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

Now, let us consider a ‘zyx’ transformation, meaning, the first rotation is about the ‘z-axis’, the second
rotation is about the ‘y-axis’ and the final rotation is about the ‘x-axis’.

This is mathematically represented as follows:


x
Rotation Rotation Rotation y
y
x
z
z
x x′
= R R R y = y′
y
x
z
z z′
1 0 0 cosθ 0 −sinθ cos θ sin θ 0 x x ′
′
= 0 cos θ sinθ 0 1 0 − sinθ cos θ 0 y = y - - - - (1.59)
0 − sin θ cos θ sinθ 0 cosθ 0 0 1 z z ′

Equation (1.59) could be used to transform the basis vectors as well:


1 0 0 cos θ 0 −sinθ cos θ sin θ 0 e 1 e ′ 1
′

0 cosθ sinθ 0 1 0 − sinθ cos θ 0 e 2 = e - - - - (1.60)
2
0 − sin θ cos θ sin θ 0 cos θ 0 0 1 e 3 e ′
3
Note: the above transformation matrices are valid, given that, the vector length is preserved.


Let us now consider a situation that involves shear distortion transformation.

šƒ’Ž‡ ͳǣ †‡”‹˜‡ –Š‡ –”ƒ•ˆ‘”ƒ–‹‘ ƒ–”‹š ”‡“—‹”‡† –‘ –”ƒ•ˆ‘” –Š‡ „Ž‘… •Š‘™ ‹ ȏ ‹‰
ͳǤͷͺȐ –‘ –Š‡ ‘‡ •Š‘™ ‹ ȏ ‹‰ ͳǤͷͻȐǤ Ž•‘ …‘‡– ‘ –Š‡ †‡–‡”‹ƒ– ‘ˆ –Š‡ –”ƒ•ˆ‘”ƒ–‹‘
ƒ–”‹šǤ
















[Fig 1.58: Square block] [Fig 1.59: Deformed square block]

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 54

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Consider the initial vector joining the origin and point (0,1) as shown in [Fig 1.58]

Let us transform this vector by an angle ‘γ’ as shown in [Fig 1.59] as follows:


a b 0 p
= - - - - (1.61)
c d 1 1
a b
Let the transformation matrix in equation . to be:
c d

a ∗ 0 + (b ∗ 1) p b p
Upon multiplication, we get: = = = - - - - (1.62)
c ∗ 0 + (d ∗ 1) 1 d 1

It can be observed, as discussed earlier, that, when a matric is multiplied with vector we get back a
vector. Therefore, we get: [b = p] and [d = 1].

Let us now transform the vector on the right, that is, the line joining the coordinates (1,0) and (1,1).


a b 1 p + 1
= , replacing b and d with P and 1 from equation . , we get:
c d 1 1

a p 1 p + 1
= - - - - (1.63)
c 1 1 1

a ∗ 1 + (p ∗ 1) p + 1 (a + p) p + 1
Simplifying equation . , we get: = = =
c ∗ 1 + (1 ∗ 1) 1 (c + 1) 1

Therefore, we get: [(a + p) = (p + 1)] and [(c + 1) = 1]. By where, we get, (a = 1) and (c = 0).

The transformation matrix in equation (1.61) can now be rewritten as:

a b 1 p
= - - - - (1.64)
c d 0 1

The determinant of this matrix is [[(1*1) – (p*0)] = 1].

This means that, the area of the transformed block in [Fig 46] does not change, that is, the block
only distorts and does not scale.

Observe that, the principal diagonal in equation (1.64) has elements (1,1), meaning, the scaling
factor is 1. The anti-principal diagonal has elements (p,0), meaning, the element is distorted, and
the distortion/translation is only along the ‘x-axis’.









QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 55

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

˜‡”› ”‡ƒŽ ƒ–”‹š ȋ™Š‘•‡ ‡Ž‡‡–• ƒ† †‡–‡”‹ƒ–• ƒ”‡ ”‡ƒŽȌ Šƒ• ”‡ƒŽ ”‘‘–•ǡ ™Š‹…Š †‡’‡†• ‘
–Š‡ ‘”†‡” ‘ˆ –Š‡ ƒ–”‹šǤ

šƒ’Ž‡ǣ ƒ ȋ͵ š ͵Ȍ ƒ–”‹š Šƒ• ͵ ”‡ƒŽ ”‘‘–•Ǥ

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

Mathematically, for a matrix, say, ‘A’, assuming ‘λ’ to be the root/Eigen value, we have:

Eigen value equation = det A − λI = 0 , where I is the identity matrix

Example 1: find the Eigen value of the matrix ‘A’ in equation (1.65).

50 10
A = - - - - (1.65)
10 25

50 10 λ 0 1 0
The Eigen value equation is given as: det − - - - - (1.66)
10 25 0 λ 0 1
50 − λ 10
Simplifying the equation . , we get: det - - - - (1.67)
10 25 − λ
The determinant of equation (1.67) is given as:


2
50 − λ 25 − λ − 100 = 0 = [1250 − 50λ − 25λ + λ − 100 = 0]
1150 − 75 + = 0 - - - - (1.68)
2
Rearranging the quadratic equation . we get: λ − 75λ + 1150
2
The roots of the quadratic equation are got as follows:


2
−b ± b − 4ac
, where, a = 1 , b = −75 and (c = 1150)
2a
2
2
−b + b − 4ac 75 + 75 − 4 1150
First root = = = λ = 53.51 - - - - (1.69)
1
2a 2
2
2
−b − b − 4ac 75 − 75 − 4 1150
Second Root = = = λ = 21.49 - - - - (1.70)
2a 2 2






QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 56

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Š‡ ‹‰‡ ˜ƒŽ—‡• ‹ ‡“—ƒ–‹‘• ȋͳǤ͸ͻȌ ƒ† ȋͳǤ͹ͲȌ Šƒ˜‡ ƒ …‘””‡•’‘†‹‰ —‹– ˜‡…–‘” …ƒŽŽ‡† Ǯ ‹‰‡
˜‡…–‘”ǯǤ Š‹• ‹• ƒ —‹– ˜‡…–‘” „‡…ƒ—•‡ǡ ‹– ‹• †‡”‹˜‡† ˆ”‘ ƒ •‡– ‘ˆ Š‘‘‰‡‘—• ‡“—ƒ–‹‘•Ǥ Š‹• ‹•
ƒ–Š‡ƒ–‹…ƒŽŽ› —†‡”•–‘‘† ƒ• ˆ‘ŽŽ‘™•Ǥ

‡ ‘™ –Šƒ–ǡ ™Š‡ ƒ ƒ–”‹š ‘’‡”ƒ–‡• ‘ ƒ ˜‡…–‘”ǡ ™‡ ‰‡– ƒ ‡™ ˜‡…–‘”Ǥ Š‡”‡ˆ‘”‡ǡ ™‡ Šƒ˜‡ǣ




Eigen Vector Equation = AX = λX = A − λI X = 0 , where X is the Eigen vector.

X X ′
Here, there are two Eigen vectors: 1 and 1
X 2 X ′ 2
Let us now find the first Eigen vector. Expanding the Eigen vector equation, we have:

50 − λ 10 X 0
= 1 1 = - - - - (1.71)
10 25 − λ 1 X 2 0

where X and X are components of the first Eigen vector.
1
2
Here, the values of X and X can not be determined, as, the equations are homogenous, that is,
1
2
RHS of the equations are zero. Only the relationship between X and X can be got by solving
1
2
either equations (1.72) or (1.73) that is got by simplifying equation (1.71).
50 − λ X + 10 ∗ X = 0 - - - - (1.72)
1
2
1
(10 ∗ X ) + (25 − λ )X = 0 - - - - (1.73)
1
1
2
Let us simplify equation (1.72) to compute a relationship between X and X .
1
2
10 ∗ X 10 ∗ X
2
2
X = − = X = − = X = 2.849 X
1
1
1
2
50 − λ 50 − 53.51
1
2.849
Substituting (X = 1), we get the components of the first Eigen vector to be: 1
2
The second Eigen vector is got by replacing (λ ) with (λ ) and the first Eigen vector with the second
1
2
in equations (1.71).
50 − λ 10 X ′ 0
= 2 1 = - - - - (1.74)
10 25 − λ 2 X ′ 2 0
Simplifying equation (1.74), we get:
′
′
50 − λ X + 10 ∗ X = 0 - - - - (1.75)
2
1
2
′
′
(10 ∗ X ) + (25 − λ )X = 0 - - - - (1.76)
1
2
2


