Download Example (1) on Journal Bearing

Example (1) on Journal Bearing
ME101 - Division III
Kaustubh Dasgupta
1
Example (1) on Journal Bearing
• Impending motion
ME101 - Division III
Kaustubh Dasgupta
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Example (1) on Journal Bearing
ME101 - Division III
Kaustubh Dasgupta
3
Example (2) on Journal Bearing
(a) For equal tension on both sides, contact point
is A; for slight rotation of the pulley, under
increased P, the contact point shifts to B.
Friction circle radius,
ME101 - Division III
Kaustubh Dasgupta
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Example (2) on Journal Bearing
(b) With reduction of P, contact point shifts to C.
Free body diagram of pulley with moment @ C,
ME101 - Division III
Kaustubh Dasgupta
5
Example (2) on Journal Bearing
(c) P, W and R must be concurrent.
R is also the tangent to the friction circle
ME101 - Division III
Kaustubh Dasgupta
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Applications of Friction in Machines
Belt Friction
Impending slippage of flexible cables, belts, ropes
over sheaves, wheels, drums
 It is necessary to estimate the frictional forces developed
between the belt and its contacting surface.
Consider a drum subjected to two belt tensions (T1 and T2)
M is the torque necessary to prevent rotation of the drum
R is the bearing reaction
r is the radius of the drum
β is the total contact angle
between belt and surface
(β in radians)
T2 > T1 since M is clockwise
ME101 - Division III
Kaustubh Dasgupta
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Applications of Friction in Machines
Belt Friction: Relate T1 and T2 when belt is about to slide to left
Draw FBD of an element of the belt of length r dθ
Frictional force for impending motion = μ dN
Equilibrium in the t-direction:
 μdN = dT (cosine of a differential quantity is unity in the limit)
Equilibrium in the n-direction:
(For the fig.
T1 > T2)
 dN = 2Tdθ/2 = Tdθ (sine of a differential in the limit equals the angle, and
product of two differentials can be neglected)
Combining two equations:
Integrating between corresponding limits:

•
•
•
 T2 = T1 e μβ
(T2 >T1; e = 2.718…; β in radians)
Rope wrapped around a drum n times  β = 2πn radians
r not present in the above eqn  eqn valid for non-circular sections as well
In belt drives, belt and pulley rotate at constant speed  the eqn describes condition of
impending slippage.
ME101 - Division III
Kaustubh Dasgupta
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Example (1) on Belt Friction
Examples: Belt Friction
A force P is reqd to be applied on a flexible cable that
supports 100 kg load using a fixed circular drum.
μ between cable and drum = 0.3
(a) For α = 0, determine the max and min P in order
not to raise or lower the load
(b) For P = 500 N, find the min α before the load begins to slip
Solution: Impending slippage of the cable over the
fixed drum is given by: T2 = T1 e μβ
Draw the FBD for each case
(a) μ = 0.3, α = 0, β = π/2 rad
For impending upward motion of the load: T2 = Pmax; T1 = 981 N
Pmax/981 = e0.3(π/2)  Pmax = 1572 N
For impending downward motion: T2 = 981 N; T1 = Pmin
981/Pmin = e0.3(π/2)  Pmin = 612 N
(b) μ = 0.3, α = ?, β = π/2+α rad, T2 = 981 N; T1 = 500 N
981/500 = e0.3β  0.3β = ln(981/500)  β = 2.25 rad
 β = 2.25x(360/2π) = 128.7o
 α = 128.7 - 90 = 38.7o
ME101 - Division III
Kaustubh Dasgupta
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Example (2) on Belt Friction
ME101 - Division III
Kaustubh Dasgupta
10
Example (2) on Belt Friction
(a) Impending slippage of the hawser
gives the application of the equation
T2 = T1 e μβ
T1 = 150 N, T2 = 7,500 N,
β = 22π rad = 12.57 rad
 μ = 0.311
(b) For 3 turns of the hawser, β = 32π rad = 18.85 rad
T1 = 150 N, μ = 0.311
Using T2 = T1 eμβ,
T2 = 52.73 kN
ME101 - Division III
Kaustubh Dasgupta
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Example (3) on Belt Friction
ME101 - Division III
Kaustubh Dasgupta
12
Example (3) on Belt Friction
Slippage will first occur for pulley B
since the angle β is smaller as
compared to pulley A
(for the same μ)
For pulley B,
T2 = 600 lb, β = 120 = 2π/3 rad
μ = 0.25
 T1 = 355.4 lb
ME101 - Division III
Kaustubh Dasgupta
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Example (3) on Belt Friction
Free body diagram of pulley A
ME101 - Division III
Kaustubh Dasgupta
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Friction in Machines :: Wheel Friction
Steel is very stiff
Significant Rolling Resistance Large Rolling Resistance
 Low Rolling Resistance between rubber tyre and tar road
due to wet field
Wheel Friction or Rolling Resistance
Resistance of a wheel to roll over a surface is caused by deformation
between two materials of contact.
This resistance is not due to tangential frictional forces
Entirely different phenomenon from that of dry friction
ME101 - Division III
Kaustubh Dasgupta
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Friction in Machines :: Wheel Friction
Rigid cylinder rolling at a constant velocity along a
rigid surface
Rigid
Rigid
- Normal force exerted by the surface on the
cylinder acts at the tangent point of contact
- No Rolling Resistance
Actually materials are not rigid
W
θ
- Deformation occurs at the contact region
- Reaction of surface on the cylinder consists of a
distribution of contact pressure.
ME101 - Division III
Kaustubh Dasgupta
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Applications of Friction in Machines
θ
W
Consider a wheel under action of a load W on axle
and a force P applied at its center to produce rolling
Deformation of wheel and supporting surface
Resultant R of the distribution of normal pressure must pass through wheel
center for the wheel to be in equilibrium (i.e., rolling at a constant speed)
R acts at point A on right of wheel center for rightwards motion
Force P reqd to maintain rolling at constant speed can be appx estimated as:
∑MA = 0  Wa = Prcosθ (cosθ ≈ 1  deformations are very small compared to r)

