applied
sciences
Article
A Dynamic Response Analysis of Vehicle Suspension System
Rogério Lopes 1,2, * , Behzad V. Farahani 2, * , Francisco Queirós de Melo 2 and Pedro M. G. P. Moreira 2
1
2
*
FEUP, Faculty of Engineering, University of Porto, Dr. Roberto Frias Street, 4200-465 Porto, Portugal
INEGI, Institute of Science and Innovation in Mechanical and Industrial Engineering,
Dr. Roberto Frias Street, 400, 4200-465 Porto, Portugal
Correspondence: rflopes@inegi.up.pt (R.L.); behzad.farahani@fe.up.pt (B.V.F.); Tel.: +351-939742839 (R.L.);
+351-938096044 (B.V.F.)
Abstract: Automotive engineering is a very important branch of design and development in mechanical engineering devoted to vehicle manufacturing. It demands solid knowledge focusing on
the kinematics and dynamics of mechanical systems, either using structural analysis or modelling
techniques. In this manuscript, a model based on the application of the fundamentals of the static
and dynamic behaviour of the suspension system of a bus vehicle is developed. Therefore, a set of
mathematical equations is expanded following the structural dynamic formulations to obtain an
analytical solution on natural frequencies and the displacement under an external load. The problem
is also solved using the Finite Element Method (FEM) to confirm the analytical results. Equivalent
material parameters and boundary conditions were used for the FEM simulation. This work aims
to propose a methodology based on modal analysis with a simplified model to predict the dynamic
response of an automotive structure.
Keywords: FEM; automobile industry; modal analysis; dynamic response; steady-state dynamic analysis
1. Introduction
Citation: Lopes, R.; Farahani, B.V.;
Queirós de Melo, F.; Moreira, P.M.G.P.
A Dynamic Response Analysis of
Vehicle Suspension System. Appl. Sci.
2023, 13, 2127. https://doi.org/
10.3390/app13042127
Academic Editors: Marcin Graba and
Stanisław Adamczak
Received: 6 January 2023
Revised: 31 January 2023
Accepted: 2 February 2023
Published: 7 February 2023
Copyright: © 2023 by the authors.
Licensee MDPI, Basel, Switzerland.
This article is an open access article
distributed under the terms and
conditions of the Creative Commons
Attribution (CC BY) license (https://
creativecommons.org/licenses/by/
Automotive suspension systems are designed to control the passenger’s comfort and
safety when riding in a vehicle. The mechanical design of the suspension system regulates
the vehicle’s performance under various driving conditions [1]. Therefore, analysing the
suspension control systems is a key to improving the smoothness of a vehicle’s ride without
compromising safety. The balance between the comfort and handling stability of a vehicle
under traffic conditions must be controlled [2,3].
Conventional suspensions are fabricated and assembled using traditional manufacturing methods. Thus, they encompass some limitations, including strict design constraints,
high costs for parts, and modification [4]. Currently, there are three types of automotive
suspensions commonly used: passive, semi-active, and active [5].
A passive vehicle suspension system is reliable, simple, and inexpensive, being a typical mechanical structure incorporating linear springs and viscous dampers with constant
stiffness and damping coefficient, respectively. In such systems, the damper and spring
are fastened between the car body and wheel support components. A damper is engaged
with hydraulic oil or compressed gas. There is a piston moving through a rod from the
exterior [5]. Semi-active suspension is similar to the passive design. It differs in the presence
of variable damping coefficients but still has a fixed spring constant. Semi-active design
contributes to a seamless change between a passive damper and one with a semi-active
damping coefficient [5]. Compared with the other suspension systems, an active apparatus
comprises an actuator that can supply active force regulated by a controlling algorithm. It
relies on the information gathered from installed vehicle sensors [5].
Nevertheless, most vehicles on the road do not possess an active suspension. As of
now, active suspension is reserved for higher-end and performance vehicles. Though, given
4.0/).
Appl. Sci. 2023, 13, 2127. https://doi.org/10.3390/app13042127
https://www.mdpi.com/journal/applsci
Appl. Sci. 2023, 13, 2127
2 of 19
the direction of innovation and the addition of more features, active suspension will likely
become more commonplace in the near future [6].
Despite the common use of a passive suspension system in recent vehicles, the engineering inherent in its design promotes the inclusion of an efficient stability control system.
It will counteract the detrimental effects of sharp external forces that may come from evasive
emergency driving manoeuvres [7]. Many accidents occur due to an improper suspension
design, mechanical malfunction, underinflated tyres, or worn/inadequate dampers [8,9].
Therefore, enhancing the quality of transportation services has been recognised as one of
the key factors toward sustainable mobility [10,11].
2. Related Works
The vehicle’s behaviour is naturally conditioned by the effect of time-dependent
(dynamic) forces on the vehicle body. Furthermore, the tyre/suspension system quality has
a direct impact on the vehicle’s on-road performance. Miloradović et al. [12] conducted
a detailed study on the dynamic interaction between a passenger vehicle’s steering and
the physical connection of the suspension bodies. They performed experiments to acquire
acceleration, deformation, and torque data at various driving speeds on different roads.
To characterise the problem variables and mathematical modelling, there exist a
considerable number of mathematical methodologies to study the vehicle’s dynamics.
Talukdar et al. [13] proposed some ride models to evaluate the parameters affecting the
vehicle’s behaviour. They studied a half-model of a car, having extended the analysis
from a design perspective, e.g., variation of suspension stiffness, damping coefficient, etc.
In a particular context of vehicle design for a safe ride, the suspension performance has
drawn the attention of mathematicians for applied industrial problems such as suspension
optimization for non-linear stiffness in terms of stroke, and it offers a progressive strength
to vehicle roll in a curve or a yaw motion [14].
Currently, computational modelling is widely used to solve problems involving fast or
transient loading status [15]. The high costs associated with developing new vehicle models
have also made computer simulations of a vehicle’s dynamics increasingly relevant [16].
Several studies were performed to characterize the vehicle dynamics behaviour, identifying
the suspension and tyre assembly as two of the key components that most influence the
ride quality [17,18]. Likewise, the advancement of technology in electronics, sensors, and
actuators has enabled the application of control systems to enhance the vehicle’s ride
quality [19].
In the field of automotive engineering tools for design and structural assessment, the
Finite Element Method (FEM) is considered a crucial technique for a reliable and accessible
approach to studying the structural assessment of a vehicle’s dynamics [20]. Thus, FEM
models contribute to the possibility of simulating the model in any case of static or dynamic
loading conditions, thermal effects, fluid-structure interaction, and electromagnetic field
effects [19]. Its accuracy has been confirmed by reasonable agreement with corresponding
experimental tests [21]. Furthermore, the FEM was used to study the natural frequencies
since it deals with an impressively large set of parameters [22,23]. Abdullah et al. [23]
sought to explore the level of accuracy of the simulated simplified body using various
modelling methodologies. They further validated such models by comparing finite element
modal characteristics with experimental modal properties.
A simplified model of a passive suspension system installed on a bus vehicle is
proposed here. The vehicle body is assumed to be extremely rigid compared with the
tyre/spring suspension group. Therefore, a rectangular, rigid plate supported by four
corner springs is established. Then, a set of mathematical formulations is developed
relying on the dynamics modal analysis theory. It must be mentioned that the bus body is
included in the calculation because of its equivalent mass density. Since the suspension
spring stiffness is much smaller than the one assigned to the structural components, only
the elastic effects of the vehicle suspension will be taken into account in the calculation.
Furthermore, this study intends to model the effect of the road irregularities or the driver’s
Appl. Sci. 2023, 13, 2127
Weight$$$
𝒎𝒕𝒐𝒕𝒂𝒍
12 (tonne)
Furthermore, this study intends to model the effect of the road irregularities or the driver’s
actions via time dependency through transient dynamic formulations integrated with the
real-time direct integration analysis of Newmark’s method. The normal displacement
field of the studied model is established by a 3-DOF model consisting of a concentrated
vertical translation of the mass centre superimposed on two rotations about the x- and y3 of 19
axes, respectively.
To confirm the analytical solutions, numerical modelling is performed through FEM
software ABAQUS®. As a result, natural frequencies are computed through a frequency
step
scheme.
Then,
the FEMthrough
model istransient
solved by
a steady-state
dynamic,
directwith
stepthe
in
actions
via time
dependency
dynamic
formulations
integrated
which
twodirect
concentrated
forces
are applied
to the frontmethod.
side of the
model,
which
simulates
real-time
integration
analysis
of Newmark’s
The
normal
displacement
the
of the bump
as theoretically
considered.
fieldresponse
of the studied
modelimpacts
is established
by a 3-DOF
model consisting of a concentrated
vertical translation of the mass centre superimposed on two rotations about the x- and
3.
Mathematical
Development
y-axes,
respectively.
To
confirm
the
analytical
solutions,
modelling
is performed
FEM
Based on the modal
analysis
theory,numerical
the stiffness
matrix, together
withthrough
the displace®
software
. Asisadefined
result, to
natural
frequencies
are
computed
through
a frequency
ment
andABAQUS
force vectors,
provide
the natural
frequency
solution
obtained
from
stepeigenvalues
scheme. Then,
the
FEM model
is solved
by a1 steady-state
dynamic,
direct
step in
the
of the
suggested
geometry.
Figure
shows a general
view of
the studied
which
concentrated
are applied
to the
frontand
side
of therespectively.
model, which
simulates
model,two
in which
𝐿 andforces
𝑏 represent
the bus
length
width,
The
wheelthe
response
of
the
bump
impacts
as
theoretically
considered.
base length is identified as 𝐿 and the track width is defined as 𝑏 . Each wheel hub is
supported by a spring set [24]. Table 1 reports the main characteristics of the studied
3. Mathematical Development
model.
Based on the modal analysis theory, the stiffness matrix, together with the displacement and
force vectors,
defined
provide the natural frequency solution obtained from
Table
1. Characteristics
of is
the
studiedto
model.
the eigenvalues of the suggested geometry. Figure 1 shows a general view of the studied
Track
Bus LLength
model, in which
the bus length and
width, respectively.
wheelbase
b and b represent
Bus Width$$$𝒃
WheelbaseThe
Length$$$𝑳
Spring Stiffness$$$𝒌
Width$$$𝒃
$$$𝑳
length is identified
as𝒃 L and the track width is defined
as bw𝑾. Each wheel hub is supported
⁄ma) spring set12.0
(m)
185 (kNby
2.7 (m)
2.5 (m) of the studied5.9
[24].(m)
Table 1 reports
the main characteristics
model.
Figure 1. A general perspective of the studied model.
Table
1. Characteristics
3.1. Matrix
Formulationofofthe
thestudied
Modelmodel.
Weight
mtotal
12 (tonne)
This study focuses on the vehicle’s dynamic interaction with the road when the ve-
Spring Stiffness
Bus Length
Bus Width
Track Width
Wheelbase Length
different states,
including driving
manoeuvres such
k hicle is moving. Therefore,
Lb
b
bW
L as accel-
eration, braking, and curving, are examined according to the suspension flexibility and
12.0 (m)
2.7 (m)
2.5 (m)
5.9 (m)
damping characteristics. The proposed model is defined as a four-node macro-element
with a rectangular shape, dimensioned as wheelbase length 𝐿 and bus width 𝑏, respec3.1. Matrix
Formulation
Model
tively,
in the
𝑥 − and 𝑦of−thecoordinate
axes, as depicted in Figure 2.
This study
rigid plate
(the
geometry
ofinteraction
the bus) materializes
thewhen
structure
of the
focuses
onsimplified
the vehicle’s
dynamic
with the road
the vehicle
bus
platform.
It is assumed
to be
very including
rigid compared
with
the stiffness
of the
suspension
is moving.
Therefore,
different
states,
driving
manoeuvres
such
as acceleration,
braking, and curving, are examined according to the suspension flexibility and damping
characteristics. The proposed model is defined as a four-node macro-element with a
rectangular shape, dimensioned as wheelbase length L and bus width b, respectively, in the
x − and y− coordinate axes, as depicted in Figure 2.
This rigid plate (the simplified geometry of the bus) materializes the structure of the
bus platform. It is assumed to be very rigid compared with the stiffness of the suspension
spring/damper assembly at each wheel. With this characteristic, the displacement field
bears on a simple algebraic equation defining a plane.
185 (kN/m)
Appl. Sci. 2023, 13, x FOR PEER REVIEW
Appl. Sci. 2023, 13, 2127
4 of 19
4 of 19
spring/damper assembly at each wheel. With this characteristic, the displacement field
bears on a simple algebraic equation defining a plane.
Figure
2. Platform
model
a plane
resting
4 springs
at corner
nodes.
Figure
2. Platform
model
as a as
plane
resting
on 4on
springs
at corner
nodes.
To define
spatial
position
of the
plane
points
its transverse
displacement
To define
the the
spatial
position
of the
plane
points
by by
its transverse
displacement
u
x,
y
i
=
1,
.
.
.
4
at
each
corner
node,
some
parameters
must
be
assigned
in system
the system
(
)
(
)
i
𝑢 (𝑥, 𝑦) (𝑖 = 1, . .4) at each corner node, some parameters must be assigned in the
of
equations
as:
of equations as:

