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 • Compute the mass moment of inertia of rotational systems • 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 • Determine the stiffness of fluid, gas, and pendulum elements • Determine the potential energy of stiffness elements • Determine the damping for systems that have different sources of dissipation: viscosity, dry friction, fluid, and material • Construct models of vibratory systems Vibrations 2.02 Modeling of Vibratory Systems HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien § 1 . Introduction - Three elements that comprise a vibrating system • Inertia elements: stores and releases kinetic energy • Stiffness elements: stores and releases potential energy • Dissipation elements: express energy loss in a system - Components comprising a vibrating mechanical system • Translational motion • Rotational motion Mass, () Mass moment of inertia, (2 ) Stiffness, (/) Stiffness, (/) Damping, (/) Damping, (/) External force, () External moment, () Vibrations 2.03 Modeling of Vibratory Systems HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien § 2 . Inertia Elements - Translational motion - Rotational motion Slender bar = 1 12 2 Circular disk = 1 2 2 Sphere = 2 5 2 Circular cylinder = = 1 12 (3 2 +ℎ 2 ) = 1 2 2 = + 2 , : distance from the center of gravity to point Vibrations 2.04 Modeling of Vibratory Systems - Parallel-axes theorem HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien § 2 . Inertia Elements - For a mass translating with a velocity of magnitude in the − plane under the driving force • The equation governing the motion of the mass = ( ) when and are independent of time = (2.2) • The kinetic energy, , of mass = 1 2 ∙ = 1 2 2 Vibrations 2.05 Modeling of Vibratory Systems HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien § 2 . Inertia Elements - For a rigid body undergoing only rotation in the plane with an angular speed • The equation governing the rotation of the mass of inertia = (2.6) : the moment acting about the center of mass or a fixed point along the direction normal to the plane of motion : the associated mass moment of inertia • The kinetic energy of the system = 1 2 2 Vibrations 2.06 Modeling of Vibratory Systems HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
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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
• Compute the mass moment of inertia of rotational systems
• 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
• Determine the stiffness of fluid, gas, and pendulum elements
• Determine the potential energy of stiffness elements
• Determine the damping for systems that have different sources
of dissipation: viscosity, dry friction, fluid, and material
