← MES Jr. Grade
Past Questions — 2024
In a disc clutch, the clutch disc acts as a
A)
Driving member
B)
Driven member
C)
Neutral member
D)
All of the above
In a standard single-plate disc clutch assembly, there are three primary components: the flywheel, the clutch plate (disc), and the pressure plate.
1. The Flywheel: This is bolted to the engine crankshaft and rotates with the engine. It acts as the 'Driving Member'.
2. The Clutch Disc (Plate): This is splined to the transmission input shaft. Its purpose is to receive the torque from the flywheel and transmit it to the gearbox. Therefore, it acts as the 'Driven Member'.
3. The Pressure Plate: This is used to press the clutch disc against the flywheel to create friction.
When the clutch is engaged, the pressure plate pushes the clutch disc against the flywheel. The friction between the flywheel (driving member) and the disc (driven member) causes the disc to rotate, thereby driving the transmission shaft. Thus, the clutch disc is the member being driven.
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.
Two tapering bars of the same material are subjected to a tensile load P. The lengths of both the bars are the same. The larger diameter of each of the bars is D. The diameter of the bar A at its smaller end is D/2 and that of the bar B is D/3. What is the ratio of elongation of the bar A to that of the bar B?
A)
3:2
B)
2:3
C)
4:9
D)
1:3
Option A is correct: 3:2.
Hardness is which kind of property?
A)
Isotropic
B)
Anisotropic
C)
Homogenous
D)
Non-homogenous
Hardness is defined as the resistance of a material to localized plastic deformation (such as a small dent or a scratch). In material science, specifically regarding crystalline materials (like metals and ceramics), hardness is an anisotropic property. This is because the arrangement of atoms and the slip planes within a crystal lattice vary depending on the crystallographic direction. Consequently, the resistance to deformation (hardness) differs depending on the direction in which the force is applied relative to the crystal orientation.
Breakdown of other options:
- Isotropic: A material is isotropic if its properties are the same in all directions. While some amorphous materials (like glass) may exhibit isotropic hardness, in the general context of material science and engineering exams, hardness is recognized as anisotropic due to the nature of crystalline structures.
- Homogenous: This refers to a material having the same composition and properties at every point throughout its volume. Homogeneity refers to spatial distribution, not directional dependence.
- Non-homogenous: This refers to a material whose properties vary from one location to another (e.g., a composite material).
A higher pair has
A)
Point contact
B)
Surface contact
C)
No contact
D)
None of these
In the Theory of Machines, kinematic pairs are classified based on the type of contact between the two mating elements:
1. **Lower Pair**: A lower pair occurs when the two mating elements have surface contact. Examples include sliding pairs (prismatic joints), turning pairs (revolute joints), and screw pairs.
2. **Higher Pair**: A higher pair occurs when the two mating elements have point or line contact. Because the contact area is limited to a point or a line, these pairs typically experience higher stress concentrations. Examples include gear teeth in mesh, a ball bearing, or a cam and follower.
Analyzing the options:
- **Option (a) Point contact**: Correct. Higher pairs are characterized by point or line contact.
- **Option (b) Surface contact**: Incorrect. This is the defining characteristic of a Lower Pair.
- **Option (c) No contact**: Incorrect. A kinematic pair, by definition, requires contact to constrain the relative motion of the parts.
- **Option (d) None of these**: Incorrect, as option (a) is the correct technical definition.
The balancing of a rigid rotor can be achieved by appropriately placing balancing masses in
A)
A single plane
B)
Two planes
C)
Three planes
D)
Four planes
In the theory of machines, balancing a rigid rotor involves eliminating both static and dynamic imbalance.
1. **Static Imbalance**: This occurs when the center of gravity of the rotor is not on the axis of rotation. It can be corrected by adding a balancing mass in a single plane.
