BPSC AE Paper 6 (Mechanical) 14 Oct 2022 Official Paper (Previous Year Paper)

Explanation:

• An Automated Guided Vehicle (AGV) is a mobile robot that is designed to move materials or products in a manufacturing or distribution facility without the need for a human operator. AGVs are often used to automate repetitive and labor-intensive tasks such as material handling, transportation, and delivery.

• AGVs can be programmed to follow a specific path or route to reach their destination. However, some advanced AGVs are equipped with sensors and software that allow them to adapt to changes in the environment and navigate around obstacles.

• The flexibility of an AGV system depends on the design and programming of the system. Some AGV systems may require significant reprogramming to accommodate changes in product or production, while others may be designed with more flexibility in mind.

• AGVs are often used as part of a larger Flexible Manufacturing System (FMS) and can be interfaced with other modules such as robots, automatic storage and retrieval systems, and CNC machines. This allows for a highly automated and efficient manufacturing process.

• Like any mechanical or automated system, AGVs can experience downtime due to technical issues or maintenance needs. However, proper maintenance and planning can help minimize downtime and ensure reliable operation.

• AGVs use a variety of guidance systems to navigate through a facility, such as lasers, magnets, or vision systems. They can be programmed to follow specific routes, pick up and deliver materials, and even interact with other machines and systems within a manufacturing or distribution facility.

• AGVs offer a number of benefits to manufacturing and distribution facilities, including increased efficiency, reduced labor costs, improved safety, and greater flexibility in operations. They are also highly customizable, with a variety of sizes, shapes, and capabilities to meet the specific needs of different industries and applications.

• Overall, AGVs are a key technology in the ongoing trend towards greater automation and efficiency in modern manufacturing and distribution operations.

Explanation:

Production planning:

• Production planning is a pre-production activity.
• Production planning is used to pre-determine the requirements of machines, materials, manpower, and other items used in the manufacturing of any product.
• The aim of production planning is:
• During the manufacturing of a product, production planning ensures that the sufficient quantity and right quality of raw materials are available.
• To ensure that capacity usage always aligns with forecasted demand.

Characteristics of production planning:

• It is a universal pre-production activity.
• It is the basis and pre-requisite for control of production.
• It consists of routing and layout of production activity.
• It includes planning, directing, and controlling the manufacturing method.

Levels of production planning:

• Strategic planning
• Tactical planning
• Operational planning

Thus, option (1) is the correct answer.

Additional Information Production control:

• To achieve the aim of production planning, various controlling methods are used for the optimum performance of the production system, this is known as production control.
• It includes dispatching, inspection, expediting, and evaluation.

Explanation:

Linear programming (LP)

• Linear programming (LP) in industrial engineering is used for the optimization of our limited resources when there is a number of alternate solutions possible for the problem like material selection.
• The real-life problems can be written in the form of a linear equation by specifying the relation between its variables.
• Linear programming can be applied effectively only if resources can be measured as quantities.
• Linear programming is used for obtaining the most optimal solution for a problem with given constraints.
• Using linear programming requires defined variables and constraints, to find the largest objective function (maximization).
• In some cases, linear programming is instead used for the smallest possible objective function value (minimization).
• Linear programming requires the creation of inequalities and then graphing those to solve problems.
• Some linear programming can be done manually.
• When the variables and calculations become too complex and require the use of computational software.

Areas of applications of Linear Programming (LP):

​Industrial Application:

• Product mix problem
• Blending problem
• Procurement problem
• Production scheduling problem
• Assembly line balancing
• Make-or-Buy problem

Management applications:

• Media selection problems
• Portfolio selection problems
• Profit planning problems
• Transportation problems

Miscellaneous applications:

• Diet problem
• Agriculture problem
• Flight scheduling problem

Concept:

The mean waiting time of a patient, needing medical checkup facility in the queue, is,

${W_q} = \frac{\lambda }{{\mu \left( {\mu - \lambda } \right)}}$

Calculation:

λ = 32 per hour

$\mu = \frac{{60; \times ;60}}{{90}} = 40;{\rm{per;hour}}$

$\therefore {W_q} = \frac{{32}}{{40\left( {40 - 32} \right)}}$

Wq = 0.1 hour

∴ Wq = 6 minute

Explanation:

Work study:

• The basic form of work system design is work study, which is used to achieve the desired function by using effective methods.
• The word "work-study" refers generally to the techniques, method studies, and work measurement used to examine human work in all of its circumstances.
• In order to make improvements, all the aspects that affect the effectiveness and economy of the working condition are methodically investigated.

Importance of work-study:

• Improve the productivity or efficiency of the organization by removing unnecessary or waste operations.
• Remove the non-value-adding processes affecting the job.
• Helps in establishing an accurate and systematic time standard.
• It also contributes to the organization's profit.

Thus, option (1) is the correct answer.

Additional Information Procedure of work-study:

1. Selection of job or process
2. Record: recording data related to the job
3. Examine: examine recorded data
4. Develop: develop an effective method
5. Measure: estimate the standard time required for the job
6. Define: new process and time standard
7. Install: a new process for practice
8. Maintain the new process

Advantages of work-study:

• Helps in achieving smooth production operations
• Minimize the product cost by removing the waste operations
• Helps in maintaining the good working condition
• helps in developing accurate standard time for operation
• Better work-layout.

Explanation:

Method study:

• Method study is the systematic recording and analysis of current and planned work processes with the goal of creating and implementing simpler, more efficient processes while also lowering costs.
• The main objectives of the method study are:
• provide the exact facts related to the working condition
• critically analyze the facts
• provide the best possible solution after examining the facts.
• The procedure involved in the method study are:

1. SELECT
2. RECORD
3. EXAMINE
4. DEVELOP
5. DEFINE
6. INSTALL
7. MAINTAIN

Thus, option (4) is the correct answer.

Additional Information Advantages of method study:

• Improve the material and plant utilization
• Increase the efficiency of labor by providing better working conditions
• Improve work layout by providing a smooth flow of material
• helps in selecting the best possible sequence of doing work
• Improve work method and work procedure.

Explanation:

Batch production:

• Batch production is suited for medium-volume lot with moderate product variety.
• The production order is repeated at regular intervals.
• Usually, the equipment of production is general purpose and suited for a higher volume of production.
• Special types of jigs and fixtures are used to minimize the time of setup and increase the rate of production.
• Industrial equipment, textbooks, furniture, household appliances, lawn movers, etc are items made by batch production shops.

Advantages of batch production:

• Better use of machine and plant
• Low cost per unit
• The investment cost is low
• Flexible to process a number of products.

Limitations of batch production:

• Complex production planning and control
• Complex material handling due to irregular flow of material
• High work-in-process inventory.

Thus, option (1) is the correct answer.

Additional Information

 Mass production:

• Mass production is suited for a large volume of production with a low variety of products.
• Generally, mass production is also known as continuous production or line production.
• In mass production, the equipment used is only dedicated to a particular product i.e., special-purpose equipment is used.
• It has a high investment cost for machines and tools.

Job shop production:

• Job shop production is suited for low volume of production with a small size of lot.
• Generally, it is used for specific customer orders.
• Aircraft, machine tools, space vehicles, etc are the products manufactured in a job shop.

Explanation:

Cellular layout:

• Cellular Manufacturing Layout (Group Technology or GT layout): This layout is based on the Group Technology (GT) principle.
• The idea of GT is to divide the manufacturing facility into small groups or cells, where each cell functions as a stand-alone system and is dedicated to manufacturing a family of parts.
• Each part in the family will require similar machine types and operations. This layout reduces the movement of material, thus decreasing production time and costs.

