With my personal experience, I have observed that, usually, questions in the interviews are taken keeping in mind for certain aspects and reasons,Company ProfileRequirement of ProfileRelevance with past experiencePossible relation between Experience and Profile.Expected knowledge to suit profile requirements.Personal Interest and relevant Knowledge.If freshers are encountered, then company looks for flexibility in knowledge and adaptability.Frequently Asked Questions (FAQs) in Interviews

1) How to decide the power of Hydraulic Cylinder when it has to lift a load of 200 kg vertically up. Mention the factors involved to calculate.Answer.2) Basic values of Sheet and Wire Gauge (SWG) to millimetre (mm).Answer. 8 SWG 4mm 10 SWG 3.2 mm 14 SWG 2.01 mm 15 SWG 1.8 mm 16 SWG 1.6 mm 17 SWG 1.4 mm 18 SWG 1.2 mm 19 SWG 1.02 mm 20 SWG 0.9 mm 21 SWG 0.8 mm 22 SWG 0.7 mm 23 SWG 0.6 mm3) Load Calculation (Stress and Strain) of Mechanical Structures (Point Loads and Truss Members)Answer.Usually we have load calculations done on regular formulae, = PL / AE or = P/ AWhere, = Stress induced on the member.P = Load mounted overL = Span between supports.A = Cross-sectional Area.E = Youngs Modulus of the Material.Now, certain basic Conditions for to calculate the deflections under various loading conditions:Here,Y = Deflection in beam.P = Load mounted over.I = Inertia of the body.x = Length between one support and the load concentration. Rigid support, Cantilever, Point Loading.Y = P / 6EI (2L3 3L2 x + x3)Ymax = PL3 / 3EI at x = 0.Where, P = Point Load., L = Length of Span.

Rigid support, Cantilever, Uniform Loading.Y = W / 24EI (3L4 4L3 x + x4)Ymax = WL4 / 8EI at x = 0.Where, W = Product of total load(w) and length(L), I = Moment of Inertia.

Simply supported, Point Loading, Loading at centre.Y = P / 48EI (3L2 x - 4x3)Ymax = PL3 / 48EI at x = L/2.

Simply supported, Uniform Loading.Y = - Wx / 24EI (L3 2L x2 + x3)Ymax = - 5WL3 / 384EI at x = L/2.

Rigid supported, Centre point Loading.Y = - P / 48EI (3Lx2 - 4x3)Ymax = - PL3 / 192EI at x = L.

Rigid supported, Uniform Loading.Y = Wx2 / 24EI (2Lx L2 x2)Ymax = WL3 / 384EI at x = L/2.4) Moment of Inertia for basic shapes. Answer.

5) Calculation for Cutting / Shearing / Stripping / punching Force in metals.Answer.If,S = Thickness of Material.L = Perimeter of cutting A = L x S = Area of Shearing.B = Shear Strength (Constant value for every material).Then, Force (in Kg) = A x B6) Calculation for unbent / development length of sheet Metal.Answer.

For a condition above we follow an easy and handy method to calculate the unbent length required for the sheet. This method is only applicable for right angle bends with same radius throughout and exterior dimensions mentioned. This is always clear that, the inner radius of the bend is equal to the thickness of the sheet. Similarly, outer radius of the bend is always the sum of the inner radius and the thickness of the sheet.Now, for this method,Unbent length = Total outer linear Length + (Product of Correction Factor and no. of bends) (Product of Outer radius and no. of bends).Table for Correction Factor:Ratio (R/s)5.03.02.01.20.80.5

Correction Factor1.00.90.80.70.60.5

Firstly, the ratio of Outer radius(R) of Bend and thickness of sheet(s) has to be calculated, which will give ratio value to follow. Then corresponding to the ratio we can pick the correction factor from the table above.For the case above,Total Linear Length = 50 + 100 + 50 = 200mm.Outer Radius (R) = 20mmNo. of Bends = 2Ratio = 20 / 10 = 2Correction factor (q) corresponding to Ratio = 0.8Therefore,Unbent Length = 300 + (0.8 x 2) (20 x 2) = 261.6mm.Note: - if the correction factor is not known, then a tolerance of 0.5mm should be added at both the end of the sheet.The other way is more flexible and can be used widely.

