Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]
Ǯ-ǯǡ ȏ ʹǤ͵ͺȐǡ ǡ
ǣ
[Fig 2.38: an inclined rod accelerated about the 'z-axis']
1) dmαx = α dmx
2
2
2) dmαxz = α dmxz
If the angular acceleration were given about the ‘x-axis’, in the ‘zx-plane’, as shown in [Fig 2.39],
then, we get the inertia terms to be:
[Fig 2.39: an inclined rod accelerated about the 'x-axis']
2
2
1) dmαz = α dmz
2) dmαzx = α dmzx
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Ǯ-ǯǡ Ǯ-ǯǡ ȏ
ʹǤͶͲȐǡ ǡ ǣ
[Fig 2.40: an inclined rod accelerated about the 'y-axis']
2
2
1) dmαz = α dmz
2) dmαzy = α dmzy
If the angular acceleration were given about the ‘z-axis’, in the ‘yz-plane’, as shown in [Fig 2.41],
then, we get the inertia terms to be:
[Fig 2.41: an inclined rod accelerated about the 'z-axis']
2
2
1) dmαy = α dmy
2) dmαyz = α dmyz
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Axis of Rotation Plane of the Rod Inertia Terms
‘y-axis’ ‘xy-plane’ α dmx and α dmxy
2
‘x-axis’ ‘xy-plane’ α dmy and α dmyx
2
‘z-axis’ ‘zx-plane’ α dmx and α dmxz
2
‘x-axis’ ‘zx-plane’ α dmz and α dmzx
2
2
‘y-axis’ ‘yz-plane’ α dmz and α dmzy
‘z-axis’ ‘yz-plane’ α dmy and α dmyz
2
The inertia values tabulated above are derived algebraically. Hence, the signs are not given
emphasis, only the magnitudes are considered.
The important reckoning from the above exercise is, whenever a component is inclined to all the
three axes of a cartesian coordinate system and given acceleration about one of the axes, then, the
component sees inertia about the axis of rotation and cross product of inertia about the other two
axes.
Let us now derive the mass moment of inertia tensor for a general case using vectors, with the
component located in the cartesian space.
From Newton’s law, for linear motion, we have F = ma and for angular motion, we have
T = r × F .
The expression for torque (T) could be rewritten as:
T = r × ma = [r × m α × r - - - - (2.33)
For an elemental mass, equation (2.33) could be rewritten as: dT = r × dm α × r
= dT = dm r × α × r - - - - (2.34)
Therefore, we have a general expression for torque (T) in terms of angular acceleration α and
radius/position vector (r).
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Ǥ
ǣ
ǡ ȋȌ
ȋǡ ǡ Ȍ ȏ
ʹǤͶʹȐǤ
[Fig 2.42: an elemental mass in a 3D space given acceleration about all 3 axes]
Let us now use the matrix form of vector product to find α × r .
i j k
α × r = α x α y α , where r = xi + yj + zk
z
x y z
Expanding the matrix, we get:
α × r = i α z − α y − j α z − α x + k α y − α x
x
y
z
y
z
x
Substituting the above expression of α × r in equation (2.34), we get:
dT = dm r × i α z − α y − j α z − α x + k α y − α x
y
x
x
z
z
y
Rewriting the above vector product in the matrix form, we get:
i j
k
dT = dm x y z
α z − α y − α z − α x α y − α x
x
z
y
z
y
x
Expanding the above matrix, we get:
dm i y α y − α x − z α x − α z − j x α y − α x − z α z − α y
y
x
x
x
z
y
y
z
+ k x α x − α z − y α z − α y
z
y
x
z
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collecting the i , j , k terms from the above expression, we get:
dm i α y − α xy − α xz + α z - - - - (2.35)
2
2
z
y
x
x
dm j −α xy + α x + α z − α yz - - - - (2.36)
2
2
x
y
y
z
dm k α x − α zx − α zy + α y - - - - (2.37)
2
2
z
x
y
z
Casting the equations (2.35), (2.36) and (2.37) in the matrix form we have:
2
2
dT x dm y + z dm(−xy) dm(−xz) α x
2
2
dT = dm(−xy) dm(x + z ) dm(−yz) α
y
y
dT z dm(−zx) dm(−zy) dm x + y α z
2
2
Integrating the above expression on both sides, we have:
dT dm y + z dm(−xy) dm(−xz)
2
2
x α
2 2 x
y
dT = dm(−xy) dm(x + z ) dm(−yz) α
y
α z
2
2
dT z dm(−zx) dm(−zy) dm x + y
dm y + z dm(−xy) dm(−xz)
2
2
T α
x x
2
2
= T = dm(−xy) dm(x + z ) dm(−yz) α
y
y
T z α z
2
2
dm(−zx) dm(−zy) dm x + y
The inertia tensor for an object in a 3 dimensional space is given as:
dm y + z dm(−xy) dm(−xz)
2
2
I −I −I
xx I xy xz 2 2
yz
Inertia tensor = −I yx yy −I = dm(−xy) dm(x + z ) dm(−yz)
−I zx −I zy I zz
2
2
dm(−zx) dm(−zy) dm x + y
The inertia tensor has perfect symmetry about the principal diagonal. Therefore, it is a symmetric
matrix and has 3 Eigen values, meaning, 3 principal inertias.
The non-diagonal terms are called cross product of inertia. These terms come about either due to
the orientation as seen in the above discussion, or, due to asymmetry of the component.
For symmetric components, such as, discs, shafts, cubes, etc., the principal diagonal terms are the
principal inertia terms, as the non-diagonal terms are zero (attributed to symmetry).
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The principal inertias have maximum, minimum and intermediate values for inertia for a given
component. That is, if I , I and I are the maximum, intermediate and minimum values of
3
2
1
principal inertias respectively, then, I > I > I .
3
1
2
For spinning objects, the inertia directions I and I that is, the maximum and minimum inertia
1
3
directions, become the axes of stability.
Example 1: compute the 3 principal inertias for a cuboid as shown in [Fig 2.43] and write the
inertia tensor.
[Fig 2.43: 3 principal inertia axes of a cuboid]
Solution: The inertia about each axis can be found by the typical procedure followed earlier. Here,
the inertia about the principal axis is computed along with the inertia about the axis of each section.
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Let us begin with the computation of I .
xx
2
The MMOI contribution of (I element ) about the ‘x-axis’ is given by: L ∗ H ∗ dz ∗ ρ ∗ z , as shown
in [Fig 2.44].