​

Thus, option (2) is the correct answer.

Additional Information Process layout:

• It is also known as a functional layout.
• In the process layout, machines of the same types are placed together.
• The process layout is used for a low volume of products with a high variety.
• Example of process layout is a machine shop, assembly line, etc.

Product layout:

• It is also known as a flow-shop layout.
• Flow shops manufacture high-volume and highly standardized products.
• Resources are allocated sequentially according to the routing of the goods in a product layout.
• An example of a product layout is automobile manufacturing.

Fixed-positioned layout:

• The fixed-positioned layout is suited for a low volume of products having low variety.
• The fixed-positioned layout is the opposite of the process layout.
• In a fixed-positioned layout, the products are placed at a fixed position, and the machining equipment is moved to the products.
• Ships, submarines, and rockets are products made by using the fixed-positioned layout.

Explanation:

Chip thickness ratio / Cutting ratio (r):

It is the ratio of chip thickness before cut (t1) to the chip thickness after cut (t2).

$r=\frac{Chip;thickness;before;cut;(t_1)}{Chip;thickness;after;cut;(t_2)}\Rightarrow \frac{uncut;chip;thickness}{chip;thickness}$

chip thickness after the cut (t2) is always greater than the chip thickness before the cut (t1), ∴ r is always < 1, i.e. the uncut chip thickness value is less than the chip thickness value.

Assuming volume to be constant:

t1b1L1 = t2b2L2

$\frac{t_1}{t_2}=\frac{L_2}{L_1};;;;(\because b_1=b_2)=\left(\frac{\text {length of chip}}{\text {length of uncut chip}}\right)$

as t2 > t1, ∴ L2 < L1 i.e. length after cut is less than the length before the cut.

Assuming discharge to be constant:

t1b1V = t2b2Vc

$\frac{t_1}{t_2}=\frac{V_c}{V};;;;(\because b_1=b_2)=\frac{Chip \ Velocity}{Cutting \ Velocity}$

as t2 > t1, ∴ V > Vc i.e. cutting velocity is greater than the chip velocity.

Explanation:

Grinding:

• Grinding is a process that is intermittent in nature and produces discontinuous chips.
• It involves removing material from a workpiece using an abrasive grinding wheel or belt. The wheel rotates and grinds the surface of the workpiece, producing small chips that are typically discontinuous.
• Grinding is used to finish workpieces that must show high surface quality and high accuracy of shape and dimension.
• The accuracy in dimensions in grinding is of the order of 0.000025 mm.
• According to the accuracy of the work to be done on a grinding machine, they are classified as Rough grinding machines and precision grinding machines.
• For grinding a job, the right grinding wheel is to be selected. The selection of a grinding wheel will depend on the following factor:
• Material to be ground
• Amount of stock to be removed
• Area of contact
• Finish required
• Wheel speed
• Work speed
• For grinding hard material, a soft wheel and for grinding soft material, a hard wheel is selected
• Specific energy requirements in a grinding process are more than those in turning for the same removal rate because of the size effect of the larger contact areas between wheel and work.
• A grinding wheel is a multi-point cutting tool which consists of a large number of abrasives.
• All the grains are not involved in cutting during the operation and few grains become blunt during the process and will only be ploughing against the workpiece.
• This process repeats as the blunt grains breaks / falls apart exposing fresh and sharp cutting edge , while the the other grains which were involved in cutting get blunt and start to plough until they break to expose sharp edges.
• Since this is a continuous process , and at all points of time there are a number of grains that are not involved in cutting and are just rubbing against the surface will increase the specific energy consumption to overcome the friction in order to break the blunt grain.

Power for Grinding operation of metal:

Power, P = Fc × V = Fc × \(\frac{\pi~×D×~N}{60}\) = e × MRR

Where Fc = Tangential cutting force, V = Grinding speed, N = Speed of wheel in rpm, e = Specific energy consumption, MRR = Material removal rate, D == Grinding wheel diameter.

Major advantages and applications of grinding:

• dimensional accuracy
• good surface finish
• good form and locational accuracy
• applicable to both hardened and unhardened material

Applications:

• It is used for slitting and parting.
• finishing of flat as well as cylindrical surfaces.
• grinding of tools and cutters and resharpening of the same..
• stock removal (abrasive milling).

Explanation:

Boring:

• Boring is the process of enlarging a pre-existing hole, which could have been formed by a drill or by a casting's core using a single-point cutting tool.
• Making the hole concentric with the work piece's axis of rotation may be an equally significant and concurrent objective of boring. This would correct any eccentricity that might have occurred as a result of the drill diverting off the centerline. One crucial characteristic of drilled holes is concentricity.
• The work is often held in a chuck or on a faceplate when boring is done in a lathe.
• Generally, boring is the internal turning process in which feed to the tool is provided parallel to the rotational axis of the work-piece.
• The principle of boring is similar to turning.
• The instrument must be positioned precisely at correct position with the axis of rotation.
• Sometimes it's necessary to utilize somewhat larger end clearance angles to avoid having the tool's heel scrape against the interior of the hole.

Thus, option (1) is the correct answer.

Explanation:

Herringbone Gear:

• Herringbone gears, also known as double helical gears, are a type of gear that consists of two mirrored helical gears that are placed side by side. The teeth on each gear are angled in opposite directions, which creates a "V" shape. This design helps to reduce noise and vibration that can occur in standard helical gears.

​

• The main advantage of herringbone gears is their ability to transmit power smoothly and quietly. The opposing angles of the teeth on the two gears cancel out any axial thrust that would occur in a standard helical gear, which can result in smoother operation and less wear and tear on the gears.
• Another advantage of herringbone gears is their ability to handle higher loads than other types of gears. The design of the gear teeth provides a larger surface area for contact between the gears, which allows for greater torque transmission and higher power transfer.
• However, herringbone gears can be more difficult and expensive to manufacture than other types of gears, due to the complexity of their design. The two helical gears must be precisely aligned and positioned to ensure that the teeth mesh properly and the opposing angles cancel out the axial thrust. This requires greater manufacturing precision and accuracy, which can drive up the cost of production.

Explanation:

Embossing:

• It is a very shallow drawing operation where the depth of the draw is limited to one to three times the thickness of the metal, and the material thickness remains largely unchanged.
• Embossing is a metal-forming operation, which is used to make an impression of a figure, letter, or any design on a metal sheet.
• The embossing process involves drawing as well as bending of the metal sheets.
• It consists of a punch and dies having the required design/contour.
• The clearance between the punch and die is equal to the thickness of the metal sheet.

Thus, option (1) is the correct answer.

Additional Information Coining:

• Coining is similar to cold forging, with the exception that the metal only flows through the top layers and not the full volume.
• Coining is a pressing operation.
• Coining is used for making coins, medals, and similar articles.

Shot peening:

• Shot peening is a cold working process,
• It is mainly employed to improve the fatigue resistance of metal by setting up compressive stresses on the surface.
• In this process, small spherical balls are bombarded on the surface of the part.

Spinning:

• Spinning is a cold-forming operation in which the rotating sheet is shaped over a mandrel.
• This process is done on a lathe machine.
• In this process, the sheet metal disc is rotating at a very high speed and the roller is pushed against the rotating disc which pushes the blank against the mandrel, and the desired shape is formed.
• This process is used to manufacture cookware, musical instruments, rocket nose cones, etc.