• Construct models of vibratory systems
Vibrations 2.02 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§1.Introduction
- Three elements that comprise a vibrating system
• Inertia elements: stores and releases kinetic energy
• Stiffness elements: stores and releases potential energy
• Dissipation elements: express energy loss in a system
- Components comprising a vibrating mechanical system
• Translational motion • Rotational motion
Mass, 𝑚 (𝑘𝑔) Mass moment of inertia, 𝐽 (𝑘𝑔𝑚2)
Stiffness, 𝑘 (𝑁/𝑚) Stiffness, 𝑘𝑡 (𝑁𝑚/𝑟𝑎𝑑)
Damping, 𝑐 (𝑁𝑠/𝑚) Damping, 𝑐𝑡 (𝑁𝑚𝑠/𝑟𝑎𝑑)
External force, 𝐹 (𝑁) External moment, 𝑀 (𝑁𝑚)
Vibrations 2.03 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§2.Inertia Elements
- Translational motion 𝑚
- Rotational motion
Slender bar 𝐽𝐺 =1
12𝑚𝐿2
Circular disk 𝐽𝐺 =1
2𝑚𝑅2
Sphere 𝐽𝐺 =2
5𝑚𝑅2
Circular cylinder 𝐽𝑥 = 𝐽𝑦 =1
12𝑚(3𝑅2 + ℎ2)
𝐽𝑧 =1
2𝑚𝑅2
𝐽𝑂 = 𝐽𝐺 + 𝑚𝑑2, 𝑑: distance from the center of gravity to point 𝐺
Vibrations 2.04 Modeling of Vibratory Systems
- Parallel-axes theorem
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§2.Inertia Elements
- For a mass 𝑚 translating with a
velocity of magnitude 𝑥 in the 𝑋−𝑌
plane under the driving force 𝐹
• The equation governing the motion of the mass 𝑚
𝐹 𝑖 =𝑑
𝑑𝑡(𝑚 𝑥 𝑖)
when 𝑚 and 𝑖 are independent of time
𝐹 = 𝑚 𝑥 (2.2)
• The kinetic energy, 𝑇, of mass 𝑚
𝑇 =1
2𝑚 𝑥 𝑖 ∙ 𝑥 𝑖 =
1
2𝑚 𝑥2
Vibrations 2.05 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§2.Inertia Elements
- For a rigid body undergoing only
rotation in the plane with an angular
speed 𝜃
• The equation governing the rotation of the mass of inertia
𝑀 = 𝐽 𝜃 (2.6)
𝑀 : the moment acting about the center of mass 𝐺 or a
fixed point 𝑂 along the direction normal to the plane of
motion
𝐽 : the associated mass moment of inertia
• The kinetic energy of the system
𝑇 =1
2𝐽 𝜃2
Vibrations 2.06 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
2
§2.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
• Uniform Disk
The mass moment of inertia about the point 𝑂 ,
which is located a distance 𝑅 from point 𝐺
𝐽𝑂 = 𝐽𝐺 + 𝑚𝑅2 =1
2𝑚𝑅2 + 𝑚𝑅2 =
3
2𝑚𝑅2
• Uniform Bar
The mass moment of inertia about the point 𝑂
𝐽𝑂 = 𝐽𝐺 + 𝑚𝐿
2
2
=1
12𝑚𝐿2 +
1
4𝑚𝐿2 =
1
3𝑚𝐿2
Vibrations 2.07 Modeling of Vibratory Systems
𝐽𝐺 =1
2𝑚𝑅2
𝐽𝐺 =1
12𝑚𝐿2
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§2.Inertia Elements
- Ex.2.2 Slider mechanism: system with varying inertia property
A slider of mass 𝑚𝑠 slides along a uniform bar of
mass 𝑚𝑙 with a pivot at point 𝑂. Another bar,
which is pivoted at point 𝑂′, has a portion of length
𝑏 that has a mass 𝑚𝑏 and another portion of
length 𝑒 that has a mass 𝑚𝑒. Determine the rotary
inertia 𝐽𝑂 of this system and show its dependence
on the angular displacement coordinate 𝜑
From geometry, 𝑟, 𝑎𝑏, 𝑎𝑒 can be described in terms of 𝜑