[Fig 2.44: inertia of element about the 'x-axis' through the CG]
Now summing up the contribution of all the sections along the ‘z-axis’ we get:
W
2
2 3 2
I
xx,elem = L ∗ H ∗ dz ∗ ρ ∗ z = L ∗ H ∗ ρ - - - - (2.38)
3
− W −
2
2
Simplifying equation (2.38), we get:
W
z 3 2 W 3 W 3 W 3
L ∗ H ∗ ρ = L ∗ H ∗ ρ + = L ∗ H ∗ ρ
3 W 24 24 12
−
2
W 2 mW 2
= L ∗ H ∗ ρ ∗ W = - - - - (2.39)
12 12
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Let us now compute the MMOI contribution of all the elements about their own axis as shown in
[Fig 2.45].
[Fig 2.45: inertia of element about its own axis]
The contribution of a single element is computed as follows:
H
H
2
2
2 y 3
I
element = L ∗ dy ∗ dz ∗ ρ ∗ y = L ∗ dz ∗ ρ 3 H - - - - (2.40)
− −
2
H
2
Simplifying equation (2.40), we get:
H
y 3 L ∗ dz ∗ ρ H 3
2
L ∗ dz ∗ ρ =
3 H 12
−
2
Integrating the single element’s contribution across the width (W), as shown in [Fig 2.46], we get
the total MMOI contribution of all the elements about their own axis.
[Fig 2.46: MMOI contribution of elements across the width (W)]
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W
L ∗ dz ∗ ρ H 3 L ∗ W ∗ ρ H 3 L ∗ W ∗ H ∗ ρ H 2 mH 2
= = = - - - - (2.41)
12 12 12 12
0
The total MMOI about the ‘x-axis’ is given by the sum of equations (2.39) and (2.41):
I xx = Rotation of all elements about the x − axis
+ Rotation of all elements about their own axis
mW 2 mH 2
I xx = +
12 12
Similarly, the MMOI about the ‘y-axis’ is as shown:
[Fig 2.47: Inertia of element about the 'y-axis' through the CG]
mL 2
From . , we have: I yy,elem = - - - - (2.42)
12
[Fig 2.48: inertia of element about its own axis] [Fig 2.49: MMOI contribution of elements
across the length (L)]
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mW 2
From . & . , we have: I element = - - - - (2.43)
12
The total MMOI about the ‘x-axis’ is given by the sum of equations (2.42) and (2.43):
I yy = Rotation of all elements about the y − axis
+ Rotation of all elements about their own axis
mL 2 mW 2
I = +
yy
12 12
And, the MMOI about the ‘z-axis’ is as shown:
[Fig 2.50: inertia of element about the 'z-axis' through the CG]
mH 2
From . , we have: I zz,elem = - - - - (2.44)
12
[Fig 2.51: inertia of element about its own axis] [Fig 2.52: MMOI contribution of elements across
the height (H)]
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mL 2
From . & . , we have: I = - - - - (2.45)
element
12
The total MMOI about the ‘x-axis’ is given by the sum of equations (2.44) and (2.45):
I zz = Rotation of all elements about the z − axis
+ Rotation of all elements about their own axis
mH 2 mL 2
I = +
zz
12 12
ʹǤ
Ǥ
Ǥ
ǡ
Ǥ
ǡ
ǡ
Ǥ
ǡ
ǡ ǡ
ǡ
ǡ
Ǥ
ǡ
ǡ
Ǥ
Ǥ
ǡ
Ǥ ǡ
Ǥ
ȏ ʹǤͷ͵Ȑ
[Fig 2.53: Variation of friction force with applied force]
The friction force can be represented as, a product of normal contact force between the contact
surfaces and coefficient of friction, indicated by Greek symbol ‘μ’.
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Therefore, friction force is given by:
Friction force = μ ∗ N
For a static situation Friction force = μ static ∗ N
It is important to note that, the coefficient of friction is operative only to ensure static equilibrium.
In situations where the motion is impending, the static coefficient of friction is fully operative. For
dynamic situations, that is, when the motion is sustained, we have:
Friction force = μ dynamic ∗ N
However, from the discussions so far, we could conclude that the dynamic coefficient of friction is
lesser that the static coefficient of friction.
If the difference between static and dynamic friction is large, then, a phenomenon called “stick-slip”
friction is observed.
Example, the chattering motion of chalk against the board, viper blades against the windshield,
violin string against the bow, etc.
Friction is an important design parameter, be it, joints or surfaces in contact, say, piston and
cylinder or gear meshing or bearing friction. In all the aforementioned cases, friction must be
designed for optimum functionality.
Example, a typical rear wheel of an automobile is attached to the hub with bolts whose pretension
provides the normal force and the available friction coefficient ensures that the wheel does not slip
when accelerated (traction developed against the road surface).
Therefore, mathematically, number of bolts used to assemble the rear wheel to the hub is given by:
maximum traction
Number of bolts n =
preload ∗ μ static
Facts of Static and Dynamic Friction:
1) Friction force does not depend on the area of contact but depends only on the normal
contact force.
2) Friction force slightly reduces at high velocities. However, it remains constant for
moderate velocities.
3) Static friction is fully operative (maximum) only in situations where the motion is
impending.
4) For metals, friction increases with increase in temperature.
5) Friction is reduced via lubrication and high quality surface finish (low height of
irregularity).
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6) Certain metals are self-lubricating (example – cast iron), wherein, the friction is reduced
as the surface wears.
7) Friction coefficient for rubber components and joints are susceptible to variation and
hence must be carefully assumed.
8) Friction coefficient varies significantly in vibratory environments due to variation in
contact force.
9) Presence of oil in a bolted joint could lead to 30% variation in preload. Incidentally, about
80 to 90% of the preload provided to a joint is consumed by friction.
10) Super polishing of surfaces leads to an increase in friction coefficient
(example – slip gauges).
Example 1: and L-bracket, as shown in [Fig 2.54], is held against the wall with a normal force
‘N’. The available coefficient of friction is 0.2, the weight of the L-bracket is 10N. Compute the
value of ‘N’ for equilibrium. Assume the contact pressure is uniform across the contact surface.
[Fig 2.54: Bracket held against a wall]
Solution: for equilibrium of the bracket, two criteria have to be met:
1) Weight (W) of the bracket must be balanced by the friction force.