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 57

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
′
Let us simplify equation (1.75) to compute a relationship between X and X .
′
2
1
′
′
10 ∗ X 10 ∗ X
′
′
′
′
X = − 2 = X = − 2 = X = −0.351 X
1
50 − λ 1 50 − 21.49 2 2
2
′
Substituting (X = 1), we get the components of the second Eigen vector to be: −0.351
2
1
The Eigen vectors are invariant (unchanged) under transformation.
ͳǤͳͳ ‰‹‡‡”‹‰ —”˜‡•
‰‹‡‡”‹‰ …—”˜‡• Šƒ˜‡ ƒŽ™ƒ›• „‡‡ …”‹–‹…ƒŽ –‘ –Š‡ ‹‡ƒ–‹…• ‘ˆ •‡˜‡”ƒŽ ‡…Šƒ‹••ǡ •—…Š ƒ•ǡ
‰‡ƒ”•ǡ …ƒǯ•ǡ –‡Ž‡•…‘’‡•ǡ …‘‘Ž‹‰ –‘™‡”• ‡–…Ǥ
 –Š‹• •‡…–‹‘ǡ ™‡ ™‘—Ž† Ž‘‘ ƒ– –Š”‡‡ ‹’‘”–ƒ– …—”˜‡•ǡ ƒ‡Ž›ǣ

ͳǤ ›…Ž‘‹†
ʹǤ ˜‘Ž—–‡
͵Ǥ ‘‰ƒ”‹–Š‹… •’‹”ƒŽ

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

ͳǤͳͳǤͳ ›…Ž‘‹†

…›…Ž‘‹† ‹• ‰‡‡”ƒ–‡† „› –”ƒ…‹‰ ƒ ’ƒ”–‹…—Žƒ” ’‘‹– ‘ ƒ …‹”…Ž‡ –Šƒ– ”‘ŽŽ• ’‡”ˆ‡…–Ž› ‘ ƒ •–”ƒ‹‰Š– Ž‹‡ǡ
™‹–Š‘—– •Ž‹’’‹‰ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ͲȐǤ















[Fig 1.60: Cycloidal Motion]


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

Š‡ …›…Ž‘‹† ‹• †‡ϐ‹‡† „› –Š‡ ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ Ǯšǯ ƒ† Ǯ›ǯ …‘‘”†‹ƒ–‡•Ǥ Š‡”‡ˆ‘”‡ǡ ƒ ‡š’”‡••‹‘
ˆ‘” Ǯšǯ ƒ› Ǯ›ǯ ‹ –‡”• ‘ˆ Ǯ ǯ ƒ† ǮɅǯǡ ™‘—Ž† ‰‹˜‡ ƒ ‡“—ƒ–‹‘ ˆ‘” …›…Ž‘‹†Ǥ

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 58

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
”‘ –Š‡ ȏ ‹‰ ͳǤ͸ͲȐǡ‹– …ƒ „‡ ‘„•‡”˜‡† –Šƒ–ǣ ȏšαȋ Ʌ - •‹ɅȌȐ ƒ† ȏ›αȋ - …‘•ɅȌȐ

Š‡”‡ˆ‘”‡ǡ ™‡ Šƒ˜‡ ˆ‘—† –Š‡ ‡“—ƒ–‹‘ ‘ˆ ƒ …›…Ž‘‹†ǡ ™Š‹…Š …ƒ „‡ ’Ž‘––‡† „› ˜ƒ”›‹‰ ǮɅǯ ˆ”‘
ȋͲ –‘ ʹɎȌ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ͳȐǤ















[Fig 1.61: Cycloidal curve (Assuming (R = 1))]


ͳǤͳͳǤʹ ˜‘Ž—–‡

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

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
























QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 59

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
‡– —• ϐ‹† ƒ ‡š’”‡••‹‘ ˆ‘” ȋšǡ›Ȍǡ ‹ –‡”• ‘ˆ –Š‡ „ƒ•‡ …‹”…Ž‡ ”ƒ†‹—• ȋ „Ȍ ƒ† –Š‡ ”‘ŽŽ ƒ‰Ž‡ ȋɅȌ ƒ•
•Š‘™ ‹ ȏ ‹‰ ͳǤ͸ʹȐǤ




























[Fig 1.62: Involute]
Let the involute originate at (O) and (α) be the initial lean angle of the line joining (O) to the center
of the base circle. It should be noted that, (O’O), which is the arc length (R θ), is equal to the length
b
of the tangent (O’O’’), that is, O O = R θ ≅ O O
′
′′
′
b
Let (x,y) be a point on the involute generated by a tangent drawn to the base circle at (O’).
Let (O O’’) subtend an angle (ϕ) at the center of the base circle.

The radius at (O’) makes an angle (θ + α) with the horizontal.

Let us start by computing the coordinates (x , y ). From the [Fig 1.62], we have:
1
1
x = R cos θ + α and y = R sin θ + α
1
b
b
1
Similarly (x, y) is given as:
x = R cos θ + α + R θ sin θ + α and y = R sin θ + α − R θ cos θ + α
b
b
b
b

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 60

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
If the involute were to originate from the horizontal instead of an arbitrary angle (α), meaning

(α = 0) , then, the equations for (x) and (y) are given as:

x = R cosθ + R θ sinθ and y = R sinθ − R θ cos θ
b
b
b
b
The coordinates (x, y) go to form the expression for the involute curve.




ͳǤͳͳǤ͵ ‘‰ƒ”‹–Š‹… ’‹”ƒŽ

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





















[Fig 1.63: Logarithmic spiral]

The rate of change of radius with respect to the angle (θ) is proportional to the instantaneous
radius (r).

dr dr dr
Mathematically, this is given as: ∝ r or = cr or = c dθ - - - - (1.77)
dθ dθ r
To find a relationship between (r) and (θ), let us integrate equation (1.77) on both sides.








QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 61

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Let us assume the initial radius to be (r ) and the radius at any instant be (r). Integrating (r)
0
between the limits (r and r) and (θ) between the limits (0 to θ), we get:
0
r θ
dr r
cθ
= c dθ = log = cθ = r = r e - - - - (1.78)
0
r r 0
r 0 0
The constant (c) in equation (1.78) can assume a positive value if the motion is inward to
outward or a negative value if the motion is outward to inward.




ͳǤͳʹ ‘…‡’– ‘ˆ ”ƒ†‹‡–

”ƒ†‹‡– ’Š›•‹…ƒŽŽ› ‹• –Š‡ †‹”‡…–‹‘ ‘ˆ ‰”‡ƒ–‡•– ˜ƒ”‹ƒ–‹‘ ‘ˆ ƒ ˆ—…–‹‘Ǥ Š‡ ‡‰ƒ–‹˜‡ ‘ˆ ‰”ƒ†‹‡–
‰‹˜‡• –Š‡ †‹”‡…–‹‘ ‘ˆ •–‡‡’‡•– †‡…”‡‡–Ǥ

‘” ƒ …‹”…Ž‡ǡ ‹– ‹• ‹–—‹–‹˜‡ –Šƒ–ǡ –Š‡ ‰”ƒ†‹‡– ‹• ƒŽ‘‰ –Š‡ ”ƒ†‹ƒŽ †‹”‡…–‹‘ ‘” ‘”ƒŽǡ –Šƒ– ‹• ™Š‡”‡
–Š‡ ”ƒ†‹—• …ƒ ˜ƒ”› ƒ– –Š‡ ƒš‹— ”ƒ–‡Ǥ

‡– —• …‘•‹†‡” ƒ …‹”…Ž‡ ‘ˆ ”ƒ†‹—• ȋ” α ͵Ȍ ƒ† ‘”‹‰‹
ȋͲǡͲȌ ƒ• …‡–‡” ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ͶȐǤ

















[Fig 1.64: Gradient of a circle]

2
2
2
Equation of the circle is given by: x + y − 3 = 0 - - - - (1.79)
∂f ∂f
∂x i + ∂y j
The direction of the gradient of a function f =
∂f 2 ∂f 2
∂x + ∂y



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 62

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Differentiating equation . , we get the gradient to be: 2xi + 2yj

Substituting (x = 1) and (y = 2 2), we get the gradient to be: 2i + 4 2 j


2i + 4 2 j 2i + 4 2 j i 2 2 j
The direction of the gradient is given by: = = + - - - - (1.80)
4 + 16 ∗ 2 6 3 3

It can be observed that equation (1.80) is the direction of the normal.

Therefore, the above property of the gradient is used in optimization to find minimum or maximum
value of a function by continuous search along the local gradient directions.





ͳǤͳ͵ ƒ‰”ƒ‰‹ƒ —Ž–‹’Ž‹‡”•


Example: locating the hottest point in a block which must lie on a circle of certain radius.

In general, if a function (f(x,y)) is to be optimized with a constraint [ax + by =c], that the
Lagrange (L) is given by:

L = f x, y + λ ax + by − c - - - - (1.81)

The above equation is solved by taking the partial derivative of (L) with respect to (x, y and λ),
where (λ) is the Lagrangian multiplier. The values of (x, y and λ) are got by simultaneously
solving the partial derivatives of (x, y and λ), as follows:

∂L
= 0 - - - - (1.82)
∂x
∂L
= 0 - - - - (1.83)
∂y

∂L
= 0 - - - - (1.84)
∂λ
Simultaneously solving equation (1.82), (1.83) and (1.84) we get the values of (x, y and λ),
which on substitution with equation (1.81), would give the optimum value of the function.