2. **Dynamic Imbalance**: This occurs when the axis of rotation is not a principal axis of inertia, causing a centrifugal couple (moment) to act on the bearings during rotation. To eliminate this couple, a second balancing mass must be placed in a different plane to create an opposing couple.
Since a rigid rotor can have both static and dynamic imbalances, a minimum of two planes (separated by some distance along the axis) are required to ensure that both the net centrifugal force and the net centrifugal couple are zero. Therefore, balancing in two planes is the standard requirement for complete balancing of a rigid rotor.
The minimum number of links in a single degree-of-freedom planar mechanism with both higher and lower kinematic pairs is
A)
2
B)
3
C)
4
D)
5
To determine the minimum number of links, we use Kutzbach's criterion for the mobility (degree of freedom) of a planar mechanism:
F = 3(n - 1) - 2j - h
Where:
- F = Degree of Freedom (given as 1)
- n = Number of links
- j = Number of lower pairs (joints with 1 degree of freedom, like pins or sliders)
- h = Number of higher pairs (joints with 2 degrees of freedom, like cams or gears)
The question specifies that the mechanism must contain BOTH higher and lower pairs. Therefore, h ≥ 1 and j ≥ 1.
Let's test the options:
1. If n = 3:
F = 3(3 - 1) - 2j - h = 6 - 2j - h.
For F = 1, 2j + h = 5.
If h = 1, then 2j = 4 → j = 2.
However, for 3 links to have 3 joints (2 lower + 1 higher), they must form a closed loop. In a 3-link chain, if you have 3 joints, the links are connected in a triangle. But a higher pair (like a cam-follower) typically requires a specific geometry that, when combined with two lower pairs in a 3-link system, often results in a structure that is either a linkage (if all are lower pairs) or doesn't satisfy the 'mechanism' definition of constrained motion without more links. More importantly, Gruebler's rule for a mechanism with higher pairs usually requires more links to maintain a single DOF without becoming a structure or having redundant constraints.
2. If n = 4:
F = 3(4 - 1) - 2j - h = 9 - 2j - h.
For F = 1, 2j + h = 8.
Possible combinations:
- If h = 2, then 2j = 6 → j = 3. (Total joints = 5)
- If h = 4, then 2j = 4 → j = 2. (Total joints = 6)
A common example of a 4-link mechanism with both pair types is a cam-follower system where the cam is driven by a crank (link 1: frame, link 2: crank, link 3: cam, link 4: follower). Here, the crank-cam connection is a lower pair, and the cam-follower contact is a higher pair.
According to the general theory of mechanisms, the minimum number of links for a constrained planar mechanism (a kinematic chain) is 4 (Gruebler's criterion). While a 3-link chain can exist, it cannot form a mechanism with a single degree of freedom that incorporates both higher and lower pairs while remaining a constrained kinematic chain. Thus, 4 is the minimum number of links required.
Resilience can also be termed as
A)
Stress energy
B)
Strain energy
C)
Modulus
D)
Tenacity
Resilience is defined as the capacity of a material to absorb energy when it is deformed elastically and then release that energy upon unloading. In the context of Strength of Materials, this absorbed energy is known as 'Strain Energy'. Specifically, the 'Modulus of Resilience' is the strain energy per unit volume stored in a material up to the elastic limit.
Analysis of options:
- (a) Stress energy: This is not a standard technical term used to describe the energy stored in a material.
- (b) Strain energy: Correct. Resilience is the ability of a material to store strain energy within the elastic range.
- (c) Modulus: This is a general term for a constant (like Young's Modulus) that describes a material's stiffness, not the energy stored.
- (d) Tenacity: This refers to the ability of a material to resist fracture; it is the total strain energy absorbed up to the point of rupture (the area under the entire stress-strain curve), whereas resilience only considers the elastic region.