2) Moment due to weight (W) of the bracket must be balanced by moment due to
normal force (N).
W 20
Therefore, we have: W = μN = = N = = N = N(Normal Force) = 200N
μ 0.2
The moment contributed by the weight of the bracket is given by:
M weight = W ∗ 0.25 = 20 ∗ 0.25 = 5 Nm
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Moment due to normal force is given by:
M normal force = N ∗ 0.25 = 200 ∗ 0.25 = 50 Nm
It can be observed that the moment due to weight of the bracket is far less compared to moment
due to normal force. However, if it were the other way around, then, the bracket would have rotated
(pivoting at the corner) or the contact pressure with the wall would have varied linearly. In the
worst case, the bracket would have lost contact with the wall.
In this case, the bracket is in equilibrium and maintains uniform contact pressure with the wall.
Example 2: a block is given a push on a rough table, as shown in [Fig 2.55], with an initial
velocity of 20m/s. The friction coefficient between the block and the table is 0.5. Find the length of
2
the table to keep the block from falling off. Assume the acceleration due to gravity to be 10 m s .
[Fig 2.55: Block travelling on a table]
Solution: the deceleration of the block due to friction is given by:
Friction force μmg
2
deceleration (−a) = = = μg = 0.5 ∗ 10 = 5 N/m
mass of the block m
The distance travelled by the block is given by:
v = U + 2as = 0 = 20 + 2 −5 ∗ s = s (length of table) = 40m
2
2
2
Neglecting the dimensions of the block, the length of the table must be at least 40m to prevent the
block from falling off.
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͵ǣ ϐ ǡ ȏ ʹǤͷȐǡ ͳʹ
ǡ ͳͲͲ Ǥ ϐ ͲǤʹǤ
ϐ
ϐ ͲǤ͵Ǥ ǡ
ϐ
Ǥ
[Fig 2.56: typical flange with bolts]
Solution: the maximum torque that the joint could transfer is determined by the friction
coefficient between the flanges and the radius of the flanges.
In order to find the friction torque, let us assume that the contact pressure between the flanges is
uniform and is given by:
force
Contact pressure = , where force = preload ∗ number of bolts
area
Let us now evaluate the friction torque.
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Ǯǯ ȏ ʹǤͷȐǤ
ǡ ǣ
[Fig 2.57: flange dimensions for torque computation]
total force on the flange area F total
Pressure force = ∗ elemental area = ∗ 2πr ∗ dr
total area of the flange πR 2
The friction torque due to elemental area is given by:
F total
Friction torque = pressure force ∗ radius ∗ μ = ∗ 2πr ∗ dr ∗ r ∗ μ
πR 2
2F total
2
= Friction torque = ∗ r ∗ dr ∗ μ
R 2
The total friction torque the joint could support is got by integrating the friction torque due to
elemental are across the radius (R) of the flange.
R R
2F total 2F total μ
2
2
Friction torque total = ∗ r ∗ dr ∗ μ = r ∗ dr
R 2 R 2
0 0
2F total μ r 3 R 2F total μR
= Friction torque = =
total
R 2 3 3
0
Equating the total friction torque with the total torque (100 Nm) to be supported, we get:
2F μR
Friction torque total = Torque to be transmitted = total = 100
3
300 300
Therefore, we have: F total = = = 2500N
2μR 2 ∗ 0.3 ∗ 0.2
2500
Since there are 12 bolts, each bolt should carry a clamping or preload of = 208.33N
12
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ʹǤǤͳ
ǡ ǡ
Ǥ
ǡ
ǡ ǡ
Ǥ
Ǥ
ǡ ȏ ʹǤͷͺȐ
ǣ
[Fig 2.58: pure rolling]
1) Since the cylinder is rigid, all the points on the cylinder have the same velocity as that of
V CM .
2) The angular velocity ω contributes to the second component, which is ω × r , where ‘r ’
is the position vector and is always directed from the center of rotation to the point of
interest.
Hence, the velocity of the instantaneous center of rotation is given by:
V = V CM + ω × r = [V CM i + (ω(−k) × r −j = [V CM i + ω −k × r −j = V CM − ωr i
Ic
Since the velocity at the instantaneous center is zero, as have: V CM − ωr i = 0 = V CM = ωr
In an ideal rolling situation, where ω and (V) are kept constant, the static friction at the point of
contact is zero.
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ǡ ȏ ʹǤͷͻȐ
ǣ
[Fig 2.59: rolling of rigid body]
The velocity at ‘B’ perpendicular to ‘OB’ is given by: [ω ∗ OB] = ω ∗ 2r cos θ
This could also be proved by the law of parallelogram of vectors, given by:
2
2
2
2
Resultant = V cm + ωr + 2 ωr ∗ V cm ∗ cos 2θ = ωr + ωr + 2 ωr ∗ ωr ∗ cos 2θ
Resultant = ωr ∗ 2 + 2 cos 2θ = ωr ∗ 2(1 + cos θ − sin θ = ωr ∗ 4 cos θ
2
2
2
Therefore, we have: Resultant = ωr ∗ 2 cosθ
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ʹǤǤʹ
Ǥ
ǡ
Ǥ ǡ
ȋ Ȍǡ
Ǥ ǡ ȏ ʹǤͲȐǡ
Ǥ
[Fig 2.60: rolling of deformable body]
The torque produced is given by:
[T = (W x d)] Resisting torque
Unlike rigid bodies, for deformable bodies, the reaction force from the ground and the CG do not
coincide, producing a torque.
It should be noted that, the ground may be flexible, and a small pit could form due to the weight (W)
of the rolling element. A torque is needed to pull the rolling element out of the pit.
Therefore, it is more appropriate to call this effect rolling resistance rather than rolling friction.
ʹǤǤ͵
ǡ
Ǥ ǡ
ǡ
ǡ
ǡ
Ǥ
ǡ
Ǥ
Ȁ
Ǥ
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ǡ ȏ ʹǤͳȐǤ
Ǥ
ϐ
Ǥ
[Fig 2.61: cylinder rolling down an inclined plane]
The torque required to provide angular inertia for the cylinder is: T = Iα
From [Fig 2.62], we could observe the following:
[Fig 2.62: maximum velocity (v) and acceleration (a) of a rolling cylinder]
Ø The point of contact of cylinder with the inclined surface is the instantaneous center of
rotation and has to be characterized by zero velocity and zero acceleration.
