Beams Transverse Vibration

Solution of Euler Bernoulli equation

Longitudinal vibration of rod and torsional vibration of shafts were carried out using Newton's law and Hamilton's principle. In all these three cases the equation of motion of the system reduces to that of Wave equation, which can be given by



To find the response of the system one may use the variable separation method by using the following equation.

is known as the mode shape of the system and q(t) is known as the time modulation. Now equation (1) reduces to


Or



Since the left side of equation (4) is independent of time t and the right side is independent of x the equality holds for all values of t and x . Hence each side must be a constant. As the right side term equals to a constant implies that the acceleration   is proportional to displacement q(t), one may take the proportionality constant equal to – ω²  to have simple harmonic motion in the system. If one take a positive constant, the response will grow exponentially and make the system unstable. Hence one may write equation (4) as

The above equation can be written as



The solution of equation (6) and (9) can be given by


 
Hence,



Here constants C₁ and C₂ can be obtained from the initial conditions and constants A, B, C, D can be obtained from the boundary conditions. Let us now determine the mode shapes of simply supported beam, fixed-fixed beam, cantilever beam and free-free beams.

For beam simply supported at both the end the boundary conditions can be given by
 
Figure 1



Using equation 2 in 13(a-d) the boundary conditions reduces to



Now using the expression for the mode shape (equation 11) and its second order derivative can be given by the following expression.



From 14(a) B+D = 0

From 14(b) B-D = 0                                                            

Hence both B-D = 0

Now from 14(c) and 14(d)

From equation 17,

As shown in Figure 1It may be noted that the hyperbolic function   is not equal to zero. Hence A=0.

Now as out of the four constants three constants A , B and D are zero, the other remaining constant C should not be equal to zero. It may be noted that C = 0 correspond to the trivial solution i.e, u = 0 of the system. As we are studying the vibration of the system i.e, about the nontrivial solution of the system, from equation (18) one may obtain.




 


Now from equation 8 and 20 one may write the expression for the frequency as





and the mode shape can be given by



Hence from equation (22) and (23) it may be noted that the simply supported system has a large number of frequency and corresponding mode shapes. As E, I, ρ, L are the system parameters and are fixed for a particular system, the frequency of the nth mode is n 2 times the fundamental frequency. For example, the second mode frequency is 4 times the fundamental or first mode frequency and the third mode frequency is 9 times the fundamental frequency.

     
Cantilever beam

In case of cantilever beam the boundary conditions are

At left end i.e.,



At the free end i.e.,





Substituting these boundary conditions in the general solution





From (27 and 30) one may have

 

Hence one may solve the frequency equation   to obtain frequencies of different modes. For the first five modes the values of   are calculated as 1.875, 4.694, 7.855, 10.996, 14.137. Since for higher roots will be quite high, so  . Hence for more than 5th mode one may write

In this case the first 4 modes are shown in the following figures 4. The points with zero displacements i.e., the node points are marked by circles.
 

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