MES Jr. Grade • 2024 • Paper-III
With the increase of percentage of carbon in the steel, which one of the following properties does increase?
A)
Modulus of elasticity
B)
Ductility
C)
Toughness
D)
Hardness
In material science, steel is an alloy of iron and carbon. The properties of steel are heavily influenced by the carbon content:
1. **Hardness and Strength**: As the percentage of carbon increases, more cementite (Fe3C) is formed. Cementite is a very hard and brittle phase. Therefore, increasing carbon content increases the overall hardness and tensile strength of the steel.
2. **Ductility**: Ductility is the ability of a material to undergo permanent deformation before rupture. Carbon atoms distort the iron lattice and hinder the movement of dislocations. Thus, as carbon content increases, ductility decreases.
3. **Toughness**: Toughness is the ability of a material to absorb energy and deform plastically before fracturing. While strength increases with carbon, the loss of ductility leads to a decrease in overall toughness (the material becomes more brittle).
4. **Modulus of Elasticity**: The Modulus of Elasticity (Young's Modulus) is a property of the atomic bonds of the base metal (Iron). It is largely independent of the carbon content within the typical range of steel (up to 2%).
Therefore, among the given options, only hardness increases with an increase in carbon percentage.
MES Jr. Grade • 2024 • Paper-III
In orthogonal cutting, shear angle is the angle between
A)
shear plane and the cutting velocity
B)
shear plane and the rake plane
C)
shear plane and the vertical direction
D)
shear plane and the direction of elongation of crystals in the chip
In orthogonal cutting (a simplified 2D model of metal cutting), the shear angle (denoted by φ) is defined as the angle formed between the shear plane and the direction of the cutting velocity (or the surface of the workpiece).
1. **Shear Plane**: This is the plane along which the metal plastically deforms and shears to form the chip. It extends from the tool tip to the free surface of the workpiece.
2. **Cutting Velocity**: This is the vector representing the direction in which the tool moves relative to the workpiece.
3. **Geometric Definition**: The shear angle is the angle that the shear plane makes with the horizontal plane (the direction of tool travel/cutting velocity). A larger shear angle generally indicates a thinner chip and lower cutting forces.
Why other options are incorrect:
- Option (b): The angle between the shear plane and the rake plane is not the shear angle; the rake angle is the angle between the rake face and a plane perpendicular to the cutting velocity.
- Option (c): The vertical direction is perpendicular to the cutting velocity; the shear angle is measured from the horizontal, not the vertical.
- Option (d): While crystals elongate in the chip, the shear angle is a geometric parameter of the cutting process, not a crystallographic measurement.
MES Jr. Grade • 2024 • Paper-III
Which one of the following factors is more relevant to represent complete solubility of two metals in each other?
A)
Chemical affinity
B)
Valency factor
C)
Crystal structure factor
D)
Relative size factor
The question refers to the conditions required for the formation of a substitutional solid solution, specifically the Hume-Rothery rules. For two metals to exhibit complete solubility in each other, four primary conditions must be met:
1. Atomic Size Factor: The difference in atomic radii of the solute and solvent atoms should be less than approximately 15%. This is the most critical factor because a large difference in size creates significant lattice strain, limiting solubility.
2. Crystal Structure: The metals must have the same crystal structure (e.g., both FCC or both BCC).
3. Valency: Metals with similar valencies are more likely to be soluble.
4. Electronegativity: The metals should have similar electronegativities to avoid the formation of intermetallic compounds.
Among the given options, the 'Relative size factor' (Atomic Size Factor) is considered the most fundamental and relevant constraint. If the size difference is too great, solubility is severely restricted regardless of the other factors. Therefore, option (d) is the most appropriate answer.
MES Jr. Grade • 2024 • Paper-III
By which one of the following machines the teeth of an internal spur gear can be cut accurately?
A)
Millingmachine
B)
Slotting machine
C)
Hobbing machine
D)
Gear-shaping machine
To cut the teeth of an internal spur gear, the cutting tool must be able to enter the internal diameter of the gear blank and perform a reciprocating motion.
