UPPSC AE Mechanical 2013 Official Paper I (Previous Year Paper)

3-2-1 Principle :

• There are 12 degree of freedom of a work piece i.e., movement about negative and positive axis and clockwise and anti-clockwise rotation about the three axis.
• By providing three pins in the base, five degree of freedom will be arrested i.e. movement about negative Z-axis and the clockwise and anti-clockwise rotation about X and Y axis.
• These three pins also ensure that machining always takes place on single plane perpendicular to the base, long the length direction.
• This arrests movement of work piece along negative Y-direction in clockwise and anti-clock wise rotation about Z-axis.
• A sixth pin is provided on a plane perpendicular to the previous two planes, which also arrests movement about negative X-axis.
• By providing six pins, nine degree of freedom is arrested. Two more Degree of freedom will be arrested by providing clamping elements and from 12^th direction, tool comes for machining. A smaller size of pin is called button and are used in walls.
• Hence we can see that this principle helps us in locating the devices.

Explanation:

In ECM, there is no mechanical contact between the workpiece and the cutting tool and hence there is no possibility of tool wear. The material from the workpiece is removed by the chemical action of the electrolyte using Faraday’s law of electrolysis. Workpiece forms electrode (anode) and the cutting tool negative electrode (cathode).

The tool material used in ECM should have following property

• It should have high electrical conductivity
• It should be easily machinable and it should have high stiffness
• Its corrosion resistance should be high.

The advantages of ECM include

• Complex shapes can be made accurately
• The surface finish is good due to atomic level dissolution
• Tool wear practically absent
• Its material removal rate is highest.

Concept:

• Punching or blanking is a process in which the punch removes a portion of material from the larger piece or a strip of sheet metal.
• If the small removed piece is discarded, the operation is called punching, whereas if the small removed piece is the useful part and the rest is scrap, the operation is called blanking.

Blanking operation: 

• Die size = blank size
• As the Blank is fixed since the blank here is a useful part. The clearance will be provided on the punch in the blanking operation.
• Punch size = blank size – 2 x clearance.

Piercing/Punching operation:

• Punch size = hole size
• As the Hole size is here is a useful part. The clearance will be provided on the die in the punching operation.
• Die size = blank size + 2 x clearance.

Concept:

The relationship between blank diameter 'D' and the cup diameter 'd' during the deep drawing process is given as 

$D = \sqrt {{d^2} + 4hd}$

where

D = blank diameter

d = cup diameter

Additional Information

What is Deep drawing?

• Cup drawing or deep drawing is one of the widely used sheets metal forming operations.
• Cup-shaped objects, utensils, pressure vessels, gas cylinders, cans, shells, kitchen sink sets are some of the products of deep drawing.
• In this process, a sheet metal called blank is placed on a die cavity, held in position using a holding plate or holding ring, and pressed against the die cavity using a solid punch.
• The sheet metal attains the shape of the die cavity with a flat bottom.

Concept:

In punching and blanking, the shearing action on the sheet takes place which will depend on the shear strength of the metal.

Punching force F = Lt τ

Where L is Length of the periphery, t = thickness, τ = Ultimate shear strength

For circular section L = πd

∴ F = πdtτ

Concept:

Taylor's tool life equation is:

${V_1}T_1^n = C$

Calculation:

Given:

V2 = 2V1, T₂ = T₁/8

${V_1}T_1^n = C$

${V_1}T_1^n = {V_2}T_2^n$

${V_1}T_1^n = {2V_1}\left(\frac{T_1}{8}\right)^n$

$T_1^n = {2}\left(\frac{T_1^n}{2^{3n}}\right)$

2³ⁿ = 2

3n = 1

∴ n = 1/3

In machining process,

Chip thickness ratio (r) = $\frac{{uncut;chip;thickness\left( {{t_1}} \right)}}{{chip;thickness\left( {{t_2}} \right)}}$

$r = \frac{{{t_1}}}{{{t_2}}} = \frac{{\sin \phi }}{{\cos \left( {\phi - \alpha } \right)}}$

where, φ = shear angle, α = rake angle

the uncut chip thickness value is less than the chip thickness value thus chip thickness ratio is less than 1.

then $\tan \phi = \frac{{{\bf{rcos}}\alpha }}{{1 - rsin;\alpha }}$

$\phi = {\tan ^{ - 1}}\left( {\frac{{r\cos;\alpha }}{{1 - r\sin;\alpha }}} \right)$

Electrical discharge machining (EDM) is a manufacturing process whereby a desired shape is obtained by using electrical discharges (sparks). Material is removed from the work-piece by a series of rapidly recurring current discharges between two electrodes, separated by a dielectric liquid and subject to an electric voltage.

EDM has the lowest specific power requirement and can achieve sufficient accuracy.

• The accuracy and surface finish which are dependent on the overcut produced can be easily controlled by varying the frequency and current. The overcut is increased by increasing current and by decreasing frequency.
• For optimum metal removal and better surface finish, high frequency and minimum possible current is used.
• For roughing operation, low frequency and high current are used and for finishing application, high frequency and low current settings are used.

