← MES Jr. Grade
Past Questions
The type of stresses developed in the key is/are-
A)
bearing stress only
B)
both bearing and shear stresses
C)
shear stress only
D)
shear, bearing and bending stresses
A key is a machine element used to connect a rotating machine element (like a gear or pulley) to a shaft to transmit torque. When torque is applied, the key is subjected to several types of stresses:
1. **Shear Stress**: The torque creates a shearing force across the plane where the key contacts the shaft and the hub. This tends to 'cut' the key in half along the length of the shaft.
2. **Bearing (Crushing) Stress**: The sides of the key are pressed against the keyway of the shaft and the hub. This compressive stress is known as bearing or crushing stress.
3. **Bending Stress**: In practical applications, the key does not fit perfectly tight in both the shaft and the hub simultaneously. This creates a small gap, causing the key to act like a cantilever beam or a beam with partial support, which induces bending stresses.
While basic textbook problems often simplify calculations by considering only shear and bearing stresses, a comprehensive engineering analysis of a key must include bending stresses to ensure the key does not fail due to fatigue or deformation. Therefore, the most complete answer is that shear, bearing, and bending stresses are all developed.
The working depth of an involute gear is given by
A)
Addendum
B)
Dedendum
C)
Addendum + Dedendum
D)
2 times of Addendum
In gear terminology, the 'Working Depth' (also known as the Whole Depth) refers to the total radial distance from the top of the tooth (addendum circle) to the bottom of the tooth space (dedendum circle).
1. **Addendum**: The radial distance from the pitch circle to the top of the tooth.
2. **Dedendum**: The radial distance from the pitch circle to the bottom of the tooth space.
3. **Whole Depth (Working Depth)**: This is the sum of the addendum and the dedendum.
Mathematically: Whole Depth = Addendum + Dedendum.
Comparing this to the options:
- Option (a) is only the top portion.
- Option (b) is only the bottom portion.
- Option (d) is incorrect because the dedendum is typically slightly larger than the addendum to provide clearance at the root, so it is not exactly 2 times the addendum.
In the Euler’s load expression P = pi^2 EI / l^2, the value of l for both the fixed ends is
A)
0.5
B)
0.33
C)
2
D)
1.5
The general formula for the Euler critical buckling load is P = (π² EI) / (L_e²), where L_e is the 'effective length' of the column. The effective length depends on the end conditions of the column. For a column with both ends fixed (fixed-fixed), the column is constrained against rotation and translation at both ends. This increases the stiffness of the column, effectively reducing the length over which it buckles. The theoretical effective length (L_e) for both ends fixed is 0.5 times the actual length (L). Therefore, L_e = 0.5L. In the context of the provided options, the value of the multiplier for the length (l) is 0.5.
Comparison of other end conditions:
- Both ends hinged: L_e = 1.0L
- One end fixed, one end hinged: L_e = 0.7L
- One end fixed, one end free: L_e = 2.0L
The number instantaneous centres for 8-link kinematic chain is
A)
16
B)
8
C)
28
D)
24
In the study of Kinematics of Machinery, the number of instantaneous centres (I-centres) for a kinematic chain with 'n' links is determined by the formula: N = n(n - 1) / 2. This formula is derived from the fact that an instantaneous centre exists for every possible pair of links in the mechanism. Given that the number of links (n) = 8, we substitute this value into the formula: N = 8(8 - 1) / 2 = 8(7) / 2 = 56 / 2 = 28. Therefore, an 8-link kinematic chain has 28 instantaneous centres. Option (a) 16, (b) 8, and (d) 24 are incorrect as they do not satisfy the mathematical relationship for the number of pairs in a set of 8 elements.
The tensile strength of the butt joint under tensile load is obtained as where, o = permissible tensile stress, l = length of weld, t = throat thickness
A)
2olt
B)
2o l^2
C)
olt
D)
o l^2
In machine design, the strength of a welded joint is calculated based on the minimum area of the weld that resists the applied load. For a butt joint, the load is distributed over the throat area of the weld.
1. **Throat Thickness (t):** This is the minimum distance from the root of the weld to its face, representing the weakest section of the weld.
2. **Length of Weld (l):** This is the total length along which the two plates are joined.
3. **Throat Area (A):** The area resisting the tensile load is the product of the length of the weld and the throat thickness: Area = l × t.
4. **Tensile Strength (P):** According to the basic stress formula (Stress = Force / Area), the force (strength) is the product of the permissible stress (σ) and the area (A).
Therefore, P = σ × (l × t) = σlt.
Comparing this to the options:
- Option (a) 2σlt: This would apply to a double-sided fillet weld or a specific joint configuration with two throat areas, not a standard single butt joint.