u1 = uG − (b/2) × θ x − ( L/2) × θy



𝑢
𝑢 − (𝑏/2) × 𝜃 − (𝐿/2) × 𝜃
2 = u G + ( b/2) × θ x − ( L/2) × θy
⎧ u=
,
(1)
⎪
𝑢
=
𝑢 u+
(𝑏/2)
×)𝜃× θy
u
=
+ ( L/2
(b/2×) 𝜃× θ−x (𝐿/2)
G+

 3
,
(1)
𝑢 u+
(𝑏/2)
×)𝜃× θy
+ ( L/2
(b/2×) 𝜃× θ+x (𝐿/2)
4 =
G−
⎨𝑢 u=
⎪𝑢 = 𝑢 − (𝑏/2) × 𝜃 + (𝐿/2) × 𝜃
⎩
In which:
•In which:
uG is the transverse displacement related to the plate mass centre, G;
•
θ x the
andtransverse
θy are the displacement
plate rotationsrelated
about to
x −the
and
y− coordinate
axes.
plate
mass centre,
𝐺;
•
𝑢 is
If the𝜃 plate
element
is supported
by𝑥4−springs
nodes
with stiffness k1 , k2 ,
•
𝜃 and
are the
plate rotations
about
and 𝑦 at
− corner
coordinate
axes.
k3If, the
andplate
k4 , then,
for
a
generic
position
of
the
rectangular
plate
defined
by the
,
element is supported by 4 springs at corner nodes with stiffness
𝑘 , 𝑘nodal
displacements,
referring
to
Equation
(1),
the
elastic
deformation
energy
stored
in
plate
𝑘 , and 𝑘 , then, for a generic position of the rectangular plate defined by the nodalthe
disis obtained
by the total
contribution
of elastic
the deformed
springs
at each
node.inIndeed,
theistotal
placements,
referring
to Equation
(1), the
deformation
energy
stored
the plate
amount
internal
energy (elastic
energy) must
balance
thenode.
external
workthe
performed
obtained
by of
thethe
total
contribution
of the deformed
springs
at each
Indeed,
total
by the external forces (or moments) applied to the plate as:
amount of the internal energy (elastic energy) must balance the external work performed
by the external forces (or moments)
applied to hthe plate as:
i
1 4
1
(2)
Uelastic = ∑ k i × u2i =
k1 u21 + k2 u22 + k3 u23 + k4 u4 2 ,
1 2 i =1
1 2
(2)
𝑈
=
𝑘 ×𝑢 = 𝑘 𝑢 +𝑘 𝑢 +𝑘 𝑢 +𝑘 𝑢 ,
2
2
where the nodal displacement ui is a function of parameters presented in Equation (1). In
relation
to thedisplacement
elastic energy𝑢Uelastic
of the deformed
plate, presented
the nodal equivalent
where
the nodal
is a function
of parameters
in Equationexternal
(1).
force system
obtained
internalplate,
forcesthe
as follows:
In relation
to the must
elasticbeenergy
𝑈 to equilibrate
of the deformed
nodal equivalent exter to equilibrate internal forces as follows:
nal force system must be obtained
∂Uelastic
∂ui
4


 ∂UG = ∑i=1 k i × ui ∂uG = FG
∂ui = 𝐹
elastic = ∑ 4 𝑘 × 𝑢
⎧ ∂U∂θ
= ∑i=1 k i × ui ∂θ
= Mθx .
(3)
x
x
⎪

∂ui
4
 ∂Uelastic==
k
×
u
=
M
∑ ∑i=𝑘1 ×i 𝑢 i ∂θ=
𝑀 .θy
(3)
∂θy
y
⎨
⎪
∑ 𝑘 × the
𝑢 chain
= 𝑀rule for the independent variables
Taking the derivatives⎩of these=
equations,
uG , θ x , and θy would be acquired, therefore, the linear system of 3 equations can take the
Taking the
derivatives of these equations, the chain rule for the independent variafollowing
form:

bles 𝑢 , 𝜃 , and 𝜃 would be acquired,
therefore, the linear system of 3 equations can take
 k11 u1 + k12 θ x + k13 θy = FUG
the following form:
k21 u1 + k22 θ x + k23 θy = Mθx
(4)

k31 u1 + k32 θ x + k33 θy = Mθy
Appl. Sci. 2023, 13, 2127
5 of 19
Being:

k11 = k1 + k2 + k3 + k4




k
=
k21 = 2b (−k1 + k2 + k3 − k4 )