1. **Gear-shaping machine (Correct):** This machine uses a reciprocating cutter (similar to a shaper) that moves linearly. Because the cutter is a bar-like tool that can be inserted into the bore of a gear blank, it is the primary and most accurate method for cutting internal gears. The tool and the workpiece rotate in synchronization to generate the involute profile.
2. **Hobbing machine (Incorrect):** Hobbing is the most efficient method for external gears. However, a hobbing cutter is a rotating worm-like tool that cannot enter the internal cavity of a gear blank to cut internal teeth due to the geometry of the tool and the workpiece.
3. **Milling machine (Incorrect):** While a milling machine can cut gear teeth using a dividing head and a form cutter, it is generally used for external gears or very simple internal slots. It lacks the specialized synchronization required to cut accurate internal involute gear teeth efficiently.
4. **Slotting machine (Incorrect):** A slotting machine is used for cutting internal keyways, splines, or square holes. While it uses a reciprocating motion, it does not have the specialized gear-cutting mechanism (synchronous rotation of tool and workpiece) required to produce the precise involute curve of a gear tooth.
MES Jr. Grade • 2024 • Paper-III
The angle between side cutting edge and end cutting edge is called as
A)
Approach angle
B)
Nose angle
C)
Sidereliefangle
D)
End reliefangle
In single-point cutting tool geometry, the 'Nose' is the intersection of the side cutting edge and the end cutting edge. The angle formed between these two edges is known as the Nose Angle (or Nose Radius angle). This angle is critical because it prevents the tool from chipping at the tip and influences the surface finish of the workpiece.
Why other options are incorrect:
- a. Approach angle: This is the angle between the side cutting edge and the direction of the feed.
- c. Side relief angle: This is the angle provided to prevent the flank of the tool from rubbing against the workpiece side.
- d. End relief angle: This is the angle provided to prevent the end flank of the tool from rubbing against the workpiece end.
MES Jr. Grade • 2024 • Paper-III
The unique property of cast iron is its high-
A)
Malleability
B)
Ductility
C)
Surface finish
D)
Damping characteristics
Cast iron, particularly Grey Cast Iron, is renowned for its exceptional damping capacity. This is primarily due to the presence of graphite flakes within the ferrite or pearlite matrix. These graphite flakes act as internal energy absorbers that dissipate vibrational energy (mechanical vibrations) into heat, preventing the resonance of the material. This makes cast iron the ideal material for machine tool beds, engine blocks, and heavy foundations where vibration suppression is critical.
Why other options are incorrect:
- a. Malleability: Cast iron is generally brittle. While 'Malleable Cast Iron' exists, it is a specially heat-treated version; standard cast iron lacks malleability.
- b. Ductility: Cast iron has very low ductility (it cannot be drawn into wires or deformed plastically without fracturing) compared to steel.
- c. Surface finish: While cast iron has excellent machinability, 'surface finish' is a result of the machining process, not an inherent material property like damping.
MES Jr. Grade • 2024 • Paper-III
Which of the following constituents of steels is softest and least strong-
A)
Austenite
B)
Pearlite
C)
Ferrite
D)
Cementite
To determine the softest and least strong constituent of steel, we must analyze the mechanical properties of the common phases in the Iron-Carbon equilibrium diagram:
1. **Ferrite (α-iron):** This is a solid solution of carbon in BCC (Body-Centered Cubic) iron. It has a very low solubility for carbon, making it the softest and most ductile phase among the options. Because it lacks significant alloying or structural barriers to dislocation movement, it has the lowest strength.
2. **Austenite (γ-iron):** This is a solid solution of carbon in FCC (Face-Centered Cubic) iron. It is generally stronger and harder than ferrite, though it is still ductile. It typically exists at high temperatures.
3. **Pearlite:** This is a lamellar (layered) mixture of ferrite and cementite. Because it contains hard cementite plates that act as barriers to dislocation movement, it is significantly stronger and harder than pure ferrite.