P
a
W   rW μr is called the Coefficient of Rolling Resistance
r
•μr is the ratio of resisting force to the normal force  analogous to μs or μk
•No slippage or impending slippage in interpretation of μr
ME101 - Division III
Kaustubh Dasgupta
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Applications of Friction in Machines
Examples: Rolling Resistance
A 10 kg steel wheel (radius = 100 mm) rests on an inclined
plane made of wood. At θ=1.2o, the wheel begins to roll-down
the incline with constant velocity.
Determine the coefficient of rolling resistance.
Solution: When the wheel has impending motion, the
normal reaction N acts at point A defined by the
dimension a.
Draw the FBD for the wheel:
r = 100 mm, 10 kg = 98.1 N
a
Using simplified equation directly: P  W   rW
r
Here P = 98.1(sin1.2) = 2.05 N
W = 98.1(cos1.2) = 98.08 N
 Coeff of Rolling Resistance μr = 0.0209
Alternatively, ∑MA = 0
 98.1(sin1.2)(r appx) = 98.1(cos1.2)a
(since rcos1.2 = rx0.9998 ≈ r)
 a/r = μr = 0.0209
ME101 - Division III
Kaustubh Dasgupta
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Center of Mass and Centroids
Concentrated Forces: If dimension of the contact area is negligible
compared to other dimensions of the body  the contact forces may be treated as
Concentrated Forces
Distributed Forces: If forces are applied over a region whose dimension
is not negligible compared with other pertinent dimensions  proper distribution of
contact forces must be accounted for to know intensity of force at any location.
Area Distribution
Ex: Water Pressure
Line Distribution
(Ex: UDL on beams)
ME101 - Division III
Kaustubh Dasgupta
Body Distribution
(Ex: Self weight)
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Center of Mass and Centroids
Center of Mass
A body of mass m in equilibrium under
the action of tension in the cord, and
resultant W of the gravitational forces
acting on all particles of the body.
- The resultant is collinear with the cord
Suspend the body from different points
on the body
- Dotted lines show lines of action of the resultant force in each case.
- These lines of action will be concurrent at a single point G
As long as dimensions of the body are smaller compared with those of the earth.
- we assume uniform and parallel force field due to the gravitational attraction of
the earth.
The unique Point G is called the Center of Gravity of the body (CG)
ME101 - Division III
Kaustubh Dasgupta
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Center of Mass and Centroids
Determination of CG
-
Apply Principle of Moments
Moment of resultant gravitational force W about
any axis equals sum of the moments about the
same axis of the gravitational forces dW acting
on all particles treated as infinitesimal elements.
Weight of the body W = ∫dW
Moment of weight of an element (dW) @ x-axis = ydW
Sum of moments for all elements of body = ∫ydW
From Principle of Moments: ∫ydW = ӯ W
x
 xdW
W
y
 ydW
W
z
 zdW
W
 Numerator of these expressions represents the sum of the moments;
Product of W and corresponding coordinate of G represents
the moment of the sum  Moment Principle.
ME101 - Division III
Kaustubh Dasgupta
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Center of Mass and Centroids
Determination of CG
 xdW
x
W
y
 ydW
W
z
 zdW
W
Substituting W = mg and dW = gdm

x
 xdm
m
y
 ydm
m
z
 zdm
m
In vector notations:
Position vector for elemental mass: r  xi  yj  zk
Position vector for mass center G: r  xi  yj  zk

rdm

r
m
The above equations are the
components of this single vector equation
Density ρ of a body = mass per unit volume
 Mass of a differential element of volume dV  dm = ρdV
 ρ may not be constant throughout the body
x
 xdV
 dV
ME101 - Division III
y
 ydV
 dV
z
 zdV
 dV
Kaustubh Dasgupta
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Center of Mass and Centroids
Center of Mass: Following equations independent of g
xdm

x




ydm

y
zdm

z
rdm

r
xdV

x
 dV
ydV

y
 dV
zdV

z
 dV
m
m
m
m
They define a unique point, which is a function of distribution of mass
This point is Center of Mass (CM)
CM coincides with CG as long as gravity field is treated as uniform and parallel
CG or CM may lie outside the body
CM always lie on a line or a plane of symmetry in a homogeneous body
Right Circular Cone
CM on central axis
ME101 - Division III
Half Right Circular Cone
CM on vertical plane of symmetry
Kaustubh Dasgupta
Half Ring
CM on intersection of two planes of symmetry
(line AB)
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