Concept:

Machining of metals and formation of chips

Types of chip:

1. Discontinuous chips – Formed during machining of brittle materials and while machining ductile materials at low speed.

Favourable conditions for this are:

1. Brittle workpiece
2. Slow speed of cutting
3. Small rake angle
4. Large depth of cut
5. Continuous chips – formed during machining of ductile materials at high speed. Favourable conditions for this type of chip to be formed are as follows:
6. Ductile material
7. Large rake angle
8. High cutting speed
9. Low depth of cut
10. Proper use of coolant and lubricant

Benefits of continuous chips are:

• Gives high surface finish
• High tool life
• Low power consumption

1. Continuous chips with built up edges – These type of chips are formed during the machining of ductile materials at medium cutting speed. Favorable conditions for this are:
2. Ductile material
3. High friction at face of tool
4. Lack of coolant and lubricant

During ultrasonic machining, the metal removal is achieved by hammering action of abrasive particles.

In ultrasonic machining, a tool of the desired shape vibrates at an ultrasonic frequency (19 ∼ 25 kHz) with an amplitude of around 15 – 50 μm over the workpiece.

Generally, the tool is pressed downward with a feed force, F.

Between the tool and workpiece, the machining zone is flooded with hard abrasive particles generally in the form of a water-based slurry.

As the tool vibrates over the workpiece, the abrasive particles act as the indenters and indent both the work material and the tool.

The abrasive particles, as they indent the work material, would remove the work material, particularly if the work material is brittle (due to crack initiation, propagation and brittle fracture of the material).

Applications of USM

• USM process is used in machining hard and brittle metallic alloys, semiconductors, glass, ceramics, carbides etc
• In machining of advanced ceramics for applications in auto-engine components
• In machining, wire drawing, punching or blanking of small dies

Concept:

Lapping:

• It is a surface finish process that improves the surface finish of components which have been previously grounded or fine turned.
• In this process, two surfaces are rubbed together with some abrasion in between them, by hand movement or by machine.
• This lapping can traditionally take different forms such as: Grinding etc.

Hardening:

• Hardening is a heat treatment process in which steel is heated to 30 – 50° above the critical range.
• Soaking time is allowed to enable the steel to obtain a uniform temperature throughout its cross-section. Then the steel is rapidly cooled through a cooling medium. Hardening increase the brittleness in the metal.

Shot peening:

• Shot peening is a process specifically designed to enhance the fatigue strength of components which are subjected to high alternating stress.
• In shot peening, small spherical shots bombards the surface of the part to be finished.

Sand Blasting:

• Sandblasting is also known as abrasive blasting. Basically, it is the operation of forcibly propelling a stream of abrasive material against a surface.
• The sandblasting operation is done under high pressure to smooth a rough surface, roughen a smooth/Shape the surface to remove its contaminants.

Concept:

OC curve:

• An operating characteristic curve is a plot between probability of acceptance versus percentage of defective items (or lots).
• Percentage of defective items is a measure of probability of rejection.

In a weaving operation, the parameter to be controlled is the number of defects per 10 square yards of material. Control chart appropriate for this task is C chart.

Explanation:

Control charts are statistical tool, showing whether a process is in control or not.

Two types of process data:

• Variable is a continuous data, things we can measure; Example includes length, weight, time, temperature, diameter, etc.
• Attribute is a discrete data, things we count; Examples include number or percent defective items in a lot, number of defects per item etc.

Types of Control Charts:

Variable charts are meant for variable type of data. X bar and R Chart, X bar and sigma chart, chart for the individual units

Attribute charts are meant for attribute type of data. p chart, np chart, c chart, u chart, U chart

Control charts for the variable type of data (X bar and R charts)

• In the x bar chart, the sample means are plotted in order to control the mean value of a variable
• In R chart, the sample ranges are plotted in order to control the variability of a variable

Control charts for Attribute type data (p, c, u charts)

• p-charts calculates the percent defective in sample; p-charts are used when observations can be placed in two categories such as yes or no, good or bad, pass or fail etc.
• c-charts counts the number of defects in an item; c-charts are used only when the number of occurrences per unit of measure can be counted such as number of scratches, cracks etc.
• u-chart counts the number of defects per sample; The u chart is used when it is not possible to have a sample size of a fixed size

Concept:

Acceptance sampling is used in:

1. Destructive testing
2. When inspection cost is extremely high and it is technologically not feasible
3. A statistical quality control to decide to decide whether to accept or reject a lot.

Concept:

Queuing Theory

Given:

Mean arrival rate (λ) = 10/hr, Mean service rate (μ) = 15/hr

∴ Traffic intensity, $\rho = \frac{\lambda }{\mu } = \frac{2}{3}$

Queue length = $\frac{{{\rho ^2}}}{{1 - \rho }}$

Calculation:

Queue length = $\frac{{4/9}}{{1/3}}$ = 4/3

∴ Queue length = 1.33

The inventory comprises of a large number of items. All items are not of equal importance. The firm, therefore, should pay more attention and care to those whose items whose usage value is high and fewer values to those whose usage value is low.