4. **Cementite (Fe3C):** This is an intermetallic compound. It is extremely hard and brittle, making it the strongest (in terms of hardness/resistance to deformation) but least ductile constituent.
Comparing these, Ferrite is clearly the softest and least strong phase.
MES Jr. Grade • 2024 • Paper-III
In CNC machine tool, the part program entered into the computer memory
A)
Can be used only once
B)
Can be used again and again
C)
Can be used again but it has to be modified every time
D)
Cannot say
In Computer Numerical Control (CNC) machining, a part program is a sequence of instructions (G-codes and M-codes) stored in the machine's control unit memory. One of the primary advantages of CNC over manual machining is reproducibility. Once a program is written, debugged, and stored, it can be executed repeatedly to produce identical parts with high precision and consistency.
- Option (a) is incorrect because the purpose of digital storage is to avoid rewriting the program for every single part.
- Option (c) is incorrect because if the part geometry remains the same, no modification is required; the program is designed for repeatability.
- Option (d) is incorrect as the functionality of CNC memory is a well-defined engineering standard.
MES Jr. Grade • 2024 • Paper-III
In grey cast iron, carbon is present in the form of-
A)
Cementite
B)
Free carbon
C)
Flakes
D)
Spheroids
Grey cast iron is characterized by the presence of graphite in the form of thin, interconnected flakes. This occurs because the carbon in the molten metal precipitates as graphite during slow cooling. These graphite flakes act as internal stress raisers, which gives grey cast iron its characteristic brittle nature and excellent vibration-damping properties.
Analysis of other options:
- a. Cementite (Fe3C) is a hard, brittle intermetallic compound found in white cast iron or steel, but in grey cast iron, most of the carbon is converted to graphite.
- b. 'Free carbon' is a general term for graphite, but 'Flakes' is the specific morphological form that defines 'Grey' cast iron.
- d. Spheroids (or nodules) are characteristic of Ductile Cast Iron (Nodular Cast Iron), where additives like magnesium or cerium are used to force the graphite into spherical shapes to improve toughness.
MES Jr. Grade • 2024 • Paper-III
Drop forging is used to produce
A)
small components
B)
large components
C)
identical Components in large numbers
D)
medium-size components
Drop forging is a closed-die forging process where a workpiece is placed between two dies (upper and lower) and the upper die 'drops' with a heavy blow to force the metal to fill the die cavity.
1. Why Option (c) is correct: Because the dies are precision-machined to a specific shape, every part produced using the same set of dies will be identical. This makes it the ideal process for mass production of components requiring high strength and dimensional consistency (e.g., connecting rods, crankshafts).
2. Why Options (a), (b), and (d) are incorrect: While drop forging can be used for small, medium, or large components depending on the size of the hammer, the defining characteristic and primary purpose of using 'drop' (die) forging over 'open-die' forging is the ability to produce standardized, identical parts in large quantities. Size is a variable of the machine capacity, but 'identicality' and 'volume' are the characteristics of the process itself.
MES Jr. Grade • 2024 • Paper-III
What do Flexible Manufacturing systems (FMS) do?
A)
Co-ordinates the whole process of manufacturing and manufactures a part, component or product
B)
Completely manufactures a range of components without significant human intervention during the processing
C)
Moves and manipulates products, parts or tools
D)
Moves materials between operations
A Flexible Manufacturing System (FMS) is an integrated system of computer-controlled machining centers (CNC machines), automated material handling systems, and a central control computer. Its primary purpose is to coordinate the entire production process to allow the manufacturing of a variety of different parts in varying quantities with minimal downtime for changeovers.
Analysis of options:
- Option (a) is correct because FMS acts as an umbrella system that coordinates the various elements (machines, transport, and control) to complete the manufacturing of a product.
- Option (b) describes 'Automated Manufacturing' or 'Lights-out Manufacturing' more specifically, but FMS is defined by its 'flexibility' to handle different part types, not just the absence of humans.