Example 1: consider the perimeter of a rectangle to be 12m. Find the optimum area of the
rectangle using Lagrangian multipliers.
The function to be optimized is the area of the rectangle. Let ‘x’ and ‘y’ be the sides of the
rectangle and the perimeter is given by (2x + 2y = 12).

The Lagrangian L is given by: xy + λ 2x + 2y − 12 - - - - (1.85)



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 63

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Differentiating equation (1.85) with respect to (x, y and λ), we get:

∂L
= y + 2λ = 0 = y = −2λ - - - - (1.86)
∂x

∂L
= x + 2λ = 0 = x = −2λ - - - - (1.87)
∂y
∂L
= 2x + 2y − 12 = 0 = x + y = 6 - - - - (1.88)
∂λ

Substituting the values of ‘x’ and ‘y’ in equation (1.88), we get:
3
−2λ − 2λ = 6 = −4λ = 6 or λ = −
2

Substituting the value of ‘λ’ in equations (1.86) and (1.87) we get the values of ‘x’ and ‘y’ to be 3.

The optimum area of the rectangle is got by substituting the value of (x, y and λ) in equation
(1.85):

3
L = 3 ∗ 3 − 6 + 6 − 12 = 9
2

It can be observed that, when the sum of ‘n’ numbers is given, each number is given by (sum/n)
for the product to be maximum.


Š‹• ‹• ƒ‘–Š‡” ‘’–‹‹œƒ–‹‘ –‡…Š‹“—‡ ™Š‡”‡ ϐ‹†‹‰ –Š‡ ƒš‹— ƒ† ‹‹— ˜ƒŽ—‡•
‹• …‘•–”ƒ‹‡†Ǥ
























QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 64

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
ͳǤͳͶ ’–‹‹œƒ–‹‘

‡•‹‰ ’–‹‹œƒ–‹‘

‡•‹‰ ‹• ƒ –”ƒ†‡ „‡–™‡‡ ƒ› ’ƒ”ƒ‡–‡”• –‘ ƒ…Š‹‡˜‡ ƒ •‹‰Ž‡ ‘” —Ž–‹’Ž‡ ‘„Œ‡…–‹˜‡•Ǥ Š‡
’ƒ”ƒ‡–‡”• ƒ› „‡ ”‡Žƒ–‡† ‘” ‹ϐŽ—‡…‡ ‡ƒ…Š ‘–Š‡” Ž‹‡ƒ”Ž› ‘” ‘- Ž‹‡ƒ”Ž›Ǥ

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

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

’–‹‹œƒ–‹‘ ‡–Š‘†• ƒ”‡ „”‘ƒ†Ž› …Žƒ••‹ϐ‹‡† ‹–‘ –Š‡ ˆ‘ŽŽ‘™‹‰ǣ
ͳǤ ”ƒ†‹–‹‘ƒŽ ‰”ƒ†‹‡– „ƒ•‡† ‡–Š‘†•
ʹǤ –‘…Šƒ•–‹… ‡–Š‘† ȋ”‘„—•– ‘’–‹‹œƒ–‹‘Ȍ
͵Ǥ ‘ ‰”ƒ†‹‡– ‡–Š‘†

”ƒ†‹–‹‘ƒŽ ‰”ƒ†‹‡– „ƒ•‡† ‡–Š‘†ǣ ™Š‡ ƒ ’”‘„Ž‡ ‹• ™‡ŽŽ †‡ϐ‹‡† ƒ† –Š‡”‡ ‹• ƒ ”‡Žƒ–‹‘•Š‹’
„‡–™‡‡ †‡•‹‰ ‘„Œ‡…–‹˜‡ ƒ† †‡•‹‰ ˜ƒ”‹ƒ„Ž‡•Ǥ

‘” ‡šƒ’Ž‡ǡ ‹ˆ †‡ϐŽ‡…–‹‘ ‘ˆ ƒ …ƒ–‹Ž‡˜‡” ‹• –‘ „‡ ‘’–‹‹œ‡†Ȁ‹‹‹œ‡†ǡ –Š‡ǡ –Š‡”‡ ‹• ƒ †‡ϐ‹‹–‡ ”‡Ǧ
Žƒ–‹‘•Š‹’ „‡–™‡‡ –Š‡ †‡ϐŽ‡…–‹‘ǡ †‡•‹‰ ‘„Œ‡…–‹˜‡ ƒ† ‘–Š‡” ’ƒ”ƒ‡–‡”• •—…Š ƒ•ǡ ˆ‘”…‡ǡ ‰‡‘‡–”›
ƒ† ƒ–‡”‹ƒŽǡ –Šƒ– ‹•ǣ

FL 3
δ =
3EI

Where, F = is the applied force

L = is the length of the beam (geometric parameter)

E = is the material modulus

I = is the area moment of inertia (geometric parameter)

δ = is the deflection of the cantilever



Š‡ •…ƒŽƒ” —Ž–‹’Ž‹‡” Ǯ͵ǯ ‹• ƒ …‘•–ƒ– ƒ† ‹• †‡’‡†‡– ‘ –Š‡ ‡† …‘†‹–‹‘• ‘ˆ –Š‡ ‡„‡”Ǥ
Š‡ †‡ϐŽ‡…–‹‘ ‘ˆ –Š‡ …ƒ–‹Ž‡˜‡” …‘—Ž† „‡ ‹‹‹œ‡† „› •‡Ž‡…–‹‰ –Š‡ ƒ’’”‘’”‹ƒ–‡ ˜ƒŽ—‡• ‘ˆ Ǯ ǯǡ Ǯ ǯǡ
Ǯ ǯ ƒ† Ǯ ǯǤ Š‹• …‘—Ž† „‡ †‘‡ ‡‹–Š‡” „› –”‹ƒŽ ƒ† ‡””‘”ǡ ‘”ǡ „› ”‡Žƒ–‹‰ –Š‡ ˜ƒ”‹ƒ–‹‘ ‘ˆ –Š‡•‡
’ƒ”ƒ‡–‡”• –‘ –Š‡ ˜ƒ”‹ƒ–‹‘ ‹ ǮɁǯǤ
Š‹• ”‡Žƒ–‹‘•Š‹’ …‘—Ž† „‡ ƒ…Š‹‡˜‡† —•‹‰ —Ž–‹˜ƒ”‹ƒ„Ž‡ …ƒŽ…—Ž—•Ǥ ‘™‡˜‡”ǡ ‹ˆ –Š‡ ƒˆ‘”‡‡–‹‘‡†
’ƒ”ƒ‡–‡”• ™‡”‡ –‘ ˜ƒ”› •–ƒ–‹•–‹…ƒŽŽ›ǡ –Š‡ǡ ƒ –”ƒ†‹–‹‘ƒŽ ‰”ƒ†‹‡– „ƒ•‡† ‡–Š‘† …ƒ‘– „‡ ƒ’’Ž‹‡†Ǥ
Š‡”‡ˆ‘”‡ǡ Ž‡– —• Ž‘‘ ƒ– ƒ‘–Š‡” ‡–Š‘† ‘ˆ ‘’–‹‹œƒ–‹‘ …ƒŽŽ‡† Dz•–‘…Šƒ•–‹… ‡–Š‘†dz ‘” Dz”‘„—•– ‘’Ǧ
–‹‹œƒ–‹‘dzǤ





QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 65

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
”‘„ƒ„‹Ž‹•–‹…Ȁ•–‘…Šƒ•–‹… †‡•‹‰ ‹ –Š‹• ‡–Š‘†ǡ –Š‡ •–ƒ–‹•–‹…ƒŽ ˜ƒ”‹ƒ–‹‘ ‹ ‡ƒ…Š ’ƒ”ƒ‡–‡”
‹• •–—†‹‡†Ǥ ƒ…Š ’ƒ”ƒ‡–‡” …‘—Ž† ƒ••—‡ ƒ …‡”–ƒ‹ ˜ƒ”‹ƒ–‹‘ǡ ˆ‘” ‡šƒ’Ž‡ǡ –Š‡ ‘—‰ǯ•
‘†—Ž—• …‘—Ž† ˆ‘” ƒ ƒ—••‹ƒȀ‘”ƒŽ †‹•–”‹„—–‹‘Ǥ


‘” ‡šƒ’Ž‡ǡ …‘•‹†‡” –Š‡ ‘”ƒŽ †‹•–”‹„—–‹‘• ƒ† –Š‡‹” ”‡•’‡…–‹˜‡ •–ƒ†ƒ”† †‡˜‹ƒ–‹‘• ˆ‘”
‘—‰ǯ• ‘†—Ž—• ȋ Ȍ ƒ† ƒ”‡ƒ ‘‡– ‘ˆ ‹‡”–‹ƒ ȋ Ȍǡ •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ͷȐǤ Š‡ ‡“—‹˜ƒŽ‡– •–ƒ†ƒ”†
†‡˜‹ƒ–‹‘ ˆ‘” –Š‡ ‡–‹‘‡† ’ƒ”ƒ‡–‡”• ‹• …‘’—–‡† ƒ• ˆ‘ŽŽ‘™•ǣ





















[Fig 1.65: Normal distribution of 'E' and 'I' and their respective standard deviations]

Let the maximum bending stress s = 32 with terms with usual meaning. Now here d needs to
3.14 3
be found optimally.

