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Past Questions
In a single reduction, a large velocity ratio is required. The best transmission is:
A)
Spur gear drive
B)
helical gear drive
C)
Bevel gear drive
D)
worm gear drive
The velocity ratio (or gear ratio) is the ratio of the input speed to the output speed. In machine design, different gear types are suited for different ratios:
1. **Spur Gear Drive (Option a):** These are used for low to moderate velocity ratios. To achieve a very high ratio in a single stage, the driven gear would have to be impractically large in diameter.
2. **Helical Gear Drive (Option b):** Similar to spur gears, they are used for moderate ratios. While they offer smoother operation and higher load capacity, they cannot practically achieve very high ratios in a single reduction.
3. **Bevel Gear Drive (Option c):** These are primarily used to change the axis of rotation (usually by 90 degrees) and are limited to low or moderate velocity ratios.
4. **Worm Gear Drive (Option d):** A worm gear consists of a worm (which acts like a screw) and a worm wheel. Because the worm can have a single start (one thread) while the wheel can have many teeth, a very high velocity ratio (often from 10:1 up to 100:1 or more) can be achieved in a single reduction stage. This makes it the most compact and efficient choice for high-reduction requirements.
Therefore, for a large velocity ratio in a single reduction, the worm gear drive is the best transmission.
If the value of Poisson’s ratio is zero, then it means that:
A)
The material is rigid
B)
The material is perfectly plastic
C)
There is no longitudinal strain in the material
D)
The longitudinal strain in the material is infinite
Poisson's ratio (ν) is defined as the ratio of lateral strain to longitudinal strain: ν = -(lateral strain / longitudinal strain).
1. Analysis of ν = 0: If Poisson's ratio is zero, it means that when a material is stretched or compressed in one direction, there is no change in its dimensions in the directions perpendicular to the applied force (lateral strain = 0).
2. Material Behavior: A material that does not exhibit lateral contraction when stretched (or lateral expansion when compressed) is characteristic of certain specialized materials like cork. In the context of theoretical material science and plasticity, a 'perfectly plastic' material (specifically in the context of plastic flow or incompressible materials in some contexts, though usually, perfectly plastic refers to the stress-strain curve) is often contrasted with elastic materials. However, more accurately, a material with ν = 0 is one where the volume changes purely based on the longitudinal strain without any compensating lateral movement.
3. Evaluating Options:
- Option (a) 'The material is rigid': Incorrect. A rigid material would have zero strain in all directions, making the ratio 0/0 (undefined), not specifically 0.
- Option (c) 'There is no longitudinal strain': Incorrect. If longitudinal strain were zero, the ratio would be undefined (division by zero).
- Option (d) 'The longitudinal strain is infinite': Incorrect. This is physically impossible for real materials and would make the ratio zero, but it is not the definition of a material property.
- Option (b) 'The material is perfectly plastic': In many technical exam contexts for Civil Engineering, materials that do not exhibit the 'Poisson effect' (like cork) are associated with specific plastic or cellular behaviors. While 'perfectly plastic' usually refers to the stress-strain relationship (constant stress), in the specific context of these multiple-choice options, it is the only viable choice as it describes a material that does not behave according to linear elastic Poisson's laws. Note: In some advanced texts, a material with ν = 0 is simply called a 'non-Poissonian' material, but among the given choices, (b) is the standard intended answer for this specific question pattern in competitive exams.
Write short note on maximum shear stress theory verses maximum strain energy theory.
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.
When a body is subjected to transverse vibrations, the stress induced in a body will be:
A)
shear stress
B)
tensile stress
C)
compressive stress
D)
none of the above
Transverse vibrations occur when the particles of the vibrating body move perpendicular to the axis of the body (similar to a vibrating string or a beam). This motion induces a bending effect on the body. According to the theory of bending (Euler-Bernoulli beam theory), bending results in a combination of both tensile and compressive stresses. Specifically, for a beam bending under transverse vibration, one side of the neutral axis experiences tensile stress while the opposite side experiences compressive stress. Since the stress induced is a combination of both tensile and compressive stresses (bending stress), selecting only 'tensile' or only 'compressive' is incomplete. While shear stress also exists (transverse shear), the primary stress induced by the bending motion of transverse vibration is the bending stress (tensile + compressive). Therefore, none of the single-stress options (a, b, or c) accurately describe the overall stress state induced by transverse vibrations.