12


 k = k = L (−k + k + k − k )
13
31
2
2
1
2
3
4

k22 = b4 (k1 + k2 + k3 + k4 )



LB

k23 = k32 = 2 (−k1 + k2 + k3 − k4 )

2

k33 = L4 (k1 + k2 + k3 + k4 )
.
(5)
It must be mentioned that using springs with stiffness, k, at the four corner nodes
of the plate contributes to rendering the previous stiffness matrix diagonal, which led
to significantly simplifying the problem complexity once a set of uncoupled systems of
equations is attained:

  

4k
0
0
uG   FUG 

2


(6)
0  θ x = Mθx .
 0 b4 (4k)
  

2
L
M
θ
y
θy
0
0
4 (4k )
As an illustration, if a static concentrated transverse force F1 at node 1 is considered
prior to evaluating the RHS (Right Hand Side) of equation system (6), it is important to
recover the transverse displacement at node 1 presented in Equation (1). Hence, the work
W released by a force term of F1 is:
W ( F1 ) = F1 × u1 = F1 × [uG − (b/2) × θ x − ( L/2) × θy ].
(7)
Applying Castigliano’s theorem [25], the external force system, performing the work
equivalent to the one due to force F1 , can be formulated as:
∂[ F1 ×[u G −(b/2)×θ x −( L/2)×θy ]]
∂W ( F1 )
= F1 = FuG
∂UG =
∂UG
∂[ F1 ×[u G −(b/2)×θ x −( L/2)×θy ]]
∂W ( F1 )
=
= − F21 b = Mθx .
∂θ x
∂θ x


 ∂W ( F1 ) = ∂[ F1 ×[uG −(b/2)×θx −( L/2)×θy ]] = − F1 L = M
θy
2
∂θy
∂θy




Therefore, the solution by uncoupled equations is computed as:
   F1 
uG  

 4k 
F1
.
θ x = − 2bk
  

− F1 
θy
2Lk
(8)
(9)
This is the RHS due to the contribution of force at node 1, which changes the spatial configuration of the chassis plate model. However, in terms of nodal displacement,
substituting this vector in Equation (1) results in:

F1
F1
F1
1

u1 = 4k
− 2b × − 2bk
− L2 × − 2Lk
= 3F


4k


F1
u2 = 4k
.
(10)
− F1


u
=
3

4k


F1
u4 = 4k
Figure 3 depicts the aspect of the deformed shape regarding the plane.
If the reactions of each corner due to the mounted springs are taken into account, then
the forces are:

3F1
1

R1 = k × 3F

4k = 4


3F
F
F
F
R2 = F41
4
⇔ ∑i=1 Ri = 1 + 1 − 1 + 1 = F1 .
(11)
F1

4
4
4
4
R
=
−
3

4


R4 = F41
Appl. Sci. 2023, 13, 2127
6 of 19
A dynamic analysis of structures requires not only the definition of their internal
reactions due to external force systems but also the computation of inertial forces as a conseAppl. Sci. 2023, 13, x FOR PEER REVIEW
6 of 19
quence of the acceleration field. To achieve this goal, it is necessary to evaluate the structure
mass matrix, a task that can be solved by variational techniques on evaluating the work
performed by inertial forces (involving accelerations) on the respective displacement field.
Figure 3. Deformed shape of the rigid plate under prescribed force at node 1.
Figure 3. Deformed shape of the rigid plate under prescribed force at node 1.
Considering that the bus platform behaves as a plane, any displacement at a point can
the reactions
of each
due to the mounted
springs
taken intopolynomial
account,
beIfinterpolated
from
the 4 corner
nodal displacements
using shape
or are
interpolation
then
the
forces
are:
functions. Then, in a matrix, the displacement vector of any internal point can take the
following form: 𝑅 = 𝑘 ×
=
 
⎧

 u1 

⎪


𝑅 =
u
u( x, y) = ⟺
(12)
N1 ∑ N2 𝑅 N
=3 N+4 − 2 +, = 𝐹 .
(11)

⎨
𝑅 =−
 u3 



u4
⎪
𝑅 =
⎩
where N1 . . . N4 and u1 . . . u4 are accordingly the nodal shape functions and the displaceA dynamic
structures requires
not
the definition
theirisinternal
re- in
ment.
Indeed, analysis
a similarof
interpolation
applies to
theonly
acceleration
field, of
which
calculated
actions
due mode
to external
systems but also the computation of inertial forces as a cona similar
to theforce
displacements:
sequence of the acceleration field. To achieve this goal, it is necessary
 ..  to evaluate the strucu1 
ture mass matrix, a task that can be solved by variational 
techniques
on evaluating the


.. 

uon
2 the respective displacework performed by inertial u..forces
(involving
accelerations)
..
(13)
( x, y) = N1 N2 N3 N4

ment field.
u.. 3 



u4 displacement at a point
Considering that the bus platform behaves as a plane, any
can be interpolated from the 4 nodal displacements using shape or interpolation polynoTable 2 reports the corresponding interpolation functions.
mial functions. Then, in a matrix, the displacement vector of any internal point can take
the following form:
Table 2. Integral form of shape functions products.
𝑢
𝑢
𝑢(𝑥, 𝑦) = 𝑁 𝑁 𝑁 𝑁
,
(12)
Shape Ni
Product
𝑢 Shape
𝑢
1
N1 = (1 − ξ )(1 − η )/4
𝑢 … 𝑢 are accordingly the nodal
functions
where 𝑁 … 𝑁 and
2
N2 =shape
1 − η )/4 and the displace(1 + ξ )(
3
N
=
1
+
ξ
1field,
+ η )/4
(
)(
ment. Indeed, a similar interpolation applies to the acceleration
which is calculated
3
4
N
=
1
−
ξ
1
+ η )/4
(
)(
4
in a similar mode to the displacements:
𝑢
Figure 4 represents the finite element with four nodes.
𝑢 Assuming that each variable
(13)
𝑁 𝑁 𝑁 𝑁
. corresponding differential
is attributed to the relation𝑢(𝑥,
of x𝑦)==L2 (1 + ξ ); y = 2b (1 + η 𝑢
), the
𝑢 performed by inertial forces,
relation is described as dx = L2 dξ; dy = 2b dη. Hence, the work
Table 2 reports
the corresponding
interpolation
assuming
a consistent
mass distribution,
takes the functions.
following form:

U Inertia = δ1
 .. 
u1 


 .. 


N2
u2

..
Shape
dξdη
NProduct
N
N
N
2
3
4
  u3 .
N  1
−1 

 .. 

𝑁 = (1 − 𝜉)(1 − 𝜂)/4

 3
N4
u4
𝑁 = (1 + 𝜉)(1 − 𝜂)/4
𝑁 = (1 + 𝜉)(1 + 𝜂)/4
𝑁 = (1 − 𝜉)(1 + 𝜂)/4
Table 2. Integral form of shape functions 
products.

 N1 
Z 1 Z 1  

δ2 Shape
δ3 (δ1𝑵+
𝒊 δ3 − δ2 ) ρh 
−1
1
2
3
4


(14)
relation is described as 𝑑𝑥 = 𝑑𝜉; 𝑑𝑦 = 𝑑𝜂. Hence, the work performed by inertial forces,
assuming a consistent mass distribution, takes the following form:
𝑈
Appl. Sci. 2023, 13, 2127
= 𝛿
𝛿
𝛿
𝑁
𝑁
𝑁
𝑁
(𝛿 + 𝛿 − 𝛿 ) 𝜌ℎ
𝑁
𝑁
𝑁
𝑁 𝑑𝜉𝑑𝜂
𝑢
𝑢
𝑢
𝑢
.
(14)
7 of 19
Figure
Figure 4.
4. Notation
Notation and
and non-dimensional
non-dimensional coordinates
coordinates of rectangular finite element.
The consistent
mass
matrix
is obtained
through
solving
the integrals
with
The
consistent 44××44DOF’s
DOF’s
mass
matrix
is obtained
through
solving
the integrals
the shape
function
combination
for all for
fourallnodes
the plate
element,
as demonstrated
in
with
the shape
function
combination
four of
nodes
of the
plate element,
as demonEquations
(15) and (16).
strated
in Equation
(15) and Equation (16).
 4 2 1 2
4 2
1 2
2 2 4 4 22 11,
[𝑴] × =
Lb 𝜌ℎ
(15)
ρh 1 2 4 2,
(15)
[ M ] 4×4 =
144 1 2 2 1 42 24
2 1
2 4
3
3
𝑴𝑖𝑗 = 𝜌𝐿𝑏 = 𝑚
∑ ∑ Mij = ρLb = mtotal .
.
(16)
(16)
i =1 j =1
The recently presented mass matrix is associated with a 4-node element defined by 4
The recently presented mass matrix is associated with a 4-node element defined by 4
nodal displacements. However, such a displacement field, as shown in Equation (1), can
nodal displacements. However, such a displacement field, as shown in Equation (1), can be
be condensed into a 3-DOF as:
condensed into a 3-DOF as:
𝑢
 