When two shafts are neither parallel nor intersecting, power can be transmitted by using
A)
a pair of spur gears
B)
a pair of helical gears
C)
an Oldham’s coupling
D)
a pair of spiral gears
To solve this, we must analyze the relative positions of the shafts for each gear/coupling type:
1. **Spur Gears (Option a):** These are used for transmitting power between two parallel shafts. Their teeth are straight and parallel to the axis of the shaft.
2. **Helical Gears (Option b):** These are primarily used for parallel shafts (though they can be used for intersecting shafts in some configurations), but they are not the standard choice for non-parallel, non-intersecting shafts.
3. **Oldham’s Coupling (Option c):** This is used to connect two parallel shafts that have a small lateral offset (misalignment). It cannot transmit power between non-parallel shafts.
4. **Spiral Gears / Hypoid Gears (Option d):** When shafts are neither parallel nor intersecting, they are known as 'skew shafts'. Spiral bevel gears or specifically Hypoid gears (a type of spiral gear) are designed specifically to transmit power between skew shafts. They allow the axis of the pinion to be offset from the axis of the gear.
Therefore, for shafts that are neither parallel nor intersecting, a pair of spiral (or hypoid) gears is the correct mechanical solution.
A mechanism where there are no restrictions on the relative motion between the particles, the mechanism is called
A)
Planar mechanism
B)
Flexure mechanism
C)
Spherical mechanism
D)
Spatial mechanism
In the study of Theory of Machines, mechanisms are classified based on the constraints imposed on the motion of their links:
1. **Planar Mechanism**: All links move in parallel planes, and all relative motions are restricted to these planes. There are significant restrictions on motion (no motion out of the plane).
2. **Spherical Mechanism**: All links move such that a common point (the center of the sphere) remains stationary. The motion is restricted to the surface of a sphere.
3. **Flexure Mechanism**: These rely on the elastic deformation of members rather than sliding or turning joints. Motion is highly restricted by the stiffness of the material.
4. **Spatial Mechanism**: Also known as a three-dimensional mechanism, these mechanisms allow motion in all three dimensions of space. Unlike planar or spherical mechanisms, there are no restrictions limiting the relative motion to a single plane or a spherical surface. Therefore, it is the most general form of mechanism where particles can move freely in 3D space relative to each other (subject only to the joint constraints).
Given the phrasing 'no restrictions on the relative motion' (meaning no restriction to a specific plane or sphere), the correct classification is a Spatial Mechanism.
To determine hoop stress, efficiency of is to be considered.
A)
Construction joint
B)
Transverse joint
C)
Longitudinal joint
D)
Rivetjoint
Hoop stress (also known as circumferential stress) occurs in cylindrical pressure vessels due to internal pressure. The stress acts along the circumference of the cylinder, which tends to pull the cylinder walls apart along its length. In a manufactured pressure vessel, the cylinder is typically formed from a flat plate rolled into a cylinder and joined along its length. This joint is called the 'longitudinal joint'. Because the hoop stress is the primary stress acting to pull this joint apart, the efficiency of the longitudinal joint (the ratio of the strength of the joint to the strength of the solid plate) must be considered to determine the safe working pressure or the actual stress the vessel can withstand. Transverse joints (circumferential joints) are subjected to longitudinal stress, which is only half the magnitude of the hoop stress, making the longitudinal joint the critical factor for hoop stress calculations.
A T-section beam is simply supported and subjected to a uniform distributed load over its whole span. Maximum longitudinal stress at
A)
Top fibre of the flange
B)
The junction of web and flange
C)
The mid-section of the web
D)
The bottom fibre of the web
1. **Bending Stress Formula**: The longitudinal stress (bending stress) in a beam is given by the formula σ = (M * y) / I, where M is the bending moment, y is the distance from the neutral axis (NA), and I is the moment of inertia.
2. **Neutral Axis Position**: In a T-section, the area is concentrated at the top (the flange). Consequently, the centroid (Neutral Axis) shifts upwards toward the flange.