Explanation:

Strain hardening:

• When yielding has ended, a further load can be applied to the specimen, resulting in a curve that rises continuously but becomes flattered until it reaches maximum stress referred to as ultimate stress. The rise in the curve is known as strain hardening.

• Strain hardening is the strengthening of a metal or polymer by plastic deformation. The strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material.

• Strain hardening increases the strength of the metal, including its hardness and tensile strength. It also decreases the ductility of the metal, meaning that it becomes less able to deform without cracking or breaking. However, it does not decrease the yield strength of the metal.

• In a strain-hardening material stiffening effect is felt at some stage. The strain hardening region is shown in the figure above. The yield plateau describes the amount of plastic flow.

• Strain hardening in the case of steel occurs between yield strength and ultimate strength.

​

Concept:

High Speed Steel:

• High-speed steels are used for cutting metals at a much higher cutting speed than ordinary carbon tool steels.
• The high-speed steels have the valuable property of retaining their hardness even when heated to red heat.
• Most of the high-speed steels contain tungsten as the chief alloying element, but other elements like cobalt, chromium, vanadium, etc. may be present in some proportion.
• 18 - 4 - 1 High-speed steel contains 18 % tungsten, 4 % chromium, and 1 % vanadium.

Explanation:

Normalizing:

• In normalizing of steel the steel specimen is heated from 30°C to 50°C above its upper critical temp, then it is held there for about fifteen minutes and then allowed to cool down in still air.

Effects of Normalizing:

• Refine the grain structure to fine equiaxial pearlite.
• Strain caused due to cold working is removed.
• Remove dislocations caused in the internal structure due to hot working.
• It restores the ductility of cold and hot worked material without appreciable alteration in their properties.

Important Points

Other heat treatment processes are:

 Full annealing: Metal is heated above the upper critical temperature & held there until the temperature of the work piece is uniform throughout, and finally cooling the workpiece at a slow controlled rate in the furnace so that the temperature of the surface and that of the centre of the workpiece is approximately the same.

Benefits of annealing are:

• Relieve stresses
• Increase softness, ductility and toughness
• Produce a specific microstructure

Process annealing

After cold working, the metal can be softened by process annealing or "recrystallization“ to reduce the distortions of the crystal lattice produced by cold working.

Spheroidising

Heat them to slightly above the critical temperature, hold them at this temp for a period of time, and then letting them cool in the furnace. Spheroidizing produces a rounded or globular form of carbide. It improves abrasion resistance.

Explanation:

Eutectoid reaction:

• In a eutectoid reaction, one solid is decomposed into two different solids on cooling i.e., austenite decomposes into ferrite (α-iron) and cementite on cooling.
• The eutectoid reaction occurs at 0.8% carbon and 725^oC.
• The composition of carbon in ferrite is 0.025% and in cementite is 6.67%.
• The eutectoid mixture of ferrite and cementite is known as pearlite.
• In the case of cooling, the eutectoid reaction is:

Austenite → ferrite (α-iron) + cementite

Thus, option (4) is the correct answer.

Additional InformationFerrite (α-iron):

• A ferrite is a structural form of pure iron at room temperature.
• It is ductile and soft in nature.
• The structure of ferrite is body-centered cubic (BCC).
• At 723^oC, the maximum amount of carbon in ferrite is 0.05%.

Austenite (γ-iron):

• Austenite is a stable form of pure iron between 910^oC to 1400^oC.
• The structure of austenite is face-centered cubic (FCC).
• At 1130^oC, austenite has a maximum carbon content of 2.1%.
• Above 1400^oC, the structure of austenite changes into BCC, and the form of iron is known as δ-iron.

Cementite (Fe₃C):

• Cementite has a maximum amount of carbon content (6.67%).
• It has an orthorhombic unit cell having 12 iron atoms and 4 carbon atoms per cell.
• Cementite is very hard and brittle.

Phase diagram of Fe and Fe₃C:

• It consists of five phases and four invariant reactions.
• The five phases are α-iron (ferrite), austenite (γ-iron), δ-iron, cementite, and liquid Fe-C solution.
• The four invariant reactions are peritectic reaction, eutectic reaction, eutectoid reaction, and monotectic reaction.
• Peritectic reaction: This reaction occurs at 1495^oC and at 0.16%C and the reaction is:

δ-iron + L → austenite (γ-iron)

• Eutectic reaction: This reaction occurs at 1147^oC and 4.3%C and the reaction is:

L → γ-iron + cementite

• Monotectic reaction: This reaction occurs at 1495^oC and 0.51%C, and the reaction is:

L → L + γ-iron

Explanation:

Thermoplasts:

• Polymers that soften when heated and stiffen when cooled, with both processes being completely reversible are known as a thermoplastic polymers.
• Thermoplastic polymers can be recycled.
• Thermoplastic polymers consist of long linear molecular chains held together by weak secondary bonds or by inter-winding.
• In thermoplastic polymers, cross-linking is absent between the molecular chains.
• Examples of thermoplastic polymers are Acrylics, PVC, Nylons, etc.
• The thermoplastics are synthesized by addition polymerization.
• The melting point and tensile strength of thermoplastic are low.

Thus, option (2) is the correct answer.

Additional Information Thermosetting polymer:

• Plastics that are ‘set’ under the application of heat and/or pressure is known as a thermosetting polymer.
• The thermosetting polymers can not be recycled.
• The thermosetting polymers consist of three-dimensional structures having strong covenant bonds to make rigid solid.
• The thermosetting polymers are synthesized by condensation polymerization.
• When compared with thermoplastic, thermosetting has high modulus, rigidity, and dimensional stability.
• Examples of thermosetting polymers are Epoxies, Amino resins, etc.

Explanation:

Types of chip:

• The chips produced while machining any part depend upon the material of the part and the machining condition.
• The types of chips formed during machining are:
• Discontinuous chip
• Continuous chip (without built-up edge)
• Continuous chip (with a built-up edge)
• Serrated chip

Discontinuous chips:

• The conditions for forming discontinuous chips are:
• machining brittle material like grey cast iron (irregular shape and size)
• ductile but hard and work hardenable (regular shape and size)
• Large feed rate
• large depth of cut
• negative rake angle
• low cutting speed
• inadequate cutting fluid.

Continuous chips (without built-up edge):

• The conditions for forming continuous chips without built-up edges are:
• ductile work material like steel
• high cutting velocity
• low feed rate
• positive rake angle
• low depth of cut

Continuous chips (with a built-up edge):

• The conditions for forming continuous chips with built-up edges are:
• Ductile work material
• medium cutting velocity
• large feed
• small rake angle

Serrated chips:

• It is also known as segmented or non‐homogeneous chips which are semi‐continuous chips having low and high shear strain zones.
• Metals with low thermal conductivity and strength that decreases sharply with temperature, such as titanium exhibit this behavior.
• The chips have a sawtooth-like appearance.

Thus, option (1) is the correct answer.

Explanation:

Single point cutting tool

Face:

• Face is the surface of the tool on which chip impinges when separated from work-piece

Side Cutting Edge Angle:

• It is also known as lead angle, is the angle between the side cutting edge and the side of the tool shank.
• The complementary angle of SCEA is called the "Approach angle**.**
• It is also called lead angle or principle cutting angle.