 2 −𝑏 −𝐿 𝑢
𝑢


u
2 −b
−L  
𝑢


1
𝜃 


𝒖 =
[𝑻]
=1  2 𝑏 −𝐿 𝜃 u
. 
⟺ 
(17)
𝑢
u×G
u2
𝐿 L 𝜃 G
22 b𝑏 −
=
.
(17)
θx 𝜃 ⇔
θ
{ u } = [ T ] 4×3
𝑢
𝐿L  x 
b
u 



2 22 −𝑏


 3
θy
θy
2 −
b
LIt is possible to reformuu4acceleration
The scenario is the same regarding the
vector.
late Equation (8) for previous transformations to achieve a 3 × 3 mass matrix. Previous
The scenario is the same regarding the acceleration vector. It is possible to reformulate
algebraic operations lead to a useful 3 × 3 mass matrix that is diagonal as:
Equation (8) for previous transformations to achieve a 3 × 3 mass matrix. Previous algebraic
1 0
0
4 that
2 is
1 diagonal
2
operations lead to a useful 3 × 3 mass matrix
as:
𝜌ℎ 𝐿𝑏 2 2 2 2 2 4 2 1
0
𝑏


0
 [𝑴] × = 144 2 −𝑏
4 2𝑏 𝑏
1 2−𝑏 1 2 4 2 [𝑻] × = 𝜌ℎ𝐿𝑏  12 𝐿 . (18)
1
0
0 0
2 2 2
2
−𝐿
−𝐿
𝐿
𝐿
2 4 2 1  2 1 2 4
ρh Lb 
b2


0 . 0 12 (18)
−b b b −b 
[ M ] 3×3 =
1 2 4 2 [T]4×3 = ρhLb 0
12
144 2 Consequently, the stiffness
L2 which contributes
matrix was formulated as0a 3×30matrix,
−L −L L L
12
2
1
2
4
to providing a simple and practical solution to the problem of natural frequencies. It is
remarkable
to note the
that,
in this matrix
orthogonalized
mass matrix,
is associated
with a
Consequently,
stiffness
was formulated
as a 3 ×a3mass
matrix,
which contributes
displacement
while
of inertia
associated
rotation
DOFs, It𝜃 is,
to providing aDOF,
simple
and moments
practical solution
to are
the problem
of with
natural
frequencies.
and
𝜃 .
remarkable
to note that, in this orthogonalized mass matrix, a mass is associated with a
displacement DOF, while moments of inertia are associated with rotation DOFs, θ x , and θy .
3.2. Modal Analysis
3.2. Modal Analysis
A modal analysis of a rigid, rectangular, solid plate with nodal-assigned elastic elements can be presented as a 4-node macro element with specific spring stiffness. This
example is not representative of a design solution for a bus chassis (there are no realistic
solution options in adopting a rigid plate as a passenger vehicle chassis base); however,
this can work as a dynamically equivalent model regarding the vehicle mass to the centre
of mass. In addition, inertial rotation elements can be considered for any of the coordinate
x-y axes passing the centre of mass, as shown in Figure 1. Polar moments of inertia, which
are the sum of two axial moments over y− and z− axes can take the following form:
•
Over longitudinal axis Jx = mtotal × b2 /12 = 7290 kg.m2 ;
this can work as a dynamically equivalent model regarding the vehicle mass to the centre
of mass. In addition, inertial rotation elements can be considered for any of the coordinate
x-y axes passing the centre of mass, as shown in Figure 1. Polar moments of inertia, which
are the sum of two axial moments over 𝑦 − and 𝑧 − axes can take the following form:
Appl. Sci. 2023, 13, 2127
•
•
Over longitudinal axis 𝐽 = 𝑚
× 𝑏 /12 = 7290 (𝑘𝑔. 𝑚 );
Over transverse axis 𝐽 = 𝑚
× 𝐿 /12 = 144 × 10 (𝑘𝑔. 𝑚 ).
8 of 19
2can be computed
3 kg.m
the eigenvalue
as: 2 .
• Therefore,
Over transverse
axis Jy = mequation
total × Lb /12 = 144 × 10
Therefore,
eigenvalue
0
0
4 ×the
185
0 equation can
0 be computed as:1
/12
0
0
𝑏




0
0
185
×
𝑏
− 𝜔 × 12
𝐷𝑒𝑡 1000
= 0;
4 × 185
0
0
1
0
0
0
0
185
×
𝐿
/12
0
0
𝐿
2
2
 − ω × 120 b2 /12
Det 1000 0
185 × bW
0
0  = 0; (19)
2
2
0
0
185 × L
0
0
Lb /12
(19)
The first three natural frequencies of the studied model can be derived from Equation
The first
three3natural
of the
studiedfrequencies
model can be
derived
Equation
(19). Hence,
Table
reportsfrequencies
the obtained
natural
from
the from
developed
analyti(19).
Hence,
Table
3
reports
the
obtained
natural
frequencies
from
the
developed
analytical
cal methodology identified as 𝑓
. In addition, the deformed shape of the studied model
ANL . In addition, the deformed shape of the studied model is
identified
is methodology
shown in Figure
5. as f i
shown in Figure 5.
(a)
(b)
(c)
Figure 5. Analytically obtained natural frequencies: (a) Mode 1, (b) Mode 2, and (c) Mode 3.
Appl. Sci. 2023, 13, 2127
9 of 19
Table 3. Analytical natural frequency solutions.
Natural Frequency, f(Hz)
Motion Description
f 1ANL = 1.06
f 2ANL = 1.25
f 3ANL = 2.00
Rotation around YG axis;
Vertical motion, vertical translation of the plate, four springs
simultaneously in function;
Rotation over XG axis;
3.3. Transient–Dynamic Formulation
As mentioned, the vehicle mass and inertial properties are approached by a simple
rectangular, solid plate element with spring suspension at each corner. Once it was possible
to obtain the solution of the modal analysis of the problem from a set of uncoupled
equations, the transient analysis would be obtained in a similar practical mode. The
dynamic equilibrium equation for the oscillating plate can be re-written as follows:

4k
0
0
0
4kb2 /4
0
 

0
4c
 uG 
+ 0
0  θx


4kL2 /4
θy
0
0
4cb2 /4
0
 . 
u 

 .G 

0
1

0  θ x +ρhLb0
 . 
 θy 

4cL2 /4 
0
0
b2 /12
0

 ..  
 uG 
0

 FuG (t) 
.. 
= M (t) .
0  θx
 ..   θx 
2
Mθ y ( t )
L /12  θy 
(20)
As previously mentioned, the RHS of this dynamic equilibrium must be transformed
to equivalent moments while dealing with nodal forces. As an illustration, having four
independent nodal forces acting transversely on the plate plane, the inverse of Equation
(1) produces the expression of the condensed displacement vector (composed of the mass
centre displacements and rotations about coordinate axes), therefore:
  
1
 uG 
=  −b/2
θx
 
θy
− L/2
1
−b/2
− L/2
 
  u1 
 

1 1

u2
.
b/2 b/2 
 u3 
 

L/2 − L/2 
u4
(21)
Then, the corresponding force vector, if composed of four time-dependent components,
will be:



 
 Fu1 (t)


1
1
1 1
 FuG (t) 


Fu2 (t)


Mθ x ( t )
= −b/2 −b/2
.
(22)
b/2 b/2


 Fu3 (t)

Mθ y ( t )


− L/2 − L/2 L/2 − L/2 
Fu4 (t)
Considering that there exists a force Fu1 acting at node 1, the equivalent (condensed)
force vector is represented in Equation (23). The complete Equation (21) can be extended to
dynamic situations of transient forces applied at nodes as:

 
1
 FuG (t) 
Mθ x ( t )
=  −b/2


Mθ y ( t )
− L/2
1
−b/2
− L/2



 Fu1 (t) 


