Ch.02 Modeling of Vibratory Systems
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Transcript of Ch.02 Modeling of Vibratory Systems
2/2/2014
1
02. Modeling of Vibratory
System
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Vibrations 2.01 Modeling of Vibratory Systems
Chapter Objectives
โข Compute the mass moment of inertia of rotational systems
โข Determine the stiffness of various linear and nonlinear elastic
components in translation and torsion and the equivalent
stiffness when many individual linear components are
combined
โข Determine the stiffness of fluid, gas, and pendulum elements
โข Determine the potential energy of stiffness elements
โข Determine the damping for systems that have different sources
of dissipation: viscosity, dry friction, fluid, and material
โข Construct models of vibratory systems
Vibrations 2.02 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง1.Introduction
- Three elements that comprise a vibrating system
โข Inertia elements: stores and releases kinetic energy
โข Stiffness elements: stores and releases potential energy
โข Dissipation elements: express energy loss in a system
- Components comprising a vibrating mechanical system
โข Translational motion โข Rotational motion
Mass, ๐ (๐๐) Mass moment of inertia, ๐ฝ (๐๐๐2)
Stiffness, ๐ (๐/๐) Stiffness, ๐๐ก (๐๐/๐๐๐)
Damping, ๐ (๐๐ /๐) Damping, ๐๐ก (๐๐๐ /๐๐๐)
External force, ๐น (๐) External moment, ๐ (๐๐)
Vibrations 2.03 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง2.Inertia Elements
- Translational motion ๐
- Rotational motion
Slender bar ๐ฝ๐บ =1
12๐๐ฟ2
Circular disk ๐ฝ๐บ =1
2๐๐ 2
Sphere ๐ฝ๐บ =2
5๐๐ 2
Circular cylinder ๐ฝ๐ฅ = ๐ฝ๐ฆ =1
12๐(3๐ 2 + โ2)
๐ฝ๐ง =1
2๐๐ 2
๐ฝ๐ = ๐ฝ๐บ + ๐๐2, ๐: distance from the center of gravity to point ๐บ
Vibrations 2.04 Modeling of Vibratory Systems
- Parallel-axes theorem
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง2.Inertia Elements
- For a mass ๐ translating with a
velocity of magnitude ๐ฅ in the ๐โ๐
plane under the driving force ๐น
โข The equation governing the motion of the mass ๐
๐น ๐ =๐
๐๐ก(๐ ๐ฅ ๐)
when ๐ and ๐ are independent of time
๐น = ๐ ๐ฅ (2.2)
โข The kinetic energy, ๐, of mass ๐
๐ =1
2๐ ๐ฅ ๐ โ ๐ฅ ๐ =
1
2๐ ๐ฅ2
Vibrations 2.05 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง2.Inertia Elements
- For a rigid body undergoing only
rotation in the plane with an angular
speed ๐
โข The equation governing the rotation of the mass of inertia
๐ = ๐ฝ ๐ (2.6)
๐ : the moment acting about the center of mass ๐บ or a
fixed point ๐ along the direction normal to the plane of
motion
๐ฝ : the associated mass moment of inertia
โข The kinetic energy of the system
๐ =1
2๐ฝ ๐2
Vibrations 2.06 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
2
ยง2.Inertia Elements
- Ex.2.1 Determination of mass moments of inertia
Illustrate how the mass moments of inertia of several different
rigid body distributions are determined
Solution
โข Uniform Disk
The mass moment of inertia about the point ๐ ,
which is located a distance ๐ from point ๐บ
๐ฝ๐ = ๐ฝ๐บ + ๐๐ 2 =1
2๐๐ 2 + ๐๐ 2 =
3
2๐๐ 2
โข Uniform Bar
The mass moment of inertia about the point ๐
๐ฝ๐ = ๐ฝ๐บ + ๐๐ฟ
2
2
=1
12๐๐ฟ2 +
1
4๐๐ฟ2 =
1
3๐๐ฟ2
Vibrations 2.07 Modeling of Vibratory Systems
๐ฝ๐บ =1
2๐๐ 2
๐ฝ๐บ =1
12๐๐ฟ2
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง2.Inertia Elements
- Ex.2.2 Slider mechanism: system with varying inertia property
A slider of mass ๐๐ slides along a uniform bar of
mass ๐๐ with a pivot at point ๐. Another bar,
which is pivoted at point ๐โฒ, has a portion of length
๐ that has a mass ๐๐ and another portion of
length ๐ that has a mass ๐๐. Determine the rotary
inertia ๐ฝ๐ of this system and show its dependence