End Cutting Edge Angle:

• The end cutting edge angle is the angle between the end cutting edge and a line normal to the shank of the cutting tool. It is an important parameter in single-point cutting tools as it affects the cutting forces, chip formation, and surface finish of the machined part.
• The end cutting edge angle is typically measured in degrees and can be positive, negative or zero. A positive end cutting edge angle means that the cutting edge is inclined towards the direction of motion of the workpiece, while a negative end cutting edge angle means that the cutting edge is inclined away from the direction of motion of the workpiece. A zero end cutting edge angle means that the cutting edge is perpendicular to the direction of motion of the workpiece.

Side Relief Angle:

• The angle between the portion of the side flank immediately below the side cutting edge and the line perpendicular to the base of the tool measured at right angles to the side flank is known as side relief angle.
• It is the angle that prevents interference, as the tool enters the work material.

End Relief Angle:

• End relief angle is the angle between the portion of the end flank immediately below the end cutting edge and the line perpendicular to the base of the tool, measured at right angles to end flank.
• It is the angle that allows the tool to cut without rubbing on the work-piece.

Back Rake Angle:

• The angle between the face of the tool and a line parallel with the base of the tool, measured in a perpendicular plane through the side cutting edge is called back rake angle.
• It is the angle that measures the slope of the face of the tool from the nose toward the rear.
• If the slope is downward toward the nose, it is a negative back rake angle. And if the slope is downward from the nose, it is a positive back rake angle. If there is not any slope, the back rake angle is zero.

Side Rake Angle:

• The angle between the face of the tool and a line parallel with the base of the tool, measured in a plane perpendicular to the base and side cutting edge is called the side rake angle.
• It is the angle that measures the slope of the tool face from cutting edge. If the slope is towards the cutting edge, it is a negative side rake angle.
• If the slope is away from the cutting edge, it is a positive side rake angle.

Concept:

In case of thin cylinders, the hoop stress is determined by assuming it to be uniform across the thickness of the cylinder but in thick cylinders, the hoop stress is not uniform across the thickness, it varies from a maximum value at the inner circumference to a minimum value at the outer circumference.

For a thick cylinder hoop stress (tangential stress) is given by,

${σ _h} = \frac{{P;{{\left( {{r_i}} \right)}^2}}}{{{{\left( {{r_0}} \right)}^2} - {{\left( {{r_i}} \right)}^2}}}\left[ {1 + \frac{{{{\left( {{r_0}} \right)}^2}}}{{{x^2}}}} \right]$

For a thick cylinder Radial stress is given by,

${σ _r} = \frac{{P;{{\left( {{r_i}} \right)}^2}}}{{{{\left( {{r_0}} \right)}^2} - {{\left( {{r_i}} \right)}^2}}}\left[ {1 - \frac{{{{\left( {{r_0}} \right)}^2}}}{{{x^2}}}} \right]$

ri ≤ x ≤ ro

Where, ri and ro are inner and outer radius, respectively

The Variation of the hoop stress and the radial stress from the figure can be concluded as 

• The hoop stress varies parabolically from the maximum value at the inner radius to the minimum value at the outer radius
• The radial stress also varies parabolically but the maximum value is at the inner radius and at the outer radius the value is zero.

Explanation:

Close-coiled helical spring:

• The close-coiled helical spring has a very small helix angle.
• The pitch between adjacent turns is also small.
• A close-coiled helical spring subjected to torque is shown in the below figure.
• The torque from the support about the longitudinal axis must be balanced by another torque created about the longitudinal axis if the top section of the spring is to remain in equilibrium.
• This would occur if internal resistive forces created on the cross-section were to grow in a way that they would cancel out, producing no net force but creating a net moment about point A.
• The torque from the upper section will be balanced by this moment of the internal resistive forces created on the c/s around A, bringing the upper section of the spring to equilibrium.
• The spring wire experiences bending stress which leads to a decrease in the coil radius.
• The bending moment remains constant through the spring and its value is equal to the applied torque.
• The maximum bending stress experiences by the spring wire are given as:

$\sigma_{max}=\frac{MY}{I_{NA}}=\frac{32T}{\pi d^3}$

Where, M = bending moment, Y = distance of any layer from the neutral axis, I_NA = area moment of inertia about the neutral axis, and d = diameter of the wire.

Bending moment (M) = T, Y = $\frac{d}{2}$, I_NA = $\frac{\pi d^4}{64}$

Thus, option (1) is the correct answer.

Explanation:

When two or more metals are mixed together to form metal, it is called an alloy.

Babbitt metal:

• Babbitt metal is a soft, white non-ferrous alloy that is used to provide a bearing surface.
• Babbitt metal also has properties that help it reduce friction which makes it a good material for use in a plain bearing.
• Babbitt metal is tin and lead-based alloy:

Tin based Babbitt metal: 

• It contains Sn (Tin) = 88%, Sb (antimony) = 8%, Cu (copper) = 4%.
• It possesses excellent anti-friction properties and sufficient mechanical strength, so most commonly used in bearing metal.

Lead-based Babbitt metal: 

• It contains Pb (Lead) = 85%, Sb (antimony) = 10%, Cu (copper) = 5%

Important Points

Here are some alloys and their properties:

Nichrome:

• Nichrome is an alloy of Manganese 1.5%, Nickel 75% to 78%, Chromium 20% to 23%, and a small percentage of Iron.
• Nichrome has low resistivity.
• Nichrome is a non-magnetic alloy that is made up of mainly chromium and nickel.
• It has a very high melting point, which makes it nichrome to make high-temperature wires.
• It is used in heating elements such as hair dryers, toasters, and ovens.
• Nichrome is used to make the coils used in water heaters.

Additional Information An alloy is a substance made by melting two or more elements together, at least one of the metals.

Explanation:

Duralumin:

• Duralumin is an alloy of aluminium that is extensively used for making automobile and aircraft components.
• The composition of duralumin is: 
• copper - 3 to 4%
• manganese - 0.4 to 0.7%
• magnesium - 0.4 to 0.7%
• remaining aluminium
• It has a maximum tensile strength (up to 400 Mpa) after age hardening and heat treatment.
• It is widely used in forging, stamping, bars, sheets, and tubes.

Thus, option (1) is the correct answer.

Additional Information Monel metal:

• It is an alloy of copper and nickel.
• The compositions in Monel metal are 68% nickel, 29% copper, and 3% other constituents (like iron, silicon, etc).
• It has superior corrosion-resisting properties as compared to bronze and brass.
• It is used in the manufacturing of pump fittings, propellers, turbine blades, etc.

Brass:

• It is an alloy of copper and zinc.
• It contains 50% of copper and 50% of zinc.
• It is used in the manufacturing of springs, bushes, electronic equipment, musical and dental instruments, etc.

Explanation:

Hastelloy:

• Hastelloy is an extremely high corrosion-resistant nickel superalloy.
• Its primary components are nickel, chromium, and molybdenum.
• Hastelloy metal resists high temperatures in addition to corrosion.
• The use of Hastelloy material is growing in the chemical, petrochemical, pharmaceutical, and oil & gas industries.
• The melting range of Hastelloy is 1323^oC to 1371^oC.

Constantan:

• Constantan is an alloy of copper and nickel.
• The composition of constituents are:
• Copper - 60%
• Nickel - 40%
• Its high specific heat resistance and great ductility allow it to be easily stretched out into wires and make it a good insulator.
• It has a low-temperature coefficient of resistance, which keeps its resistivity constant across a large temperature range.
• It is used for manufacturing thermocouples, Wheatstone bridge circuit, low-temperature heaters, and resistance.