- Option (c) describes the function of 'Industrial Robots'.
- Option (d) describes the function of 'Automated Material Handling Systems' (AMHS) or conveyors, which are only one component of an FMS, not the FMS itself.
MES Jr. Grade • 2024 • Paper-III
A built-up-edge is formed while machining
A)
ductile materials at high speed
B)
ductile materials at low speed
C)
brittle materials at high speed
D)
brittle materials at low speed
A Built-Up Edge (BUE) is a phenomenon in machining where fragments of the workpiece material weld themselves to the cutting edge of the tool. This occurs due to high pressure and friction at the tool-chip interface.
1. **Material Property (Ductility):** BUE occurs primarily in ductile materials (like aluminum, mild steel, or copper) because these materials have a tendency to undergo plastic deformation and adhere to the tool surface. Brittle materials (like cast iron) fracture rather than flow, making BUE formation unlikely. Therefore, options (c) and (d) are incorrect.
2. **Cutting Speed:** BUE is most prominent at low to medium cutting speeds. At low speeds, the temperature at the tool-chip interface is high enough to cause adhesion but not high enough to soften the material sufficiently to allow it to flow smoothly over the tool. As the cutting speed increases, the temperature rises, which reduces the strength of the adhered material and increases its flowability, eventually causing the BUE to disappear. Therefore, option (a) is incorrect.
Conclusion: BUE is characteristic of machining ductile materials at low speeds.
MES Jr. Grade • 2024 • Paper-III
In a blanking operation to produce steel washer, the maximum punch load used in 2 x 10^5 N. The plate thickness is 4 mm and percentage penetration is 25. The work done during this shearing operation is :
A)
200J
B)
400J
C)
600J
D)
800J
Step-by-step calculation:
1. **Identify Given Data**:
- Maximum punch load ($F_{max}$) = $2 \times 10^5$ N
- Plate thickness ($t$) = 4 mm = $0.004$ m
- Percentage penetration ($p$) = 25% = 0.25
2. **Understand the Shearing Process**:
In a blanking operation, the force does not remain constant. It increases linearly from zero to a maximum value ($F_{max}$) as the punch penetrates the material, and then drops sharply once the material fractures.
3. **Calculate the Distance of Penetration**:
The punch does not travel the full thickness of the plate to complete the cut. The fracture occurs after a certain penetration.
The distance over which the force acts is the thickness minus the penetration (or the distance until fracture).
However, in standard manufacturing technology problems for work done in shearing, the work is calculated as the area under the force-displacement curve.
The displacement ($d$) until fracture is: $d = t \times (1 - p)$ is not the standard approach; rather, the penetration $p$ refers to the distance the punch travels *before* the material shears.
Actually, the work done in shearing is typically approximated as the area of the triangle formed by the force-displacement curve:
$ ext{Work Done} = \frac{1}{2} \times F_{max} \times t \times (1 - p)$ is incorrect. The correct formula for work done in a shearing operation where the force increases linearly to $F_{max}$ and then drops is:
$ ext{Work Done} = \frac{1}{2} \times F_{max} \times t \times (1 - p)$? No, the standard formula is $\text{Work Done} = \frac{1}{2} \times F_{max} \times t \times (1 - p)$ if $p$ is the percentage of thickness remaining.
Let's re-evaluate: The punch travels a distance $t \times (1 - p)$ before the material fractures.
Distance $s = 4 \text{ mm} \times (1 - 0.25) = 4 \times 0.75 = 3 \text{ mm} = 0.003 \text{ m}$.
$ ext{Work Done} = \frac{1}{2} \times F_{max} \times s = \frac{1}{2} \times (2 \times 10^5) \times 0.003 = 10^5 \times 0.003 = 300 \text{ J}$.
Wait, let's check the definition of 'percentage penetration'. In many textbooks, penetration refers to the distance the punch travels *before* the crack propagates. If penetration is 25%, the punch travels $0.25 \times t$ before the force drops.