on the angular displacement coordinate ๐
From geometry, ๐, ๐๐, ๐๐ can be described in terms of ๐
๐2 ๐ = ๐2 + ๐2 โ 2๐๐๐๐๐ ๐
๐๐2 ๐ = (๐/2)2+๐2 โ ๐๐๐๐๐ ๐
๐๐2 ๐ = (๐/2)2+๐2 โ ๐๐๐๐๐ (๐ โ ๐)
๐๐ : the distance from the midpoint of bar of mass ๐๐ to ๐
๐๐ : the distance from the midpoint of bar of mass ๐๐ to ๐
Vibrations 2.08 Modeling of Vibratory Systems
Solution
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง2.Inertia Elements
๐2 ๐ = ๐2 + ๐2 โ 2๐๐๐๐๐ ๐
๐๐2 ๐ = (๐/2)2+๐2 โ ๐๐๐๐๐ ๐
๐๐2 ๐ = (๐/2)2+๐2 โ ๐๐๐๐๐ (๐ โ ๐)
The rotary inertia ๐ฝ๐ of this system
๐ฝ๐ = ๐ฝ๐๐+ ๐ฝ๐๐
(๐) + ๐ฝ๐๐(๐) + ๐ฝ๐๐
(๐)
where
๐ฝ๐๐=
1
3๐๐๐
2
๐ฝ๐๐ (๐) = ๐๐ ๐
2(๐)
๐ฝ๐๐๐ = ๐๐
๐2
12+ ๐๐๐๐
2 = ๐๐
๐2
3+ ๐2 โ ๐๐๐๐๐ ๐
๐ฝ๐๐๐ = ๐๐
๐2
12+ ๐๐๐๐
2 = ๐๐
๐2
3+ ๐2 โ ๐๐๐๐๐ (๐ โ ๐)
Vibrations 2.09 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
1.Introduction
- Stiffness elements are manufactured from different materials
and they have many different shapes
- Application
โข to minimize vibration transmission from machinery to the
supporting structure
โข to isolate a building from earthquakes
โข to absorb energy from systems subjected to impacts
Vibrations 2.10 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Some representative types of stiffness elements that are
commercially available along with their typical application
Vibrations 2.11 Modeling of Vibratory Systems
Building or highway
base isolation for
lateral motion using
cylindrical rubber
bearings
Wire rope isolators to isolate vertical motions of machinery
Steel cable
springs
used in a
chimney
tuned
mass
damper to
suppress
lateral
motions
Air springs used in
suspension systems to
isolate vertical motions
Typical steel
coil springs
for isolation
of vertical
motions
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- The stiffness elements store and release the potential energy
of a system
- Consider a spring under acting force of magnitude ๐น is
directed along the direction of the unit vector ๐
๐น๐ = โ๐น ๐
๐น๐ tries to restore the stiffness element to its undeformed
configuration, it is referred to as a restoring force
Vibrations 2.12 Modeling of Vibratory Systems
Stiffness element with a force acting on it Free-body diagram
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
3
ยง3.Stiffness Elements
- As the stiffness element is deformed, energy is stored in this
element, and as the stiffness element is undeformed, energy
is released
- The potential energy ๐ is defined as the work done to take
the stiffness element from the deformed position to the
undeformed position; that is, the work needed to undeform
the element to its original shape
๐ ๐ฅ = ๐๐๐๐๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐
๐๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๐ ๐๐๐ ๐๐ก๐๐๐
๐น๐ ๐๐ฅ
=
๐ฅ
0
๐น๐ ๐๐ฅ =
๐ฅ
0
โ๐น ๐ โ ๐๐ฅ ๐ =
๐ฅ
0
๐น๐๐ฅ
Vibrations 2.13 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
2.Linear Springs
- Translation Spring
โข Deformation
๐น ๐ฅ = ๐๐ฅ (2.9)
๐น : the applied force
๐ : the spring constant
๐ฅ : the spring deflection
โข The potential energy ๐ stored in the spring
๐ ๐ฅ =
0
๐ฅ
๐น ๐ฅ ๐๐ฅ =
0
๐ฅ
๐๐ฅ๐๐ฅ = ๐
0
๐ฅ
๐ฅ๐๐ฅ =1
2๐๐ฅ2
Vibrations 2.14 Modeling of Vibratory Systems
(2.10)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Torsion Spring
โข Deformation
๐ ๐ = ๐๐ก๐ (2.11)
๐ : the applied moment
๐๐ก : the spring constant
๐ : the spring deflection
โข The potential energy ๐ stored in the spring
๐ ๐ =
0
๐
๐ ๐ ๐๐ =
0
๐
๐๐ก๐๐๐ =1
2๐๐ก๐
2
Vibrations 2.15 Modeling of Vibratory Systems
(2.12)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Combinations of Linear Springs
โข Parallel Springs
Translation springs
Total force
