Question 2.1: The system shown in Fig. 2.2(a) consists of a pivoted, rigid......

(a) Method  1:  using  the  principle  of  virtual  work
The system has only one degree of freedom, and its motion must be represented by only one coordinate. However, the choice of which coordinate to use is arbitrary.

Possible candidates are the angle of rotation about the pivot, or the vertical displacement of one of the masses. Let us choose to use the displacement, z , of the mass m_{2} , which is defined as positive in the downward direction. It follows that the velocity, \dot{z} , and the acceleration, \ddot{z} , will also be positive downwards at that point. The inertia, stiffness, damping and external forces acting are shown in Fig. 2.2(b). These forces are now assumed not to change when an infinitesimal additional downward virtual displacement, \delta z , is applied to m_{2}. Figure 2.2(c) shows the resulting virtual displacements at this and other points in the system. The virtual displacement at the left end of the beam, for example, is \frac{1}{2} \delta z , and upwards. At the point where the spring and damper are attached, the virtual displacement is \frac{1}{2} \delta z , and downwards.
The total work done by the forces in Fig. 2.2(b), acting on the corresponding virtual displacements in Fig. 2.2(c) (counting the work as positive when the force and virtual displacement are in the same direction, and negative when they are in opposite directions), is, by the principle of virtual work, equal to zero, so:

–  m_{1} \left(\frac{1}{2}\ddot{z} \right) \left(\frac{1}{2}\delta z \right)  –  m_{2} \ddot{z} \cdot \delta z-k_{1}\left(\frac{1}{2} z \right)\left(\frac{1}{2}\delta z \right)  –  c_{1} \left(\frac{1}{2}\dot{z} \right) \left(\frac{1}{2}\delta z \right) + \overline{F}\cdot \delta z = 0        (A)

Dividing Eq. (A) through by \delta z , and simplifying, gives the required equation of motion:

\left(\frac{1}{4}m_{1} + m_{2} \right) \ddot{z} + \left(\frac{1}{4}c_{1} \right) \dot{z} + \left(\frac{1}{4} k_{1} \right)z = \overline{F}        (B)

It is seen that the above process has ‘lumped’ the two masses m_{1} and m_{2} into a single effective   mass of \left(\frac{1}{4}m_{1} + m_{2} \right), considered to be located at the point where z and \overline{F} are defined, at the right end of the beam. Similarly, the stiffness and damping have changed because they have been referred to the new location.
(b) Method  2:  using  Lagrange’s  equations
If a single-DOF system has viscous damping, and also the kinetic energy depends only upon the velocities (i.e. not the displacements) of the masses, as in this case, Lagrange’s equation is given by Eq. (1.36),

\frac{d}{dt}\left(\frac{∂T}{∂\dot{q}_{i} } \right) – \frac{∂T}{∂ q_{i} } + \frac{∂U}{∂ q_{i} } + \frac{∂D}{∂\dot{q}_{i} } = Q_{i},       i=1,2,3,…,n.        (1.36)

Chapter 1, noting that ∂T/∂q = 0:

\frac{d}{dt}\left(\frac{∂T}{∂\dot{q} } \right) + \frac{∂U}{∂ q } + \frac{∂D}{∂\dot{q} } = Q         (C)

where q is the single generalized coordinate, taken as equal to z in this case; T the total kinetic energy in the system; U the total potential energy in the system; D a dissipation function, included to represent the energy lost in damping and Q the generalized external force. This last quantity, Q, must be defined so that when it acts upon q, the correct expression for the work done by the external force is obtained, so that in this case, since q = z, then Q = \overline{F}, since clearly \overline{F} z is the work done by the external force \overline{F}, and this must be equal to Qq.
The kinetic energy, T, the potential energy, U, and dissipation energy, D, of the system can now be written down. We know that the kinetic energy of a single mass particle, m, with velocity \dot{z} is equal to \frac{1}{2} m \dot{z}^{2}; the potential energy stored in a spring of stiffness k, with displacement z, is \frac{1}{2}kz^{2} , and the dissipation energy lost in the damper c with closure velocity \dot{z} is \frac{1}{2} c \dot{z}^{2} . Using these simple relationships, we have, by inspection of Fig. 2.2(a):

T= \frac{1}{2} m_{1} \left(\frac{\frac{1}{3}L }{\frac{2}{3}L } \dot{z} \right)^{2} + \frac{1}{2}m_{2}\dot{z}^{2} = \frac{1}{2}m_{1}\left(\frac{1}{2}\dot{z} \right)^{2} + \frac{1}{2} m_{2}\dot{z}^{2}           (D)

U = \frac{1}{2} k_{1} \left(\frac{\frac{1}{3}L }{\frac{2}{3}L } z \right)^{2} = \frac{1}{2} k_{1} \left(\frac{1}{2} z \right)^{2}            (E)

D = \frac{1}{2} c_{1} \left(\frac{\frac{1}{3}L }{\frac{2}{3}L } \dot{z} \right)^{2} = \frac{1}{2} c_{1}\left(\frac{1}{2}\dot{z} \right)^{2}          (F)

Applying Eq. (C), term by term:

\frac{d}{dt}\left(\frac{∂T}{∂\dot{q} } \right) = \frac{d}{dt}\left(\frac{∂T}{∂\dot{z} } \right) = \left(\frac{1}{4}m_{1} + m_{2} \right) \ddot{z}          (G)

\frac{∂U}{∂q} = \frac{∂U}{∂z} = \frac{1}{4} k_{1} z         (H)

\frac{∂D}{∂\dot{q} } = \frac{∂D}{∂\dot{z} } = \frac{1}{4} c_{1} z        (I)

Q = \overline{F}        (J)

Substituting Eqs (G), (H), (I), and (J) into Eq. (C) gives the equation of motion:

\left(\frac{1}{4}m_{1} + m_{2} \right) \ddot{z} + \left(\frac{1}{4}c_{1} \right) \dot{z} + \left(\frac{1}{4} k_{1} \right) z = \overline{F}          (K)

which is seen to agree with Eq. (B), derived by the principle of virtual work.