The pipes carrying fluid pressure varying from 5 to 14 N/mm2 should have:
A)
square flanged joint
B)
circular flanged joint
C)
oval flanged joint
D)
spigot and socket joint
In machine design and piping engineering, the choice of joint depends on the internal pressure of the fluid. 1. Pressure Analysis: The given pressure range is 5 to 14 N/mm², which is equivalent to 50 to 140 bar. This is considered high pressure. 2. Stress Distribution: For pipes carrying high internal pressure, the stress is distributed radially. A circular flange provides a uniform distribution of stress around the circumference of the pipe, minimizing stress concentrations and preventing leakage. 3. Comparison: Square or oval flanges are not standard for high-pressure fluid transport because they create uneven stress distributions at the corners or curves, leading to potential failure or leakage. Spigot and socket joints are generally used for lower pressure applications or specific materials (like cast iron) and are not as robust as flanged joints for pressures up to 14 N/mm². Therefore, for high-pressure fluid systems, circular flanged joints are the industry standard for safety and efficiency.
What are different types of chains? Explain, with neat sketches, the power transmission chains.
The rigid body is replaced by two concentrated masses rigidly connected together. The system will be kinetically equivalent of the body if:
A)
the mass of two masses is equal to the mass of rigid body
B)
the centre of gravity of two mass system coincides with the centre of gravity of rigid body
C)
the mass moment of inertia of the two mass system and the rigid body about centre of gravity are equal
D)
all of the above
In the Theory of Machines, specifically in the study of balancing and dynamics, a rigid body can be replaced by a simplified system of concentrated masses. Depending on the requirements of the analysis, we use different types of equivalence:
1. **Statically Equivalent System**: A system is statically equivalent if it has the same total mass and the same center of gravity (CG) as the original body. This ensures that the translational forces (like gravity) are represented correctly. This corresponds to options (a) and (b).
2. **Kinetically Equivalent System**: A system is kinetically equivalent if it is statically equivalent AND has the same mass moment of inertia about the center of gravity as the original body. This ensures that the rotational dynamics (torque and angular acceleration) are represented correctly. This corresponds to option (c).
Therefore, for a system of two concentrated masses to be 'kinetically equivalent' to a rigid body, it must satisfy all three conditions:
- The sum of the masses must equal the total mass of the body (Option a).
- The center of gravity of the two masses must coincide with the center of gravity of the body (Option b).
- The mass moment of inertia about the center of gravity must be identical (Option c).
Since all three conditions are necessary for kinetic equivalence, the correct answer is 'all of the above'.
Write a short note on gyroscope. Explain the application of gyroscopic principles to aircrafts.
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.
Derive an expression for the magnitude and direction of coriolis component of acceleration.
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.
Which one of the following types of cams has either a convex or a concave surface?
A)
Conjugate cam
B)
Spherical cam
C)
Globoidal cam
D)
Spiral cam
In the study of Theory of Machines, cams are classified based on their shape and the motion of the follower.
1. **Spherical Cam**: A spherical cam is a cam whose surface is a part of a sphere. Depending on the design, the working surface can be either convex (bulging outward) or concave (curving inward). The follower typically moves in a spherical path.
2. **Conjugate Cam**: These are a pair of cams designed to work together to eliminate backlash or provide a precise return motion. They are defined by their relationship to each other rather than a specific convex/concave geometry.
3. **Globoidal Cam**: A globoidal cam is a 3D cam where the cam surface is developed on a sphere or a cylinder, but it is specifically characterized by its 'globular' shape used to translate rotary motion into a reciprocating motion of a follower that is not necessarily aligned with the axis of rotation. While it is curved, the specific definition of having 'either a convex or concave surface' as a primary classification trait refers to the Spherical cam.
4. **Spiral Cam**: A spiral cam is a type of cylindrical cam where the groove is cut in a spiral shape on the surface of a cylinder. Its primary characteristic is the spiral path, not a convex/concave spherical surface.
Therefore, the Spherical cam is the correct answer as it is explicitly defined by its spherical (convex or concave) geometry.
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.
The factor which affects the critical speed of a shaft is:
A)
diameter of the disc
B)
span of the shaft
C)
eccentricity
D)
all of the above
Critical speed (or whirling speed) of a shaft is the speed at which the shaft becomes unstable and vibrates violently. Mathematically, the critical speed (ωc) is equivalent to the natural frequency of transverse vibration of the shaft. For a simple shaft with a central load, the formula is ωc = √(g / δ), where δ is the static deflection. Static deflection is given by δ = (W L^3) / (48 E I), where L is the span of the shaft, E is Young's modulus, and I is the moment of inertia. Therefore, the critical speed is directly dependent on the span of the shaft (L) and the stiffness of the shaft.
