xwd nm mdxx m xwd n n nn n m      
17. 17. Aero631 Dr. Eng. Mohammad Tawfik A mode shape             t c xtxw d ndxx n xd ncxw nxxw n n          cos)sin(),( 1 3,2,0)sin()sin( 2 ,0,0)( 1)=(ioneigenfunctfirsttheiswhich,sin)( 1 0 0 0 Causes vibration in the first mode shape
18. 18. Aero631 Dr. Eng. Mohammad Tawfik Plots of mode shapes 0 0.5 1 1.5 2 1 0.5 0.5 1 X ,1 x X ,2 x X ,3 x x sin n 2 x     nodes
19. 19. Aero631 Dr. Eng. Mohammad Tawfik Homework #3 1. Solve the cable problem with one side fixed and the other supported by a flexible support with stiffness k N/m 2. Solve the cable problem for a cable that is hanging from one end and the tension is changing due to the weight  N/m
20. 20. Aero631 Dr. Eng. Mohammad Tawfik Bar Vibration
21. 21. Aero631 Dr. Eng. Mohammad Tawfik Bar Vibration • The bar is a structural element that bears compression and tension loads • It deflects in the axial direction only • Examples of bars may be the columns of buildings, car shock absorbers, legs of chairs and tables, and human legs!
22. 22. Aero631 Dr. Eng. Mohammad Tawfik Vibration of Rods and Bars • Consider a small element of the bar • Deflection is now along x (called longitudinal vibration) • F= ma on small element yields the following: x x +dx w(x,t) x dx F+dFF Equilibrium position Infinitesimal element 0 l
23. 23. Aero631 Dr. Eng. Mohammad Tawfik 0 ),( :endfreeAt the ,0),0(:endclampedAt the ),(),( constant)( ),( )( ),( )( ),( )( ),( )( ),( )( 2 2 2 2 2 2 2 2                  xx txw EA tw t txw x txwE xA t txw xA x txw xEA x dx x txw xEA x dF x txw xEAF t txw dxxAFdFF                        Force balance: Constitutive relation:
24. 24. Aero631 Dr. Eng. Mohammad Tawfik Note • The equation of motion of the bar is similar to that of the cable/string  the response should have similar form • The bar may have different boundary conditions
25. 25. Aero631 Dr. Eng. Mohammad Tawfik Homework #4 • Solve the equation of motion of a bar with constant cross-section properties with 1. Fixed-Fixed boundary conditions 2. Free-Free boundary conditions • Compare the natural frequencies for all three cases
26. 26. Aero631 Dr. Eng. Mohammad Tawfik Beam Vibration
27. 27. Aero631 Dr. Eng. Mohammad Tawfik Beam Vibration • The beam element is the most famous structural element as it presents a lot of realistic structural elements • It bears loads normal to its longitudinal axis • It resists deformations by inducing bending stresses
28. 28. Aero631 Dr. Eng. Mohammad Tawfik Bending vibrations of a beam 2 2 ),( )(),( aboutinertia ofmomentareasect.-cross)( modulusYoungs )(stiffnessbending x txw xEItxM z xI E xEI       Next sum forces in the y- direction (up, down) Sum moments about the point Q Use the moment given from stenght of materials Assume sides do not bend (no shear deformation) f (x,t) w (x,t) x dx A(x)= h1h2 h1 h2 M(x,t)+Mx(x,t)dx M(x,t) V(x,t) V(x,t)+Vx(x,t)dx f(x,t) w(x,t) x x +dx ·Q
29. 29. Aero631 Dr. Eng. Mohammad Tawfik Summing forces and moments 0)( 2 ),(),( ),( ),( 0 2 ),( ),( ),(),( ),( ),( ),( )(),(),( ),( ),( 2 2 2                                   dx txf x txV dxtxVdx x txM dx dxtxf dxdx x txV txVtxMdx x txM txM t txw dxxAdxtxftxVdx x txV txV              0