Distance $s = 4 \text{ mm} \times 0.25 = 1 \text{ mm} = 0.001 \text{ m}$.
$ ext{Work Done} = \frac{1}{2} \times F_{max} \times s = \frac{1}{2} \times (2 \times 10^5) \times 0.001 = 10^5 \times 0.001 = 100 \text{ J}$.
Let's consider the most common interpretation for this specific exam question: The work done is often calculated as the area of the trapezoid or triangle. If the force increases linearly to $F_{max}$ and the fracture occurs at $p \times t$ penetration:
$ ext{Work Done} = \frac{1}{2} \times F_{max} \times (p \times t) = \frac{1}{2} \times 200,000 \times (0.25 \times 0.004) = 100,000 \times 0.001 = 100 \text{ J}$.
However, if the 'penetration' refers to the amount of material *not* sheared (the burr height), the punch travels $t - (p \times t)$.
If the work is calculated as $ ext{Work} = F_{avg} \times t$, where $F_{avg} = F_{max}/2$ and we account for the penetration:
$ ext{Work} = \frac{1}{2} \times F_{max} \times t \times (1 - p) = 300 \text{ J}$.
Let's look at the options: 200, 400, 600, 800.
If we use the formula $ ext{Work} = F_{max} \times t \times (1 - p) / 2$ it doesn't match.
If we use $ ext{Work} = F_{max} \times t \times p / 2$ it doesn't match.
What if the force is constant? $ ext{Work} = F_{max} \times t \times (1-p) = 200,000 \times 0.003 = 600 \text{ J}$.
What if the force is constant and penetration is 25%? $ ext{Work} = 200,000 \times 0.001 = 200 \text{ J}$.
In many competitive exams for ME, the work done in blanking is simplified to $ ext{Work} = F_{max} \times t \times p$ (where $p$ is the penetration fraction).
$ ext{Work} = 200,000 \times 0.004 \times 0.25 = 200 \text{ J}$.
This matches option (a). This assumes the average force is effectively $F_{max}$ over the penetration distance or a specific energy model used in the PHED/MES syllabus.
MES Jr. Grade • 2024 • Paper-III
For 50% penetration of work material, a punch with single shear equal to thickness will
A)
reduce the punch load to half the value
B)
increase the punch load by half the value
C)
maintain the same punch load
D)
reduce the punch load to quarter load
In a punching or blanking operation, the force (punch load) required to shear the material is calculated by the formula: F = L × t × τ, where 'F' is the punch load, 'L' is the length of the cut (perimeter), 't' is the thickness of the material, and 'τ' is the shear strength of the material. In a standard single shear operation, the punch must penetrate the full thickness (t) of the material to complete the cut. If the operation is modified such that only 50% penetration of the work material is required (or if the effective thickness being sheared at any given moment is reduced by half), the value of 't' in the equation is effectively halved. Since the force F is directly proportional to the thickness t (F ∝ t), reducing the penetration/thickness by 50% results in the punch load being reduced to half of its original value. Therefore, option (a) is correct.
MES Jr. Grade • 2024 • Paper-III
In ultrasonic machining process, the material removal rate will be higher for materials with
A)
higher toughness
B)
higher ductility
C)
lower toughness
D)
higher fracture strain
Ultrasonic Machining (USM) is a mechanical material removal process where a tool vibrates at high frequency, driving abrasive particles into the workpiece. The primary mechanism of material removal is brittle fracture (micro-chipping) caused by the repeated impact of the abrasive grains.
1. **Brittle vs. Ductile Materials**: USM is most effective on hard and brittle materials (like glass, ceramics, and quartz). In brittle materials, the impact of the abrasive causes cracks to propagate quickly, leading to the removal of small chips.
2. **Toughness**: Toughness is the ability of a material to absorb energy and deform plastically before fracturing. A material with high toughness (or high ductility) will absorb the energy of the abrasive impact through plastic deformation rather than fracturing. This results in a significantly lower Material Removal Rate (MRR).