𝑟2 𝜑 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝜑
𝑎𝑏2 𝜑 = (𝑏/2)2+𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑
𝑎𝑒2 𝜑 = (𝑒/2)2+𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)
𝑎𝑏 : the distance from the midpoint of bar of mass 𝑚𝑒 to 𝑂
𝑎𝑒 : the distance from the midpoint of bar of mass 𝑚𝑏 to 𝑂
Vibrations 2.08 Modeling of Vibratory Systems
Solution
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§2.Inertia Elements
𝑟2 𝜑 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠𝜑
𝑎𝑏2 𝜑 = (𝑏/2)2+𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑
𝑎𝑒2 𝜑 = (𝑒/2)2+𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)
The rotary inertia 𝐽𝑂 of this system
𝐽𝑂 = 𝐽𝑚𝑙+ 𝐽𝑚𝑠
(𝜑) + 𝐽𝑚𝑏(𝜑) + 𝐽𝑚𝑒
(𝜑)
where
𝐽𝑚𝑙=
1
3𝑚𝑙𝑗
2
𝐽𝑚𝑠(𝜑) = 𝑚𝑠𝑟
2(𝜑)
𝐽𝑚𝑏𝜑 = 𝑚𝑏
𝑏2
12+ 𝑚𝑏𝑎𝑏
2 = 𝑚𝑏
𝑏2
3+ 𝑎2 − 𝑎𝑏𝑐𝑜𝑠𝜑
𝐽𝑚𝑒𝜑 = 𝑚𝑒
𝑒2
12+ 𝑚𝑒𝑎𝑒
2 = 𝑚𝑒
𝑒2
3+ 𝑎2 − 𝑎𝑒𝑐𝑜𝑠(𝜋 − 𝜑)
Vibrations 2.09 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
1.Introduction
- Stiffness elements are manufactured from different materials
and they have many different shapes
- Application
• to minimize vibration transmission from machinery to the
supporting structure
• to isolate a building from earthquakes
• 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
§3.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
§3.Stiffness Elements
- The stiffness elements store and release the potential energy
of a system
- Consider a spring under acting force of magnitude 𝐹 is
directed along the direction of the unit vector 𝑗
𝐹𝑠 = −𝐹 𝑗
𝐹𝑠 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
§3.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 𝑉 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
𝑉 𝑥 = 𝑑𝑒𝑓𝑜𝑟𝑚𝑒𝑑 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑜𝑟 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛
𝐹𝑠𝑑𝑥
=
𝑥
0
𝐹𝑠𝑑𝑥 =
𝑥
0
−𝐹 𝑗 ∙ 𝑑𝑥 𝑗 =
𝑥
0
𝐹𝑑𝑥
Vibrations 2.13 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
2.Linear Springs
- Translation Spring
• Deformation
𝐹 𝑥 = 𝑘𝑥 (2.9)
𝐹 : the applied force
𝑘 : the spring constant
𝑥 : the spring deflection
• The potential energy 𝑉 stored in the spring
𝑉 𝑥 =
0
𝑥
𝐹 𝑥 𝑑𝑥 =
0
𝑥
𝑘𝑥𝑑𝑥 = 𝑘
0
𝑥
𝑥𝑑𝑥 =1
2𝑘𝑥2
Vibrations 2.14 Modeling of Vibratory Systems
(2.10)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
- Torsion Spring
• Deformation
𝜏 𝜃 = 𝑘𝑡𝜃 (2.11)
𝜏 : the applied moment
𝑘𝑡 : the spring constant
𝜃 : the spring deflection
• The potential energy 𝑉 stored in the spring
𝑉 𝜃 =
0
𝜃
𝜏 𝜃 𝑑𝜃 =
0
𝜃
𝑘𝑡𝜃𝑑𝜃 =1
2𝑘𝑡𝜃
2
Vibrations 2.15 Modeling of Vibratory Systems
(2.12)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
- Combinations of Linear Springs
• Parallel Springs
Translation springs
Total force
𝐹 𝑥 =𝐹1 𝑥 +𝐹1 𝑥 =𝑘1𝑥+𝑘2𝑥= 𝑘1 +𝑘2 𝑥 =𝑘𝑒𝑥
Equivalent spring
𝑘𝑒 = 𝑘1 + 𝑘2
Torsion springs
Total moment
