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Aero631 continuous Structures

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  • 1. Vibration of Continuous Structures Aero631 Dr. Eng. Mohammad Tawfik
  • 2. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 3. String and Cables Aero631 Dr. Eng. Mohammad Tawfik
  • 4. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 5. The string/cable equation • Start by considering a uniform string stretched between two fixed boundaries f(x,t) • Assume constant, axial tension  in string • Let a distributed force y  f(x,t) act along the x string Aero631 Dr. Eng. Mohammad Tawfik
  • 6. Examine a small element of string f (x,t) 2  2 w( x, t ) 1 2  Fy  x t 2  w(x,t) 1   1 sin 1   2 sin  2  f ( x, t )x x1 x2 = x1 +x • Force balance on an infinitesimal element • Now linearize the sine with the small angle approximate sinx=tanx=slope of the string Aero631 Dr. Eng. Mohammad Tawfik
  • 7.  w( x, t )   w( x, t )   2 w( x, t )       f ( x, t )x   x  x  x  2 x  x 1 t 2 Recall the Taylor series of w/x about x1 :  w   w    w         x    O(x 2 )    x  x2  x  x1 x  x  x1   w( x, t )   2 w( x, t )   x  f ( x, t )x   x x  x  x 1 t 2   w( x, t )   2 w( x, t )    f ( x, t )   x  x  t 2 Aero631 Dr. Eng. Mohammad Tawfik
  • 8. Since  is constant, and for no external force the equation of motion becomes:  2 w( x, t )  2 w( x, t )  , wave speed  , c c t 2 2 x 2  Second order in time and second order in space, therefore 4 constants of integration. Two from initial conditions: w( x,0)  w0 ( x), wt ( x,0)  w0 ( x) at t  0  And two from boundary conditions: w(0, t )  w(, t )  0, t 0 Aero631 Dr. Eng. Mohammad Tawfik
  • 9. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 10. Modes and Natural Frequencies Solve by the method of separation of variables: w( x, t )  X ( x)T (t )  Substitute into the equation of motion to 2get: d d2 c 2 X ( x)T (t )  X ( x)T(t ) where = 2 and = 2  dx dt X ( x) T(t )  d  X ( x)  X ( x)   2 ,  0   2 X ( x) c T (t ) dx  X ( x)    X ( x) Results in two second order equations T(t )   2   2 coupled only by a constant: c T (t ) Aero631 Dr. Eng. Mohammad Tawfik
  • 11. Solving the spatial equation: X ( x)   2 X ( x)  0  A second order equation with solution of the form: X ( x)  a1 sin  x  a2 cos  x, a1 and a2 are constants of integratio n Next apply the boundary conditions: X (0)T (t )  0, X ()T (t )  0  Since T(t) is not zero X (0)  0, X ()  0, characteri stic equation  X (0)  a2  0     n   sin   0   n   X ()  a1 sin   0  an infinite number of values of  Aero631 Dr. Eng. Mohammad Tawfik
  • 12. Also an eigenvalue problem The spatial solution becomes:  n  For n = 1,2,3, X n ( x)  an sin  x    Here the index n results because of the indexed value of  The spatial problem also can be written as: 2  2 ( X n )  n X n , X n ( x)  0, X n (0)  X n ()  0 x Which is also an eigenvalue, eigenfunction problem where  =2 is the eigenvalue and Xn is the eigenfunction. Aero631 Dr. Eng. Mohammad Tawfik
  • 13. Analogy to Matrix Eigenvalue Problem 2 A   2 , plus boundary conditions , matrix  operator x u i  X n ( x), eigenvecto r  eigenfunct ion X n ( x) also and eigenvecto r orthoganal ity also results and the condition of normalizin g plays the same role eigenvecto rs become mode shapes, eigenvalue s frequencie s modal expansion will also happen Aero631 Dr. Eng. Mohammad Tawfik
