5Base excitation and vibration isolation

Often the force does not act on the mass directly; instead the support moves. Think of a car on a bumpy road or an instrument on a vibrating floor. If the base moves as $y(t)$, the spring and damper respond to the relative motion:

Read the right-hand side as a force: moving the base pushes on the mass exactly as an applied force $c\dot y+ky$ would. If the base is connected only through springs, the equivalent force is just a spring constant times the base motion. For example, with a spring $k_1$ to the moving base and a spring $k_2$ to a fixed wall, $m\ddot x+(k_1+k_2)x=k_1y$: the base acts like a force of amplitude $k_1Y$, and everything from Section 4 applies with that $F_0$.

For $y=Y\sin\omega t$, the steady-state amplitude ratio is the displacement transmissibility:

The same expression gives the force transmissibility: the fraction of a harmonic force on a machine that the mounts pass to the floor. Every curve crosses $TR=1$ at $r=\sqrt2$. Isolation ($TR<1$) only happens for $r>\sqrt2$, which means soft mounts: a natural frequency well below the operating frequency.

Force transmissibility. Turn the problem around: a machine with an internal harmonic force of fixed amplitude, $F_0\sin\omega t$, shakes on its mounts. The mounts pass a force $kx+c\dot x$ to the floor, and its amplitude relative to the shaking force is the same $TR$:

5bRotating imbalance

Many machines (motors, fans, pumps, a washing machine on spin) excite themselves. If the center of mass G of a spinning rotor sits a small distance $e$ (the eccentricity) from the spin axis O, then at constant speed $\omega$ the point G travels in a circle and accelerates toward the axis with magnitude $e\omega^2$. The shaft must pull the rotor inward with force $me\omega^2$, so the rotor pulls on its bearing with $me\omega^2$ outward, along the line from O to G, rotating with the shaft.

Only $m$ and $e$ matter, not the rotor's shape, so the rotor can be modeled as a point mass $m$ at radius $e$. Now mount it in a frame of mass $M$ that slides on a spring $k$ and dashpot $c$, and measure the frame's position $x$. The point mass is at $x+e\cos\omega t$, so its acceleration is $\ddot x-e\omega^2\cos\omega t$. Newton's law for the frame and rotor together gives

• It is a harmonic force at the shaft speed, but its amplitude $me\omega^2$ grows with the square of the speed.
• The natural frequency uses the total mass: $\omega_n=\sqrt{k/(M+m)}$, and $\zeta=c/2\sqrt{k(M+m)}$.
• The steady amplitude is $X=\dfrac{me\omega^2/k}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}$, or in dimensionless form $\dfrac{X(M+m)}{me}=\dfrac{r^2}{\sqrt{(1-r^2)^2+(2\zeta r)^2}}$.

This curve behaves differently from the fixed-amplitude case: it starts at zero (a slowly turning rotor barely shakes anything), peaks near resonance, and at high speed levels off at 1, i.e. $X\to me/(M+m)$. At high speed the frame and rotor simply circle around their combined center of mass, which stays still.