Ø Any other point away from the instantaneous center of rotation will have velocity
proportional to ‘ω’ and acceleration proportional to ‘α’.
Ø The highest point will have twice the velocity and acceleration than that of the center of
mass.
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A rolling member has both linear and angular degrees of freedom.
We know that: T = Iα , where [T = torque], [I = Mass moment of inertia] and
[α = angular acceleration]
The Newton’s second law, since friction provides angular inertia, we have:
T = Iα = μ mg cos θ r = Iα ,
a
We know that α in terms of its linear accelerations is
r
a
Therefore μ mg cos θ r = I - - - - (2.46)
r
Equation (2.46) gives the friction torque about the CM.
Simplifying equation (2.46), we get:
μ mg cos θ r 2
2
μ mg cos θ r = I a , or a = - - - - (2.47)
I
Acceleration (a) can be found by balancing the horizontal forces:
ma + μ mg cosθ = mg sinθ or a = g sin θ − μ g cos θ
Substituting the value of ‘a’ in equation (2.47), we get:
μ mg cos θ r 2 μ mg cos θ r 2
g sin θ − μ g cos θ = = g sin θ = + μ g cos θ
I I
μ mg cos θ r 2 mr 2
g sin θ = μg + cos θ = g sin θ = μg cos θ + 1 - - - - (2.48)
I I
Simplifying equation (2.48), we get minimum friction coefficient required to prevent the cylinder
from slipping.
tanθ
μ = mr 2
+ 1
I
It is to be noted that, the friction provides angular inertia for the cylinder. Therefore, the friction
torque must be greater than or equal to (Iα).
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tanθ
condition for perfect rolling − μ ≥
mr 2
I + 1
tanθ
condition for slipping − μ <
mr 2
I + 1
Example 1: find the time taken by a solid cylinder to roll down an inclined plane if it is released
with zero initial velocity.
[Fig 2.63: cylinder rolling down an inclined plane]
Solution: considering the conservation of energy, as there is no loss of energy.
Potential Energy = Transational Kinetic Energy + Rotational Kinetic Energy
1 1 1 1 mr 2 v 2 3
2
2
2
2
mgH = mv + Iω = mv + ∗ ∗ = mv
2 2 2 2 2 r 2 4
Further simplifying the above expression, we get:
4 4gH
2
v = gH = v = - - - - (2.50)
3 3
The acceleration of the cylinder could be computed using equations (2.47) and (2.49), we have:
μ mg cos θ r 2
a = - - - - (2.47)
I
tanθ
μ = 2 - - - - (2.49)
mr + 1
I
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Substituting the value of ‘μ’ from equation (2.49) in equation (2.47), we get:
tanθ 2
mr 2 mg cos θ ∗ r
I + 1
g sin θ mr 2 1 g sin θ mr 2 1
a = I = mr 2 ∗ = I ∗ mr 2
I
I + 1 I + 1
mr 2
To simplify the above expression, let us assume that, = x . Therefore, we have:
I
1
a = g sinθ ∗ x ∗
x + 1
Let us simplify the above expression by dividing the numerator and denominator by ‘x’.
1 g sin θ g sinθ g sin θ 2 g sinθ - - - - (2.51)
a = g sin θ 1 = I = mr 2 = 1 = 3
1 + 1 + mr 2 2 1 + 2
x
1 + 2
mr
To compute the time of travel, let us use the expression [v = U+at], where, the initial velocity ‘U’ is
zero.
Substituting the value of ‘v’ and ‘a’ in the expression [v = U+at], we get:
4gH 2 g sinθ 4gH 3 4gH 9
= 0 + ∗ t = t = ∗ = t = ∗
2
2
3 3 3 2 g sinθ 3 4 g sin θ
3H
Therefore, we have the time of travel to be: t =
2
g sin θ
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ʹǣ ϐ ǡ
ǡ ȏ ʹǤͶȐǤ
[Fig 2.64 : Hoop rolling down an inclined plane]
Solution: for the hoop rolling down the inclined plane, the acceleration is given by:
g sin θ g sinθ g sin θ
a = I = mr 2 2
1 + 2 1 + 2
mr mr
considering the conservation of energy, as there is no loss of energy.
Potential Energy = Transational Kinetic Energy + Rotational Kinetic Energy
1 1 1 1 v 2
2
2
2
2
2
mgH = mv + Iω = mv + ∗ mr ∗ = mv
2 2 2 2 r 2
Further simplifying the above expression, we get:
v = gH = v = gH
2
To compute the time of travel, let us use the expression [v = U+at], where, the initial velocity ‘U’ is
zero.
Substituting the value of ‘v’ and ‘a’ in the expression [v = U+at], we get:
g sin θ 2 4
gH = 0 + ∗ t = t = gH ∗ = t = gH ∗
2
2
2 g sin θ g sin θ
H
Therefore, we have the time of travel to be: t = 2 - - - - (2.52)
2
g sin θ
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For a hoop sliding down the inclined plane, the acceleration is given by:
g sin θ
a =
1 + I
mr 2
As for sliding, there is no rotational inertia, the mass moment of inertia (I) is zero. Therefore, the
acceleration for a hoop sliding down an inclined plane is given by:
g sinθ
a = = g sin θ
1 + 0
considering the conservation of energy, as there is no loss of energy.
Potential Energy = Transational Kinetic Energy + Rotational Kinetic Energy
1 1 1 1 v 2 1
2
2
2
2
mgH = mv + Iω = mv + ∗ 0 ∗ = mv
2 2 2 2 r 2 2
Further simplifying the above expression, we get:
1
v = gH = v = 2gH
2
2
To compute the time of travel, let us use the expression [v = U+at], where, the initial velocity ‘U’ is
zero.
Substituting the value of ‘v’ and ‘a’ in the expression [v = U+at], we get:
1 2H
2gH = 0 + g sinθ ∗ t = t = 2gH ∗ = t = - - - - (2.53)
2
g sinθ g sin θ
Dividing equation (2.53) by equation (2.52), gives us the ratio of time taken by the hoop slide and
roll down the inclined plane. Therefore, we have:
2 H
2
t sliding g sin θ 2 H 2
t = 2H = 2 H = 1 - - - - (2.54)
rolling 2
g sin θ
It can be seen from equation (2.54) that, the sliding time is [ 2 = 1.414] times more than that of
rolling time.
