Question 4.7.5: A centrifugal pump is supported on a rigid foundation of mas......

a. Given parameters
Given that the speed of the pump is 1200 rpm, the applied frequency ω =2 π(1200)/60 rad/s = 125.66 rad/s.
The applied force due to unbalance,

f_{1} =  meω²sinωt
= 0.25 (0.1525) (125.66)² sin 125.66 tN
= 602 sin 125.66 tN

The stiffness of the isolator springs, k_{1}  = 230 MN/m =   2.3  ×  10^{8} N/m.
The stiffness of the soil, k_{2}  =  115   MN/m   =   1.15  ×  10^{8} N/m.
The mass of centrifugal pump, m_{1} = 364 kg, and the mass of the foundation, m_{2} = 909 kg.
The damping coefficient of soil, c_{2}  =  0.25   MNs/m  = 0.25  ×  10^{6}  Ns/m.

b. Equations of motion

Let x_{1}  and   x_{2} be the absolute equilibrium vertical displacements of the centrifugal pump (not including the unbalance mass m) and foundation, respectively.
With reference to Figure 4E5c, the vertical displacement of m is x_{1} + e sin ωt. With reference to this equation and the FBD in Figure 4E5d, by applying Newton’s second law of motion for the centrifugal pump of mass m_{1} and the unbalance mass m, one has

Re-arranging terms, it becomes

( m_{1}  +  m )  \ddot{ x_{1}}    +  k_{1} x_{1}  −  k_{1} x_{2}   =  f_{1}   =  F_{1} sinωt,    F_{1}  = meω² .

As m is much smaller than m_{1} , one can simply write

Similarly, applying Newton’s second law of motion for the foundation m_{2} , the equation of motion is given by

After re-arranging terms, it leads to

For subsequent analysis, one can express the two equations of motion in matrix form.
Thus, the above two equations of motion can be written as

\begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix}  \begin{pmatrix} \ddot{ x_{1}}\\ \ddot{ x_{2}} \end{pmatrix}   +    \begin{pmatrix}0 & 0 \\ 0 & c_{2} \end{pmatrix}    \begin{pmatrix} \dot{ x_{1}}\\ \dot{ x_{2}} \end{pmatrix}    +    \begin{bmatrix} k_{1}  & –  k_{1}  \\ –  k_{1}  & k_{22} \end{bmatrix}    \begin{pmatrix}  x_{1}\\ x_{2} \end{pmatrix}   =    \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}  sinωt                    (i)
in which k_{22}   =   k_{1}   +   k_{2} .

c. Solution for responses
For convenience, one can write sin ωt as Im  \left\{  e^{iωt} \right\}    such   that   x_{i}   =   X_{i} sin ωt, i = 1, 2, can be written as

   x_{i}   =  Im  \left\{  X_{i}   e^{iωt} \right\}  ,

where Im{.} denotes the imaginary part of the enclosing complex variable.
Without loss of generality, one can disregard Im{.} for the time being so that the displacements can be written as    x_{i}   =   X_{i}   e^{iωt}  . Then, the velocity and acceleration are given by   \dot{x_{i}}  =  iωX_{i} e^{iωt} ,   \ddot{x_{i}}  =  − ω²X_{i} e^{iωt} .

Substituting the above expressions into Equation (i) and canceling e^{iωt}  on both sides of the resulting matrix equation of motion, one has

\begin{bmatrix} − ω²m_{1}  & 0 \\ 0 &  − ω² m_{2} \end{bmatrix}    \begin{pmatrix} X_{1}\\ X_{2} \end{pmatrix}  +    \begin{pmatrix} 0 & 0 \\ 0 & iωc_{2} \end{pmatrix}      \begin{pmatrix} X_{1}\\ X_{2} \end{pmatrix}   +  \begin{bmatrix} k_{1}  & –  k_{1}  \\ –  k_{1}  & k_{22} \end{bmatrix}  \begin{pmatrix} X_{1}\\ X_{2} \end{pmatrix}   =    \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}  .

