About this previous year paper
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(x2x1)
(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=2π1mk
(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)2meω2/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=ccc=41640=0.096
Logarithmic decrement
δ=2πζ1−ζ2≈0.61\delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \approx 0.61δ=1−ζ22πζ≈0.61
Ratio of successive amplitudes
x1x2=eδ=1.84\frac{x_1}{x_2} = e^\delta = 1.84x2x1=eδ=1.84
(c) Frequency of free damped vibration
Damped natural frequency:
ωd=ωn1−ζ2\omega_d = \omega_n\sqrt{1-\zeta^2}ωd=ωn1−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=2π1mk
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=∑ωi21
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=IeqK
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=ωnx
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=n1ln(xnx1)
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.