Here, the formula is,L = { a + (R1 + q1 x s/2)(1/180) } + { b + (R2 + q2 x s/2)(2/180) } + ..... + ending linear length.Here till 120, we take smallest angle in the bend. Bends greater than 120, we take exterior angles.a = 70mm, b = 100mm, c = 100mm, R1 = 30mm, R2 = 20mm, s = 10mm, q1 = 0.9, q2 = 0.8, 1 = 2 = 90.Hence, Unbent Length (L) = {70 + (30 + 0.9 x 10/2)( x 90/180)} + {100 + (20 + 0.8 x 10/2)( x 90/180)} + 100 = 124.19 + 194.77 + 100 = 418.96mmNote: - In the above method there is a possibility of wastage of material if the bends are not equal to 90 throughout.7) Calculation for Radial Centripetal Force (RCF) for rorating Equipments.Answer.The formula is as under;RCF (g force) = {1.118 x (RPM) 2 x (R/100)} / 105Where,RCF = Centripetal force in Newtons.RPM = Revolutions per minute.R = Radius of string from the centre.8) Specification of Pipes, tubes and their Uses used in petrochemical / oil / process industries.Answer.9) Highlights of ASME Y 14.5M Standard for GD&T.Answer.

10) Highlights of ASME Standards of 31.1 and 31.3.Answer.11) Types of Measuring System and their least counts.Answer. Verneir Calliper :- 0.02mm Screw Gauge :- 0.01mm Dial Verneir :- 0.02mm Micrometer :- 0.01mm Dial Bore Gauge :- 0.001 to 0.01mm12) Classify Boilers with specifications.Answer.The classification of boilers is widely based on; According to the flow of water and hot gasesi. Fire (smoke) Tube Boilers.In fire tube boilers, hot gases pass through tubes which are surrounded with water. Examples: Vertical, Cochran, Lancashire and Locomotive boilers.ii. Water Tube Boilers.In water tube boilers, water circulates through a large number of tubes and hot gases pass around them. Eg, Bobcock & Wilcox boiler. According to axis of Shelli. Vertical Boilers.The vertical boilers are boilers which are mounted vertically up-straight.ii. Horizontal Boilers.The horizontal boilers are boilers which are laid down horizontally rigidly supported by a boiler bed. According to position of furnace.i. Internally Fired Boilers.In internally fired boilers, the furnace forms an integral part of the boilers structure. The vertical tubular, locomotive and the scotch marine boilers are well known examples.ii. Externally Fired Boilers.Externally fired boilers have a separate furnace built outside the boiler shell and usually below it. The horizontal return tube (HRT) boiler is probably the most widely known example of this type. According to the application and use.i. Stationary Boilers.Stationary boilers are one of which is installed permanently on a land installation.ii. Mobile Boilers.A marine boiler is a mobile boiler meant for ocean cargo and passenger ships with an inherent fast steaming capacity. According to the steam pressure.i. Low pressure Boilers.ii. Medium pressure Boilers.iii. High pressure Boilers.13) Mention Auxiliaries used in Boilers.Answer.14) Types of Power generation.Answer. Thermal Power Generation1. Fossil FuelsI. CoalII. Petroleum III. Natural Gas2. Solar Thermal Energy3. Geothermal Energy4. Nuclear Energy Hydroelectric Power Generation Wind Power Generation Solar Power Generation Chemical Power Generation1. Fuel Cells2. Batteries15) Relation between Current, Voltage and Resistance.Answer.The relation between current, voltage and resistance is well explained in Ohms law. According to Ohms Law, Voltage (V) is directly proportional to Current (I), where Resistance (R) remains constant. In other words, Voltage is a product of Current and the Resistance. Mathematically,V = IR16) Classification of Turbine. Answer.17) How to classify and measure