Phosphor bronze:

• According to Indian standards, a typical type of phosphor bronze has the following composition:
• Tin = 9–10%,
• copper = 8–70%,
• phosphorus = 0.1–3%.
• It has a high degree of flexibility and excellent wear resistance.
• Castings become stronger, more ductile, and more sound when phosphorus is added.
• It is used for manufacturing pump components, gears, springs, power screw nuts, and bearings.

Inconel:

• The main components of Inconel are nickel, chromium, and iron.
• The composition of constituents are:
• Nickel - 80%
• Chromium - 14%
• Iron - 6%
• Even at high temperatures, this alloy has good mechanical characteristics.
• It is utilized to create springs that must tolerate extreme temperatures and corrosive action.
• Moreover, it is utilized for aviation engine exhaust manifolds.

Thus, option (2) is the correct answer.

Explanation:

<br>

In Mild steel percentage of carbon is in the range of 0.05% to 0.30%. Hence it comes under the category of low carbon steel.

Concept:

Mohr's circle is a graphical representation used in engineering mechanics to visualize the state of stress at a point in a material.

Analysis of Options:

1. State of stress at a point with reference to any arbitrary set of orthogonal axes passing through that point.
   

This is true because Mohr's circle can represent the stress state at any point with reference to any set of orthogonal axes.

2. Principal stresses at a point.
   

The principal stresses are represented by the points on the horizontal axis (where the shear stress is zero) on Mohr's circle. Thus, Mohr's circle can be used to determine the principal stresses.

3. One of the two direct stresses and shearing stress at a point.
   

This is true because any point on Mohr's circle represents a combination of normal stress (direct stress) and shear stress at a particular orientation.

4. All the above.
   

Since all the above statements are correct, this is the most inclusive and correct answer.

Conclusion:

The correct answer is:

4. All the above.
   

Additional Information

• Principal stress: The normal stresses acting on the plane of zero shear stress are known as principal stress.
• Direct stress: When the force applied on any body is parallel or co-linear to the axis, then the stress developed in the body is known as direct stress.
• Shear stress: The stress developed in the body, when two equal and opposite forces are acting tangentially on the body is known as shear stress.

Explanation:

Bulk modulus:

• When equal and like direct stresses acting on the body then the ratio of direct stress to the volumetric strain of the body is known as bulk modulus.
• Generally, the bulk modulus is represented by 'K' and is given as:

$K = \frac{Direct\space stress}{Volumetric\space strain}=-\frac{V\delta P}{dV}$

• The SI unit of the bulk modulus is N/m² or Pascal (Pa).
• The dimensional formula of the bulk modulus is [M¹L^-1T^-2].
• The (-)ve sign in the expression denotes that, with an increase in direct stress the volume of the body decreases.
• Thus, in equilibrium conditions, the bulk modulus is always positive and it depends upon applied direct stress and volumetric stress.
• The bulk modulus is also a function of Poisson's ratio.
• The bulk modulus of incompressible material is infinite.
• The relation between bulk modulus and Poisson's ratio is:

$K = \frac{E}{3(1-2μ)}$

Where, E = young's modulus, μ = Poisson's ratio

Thus, option (1) is the correct answer.

Additional InformationElastic constant:

• Elastic constants are used to determine the stress and strain relationships.
• For a homogeneous and isotropic material, the number of elastic constants is 4 i.e., young's modulus (E), bulk modulus (K), Poisoon's ratio (μ), and modulus of rigidity (G).

Poisson's ratio:

• It is defined as the ratio of lateral strain to longitudinal strain and denoted by:

$μ = \frac{Lateral\space strain}{Longitudinal\space strain}$ = $\frac{\delta d/d}{\delta l/l}$

Explanation:

Concept:

Simply supported beam:

• A simply supported beam is a beam that is supported at its two extreme ends by either two rollers, two hinges, or one roller and one hinge support.

Calculation:

Given:

Span/Length of the beam (l) = 4 m, total uniformly distributed load (W) = 20 kN

R_A + R_B = 20

Take moment about point A, then

(20×2) + (R_B×4) = 0

R_B = 10 kN and R_A = 10 kN

The maximum bending moment (BM_max) =  occurs at the midpoint of the simply supported beam.

In question, the total load is given which means, w = $\frac{W}{l}$

Then, the maximum bending moment is:

BM_max = $\frac{Wl}{8}$ = $\frac{20\times4}{8}$ = 10 kN-m​

Thus, option (4) is the correct answer.

Explanation:

Neutral Plane:

• When loaded by a bending force, the beam bends so that the inner surface is in compression and the outer surface is in tension.
• The neutral plane is the surface within the beam between these zones, where the material of the beam is not under stress 

Explanation:

Neutral Plane:

• When loaded by a bending force, the beam bends so that the inner surface is in compression and the outer surface is in tension.
• The neutral plane is the surface within the beam between these zones, where the material of the beam is not under stress (either compression or tension).
• As there is no lengthwise stress force on the neutral plane, there is no strain so when the beam bends, the length of the neutral plane remains constant.
• The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains.
• If the section is symmetric, isotropic, and is not curved before a bend occurs, then the neutral axis is at the geometric centroid.
• All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.
• As there is no lengthwise stress force on the neutral plane, there is no strain so when the beam bends, the length of the neutral plane remains constant.
• The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains.
• If the section is symmetric, isotropic, and is not curved before a bend occurs, then the neutral axis is at the geometric centroid.
• All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.

Concept:

A circular cross-sectional solid shaft is rigidly held at one end, and torque is applied at the free end. The material of the shaft is isotropically elastic and homogeneous. The circular section remains circular after loading, and the plane cross-section remains plane after loading. Each cross-section rotates as if rigid.

Shear Stress Distribution in a Circular Shaft:

When a circular shaft is subjected to a torque, the shear stress (ττ

) at a distance rr

from the center is given by:

$\tau = \frac{T \times r}{J}$

where T is the applied torque, r is the radial distance from the center, and J is the polar moment of inertia of the cross-section.

Key Points about Shear Stress in a Shaft:

The shear stress varies linearly with the radius rr

. The maximum shear stress occurs at the outer surface of the shaft (i.e., at the maximum radius r=Rr=R

). The shear stress is zero at the center of the shaft (r=0r=0

).

Given Options:

1. Shear stress will have the lowest value on the surface.
   

2. Shear stress will have different values on points lying on the circle of given radius.
   

3. Shear stress will be maximum at the center of the section.
   

4. Shear stress will have the same value at the point lying on the circle of given radius.

Correct Answer Analysis:

Option 1: This is incorrect. The shear stress is maximum at the surface.

Option 2: This is incorrect. For a given radius, the shear stress is the same at all points on the circle of that radius.

Option 3: This is incorrect. The shear stress is zero at the center of the section.

Option 4: This is correct. The shear stress will have the same value at all points lying on the circle of a given radius.

Conclusion:

The correct answer is:

4. Shear stress will have the same value at the point lying on the circle of given radius.

Concept:

Open Belt Drive:

• The open belt drive is used in shaft arranged parallel and rotating in the same direction.
• In this case, the drivers pulls from one side i.e. lower side and delivers it to the other side i.e. upper side.
• Thus the tension in the lower side belt will be more than in the upper side belt.
• Thus the lower side is known as tight side and the upper side is known as slack side.