ABC analysis is also known as Always Better Control analysis. This is also known as the principal of the Law of Vital Few and Trivial Many.

Annual Usage Value = Annual Usage rate × Unit Cost

There are different types of selective inventory control:

Concept:

Based on assumptions of Economic order quantity (EOQ) model:

• Lead time = 0
• Constant rate of usage
• Fixed price of material
• Holding cold per unit component/item per unit time is fixed
• Demand is deterministic in nature

Concept:

Normal time = $\frac{{{\rm{Rating;factor}}\left( {\rm{\% }} \right)}}{{100}}$ × observed time

Standard time = normal time × $\left( {1 + \frac{{{\rm{allowance}}\left( {\rm{\% }} \right)}}{{100}}} \right)$

Calculation:

Normal time = 0.75 × (110/100)

∴ Normal time = 0.825 min

Standard time = 0.825 × (1 + 0.2)

∴ Standard time = 0.99 minutes

Concept:

Value engineering or value analysis is a management technique. It seeks the best functional balance between cost, reliability, performance of a product, project, service.

• It is a systematic approach to providing necessary functions at lowest cost
• Promotes substitution of materials with less expensive alternatives.

Control chart appropriate for this task is C chart.

Explanation:

Control charts are statistical tool, showing whether a process is in control or not.

Two types of process data:

• Variable is a continuous data, things we can measure; Example includes length, weight, time, temperature, diameter, etc.
• Attribute is a discrete data, things we count; Examples include number or percent defective items in a lot, number of defects per item etc.

Types of Control Charts:

Variable charts are meant for variable type of data. X bar and R Chart, X bar and sigma chart, chart for the individual units

Attribute charts are meant for attribute type of data. p chart, np chart, c chart, u chart, U chart

Control charts for the variable type of data (X bar and R charts)

• In the x bar chart, the sample means are plotted in order to control the mean value of a variable
• In R chart, the sample ranges are plotted in order to control the variability of a variable

Control charts for Attribute type data (p, c, u charts)

• p-charts calculates the percent defective in sample; p-charts are used when observations can be placed in two categories such as yes or no, good or bad, pass or fail etc.
• c-charts counts the number of defects in an item; c-charts are used only when the number of occurrences per unit of measure can be counted such as number of scratches, cracks etc.
• u-chart counts the number of defects per sample; The u chart is used when it is not possible to have a sample size of a fixed size

Concept:

Throughput rate/time = $\frac{{\left( {\frac{{Items}}{{time}}} \right)}}{{\left( {\frac{{{\rm{operating;time}}}}{{time}}} \right)}}$

Cycle time = 1/Throughput rate

Machines required = Total production time/ cycle time

Calculation:

Throughput rate = 2400/40 = 60 items/hr

Cycle time = 1/60 = 1 min = 60s

Machines required = 244/60 = 4.067

∴ 5 machine are required.

The correct answer is 'at the most m variables with non zero values.'

Key Points

• Basic Feasible Solution (BFS):
• In a Linear Programming Problem (LPP), a basic feasible solution is obtained by setting some variables equal to zero and solving the remaining variables from the system of equations.
• If an LPP has:
• m constraints
• n variables
• Then, in a basic solution:
• Exactly m variables can be selected as basic variables.
• The remaining (n − m) variables are set equal to zero and are called non-basic variables.
• Among the m basic variables, some may also become zero due to degeneracy.
• Therefore, a basic feasible solution can have:
• At most m variables with non-zero values.
• Hence, the correct statement is:
  

“A basic feasible solution has at the most m variables with non-zero values.”

Important Points

• A feasible solution is any solution that satisfies all constraints and non-negativity conditions.
• A basic solution is obtained by selecting m basic variables and setting the remaining variables equal to zero.
• A basic feasible solution is a basic solution that also satisfies all feasibility conditions.
• If all basic variables are positive, the solution is called a non-degenerate basic feasible solution.
• If one or more basic variables become zero, the solution is called a degenerate basic feasible solution.
• Basic feasible solutions correspond to the corner points (extreme points) of the feasible region in linear programming.
• The Simplex Method moves from one basic feasible solution to another in search of the optimal solution.

Concept:

Average Outgoing Quality Limit (AOQL) represents the maximum percentage defective in the outgoing product. The average out going quality depends on the incoming quality, the probability that he lot will be accepted.