𝜏 𝜃 = 𝜏1 𝜃 + 𝜏2 𝜃
= 𝑘𝑡1𝜃 + 𝑘𝑡2
𝜃 = 𝑘𝑡1+ 𝑘𝑡2
𝜃 = 𝑘𝑡𝑒𝜃
Equivalent spring
𝑘𝑡𝑒= 𝑘𝑡1
+ 𝑘𝑡2
Vibrations 2.16 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
• Series Springs
Translation springs
𝑥 = 𝑥1 + 𝑥2 =𝐹
𝑘1+
𝐹
𝑘2=
1
𝑘1+
1
𝑘2𝐹 =
𝐹
𝑘𝑒
Equivalent spring
𝑘𝑒 =1
𝑘1+
1
𝑘2
−1
=𝑘1𝑘2
𝑘1 + 𝑘2
Torsion springs
𝜃 = 𝜃1 + 𝜃2 =𝜏
𝑘𝑡1
+𝜏
𝑘𝑡2
=1
𝑘𝑡1
+1
𝑘𝑡2
𝜏 =𝜏
𝑘𝑡𝑒
Equivalent spring
𝑘𝑡𝑒=
1
𝑘𝑡1
+1
𝑘𝑡2
−1
=𝑘𝑡1
𝑘𝑡2
𝑘𝑡1+ 𝑘𝑡2
Vibrations 2.17 Modeling of Vibratory Systems
Displacement
Displacement
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
- Spring Constants for Some Common Elastic Elements
1. Axially loaded rod or cable
𝑘 =𝐴𝐸
𝐿
𝐴 : cross-sectional area, 𝑚2
𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2
𝐿 : length of the rod, 𝑚
2. Axially loaded tapered rod
𝑘 =𝜋𝐸𝑑1𝑑2
4𝐿
𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2
𝑑1 : rod diameter, 𝑚
𝑑2 : rod diameter, 𝑚
𝐿 : length of the rod, 𝑚
Vibrations 2.18 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
2/2/2014
4
§3.Stiffness Elements
3. Hollow circular rod in torsion
𝑘𝑡 =𝐺𝐼
𝐿
𝐺 : shear modulus of elasticity, 𝑁/𝑚2
𝐼 : the torsion constant (polar moment of inertia), 𝑚4
For the concentric circular tubes,
𝐿 : length of the rod, 𝑚
𝑑𝑜 : outside rod diameter, 𝑚
𝑑𝑖 : inside rod diameter, 𝑚
4. Cantilever beam
𝑘 =3𝐸𝐼
𝑎3, 0 < 𝑎 ≤ 𝐿
𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2
𝐼 : the area moment of inertia about the bending axis, 𝑚4
𝑎 : position of applied force, 𝑚
𝐿 : length of the beam, 𝑚
Vibrations 2.19 Modeling of Vibratory Systems
𝐼 =𝜋(𝑑𝑜
4 − 𝑑𝑖4)
32
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
5. Pinned-pinned beam (Hinged, simply supported)
𝑘 =3𝐸𝐼(𝑎 + 𝑏)
𝑎2𝑏2
𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2
𝐼 : the area moment of inertia about the bending axis, 𝑚4
𝑎,𝑏: position of applied force, 𝑚
6. Clamped-clamped beam (Fixed-fixed beam)
𝑘 =3𝐸𝐼(𝑎 + 𝑏)3
𝑎3𝑏3
𝐸 : Young’s modulus of elasticity, 𝑁/𝑚2
𝐼 : the area moment of inertia about the bending axis, 𝑚4
𝑎,𝑏: position of applied force, 𝑚
Vibrations 2.20 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
7. Two circular rods in torsion
𝑘𝑡𝑒= 𝑘𝑡1
+ 𝑘𝑡2, 𝑘𝑡𝑖
=𝐺𝑖𝐼𝑖𝐿𝑖
𝐺𝑖 : modulus of elasticity, 𝑁/𝑚2
𝐼𝑖 : the torsion constant (polar moment of inertia), 𝑚4
𝐿𝑖 : position of applied force, 𝑚
8. Two circular rods in torsion
𝑘𝑡𝑒=
1
𝑘𝑡1
+1
𝑘𝑡2
−1
, 𝑘𝑡𝑖=
𝐺𝑖𝐼𝑖𝐿𝑖
𝐺𝑖 : modulus of elasticity, 𝑁/𝑚2
𝐼𝑖 : the torsion constant (polar moment of inertia), 𝑚4
𝐿𝑖 : position of applied force, 𝑚
Vibrations 2.21 Modeling of Vibratory Systems
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
§3.Stiffness Elements
9. Coil springs
𝑘 =𝐺𝑑4
8𝑛𝐷3
𝐺 : modulus of elasticity, 𝑁/𝑚2
𝑑 : wire diameter, 𝑚
𝑛 : number of active coil
𝐷 : mean coil diameter, 𝑚
10. Clamped rectangular plate, constant thickness, force at center