๐น ๐ฅ =๐น1 ๐ฅ +๐น1 ๐ฅ =๐1๐ฅ+๐2๐ฅ= ๐1 +๐2 ๐ฅ =๐๐๐ฅ
Equivalent spring
๐๐ = ๐1 + ๐2
Torsion springs
Total moment
๐ ๐ = ๐1 ๐ + ๐2 ๐
= ๐๐ก1๐ + ๐๐ก2
๐ = ๐๐ก1+ ๐๐ก2
๐ = ๐๐ก๐๐
Equivalent spring
๐๐ก๐= ๐๐ก1
+ ๐๐ก2
Vibrations 2.16 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข Series Springs
Translation springs
๐ฅ = ๐ฅ1 + ๐ฅ2 =๐น
๐1+
๐น
๐2=
1
๐1+
1
๐2๐น =
๐น
๐๐
Equivalent spring
๐๐ =1
๐1+
1
๐2
โ1
=๐1๐2
๐1 + ๐2
Torsion springs
๐ = ๐1 + ๐2 =๐
๐๐ก1
+๐
๐๐ก2
=1
๐๐ก1
+1
๐๐ก2
๐ =๐
๐๐ก๐
Equivalent spring
๐๐ก๐=
1
๐๐ก1
+1
๐๐ก2
โ1
=๐๐ก1
๐๐ก2
๐๐ก1+ ๐๐ก2
Vibrations 2.17 Modeling of Vibratory Systems
Displacement
Displacement
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Spring Constants for Some Common Elastic Elements
1. Axially loaded rod or cable
๐ =๐ด๐ธ
๐ฟ
๐ด : cross-sectional area, ๐2
๐ธ : Youngโs modulus of elasticity, ๐/๐2
๐ฟ : length of the rod, ๐
2. Axially loaded tapered rod
๐ =๐๐ธ๐1๐2
4๐ฟ
๐ธ : Youngโs modulus of elasticity, ๐/๐2
๐1 : rod diameter, ๐
๐2 : rod diameter, ๐
๐ฟ : length of the rod, ๐
Vibrations 2.18 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
4
ยง3.Stiffness Elements
3. Hollow circular rod in torsion
๐๐ก =๐บ๐ผ
๐ฟ
๐บ : shear modulus of elasticity, ๐/๐2
๐ผ : the torsion constant (polar moment of inertia), ๐4
For the concentric circular tubes,
๐ฟ : length of the rod, ๐
๐๐ : outside rod diameter, ๐
๐๐ : inside rod diameter, ๐
4. Cantilever beam
๐ =3๐ธ๐ผ
๐3, 0 < ๐ โค ๐ฟ
๐ธ : Youngโs modulus of elasticity, ๐/๐2
๐ผ : the area moment of inertia about the bending axis, ๐4
๐ : position of applied force, ๐
๐ฟ : length of the beam, ๐
Vibrations 2.19 Modeling of Vibratory Systems
๐ผ =๐(๐๐
4 โ ๐๐4)
32
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
5. Pinned-pinned beam (Hinged, simply supported)
๐ =3๐ธ๐ผ(๐ + ๐)
๐2๐2
๐ธ : Youngโs modulus of elasticity, ๐/๐2
๐ผ : the area moment of inertia about the bending axis, ๐4
๐,๐: position of applied force, ๐
6. Clamped-clamped beam (Fixed-fixed beam)
๐ =3๐ธ๐ผ(๐ + ๐)3
๐3๐3
๐ธ : Youngโs modulus of elasticity, ๐/๐2
๐ผ : the area moment of inertia about the bending axis, ๐4
๐,๐: position of applied force, ๐
Vibrations 2.20 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
7. Two circular rods in torsion
๐๐ก๐= ๐๐ก1
+ ๐๐ก2, ๐๐ก๐
=๐บ๐๐ผ๐๐ฟ๐
๐บ๐ : modulus of elasticity, ๐/๐2
๐ผ๐ : the torsion constant (polar moment of inertia), ๐4
๐ฟ๐ : position of applied force, ๐
8. Two circular rods in torsion
๐๐ก๐=
1
๐๐ก1
+1
๐๐ก2
โ1
, ๐๐ก๐=
๐บ๐๐ผ๐๐ฟ๐
๐บ๐ : modulus of elasticity, ๐/๐2
๐ผ๐ : the torsion constant (polar moment of inertia), ๐4
๐ฟ๐ : position of applied force, ๐
Vibrations 2.21 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
9. Coil springs
๐ =๐บ๐4
8๐๐ท3
๐บ : modulus of elasticity, ๐/๐2
๐ : wire diameter, ๐
๐ : number of active coil
๐ท : mean coil diameter, ๐
10. Clamped rectangular plate, constant thickness, force at center
๐ =๐ธโ3
12๐ผ๐2(1 โ ๐2)
๐ธ : Youngโs modulus of elasticity, ๐/๐2
โ : thickness of plate, ๐
๐ผ : coefficient
๐ : width of the plate, ๐
๐ : poison ratio
Vibrations 2.22 Modeling of Vibratory Systems
๐/๐ ๐ผ1.0 0.005601.2 0.006471.4 0.006911.6 0.007121.8 0.007202.0 0.00722
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Force-displacement relationships may also be used to
determine parameters such as ๐ that characterize a stiffness
element
Vibrations 2.23 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.3 Equivalent stiffness of a beam-spring combination
Consider a cantilever beam that has a spring attached at its free end
โข The force is applied to the free end of the spring
๐๐ =1
๐๐๐๐๐+
1
๐1
โ1
๐๐๐๐๐ =3๐ธ๐ผ
๐ฟ3
โข The force is applied simultaneously to the free end of the
cantilever beam
๐๐ = ๐๐๐๐๐ + ๐1
