C:\Users\trainee\AppData\Local\Temp\msoEAA3.tmp

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]









Engineering Mechanics











Contents

Introduction ..................................................................................................... 2

Units and dimensions ...................................................................................... 4

Equations of motion ....................................................................................... 9

Newton’s laws of motion ............................................................................... 37

Concept of friction .......................................................................................... 61
Curvilinear motion .......................................................................................... 77

Gyroscopic couple ........................................................................................... 89

Free Body Diagram ......................................................................................... 92

































QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 1 age 1
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Šƒ’–‡” ʹ - ‰‹‡‡”‹‰ ‡…Šƒ‹…•


ʹǤͳ –”‘†—…–‹‘

Š‡ ”‡ƒ•‘ ‡‰‹‡‡”‹‰ ‡…Šƒ‹…• ‹• ‰‹˜‡ –Š‡ ‘•– ”—†‹‡–ƒ”› •–ƒ–—• ‹• „‡…ƒ—•‡ ‹– ’”‘˜‹†‡• –Š‡
’‡†‡•–ƒŽ ˆ‘” ‡‰‹‡‡”‹‰ †‡•‹‰ ƒ…”‘•• †‹•…‹’Ž‹‡•Ǥ
‡…Šƒ‹…• ‹• ƒŽŽ ƒ„‘—– ϐ‹†‹‰ –Š‡ ‹–‡”ƒ…–‹‘ „‡–™‡‡ …‘’‘‡–•ǡ ƒ‡Ž›ǡ ƒ…–‹‘ ƒ† ”‡ƒ…–‹‘ǡ ‹
–Š‡ •‹’Ž‡•– –‡”•Ǥ Š‹• …Šƒ’–‡” ’”‹ƒ”‹Ž› ˆ‘…—•‡• ‘ •‘Ž˜‹‰ •–ƒ–‹… ƒ† †›ƒ‹… •‹–—ƒ–‹‘• ˆ‘”
˜ƒ”‹‘—• ’”‘†—…–• •‘ –Šƒ– –Š‡ ’”ƒ…–‹…ƒŽ‹–› ‘ˆ ‡…Šƒ‹…• ‹• ™‡ŽŽ ƒ’’”‡…‹ƒ–‡† „› ›‘—‰ ‡‰‹‡‡”•Ǥ Š‹•
…Šƒ’–‡” ƒŽ•‘ Žƒ›• ƒ •–”‘‰ ˆ‘—†ƒ–‹‘ ˆ‘” ƒ–‡”‹ƒŽ ‡…Šƒ‹…•ǡ ‡…Šƒ‹••ǡ ˜‹„”ƒ–‹‘• ƒ†
‡‰‹‡‡”‹‰ †‡•‹‰Ǥ
Š‡ ‡…Šƒ‹…ƒŽ ƒ† •’ƒ–‹ƒŽ ˜‹•—ƒŽ‹œƒ–‹‘ –Šƒ– ‹• „—‹Ž– –Š”‘—‰Š –Š‹• …Šƒ’–‡” ‰‘‡• ƒ Ž‘‰ ™ƒ› ‹
”‡ϐ‹‹‰ ‡‰‹‡‡”‹‰ –Š‹‹‰Ǥ
‡…Šƒ‹•• ‹• •ƒ‹† –‘ „‡ ‘‡ ‘ˆ –Š‡ …‘”‡ …‘’‡–‡…‹‡• ˆ‘” ƒ ‡…Šƒ‹…ƒŽ ‡‰‹‡‡”Ǥ Š‹• …Šƒ’–‡”
ˆ‘…—•‡• ‘ –Š‡ •…‹‡…‡ ‘ˆ ‡…Šƒ‹•• ƒ† ‘– ‘ –Š‡ –‡…Š‘Ž‘‰›Ǥ
Š‡ ˆ‘ŽŽ‘™‹‰ ‡‰‹‡‡”‹‰ ™Š‡‡Ž ‹• —•‡† –‘ ‡’Šƒ•‹œ‡ –Š‡ •‹‰‹ϐ‹…ƒ…‡ ‘ˆ ‡…Šƒ‹…•ǣ























Š‡ ‡‰‹‡‡”‹‰ ™Š‡‡Ž ‰‹˜‡• ƒ „‹”†ǯ• ‡›‡ ˜‹‡™ ‘ˆ ˜ƒ”‹‡† ƒ’’Ž‹…ƒ–‹‘• –Šƒ– ƒ”‡ ‡ƒ„Ž‡† „› –Š‡
’”‹…‹’Ž‡• ‘ˆ ‡…Šƒ‹…•Ǥ Š‹• „‘‘ Ž‹‹–• ‹–• †‹•…—••‹‘ –‘ ‡™–‘‹ƒ ‡…Šƒ‹…• ȋ ƒ‰”ƒ‰‹ƒ
‡…Šƒ‹…• ‹• ‘– †‹•…—••‡†ȌǤ


ʹǤʹ „Œ‡…–‹˜‡

Š‡ ‹–‡–‹‘ ‘ˆ –Š‡ …Šƒ’–‡” ‹• –‘ ‹–”‘†—…‡ ˆ—†ƒ‡–ƒŽ• ‘ˆ ‡…Šƒ‹…• ˆ‘” •–”‡•• ƒ† ˜‹„”ƒ–‹‘
ƒƒŽ›•–•Ǥ ‘ŽŽ‘™‹‰ ƒ”‡ –Š‡ ‡› –ƒ‡ƒ™ƒ›• ˆ”‘ –Š‹• …Šƒ’–‡”ǣ
ͳǤ ‘ ’”‘˜‹†‡ …Žƒ”‹–› ‘ ‘”‹‰‹ƒ–‹‘ ‘ˆ ƒŽŽ ‡…Šƒ‹…ƒŽ ƒ† ‹‡”–‹ƒŽ ˆ‘”…‡•Ǥ
ʹǤ …—Ž…ƒ–‡ –Š‡ …—Ž–—”‡ ‘ˆ ˆ‘” …‘’‘‡– ‹–‡”ƒ…–‹‘ ƒ† Ž‘ƒ† ’ƒ–Š …‘’”‡Š‡•‹‘Ǥ
͵Ǥ ‡˜‡Ž‘’ –Š‡ ƒ„‹Ž‹–› –‘ ƒ’’Ž› ‡™–‘ǯ• ʹ Žƒ™ ƒ† ‹‡ƒ–‹… ”‡Žƒ–‹‘• –‘ ’”ƒ…–‹…ƒŽ
†
•‹–—ƒ–‹‘•


Page 2 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

2.3 Practical Value Addition

Ø Units and dimensions
This chapter helps establish equations of physics for various engineering situations.
This chapter also emphasizes the need for conversion of units and its correctness from
the engineering simulation perspective.
Significant Practical Application: Analysis of windmill

Ø Equations of motion
This chapter treats equations of motion from an engineering perspective. Example, the fact
that, the equations of motion are fundamental in the explicit and implicit dynamic analyses,
is given a detailed discussion. Also, facts such as, the effect of Coriolis force on the target of
a missile, the analysis of stopping distance of a vehicle, optimization of projectile path and
others are discussed in detail.
Significant Practical Application: Optimization of projectile path and analysis of
stopping distance of cars.

Ø Dynamic analysis
In this chapter basic conservation laws of energy and momentum are well exploited.
Concepts of moment of momentum, rate of change of angular momentum and resulting Gyro
and Coriolis effects are given a detailed treatment. The chapter provides enough clarity on
rolling and sliding friction. Concept of rolling is covered in its entirety. Curvilinear motion is
given a detailed discussion via vectors. The highlight of the chapter is the clarity and
visualization given to the concept of mass moment of inertia matrix for 3D objects. This is
significant not just for mechanics, but, even vibrations, rotor-dynamics and other
applications.
Significant Practical Application: Generalized treatment of a toy car subjected to
several dynamic forces.

Ø Free body diagram
The chapter deals with all the rules of constructing FBD for both dynamic and static
situations. The rules cover, contact forces, Pseudo forces and Joints (welded, bolted and
hinged).
Significant Practical Application: Rotating overhung shaft system with Gyro effect.













QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 3 age 3
Copyright Diary No – 9119/2018-CO/L

Page 4
Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

ʹǤͶ ‹–• ƒ† ‹‡•‹‘•

‹–• ƒ† †‹‡•‹‘• ‹• ‘‡ ‘ˆ –Š‡ ‘•– ’‘™‡”ˆ—Ž –‘‘Ž• –Šƒ– …‘—Ž† „‡ —•‡† ‹ ƒ› •‹–—ƒ–‹‘ǡ
‹””‡•’‡…–‹˜‡ ‘ˆ –Š‡ ‡‰‹‡‡”‹‰ †‹•…‹’Ž‹‡ǡ –‘ ‡•–ƒ„Ž‹•Š –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ –Š‡ ’Š›•‹…ƒŽ
’ƒ”ƒ‡–‡”• ‹ “—‡•–‹‘Ǥ
šƒ’Ž‡ǣ ‹– …‘—Ž† „‡ ƒ •‹’Ž‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ǣ
· ‹•’Žƒ…‡‡–ǡ ƒ……‡Ž‡”ƒ–‹‘ ƒ† –‹‡Ǥ
· ‘™‡”ǡ ƒ‰—Žƒ” ˜‡Ž‘…‹–› ƒ† –‘”“—‡ Ǥ

· Š”—•–ǡ †‡•‹–›ǡ ™‹†‹ŽŽ ǡ ”ƒ†‹—• ‘ˆ –Š‡ ƒ‡”‘ ˆ‘‹Ž ƒ† ™‹† ˜‡Ž‘…‹–›Ǥ
Š‡ ’Š‹Ž‘•‘’Š› ‘ˆ —‹–• ƒ† †‹‡•‹‘• ‹• –‘ ‹†‡–‹ˆ› ƒŽŽ –Š‡ ’Š›•‹…ƒŽ ’ƒ”ƒ‡–‡”• ‹ ƒ ‰‹˜‡
•‹–—ƒ–‹‘ǡ •—…Š –Šƒ–ǡ –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ ƒ› –™‘ ’ƒ”ƒ‡–‡”• …‘—Ž† „‡ ‡•–ƒ„Ž‹•Š‡†Ǥ ‘™‡˜‡”ǡ

‘‡ ‘ˆ –Š‡ Ž‹‹–ƒ–‹‘• ‘ˆ —‹–• ƒ† †‹‡•‹‘• ‹• –Šƒ–ǡ ‹– …ƒ‘– „‡ —•‡† –‘ ϐ‹† ‘—– –Š‡
…‘•–ƒ–• ‘ˆ ’”‘’‘”–‹‘ƒŽ‹–›Ǥ

Example1: Consider the relationship between displacement, acceleration and time. This is
y z
x
mathematically given by: s = a t - - - - (2.1)
Where, (s) is the displacement, (a) is the acceleration and (t) is time, and (x, y and z) are their
respective exponents that need to be computed.