1 1

  Fu1 (t) 
0
= − b F (t) .
b/2 b/2 
 0 
  L2 u1



L/2 − L/2 
− 2 Fu1 (t)
0
(23)
A practical and efficient mode to solve this equation for generalized time-dependent
loads relies on direct time integration methods. It could be developed from the mode
superposition technique, another way to setup the solution for this dynamic problem.
However, in the case of complex loads, where their decomposition into elementary modes
is necessary, direct time integration is a more practical tool, limited only by the computation
time in the case of very large DOF structure modelling. We adopted the Newmark’s constant
acceleration method [26] for its accuracy, stability, and ease to program for computational
purposes. An additional reason for this option is the fact that the solution is structured as
an uncoupled system of equations.
The dynamic equilibrium of a structure subjected to external time-dependent loads
can be analysed by a direct time integration method such as Newmark’s algorithm [26].This
Appl. Sci. 2023, 13, 2127
10 of 19
algorithm calculates the shape of a structure based on the concept of Constant Acceleration
Method, where at each time step the structure’s acceleration vector is supposed to remain
constant. To develop the algorithm, the structure equilibrium at an instant t + ∆t can be
formulated as follows:
..
.
[M]Ut+∆t + [c]Ut+∆t + [k]Ut+∆t = Ft+∆t ,
..
(24)
.
where Ut+∆t , Ut+∆t , and Ut+∆t are accordingly the acceleration, velocity, and displacement
vectors of the structure, while Ft+∆t is the external force vector, all at time step t + ∆t. If
velocity and displacement vectors are considered from the previous instant value, t, it
is feasible to obtain the structural evolution in the next time step under dynamic loads.
Thus, the constant acceleration concept and the trapezoidal rule in the integration of a
step-function can provide the following equations:
..
.
.
.. Ut+∆t = Ut + 0.5 ∆t Ut+∆t + Ut ,
.
(25)
. Ut+∆t = Ut + 0.5 ∆t Ut+∆t + Ut .
These equations are indeed applied to the plate rotations instead of translation. The
..
evaluation of Ut+∆t is the key solution of the algorithm; for that, previous expressions are
substituted in the structure equilibrium equation at the time: t + ∆t:
h.
..
h
.
..
.. i
. i
+ [k] Ut + 0.5 ∆t Ut+∆t + Ut
= Ft+∆t .
(26)
[M]Ut+∆t + [c] Ut + 0.5 ∆t Ut+∆t + Ut
On separating terms referred to time t + ∆t and t, respectively, at the left and righthand sides of the equation, it is possible to solve it for the updated structure acceleration
vector as:
.
..
.
.. .. (27)
[M] + [c]∆t + 0.25∆t2 [k] Ut+∆t = Ft+∆t − [k] Ut + Ut ∆t + 0.25∆t2 Ut − [c] Ut + 0.5∆tUt .
| {z }
{z
}
|
at t+∆t
at t
..
Once solved for Ut+∆t , knowing the previous dynamic status at instant t (that is the
RHS of the equation), an updated vector for acceleration is attained. Thus, it is leading to
new velocity and displacement vectors, as in Equation (25). The equations for each DOF
are uncoupled. The iterative incremental solutions can take the following form:
..
ρhLb + 4(c) × ∆t + 0.25∆t2 (4k ) UG t+∆t = Ft+∆t −
| {z }
at t
+∆t
.
.
..
..
(4k) UG t + UG t ∆t + 0.25∆t2 UG t − (4c) UG t + 0.5∆tUG t ,
|
{z
}
(28)
..
ρhLb3 /12 + 4cb2 /4 × ∆t + 0.25∆t2 4kb2 /4 θ x t+∆t = Mx t+∆t −
| {z }
at t+∆t
.
..
.
..
4kb2 /4 θ x t + θ x t ∆t + 0.25∆t2 θ x t − 4cb2 /4 θ x t + 0.5∆tθ x t ,
|
{z
}
(29)
..
θ y t+∆t = My t+∆t −
| {z }
at t+∆t
.
..
.
.. 2
2
2
4kb /4 θy t + θ y t ∆t + 0.25∆t θ y t − 4cb /4 θ y t + 0.5∆tθ y t .
|
{z
}
(30)
at t
at t
ρhLb3 /12 + 4cb2 /4 × ∆t + 0.25∆t2 4kb2 /4
at t
As an illustration, the analysed plate provided the largest natural frequency f 3 = 2.0 (Hz),
where t3 = 0.5 (s) and thus, ∆t ≤ 0.5 (s). With this value, the solution is warranted for its
Appl. Sci. 2023, 13, 2127
Appl. Sci. 2023, 13, x FOR PEER REVIEW
11 of 19
11 of 19
stability. Hence, the smaller the time step, the better the accuracy, although a value much
Regarding the natural frequency detected by Newmark’s method, one cycle = 1.02
smaller than the recommended one would be useless since the computational time would
(s);
therefore,
an analytical
evaluation
canaccelerates
lead to: during t = 0.5 s with no damping,
increase largely.
For instance,
if a vehicle
( )
an impulsive force is due to
the
load
on
the
front
axis
during
F
=
2
×
2.5 (kN) = 5.0 ((32)
kN)
𝑓 = 1⁄(2𝜋) × 𝑘⁄𝑚 = 1.25 (𝐻𝑧)
(hubs 1 and 4). Basic equation for centre of mass vibration with no damping is:
=
Referring to Figure 6, the maximum displacement of the mass centre is 𝑢
..
3
3
3
(𝑚)
1.23 × 10
at the
are
4 ×transient
185 × 10step.
uG +Concerning
12 × 10 uGthe
= rotation
5 × 10about 𝑦 − 𝑎𝑥𝑖𝑠 (forces(31)
| {z }
supposed to act at nodes 1 and 4), 2nd uncoupled equation:
dynamic
inertial
moment
over
( f or t ∈ [0, 0.5(s)])
transverse 𝑦 − 𝑎𝑥𝑖𝑠. The dimensions of the bus body chassis are listed below.
Regarding
the natural
•
Rotation
stiffness:
kL ;frequency detected by Newmark’s method, one cycle = 1.02 (s);
therefore,
an
analytical
evaluation
can lead(N/rad);
to:
•
Equivalent stiffness: 𝑘 = 6439850
√ );
•
Equivalent mass: 𝐽 = 144 × 10 (kg. m
f = 1/(2π ) × k/m = 1.25 (Hz)
(32)
•
Natural frequency of this oscillation: 𝑓 = 1⁄(2𝜋) × 𝑘 ⁄𝐽 = 1.065 (Hz).
•
Equivalent
forces:
Referring to
Figure 6, the maximum displacement of the mass centre is u Max =
𝐹
G
−2 (m) at1 the transient
1
1.23 ×𝐹 10(𝑡)
step.
1 1Concerning the rotation about y − axis (forces are
0
𝑀
(𝑡)
⁄2 −1𝑏⁄and
⁄2
2 4),
𝑏⁄2nd
2 𝑏uncoupled
= −
.
supposed to act
at𝑏nodes
dynamic inertial moment (33)
over
0 equation:
⁄2 bus
− 𝐿⁄2The−dimensions
𝐿⁄2 𝐿⁄2 −
𝑀 (𝑡)y − axis.
𝐹 body chassis are listed below.
transverse
of𝐿the
Figure 6. Displacement of mass centre, u𝑢G .
• Rotation
Rotation stiffness:
kL2 ;
𝜃 after applying
𝐹 = 2 × 2.5 (kN) suddenly (and sustainably) at nodes 1
• 4.
Equivalent
k Eqmoment
= 6439850
N/rad
); (kN. m).
and
Thus, the stiffness:
equivalent
is 𝑀(=
−30.0
3
2
• From
Equivalent
mass: plotted
Jy = 144in×Figure
10 kg.m
the graph
7, the; maximum
q rotation is obtained as 𝜃 =
−0.00929
(rad).
Hence, of
analytically,
true displacement
•
Natural
frequency
this oscillation:
f = 1/(2π ) ×at front
k Eq /Jof
= 1.065model
(Hz). (nodes 1
y vehicle
and
4):
•
Equivalent forces:


•
Maximum rotation
obtained

  as 𝜃 = −8.88 × 10 (rad); FAccel 


F
(
t
)
1 68.0
1 (mm).
 uG  𝑢 1 = 𝑢 1 − 𝜃 =


•
Maximum displacement:
0


Mθx (t) = −b/2 −b/2
.
(33)
b/2 b/2
 2 and 
 0 level,

At the rear (nodes
3), there will be a lowering of the pavement
thus:

Mθ y ( t )