3. **Lower Toughness**: Materials with lower toughness are more prone to brittle fracture. Since USM relies on fracturing the material to remove it, a lower toughness ensures that the abrasive particles can easily create cracks and remove material efficiently, thereby increasing the MRR.
Therefore, options (a) higher toughness, (b) higher ductility, and (d) higher fracture strain all describe properties that resist brittle fracture, which would decrease the efficiency of the USM process.
MES Jr. Grade • 2024 • Paper-III
Poor machinability of centrifugally cast-iron pipe is due to
A)
chilling
B)
segregation
C)
dense structure
D)
high mould rotation speed
In centrifugal casting of cast-iron pipes, the molten metal is poured into a rotating mold. Because the mold wall is cold, the metal in direct contact with the mold surface cools very rapidly. This rapid cooling leads to a phenomenon called 'chilling,' where the carbon (which would normally form soft graphite flakes in gray cast iron) instead forms hard iron carbides (cementite). These carbides are extremely hard and abrasive, creating a 'hard skin' or 'chill zone' on the inner and outer surfaces of the pipe. This hard layer significantly increases tool wear and makes the material very difficult to machine, thereby resulting in poor machinability.
Regarding other options:
- b. Segregation: While centrifugal casting causes segregation (heavier elements move outward), this generally affects mechanical properties rather than causing the specific 'poor machinability' associated with the surface skin.
- c. Dense structure: A dense structure is generally a goal of centrifugal casting to eliminate porosity; it does not inherently cause poor machinability unless it is accompanied by chilling.
- d. High mould rotation speed: Rotation speed affects the distribution of impurities and the thickness of the wall, but the primary cause of the hard, unmachinable surface is the thermal gradient (chilling), not the speed of rotation itself.
MES Jr. Grade • 2024 • Paper-III
The fixed costs for a year is Rs. 8 lakhs, variable cost per unit is Rs. 40/- and the selling price of each unit is Rs. 200/-. If the annual estimated sales is Rs. 20,00,000/-, then the break-even volume is :
A)
2000
B)
3000
C)
3333
D)
5000
To find the break-even volume (BEP in units), we use the standard Break-Even Analysis formula.
1. **Identify Given Data:**
- Fixed Costs (FC) = Rs. 8,00,000
- Variable Cost per unit (VC) = Rs. 40
- Selling Price per unit (P) = Rs. 200
- Annual Estimated Sales = Rs. 20,00,000 (Note: This value is provided as context but is not required to calculate the break-even point itself, as BEP depends on costs and price per unit).
2. **Formula for Break-Even Volume (Units):**
Break-Even Point (Units) = Fixed Costs / (Selling Price per unit - Variable Cost per unit)
The term (Selling Price - Variable Cost) is also known as the 'Contribution Margin per unit'.
3. **Calculation:**
- Contribution Margin = 200 - 40 = Rs. 160 per unit
- BEP (Units) = 8,00,000 / 160
- BEP (Units) = 80,000 / 16
- BEP (Units) = 5,000 units
4. **Verification of Options:**
- Option (a) 2000: Incorrect.
- Option (b) 3000: Incorrect.
- Option (c) 3333: Incorrect.
- Option (d) 5000: Correct.
Therefore, the break-even volume is 5,000 units.
MES Jr. Grade • 2024 • Paper-III
Light and intricate parts with close dimensional tolerances of the order of + 0.005 mm are produced by
A)
investment casting
B)
die casting
C)
centrifugal casting
D)
shell mould casting
Step-by-step reasoning:
1. **Analysis of Requirements**: The question specifies three key requirements: 'light and intricate parts', 'close dimensional tolerances', and a specific precision order of approximately ±0.005 mm (though typically investment casting is cited for tolerances in the range of ±0.05 to ±0.1 mm, in the context of these specific options, it is the most precise).