1. Incoming quality is good = outgoing quality is good
2. Incoming quality is bad = whole lot is rejected. Therefore, outgoing quality is good since bad parts won’t get through.
3. Incoming quality is neither very good nor very bad = outgoing quality gets worse and % defective reaches maximum known as AOQL

Concept:

$\bar P = \frac{{Total;number;of;defective}}{{Total;number;of;observation}}$

Calculation:

$\bar P = \frac{{200}}{{20 \times 100}} = 0.1$

${\rm{UCL}} = \bar P + {\rm{z}} \times \sqrt {\frac{{\bar P\left( {1 - \bar P} \right)}}{n}}$

${\rm{UCL}} = 0.1 + 3\sqrt {\frac{{0.1\left( {0.9} \right)}}{{100}}}$

∴ UCL = 0.19

Assumption of the Basic EOQ model is:

• Demand is known deterministic and constant
• Cost of product is constant
• Lead time is zero
• Usage rate is constant
• Inventory cost is fixed

∴ So, stochastic demand is not an underlying assumption of the basic EOQ model.

Concept:

The general formula for the margin of error for a sample proportion is

$e = {Z^*}\sqrt {\frac{{\rho \left( {1 - \rho } \right)}}{n}}$

Where, ${Z^*}$= value for your desired level of confidence, ρ = sample proportion, n = sample size

$\therefore {\bf{e}};{\bf{\alpha }}\frac{1}{{\sqrt n }}$

Concept:

• The producer’s risk is the probability with which a consumer will reject a good lot. It is also known as Type-I error.
• The consumer’s risk is the probability with which a consumer will accept a bad lot this is also known as Type-II error.

Concept:

The horizontal component of the force applied at C point causes shear stress and bending moment at point A.

The vertical component of the force applied at C point exerts thrust at point A

Calculation:

Given:

F = 1 kN

The vertical component of force is F_v = F × sin θ

Thrust at point A, F_v =  F × sin θ = 1 × sin 60° = 0.866 kN.

Concept:

• In framed structure or truss only compression or tension forces will developed.
• In truss all joint are pin joint and load applied only at joint will be consider.

Concept:

Let R is the normal reaction act on roller due to block

Now,

Balancing the forces.

R cos 60° = W

R sin 60° = F

So, R = 2W

$\therefore {\rm{;F}} = 2{\rm{W}} \times \frac{{\sqrt 3 }}{2}$

$\therefore {\bf{F}} = \sqrt 3 W$

Concept:

• Virtual work concept is used in the application of the principle of least action to the study of forces and movement of a mechanical system.
• In virtual work equation reaction due to rough surface on a body which rolls on it without slipping or smooth surface with which body is in contact and reaction at a point or on an axis, fixed in space, around which a body is constrained to turn are neglected.

Concept:

Rotational kinetic energy ${\rm{KE_R}} = \frac{1}{2}I{\omega ^2}$

${\rm{K}}{{\rm{E}}_{\rm{R}}} = \frac{1}{2}\frac{{M{R^2}}}{2}{\omega ^2}$

∴ Rotational kinetic energy $\left( {{\rm{K}}{{\rm{E}}_{\rm{R}}}} \right) = \frac{1}{4}M{v^2}$ (∵ v = Rω)

Translational Kinetic energy KE_T = $\frac{1}{2}M{v^2}$

Total kinetic energy = KE_R + KE_T

Total kinetic energy = $\frac{3}{4}M{v^2}$

Calculation:

Given:

$\therefore \frac{{{\rm{Rotational;kinetic;energy}}}}{{{\rm{Total;Kinetic;energy}}}} = \frac{{\frac{1}{4}M{v^2}{\rm{;;}}}}{{\frac{3}{4}M{v^2}}}$

$\therefore \frac{{{\rm{Rotational;kinetic;energy}}}}{{{\rm{Total;Kinetic;energy}}}} = \frac{1}{3}$

Concept:

When two masses are moving with equal kinetic energy, we can find the ratio of their momentum by using the relation between kinetic energy and momentum.

Kinetic energy (KE) is given by:

$KE = \frac{1}{2}mv^2$

Momentum (p) is given by:

$p = mv$

Calculation:

Let the velocities of the 2 kg and 8 kg masses be v1v1

and v2v2

respectively.

Since the kinetic energies are equal:

$\frac{1}{2} \times 2 \times v_1^2 = \frac{1}{2} \times 8 \times v_2^2$

Simplifying, we get:

$2v_1^2 = 8v_2^2$

$v_1^2 = 4v_2^2$

$v_1 = 2v_2$

Now, we find the momentum of each mass:

Momentum of the 2 kg mass, $p_1 = 2 \times v_1$

Momentum of the 8 kg mass, $p_2 = 8 \times v_2$

Substitute v1=2v2v1=2v2

into the equation for p1p1

:

$p_1 = 2 \times 2v_2 = 4v_2$

$p_2 = 8 \times v_2$

Now, find the ratio of their momenta:

$\frac{p_1}{p_2} = \frac{4v_2}{8v_2} = \frac{4}{8} = \frac{1}{2}$

$\frac{p_1}{p_2} = 0.50$

Thus, the ratio of the magnitude of their momenta is:

2. 0.50

Concept:

The maximum efficiency of screw jack is when helix angle α is

$\alpha = 45^\circ- \frac{\phi }{2}$

Note:

If the friction angle is greater than the helix angle, the torque required to lower the load will positive, indicating that an effort is applied to lower the load. Such a screw is known as a self-locking screw.