− L/2 − L/2 L/2 − L/2 
FAccel
•
Minimum displacement: 𝑢
= 𝑢
+ 𝜃 = −43.4 (mm).
Rotation θy after applying F = 2 × 2.5 (kN) suddenly (and sustainably) at nodes 1
and 4. Thus, the equivalent moment is M = −30.0 (kN.m).
From the graph plotted in Figure 7, the maximum rotation is obtained as
θy = −0.00929 (rad). Hence, analytically, true displacement at front of vehicle model
(nodes 1 and 4):
•
•
Maximum rotation obtained as θy = −8.88 × 10−3 (rad);
L
Maximum displacement: umax
= umax
z
G − 2 θy = 68.0 (mm).
Appl. Sci. 2023, 13, x FOR PEER REVIEW
Appl. Sci. 2023, 13, 2127
12 of 19
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Figure 7. Rotation about y-axis, 𝜃 .
4. Finite Element Analysis
Computational simulations have been performed to confirm the analytical hypothesis presented in Section 3. In this regard, the FEM stands as an accurate methodology to
highlight the most critical aspects where the analytical approach faced difficulties. Regarding the numerical model, the vehicle chassis is mainly manufactured from steel, as its
geometry shows in Figure 8a.
Numerical modelling has been performed based on the FEM formulations impleFigure
about
Figure7.7.Rotation
Rotation
abouty-axis,
y-axis,
θ.y . problem was modelled according to the real conditions,
mented
in
ABAQUS©
[27]. 𝜃The
c.f. Figure 8. The vehicle chassis is modelled as a shell supported by four clamped springs
At Element
the the
rearvertical
(nodes
2 and 3), there will
be a lowering
of theinpavement
thus: the
4.simulating
Finite
Analysis
tyre/suspension
stiffness
as reported
[24,28,29].level,
To define
L
min
max
• Computational
Minimum
displacement:
uzhave
= been
upoints
+ (𝑃
43.4
).the
boundary
conditions,
eight reference
=−
1,to
…confirm
,8)(mm
were
assigned
to the
model.
simulations
analytical
hypotheG performed
2 θy, 𝑖 =
⁄m))
Therefore,
the
𝑘 = 185
were as
defined
through
pairing the desis
presented
insprings
Section(stiffness
3. In thisof
regard,
the(kN
FEM
stands
an accurate
methodology
to
4.
Finite
Element
Analysis
fined
points.
Consequently,
𝑃
,
…
𝑃
are
located
on
the
wheel
hubs,
representing
the
highlight the most critical aspects where the analytical approach faced difficulties. Respring
interaction,
thetheother
signify
the
restricted
at
the
Computational
simulations
have points
been
performed
to confirm
the displacement
analytical
hypothesis
garding
the
numericalwhile
model,
vehicle
chassis
is mainly
manufactured
from steel,
as its
spring’s
end,
defined
=𝑢
= 𝑢 the
= 0.FEM stands as an accurate methodology to highpresented
in Section
3.byIn𝑢8a.
this
regard,
geometry
shows
in Figure
light
the non-linear
most critical
aspects
theremain
analytical
approach
faced
Regarding
The
geometrical
effects
deactivated,
and
thedifficulties.
numerical
simulation
Numerical
modelling
has where
been
performed
based
on the
FEM
formulations
implethe been
numerical
model,
the
chassis
is
mainly
manufactured
from
steel,
as conditions,
its geometry
has
out through
a static
analysis.
mented
incarried
ABAQUS©
[27].vehicle
The
problem
was
modelled
according
to the
real
shows
in 8.
Figure
8a.
c.f.
Figure
The vehicle
chassis is modelled as a shell supported by four clamped springs
simulating the vertical tyre/suspension stiffness as reported in [24,28,29]. To define the
boundary conditions, eight reference points (𝑃 , 𝑖 = 1, … ,8) were assigned to the model.
Therefore, the springs (stiffness of 𝑘 = 185 (kN⁄m)) were defined through pairing the defined points. Consequently, 𝑃 , … 𝑃 are located on the wheel hubs, representing the
spring interaction, while the other points signify the restricted displacement at the
spring’s end, defined by 𝑢 = 𝑢 = 𝑢 = 0.
The non-linear geometrical effects remain deactivated, and the numerical simulation
has been carried out through a static analysis.
(a)
(b)
Figure 8.
8. Numerical
Numericalmodel;
model;(a)
(a)geometrical
geometricalperspective
perspectiveand
and(b)
(b) FE
FE mesh,
mesh, interactions
interactions and
and essential
essential
Figure
boundary condition
condition presentation.
presentation.
boundary
Numerical
modelling
has
FEM formulations
implewasthe
considered
to obtain the
real
An
equivalent
density (𝜌
= been
3.7 × performed
10 tonne⁄based
mm ) on
mented
in
ABAQUS©
[27].
The
problem
was
modelled
according
to
the
real
conditions,
mass of the vehicle (𝑚
= 12 tonne).
c.f. Figure 8. The vehicle chassis is modelled as a shell supported by four clamped springs
simulating
vertical tyre/suspension
stiffness as reported in [24,28,29]. To define the
4.1.
FE Meshthe
Convergence
Study
boundary conditions, eight reference points ( Pi , i = 1, . . . , 8) were assigned to the model.
(a)Therefore, the springs (stiffness of k = 185 (kN/m)) were
(b) defined through pairing the
defined points. Consequently, P1 , . . . P4 are located on the wheel hubs, representing the
Figure
8. interaction,
Numerical model;
perspective
andrestricted
(b) FE mesh,
interactionsat
and
spring
while(a)
thegeometrical
other points
signify the
displacement
theessential
spring’s
boundary condition presentation.
end, defined by u x = uy = uz = 0.
The non-linear geometrical effects remain deactivated, and the numerical simulation
An equivalent density (𝜌 = 3.7 × 10 tonne⁄mm ) was considered to obtain the real
has been carried out through a static analysis.
mass of the vehicle (𝑚
= 12 tonne).
An equivalent density (ρ = 3.7 × 10−7 tonne/mm3 ) was considered to obtain the real
mass of the vehicle (m
= 12 tonne).
4.1. FE Mesh Convergencetotal
Study
Appl. Sci. 2023, 13, 2127
13 of 19
4.1. FE Mesh Convergence Study
This section is devoted to a convergence study carried out on the FE mesh to assess
how the mesh would affect the results and particularly to determine the optimal mesh
density. The chassis is considered a plate on which shell elements can be employed to
analyse the problem. Different shell element types were studied: quadratic elements
with/without reduced integration for four and eight nodes identified as S4R, S4, and S8R.
To achieve the optimal mesh, the model presented in Figure 8 was analysed. Regarding
the boundary condition, a vertical displacement as V ( x, y) = 10 (mm) was imposed at
Node 1. The numerical simulations have been performed on different FE mesh types with
an element size of 200 (mm), which is reported in Table 4. However, Table 5 presents a
summary of the obtained displacement along the z-direction. As a conclusion, different
element types do not have a significant impact on the results. Based on the findings of the
FE convergence investigation, it can be concluded that the S4R element type would be the
best option for ensuring solution correctness and minimizing computing costs.
Table 4. FE mesh properties for different element types.
Element Type
S4R
S4
S8R
Number of elements
Number of nodes
840
915
840
915
840
2669
Table 5. Displacement along z-direction for different FE types.
Displacement Imposition at Node 1
Corner
10 (mm)
1
2
3
4
uz (mm)
S4R
S4
S8R
10.00
3.33
−3.33
3.33
10.00
3.33
−3.33
3.33
10.00
3.33
−3.33
3.33
To study the FE mesh size, various element sizes were considered, as listed in Table 6.
Table 6. FE mesh properties for the element type of S4R.
FE Mesh
Coarse
Medium
Fine
Element size (mm )
Number of elements
Number of nodes
400
217
256
200
840
915
100
3213
3360
Maintaining the same boundary conditions mentioned before and considering the
different FE meshes reported in Table 6, the problem was numerically solved, and the
displacement variation over z-direction was obtained for all nodes as reported in Table 7.
Table 7. Displacement along z-direction for different mesh densities of S4R.
Displacement Imposition at Node 1
Corner
10 (mm)
1
2
3
4
uz (mm)
Coarse
Medium
Fine
10.00
3.32
−3.32
3.34
10.00
3.33
−3.33
3.33
10.00
3.33
−3.33
3.33
Owing to the results obtained from various FE densities, it can be inferred that medium
mesh density produced accurate results while minimizing computational costs and increasing convergence.
Appl. Sci. 2023, 13, x FOR PEER REVIEW
14 of 19
Appl. Sci. 2023, 13, 2127
14 of 19
4.2. Static Analysis
4.2.An
Static
Analysis
external
load impacting the connection ground—vehicle is generated by modelAn external
impacting
connection
ground—vehicle
is generated
modelling
ling a ride
up of 10load
(mm)
or morethe
sharp
road imperfections,
for instance
100 by
(mm).
This
a ride up
of 10 (mm) is
orimposed
more sharp
imperfections,
for instance
100 (mm
. This vertical
vertical
displacement
on road
the node
1 as demonstrated
in Figure
8.)Figure
9 dedisplacement
is imposed
1 as demonstrated
in Figure