Parameters Units Dimensions


Displacement (s) Meters (m) Length (L)


m Length ∗ Time LT −2
−2
2
Acceleration (a) meter/second
s 2
Time (t) Seconds(t) Time(T)


Writing the dimension of parameters in equation (2.1), we get:

L y
x
z
y
y
x
z
y z
x
x
s = a t = L = T = L = L T −2y T = L = L T −2y+z - - - - (2.2)
T 2
By equating the exponents of the respective dimensions of LHS and RHS from equation (2.2),
we get:
(x = y) : powers of ‘L’ on both sides
(-2y + z = 0) : comparing powers of time (T) on both sides
Therefore, we have, (z = 2y=2x)
Since we seek displacements relation with acceleration and time, let the power of displacement

(x) be 1.

Page 4 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Since (x = 1), we have (z = 2)

2
This given us the relationship: s = at - - - - (2.3)
It is to be noted that, the actual relationship between displacement, acceleration and time, for a
straight line motion, is given by:

1
2
s = at
2
The limitation of this approach is that, the constant (1/2) cannot be computed.


šƒ’Ž‡ ʹǣ ‹– …ƒ „‡ ƒ”‰—‡† …‘‘•‡•‹…ƒŽŽ› –Šƒ–ǡ –Š‡ –Š”—•– †‡˜‡Ž‘’‡† „› ƒ ™‹†‹ŽŽ
†‡’‡†• ‘ ™‹† ˜‡Ž‘…‹–›ǡ †‡•‹–› ‘ˆ –Š‡ ƒ‹”ǡ ‰‡‘‡–”› ‘ˆ –Š‡ ƒ‡”‘ ˆ‘‹Ž ȋ”ƒ†‹—• ‘ˆ –Š‡ ƒ‡”‘ ˆ‘‹ŽȌ ƒ†
‘ˆ –Š‡ ™‹†‹ŽŽǤ

a b
c d
Mathematically, thrust is given by: T = v ρ N r - - - - (2.4)
Š‡”‡ǡ ȋƒǡ „ǡ … ƒ† †Ȍ ƒ”‡ ‡š’‘‡–• ‘ˆ –Š‡‹” ”‡•’‡…–‹˜‡ –‡”•Ǥ


Parameters Units Dimensions


Thrust (T) Newton (N) Mass ∗ Acceleration(MLT −2 )

m −1
Velocity (v) meter/second Length/time LT
s

−3
kg Mass/Volume ML
Density (ρ) kilogram/meter
3
m 3
Radius (r) Meters (m) Length (L)

1 (T −1 )
RPS (N) revolutions/second
s


Writing the dimension of parameters in equation (2.4), we get:

b
c d
a b
L = MLT
ML T
T = v ρ N r = MLT −2 = LT −2 a 3 b −1 c d −2 = L a+d−3b T −a−c M - - - - (2.5)
By equating the exponents of the respective dimensions of LHS and RHS from equation (2.5), we get:
(b = 1) : exponents of mass (M)
(a + c = 2) : exponents of time (T)

(a + d – 3b = 1) : exponents of length (L)

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 5 age 5
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

Putting (b = 1) in the exponents of length, we get: (a + d = 4)

Subtracting (a + c = 2) from (a + d = 4), we get: (d – c = 2), by where, (d = 2 + c) and also (a = 2 – c).

Rewriting equation (2.4) in terms of exponent ‘c’, we get:

2
c
c
T = v (2−c) ρ N r (2+c) = T = v 2−c ρ Nr r

(Nr) is equivalent to the tangential velocity of the blade (ωr). Therefore, combining tangential
velocity (Nr) with velocity (v), assuming that air hugs the aero foil perfectly (no slip), we get:
T = v 2−c ρ Nr r = T = v 2−c ρ v r = T = v ρ r
2
c
2
2
c
2
Recasting the above expression in terms of RPM (N), we get:
Or T = (Nr) 2−c ρ Nr r = T = N r ρ
c
2
2 4
Further, for the power generated, we get:
2
3
Power = Thrust ∗ Velocity = P = T ∗ v = P = v ρ r
The above expression for power (P) can also be written as: P = N ρ r
5
3
Where, [N (RPM) * r (radius) = v (velocity)]

This clearly establishes that, the typical fan power is equal to cube of the RPM.



Example 3: let us relate power to torque and angular velocity.

Mathematically, power is given by: P = T ω - - - - (2.6)
a
b

Where, (a and b) are exponents of their respective terms.


Parameters Units Dimensions

J 2 −3
Power (P) Joules/second newton meter/second ( T )
s

Torque (T) newton meter(Nm) (ML T )
2 −2
1 −1
Angular velocity (ω) radians/second T
t






Page 6 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Writing the dimension of parameters in equation (2.6), we get:


b
a
2a
a
2 −3
2 −2 a
P = T ω = ML T = ML T T −1 b 2 −3 = M L T −2a T −b - - - - (2.7)
= ML T

By equating the exponents of the respective dimensions of LHS and RHS from equation (2.7), we

get:

(a = 1) : exponent of mass (M)
(2a = 2) : exponent of length (L)
(-2a – b = -3) : exponent of time (T)
Substituting (a = 1) in the exponent of time (T), we get: (b = 1).
Substituting the exponent (a and b) in equation (2.7), we get:

a
P = T ω = P = Tω
b

It can be observed that, as the torque increases, for a given power, the RPM decreases.

This is equivalent to the first gear of an automobile providing maximum torque with minimum

RPM. Whereas, the maximum gear, say, the fifth gear, provides maximum RPM with minimum
torque.



2.4.1 Unit Conversion


In the simulation of components, the units play a major role, and in many situations, they need to

be converted from one system of units to another.

Example 1: convert the British unit of density to SI unit, using the conversion factor to find the
density of steel, which is given to be, 498.3 pounds per cubic feet.

The conversion factor is got as follows:

mass m kg
density = = ρ = =
volume V L 3

Let (M L ) be the SI unit dimensions and (M L ) be the British unit dimensions.
2 2
1 1

The conversion factor is given by:

ρ SI M L −3 M 1 L 1 −3
1 1
= =
ρ British M L −3 M 2 L 2
2 2

We know that, 1 pound is approximately equal to 0.45 kg, and 1 foot is equal to 0.3048 meters.



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 7 age 7
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Substituting these values for (M and L ) and a value of 1 for (M and L ) we get:
1
2
2
1
M 1 L 1 −3 1 1 −3
= = 0.06286
M 2 L 2 0.45 0.3048


The density in SI unit is given as follows:
ρ = ρ British = ρ = 498.3 ≈ 7909.52 kg
SI
conversion factor SI 0.06286 m 3

Example 2: A component modeled in millimeters, whose vibration or dynamic

analysis is carried out. Determine the appropriate unit of density for numerical
simulation.

Solution: consider a dynamic force, say, centrifugal force, which is given by mω r .
2

therefore, we have:

2
2
F = mω r , this can be rewritten as F = volume ∗ density ∗ ω r - - - - (2.8)
Let us rewrite equation (2.8) in terms of their respective units.
kg 1
3
N = mm ∗ 3 ∗ 2 ∗ mm - - - - (2.9)
mm s
3
The unit of density is chosen to be kg mm , as the model is in mm.


Simplifying equation (2.9), we get:

kg
N = mm - - - - (2.10)
s 2
Since newton N has the units kg m , we could conclude that the RHS of equation . is not in
s 2
newton.

This shows, that, the density must be entered in tonne mm . Therefore, substituting
3

the new unit of density in equation (2.9), we get:

tonne 1 kg × 10 3 kg m
3
N = mm ∗ ∗ ∗ mm = mm =
mm 3 s 2 s 2 s 2

One could test the fact that, for a dynamic analysis, the computed natural frequencies go
3
completely incorrect if the density is input in terms of kg mm instead of


3
tonne mm .




Page 8 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Let us also verify the fact discussed, for [F = ma]. Writing the units for [F = ma], we get:

mm tonne mm kg × 10 3 kg
3
N = volume ∗ density ∗ = mm ∗ ∗ = mm = m
s 2 mm 3 s 2 s 2 s 2
Therefore, for all inertial forces, where density is involved, its unit must be in tonne mm . This
3

fact only holds for engineering simulation or FEM analysis.

2.5 Equations of Motion


2.5.1 Straight Line Motion

The analysis of straight line motion is useful, not just for vehicles, but also for mechanisms and
mathematical schemes, used for transient analyses. These equations are derived based on
conservation of energy and work done. The set of equations that are going to be derived for linear
motion are also going to be used for angular motion.

For curvilinear motion, we need new acceleration terms, to define the motion, which would be
taken up subsequently.



2.5.2 Energy and Work done























[Fig 2.1: mass(m) travelling on a frictionless platform]





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 9 age 9
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Š‡ ƒ ‡š–‡”ƒŽ ˆ‘”…‡ ƒ…– ‘ –Š‡ ‘„Œ‡…–ǡ –Š‡ ‘„Œ‡…– ‡‹–Š‡” †‡ˆ‘”• ‘” ƒ……‡Ž‡”ƒ–‡•Ǥ Š‹• ”‡•—Ž–• ‹
™‘” †‘‡ „› –Š‡ ‡š–‡”ƒŽ ˆ‘”…‡ ‘ –Š‡ ‘„Œ‡…–Ǥ

 …ƒ•‡ ‘ˆ ‡Žƒ•–‹… †‡ˆ‘”ƒ–‹‘ǡ –Š‡ „‘†› •–‘”‡• ’‘–‡–‹ƒŽ ‡‡”‰›Ǥ –Š‡”™‹•‡ǡ –Š‡ „‘†› †‡˜‡Ž‘’•
‹‡–‹… ‡‡”‰› „‡…ƒ—•‡ ‘ˆ –Š‡ ˜‡Ž‘…‹–› ‹†—…‡†Ǥ ‹‡–‹… ‡‡”‰› ‹• …ƒ—•‡† †—‡ –‘ ‘–‹‘ ƒ••‘…‹ƒ–‡†
™‹–Š –Š‡ „‘†›Ǥ Ž•‘ ‘–‡ –Šƒ–ǡ ‘•– ‘ˆ –Š‡ „‘†‹‡• ™‡ †‡ƒŽ ™‹–Šǡ ‹ –Š‹• „‘‘ǡ ƒ”‡ ”‹‰‹†ǡ ‡ƒ‹‰ǡ
–Š‡› Šƒ˜‡ œ‡”‘ ‡Žƒ•–‹…‹–›Ǥ ‘™‡˜‡”ǡ –‘ ”‡’”‡•‡– –Š‡ ‡Žƒ•–‹…‹–›ǡ •’”‹‰• ƒ”‡ —•‡†ǡ ™Š‹…Š ‘
…‘’”‡••‹‘ǡ •–‘”‡ ’‘–‡–‹ƒŽ ‡‡”‰›ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤͳȐǤ

‘–Š‡” ˆ‘” ‘ˆ ’‘–‡–‹ƒŽ ‡‡”‰›ǡ ‹• –Š‡ ‡‡”‰› †—‡ –‘ ‰”ƒ˜‹–›ǡ –Šƒ– ‹•ǡ ™Š‡ –Š‡ „‘†› ‹• ‘˜‡†
ˆ”‘ ’‘•‹–‹‘ Ǯ ǯ –‘ Ǯ ǯ ƒ‰ƒ‹•– ‰”ƒ˜‹–›ǡ ™‘” ‹• •–‘”‡† ƒ• ‰”ƒ˜‹–ƒ–‹‘ƒŽ ’‘–‡–‹ƒŽ ‡‡”‰›ǡ ƒ• •Š‘™
‹ ȏ ‹‰ ʹǤʹȐǤ












[Fig 2.2: work done in displacing the car from position ‘A’ to position ‘B’]

‘–‡ –Šƒ–ǡ ‰”ƒ˜‹–ƒ–‹‘ƒŽ ϐ‹‡Ž† ‹• …‘•‡”˜ƒ–‹˜‡ ƒ† –Š‡ ™‘” †‘‡ ‹• ‹†‡’‡†‡– ‘ˆ –Š‡ ’ƒ–ŠǤ Š‡
™‘” †‘‡ ‹• …ƒŽ…—Žƒ–‡† „› –Š‡ •Š‘”–‡•– †‹•–ƒ…‡ǡ –Šƒ– ‹•ǡ –Š‡ †‹•’Žƒ…‡‡– „‡–™‡‡ –™‘ ’‘‹–• ‹
–Š‡ ˜‡”–‹…ƒŽ †‹”‡…–‹‘ ȋƒ‰ƒ‹•– ‰”ƒ˜‹–›ȌǤ
‡– —• ‘™ †‡”‹˜‡ –Š‡ ‡š’”‡••‹‘ ˆ‘” ‹‡–‹…ǡ ’‘–‡–‹ƒŽ ƒ† •’”‹‰ ‡‡”‰‹‡•Ǥ

Kinetic Energy: A force (F) acts on a mass (m) for (t-seconds), as shown in [Fig 2.1]. Find the
kinetic energy of the mass.