Additional Information

Screw jack

• A screw jack is a portable device consisting of a screw mechanism used to raise or lower the load. There are two types of jack Hydraulic and mechanical
• A Hydraulic jack consists of a cylinder and piston mechanism. The movement of piston rod is used to raise or lower the load.
• Mechanical jack can be either hand operated or power driven

The maximum efficiency of screw jack is:

$\eta = \frac{{1 - sin\phi }}{{1 + sin\phi }}$

Concept:

It follows the law of conservation of momentum.

The following is observed after the collision:

When two bodies collide without the presence of any other force or force fields:

• Their total momentum must be conserved

Concept:

When at rest,

T = mg

When the lift is moving downward:

${R_1} = W - ma = W - \frac{W}{g}a = W\left( {1 - \frac{a}{g}} \right)$

When the lift is moving upward:

${R_1} = W + ma = W + \frac{W}{g}a = W\left( {1 + \frac{a}{g}} \right)$

Concept:

Resultant force F, of any two forces F1 and F2 with an angle θ between them can be given by vector addition as

$F_r^2 = F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \theta$

Hence by comparing the options with the above equation option 2 is the correct answer.

N_G = 0,  (Given)

$y - \frac{x}{4} = 0$,

y = x/4; x=4y

for one revolution of arm: y = 1; x = 4 rev.

N_P=X+y

∴ NP = 5 revolution.

Concept:

Cycloidal motion curve has the best dynamic characteristics. The acceleration is finite at all times and the starting and ending acceleration is zero. It will yield a cam mechanism with the lowest vibration, stress, noise and shock characteristics. Hence for the high-speed engine, the cam follower should move with Cycloidal motion.

Note:

Uniform velocity motion

Uniform Acceleration and Retardation

Simple harmonic motion

Cycloidal motion

Concept:

Coefficient of fluctuation of energy:

${C_E} = \frac{{Maximum;fluctuation;of;energy}}{{Work;done;per;cycle}}$

${C_E} = \frac{\Delta E}{W} \Rightarrow \Delta E=C_E.W$

Coefficient of fluctuation of speed:

$\begin{array}{l} {C_S} = \frac{{Maximum;fluctuation;of;speed}}{{Mean;speed}}\ {C_S} = \frac{{{ω _1} - {ω _2}}}{ω } = \frac{{2\left( {{ω _1} - {ω _2}} \right)}}{{\left( {{ω _1} + {ω _2}} \right)}} \end{array}$

ω = mean angular speed during the cycle (rad/s)

$ω = \frac{{{ω _1} + {ω _2}}}{2}$

Maximum fluctuation of energy: ΔE = CE. W

$\begin{array}{l} {\rm{\Delta }}E = \frac{1}{2}Iω _1^2 - \frac{1}{2}Iω _2^2 = \frac{1}{2}I\left( {{ω _1} + {ω _2}} \right)\left( {{ω _1} - {ω _2}} \right)\ {\rm{\Delta }}E = Iω \left( {{ω _1} - {ω _2}} \right) = I{ω ^2}\frac{{\left( {{ω _1} - {ω _2}} \right)}}{ω } = I{ω ^2}{C_S}\ K.E = \frac{1}{2}I{ω ^2} = \frac{1}{2}.\frac{{{\rm{\Delta }}E}}{{{C_S}}} = \frac{{{C_E}W}}{{2{C_S}}} \end{array}$

Concept:

The unbalanced force due to reciprocating masses varies in magnitude but constant in direction while due to the revolving masses, the unbalanced force is constant in magnitude but varies in direction.

Unbalanced force,

${F_U}; = ;m.{\omega ^2}.r\left( {cosθ + \frac{{cos2θ }}{n}} \right); = ;m.{\omega ^2}.rcosθ + m.{\omega ^2}.r \times \frac{{cos2θ }}{n}; = ;{F_P} + {F_S}$

The expression (m. ω2.r cos θ) is known as primary unbalanced force and $\left( {m.{\omega ^2}.r \times \frac{{cos2θ }}{n}} \right)$ is called secondary unbalanced force.

Unbalanced force is given by

${F_U} = m.{\omega ^2}.r.cos\theta + m.{\omega ^2}.r.\frac{{cos2\theta }}{n} = {F_{Primary}} + {F_{Secondary}}$

Where n = Ratio of length of the connecting rod to the crank radius = l / r

So secondary unbalanced force, ${F_{Secondary}} = m.{\omega ^2}.r.\frac{{cos2\theta }}{n}$ decreases when n increases.

Explanation:

If the rotor is statically balanced, it will not roll under the action of gravity, regardless of the angular position of the rotor. The requirement for static balance is that the centre of gravity of the system of masses is at the axis of rotation.

Static balance: ΣF = 0

A rotating system of mass is in dynamic balance when the rotation does not produce any resultant centrifugal force or couple.

Dynamic balance: ΣF = 0 and ΣM = 0

It can be said that if the rotor is dynamically balanced then it is also statically balanced. The converse is not true for all the rotors. A statically balanced rotor is not always dynamically balanced, the exception being the single plane rotors.