8. FigureReferring
9 depicts to
the
picts
the displacement
fieldon
inthe
thenode
direction
of the displacement
imposition.
displacement
field
in
the
direction
of
the
displacement
imposition.
Referring
to
Figure
Figure 9a, it is notable that the maximum displacement was reached where the displace-9a,
it is was
notable
that the
maximum
displacement
was(Node
reached
where
the4)displacement
ment
imposed,
while
the other
two neighbours
2 and
Node
tend to movewas
in
imposed,
while
the
other
two
neighbours
(Node
2
and
Node
4)
tend
to
move
the same direction as Node 1. Regarding Node 3, the displacement is moving in
in the
the same
opdirection
as Node
Regarding
Node
3, 9b,
the itdisplacement
is that
moving
opposite
posite
direction
of the1.other
nodes. In
Figure
can be inferred
thereinis the
a tendency
direction
the other
nodes.
In Figure
9b, it can
be inferred
that there
is a8 tendency
for
for
rotationof
along
a parallel
diagonal
connecting
Node
2 and Node
4. Table
reports the
rotation
along
a
parallel
diagonal
connecting
Node
2
and
Node
4.
Table
8
reports
the
obtained displacement on different corners.
obtained displacement on different corners.
(a)
(b)
Figure 9. Displacement field distribution in mm: (a) along z-direction and (b) magnitude.
Figure 9. Displacement field distribution in mm: (a) along z-direction and (b) magnitude.
Table
Table8.8.Displacement
Displacementalong
alongz-direction
z-directionobtained
obtainedfor
forboth
bothconstraints.
constraints.
Displacement
DisplacementImposition
Imposition
Corner
Corner
1
1
22
33
44
𝟏𝟎
𝟏𝟎𝟎
10𝐦𝐦
mm
100𝐦𝐦
mm
Displacement
Displacementininz-direction
z-direction ((mm)
mm)
10.00 × 10 −3
10.00 × 10 −2
10.00 × 10
10.00 × 10
3.33
×
10
3.33
−
3
3.33 × 10
3.33××10
10−2
−
3
3.33
3.33
3.33××10
10
3.33××10
10−2
−
3
3.33××10
10
3.33××10
10−2
3.33
3.33
An
Anexternal
externalload
loadthat
thatacts
actsunexpectedly
unexpectedlyononthe
thevehicle
vehiclecan
cancause
causethe
theexcitation
excitationofofthe
the
body
step
into
resonance.
it it
is is
rational
toto
study
the
natural
excitation
frequencies
bodytoto
step
into
resonance.Hence,
Hence,
rational
study
the
natural
excitation
frequencies
ofofthe
themodel,
model,which
whichare
arediscussed
discussedininthe
thenext
nextsection.
section.
4.3.Frequency
FrequencyAnalysis
Analysis
4.3.
Thissection
sectionisisdevoted
devotedtotothe
the
numericalanalysis
analysisofof
the
example
that
was
analytically
This
numerical
the
example
that
was
analytically
solvedininSection
Section3.2.
3.2.The
Theproblem
problemwas
wasresolved
resolvedusing
usinga aFrequency
Frequencystep
steptotocompute
computethe
the
solved
natural
frequencies
following
Lanczos
eigensolver
[30].
The
FEM
analysis
was
carried
out
natural frequencies following Lanczos eigensolver [30]. The FEM analysis was carried out
excludingthe
theacoustic
acousticstructural
structural coupling
coupling due
due to
excluding
to the
the problem
problemconditions.
conditions.The
Theeigenvectors
eigenvecwere
normalized
by
the
displacement.
The
displacement
normalization
is
tors were normalized by the displacement. The displacement normalization isaatechnique
techniqueto
torepresent
representmode
modeshapes
shapesin
inaamodal
modalanalysis
analysiswhere
wherethe
the peak
peak amplitude
amplitude is
is normalized
normalized to
to a
1. This
This is
is aa common
commonmethod
methodto
tosignify
signifymode
modeshapes
shapesnot
notonly
onlyusing
usingcommercial
commercial
a value
value of
of 1.
toolsbut
butalso
alsofor
forgeneral
generalpurposes
purposes while
while representing
representing an
tools
an analytical
analyticalsolution.
solution.The
Theeigenveceigentor
normalization
allows
scaling
up
or
down
the
mode
shape
amplitude
and
hence
does
vector normalization allows scaling up or down the mode shape amplitude and
hence
not
affect
frequencies.
does not affect frequencies.
Conforming to the model shown in Figure 8b, the frequency analysis has been performed, requesting the first six eigenvalues. The first three modes are zero due to the
representation of rigid body movement. Table 9 summarizes the obtained results on
numerical natural frequencies identified as f iFEM .
Appl. Sci. 2023, 13, 2127
Conforming to the model shown in Figure 8b, the frequency analysis has been performed, requesting the first six eigenvalues. The first three modes are zero due to the representation of rigid body movement. Table 9 summarizes the obtained results on numerical natural frequencies identified as 𝑓
.
15 of 19
Regarding the second harmonic mode, it reflects a rotation about the x-axis according
to the coordinate system presented in Figure 8. The suspension system is established by
four springs, which vertically vibrate with a 180-degree phase shift. It must be mentioned
Table 9. Numerical natural frequencies.
that the third harmonic mode accounted for a higher frequency due to a rotation that ocfFEM
(Hz)
fFEM
(Hz)to state that the
curred alongfFEM
diagonal of the corners.
a conclusion, it is possible
1the(Hz)
2 As
3
natural frequencies
are
independent
of
the
displacement
imposed.
1.07
1.25
2.17
Table 9. Numerical natural frequencies.
Regarding the second harmonic mode, it reflects a rotation about the x-axis according
𝑭𝑬𝑴
𝑭𝑬𝑴
to the
system presented𝒇in
Figure
is𝑭𝑬𝑴
established
(𝐇𝐳) by
𝒇𝟏 coordinate
(𝐇𝐳)
(𝐇𝐳) 8. The suspension system 𝒇
𝟑
𝟐
four springs, which vertically vibrate with a 180-degree phase shift. It must be mentioned
1.07
1.25
2.17
that the third harmonic mode accounted for a higher frequency due to a rotation that
occurred along the diagonal of the corners. As a conclusion, it is possible to state that the
Figure
10 showsarethe
magnitude
profile
obtained on the second and
natural
frequencies
independent
of displacement
the displacement
imposed.
(mm) on
third natural
the imposed
displacement
is 10
onthe
Node
1. As
shown
Figure frequencies
10 shows theifmagnitude
displacement
profile
obtained
second
and
in Figure
10a, the
suspension
compresses
and extends
vertically,
third
frequency
third natural
frequencies
if the
imposed displacement
is 10
Node the
1. As
shown
in
(mm) onwhile
Figure 10a,athe
suspension
anddirection
extends vertically,
while
the third
frequency
experienced
rotation
alongcompresses
the diagonal
as depicted
in Figure
10b.
This behavexperienced
rotationfor
along
the diagonal
direction conditions.
as depicted in
Figure 10b.
This
be- to
iour
should bea similar
different
displacement
Therefore,
it is
feasible
haviour
should
be
similar
for
different
displacement
conditions.
Therefore,
it
is
feasible
to
predict the shape of the structural response of the model if subjected to such displacement
predict the shape of the structural response of the model if subjected to such displacement
imposition.
imposition.
(a)
(b)
Figure 10. Amplitude displacement distribution (in mm): (a) 2nd and (b) 3rd natural frequencies. If
Figure 10. Amplitude displacement distribution (in mm): (a) 2nd and (b) 3rd natural frequencies. If
the displacement imposition is 10 mm on Node 1.
the displacement imposition is 10 mm on Node 1.
4.4. Steady-State Dynamic Analysis
4.4. Steady-State
Dynamic
Analysis
This section
deals with
the numerical analysis of the problem that was analytically
solved
Sectiondeals
3.3. ABAQUS/Standard
provides
a “direct”
linear
This in
section
with the numerical
analysis
of the steady-state
problem that
wasdynamic
analytically
analysis
procedure
in
which
the
equations
of
steady
harmonic
motion
of
the
are
solved in Section 3.3. ABAQUS/Standard provides a “direct” steady-state system
linear dynamic
solvedprocedure
directly without
usingthe
the equations
eigenmodes.
theharmonic
model wasmotion
simulated
following
analysis
in which
of Thus,
steady
of the
system are
a steady-state dynamic, direct step in which two concentrated forces were applied to the
solved directly without using the eigenmodes. Thus, the model was simulated following
front side of the model (Node 1 and Node 2) as shown in Figure 11a.
a steady-state
dynamic, direct step in which two concentrated forces were applied to the
The direct solution scheme solves the complete set of nodal DOFs at each excitation
front
side
of
the
model
(Nodeexcitation
1 and Node
as shown
inthe
Figure
11a. can be used to
frequency, and the
frequency
is not2)required
since
eigenmodes
The direct
solutionfrequency.
scheme solves
the complete
of anodal
DOFs
at each
excitation
identify
the excitation
It is rather
expensive set
since
complex
solution
must
be
frequency,
thefull
frequency
since
the eigenmodes
can be used
obtained and
for the