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͵ǣ ǡ Ǥ
ǣ
ǡ ǡ Ǥ
ǡ
ǡ
Ǥ
ϐ
ǡ ȏ ʹǤͷȐǤ
[Fig 2.65 : a composite cylinder rolling down an inclined plane]
The acceleration term that determines the time of travel is given as follows:
g sin θ
a =
1 + I
mr 2
I
The term , is independent of mass for a homogenous cylinder.
mr 2
Let us verify, if the same holds for a composite cylinder.
The mass moment of inertia of the composite cylinder is given as follows:
2
m r 2 ρ πr ∗ L ∗ r 2
c
c
2
2
I composite = I cylinder + I hoop = + m r = + ρ 2πr ∗ t ∗ L ∗ r
h
h
2 2
Now that we have found ‘I’ for the composite cylinder, let us compute ‘mr ’.
2
2
2
2
mr 2 composite = ρ πr ∗ L ∗ r + ρ 2πr ∗ t ∗ L ∗ r
c
h
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Taking ratio of ‘I composite ’ and ‘mr 2 composite ’, we get:
4
2
ρ πr ∗ L ∗ r 2 + ρ 2πr ∗ t ∗ L ∗ r ρ πr ∗ L + ρ 2πr ∗ t ∗ L
3
c
c
2
I composite 2 h 2 h
mr 2 = ρ πr ∗ L ∗ r + ρ 2πr ∗ t ∗ L ∗ r = ρ πr ∗ L + ρ 2πr ∗ t ∗ L
3
2
2
4
2
composite c h c h
4
Dividing the numerator and denominator by the common terms (r πL), we get:
I ρ c + 2ρ ∗ t
h
composite 2 r - - - - (2.55)
mr 2 composite ρ + 2ρ ∗ t
h
c r
I
It can be observed that, is a function of the densities of materials in the composite cylinder.
mr 2
Therefore, the acceleration ‘a’ is also a function of densities and so is the time of travel ‘t’.
As long as the hoop thickness remains small, the time of travel does not alter significantly. However,
as the thickness increases, it significantly changes the time of travel.
2.8 Curvilinear Motion
In a curve, every parameter changes from point to point, such as:
ü Radius of curvature
ü Direction of path
ü Speed of travel
[Fig 2.66: Curvilinear motion]
ǡ ǡ ϐ
ǡ
ǡ ǡ
ǡ
ǡ
ǡ
Ǥ
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However, a fourth acceleration is produced due the radial and angular velocities. This acceleration is
called the “Coriolis” acceleration. The Coriolis acceleration is responsible for several natural and
mechanical phenomena. Example, cyclones, missiles missing the target, ocean currents, turbulence
in flows, mechanisms (quick return motion mechanism) etc.
Therefore, finding the acceleration becomes a challenge, these accelerations could be:
ü Radial acceleration d r dt due to change of radius with respect to time.
2
2
ü Tangential acceleration due to rate of change of angular velocity dω dt .
2
ü The centripetal acceleration (V r) is variable, as both the radii and velocity change from
point to point.
ü Since there is a rate of change of radius and the radius itself is rotating (as there is angular
motion), the Coriolis 2Vω component comes into play.
Let us now derive the four accelerations for curvilinear motion.
Consider a point on the curvilinear motion path, which has an instantaneous radius vector r , as
shown in [Fig 2.67].
[Fig 2.67: Curvilinear motion of a particle]
Mathematically, the radius vector is given by: r = rr .
The first step is to find the rate of change of the radius vector along the curvilinear path. This is
given by, the derivative of the radius vector with respect to time:
dr dr dr
= r + r - - - - (2.56)
dt dt dt
The unit radius vector r is rotating at a certain angular rate dθ dt . Therefore, the rate of
change of the unit radius vector is represented as follows.
dr dr dθ
= - - - - (2.57)
dt dθ dt
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Substituting equation (2.57) in equation (2.56), we get:
dr dr dr dθ
= r + r
dt dt dθ dt
From the discussion on circular motion, remember that the radius unit vector r when
differentiated with respect to θ , we get unit tangent vector T . Also, we know that (dθ dt) is the
instantaneous angular velocity (ω). Substituting these facts in the above expression, we get:
dr dr
= r + Tω r - - - - (2.58)
dt dt
Since we seek accelerations, we differentiate equation (2.58) again:
dr
Let us consider the first term in equation . , which is : r
dt
Differentiating the above term by chain rule, we get:
2
d dr d r dr dr
r = r +
dt dt dt 2 dt dt
Recalling the derivative of unit radius vector, which is dr dt = Tω and substituting this, in the
above expression, we get:
2
d dr d r dr
Tω
r = r + - - - - (2.59)
dt dt dt 2 dt
Let us now consider the second term in equation . , that is: Tω r
We shall differentiate the above term using chain rule of differentiation. However, the chain rule
must applied once again, to (ωr). Mathematically, we have,
d dT d ωr dT
Tω r = ωr + T , from circular motion, we have: = −r ω
dt dt dt dt
d dω dr dω dr
2
= Tω r = −r ω ωr + r + ω T = −ω rr + r T + ω T - - - - (2.60)
dt dt dt dt dt
Adding equations (2.59) and (2.60), we get the rate of, rate of change of the radius vector, which is:
2
2
d r d r dr dω dr
2
= r + ωT − ω rr + r T + ω T - - - - (2.61)
dt 2 dt 2 dt dt dt
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Rearranging the like terms in equation (2.61), we get:
2
2
d r d r dr dω
2
= r + 2 ωT − ω rr + rT
dt 2 dt 2 dt dt
The above expression can also be written to get the four accelerations as:
2
2
2
d r d r dr dθ dθ 2 d θ
= r + 2 rr + rT
T −
dt 2 dt 2 dt dt dt dt 2
In summary the accelerations for curvilinear motion is tabulated as follows:
Accelerations In the form of derivatives In the form of vectors
Radial acceleration d r d r
2
2
dt 2 r dt 2 r
Coriolis dr dθ 2 ω × V
2 T
dt dt
Centripetal acceleration dθ 2 ω × ω × r
rr
dt
2
Tangential acceleration d θ
dω
rT × r
dt 2 dt
Example 1: find the ratio of maximum to minimum centripetal acceleration for a particle
negotiating an elliptical curve at constant velocity, as shown in [Fig 2.68].