The equivalent standard deviation is given by:


∂σ 2 ∂σ 2 ∂σ 2

2
2
2
Equivalent standard deviation σ eqv = σ + σ + σ
d
L
F
∂L ∂F ∂d
Where,
∂s ∂s ∂σ
and are the sensitivity coefficients which are multiplied by their respective
∂L ∂F ∂d
standard deviations. Since we seek to optimize the diameter σ eqv is in terms of the mean
variable diameter.
Let mean stress be computed by putting all mean parameters including variable mean
diameter.

Let the mean yield strength be (Y) and standard deviation be σ
Y
Now assuming a failure probability (recommended is less than 0.0001) from the stress
distribution subtracted from strength distribution, we get


Y − S
3.1 = 0 − mean
σ 2 + σ 2
eqv Y
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 66

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
˜ƒŽ—‡ ȋ͵Ǥͳ ‹• ˆ‘—† ˆ”‘ ‘”ƒŽ †‹•–”‹„—–‹‘ –ƒ„Ž‡ ˆ‘” ƒ ˆƒ‹Ž—”‡ ’”‘„ƒ„‹Ž‹–› Ž‡•• –Šƒ ͲǤͲͲͳȌǤ ›
™Š‡”‡ –Š‡ ˜ƒŽ—‡ ‘ˆ †‹ƒ‡–‡” ‹• …‘’—–‡†Ǥ
‘–‡ǣ ‡•‹–‹˜‹–› ‹ †‡•‹‰ …‘–‡š– …‘—Ž† ƒŽ•‘ „‡ ™‹–Š ”‡•’‡…– –‘ –‘Ž‡”ƒ…‡ •–ƒ… —’ ™Š‡”‡ …‡”–ƒ‹
–‘Ž‡”ƒ…‡• ȋ‰‡‘‡–”‹… ‘” †‹‡•‹‘ƒŽȌǡ ˆ‘” ‹•–ƒ…‡ ƒˆˆ‡…– –Š‡ ˆ—…–‹‘ƒŽ‹–› ‘” ’‡”ˆ‘”ƒ…‡ ‘ˆ ƒ
‡…Šƒ‹• ƒ›„‡ •‡•‹–‹˜‡ –‘ ˆ‡™ –‘Ž‡”ƒ…‡•Ǥ ”‘„—•– †‡•‹‰ •Š‘—Ž† „‡ ƒ†‡ ‹•‡•‹–‹˜‡ –‘ –Š‡
–‘Ž‡”ƒ…‡•Ǥ
‘-‰”ƒ†‹‡– ‡–Š‘† ‘” †‡”‹˜ƒ–‹˜‡ ˆ”‡‡ ȋ‘– †‹•…—••‡† ‹ –Š‹• „‘‘Ȍǣ

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

‘” ‡šƒ’Ž‡ǡ …ƒ”„‘ ‹†—…–‹‘ ‹ ‰‡ƒ”• •ƒ› –‘ „‡ …‘–”‘ŽŽ‡† ™‹–Š ‹ ͲǤͳ ‘ˆ ƒ……—”ƒ…› ‡‡†• †ƒ–ƒ
‘ˆ –‘‘ ƒ› ˜ƒ”‹ƒ„Ž‡•Ǥ ‡ ‹– ‰”‡‡ †‹‡•‹‘•ǡ –‡’‡”ƒ–—”‡ǡ –‹‡ ‘ˆ ‡š’‘•—”‡ǡ ˆ—”ƒ…‡ ‡ˆϐ‹…‹‡…›
ƒ† ƒ› ‘”‡Ǥ Š‡ ‘’–‹— ’ƒ”ƒ‡–‡”• ‡‡†‡† …‘—Ž† „‡ Ž‘…ƒ–‡† ‘Ž› „› •–”—…–—”‡† ”ƒ†‘
•‡ƒ”…ŠǤ ‡‡”ƒŽŽ›ǡ •‹–—ƒ–‹‘• …Š‘•‡ ˆ‘” †‡”‹˜ƒ–‹˜‡ ˆ”‡‡ •‹–—ƒ–‹‘•ǡ Šƒ˜‡ —Ž–‹’Ž‡ Ž‘…ƒŽ ‹‹ƒǤ
Š‘—‰Š •‘‡ †ƒ–ƒ •—”ˆƒ…‡• ƒ› Ž‘‘ •‘‘–Š ‹– ƒ› Šƒ˜‡ ƒ› †‹•…”‡–‡ ˆ‡ƒ•‹„Ž‡ •’ƒ…‡•Ǥ Š‘—‰Š
–Š‡ •‡ƒ”…Š ƒ› ‘– ‰‹˜‡ ƒŽ™ƒ›• –”—‡ ‰Ž‘„ƒŽ ‘’–‹ƒŽ •‘Ž—–‹‘ „—– ’”‘˜‹†‡• ’”ƒ…–‹…ƒŽ •‘Ž—–‹‘Ǥ

šƒ’Ž‡ ͳǣ ‘’–‹‹œ‡ –Š‡ •‡…–‹‘ ’”‘’‡”–› ȋš› Ȍ ‰‹˜‡ –Šƒ– ‹– ‹• ‹•…”‹„‡† ‹ ƒ …‹”…Ž‡ǡ ƒ• •Š‘™ ‹
͵
ȏ ‹‰ ͳǤ͸͸ȐǤ














[Fig 1.66: a section inscribed in a circle]


3
Solution 1: The function to be optimized is xy
2
2
2
For a circle, by Pythagoras theorem, we have: x + y = d
Simplifying the above expression we get y = d − x
2
2
3
In order to optimize for xy let us substitute the value of ‘y’ in xy . Therefore, we get:
3
3
3
xy = d − x 2 ∗ x
2
2
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 67

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Taking ‘x’ inside the root we get:

3 3
2 2 2 2 8 2
3
3 = xy = x 3 d − x 3
2
xy = x 3 d − x 2+ 3 2


2 8

2
Therefore the term under the root is: x 3 d − x 3

Differentiating with respect to x we get:

3
d y x 8 5 2 1
−
2
= − ∗ x 3 + d ∗ ∗ x 3 = 0
dx 3 3

d
Therefore, we get x =
2
2
Substituting the value of ‘x’ in the equation x + y = d , we get:
2
2
d
y = 3
2

The optimum or maximum value of xy is given by:
3

d d 3 d 4
3
xy = ∗ 3 = 3 3
2 2 16



Example 2: Find the minimum value of f x, y = x + y + 2xy , using gradient approach.
2
2
2

Note: Though this is a simple case, the entire procedure of using the Hessian matrix is well
demonstrated.

Solution 2: in order to solve this situation we would use the Hessian matrix, which is given as
follows:

2
2
∂ f ∂ f
∂x 2 ∂x ∂y
2
Hessian Matrix = ∂ f ∂ f
2
2
∂y ∂x ∂y
Š‹• ƒ–”‹š ‹• —•‡† –‘ …Š‡… –Š‡ ƒš‹— ƒ† ‹‹— ˜ƒŽ—‡ ‘ˆ ƒ ˆ—…–‹‘ ™Š‡ –Š‡”‡ ‹• ‘”‡
–Šƒ ‘‡ ˜ƒ”‹ƒ„Ž‡Ǥ  –Š‹• ‡šƒ’Ž‡ǡ –Š‡”‡ ƒ”‡ ʹ ˜ƒ”‹ƒ„Ž‡• Ǯšǯ ƒ† Ǯ›ǯǤ ˆ –Š‡ †‡–‡”‹ƒ– ‘ˆ –Š‹• ƒ–”‹š
‹• ’‘•‹–‹˜‡ǡ –Š‡ǡ –Š‡ ˆ—…–‹‘ ƒ••—‡• ‹‹— ˜ƒŽ—‡Ǥ ˆ –Š‡ †‡–‡”‹ƒ– ‹• ‡‰ƒ–‹˜‡ǡ –Š‡ǡ –Š‡
ˆ—…–‹‘ ƒ••—‡• ƒš‹— ˜ƒŽ—‡Ǥ Š‡”‡ˆ‘”‡ǡ ‹ ‘”†‡” –‘ ‰‡– –Š‡ ‹‹— ˜ƒŽ—‡ ‘ˆ –Š‡ ˆ—…–‹‘ǡ


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 68

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
‘‡ —•– ϐ‹”•– Ž‘…ƒ–‡ ȋšǡ›Ȍ „› ‡“—ƒ–‹‰ –Š‡ †‡”‹˜ƒ–‹˜‡ ‘ˆ ˆ—…–‹‘ ™‹–Š ”‡•’‡…– –‘ Ǯšǯ ƒ† Ǯ›ǯ –‘ œ‡”‘Ǥ
—„•‡“—‡–Ž›ǡ –Š‡ ƒ–”‹š —•– „‡ …‘•–”—…–‡† –‘ —‡”‹…ƒŽŽ› ‡˜ƒŽ—ƒ–‡ –Š‡ †‡–‡”‹ƒ–Ǥ
‹ˆˆ‡”‡–‹ƒ–‹‰ –Š‡ ˆ—…–‹‘ ™‹–Š ”‡•’‡…– –‘ Ǯšǯ ™‡ ‰‡–ǣ

∂f x, y
2
= 2x + 2y = 0
∂x
Simplifying the above expression we get [x = −y ]
2
Differentiating the function with respect to ‘y’ we get:

∂f x, y
= 2y + 4xy = 0
∂y

1
Simplifying the above expression we get x = − and y = 0
2

Substituting the values [y = 0] in the expression [x = −y ], we get [x = 0]. Therefore we have:
2
1
x = 0 and x = −
2

In order to find the minimum value of the function let us choose the coordinates (x = 0) and (y = 0).