  • 14. The temporal solution T (t )  c 2 n Tn (t )  0, n  1,2,3  n 2 Again a second order ode with solution of the form: Tn (t )  An sin  n ct  Bn cos  n ct An , Bn are constants of integratio n (get from initial conditions ) Substitution back into the separated form X(x)T(t) yields: wn ( x, t )  cn sin  n ct sin  n x  d n cos  n ct sin  n x n n n n  cn sin( ct ) sin( x)  d n cos( ct ) sin( x)  The total solution becomes:     n n n n w( x, t )   cn sin( ct ) sin( x)  d n cos( ct ) sin( x) n 1     Aero631 Dr. Eng. Mohammad Tawfik
  • 15. Using orthogonality to evaluate the remaining constants from the initial conditions  n m  2 , n  m   sin(  x) sin(  x)dx  0, n  m  2  nm 0  From the initial conditions :  n w( x,0)  w0 ( x)   d n sin( x) cos(0)  n 1  m n m     w0 ( x) sin(  x)dx   d n  sin(  x) sin(  x)dx  0 n 1 0 Aero631 Dr. Eng. Mohammad Tawfik
  • 16. m  2 d m   w0 ( x) sin( x)dx, m  1,2,3 0  mn n  2 d n   w0 ( x) sin( x)dx, n  1,2,3 0   n w0 ( x)   cn n c sin(  x) cos(0) n 1  n  2 cn   w0 ( x) sin(  x)dx, n  1,2,3 nc 0  Aero631 Dr. Eng. Mohammad Tawfik
  • 17. A mode shape  w0 ( x)  sin x, which is the first eigenfunct ion (n = 1)  w0 ( x)  0,  cn  0, n  2   n d n   sin( x) sin( x)dx  0, n  2,3 0   d1  1    c  Causes vibration in the first mode w( x, t )  sin( x) cos t      shape Aero631 Dr. Eng. Mohammad Tawfik
  • 18. Plots of mode shapes n  sin x  2 1 0.5 X 1, x X 2, x 0 0.5 1 1.5 2 X 3, x 0.5 nodes 1 x Aero631 Dr. Eng. Mohammad Tawfik
  • 19. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 20. Bar Vibration Aero631 Dr. Eng. Mohammad Tawfik
  • 21. 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! Aero631 Dr. Eng. Mohammad Tawfik
  • 22. Vibration of Rods and Bars Equilibrium position x x +dx Infinitesimal dx • Consider a small element element of the bar F F+dF • Deflection is now along x (called longitudinal x vibration) • F= ma on small element w(x,t) yields the following: 0 l Aero631 Dr. Eng. Mohammad Tawfik
  • 23. Force balance:  2 w( x, t ) F  dF  F  A( x)dx Constitutive relation:  t2 w( x, t )   w( x, t )  F  EA( x)  dF   EA( x) dx  x  x  x    w( x, t )   2 w( x, t )  EA( x)   A( x)  x  x   t2 E  2 w( x, t )  2 w( x, t ) A( x)  constant     x 2  t2 At theclamped end : w(0, t )  0, w( x, t ) At thefree end : EA 0  x x  Aero631 Dr. Eng. Mohammad Tawfik
  • 24. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 25. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 26. Beam Vibration Aero631 Dr. Eng. Mohammad Tawfik
  • 27. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 28. Bending vibrations of a beam w (x,t) f (x,t) M(x,t)+Mx(x,t)dx f(x,t) h1 M(x,t) w(x,t) V(x,t)+Vx(x,t)dx x h2 V(x,t) ·Q dx A(x)= h1h2 bending stiffness  EI ( x) x x +dx Next sum forces in the y - direction (up, down) E  Youngs modulus Sum moments about the point Q I ( x)  cross - sect. area moment of Use the moment given from inertia about z stenght of materials  2 w( x, t ) Assume sides do not bend M ( x, t )  EI ( x) x 2 (no shear deformation) Aero631 Dr. Eng. Mohammad Tawfik