Let the displacement of the body, during (t-seconds) be ‘d’. therefore, the work done by the force is
given by: work done = Force ∗ average displacement
Average displacement is considered, as the body is accelerating uniformly, and travels different
displacements in different intervals of time. Mathematically, we have:

s
Work done = F ∗ s avg = ma ∗ - - - - (2.11)
2
Let us assume that the velocity of the body is ‘v’, after an interval of time ‘t’. Therefore, equation
(2.11) can be rewritten as:


velocity mv (v + 0) 1
2
Work done = ma ∗ s avg = m ∗ velocity avg ∗ time = ∗ t = mv
time t 2 2






Page 10 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


”ƒ’Š‹…ƒŽŽ›ǡ ‹‡–‹… ‡‡”‰› …‘—Ž† „‡ ”‡’”‡•‡–‡† ƒ• ˆ‘ŽŽ‘™•ǣ










[Fig 2.3: work done by a force (F) over a displacement (s)]



”ƒ˜‹–ƒ–‹‘ƒŽ ‘–‡–‹ƒŽ ‡”‰›ǣ –Š‹• ‹• ‰‹˜‡ „› –Š‡ ’”‘†—…– ‘ˆ ‰”ƒ˜‹–ƒ–‹‘ƒŽ ’—ŽŽ ƒ†
†‹•’Žƒ…‡‡–Ǥ Š‹• ‹• ƒ–Š‡ƒ–‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• ˆ‘ŽŽ‘™•ǣ

work done = force ∗ displacement = F ∗ s = mg ∗ s













[Fig 2.4: work done against gravity/self weight]



‘‹– –‘ ’‘†‡”ǣ ™Š‡ ƒ ‘„Œ‡…– ‹• Ž‹ˆ–‡†ǡ –Š‡ ˆ‘”…‡ ƒ’’Ž‹‡† ‹ Ž‹ˆ–‹‰ –Š‡ ‘„Œ‡…– Šƒ• –‘ „‡
‰”‡ƒ–‡” –Šƒ –Š‡ ™‡‹‰Š– ‘ˆ –Š‡ ‘„Œ‡…–ǡ ‡ƒ‹‰ǡ –Š‡ ‘„Œ‡…– ‹• ƒ……‡Ž‡”ƒ–‹‰Ǥ ‘™‡˜‡”ǡ –Š‡ ‘„Œ‡…– ‹•
‰‘‹‰ –‘ •–‘’ ƒ– ƒ …‡”–ƒ‹ Š‡‹‰Š–ǡ ‡ƒ‹‰ǡ –Š‡ ‘„Œ‡…– ‹• †‡…‡Ž‡”ƒ–‡† ȋŠ‡”‡ǡ –Š‡ ˆ‘”…‡ ‡š‡”–‡† ‹• Ž‡••
–Šƒ –Š‡ ™‡‹‰Š–ȌǤ Š‡”‡ˆ‘”‡ǡ –Š‡ ˆ‘”…‡ ƒ’’Ž‹‡† ‹• ‘”‡ –Šƒ –Š‡ •‡Žˆ™‡‹‰Š– ‘ˆ –Š‡ ‘„Œ‡…– ˆ‘” ‘‡ ŠƒŽˆ
‘ˆ –Š‡ ‘–‹‘ ƒ† Ž‡•• –Šƒ –Š‡ •‡Žˆ™‡‹‰Š– ˆ‘” –Š‡ •‡…‘† ŠƒŽˆ ‘ˆ –Š‡ ‘–‹‘Ǥ Š‡ ƒ˜‡”ƒ‰‡ ˆ‘”…‡
‡š‡”–‡†ǡ ‹• –Š‡ •‡Žˆ™‡‹‰Š– ‹–•‡ŽˆǤ











QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 11 age 11
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Spring Potential Energy: this is given by:

work done = Spring Force ∗ average displacement

This is because, the spring force is growing linearly with displacement.

x 1
2
work done = kx ∗ s = kx ∗ = kx
avg
2 2











[Fig 2.5: work done by a spring force (F) over a stretch (x)]


šƒ’Ž‡ǣ ƒ ’‘‹– ƒ•• ‹• ”‡Ž‡ƒ•‡† ˆ”‘ ƒ Š‡‹‰Š– ȋ Ȍ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤ͸ȐǤ ‹† –Š‡ …‘’”‡••‹‘
‹ –Š‡ •’”‹‰ǡ ƒ••—‹‰ ƒ ‹‹–‹ƒŽ …‘†‹–‹‘ –Šƒ–ǡ –Š‡ •’”‹‰ ‹• ‹ ‹–• ƒ–—”ƒŽ Ž‡‰–ŠǤ

‘Ž—–‹‘ǣ –Š‹• ’”‘„Ž‡ ‹• „ƒ•‡† ‘ –Š‡ ’”‹…‹’Ž‡ ‘ˆ ‡‡”‰› ƒ† ™‘” ”‡Žƒ–‹‘Ǥ ‹–‹ƒŽŽ›ǡ –Š‡ ’‘‹–
ƒ••ǡ Šƒ• –‘–ƒŽ ‰”ƒ˜‹–ƒ–‹‘ƒŽ ’‘–‡–‹ƒŽ ‡‡”‰›ǡ ƒ ’ƒ”– ‘ˆ ™Š‹…Š ‹• •’‡– ƒ‰ƒ‹•– ˆ”‹…–‹‘Ǥ Š‡
”‡ƒ‹‹‰ ‹• …‘˜‡”–‡† –‘ ‰”ƒ˜‹–ƒ–‹‘ƒŽ ’‘–‡–‹ƒŽ ‡‡”‰› ƒ† ‡Žƒ•–‹… •’”‹‰ ‡‡”‰›Ǥ
‡– —• †‡˜‡Ž‘’ ƒ ‡š’”‡••‹‘ ˆ‘” Š‡‹‰Š– ȋ Ȍ –Šƒ– ƒŽŽ‘™• …‡”–ƒ‹ •’”‹‰ …‘’”‡••‹‘Ǥ

















[Fig 2.6: a point mass negotiating an inclined track]





Page 12 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


‘‹– –‘ ’‘†‡”ǣ ™‘” †‘‡ ˆ‘” ‘–‹‘ ƒŽ‘‰ ƒ …—”˜‡ ‘” ƒ ‹…Ž‹‡† •–”ƒ‹‰Š– Ž‹‡ „› ˆ”‹…–‹‘
ƒŽ™ƒ›• †‡’‡†• ‘ –Š‡ Š‘”‹œ‘–ƒŽ †‹•’Žƒ…‡‡–ǡ ˆ”‹…–‹‘ …‘‡ˆϐ‹…‹‡– ƒ† ™‡‹‰Š–Ǥ ‡– —• ƒƒŽ›œ‡ –Š‡
•ƒ‡ǣ
‘•‹†‡” ƒ ’‘‹– ƒ•• ‡‰‘–‹ƒ–‹‰ ƒ …—”˜‡† ˆ”‹…–‹‘ƒŽ –”ƒ…ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤ͹ȐǤ


















- - - - (2.12)







[Fig 2.7: a point mass negotiating a frictional curved track]

The frictional work done is given as:

Frictional work done = N ∗ μ ∗ ds arc length = mg cosθ ∗ μ ∗ ds

We know from [Fig 2.7] that, ds cos θ = dx

Therefore, the frictional work done is given by: mg ∗ μ ∗ dx

Integrating the frictional work done for the entire curve between the limits of ‘x’ (0 to L), we get:

L
Total friction work done = μmg dx = μmg L
0






QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 13 age 13
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Š‹• ƒ’’Ž‹‡• ‡˜‡ –‘ ƒ ‹…Ž‹‡† •–”ƒ‹‰Š– Ž‹‡Ǥ ‘™‡˜‡”ǡ ‘‡ ƒ••—’–‹‘ –Šƒ– ™‡ Šƒ˜‡ ƒ†‡ ‹• –Šƒ–ǡ
–Š‡ …‡–”‹ˆ—‰ƒŽ ˆ‘”…‡ǡ –Šƒ– ‹…”‡ƒ•‡• –Š‡ ‘”ƒŽ ”‡ƒ…–‹‘ ‹• ‘– ƒ……‘—–‡† ˆ‘”Ǥ Š‹• ‹• –‘ ƒ˜‘‹†
ƒ–Š‡ƒ–‹…ƒŽ …‘’Ž‡š‹–›Ǥ Š‹• ƒ••—’–‹‘ †‘‡• ‘– •‹‰‹ϐ‹…ƒ–Ž› ƒŽ–‡” –Š‡ ”‡•—Ž– ˆ‘” …—”˜‡• ™‹–Š
Žƒ”‰‡ ”ƒ†‹—• ƒ† ‘†‡”ƒ–‡ ˜‡Ž‘…‹–›Ǥ

‘Ž—–‹‘ ȋ…‘–ǥȌǣ „› ‡‡”‰› „ƒŽƒ…‡ ȋ‡‰Ž‡…–‹‰ ‡‡”‰› Ž‘•– ‹ …‘ŽŽ‹•‹‘Ȍǡ ™‡ ‰‡–ǣ

Initial potential energy = frictional work done + final potential energy + spring work done

1
2
= mgH = μmg L + mg h + x sinθ + kx - - - - (2.12)
2
The final potential energy mg h + x sin θ , is got as follows:

Ø Initially the mass climbs a height (h).
Ø Further, as the momentum of the mass compressed the spring, the mass gains further
height, which is sinθ times the spring compression (x), as shown in [Fig 2.8]



















[Fig 2.8: compression of the spring (x) due to momentum of the mass]


Simplifying equation (2.12) we get:

1
2
H = μ L + h + x sinθ + kx - - - - (2.13)
2mg
k
For ease of simplification, let us assume to be C and (H − μL − h = C )
1
2mg
Rearranging the expression (2.13) and applying the assumption, we get:

1
2
2
2
−(H + μ L + h) + x sin θ + kx = 0 = −C + x sinθ + Cx = 0 or Cx + x sinθ − C = 0
1
1
2mg


Page 14 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Solving the above quadratic equation, we get:


2
− sinθ ± sin θ + 4CC 1
x =
2C
For the spring compression (x) to be positive, ‘C ’ must be greater than zero, that is:
1
H − μL − h > 0 , therefore, we get: H > μL + h
The spring compression is now given by the positive root, as follows:


2
sin θ + 4CC − sinθ
x = 1
2C


2.5.3 The 3 Equation of motion

Let us consider the energy and work relation between two positions of a particle in motion. The
work done on the particle is the change in its kinetic energy, assuming that, the potential energy
change is zero.