Let us begin by computing the remaining component of the Hessian matrix.

2
∂ f x, y
= 2
∂x 2
2
∂ f x, y
= 2 + 4x = 0 , since we have chosen the coordinates x = 0 and y = 0 , we get:
∂y 2
2
∂ f x, y
= 2 + 4 0 = 2
∂y 2
The cross derivatives are computed as follows:


∂ ∂f x, y
= 4y
∂x ∂y

∂ ∂f x, y
= 4y
∂x ∂x

Substituting the values (x = 0) and (y = 0) in the cross derivatives, we get:

2
2
∂ f x, y ∂ f x, y
= = 0
∂x ∂y ∂y ∂x
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 69

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Substituting the values of the derivatives in the Hessian matrix, we get:


2
2
∂ f ∂ f
∂x 2 ∂x ∂y 2 0

2
Hessian Matrix = ∂ f ∂ f = 0 2
2
2
∂y ∂x ∂y

The determinant of the Hessian matrix is ‘4’ which is positive. Hence, at (x = 0) and (y = 0) the
function assumes a minimum value, and is given by:

f x, y = x + y + 2xy = 0
2
2
2

ͳǤͳͷ ‘—”‹‡” ‡”‹‡• ƒ† ”ƒ•ˆ‘”•

› ‰‡‡”ƒŽ ™ƒ˜‡ ˆ‘” ‹• ƒ …‘„‹ƒ–‹‘ ‘ˆ •‡˜‡”ƒŽ •‹—•‘‹†ƒŽ ‘” …‘•‹—•‘‹†ƒŽ ™ƒ˜‡•ǡ
‡ƒ‹‰ǡ ƒ ™ƒ˜‡ …‘—Ž† „‡ …‘•–”—…–‡† „› •‡˜‡”ƒŽ ™ƒ˜‡• ‘ˆ †‹ˆˆ‡”‡– ˆ”‡“—‡…‹‡•Ǥ












[Fig 1.67: Addition of two harmonics]



‘” ‡šƒ’Ž‡ǡ Ž‡– —• …‘•‹†‡” –Š‡ ƒ††‹–‹‘ ‘ˆ –™‘ •‹—•‘‹†ƒŽ ™ƒ˜‡•ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸͹ȐǤ















QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 70

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Š‡ …‘„‹‡† ™ƒ˜‡ ˆ‘” ‘ˆ Šƒ”‘‹…• ͳ ƒ† ʹ ‹• ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͸ͺȐǤ





















[Fig 1.68: Combined wave form of harmonics 1 and 2]

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

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


n=∞
1 1 1
f x = − sin nπx
2 π n
n=1
‘n’ could assume a large value, but for simplicity of calculations, we have chosen n=4. Plotting
addition of 4 harmonics we get:

n=4
1 1 1
f x = − sin nπx
2 π n
n=1
1 1 1 1 1 1 1 1
= − sin 1πx + − sin 2πx + − sin 3πx + − sin 4πx
2 π 2 2π 2 3π 2 4π

The above function could be graphically represented as shown in [Fig 1.69].








QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 71

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


































[Fig 1.69: Summation of the first four harmonics leading to sawtooth wave]


Š‡ •ƒ™–‘‘–Š ™ƒ˜‡ Šƒ• ƒ ”‡•‡„Žƒ…‡ –‘ ‡›„‘ƒ”† ™ƒ˜‡ˆ‘” ™Š‹…Š Šƒ• •‡˜‡”ƒŽ ‡›• ‰‹˜‹‰ …‘Ǧ
„‹ƒ–‹‘ ‘ˆ •‡˜‡”ƒŽ Šƒ”‘‹…•Ǥ Š‡ …‘„‹ƒ–‹‘ ‘ˆ •‡˜‡”ƒŽ Šƒ”‘‹…• ‹• ƒŽ•‘ …ƒŽŽ‡† Šƒ”‘‹… †‹•Ǧ
–‘”–‹‘Ǥ Š‹• ‘……—”• ‹ ‡Ž‡…–”‹…ƒŽ …‹”…—‹–• †—‡ –‘ ‘-Ž‹‡ƒ” ”‡Žƒ–‹‘•Š‹’ „‡–™‡‡ …—””‡– ƒ†
˜‘Ž–ƒ‰‡Ǥ
‘”ƒŽŽ› –Š‡ ‘—”‹‡” •‡”‹‡• ‹• ƒ–Š‡ƒ–‹…ƒŽŽ› ‰‹˜‡ „› ƒ› ’‡”‹‘†‹… ‡š…‹–ƒ–‹‘ ™‹–Š ˆ”‡“—‡…› ™
”ƒ†Ȁ ǡ ƒ• ƒ •—ƒ–‹‘ ‘ˆ ƒ –”‹‰‘‘‡–”‹… •‡”‹‡•Ǥ ‡…‡ Ǯ™ǯ ‹• –Š‡ ˆ—†ƒ‡–ƒŽ ˆ”‡“—‡…›ǡ ‹–
”‡’”‡•‡–• –Š‡ ƒš‹— ‡‡”‰›Ǥ ‹–Š Š‹‰Š‡” Šƒ”‘‹…• ‡‡”‰› ”—• †‘™Š‹ŽŽǤ ‘™‡˜‡”ǡ –Š‡ ϐ‹”•–
ˆ‡™ Šƒ”‘‹…• …‘–ƒ‹ ƒš‹— ‡‡”‰›

n=∞
a
f t = + a cosnωt + b sinnωt
2 n n
n=1
π
1
a = f x cos nωt dt
n
π
−π
π
1
b = f x sin nωt dt
n
π
−π

π
1
a = f t dt
π
−π

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 72

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
ͳǤͳͷǤͳ ‘—”‹‡” ”ƒ•ˆ‘”

 ‡‰‹‡‡”‹‰ǡ ƒ› •‹–—ƒ–‹‘• ‡š‹•– ™Š‡”‡ –‹‡ †‘ƒ‹ †ƒ–ƒ ȋƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹ͲȐȌ †‘‡• ‘–
ƒ‡ —…Š •‡•‡ǡ ”ƒ–Š‡” –Š‡ ˆ”‡“—‡…› †‘ƒ‹ †ƒ–ƒ †‘‡• ȋƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹ͳȐȌǤ


















[Fig 1.70: Force Vs Time] [Fig 1.71: Force Vs Frequency]



šƒ’Ž‡ǣ ‘•‹†‡” ƒ …‘„‹‡† ˜‹„”ƒ–‘”› Ž‘ƒ†‹‰ ‘ˆ ƒ …‘’‘‡– †—‡ –‘ –Š”‡‡ •‘—”…‡• ‘ˆ ‡š…‹Ǧ
–ƒ–‹‘Ǥ
‡”‡ ƒ ‡‰‹‡‡” ‹• ‹–‡”‡•–‡† ‹ –Š‡ ƒ‰‹–—†‡ ‘ˆ –Š‡ ‡š…‹–ƒ–‹‘ ˆ‘”…‡ ƒ– ˜ƒ”‹‘—• ˆ”‡“—‡…‹‡• –‘
…ƒ””› ‘—– †›ƒ‹… •–”‡•• ƒƒŽ›•‹•Ǥ ‘” ‡šƒ’Ž‡ǡ ƒ Šƒ‡” –‡•– ǡ –Š‡ ”‡•’‘•‡ ‹• …ƒ’–—”‡† ‹ –Š‡
–‹‡ †‘ƒ‹Ǥ Š‡ ƒ–—”ƒŽ ‘” ”‡•‘ƒ– ˆ”‡“—‡…› ‹• ‹†‡–‹ϐ‹‡† „› Ž‘…ƒ–‹‰ –Š‡ ’‘‹– ‘ˆ ƒš‹—
”‡•’‘•‡ ‹ –Š‡ ˆ”‡“—‡…› †‘ƒ‹Ǥ
‘—”‹‡” –”ƒ•ˆ‘” Šƒ• •‡˜‡”ƒŽ ƒ’’Ž‹…ƒ–‹‘•Ǥ ‡ ‹– ‹ƒ‰‡ ’”‘…‡••‹‰ǡ ‘‹•‡ ϐ‹Ž–‡”‹‰ǡ •‘Ž˜‹‰ ’ƒ”–‹ƒŽ
†‹ˆˆ‡”‡–‹ƒŽ ‡“—ƒ–‹‘• ‡–…Ǥ  ‡…Šƒ‹…ƒŽ ‡‰‹‡‡”‹‰ ™‡ —•‡ ȋ ƒ•– ‘—”‹‡” ”ƒ•ˆ‘”•Ȍ –‘
•–—†› •‹‰‹ϐ‹…ƒ– ȋϐ‹Ž–‡”‡†Ȍ ”‡•’‘•‡ ‘ˆ •›•–‡• •ƒ› ’—’ ”‹’’Ž‡ǡ –‘ …‘˜‡”– –Š‡ ˜‹„”ƒ–‹‘ ‡ƒ•—”‡Ǧ
‡– ‹ –Š‡ –‹‡ †‘ƒ‹ –‘ ˆ”‡“—‡…› †‘ƒ‹ ƒ† •‘ ‘Ǥ
ƒ–Š‡ƒ–‹…ƒŽŽ›ǡ –Š‡ ‘—”‹‡” –”ƒ•ˆ‘” ‹• ”‡’”‡•‡–‡† ƒ• ˆ‘ŽŽ‘™•ǣ