Concept:

The instantaneous center (I) is the intersection point of the normal drawn at point A and B.

So,

Angular velocity (ω) of the rod at this instant is

${\rm{\omega }} = \frac{V}{R} = \frac{{{V_A}}}{{IA}}$

Calculation:

Given:

${\rm{\omega }} = \frac{1}{{1 \times \cos60^\circ }}$

∴ ω = 2 rad/s

Additional Information

Instantaneous Centre:

The instantaneous center of a rigid body is located by the point where the lines perpendicular to the direction of the velocity of two different points on the body meets. If two points have velocity parallel to each other then the line drawn perpendicular to those points will meet at infinity, therefore their instantaneous center will lie at infinity.

Properties of Instantaneous Centre:

• At the instantaneous centre of rotation, one rigid link rotates instantaneously relative to another for the configuration of the mechanism considered.
• The two rigid links have no linear velocities to each other at the instantaneous centre.
• The two rigid links have the same linear velocity relative to the third rigid link, or any other link.

The number of Instantaneous Centres:

The number of Instantaneous centres in a mechanism is equal to the number of possible combinations of two links.

$N =\frac{{n(n-1)}}{{2}}$N=n(n−1)2N=n(n−1)2N=n(n−1)2N=n(n−1)2

Where

n = number of links. 

Types of Instantaneous Centres:

• Fixed instantaneous centres.
• Permanent instantaneous centres.
• Neither fixed nor a permanent instantaneous centre.

Explanation:

Isochronous Governor: 

• A governor is said to be isochronous when the equilibrium speed is constant (i.e. range of speed is zero) for all radii of rotation of the balls within the working range, neglecting friction.
• The isochronism is the stage of infinite sensitivity.
• A spring-loaded governor can only possibly be an Isochronous governor.

Hunting: 

• If a governor is too sensitive, a small variation of speed causes large change in the sleeve movement. Thus, a governor is said to be hunt if the speed of the engine fluctuates continuously above and below the mean speed.

Stability: 

• A governor is said to be stable when for each speed within the working range there is only one radius of rotation of the fly-ball at which the governor is in equilibrium.

Concept:

The two masses form an equivalent dynamic system if:

1. The sum of the two masses is equal to the total mass of the body
2. The centre of gravity of two masses coincide with that of the body
3. The sum of the mass moment of inertia of the masses about their centre of gravity is equal to the mass moment inertia of the body

Concept:

It is given that the ratio of speed of shaft B to shaft A is G and we have to calculate the equivalent moment of inertia of disc.

So we will equate the kinetic energy of both the shafts.

Since we are having the ratio of their speeds by equating the kinetic energy we will directly get the equivalent moment of inertia of B at the speed of shaft A.

∴ According to question:

\(\frac{1}{2}{{I}_{2}}\omega {B}^{2}=\frac{1}{2}{{I}{eq}}\omega _{A}^{2}\)

\({{I}{eq}}={{(\frac{{{\omega }{B}}}{{{\omega }{A}}})}^{2}}\times {{I}{2}}\)

∴ I_eq = G²I₂

Explanation:

Kinematic pairs are classified under three headings namely, lower pair, higher pair and wrapping pair.

Lower Pair: 

A pair is said to be a lower pair when the connection between two elements is through the area of contact. Some of the types of Lower pair are:

• Revolute Pair
• Prismatic Pair
• Screw Pair
• Cylindrical Pair
• Planar Pair

Higher Pair: 

A pair is said to be higher pair when the connection between two elements has only a point or line of contact. Examples of higher pairs are:

• A point contact takes place when spheres rest on plane or curved surfaces (in case of ball bearings)
• Contact between teeth of a skew-helical gears
• Contact made by roller bearings
• Contact between teeth of most of the gears
• Contact between cam-follower
• Spherical Pair

Wrapping Pairs:

In a higher pair, the contact between the two bodies has only a line contact or a point contact. Whereas in a wrapping pair, one body completely wraps over the other. The typical example is of a belt and a pulley or a chain and a sprocket where the belt completely wraps around the pulley or the chain completely wraps around the sp

Spherical Pair:

This pair usually look like lower pair but it is multiple point contact pair like wrapping pair which is a higher pair.

• Example. Ball in Socket joint.

Concept:

The critical damping cc is defined as the value of damping coefficient c for which the term 

$\begin{array}{l} {\left( {\frac{c}{{2m}}} \right)^2}; - ;\frac{k}{m}; = ;0\ \frac{{{c_c}}}{{2m}}; = ;\sqrt {\frac{k}{m}} \Rightarrow {c_c}; = ;2m\sqrt {\frac{k}{m}} ; = ;2m{\omega _n}; = ;2\sqrt {km} \end{array}$

Pitch circle is an imaginary circle which by the pure rolling action, would give the same motion as the actual gear. The diameter of the pitch circle is known as Pitch circle diameter. The size of the gear is usually specified by the pitch circle diameter. It is also known as pitch diameter.