[Fig 2.9: a particle in uniform straight-line motion]

1 1
2
2
Work done = final kinetic energy − initial kinetic energy = F ∗ s = mv − mU
2 2
1 1
= (ma) ∗ s = mv − mU = 2(a ∗ s) = v − U = v = U + 2as - - - - (2.14)
2
2
2
2
2
2
2 2
Another method to derive equation (2.14) is as follows:
For a uniform straight line motion: Distance travelled = average speed ∗ time
v + U
Therefore, we have: s = t - - - - (2.15)
2
change in velocity v − U
t is got by the definition of acceleration, which is: a = = - - - - (2.16)
time t




QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 15 age 15
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

Substituting the value of (t) in equation (2.15), we get:

v + U v − U
2
2
2
2
s = ∗ = 2as = v − vU + Uv − U = v = U + 2as
2 a
Equation (2.14) can be graphically represented as shown in [Fig 2.10]














[Fig 2.10: Graphical representation of [v = u + 2as]]
2
2
The equation (2.16) could be recast to give: [v = U+at] - - - - (2.17)

Equation (2.17) can be graphically represented as shown in [Fig 2.11]
















[Fig 2.11: Graphical representation of [v = U + at]


Substituting the value of (v) from equation (2.17) in equation (2.14), we get:

1
2
2 2
2
2
(U + at) = U + 2as = U + a t + 2Uat = U + 2as = s = ut + at - - - - (2.18)
2
2
2

Page 16 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


“—ƒ–‹‘ ȋʹǤͳͺȌ …ƒ „‡ ‰”ƒ’Š‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤͳʹȐ















1
2
[Fig 2.12: Graphical representation of v = ut + at ]
2
Practically, mechanical systems are characterized by constant velocity, constant acceleration and
variable acceleration.

Example:

Ø The motion of a rocket is characterized by variable acceleration.
Ø A cruising, level flight, is characterized by zero acceleration, as thrust is fully balanced by
drag.
Ø The rate of change of acceleration (jerk) of a vehicle, experienced during sudden braking.
These variations in kinematic parameters, such as, velocity and acceleration are either due to the
characteristics of the power plant, brake system or external factors, such as, drag, friction, etc.

Therefore, it is important to represent such motions mathematically.


Constant velocity: this is a result of zero net force acting on the system, meaning, the system has
no acceleration. Constant velocity could be graphically represented as shown in [Fig 2.13]















[Fig 2.13: constant velocity motion]

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 17 age 17
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


‘•–ƒ– ƒ……‡Ž‡”ƒ–‹‘ǣ –Š‹• ‹• ƒ ”‡•—Ž– ‘ˆ …‘•–ƒ– ˆ‘”…‡ ƒ…–‹‰ ‘ –Š‡ •›•–‡Ǥ Š‡ ƒ•• ‘ˆ –Š‡
•›•–‡ ƒŽ•‘ ”‡ƒ‹• …‘•–ƒ–Ǥ ‘•–ƒ– ƒ……‡Ž‡”ƒ–‹‘ …‘—Ž† „‡ ‰”ƒ’Š‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• •Š‘™ ‹
ȏ ‹‰ ʹǤͳͶȐ














[Fig 2.14: constant acceleration motion]



ƒ”‹ƒ„Ž‡ ƒ……‡Ž‡”ƒ–‹‘ǣ –Š‹• ”‡•—Ž–• •‹…‡ǡ –Š‡ ˆ‘”…‡ ƒ…–‹‰ ‘ –Š‡ •›•–‡ ‹• ˜ƒ”‹ƒ„Ž‡ǡ ‘” –Š‡
ƒ•• ‘ˆ –Š‡ •›•–‡ ‹• ˜ƒ”‹ƒ„Ž‡Ǥ ƒ”‹ƒ„Ž‡ ƒ……‡Ž‡”ƒ–‹‘ …‘—Ž† „‡ ‰”ƒ’Š‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• •Š‘™ ‹
ȏ ‹‰ ʹǤͳͷȐ













[Fig 2.15: variable acceleration motion]














Page 18 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Example 1: a ball is thrown into the water, with an initial velocity U (as shown in [Fig 2.16]).
0
The drag experienced by the ball in water is proportional to square of the instantaneous velocity.
Find the time taken by the ball, to travel a depth of ‘d’, where the velocity of the ball becomes v .
0



















[Fig 2.16: ball travelling in water]

Solution: the factors affecting the motion of the ball are:

Ø Drag due to water
Ø Gravity

Here, we neglect viscous drag. Therefore, the acceleration of the ball at any instant is given by:

net force gravity force − drag force
Instantaneous acceleration (a) = =
mass of ball mass of ball
The drag force is given by kv , where, ‘k’ is the constant of proportionality. Therefore, we have:
2
mg − kv 2 kv 2
a = = g −
m m

Writing acceleration in terms of velocity, we get:

dv kv 2
= g − - - - - (2.19)
dt m

Equation (2.19) represents the motion of the ball and its instantaneous acceleration. We would
use this equation to find the time of journey, as, we know the initial and final velocities.







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 19 age 19
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Separating the variables of velocity and time in equation (2.19), we get:


dv
kv 2 = dt , the time of travel is got by integrating both sides.
g −
m

v 0 t
dv k
= 2 = dt , taking in the LHS common, we get:
g − kv m
U 0 m 0

v 0
m dv
= t - - - - (2.20)
k gm − v
2
U 0 k
To reduce equation (2.20) to a standard integral form, it is rewritten as follows:



v 0
m dv dx 1 −1 x

k 2 = t , this is a standard integral of the form: a − x = tanh
2
2
a
a
U 0 gm 2 1
2
k − v

Using the standard formula for the integral, we get:

v 0
m dv m 1 −1 v 0 −1 U 0
tanh

k 1 2 = t = k ∗ mg mg − tanh mg = t

U 0 gm 2 2
k − v k k k




















Page 20 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Example 2: a cart filled with water, is being pulled with a constant force by a car. If the cart has
a small hole and water starts leaking form it [Fig 2.17], at the rate of m kg s , find the velocity of
the cart at any time (t), assuming that the cart starts with zero velocity at (t = 0) and the initial mass
is ‘M’.

Solution: the pulling force, though constant, the mass of the cart is continuously changing (as
shown in [Fig 2.17]). This is a case of variable acceleration, and is given by:

dv force F
acceleration dt = mass of cart at any instant = M − m t - - - - (2.21)











[Fig 2.17: water cart being pulled by a car]


Separating the variables (dv) and (dt) in equation (2.21), we get:

F dt
dv = , integrating both sides, we get:
M − m t
v t
F dt
dv = - - - - (2.22)
M − m t
0 0

We know that, when the denominator of the integrand is differentiated, yielding the numerator,
then, the value of the integral is the log of the denominator. To arrive at this condition, let us divide

and multiply the integrand on the RHS by −m in equation (2.22).

v t v
F −m dt F F
t
dv = − m M − m t = dv = − m log⁡(M − m t = v = − m log M − m t − log M

0
0 0 0

Therefore, we have the instantaneous velocity of the cart at ‘t’ seconds, to be:

F M − m t F M
v = − log = log
m M m M − m t





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 21 age 21
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


šƒ’Ž‡ ͵ǣ ‘’—–‡ –Š‡ –‹‡ –‘ •–‘’ ˆ‘” ˆ‘—” …ƒ”•Ǥ ••—‡ –Šƒ– –Š‡ …ƒ”• „”ƒ‡ ƒ– ͳͲͲ ȀŠ‘—”Ǥ
”ƒ™ ƒ ‰”ƒ’Š ‘ˆ •’‡‡† ˜• •–‘’’‹‰ †‹•–ƒ…‡Ǥ Š‡ •–‘’’‹‰ †‹•–ƒ…‡ ˆ‘” –Š‡ ˆ‘—” …ƒ”• ƒ– ͳͲͲ Ȁ
Š‘—”ǡ ‹• –ƒ„—Žƒ–‡† ƒ• ˆ‘ŽŽ‘™•ǣ

Car Breaking speed Stopping distance


Car 1 100 km/hr 30 meters


Car 2 100 km/hr 32 meters


Car 3 100 km/hr 34 meters


Car 4 100 km/hr 36 meters


Solution: let us now compute the stopping distance for each car. Since we seek a relationship
between velocity and distance travelled, the following equation of motion could be used.
v = U + 2as
2
2
We know that, the final velocity (v) is zero and the initial velocity is given to be 100 km/hr.

The stopping distance (s) is therefore given by:

2
0 = U + 2(−a)s we consider a to be negative as the car is decelerating.
U 2
s =
2a

Since ‘U’ is in km/hr and we have the stopping distance in meters, we multiply ‘U’ with (5/18) to
convert its units to m/s.


5U 2
18 25U 2
s = = - - - - (2.23)
2a 648 a


25U 2
Deceleration is given by: a = - - - - (2.24)
648 s

The stopping time is computed using the equation: [v = U + (-a) t]

U km S 2 5U m s 2
Since the final velocity is zero, we have: 0 = U − at = t = ∗ = t = ∗ = s
a hr m 18a s m





Page 22 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


‡– —• ‘™ –ƒ„—Žƒ–‡ –Š‡ †‡…‡Ž‡”ƒ–‹‘ ƒ† •–‘’’‹‰ –‹‡ ˆ‘” –Š‡ ˆ‘—” …ƒ”•Ǥ


Car Stopping distance Deceleration Time


Car 1 30 meters 12.86 m/s 2.16 s
2

Car 2 32 meters 12.06 m/s 2.304 s
2
Car 3 34 meters 11.35 m/s 2.448 s
2

Car 4 36 meters 10.72 m/s 2.592 s
2





– …ƒ „‡ ‘„•‡”˜‡† –Šƒ–ǡ ƒ •‹‰‹ϐ‹…ƒ– …Šƒ‰‡ ‹ †‡…‡Ž‡”ƒ–‹‘ †‘‡• ‘– ƒŽ–‡” –Š‡ •–‘’’‹‰
–‹‡ ƒ• •‹‰‹ϐ‹…ƒ–Ž›Ǥ
‡– —• ‘™ –ƒ„—Žƒ–‡ –Š‡ •–‘’’‹‰ †‹•–ƒ…‡ ˆ‘” †‹ˆˆ‡”‡– ˜‡Ž‘…‹–‹‡• ˆ‘” ƒŽŽ –Š‡ ˆ‘—” …ƒ”•Ǥ



Stopping Stopping Stopping Stopping
Velocities distance for distance for distance for distance for
car 1 car 2 car 2 car 2


200 km/hr 120 meters 127.96 meters 135.97 meters 143.96 meters


150 km/hr 67.5 meters 71.97 meters 76.48 meters 80.98 meters


120 km/hr 43.2 meters 46 meters 48.95 meters 51.82 meters

90 km/hr 24.3 meters 25.91 meters 27.53 meters 29.15 meters


60 km/hr 10.8 meters 11.5 meters 12.24 meters 12.96 meters


30 km/hr 2.7 meters 2.88 meters 3.06 meters 3.24 meters










QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 23 age 23
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


’‡‡† ˜• †‹•–ƒ…‡ –”ƒ˜‡ŽŽ‡† ˆ‘” –Š‡ ˆ‘—” …ƒ”• ‹• ‰”ƒ’Š‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤͳͺȐ






























[Fig 2.18: Graph of speed Vs stopping distance for 4 cars]



– …ƒ „‡ ‘„•‡”˜‡† ˆ‘” ȏ ‹‰ ʹǤͳͺȐ –Šƒ–ǡ ƒ– Š‹‰Š •’‡‡†•ǡ –Š‡ •–‘’’‹‰ †‹•–ƒ…‡ ƒ––‡”•ǡ ƒ† ‹•
…‘•‹†‡”ƒ„Ž› †‹ˆˆ‡”‡– ˆ‘” …ƒ” ͳ ƒ† Ͷǡ ™Š‹…Š Šƒ˜‡ ƒ †‡…‡Ž‡”ƒ–‹‘ †‹ˆˆ‡”‡…‡ ‘ˆ ʹ Ȁ• Ǥ
ʹ
…‹†‡–ƒŽ ‡‰‹‡‡”‹‰ ˆƒ…–ǣ ……‡Ž‡”ƒ–‹‘ ƒ† †‡…‡Ž‡”ƒ–‹‘ ‘ˆ ƒ …ƒ” ’”‹ƒ”‹Ž› †‡’‡† ‘ –Š‡
”‘ƒ† ˆ”‹…–‹‘ ™‹–Š –›”‡•Ǥ ˆƒ…– Š‡ƒ–‡† –›”‡• Šƒ˜‡ „‡––‡” ˆ”‹…–‹‘ ™‹–Š ”‘ƒ† –Šƒ …‘Ž† –›”‡•Ǥ ƒ…‡
†”‹˜‡”ǯ•ǡ ˆ”‹…–‹‘ Š‡ƒ– –Š‡‹” –›”‡• „› ƒ……‡Ž‡”ƒ–‹‰ ‹ ’Žƒ…‡ǡ –‘ ‰ƒ‹ –Š‡ ˆ”‹…–‹‘ ƒ†˜ƒ–ƒ‰‡Ǥ
Š‡”‡ˆ‘”‡ǡ –Š‡ ‡“—ƒ–‹‘• ‘ˆ ‘–‹‘ ƒ”‡ ’‘™‡”ˆ—ŽŽ› ƒ’’Ž‹‡† –‘ •‡˜‡”ƒŽ •‹–—ƒ–‹‘•ǡ „‡ ‹–ǣ
· ‹‡ƒ–‹…• ‘ˆ ƒ ˜‡Š‹…Ž‡
· ‹‡ƒ–‹…• ‘ˆ ƒ ‡…Šƒ‹•
· ƒ–Š‡ƒ–‹…ƒŽ ‘†‡Ž• ˆ‘” –”ƒ•‹‡– ƒ† ‘-Ž‹‡ƒ” ƒƒŽ›•‡•
Ž•‘ ‘–‡ –Šƒ–ǡ –Š‡ ƒ‰—Žƒ” “—ƒ–‹–‹‡•ǡ †‘ „‡ƒ” –Š‡ •ƒ‡ ”‡Žƒ–‹‘•Š‹’ ƒ• †‘ –Š‡ Ž‹‡ƒ” “—ƒ–‹–‹‡•Ǥ














Page 24 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


‘ŽŽ‘™‹‰ ‹• ƒ •—ƒ”‹œ‡† –ƒ„—Žƒ–‹‘ ˆ‘” Ž‹‡ƒ” ƒ† ƒ‰—Žƒ” ‡“—ƒ–‹‘• ‘ˆ ‘–‹‘Ǥ

Linear quantities Angular quantities


2
2
V = u + 2as Ω = ω + 2αθ
2
2
V = u + at Ω = ω + at

1 1
2
s = ut + at θ = ωt + αt
2
2 2
Where,

V Final velocity Ω Final angular velocity


u Initial velocity ω Initial Angular velocity


a Acceleration α Angular acceleration

t Time θ Angular displacement


s Displacement




Example 4 (Basic Optimization): Consider a cannon firing a ball, as shown in [Fig 2.19].
What is the optimum angle θ and height (H), with which, the ball must be fired such that it hits
the target?
Parameters of the problem are as tabulated.

Parameters Magnitude


Velocity of fire (V) 25 m/s

Limits of variable height (H) 40m to 80m


Limits of angle of projection (θ) 30° to 40°

Horizontal displacement of bullet (L) 100 m


Acceleration due to gravity (g) 10 m/s
2
0
The accuracy of angle in the final solution must be within ±0.1 .

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 25 age 25
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics






















[Fig 2.19: Canon firing at the target]


‘Ž—–‹‘ǣ •‹…‡ –Š‡ …ƒ‘ ‹• ϐ‹”‡† ƒ– ƒ ƒ‰Ž‡ ȋɅȌǡ Š‘”‹œ‘–ƒŽ ƒ† ˜‡”–‹…ƒŽ …‘’‘‡–• ‘ˆ ˜‡Ž‘…‹–›
ȋ Ȍ ‡š‹•–ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤʹͲȐǤ









[Fig 2.20: components of velocity]

The horizontal displacement (L) of the bullet is given by: [L = V cos θ ∗ time of flight t - - - - (2.25)

The vertical displacement (H) is negative, as the target is H meters below the point of fire. Further,
the initial velocity is vertically upwards and the acceleration due to gravity for an upward motion is
negative.

1 1
2
2
Using the expression: s = Ut + at , we get: −H = V sinθ t − gt - - - - (2.26)
2 2
Substituting the value of ‘t’ from equation (2.25) in equation (2.26), we get:
L 1 L 2 1 gL 2
−H = V sin θ − g = −H = L tan θ −
2
2
V cos θ 2 V cos θ 2 V cos θ


Page 26 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Substituting the values of ‘L’, ‘V’ and ‘g’ in the above expression, we get:

1 10 ∗ (100) 2 80
−H = 100 tan θ − = H = −100 tan θ +
2
2
2
2 25 cos θ cos θ
Since, ‘H’ and ‘θ’ are variable, they must be plotted as dependent and independent variables
respectively. The zone for optimization is marked to select the optimum point, using the ranges for
‘H’ and ‘θ’ specified in the problem, as shown in [Fig 2.21].

























[Fig 2.21: Minimum value of Height and Theta]

The minimum point is located in the area determined by the ranges of ‘H’ and ‘θ’. This is the
optimum value which minimizes (H).
The minimum value of ‘H’ could also be got more accurately via calculus, as follows:

80
We have, from equation . : H = −100 tan θ +
2
cos θ
Differentiating ‘H’ with respect to ‘θ’, we get:

dH
2
H = −100 tan θ + 80 cos −2 θ = = −100 sec θ + 80 ∗ −2 cos −3 θ ∗ (− sinθ)
dθ
dH 2 sinθ
2
= = −100 sec θ + 80 ∗
3
dθ cos θ


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 27 age 27
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


From [Fig 2.21], it can be seen that, the slope of the curve is zero at minimum ‘H’, therefore, we
have:

2 sinθ 100 sin θ 5
2
0 = −100 sec θ + 80 ∗ = = 160 = = tan θ
2
3
3
cos θ cos θ cos θ 8
Therefore, we have θ = 32.005
0
It can be observed that the value of ‘θ’ for minimum ‘H’ via XLS, as shown in [Fig 2.21], is close to
the exact value.
L 100
The time of flight for equation . is: t = = = 4.72 s
V cos θ 25 cos 32.005

It can be shown that, the range of 100 m (given) is maximum for the computed ‘H’ and ‘θ’.


Example 5: For (Example 4) establish a general expression for maximum range (R) of the
projectile.

Solution 5: For the given situation, the following example computes the expression for
maximum range:

The canon is at a height ‘h’ from the ground. The bullet traces a parabolic path, reaching a
maximum height of (h+h’) from the ground. The total time the bullet takes to reach the ground,
which is the range of travel for the bullet, is a summation of the time taken by the bullet to reach
the maximum height (h+h’) and the time to reach the target, as shown in [Fig 2.22].





















[Fig 2.22: Canon firing at the target]



Page 28 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


The time taken to reach the maximum vertical height is given by the expression: [v = U+at]

v sinθ
0 = v sinθ − gt , rearranging for time t , we get: t =
g

Where, ‘t’ is the time taken to reach the highest point of the trajectory.

The time taken by the bullet to travel a vertical height of (h+h’) in the downward direction
after reaching the highest point is given by the expression:
1 1
2
2
s = Ut + at = 0 + gt , rearranging for time t , we get:
2 2
2 s 2 h + h
′
t = = - - - - (2.28)
g g

2
The distance (h’) could be computed using the expression: v = U + 2as
2
2
2
v sin θ
2
′
′
= v = 0 + 2ah′ , rearranging for height h , we get: h =
2g

Substituting the value of (h’) in equation (2.28) we get,

2 h + h 2h v sin θ
′
2
2
t = = +
g g 2g

The range ‘R’ of travel is given by:

Range R = horizontal component of velocity ∗ total time of travel

2
v sinθ 2h v sin θ
2
= R = v cos θ ∗ + +
g g 2g

Taking (v/g) common in the above equation, we get:

v 2 2hg
2
R = cos θ ∗ sinθ + + sin θ
g v 2
2hg
For ease of simplification, let us assume the expression 2 = k . Therefore, we have:
v

v 2
2
R = cos θ ∗ sinθ + k + sin θ
g


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 29 age 29
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Differentiating the above expression with respect to theta we get:

dR v 2 1 2 ∗ sinθ ∗ cos θ
2
= − sin θ sinθ + k + sin θ + cos θ cos θ + ∗ = 0
2
dθ g 2 k + sin θ
Let us simplify the terms inside the bracket first and then combine the rest, as follows:


1 2 ∗ sin θ ∗ cosθ
2
− sinθ sin θ + k + sin θ + cos θ cosθ + ∗ = 0
2 k + sin θ
2

sinθ
2
2
− sinθ sin θ + k + sin θ + cos θ 1 + = 0
2
k + sin θ

k + sin θ + sinθ
2
2
2
− sinθ sinθ + k + sin θ + cos θ = 0
2
k + sin θ

2
2
2
2
− sin θ k + sin θ sinθ + k + sin θ + cos θ ( k + sin θ + sinθ)
= 0
k + sin θ
2

Taking k + sin θ + sin θ common in the above expression, we get:
2

2
2
− sin q k + sin θ + cos q]
2
sinθ + k + sin θ = 0
2
k + sin θ

Since the first term of the above expression gives sinθ = − k + sin θ , and the sign of the
2
acute angle (angle of launch of the bullet) cannot be negative, only the second term is equated
to zero. Therefore, we have:

2
2
− sin q k + sin θ + cos q] 2 2
k + sin θ = 0 = − sinq k + sin θ + cos θ = 0
2

Upon further simplification we have:

2
2
cos q = sinq k + sin θ

Squaring both sides and simplifying we get

(1 − sin θ) = sin θ (k + sin θ)
2
2
2
2
Expanding the above expression, we get:
1 + sin θ − 2 sin θ = k sin θ + sin θ
2
4
2
4
Page 30 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

1
2
= 1 = sin θ k + 2 = sinθ =
k + 2

2
adjacent hypotenuse − opposite 2 k + 2 − 1 k + 1
Therefore, we have cosθ = = = =
hypotenuse hypotenuse k + 2 k + 2

Substituting the value of sin θ and cos θ in the equation for range we get:

2 2
v 2 v k + 1 1 1

R = g cos θ ∗ sinθ + k + sin θ = g ∗ + K + K + 2
k + 2 k + 2



v 2 k + 1 1 k + 1 2 v 2 k + 1 1 k + 1
= ∗ + = ∗ +
g k + 2 k + 2 K + 2 g k + 2 k + 2 K + 2


v 2 k + 1 ∗ (k + 2) v 2
= = k + 1
g k + 2 2 g


2gh
Substituting k = in the above expression, we get the maximum range of the projectile to be:
v 2

v 2 2hg
R = + 1
g v 2
























QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 31 age 31
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


šƒ’Ž‡ ͸ǣ ‘•‹†‡” ƒ ‹••‹Ž‡ Žƒ—…Š‡† ˆ”‘ –Š‡ •—”ˆƒ…‡ ‘ˆ –Š‡ ‡ƒ”–ŠǤ ‡– –Š‡ Ǯœ-†‹”‡…–‹‘ǯ „‡
ƒŽ‘‰ –Š‡ ˜‡”–‹…ƒŽ ȋ…‘‹…‹†‡• ™‹–Š –Š‡ ”ƒ†‹—• ‘ˆ –Š‡ ‡ƒ”–ŠȌǤ ‡– –Š‡ Ǯš-†‹”‡…–‹‘ǯ „‡ ƒŽ‘‰ ƒ•– ƒ† Ǯ›-
†‹”‡…–‹‘ǯ „‡ ƒŽ‘‰ ‘”–ŠǤ Š‡ ‹••‹Ž‡ –”ƒ˜‡Ž• ‹ –Š‡ Ǯœšǯ ’Žƒ‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤʹ͵ȐǤ ••—‡ –Š‡
ƒŽ–‹–—†‡ ‘ˆ –Š‡ ƒ”–Š –‘ „‡ ȋͻͲ - Ʌ Ȍ ƒ† ˜‡Ž‘…‹–› ƒ† ƒ‰Ž‡ ‘ˆ Žƒ—…Š ˆ‘” –Š‡ ’”‘Œ‡…–‹Ž‡ ƒ”‡ Ǯ ǯ ƒ† ǮɅǯ
”‡•’‡…–‹˜‡Ž›Ǥ Š‡ ”‘–ƒ–‹‘ƒŽ •’‡‡† ‘ˆ –Š‡ ‡ƒ”–Š ‹• ǮɘǯǤ š’Žƒ‹ Š‘™ –Š‡ ‘”‹‘Ž‹• ˆ‘”…‡ ƒˆˆ‡…–• –Š‡
–”ƒ˜‡Ž ‘ˆ ƒ ‹••‹Ž‡Ǥ






















[Fig 2.23: local axis of the missile on the surface of the earth]
























Page 32 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]
‘Ž—–‹‘ ͸ǣ •‹…‡ –Š‡ ’”‘Œ‡…–‹Ž‡ Šƒ• Ž‹‡ƒ” ˜‡Ž‘…‹–› …‘’‘‡–• ‹ –Š‡ Ǯœǯ ƒ† Ǯšǯ †‹”‡…–‹‘•ǡ –Š‡
‘”‹‘Ž‹• …‘’‘‡– …‘‡• ‹–‘ ’‹…–—”‡ „‡…ƒ—•‡ ‘‡ ‘ˆ –Š‡ …‘’‘‡–• ‘ˆ –Š‡ ƒ”–Šǯ• ƒ‰—Žƒ”
˜‡Ž‘…‹–› ‹• ’‡”’‡†‹…—Žƒ” –‘ –Š‡ ’Žƒ‡ ‘ˆ –Š‡ ’”‘Œ‡…–‹Ž‡ ƒ† –Š‡ ‘–Š‡” …‘’‘‡– ‹• ‹ –Š‡ ’Žƒ‡ ‘ˆ
–Š‡ ’”‘Œ‡…–‹Ž‡ „—– ’‡”’‡†‹…—Žƒ” –‘ –Š‡ Š‘”‹œ‘–ƒŽ …‘’‘‡– ‘ˆ –Š‡ ’”‘Œ‡…–‹Ž‡ǡ ƒ• •Š‘™ ‹
ȏ ‹‰ ʹǤʹͶȐǤ

















[Fig 2.24: local axis of the missile on the surface of the earth]


Š‡ ƒ”–Šǯ• ƒ‰—Žƒ” ˜‡Ž‘…‹–› ‹• •’Ž‹– ƒŽ‘‰ –Š‡ Ǯœǯ ƒ† Ǯ›ǯ ƒš‡• ”‡•’‡…–‹˜‡Ž›ǡ ƒ† –Š‡ ˜‡Ž‘…‹–›
…‘’‘‡–• ‘ˆ –Š‡ ’”‘Œ‡…–‹Ž‡ ƒ”‡ •Š‘™ ‹ ȏ ‹‰ ʹǤʹͷ ƒ Ƭ „ȐǤ













[Fig 2.25: components of angular and projectile velocities]

From the [Fig 2.24 & 2.25] the Coriolis force is given by:


i j k
2m v × ω = 2m v cosθ 0 v sin θ
0 ω sinθ ω cos θ





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 33 age 33
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


ʹǤͷǤͶ ‘Ž‡ ‘ˆ “—ƒ–‹‘• ‘ˆ ‘–‹‘ ‹ ”ƒ•‹‡– ƒ† ‘-Ž‹‡ƒ” ƒŽ›•‡•

“—ƒ–‹‘• ‘ˆ ‘–‹‘
Š‡ †›ƒ‹… ”‡•’‘•‡ ‘ˆ •–”—…–—”ƒŽ •›•–‡• ‹• …‘’—–‡† „› †‹”‡…– —‡”‹…ƒŽ ‹–‡‰”ƒ–‹‘ ‘ˆ –Š‡
†›ƒ‹… ‡“—‹Ž‹„”‹— ‡“—ƒ–‹‘•Ǥ

Š‡ ‡“—ƒ–‹‘ ‘ˆ ‘–‹‘ ‹• ‹–‡‰”ƒ–‡† •–‡’ „› •–‡’ ™‹–Š‘—– –”ƒ•ˆ‘”‹‰ –Š‡ ‡“—ƒ–‹‘•Ǥ Š‡ ‹†‡ƒ ‹•
–Šƒ–ǡ ƒ– ‡ƒ…Š –‹‡ •–‡’ǡ –Š‡ ‡“—‹Ž‹„”‹— ‹• •ƒ–‹•ϐ‹‡† —†‡” –Š‡ ƒ…–‹‘ ‘ˆ ‹‡”–‹ƒǡ †ƒ’‹‰ǡ •–‹ˆˆ‡••
ƒ† ‡š–‡”ƒŽ ˆ‘”…‡•Ǥ

Šƒ– ‹• š’Ž‹…‹– ƒƒŽ›•‹•ǫ
Explicit methods do not involve the solution of a set of linear equations at each step. Basically, these
methods use the differential equation at time “t” to predict a solution at time “t+∆t”.

The equation of motion is given by, Newton’s II law.

F = mu + cu + ku

The equation of motion is solved in several time steps till the solution at time “t+∆t” is got. These
time steps have to be carefully determined, as they in turn determine the accuracy of the solution.

There are many direct integration methods available to solve the equation of motion. A few popular
ones are listed below.

Ø Central difference scheme
Ø The Newmark family of methods
Ø The Hilber, Hughes and Taylor method.
Ø The Houbolt method.
Ø The Wilson method.

By average slope/central difference method, the velocity and acceleration in the equation of motion
could be expressed in terms of displacement as follows:
Change in displacement
Velocity =
time interval ∆t
If we consider the time intervals to be t + ∆t and t − ∆t , we get:


u t+∆t − u t−∆t u t+∆t − u t−∆t
Velocity u = =
t + ∆t − t − ∆t 2∆t

Let us now find acceleration by considering the rate of change of velocity as follows:


u t+∆t − u t−∆t
Acceleration u =
t + ∆t − t − ∆t



Page 34 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


The velocity component in the above expression could be rewritten as change in displacement
divided by time;

u (t+∆t) − u t u − u t−∆t
∆t − t ∆t u − 2u − u t−∆t
Acceleration u = = (t+∆t) t 2
t + ∆t − t − ∆t 2 ∆t

Since we have computed velocity, acceleration and displacement, their respective expression
could be substituted in the equation of motion and simplified to get the value of u t+∆t , using

the displacement data of the previous step. The accuracy of this method depends on square of
the time step ∆t chosen.
2
Newmark (Implicit time step)

This method, the displacement computation for average acceleration and average velocity, say,
from step 1 (s) to step 2 (s+1), is given by:

1
2
2
Displacement = u s+1 = u + u ∆t + ∆t − β u + ∆t β u s+1
s
s
2 s

Where, β is a constant of value (1/4) or 0.25. Substituting the value of β in the above
expression, we get:

1
1
1
Displacement = u s+1 = u + u ∆t + ∆t − u + ∆t u s+1
2
2
s
s
2 4 s 4

∆t 2 u ∆t 2 u s+1
s
Displacement = u s+1 = u + u ∆t + +
s
s
2 2 2 2

∆t 2 u + u s+1
s
Displacement = u s+1 = u + u ∆t + 2 2

s
s

The above expression is comparable to:
1
Initial displacement u + ut + at
2
0
2
It can be observed that, the Newmark’s equation contains average acceleration, given by:

s
u + u s+1
Average acceleration =
2



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 35 age 35
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics



Similarly the velocity is given by:

2
Velocity = u s+1 = u + u ∆t 1 − γ + ∆t γ u s+1
s
s

Where, γ is a constant of value (1/2) or 0.5. Substituting the value of γ in the above
expression, we get:

Velocity = u s+1 = u + u ∆t 1 − 0.5 + ∆t 0.5 u s+1
s
s

1
Velocity = u s+1 = u + u ∆t + ∆t u s+1 = u + (average acceleration ∗ ∆t)
s 2 s s

The above expression is comparable to [v = U + at]

Since, this scheme is for an implicit analysis, the choice of a larger time step does not affect the
accuracy of the solution. This scheme is unconditionally stable.