30. 30. Aero631 Dr. Eng. Mohammad Tawfik A EI c x txw c t txw txf x txw xEI xt txw xA t txw dxxAdxtxfdx x txM x txM txV                             ,0 ),(),( ),( ),( )( ),( )( ),( )(),( ),( ),( ),( 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 Substitute into force balance equation yields: Dividing by dx and substituting for M yields Assume constant stiffness to get:
31. 31. Aero631 Dr. Eng. Mohammad Tawfik Boundary conditions (4) 0forceshear 0momentbending endFree 2 2 2 2         x w EI x x w EI       0slope 0deflection endfixed)(orClamped   x w w   0momentbending 0deflection endsupported)simply(orPinned 2 2   x w EI w   0forceshear 0slope endSliding 2 2         x w EI x x w      
32. 32. Aero631 Dr. Eng. Mohammad Tawfik Solution of the time equation: )()0,(),()0,( :conditionsinitialTwo cossin)( 0)()( )( )( )( )( 00 2 22 xwxwxwxw tBtAtT tTtT tT tT xX xX c t           
33. 33. Aero631 Dr. Eng. Mohammad Tawfik Spatial equation (BVP) xaxaxaxaxX AexX EI A c xX c xX x      coshsinhcossin)( :getto)(Let Define .0)()( 4321 22 4 2                 Apply boundary conditions to get 3 constants and the characteristic equation
34. 34. Aero631 Dr. Eng. Mohammad Tawfik Example: compute the mode shapes and natural frequencies for a clamped-pinned beam. 0)coshsinhcossin( 0)( 0coshsinhcossin 0)( andend,pinnedAt the 0)(0)0( 00)0( and0endfixedAt 4321 2 4321 31 42                 aaaa XEI aaaa X x aaX aaX x
35. 35. Aero631 Dr. Eng. Mohammad Tawfik            tanhtan 0)det(, 0 0 0 0 coshsinhcossin coshsinhcossin 00 1010 4 3 2 1 2222                                         BB a a a a B 0a0a a The 4 boundary conditions in the 4 constants can be written as the matrix equation: The characteristic equation
36. 36. Aero631 Dr. Eng. Mohammad Tawfik Solve numerically to obtain solution to transcendental equation 4 )14( 5 493361.16351768.13 210176.10068583.7926602.3 54 321          n n n   Next solve Ba=0 for 3 of the constants:
37. 37. Aero631 Dr. Eng. Mohammad Tawfik Solving for the eigenfunctions:                  xxxxaxX aa aa aa aa B nnnn nn nn nn nn nn nnnn             coscosh)sin(sinh sinsinh coscosh )()( sinsinh coscosh :yieldsSolving equationfourth)(orthirdthefrom 0)cos(cosh)sinsinh( equationsecondthefrom equationfirstthefrom :4ththeofin termsconstants3yields 4 43 43 42 31 0a
38. 38. Aero631 Dr. Eng. Mohammad Tawfik Mode shapes X ,n x .cosh n cos n sinh n sin n sinh .n x sin .n x cosh .n x cos .n x 0 0.2 0.4 0.6 0.8 1 2 1.5 1 0.5 0.5 1 1.5 X ,3.926602 x X ,7.068583 x X ,10.210176 x x Mode 1 Mode 2 Mode 3 Note zero slope Non zero slope
39. 39. Aero631 Dr. Eng. Mohammad Tawfik Summary of the Euler-Bernoulli Beam • Uniform along its span and slender • Linear, homogenous, isotropic elastic material without axial loads • Plane sections remain plane • Plane of symmetry is plane of vibration so that rotation & translation decoupled • Rotary inertia and shear deformation neglected
40. 40. Aero631 Dr. Eng. Mohammad Tawfik Homework #5 • Get an expression for  for the cases of a uniform Euler-Bernoulli beam with BC’s as follows: – Clamped-Clamped – Pinned-Pinned – Clamped-Free – Free-Free