values for Wire, Cable, and Bus Bar.Answer.18) Formulas of Series and Parallel Connection.Answer.19) Define Alpha and Delta Connection and differentiate them.Answer.20) Difference between Relay and Switch.Answer.Relays are a form of a switch that can be actuated without any human intervention. They most often are active devices that must be plugged in to operate. Relays are used on high current values with no limits.Switches are rated for certain load limits, so it has to be sure about use of appropriate switch especially if the switch is being used on an amplified level circuit. Switches are bound to current values to operate.21) Types of Motors and their Specification.Answer.Electric Motors are broadly classified as under;i. DC (Direct Current) Motors.Disadvantages of DC Motor are, Since they need brushes to connect the rotor winding. Brush wear occurs, and it increases dramatically in lowpressure environment. So they cannot be used in artificial hearts. If used on aircraft, the brushes would need replacement after one hour off operation. Sparks from the brushes may cause explosion if the environment contains explosive materials. RF noise from the brushes may interfere with nearby television sets, or electronic devices, etc.a. Shunt Motor.In Shunt Motor the rotor and stator windings are connected in parallel.b. Separately Excited Motor.In Separately Excited Motor the rotor and stator are each connected from a different power supply; this gives another degree of freedom for controlling the motor over the shunt.c. Series Motor.In series motor the stator and rotor windings are connected in series. Thus the torque is proportional to square of current, so it gives the highest torque per current ratio over all other dc motors. It is therefore used in starter motors of cars and elevator motors.d. Permanent Magnet DC Motor (PMDC).In PMDC motors, the stator is a permanent magnet, so the motor is smaller in size. But, it can be only used for low torque applications.e. Compound Motor.In compound Motors, the stator is connected to the rotor through a compound of shunt and series windings, if the shunt and series windings add up together, the motor is called cumulatively compounded. If they subtract from each other, then a differentially compounded motor results, which is unsuitable for any application.ii. AC (Alternating Current) Motors.a. Induction Motor.The Induction Motor is so called because voltage is induced in the rotor (thus no need for brushes), but for this to happen, the rotor must rotate at a lower speed than the magnetic field to allow for the existence of an induced voltage. Therefore a new term is needed to describe the induction motor: The slip. b. Synchronous Motor.The Synchronous Motor is so called because rotor tries to line up with the rotating magnetic field in the stator. It has the stator of an induction motor, and the rotor of a dc motor.iii. Other Motors. a. Reluctance motorA Reluctance Motor is like a synchronousinduction motor. The rotor has salient poles and a cage so that it starts like an induction motor, and runs like a synchronous motor.b. Hysteresis motorThe Hysteresis motor uses its own hysteresis producing the torque, which can be can be very tiny and are used as the driver for electric clocks.c. Stepper Motor.A Stepper motor is a special type of synchronous motors which rotates a number of degrees with each electric pulse.d. Brushless DC Motor.A Brushless DC Motor is a close cousin of a permanent magnet stepper motor with electronic controllers.e. Universal Motor.If a series DC motor has a