Vibrations 2.24 Modeling of Vibratory Systems
โน parrallel
โน series
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
5
ยง3.Stiffness Elements
- Ex.2.4 Equivalent stiffness of a cantilever beam with a
transverse end load
Acantilever beam:๐ธ = 72ร109๐/๐2, ๐ = 750๐๐, ๐๐ = 110๐๐,
๐๐ = 120๐๐. Determine the equivalent stiffness of this beam
Solution
The area moment of inertia ๐ผ about the bending axis
๐ผ =๐
32๐๐
4 โ ๐๐4
=๐
32120 ร 10โ3 4 โ 110 ร 10โ3 4
= 5.98 ร 10โ6๐4
The equivalent stiffness of the cantilever beam
๐ =3๐ธ๐ผ
๐ฟ3 =3 ร 72 ร 109 ร 5.98 ร 10โ6
750 ร 10โ3 3 = 3.06 ร 106๐/๐
Vibrations 2.25 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.5 Equivalent stiffness of a beam with a fixed end and a
translating support at the other end
Consider a uniform beam of length ๐ฟ with flexural rigidity ๐ธ๐ผ. When the
beam is subjected to a transverse loading ๐นat the translating support end, determine
the equivalent stiffness of this beam
Solution
By observation
โบ
Vibrations 2.26 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
To this end, we use Case 6 of Table 2.3 and set ๐ = ๐ = ๐ฟ and
obtain
๐๐๐๐ฅ๐๐ = 3๐ธ๐ผ ๐ + ๐ 3
๐3๐3
๐=๐=๐ฟ
=3๐ธ๐ผ ๐ฟ + ๐ฟ 3
๐ฟ3๐ฟ3 =24๐ธ๐ผ
๐ฟ3
Recognizing that the equivalent stiffness of a fixed-fixed beam
of length 2๐ฟ loaded at its middle is equal to the total equivalent
stiffness of a parallel spring combination of two end loaded
beams, we obtain that
๐๐ =1
2๐๐๐๐ฅ๐๐ =
1
2
24๐ธ๐ผ
๐ฟ3 =12๐ธ๐ผ
๐ฟ3
Vibrations 2.27 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.6 Equivalent stiffness of a micro-electromechanical
system (MEMS) fixed-fixed flexure
A micro-electromechanical sensor
system (MEMS) consisting of four
flexures. Each of these flexures is
fixed at one end and connected to a
mass at the other end. The length of each flexure is๐ฟ =100๐๐, the thickness of each flexure is โ = 2๐๐, and the
width of each flexure is ๐ = 2๐๐. A transverse loading acts on
the mass along the ๐-direction, which is normal to the ๐ โ ๐plane. Each flexure is fabricated from a poly-silicon material,
which has a Youngโs modulus of elasticity ๐ธ = 150๐บ๐๐
Determine the equivalent stiffness of the system
Vibrations 2.28 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
Solution
Each of the four flexures can be treated as a beam that is fixed
at one end and free to translate only
at the other end, similar to the
system in Ex.2.5
The equivalent stiffness of each
flexure is given by
๐๐๐๐๐ฅ๐ข๐๐ =12๐ธ๐ผ
๐ฟ3 , ๐ผ =๐โ3
12The equivalent stiffness of the system
๐๐ = 4 ร ๐๐๐๐๐ฅ๐ข๐๐ = 4 ร12๐ธ ร
๐โ3
12๐ฟ3 = 4
๐ธ๐โ3
๐ฟ3
= 4150 ร 109 ร 2 ร 10โ6 ร 2 ร 10โ6
100 ร 10โ6 = 9.6๐/๐
Vibrations 2.29 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.7 Equivalent stiffness of springs in parallel: removal of a restriction
Determine of the equivalence spring constant for
the parallel springs subjected to unequal forces
Solution
From similar triangles
๐ฅ = ๐ฅ2 +๐
๐ + ๐๐ฅ1 โ ๐ฅ2 =
๐
๐ + ๐๐ฅ1 +
๐
๐ + ๐๐ฅ2
Consider the bar
๐น = ๐น1 + ๐น2
๐๐น2 = ๐๐น1
Therefore
๐ฅ1 =๐น1
๐1=
๐๐น
๐1 ๐ + ๐, ๐ฅ2 =
๐น2
๐2=
๐๐น
๐2 ๐ + ๐
Vibrations 2.30 Modeling of Vibratory Systems
โน ๐น1 =๐๐น
๐ + ๐, ๐น2 =
๐๐น
๐ + ๐
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
6
ยง3.Stiffness Elements
โน ๐ฅ =๐
๐ + ๐๐ฅ1 +
๐
๐ + ๐๐ฅ2
=๐
๐ + ๐
๐๐น
๐1 ๐ + ๐+
๐
๐ + ๐
๐๐น
๐2 ๐ + ๐
=๐น
๐ + ๐ 2
๐1๐2 + ๐2๐
2
๐1๐2
For the equivalent system
๐น = ๐๐๐ฅ
โน ๐๐ =๐น
๐ฅThe equivalence spring constant for the parallel
springs subjected to unequal forces
๐๐ =๐1๐2 ๐ + ๐ 2
๐1๐2 + ๐2๐
2
Vibrations 2.31 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
3.Nonlinear Springs
- Nonlinear stiffness elements appear in many applications,
including leaf springs in vehicle suspensions and uniaxial