- Option (b) 2σl²: Incorrect units and formula.
- Option (c) σlt: Correct formula for the strength of a butt joint.
- Option (d) σl²: Incorrect units and formula.
A shaft carrying two rotors as its end will have
A)
No node
B)
One node
C)
Two nodes
D)
Three nodes
In the context of rotor dynamics and vibrations, a 'node' is a point on the shaft that remains stationary (zero amplitude) during a specific mode of vibration. For a shaft carrying two rotors (masses) at its ends, we consider the fundamental modes of vibration. In the first mode (fundamental mode), the two masses typically move in opposite directions (out of phase). Because the masses are moving in opposite directions, there must be a point somewhere between them where the displacement is zero. This point is called the node. For a system with two concentrated masses at the ends, the first mode of vibration results in exactly one node located between the two rotors. Higher modes would introduce more nodes, but in standard engineering exam contexts regarding basic rotor configurations, the primary mode is considered unless specified otherwise.
A bar of 12 mm diameter is under an axial load of 20 KN. What will be the change in diameter, if the Poisson’s ratio is v = 0.2923 and modulus of rigidity is G = 80 GPa?
A)
0.001
B)
0.003
C)
0.0003
D)
0.0001
Step-by-step solution:
1. **Given Data:**
- Diameter (d) = 12 mm
- Axial Load (P) = 20 kN = 20,000 N
- Poisson's ratio (v) = 0.2923
- Modulus of Rigidity (G) = 80 GPa = 80 x 10^3 MPa (or N/mm²)
2. **Find Young's Modulus (E):**
The relationship between E, G, and v is given by: E = 2G(1 + v)
E = 2 * 80,000 * (1 + 0.2923)
E = 160,000 * 1.2923 = 206,768 MPa
3. **Calculate Axial Stress (σ):**
Cross-sectional area (A) = (π/4) * d² = (π/4) * (12)² ≈ 113.097 mm²
Stress (σ) = P / A = 20,000 / 113.097 ≈ 176.839 MPa
4. **Calculate Longitudinal Strain (ε_long):**
ε_long = σ / E = 176.839 / 206,768 ≈ 0.0008553
5. **Calculate Lateral Strain (ε_lat):**
Poisson's ratio (v) = - (Lateral Strain / Longitudinal Strain)
ε_lat = -v * ε_long = -0.2923 * 0.0008553 ≈ -0.00025
6. **Calculate Change in Diameter (Δd):**
Δd = ε_lat * d = -0.00025 * 12 mm ≈ -0.003 mm
Wait, let's re-verify the calculation.
Δd = (v * P * d) / (E * A)
Δd = (0.2923 * 20000 * 12) / (206768 * 113.097)
Δd = 70152 / 23385125 ≈ 0.003 mm.
Looking at the options provided: a. 0.001, b. 0.003, c. 0.0003, d. 0.0001.
Based on the calculation, the magnitude is 0.003 mm. However, in many competitive exam keys for this specific problem, if the options are shifted or a decimal error occurred in the question paper's source, 0.0003 is often cited. Let's re-check the math:
E = 206.7 GPa.
σ = 176.8 MPa.
ε_long = 0.000855.
ε_lat = 0.2923 * 0.000855 = 0.00025.
Δd = 0.00025 * 12 = 0.003 mm.
Comparing the calculated result (0.003) with the options, option (b) is the mathematically correct result. However, if the question intended a different load or diameter, the result would change. Given the standard options for this specific problem in engineering test banks, 0.003 is the correct numerical value.
Correction: Re-calculating carefully:
Δd = (0.2923 * 20000 * 12) / (2 * 80000 * (1 + 0.2923) * (π/4 * 12^2))
Δd = 70152 / (160000 * 1.2923 * 113.097)
Δd = 70152 / 23385125 = 0.002999... ≈ 0.003 mm.
Therefore, option (b) is the correct answer.
The ability of a material to regain its original shape on removal of the applied load within elastic limit is called as-
A)
Toughness
B)
Resilience
C)
Absorption energy
D)
Distortion energy
Step-by-step reasoning:
1. **Definition of Resilience**: Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. In simpler terms, it is the ability of a material to regain its original shape after the removal of an applied load, provided the load was within the elastic limit.
2. **Analysis of Options**:
- **a. Toughness**: This is the ability of a material to absorb energy and deform plastically before fracturing. It involves both elastic and plastic deformation (the total area under the stress-strain curve up to the point of fracture).
- **b. Resilience**: This specifically refers to the energy stored during elastic deformation, which is fully recovered upon unloading. This matches the question's criteria perfectly.
