THEORY EXAMINATION (SEM–VI) 2016-17 MECHANICAL VIBRATION

MECHANICAL VIBRATION (NME013)

SECTION – A

(10 × 2 = 20 Marks | Short Answers)

(a) Causes and effects of vibration

Causes: unbalanced masses, misalignment, looseness, gear meshing errors, external excitation.
Effects: noise, fatigue failure, wear, reduced efficiency, discomfort, and structural damage.

(b) Logarithmic decrement

Logarithmic decrement (δ) is a measure of damping in a system and is defined as the natural logarithm of the ratio of two successive amplitudes:

δ=ln⁡(x1x2)\delta = \ln\left(\frac{x_1}{x_2}\right)δ=ln(x2​x1​​) 

(c) Transmissibility

Transmissibility is the ratio of transmitted force (or motion) to the applied force (or motion). It indicates how effectively vibrations are isolated.

(d) Basic elements of vibratory system & Degree of freedom

Elements: mass, stiffness (spring), damping.
Degree of freedom (DOF): number of independent coordinates required to describe the motion of the system.

(e) Steady-state response in forced vibration

It is the constant-amplitude response of a system after transient vibrations have died out, occurring at the forcing frequency.

(f) Whirling of shaft and critical speed

Whirling is the lateral vibration of a rotating shaft.
Critical speed is the speed at which the natural frequency equals the rotational frequency, causing large deflections.

(g) Multifilar system and applications

A multifilar system consists of multiple strings or wires supporting a mass.
Applications: vibration isolation, torsional pendulums, suspended platforms.

(h) Natural frequency of mass–spring system

For a mass m suspended from a spring of stiffness k:

fn=12πkmf_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}}fn​=2π1​mk​​ 

(i) Under, over and critical damping

Underdamping: oscillatory motion with decreasing amplitude

Critical damping: fastest return to equilibrium without oscillation

Overdamping: slow return to equilibrium without oscillation

(j) Vibration isolation

Vibration isolation is the technique of reducing transmission of vibration from source to receiver using springs, dampers, or isolators.

SECTION – B

(Attempt Any Five | 5 × 10 = 50 Marks)

(a) Steady-state amplitude due to unbalance

For a rotating unbalanced mass, steady-state amplitude:

X=meω2/k(1−r2)2+(2ζr)2X = \frac{me\omega^2/k}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}X=(1−r2)2+(2ζr)2​meω2/k​

where
r = frequency ratio, ζ = damping factor

Plots:

Amplitude vs frequency ratio                Peak occurs near resonance

Damping reduces peak amplitude        (b) Damped vibrating system (Numerical)

Given:                                                     m = 8 kg
k = 5.4 N/mm = 5400 N/m                    Damping force = 40 N at 1 m/s → c = 40 Ns/m

Critical damping coefficient

cc=2km=25400×8=416 Ns/mc_c = 2\sqrt{km} = 2\sqrt{5400 \times 8} = 416 \text{ Ns/m}cc​=2km​=25400×8​=416 Ns/m

Damping factor

ζ=ccc=40416=0.096\zeta = \frac{c}{c_c} = \frac{40}{416} = 0.096ζ=cc​c​=41640​=0.096

Logarithmic decrement

δ=2πζ1−ζ2≈0.61\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \approx 0.61δ=1−ζ2​2πζ​≈0.61

Ratio of successive amplitudes

x1x2=eδ=1.84\frac{x_1}{x_2} = e^\delta = 1.84x2​x1​​=eδ=1.84 

(c) Frequency of free damped vibration

Damped natural frequency:

ωd=ωn1−ζ2\omega_d = \omega_n\sqrt{1-\zeta^2}ωd​=ωn​1−ζ2​

where

ωn=km\omega_n = \sqrt{\frac{k}{m}}ωn​=mk​​ 

(d) Spring-mass system with changed mass

Given:

f1=3.56 Hz,f2=2.9 Hzf_1 = 3.56\text{ Hz}, \quad f_2 = 2.9\text{ Hz}f1​=3.56 Hz,f2​=2.9 Hz

Using:

f=12πkmf = \frac{1}{2\pi}\sqrt{\frac{k}{m}}f=2π1​mk​​

Solving simultaneous equations gives:
 

Original mass = 4.8 kg
Spring constant = 2400 N/m

(e) Natural frequency of simply supported beam

(a) Energy method:
Equates maximum strain energy to maximum kinetic energy.

(b) Dunkerley’s method:

1ω2=∑1ωi2\frac{1}{\omega^2} = \sum \frac{1}{\omega_i^2}ω21​=∑ωi2​1​

Used for approximate lower bound frequency.

(f) Torsional vibration of two-rotor system

Natural frequency obtained using:

ωn=KIeq\omega_n = \sqrt{\frac{K}{I_{eq}}}ωn​=Ieq​K​​

where K is torsional stiffness and I is mass moment of inertia.

(g) Shaft torsional vibration (Numerical)

Given:
d = 100 mm, L = 1 m
m = 500 kg, k = 0.45 m
G = 80 GN/m²

K=GJLK = \frac{GJ}{L}K=LGJ​ I=mk2I = mk^2I=mk2

Substituting values:

fn=2.1 Hzf_n = 2.1 \text{ Hz}fn​=2.1 Hz 

(h) Gun recoil problem

Given:
m = 300 kg
k = 250 N/mm = 250000 N/m
x = 0.8 m

Critical recoil velocity

vc=ωnxv_c = \omega_n xvc​=ωn​x

Critical damping coefficient

cc=2kmc_c = 2\sqrt{km}cc​=2km​ 

SECTION – C

(Attempt Any Two | 2 × 15 = 30 Marks)
 

Whirling speed of shaft with distributed mass

Steps:

Calculate shaft mass using density and volume

Use Rayleigh’s method:

ω2=g∑Wy∑Wy2\omega^2 = \frac{g\sum Wy}{\sum Wy^2}ω2=∑Wy2g∑Wy​

Include mass of shaft and wheels

Lowest whirling speed obtained from minimum ω

 Damped oscillations of spring–mass system

Given:
60 oscillations in 35 s → damped frequency

fd=6035f_d = \frac{60}{35}fd​=3560​

Using logarithmic decrement from displacement ratio:

δ=1nln⁡(x1xn)\delta = \frac{1}{n}\ln\left(\frac{x_1}{x_n}\right)δ=n1​ln(xn​x1​​)

From this:

Spring stiffness

Damping resistance

Critical damping resistance

 Three-rotor vibratory system

A three-rotor system consists of three inertias connected by shafts.

Steps:

Write torsional stiffness equations

Assume harmonic motion

Solve characteristic equation

Determine amplitude ratios from mode shapes

Used in turbines, generators, and multi-disc systems.