micro-electromechanical devices in the presence of
electrostatic actuation
- The spring force ๐น(๐ฅ)
๐น ๐ฅ = ๐๐ฅ๐๐๐๐๐๐ ๐ ๐๐๐๐๐ ๐๐๐๐๐๐๐ก
+ ๐ผ๐๐ฅ3
๐๐๐๐๐๐๐๐๐ ๐ ๐๐๐๐๐ ๐๐๐๐๐๐๐ก
(2.23)
๐ผ : the stiffness coefficient of the nonlinear term
๐ผ > 0 hardening spring ๐ผ < 0 softening spring
๐ : the linear spring constant
- The potential energy ๐
๐ ๐ฅ = 0
๐ฅ
๐น ๐ฅ ๐๐ฅ =1
2๐๐ฅ2 +
1
2๐ผ๐๐ฅ4
Vibrations 2.32 Modeling of Vibratory Systems
(2.24)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
๐น ๐ฅ = ๐๐ฅ + ๐ผ๐๐ฅ3 (2.23)
ยง3.Stiffness Elements
- For a nonlinear stiffness element described by Eq. (2.23), the
graph is no longer a straight line. The slope of this graph at a
location ๐ฅ = ๐ฅ๐ is given by
๐๐น
๐๐ฅ๐ฅ=๐ฅ๐
= ๐ + 3๐ผ๐๐ฅ2
๐ฅ=๐ฅ๐
= ๐ + 3๐ผ๐๐ฅ๐2
โน in the vicinity of displacements in a neighborhood of ๐ฅ =๐ฅ๐, the cubic nonlinear stiffness element may be replaced by
a linear stiffness element with a stiffness constant (2.25)
- The constant of proportionality ๐ผ๐ for the nonlinear cubic
spring is determined experimentally
Vibrations 2.33 Modeling of Vibratory Systems
(2.25)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Experimentally obtained data used to determine the
nonlinear spring constant ๐ผ๐
Vibrations 2.34 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.8 Nonlinear stiffness due to geometry
a. Nonlinear stiffness due to geometry
โข The initial tension force
๐0 = ๐น๐ ๐พ=0
= ๐๐ฟ0
โข The force in the spring
๐น๐ ๐ฅ = ๐๐ฟ0 + ๐ ๐ฟ2 + ๐ฅ2 โ ๐ฟ
โข The force in the ๐ฅ-direction is obtained
๐น๐ฅ ๐ฅ = ๐น๐ ๐ ๐๐๐พ =๐น๐ ๐ฅ
๐ฟ2 + ๐ฅ2=
๐ฅ๐๐ฟ0
๐ฟ2 + ๐ฅ2+
๐๐ฅ ๐ฟ2 + ๐ฅ2 โ ๐ฟ
๐ฟ2 + ๐ฅ2
โน the spring force opposing the motion is a nonlinear function
of the displacement ๐ฅ . Hence, this vibratory model of the
system will have nonlinear stiffness
Vibrations 2.35 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Binomial expansion 1 + ๐ฅ ๐ = 1 + ๐๐ฅ +1
2๐(๐ โ 1)๐ฅ2 + โฏ
ยง3.Stiffness Elements
Cubic Springs and Linear Springs
Assume that |๐ฅ/๐ฟ| โช 1, using binomial expansion
1 +๐ฅ
๐ฟ
2
= 1 +๐ฅ
๐ฟ
2 1/2
= 1 +1
2
๐ฅ
๐ฟ
2
+1
8
๐ฅ
๐ฟ
4
+ โฏ
โน ๐น๐ฅ ๐ฅ = ๐๐ฟ0
๐ฅ/๐ฟ
1 + ๐ฅ/๐ฟ 2+
๐๐ฅ 1 + ๐ฅ/๐ฟ 2 โ 1
1 + ๐ฅ/๐ฟ 2
= ๐๐ฟ0
๐ฅ/๐ฟ
1 +12
๐ฅ๐ฟ
2 +๐๐ฅ 1 +
12
๐ฅ๐ฟ
2โ 1
1 +12
๐ฅ๐ฟ
2
= ๐๐ฟ0
๐ฅ
๐ฟ+
๐
2๐ฟ
๐ฅ
๐ฟ
3
Vibrations 2.36 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
7
ยง3.Stiffness Elements
๐น๐ฅ ๐ฅ = ๐๐ฟ0
๐ฅ
๐ฟ+
๐
2๐ฟ
๐ฅ
๐ฟ
3
When the nonlinear term is negligible
๐น๐ฅ ๐ฅ = ๐๐ฟ0
๐ฅ
๐ฟ= ๐0
๐ฅ
๐ฟand the spring constant is proportional to the initial tension in
the spring
b. Nonlinear spring composed of a set of linear springs
Another example of a nonlinear spring is one that is piecewise
linear as shown in figure
Vibrations 2.37 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
4.Other Forms of Potential Energy Elements
Consider other stiffness elements in which there is a
mechanism for storing and releasing potential energy. The
source of the restoring force is a fluid element or a gravitational
loading
- Fluid Element
โข The magnitude of the total force of the
displaced fluid acting on the rest of the fluid
๐น๐ ๐ฅ = 2๐๐๐ด0๐ฅ
๐ : the mass density of the fluid, ๐๐/๐3
๐ : gravitational constant, ๐/๐ 2
๐ด0 : the manometer cross-sectional area, ๐2
๐ฅ : the fluid displacement, ๐
Vibrations 2.38 Modeling of Vibratory Systems
Manometer
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข The equivalent spring constant of this fluid system
๐๐ =๐๐น๐
๐๐ฅ= 2๐๐๐ด0
โข The potential energy
๐ ๐ฅ =1
2๐๐๐ฅ
2 = ๐๐๐ด0๐ฅ2
Alternatively, the potential energy can also be obtained
directly from the work done
๐ ๐ฅ = 0
๐ฅ
๐น๐ ๐ฅ ๐๐ฅ
= 2๐๐๐ด0 0
๐ฅ
๐ฅ๐๐ฅ
= ๐๐๐ด0๐ฅ2
Vibrations 2.39 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Compressed Gas