3. **Distance 'y'**: The maximum stress occurs at the fiber furthest from the neutral axis. Because the NA is shifted upwards, the distance from the NA to the bottom fiber of the web (y_bottom) is significantly greater than the distance from the NA to the top fiber of the flange (y_top).
4. **Comparison**: Since σ is directly proportional to y, and y_bottom > y_top, the maximum longitudinal stress will occur at the bottom-most fiber of the beam.
5. **Conclusion**: For a simply supported beam under UDL, the maximum bending moment occurs at the center, and the maximum stress at that section occurs at the fiber furthest from the NA, which is the bottom fiber of the web.
The ductility of a material is to the increase in percentage reduction in an area.
A)
inversely proportional
B)
directly proportional
C)
equal
D)
uniform
Ductility is a mechanical property of a material that allows it to undergo significant plastic deformation under tensile stress before rupture. In material science, ductility is quantitatively measured using two primary parameters: 1) Percentage Elongation and 2) Percentage Reduction in Area. The 'Percentage Reduction in Area' is calculated as [(Original Area - Final Area) / Original Area] * 100. A material that is highly ductile will experience a significant 'necking' effect, leading to a large reduction in the cross-sectional area at the point of fracture. Therefore, as the ductility of a material increases, the percentage reduction in area also increases. This positive correlation means that ductility is directly proportional to the increase in percentage reduction in area.
Which of the following methods is also known as individual wall method?
A)
Centre line method
B)
Alignment method
C)
Long wall and short wall method
D)
Voluminous method
In civil engineering estimation and costing, there are two primary methods used for calculating the quantities of masonry and concrete work in buildings: the Centre Line Method and the Long Wall-Short Wall Method.
1. **Long Wall and Short Wall Method**: This method is also known as the **Individual Wall Method** or the Out-to-Out and In-to-In method. In this approach, the lengths of the long walls and short walls are calculated separately. The long walls are measured from the outer face to the outer face, and the short walls are measured between the inner faces of the long walls. This ensures that no volume is counted twice and each wall is treated as an individual entity.
2. **Centre Line Method**: This method involves calculating the total length of the center line of all walls and multiplying it by the width and height. Deductions are then made at junctions (T-junctions or cross-junctions) to avoid double counting. It is faster but not an 'individual wall' approach.
3. **Alignment and Voluminous methods**: These are not standard technical terms for calculating masonry quantities in building estimation.
Therefore, the Long Wall and Short Wall method is the correct answer as it treats walls individually.
A weight falls on a plunger fitted in a container filled with oil thereby producing a pressure of 1.5 N/mm² in the oil. The Bulk Modulus of oil is 2800 N/mm². Given this situation, the volumetric compressive strain produced in the oil will be:
A)
400 x 10⁻⁶
B)
800 x 10⁻⁶
C)
268 x 10⁻⁶
D)
535 x 10⁻⁶
To find the volumetric compressive strain, we use the definition of the Bulk Modulus of elasticity.
1. **Given Data:**
- Pressure produced (Change in pressure, ΔP) = 1.5 N/mm²
- Bulk Modulus of oil (K) = 2800 N/mm²
2. **Formula:**
The Bulk Modulus (K) is defined as the ratio of the change in pressure (ΔP) to the volumetric strain (εᵥ):
K = ΔP / εᵥ
3. **Calculation:**
Rearranging the formula to solve for volumetric strain (εᵥ):
εᵥ = ΔP / K
εᵥ = 1.5 / 2800
εᵥ ≈ 0.000535714
4. **Converting to Scientific Notation:**
εᵥ = 535.714 x 10⁻⁶
Comparing this result to the given options, it aligns most closely with option (d) 535 x 10⁻⁶.
- Option (a) 400 x 10⁻⁶ is incorrect as it would require a pressure of ~1.12 N/mm².
- Option (b) 800 x 10⁻⁶ is incorrect as it would require a pressure of ~2.24 N/mm².
- Option (c) 268 x 10⁻⁶ is incorrect as it would require a pressure of ~0.75 N/mm².
