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

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Presentation on theme: "Hugh Hunt, Trinity College, Cambridgewww.hughhunt.co.uk mode 1 mode 2 mode 3 mode 4 EULER BENDING VIBRATION."— Presentation transcript:

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

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

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

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

5 Hugh Hunt, Trinity College, Cambridgewww.hughhunt.co.uk From aL the frequencies of free vibration are found using a j = 22.37, 61.67, 120.90, 199.86,... or a j The corresponding mode shapes are obtained by substituting  j 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

6 Hugh Hunt, Trinity College, Cambridgewww.hughhunt.co.uk 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


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