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EULER BENDING VIBRATION. mode 4. mode 3. mode 2. mode 1. Free vibration of a beam. mass per unit length m flexural rigidity EI , length L. y. z. Equation of motion: For vibration, assume y ( x,t )= Y ( x )cos( w t ), so This has general solution

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  1. EULER BENDING VIBRATION mode 4 mode 3 mode 2 mode 1 Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

  2. Free vibration of a beam mass per unit length m flexural rigidity EI, length L y z Equation of motion: For vibration, assume y(x,t)=Y(x)cos(wt), so This has general solution Boundary condition for a fee end at z=0: Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

  3. so i.e.C=A and D=B Boundary condition for a free end at z=L: so and or, in matrix form, Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

  4. For a non-trivial solution, the determinant must be zero, so 1 0 Exact solutions for aL: 4.730 7.853 10.996 14.137 Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

  5. From aL the frequencies of free vibration are found using aj= 22.37, 61.67, 120.90, 199.86, ... or aj The corresponding mode shapes are obtained by substituting aj into the matrix equation to find the ratio between A and B so that The location of nodal points is then found by looking for where Y(z)=0 Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

  6. The location of the nodal points needs to be computed numerically, and the values are: Position of nodal points for a beam of L=1000mm (measured in mm from one end) mode 1: 224 776 mode 2: 132 500 868 mode 3: 94 356 644 906 mode 4: 73 277 500 723 927 mode 5: 60 226 409 591 774 940 mode 6: 51 192 346 500 654 808 949 mode 7: 44 166 300 433 567 700 834 956 mode 8: 39 147 265 382 500 618 735 853 961 Hugh Hunt, Trinity College, Cambridge www.hughhunt.co.uk

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