A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to
A)
41 MPa
B)
82 MPa
C)
164 MPa
D)
204 MPa
Step-by-step solution:
1. **Given Data:**
- Diameter of shaft (d) = 100 mm = 0.1 m
- Axial stress (σₓ) = 50 MPa = 50 × 10⁶ N/m²
- Torque (T) = 10 kNm = 10 × 10³ Nm
2. **Calculate Torsional Shear Stress (τ):**
The shear stress induced by torque in a solid circular shaft is given by:
τ = (16T) / (πd³)
τ = (16 × 10,000) / (π × 0.1³)
τ = 160,000 / (π × 0.001)
τ = 160,000,000 / π ≈ 50.93 MPa
3. **Calculate Maximum Principal Stress (σ₁):**
The shaft is subjected to a combined state of axial stress (σₓ) and shear stress (τ). The formula for the maximum principal stress is:
σ₁ = (σₓ / 2) + √((σₓ / 2)² + τ²)
σ₁ = (50 / 2) + √((50 / 2)² + 50.93²)
σ₁ = 25 + √(25² + 50.93²)
σ₁ = 25 + √(625 + 2593.86)
σ₁ = 25 + √3218.86
σ₁ = 25 + 56.73
σ₁ = 81.73 MPa
4. **Conclusion:**
The calculated maximum principal stress is approximately 81.73 MPa, which is closest to 82 MPa.
Comparing with options:
- (a) 41 MPa: Incorrect.
- (b) 82 MPa: Correct (closest value).
- (c) 164 MPa: Incorrect.
- (d) 204 MPa: Incorrect.
The following is not a friction clutch
A)
Centrifugal clutch
B)
Cone clutch
C)
Disc clutch
D)
Fluid clutch
A friction clutch is a device that transmits torque from one shaft to another through the application of friction between two surfaces.
1. Centrifugal clutch (Option a): Uses centrifugal force to push friction pads against a drum to engage the drive. It is a friction clutch.
2. Cone clutch (Option b): Uses conical surfaces that press against each other to transmit torque via friction. It is a friction clutch.
3. Disc clutch (Option c): Uses flat friction plates (discs) pressed together to transmit torque. It is a friction clutch.
4. Fluid clutch (Option d): Also known as a fluid coupling, it transmits torque through the hydrodynamic action of a fluid (usually oil) between an impeller and a runner. There is no physical contact or sliding friction between the driving and driven members. Therefore, it is a hydraulic/fluid coupling, not a friction clutch.
Which gear is used for connecting two coplanar and intersecting shafts?
A)
Spur gear
B)
Helical gear
C)
Bevel gear
D)
None of these
To determine the correct gear type, we must analyze the spatial relationship between the two shafts: 1. Coplanar: The axes of the shafts lie in the same plane. 2. Intersecting: The axes meet at a specific point.
- Spur gears (Option a) are used for parallel shafts.
- Helical gears (Option b) are primarily used for parallel shafts (though crossed helical gears exist for non-parallel, non-intersecting shafts).
- Bevel gears (Option c) are specifically designed with conical shapes to transmit power between shafts that intersect, typically at a 90-degree angle, though other angles are possible. Because they are conical, their axes intersect at the apex of the cones, and they operate in a single plane relative to that intersection point.
Therefore, the Bevel gear is the only option that satisfies the condition of connecting intersecting shafts.
Belt slip may occur due to
A)
Heavy load
B)
Loose belt
C)
Driving pulley too small
D)
All of these
Belt slip occurs when the frictional force between the belt and the pulley is insufficient to transmit the required torque. The condition for no-slip is governed by the belt tension ratio formula: (T1/T2) ≤ e^(μθ), where T1 is the tight side tension, T2 is the slack side tension, μ is the coefficient of friction, and θ is the angle of wrap.