2. **Evaluating Investment Casting**: Also known as the 'lost-wax process', it is specifically designed for producing complex, intricate geometries with high surface finish and excellent dimensional accuracy. Because the mold is a ceramic shell created around a wax pattern, there is no parting line, allowing for extreme intricacy.
3. **Evaluating Die Casting**: While die casting offers good tolerances and high production rates, it is generally used for non-ferrous metals and cannot achieve the same level of intricate internal detail or the extreme precision associated with high-end investment casting for specialized parts.
4. **Evaluating Centrifugal Casting**: This is used primarily for cylindrical parts (like pipes or bushings) to ensure a dense, grain-refined structure. It is not used for 'intricate' parts.
5. **Evaluating Shell Mould Casting**: This provides better accuracy than sand casting but is significantly less precise than investment casting and cannot produce the same level of intricate detail.
6. **Conclusion**: Investment casting is the industry standard for high-precision, intricate components (such as turbine blades or jewelry) that require the tightest tolerances among the listed casting methods.
MES Jr. Grade • 2024 • Paper-III
The process of improving the cutting action of grinding wheel is called............
A)
Truing
B)
Dressing
C)
Facing
D)
Clearing
In grinding operations, the cutting action of the wheel degrades over time due to two main phenomena: 'loading' (where chips of the workpiece material clog the pores of the wheel) and 'glazing' (where the abrasive grains become dull and flat). Dressing is the process of removing these clogged particles and stripping away the dull grains to expose fresh, sharp abrasive grains, thereby restoring the cutting action and efficiency of the wheel.
Comparison with other options:
- Truing (a): This is the process of shaping the grinding wheel to ensure it is perfectly concentric with the spindle axis and has the desired geometric profile. While dressing often happens during truing, truing focuses on geometry/concentricity, whereas dressing focuses on cutting sharpness.
- Facing (c): This is a machining process used in turning or milling to create a flat surface on the end of a workpiece; it is not a grinding wheel maintenance process.
- Clearing (d): This is a general term and not a technical term used for the restoration of a grinding wheel's cutting ability.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
The shapes of the bending moment diagram for a uniform cantilever beam carrying a uniformly distributed load over its length is:
A)
A straight line
B)
A hyperbola
C)
An ellipse
D)
A parabola
To determine the shape of the Bending Moment (BM) diagram, we analyze the relationship between the bending moment and the position along the beam.
1. **Setup**: Consider a cantilever beam of length 'L' fixed at one end and free at the other, carrying a uniformly distributed load (UDL) of intensity 'w' per unit length.
2. **Moment Equation**: Let 'x' be the distance from the free end. The total load acting on the segment of length 'x' is (w * x). This load acts at the center of gravity of the segment, which is at a distance of (x/2) from the section being considered.
3. **Calculation**: The bending moment M at distance 'x' is given by:
M = (Load) * (Distance) = (w * x) * (x / 2) = (w / 2) * x².
4. **Analysis of the Equation**: The equation M = (w/2)x² is a second-degree polynomial equation. In mathematics, an equation where the dependent variable (M) is proportional to the square of the independent variable (x) represents a parabola.
**Comparison with other options:**
- **Straight line**: This occurs when the load is a point load (M is proportional to x).
- **Hyperbola/Ellipse**: These shapes do not occur in standard beam bending moment diagrams for common loading conditions like UDL or point loads.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
Aircraft body is usually fabricated by
A)
Welding
B)
Precoating
C)
Riveting
D)
Casting
Aircraft bodies are primarily fabricated using riveting for several critical engineering reasons: 1. Material Properties: Aircraft are mostly made of aluminum alloys. Many of these alloys lose their heat-treatment properties (strength) when subjected to the high temperatures of welding. 2. Fatigue Resistance: Riveted joints are generally better at resisting fatigue and vibration compared to welded joints in thin-sheet structures. 3. Inspection and Maintenance: Riveted joints are easier to inspect for cracks and can be disassembled for repairs or replacements. 4. Weight and Stress: Riveting allows for the joining of thin sheets without introducing the thermal stresses or distortion associated with welding.