โข When the piston moves by an amount ๐ฅ along
the axis of the piston, ๐0 decreases to a
volume ๐๐
๐๐ ๐ฅ = ๐0 โ ๐ด0๐ฅ = ๐ด0๐ฟ0 1 โ ๐ฅ/๐ฟ0
โน ๐๐ ๐ฅ = ๐0 1 โ ๐ฅ/๐ฟ0
๐ด0: the piston cross-sectional area, ๐2
โข The equation of state for the gas
๐๐๐๐ = ๐0๐0
๐ = ๐0 = ๐๐๐๐ ๐ก โน ๐ = ๐0๐๐โ๐
๐ : gas pressure, ๐/๐2 ๐ : gas volume, ๐3
๐ : the ratio of specific heats of the gas, when compressed
- slowly, the compression is isothermal, ๐ = 1
- rapidly, the compression is adiabatic, ๐ = ๐๐/๐๐ฃ = 1.4
Vibrations 2.40 Modeling of Vibratory Systems
Gas compression
with a piston
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข The magnitude of the force on the piston
๐น ๐ฅ = ๐ด0๐
= ๐ด0๐0๐๐โ๐
= ๐ด0๐0๐0โ๐ 1 โ ๐ฅ/๐ฟ0
โ๐
โน ๐น ๐ฅ = ๐ด0๐0 1 โ ๐ฅ/๐ฟ0โ๐ (2.33)
The Eq. (2.33) describes a nonlinear force
versus displacement relationship
At the vicinity of ๐ฅ = ๐ฅ๐, the stiffness of an equivalent linear
stiffness element
๐๐ = ๐๐น
๐๐ฅ๐ฅ=๐ฅ๐
=๐๐ด0๐0
๐ฟ01 โ ๐ฅ๐/๐ฟ0
โ๐โ1
For ๐ฅ๐/๐ฟ0 โช 1,
Vibrations 2.41 Modeling of Vibratory Systems
๐๐ =๐๐ด0๐0
๐ฟ0
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข For arbitrary ๐ฅ/๐ฟ0, the potential energy
๐ ๐ฅ = 0
๐ฅ
๐น ๐ฅ ๐๐ฅ
= ๐ด0๐0 0
๐ฅ
1 โ ๐ฅ/๐ฟ0โ๐๐๐ฅ
โน ๐ ๐ฅ = โ๐ด0๐0๐ฟ0๐๐ 1 โ ๐ฅ/๐ฟ0 ๐ = 1
๐ด0๐0๐ฟ0
๐ โ 11 โ ๐ฅ/๐ฟ0
โ๐ โ 1 ๐ โ 1
Vibrations 2.42 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
8
ยง3.Stiffness Elements
- Pendulum System
โข Pendulum systems: bar with uniformly
distributed mass
At ๐, the vertical distance through which the
center of gravity of the bar moves up from the
reference position
๐ฅ =๐ฟ
2โ
๐ฟ
2๐๐๐ ๐ =
๐ฟ
21 โ ๐๐๐ ๐
The increase in the potential energy
๐ ๐ฅ = 0
๐ฅ
๐น ๐ฅ ๐๐ฅ = 0
๐ฅ
๐๐๐๐ฅ = ๐๐๐ฅ
or in ๐
๐ ๐ =1
2๐๐๐ฟ(1 โ ๐๐๐ ๐)
Vibrations 2.43 Modeling of Vibratory Systems
Pendulum systems: bar
with uniformly distributed
mass
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Taylor series approximation ๐๐๐ ๐ = 1 โ๐2
2+ โฏ
ยง3.Stiffness Elements
๐ ๐ =1
2๐๐๐ฟ 1 โ ๐๐๐ ๐
โ1
2
๐๐๐ฟ
2๐2
=1
2๐๐๐
2
where the equivalent spring constant
๐๐ =๐๐๐ฟ
2
Vibrations 2.44 Modeling of Vibratory Systems
Pendulum systems: bar
with uniformly distributed
mass
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข Pendulum systems: mass on a weightless rod
The increase in the potential energy
๐ ๐ โ1
2๐1๐๐ฟ๐2 =
1
2๐๐๐
2
where the equivalent spring constant
๐๐ = ๐1๐๐ฟ
If the weightless bar is replaced by one that
has a uniformly distributed mass ๐, then the
total potential energy of the bar and the mass
๐ ๐ โ1
4๐๐๐ฟ๐2 +
1
2๐1๐๐ฟ๐2 =
1
2
๐
2+ ๐1 ๐๐ฟ๐2 =
1
2๐๐๐
2
where the equivalent spring constant
๐๐ =๐
2+ ๐1 ๐๐ฟ
Vibrations 2.45 Modeling of Vibratory Systems
Pendulum systems: mass
on a weightless rod
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
โข Pendulum systems: inverted mass on a
weightless rod
The decrease in the potential energy
๐ ๐ โ โ1
2๐1๐๐ฟ๐2
Vibrations 2.46 Modeling of Vibratory Systems
Pendulum systems:
inverted mass on a
weightless rod
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง3.Stiffness Elements
- Ex.2.9 Equivalent stiffness due to gravity loading
For โsmallโ rotations about the upright position ฮธ = 0,
the potential energy
๐ ๐ =1
2๐1๐๐๐2 โ
1
2๐2๐๐๐2
=1
2๐1๐ โ ๐2๐ ๐๐2
There is a gain or loss in potential energy depending
on whether ๐1๐ > ๐2๐ or vice versa
โข When the bar has a uniformly distributed mass ๐
๐1 = ๐๐ฟ1/๐ฟ, ๐2 = ๐๐ฟ2/๐ฟ, ๐ฟ = ๐ฟ1 + ๐ฟ2, ๐ = ๐ฟ1/2, ๐ = ๐ฟ2/2
๐ ๐ =๐ฟ12 โ ๐ฟ2
2
4(๐ฟ1 + ๐ฟ2)๐๐๐2 =
1
2
๐ฟ1 + ๐ฟ2
2๐๐๐2 =
1
2๐๐๐
2
where the equivalent stiffness ๐๐ = ๐ฟ1 + ๐ฟ2 ๐๐/2
Vibrations 2.47 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
- Damping elements are assumed to have neither inertia nor the