laminated stator frame, it can run effectively from an ac supply as well as dc, this is the Universal Motor.22) Classify Pumps.Answer.Pumps can be broadly classified as under, Positive Displacement Pumps.In a positive displacement pumps, energy is periodically added to the liquid by the direct application of force to one or more movable volumes of liquid. This cause an increase in pressure up to the value required to move the liquid through ports in the discharge line. This is more visible in Reciprocating Pumps.i. Reciprocating Pumps.ii. Blow Case Pumps.iii. Rotary Pumps. Kinetic / Dynamic Pumps.In a kinetic pump, energy is continuously added to the liquid to increase its velocity. When the liquid velocity is subsequently reduced, this produces a pressure increase. Although there are several special types of pumps that fall into this classication, for the most part this classication consists of centrifugal pumps.i. Centrifugal Pumps.ii. Regenerative Pumps.iii. Special Effect. Others, viz; Buoyancy, impulse, etc.23) Classify Reciprocating Pumps.Answer.Reciprocating Pumps are broadly classified as under;iv. Air operated reciprocating Pumps. f. Diaphram Air operated reciprocating Pumps.g. Bellows Air operated reciprocating Pumps.h. Piston Air operated reciprocating Pumps.v. Steam reciprocating Pumps.c. Horizontal Steam reciprocating Pumps.d. Vertical Steam reciprocating Pumps.vi. Power reciprocating Pumps.a. Horizontal Power reciprocating Pumps.b. Vertical Power reciprocating Pumps.vii. Controlled Volume reciprocating Pumps.a. Horizontal Controlled Volume reciprocating Pumps.b. Vertical Controlled Volume reciprocating Pumps.24) Classify Rotary Pumps.Answer.Rotary Pumps Rotary pumps are pumps in which displacement is done by gears, cam or vanes in a chamber of diaphragm in a fixed casing. Rotary pumps are classified as under. Vane Rotary Pumps. i. Blade, bucket roller Vane Rotary Pumps. ii. Slipper Vane Rotary Pumps. Piston Rotary Pumps. i. Axial Piston Rotary Pumps. ii. Radial Piston Rotary Pumps. Flexible member Rotary Pumps. i. Flexible tube Rotary Pumps. ii. Flexible vane Rotary Pumps.iii. Flexible liner Rotary Pumps Lobe Rotary Pumps. i. Single Lobe Rotary Pumps. ii. Multiple Lobe Rotary Pumps. Gear Rotary Pumps. i. Internal Gear Rotary Pumps. ii. External Gear Rotary Pumps. Circumferential Piston Rotary Pumps. i. Single Circumferential Piston Rotary Pumps. ii. Multiple Circumferential Piston Rotary Pumps. Screw Rotary Pumps. i. Single Screw Rotary Pumps. ii. Multiple Screw Rotary Pumps.25) Classify Centrifugal Pumps.Answer.Centrifugal Pumps - Centrifugal pumps are used in more industrial applications than any other kind of pump. This is primarily because these pumps offer low initial and upkeep costs. Traditionally these pumps have been limited to low-pressure-head applications, but modern pump designs have overcome this problem unless very high pressures are required.The single-stage, horizontal, overhung, centrifugal pump is by far the most commonly type used in the chemical process industry. Centrifugal Pumps were classified perfectly in (American Petroleum Institute) API 610 technically equivalent to ISO 13709. They are broadly classified as: Overhung Centrifugal Pumps. A pump with the impeller(s) cantilevered from its bearing assemblies is classified as overhung pumps.a) Flexibly Coupled(1) Horizontal i. Foot Mounted OH1ii. Centreline Supported OH2(2) Vertical In-line with Bearing Bracket OH3 b) Rigidly Coupled(1) Vertical In-line OH4c) Close Coupled(1) Vertical In-line OH5(2) High Speed integrally geared OH6

