Mechanical Vibrations lecture notes ( Undamped Single Degree-of-Freedom System) (Chapter Two)

CHAPTER TWO Free Vibrations Undamped Single Degree-of-Freedom System A system is said to undergo free vibration when it oscillates only under an initial disturbance with no external forces acting afterward. Some examples are the oscillations of the pendulum of a grandfather clock, the vertical oscillatory motion felt by a bicyclist after hitting a road bump, and the motion of a child on a swing after an initial push.

Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

2-1

Undamped Single Degree-of-Freedom System Single-degree-of-freedom (SDOF) system is a system whose motion is defined just by a single independent co-ordinate (or function). SDOF systems are often used as a very crude approximation for a generally much more complex system. Newton's second law is the first basis for examining the motion of the system. As shown in Figure below the deformation of the spring in the static equilibrium position is ∆ , and the spring force 𝑘∆ is equal to the gravitational force 𝑊 acting on mass 𝑚:

Static

∑ 𝐹𝑦 = 0

Dynamic

(𝑆𝑡𝑎𝑡𝑖𝑐 𝑠𝑡𝑎𝑡𝑒)

𝑘∆= 𝑊 = 𝑚𝑔 By measuring the displacement 𝑥 from the static equilibrium position, the forces acting on mare 𝑘(∆ + 𝑥) and 𝑊. With 𝑥 chosen to be positive in the downward direction, all quantities-force, velocity, and acceleration-are also positive in the downward direction. We now apply Newton's second law of motion to the mass m:

∑ 𝐹𝑦 = 𝑚𝑎

(Dynamic state)


2-2

∑ 𝐹𝑦 = 𝑚𝑥̈ 𝑚𝑥̈ = ∑ 𝐹𝑦 = 𝑚𝑔 − 𝑘(∆ + 𝑥) 𝑚𝑥̈ = 𝑚𝑔 − 𝑘∆ − 𝑘 𝑥 𝑘∆= 𝑊 = 𝑚𝑔 𝑚𝑥̈ = 𝑚𝑔 − 𝑚𝑔 − 𝑘 𝑥 𝑚𝑥̈ + 𝑘 𝑥 = 0

(Equation of motion) 𝑥 = 𝐴 sin 𝜔𝑡

𝑚(−𝜔2 𝑥) + 𝑘 𝑥 = 0

𝑥̇ = 𝜔𝐴 cos 𝜔𝑡 𝑥̈ = −𝜔2 𝐴 sin 𝜔𝑡

2

𝑚𝜔 𝑥 = 𝑘 𝑥 𝜔2 =

𝑥̈ = −𝜔2 𝑥

𝑘 𝑚

𝜔𝑛 = √

𝑘 𝑚

𝑚𝑥̈ + 𝑘 𝑥 = 0

(÷ m)

𝑥̈ + 𝜔2𝑛 𝑥 = 0 Equation is a homogeneous second-order linear differential equation, has the following general solution: Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

2-3

Equation of motion 𝑚𝑥̈ + 𝑘 𝑥 = 0

Forms of solution:

:

𝑚𝑥̈ + 𝑘 𝑥 = 0 2nd order Differential equation Homogenous Constant coefficients Linear 𝑥(𝑡) = 𝑋 sin(𝜔𝑡 + Φ) 𝑥(𝑡) = 𝑋 cos(𝜔𝑡 − Φ) 𝑥(𝑡) = 𝐶𝑒 𝑠𝑡

We will use this form

𝑚𝑥̈ + 𝑘 𝑥 = 0 Assume : 𝑥(𝑡) = 𝐶𝑒 𝑠𝑡 𝑥̇ (𝑡) = 𝑠𝐶𝑒 𝑠𝑡 𝑥̈ (𝑡) = 𝑠 2 𝐶𝑒 𝑠𝑡

𝑚𝑠2 𝐶𝑒𝑠𝑡 + 𝑘 𝐶𝑒𝑠𝑡 = 0 (𝑚𝑠2 + 𝑘 )𝐶𝑒𝑠𝑡 = 0

for a non-trivial solution (𝐶𝑒 𝑠𝑡 ≠ 0) 𝑚𝑠2 + 𝑘 = 0

𝑠1,2 = ±𝑗√

𝑘 𝑚

𝑥(𝑡) = 𝐶1 𝑒 𝑠1𝑡 + 𝐶2 𝑒 𝑠2𝑡


2-4

𝑥(𝑡) = 𝐶1 𝑒

𝑗√

𝑘 𝑚𝑡

+ 𝐶2 𝑒

𝑘 𝑚𝑡

−𝑗√

𝐶1 𝑎𝑛𝑑 𝐶2 𝑎𝑟𝑒 𝑎𝑟𝑏𝑖𝑡𝑟𝑎𝑟𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠 𝑡𝑜 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 𝑖𝑛𝑖𝑡𝑖𝑜𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠 𝑒 ±𝑗𝜃 = cos(𝜃) ± 𝑗𝑠𝑖𝑛(𝜃)