Formulas in Belt drive:

Velocity ratio: \(\frac{N_2}{N_1}=\frac{d_1}{d_2}\)

Where d₁ = Diameter of the driver, d₂ = Diameter of the follower, N₁ = Speed of the driver, N₂ = Speed of the follower

Angle of lap, θ = 180° - 2α

α = $\frac{r_2~-~r_1}{x}$

Where r1 = radius of the driver, r2 = radius of the follower, x = centre distance between the driver and the follower

Power, P = (T₁ - T₂) × V

Where T₁ = Tension in the tight side, T₂ = Tension in the slack side

Ratio of driving tensions:

$\frac{T_1}{T_2}~=~e^{μ × \theta}$

Where μ = Coefficient of friction between the belt and the pulley

Tension in the tight side, T₁ = σ × b × t

Where, σ = safe stress in the belt, b = width of the belt, t = thickness of the belt

Calculation:

Given:

safe stress in the belt = 2.5 N/mm2, t = 10 mm, tight side tension (T₁) = 2400 N

∵ T₁ = σ × b × t

⇒ 2400 = 2.5 × b × 10

⇒ b = $\frac{2400}{2.5~\times~10}$

⇒ b = 96 mm

∴ Width of the belt = 96 mm

Note: In the revised official answer, this question was deleted because in the official question effect of centrifugal force was given to consider but velocity was not given so we neglected the centrifugal force while calculating.

Explanation:

Chain drive:

• A chain drive is a way of transmitting mechanical power from one place to another.
• It is often used to convey power to the wheels of a vehicle, particularly bicycles and motorcycles.
• It is also used in a wide variety of machines besides vehicles.
• Chains provide a constant velocity ratio and high transmission efficiency.
• Most often, the power is conveyed by a roller chain, known as the drive chain or transmission chain, passing over a sprocket gear, with the teeth of the gear meshing with the holes in the links of the chain.
• The gear is turned, and this pulls the chain putting mechanical force into the system.
• The gear is rotated, and this pulls the chain, exerting mechanical force in the mechanism.
• The chain is made up of several rigid links that are held together by pin joints to provide the necessary flexibility to wrap the driving and driven wheels.

Types of Chain Drives:

• Roller Chain
• Silent Chain
• Leaf Chain
• Flat Top Chain
• Engineering Steel Chain

Explanation:

Coupling:

• In mechanical system, coupling can be defined as a kind of linkage between two rotating shafts which joins driving and driven shafts together.
• The joint between two shafts may be permanent or temporary.
• In simple words we can say that coupling is used to join input and output shafts in any power transmission system such as in machine tools gear box shaft is connected to the input engine shaft by means of coupling; engine shaft with pump or compressor shafts etc.
• The function of the coupling is almost the same as the clutches but clutches are temporary joints while the coupling joints are permanent connections.

Types of Couplings:

• Oldham’s Coupling
• Hook’s Coupling
• Double Hook’s coupling
• Rigid Coupling
• Flexible coupling

Rigid Coupling:

• Rigid coupling is used where the axes of both shafts are collinear.
• Rigid couplings are simple and less costly, but the major drawback of this type of coupling is that they cannot tolerate any kind of small misalignment between the axis of the shafts.
• Different types of coupling come under this category some are them are as follow:
• Muff coupling
• Clamp coupling
• Flange coupling

Flexible coupling:

• This is another category of coupling.
• As name suggest it has flexible connection between the input and the output shaft because it is very difficult to make a proper aligned connection between the two mating shafts and there are various reasons for the misalignment between the two shafts such as deflection of shafts, thermal expansions etc. so to remove all such causes flexible coupling comes into the play.
• Flexible coupling able to tolerate small eccentricity between the two shafts and in flexible couplings some kind of flexible element know as bush are used as extra part which helps in reduction of vibrations and jerks so which make it to use in little vibration conditions.
• We can say that flexible couplings are the best for the power transmission between two shafts.
• The cost and the complex construction is the main disadvantage of this coupling.
• Oldham coupling, universal joint and bush pin type coupling comes under this category.

Additional InformationFlanged coupling:

• These are the standard forms of couplings, most extensively used. In a flanged coupling, flanges are either fitted or provided at the ends of shafts. The flanges are fastened together by means of a number of bolts and nuts. The number and size of the bolts depend upon the power to be transmitted and hence, the shaft diameter.

Universal Coupling:

• Coupling which allows a large angle between drive halves (20-30°). Generally, it is based on a yoke mounted on each shaft. Between to yokes in mounted a trunnion cross. Needle bearings are used at the bearing points between the cross and the yokes. These type or units are used in pairs on carden shafts. It is used widely on rear-wheel drive vehicle prop shafts. The other name of universal coupling is hook coupling.

Concept:-

Bearing modulus:

• Bearing modulus is a term used in the design of journal bearings, specifically hydrodynamic journal bearings.
• It's a dimensionless number that helps analyze the stability and operating conditions of the bearing.
• It is given by:

C = $\frac{ZN}{p}$

Where, Z = absolute viscosity (in N-s/m²), N = RPM, p = bearing pressure (N/mm²)

Calculation:-

Given:-

diameter of journal, D = 0.1 m = 100 mm, Length of the bearing, L = 0.15 m = 150 mm, Z = 0.02 N-s/m2, N = 700 rpm, Load, W = 15 kN

∵ p = \(\frac{W}{L~\timesD}\) = \(\frac{15\times10^3}{100\times150}=1N/mm^2\)

∴ C = $\frac{ZN}{p}$ = \(\frac{0.02~\times700}{1}=~14\)

Explanation:

Bearings:

• Bearing is a mechanical element that permits relative motion between two parts, such as the shaft and the housing, with minimum friction. The functions of the bearing are as follows:
  

(i) The bearing ensures free rotation of the shaft or the axle with minimum friction.

(ii) The bearing supports the shaft or the axle and holds it in the correct position.

(iii) The bearing takes up the forces that act on the shaft or the axle and transmits them to the frame or the foundation.

Depending upon the type of friction, bearings are classified into two main groups: 

• sliding contact bearings
• rolling contact

<br>

Roller bearings:

• Roller bearings, also known as rolling-element bearings, are similar to ball bearings in that they are designed to carry a load while minimizing friction.
• However, roller bearings transmit loads using cylinder rolling elements, rather than balls, to maintain the separation between moving parts of the bearing.
• These versatile bearings can contain single or multiple rows of rolling elements; multiple rows can significantly improve radial load capacity.
• Also, the use of different roller shapes can further reduce friction and support both radial and axial loads.
• While roller bearings can handle higher loads than conventional ball bearings, their applications are generally limited to low-speed operations.
• Many types of roller bearings are self-aligning, and are easily able to overcome misalignment and mounting issues, cutting down on maintenance, repair, and labor needs.
• Roller bearings come in a wide range of shapes and sizes, and can be customized for specialized situations.
• Also, the use of flanges, cages, and multiple bearing rows can allow for higher performance to meet specific application needs.

Roller Bearing Types:

• Cylindrical roller bearings
• Spherical roller bearings
• Needle roller bearings
• Taper roller bearings

Explanation:

Axial load:

• When a bar of uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings.
• In general terms, a load is an external force acting on a component.
• An axial load acts perpendicular to the geometric cross-section of the component.
• Based on the direction of the application of the load, an axial load can be tensile or compressive.
• The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.
• If a body is loaded in such a way that the force passes through the centroids of all resisting cross-sections, then the body is loaded under axial load.

Torsion:

• A torsion force is a load that is a applied to a material through torque.
• The torque that is applied creates a shear stress.
• If a torsion force is large enough, it can cause a material to undergo a twisting motion during elastic and plastic deformation.