[Fig 2.68: Point traversing an elliptical path]
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Solution: for a general curve the radius of curvature varies from point to point, as shown in
[Fig 2.66] and is attributed to the fact that, both slope and the rate of change of slope vary form
point to point.
For the ellipse, inspection reveals that, points (A, B, C, D) are the points of maximum and minimum
radii of curvature. The centripetal acceleration depends on the square of the speed and the radius
of curvature, and hence, can assume extreme values at (A, B, C, D), given that, the speed is constant.
Let us evaluate the maximum and minimum curvature/radius and hence the centripetal
acceleration ratio.
In order to find the maximum and minimum radius of curvature, let us derive a general equation
by parameterizing the equation of the ellipse. The equation of the ellipse is given by:
x 2 y 2
+ = 1
a 2 b 2
The ‘x’ and ‘y’ coordinates can be parametrically rewritten as:
x = a cos θ and y = b sinθ , as shown in .
2
2
In order to determine the radius of curvature, we need (dy/dx) and d y dx , which in turn have
to be substituted in the equation for radius of curvature, given by:
3
dy 2 2
1 + dx
R = - - - - (2.62)
2
d y
dx 2
[Fig 2.69: parametric coordinates of the ellipse]
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From the equations x = a cosθ and y = b sinθ , we have:
dx dy
= −a sinθ and = b cos θ - - - - (2.63)
dθ dθ
dy b cos θ b
Therefore, we have: = − = − cot θ
dx a sin θ a
2
d y b dθ b 1 1 b
2
And = − cosec θ ∗ = − − = - - - - (2.64)
3
2
2
dx 2 a dx a sin θ a sinθ a sin θ
Substituting equations (2.63) and (2.64) in equation (2.62) we get:
3
2 2
b
1 + − cot θ 3 3
2
3
2
3
2
2
2
2
2
2
a a sin θ b cos θ 2 a sin θ a sin θ + b cos θ 2
R = b = b 1 + a sin θ = b ∗ 3
2
2
2
2
a sin θ a cos θ 2
2
3
Simplifying the above expression further, we have:
3 3
2
2
3
2
2
2
2
2
2
2
a sin θ a sin θ + b cos θ 2 a sin θ + b cos θ 2
R = ∗ = - - - - (2.65)
3
3
b a cos θ ab
Equation (2.65) gives us the radius of curvature at any point on the ellipse and the radius of
curvature at points (A, B, C and D) is tabulated as follows:
Points on the Ellipse Angle Radius of Curvature (R)
a 3 a 2
A 90 =
0
ab b
b 3 b 2
0
B 0 =
ab a
a 3 a 2
C 270 =
0
ab b
b 3 b 2
D 180 =
0
ab a
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Since we know that, ‘a’ is greater than ‘b’ (‘a’ being the major axis), the radius of curvature is
maximum at ‘B’ and ‘D’ and minimum at ‘A’ and ‘C’.
Hence, the ratio of minimum and maximum centripetal acceleration given by:
v 2 b 2
Centripetal acceleration minimum R R max a b 2 b b 3
= min = = = ∗ =
2
Centripetal acceleration maximum v 2 R min a a 2 a 3
a
R max b
Example 2: a cylinder when at the top of a hemisphere, is given a small perturbance, to which,
the cylinder starts rolling down. Compute the angle ‘θ’ at which the cylinder loses contact with the
hemisphere. Assume that, friction is absent, and the hemisphere is firmly fixed to the ground.
[Fig 2.70: cylinder rolling down a hemisphere]
Solution: as the cylinder rolls down, the kinetic energy increases and so does the centrifugal
force, as shown in [Fig 2.71]. At one point, the contact is lost, where, the weight component of the
cylinder, which provides the contact reaction is reduced to zero.
[Fig 2.71: cylinder rolling down a hemisphere]
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ȏ ʹǤʹȐ
[Fig 2.72: vertical displacement of the CG of the cylinder]
The potential energy lost is given by:
Potential Energy = Transational Kinetic Energy + Rotational Kinetic Energy
1 1 1 1 mr 2 3
2
2
2 2
2
2 2
mg R + r 1 − cos θ = mv + Iω = m ω r + ∗ ∗ ω = mω r
2 2 2 2 2 4
4g R + r 1 − cos θ
2
By where, ω =
3r 2
Since mg cos θ is just balanced by the centrifugal force, as shown in [Fig 2.71], we have:
4g R + r 1 − cos θ
mg cos θ = ∗ m(R + r)
3r 2
2
4g R + r 1 − cos θ
By where, we get cos θ = - - - - (2.66)
3r 2
4g R + r 2
For ease of simplification, let as assume = k , and rewritting equation . , we get:
3r 2
cosθ = k 1 − cos θ = k − k cos θ
k k
Upon further simplification, we get: cosθ = , therefore, θ = cos −1
1 + k 1 + k
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͵ǣ ǮɅǯǡ ȏ ʹǤ͵ȐǤ
Ǥ
[Fig 2.73: train travelling down south at a latitude 'θ']
Solution: the Coriolis force is given by:
Coriolis force = 2m v × ω (this expression is for reactive Coriolis force, which is the exact
opposite to active Coriolis force [2m ω × v ]. This is similar to the relation between centrifugal
and centripetal acceleration)
The ‘x-component’ of the velocity is perpendicular to the rotation vector of the earth. Therefore,
the Coriolis force experienced by the train is given by:
[(Coriolis force) train = 2m vsin θ i × ω j = 2m v sinθ ω k
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Ͷǣ
ǡ ȏ ʹǤͶȐǡ
Ǯɘǯ
Ǥ
ǮǯǤ ǡ
Ǥ Ǯǯ
Ǯǯ
Ǥ
[Fig 2.74: toy car travelling on a turn table]
Solution: the toy car experiences three major force components, as shown in [Fig 2.75]:
1) Gravitational force
2) Coriolis force
3) Centrifugal force
[Fig 2.75: forces acting on CG of the toy car]
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mg
Each wheel sees a reaction due to self weight gravitational force .
4
The centrifugal force causes an additional load on the front wheels and reduces the pressure on
the rear wheels. This is because, the centrifugal force, through the CG of the toy car, causes a
couple in the clockwise direction about the ‘z-axis’. This couple is balanced by the couple due to
reaction on the wheels, as shown in [Fig 2.76]. Mathematically we have:
2
mω r CG ∗ h CG
2
R ∗ b = mω r ∗ h , therefore, we have: R =
CG
CG
b
This reaction adds on to the gravity component of the front wheel and is subtracted from that of
the rear wheel.
