2.0 Mechanical Systems
Mechanical systems are composed of three types of elements: inertial, stiffness and damping. A
mechanical system may have translational elements, rotational elements or some combination of
both. In this section we review the procedure for finding equations of motion of mechanical
systems.
Inertial Elements
For a translational system, an inertial element is simply a “lumped mass”. A lumped mass is
assumed to have all of its mass concentrated at one point (its center of mass) and to be rigid. The
equation of motion for the lumped mass is Newton’s law
(1)
𝑓(𝑡) = 𝑚𝑥̈
m
f
J
T
x
We will also encounter rotational inertial elements whose equation of motion is given by
𝜏(𝑡) = 𝐽𝜃̈
(2)
where J is the polar moment of inertia.
Stiffness Elements
The most common stiffness element is the linear spring, shown below
f
f
x2
x1
The force in the spring is proportional to the deflection of the spring such that
𝑓𝑘 = 𝑘∆
(3)
where k is the “spring constant” and Δ is the deflection. For the spring shown above the deflection
is given by
1
(4)
∆= 𝑥1 − 𝑥2
How do we know that Δ is x1 – x2, and not x2 – x1? Here is a simple rule of thumb to help you
remember. A positive deflection should produce a positive (tensile) force. Physically, if x1 is greater
than x2 the spring is stretched, and under tension. The quantity x1 – x2 is also positive. This rule
will be extremely useful when analyzing more complicated systems with several springs.
There are also rotational (torsional) springs that are governed by the following equation
(5)
𝜏 = 𝑘𝑡 (𝜃1 − 𝜃2 )
Damping Elements
Damping in mechanical systems is most often created by friction between elements, or between an
element and “ground”. The most common example of a damping element is the resistance to
motion caused by friction. Another damping element is a shock absorber in a car. Damping force is
proportional to velocity, as shown
(6)
𝑓 = 𝑏(𝑥̇ 1 − 𝑥̇ 2 )
The symbol for a damping element is shown below.
f
f
x2
x1
A torsional damper is governed by the following equation.
𝜏 = 𝑏𝑡 (𝜃̇1 − 𝜃̇2 )
(7)
Example Problems
Over the next few days we will analyze several example systems. In each problem we will use the
governing equation for each element to derive a differential equation (or set of differential
equations) for the system. Once we have the differential equation(s), we can use the Laplace
transform to obtain the solution. For these problems, however, we will stop once we have the
governing differential equation.
We will also introduce the concept of natural frequency – the frequency at which systems tend to
oscillate when given an initial disturbance. We won’t be very rigorous in our initial discussion of
natural frequency – we’ll save that for a later section on vibration. For the present, we will
demonstrate the method for finding natural frequency in an SDOF system.
2
x
k
m
b
Mass/Spring/Damper
Example 1: Spring-Mass-Damper System
The figure above shows the classic spring-mass-damper system, which has a single degree of
freedom, the displacement x. This is a decent first-pass model of an automotive suspension system,
with the stiffness and damping of the tire combined into single quantities. To analyze this system
we will first construct the free-body diagram
x
fk
m
fb
where
𝑓𝑘 = 𝑘𝑥
(8)
𝑓𝑏 = 𝑏𝑥̇
(9)
∑ 𝐹 = 𝑚𝑥̈ = −𝑏𝑥̇ + 𝑘𝑥
(10)
𝑚𝑥̈ + 𝑏𝑥̇ + 𝑘𝑥 = 0
(11)
Sum forces in the x direction
Or, rearranging
To find the natural frequency of the system, divide through by m so that the acceleration term is
isolated.
𝑥̈ +
𝑏
𝑘
𝑥̇ + 𝑥 = 0
𝑚
𝑚
(12)
We state here without proof that the natural frequency of an SDOF system is the square root of the
coefficient of displacement.
3
𝜔𝑛 = √
𝑘
𝑚
(13)
Of course, we will prove this in a later section. For now, let us use this formula to find the natural
frequencies of other SDOF systems.
k1
k2
f(t)
m2
m1
b1
b2
x1
x2
Example 2: Double mass-spring-damper system
The figure above shows a double mass-spring-damper system, which might be used as a second
approximation to the automotive suspension system. In this case, m2 is the mass of the chassis, m1 is
the mass of the wheel hub, k2 and b2 are the stiffness and damping of the shock absorber and k1 and
b1 are the internal stiffness and damping of the tire. To analyze the system, we will construct freebody diagrams as before
f k1
f k2
f k2
f b1
f(t)
m2
m1
f b2
f b2
x1
x2
Note that the direction that the forces point is very important for deriving the equations of motion –
forces in the positive x-direction will be written as positive and vice versa. It is worth pausing a
moment here to consider how the direction of the forces was arrived at in constructing the diagram.