- **c. Absorption energy**: This is a general term and not a specific material property. While resilience and toughness both involve energy absorption, 'absorption energy' is not the technical term for the ability to regain shape.
- **d. Distortion energy**: This refers to the energy associated with the change in shape (shear strain) of a material, often used in failure theories (like Von Mises), but it does not describe the ability to regain original shape.
Therefore, the correct technical term for the ability to regain original shape within the elastic limit is Resilience.
The mould in shell moulding process is made up of which of the following?
A)
Gypsum + setting agents
B)
Green sand + clay
C)
Sodium silicate + dried sand
D)
Dried silica + phenolic resin
Shell moulding is a precision casting process. In this process, a metal pattern is heated and then coated with a mixture of sand (typically silica sand) and a thermosetting resin binder (most commonly phenolic resin). The heat from the pattern causes the resin to cure (polymerize), forming a hard, thin 'shell' around the pattern. Once the shell is sufficiently hardened, it is removed from the pattern and joined to form the final mould.
Analysis of options:
- Option (a) Gypsum is used in investment casting or plaster moulds, not shell moulding.
- Option (b) Green sand and clay are used in traditional green sand casting.
- Option (c) Sodium silicate is used in the CO2 process (chemical bonding), not shell moulding.
- Option (d) Dried silica (sand) combined with a phenolic resin (thermosetting binder) is the standard composition for shell moulding.
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.
If the surface of a component is heavily stressed while the stresses in the core are of comparative small magnitude, which one of the following heat treatment methods is employed?
A)
Annealing
B)
Tempering
C)
Quenching
D)
Case hardening
The question describes a scenario where a component requires high surface hardness (to resist wear and heavy surface stress) while maintaining a tough, ductile core (to absorb shocks and resist impact). This specific combination of properties is achieved through 'Case Hardening'.
1. Case Hardening: This is a thermochemical process that diffuses carbon or other elements into the surface layer (the 'case') of a low-carbon steel part. The result is a hard, wear-resistant outer shell and a soft, ductile inner core. Examples include carburizing and nitriding.
Why other options are incorrect:
- Annealing: This is a process used to soften a material, relieve internal stresses, and improve ductility. It makes the entire component softer, not just the core.
- Tempering: This is performed after quenching to reduce brittleness and relieve internal stresses. It affects the bulk properties of the material rather than creating a surface-core gradient.
- Quenching: This involves rapid cooling of a material to increase hardness. If a component is quenched throughout, it becomes hard and brittle across the entire cross-section, which would make the core too brittle for applications requiring toughness.
The linear programming techniques can be applied successfully to industries like
A)
iron and steel
B)
food processing
C)
banking
D)
all of the above
Linear Programming (LP) is a mathematical optimization technique used to achieve the best outcome (such as maximum profit or lowest cost) in a mathematical model whose requirements are represented by linear relationships. Its versatility allows it to be applied across virtually all sectors of industry:
1. Iron and Steel (Manufacturing): LP is used for 'blending problems' (mixing raw materials to achieve a specific grade of steel at minimum cost) and production scheduling to optimize furnace usage.
2. Food Processing: LP is used for 'least-cost feed formulation' (mixing ingredients to meet nutritional requirements at minimum cost) and optimizing the supply chain for perishable goods.
3. Banking (Finance): LP is used for portfolio optimization, managing asset allocation to maximize returns while minimizing risk, and optimizing loan approvals based on capital constraints.
Since LP is a general-purpose optimization tool applicable to any system with linear constraints and a linear objective function, it is successfully applied to all the listed industries.
A company sells 14,000 units of its product. It has a variable cost of Rs. 15 per unit. Fixed cost is Rs. 47,000 and the required profit is Rs. 23,000 Per unit product price (in Rs.) will be:
A)
60
B)
40
C)
30
D)
20
Option D is correct: 20.
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.
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.
Choose the basic element for an automated machine tool
A)
Logic
B)
NC tape programming
C)
Software
D)
Workstation
In the context of automation and automated machine tools, the 'basic element' refers to the decision-making or control mechanism that governs the sequence of operations.
1. **Logic (Correct):** Automation is fundamentally based on logic (Boolean logic, PLC logic, or control logic). Logic is the core element that determines when a specific action should occur based on inputs (sensors) and outputs (actuators). Without a logic system, a machine cannot be 'automated' as it would lack the ability to make decisions or follow a programmed sequence independently.
2. **NC tape programming (Incorrect):** This is a method of providing instructions to a Numerical Control machine. While it is a tool for automation, it is a medium of data storage/input, not the basic functional element of the machine tool itself.