means to store or release potential energy
- The mechanical motion imparted to these elements is
converted to heat or sound and, hence, they are called non-
conservative or dissipative because this energy is not
recoverable by the mechanical system
- There are four common types of damping mechanisms used to
model vibratory systems
โข Viscous damping
โข Coulomb or dry friction damping
โข Material or solid or hysteretic damping
โข Fluid damping
In all these cases, the damping force is expressed as a
function of velocity
Vibrations 2.48 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
9
ยง4.Dissipation Elements
1.Viscous damping
- When a viscous fluid flows through a slot or around a piston in
a cylinder, the damping force generated is proportional to the
relative velocity between the two boundaries confining the fluid
- A common representation of a viscous damper is a cylinder
with a piston head
- Depending on the damper construction and the velocity
range, the magnitude of the damper force ๐น( ๐ฅ) is a nonlinear
function of velocity or can be approximated as a linear
function of velocity
Vibrations 2.49 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
- In the linear case, the damper force is expressed as
๐น ๐ฅ = ๐ ๐ฅ (2.46)
๐ : the damping coefficient, ๐/(๐/๐ )
Viscous damping of the form given by Eq. (2.46) is also
called slow-fluid damping
- In the case of a nonlinear viscous damper described by a
function ๐น ๐ฅ , the equivalent linear viscous damping around
an operating speed ๐ฅ = ๐ฅ๐ is determined as follows
๐๐ = ๐๐น ๐ฅ
๐ ๐ฅ ๐ฅ= ๐ฅ๐
- Linear viscous damping elements can be combined in the
same way that linear springs are, except that the forces are
proportional to velocity instead of displacement
Vibrations 2.50 Modeling of Vibratory Systems
(2.47)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
โข Energy Dissipation
The energy dissipated by a linear viscous damper
๐ธ๐ = ๐น๐๐ฅ = ๐น ๐ฅ๐๐ก = ๐ ๐ฅ2๐๐ก = ๐ ๐ฅ2๐๐ก
โข Parallel-Plate Damper
An example of a viscous damper is shown in the figure
The shear force acting on the
bottom plate
๐น ๐ฅ =๐ ๐ฅ
โ๐ด =
๐๐ด
โ ๐ฅ
The damping coefficient ๐ for the parallel-plate construction
๐ =๐๐ด
โ
Vibrations 2.51 Modeling of Vibratory Systems
(2.48)
(2.49)
(2.50)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
๐ =๐๐ด
โ(2.50)
ยง4.Dissipation Elements
- Ex.2.10 Design of a parallel-plate damper
A parallel-plate damper with a top plate of dimensions100๐๐ร 100๐๐ is to be pulled across an oil layer of thickness
0.2๐๐, which is confined between the moving plate and a
fixed plate. We are given that this oil is SAE30 oil, which has a
viscosity of 345๐๐๐๐ (345 ร 103๐๐ /๐2)
Determine the viscous damping coefficient of this system
Solution
To this end, using Eq. (2.50) and substitute the given values
into this expression and find that
๐ =๐๐ด
โ=
345ร10โ3 100ร10โ3 ร100ร10โ3
0.2 ร 10โ3 = 17.25๐๐ /๐
Vibrations 2.52 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
- Ex.2.11 Equivalent damping coefficient and
equivalent stiffness of a vibratory system
Consider the vibratory system in which the motion of mass ๐is restrained by a set of linear springs and linear viscous
dampers. Determine ๐๐ and ๐๐
Solution
The equivalence system
๐๐ = ๐1 +๐2๐3
๐2 + ๐3, ๐๐ = ๐1 + ๐2
Vibrations 2.53 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
๐๐ = ๐๐น ๐ฅ
๐ ๐ฅ ๐ฅ= ๐ฅ๐
(2.47)
ยง4.Dissipation Elements
- Ex.2.12 Equivalent linear damping coefficientof a nonlinear damper
It has been experimentally determined that the damper force-
velocity relationship is given by the function
๐น ๐ฅ = 4๐๐ /๐ ๐ฅ + (0.3๐๐ 3/๐) ๐ฅ3
Determine the equivalent linear damping coefficient around an