To begin, let us imagine moving both masses in the positive x-direction by a small amount. If we
move the mass m2 a little more than we move mass m1, then the quantity x2-x1 will be positive.
𝑥2 − 𝑥1 > 0 → tension in 𝑘2
(14)
and the spring k2 will be put into tension, since tensile forces are taken to be positive. Tension in
the spring k2 results in the forces shown in the diagram above, and a similar result obtains for the
damper b2. Thus
4
𝑓𝑏2 = 𝑏2 (𝑥̇ 2 − 𝑥̇ 1 )
𝑓𝑘2 = 𝑘2 (𝑥2 − 𝑥1 )
Similarly, if we move mass m1 in the positive direction, we will put the spring k1 under tension,
resulting in the force shown in the diagram above. Thus:
𝑓𝑘1 = 𝑘1 𝑥1
𝑓𝑏1 = 𝑏1 𝑥̇ 1
We can now sum forces on both masses
∑ 𝐹1 = 𝑚1 𝑥̈ 1 = −𝑓𝑘1 − 𝑓𝑏1 + 𝑓𝑘2 + 𝑓𝑏2
(15)
∑ 𝐹2 = 𝑚2 𝑥̈ 2 = −𝑓𝑘2 − 𝑓𝑏2 + 𝑓(𝑡)
𝑚1 𝑥̈ 1 + 𝑘1 𝑥1 + 𝑏1 𝑥̇ 1 − 𝑘2 (𝑥2 − 𝑥1 ) − 𝑏2 (𝑥̇ 2 − 𝑥̇ 1 ) = 0
𝑚2 𝑥̈ 2 + 𝑘2 (𝑥2 − 𝑥1 ) + 𝑏2 (𝑥̇ 2 − 𝑥̇ 1 ) = 𝑓(𝑡)
(16)
𝑚1 𝑥̈ 1 + (𝑏1 + 𝑏2 )𝑥̇ 1 − 𝑏2 𝑥̇ 2 + (𝑘1 + 𝑘2 )𝑥1 − 𝑘2 𝑥2 = 0
𝑚2 𝑥̈ 2 − 𝑏2 𝑥̇ 1 + 𝑏2 𝑥̇ 2 − 𝑘2 𝑥1 + 𝑘2 𝑥2 = 𝑓(𝑡)
(17)
It is worthwhile to put this into matrix form, since we would probably model this using engineering
software (such as Matlab) as a next step.
[
𝑚1
0
0 𝑥̈ 1
𝑏 + 𝑏2
]{ } + [ 1
𝑚2 𝑥̈ 2
−𝑏2
−𝑏2 𝑥̇ 1
𝑘 + 𝑘2
]{ } + [ 1
𝑏2 𝑥̇ 2
−𝑘2
−𝑘2 𝑥1
0
] {𝑥 } = {
}
𝑓(𝑡)
𝑘2
2
(18)
We don’t have the tools yet to find the natural frequencies of a 2DOF system, so we finish the
example here.
kt
m
m
bt from friction in bearing
Example 3: Torsional Pendulum
The figure above shows a torsional pendulum, as might be found inside an old-style mechanical
watch (or chronometer). The mechanism pivots at the center, and two masses, m, are suspended a
distance r from the center. The composite moment of inertia of the two masses is:
5
(19)
𝐽 = 2𝑚𝑟 2
m
m
Tk Tb
To create the free-body diagram of the system, we employ the same technique as earlier. First, turn
the pendulum in the positive θ direction (counterclockwise) and record the resulting torques from
the spring and damper (counterclockwise, in this case). Summing torques about the pivot gives
∑ 𝑇 = 𝐽𝜃̈ = −𝑇𝑘 − 𝑇𝑏
(20)
𝐽𝜃̈ + 𝑏𝑡 𝜃̇ + 𝑘𝑡 𝜃 = 0
(21)
This is the same equation as for the mass-spring-damper system! The natural frequency is then
𝑘𝑡
𝜔𝑛 = √
𝐽
(22)
which looks pretty similar to the one for the mass-spring-damper system, where linear stiffnesses
and inertias have been replaced by torsional stiffnesses and inertias.
k2 stiffness of rim
k1 stiffness of rim
Example 4: Tire and Rim Stiffness
The figure above shows a wheel from a car, with tire and rim. We wish to include the stiffness of
the rim and the stiffness of the tire in our model, to make it as accurate as possible. Vibrations
between the ground and the hub will pass through each stiffness in series, so we make the model of
the system shown below:
6
k1
k2
m
x'
f(t)
x
It might not be obvious that the tire and rim should be placed in series, rather than in parallel. To
understand this, consider the “intermediate node”, x′. This is the displacement of the interface
between the tire and the rim (the “bead” on the tire), which we assume moves as a unit. Because of
this intermediate node, we must place the tire and rim in series.
f k1
f k2
f k2
f(t)
m
x'
x
The free body diagram of the system is shown above. We use the same technique as before to find
the direction of the forces.
𝑓𝑘2 = 𝑘2 (𝑥 − 𝑥′)
𝑓𝑘1 = 𝑘1 𝑥′
So that
∑ 𝐹1 = 𝑓𝑘2 − 𝑓𝑘1
(23)
∑ 𝐹2 = 𝑚𝑥̈ = 𝑓(𝑡) − 𝑓𝑘2
𝑘2 (𝑥 − 𝑥 ′ ) − 𝑘1 𝑥 ′ = 0
𝑚𝑥̈ + 𝑘2 (𝑥 − 𝑥′) = 𝑓(𝑡)
(24)
solve the first equation to eliminate x′.
𝑥′ =
𝑘2 𝑥
𝑘1 + 𝑘2
And substitute this into the second equation
𝑚𝑥̈ + 𝑘2 (𝑥 −
𝑘2 𝑥
) = 𝑓(𝑡)
𝑘1 + 𝑘2
𝑘1 𝑘2
𝑚𝑥̈ + (
) 𝑥 = 𝑓(𝑡)
𝑘1 + 𝑘2
(25)
(26)
7
Thus, we have arrived at the equation of motion for the mass-spring system, but with a slightly more
complicated expression for the spring stiffness. In fact, the expression for the stiffness of two
springs in series is:
𝑘1 𝑘2
(27)
𝑘𝑠𝑒𝑟𝑖𝑒𝑠 =
𝑘1 + 𝑘2
which the reader will recognize as similar to the formula for two resistors in parallel.
x
A,
l
Example 5: Fluid Manometer
A simple method for measuring air pressure is the use of a fluid manometer, as shown above.
Pressure on one of the openings forces the fluid down at one end, and up at the other. A scale on
one side of the manometer can be calibrated to indicate air pressure at the other side. For this
example, we are interested in the dynamic behavior of the manometer. The total mass of the fluid
inside the tube is
(28)
𝑚 = 𝐴𝑙𝜌
where A is the cross-sectional area, l is the total length of the fluid (not the tube) and ρ is the density
of the fluid. The dotted line in the figure above shows the equilibrium position of the fluid, and x
gives the displacement of the top of the fluid above or below the equilibrium line.
2xA g
2x
x
8
Let us now displace one end of the fluid a distance x above the equilibrium line, as shown in the
figure above. The height of the fluid at one end of the tube above the fluid at the other end of the
tube is 2x, as seen in the figure, and the weight of this fluid is 2xAρg – this is the restorative force
on the fluid.
∑ 𝐹 = 𝑚𝑥̈ = −2𝑥𝐴𝜌𝑔
(29)
𝐴𝑙𝜌𝑥̈ + 2𝐴𝜌𝑔𝑥 = 0
(30)
𝑥̈ +
2𝑔
𝑥=0
𝑙
(31)
Thus, the natural frequency of the fluid in the manometer is
2𝑔
𝜔𝑛 = √
𝑙
(32)
9
Energy Methods
Some problems are easier to solve using the principle of conservation of energy. This principle
states that
Change in 𝐾𝐸 + Change in 𝑃𝐸 = Work added to system
or mathematically
∆(𝑇 + 𝑈) = ∆𝑊
(33)
If no work is added to a system it is called conservative and the conservation of energy principle can be
written
∆(𝑇 + 𝑈) = 0
(34)
This implies that the amount of kinetic and potential energy remains constant with respect to time,
or
𝑇 + 𝑈 = constant
(35)
Differentiating this with respect to time gives
𝑑
(𝑇 + 𝑈) = 0
𝑑𝑡
(36)
We’ll use this equation to solve a few example problems.
10
l
m
g
Example 1: The Simple Pendulum
Consider the simple pendulum shown above. For this example, we assume that the deflection angle,
θ, is small, such that
sin 𝜃 ≈ 𝜃
cos 𝜃 ≈ 1
The kinetic energy of a rotating body is given by
𝑇=
1 2
𝐽𝜃̇
2
(37)
where J is the moment of inertia of the body. For a mass on a thin rod, the moment of inertia is
𝐽 = 𝑚𝑙 2
(38)
1
𝑇 = 𝑚𝑙 2 𝜃̇ 2
2
(39)
so that the kinetic energy is
l
l cos( )
l
m
h
g
The potential energy of a mass in a gravitational field is given by
𝑉 = 𝑚𝑔ℎ = 𝑚𝑔(𝑙 − 𝑙 cos 𝜃)
(40)
11
Adding the energies together gives
𝑇+𝑉 =
1 2 2
𝑚𝑙 𝜃̇ + 𝑚𝑔(𝑙 − 𝑙 cos 𝜃)
2
(41)
and taking the derivative gives
𝑑
𝑑 1
(𝑇 + 𝑉) = ( 𝑚𝑙 2 𝜃̇ 2 + 𝑚𝑔(𝑙 − 𝑙 cos 𝜃)) = 0
𝑑𝑡
𝑑𝑡 2
(42)
𝑚𝑙 2 𝜃̇ ∙ 𝜃̈ + 𝑚𝑔𝑙𝜃̇ sin 𝜃 = 0
(43)
𝑙𝜃̈ + 𝑔 sin 𝜃 = 0
(44)
and using the small angle approximation
𝑔
𝜃=0
𝑙
𝜃̈ +
(45)
The natural frequency of the pendulum is then
𝑔
𝑙
(46)
𝜔𝑛 = √
which should be a familiar formula from physics class.