Recall Euler’s identity:

𝑘 𝑘 𝑘 𝑘 𝑡) + 𝑗𝑠𝑖𝑛 (√ 𝑡)] + 𝐶2 [cos (√ 𝑡) − 𝑗𝑠𝑖𝑛 (√ 𝑡)] 𝑚 𝑚 𝑚 𝑚

𝑥(𝑡) = 𝐶1 [cos (√

𝑘 𝑘 𝑡)] + 𝑗(𝐶1 − 𝐶2 ) [𝑠𝑖𝑛 (√ 𝑡)] 𝑚 𝑚

𝑥(𝑡) = (𝐶1 + 𝐶2 ) [cos (√

𝑘 𝑘 𝑡)] + 𝐵 [𝑠𝑖𝑛 (√ 𝑡)] 𝑚 𝑚

𝑥(𝑡) = 𝐴 [cos (√

A& B area always real , since 𝐶1 𝑎𝑛𝑑 𝐶2 are complex conjugates


2-5

𝑘 𝜔𝑛 = √ 𝑚

∴ 𝑚𝑥̈ + 𝑘 𝑥 = 0 Equation is a homogeneous second-order linear differential equation, has the following general solution:

𝑥 = 𝐴 sin 𝜔𝑛 𝑡 + 𝐵 cos 𝜔𝑛 𝑡

(General solution)

where 𝐴 and 𝐵 are the two necessary constants. These constants are evaluated from initial conditions 𝑥(0) and 𝑥̇ (0) t=0

𝑥(𝑡) = 𝑥(0) 𝑥(0) = 𝐴 sin 𝜔𝑛 (0) + 𝐵 cos 𝜔𝑛 (0) 𝑥(0) = 0 + 𝐵(1)

𝐵 = 𝑥(0) 𝑥̇ = 𝐴 𝜔𝑛 cos 𝜔𝑛 𝑡 − 𝐵 𝜔𝑛 sin 𝜔𝑛 𝑡 𝑥̇ (0) = 𝐴 𝜔𝑛 cos 𝜔𝑛 (0) − 𝐵 𝜔𝑛 sin 𝜔𝑛 (0) 𝑥̇ (0) = 𝐴 𝜔𝑛 (1) − 𝐵 𝜔𝑛 (0) 𝐴=

𝑥̇ (0)

𝜔𝑛

𝑥(0) = 𝐴 sin 𝜔𝑛 (0) + 𝐵 cos 𝜔𝑛 (0) Sub A and B in general equations

𝑥=

𝑥̇ (0)

𝜔𝑛

sin 𝜔𝑛 𝑡 + 𝑥(0) cos 𝜔𝑛 𝑡


2-6

The natural period of the oscillation is established from 𝜔𝑛 𝜏 = 2𝜋, or 𝜏 = 2𝜋√

𝑚 𝑘

and the natural frequency is 𝑓𝑛 =

1 1 𝑘 √ = 𝜏 2𝜋 𝑚

𝑓𝑛 = ∆ ∶Statical deflection

1 𝑔 √ 2𝜋 ∆

Harmonic motion : Different forms of the same solution

𝑥(𝑡) = 𝑋 sin(𝜔𝑡 + Φ) 2

𝑥̇ (0) 𝑋 = √𝐴2 + 𝐵2 = √𝑋(0)2 + ( )

𝜔𝑛

Φ = tan−1

𝑥(0)𝜔𝑛 𝑥̇ (0)


Φ: phase

2-7

 Several mechanical and structural systems can be idealized as singledegree-of-freedom systems. In many practical systems, the mass is distributed, but for a simple analysis, it can be approximated by a single point mass. Similarly, the elasticity of the system, which may be distributed throughout the system, can also be idealized by a single spring.