Between Bearings Centrifugal Pumps. A pump with the impeller(s) located between the bearings is classified as between bearing pumps. A pump may be single stage, two stage or multistage. It can be axially (horizontally) split or radially split. (1) 1st and 2nd stagei. Axially Split BB1ii. Radially split BB2 Vertically Suspended Centrifugal Pumps. A pump with the impeller(s) cantilevered vertically and the suction nozzle typically submerged is classified as vertically suspended pump.(1) Single Casingi. Discharge through column Diffuser VS1 Volute VS2 Axial Flow VS3ii. Separate Discharge Line-shaft VS4 Cantilever VS5(2) Single Casingi. Diffuser VS6ii. Volute VS7 Seal-less Centrifugal Pumps. Seal-less pumps are special pumps which do not require shaft seals. Construction for seal-less pumps is driven by canned motors or magnetic couplings. It is normally used in process involving extreme hazardous fluid, where leakage cannot be tolerated. Submersible Centrifugal Pump. Submersible Pumps are designed to prevent cavitations. The driver components inside are completely surrounded by the pumped fluid. Horizontal self-priming Centrifugal pump. Horizontal self-priming pump are designed to create a vacuum at the pump inlet. This enables the pump to suck fluid into its casing. The suction nozzle of the pump can therefore be located above the level of liquid being pumped.Pumps shall be capable of operating at the maximum continuous speed. The maximum continuous speed is defined as, Equal to synchronous speed at maximum supply frequency of electrical motors. At least 105% of rated speed for variable speed pumps capable of exceeding rated speed.26) Highlights / similarities in API 610 and ISO 13709.Answer.The deflection () is measured in terms of Shaft Flexibility Index (SFI), occurred by load (w),