l
k
l sin
m
g
Example 2: Pendulum with spring
We now add a spring to the pendulum, to show how easy it is to revise the equation of motion using
the energy method. The potential energy stored in a spring is
12
1
𝑈𝑠𝑝𝑟𝑖𝑛𝑔 = 𝑘𝑥 2
2
(47)
where x is the deflection of the spring. As seen in the diagram above, the deflection of the spring is
𝑥 = 𝑙 sin 𝜃
(48)
𝑑
𝑑 1
1
(𝑇 + 𝑉) = ( 𝑚𝑙 2 𝜃̇ 2 + 𝑚𝑔(𝑙 − 𝑙 cos 𝜃) + 𝑘𝑙 2 sin2 𝜃) = 0
𝑑𝑡
𝑑𝑡 2
2
(49)
𝑚𝑙 2 𝜃̇ ∙ 𝜃̈ + 𝑚𝑔𝑙𝜃̇ sin 𝜃 + 𝑘𝑙 2 𝜃̇ sin 𝜃 cos 𝜃 = 0
(50)
𝑚𝑙𝜃̈ + 𝑚𝑔 sin 𝜃 + 𝑘𝑙 sin 𝜃 cos 𝜃 = 0
(51)
so that
Using the small angle approximation yields
𝑚𝑙𝜃̈ + (𝑚𝑔 + 𝑘𝑙)𝜃 = 0
𝜔𝑛 = √
𝑚𝑔 + 𝑘𝑙
𝑚𝑙
(52)
(53)
k
m
g
l
Example 3: The inverted pendulum
Let us turn the pendulum upside down, as shown in the figure above. The kinetic energy remains
the same as before, as does the expression for the deflection of the spring. The only major
difference is the expression for potential energy due to gravity, which we must examine more
closely.
13
h
l
m
l cos( )
As shown in the figure above, the height variable, h, is measured from the equilibrium position,
when the pendulum is at the top of its swing. Therefore, h will be either zero (at the top) or
negative. Thus
ℎ = 𝑙 cos 𝜃 − 𝑙
(54)
and the total energy expression becomes
1 2 2
1
𝑚𝑙 𝜃̇ + 𝑚𝑔(𝑙 cos 𝜃 − 1) + 𝑘𝑙 2 sin2 𝜃
2
2
(55)
𝑑 1 2 2
1
( 𝑚𝑙 𝜃̇ + 𝑚𝑔(𝑙 cos 𝜃 − 1) + 𝑘𝑙 2 sin2 𝜃) = 0
𝑑𝑡 2
2
(56)
𝑚𝑙 2 𝜃̇ ∙ 𝜃̈ − 𝑚𝑔𝑙𝜃̇ sin 𝜃 + 𝑘𝑙 2 𝜃̇ sin 𝜃 cos 𝜃 = 0
(57)
𝑚𝑙𝜃̈ − 𝑚𝑔 sin 𝜃 + 𝑘𝑙 sin 𝜃 cos 𝜃 = 0
(58)
𝑇+𝑉 =
The time derivative is
and taking the small angle approximation yields
with the natural frequency
𝑚𝑙𝜃̈ + (𝑘𝑙 − 𝑚𝑔)𝜃 = 0
𝜔𝑛 = √
𝑘𝑙 − 𝑚𝑔
𝑚𝑙
(59)
(60)
What happens if mg > kl? It appears as though we are left with a negative quantity inside the radical,
which would result in an imaginary natural frequency! Obviously, this is physically unrealizeable – a
natural frequency must be a real quantity. We conclude that, if mg > kl, the system is unstable, which
means that the pendulum will simply fall over, rather than returning to equilibrium. An unstable
system occurs when the restoring force is insufficient to return a system to equilibrium, as would be
the case if the spring were too weak in the inverted pendulum. The method shown above is very
commonly used to investigate the stability of systems; where imaginary natural frequencies are found
the system is said to be unstable.
14