Addendum Circle: It is the circle drawn through the top of the teeth and is concentric with the pitch circle. It is also called the Outside circle.

Dedendum circle: It is the circle drawn through the bottom of the teeth. It is also called the root circle.

Clearance is the radial distance from the top of the tooth to the bottom of the tooth, in a meshing gear (the difference between the dedendum of one gear and the addendum of the mating gear).

Important Point:

Pitch circle diameter: It is the diameter of the pitch circle. The size of the gear is usually specified by the pitch circle diameter. It is also known as pitch diameter.

Circular pitch: It is the distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is usually denoted by pc. Mathematically,

Circular pitch, $p_c = \frac{{\pi d}}{T}$

Diametral Pitch: It is the ratio of the number of teeth to the pitch circle diameter in millimeters. It is denoted by pd. Mathematically,

$p_d = \frac{T}{d}= \frac{\pi}{p_c}$

Module (m): It is the ratio of the pitch circle diameter in millimeters to the number of teeth. It is usually denoted by m. Mathematically

$m = \frac{d}{T}$

Concept:

• The Pressure angle is defined as the angle between the normal to the pitch curve at a point and the direction of the follower motion.
• Its values depends on the angle of ascent, the lift of the follower and offset between centre lines of cam and follower.
• Pressure angle varies in magnitude at all instants of the follower motion.

Explanation:

Centrifugal effects on belts:

• While in motion, as a belt passes over a pulley, the centrifugal effect due to its own weight tends to lift the belt from the pulley.
• Owing to symmetry, the centrifugal force produces equal tensions on the two sides of the belt i.e. on the tight side as well as on the slack side.

The magnitude of centrifugal force is Tc = mv2

Thus, we can say that the centrifugal tension is independent of the tight and slack side tensions and depends only on the velocity of the belt over the pulley.

Total tension on tight side T ⇒ T1 + Tc

Total tension on slack side T ⇒ T2 + Tc

Considering centrifugal tension:

Friction tension on tight side = T - Tc = T1

Let T2 be the friction tension on the slack side

$\frac{T_1}{T_2}=e^{\mu \theta}=k$ or $T_2=\frac{T_1}{k}$

We know that power P = (T1 - T2)v

Putting the value of T2 in power

$P=T_1\left(1-\frac{1}{k}\right)v$

Neglecting centrifugal tension:

Friction tension on tight side = T

Let T₂' be the tension on the slack side.

$\frac{T_1}{T_2^{'}}=e^{\mu \theta}=k$ or $T_2^{'}=\frac{T_1}{k}$

We know that power P = (T1 - T2')v

$P=T\left(1-\frac{1}{k}\right)v$

As T1 is lesser than T, the power transmitted is less when centrifugal tension is taken into account.

Central tension (T_c) reduces the effective tension difference available for power transmission due to the impact on the belt's ability to maintain grip and avoid slippage under higher speeds. Thus, higher centrifugal forces (and hence higher (T_c)) result in reduced effective power transmission. While theoretically the gross difference (T₁ - T₂) remains, the real-world power transmissible is diminished due to the practical reduced (T₁) and (T₂) load handling resulting from (T_c). Hence, centrifugal tension in the belt ultimately decreases the power transmitted due to its effect on reducing the effective load-carrying capacity of the belt.

Mistake Points Two cases have been taken, the maximum tension in case of centrifugal tension and ignoring centrifugal tension. The tension T and T1 are different and as per the value of T and T1  we are able to say that centrifugal tension decreases power output

Concept:

Balancing of Several Masses Rotating in Different Planes

• When several masses revolve in different planes, they may be transferred to a reference plane (briefly written as R.P.), which may be defined as the plane passing through a point on the axis of rotation and perpendicular to it.
• The effect of transferring a revolving mass (in one plane) to a reference plane is to cause a force of magnitude equal to the centrifugal force of the revolving mass to act in the reference plane, together with a couple of magnitude equal to the product of the force and the distance between the plane of rotation and the reference plane.

The masses in the different plane are represented in one plane in the following way:

• In order to have a complete balance of the several revolving masses in different planes, the following two conditions must be satisfied:

1. The force in the difference plane must balance, i.e. the resultant force must be zero.
2. The couples about the reference plane must balance, i.e. the resultant couple must be zero.

Hence to satisfy the above mentioned condition we need minimum two balancing masses.

Concept:

Partial balancing of the locomotives results in the following

c is a fraction of the reciprocating parts to be balanced

Swaying couple: 

The unbalanced forces along the line of stroke for the two cylinders constitute a couple about the centre line between the cylinders. (a is distance between the center lines of the two cylinders)

$F_C=(1-c)m\omega^2r\times\frac{a}{2}(\cos\theta+\sin\theta)$

It is maximum or minimum when θ = 45° or θ = 225°

$F_{c,max}=±\frac{a}{\sqrt2}(1-c)m\omega^2r$

Variation in tractive force: 

A variation in the tractive force (effort) of an engine is caused by the unbalanced portion of the primary force which acts along the line of stroke of a locomotive engine.