ISF = {(L13 / D14) + (L12 L2 / D14)}But, since the second part of equation above contributes only 20% of the actual value of SFI (Shaft Flexibility Index) abbreviated as ISF. Hence, ISF = (L13 / D14)Generally, the lower the SFI number for an overhung process, better is the shaft stiffness, thus giving higher Mean time between Repair (MTBR).There is other factor called pump sizing factor,Kt = QH / NWhere,Q = Flow RateH = Head at flow rateN = Rotating speedFor efficient motor;Higher point of graph should lie between 10 < ISF < 102 1 < Kt < 10Lower point of graph should lie between 10-3 < ISF < 10-2 104 < Kt < 105

Applying the new formula with standards, it is found that the system life of bearings is shorter than the shortest life of individual bearing.Three ways of torsional analysis;i. Undamped Natural Frequency.ii. Steady State damped response analysisiii. Transient TorsionalA head drop of 3% is allowable at the maximum at the first stage termed as NPSH3.Note: - Old overhung rotor designs have SFI value of 300 or even 1000. SFI of 50 is really fine.The bearing life as per API should be at 25000 hours at continuous run and if rated for conditions, bearing life should be 16000 hours at maximum axial and radial load at rated speed. This bearing life applies to the system bearing life.Formula, for bearing system life,L10 = {(1 / L10hA) 3/2 + (1 / L10hB ) 3/2 + ..... + (1 / L10hN) 3/2}-2/3 Where,L10hA = Basic rating life L10h as per ISO 281 for bearing A. L10hB = Basic rating life L10h as per ISO 281 for bearing B.L10hN = Basic rating life L10h as per ISO 281 for bearing N.N = Number of Bearings.Note:- Non-metallic cages shall not be used.27) Specification of Turbines.Answer.28) Types of Flow with relation.Answer.29) Explain terms used in Psychrometric Chart.Answer.30) Transformer types and Specification. Answer.A transformer is an apparatus for converting electrical power in an ac system at one voltage or current into electrical power at some other voltage or current without the use of rotating parts.In general, transformers are used for two purposes: signal matching and power supplies. Power TransformersPower transformers are used to convert from one voltage to another, at significant power levels. Step-up TransformersA "step-up transformer" allows a device that requires a high voltage power supply to operate from a lower voltage source. The transformer takes in the low voltage at a high current and puts out the high voltage at a low current. Step-down TransformersA "step-down transformer" allows a device that requires a low voltage power supply to operate from a higher voltage. The transformer takes in the high voltage at a low current and puts out a low voltage at a high current. Isolation TransformersAn "isolation transformer" does not raise or lower a voltage; whatever voltage comes in is what goes out. An isolation transformer prevents current from flowing directly from one side to the other. This usually serves as a safety device to prevent electrocution. Variable Auto-TransformersA "variable auto-transformer" (variac) can act like a step-up transformer or step-down transformer. It has a big knob on top that allows you to dial in whatever output voltage you want. Poly-phase TransformersThis type of transformer is commonly associated with three phase electric power, which is a common method of transmitting large amounts of high voltage power, such as the national power grid. In this system, three separate wires carry alternating currents of the same frequency, but they reach their peak at different times, thus resulting in a continuous power flow. Occasionally these three-phase systems have a neutral wire, depending on the application. Other times, all three phases can be incorporated into one, multiphase transformer. Winding patterns can vary and so can the phases of a poly-phase transformer. Leakage TransformerLeakage transformers have a loose binding between the primary and secondary winding, which leads to a large increase in the amount of inductance leakage. All currents are kept low with leakage transformers, which help prevent overload. They are useful in applications such as arc welding and certain high-voltage lamps, as well as in the extremely low-voltage applications found in some childrens toys. Resonant TransformerAs a type of leakage transformer, resonant transformers depend on the loose pairing of the primary and secondary winding, and on external capacitors to work in combination with the second winding. They can effectively transmit high voltages, and are useful in recovering data from certain radio wave frequency levels. Audio TransformerOriginally found in early telephone systems, audio transformers help isolate potential interference and send one signal through multiple electrical circuits. Modern telephone systems still use audio transformers, but they are also found in audio systems where they transmit analog signals between systems. Because these transformers can serve multiple functions, such as preventing interference, splitting a signal, or combining signals, they are found in numerous applications. Amplifiers, loudspeakers, and microphones all depend on audiotransformersin order to properly perform.31) Classify Transformers. Answer. According to method of Coolinga) Self-aircooled (dry type)b) Air-blastcooled (dry type)c) Liquid-immersed, self-cooledd) Oil-immersed, combination self-cooled and air-blaste) Oil-immersed, water-cooledf) Oil-immersed, forced-oilcooledg) Oil-immersed, combination self-cooled and water-cooled According to insulation between windingsa) Windings insulated from each otherb) Auto transformers According to No. of Phasesa) Single-phaseb) Poly-phase According to Method of Mountinga) Pole and platformb) Subwayc) Vaultd) Special According to purposea) Constant-voltageb) Variable-voltagec) Currentd) Constant-current According to servicea) Large powerb) Distributionc) Small powerd) Sign lightinge) Control and signallingf) Gaseous-discharge lamp transformersg) Bell ringingh) Instrumenti) Constant-currentj) Series transformers for street lighting32) Cost Estimation and Analysis.Answer.33) Areas, Volumes and Perimeter formulas of all basic Geometric Shapes.Answer.34) Hole Fit Precision to define tolerances.Answer. H5 Grinding, Honing (Fine Surface Finish) H6 Grinding, Boring (Fine Surface Finish) H7 Internal Operations like reaming, drilling, boring, etc (Normal Surface Finish) H8 Machining (Normal Surface Finish) H9 Boring and Reaming (Coarse Surface Finish) H10 / H11 Coarse Running (No Diameters)35) Running Fit tolerances (Tolerances between two running machine parts).Answer.