The net reaction is therefore given by:
2
mg mω r ∗ h CG
CG
Net reaction on each of the rear wheels = −
4 2b
2
mg mω r CG ∗ h CG
Net reaction on each of the front wheels = +
4 2b
[Fig 2.76: forces acting on CG of the toy car]
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The Coriolis force is given by: 2m v × ω = [2m v i × ω j = 2mvω k
The Coriolis force leads to another couple about the positive ‘x-axis’, where, the reactions on the
inner wheels are increased and that of the outer wheels are decreases. Mathematically, from
[Fig 2.77], we have:
[Fig 2.77: Force on the CG of the toy car]
Reaction on (wheel 1) is given by:
2
mg mω r ∗ h 2mvω ∗ h
= − CG CG + CG
4 2b 2a
Reaction on (wheel 2) is given by:
2
mg mω r ∗ h CG 2mvω ∗ h CG
CG
= + +
4 2b 2a
Reaction on (wheel 3) is given by:
2
mg mω r ∗ h 2mvω ∗ h
= + CG CG − CG
4 2b 2a
Reaction on (wheel 4) is given by:
2
mg mω r ∗ h 2mvω ∗ h
= − CG CG − CG
4 2b 2a
It can be observed form the wheel reactions that the (wheel 2) is the worst loaded and (wheel 4) is
the least loaded.
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2.9 Gyroscopic Couple
Let us consider a pseudo effect, which is generally called the gyroscopic effect. This effect is brought
about, whenever the direction of the angular momentum changes.
The gyroscopic effect is present in several situations, be it:
Ø Over hung rotor system
Ø Solid axels of a car
Ø Turbo charger system, and more.
Let us understand mathematically and physically the consequences of the gyroscopic effect.
Case 1: Consider a rigid shaft-disc system with a massless shaft and an overhung disc mounted onto
a rotating platform, as shown in [Fig 2.78]. The shaft is given and angular velocity ω , about the
positive ‘x-axis’. The system undergoes an angular precision Ω about the positive ‘y-axis’. Let the
polar mass moment of inertia be (I ) and the diametral mass moment of inertia be (I ).
p
d
[Fig 2.78: shaft disc system spinning and precising about 2 perpendicular axes]
The total angular momentum [Fig 2.79] of the system is given by: I ω i + I Ω j
p
d
[Fig 2.79: direction of spin and precision of the system]
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The angular momentum along the ‘y-axis’ is much smaller than the angular momentum along the
‘x-axis’. Therefore, we get:
Net angular momentum = I ω i + I Ω j
p
d
= I ω i (neglecting I assuming the disc to be thin)
p
d
Differentiating the net angular momentum, we get:
di
I ω , where i is the radius vector.
p
dt
Since the tangent vector is along the negative ‘z − axis’, we get:
di dθ
I ω = I ω −k Ω = −I ωΩ k - - - - (2.67)
p
p
p
dθ dt
Equation (2.67) can be vectorially represented as follows:
−I ωΩ k = Ω j × I ω
p
p
Since the gyroscopic torque is an inertia torque, the reaction must be considered on the shaft, which
is along the positive ‘z-axis’ k .
This means that, the gyroscopic torque goes to balance the gravity torque on the shaft, as shown in
[Fig 2.80], which is along the negative ‘z-axis’ −k .
[Fig 2.80: Forces on the disc shaft system]
Vectorially, we have:
Gravity torque = r × mg = r i × mg −j = −r mg k
For equilibrium, the bearing also experiences a vertical upward reaction equal to ‘mg’
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ʹǣ
ǮɅǯ
ǡ Ǯɘǯ ǡ ȏ ʹǤͺͳȐǤ
[Fig 2.80: motion of symmetric spinning top]
Let us represent the angular velocity of the system vectorially. Splitting the spin component of
angular velocity ‘Ω’ along the ‘x’ and ‘y’ directions and adding the vertical and horizontal
components to get the total angular velocity of the system, we have:
Ω cosθ + ω + Ω sin θ = Ω cosθ + ω j + Ω sin θ i
In order to get the angular acceleration, we differentiate the above expression with respect to time.
Also note that, the vertical axis is stationary and hence only the second term has to be
differentiated.
di dθ
= Ω sin θ ∙ = Ω sin θ kω - - - - (2.68)
dθ dt
Note that, differentiating the i unit vector with respect to ‘θ’, we get the tangent unit vector,
th
which is k in our case (Refer to the discussion on vector differentiation).
Multiplying equation (2.68) with the polar mass moment of inertia of the top, we get:
I Ω sin θ kω
p
This is the gyroscopic torque that prevents the top from falling off against the gravity torque.
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Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics
2.10 Free Body Diagram
The concept of FBD is fundamental to finding loads experienced by components in an assembly
and also their load path.
In this chapter, examples that are static and dynamic are considered. To illustrate the fundamentals,
let us begin with the concept of equilibrium.
Ø Equilibrium could be static or dynamic, wherein, the sum all forces and moments about
any point is zero.
Ø In static equilibrium, the component is at rest and may suffer elastic or plastic deformation.
Ø In dynamic equilibrium, the forces and moments are balanced in such a way that the body
has constant dynamic characteristics.
Ø In this chapter, all the bodies are considered to be absolutely rigid.
Ø The elastic deformations is sometimes represented by a linear or a torsional spring.
Example 1: find the instantaneous tension in the rope and acceleration of the bob.
[Fig 2.82: Simple pendulum]
Solution: let us write the free body diagram for the bob, as shown in [Fig 2.83]
[Fig 2.83: FBD of bob accelerating towards the mean position]
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Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]
It is to be observed that, the position of the bob in [Fig 2.83] is instantaneous and the tangential
velocity ‘v’ does not remain constant, rather, it is an accelerating motion.
Further, though the bob is moving away from the mean position, the acceleration is always directed
towards the mean position. This could be explained as follows:
The equation of SHM for displacement at any point is given by: y = a sin ωt
[Fig 2.84: Graph of displacement vs time]
dy y 2
2
2
2
The velocity is given by: v = = aω cosωt = aω 1 − sin ωt = aω 1 − = ω a − y
dt 2
[Fig 2.85: Graph of velocity vs time]
2
d y
2
2
And the acceleration is given by: a = = −aω sinωt = −ω y
dt 2
[Fig 2.86: Graph of Acceleration vs time]
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Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics
Resolving all the forces along the rope, we have:
v 2
T = m + mgcosθ
r
Resolving all the forces perpendicular to the rope, we have:
mgsinθ = ma ,therefore,we get: a = gsinθ
Here, the acceleration is positive as we seek only the magnitude and is directed towards the mean
position.
