Overview

Part 4 - Models for mechanical systems

Introduction

In this part we’ll see how we can model mechanical systems with controls.

Models for mechanical systems

In this series we’ll mainly deal with springs and dampers. Mainly translation motion.

For springs, we’ll of course use Hooke’s law:

Fs=ky.F_s = ky.

Where kk is the spring constant in [N/m][\mathrm{N/m}].

For dampers, we’ll use:

Fy=by˙.F_y = b \dot{y}.

Where bb is the damping constant in [Ns/m][\mathrm{Ns/m}].

Example 1 (Mass-spring-damper model)

Imagine we have a mass attached to both a spring and damper.

We have a downward force called FdF_d. Assume the system is rest at the start, using Newton’s second law:

F=my¨Fd−ky−by˙=my¨Fd=my¨+ky+by˙.\begin{align*} F & = m\ddot{y} \newline F_d - ky - b\dot{y} & = m\ddot{y} \newline F_d & = m\ddot{y} + ky + b\dot{y}. \end{align*}

Now we have a second-order differential equation that describes the system. Let’s Laplace transform it and find its transfer function from Fd→yF_d \to y.

Meaning GFdy(s)=Y(s)Fd(s)G_{F_dy}(s) = \dfrac{Y(s)}{F_d(s)}

Fd(s)=ms2Y(s)+kY(s)+bsY(s)=Y(s)(ms2+bs+k).\begin{align*} F_d(s) & = ms^2 Y(s) + kY(s) + bs Y(s) \newline & = Y(s) (ms^2 + bs + k). \end{align*}GFdy(s)=Y(s)Fd(s)=1ms2+bs+k.\boxed{G_{F_dy}(s) = \dfrac{Y(s)}{F_d(s)} = \dfrac{1}{ms^2 + bs + k}}.
Example 2 (Tank system)

Given a tank system, we can write using Bernoulli’s equation as:

P1+ρgh1+ρv122=P2+ρgh2+ρv222.P_1 + \rho g h_1 + \dfrac{\rho v_1^2}{2} = P_2 + \rho g h_2 + \dfrac{\rho v_2^2}{2}.

Let qoutq_{out} and qinq_{in} be volume flow. Let h2=0h_2 = 0, P1=P2=PP_1 = P_2 = P and that v1≈0v_1 \approx 0.

This yields:

ρgh1=ρv222  ⟹  v2=2gh1.\rho g h_1 = \dfrac{\rho v_2^2}{2} \implies v_2 = \sqrt{2gh_1}.

We define qoutq_{out} with qout=a⋅vq_{out}= a \cdot v, where aa is the area.

qout=a2gh1=a2g⋅h1,non-linear.q_{out} = a \sqrt{2gh_1} = a \sqrt{2g} \cdot \sqrt{h_1}, \quad \text{non-linear}.

Let V=A⋅h1V = A \cdot h_1. This means that V˙\dot{V} is the rate of change of the volume per second. V˙=qin−qout\dot{V} = q_{in} - q_{out}.

Which means:

Ah˙=qin−qouth˙=1A(qin−qout)=1A(qin−a2g⋅h1).\begin{align*} A\dot{h} & = q_{in} - q_{out} \newline \dot{h} & = \dfrac{1}{A}(q_{in} - q_{out}) \newline & = \dfrac{1}{A}(q_{in} - a \sqrt{2g} \cdot \sqrt{h_1}). \end{align*}

In the next part, we’ll see how we can linearize this function to obtain the so-called space-state representation of this system.

Search

Essays, lecture notes, and research

Type to search the site.