3. **Software (Incorrect):** Software is the implementation of logic. While modern machines use software, the fundamental engineering principle behind automation is 'Logic' (which can be implemented via hardware like relays/hard-wired logic or software).
4. **Workstation (Incorrect):** A workstation is a physical location or a computer setup where programming or monitoring occurs; it is an external support system, not a basic element of the machine tool's internal automated operation.
A strip is to be rolled from a thickness of 30 mm to 15 mm using a two-high mill having rolls of diameter 300 mm. The coefficient of friction for unaided bite should nearly be
A)
0.35
B)
0.5
C)
0.25
D)
0.07
Option C is correct: 0.25.
Gamma iron exists at following temperature-
A)
Room temperature
B)
Near melting point
C)
Between 1400°C and 1539°C
D)
Between 910°C and 1400°C
Pure iron undergoes allotropic transformations (changes in crystal structure) as temperature increases.
1. Alpha-iron (α-iron): This is the Body-Centered Cubic (BCC) phase, stable from room temperature up to 910°C.
2. Gamma-iron (γ-iron): Also known as Austenite, this is the Face-Centered Cubic (FCC) phase. It is stable between 910°C and 1394°C (approximately 1400°C).
3. Delta-iron (δ-iron): This returns to a Body-Centered Cubic (BCC) structure and is stable between 1394°C (1400°C) and the melting point of iron (1538°C/1539°C).
Analyzing the options:
- Option (a) is incorrect because room temperature is the domain of Alpha-iron.
- Option (b) is incorrect because near the melting point, iron exists as Delta-iron.
- Option (c) is incorrect because this range describes Delta-iron.
- Option (d) is correct as it aligns with the stability range of Gamma-iron (Austenite).
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.
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.
In a single-point turning operation of steel with a cemented carbide tool, Taylor’s tool life exponent is 0.25. If the cutting speed is halved, the tool life will increase by
A)
two times
B)
four times
C)
eight times
D)
sixteen times
The relationship between cutting speed and tool life is governed by Taylor's Tool Life Equation: V * T^n = C, where V is the cutting speed, T is the tool life, n is the tool life exponent, and C is a constant.
Given:
- Tool life exponent (n) = 0.25
- Initial speed = V1
- Initial tool life = T1
- Final speed (V2) = V1 / 2
- Final tool life = T2
According to the equation:
V1 * (T1)^n = V2 * (T2)^n
Substitute the values:
V1 * (T1)^0.25 = (V1 / 2) * (T2)^0.25
Divide both sides by V1:
(T1)^0.25 = (1 / 2) * (T2)^0.25
Rearrange to solve for the ratio T2/T1:
(T2 / T1)^0.25 = 2
To solve for T2/T1, raise both sides to the power of (1 / 0.25), which is the power of 4:
(T2 / T1) = 2^(1 / 0.25)
(T2 / T1) = 2^4
(T2 / T1) = 16
Therefore, the tool life increases by 16 times.
Comparing with options:
- a. two times (Incorrect)
- b. four times (Incorrect)
- c. eight times (Incorrect)
- d. sixteen times (Correct)
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.
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.
Which one of the following gating systems is best suited to obtain directional solidification?
A)
Top grating
B)
Part-line grating
C)
Bottom grating
D)
Stepped grating
Directional solidification is a process where the casting solidifies in a controlled direction, typically from the furthest point from the riser toward the riser, to prevent shrinkage cavities and porosity.
1. **Top Gating (Top grating):** In a top gating system, the molten metal enters from the top. The hottest metal remains at the top of the mold cavity, while the coldest metal settles at the bottom. Since the riser is also located at the top, the temperature gradient is established such that the bottom solidifies first and the top (near the riser) solidifies last. This creates an ideal condition for directional solidification.
2. **Bottom Gating (Bottom grating):** In bottom gating, the hottest metal ends up at the top of the casting, but the metal entering the bottom cools as it rises. While this reduces turbulence, it often leads to a temperature profile where the bottom may remain liquid longer or the gradient is not as steep as top gating, making it less ideal for strict directional solidification compared to top gating.
3. **Part-line Gating:** This is a compromise between top and bottom gating, typically used to reduce turbulence and dross, but it does not optimize the temperature gradient for directional solidification as effectively as top gating.
4. **Stepped Gating:** This is used for very thick sections to ensure the mold fills uniformly and to reduce turbulence, but it is not the primary choice for establishing a simple directional solidification gradient.
Therefore, Top Gating is the best suited for obtaining directional solidification because it ensures the highest temperature of the metal is maintained at the top, where the riser is located.
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.