operating speed of 3๐/๐
Solution
Using the Eq. (2.47)
๐๐ = ๐๐น ๐ฅ
๐ ๐ฅ ๐ฅ=3๐/๐
= 4 + 0.9 ๐ฅ2
๐ฅ=3๐/๐ = 4 + 0.9 ร 32
โน ๐๐ = 12.1๐๐ /๐
Vibrations 2.54 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
10
ยง4.Dissipation Elements
2.Other Forms of Dissipation
- Coulomb Damping or Dry Friction
This type of damping is due to the force caused by friction
between two solid surfaces
The friction force acting on the system
๐น ๐ฅ = ๐๐๐ ๐๐( ๐ฅ) (2.51)
๐ : the kinetic coefficient of friction
๐ : the force compressing the surfaces, ๐
๐ ๐๐: the signum function, ๐ ๐๐ ๐ฅ = +1 ๐ฅ > 0โ1 ๐ฅ < 0
0 ๐ฅ = 0
Vibrations 2.55 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
If the normal force is due to the system weight, then ๐ = ๐๐
๐น ๐ฅ = ๐๐๐๐ ๐๐( ๐ฅ) (2.52)
The energy dissipated in this case
(2.53)
- Fluid Damping (Velocity-Squared Damping)
This type of damping is associated with a system whose mass is
vibrating in a fluid medium. The magnitude of the damping force
๐น ๐ฅ = ๐๐ ๐ฅ2๐ ๐๐ ๐ฅ = ๐๐| ๐ฅ| ๐ฅ (2.54)
๐๐ : friction coefficient, ๐๐ = ๐ถ๐๐ด/2
๐ถ : drag coefficient
๐ : the mass density of the fluid
๐ด : the projected area of the mass in a direction normal to ๐ฅ
Fluid damping of (2.54) is often referred to as fast-fluid damping
Vibrations 2.56 Modeling of Vibratory Systems
๐ธ๐ = ๐น๐๐ฅ = ๐น ๐ฅ๐๐ก = ๐๐๐ ๐ ๐๐ ๐ฅ ๐ฅ๐๐ก
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง4.Dissipation Elements
The energy dissipated
๐ธ๐ = ๐น๐๐ฅ = ๐น ๐ฅ๐๐ก = ๐๐ ๐ ๐๐( ๐ฅ) ๐ฅ3๐๐ก
- Structural or Solid or Hysteretic Damping
This type of damping describes the losses in materials due to
internal friction. The damping force is a function of
displacement and velocity and is of the form
๐น = ๐๐๐ฝโ๐ ๐๐ ๐ฅ |๐ฅ| (2.57)
๐ฝโ : an empirically determined constant
The energy dissipated
๐ธ๐ = ๐น๐๐ฅ = ๐น ๐ฅ๐๐ก = ๐๐๐ฝโ ๐ ๐๐( ๐ฅ)|๐ฅ|๐๐ก
Vibrations 2.57 Modeling of Vibratory Systems
(2.54)
(2.58)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง5.Model Construction
1.Introduction
- In this section, four examples are provided to illustrate how
the previously described inertia, stiffness, and damping
elements are used to construct system models
- Modeling is an art, and often experience serves as a guide in
model construction
- In this section, the examples are drawn from different areas,
and are presented in a progressive order proceeding from
the use of discrete inertia, stiffness, and damping elements in
a model to distributed elements, and finally, to a combination
of distributed and discrete elements
- As discussed in the subsequent chapters, the mass,
stiffness, and damping of a system appear as parameters in
the governing equations of the system
Vibrations 2.58 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง5.Model Construction
2.A Micro-electromechanical System
Micro-electromechanical accelerometer and a vibratory model
of this sensor
Vibrations 2.59 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง5.Model Construction
3.The Human Body
Human body and a vibratory model
Vibrations 2.60 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
11
ยง5.Model Construction
4.A Ski
Cross-country ski, which is a physical system with distributed
stiffness and inertia properties, and its vibratory model
Vibrations 2.61 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง5.Model Construction
5.Cutting Process
Work-piece-tool turning system and vibratory model of this
system
Vibrations 2.62 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
ยง6.Design for Vibration
Vibrations 2.63 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Exercises
Vibrations 2.66 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien