Question 6.1: The equipment box shown in Fig. 6.2, to be fitted into an ai......

Since there are no external forces or damping, the equations of motion will be of the form:

\left[M\right] \left\{\ddot{\underline{z}} \right\} + \left[K\right] \left\{\underline{z} \right\} = 0               (A)

where \left\{\underline{z} \right\} = \begin{Bmatrix} z \\ \theta \end{Bmatrix}. It is now required to find the 2×2 matrices [M] and [K].

The mass matrix, [M], is given by Eq. (6.17):

[M] = [X_{m}]^{T} [\overline{m}] [X_{m}]             (B)

where

i.e. a diagonal matrix of the individual mass or inertia elements. Note that there are no cross terms between the mass of the empty box, m, and its moment of inertia, I, because the latter is defined about the mass center, G.

The matrix [X_{m}] is defined by Eq. (6.11), \left\{r\right\} = [X_{m}] \left\{\underline{z} \right\},

\left\{r\right\} = [X_{m}] \left\{z \right\}           (6.11)

and is the transformation matrix, for small angles, between the five displacements, r_{m}, r_{I}, r_{m_{1}}, r_{m_{2}}, r_{m_{3}} of the five ‘mass’ elements m, I, m_{1} ,m_{2}, m_{3} and the two global coordinates z and θ, as shown in Fig. 6.2. It is therefore of size 5×2, i.e. five rows and two columns. To derive the matrix [X_{m}], the first column consists of the displacements r_{m}, r_{I}, r_{m_{1}}, r_{m_{2}}, r_{m_{3}}, when the global coordinate z is increased from 0 to 1 unit, i.e. displaces one unit upwards.

The second column of [X_{m}] consists of the displacements r_{m}, r_{I}, r_{m_{1}}, r_{m_{2}}, r_{m_{3}} when the global coordinate θ is increased by one unit (one radian) clockwise.

The two columns of [X_{m}], and the complete matrix [X_{m}], are therefore as follows:

\begin{Bmatrix} r_{m} \\ r_{I} \\ r_{m_{1}} \\ r_{m_{2}} \\ r_{m_{3}} \end{Bmatrix}_{1} = \begin{Bmatrix} 1 \\ 0 \\ 1 \\ 1 \\ 1 \end{Bmatrix}             (D)

\begin{Bmatrix} r_{m} \\ r_{I} \\ r_{m_{1}} \\ r_{m_{2}} \\ r_{m_{3}} \end{Bmatrix}_{2} = \begin{Bmatrix} -f \\ 1 \\ a \\ b \\ -c \end{Bmatrix}             (E)

and

[X_{m}] = \left[\begin{matrix} 1 & -f \\ 0 & 1 \\ 1 & a \\ 1 & b \\ 1 & -c \end{matrix} \right]             (F)

The displacement r_{I} in the first column is zero, because it is defined as a rotation, and displacement has no effect.

The small angle approximations are justified here because the expected vibration displacements are all very small compared with the dimensions of the box. The assumed unit displacements of one meter and one radian (say) should not, of course, be taken literally.

The mass matrix, in global coordinates, is now given by Eq. (B):

which multiplies out to:

The stiffness matrix in global coordinates is given by Eq. (6.23):

[K] = [X_{s}]^{T} [\overline{k}] [X_{s}]                (I)

where [\overline{k}] is a diagonal matrix of spring elements, in this case \left[\begin{matrix} k_{1} & 0 \\ 0 & k_{2} \end{matrix} \right], and [X_{s}] a matrix of individual spring displacements, when unit displacements of the global coordinates z and θ are applied. Either compression or extension may be taken as positive, but the same convention must be used throughout. Taking the spring extensions s_{1}  \mathrm{and}  s_{2}, as shown in Fig. 6.2, as positive, Eq. (6.20), \left\{s\right\} = [X_{s}]\left\{\underline{z} \right\}, becomes:

\left\{s\right\} = [X_{s}]\left\{z \right\}            (6.20)

\begin{Bmatrix} s_{1} \\ s_{2} \end{Bmatrix} = \left[\begin{matrix} 1 & d \\ 1 & -e \end{matrix} \right]\begin{Bmatrix} z \\ \theta \end{Bmatrix}               (J)

Then from Eq. (I):

Now that we have the mass matrix [M] and the stiffness matrix [K] in global coordinates, the equations of motion, in the form of Eq. (A):

are simply: