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

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ͳǤͶ ’’Ž‹‡† ƒŽ…—Ž—•

ƒŽ…—Ž—• ™ƒ• ˆ—†ƒ‡–ƒŽŽ› †‡˜‡Ž‘’‡† –‘ †‡ƒŽ ™‹–Š ‘-Ž‹‡ƒ” •›•–‡•ǡ ™Š‡”‡‹ǡ …Šƒ‰‡• ‘ˆ ƒ
“—ƒ–‹–› ™‹–Š ”‡•’‡…– –‘ ƒ‘–Š‡” “—ƒ–‹–› ‹• ˜ƒ”›‹‰ ‹ ƒ ‘-Ž‹‡ƒ” ˆƒ•Š‹‘Ǥ

‘ ‹ŽŽ—•–”ƒ–‡ –Š‹• Ž‡– —• …‘•‹†‡” ƒ •–”ƒ‹‰Š– Ž‹‡ ƒ† ƒ …—”˜‡ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳ Ƭ ͳǤʹȐǤ
















[Fig 1.1: slope for a straight line] [Fig 1.2: slope for a curve]




 ’”ƒ…–‹…‡ǡ –Š‡”‡ ƒ”‡ ˆ‡™ •›•–‡• –Šƒ– ƒ”‡ Ž‹‡ƒ”ǡ ƒ† ‘–Š‡”• ƒ”‡ ‘-Ž‹‡ƒ”Ǥ


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

ƒŽ›œ‹‰ Ž‹‡ƒ” •›•–‡• ȏ ‹‰ ͳǤͳȐ ‹• ‡ƒ•›ǡ ƒ•ǡ –Š‡ ”ƒ–‡ ‘ˆ …Šƒ‰‡ ‘ˆ †‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ ȋ›Ȍ ™‹–Š
”‡•’‡…– –‘ –Š‡ ‹†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ ȋšȌ ‹• …‘•–ƒ– ƒŽŽ –Š‡ ™Š‹Ž‡Ǥ Š‡ †‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ ‹• …ƒŽŽ‡†
•‘ǡ „‡…ƒ—•‡ǡ ‹– ‹• ƒ ˆ—…–‹‘ ‘ˆ –Š‡ ‹†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ ȏ› α ˆȋšȌȐǤ Š‡”‡ˆ‘”‡ǡ –Š‡ …Šƒ‰‡• ‹ –Š‡
†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ †‡’‡† ‘ –Š‡ …Šƒ‰‡• ‹ –Š‡ ‹†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡Ǥ
ƒŽ›œ‹‰ ‘-Ž‹‡ƒ” •›•–‡• ‹• †‹ˆϐ‹…—Ž–ǡ ƒ•ǡ ƒ– ‡ƒ…Š ’‘‹– –Š‡ ”ƒ–‡ ‘ˆ …Šƒ‰‡ ‘ˆ †‡’‡†‡– ˜ƒ”‹ƒ„Ž‡
™‹–Š ”‡•’‡…– –‘ –Š‡ ‹†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡ ‹• ‹…”‡ƒ•‹‰ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤʹȐǤ

’Š›•‹…ƒŽ ‹–‡”’”‡–ƒ–‹‘ ‘ˆ ƒ ‘-Ž‹‡ƒ” •›•–‡ǡ •ƒ›ǡ ƒ …ƒ” ƒ……‡Ž‡”ƒ–‹‰ ƒ– ƒ …‘•–ƒ– ”ƒ–‡ǡ ‹• ƒ•
ˆ‘ŽŽ‘™•Ǥ










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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
‘•‹†‡” ƒ †‹•–ƒ…‡ ˜• –‹‡ ‰”ƒ’Š ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͵Ȑ











[Fig 1.3: distance vs time]

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

‡– —• …‘•‹†‡” –Š‡ ˆ‘ŽŽ‘™‹‰ ’”‘…‡†—”‡ǣ


‘ †‡–‡”‹‡ –Š‡ •Ž‘’‡ǡ ™‡ ‡‡† –™‘ ’‘‹–• ‘ –Š‡ …—”˜‡ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͶȐǤ ˆ –Š‡ ’‘‹–• ƒ”‡
ˆƒ” ƒ’ƒ”–ǡ –Š‡ •Ž‘’‡ ‹• ƒ˜‡”ƒ‰‡† ƒ† ‘– ƒ…–—ƒŽ ƒ– –Š‡ †‡•‹”‡† ’‘‹–Ǥ Š‡ ‡””‘” …‘—Ž† „‡ •‹‰‹ϐ‹…ƒ–
†‡’‡†‹‰ ‘ –Š‡ …—”˜‡ ƒ† –Š‡ ’‘‹–• …Š‘•‡Ǥ














[Fig 1.4: slope of a curve]

‘•‹†‡” –Š‡ •Ž‘’‡ ‘ˆ …—”˜‡ ƒ– ’‘‹– ȋ͵ȌǤ Š‡ •Ž‘’‡ ƒ– ’‘‹– ȋ͵Ȍ ‹• †‡–‡”‹‡† „› …Š‘‘•‹‰ –™‘
‡ƒ”„› ’‘‹–• ȋͳȌ ƒ† ȋʹȌǤ – …‘—Ž† „‡ ‘„•‡”˜‡† ˆ”‘ ȏ ‹‰ ͳǤͶȐ –Šƒ–Ǥ –Š‡ •Ž‘’‡ ‘ˆ –Š‡ …—”˜‡ ƒ– ’‘‹–•
ȋͳȌ ƒ† ȋʹȌ ‹• †‹ˆˆ‡”‡– ˆ”‘ –Š‡ •Ž‘’‡ ‘ˆ –Š‡ …—”˜‡ ƒ– ’‘‹– ȋ͵ȌǤ

‡– —• ‘™ †‡–‡”‹‡ –Š‡ ’”‘š‹‹–› ‘ˆ ’‘‹–• ȋͳȌ ƒ† ȋʹȌ –‘ ‹‹‹œ‡ ‡””‘” ‹ –Š‡ †‹ˆˆ‡”‡…‡ ‘ˆ
•Ž‘’‡•Ǥ – ‹• ‹–—‹–‹˜‡ –Šƒ– –Š‡ ‡””‘” ‹• ‹‹— ™Š‡ –Š‡ ’‘‹–• ȋͳȌ ƒ† ȋʹȌ ƒ”‡ ‹ –Š‡ …Ž‘•‡•–
ƒ–Š‡ƒ–‹…ƒŽ ’”‘š‹‹–› –‘ ‡ƒ…Š ‘–Š‡”Ǥ Š‹• ‡ƒ• –Šƒ–ǡ –Š‡ †‹ˆˆ‡”‡…‡ ‹ –Š‡‹” ‹†‡’‡†‡–
…‘‘”†‹ƒ–‡ ȋšȌ ‹• ˜‡”› •ƒŽŽ ƒ† ‹• –‡†‹‰ –‘ œ‡”‘Ǥ





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 5

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics


Š‡”‡ˆ‘”‡ǡ –Š‡ •Ž‘’‡ ‹• ‰‹˜‡ „›ǣ
∆y
slope = lim - - - - (1.1)
∆x→0 ∆x

The expression in equation (ͳǤͳ) is called the derivative of (y) with respect to (x)
dy
and is represented as: , which is also used to find the slope of the curve at a given point.
dx
dy dy
−
2
dx dx d dy d y
Further, the rate of change of slope: lim 2 1 = = - - - - (1.2)
∆x→0 ∆x dx dx dx 2
dy
is got by appying the concept of limit, to change in , rather than (y).
dx



‘–‡ǣ Š‡ •Ž‘’‡ ‘ˆ –Š‡ †‹•–ƒ…‡ ˜• –‹‡ ‰”ƒ’Š ›‹‡Ž†• ˜‡Ž‘…‹–›ǡ ƒ† –Š‡ ”ƒ–‡ ‘ˆ …Šƒ‰‡ ‘ˆ •Ž‘’‡
›‹‡Ž†• ƒ……‡Ž‡”ƒ–‹‘Ǥ

ͳǤͶǤͳ –ƒ†ƒ”† ‡”‹˜ƒ–‹˜‡•

‘ŽŽ‘™‹‰ ƒ”‡ ƒ ˆ‡™ •–ƒ†ƒ”† †‡”‹˜ƒ–‹˜‡• …‘‘Ž› —•‡† ‹ –Š‡ ‡‰‹‡‡”‹‰ …‘–‡š–Ǥ

ˆȋšȌ ˆ ǯȋšȌ ˆȋšȌ ˆ ǯȋšȌ

-ͳ
•‹ š …‘• š š š
…‘• š -•‹ š −2x

a − x 2 2
2
2
–ƒ š •‡… š 2 a − x
ʹ
š š
•‡… š –ƒ š •‡… š ‡ ‡
ƒš
ƒš
…‘•‡… š -…‘– š …‘•‡… š ‡ ƒ‡
…‘– š -…‘•‡… š Ž‘‰‡ š 1
ʹ
x
cos x
log sin x = cot x
e
sin x

′
f x ∗ g x [Chain Rule] f x ∗ g x + g x ∗ f(x)
′

g(x) g x ∗ f x − f x ∗ g(x)
′
′


f(x) [Division Rule] (f(x)) 2



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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
ͳǤͶǤʹ ‡”‹˜ƒ–‹˜‡ –‘ ϐ‹† ƒš ƒ† ‹ ‘ˆ ƒ ˆ—…–‹‘

‘•‹†‡” –Š‡ ˆ‘ŽŽ‘™‹‰ ‰”ƒ’Š•ǣ














[Fig 1.5: slope at peaks]
dy
The . helps us conclude that, is zero when the curve is either at maximum or
dx

minimum value


‘ ƒ•…‡”–ƒ‹ –Š‡ ƒš ‘” ‹ ˜ƒŽ—‡•ǡ –Š‡ ˆ‘ŽŽ‘™‹‰ …‘†‹–‹‘• ƒ’’Ž›ǣ

2
d y
1) when < 0 , the value of y is maximum
dx 2
2
d y
2) when > 0 , the value of y is minimum
dx 2
2
d y
3) when = 0 , this is called a saddle point stationary point , where the rate of change
dx 2


‘ˆ •Ž‘’‡ ‹• œ‡”‘ǡ ‡ƒ‹‰ǡ –Š‡ †‹”‡…–‹‘ ‘ˆ •Ž‘’‡ …Šƒ‰‡ …‘—Ž† „‡ ’‘•‹–‹˜‡ ‘” ‡‰ƒ–‹˜‡Ǥ Š‹• ‹• ƒŽ•‘
…ƒŽŽ‡† –Š‡ •–ƒ–‹‘ƒ”› ‘” •ƒ††Ž‡ ’‘‹–ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ȐǤ













[Fig 1.6: max, min and saddle points]

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

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
2
šƒ’Ž‡ ͳǣ ϐ‹† –Š‡ ƒš ‘” ‹ ˜ƒŽ—‡ ˆ‘” –Š‡ ˆ—…–‹‘ [y = 4x + 2x − 16] - - - - (1.3)

‘Ž—–‹‘ ͳǣ

dy
Differentiating y , and equating the slope to zero, we get: = 8x + 2 = 0
dx
2
Therefore, we have x = − = −0.25
8


‘™ǡ –‘ —†‡”•–ƒ† ‹ˆ ƒ ‹‹— ‘” ƒš‹— ‡š‹•–• ƒ– ȋš α -ͲǤʹͷȌǡ Ž‡– —• …‘’—–‡ ‹ˆ –Š‡ ”ƒ–‡ ‘ˆ
…Šƒ‰‡ ‘ˆ •Ž‘’‡ ‹• ’‘•‹–‹˜‡ ‘” ‡‰ƒ–‹˜‡Ǥ
2
d y
Rate of change of slope = = 8
dx 2
2
d y
Since the rate of change of slope is positive > 0 , y assumes a minimum value.
dx 2
Š‡ ˜ƒŽ—‡ ‘ˆ ȋ›Ȍ ‹• ‰‘– „› •—„•–‹–—–‹‰ –Š‡ ˜ƒŽ—‡ ‘ˆ ȋšȌ ‹ ‡“—ƒ–‹‘ ȋͳǤ͵Ȍ


2
That is: y = 4 0.25 + 2 ∗ −0.25 − 16 = −16.25
Š‡ ‹‹— ˜ƒŽ—‡ ‘ˆ –Š‡ ˆ—…–‹‘ ‹• ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹ȐǤ
















[Fig 1.7: minimum value of function ( = 4 + 2 − 16)]
2











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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
šƒ’Ž‡ ʹǣ

…‘’—–‡ –Š‡ ‘’–‹— ƒ”‡ƒ ‘ˆ ƒ ‹•‘•…‡Ž‡• –”‹ƒ‰Ž‡ ȋȏ ‹‰ ͳǤͺȐȌ ˆ‘” ƒ ‰‹˜‡ ’‡”‹‡–‡” ȋ ȌǤ


‘Ž—–‹‘ ʹǣ ‘•‹†‡” –Š‡ ‹•‘•…‡Ž‡• –”‹ƒ‰Ž‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͺȐǤ

‡– –Š‡ Š‡‹‰Š– ‘ˆ –Š‡ –”‹ƒ‰Ž‡ „‡ ǮŠǯǡ „ƒ•‡ „‡ Ǯšǯ ƒ† –Š‡ ‡“—ƒŽ •‹†‡• „‡ Ǯ›ǯǤ

P − x
The perimeter of the triangle is given by P = x + 2y , therefore y = - - - - (1.4)
2


Š‡ Š‡‹‰Š– ǮŠǯ …‘—Ž† „‡ ‰‘– „› ƒ’’Ž›‹‰ ›–Šƒ‰‘”ƒ• –Š‡‘”‡ǣ



P − x 2 x 2
h = −
2 4
















[Fig 1.8: Isosceles triangle of base ‘x’ and height ‘h’]



‘™ǡ –Š‡ ƒ”‡ƒ ‘ˆ –Š‡ –”‹ƒ‰Ž‡ ‹• ‰‹˜‡ „›ǣ


1 1 P − x 2 x 2 1 P 2 x 2 2Px x 2

Area A = ∗ x ∗ h = ∗ x ∗ − = ∗ x ∗ + − −
2 2 2 4 2 4 4 4 4

2 2
1 P x Px 3
Simplifying the above expression, we get: Area A = − - - - - (1.5)
2 4 2

Š‡ ‘’–‹— ˜ƒŽ—‡ ‘ˆ ƒ”‡ƒ ȋ Ȍ ‹• ‰‘– „› †‹ˆˆ‡”‡–‹ƒ–‹‰ ‡“—ƒ–‹‘ ȋͳǤͷȌ ™‹–Š ”‡•’‡…– –‘ ȋšȌ ƒ†
‡“—ƒ–‹‰ –Š‡ †‡”‹˜ƒ–‹˜‡ –‘ œ‡”‘Ǥ Š‹• ‰‹˜‡• –Š‡ ˜ƒŽ—‡ ‘ˆ ȋšȌ ƒ– ™Š‹…Š –Š‡ ˆ—…–‹‘ ‹ ‡“—ƒ–‹‘ ȋͳǤͷȌ
ƒ••—‡• –Š‡ ‘’–‹— ˜ƒŽ—‡Ǥ



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 9

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics

′
2 2
1 f x P x Px 3
′
Equation . is of the form: f x = = 0 or f x = 0 , where f x = −
2 f x 4 2
2
d f x P x 3Px 2 P
′
Therefore, we have: f x = = − = 0 , by where, we get: x =
dx 2 2 3
P
P − P
3
Substituting the value of x in the expression . , we get: y = =
2 3
‹…‡ ȋšȌ ‹• ‡“—ƒŽ –‘ ȋ›Ȍǡ –Š‡ –”‹ƒ‰Ž‡ ‹• ‡“—‹Žƒ–‡”ƒŽ ˆ‘” ‘’–‹— ƒ”‡ƒǤ

Š‡ ‘’–‹— ƒ”‡ƒ ȋ Ȍ ‹• ‰‘– „› •—„•–‹–—–‹‰ –Š‡ ˜ƒŽ—‡ ‘ˆ ȋšȌ ‹ ‡“—ƒ–‹‘ ȋͳǤͷȌǣ


P 2 P 3
2
1 P P 1 P 4 P 4 P 2 18 P 2
3
3
Area A = 2 4 − 2 = − 54 = 2 36 ∗ 54 = 12 3
2 36


P 2
Therefore, we have the optimum area of the triangle to be:
12 3

– …ƒ ƒŽ•‘ „‡ •Š‘™ –Šƒ– –Š‡ ƒ”‡ƒ ȋ Ȍ ‹• –Š‡ ƒš‹— ˜ƒŽ—‡ǡ ƒ• ˆ‘ŽŽ‘™•ǣ

2
2
d f x P x 3Px 2 d f x P 2
′
f x = = − , therefore, = − 3Px
dx 2 2 dx 2 2
—„•–‹–—–‹‰ –Š‡ ˜ƒŽ—‡ ‘ˆ ȋšȌ ‹ –Š‡ ƒ„‘˜‡ ‡š’”‡••‹‘ǡ ™‡ ‰‡–ǣ

2
2
d f x P 2 P P 2 d y
= − 3P = − , which means < 0 , indicating maximum value.
dx 2 2 3 2 dx 2

‘–‡ǣ ™Š‡‡˜‡” –Š‡ •— ‘ˆ –™‘ ˜ƒ”‹ƒ„Ž‡• ‹• ϐ‹š‡†ǡ –Š‡ǡ –Š‡‹” ’”‘†—…– ™‘—Ž† „‡ ƒš‹— ‘Ž›
™Š‡ –Š‡› ƒ”‡ ‡“—ƒŽǤ

šƒ’Ž‡ǣ ‹ˆ –Š‡ •— ‘ˆ –™‘ —„‡”• ‹• ʹͷǡ –Š‡ ƒš‹— ˜ƒŽ—‡ ‘ˆ –Š‡‹” ’”‘†—…– ‹• ‰‹˜‡ „›ǣ
ȋͳʹǤͷȗͳʹǤͷȌǡ ™Š‹…Š ‹• ͳͷ͸ǤʹͷǤ

• ƒ ’”ƒ…–‹…ƒŽ …ƒ•‡ǡ ™Š‡ –Š‡ •— ‘ˆ –Š”‡‡ •‹†‡• ‘ˆ ƒ …—„‘‹† ‹• ‰‹˜‡ǡ –Š‡ ˜‘Ž—‡ …‘—Ž† „‡
ƒš‹‹œ‡† ™Š‡ –Š‡ •‹†‡• ƒ”‡ ‡“—ƒŽ –‘ ‡ƒ…Š ‘–Š‡”ǡ –Šƒ– ‹•ǡ ‹– „‡…‘‡• ƒ …—„‡Ǥ


Š‹• Ž‘‰‹… …‘—Ž† „‡ ‡š–‡†‡† –‘ ƒ› —„‡” ‘ˆ ˜ƒ”‹ƒ„Ž‡•Ǥ




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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
šƒ’Ž‡ ͵ǣ ϐ‹† –Š‡ ƒš‹— ƒ† ‹‹— ˜ƒŽ—‡• ‘ˆ –Š‡ ˆ—…–‹‘ a cos θ + b sin θ .

‘Ž—–‹‘ ͵ǣ ‹ ‘”†‡” –‘ ϐ‹† –Š‡ ƒš‹— ƒ† ‹‹— ˜ƒŽ—‡ǡ ™‡ ™‘—Ž† ƒ‡ —•‡ ‘ˆ –Š‡ ˆƒ…–
–Šƒ– Ǯ•‹Ʌǯ ˜ƒ”‹‡• „‡–™‡‡ ƒš‹— ȋΪͳȌ ƒ† ‹‹— ȋ-ͳȌǤ

‘ ƒ’’Ž› –Š‡ ƒ„‘˜‡ ˆƒ…– Ž‡– —• †‹˜‹†‡ ƒ† —Ž–‹’Ž› –Š‡ ‡š’”‡••‹‘ ȋƒ …‘• Ʌ Ϊ „ •‹ ɅȌ ™‹–Š a + b ,
2
2
—…Š –Šƒ–ǡ ™‡ …‘—Ž† —•‡ –Š‡ ˆ‘”—Žƒ ˆ‘” •‹ȋ Ϊ ȌǤ Š‡”‡ˆ‘”‡ǡ ™‡ Šƒ˜‡ǣ


a cos θ + b sin θ a b
2
2
2
2
∗ a + b = cos θ + sinθ a + b
2
2
2
a + b 2 a + b 2 a + b 2
”‘ –Š‡ ȏ ‹‰ ͳǤͻȐǡ ™‡ Šƒ˜‡ǣ








[Fig 1.9: by Pythagoras theorem]

= sinΦ cos θ + cos Φ sin θ a + b = sin Φ + θ ∗ a + b - - - - (1.6)
2
2
2
2
Since we know that the value of sin θ varies from (+1) to (-1),
The maximum value of equation (1.6) = 1 ∗ a + b
2
2
The minimum value of equation (1.6) = −1 ∗ a + b
2
2



















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 11

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
1.4.3 Concept of Curvature

The expression for curvature is highly useful, be it, for mechanics of a vehicle negotiating
undulations or curves, or, curvature of a bent beam. Therefore, it is important to derive a
general expression which could be used for any curve as shown in [Fig 1.10].
Point of the curve







[Fig 1.10: a general curve with varying curvature]




 ‘”†‡” –‘ ϐ‹† –Š‡ ”ƒ†‹—• ƒ– ƒ ‰‹˜‡ ’‘‹– ‘ ƒ ‰‡‡”ƒŽ …—”˜‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳͲȐǡ ƒ ƒ”… ‹•
†”ƒ™ •—…Š –Šƒ– –Š‡ …—”˜‡ ƒ– –Šƒ– ’‘‹– ‹• –ƒ‰‡–‹ƒŽ –‘ –Š‡ ƒ”… ‘ˆ ”ƒ†‹—• ȋ”ȌǤ

• „‘–Š –Š‡ •Ž‘’‡ ƒ† ‹–• ”ƒ–‡ ‘ˆ …Šƒ‰‡ ˜ƒ”› ˆ‘” ƒ …—”˜‡ǡ Ž‡– —• ϐ‹† –Š‡ ‰‡‡”ƒŽ ‡š’”‡••‹‘ ˆ‘”
…—”˜ƒ–—”‡ ‘” ”ƒ†‹—• ‹ –‡”• ‘ˆ ȋ†›Ȁ†šȌ ƒ† ȋ† ›Ȁ†š Ȍ
ʹ
ʹ
‘•‹†‡” ƒ …—”˜‡ ‹ –Š‡ Ǯš-›ǯ ’Žƒ‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳͳȐǤ ”ƒ™ –ƒ‰‡–• ƒ– –™‘ ’‘‹–• •‡’ƒ”ƒ–‡†
„› ƒ ‡Ž‡‡–ƒŽ ƒ”… Ž‡‰–Š ȋ†•ȌǤ Š‡ ƒ”…
Ž‡‰–Š •—„–‡†• ƒ ƒ‰Ž‡ ȋ†ɅȌ ƒ– –Š‡
…‡–‡”Ǥ Š‹• ‹• ‡šƒ…–Ž› –Š‡ ƒ‰Ž‡ „‡–™‡‡
–Š‡ –ƒ‰‡–• ƒ– ’‘‹– ȋͳȌ ƒ† ȋʹȌǤ














[Fig 1.11: radius of curvature]




dy 2

2
2
Therefore, we have: Rdθ = dx + dy or Rdθ = dx 1 +
dx

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Page 12

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]


dθ dy 2
Further simplifying, we get: R = 1 + - - - - (1.7)
dx dx

2
dy d y dθ
2
We know that = tanθ , differentiating again we get: = sec θ - - - - (1.8)
dx dx 2 dx
dy 2
2
2
We also know that, sec θ = 1 + tan θ = 1 +
dx
2
d y dy 2 dθ
Therefore, equation . could be rewritten as: = 1 + - - - - (1.9)
dx 2 dx dx




1 2
2 2
2
d y 2 1 + dy ∗ 1 + dy
dx 2 dy dx dx
R 2 = 1 + dx = R = d y
2

1 + dy 2
dx dx

3
dy 2 2
1 + dx
Simplifying the above expression, we get: R = - - - - (1.10)
2
d y
dx 2

1
Expression . is that of radius of curvature R and curvature is given by .
R

2
d y
is positive for a concave curve and negative for a convex curve.
dx 2



”‘ ȏ ‹‰ ͳǤͳͳȐǡ ™‡ Šƒ˜‡ǣ ȏ †Ʌ α †•Ȑ
3
dy 2 2
1 + dx
Radius of Curvature R = - - - - (1.11)
2
d y
dx 2


“—ƒ–‹‘ ȋͳǤ͹Ȍ …ƒ ‘™ „‡ ”‡™”‹––‡ ƒ•ǣ





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 13

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
dy
If the curve has a small slope ≈ 0 at the point where the radius of curvature is sought, then,
dx


1
the radius of curvature reduces to: R = d y - - - - (1.12)
2

dx 2
2
1 1 d y
And the curvature is given by: = - - - - (1.13)
R R dx 2
dy
It can be observed that, for a general curve, the slope of the tangent is
dx


1
And slope of the normal is − dy , as shown in [ . ]

dx




















[Fig 1.12: slope for tangent and normal of a curve]
‹…‡ ™‡ ‘Ž›
•‡‡ –Š‡ ƒ‰‹–—†‡ ‘ˆ ”ƒ–‡ ‘ˆ …Šƒ‰‡ ‘ˆ •Ž‘’‡ ƒ† ‘– ‹–• †‹”‡…–‹‘ǡ ‡“—ƒ–‹‘ ȋͳǤͳͲȌ …ƒ „‡
”‡™”‹––‡ ƒ• ˆ‘ŽŽ‘™•ǣ



















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Page 14

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
ͳǤͶǤͶ –‡‰”ƒ–‹‘

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

–‡‰”ƒ–‹‘ ‘ˆ ƒ› ˆ—…–‹‘ ‹• —†‡”•–‘‘† ƒ• ˆ‘ŽŽ‘™•ǣ













[Fig 1.13: function f(x) between limits 'a' and 'b']

Š‡ ‹–‡‰”ƒ–‹‘ „‡–™‡‡ Ž‹‹–• ȋš α ƒȌ ƒ† ȋš α „Ȍ ‘ˆ –Š‡ ˆ—…–‹‘ ˆȋšȌǡ ™Š‹…Š ‹• ƒ …—”˜‡ ƒ• •Š‘™ ‹
ȏ ‹‰ ͳǤͳ͵Ȑ ‹• –Š‡ ƒ”‡ƒ —†‡” –Š‡ …—”˜‡Ǥ Š‡ ˆ—†ƒ‡–ƒŽ ƒ’’”‘ƒ…Š –‘ ϐ‹†‹‰ ƒ”‡ƒ —†‡” •—…Š ƒ
‰‡‡”ƒŽ …—”˜‡ ‹• ƒ• ˆ‘ŽŽ‘™•ǣ

‘•‹†‡” –Š‡ ƒ”‡ƒ —†‡” –Š‡ …—”˜‡ „‡–™‡‡ –Š‡ Ž‹‹–• ȋš α ƒȌ ƒ† ȋš α „ȌǤ

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

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

x=b
Total area under the curve = f x dx
x=a
b
The above expression is symbolically given to be: f x dx
a


Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ ǮˆȋšȌǯ ƒ† Ǯšǯ …‘—Ž† „‡ ‘ˆ ƒ› ƒ–—”‡ǡ ‹ˆ ‹– ‹• …‘–‹—‘—•Ǥ Š‡ ‹–‡‰”ƒŽ ‘ˆ
ǮˆȋšȌǯ ‹• ƒ ƒ–‹-†‡”‹˜ƒ–‹˜‡ǡ ‡ƒ‹‰ǡ †‹ˆˆ‡”‡–‹ƒ–‹‰ –Š‡ ‹–‡‰”ƒŽ ™‘—Ž† Ž‡ƒ† —• –‘ –Š‡ ˆ—…–‹‘ ‹–•‡ŽˆǤ
Š‹• …ƒ „‡ ƒ–Š‡ƒ–‹…ƒŽŽ› ”‡’”‡•‡–‡† ƒ• ˆ‘ŽŽ‘™•ǣ


d
f x dx = f x
dx

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 15

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
šƒ’Ž‡ ͳǣ ϐ‹† –Š‡ ƒ”‡ƒ —†‡” –Š‡ …—”˜‡ •Š‘™ ‹ ȏ ‹‰ ͳǤͳͶȐǤ














2
[Fig 1.14: area under a parabola [y = x ] ]

b 2 2
x 3 8 8
2
The area under the curve is given by: f x dx = x dx = = − 0 =
3 3 3
a 0 0

šƒ’Ž‡ ʹǣ ϐ‹† –Š‡ •—”ˆƒ…‡ ƒ”‡ƒ ƒ† ˜‘Ž—‡ ‘ˆ ƒ •’Š‡”‹…ƒŽ …ƒ’ ™Š‹…Š •—„–‡†• ƒ ƒ‰Ž‡ ǮʹɅͳǯ ƒ–
–Š‡ …‡–‡” ‘ˆ –Š‡ •’Š‡”‡ ƒ† ‹ –Š‡ ’Žƒ‡ ‘ˆ –Š‡ ‹ƒ‰‡ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳͷȐǤ



































[Fig 1.15: a spherical cap]





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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
‘•‹†‡” ƒ •ƒŽŽ •–”‹’ ‘ˆ ™‹†–Š Ǯ †Ʌǯ ƒ† Ž‡‰–ŠȀ…‹”…—ˆ‡”‡…‡ ǮʹɎ” α ʹɎ •‹ɅǯǤ Š‡ ƒ”‡ƒ ‘ˆ –Š‡
•–”‹’ ‹• ‰‹˜‡ „›ǣ ȏʹɎ •‹Ʌ ȗ †ɅȐǤ

‡– —• ‹–‡‰”ƒ–‡ –Š‡ ƒ”‡ƒ ‘ˆ –Š‡ ‡Ž‡‡–ƒŽ •–”‹’ ˆ”‘ ǮͲ –‘ ɅͳǯǤ



θ 1 θ 1
2
2
2π Rsinθ ∗ Rdθ = 2πR sinθ dθ = 2πR 1 − cosθ - - - - (1.14)
1
0 0
“—ƒ–‹‘ ȋͳǤͳͶȌ ‹• –Šƒ– ‘ˆ –Š‡ •—”ˆƒ…‡ ƒ”‡ƒ ‘ˆ –Š‡ •’Š‡”‹…ƒŽ …ƒ’Ǥ

‡– —• ˜‡”‹ˆ› –Š‡ ƒ……—”ƒ…› ‘ˆ ‡“—ƒ–‹‘ ȋͳǤͳͶȌǤ ˆ ™‡ •—„•–‹–—–‡ ˆ‘” ǮɅͳ α ͻͲ ǯǡ ™‡ ‰‡– ǮʹɎ ǯ ™Š‹…Š ‹•
ʹ
–Š‡ •—”ˆƒ…‡ ƒ”‡ƒ ‘ˆ –Š‡ Š‡‹•’Š‡”‡Ǥ ˆ ™‡ •—„•–‹–—–‡ ˆ‘” ǮɅͳ α ͳͺͲǯǡ ™‡ ‰‡– ǮͶɎ ǯ ™Š‹…Š ‹• –Š‡ ƒ”‡ƒ ‘ˆ
ʹ
–Š‡ •’Š‡”‡Ǥ
Š‡ ˜‘Ž—‡ ‘ˆ –Š‡ •’Š‡”‡ ‹• …‘’—–‡† ƒ• ˆ‘ŽŽ‘™•ǣ

ȋ ‘Ž—‡ ‘ˆ –Š‡ ‡Ž‡‡–ƒŽ •–”‹’ α ȏƒ”‡ƒ ‘ˆ –Š‡ ‡Ž‡‡–ƒŽ •–”‹’ ȗ Š‡‹‰Š– ȋ†ŠȌȐȌ


2
V strip = π Rsinθ ∗ Rdθ sinθ physically, this the volume of the plate
Š‡ ˜‘Ž—‡ ‘ˆ –Š‡ •’Š‡”‹…ƒŽ …ƒ’ ‹• ‰‹˜‡ „›ǣ

θ 1 θ 1 θ 1
π Rsinθ ∗ Rdθ sinθ = πR sin θ dθ = πR sin θ 1 − cos θ dθ - - - - (1.15)
2
3
3
3
2
0 0 0
‘ ˆƒ…‹Ž‹–ƒ–‡ ‹–‡‰”ƒ–‹‘ǡ Ž‡– —• ƒ‡ –Š‡ ˆ‘ŽŽ‘™‹‰ •—„•–‹–—–‹‘Ǥ
dt
Assume cosθ = t , differentiating, we get = − sinθ or dt = − sin θ dθ
dθ


Ž•‘ ™Š‡ (θ = θ ), t = cosθ , and, when θ = 0 , t = 1 .
1
1
Š‡”‡ˆ‘”‡ǡ ‡“—ƒ–‹‘ ȋͳǤͳͷȌ …ƒ „‡ ”‡™”‹––‡ ƒ•ǣ

cos θ 1 cos θ 1
t 3 cos θ 1
2
3
3
2
3
πR − 1 − t dt = πR t − 1 dt = πR − t cos θ 1
3 1
1 1 1
—„•–‹–—–‹‰ –Š‡ —’’‡” ƒ† Ž‘™‡” Ž‹‹–•ǡ ™‡ ‰‡–ǣ
3
3
cos θ 1 1 cos θ 1 2
3
3
πR − − cosθ + 1 = πR + − cosθ - - - - (1.16)
1
1
3 3 3 3
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Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
2
3
: if θ = 90 then, we get πR , which is the volume of a hemisphere
1
3
4
3
if θ = 180 then,we get πR , which is the volume of a sphere
1
3
ͳǤͶǤͷ –ƒ†ƒ”† –‡‰”ƒŽ•



( ) ( )



x x n+1
+ c
n + 1
n
x (when( n = −1)) log e x + c
x e e x + c


e ax
e ax + c
a

sin x − cos x + c

cos x sin x + c

x + c
tan x log sec
e
log e x (log x + 1)x + c
e

a 2 x x x 2
2
a 2 − x sin −1 + 1 − + c
2 a a a 2

f x ∗ g(x) [integr ation by parts] g x f x − f x ∗ g x dx
′


ͳǤͶǤ͸ ͳ ‘”†‡” ‹ˆˆ‡”‡–‹ƒŽ “—ƒ–‹‘•
•–
Š‡ ‘”†‡” ‘ˆ ƒ †‡”‹˜ƒ–‹˜‡ †‡–‡”‹‡• –Š‡ ‘”†‡” ‘ˆ –Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘Ǥ šƒ’Ž‡ǡ ‹ˆ ‘”†‡” ‘ˆ –Š‡
†‡”‹˜ƒ–‹˜‡ ‹• Ǯ͵ǯǡ –Š‡ǡ ‹– ‹• ƒ ͵ ‘”†‡” †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘Ǥ
Ӡ
ˆ –Š‡ ‘”†‡” ‘ˆ –Š‡ †‡”‹˜ƒ–‹˜‡ ‹• Ǯǯǡ –Š‡ǡ ‹– ‹• ƒ  ‘”†‡” †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘Ǥ Š‡ †‡‰”‡‡ ‘ˆ –Š‡
–Š
†‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ‹• –Š‡ ’‘™‡” ‘ˆ –Š‡ †‡”‹˜ƒ–‹˜‡Ǥ







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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
n
d y n
Example: has order n and degree n.
dx n


šƒ’Ž‡ ͳǣ ‡– —• …‘•‹†‡” ƒ •‹–—ƒ–‹‘ ™Š‡”‡ ƒ …‘’‘‡– ‹• …‘‘Ž‹‰ǡ ƒ† –Š‡ ”ƒ–‡ ‘ˆ …‘‘Ž‹‰ ‹•
’”‘’‘”–‹‘ƒŽ –‘ –Š‡ ‹•–ƒ–ƒ‡‘—• –‡’‡”ƒ–—”‡ ‘ˆ –Š‡ …‘’‘‡– ‹–•‡ŽˆǤ ••—‹‰ ƒ
’”‘’‘”–‹‘ƒŽ‹–› …‘•–ƒ– ȋ…Ȍǡ ‡•–ƒ„Ž‹•Š ƒ –‹‡ ȋ–Ȍ – –‡’‡”ƒ–—”‡ ȋ Ȍ ”‡Žƒ–‹‘•Š‹’ ƒ– ȋ–ͳ α ͳȌ ƒ†
ȋ–ʹ α ʹȌǤ

ƒ–Š‡ƒ–‹…ƒŽŽ›ǡ ™‡ Šƒ˜‡ǣ

Rate of change of temprature ∝ Instantaneous temperature


dT dT
= ∝ T = = −cT - - - - (1.17)
dt dt
Š‡ ‡‰ƒ–‹˜‡ •‹‰ ‹ ‡“—ƒ–‹‘ ȋͳǤͳ͹Ȍ ‹†‹…ƒ–‡• –Šƒ– –Š‡ –‡’‡”ƒ–—”‡ ‹• †‡…”‡ƒ•‹‰ ™‹–Š –‹‡Ǥ

‡™”‹–‹‰ ‡“—ƒ–‹‘ ȋͳǤͳ͹Ȍ ƒ† ‹–‡‰”ƒ–‹‰ǡ ™‡ ‰‡–ǣ


T 2 t 2
dT T 2 T 2
= −c dt = log = −c t − t = = e −c t 2 −t 1 or T = T e −c t 2 −t 1 - - - - (1.18)
1
2
2
1
T T 1 T 1
T 1 t 1

– …ƒ „‡ ‘„•‡”˜‡† –Šƒ–ǡ –‡’‡”ƒ–—”‡ ȋ ʹȌ ‹• ‰‘– „› —Ž–‹’Ž›‹‰ ȋ ͳȌ „› ƒ ˆƒ…–‘” e c t 2 −t 1


šƒ’Ž‡ ʹǣ †‡”‹˜‡ ƒ† ’Ž‘– –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯ›ǯ ƒ† Ǯšǯǡ ’”‘˜‹†‡†ǡ ™Š‡ ȋš α ͳȌ ƒ– ȋ›α ͳȌǤ
Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ‹• ‰‹˜‡ „›ǣ


dy y
= k - - - - (1.19)
dx x

dy dx
Equation . can be rewritten as: = k = log y = k log x + c
e
e
y x
y
Simplifying the above expression, we get: log y − k log x = c = log = c
e
e
e
x k
Upon further simplification, we get: y = x e - - - - (1.20)
k c
’’Ž›‹‰ –Š‡ ‹‹–‹ƒŽ …‘†‹–‹‘• ‘ˆ ȋš α ͳȌ ƒ† ȋ› α ͳȌ –‘ ‡“—ƒ–‹‘ ȋͳǤʹͲȌǡ ™‡ ‰‡–ǣ

0
c
1 = e ,we know that e = 1 , therefore c = 0 .


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 19

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
—„•–‹–—–‹‰ –Š‡ ˜ƒŽ—‡ ‘ˆ Ǯ…ǯ ‹ ‡“—ƒ–‹‘ ȋͳǤʹͲȌǡ ™‡ ‰‡– –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯ›ǯ ƒ† Ǯšǯ –‘ „‡ǣ

k
y = x − this is parabolic equation of degree k .
‘ ’Ž‘– –Š‡ …—”˜‡ǡ Ž‡– —• ƒ••—‡ ȋ α ͵Ȍ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳ͸Ȑ














3
[Fig 1.16: parabola of form [y = x ]]


‘–‡ǣ –Š‡ •‘Ž—–‹‘ –‘ ƒ› †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ‹• ƒŽŽ ƒ„‘—– ‡•–ƒ„Ž‹•Š‹‰ ƒ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ –Š‡
†‡’‡†‡– ƒ† ‹†‡’‡†‡– ˜ƒ”‹ƒ„Ž‡•Ǥ



ͳǤͶǤ͹ ʹ ‘”†‡” ‹ˆˆ‡”‡–‹ƒŽ “—ƒ–‹‘•
†

 •‡˜‡”ƒŽ ’”ƒ…–‹…ƒŽ ‡‰‹‡‡”‹‰ •‹–—ƒ–‹‘•ǡ •—…Š ƒ•ǡ ˆ”‡‡ ˜‹„”ƒ–‹‘ǡ „—…Ž‹‰ǡ ‡–…Ǥǡ –Š‡ ʹ ‘”†‡”
†
†‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ‹• …‘ˆ”‘–‡†Ǥ

šƒ’Ž‡ ͳǣ ϐ‹† –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯ›ǯ ƒ† Ǯ–ǯ ˆ”‘ –Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ȋͳǤʹͳȌǡ ‰‹˜‡
–Š‡ ‹‹–‹ƒŽ …‘†‹–‹‘ ȋ› α ͲȌ ƒ– ȋ– α ͲȌǤ
2
d y
+ ky = 0 - - - - (1.21)
dt 2
‹…‡ –Š‡ ‡“—ƒ–‹‘ ‹• ƒ ʹ ‘”†‡” Š‘‘‰‡‘—• †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ǡ ‹– ‹• ‡ƒ•› –‘ ’”‘’‘•‡ ƒ •‘Ž—–‹‘Ǥ
†
– …ƒ „‡ ‘„•‡”˜‡† –Šƒ–ǡ —’‘ †‹ˆˆ‡”‡–‹ƒ–‹‰ ƒ •‹—•‘‹†ƒŽ ‘” ƒ …‘•‹—•‘‹†ƒŽ ˆ—…–‹‘ –™‹…‡ǡ ‹ –Š‡
‰‹˜‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ǡ ™‡ ‰‡– „ƒ… –‘ –Š‡ •ƒ‡ ˆ—…–‹‘ǡ „—– ™‹–Š ƒ ‡‰ƒ–‹˜‡ •‹‰Ǥ

‡– —• ƒ••—‡ –Š‡ •‘Ž—–‹‘ –‘ –Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ –‘ „‡ ‘ˆ –Š‡ ˆ‘”ǣ y = a sinct - - - - (1.22)
Š‡ …Š‘‹…‡ ‘ˆ •‘Ž—–‹‘ ƒŽ•‘ ‹• ‹ Ž‹‡ ™‹–Š –Š‡ ‹‹–‹ƒŽ …‘†‹–‹‘ •—‰‰‡•–‡†Ǥ

dy
Differentiating equation . , we get: = ca cos ct
dx

2
d y
2
Differentiating again, we get: = − c a sinct
dx 2
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

2
d y
Substituting the values of ‘y’ and in equation . , we get:
dx 2
2
2
− c a sinct + k a sin ct = 0 = −c + k asinct = 0
‹…‡ ȋ•‹ …–Ȍ ‹• œ‡”‘ ‘Ž› ˆ‘” •’‡…‹ϐ‹… ‹•–ƒ…‡• ‘ˆ ȋ–Ȍǡ ȋ-… Ϊ Ȍ ‰‘‡• –‘ œ‡”‘Ǥ
ʹ
2
Therefore,we have: k − c = 0 = c = K
—„•–‹–—–‹‰ –Š‡ ˜ƒŽ—‡ ‘ˆ Ǯ…ǯ ‹ ‡“—ƒ–‹‘ ȋͳǤʹʹȌǡ ™‡ ‰‡– –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯ›ǯ ƒ† Ǯ–ǯ –‘ „‡ǣ

y = a sin kt

Š‡ …—”˜‡ ‹• •‹—•‘‹†ƒŽ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳ͹Ȑ








[Fig 1.17: sine curve of the form y = a sin kt]


šƒ’Ž‡ ʹǣ ‘•‹†‡” ‡“—ƒ–‹‘ ȋͳǤʹ͵Ȍǡ ™Š‹…Š ‹• –Š‡ –›’‹…ƒŽ ‡š’”‡••‹‘ ˆ‘” †ƒ’‡† ˜‹„”ƒ–‹‘Ǥ
•–ƒ„Ž‹•Š –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯ›ǯ ƒ† Ǯ–ǯ ˆ‘” –Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘ ȋͳǤʹ͵ȌǤ


2
d y dy
c 1 + c 2 + c y = 0 - - - - (1.23)
3
dt 2 dt
bt
The solution for equation . is of the form: y t = pe + qe
at
Š‡”‡ǡ Ǯƒǯ ƒ† Ǯ„ǯ ƒ”‡ –Š‡ ”‘‘–• ‘ˆ –Š‡ “—ƒ†”ƒ–‹… ‡“—ƒ–‹‘ ȋͳǤʹͶȌǣ

2
c m + c m + c = 0 - - - - (1.24)
2
3
1
Š‡ •‘Ž—–‹‘ ’”‘’‘•‡† ‹• ‹ –Š‡ ‡š’‘‡–‹ƒŽ ˆ‘” ƒ† †‹ˆˆ‡”‡–‹ƒ–‹‰ –Š‡ ‡š’‘‡–‹ƒŽ ˆ‘” ‰‹˜‡•
–Š‡ ‡š’‘‡– –‡” —Ž–‹’Ž‹‡† „› ƒ …‘•–ƒ–ǡ ‡ƒ‹‰ǡ •—„•–‹–—–‹‰ –Š‡ •‘Ž—–‹‘ ‹ –Š‡ †‹ˆˆ‡”‡–‹ƒŽ
‡“—ƒ–‹‘ ȋͳǤʹ͵Ȍǡ ™‡ —•– ‰‡– œ‡”‘ǡ •ƒ–‹•ˆ›‹‰ –Š‡ †‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘Ǥ

‡– —• ‘™ ϐ‹† –Š‡ ”‘‘–• ‘ˆ –Š‡ “—ƒ†”ƒ–‹… ‡“—ƒ–‹‘ ȋͳǤʹͶȌǤ

2
2
2
c 2 c − 4c c c 2 c − 4c c c 2 c − 4c c
3 1
3 1
3 1
− ± 2 , where, a = − + 2 and b = − − 2
2c 1 2c 1 2c 1 2c 1 2c 1 2c 1

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Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics


2
2

− c 2 c −4c 3 c 1 t − c 2 c −4c 3 c 1
2
2
t
2c 1 + 2c 1 2c 1 − 2c 1
The solution is given by: y t = pe + qe



‘ ’Ž‘– –Š‡ ‰”ƒ’Šǡ ™‡ ƒ••—‡ –Š‡ ˆ‘ŽŽ‘™‹‰ǣ

’ ʹ

“ ͵

…ͳ ʹ

…ʹ ͵

…͵ ͳ


Š‡ ‰”ƒ’Š ‹• ‡š’‘‡–‹ƒŽ ‹ ƒ–—”‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͳͺȐ
















at
bt
[Fig 1.18: exponential curve of the form [y t = pe + qe ]]














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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
ͳǤͷ ‡…–‘” Ž‰‡„”ƒ


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

ͳǤͷǤͳ ‡ϐ‹‹–‹‘ ‘ˆ ˜‡…–‘”• ƒ† •…ƒŽƒ”•

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

‡…–‘” “—ƒ–‹–‹‡• ƒ”‡ –Š‘•‡ –Šƒ– ƒ”‡ …Šƒ”ƒ…–‡”‹œ‡† „› „‘–Š †‹”‡…–‹‘ ƒ† ƒ‰‹–—†‡Ǥ ‡…–‘”•
ˆ‘ŽŽ‘™ǡ Žƒ™ ‘ˆ ˜‡…–‘” ƒ††‹–‹‘ǡ ˜‡…–‘” —Ž–‹’Ž‹…ƒ–‹‘ ƒ† —Ž–‹’Ž‹…ƒ–‹‘ „› ƒ •…ƒŽƒ”Ǥ ‡– —•
—†‡”•–ƒ† ‡ƒ…Š ‘‡ ‘ˆ –Š‡ ‹ †‡–ƒ‹ŽǤ
ͳǤͷǤʹ –”‘†—…–‹‘ –‘ ‡…–‘”• ƒ† ƒ’’Ž‹…ƒ–‹‘•


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

ƒŽ‘‰ ƒ •–”ƒ‹‰Š– Ž‹‡ –‘™ƒ”†• ƒ•– ˆ”‘ ‡•–ǡ ‹• ‘„˜‹‘—• ƒ† ‹• ‹†‹…ƒ–‡† „›ǣ V = vn ,
™Š‡”‡ –Š‡ —‹– ˜‡…–‘” (n ) ‹†‹…ƒ–‡• –Š‡ †‹”‡…–‹‘ ˆ”‘ ‡•– –‘ ƒ•– ƒ† Ǯ˜ǯ ‹• –Š‡ ƒ‰‹–—†‡ ‘ˆ

–Š‡ ˜‡Ž‘…‹–› ˜‡…–‘” –Šƒ– ‹•ǡ •’‡‡†Ǥ ‡…‡ ‡šƒ’Ž‡• ˆ‘” ’‘Žƒ” ˜‡…–‘”• ƒ”‡ Ž‹‡ƒ” ˜‡Ž‘…‹–›ǡ Ž‹‡ƒ”
V,
ƒ……‡Ž‡”ƒ–‹‘ǡ Ž‹‡ƒ” ‘‡–—ǡ ˆ‘”…‡ ƒ† ‘–Š‡” Ž‹‡ƒ” ˜‡…–‘”•Ǥ


š‹ƒŽ ˜‡…–‘”• ƒ”‡ –Š‘•‡ ™Š‘•‡ †‹”‡…–‹‘ …‘‹…‹†‡• ™‹–Š –Š‡ ƒš‹• ‘ˆ ”‘–ƒ–‹‘Ǥ ‘•‹†‡” ƒ ™Š‡‡Ž
”‘ŽŽ‹‰ ‘ ƒ ”‘—‰Š Š‘”‹œ‘–ƒŽ •—”ˆƒ…‡ ™‹–Š ƒ‰—Žƒ” ˜‡Ž‘…‹–›Ǥ












[Fig a: Polar and Axial vectors]



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






QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 23

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ˆ –Š‡ •…”‡™ ‹• ‰‹˜‡ ƒ …Ž‘…™‹•‡ ”‘–ƒ–‹‘ǡ –Š‡ǡ –Š‡ •…”‡™ ƒ†˜ƒ…‡• ‹ –Š‡ ‡‰ƒ–‹˜‡ †‹”‡…–‹‘Ǥ
‡…‡ ‰‘‹‰ „› –Š‡ ”—Ž‡ǡ –Š‡ Ǯwǯ ˆ‘” –Š‡ •…”‡™ •Š‘™ ‹ –Š‡ ȏ ‹‰ „Ȑ ‹• ˜‡…–‘”‹ƒŽŽ› ”‡’”‡•‡–‡† ƒ•ǣ


w = w −k , indicating that the direction is along negative Z axis (clock wise)
w = w k , indicating that the direction is along positive Z axis (anti - clock wise)


Š‡ ˜‡…–‘”‹ƒŽ ”‡’”‡•‡–ƒ–‹‘ ‘ˆ –Š‡ ’Š›•‹…ƒŽ •‹–—ƒ–‹‘ ‹• ƒŽ™ƒ›• ‡š–”‡‡Ž› —•‡ˆ—Ž ‡•’‡…‹ƒŽŽ› ‹ –™‘
†‹‡•‹‘ƒŽ ‘” –Š”‡‡-†‹‡•‹‘ƒŽ •‹–—ƒ–‹‘•Ǥ Š‡ ‡šƒ’Ž‡• ˆ‘” ƒš‹ƒŽ ˜‡…–‘”• ƒ”‡ ƒ‰—Žƒ” ˜‡Ž‘…‹–›ǡ
ƒ‰—Žƒ”-‘‡–—ǡ ƒ‰—Žƒ” ƒ……‡Ž‡”ƒ–‹‘ǡ –‘”“—‡ ƒ† ‘–Š‡” ƒ‰—Žƒ” ˜‡…–‘”•Ǥ












[Fig b: Right Handed Screw]


ͳǤͷǤ͵ …ƒ”–‡•‹ƒ ˜‡…–‘”


‡– —• …‘•‹†‡” ƒ ˜‡…–‘” …‘‡…–‹‰ –Š‡ ‘”‹‰‹ ȋͲǡͲǡͲȌ ƒ† ƒ ƒ”„‹–”ƒ”› ’‘‹– ‹ •’ƒ…‡ ȋšǡ ›ǡ œȌǤ

˜‡…–‘” ‹ ƒ ͵ …ƒ”–‡•‹ƒ •’ƒ…‡ ‹• ”‡’”‡•‡–‡† ƒ• ˆ‘ŽŽ‘™•ǣ


Vector V = [(magnitude along ‘x’ ∗ unit vector along ‘x’) + (magnitude along ‘y’
∗ unit vector along ‘y’) + (magnitude along ‘z’ ∗ unit vector along ‘z’)



That is: [V = x − x i + y − y j + (z − z )k]
1
2
1
2
1
2
Š‡”‡ǡšͳǡ ›ͳ ƒ† œͳ ƒ”‡ …‘‘”†‹ƒ–‡• ‘ˆ –Š‡ ‘”‹‰‹Ǥ Š‡”‡ˆ‘”‡ǡ
™‡ Šƒ˜‡ǣ


[V = x − 0 i + y − 0 j + (z − 0)k]
2
2
2
Let us analyze a vector V in a 3D space as shown in [ . ].
,




[Fig 1.19: Vector in a 3D space]

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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]


Assume the magnitude of vector V to be′v′.
Let the inclination of the vector V with respect to the vertical axis ‘z’, be ‘θ’. Therefore,

its inclination to the ‘x − y’ plane is 90 – θ . The vector is contained in the plane ‘ABCD’ as shown.


Projection of vector V on the ‘z − axis’ is ‘AD’ . , and, projection of vector V on the

‘x − y’ plane is ‘DC’ [ . ].
















[Fig 1.20: Vector projection onto 'z-axis'] [Fig 1.21: vector projection onto 'x-y'plane]


ͳǤͷǤͶ †‡”•–ƒ†‹‰ –Š‡ ‘…‡’– ‘ˆ ”‘Œ‡…–‹‘


Consider a vector ( x ) being projected onto the horizontal axis, as shown in [ . ].







[Fig 1.22: Vector resolution]

By trigonometry, the magnitude of ((x )) is given by:
1

x
cosϕ = 1 , therefore, we have x = x cosϕ
x 1









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Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics


To summarize: The vector ( V ) in a 3D space as shown in [Fig 1.19] can be resolved along
the ‘x’, ‘y’ and ‘z’ axes.


The magnitude of vector ( V ) projected on to the ‘z’ axis is: |V |cosθ as shown in [Fig 1.23]













[Fig 1.23: Vector Projection onto ‘z’ axis]

To get the magnitude of vector ( V ) projected on to the ‘x’ and ‘y’ axes, we first need to project
the vector (V ) onto the ‘x-y’ plane, as shown in [Fig 1.24]. We then project the vector on the ‘x-
y’ plane onto the ‘x’ and ‘y’ axes, as shown in [Fig 1.25] and [Fig 1.26] respectively.



Magnitude of vector (V ) projected onto the ‘x-y’ plane is: (|V|sinθ) as shown in [Fig 1.24]















[Fig 1.24: Vector Projection onto ‘x-y’ plane]









QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Vector (|V |sinθ) projected on to the ‘y’ axis is: (|V |sinθ * cosϕ) as shown in [Fig 1.25]
















[Fig 1.25: Vector Projection onto ‘y’ axis]

Vector (|V |sinθ) projected on to the ‘x’ axis is: (|V |sinθ * sinϕ) as shown in [Fig 1.26]
















[Fig 1.26: Vector Projection onto ‘x’ axis]


















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 27

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ͳǤͷǤͷ ‡…–‘” ††‹–‹‘ – ”ƒ’Š‹…ƒŽ ’’”‘ƒ…Š


when we add two vectors, say, a + b we add their respective cartesian components.
let us illustrate the same:














[Fig 1.27: Vector Addition]

It is easy to see that the ‘x’ and ‘y’ components of the two vectors, algebraically add up to give the
‘x’ and ‘y’ components of the resultant vector.

Incidentally, the longer diagonal of the parallelogram represents the resultant vector, which is the

addition of vectors a and b .

ͳǤͷǤ͸ ‡…–‘” ††‹–‹‘ – ƒŽ›–‹…ƒŽ ’’”‘ƒ…Š


From [Fig 1.27], the ‘x’ and ‘y’ components of vectors ( a ) and ( b ) can be vectorially

represented as follows:


a + b i --------- ‘x’ component of vectors ( a ) and ( b )
x
x

a + b j --------- ‘y’ component of vectors ( a ) and ( b )
y
y

The magnitude of vectors ( a ) and ( b ) is given as:
2

a + b = x − comp + y − comp = a + b + a + b
2
2
2
y
x
x
y

= a + b + a + b + 2 a b + a b - - - - (1.25)
2
2
2
2
y
x
x
x x
y y
y
Equation (1.25) is the magnitude of the longer diagonal of the parallelogram, as shown in
[Fig 1.27]

Considering the graphical representation of vector addition, as shown in [Fig 1.27], let us derive
an expression for the longer diagonal of the parallelogram.

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]














[Fig 1.28: Parallelogram law of Vectors]


From [Fig 1.28], the longer diagonal of the parallelogram is given by:


2
2
2
2





a + b cosθ + b sinθ = a + b cos θ + 2 a b cosθ + b sin θ

2
2
2
Applying the fact that sin θ + cos θ = 1) in the above equation, we get:
2
2
2


a + b + 2 a b cosθ - - - - (1.26)
2
From the above discussion, we have the magnitudes of vectors [ a ] and [ b ] to be:

a = a + a and b = b + b

2
2
2
2
x
y
x
y
Substituting the magnitudes of the vectors in equation (1.26), we get:
a + a + b + b + 2 a + a b + b cosθ - - - - (1.27)
2
2
2
2
2
2
2
2
x
y
x
y
y
y
x
x
Equating, expression (1.27) with (1.25), we have:
2
2
2
2
a + b + a + b + 2 a b + a b = a + a + b + b + 2 a + a b + b cosθ
2
2
2
2
2
2
2
2
y
y
x
y
y
x
x
x x
x
y y
y
y
x
x
Cancelling the common terms in the above expression, we have:
a b + a b
cosθ = x x y y
a + a b + b 2
2
2
2
x y x y
The denominator a + a b + b , is the product of magnitude of two vectors.
2
2
2
2
y
y
x
x
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 29

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics

Let us interpret the numerator (ax bx + ay by) , by relating vectors to physical quantities as follows:



Consider the displacement and force vectors to be ( d ) and ( F ) respectively, as shown in [Fig 1.29].









[Fig 1.29: Scalar product]



Š‡‹” …‘’‘‡–• ƒ”‡ ‰‹˜‡ ƒ•ǣ



d = d i + d j and F = F i + F j
x
y
x
y
The work done in the ‘x’ direction is: W = F ∙ d
x
x
x
The work done in the ‘y’ direction is: W = F . d
y
y
y
The total work done is given by:
W + W = F . d + F . d
x
y
y
y
x
x


If we were to assume, vectors ( F ) and ( d ) are equivalent to vectors ( a ) and ( b ) (as discussed in

addition and subtraction of vectors), we get:
W + W = a . b + a . b
y
y
x
x
y
x

Further by projecting ( F ) onto ( d ) as shown in [Fig 1.30], the work done is given by:


Work done = F d cosθ Force in the direction of displacement ∗ displacement









QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]














[Fig 1.30: scalar product - vector resolution]

By similar argument, work done for vectors ( a ) onto ( b ) is also given by:

Work done = a b cosθ

Expanding the above expression, we get:


Work done = a . b + a . b = a b cosθ
y
x
x
y
The above expression is a Scalar quantity.
Therefore, when two vectors are multiplied in a certain way, we get a scalar quantity and is called
‘Scalar product’ of vectors. This is formally represented as:



a . b = a b cosθ ----- Scalar product of Vectors
To summarize: The Scalar/Dot product is a product of vectors, that gives us a scalar quantity.

This product is used while computing scalar quantities like energy, power, work done, angular
displacement, etc.



Example: let us consider two vectors ( a ) and ( b ), in the ‘x-y’ plane, as shown in [Fig 1.31] .

Find the magnitude and direction of the vector resulting from addition of vector ( b ) from ( a ).











[Fig 1.31: Vectors on an ‘x-y’ plane]



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 31

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
‹…‡ –Š‡ ˜‡…–‘”• ƒ”‡ ˆ—ŽŽ› †‡ϐ‹‡†ǡ –Š‡ ”‡•—Ž–ƒ– ˜‡…–‘”ǡ ‹• –Š‡ ƒ††‹–‹‘ ‘ˆ –Š‡ ”‡•’‡…–‹˜‡ …‘’‘‡–• ‘ˆ
–Š‡ ˜‡…–‘”•ǡ ƒ† ‹• ‰‹˜‡ „›ǣ
a + b = 4i + 5j + 7i + 2j = 11i + 7j

The magnitude of the resultant vector is: 11 + 7 = 13.03
2
2
The direction of the resultant vector is given by:
7
0
−1
tan −1 = tan 0.6363 = 32.47 , with respect to the positive (x − axis)
11
The magnitude and direction of the resultant is graphically represented as shown in [Fig 1.32]:
















[Fig 1.32: Magnitude and direction of the resultant vector]



ͳǤͷǤ͹ ‡…–‘” —„–”ƒ…–‹‘ – ”ƒ’Š‹…ƒŽ ’’”‘ƒ…Š



when we subtract two vectors, say, a − b , we subtract their respective cartesian components.
Let us illustrate the same, [Fig 1.33]:


















[Fig 1.33: Vector subtraction


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

It is easy to see that the ‘x’ and ‘y’ components of the two vectors, algebraically subtract each other to
give the ‘x’ and ‘y’ components of the resultant vector.

Incidentally, the shorter diagonal of the parallelogram represents the resultant vector.



ͳǤͷǤͺ ‡…–‘” —„–”ƒ…–‹‘ – ƒŽ›–‹…ƒŽ ’’”‘ƒ…Š

”‘ ȏ ‹‰ ͳǤ͵͵Ȑǡ ‹– ‹• ‘„˜‹‘—• –Šƒ– –Š‡ –Š‹”† •‹†‡ ‘ˆ –Š‡ –”‹ƒ‰Ž‡ ”‡’”‡•‡–• –Š‡ †‹ˆˆ‡”‡…‡ ‘ˆ –Š‡ ˜‡…–‘”•


( a ) and ( b ), represented as follows:

a − b i --------- ‘x’ component of vectors ( a ) and ( b )
x
x

a − b j --------- ‘y’ component of vectors ( a ) and ( b )
y
y

The magnitude of vectors ( a ) and ( b ) is given as:
2
a − b = x − comp + y − comp = a − b + a − b

2
2
2
y
y
x
x
= a + b + a + b − 2(a b + a b )
2
2
2
2
y y
x x
y
y
x
x
- - - - (1.28)
From the discussion on scalar products, we have:
a b + a b
cosθ = x x y y , or a b + a b = cosθ a + a b + b
2
2
2
2
y
y y
x
x
x x
y
a + a b + b 2
2
2
2
x y x y
Equation (1.28) is now rewritten as:
2


= a + b + a + b − 2 cosθ a + a b + b = a + b − 2 a b cosθ

2
2
2
2
2
2
2
2
2
y
y
y
x
x
x
x
y
The above expression is got by the application of cosine rule to the triangle.
For any planar vector in the ‘x-y’ plane, its orientation with respect to the positive ‘x-axis’ is given
by:
y − comp a − b
y
y
θ = tan −1 = tan −1
x − comp a − b
x
x

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 33

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics


Example: let us consider two vectors ( a ) and ( b ), in the ‘x-y’ plane. Find the magnitude and

direction of the vector resulting from subtraction of vector ( b ) from ( a )













[Fig 1.31: Vectors on an ‘x-y’ plane]

Since the vectors are fully defined, the resultant vector, is the difference of the vectors, and is given
by:


a − b = 4i + 5j − 7i + 2j = −3i + 3j
The magnitude of the resultant vector is: 3 + 3 = 3 2
2
2
The direction of the resultant vector is given by:

3
−1
0
tan −1 = tan −1 = 135 , with respect to the positive (x − axis)
−3
Let us now represent the magnitude and direction of the resultant graphically, [Fig 1.34]:



















[Fig 1.34: Magnitude and direction of the resultant vector]



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Š‹• ‡ƒ• –Šƒ–ǡ –Š‡ ˜‡…–‘”• …ƒ ‡š‹•– ƒ– ƒ› ’‘‹– ‘ –Š‡ …‘-‘”†‹ƒ–‡ ’Žƒ‡Ǥ Š‡ ˜‡…–‘”• ‡‡† ‘–
‘”‹‰‹ƒ–‡ ˆ”‘ –Š‡ ‘”‹‰‹ „—– …‘—Ž† Ž‹‡ ‘ ƒ› ’ƒ”– ‘ˆ –Š‡ š-› ’Žƒ‡Ǥ




ͳǤͷǤͻ —Ž‡• ‘ˆ ‡…–‘” ††‹–‹‘ ƒ† —„–”ƒ…–‹‘

Ø ††‹–‹‘ ƒŽ™ƒ›• ’”‘†—…‡• ƒ ˜‡…–‘” ‘ˆ ‰”‡ƒ–‡” ƒ‰‹–—†‡ ‹ˆ „‘–Š ˜‡…–‘”• ƒ”‡
†‹˜‡”‰‹‰ ȏ ‹‰ ͳǤ͵ͷȐ ‘” …‘˜‡”‰‹‰ ȏ ‹‰ ͳǤ͵͸Ȑ ’”‘˜‹†‡† –Š‡ ƒ‰Ž‡ „‡–™‡‡ –Š‡ –™‘
˜‡…–‘”• ‹• ƒ…—–‡Ǥ






[Fig 1.35: Diverging vectors] [Fig 1.36: Converging vectors]



Ø ˆ ‘‡ ‘ˆ –Š‡ ˜‡…–‘” ‹• …‘˜‡”‰‹‰ ƒ† –Š‡ ‘–Š‡” ‹• †‹˜‡”‰‹‰ǡ –Š‡ǡ ƒ††‹‰ –Š‡ –™‘
˜‡…–‘”•ǡ ”‡•—Ž–• ‹ ‡“—‹˜ƒŽ‡– •—„–”ƒ…–‹‘ ȏ ‹‰
ͳǤ͵͹ȐǤ








[Fig 1.37: Vector subtraction]






Ø ˆ –Š‡ ˜‡…–‘”• ”‡’”‡•‡– –Š‡ –™‘ •‹†‡• ‘ˆ ƒ –”‹ƒ‰Ž‡ǡ –Š‡ –Š‹”† •‹†‡ ”‡’”‡•‡–• –Š‡
†‹ˆˆ‡”‡…‡ ‘ˆ –Š‡ –™‘ ˜‡…–‘”• ȏ ‹‰ ͳǤ͵ͺȐǡ ’”‘˜‹†‡†ǡ ˜‡…–‘”• ‡‹–Š‡” …‘˜‡”‰‡ ‘”
†‹˜‡”‰‡Ǥ





[Fig 1.38: Diverging Vectors]





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 35

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ͳǤ͸ ‡…–‘” ’”‘†—…–


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

‡…‘ŽŽ‡…– ‘—” †‹•…—••‹‘ ‘ –Š‡ ”‹‰Š–-Šƒ†‡† •…”‡™ ’”‹…‹’Ž‡ ȏ ‹‰ „ȐǤ Š‡ –Š‡ Š‡ƒ† ‘ˆ –Š‡ •…”‡™
‹• ‰‹˜‡ ƒ …Ž‘…™‹•‡ ”‘–ƒ–‹‘ǡ ‹– ‘˜‡• ‹–‘ –Š‡ •—”ˆƒ…‡ǡ ƒ† ™Š‡ ‰‹˜‡ ƒ ƒ–‹-…Ž‘…™‹•‡ ”‘–ƒ–‹‘ǡ
‹– ‘˜‡• ‘—– ‘ˆ –Š‡ •—”ˆƒ…‡Ǥ

Ž‡– —• …‘•‹†‡” ƒ ’”ƒ…–‹…ƒŽ •‹–—ƒ–‹‘ ‘ˆ ƒ •’ƒ‡” ƒ† —– ƒ””ƒ‰‡‡–ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͵ͻȐǤ










[Fig 1.39: A right handed system]
Let the length of the spanner be ‘L’. When we apply a force ‘F’, perpendicular to the length of the
spanner, the nut rotates. If the direction of force is along the negative ‘y-axis’ (acting downwards
[−j ]) and the spanner length is along the ‘x-axis’ [i ] , then the torque experienced by the nut is

along the negative ‘z-axis’ [−k] (clockwise).
Let us prove the same using vector multiplication.

The direction of torque experienced by the nut can be computed using vector product. Let us use


the vector product ( L x F ), where ( L ) is the position vector/length and ( F ) is the force vector.

Therefore, the direction of torque is vectorially got as follows:

T = L i x F −j = T = LF −k - - - - (1.29)

The resultant direction −k , and this is practically correct.
If we were to reverse the direction of force, then the resultant direction would have been k .

It is now evident that, when we multiply two unit vectors, we get a third unit vector perpendicular
to the plane of vectors being multiplied. This can be applied to any two vectors, as, every vector is
represented using its magnitude and direction (unit vector).

Considering many such cases, we conclude the following:










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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]














[Fig 1.40: Vector product rule]


Thus, two vectors can be multiplied to yield another vector. The fundamental advantage of vector
product is that, the direction of the resultant quantity is readily got, which helps in complex
situations.

A vector product is given by the following expression:



a x b = a b sinθ - - - - (1.30)
This can be visualized as follows:

Let us consider two vectors a and b represented on a cartesian plane as shown in [Fig 1.41].




















[Fig 1.41: Cartesian components of vectors a and b ]


Their product of vectors a and b in terms of their components could be computed as follows:

Components of a = a i + a j
1
2

Components of b = b i + b j
2
1

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 37

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics


The product of vectors a and b is given as:
a × b = a i + a j × b i + b j

2
1
1
2
= a i × b i + a i × b j + a j × b i + a j × b j - - - - (1.31)
1
1
2
2
2
2
1
1
It can be observed from the example of spanner and nut, that, no torque is produced if force and
the position vector (length of spanner) are along the same axis. Hence, vector products of identical
unit vectors are zero. Therefore, the first and last terms of equation (1.31) go to zero. Also, we


know that, j × i = −k and i × j = k . Using these facts, we get:



= a i × b j + a j × b i = a b k − a b k = a b − a b k - - - - (1.32)
1 2
2
2
1
2 1
2 1
1 2
1
It can be proved that a b − a b is double the area of the triangle ‘ACD’ as shown in [Fig 1.42]
1 2
2 1








[Fig 1.42: Area of triangle formed by vectors a and b ]

The area of the triangle ‘ACD’ is equal to the sum of area of the rectangle ‘ABCF’ area of trapezium
‘FCDE’, subtracted by the area of triangles ‘ABC’ and ‘ADE’.

(b + a 2 1
2
Area of Triangle ACD = a a + b − a − b b + a a
1 2
1 2
1
1
2 1
2 2
1
= 2 a a + b b + a b − a b − a a − b b − a a - - - - (1.33)
1 2
1 2
1 2
2 1
2 1
2 1
1 2
2
Simplifying and rearranging equation (1.33) we get:
1
Area of Triangle ACD = a b − a b - - - - (1.34)
1 2
2 1
2
It can be observed that, the area of the triangle in equation (1.34) is exactly half the magnitude of
vector product in equation (1.32).

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

However, from elementary trigonometry, we know that, the area of a triangle whose sides and
included angle are known is given by:

1

Area of the triangle ACD = a b sinθ
2
This proves equation (1.30).

Vector products could also be represented in the matrix form, as follows:


i j k

a × b = a x a y a
z
b x b y b z
Taking determinant of the above matrix, we get:

k
i j
det a x a y a
z
b x b y b z

= i a b − a b − j a b − a b + k a b − a b
y z
y x
z y
z x
x z
x y
The above expression is the vector whose magnitude is given by: a b sinθ


Which is the area of the parallelogram, whose two sides the vectors a and b themselves.

Example 1: a force applied in a 3D space
has 3 components, one in each direction
(x, y, and z). The position vector ‘r’ also has
3 components (x, y, and z).

Let the 3 components of force be F , F and F
y
x
z
and the distances be r , r and r as
z
x
y
shown in [Fig 1.43].

Each component of force produces two
moments as one of the position vectors
coincides with its own direction, which
cannot produce a moment.




[Fig 1.43: Forces on a Point in a 3D space]





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 39

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics

Therefore, from [Fig 1.44], we have moments due to force ‘F ’ to be:

x

















[Fig 1.44: Moments due to force 'Fx']




Moment due to F at a distance r = r k × F i = r F j
x
x
z x
z
z


Moment due to F at a distance r = r j × F i = r F −k
z x
x
y
x
y


Similarly, from [Fig 1.45], we have moments due to force ‘F ’ as follows:
y













[Fig 1.45: Moments due to force 'Fy']



Moment due to F at a distance r = r k × F j = r F −i
z
y
z
z x
y

Moment due to F at a distance r = r i × F j = r F k

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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

And, from [Fig 1.46], we have the moments due to force ‘F ’ to be:

y















[Fig 1.46: Moments due to force 'Fz']




Moment due to F at a distance r = r i × F k = r F −j
z
z
x
x
x z

Moment due to F at a distance r = r j × F k = r F i

y
y
y z
z
z
Collecting and adding the F , F and F terms, we get:
x
y
z


F r j − r k + F r k − r i + F r i − r j - - - - (1.35)
y
x
z
z
y
x
x
z
y
Let us now cross verify equation (1.35) using the matrix form of vector product, as follows:

i j k
det r x r y r
z
F x F y F z

= i r F − r F − j r F − r F + k r F − r F
z y
z x
y z
x y
x z
y x


= i r F − i r F − j r F + j r F + kr F − kr F
x y
y x
y z
z x
z y
x z
Collecting the F , F and F terms, we get:
z
x
y


F r j − r k + F r k − r i + F r i − r j - - - - (1.36)
y
x
x
z
y
z
x
y
z




QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 41

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ͳǤ͹ ‘™ –‘ ‹†‡–‹ˆ› ƒ ‹‰Š–- ƒ†‡† …‘‘”†‹ƒ–‡ •›•–‡


‘•‹†‡” –Š‡ ˆ‘ŽŽ‘™‹‰ ‡š’”‡••‹‘• ‘ˆ ˜‡…–‘” ’”‘†—…–ǣ

i × j = k - - - - (1.37)
j × k = i - - - - (1.38)



k × i = j - - - - (1.39)
Let us graphically understand equation (1.37), (1.38) and (1.39) and verify whether the resultant
vector matches with the coordinate system, by the definition of vector product/ right-handed screw
principle.












[Fig 1.47: vector products for anti-clockwise rotations of coordinate axes]


‹‹Žƒ”Ž›ǡ ˆ‘” …Ž‘…™‹•‡ ”‘–ƒ–‹‘•ǡ ™‡ Šƒ˜‡ǣ













[Fig 1.48: vector products for clockwise rotations of coordinate axes]

‡…–‘”• ‘ˆ †‹ˆˆ‡”‡– ‹†•ǡ ™Š‡ —Ž–‹’Ž‹‡† Ž‡ƒ† –‘ ‹’‘”–ƒ– ”‡•—Ž–• ƒ† –Š‡ ˆ‘ŽŽ‘™‹‰ ‡“—ƒ–‹‘•
ƒ”‡ ‡‡†‡† –‘ •‘Ž˜‡ ƒ› ˆ—†ƒ‡–ƒŽ •‹–—ƒ–‹‘• ‹ ‡…Šƒ‹…•Ǥ







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

Vector Product Definition


(ω × r ) Velocity (‘r’ is the position vector, ‘ω’ is the angular velocity.

Angular momentum/moment of momentum (‘r’ is the position

(mr × V) vector, ‘mV’ is linear momentum) – widely used in turbomachines,
rigid body dynamics, fluid mechanics, etc.…

r × F = τ Torque (‘r’ is the position vector)


Coriolis force (‘v’ is the linear velocity of the element subjected to

−2m(ω × V) angular velocity ‘ω’) – widely used in mechanisms, rotor dynamics,
fluid mechanics, cyclones…


Gyroscopic couple (‘I’ is the moment of inertia about ‘Ω’, ‘ω’ is the

(ω × IΩ) velocity of precession) – widely used in rigid body dynamics, vehicle
dynamics, rotor dynamics, etc.


ω × (ω × r ) Centripetal acceleration (negative of which gives the centrifugal
acceleration)


Scalar Product Definition


Work done (where (F) is the applied force and (S) is the


(F. S) displacement.
(F. V) Power (where (F) is the applied force and (V) is the velocity.






(T. ω ) Power (where (T) is the torque and (ω ) is the angular velocity.

(T. θ) Work done, where (θ) is the angular displacement.



1 (L. ω ) Rotational kinetic energy (‘L’ is the angular momentum) – Note that
the direction of angular momentum and angular velocity need not

2 necessarily co-inside.


1 Kinetic energy, where (p ) is the linear momentum (mV) and (V) is
2 (p . V) the velocity vector.



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 43

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
ͳǤͺ ‡…–‘” †‹ˆˆ‡”‡–‹ƒ–‹‘


Vector differentiation is found to be naturally needed in several cases, such as, rate of change
of velocity, momentum, angular momentum, and many other vectors.
Let us consider a particle negotiating a circle and the direction of the velocity keeps changing
meaning the tangent is continuously changing direction. Let us compute the rate of change of the
tangent vector with respect to time.
Before we dive into the crux, let us build a little background.
A vector is characterized by direction and magnitude. Mathematically, we have:

Vector V = f magnitude, direction
For a real-time situation, say, a car negotiating a curvy road, both magnitude and direction of
velocity are functions of time. Therefore, both magnitude and direction can be differentiated with
respect to time, assuming their variations to be continuous.
The direction of a vector is always a unit vector. The unit vector can be conveniently chosen with a
coordinate system. Since we are already familiar with a cartesian coordinate system, any vector
with any orientation in the cartesian space, could be represented in terms of unit vectors
i, j and k .

If this vector undergoes a rotation, then, the new direction can be expressed in terms of
i, j and k using the angle of rotation.

In a real-time situation, the angle is a function of some physical quantity, say, angular speed.
Therefore, the rate of change of direction is got by differentiating the function of the angle.


Let us qualify the above discussion with a practical example.














[Fig 1.49: Uniform circular motion]


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




QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Let us compute the rate of change of the tangent unit vector T .



