Dynamics of mechanical system - prima parte

DYNAMICS OF

MECHANICAL SYSTEMS

Theory, Vibrations, and Control

k

c

x

o

e

q

w

F(t)

c^x + kx = 0

ΣF

x

ẍ + c ẋ + kx = 0

J₀ k

m

m

ω

POLITECNICO DI MILANO

Cae running on even surface: typical mechanical system full of electronics on mechatronics

16.09.2025

We want to study how the body reacts to vibrations and which frequencies are worst for it.

One simple choice is the quarter-car model:

This is the lumped parameter model

We can say we miss the vertical or pitch movement with are not considered by the quarter car model so we represent the overall effect:

lumped parameter model

Now we deal with internal noise: cars can hear waves at least 5 kHz, the motion becomes a deformable motion which makes vibrations. We will have deformability effects.

• LP
• DP
• DISCRETIZATION (FEM)

large motion

small motion

stability

1. Dynamics of lumped parameter system

where does it come from?

1 dof. Why? 1 r b in plane motion has 3 dof

cost constants: prevent any displacement along the x of the mass

2 costs: 2 x 1 dof

So the remaining dof is 1 which is the number of independents coords.

Introduce 4 methods:

1a) Newton's 3rd law

sum of forces applied to a point

For rigid body we can generalize with

acceleration of centre of mass c.o.m.

Scheme of forces acting on the system:

1b) D'Alembert's principle

rewriting the system in the form of a static state simpler and more convenient: makes clear that any acceleration imposed to a mass will affect the balance of forces.

2a) Principle of virtual work

Virtual displacement (box def.)

With the principle of virtual work we automatically rule out the additione unknowns related to the constant forces.

2b) Lagrange's equations (different form of the law)

in case of non dissipative constraints, the constraint forces do not appear in the equations of motions

δW_con = -∂V/∂x dx

δW_diss = -∂D/∂x dx

→ d/dt(∂T/∂x) - ∂T/∂x + ∂D/∂x + ∂V/∂x = dW_b/dx

example: spring

T = 1/2 m x²

V = 1/2 k x²

D = 1/2 c x²

δW_b = P(t) δx

∂T/∂x = m xdot → d/dt(∂T/∂x) = m xddot

∂T/∂x = 0

∂V/∂x = kx

∂D/∂x = c xdot

→ m xddot + c xdot + kx = P(t)

example: slider-crank mechanism

connecting rod: massless and lumped to two points

slider mass

17.09.2025

Instead of force balance, we prefer energy balance

I want to write a single equation of motion and I can choose a single kine=matic parameter: if I chose α I know exactly the position of A

m_s is not only representing the mass of the piston, but also contains part of the mass of the rod; A is an eccentric rotating mass, so for simplifying we position the com (center of mass) of the crank not in O so that the total com is positioned in O. The distance OA is fixed.

3 bodies -> 9 independent equations

Using the approach with Newtons Dynamics on d'Alembert principle:

I'm using the convention for the positive sign, also if the actual direction could be the opposite.

More than one body => I can write the equation considering one body at time: up to 3 scalar equations with forces or moments

While using force method, we have to consider V, H, M's: so many forces than necessary

Solution with Lagrange

virtual displacement

Considering β:

x_B = r cos α + e cos β

r sin α = e sin β → β = arcsin(r sin α / e)

λ = r/e < 1 necessary, otherwise the mechanism can't move. (better if λ << 1)

→ x_B = r cos α(t) + e cos β

→ x_B = -α r sin α(t)

Now I can follow with Lagrange:

Kinetic energy T = 1/2 J αdot² + 1/2 m_s xdot_B² = 1/2 (J + m_sr² sin²α) αdot² = 1/2 m*(α) αdot²

Potential energy V = 0 (no conservative forces)

D = 0 (no viscous forces or dissipative elements -> frictionless)

δW_b = M δα + f δx_B = (M - f r sin α) δα in which δx_B = ∂x_B/∂α δα = -r sin α δα

∂T/∂α = (J + m_sr² sin²α) αdot → d/dt(∂T/∂α) = (J + m_sr² sin²α) αddot + m