To get the frequency domain F ω , the time domain function [f(t)] is input to the Fourier
transformation function (ℱ). Therefore, we get:

∞
F ω = ℱ f t = f t e −iωt dt
−∞

The above equation is used to get frequency domain data and the inverse could be taken to get
back to time domain as follows:


f t = ℱ −1 F ω = 1 ∞ f ω e iωt dω
2π −∞

”ƒ…–‹…‹‰ ‡…Šƒ‹…ƒŽ ‡‰‹‡‡”• ‰‡‡”ƒŽŽ› —•‡ ȋ ƒ•– ‘—”‹‡” ”ƒ•ˆ‘”•Ȍ –‘ …ƒ’–—”‡ ‘Ž› –Š‡
•‹‰‹ϐ‹…ƒ– ”‡•’‘•‡• —•‹‰ –‹‡ †‘ƒ‹ †ƒ–ƒǤ ‘” ‡šƒ’Ž‡ǡ ƒ Šƒ‡” ƒ› Š‹– ƒ …ƒ–‹Ž‡˜‡”
…‘–‹—‘—•Ž› ƒ– ˜ƒ”‹‘—• ƒ‰‹–—†‡•Ǥ Š‡ †‡ϐŽ‡…–‹‘ ‘ˆ –Š‡ …ƒ–‹Ž‡˜‡” ™‹–Š –‹‡ …ƒ „‡ ”‡…‘”†‡†Ǥ

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 73

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Š‡ •ƒ‡ …‘—Ž† „‡ ˆ‡† –‘ ƒ ‘Ž‹‡ …ƒŽ…—Žƒ–‘”Ǥ Š‡ …ƒŽ…—Žƒ–‘” ™‘—Ž† …‘’—–‡ •‹‰‹ϐ‹…ƒ–
”‡•’‘•‡• ƒ– ˜ƒ”‹‘—• ˆ”‡“—‡…‹‡•ǡ ™Š‹…Š ‹-–—” ™‘—Ž† Š‡Ž’ Ž‘…ƒ–‡ –Š‡ ”‡•‘ƒ– ˆ”‡“—‡…‹‡• ƒ–
™Š‹…Š –Š‡ ”‡•’‘•‡ ‹• ”‡Žƒ–‹˜‡Ž› ƒš‹—Ǥ

ͳǤͳ͸ ‘—–‹‡ ‰‹‡‡”‹‰ ƒŽ…—Žƒ–‹‘•

 ƒ› ‡‰‹‡‡”‹‰ •‹–—ƒ–‹‘•ǡ –Š‡ …ƒŽ…—Žƒ–‹‘ ‘ˆ –Š‡ ‡Ž‡˜ƒ–‹‘ ‘ˆ ƒ ‰‡‘‰”ƒ’Š‹…ƒŽ Ž‘…ƒ–‹‘ ™‹–Š ”‡Ǧ
•’‡…– –‘ ƒ‘–Š‡” Ž‘…ƒ–‹‘ǡ ˆ‘” ϐŽ‘™ ƒƒŽ›•‹• ‘ˆ ‰ƒ•‡• ‘” †‹•–ƒ…‡ …ƒŽ…—Žƒ–‹‘•ǡ ‡…‡••‹–ƒ–‡ –Š‡ …ƒŽ…—Ǧ
Žƒ–‹‘ ‘ˆ ȋšǡ › ƒ† œȌ …‘‘”†‹ƒ–‡•ǡ ˆ”‘ –Š‡ ƒ˜ƒ‹Žƒ„Ž‡ ˜ƒŽ—‡• ‘ˆ Žƒ–‹–—†‡ ƒ† Ž‘‰‹–—†‡ ‘ˆ –Š‡ ‰‡‘Ǧ
‰”ƒ’Š‹…ƒŽ Ž‘…ƒ–‹‘Ǥ
šƒ’Ž‡ ͳǣ …ƒŽ…—Žƒ–‡ –Š‡ •Š‘”–‡•– –—‡Ž †‹•–ƒ…‡ „‡–™‡‡ ƒ‰ƒŽ‘”‡ ƒ† ‘‹„ƒ–‘”‡Ǥ Ž•‘ …ƒŽǦ
…—Žƒ–‡ –Š‡ †‹•–ƒ…‡ ƒŽ‘‰ –Š‡ ‡ƒ”–Šǯ• •—”ˆƒ…‡ ȋ–Š‡ ‰”‡ƒ– …‹”…Ž‡Ȍǡ ƒ••—‹‰ –Š‡ ‡ƒ”–Š –‘ „‡ ƒ ’‡”ˆ‡…–
•’Š‡”‡ ‘ˆ ”ƒ†‹—• Ǯ ǯǤ Ž•‘ ‡•–‹ƒ–‡ –Š‡ ‡ƒ”–Šǯ• •—”ˆƒ…‡ ˜‡Ž‘…‹–› ƒ† …‡–”‹ˆ—‰ƒŽ ˆ‘”…‡ ƒ– ƒ‰ƒŽ‘”‡Ǥ
Š‡ ‰‡‘‰”ƒ’Š‹… …‘‘”†‹ƒ–‡• ‘ˆ ƒ‰ƒŽ‘”‡ ƒ† ‘‹„ƒ–‘”‡ ‹• ƒ• –ƒ„—Žƒ–‡†ǣ


‹–› ƒ–‹–—†‡ ȋȽȌ ‘‰‹–—†‡ ȋȾȌ
ƒ‰ƒŽ‘”‡ ͳ͵-†‡‰”‡‡ ‘”–Š ͹ͺ-†‡‰”‡‡ ƒ•–
‘‹„ƒ–‘”‡ ͳͳ-†‡‰”‡‡ ‘”–Š ͹͹-†‡‰”‡‡ ƒ•–



Š‡ Žƒ–‹–—†‡ ƒ† Ž‘‰‹–—†‡ ƒ‰Ž‡• –Šƒ– ƒ”‡ ȋȽ ƒ† ȾȌ ”‡•’‡…–‹˜‡Ž› ƒ”‡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹ʹȐǤ





















[Fig 1.72: A location on the surface of the earth with geographic coordinates]







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 74

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
– ‹• …Ž‡ƒ” –Šƒ–ǡ –Š‡ ȋšǡ › ƒ† œȌ …‘‘”†‹ƒ–‡• ‘ˆ –Š‡ Ž‘…ƒ–‹‘ …ƒ „‡ †‡”‹˜‡† ‹ –‡”• ‘ˆ ǮȽǯǡ ǮȾǯ ƒ† ”ƒǦ
†‹—• Ǯ ǯ ‘ˆ –Š‡ ‡ƒ”–ŠǤ Š‡ ”ƒ†‹—• Œ‘‹‹‰ –Š‡ Ž‘…ƒ–‹‘ ‘ˆ ‹–‡”‡•– ƒ† –Š‡ ‘ˆ –Š‡ ‡ƒ”–Š ‹• ’”‘Œ‡…–‡†
‘–‘ –Š‡ ȋšǡ › ƒ† œȌ ƒš‡• –‘ ‰‡– –Š‡ ȋšǡ › ƒ† œȌ …‘‘”†‹ƒ–‡•ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹͵Ȑ





















[Fig 1.73: (x, y and z) coordinates of the location of interest]

The (x, y and z) coordinates are tabulated as follows:

z R sin α

x R cos α cosβ


y R cos α sinβ



Substituting the values of (α and β) for Bangalore and Coimbatore, and assuming the radius of the
earth to be 6700 kms, their locations are as follows:

Coordinates Bangalore (from Center of earth) Coimbatore (from Center of earth)

x (6700 cos 13 cos78) = 1357 Kms (6700 cos 11 cos77) = 1479.5 Kms

y 6700 cos 13 sin78 = 6385 kms 6700 cos 11 sin77 = 6408 kms

z 6700 sin 13 = 1507 kms 6700 sin 11 = 1278 kms







QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 75

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Let us now find the shortest distance, using the distance formula, as follows:


x − x + y − y + z − z
2
2
2
1
2
1
2
2
1
= 1357 − 1479.8 + 6385 − 6408 + 1507 − 1278 = 260.2 kms
2
2
2
Now let us find the distance between Bangalore and Coimbatore in the great circle as shown in
[Fig 1.74].




