1. Heavy load (Option a): An increase in load increases the required torque, which increases the difference between T1 and T2. If T1/T2 exceeds e^(μθ), the belt will slip.
2. Loose belt (Option b): A loose belt reduces the initial tension (T0), which directly lowers the normal force acting on the pulley surface. Since frictional force is proportional to the normal force (F = μN), a loose belt significantly reduces the grip, leading to slip.
3. Driving pulley too small (Option c): A smaller pulley diameter reduces the angle of wrap (θ) if the center distance remains constant, or it increases the bending stress and reduces the effective contact area/pressure distribution. A smaller θ decreases the value of e^(μθ), making the system more prone to slipping.
Since all three factors contribute to the occurrence of belt slip, the correct option is 'All of these'.
Which of the following is a spring-controlled governor?
A)
Hartnell governor
B)
Hartung governor
C)
Wilson-Hartnell governor
D)
All of these
Governors are devices used to maintain the speed of an engine within specified limits regardless of the load. They are broadly classified into two types: Centrifugal Governors (which are further divided into Pendulum type and Fly-ball type) and Spring-controlled Governors.
1. Hartnell Governor: This is a classic example of a spring-loaded governor where a central spring provides the controlling force to counteract the centrifugal force of the rotating balls.
2. Hartung Governor: This is a specific type of spring-controlled governor where the spring is arranged such that the controlling force remains constant regardless of the lift of the sleeve.
3. Wilson-Hartnell Governor: This is a modification of the Hartnell governor that incorporates an additional spring to provide a more linear relationship between speed and lift, improving the sensitivity and stability of the governor.
Since all three listed governors (Hartnell, Hartung, and Wilson-Hartnell) utilize a spring to provide the controlling force, the correct option is 'All of these'.
The governor used in gramophone is of the following type
A)
Pickering
B)
Porter
C)
Hartnell
D)
Watt
A governor is a device used to maintain a constant speed of a machine regardless of the load. In the context of a gramophone (record player), it is critical to maintain a very precise and constant angular velocity to ensure the music plays at the correct pitch and tempo.
1. **Pickering Governor**: This is a centrifugal governor specifically designed for high precision and sensitivity in a compact space. It uses a spring-loaded mechanism that can react quickly to small changes in speed, making it ideal for the speed regulation required in gramophones.
2. **Porter Governor**: This is a dead-weight governor (a modification of the Watt governor) used primarily in large steam engines. It is too bulky and lacks the sensitivity required for a small electronic/mechanical device like a gramophone.
3. **Hartnell Governor**: This is a spring-loaded governor used in industrial applications where a wide range of speed control is needed. While it uses a spring, its design is intended for larger machinery rather than the precision instrumentation of a record player.
4. **Watt Governor**: This is the simplest form of a centrifugal governor (centrifugal balls on a rotating arm). It is highly inefficient at high speeds and lacks the precision needed for audio playback devices.
Therefore, the Pickering governor is the correct choice for the specific application of a gramophone.
In the case of an involute toothed gear, involute starts from
A)
addendum circle
B)
dedendum circle
C)
pitch circle
D)
base circle
An involute curve is defined as the path traced by a point on a straight line that rolls without slipping on a circle. In gear geometry, this 'rolling circle' is known as the base circle. Therefore, the involute profile of a gear tooth physically begins at the base circle and extends outward toward the addendum circle.
Analysis of options:
- a. Addendum circle: This is the outermost circle of the gear; the involute ends here.
- b. Dedendum circle: This is the bottom-most circle of the tooth space; the profile here is usually a radial fillet curve, not an involute.
- c. Pitch circle: This is the theoretical circle used to define the gear ratio and kinematics; while the involute passes through it, it does not start there.
- d. Base circle: This is the fundamental circle from which the involute is generated. The involute profile starts exactly at the base circle.
If the shear force acting at every section of a beam is of the same magnitude and of the same direction then it represents a
A)
Simply supported beam with a concentrated load at the centre.
