← MES Jr. Grade
Past Questions
The properties of material to resist penetration by another material is known as:
A)
Impact strength
B)
Hardness
C)
Fatigue
D)
Resilience
Step-by-step reasoning:
1. Definition of Hardness: In material science, hardness is defined as the resistance of a material to localized plastic deformation, such as indentation, scratching, or penetration by a harder object. Common tests for hardness include the Brinell, Rockwell, and Vickers tests, which all measure the depth or area of penetration by a standardized indenter.
2. Analysis of other options:
- Impact strength (a): This refers to the ability of a material to absorb energy and plastically deform before fracturing under a high-rate load (sudden impact). It is measured by tests like the Charpy or Izod tests.
- Fatigue (c): This is the tendency of a material to break or fail under repeated or cyclic loading, even if the stress is below the yield strength.
- Resilience (d): This is the capacity of a material to absorb energy when it is deformed elastically and then release that energy upon unloading.
3. Conclusion: Since the question specifically asks for the property to resist 'penetration', the correct term is Hardness.
What is a surface imperfection, which separates crystals of different orientations a polycrystalline aggregate, called?
A)
Edge dislocation
B)
Stacking fault
C)
Grain boundary
D)
Screw dislocation
In material science, most engineering materials are polycrystalline, meaning they consist of many small crystals called 'grains'. Each grain has a specific crystallographic orientation. A grain boundary is the interface or surface imperfection that separates two grains of different orientations. These boundaries are considered 2D (surface) defects.
Why other options are incorrect:
- a. Edge dislocation: This is a 1D (line) defect involving an extra half-plane of atoms inserted into the crystal lattice.
- b. Stacking fault: This is a 2D (surface) defect, but it occurs within a single crystal due to an error in the sequence of stacking atomic planes (e.g., ABCABC becoming ABCBCA), not as a boundary between crystals of different orientations.
- d. Screw dislocation: This is a 1D (line) defect where the lattice is distorted in a helical path around the dislocation line.
How does the distribution of heat between the tool, chip, and workpiece affect machining performance?
Gunmetal, which is used in journal bearings, contains :
A)
88% Cu, 10% Sn, 2% Zn
B)
80% Cu, 10% Zn, 10% Al
C)
85% Cu, 5% Mg, 10% Al
D)
85% Cu, 5% Sn, 10% Pb
Gunmetal is a type of bronze, which is primarily an alloy of copper (Cu) and tin (Sn). It is specifically designed for applications requiring high strength, corrosion resistance, and low friction, making it ideal for journal bearings, valves, and pump components.
1. **Composition Analysis**: Standard gunmetal typically consists of approximately 88% Copper, 10% Tin, and 2% Zinc. The addition of zinc improves the casting properties and strength of the alloy.
2. **Evaluation of Options**:
- **Option (a)**: Matches the standard composition of gunmetal (88% Cu, 10% Sn, 2% Zn).
- **Option (b)**: This composition (Cu-Zn-Al) describes an aluminum bronze or a modified brass, not gunmetal.
- **Option (c)**: This composition (Cu-Mg-Al) is not a standard gunmetal formulation.
- **Option (d)**: While lead (Pb) is sometimes added to bearings for lubricity (creating 'leaded bronze'), the specific proportions listed here do not define standard gunmetal.
Therefore, option (a) is the correct technical specification for gunmetal used in engineering applications like journal bearings.
The crystal structure of austenite is:
A)
body centered cubic
B)
face centered cubic
C)
hexagonal closed packed
D)
body centered tetragonal
In the study of the Iron-Carbon equilibrium diagram, iron exists in different allotropic forms depending on the temperature.
1. **Austenite (γ-iron)**: This is the stable phase of iron between 912°C and 1394°C. It possesses a **Face Centered Cubic (FCC)** crystal structure. Because of the FCC arrangement, austenite has larger interstitial sites, allowing it to dissolve more carbon compared to ferrite.
2. **Ferrite (α-iron)**: This is the stable phase at room temperature (up to 912°C) and possesses a **Body Centered Cubic (BCC)** structure. This corresponds to option (a).
3. **Delta Ferrite (δ-iron)**: This is the stable phase at very high temperatures (above 1394°C) and also possesses a **Body Centered Cubic (BCC)** structure.
4. **Martensite**: When austenite is quenched rapidly, it transforms into a non-equilibrium phase called Martensite, which has a **Body Centered Tetragonal (BCT)** structure. This corresponds to option (d).
5. **Hexagonal Closed Packed (HCP)**: While some metals exhibit HCP structures, the standard phases of pure iron/steel (Austenite, Ferrite) do not. This corresponds to option (c).
Therefore, the crystal structure of austenite is specifically Face Centered Cubic (FCC).
In shell moulding, the resin-coated sand is hardened by:
A)
Cooling
B)
Chemical catalyst
C)
Heating to cure the resin
D)
Applying pressure
Shell moulding is a precision casting process that uses a mixture of sand and a thermosetting resin binder. The process works as follows: 1. A metal pattern is heated to a specific temperature (typically 200-300°C). 2. Resin-coated sand is dropped onto the heated pattern. 3. The heat from the pattern causes the thermosetting resin to melt and then cure (polymerize), forming a hard, thin 'shell' around the pattern. 4. Once the shell is hardened, the excess uncured sand is dumped, and the shell is removed from the pattern.
Why other options are incorrect:
- a. Cooling: Cooling would not trigger the polymerization of a thermosetting resin; heat is required for the chemical reaction.
- b. Chemical catalyst: While some resins use catalysts, shell moulding specifically relies on thermal energy (heat) from the pattern to initiate curing.
- d. Applying pressure: Pressure is not the mechanism used to harden the resin in this process; the chemical bond is formed via thermal curing.
Define ‘slenderness ratio’. How it is used to define long and short columns?
The motion of a circular shaft in a circular hole constitutes between the elements of kinematic pair:
A)
Successfully constrained motion
B)
Completely constrained motion
C)
Incompletely constrained motion
D)
None of the above
In the Theory of Machines, kinematic pairs are classified based on the nature of the constraint they impose on the relative motion between two links:
1. **Completely Constrained Motion**: When the motion is limited to a single direction regardless of the direction of the force applied. Example: A square bar in a square hole can only slide.
2. **Successfully Constrained Motion**: When the motion is constrained not by the geometry of the parts, but by some other means, such as gravity or a spring. Example: A shaft in a hole that is kept in place by the weight of a pulley.
3. **Incompletely Constrained Motion**: When the relative motion between the elements is not limited to a single direction by the geometry of the joint.
In the case of a **circular shaft in a circular hole**, the shaft can both **rotate** (about its axis) and **slide** (along its axis). Because the geometry of the circular shaft and hole allows for more than one type of motion (rotation and translation), the motion is not uniquely determined by the joint itself. Therefore, it is classified as an incompletely constrained motion.
A concentrated mass m is attached at the centre of a rod of length 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the undamped natural frequency of the system is:
A)
k/m
B)
2k/m
C)
k/2m
D)
4k/m
Option B is correct: 2k/m.
According to the Lewis equation which one of the following is correct?
A)
Pinion is always weaker than gear.
B)
Pinion is weaker than gear if both are made up of same material
C)
Gear is weaker than pinion if both are made up of same material
D)
Gear is always weaker than pinion
The Lewis equation for the beam strength of a gear tooth is given by: W = σ * b * m * Y, where W is the tangential load, σ is the allowable bending stress, b is the face width, m is the module, and Y is the Lewis form factor. The Lewis form factor (Y) depends on the number of teeth on the gear. For a given module and material, the form factor Y increases as the number of teeth increases. In a gear pair, the pinion is the smaller gear and thus has fewer teeth than the gear. Consequently, the Lewis form factor for the pinion (Yp) is smaller than that for the gear (Yg). Since the strength of the tooth is directly proportional to the form factor, the pinion tooth is weaker than the gear tooth when both are made of the same material and have the same face width and module. Therefore, the design of a gear drive is usually governed by the strength of the pinion.
