CHAPTER 3: Beam on springs.
Figure 3.1 Beam on springs. The static deflections will be omitted. We will consider all deflections relative to the equilibrium position. We have a choice over the two coordinates to use to describe the degrees-of-freedom. We could use the rotation of the beam and the vertical motion of the centre of gravity. However, we will use the vertical deflections at each end of the beam. Thus the two coordinates are x1 and x 2 . Consider the free-body diagram of the beam as shown in figure 3.2.
Figure 3.2 Free body diagram of beam. As we omit the static deflections then the mg term need not be included as the static deflection produces a resultant force that is equal and opposite to mg. Considering the deflections from the equilibrium position.
The natural frequencies and mode shapes for a range of k1 and k2 may be examined. 3.1.2 Transient motion After much maths equations (3.5) and (3.6) may be solved for a given set of initial conditions. A typical response is shown in figure 3.4.
Different initial conditions and parameter values give different transients. The effect on the motion may be examined by running the program.
3.2 Forced vibration Consider a force F(t) applied a distance b from the left hand end as shown in figure 3.5.
Figure 3.6 Free body diagram Again as we omit static deflections then the mg term need not be included. Considering the deflections from the equilibrium position as before, x = ( x1 + x 2 ) / 2 .................................................. (3.1) and for small angles, θ = (x 1 − x 2 ) / L .................................................. (3.2) Now to the kinematics.
The linear motion of the centre of mass gives, mx' ' = − k1x1 − k2x 2 + F(t) .......................................... (3.11) and the angular motion about the centre of mass gives 1 mL2θ' ' = −k1x1L / 2 + k2 x2L / 2 + F(t)( L / 2 − b) ....................... (3.12) 12 Substituting for x and q from equations (3.1) and (3.2). x1'' + x2' ' = 2(− k1x1 − k 2x2 + F(t)) / m ..................................... (3.13) x1'' −x 2' ' = 12(− k1x1 / 2 + k 2x2 / 2 + F(t)(0.5 − b / L )) / m ...................... (3.14) Adding (3.13) and (3.14) x1'' = (−4k1x1 + 2k2 x2 + F(t)(4 − 6b / L )) / m ................................. (3.15) Subtracting (3.13) and (3.14) x 2'' = (+2k1x1 − 4k2 x2 − F(t)(2 − 6b / L )) / m ................................ (3.16) Equations (3.15) and (3.16) can be solved in various ways. They have been presented in the form above as then a numerical method such as Runge Kutta may be used. 3.2.1 Sinusoidal excitation If the force Fsinωt then as found previously for other examples there will be a start up transient. The maths is significant. A typical start up transient is shown in figure 3.7.
Figure 3.7 Start up transient with sinusoidal force excitation.
The motion depends on the system parameters and may be investigated by running the program.
Some typical responses are shown in figure 3.8.
Figure 3.8 End responses for force excitation.
Note The natural frequencies are given when the denominator of equations (3.19) and (3.20) is zero (and resonance occurs). Thus when m 2ω 4 − 4mω 2 (k1 + k2 ) + 12k1k 2 = 0 This is identical to the natural frequency equation obtained previously - equation (3.10). The associated mode shapes can be found from equation (3.17) with F=0, X1 2k 2 = ................................................... (3.21) X 2 4k1 − mω 2 3.3 The effect of damping.
Figure 3.9 Beam supported on springs and viscous dampers.
Figure 3.10 Transient response of beam on damped supports.
Note that the dampers are omitted from the animation for reasons of clarity. Except for particular values of the damping coefficients it is not possible to separate the motion into modes that may be superimposed.
..... (3.31) The responses have amplitude and phase. A computer program was written to allow the amplitude of the response at each end of the beam to be calculated.
Typical responses are shown in figure 3.11.
Figure 3.11 End responses for force excitation.