Bending:

• A bending force is a force that is applied to a length of material.
• The bending force is applied to a point, area or volume that is some distance from a fixed portion of the component or structure to which the force is being applied.
• An excessive bending force can cause material failure, especially for materials that have corroded.
• In solid mechanics, a bending moment is the reaction induced in a structural element when an external force or moment is applied to the element, causing the element to bend.
• The most common or simplest structural element subjected to bending moments is the beam.

Uniformly distributed load:

• A uniformly distributed load is a type of load which acts in constant intensity throughout the span of a structural member.
• A uniformly distributed load is spread over a beam so that the rate of loading w is uniform along the length (i.e., each unit length is loaded at the same rate).
• The uniformly distributed load can act over a member in many forms, like hydrostatic force on a horizontal beam, the dead load of a beam, etc.
• The uniformly distributed load will be of the same intensity throughout the span of the beam.

Explanation:

In the case of reciprocating masses, Primary forces are partially balanced. Because in the reciprocating masses "the resultant forces" will be completely balanced but "the resultant couple" won't be balanced, that is why we say that reciprocating masses are only partially balanced.

Let F_R = Force required to accelerate the reciprocating parts,

F₁  =  Inertia force due to reciprocating parts,

F_N = Force on the sides of the cylinder walls or normal force acting on the cross-head guides, and

F_B = Force acting on the crankshaft bearing or main bearing.

Since F_R and F₁ are equal in magnitude but opposite in direction, therefore they balance each other. The horizontal component of FB (i.e. F_BH) acting along the line of reciprocation is also equal and opposite to F. This force F_BH = F_U is an unbalanced force or shaking force and required to be properly balanced.

The force on the sides of the cylinder walls (F_N) and the vertical component of F_B (i.e. F_BV) are equal and opposite and thus form a shaking couple of magnitude F_N × x or F_BV × x.

From above we see that the effect of the reciprocating parts is to produce a shaking force and a shaking couple. Since the shaking force and a shaking couple vary in magnitude and direction during the engine cycle, therefore they cause very objectionable vibrations.

22.2. Primary and Secondary Unbalanced Forces of Reciprocating Masses

Consider a reciprocating engine mechanism as shown in Fig. 22.1.

 

Let m = Mass of the reciprocating parts,

l  = Length of the connecting rod PC,

r = Radius of the crank OC,

θ = Angle of inclination of the crank with the line of stroke PO,

ω = Angular speed of the crank,

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

We have already discussed in Art. 15.8 that the acceleration of the reciprocating parts is approximately given by the expression,

$\rm a_R=ω^2.r\left(\cos θ+\frac{\cos 2θ}{n}\right)$

∴ Inertia force due to reciprocating parts or force required to accelerate the reciprocating parts,

F₁ = F₂ = Mass × acceleration = $\rm m.ω^2.r\left(\cos θ+\frac{\cos 2θ}{n}\right)$

We have discussed in the previous article that the horizontal component of the force exerted on the crank shaft bearing (1.e. F_BH) is equal and opposite to inertia force (F₁). This force is an unbalanced one and is denoted by F_U.

∴ Unbalanced force,

$\rm F_U=m.ω^2.r\left(\cos θ+\frac{\cos 2θ}{n}\right)=m.ω^2.r\cos θ+m.ω^2.r\times \frac{\cos 2θ}{n}=F_P+F_S$

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

Since reciprocating mass cannot balance fully hence partial balancing of the reciprocating mass is done while revolving masses can be balanced fully.

Primary unbalanced Force ${F_P} = mr{ω ^2}cosθ$

Secondary unbalanced Force = ${F_S} = \frac{{mr{ω ^2}cos2θ }}{n}$

Effect of Partial Balancing of Reciprocating Parts

Reciprocating parts are only partially balanced. Due to this partial balancing of the reciprocating parts, there is an unbalanced primary force along the line of stroke and also an unbalanced primary force perpendicular to the line of stroke. The effect of an unbalanced primary force along the line of stroke is to produce;

• Variation in tractive force along the line of stroke
• Swaying couple

Concept:

• For the shaft, the critical speed is given by,

${w_n} = \sqrt {\frac{9}{Δ }}$

Where Δ = The deflection of the shaft

• We know, Δ = f (Young's modulus  (E), Dimension (d), Span  (L))

Explanation:

• Thereby, the critical speed is also a function of all these parameters which means Young's modulus, Dimensions, and Span length of the shaft.
• The critical speed of the shaft will not be the function of the Eccentricity of the shaft.

Concept:-

Frequency is the number of oscillations per unit of time, and its unit is Hz.

Frequency= fn = Number of oscillations/time

For a single-degree damped vibrating system, the frequency can be expressed as,

⇒ $f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}};$      ....(1)

Where,

k = Stiffness of the spring

m = mass suspended 

Calculation:-

Given:-

m = 5 kg, No. Of oscillations = 21, time = 11 s, k = ?

The frequency of the spring-mass system is, 

fn = Number of oscillations/time = 21/11 = 1.909 Hz

Now putting this value of frequency in equation(1),

 $f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}};$

⇒ $1.909= \frac{1}{2\pi}\sqrt{\frac{k}{5}};$

⇒ k = 719.42 N/m

Or, k = .72 N/mm

Concept:

Tearing resistance of plate (per pitch length), P_t = (p - d) × t × σ_t

Where, p = pitch of the rivet, d = diameter of the rivet, t = Thickness of the plate, σ_t = safe working stress in tension

Shear resistance of plate, P_s = n × $\frac{\pi}{4}$ × d² × $\tau$

Where, n = Number of rivets, $\tau$ = safe working stress in shear

Crushing resistance of the rivet, P_c = n × d × t × σ_c

Where, d = diameter of the rivet, t = Thickness of the plate, n = Number of rivets, σ_c = safe working stress in crushing

Calculation:

Given:

n = 2, t = 10 mm, d = 20 mm, $\tau$ = 56 N/mm2 , σt = 80 N/mm2 

∵ Ps = n × $\frac{\pi}{4}$ × d2 × $\tau$

⇒ Ps = 2 × $\frac{\pi}{4}$ × 202 × 56

⇒ Ps = 35185.83 N ................................................. (1)

∵ Pt = (p - d) × t × σ_t

⇒ Pt = (p - 20) × 10 × 80 = 800 × (p - 20) ................ (2)

For the riveted joint not to fail, Tearing and shearing strength should be equated,

Equating (1)and (2):

⇒ 800 × (p - 20) = 35185.83

⇒ (p - 20) = 43.98

⇒ p = 63.98 mm

The rivet pitch will be 64 mm

Concept:

Total force or external load acting on the cylinder = Pressure × Area of the cylinder head

⇒ Total force or external load acting on the cylinder = p × \(\frac{\pi}{4}\timesd^2\)

External load on the cylinder head per bolt = $\frac{1}{n}$ × {p × \(\frac{\pi}{4}\timesd^2\)}

Where n = Number of bolts

Calculation:

Given:

p = 0.7 N/mm2, diameter of the cylinder, d = 0.2 m = 200 mm, n = 10

∵ Total force or external load acting on the cylinder = p × \(\frac{\pi}{4}\timesd^2\)

⇒ Total force or external load acting on the cylinder = 0.7 × \(\frac{\pi}{4}\times200^2\) = 21991.14 N

External load on the cylinder head per bolt = $\frac{1}{n}$ × {p × \(\frac{\pi}{4}\timesd^2\)}

⇒ External load on the cylinder head per bolt = \(\frac{1}{10}\times21991.14\) = 2199.11 N

∴ The external load on each bolt is 2200 N.