Show that the expression for Mohr’s circle for a stress element subjected at any state of stress is (σx + σy)/2 + τxy^2 = τs^2 + ((σx - σy)/2)^2.
The Poisson’s ratio for steel varies from:
A)
0.21 to 0.25
B)
0.25 to 0.33
C)
0.33 to 0.38
D)
0.38 to 0.45
Poisson's ratio (ν) is the ratio of lateral strain to longitudinal strain when a material is stretched or compressed. For most structural metals, this value typically falls between 0.25 and 0.35. Specifically for steel, the standard accepted range in strength of materials and engineering textbooks is approximately 0.27 to 0.30, which fits squarely within the range of 0.25 to 0.33.
- Option (a) 0.21 to 0.25 is generally too low for standard structural steel.
- Option (c) 0.33 to 0.38 is more characteristic of aluminum or some copper alloys.
- Option (d) 0.38 to 0.45 is characteristic of highly plastic materials or rubber (which approaches 0.5 for incompressible materials).
Every material obeys the Hooke’s law within:
A)
Elastic limit
B)
Plastic limit
C)
Limit of proportionality
D)
None of these
Option C is correct: Limit of proportionality.
A ball and socket joint forms:
A)
A turning pair
B)
Rolling pair
C)
Spherical pair
D)
Sliding pair
In the Theory of Machines, a kinematic pair is classified based on the type of relative motion between two links.
1. **Spherical Pair (Option c):** A ball and socket joint is the classic example of a spherical pair. In this joint, one link (the ball) is a sphere that fits into a hollow socket of another link. This allows the ball to rotate freely in three dimensions (three rotational degrees of freedom) about a fixed point. Therefore, it is termed a spherical pair.
2. **Turning Pair (Option a):** Also known as a revolute pair, this allows only one rotational motion around a fixed axis (e.g., a hinge or a shaft in a bearing). A ball and socket joint allows more than one axis of rotation.
3. **Rolling Pair (Option b):** This occurs when two surfaces roll over each other without slipping (e.g., a ball bearing or a wheel rolling on a road). This is a higher pair involving point or line contact.
4. **Sliding Pair (Option d):** Also known as a prismatic pair, this allows only linear translation between two links (e.g., a piston in a cylinder).
Which of the following mechanism is used to enlarge or reduce the size of a drawing?
A)
Grasshopper mechanism
B)
Watt mechanism
C)
Pantograph
D)
None of the above
A pantograph is a mechanical linkage based on the principle of similar triangles (specifically a parallelogram linkage) used to copy a drawing or a shape on a different scale. By adjusting the position of the pivot point and the lengths of the arms, the device can either enlarge (magnify) or reduce the size of the original drawing.
Analysis of other options:
- Grasshopper mechanism: This is a type of linkage used to approximate straight-line motion, often used in early beam engines.
- Watt mechanism: This is a classic linkage designed by James Watt to convert rotary motion into approximate straight-line motion for the piston rod of a steam engine.
- None of the above: Incorrect, as the pantograph is the specific tool designed for scaling drawings.
Instantaneous centre of a body rolling on a stationary curved surface, lies:
A)
at the point of contact
B)
on the common normal at the point of contact
C)
on the common tangent at the point of contact
D)
at the centre of curvature of the stationary surface
1. Definition of Instantaneous Centre (I-centre): The instantaneous centre of rotation is a point in a body (or its extension) that has zero velocity at a given instant.
2. Pure Rolling Condition: When a body rolls without slipping on a stationary surface, the point of the body that is currently in contact with the surface must have a velocity of zero relative to the stationary surface.
3. Velocity Analysis: If there were any velocity at the point of contact, it would imply slipping (sliding). Since the condition is 'rolling' (which implies pure rolling unless specified otherwise), the velocity of the point of contact is zero.
4. Conclusion: Because the point of contact has zero instantaneous velocity, it satisfies the definition of the instantaneous centre of rotation for the rolling body.
Why other options are incorrect:
- Option b: The common normal passes through the center of curvature, but the I-centre is specifically at the contact point, not just anywhere on the normal.
- Option c: The common tangent is perpendicular to the velocity vector of the center of mass, but the point of zero velocity is the contact point itself.
- Option d: The center of curvature is the center of the arc, not the point where the velocity is zero.
The effect of increasing the stiffness of springs of centrifugal clutch is:
A)
The decrease of engagement speed
B)
The increase of engagement speed
C)
The increase of frictional force at maximum speed
D)
There is no effect on the engagement speed
In a centrifugal clutch, engagement occurs when the centrifugal force acting on the clutch shoes overcomes the compressive force exerted by the engagement springs.
1. The centrifugal force (F_c) is given by the formula: F_c = m * ω² * r, where 'm' is the mass of the shoe, 'ω' is the angular velocity, and 'r' is the radius of the center of gravity of the shoe.
2. The spring force (F_s) is given by: F_s = k * x, where 'k' is the stiffness (spring constant) and 'x' is the deflection.
3. Engagement begins at the 'engagement speed' (ω_e) when the centrifugal force equals the spring force: m * ω_e² * r = k * x.
4. Rearranging for engagement speed: ω_e = √( (k * x) / (m * r) ).
From this relationship, it is clear that the engagement speed (ω_e) is directly proportional to the square root of the spring stiffness (k). Therefore, if the stiffness of the springs is increased, a higher angular velocity (speed) is required to generate enough centrifugal force to compress the springs and engage the clutch.
Regarding other options:
- Option (a) is incorrect because increasing stiffness increases the speed required for engagement, not decreases it.
- Option (c) is incorrect because the frictional force at maximum speed depends on the total centrifugal force minus the spring force; while stiffness affects the net force, the primary and most direct effect of increasing stiffness is the shift in the engagement threshold.
- Option (d) is incorrect as the spring constant is a primary variable in the engagement equation.
For a point inside a rod subjected to pure torsion with shear stress , which of the following correctly describes Mohr’s circle?
A)
Centre at (0,0) with a radius equal to the normal stress
B)
Centre at (0,0) with a radius equal to the shear stress ()
C)
Centre at (,0) with a radius equal to the shear stress ()
D)
A point along the normal stress axis
Step-by-step reasoning:
1. **State of Stress**: A rod subjected to pure torsion experiences a state of 'pure shear'. For a point inside the rod, the normal stresses in the longitudinal and transverse directions are zero (σx = 0, σy = 0), and the only non-zero stress is the shear stress (τxy = τ).
2. **Mohr's Circle Parameters**: The center of Mohr's circle is located at the average normal stress: Center = ((σx + σy) / 2, 0). Substituting the values: Center = ((0 + 0) / 2, 0) = (0, 0).
3. **Radius Calculation**: The radius (R) of Mohr's circle is given by the formula: R = sqrt(((σx - σy)/2)^2 + τxy^2). Substituting the values: R = sqrt(((0 - 0)/2)^2 + τ^2) = sqrt(τ^2) = τ.
4. **Conclusion**: The circle is centered at the origin (0,0) and has a radius equal to the magnitude of the shear stress τ.
Analysis of other options:
- Option (a) is incorrect because there is no applied normal stress in pure torsion.
- Option (c) is incorrect because the center is at the origin, not at (τ, 0).
- Option (d) is incorrect because pure shear results in a full circle representing various stress states at different planes, not a single point.
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 two primary sensors: a load cell (to measure the applied force or load) and an extensometer or crosshead displacement sensor (to measure the change in length or elongation).
1. **Load (P):** This is the direct force applied by the machine's crosshead.
2. **Elongation (ΔL):** This is the direct measurement of how much the specimen has stretched.
All other parameters mentioned in the options are 'calculated' or 'derived' values, not 'measured' values:
- **Engineering Stress (σ):** Calculated as Load / Original Area (P/A₀).
- **Engineering Strain (ε):** Calculated as Elongation / Original Length (ΔL/L₀).
- **True Stress and True Strain:** Calculated using the instantaneous area and length during the test.
- **Young's Modulus (E) and Poisson's Ratio (ν):** These are material properties derived from the slope of the stress-strain curve and the ratio of lateral to longitudinal strain, respectively.
The two links OA and OB are connected by a pin joint at O. If the link OA turns with angular velocity ω1 rad/s in the clockwise direction and the link OB turns with angular velocity ω2 rad/s in the anti-clockwise direction, then the rubbing velocity at the pin joint O is: (where r = Radius of the pin at O)
A)
ω1 . ω2 . r
B)
(ω1 – ω2) r
C)
(ω1 + ω2) r
D)
(ω1 – ω2) 2r
Rubbing velocity (also known as sliding velocity) at a pin joint is the relative velocity between the two surfaces in contact at the interface of the pin.
1. **Analyze the velocities at the surface of the pin**:
- Link OA is rotating clockwise with angular velocity $\omega_1$. The linear velocity of the surface of the pin belonging to link OA at the point of contact is $v_1 = \omega_1 \cdot r$.
- Link OB is rotating anti-clockwise with angular velocity $\omega_2$. The linear velocity of the surface of the pin belonging to link OB at the point of contact is $v_2 = \omega_2 \cdot r$.
2. **Determine the relative direction**:
Since link OA is moving clockwise and link OB is moving anti-clockwise, at the point of contact on the pin's circumference, the two surfaces are moving in opposite directions relative to each other.
3. **Calculate Rubbing Velocity**:
When two surfaces move in opposite directions, the relative velocity (rubbing velocity $V_r$) is the sum of their individual linear velocities:
$V_r = v_1 + v_2$
$V_r = (\omega_1 \cdot r) + (\omega_2 \cdot r)$
$V_r = (\omega_1 + \omega_2)r$
Therefore, option (c) is correct. Option (b) would be correct if both links rotated in the same direction. Option (d) is incorrect as it uses the diameter instead of the radius for the relative velocity calculation at the surface.
What is the difference in the analysis of thick tubes compared to that for thin tubes? State the basic equations describing stress distribution in a thick tube.
With the increase of carbon content in steel, maximum stress:
A)
Increases
B)
Decreases
C)
Remains the same
D)
None of these
In the context of material science and strength of materials, the mechanical properties of steel are heavily influenced by its carbon content. Carbon acts as an interstitial alloying element that hinders the movement of dislocations within the crystal lattice of the iron (a process known as solid solution strengthening). As the carbon content increases: 1. Hardness increases: The material becomes more resistant to plastic deformation. 2. Tensile Strength (Maximum Stress) increases: The ultimate tensile strength (the maximum stress the material can withstand before necking/failure) increases because the carbon atoms create internal stresses that make it harder for the metal to deform. 3. Ductility decreases: While the strength increases, the material becomes more brittle and less capable of plastic deformation. Therefore, with an increase in carbon content, the maximum stress (ultimate strength) the steel can sustain increases.
Derive the expression for thermal stress when expansion of a body is fully restrained.
Explain the construction and working of Whitworth quick return mechanism.
The design of shafts made of brittle materials is based on:
A)
Guest’s theory
B)
Rankine’s theory
C)
St. Venant’s theory
D)
Von Mises Theory
In machine design, different failure theories are applied based on the nature of the material (ductile vs. brittle).
1. **Rankine's Theory (Maximum Normal Stress Theory):** This theory states that failure occurs when the maximum principal stress reaches the yield point or ultimate strength of the material. It is the most accurate theory for **brittle materials** (like cast iron), as brittle materials fail due to sudden fracture caused by normal stresses rather than shear deformation.
2. **Guest's Theory (Maximum Shear Stress Theory):** This theory states that failure occurs when the maximum shear stress reaches a critical value. It is primarily used for **ductile materials**.
3. **Von Mises Theory (Maximum Distortion Energy Theory):** This is the most accurate theory for predicting the yielding of **ductile materials** under complex loading.
4. **St. Venant's Theory:** While St. Venant is famous for his principle regarding stress distribution in beams, it is not a primary failure criterion used for the general design of shafts in the same context as the others listed.
Therefore, for shafts made of brittle materials, Rankine's theory is the standard design basis.