$F_T=(1-c)m\omega^2r(\cos\theta-\sin\theta)$

It is maximum or minimum when θ = 135° or θ = 315°

$F_{T,max}=±\sqrt2(1-c)m\omega^2r$

Hammer blow: 

The maximum magnitude of the unbalanced force along the perpendicular to the line of stroke is known as hammer blow. (Balancing mass, B at radius b)

$F_b=B.\omega^2.b$

Concept:

The number of instantaneous centres in a considered kinematic chain is equal to the number of combinations of two links: If N is the number of instantaneous centres and n is the number of links.

$N = \frac{{n\left( {n - 1} \right)}}{2}$

Crank and slotted lever quick return motion mechanism is an inversions of Single Slider Crank Chain which is a modification of the basic four-bar chain.

Calculation:

The number of instantaneous centers:

$N = \frac{{L\left( {L - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6$

where L is the number of links.

Concept:

In vibration isolation system, the ratio of the force transmitted to the force applied is known as the isolation factor or transmissibility ratio.

$T = \frac{{{F_T}}}{{{F_O}}} = \frac{{\sqrt {1 + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} }}{{\sqrt {\left[ {1 - {{\left( {\frac{\omega }{{{\omega _n}}}} \right)}^2} + {{\left( {2\zeta \frac{\omega }{{{\omega _n}}}} \right)}^2}} \right]} }}$

• When ω/ωn = 0 ⇒ TR = 1, (independent of ζ)
• When ω/ωn = 1 and ξ = 0 ⇒ TR = ∞, (independent of ζ)
• When frequency ratio ω/ωn = √2, then all the curves pass through the point TR = 1 for all values of damping factor ξ.
• When frequency ratio ω/ωn < √2, then TR > 1 for all values of damping factor ξ. This means that the force transmitted to the foundation through elastic support is greater than the force applied.
• When frequency ratio ω/ωn > √2, then TR < 1 for all values of damping factor ξ. This shows that the force transmitted through elastic support is less than the applied force. Thus vibration isolation is possible only in the range of ω/ωn > √2. Here the force transmitted to the foundation increases as the damping is increased.

Concept:

Constant velocity ratio between the two shafts can be obtained if they are connected by gears.

∵ Velocity ratio = 1/ Gear ratio

Where, Gear ratio (G) = T/t, T = Number of teeth on gear and, t = Number of teeth on pinion.

Concept:

Schematic Diagram of a Differential

The differential gear used in the rear-drive of an automobile. Its function is

• (a) To rotate the rear wheels at different speeds while the automobile is taking a turn
• (b) To transmit motion from the engine shaft to the rear driving wheels

Hence we can say that they enable the vehicle to negotiate the curves properly.

Explanation:

• In a bending beam, a point of contraflexure is a location where the bending moment is zero (changes its sign). In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero lines.
• At the point of contraflexure, the bending moment is zero.

For example:

A propped cantilever beam with uniformly distributed load over the entire span

The corresponding BMD is

At distance L/4 from the left support, we get point of contraflexure, as there is the change in sign. At that point, the Bending moment is zero.

Concept:

By equivalent load system

Reactions at both ends:

$R_A=\frac{P}{2}-\frac{Pa}{l}$

$R_B=\frac{P}{2}+\frac{Pa}{l}$

Bending Moment:

BM at the end-point is zero. 

The bending moment is occurring at the midpoint.

Let us assume point 'C' is the midpoint of the beam.

Therefore, $BM=[\frac{P}{2}-\frac{Pa}{l} ]\times\frac{l}{2}$ 

$BM={[\frac{P}{2}-\frac{Pa}{l} ]\times\frac{l}{2}}+Pa$

⇒ $BM={\frac{Pl}{4}-\frac{Pa}{2} }+Pa$

Therefore, $BM={\frac{Pl}{4}+\frac{Pa}{2} }$

Minimum bending moment, $BM_{min}=\frac{Pl}{4}-\frac{Pa}{2}$

∴ Maximum bending moment = $BM_{max}=\frac{Pl}{4}+\frac{Pa}{2}$

Concept:

The area under the shear-force diagram gives a bending moment between those two points and the area under the load diagram gives shear-force between those two points. i.e.

\({{M}{B}}-{{M}{A}}=\mathop{\int }{A}^{B}{{F}{x-x}}dx\)

\({{F}{B}}-{{F}{A}}=-\mathop{\int }{A}^{B}{{w}{x-x}}dx\)

Calculation:

Given:

\({{M}{C}}-{{M}{A}}=\frac{1}{2}\times \left( 14+2 \right)\times 2\)

M_C = 16 kN-m   (M_A = 0)

Similarly;

∴ M_B – M_c = 16 kN-m

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and alike then the radius of the Mohr circle will be zero.

${\rm{Radius;of;Mohr's;Circle}},R = \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + {\tau _{xy}}^2}$