B)
Overhung beam having equal overhang at both supports and carrying equal concentrated loads acting in the same direction at the free ends.
C)
Cantilever subjected to concentrated load at the free end.
D)
Simply supported beam having concentrated loads of equal magnitude and in the same direction acting at equal distances from the supports.
To determine the correct option, we must analyze the Shear Force (SF) distribution for each case. The condition given is that the shear force must be of the same magnitude and the same direction (constant SF) across every section of the beam.
1. **Option (a): Simply supported beam with a concentrated load at the center.** The SF is positive from the left support to the center and negative from the center to the right support. The magnitude is the same, but the direction changes. Thus, it is not constant.
2. **Option (b): Overhung beam with equal overhangs and equal loads at free ends.** The SF in the overhangs will be equal to the load, but the SF in the central span between the supports will be zero (since the loads at the ends cancel each other out in the internal section). Thus, it is not constant across the entire beam.
3. **Option (c): Cantilever subjected to a concentrated load at the free end.** For a cantilever of length L with a load P at the free end, the shear force at any section x (measured from the free end) is constant and equal to P. Since the load is acting in one direction, the SF remains the same in magnitude and direction throughout the entire length of the beam. This satisfies the condition.
4. **Option (d): Simply supported beam with equal loads at equal distances from supports.** The SF will be constant between the support and the load, but it will change magnitude (drop to zero or change sign) between the two loads. Thus, it is not constant across every section.
Therefore, only the cantilever beam with a point load at the free end maintains a constant shear force throughout its length.
In a single slider four-bar linkage, when the slider is fixed, it forms a mechanism of
A)
hand pump
B)
reciprocating engine
C)
quick return
D)
oscillating cylinder
A single slider-crank mechanism consists of four links: a frame (fixed link), a crank, a connecting rod, and a slider. The inversion of a mechanism occurs when a different link is chosen as the fixed link.
1. **First Inversion (Frame fixed):** This is the standard reciprocating engine or compressor where the cylinder is fixed and the piston reciprocates. (Option b)
2. **Second Inversion (Crank fixed):** This results in a mechanism where the cylinder oscillates. Examples include the oscillating cylinder engine or the Whitworth quick return mechanism (though the quick return is a specific application of this inversion). (Option d)
3. **Third Inversion (Connecting rod fixed):** This results in a crank and slider both rotating, used in the hand pump (Option a) or the rotary engine.
In this specific question, the 'slider' is fixed. When the slider is fixed, the cylinder (which normally houses the slider) is forced to move/pivot relative to the other links. This creates the 'oscillating cylinder' mechanism. While 'quick return' mechanisms are derived from the second inversion (fixed crank), the most direct description of the physical arrangement when the slider is fixed is the oscillating cylinder mechanism.
During tensile-testing of a specimen using a Universal Testing Machine, the parameters actually measured include
A)
True stress and true strain
B)
Poisson’s ratio and Young’s modulus
C)
Engineering stress and engineering strain
D)
Load and elongation
In a tensile test conducted using a Universal Testing Machine (UTM), the machine is equipped with sensors (such as a load cell for force and an extensometer or crosshead displacement sensor for length change). Therefore, the raw data 'actually measured' by the hardware are the applied Load (P) and the resulting Elongation (ΔL).
All other parameters mentioned in the options are 'calculated' or 'derived' values:
- Engineering Stress (σ = P/A₀) and Engineering Strain (ε = ΔL/L₀) are calculated by dividing the measured load and elongation by the original cross-sectional area and original length, respectively.
- True Stress and True Strain are derived from engineering values by accounting for the instantaneous change in area and length during deformation.
- Young's Modulus (E = σ/ε) and Poisson's ratio are material properties calculated from the slope of the stress-strain curve and the ratio of lateral to longitudinal strain, respectively.
Since the question asks for parameters 'actually measured', the correct answer is Load and elongation.