It can be observed that, the tension is maximum at the mean position (where ‘θ’ is zero).
2.10.1 General forces and their directions
Ø Self-weight (mg) always acts in the down ward direction. If the weight is oriented in a
plane, it would have two components. In some situations, the platform of a machine could
have a spatial orientations leading to three components of weight.
Ø Inertia forces (inertia (ma), centrifugal force mω r , Coriolis force 2mvω and
2
gyroscopic torque I ωΩ . All these agencies have a reaction or effect in the direction
p
opposite to that of acceleration or the vectorial direction derived from a standard formula.
Example: the centripetal force is given by m ω × ω × r (action), whereas, the
centrifugal force (reaction) has a direction negative to that of centripetal force.
The centrifugal force can be observed only in a rotating frame of reference and not in a
stationary frame of reference. However, since we reduce our dynamic situations to pseudo
static, we represent pseudo force, with their real effect, in the FBD.
Ø Contact forces: any contact is characterized by two forces, namely, tangential and normal.
For non-deformable bodies in contact, the tangential force is the friction force and normal
force is the contact pressure.
Ø Hinge joints: these joints cannot take any moments but develop vertical and horizontal
reactions.
Ø Joints: bolted and welded joints are generally treated as fixed. In very few situations, the
bolted joints can behave as sliding joints (when friction between the flanges is less that the
applied load).
Ø Shock loads are generally modified using shock factors and applied as equivalent static
loads.
Ø Gear loads: except spur gear, helical and bevel gears have 3 components of loads, namely
tangential, radial and axial
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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]
Case Study: consider a shaft-disc system, as shown in [Fig 2.87]. The shaft is given an angular
velocity ‘Ω’, about the ‘x-axis’. The shaft-disc system is a part of an assembly which precesses about
the ‘y-axis’. Assume the inertia of the disc-shaft system to be ’I’ about the axis of spin and the shaft
and disc to be rigid. The platform remains perfectly horizontal. Find the vertical bearing reactions.
[Fig 2.87: a rotating shaft disc system on a precising platform]
Solution: the loads considered are as follows:
1) Gyroscopic moment
2) Weight of the shaft (Mg) and that of the disc (mg)
Let us start by writing the FBD of the shaft, as shown in [Fig 2.88].
[Fig 2.88: FBD of the disc mass system without the gyro effect]
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Copyright Diary No – 9119/2018-CO/L
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Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics
Let us find the reactions at the supports by taking moments about the bearing support 1.
L d Mg
R ∗ L = Mg + + mg L + d = R ∗ L = L + d + mg L + d
2
2 2 2 2
Mg d Mg
= R ∗ L = L + d + mg = R = 1 + + mg - - - - (2.69)
2
2 2 L 2
From vertical force balance we have:
d Mg
R = Mg + mg − R = R = Mg + mg − 1 + + mg - - - - (2.70)
1
1
2
L 2
Since the disc-shaft system is spinning and precessing, it experiences a torque about the ‘z-axis’,
called the gyroscopic torque and is given by:
Gyroscopic torque = ω j × Ω i I = ωΩI −k
Since gyroscopic moment is an inertial moment, the reactive moment is considered, which is:
Gyroscopic torque = ωΩI k - - - - (2.71)
This torque induces reactions, as shown in [Fig 2.89]
[Fig 2.89: FBD of the disc mass system with the gyro effect]
The disc-shaft system, being rigid, exert equal and opposite reactions on the bearing. Therefore, the
couple formed by the bearing reaction would go to balance the gyro torque. Therefore, we have:
ωΩI
′
′
′
R ∗ L = ωΩI = R = = −R - - - - (2.70)
1
2
1
L
Thereby, the effective reactions for the disc-shaft system is got as follows:
1) Effective reaction at left support = R = R + R
′
1
1
L
′
2) Effective reaction at right support = R = R − R
2
R
2
Page 96 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]
d Mg ωΩI
Therefore, R = Mg + mg − 1 + + mg +
L
L 2 L
d Mg ωΩI
And R = 1 + + mg −
R
L 2 L
This problem is designed based on the design situations such as:
Ø Solid rotor of a car
Ø Overhung shaft of a turbo charger
Ø Overhung disc-shaft system of a gas turbine
Ø Overhung centrifugal compressor of a helicopter, and more…
All these systems experience gyro couple, which not only affects aerodynamics, system dynamics,
vibration, but also the loads on the bearing.
2.10.2 Engineering Facts of FBD
Ø FBD helps reduce multi component analyses to a single component analysis.
Ø FBD helps compute load path and attenuation, as each component in the system could be
singled out with force interactions.
Ø FBD helps estimate friction loads, which is difficult to compute, as the friction coefficient’s
have significant variability.
Ø FBD helps in simulating components with equivalent force forces, as, it is difficult to model
inertial loads. For simulation of stress, the above case study could be idealized to a beam
with simple supports, subjected to self-weight and gyro moment, reducing significant time
and effort.
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Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
Mathematics for
Structural Mechanics
Contents
Introduction ..................................................................................................... 2
Applied calculus ............................................................................................... 4
Vector algebra .................................................................................................. 23
Vector product ................................................................................................. 36
Identifying right handed coordinates ............................................................ 42
Vector differentiation ...................................................................................... 44
Transformation ................................................................................................ 47
Eigen values and Eigen vectors ..................................................................... 56
Engineering curves .......................................................................................... 58
Concept of Gradient ....................................................................................... 62
Lagrangian Multipliers..................................................................................... 63
Concept of Optimization ............................................................................... 65
Fourier Series and Transforms ...................................................................... 70
Routine Engineering hand Calculations ....................................................... 74
QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, Page 1
Fundamentals of Stress and Vibration
[A Practical guide for aspiring Designers / Analysts] 1. Mathematics for Structural mechanics
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QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,
Page 2
Fundamentals of Stress and Vibration 1. Mathematics for Structural mechanics
[A Practical guide for aspiring Designers / Analysts]
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