Question

Describe and quantify what you can do to improve ride comfort in a multi-degree-of-freedom model. make an example

Answer 1

In the vibration analysis of mechanical systems with multiple degrees of freedom (MDOF), instead of one vibrating mass, we replace our real structure with two or more oscillating masses. If the real system has well-defined separate moving parts, we can consider it as a lumped interconnected parameter system. The degrees of freedom of a lumpedparameter system are equal to the number of vibrating mass points and this is also true for the number of resonant frequencies. A mechanical system or structure which does not have well-defined separately oscillating parts but consists of a continuously spread mass and geometry is a distributed system. Continuum or distributed parameter systems have an infinite amount of resonant frequencies and corresponding vibration shapes. It is, however, possible to discretize the system into large amounts of lumped interconnected parameters and approximate its behavior with methods commonly used for lumped parameter systems. This idea is in fact used in FEM software to extract the vibration dynamics of distributed mechanical systems defined with complex three-dimensional geometry.

Let us choose a very simple example, which has more than one degree of freedom and therefore may be considered as a MDOF system. The system of connected

Fig. 2.5

0 q1t,fe1t 0 q2t fe2t
k1 k2 k3

Mass: m1 Mass: m2
b1 b2 b3

Multiple degrees of freedom system: connected set of two masses

moving masses illustrated in Fig. 2.5 is sliding on a frictionless surface. Now instead of one coordinate defining the movement, we have two for each moving mass: q1(t)and q2(t). The two moving masses m1 and m2 are connected to each other and the surrounding fixed wall through a series of springs and dampers with the spring and damping coefficients k1, k2, k3 and b1, b2, b3. There are external force inputs associated with individual masses denoted by fe1 and fe2 .

Using a simple mechanical analysis, we may create a free body diagram for each mass and analyze the forces acting on them. After assembling the equations of motion, we obtain the following set of equations for our two masses [4, 10, 18]:

mq?+(b +b)q??bq?+(k +k)q ?kq =f (2.40)1112122121221

mq??bq?+(b +b)q??kq +(k +k)q =f (2.41)2221232212322

It is possible to rewrite the one equation for motion per moving mass into a compact set, using matrix notation [10, 37, 52]:

m0q? b+b?b q?11+1221

0mq? ?bb+bq?22 2232

+ k1+k2 ?k2 q1 = fe1?k2 k2 + k3 q2 fe2

Note the similarity between this equation of motion and the SDOF forced equation motion in (2.34). We have a matrix containing the masses, which is multiplied by a vector of accelerations. Similarly, we have matrices containing damping elements and spring constants. We can in fact use a matrix notation to create from this [21]:

Mq? + Bdq? + Ksq = fe (2.43)

where matrix M is the mass matrix,3 Bd is the structural damping matrix and Ks is the stiffness matrix. Vector q contains the displacement coordinates for each degree of freedom. For an N degree of freedom system the constant matrices M, Bd and Kswill all have N × N elements.

The solution of such systems is in fact very similar to the solution of SDOF systems. To illustrate this, let us consider a case without damping and with no outside force. Removing these effects from the equation of motion in matrix from (2.43), we get [38]:

Mq? + Ksq = 0 (2.44)

for which we have to find a solution. Similar to the SDOF systems, our solution can be expected in a form of a set of harmonic functions, which mathematically is simply an amplitude multiplied by a complex exponential:

q = q?ej?nt = q?e?t (2.45)

As introduced previously, the term ej?nt or analogously e?t is just a mathematical trick to solve differential equations by using the so-called phasor form for the solution. If we take the real part of Euler’s formula, we essentially expect a cosine function. Let us substitute this solution to our matrix equation of motion, and differentiate to get:

??n2M+Ks q?ej?nt =0