From [Fig 1.49] the velocity vector V at time t = t is inclined at an angle 90 − θ with respect
1
to the negative ‘x-axis’.
















[Fig 1.50: Uniform circular motion] [Fig 1.51: Velocity vector in the cartesian space]

From [Fig 1.51] the velocity vector vT is given by: vT = vsinθ −i + vcosθ j


By where, we get: T = −sinθ i + cosθ j


The angle swept by the radius vector from (t = 0) to (t = t ), that is angle ‘θ’ is given by ‘ωt ’. This
1
1
is because, the radius vector ‘ r ’ is rotating at a constant rate of ‘ω’ radians per second. Therefore,

the tangent vector T is given by: T = sinωt −i + cosωt j

1
1
Differentiating the tangent vector T with respect to time, we get:



dT dT
= ω cosωt −i − ω sinωt j or = −ω cosωt i + sinωt j - - - - (1.40)
1
1
1
1
dt dt
Now, let us compute the radius unit vector r at the same instant of time (t = t )
1











QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 45

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics























[Fig 1.52: the radius vect at an instant (t= t1)]

The radius vector (r ) as shown in [Fig 1.52] is given by: r = rcosθ i + rsinθ j

r rcosθ i + rsinθ j
The unit radius vector is given as: r = = r = - - - - (1.41)
r r cos θ + r sin θ
2
2
2
2
Simplifying equation (1.41) we get:
r = cosθ i + sinθ j or r = cosωt i + sinωt j - - - - (1.42)
1
1
Substituting equation (1.42) in equation (1.40), we get:



dT dT
= −ω cosωt i + sinωt j = = −ωr
1
1
dt dt
Also, upon multiplying −ωr with velocity (v), we get an acceleration term called the centripetal
acceleration. This is explained as follows:
As the particle negotiates the circle (refer to [Fig 1.49]), the direction of the velocity continuously


changes. And since velocity is a vector V , the rate of change of the magnitude (v) or direction T
results in acceleration. The acceleration resulting from the directional change is called centripetal
acceleration, and the acceleration resulting from magnitudinal change is called the tangential
acceleration.


It can also be observed that, on differentiating the radius vector r , we get the tangent vector T .
dr

r = cosθ i + sinθ j = = −sinθ i + cosθ j = T - - - - (1.43)
dθ

And also, on differentiating the tangent vector T , we get the negative radius vector r .




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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]


dT

T = −sinθ i + cosθ j = = − cosθ i + sinθ j = −r - - - - (1.44)
dθ
The facts expressed in equations (1.43) and (1.44), are used in deriving accelerations for curvilinear
motion and many other rigid body dynamic situations.


ͳǤͻ ”ƒ•ˆ‘”ƒ–‹‘


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

ͳǤͻǤͳ ‡…–‘” ”ƒ•ˆ‘”ƒ–‹‘


Consider a vector V , in the ‘x-y’ plane, passing through the origin, as shown in [Fig 1.53].





















[Fig 1.53: Rotation of a vector]



The vector V has an initial orientation of angle (θ), with respect to positive ‘x-axis’.

The vector V is rotated by an angle (α) in the anti-clockwise direction. Let us derive a matrix,

relating (x, y) to x , y .
1
1




QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 47

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics

Note that, transformation and scaling are two different mathematical operations. In transformation,
the vector is rotated and translated, but preserved in its length. During scaling, both length and
orientation would change, unless, all the components of the vector are equally scaled, which
preserves the orientation.

Let us find the relationship between the original coordinates (x, y), and new coordinates x , y , in
1
1
terms of (α), as shown in [Fig 1.53]. This gives us an important rule of linear algebra, that, when a
matrix operates on a vector, we get a new vector.
From the [Fig 1.53], assuming the magnitude of the vector to be ‘V’, we have:


x = V cos θ + α = V cosθ cosα − Vsinθ sinα
1
y = V sin θ + α = V sinθ cosα + Vcosθ sinα
1
Rewriting (V cosθ) as ‘x’ and (V sinθ) as ‘y’, we get:

x = V cos θ + α = x cosα − y sinα - - - - (1.45)
1
y = V sin θ + α = y cosα + x sinα - - - - (1.46)
1
Equations (1.45) and (1.46) can be rewritten in the matrix form to get a relationship between
original and transformed coordinates of vector V .


cosα −sinα x x 1
= - - - - (1.47)
sinα cosα y y 1

transformation matrix original vector = transformed vector

Let us now express the original coordinates in terms of the transformed coordinates.

In order to get ‘x’ in terms of ‘x ’, let us multiply equation (1.45) by ‘cosα’ and equation (1.46) by
1
‘sinα’ and add the resulting equations.

x cosα = x cos α − y cosα sinα - - - - (1.48)
2
1
2
y sinα = y cosα sinα + x sin α - - - - (1.49)
1
Adding equations (1.48) and (1.49), we get:

2
2
x cosα + y sinα = x cos α − y cosα sinα + y cosα sinα + x sin α
1
1
2
x cosα + y sinα = x cos α + sin α = x cosα + y sinα = x - - - - (1.50)
2
1
1
1
1


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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

In order to get ‘y’ in terms of ‘y ’, let us multiply equation (1.45) by ‘−sinα’ and equation (1.46) by
1
‘cosα’ and add the resulting equations.
−x sinα = −x sinα cosα + y sin α - - - - (1.51)
2
1
y cosα = y cos α + x cosα sinα - - - - (1.52)
2
1
Adding equations (1.51) and (1.52), we get:

2
2
−x sinα + y cosα = −x sinα cosα + y sin α + y cos α + x cosα sinα
1
1
−x sinα + y cosα = y cos α + sin α = −x sinα + y cosα = y - - - - (1.53)
2
2
1
1
1
1
Equations (1.50) and (1.53) can be rewritten in the matrix form to get a relationship between
original and transformed coordinates of vector V .

cosα sinα x 1 x
= - - - - (1.54)
−sinα cosα y 1 y
transformation matrix transformed vector = original vector
Let us understand the nature of the transformation matrices in equations (1.47) and (1.54).

It can be observed that the transformation matrix in equation (1.54) is the transpose or inverse of
the transformation matrix in equation (1.47).

cosα −sinα
Let the transformation matrix in equation . be ‘A’: = A
sinα cosα


cosα sinα
Let the transformation matrix in equation . be ‘B’: = B
−sinα cosα
We could now conclude the following:


1) A = B and B = A
T
T
2) B B = I (identity matrix) and (A A = I(identity matrix))
T
T

ͳǤͻǤʹ ‘–ƒ–‹‘ ƒ„‘—– ƒ ‘‘”†‹ƒ–‡ ƒš‡•

‡– —• ”‡…ƒ•– –Š‡ –”ƒ•ˆ‘”ƒ–‹‘ ƒ–”‹š ‹ –‡”• ‘ˆ –Š‡ „ƒ•‹• ˜‡…–‘”•ǡ –Šƒ– ‹•ǡ –Š‡ —‹– ˜‡…–‘”• ƒŽ‘‰
–Š‡ Ǯšǯǡ Ǯ›ǯǡ ƒ† Ǯœǯ ƒš‡•ǡ –Šƒ– ‹•ǡ Ǯ‡ͳǯǡ Ǯ‡ʹǯ ƒ† Ǯ‡͵ǯ ”‡•’‡…–‹˜‡Ž›Ǥ







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 49

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
‘•‹†‡” ƒ …ƒ”–‡•‹ƒ …‘‘”†‹ƒ–‡ •›•–‡ ™Š‘•‡ Ǯš-›ǯ ’Žƒ‡ ‹• ”‘–ƒ–‡† ƒ„‘—– –Š‡ Ǯœ-ƒš‹•ǯ ƒ• •Š‘™ ‹
ȏ ‹‰ ͳǤͷͶȐ




















[Fig 1.54: Rotation about the ‘z-axis’]

‘” ƒ ’‘‹– ‘ –Š‡ Ǯš-›ǯ ’Žƒ‡ǡ Ž‡– —• ”‡Žƒ–‡ –Š‡ ‘”‹‰‹ƒŽ ƒ† –”ƒ•ˆ‘”‡† …‘‘”†‹ƒ–‡• ƒ• •Š‘™ ‹
ȏ ‹‰ ͳǤͷͷȐǤ




















[Fig 1.55: Rotation about the ‘z-axis’]


‡– —• ϐ‹† ƒ† ‡š’”‡••‹‘ ˆ‘” ȋšǯȌ ƒ† ȋ›ǯȌ ‹ –‡”• ‘ˆ ȋšȌ ƒ† ȋ›Ȍ ˆ”‘ ȏ ‹‰ ͳǤͷͷȐǤ




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

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]

OD x
′
x = OB + BC , OB = = and BC = AB sin θ
cos θ cos θ
x
′
AB = y − x tanθ , therefore, BC = y − x tan θ sinθ and x = + y − x tan θ sin θ
cosθ
2
x sin θ
′
Simplifying for (x’) we get: x = + y sinθ − x
cosθ cos θ
x x
′
′
2
2
= x = 1 − sin θ + y sinθ = x = cos θ + y sinθ
cos θ cosθ
′
Therefore, we have: x = x cosθ + y sin θ - - - - (1.55)
sinθ
′
′
′
y = AB cos θ = y = y − x tanθ cos θ = y = y cos θ − x cos θ
cos θ
Therefore, we have: y = y cos θ − x sin θ - - - - (1.56)
′
Rewriting equation (1.55) and (1.56) in the matrix from, gives us a relationship between the
original and transformed coordinates.

cosθ sinθ x x′
= - - - - (1.57)
−sinθ cosθ y′

The equation (1.57) could be recast in terms of the basis vectors (using dot product), that is, (e – for
original coordinate system) and (e’ – for the transformed coordinate system).

′
′
e ∙ e e ∙ e x x′
1 1 1 2 = - - - - (1.58)
′
′
e ∙ e 1 e ∙ e 2 y′
2
2
In a 3D space, equation (1.57) could be represented as follows:
cosθ sinθ 0 x x ′
′
− sinθ cos θ 0 y = y
0 0 1 z z ′















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 51

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
‹‹Žƒ”Ž›ǡ ”‘–ƒ–‹‘ ƒ„‘—– –Š‡ Ǯš-ƒš‹•ǯ ‹• ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͷ͸Ȑ



















[Fig 1.56: Rotation about the ‘x-axis’]

Equation (1.58) can be recast for rotation about the ‘x-axis’ as follows:

′
′
e ∙ e 2 e ∙ e 3 y y′
2
2
=
′
′
e ∙ e 2 e ∙ e 3 z′
3
3
Representing rotation about ‘x-axis’ in the 3D space, we get:
1 0 0 x x ′
′
0 cos θ sin θ y = y
0 − sinθ cos θ z z ′




















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 52

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
‘–ƒ–‹‘ ƒ„‘—– –Š‡ Ǯ›-ƒš‹•ǯ ‹• ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤͷ͹Ȑ






















[Fig 1.57: Rotation about the ‘y-axis’]

Equation (1.58) can be recast for rotation about the ‘y-axis’ as follows:


′
′
e ∙ e e ∙ e y y′
1 1 1 3 =
′
′
e ∙ e 1 e ∙ e 3 z′
3
3
Representing rotation about ‘y-axis’ in the 3D space, we get:
cos θ 0 −sinθ x x ′
′
0 1 0 y = y
sin θ 0 cos θ z z ′


















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 53