Explanation:

Square thread: 

• In this thread, the flanks are perpendicular to the axis of the thread.
• This is used for transmitting motion or power. E.g. Fly presses, Screw jack, vice handles, cross-slide and compound slide, etc.
• Square thread has the highest efficiency as compared to other power screws.
• The efficiency of a square thread is more than that of trapezoidal thread as its profile angle is zero.
• It can transmit load and power in both direction.
• There is no radial or side thrust on the nut.
• It is difficult to manufacture.
• More transmission efficiency due to less friction.
• Load carrying capacity is low due to less thickness at core diameter than trapezoidal threads.

Trapezoidal threads: 

• These threads have a profile which is neither square nor V thread form and have a form of a trapezoid. Different forms of trapezoidal threads are ACME thread, Buttress thread, Saw-tooth thread, Worm thread
• ACME thread: It has an included angle of 29°. These are used in lathe lead screws.
• Buttress Thread: In this thread, one flank is perpendicular to the axis of the thread and another flank is at 45°. These threads are used in power presses, carpentry vices, gun breeches, ratchets, etc.
• Knuckle thread: The shape of the knuckle thread is not trapezoid, but it has a rounded shape. It has limited applications. It is used for valve spindles, railway carriage coupling, hose connections, etc.

Explanation:

Kennedy's theorem:

• Kennedy's theorem states that if three bodies have plane motion relative to one another, then their I-centers i.e. instantaneous center must lie on a straight line.
• Hence if two rigid bodies A and B are moving in the same plane relative to a third rigid body C, then the instantaneous centers of A relative to C and B relative to A all lie in a straight line.

The number of instantaneous centers (N) is given by:

N = $\frac{n~\times~(n~-~1)}{2}$ 

Where n = number of links

Additional InformationInstantaneous center of velocity:

• The instantaneous center method of analyzing the motion in a mechanism is based on the concept that any displacement of a body having motion in one plane can be considered as a pure rotational motion of a rigid link as a whole about the same center, known as instantaneous center or virtual center of motion.
• The instant center of rotation (also, instantaneous velocity center, instantaneous center, or instant center) is the point fixed to a body undergoing planar movement that has zero velocity at a particular instant of time.

Properties of instantaneous center of rotation:

• Each configuration of the link has one instantaneous center. Instantaneous center changes with the alteration of configuration of mechanism.
• Instantaneous center is one about which the body has pure rotation. Hence, for the body having straight line motion, the radius of curvature of it is at infinity and hence, instantaneous center of this lies at infinity.
• An instantaneous center helps to determine the velocity of any point of a body if the direction of velocity of two points and magnitude of velocity of one point is known.

Explanation:

Let a link AR rotate about a fixed point A on it. P is a point on a slider on the link.

At any given instant, Let:

ω = Angular velocity of the link,

α = Angular acceleration of the link,

v = Linear velocity of the slider on the link,

f = Linear acceleration of the slider on the link,

r = Radial distance of point P on the slider

In a short interval of time δt, let δθ be the angular displacement of the link and δr be the displacement of the slider in the outward direction.

After the short interval of time δt, let

ω′ = ω + α ∙ δt = angular velocity of the link.

v′ = v + f∙ δt = Linear velocity of the slider on the link.

r′ =r + δr = Radial distance of the slider.

Acceleration of P perpendicular to AR:

Initial velocity of P perpendicular to AR = ωr

Final velocity of P perpendicular to AR = v′sin δθ + ω′r′ cos δθ

Change of velocity perpendicular to AR = (v′sinδθ + ω′r′cosδθ) ‒ ωr

<br>

Acceleration of P perpendicular to AR = \(\frac{(v~+δ t)\cos δ θ-(ω+α δ t)(r+δ r)\cos δ θ-~ω r}{δ t}\)

In the limit, as δt → 0, cos δθ → 1 and sin δθ → δθ

Acceleration of P perpendicular to AR = \(v\frac{d\theta}{dt}~+ω\frac{dr}{dt}+~rα\)

⇒ Acceleration of P perpendicular to AR = vω + vω + αr

⇒ Acceleration of P perpendicular to AR = 2vω + αr

And Acceleration of P parallel to AR = f - ω2r

Pitch Curve:

• It is the curve drawn by the trace point assuming that the cam is fixed, and the trace point of the follower rotates around the cam i.e. if we hold the cam fixed and rotate the follower in a direction opposite to that of the cam, then the curve generated by the locus of the trace point is called pitch curve.
• For a knife-edge follower, the pitch curve and the cam profile are the same whereas for a roller follower they are separated by the radius of the roller.

Prime Circle:

• The prime circle is the smallest circle that can be drawn so as to be tangential to the pitch curve, with its centre at the cam centre.

Base Circle:

• It is the smallest circle, keeping the centre at the cam centre, drawn tangential to cam profile.
• The base circle decides the overall size of the cam and thus it is a fundamental feature.

Concept:

The kinetic energy of the flywheel is given by, $E = \frac{1}{2}I{\omega ^2}$

Where, I = mass moment of flywheel = m × K2, and ω is the angular velocity of the flywheel.

Angular velocity ω = ω0 + αt

Where α is the angular acceleration of the flywheel

Torque, T = Iα

Calculation:

Given,

T = 1000 N-m, radius of gyration K= 500 mm = 0.5 m and m = 1000 kg, ω0 = 0 and t = 10 sec

I = 1000 × 0.52 = 250 kg.m2

So the angular acceleration $\alpha = \frac{T}{I} = ;\frac{{1000}}{{250}} = 4;rad/{s^2}$

Angular velocity ω = ω0 + αt

⇒ ω = 0 + 4 × 10 = 40 rad/sec

The kinetic energy of the flywheel is, = $\frac{1}{2} \times 250 \times {40^2} = 200;\times10^3;J=200;kJ$

Explanation:

Governor Effort

• It is the force exerted by the governor at the sleeve as the sleeve tends to move.
• When the speed of the governor is constant, the force exerted on the sleeve is zero as the sleeve doesn't tend to move and hence at the constant speed, the effort of the governor is zero, but when the speed changes and the sleeve tends to move to new equilibrium position force is exerted on the sleeve.
• This force gradually diminishes to zero as the sleeves move to a new equilibrium position corresponding to the new speed.
• The mean force exerted on the sleeve during the given change of speed is known as the effort of the governor.
• The given change of speed is generally taken as 1%, hence effort is defined as the force exerted on the sleeve for 1% change of speed.

Concept:

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

If the circular diameter is d, and No. of teeth is T

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

Pitch line velocity (VP) $= {\rm{\omega }}.r = \frac{{2\pi N}}{{60}} × \frac{d}{2}$

where m is the module of gear. (m = D/T)

Calculation:

Given:

number of teeth in spur gear (T) = 28

m = 2 mm 

speed (N) = 250 rpm

Now,

pc = πm

∴ pc = π × 2

∴ pc = 6.28 mm

$Pitch\ line\ velocity= \frac{{2\pi N}}{{60}} × \frac{{mT}}{2};;;;;;;\left( {as;m = \frac{d}{T}} \right)$

$= \frac{{ 2 \pi× 250×2×28}}{{60×2}} = \frac{{87964.594}}{{120}} = 733.0382\frac{{mm}}{s}$

∴ Pitch line velocity = 733.0382× 60 = 43982 $\frac{mm}{min}$ $\approx$ 44000 $\frac{mm}{min}$