2-8

 Stiffness Element : 1- Spring:

𝐹 = 𝑘𝑥

2- Bar in Tension / Compression

3- Beam in bending


2-9

Beam in bending

4- Torsion


2-10

Example (1) :

A 1/4 𝑘𝑔 mass is suspended by a spring having a stiffness of 0.1533 𝑁/𝑚𝑚. Determine its natural frequency in cyc1es per second. Determine its statical deflection.

Example (2) :


2-11

Torsional Spring 𝑇 𝐺𝜃 𝜏 = = 𝐽 𝑙 𝑟 𝜃=

𝑇. 𝑙 (𝑟𝑜𝑑) 𝐺𝐽

𝐹 = 𝑘𝑥 𝑇 = 𝑘𝑡 𝜃 𝜃=

(𝑙𝑖𝑛𝑒𝑎𝑟) (𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙)

𝑘𝑡 . 𝜃. 𝑙 𝐺𝐽𝑟𝑜𝑑

(𝑟𝑜𝑑)

𝐺𝐽𝑟𝑜𝑑 𝑙

(𝑟𝑜𝑑)

𝑘𝑡 =

𝐽𝑟𝑜𝑑 : polar moment of inertia of rod ∑ 𝑇 = 𝐽0 𝜃̈

(Disk)

𝐽0 is the rotation mass moment of inertia of disk 𝑘𝑡 is the rotational stiffness 𝜃 is the angle of rotation in radians. 𝜔𝑛 is the natural frequency of oscillation Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

2-12

−𝑘𝑡 𝜃 = 𝐽0 𝜃̈ 𝜃̈ +

𝜔𝑛 = √

𝑘𝑡 𝐽0

𝑓𝑛 =

𝑓𝑛 =

𝑘𝑡 𝜃=0 𝐽0

(𝑟𝑎𝑑/𝑠𝑒𝑐)

𝜔𝑛 2𝜋

(𝐻𝑧)

1 𝑘𝑡 √ 2𝜋 𝐽0

(𝐻𝑧)

General Solution: 𝜃(𝑡) = 𝐴1 sin 𝜔𝑛 𝑡 + 𝐴2 cos 𝜔𝑛 𝑡 𝜃 (𝑡 = 0) = 𝜃0

𝑎𝑛𝑑

𝜃̇ (𝑡 = 0) =

𝑑𝜃 (𝑡 = 0) = 𝜃̇0 𝑑𝑡

𝐴1 = 𝜃0 𝜃̇0 𝐴2 = 𝜔𝑛 𝜃̇0 𝜃(𝑡) = 𝜃0 sin 𝜔𝑛 𝑡 + cos 𝜔𝑛 𝑡 𝜔𝑛 Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

2-13

EXAMPLE (3): An automobile wheel and tire are suspended by a steel rod 0.50 cm in diameter and 2 m long, as shown in Figure below. When the wheel is given an angular displacement and released, it makes 10 oscillations in 30.2 sec. Determine the polar moment of inertia of the wheel and tire. 𝐺𝑠𝑡𝑒𝑒𝑙 = 80 𝑋 109 𝑁/𝑚2

SOLUTION:

𝐽𝜃̈ = −𝑘𝑡 𝜃 10 = 2.081 𝑟𝑎𝑑/𝑠𝑒𝑐 30.2

𝜔𝑛 = 2𝜋

The torsion al stiffness of the rod is given by the equation 𝑘𝑡 = where 𝐽𝑟𝑜𝑑 =

𝜋𝑑 4 32

𝑙

,

= polar moment of inertia of the circular cross-sectional area of the rod 𝑙 = length of rod 𝐺𝑠𝑡𝑒𝑒𝑙 = 80 × 109 𝑁/𝑚2 = shear modulus of steel.

𝐽𝑟𝑜𝑑

𝑘𝑡 =

𝐺𝐽𝑟𝑜𝑑

𝜋(0.5 × 10−2 )4 = = 0.006136 × 10−8 𝑚4 32

80 × 109 × 0.006136 × 10−8

2

= 2.455 𝑁. 𝑚/𝑟𝑎𝑑

By substituting into the natural frequency equation, the polar moment of inertia of the wheel and tire (mass moment of inertia) (𝐽0 ) (𝐽𝐷𝑖𝑠𝑘 ) is 𝐽𝑤ℎ𝑒𝑒𝑙 =

𝑘𝑡 𝜔𝑛2

=

2.455 (2.081)2


= 0.567 𝑘𝑔. 𝑚2

2-14

Equivalent Spring ( Combination of springs )

Spring in Parallel


2-15

EXAMPLE: Model the system shown in figure below by a block attached to a single spring of an equivalent striffness