Precision Running Fit H6 / g5 H7 / g6 H8 / g7Close Running Fit H6 / f5 H7 / f6 H8 / f7Normal Running Fit H6 / e5 H7 / e6 H8 / e7Loose Running Fit H6 / d5 H7 / d6 H8 / d7Slack Running Fit(Fine Surface Finish) H8 / c5 H8 / b6 H9 / a8(Normal Surface Finish) H9 / c9 H11 / c9(Coarse Surface Finish) H11 / c11 H11 / b11 H11 / a1136) Non-Running Fit tolerances (Tolerances between two fixed non-running machine parts).Answer.Precision Running Fit H6 / h5 H6 / h6Close Running Fit H7 / h6 H7 / h7Normal Running Fit H8 / h7 H8 / h8Loose Running Fit H9 / h8 H9 / h9Slack Running Fit H11 / h9 H11 / h1137) Transitional Fit tolerances (Tolerances between two machine parts which are connected to stay fixed in a moving assembly).Answer.Clearance Running Fit H6 / j5 H7 / j6 H8 / j7True Transitional Fit H6 / k5 H7 / k6 H8 / k7Interface Fit H6 / m5 H7 / m6 H8 / m7Press Fit H6 / p5 H7 / p6Medium Drive Fit H6 / r5 H7 / r6Semi-Permanent drive Fit H6 / s5 H7 / s6Shrink / Permanent Fit (Shrinks are made in parts by heating) H6 / u5 H7 / u638) Types of Gears and its features.Answer.Advantages of Gears over other drives:-1. The gear has a positive drive and a constant velocity ratio. The velocity ratio is the ratio of effort used to the distance travelled.Velocity Ratio = (Effort Used to move / Distance Travelled)2. The gear drive can transmit at very low velocities which is not possible in belt drives.3. The efficiencies of the gear drives are very high.4. The gear drives consumes very less space for assembly.The gears are widely classifies as:- 1. Spur Gears; are gears in which teeth are cut parallel to the axis of shaft. These gears impose radial loads on the shaft.2. Helical Gears; are gears in which teeth are cut at an angle to the axis of shaft. . These gears impose radial and thrust loads on the shaft.3. Bevel Gears; are used where shafts are aligned to right angles. The teeth of the spiral gear are cut straight or spiral. These gears impose radial and thrust loads on the shaft.4. Worm Gears; consist of a threaded screw and a matching wheel. The thread may be single or multistart. The worm gear induces high thrust loads while the other induces high radial loads.39) Types of Gears Trains.Answer.1. Simple Gear Train; which carries only one gear in a shaft.Velocity Ratio (n1/n4) = (z4/z1)2. Compound Gear Train; Velocity Ratio (n1/n4) = (z2/z1) (z4/z3)3. Reverted Gear Train; the centre distance of gears 1 and 2 are same and gears 3 and 4 have the same centre distance.Velocity Ratio (n1/n4) = m1 (z1 + z2) = m1 (z3 + z4)4. Epicyclic Gear Train; is an arrangement with sun and planet system.Idler Gear is a gear, which operates between two main shaft gears. If the no. of idler gears in a gear train is even (even zero), then the first and last gears rotate in the opposite gears.Similarly if the no. of idler gears in a gear train is odd, then the first and the last gear rotate in the same direction.40) Basic formula for Spur Gears.Answer.Let us first understand the terms; Pitch Circle Diameter (d = mz) Circular Pitch (p = d/z) Diameteral Pitch (P = z/d) Addendum Circle Diameter {da = m(z + 2)} Addendum (A = M) Deddendum (D = 1.25M) Deddendum Circle Diameter {df = m(z - 2.5)} Tooth Thickness 1.5708M Bottom Clearance 0.25M Centre Distance {C = m(zP + zG) / 2} No. of Teeth (Z = 2 / sin2), where is the pressure angle.41) How to calculate shear stress in Bolts.Answer.Bolts and other types of connectors in structures undergo forces as the structures are loaded and unloaded. One of the forces that affect bolts is shear stress. When a bolt connects two or more parts, each of the parts can impart separate forces on the bolt, often in different directions. The result of these opposing forces on the bolt is shear stress at the plane through the bolt between the two connected components. If the shear stresses in the bolt are too high, then the bolt can break. An extreme example of shear is the use of bolt cutters on a bolt. The two blades of the cutters impart opposite forces on a single plane of the bolt, resulting in a cut bolt. Determining the shear stress in a bolt is a straightforward calculation using only a few inputs.Procedure to calculate Shear Stress in Bolts; Use the ruler or digital callipers to measure the thickness of each part of the bolted assembly. Label each thickness t1, t2, t3, and so on. Calculate the shear stress using the formula, = F/(d*(t1+t2)) if the bolt connects two plates where