For an explicit analysis, a larger time step adversely affects the accuracy of the solution. Further,

the choice of time step also determines the damping introduced numerically. Therefore, if the
analyst were to assume a physical damping, then, it must be well justified, as, the combination of
physical and numerical damping can overdamp the system, leading to inaccurate computation of
response.



























Page 36 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


ʹǤ͸ ‡™–‘ǯ• ƒ™• ‘ˆ ‘–‹‘

ƒ› ‘ˆ –Š‡ „ƒ•‹… ‡‰‹‡‡”‹‰ ’”‘„Ž‡• …ƒ „‡ •‘Ž˜‡† —•‹‰ –Š‡ ‡™–‘ǯ• Žƒ™• ‘ˆ ‘–‹‘Ǥ Š‡
‘•– …‘‘ ƒ’’Ž‹…ƒ–‹‘• ƒ”‡ǣ
· ‡Š‹…Ž‡ †›ƒ‹…•
· ‡…Šƒ‹••
Š‡ –Š”‡‡ Žƒ™• ‘ˆ ‘–‹‘ –Šƒ– ‰‘˜‡” …Žƒ••‹…ƒŽ ‡…Šƒ‹…•ǡ ‹˜‘Ž˜‡ ƒ ‹’‘”–ƒ– …‘…‡’– …ƒŽŽ‡†
‹‡”–‹ƒǤ
Newton’s laws Equations Description
(For linear motion)

An external force is needed to change the state of
1 Law F = ma (a = 0, if F = 0)
st
uniform motion or rest of a body
The rate of change of momentum of a body is
dp
2 Law = F directly proportional to the force applied and is
nd
dt in the same direction of the applied force
The force exerted on body (2) by body (1) is
3 Law [F 12 = −F ] equal and opposite to the force exerted on body
rd
21
(1) by body (2).


Newton’s laws Equations Description
(For linear motion)

An external torque is needed to change the state
1 Law T = Iα (α = 0, if T = 0)
st
of uniform motion or rest of a body
The rate of change of angular momentum of a
dH body is directly proportional to the torque
2 Law = T
nd
dt applied and is in the same direction of the
applied torque
The torque exerted on body (2) by body (1) is
3 Law [T 12 = −T ] equal and opposite to the torque exerted on
rd
21
body (1) by body (2).


Inertia: whenever a body is accelerated, the acceleration is opposed by the tendency of the body
to continue in the state of rest or uniform motion. The inertia for linear and angular acceleration
are contributed by mass and mass moment of inertia respectively.







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 37 age 37
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


ʹǤ͸Ǥͳ ƒ•• ‘‡– ‘ˆ ‡”–‹ƒ ȋ Ȍ

‡™–‘ǯ• •‡…‘† Žƒ™ ‘ˆ ‘–‹‘ ™‡ Šƒ˜‡ ȏ α ƒȐǤ • –Š‡ ƒ’’Ž‹‡† ˆ‘”…‡ Ǯ ǯ ‹…”‡ƒ•‡•ǡ –Š‡ ‹‡”–‹ƒ Ǯƒǯ
ƒŽ•‘ ‹…”‡ƒ•‡•Ǥ Š‹• ‹• „‡…ƒ—•‡ǡ ‹‡”–‹ƒ Šƒ• –‘ „‡ ’”‘˜‹†‡† „› –Š‡ ˆ‘”…‡ Ǯ ǯǤ

‹‹Žƒ”Ž›ǡ ™Š‹Ž‡ ”‘–ƒ–‹‰ ƒ „‘†›ǡ –Š‡ ƒ‰—Žƒ” ‹‡”–‹ƒ ȋ‘’’‘•‹–‹‘ –‘ ”‘–ƒ–‹‘Ȍ ‹• ’”‘˜‹†‡† „› –Š‡
ƒ’’Ž‹‡† –‘”“—‡ ǯ ǯǤ

 …ƒ•‡ ‘ˆ Ž‹‡ƒ” ‘–‹‘ǡ ƒ•• Ǯǯ ‹• –Š‡ ‹†‡š ‘ˆ ‹‡”–‹ƒǤ ‘™‡˜‡”ǡ ƒ ‡“—‹˜ƒŽ‡– –‡” Šƒ• –‘ „‡
†‡˜‡Ž‘’‡† –‘ …‘’—–‡ –Š‡ ƒ‰—Žƒ” ‹‡”–‹ƒǤ Š‹• –‡” ‹• …ƒŽŽ‡† Dzƒ•• ‘‡– ‘ˆ ‹‡”–‹ƒdzǤ

Š‡ ƒ‡ ƒ•• ‘‡– ‘ˆ ‹‡”–‹ƒ ‘”‹‰‹ƒ–‡• ˆ”‘ –Š‡ ˆƒ…– –Šƒ–ǡ ‹– ‹• –Š‡ ‘‡– ‘ˆ ƒŽŽ –Š‡ Ž‹‡ƒ”
‹‡”–‹ƒ• ȋƒȌ ‘ˆ –Š‡ ‡Ž‡‡–• –Šƒ– ˆ‘” –Š‡ ”‘–ƒ–‹‰ ‡„‡”Ǥ

‡– —• …‘•‹†‡” ȏ ‹‰ ʹǤʹ͸ȐǤ






[Fig 2.26: Uniform rot rotating with angular acceleration 'α']






[Fig 2.26: Uniform rot rotating with angular acceleration 'α']

Each element on the rod experiences an inertia dmαx . This inertia contributes to a moment about
the ‘z-axis’ as show in [Fig 2.26]. All the elements along the rod contribute to the same sense of
moment. The sum of all the moments is given by:

L
2
Moment of Inertia of the rod = dmαx
0
This inertia has to be provided by the applied torque to sustain the acceleration. Therefore, we have:


L
2
Torque T = dmαx
0
L L
By Newton’s second law of motion: Torque T = dmαx = Iα , by where I = dmx
2
2
0 0
Where (I) is the mass moment of inertia.




Page 38 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


– ‹• –‘ „‡ ‘–‡† –Šƒ–ǡ –Š‡ …‘…‡’– ‘ˆ ƒ”‡ƒ ‘‡– ‘ˆ ‹‡”–‹ƒ ȋ Ȍ ȋ–‘ „‡ ‹–”‘†—…‡† ‹ ‘Ž‹†
‡…Šƒ‹…•Ȍǡ •Š‘—Ž† ‘– „‡ …‘ˆ—•‡† ™‹–Š ƒ•• ‘‡– ‘ˆ ‹‡”–‹ƒǡ ƒ•ǡ ‹• ƒ ’”‘’‡”–› ‘ˆ –Š‡
ƒ••ǡ ™Š‡”‡ƒ•ǡ ‹• ƒ ’”‘’‡”–› ‘ˆ –Š‡ …”‘••-•‡…–‹‘ƒŽ ‰‡‘‡–”›Ǥ

—”–Š‡”ǡ –Š‡ ƒ•• ‘‡– ‘ˆ ‹‡”–‹ƒ ‹• –Š‡ ”‡•‹•–ƒ…‡ ‘ˆˆ‡”‡† ƒ‰ƒ‹•– ƒ‰—Žƒ” ”‘–ƒ–‹‘• ƒ† ƒ”‡ƒ
‘‡– ‘ˆ ‹‡”–‹ƒ ‹• –Š‡ ”‡•‹•–ƒ…‡ ‘ˆˆ‡”‡† ƒ‰ƒ‹•– ƒ‰—Žƒ” †‡ˆ‘”ƒ–‹‘•Ǥ
Š‡ ‹• ˆ‘—† ˆ‘” ˜ƒ”‹‘—• –›’‡• ‘ˆ •Šƒ’‡• ƒ„‘—– ƒ ‰‹˜‡ ƒš‹• ‘ˆ ”‘–ƒ–‹‘Ǥ ȏ ‘” ‡šƒ’Ž‡ǡ ƒ
”‘–ƒ–‹‰ •Šƒˆ–ǡ ™‹–Š ƒ †‹•… ‘—–‡† ‘ ‹–ǡ ‹• ‡“—‹˜ƒŽ‡– –‘ ƒ ”‘† ™‹–Š ƒ …›Ž‹†‡”ȐǤ Š‡”‡ˆ‘”‡ǡ ˆ‘”
ƒ› …‘’‘‡–•Ȁ‰‡‘‡–”‹‡•ǡ –Š‡ ‹• ˆ‘—† ƒ• ˆ‘ŽŽ‘™•ǣ
‹–—ƒ–‹‘ ͳǣ …‘•‹†‡” ƒ ”‘† ƒ……‡Ž‡”ƒ–‡† ƒ„‘—– ƒ ƒš‹• –Š”‘—‰Š ‘‡ ‘ˆ ‹–• ‡†• ƒ• •Š‘™ ‹ ȏ ‹‰
ʹǤʹ͹ȐǤ Š‡ ”‘† ‹• ’”‹•ƒ–‹… ƒ† —‹ˆ‘”Ǥ ‡– —• …‘’—–‡ –Š‡ ˆ‘” –Š‹• •‹–—ƒ–‹‘Ǥ










[Fig 2.27: Rod accelerating about one of its ends]



Solution 1: here, we could utilize the formula derived for a general case as shown in [Fig 2.26].

L L L
m m x 3 mL 2
2
2
I = dmx = dx x = =
L L 3 3
0 0 0
Situation 2: consider a rod accelerated about an axis passing through its CG as shown in
[Fig 2.28]. The rod is prismatic and uniform. Let us compute the MMOI for this situation.
















[Fig 2.28: Rod accelerating about its CG]

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 39 age 39
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Solution 2: using the general formula, with limits [-(L/2) to (L/2)] we have:

L L
2 2 3 2 L 2
2 m 2 m x mL
I = dmx = dx x = L = 12
L
3
L L − L 2
− 2 − 2


Situation 3: consider a thick cylinder rotating about an axis through the CG and parallel to the
length of the cylinder, as shown in [Fig 2.29]. Find its MMOI.





















[Fig 2.29: Thick cylinder rotating about an axis of symmetry
perpendicular to its cross-section]


Solution 3: the mass of the elemental cylinder at a distance ‘r’ from the CG is given by:

dm = 2πr ∗ dr ∗ L ∗ ρ and the elemental MMOI is given by: dI = dm ∗ r
2
Therefore, the total mass moment of inertia of the cylinder is given by:


R R
2
r 4 2πLρR 4 πR ∗ L ∗ ρ ∗ R 2 mR 2
3
I = 2πLρ r dr = 2πLρ = = =
4 4 2 2
0 0





Page 40 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Situation 4: consider a thin disc of thickness ‘t’. Assuming that the disc is rotating about its
vertical diametral axis, as shown in [Fig 2.30]. Find its MMOI.






















[Fig 2.30: thin disc rotation about the vertical diameter]


Solution 4: the mass of the elemental strip is given by:


dm = 2y ∗ dx ∗ t ∗ ρ , the elemental MMOI is given by: dI = dm ∗ x
2
= dI = 2y ∗ dx ∗ t ∗ ρ ∗ x
2
R
2
I total = 2y ∗ dx ∗ t ∗ ρ ∗ x - - - - (2.29)
−R

Since the integration is with respect to ‘dx’, let us represent ‘y’ in equation (2.29) in terms of ‘x’.