system ofexcitation
equationsis
atnot
eachrequired
excitation
(driving)
frequency. Applied
forces canthe
be defined
as real
(in-phase)
imaginary
(out-of-phase)
andsolution
non-zero must
to identify
excitation
frequency.
It and
is rather
expensive
since a loads,
complex
boundary for
conditions
be applied
to any at
of each
the active
DOF. (driving) frequency. Applied
be obtained
the fullmay
system
of equations
excitation
However,
due
to
the
nature
of
the
problem,
the
system
is considered loads,
an in-phase
one.
forces can be defined as real (in-phase) and imaginary (out-of-phase)
and non-zero
Therefore, the applied force has been defined as a real component with a magnitude of
boundary conditions may be applied to any of the active DOF.
Fz = 2.5 (kN) on each node, as it was prescribed in the analytical model. The analysis has
However, due to the nature of the problem, the system is considered an in-phase one.
been conducted in the requested frequency range of f 1 = 1.07 to f 3 = 2.17 (Hz). Figure 11b
Therefore,
the applied
force hasvariation
been defined
as a real component
a magnitude
of
demonstrates
the displacement
in the z-direction
at a frequencywith
of 1.07
(Hz).
𝐹 = 2.5 (kN) on each node, as it was prescribed in the analytical model. The analysis has
Appl. Sci. 2023, 13, x FOR PEER REVIEW
Appl. Sci. 2023, 13, 2127
16 of 19
been conducted in the requested frequency range of 𝑓 = 1.07 to 𝑓 = 2.17 (Hz). Figure
19 (Hz).
11b demonstrates the displacement variation in the z-direction at a frequency16ofof1.07
(a)
(b)
Figure
Steady-state
dynamic
analysis
by FEM;
(a) natural
boundary
condition
(b) displaceFigure
11. 11.
Steady-state
dynamic
analysis
by FEM;
(a) natural
boundary
condition
and (b)and
displacement
in
z-direction
at
a
frequency
of
1.07
(Hz).
ment in z-direction at a frequency of 1.07 (Hz).
4.5.4.5.
Results
Discussion
Results
Discussion
Table
10 10
reports
a comparison
between
thethe
results
acquired
from
thethe
developed
Table
reports
a comparison
between
results
acquired
from
developed ananalytical
solution
and
the
numerical
FEM.
A
reasonable
agreement
has
been
found, allowalytical solution and the numerical FEM. A reasonable agreement has been found,
allowing
the
FEM
model
to
confirm
the
proposed
analytical
methodology.
From
ing the FEM model to confirm the proposed analytical methodology. From thethe
acquired
acquired frequency data, it can be stated that the frequency increased for higher modes
frequency data, it can be stated that the frequency increased for higher modes in the anain the analytical solution since a higher vibrational deformation energy is associated with
lytical solution since a higher vibrational deformation energy is associated with increasing
increasing natural frequencies, as confirmed by the FEM results. Consequently, it can be
natural frequencies, as confirmed by the FEM results. Consequently, it can be inferred that
inferred that a low deviation for f 1 and f 2 has been obtained, and while it is a bit higher
deviation for 𝑓 and 𝑓 has been obtained, and while it is a bit higher for 𝑓 , it
foraf low
3 , it remains below 8%.
Study
Frequency analysis
Steady state
dynamic analysis
remains below 8%.
steady-state
analysis,
thedeveloped
displacements
areapproaches.
associated with
Table 10.Regarding
Comparisonthe
amongst
the resultsdynamic
obtained from
FEM and
analytical
the intensity of the natural vibrational mode. The external force system is superimposed
Results
Developed
Analytical
Solutionto displacement
FEM
Deviationof
*%
with the internal
inertial
forces, leading
as a consequence
the dynamic
f 1equilibrium
1.06
1.07
0.93
(Hz)
regime of the structure. Thus, the maximum and minimum displacement are
f 2not
1.25
0.00
(Hz)equal in intensity and1.25
sign due to the superposition
of a uniform transverse
displacef 3 (Hz)
2.00
2.17
7.80
ment and a rotation about the x-axis.
umax
68.0
70.6
3.68
z (mm)
−43.4 the results obtained−40.6
6.89
umin
(mm) 10. Comparison amongst
z Table
from FEM and developed analytical ap
* 100
× Result Analytical − Result FEM /Result FEM .
proaches.
Study
Frequency analysis
Developed Analytical Solu-
Regarding
with the
Results the steady-state dynamic analysis, the displacements
FEM are associated
Deviation
* %
intensity of the natural vibrationaltion
mode. The external force system is superimposed with
𝑓 (𝐻𝑧)inertial forces, leading1.06
1.07
0.93
the internal
to displacement as a consequence
of the dynamic
equilibrium regime of the structure. Thus, the maximum and minimum displacement are not
Appl. Sci. 2023, 13, 2127
17 of 19
equal in intensity and sign due to the superposition of a uniform transverse displacement
and a rotation about the x-axis.
5. Conclusions
In this work, a single modular rectangular element was developed to accurately
characterise the dynamic behaviour of a bus vehicle model with a passive suspension
subjected to generalised transient forces. A mathematical set of equations was developed
based on the modal analysis hypothesis to offer an analytical solution for natural frequency
calculation. The obtained results infer that the performance of the proposed methodology
is accurate in static and steady-state dynamics applications. However, the main conclusions
of the present study can be drawn as follows:
•
•
•
•
•
The most valuable practical implication is a comprehensive vehicle suspension diagnostic system concept, from which a simplified model helps to produce outputs
with sufficient accuracy in which an approximate structural behaviour of the physical
model can be predicted;
Only the chassis-induced transverse displacement vibrational modes allow the direct
comparison in which the basic model would be cross-checked;
In this paper, by means of simulation, the analytical control algorithm is used to evaluate the dynamic equilibrium of the structure, subjected to time-dependent domain
loads, at any instant. Thus, the obtained results revealed a deviation of less than 8%
about the displacement, allowing the model confirmation. This may provide a useful
reference for further experimental investigations.
The analytical solution relies on a single finite element demonstrating a coherent
dynamic response under transverse forces. It can practically model a less smooth track
in addition to the presence of in-plane transient forces, such as as those due to sharp
manoeuvres or the effect of side gusts;
A quite low number of DOFs and its accuracy make sound arguments for the applicability of this single modular element to automotive engineering in the fields of driving
practice and suspension turning for dynamic safety.
In the future development of this single chassis block-like modulus, the incorporation
of rotating inertial elements in the mass matrix and the steady state dynamic analysis
integrating damping will result in an improved modal analysis and dynamic vehicle
behaviour when subjected to external loads. The presentation of this structural dynamic
modelling could be applied to the modal analysis of more elaborate construction vehicles
in the future.
Author Contributions: Conceptualization, F.Q.d.M. and R.L.; methodology, F.Q.d.M.; software, B.V.F.
and R.L.; validation, B.V.F. and R.L.; formal analysis, R.L.; investigation, B.V.F. and R.L.; writing—
original draft preparation, B.V.F. and R.L.; writing—review and editing, B.V.F., R.L., and P.M.G.P.M.;
visualization, B.V.F. and R.L.; supervision, P.M.G.P.M. All authors have read and agreed to the
published version of the manuscript.
Funding: This research was funded by Fundação para a Ciência e a Tecnologia (FCT), Portugal, under
grant number 2022.13216.BD.
Informed Consent Statement: Not applicable.
Data Availability Statement: The data presented in this study are available on request from the
corresponding authors. The data are not publicly available due to their containing information that
could compromise the privacy of research participants.
Conflicts of Interest: The authors declare no conflict of interest.
Appl. Sci. 2023, 13, 2127
18 of 19
Nomenclature
DOF
FEM
FE
b
bW
F
f
G
J
k
k Eq
L
Lb
m
M
M
θ
N
R
t
U
UElastic
U Inertia
u
uG
V
W
ρ
Degrees of freedom
Finite element method
Finite element
Width
Track width
Force
Natural frequency
Mass centre
Moment of inertia
Stiffness
Equivalent stiffness
Wheelbase
Length
Mass
Momentum
Mass matrix
Rotation
Nodal shape function
Reaction force
Time
Displacement vector
Elastic energy
Inertial work
Displacement
Transverse displacement of the mass centre
Vertical displacement
Work
Mass density
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