each plate is subjected to a force (F) in opposite directions. This load case is called single shear. For example, if two plates with 1-inch thicknesses are connected by a bolt with a diameter (d) of 1 inch, and each plate is subjected to a force of 100 lb, then it equals to 50 psi. Calculate the shear stress using the formula = F/(2d*(t1+t2+t3)) if the bolt connects three plates, where the centre plate experiences a force in one direction and the other two plates experience a force in the other direction. This load case is considered double shear because shear occurs in two different planes in the bolt. For example, if three plates with 1-inch thicknesses are connected by a bolt with a diameter (d) of 1 inch, and the plates are subjected to a force of 100 lb, then it is equal to 16.7 psi.42) The old welding machines used electromagnetic winding to step up the current for welding rods. What is the new principle inside the welding machines?Answer.The new principle inside the welding machine is based on the metal-arc welding. When one side of the welding circuit is connected to a piece of steel, an arc will be formed when a welding electrode attached to the other side and the two brought into contact. When the electrode is transferred along the steel at an accurate speed, the metal will deposit in an even layer known as a bead. Regulating the power is important in order to make use of these principles in metal-arc welding.43) What can be the reasons for a metal gasket to leak?Answer.Metal Gaskets may leak due to high pressures, temperature deformations, chemical reactions if any, or wear outs of Rubber, metal or both in the Gasket.44) In order to ensure constant pressure and constant flow rate of compressed air what should we put between the compressor and end user?Answer.Air Receiver is associated with a storage tank is used to serve the purpose end to avoid the return stroke of air in the compressor.45) What are the different types of mechanism to engage primary moving engine and transmission (in other words how many different types of clutches are there)?Answer.Dog Clutches, Cone Clutch, Friction Clutch, Safety Clutch, Hydraulic Clutch, Electromagnetic Clutch, Centrifugal Clutch, Over-running Clutch.46) Instead of having individual transmission what is the benefit of unified transmission?Answer.It controls the active and reactive power flows in the same transmission line.47) What is a torque converter?Answer.A torque converter acts like a clutch in an automobile which separates the load(fluid pressure) from the power source.48) Oil melts, grease melts. Is there any other lubrication which is effective at 200 deg c?Answer.Materials like Nylon, Derlin, etc may be a suggestion. Apart from that synthetic lubricants and high performance thickners are an option.49) If any machinery or transmission needs periodic and regular lubrication what is the best method to give it?Answer.Automatic Lubrication methods can be done by mechanizing systems. Alternatively, we can know the oil deterioration rates and can maintain lubrication.50) In which industrial process you will use centrifugal separator?Answer.They are used in Environmental Processes, Water Purification, Oil Purification processes, Dairy Processes.51) When mesuring the flow of liquid what are the things to be known?Answer.Density, Area and Velocity.52) How to calculate shear strength in Bolts.Answer.Shear stress is defined as the internal forces acting on an object when a force acts parallel to the face of the material, whereas shear strength is the maximum force the object can handle before becoming structurally compromised. Calculating the shear strength of a bolt can be particularly important, especially during construction work, as very large stresses can cause the bolts to fail.

Procedure to calculate Shear Stress in Bolts; Multiply the failing force of the bolt (the maximum force the bolt can sustain before failure) by 4. For example, if the failing force is 2 Newtons (N), then the result will be 8 N. Call this result A. Square the diameter of the circular component of the bolt, and multiply by the constant pi (3.14159). For example, if the diameter of the bolt is 0.1 meters (m), squaring gives 0.01 meters squared (m^2), and multiplying by pi gives 0.031459. Call this result B. Divide result A by result B. In our example, dividing 8 N by 0.0314159 gives 255 Pascals (Pa). This is the shear strength of the bolt.