From [Fig 2.30], we have: y = R − x
2
2
Substituting the value of ‘y’ in equation (2.29), we get:


R
2
2
2
I total = 2 R − x ∗ t ∗ ρ ∗ x dx - - - - (2.30)
−R
Further, from [Fig 2.30], we have: x = R sinθ , upon differentiation, we get: dx = R cos θ dθ


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 41 age 41
P
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


Since ‘x’ is expressed in terms of ‘θ’, the limits of the integral also change accordingly. That is:

In the expression x = R sin θ , when, (x = R), we have: R = R sin θ = sinθ = 1 = θ = π 2

And when, (x = -R), we have: −R = R sinθ = sin θ = −1 = θ = − π 2

Substituting the values of ‘x’ and ‘dx’ in equation (2.30), and replacing the limits with that of ‘θ’,
we get:

π
2
2 2 2 2 2
I

total = 2 R − R sin θ ∗ t ∗ ρ ∗ R sin θ ∗ R cos θ dθ


− π
2
π π π
2 2 sin 2θ tρR 4 2 1 − cos4θ
2
2

4
2
4




= 2tρR cos θ sin θ dθ = 2tρR dθ = dθ
π π 4 2 π 2
− 2 − 2 − 2
π π
2 2 π
tρR 4 tρR 4 π sin4θ 2

= dθ − cos 4θ dθ = θ 2 π − π

4 − π − π 4 − 2 4 − 2
2 2
2
tρR 4 sin 2π sin2π tρR 4 πtρR 4 πR tρ R 2 mR 2
I = π − + = π − 0 = = =
total
4 4 4 4 4 4 4





















Page 42 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Situation 5: find the mass moment of inertia of a cylinder about a vertical axis perpendicular to
the axis of the cylinder, as shown in [Fig 2.31].



















[Fig 2.31: Thick cylinder rotating about an axis of symmetry
perpendicular axis of the cylinder]



Solution 5 the mass moment of inertia about the vertical axis has two components:

1) Inertia about the axis of rotation (treating the elemental section on the cylinder as a point
mass)
2) Inertia about the vertical axes of the elemental sections themselves. Therefore, from
[Fig 2.32] we have the total MMOI of the cylinder about the vertical axis of symmetry to be:





















[Fig 2.32: Rotation of an elemental section about its vertical diameter]







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 43 age 43
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


[Total MMOI = MMOI of the elemental section about the axis of rotation (treating it as a thin rod) +
MMOI about the vertical diameter of all the elemental sections]

L
L 2
2
2 2 dm ∗ R 2
I



total = πR ρ dx ∗ x + 4 Proved in situation 4
L L
−
−
2 2

L L
2 2
2
2 2 πR ρ dx ∗ R 2
I



total = πR ρ dx ∗ x + 4
L L
−
−
2 2

L L L
2 πR ρ 2 x 3 2 πR ρ L
4
4
2
2
2
I total = πR ρ x dx + dx = πR ρ + x 2



L
4 3 L 4 −
2
− L − L − 2
2 2
L 3 L 3 πR ρ L L πR ρL ∗ L 2 πR ρL ∗ R 2
2
2
4
2
I total = πR ρ + + + = +
24 24 4 2 2 12 4

mL 2 mR 2
= I total = +
12 4
Situation 6: find the mass moment of inertia for a sphere, shown in [Fig 2.33]










[Fig 2.33: spherical object rotating about its vertical axis]



Page 44 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]


Solution 6: the sphere could be deemed upon as a pile up of plates. Each plate is a thin disc with
2
MMOI of [dm r 2]. The summation of MMOI’s of all such plates about the axis of rotation gives the
MMOI of the sphere. Therefore, we have:

R R
2
dm r 2 πr ∗ dy ∗ ρ ∗ r 2
I = =
sphere
2 2
−R −R
Since the integration is with respect to ‘dy’, let us represent ‘r’ in equation (2.31) in terms of ‘y’.
Therefore, from [Fig 2.33] we have: r = R − y
2
2
2 2
R π R − y ∗ dy ∗ ρ ∗ R − y
2
2
2
2

I
sphere = 2
−R

R
2
2 2
π R − y ∗ dy ∗ ρ
I =
sphere
2
−R
R R
2 2
4
4
π R + y − 2R y ∗ dy ∗ ρ ρπ
4
2 2
4
I sphere = = + y − 2R y dy
2 2
−R −R
2 3
ρπ y 5 R 2R y R ρπ 2R 5 4R 5 1 2
4 R
5
5
I sphere = yR + − = 2R + − = ρπR 1 + −
2 −R 5 3 2 5 3 5 3
−R −R
6 2 8ρπR 5
5
Therefore, I = ρπR − = - - - - (2.32)
sphere
5 3 15
4
3
We know that the mass of a sphere is πR ρ
3
4 2 2
3
2
2
Rewriting equation . we get: I = πR ρ ∗ R = mR
sphere
3 5 5
It can be noted that the axis of rotation for a sphere could be along any diameter. Due to symmetry,
the mass moment of inertia of a sphere about any axis of rotation passing through the CG is the
same.


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 45 age 45
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics


2.6.2 Parallel and Perpendicular axis theorem

The axes of rotation need not always pass through the CG. For example, the axis of rotation for a
compound pendulum, shown in [Fig 2.34], is through the pivot, and hence, we need mass
moment of inertia about an axis through the pivot.






















[Fig 2.34: Compound pendulum oscillating about a pivot]




Parallel axis theorem: This situation, could be tackled using parallel axis theorem instead of
integrating for the entire length, assuming that I CG is known.
The mass moment of inertia of the compound pendulum about the pivot axis is given by:


2
I pivot = I CG + m mass of the compound pendulum ∗ d
This approach of evaluating MMOI about an axis parallel to the CG axis is called “parallel axis
theorem”. This theorem could be applied to an assembly of components as well.

Perpendicular axis theorem: when the mass moments of inertia of a component along two
perpendicular axes are known, then, the mass moment of inertia could be computed about the third
perpendicular axis using the perpendicular axis theorem.











Page 46 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]




‘•‹†‡” ȏ ‹‰ ʹǤ͵ͷȐ ƒ• •Š‘™ǡ





















[Fig 2.35: MMOI of a thin lamina about the cartesian axes]


For the diagram, the inertias of the point mass in the (x, y and z) directions are as follows:

2
dI xx = point mass ∗ perpendicular distance = dm ∗ y
2
dI yy = point mass ∗ perpendicular distance = dm ∗ x

2
2
dI zz = point mass ∗ perpendicular distance = dm ∗ (x + y )
2
2
Therefore, we have: dI xx + dI yy = dI and I zz = dmy + dmx
zz
Points to Ponder: Parallel and perpendicular axis theorem

1) The perpendicular axis theorem applies only to objects whose thickness is small. For
example, rectangular and circular laminae or laminae of any irregular shape.

2) Parallel axis theorem can be applied to any object provided that, the inertia at the CG is
given or known.








QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 47 age 47
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics



3) For thick objects, the inertias are computed about two axes:

Ø About the axis of rotation.
Ø About the axis of each section parallel to the axis of rotation.

4) For a general object, about the 3 perpendicular axes of symmetry, mass moment inertia
exists in such a way that, one of them is maximum and the others are intermediate and
minimum. These are called the principal mass moment of inertias.

The most commonly used MMOI’s for engineering calculations are as tabulated:


Component Description Mass moment of inertia
Cuboid: Since the cuboid is symmetric about


each axis, it has 3 principal inertias.
W

CG

H

L


Hollow shaft: The mass moment of inertia about the
axis of the cylinder can be got by


subtracting the inertia of solid cylinder
with radius with that of the

L cylinder with radius For the CG

axis it is treated as a thick rod.


Component Description Mass moment of inertia
Thin rod: The mass moment of inertia is sought
about an axis through the CG of the rod.
The inertia about the axis of the rod is
CG
not considered as the entire mass is
almost concentrated along the axis of
the rod.
Thick Cylinder:


CG
For a thin disc:
x










Page 48 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,

Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]




Component Description Mass moment of inertia

Sphere: A sphere is absolutely symmetric and Solid Sphere:
the inertia about any axis through the

CG is constant.
CG

Thin Spherical shell:

Cube: A cube has 13 axes of rotational
symmetry. Here, we have considered
a
cartesian axes, about all of which,
CG a inertia is the same.




a






ʹǤ͸Ǥ͵ ƒ•• ‘‡– ‘ˆ ‡”–‹ƒ ‡•‘”

 ‡…Šƒ‹…ƒŽ ƒ••‡„Ž‹‡•ǡ –Š‡ ”‹‰‹† ‡„‡”• …ƒ „‡ ‘”‹‡–‡† ƒ”„‹–”ƒ”‹Ž› ȋ‘– ƒŽ‹‰‡† ™‹–Š –Š‡
•–ƒ†ƒ”† ƒš‡•ȌǤ ‘ ˆƒ…‹Ž‹–ƒ–‡ ƒ †‡–ƒ‹Ž‡† —†‡”•–ƒ†‹‰ ƒ• –‘ Š‘™ –Š‡ ˜ƒŽ—‡ ‘ˆ ‹‡”–‹ƒ …‘—Ž† „‡ ˆ‘—†
ƒ„‘—– ƒ ‰‡‡”ƒŽ ƒš‹•ǡ ™‡ ‡‡† –‘ —†‡”•–ƒ† –Š‡ ’”‘’‡”–‹‡• ‘ˆ –Š‡ ‹‡”–‹ƒ –‡•‘” ƒ† –Š‡ ‹‡”–‹ƒ
–‡•‘” ‹–•‡Žˆ ˆ‘” ƒ ”‹‰‹† „‘†›Ǥ

‡– —• …‘•‹†‡” ƒ „ƒ” ‹…Ž‹‡† ƒ– ƒ ƒ‰Ž‡ –‘ –Š‡ Ǯ›-ƒš‹•ǯǡ ™Š‹…Š ‹• ƒŽ•‘ –Š‡ ƒš‹• ‘ˆ ”‘–ƒ–‹‘ǡ ƒ• •Š‘™
‹ ȏ ‹‰ ʹǤ͵͸ȐǤ















[Fig 2.36: an inclined rod accelerated about the 'y-axis']





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
Page 49 age 49
Copyright Diary No – 9119/2018-CO/L

Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

The rod is given an angular acceleration ‘α’, about the vertical axis. Consider an element on the rod
with coordinates (x, y). This element experiences a linear acceleration αx . Therefore, the linear
inertia of the element is given by dmαx .

The linear inertia has a moment about both ‘x and y-axis’, given by:

About the ‘x-axis’ = dmαx ∗ y

About the ‘y-axis’ = dmαx ∗ x

Therefore, if we sum up the contributions of all the elements along the length of the rod, we would
get two inertia terms:


2
1) dmαx = α dmx
2
2) dmαxy = α dmxy
The first term gives the inertia of the rod about the ‘y-axis’. However, the second term is called the
cross product of inertia, as, it contains two coordinates (x and y).

If the angular acceleration were given about the ‘x-axis’, as shown in [Fig 2.37], then the two inertia
terms would be:

2
2
1) dmαy = α dmy
2) dmαyx = α dmyx
These terms have the same explanations as before.


















[Fig 2.37: an inclined rod accelerated about the 'x-axis']









Page 50 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,