Mechanical Vibration, Lecture notes of Theory of Machines

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MECHANICAL VIBRATIONS-MENG 4

LECTURE 2

TUTORIALS AND CONSTANT ENERGY METHOD

Mechanical & Maintenance Engineering Dept. FBC/USL/STN-MECH VIB MENG 424

STEPWISE APPROACH TO SOLVING/ANALYSING MECHANICAL VIBRATIONS PROBLEMS USING

NEWTON’S METHOD

1. Be sure to establish the number of degrees of freedom first and formulate all

terms in those variable only.

1. Clearly adopt a sign convention with respect to the direction of motion for

each coordinate

1. Try to establish whether motion will be translational as in the case of

spring deflection under a fixed mass loading or rotational as in the case of a

pendulum system

1. Evaluate the static balance for the system in order to establish whether the

Mechanical & Maintenance Engineering Dept. FBC/USL/STN-MECH VIB MENG 424

1. Develop ONE equation for each degree of freedom of the system using

Newton’s or the energy conservation method. Do not forget to use

consistent units in the equations.

Example 1.

The motion of a particle is described by the equation 𝒙 = 𝟎. 𝟐 𝒄𝒐𝒔𝟏𝟎𝝅𝒕 +

𝝅

𝟑

), where 𝒙 is expressed in metres and t in seconds.

Determine

(a ) The period of the resulting motion.

(b) Its amplitude and

(c) Its phase angle.

Mechanical & Maintenance Engineering Dept. FBC/USL/STN-MECH VIB MENG 424

Solution:

Given the equation of motion of a particle as 𝑥 = 0. 2 𝑐𝑜𝑠10𝜋𝑡 +

1. 15 sin(10𝜋𝑡 −

𝜋

3

), let us simplify to get it into the standard form

𝑥 = 𝐴𝑠𝑖𝑛 𝜋𝑡 + 𝐵𝑐𝑜𝑠𝑡 ……………………………..(i)

Where A and B are constants.

𝑥 = 0. 2 𝑐𝑜𝑠10𝜋𝑡 + 0. 15 sin(10𝜋𝑡 −

𝜋

3

Mechanical & Maintenance Engineering Dept. FBC/USL/STN-MECH VIB MENG 424

Where X is the amplitude and  the phase angle

Expanding equation (iii) we have;

𝑥 = 𝑋𝑠𝑖𝑛 10𝜋𝑡 + 𝜑 = 𝑋𝑠𝑖𝑛10𝜋𝑡𝑐𝑜𝑠𝜑 + 𝑋𝑐𝑜𝑠10𝜋𝑡𝑠𝑖𝑛𝜑 …………………..(iv)

Comparing equations (ii and iv) by comparing coefficients of sin10лt

and cos10лt we have;

𝑋𝑠𝑖𝑛𝜑 = 0. 070 ………………………………….(v)

𝑋𝑐𝑜𝑠𝜑 = 0. 075 ………………………………….(vi)

Squaring and adding both sides of equations (v) and (iv) we have;

2

• 𝑋𝑠𝑖𝑛𝜑

2

= 0. 075

2

• 1. 070

2

2

𝑐𝑜𝑠𝜑

2

• 𝑠𝑖𝑛𝜑

2

= 0. 010525 ,

𝑋 = 0. 010525 = 0.1026 m

(a)Dividing equation (v) by (vi), we have;

𝑋𝑠𝑖𝑛𝜑

𝑋𝑐𝑜𝑠𝜑

=

1. 070

2. 075

→ 𝑡𝑎𝑛𝜑 = 0. 9333 ,

− 1

1. 9333 = 𝟒𝟑

𝒐

Solution.

Step 1 ;

Draw the free body diagram for each independent rigid body in the system.

Forces and moments acting are as follows;

(1)Tensions in spring T=kx (1)

Moment of the frictional force = M f

=F

f

r

(2)Weight of body W=mg (2)

(“Inertia of turning forces”)=I

(3)Normal reaction= R N

(4)Friction at point of contact= F f

(5)Inertia force = m xሷ

Step 2 ;

Write down all force and moment balance equations;

(i)When the body is displaced by x, the equation of motion,

σ (^) 𝐹 𝑥

𝑓

OR

σ (^) 𝐹 𝑥

𝑓

− 𝑘𝑥 = 𝑚 𝑥ሷ …………….…………..(i)

Considering the y-direction we have;

σ (^) 𝐹 𝑦

𝑁

− 𝑚𝑔 = 0 ……….…………………….…………..(ii)

(No motion in the y-direction)

𝑘

𝑚+

𝐼 𝑐𝑔

𝑟

2

Hence 𝑥ሷ + 

2

𝑥 = 0 𝜔 =

𝑘

𝑚+

𝐼 𝑐𝑔

𝑟

2

𝟐л

𝒄𝒈

𝟐

Example 3.

Using Newton’s method derive the equation of motion of the mass m for

small angular displacement , hence determine the angular frequency and

period of free oscillations of the spring loaded pendulum shown in Figure

Q 3 below with spring stiffness being k and assuming the connecting rod is

massless.

Figure Q

Solution;

For the spring loaded pendulum if an

infinitesimal angular displacement is given to the

mass m either to the left or right of the mean/ rest

position as show in the free body diagram below

we have,

σ (^) 𝑀 𝐻

𝜃 = −𝑘𝑏𝑠𝑖𝑛. 𝑏 − 𝑘𝑏𝑠𝑖𝑛. 𝑏 − 𝑚𝑔𝑠𝑖𝑛. 𝑙 ……………………..(i)

For very small , we have sin 𝜃 ≈ 𝜃 , so we have

2

𝜃 + 𝑘𝑏

2

𝜃 + 𝑚𝑔𝑙𝜃 = 0

2

• 𝑚𝑔𝑙

If 𝐼 = 𝑚𝑙

2

, we have,

2

• 𝑚𝑔𝑙

2

𝟐

 = 𝟎 ; this is the equation of motion of the mass m

Where

2

=

2

• 𝑚𝑔𝑙

2

2

• 𝑚𝑔𝑙

2

Hence the period of small oscillation is;

𝟐л

= 𝟐л

𝟐

𝟐

• 𝒎𝒈𝒍

T+V= constant. Differentiating with respect to time yields;

If the kinetic and potential energies are expressed as function of an

appropriate position coordinate (linear(x) or angular ( or )), performing the

differentiation with respect to time as indicated in equation above will yield the

required differential equation of motion.

In summary, the stepwise approach for deriving the equation of motion using

the energy method is as follows;

(1)Choose a coordinate x that describes the position of the rigid body,

noting that x must be zero in the equilibrium position.

(2)Write the expression for the potential energy in an arbitrary position.

When making small displacement approximations, we must bear in mind

that quadratic as well as linear terms in x must be kept.

(3)Write the expression for the kinetic energy in an arbitrary position.

(4)Sum the energies (potential and kinetic) and equate to a constant.

(5) Perform the differentiation on the expression for total energy in step ( 4 ).