Solution: Left of block

1 1 1 1 1 + + + 𝑘 + 2𝑘 3𝑘 𝑘 𝑘 + 2𝑘

1 1 1 1 1 + + + 3𝑘 3𝑘 𝑘 3𝑘

=

=

𝑘 2

𝑘 2

The springs attached to the right of block are in series and are replaced by a spring of stiffness:

1 1 1 + 𝑘 2𝑘

=


2𝑘 3 2-16

Thus these springs behave as if they are in parallel and can be replaced by a spring of stiffness 𝑘 2𝑘 7𝑘 + = 2 3 6


2-17

Energy Concepts : Stiffness Elements

Mass / Inertia Elements


2-18

Energy Methods for analysis: In a conservative system, the total energy is constant. Kinetic energy is stored in the mass in terms of velocity and potential energy is stored as strain energy in the spring

𝐾𝐸 + 𝑃𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 + 𝑈 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑 (𝑇 + 𝑈) = 0 𝑑𝑡 From conservation of energy, an equilibrium on state 1 and state 2 exists – and at the extremes the maximums result in 𝑇1 + 𝑈1 = 𝑇2 + 𝑈2 where 1 and 2 represent two instances of time. Let 1 be the time when the mass is passing through its static equilibrium position and choose 𝑈1 = 0 as reference for the potential energy. Let 2 be the time corresponding to the maximum displacement of the mass. At this position, the velocity of the mass is zero, and hence 𝑇2 = 0. We then have 𝑇1 + 0 = 0 + 𝑈2 If the system is undergoing harmonic motion, then 𝑇1 and 𝑈2 are maximum values, and hence 𝑇𝑚𝑎𝑥 = 𝑈𝑚𝑎𝑥 Applying this method to the case, already considered, of a body of mass 𝑚 fastened to a spring of stiffness 𝑘, when the body is displaced a distance 𝑥 from its equilibrium position, 1 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 (𝑆𝐸) 𝑖𝑛 𝑠𝑝𝑟𝑖𝑛𝑔 = 𝑘𝑥 2 2 Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

2-19

𝑘𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 (𝐾𝐸 )𝑜𝑓 𝑏𝑜𝑑𝑦 =

1 𝑚 𝑥̇ 2 2

𝑑 1 1 ( 𝑘𝑥 2 + 𝑚 𝑥̇ 2 ) = 0 𝑑𝑡 2 2

That is 𝑚𝑥̈ 𝑥̇ + 𝑘𝑥̇ 𝑥 = 0 Or 𝑥̈ +

𝑘 𝑚

𝑥=0

This is a very useful method for certain types of problem in which it is difficult to apply Newton’s laws of motion. Alternatively, assuming SHM, if 𝑥 = 𝑥0 𝑐𝑜𝑠 𝜔𝑡

𝑇ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑆𝐸, 𝑇ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑆𝐸,

𝑈𝑚𝑎𝑥 = 𝑇𝑚𝑎𝑥 =

1 𝑘𝑥02 2

1 𝑚(𝑥0 𝜔)2 2

Thus since 𝑇𝑚𝑎𝑥 = 𝑈𝑚𝑎𝑥 1 1 𝑘𝑥02 = 𝑚(𝑥0 𝜔)2 2 2 𝜔𝑛 = √

𝑘 𝑚


(rad/sec)

2-20

EXAMPLE (4) : Determine the natural frequency of the system shown in Figure

Assume that the system is vibrating harmonically with amplitude 𝜃 from its static equilibrium position. The maximum kinetic energy is

𝐾𝐸 = 𝑇𝑚𝑎𝑥 = (𝐾𝐸𝑑𝑖𝑠𝑘 + 𝐾𝐸𝑚𝑎𝑠𝑠 ) 𝑇𝑚𝑎𝑥 = (𝐾𝐸𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 + 𝐾𝐸𝑙𝑖𝑛𝑒𝑎𝑟 ) 1 1 𝑇𝑚𝑎𝑥 = [( 𝐽𝜃̇ 2 + 𝑚𝑥̇ 2 )] 2 2 𝑚𝑎𝑥 1 1 𝑇𝑚𝑎𝑥 = [( 𝐽𝜃̇ 2 + 𝑚(𝑟1 𝜃̇ )2 ] 2 2 𝑚𝑎𝑥 The maximum potential energy is the energy stored in the spring, which is:

𝑈𝑚𝑎𝑥 =

1 2 𝑘𝑥𝑚𝑎𝑥 2


2-21

𝑈𝑚𝑎𝑥 =

1 𝑘(𝑟2 𝜃)2𝑚𝑎𝑥 2

1 1 1 [( 𝐽𝜃̇ 2 + 𝑚(𝑟1 𝜃̇ )2 ] = 𝑘(𝑟2 𝜃)2𝑚𝑎𝑥 2 2 2 𝑚𝑎𝑥 Equating the two, the natural frequency is 𝑘𝑟22 𝜔𝑛 = √ 𝐽 + 𝑚𝑟12 The student should verify that the loss of potential energy of m due to position 𝑟1 𝜃 is canceled by the work done by the equilibrium force of the spring in the position 𝜃 = 0.


2-22

EXAMPLE (5): 𝐴 link 𝐴𝐵 in a mechanism is a rigid bar of uniform section 0.3 𝑚 long. It has a mass of 10 𝑘𝑔, and a concentrated mass of 7 𝑘𝑔 is attached at 𝐵. The link is hinged at 𝐴 and is supported in a horizontal position by a spring attached at the midpoint of the bar. The stiffness of the spring is 2 𝑘𝑁/𝑚. Find the frequency of small free oscillations of the system. The system is as shown below.

SOLUTION: For rotation about A the equation of motion is 𝐽𝜃̈ = −(𝑘)(𝑎𝜃)(𝑎) 𝐽𝜃̈ = −𝑘𝑎2 𝜃 That is, 𝜃̈ + (

𝑘𝑎2 )𝜃 = 0 𝐽

This is SHM with frequency


2-23

1 𝑘𝑎2 √ 2𝜋 𝐽 In this case 𝑎 = 0.15𝑚

,

𝑙 = 0.3𝑚 , 𝑘 = 2000𝑁/𝑚

and 1 𝐽 = 𝑚𝑏𝑎𝑟 𝑙2 + 𝑚𝑏𝑎𝑙𝑙 𝑙2 3 1 𝐽 = (7)(0.3)2 + (10)(0.3)2 = 0.93 𝑘𝑔. 𝑚2 3 Hence

1 2000 × (0.15)2 √ 𝑓= = 1.1 𝐻𝑧 2𝜋 0.93


2-24

EXAMPLE (6): Determine the natural frequency of the beam-spring system shown in Figure below consisting of a weight of 𝑊 = 50.0 𝐼𝑏 attached to a horizontal cantilever beam through the coil spring 𝑘2 . The cantilever beam has a thickness ℎ = 1⁄4 𝑖𝑛, a width 𝑏 = 1 𝑖𝑛. modulus of elasticity 𝐸 = 30 × 106 psi, and length 𝐿 = 12.5 𝑖𝑛. The coil spring has a stiffness 𝑘2 = 100 𝑖𝑏/𝑖𝑛

SOLUTION:

The deflection ∆ at the free end of a uniform cantilever beam acted upon by a static force 𝑃 at the free end is given by

𝑃𝑙 3 ∆= 3𝐸𝐼 The corresponding spring constant 𝑘1 is then 𝑘1 =

𝑃 3𝐸𝐼 = 3 ∆ 𝑙

Where the cross-section moment of inertia 𝐼 Lecture notes on Mechanical Vibrations website : www.abdulrahmanbahaddin.epu.edu.krd

=

𝑏ℎ3 12

( for a rectangular section) 2-25

1 1 1 = + 𝑘𝑒𝑞 𝑘1 𝑘2 Substituting corresponding numerical values ,we obtain

(1)(1⁄4)3 1 𝐼= = 𝑖𝑛4 12 768 3 × 30 × 106 𝑘1 = = 60 𝑙𝑏/𝑖𝑛 (12.5)3 × 768 And

1 1 1 = + 𝑘𝑒𝑞 60 100

𝑘𝑒𝑞 = 37.5 𝑙𝑏/𝑖𝑛 The natural frequency for this system is then equation 𝜔𝑛 = √

𝑘𝑒𝑞

𝑚=

𝑚

𝑊 𝑔

𝑔 = 386

37.5 × 386

𝜔𝑛 = √

50

= 17.01 𝑟𝑎𝑑/𝑠𝑒𝑐

𝑓 = 2.71 𝑐𝑝𝑠


2-26

www.abdulrahmanbahaddin.epu.edu.krd


2-27

Mechanical Vibrations lecture notes ( Undamped Single Degree-of-Freedom System) (Chapter Two)

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