Upon simplification, one arrives at

\begin{bmatrix} k_{1}  − ω²m_{1}  & –  k_{1} \\ –  k_{1} &  k_{1}  +  k_{2}  − ω² m_{2}  +  iωc_{2} \end{bmatrix}  \begin{pmatrix} X_{1}\\ X_{2} \end{pmatrix}   =    \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}  .      (ii)

This is identical in form to Equation (4.18). Thus, one has

\begin{bmatrix}( k_{11}  − m_{1}ω² ) &   k_{12} \\  k_{21} &  ( k_{22}  − m_{2}ω² ) \end{bmatrix}  \begin{pmatrix}  X_{1}\\ X_{2} \end{pmatrix}   =    \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}      (4.18)

[Z (ω)] =  \begin{bmatrix} k_{1}  − ω²m_{1}  &   – k_{1} \\ –  k_{1} & k_{1}  +  k_{2}  − ω² m_{2}  +  iωc_{2}  \end{bmatrix}  .                              (iii)

The form of the impedance or coefficient matrix [Z(ω)] is identical to that in Equation (4.18) for a 2-dof system. That is, the elements of the coefficient matrix are

z_{11}  =  k_{1}  −  ω²  m_{1},       z_{12}  =  z_{21}   =  −  k_{1},           z_{22}  =  k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2} .

From Equation (4.19), the amplitudes of the displacements are

\begin{pmatrix}  X_{1}\\ X_{2} \end{pmatrix}   =  [Z(ω)]^{-1} \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}   =   \frac{adj[Z(ω)]   \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}  }{\left|Z(ω)\right| } or                    (4.19a)
\begin{pmatrix}  X_{1}\\ X_{2} \end{pmatrix}   =  \frac{\begin{bmatrix}( k_{22}  − m_{2}ω² ) &   – k_{12} \\  – k_{21} &  ( k_{11}  − m_{1}ω² ) \end{bmatrix} \begin{pmatrix}  F_{1}\\ 0 \end{pmatrix}  }{\left|Z(ω)\right| } .

\left|Z(ω)\right|  =   m_{1}m_{2} (ω_{1}²  −   ω²)  (ω_{2}²  − ω²)                                             (4.19b)

therefore

X_{1}   =   \frac{ k_{1} F_{1}}{(k_{1}  −  ω²  m_{1})  (k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2})  –   k_{1}²}  ,                                   (iv)

and

X_{2}  =   \frac{-  z_{12}  F_{1}}{\left|Z(ω)\right|}  =  \frac{ (k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2})  F_{1}  }{(k_{1}  −  ω²  m_{1})  (k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2})  –   k_{1}²}   .                              (v)

Now, substituting the given data into the numerator and denominator terms in Equation (iv), one has

k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2}
=[(2.3 + 1.15)   × 10^{8} − 125.66² (909) + i (125.66  × 0.25 ) × 10^{6} ] 602
= 19.904919  × 10^{10} +i (1.891183) × 10^{10} N/m
( k_{1}   +  k_{2} −  ω²m_{2}  +  iωc_{2}) (k_{1}  −  ω²  m_{1})  –   k_{1}²
= 7.4148238  ×  10^{16} − 5.29  ×  10^{16} +  i (0.7044886) ×  10^{16}   (N/m )²
= (2.1248238  +  i 0.7044886) × 10^{16}   (N/m )² .

Therefore,

Rationalizing, one has

=  (8.6870 – i 1.9942)  × 10^{- 6} m .

From the theory of complex variables,

where

Therefore,

Thus,

  x_{1}   =  Im  \left\{  X_{1}  e^{iωt}\right\}   =  Im  \left\{  X_{10}   e^{i(ωt  +  φ_{1} )}\right\}  =  X_{10}  sin (ωt  +  φ_{1} ) .   
  x_{1}   =  8.9129   ×  10^{- 6}    sin (ωt −  12.9°)  m  .    (vi)

Similarly, applying Equation (v), one obtains

Rationalizing, one has

= (5.8660  −   i 1.9449 )  ×  10^{−6}  m .

Again, from the theory of complex variables,

where

Therefore,

X_{20}  =  6.18  ×  10^{- 6}   m,     φ_{1} = −18.3˚ .

Thus,

  x_{2}   =  Im  \left\{  X_{2}  e^{iωt}\right\}   =  Im  \left\{  X_{20}   e^{i(ωt  +  φ_{2} )}\right\}  =  X_{20}  sin (ωt  +  φ_{2} ) .   
  x_{2}   =  6.1800   ×  10^{- 6}    sin (ωt − 18.3°)  m  .    (vii)