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# 03 Vibration of string

Vibrations of strings and Cables

Vibrations of strings and Cables

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### 03 Vibration of string

1. 1. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Vibration of Continuous Structures
2. 2. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Course Contents  SDOF  M-DOF • Cables/String • Bars • Shafts • Vibration Attenuation • Beams • FEM for Vibration • Plates • Aeroelasticity
3. 3. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Derive the equation of motion for Cable/ String • Estimate the Natural Frequencies • Understand the concept of mode shapes • Apply BC’s and IC’s to obtain structure response
4. 4. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com String and Cables
5. 5. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Derive the equation of motion for Cable/String • Estimate the Natural Frequencies • Understand the concept of mode shapes • Apply BC’s and IC’s to obtain structure response
6. 6. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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
7. 7. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com The string/cable equation • Start by considering a uniform string stretched between two fixed boundaries • Assume constant, axial tension t in string • Let a distributed force f(x,t) act along the string f(x,t) t x y
8. 8. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Examine a small element of string xtxf t txw xFy   ),(sinsin ),( 2211 2 2 tt    • Where  is the mass per unit length of the cable • Force balance on an infinitesimal element • Now linearize the sine with the small angle approximate sin(x) = tan(x) = slope of the string 1 2 t2 t1 x1 x2 = x1 +x w(x,t) f (x,t)
9. 9. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com                   )( :about/ofseriesTaylortheRecall 2 1 112 xO x w x x x w x w xxw xxx   t     t   t t x t txw xtxf x txw x txw xx             2 2 ),( ),( ),(),( 12      t   t 2 2 ),( ),( ),( t txw txf x txw x      t         x t txw xtxfx x txw x x       2 2 ),( ),( ),( 1      t  
10. 10. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 0,0),(),0( 0at)()0,(),()0,( , ),(),( 00 2 2 22 2    ttwtw txwxwxwxw c x txw tc txw t    t     Since t 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
11. 11. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Physical quantities • Deflection is w(x,t) in the y-direction • The slope of the string is wx(x,t) • The restoring force is twxx(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 Subscript denotes differentiation w.r.t. to that parameter
12. 12. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Derive the equation of motion for Cable/String • Estimate the Natural Frequencies • Understand the concept of mode shapes • Apply BC’s and IC’s to obtain structure response
13. 13. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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:
14. 14. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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:
15. 15. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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:
16. 16. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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    
17. 17. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com             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      
18. 18. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Objectives • Derive the equation of motion for Cables/Strings • Estimate the Natural Frequencies • Understand the concept of mode shapes • Apply BC’s and IC’s to obtain structure response
19. 19. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com A mode shape             t c xtxw d ndxx n xd ncxw nxxw Assume 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
20. 20. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com 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
21. 21. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com String mode shapes Video 1 String mode shapes Video 2
22. 22. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Example :Piano wire: L=1.4 m, t=11.1x104 N, m=110 g. Compute the first natural frequency.   110 g per 1.4 m = 0.0786 kg/m 1  c l   1.4 t    1.4 11.1104 N 0.0786 kg/m  2666.69 rad/s or 424 Hz
23. 23. Dynamics of Continuous Structures Maged Mostafa #WikiCourses http://WikiCourses.WikiSpaces.com Assignment 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