  • 29. Summing forces and moments  V ( x, t )   2 w( x, t )  V ( x, t )  dx   V ( x, t )  f ( x, t )dx  A( x)dx  x  t 2  M ( x, t )   V ( x, t )   M ( x, t )  dx   M ( x, t )  V ( x, t )  dx  dx  x   x  dx  f ( x, t )dx 0 2 0  M ( x, t )   V ( x, t ) f ( x, t )   dx  V ( x, t ) dx    (dx) 2  0  x   x 2   Aero631 Dr. Eng. Mohammad Tawfik
  • 30. M ( x, t )  V ( x, t )   x Substitute into force balance equation yields:  2 M ( x, t )  2 w( x, t )  dx  f ( x, t )dx  A( x)dx x 2 t 2 Dividing by dx and substituting for M yields  2 w( x, t )  2   2 w( x, t )  A( x)  2  EI ( x)   f ( x, t ) t 2 x  x 2  Assume constant stiffness to get:  2 w( x, t ) 2  4 w( x, t ) EI c  0, c  t 2 x 4 A Aero631 Dr. Eng. Mohammad Tawfik
  • 31. Boundary conditions (4) Free end Clamped (or fixed) end  2w deflection  w  0 bending moment  EI 0 x 2 w shear force     2w 0 slope  0  EI x  x 2  x Pinned (or simply supported)end Sliding end deflection  w  0 w slope  0  2w x bending moment  EI 0    2w  x 2 shear force   EI 2   0 x  x  Aero631 Dr. Eng. Mohammad Tawfik
  • 32. Solution of the time equation: X ( x) T(t )  c2   2  X ( x) T (t ) T(t )   2T (t )  0  T (t )  A sin t  B cos t Two initial conditions : w( x,0)  w0 ( x), wt ( x,0)  w0 ( x)  Aero631 Dr. Eng. Mohammad Tawfik
  • 33. Spatial equation (BVP)   2 X ( x)    X ( x)  0. c   A 2 2 Define  4     c EI Let X ( x)  Aex to get : X ( x)  a1 sin x  a2 cos x  a3 sinh x  a4 cosh x Apply boundary conditions to get 3 constants and the characteristic equation Aero631 Dr. Eng. Mohammad Tawfik
  • 34. Example: compute the mode shapes and natural frequencies for a clamped-pinned beam. At fixed end x  0 and X (0)  0  a2  a4  0 X (0)  0   (a1  a3 )  0 At the pinned end, x   and X ( )  0  a1 sin   a2 cos   a3 sinh   a4 cosh   0 EIX ()  0   2 (a1 sin   a2 cos   a3 sinh   a4 cosh  )  0 Aero631 Dr. Eng. Mohammad Tawfik
  • 35. The 4 boundary conditions in the 4 constants can be written as the matrix equation:  0 1 0 1   a1  0   0  0   a  0    2      sin  cos  sinh  cosh    a3  0        cos    a4  0  sin  sinh   cosh  2 2 2 2         B a Ba  0, a  0  det(B)  0  tan   tanh  The characteristic equation Aero631 Dr. Eng. Mohammad Tawfik
  • 36. Solve numerically to obtain solution to transcendental equation 1  3.926602  2  7.068583  3  10.210176  4  13.351768  5  16.493361  n 5 (4n  1)  n  4 Next solve Ba=0 for 3 of the constants: Aero631 Dr. Eng. Mohammad Tawfik
  • 37. Solving for the eigenfunctions: Ba  0 yields 3 constants in terms of the 4th : a1  a3 from the first equation a2  a4 from the second equation (sinh  n   sin  n )a3  (cosh  n   cos  n )a4  0 from the third (or fourth) equation Solving yields : cosh  n   cos  n  a3   a4 sinh  n   sin  n   cosh  n   cos  n    X n ( x )  ( a4 ) n  (sinh  n x  sin  n x)  cosh  n x  cos  n x   sinh  n   sin  n   Aero631 Dr. Eng. Mohammad Tawfik
  • 38. Mode shapes cosh n cos n . X n, x sinh n. x sin n. x cosh n. x cos n. x sinh n sin n 1.5 1 Mode 2 Mode 3 Note zero slope 0.5 X 3.926602, x 0 0.2 0.4 0.6 0.8 1 X 7.068583, x 0.5 Non zero slope X 10.210176, x 1 1.5 2 Mode 1 x Aero631 Dr. Eng. Mohammad Tawfik
  • 39. 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 Aero631 Dr. Eng. Mohammad Tawfik
  • 40. 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 Aero631 Dr. Eng. Mohammad Tawfik