[Fig 1.74: Distance between the cities on the great circle]


Š‡ ƒ”… Ž‡‰–Š ȋ Ȍ …‘‡…–‹‰ ƒ‰ƒŽ‘”‡ ƒ† ‘‹„ƒ–‘”‡ ‹• ‡“—ƒŽ –‘ ȋ ɅȌǡ ™Š‡”‡‹ǡ ǮɅǯ ‹• –Š‡ ˜ƒŽ—‡
–Šƒ– ‡‡†• –‘ „‡ …‘’—–‡†Ǥ ‡– —• …‘’—–‡ ǮɅǯ „› •…ƒŽƒ” ’”‘†—…–Ǥ

ȋ…‘• ɅȌ ‹• ‰‹˜‡ „›ǡ •— ‘ˆ ’”‘†—…– ‘ˆ ”‡•’‡…–‹˜‡ …‘’‘‡–• ‘ˆ –Š‡ –™‘ ˜‡…–‘”•ǡ †‹˜‹†‡† „› –Š‡
’”‘†—…– ‘ˆ ƒ‰‹–—†‡ ‘ˆ –Š‡ ˜‡…–‘”•Ǥ ‡”‡ǡ –Š‡ –™‘ ˜‡…–‘”• ƒ”‡ ȋ Ȍ ƒ† ȋ Ȍǡ ™Š‘•‡ ƒ‰‹–—†‡ ‹•
–Š‡ ”ƒ†‹—• ȋ Ȍ ‘ˆ –Š‡ ‡ƒ”–ŠǤ

ƒ–Š‡ƒ–‹…ƒŽŽ›ǡ ™‡ Šƒ˜‡ǣ

x x + y y + z z 1357 ∗ 1479.8 + 6385 ∗ 6408 + 1507 ∗ 1278
1 2
1 2
1 2
cos θ = = = 0.9992

R R 6700 2
Therefore, we have: θ = cos −1 0.9992 = 2.29 degrees = 0.0407 radians
And we have, Rθ = 6700 ∗ 0.0407 = 272.69 kms


QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 76

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

The angular velocity ω of the earth is given by:

2π
ω =
T time taken by the earth to complete one rotation in seconds

2π radians
= = 7.27 × 10 −5
24 ∗ 3600 second
m
Surface velocity of the earth at the equator = ωR = 7.27 × 10 −5 ∗ 6700000 = 487
s

2
∗ 6700000] = 0.035N
For a unit mass, centrifugal force is given by = mω R = [ 7.27 × 10 −5 2


šƒ’Ž‡ ʹǣ ϐ‹† –Š‡ ƒ–‡”‹ƒŽ ƒ”‡ƒ ”‡“—‹”‡† –‘ ˆƒ„”‹…ƒ–‡ ƒ …‘‹…ƒŽ •‹ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ͳǤ͹ͷȐǤ



























[Fig 1.75: A sheet metal sink]


Š‡ ƒ‘—– ‘ˆ •Š‡‡– ‡–ƒŽ ‡‡†‡† ˆ‘” –Š‡ •‹ ‹• ƒ• ˆ‘ŽŽ‘™•ǣ

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



QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 77

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



























[Fig 1.76: cone develops to a sector of a circle]

Let the sector subtend an angle ‘θ’ at the center of the circle. For an angle ‘2π’, the area of the
sector is the area of the circle itself, for an angle ‘θ’, the area of the circle is got as follows:
2
πr ∗ θ 1
2
Area of a sector of a circle for a subtended angle θ = = r θ
2π 2
Let us apply above observation to the case in hand. Therefore, the net area of the sector in
[Fig 1.76] is given by:

1 1 1 1
2
2
2
2
Net area of sector = R θ − R θ = θ R − R = θ R + R ∗ R − R
1
2
1
2
2 1 2 2 2 1 2 2

From the [Fig 1.76], we know that R − R = S and R θ = 2πr and similarly
1
1
2
1
R θ = 2πr .
2
2
1
Therefore, we have the net area of the sector = 2πr + 2πr ∗ S = π r + r ∗ S
2 1 2 1 2








QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 78

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













[Fig 1.77: cylinder develops to a rectangle]


Area of the developed cylinder = H ∗ 2πr
2

The sum of the developed cylinder area and that of the cone will give us the total material area.

π r + r ∗ S + H ∗ 2πr
2
2
1

‘S’ is the slant height of the frustum of the cone, as shown in [Fig 1.78], and is given by:















[Fig 1.78: slant length (S) of the frustum of the cone]

We have calculated the area of the sector/frustum of a cone to be: π r + r ∗ S
2
1
Replacing the value of ‘s’ from [Fig 1.78] in the above expression, we get:

π r + r ∗ H + r − r
2
2
2
1
1
2

The total area = π r + r ∗ H + r − r + H ∗ 2πr
2
2
1
2
1
2
2

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 79

Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
Substituting the values of ‘r ’, ‘r ’ and ‘H’ in the above expression we get:
1
2
The total material area = π 12 + 3 ∗ 32 + 12 − 3 + 32 ∗ 2π ∗ 3 = 2169.655 inch
2
2
2
šƒ’Ž‡ ͵ǣ ‹ ƒ› ‡‰‹‡‡”‹‰ •‹–—ƒ–‹‘•ǡ ‘™‹‰ –‘ –Š‡ ƒ–—”‡ ‘ˆ ‡‰‹‡‡”‹‰ǡ –Š‡ ’ƒ”ƒ‡–‡”•
Šƒ˜‡ •‹‰‹ϐ‹…ƒ– ˜ƒ”‹ƒ„‹Ž‹–›Ǥ ‘ ‰‡– –Š‡ ƒ˜‡”ƒ‰‡ ˜ƒŽ—‡ ‘ˆ ƒ ’ƒ”ƒ‡–‡”ǡ ‘‡ …‘—Ž† —•‡ –Š‡ ȋ ‘‘–
‡ƒ “—ƒ”‡Ȍ ˜ƒŽ—‡Ǥ

‡– —• …‘•‹†‡” ƒ ‘–‘”ǡ ™Š‘•‡ –‘”“—‡ ”‡“—‹”‡‡– ‹• †‹ˆˆ‡”‡– ˆ‘” †‹ˆˆ‡”‡– ’‡”‹‘†• ‘ˆ –‹‡Ǥ Š‡
‘–‘” —•– „‡ …‘–”‘ŽŽ‡† ‹ •—…Š ƒ ™ƒ› –Šƒ– ‹– †‘‡• ‘– •—ˆˆ‡” ‡š…‡••‹˜‡ †ƒƒ‰‡ †—‡ –‘ ‘˜‡” Š‡ƒ–Ǧ
‹‰ ‘” †‡ƒ‰‡–‹œƒ–‹‘ ‡ˆˆ‡…–•Ǥ
Let us say, that, the torque requirements (T , T , T − − − −−, T ) for different time ranges and the
2
3
n
1
time intervals be (t , t , t − − − −−, t ).
3
1
2
n
The RMS torque value is given by:
2
2
2
2
T t + T t + T t + − − − − T t
n n
1 1 2 2 3 3
t + t + t + − − − − −t n
3
1
2
‡”‡ǡ ‡ƒ…Š –‘”“—‡ ˜ƒŽ—‡ Šƒ• –Š‡ ™‡‹‰Š– ƒ• –‹‡ ‹–‡”˜ƒŽǤ Š‡”‡ˆ‘”‡ǡ –Š‡ •— ‘ˆ –Š‡ ™‡‹‰Š–‡†
•“—ƒ”‡• ‘ˆ –‘”“—‡• ‹• †‹˜‹†‡† „› –Š‡ –‘–ƒŽ ™‡‹‰Š–Ȁ–‹‡ –‘ ‰‡– –Š‡ ™‡‹‰Š–‡† ‡ƒ •“—ƒ”‡ǡ ™Š‘•‡
•“—ƒ”‡ ”‘‘– ‰‹˜‡• –Š‡ ˜ƒŽ—‡ ‡‡†‡†Ǥ
Š‡ ˜ƒŽ—‡ Š‡Ž’• ‹ •‡Ž‡…–‹‰ –Š‡ „‡•– ‘–‘” ˆ‘” ƒ ‰‹˜‡ •‹–—ƒ–‹‘Ǥ
‡– —• …‘•‹†‡” ƒ‘–Š‡” •‹–—ƒ–‹‘Ǥ
ƒ–‡”‹ƒŽ ™‹–Š ”‡Žƒ–‹˜‡ ’”‘’‡”–› ”ƒ ƒ† ™‡‹‰Š–ƒ‰‡ǡ ‰‹˜‡ –‘ ‡ƒ…Š ’”‘’‡”–› ƒ• •Š‘™ ‹ –Š‡ –ƒ„Ž‡
„‡Ž‘™Ǥ Š‡ ™‡‹‰Š–‡† ƒ˜‡”ƒ‰‡ ‹• ˆ‘—† •—…Š –Šƒ–ǡ –Š‡ ƒ–‡”‹ƒŽ• …‘—Ž† „‡ …‘’ƒ”‡† ˆ‘” ƒ’’Ž‹…ƒ–‹‘Ǥ
Š‡ ƒ–‡”‹ƒŽ ™‹–Š Š‹‰Š‡•– ™‡‹‰Š–‡† ƒ˜‡”ƒ‰‡ ‹• –Š‡ „‡•– •—‹–‡† ƒ–‡”‹ƒŽǤ