Why other options are incorrect:
- Welding (a): While used in some specific areas (like titanium components), it is not the primary method for the main body due to heat-affected zones (HAZ) that weaken the aluminum alloy.
- Precoating (b): This is a surface treatment process to prevent corrosion, not a fabrication/joining method.
- Casting (d): Casting is used for complex engine parts or landing gear components, but it is impossible to cast the large, thin-walled skin of an aircraft fuselage.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
In V belt drive, the belt touches
A)
At the bottom
B)
At sides only
C)
Could touch anywhere
D)
None of these
A V-belt is designed with a trapezoidal cross-section and fits into a V-shaped groove on the pulley. Due to the wedge action, the belt is forced against the inclined sides of the pulley groove. Under normal operating tension, the belt makes contact with the side walls of the groove, but it does not touch the bottom of the groove. This 'wedge effect' increases the normal force between the belt and the pulley, which significantly increases the frictional grip and reduces the likelihood of slippage compared to flat belts. If the belt were to touch the bottom, it would indicate that the belt is too loose or worn out, leading to a loss of efficiency.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
If air resistance is neglected, while it is executing small oscillations the acceleration of the bob of a simple pendulum at the mid-point of its swing will be
A)
zero
B)
a minimum but not equal to zero
C)
a maximum
D)
not determinable unless the length of the pendulum and the mass of the bob are known
To determine the acceleration of the bob of a simple pendulum at the mid-point of its swing, we must analyze the forces acting on the bob.
1. **Definition of Mid-point**: The mid-point of the swing is the equilibrium position (the lowest point of the arc), where the displacement angle θ = 0.
2. **Restoring Force**: The restoring force acting on a simple pendulum is given by F = -mg sin(θ). This force is responsible for the tangential acceleration (a_t = -g sin(θ)). At the mid-point, θ = 0, therefore sin(0) = 0, and the tangential acceleration is zero.
3. **Centripetal Acceleration**: While there is a centripetal acceleration (a_c = v²/L) acting toward the pivot at the bottom of the swing, in the context of 'oscillations' and 'swing' in basic physics/mechanical vibration problems, 'acceleration' typically refers to the restoring acceleration that drives the oscillation (tangential acceleration).
4. **Analysis of Options**:
- At the extreme positions, velocity is zero and acceleration is maximum.
- At the mid-point (equilibrium), velocity is maximum and the restoring force (and thus the tangential acceleration) is zero.
Therefore, the acceleration driving the oscillation is zero at the mid-point.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
A beam AB of length 2 L having a concentrated load P at its mid-span is hinge supported at its two ends A and B on two identical cantilevers as shown in the given figure. The correct value of bending moment at A is
A)
Zero
B)
PL/2
C)
PL
D)
2PL
Option A is correct: Zero.
MES Jr. Grade • 2024 • Mechanical Engineering Paper-II
The cam follower generally used in automobile engines is
A)
Knife edge follower
B)
Flat faced follower
C)
Spherical faced follower
D)
Roller follower
In automobile engines, the cam-follower mechanism is used to operate the intake and exhaust valves. The choice of follower is based on reducing friction and wear due to high operating speeds.
1. **Roller Follower (Correct):** A roller follower uses a rolling element (bearing) at the contact point with the cam. This converts sliding friction into rolling friction, significantly reducing wear and tear and increasing the efficiency of the engine. This is the standard choice for high-speed internal combustion engines.
2. **Knife Edge Follower (Incorrect):** This has a sharp point. While theoretically precise, it creates extremely high contact stress and wears out almost instantly, making it unsuitable for any practical engine application.
3. **Flat Faced Follower (Incorrect):** These are used in some specific applications to reduce surface stress, but they involve significant sliding friction across the face, which would lead to excessive heat and wear at automotive RPMs.
4. **Spherical Faced Follower (Incorrect):** These are used to reduce stress compared to knife-edge followers, but they still involve sliding friction, which is inferior to the rolling action of a roller follower in a high-speed engine environment.