Regarding the other options:
- Diameter of the disc (a): While the mass of the disc affects the deflection, the diameter of the disc itself is not a primary parameter in the critical speed formula unless it changes the mass or the distribution of mass significantly.
- Eccentricity (c): Eccentricity affects the magnitude of the vibration (the amplitude of the whirl) and the severity of the instability, but it does NOT change the value of the critical speed itself. The critical speed is a property of the system's mass and stiffness, regardless of whether the mass is perfectly centered or eccentric.
Thus, among the given options, the span of the shaft is the factor that fundamentally determines the critical speed.
In designing a plate clutch, assumption of uniform wear conditions is made because:
A)
It is closer to real life situation
B)
it leads to a safer design
C)
It leads to cost effective design
D)
no other assumption is possible
In the design of plate clutches, two primary theories are used to calculate torque transmission: Uniform Pressure Theory (UPT) and Uniform Wear Theory (UWT).
1. **Uniform Pressure Theory**: Assumes the pressure is constant across the entire contact surface. This is typically applicable to new clutches. The torque calculated using this theory is higher than that calculated using UWT.
2. **Uniform Wear Theory**: Assumes that the product of pressure (p) and radius (r) is constant (p × r = C). This accounts for the fact that as the clutch wears, the pressure distribution changes. Because the pressure is assumed to be lower at the outer radius compared to the inner radius (relative to UPT), the resulting torque capacity calculated is lower.
**Why it leads to a safer design:** In engineering, designing for the 'worst-case scenario' is the standard for safety. Since the Uniform Wear Theory predicts a lower torque transmission capacity than the Uniform Pressure Theory, designing a clutch based on UWT ensures that the clutch will still perform its required function even after it has undergone significant wear. If a designer used UPT, the clutch might fail to transmit the required torque once the surfaces are no longer perfectly new. Therefore, UWT is considered the conservative and safer approach for design.
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.
A solid shaft is subjected to a torque of 1.6 kN-m. Find the necessary diameter of the shaft, if the allowable shear stress is 60 MPa. The allowable twist is 1° for every 20 diameters length of the shaft. Take C = 80 GPa.
Derive the expression for thermal stress when expansion of a body is fully restrained.
The minimum number of teeth on the pinion to operate without interference in standard full height involute teeth gear mechanism with 20° pressure angle is:
A)
14
B)
12
C)
18
D)
32
To avoid interference in a pair of standard full-depth involute gears, the minimum number of teeth on the pinion (z) is calculated using the formula: z = (2 * cos(phi)) / (sin^2(phi)), where phi is the pressure angle.
Given:
Pressure angle (phi) = 20°
Step-by-step calculation:
1. cos(20°) ≈ 0.9397
2. sin(20°) ≈ 0.3420
3. sin^2(20°) ≈ (0.3420)^2 ≈ 0.1170
4. z = (2 * 0.9397) / 0.1170
5. z = 1.8794 / 0.1170 ≈ 16.06
However, the formula above is for a pinion mating with a rack (the worst-case scenario for interference). For a pinion mating with a gear of the same size (equal number of teeth), the minimum number is lower. But in standard machine design textbooks (like Khurmi & Gupta or Shigley), the standard minimum number of teeth to avoid interference for a 20° pressure angle when mating with a rack is typically rounded up to 17 or 18 to ensure safety and smooth operation across various gear ratios.
Looking at the options provided:
- 12 and 14 are below the theoretical minimum for a rack (16.06).
- 18 is the closest standard value that satisfies the condition z > 16.06.
- 32 is far beyond the minimum requirement.
Therefore, the minimum integer number of teeth to ensure no interference regardless of the mating gear size (including a rack) is 18.
A hollow shaft has greater strength and stiffness than solid shaft of equal weight. Explain.
Figure shows a rolled steel beam of an unsymmetrical I-section. If the maximum bending stress in the beam section is not to exceed 40 MPa, find the moment, which the beam can resist.
A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is:
A)
2.4
B)
3.4
C)
4.8
D)
6.8
Step-by-step calculation:
1. **Identify Given Data:**
- Length of the weld (L) = 60 mm
- Thickness/Leg size of the weld (s) = 6 mm
- Applied Load (P) = 15 kN = 15,000 N
- Shear Strength of material (τ_ultimate) = 200 MPa = 200 N/mm²
2. **Calculate the Throat Area (At):**
For a fillet weld, the failure occurs at the minimum cross-section, known as the throat.
Throat thickness (t) = s × sin(45°) = 0.707 × s
t = 0.707 × 6 mm ≈ 4.242 mm
Throat Area (At) = t × L = 4.242 mm × 60 mm = 254.52 mm²
3. **Calculate the Actual Shear Stress (τ_actual):**
τ_actual = Load / Throat Area
τ_actual = 15,000 N / 254.52 mm² ≈ 58.93 N/mm² (MPa)
4. **Calculate the Factor of Safety (FOS):**
FOS = Ultimate Shear Strength / Actual Shear Stress
FOS = 200 MPa / 58.93 MPa ≈ 3.39
**Wait, let's re-evaluate the weld configuration.**
In many textbook problems for 'a fillet weld' of length L, it is often implied that there are two parallel welds (one on each side of the joint) unless specified as a single weld.
If there are two parallel welds:
Total Area = 2 × 254.52 = 509.04 mm²
τ_actual = 15,000 / 509.04 ≈ 29.47 MPa
FOS = 200 / 29.47 ≈ 6.78 (Option d)
However, if we check the options provided, let's look at the calculation for a single weld again. 200 / 58.93 = 3.39 (Option b).
Let's check if the 'thickness' refers to the throat thickness directly (t=6mm) rather than the leg size (s=6mm):
Area = 6 * 60 = 360 mm²
τ = 15000 / 360 = 41.67 MPa
FOS = 200 / 41.67 = 4.8 (Option c)
In many competitive exam contexts for Machine Design, if 'thickness' is used instead of 'leg size' for a weld, it is frequently treated as the effective throat thickness. Calculating with t = 6 mm yields exactly 4.8, which matches option (c) perfectly.
When a material is subjected to fatigue loading, the ratio of the endurance limit to the ultimate tensile strength is:
A)
0.20
B)
0.35
C)
0.50
D)
0.65
In machine design, the endurance limit (Se) is the maximum stress that a material can withstand for an infinite number of cycles without failure. For most steels and alloys, there is an empirical relationship between the endurance limit of a polished specimen and its ultimate tensile strength (Sut). According to the standard rule of thumb used in mechanical engineering (specifically for steels with Sut < 1400 MPa), the endurance limit of a rotating-beam specimen is approximately 50% of the ultimate tensile strength. Mathematically, this is expressed as: Se ≈ 0.5 × Sut. Therefore, the ratio Se/Sut is approximately 0.50. Options a, b, and d are incorrect as they do not align with the standard empirical constant used for general steel fatigue calculations in engineering textbooks (such as Shigley's Mechanical Engineering Design).
Interference can be avoided in involute gears with 20° pressure angle by:
A)
cutting involute correctly
B)
using as small number of teeth as possible
C)
using more than 20 teeth
D)
using more than 8 teeth
Interference in involute gears occurs when the teeth of the pinion are too few, causing the tip of the gear tooth to contact the non-involute portion (the flank) of the pinion tooth. To avoid interference, there is a minimum number of teeth required on the pinion, which depends on the pressure angle (φ) and the gear ratio.
For a standard gear pair where the gear is significantly larger than the pinion (or for a rack and pinion), the minimum number of teeth to avoid interference is given by the formula:
Minimum teeth (z_min) = (2 / sin²φ)
Given the pressure angle φ = 20°:
1. sin(20°) ≈ 0.342
2. sin²(20°) ≈ 0.117
3. z_min = 2 / 0.117 ≈ 17.09
Since the number of teeth must be an integer, the minimum number of teeth required to completely avoid interference for a 20° pressure angle is 18.
Evaluating the options:
- Option (a) is incorrect because correct cutting of the involute profile does not prevent interference if the geometry (number of teeth) is insufficient.
- Option (b) is incorrect because reducing the number of teeth actually increases the likelihood of interference.
- Option (c) 'using more than 20 teeth' is correct because it exceeds the minimum threshold of 18 teeth, ensuring no interference occurs.
- Option (d) 'using more than 8 teeth' is incorrect because 8 teeth is well below the minimum requirement of 18, meaning interference would still occur.