‘–‡ǣ ‡”‡ ™‡‹‰Š–• ƒ”‡ ‘”ƒŽ‹œ‡† •—…Š –Šƒ– –Š‡‹” •— ‹• ͳ

Material Property Relative Rank Weightage given



Yield strength 9 0.1


Fatigue strength 7 0.4

Young’s Modulus to density ratio 8 0.2


Cost 9 0.3




QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 80

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
The weighted average property is given by using the formula,

n a W
Weighted average = i=1 1 i where a is the relative rank and W is the weightage
n i=1 W i
Therefore the weighted average for the given situation is;


9 ∗ 0.1 + 7 ∗ 0.4 + 8 ∗ 0.2 + 9 ∗ 0.3
= 16.2
0.1 + 0.4 + 0.2 + 0.3


šƒ’Ž‡ Ͷǣ ϐ‹† –Š‡ …ƒ”„—”‹œƒ–‹‘ †‡’–Š Ž‘•–ǡ ˆ‘” ƒ ‰‡ƒ” ‹ˆ –Š‡ ‘˜‡” ’‹ †‹‡•‹‘ ‰‘‡• †‘™ „›
ͲǤͷǤ Š‡ †‹ƒ‡–‡” ‘ˆ –Š‡ ’‹ ‹• ͳͲǡ –Š‡ ’”‡••—”‡ ƒ‰Ž‡ ‘ˆ –Š‡ ‰‡ƒ” ‹• ʹͲ †‡‰”‡‡•Ǥ

ƒ…‰”‘—†ǣ –Š‡ ‰‡ƒ” –‘‘–Š •‹œ‡ ‹• ‰‡‡”ƒŽŽ› ‡ƒ•—”‡† „› ‡ƒ•—”‹‰ –Š‡ ‘˜‡” ’‹ †‹‡•‹‘ ƒ•
†‡’‹…–‡† ‹ ȏ ‹‰ ͳǤ͹ͻȐǤ Š‡•‡ ’‹•Ȁ”‘ŽŽ‡”• ƒ”‡ –ƒ‰‡–‹ƒŽ –‘ –Š‡ ‰‡ƒ” –‘‘–Šǡ ™Š‡”‡ǡ –Š‡ –ƒ‰‡–
ƒ‡• ƒ ƒ‰Ž‡ ‡“—ƒŽ –‘ –Š‡ ’”‡••—”‡ ƒ‰Ž‡ ™‹–Š –Š‡ ˜‡”–‹…ƒŽǤ Š‡ ‘˜‡” ’‹ †‹‡•‹‘ ‹• ‰‹˜‡ ƒ
–‘Ž‡”ƒ…‡Ǥ ˆ –Š‡ †‹‡•‹‘• ‰‘ ƒ„‘˜‡ ‘” „‡Ž‘™ –Š‡ •‡– –‘Ž‡”ƒ…‡ǡ –Š‡ǡ –Š‡ –‘‘–Š •‹œ‡ ‹• ‡‹–Š‡”
‘”‡ǡ ‘” Ž‡•• –Šƒ –Š‡ •–‹’—Žƒ–‡† †‹‡•‹‘Ǥ
























[Fig 1.79: over pin dimension of a gear]








QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 81

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


ˆ –Š‡ –‘‘–Š •‹œ‡ ‰‘‡• †‘™ǡ –Š‡”‡ ƒ”‡ –™‘ …‘•‡“—‡…‡•ǣ
· Š‡ „ƒ…Žƒ•Š ‹…”‡ƒ•‡•Ǥ ˆ –Š‡ –‘‘–Š •‹œ‡ ‹• Š‹‰Š‡”ǡ –Š‡ǡ –Š‡ „ƒ…Žƒ•Š †‡…”‡ƒ•‡•Ǥ 
‡š–”‡‡ …ƒ•‡•ǡ •ƒ›ǡ ƒ…Š‹‹‰ –Š‡ ™”‘‰ —„‡” ‘ˆ –‡‡–Šǡ –Š‡ ‰‡ƒ”• †‘ ‘– ”‘–ƒ–‡Ǥ
· Š‡ …ƒ”„—”‹œƒ–‹‘ †‡’–Š ƒŽ•‘ †‡…”‡ƒ•‡•Ǥ


Solution: the over pin dimension goes down by 0.5mm, meaning, for a symmetric gear about the
diameter, the dimension lost, radially, is:


0.5
= 0.25mm
2
Since, it is also known that the pin is tangential to the gear tooth (approximately at the pitch circle
radius), the radius of the pin, makes an angle Φ , which is equal to the pressure angle, with respect
to the horizontal, as shown in [Fig 1.80].

It can be observed that, the material lost on the profile is measured along the radius of the pin.


















[Fig 1.80: pressure angle subtended at the center of the pin]














QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 82

Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Š‡ ƒ–‡”‹ƒŽ Ž‘•– „› –Š‡ ‰‡ƒ” –‘‘–Šǡ •Š‘™ ‹ ȏ ‹‰ ͳǤͺͳȐ ‹• …ƒŽ…—Žƒ–‡† ƒ• ˆ‘ŽŽ‘™•









































[Fig 1.81: material lost by the gear tooth]


Material lost = Drop in over pin dimension ∗ cos 90 − Φ = [0.25 ∗ cos(90 − 20)] = 0.085mm


Observe that, as the pressure angle Φ increases, the material loss also increases.

Example 5: find the maximum weight the mechanism could lift in the orientations shown in
[Fig 1.82], so that the cart of weight (W ) is not toppled. The orientation of the lifting mechanism
1
of weight (W ) is as follows:
2
Ø The cart is aligned parallel to the ‘x-axis’.
Ø The lifting mechanism makes an angle (θ) with the ‘x-axis’.
Ø The boom is in the plane of the lifting mechanism and makes an angle (Φ) with the
horizontal.
Ø The mechanism is firmly fixed to the cart.










QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 83

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







































[Fig 1.82: weight lifting mechanism on a four wheel cart]


Solution: it can be observed that, the external weight (mg) has a moment, which must be
countered by the moment due to the weight of the cart (W ) and lifting mechanism (W ).
2
1
The major assumptions to compute the moment due to external weight are as follows:

Ø The boom is rigid.
Ø The boom length sufficiently large.
Ø The external weight does not alter the CG of the cart system.
Ø The lifting mechanism is centrally placed on the cart. The CG of the cart and the lifting
mechanism are along the same vertical axis.
Let us use vectors to calculate the moment of the external load.

Moment due to external load = r × mg













QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 84

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


















[Fig 1.83: x, y and z coordinates of a boom]

Let us find the position vector r , which is given by:

x − x i + y − y j + z − z k

2
2
1
1
1
2
Treating the CG of the lifting mechanism to be the origin, the x , y , z and x , y , z
2
1
2
2
1
1
coordinates as shown in [Fig 1.83] are given by:
x , y , z = C cos θ , 0, −c sinθ − form .
1
1
1





[Fig 1.84: top view of boom’s bottom coordinates]


x , y , z = x + L cosΦ cosθ , y + L sin Φ , (z − L cosΦ sinθ − form .
1
2
2
2
1
1









QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 85

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















[Fig 1.85: boom’s top coordinates in a 3D space]

Therefore, the position vector r , is given by:


x − x i + y − y j + z − z k
1
2
1
1
2
2

r = L cos Φ cos θ i + L sinΦ j − L cosΦ sinθ k


Therefore, the moment of the external weight about the point x , y , z is:
1
1
1


r × mg = L cos Φ cos θ i + L sinΦ j − L cosΦ sin θ k × mg −j


= L cos Φ cos θ mg i × −j + L sin Φ mg j × −j + L cos Φ sin θ mg k × −j

Moment due to external weight = L cos Φ cos θ mg −k + L cosΦ sinθ mg i
These moments are fully reacted/transferred onto the cart, which must be balanced by the
moment due to self-weight of the cart and lifting mechanism. These are computed as follows:

W + W a
Reactive moment about the z − axis = 1 2 ∗
2 2
b
Reactive moment about the x − axis = W + W 2 ∗
1
2 2

Reactive moment about the y − axis is zero

On equating the respective moments, the value of (mg) could be computed.










QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 86