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Test: Strength of Materials Level - 1 - Mechanical Engineering MCQ


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25 Questions MCQ Test - Test: Strength of Materials Level - 1

Test: Strength of Materials Level - 1 for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Test: Strength of Materials Level - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Strength of Materials Level - 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Strength of Materials Level - 1 below.
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Test: Strength of Materials Level - 1 - Question 1
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Wherever some external system of forces acts on a body, it undergoes some deformation. As the body under goes some deformation, it sets up some resistance to the deformations. This resistance per unit area to deformation, is called

Detailed Solution for Test: Strength of Materials Level - 1 - Question 1
Explanation:
The correct answer is stress.
Stress is the resistance per unit area to deformation. It is a measure of the internal forces within a material that counteract the external forces acting on it. When an external force is applied to a body, it causes the body to undergo deformation. This deformation results in internal stresses that develop within the material.
Here is a more detailed explanation:
External System of Forces:
- Whenever an external force is applied to a body, it can cause the body to deform.
- Examples of external forces include compression, tension, shear, and bending.
Deformation:
- Deformation refers to the change in shape or size of a body due to the applied external forces.
- It can be either elastic or plastic deformation, depending on the material's behavior.
Resistance to Deformation:
- When a body undergoes deformation, it sets up resistance to counteract the applied external forces.
- This resistance per unit area is known as stress.
- Stress is a measure of the internal forces that develop within the material.
Types of Stress:
- There are different types of stress, including:
1. Tensile Stress: It occurs when a body is being pulled apart.
2. Compressive Stress: It occurs when a body is being squeezed or compressed.
3. Shear Stress: It occurs when a body is subjected to forces parallel to its surface.
4. Bending Stress: It occurs when a body is bent or subjected to a combination of tension and compression.
Importance of Stress:
- Understanding stress is crucial in various fields, including engineering, mechanics, and materials science.
- It helps in designing structures and materials that can withstand the applied forces without failure.
- The study of stress and its effects on materials is essential for ensuring the safety and reliability of structures and components.
In conclusion, stress is the resistance per unit area to deformation, which develops within a material when external forces are applied. It is a crucial concept in understanding the behavior of materials and designing structures that can withstand the applied forces.
Test: Strength of Materials Level - 1 - Question 2
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Which of the following is a proper sequence ?

Detailed Solution for Test: Strength of Materials Level - 1 - Question 2
Proper Sequence of Mechanical Properties:
To determine the proper sequence of mechanical properties, we need to understand the definitions and relationships between these properties.
1. Proportional Limit: It is the maximum stress at which stress and strain remain directly proportional to each other. It represents the elastic behavior of a material and is the point where Hooke's law applies.
2. Elastic Limit: It is the maximum stress that a material can withstand without permanent deformation once the stress is removed. Beyond this limit, the material exhibits plastic behavior and undergoes permanent deformation.
3. Yielding: It is the point at which the material starts to deform plastically under stress. The stress required for yielding is higher than the elastic limit.
4. Failure: It is the point at which the material breaks or fractures under stress. Failure can occur after the yielding point or directly without any plastic deformation, depending on the material and loading conditions.
Now, let's analyze the given options:
A: Proportional limit, elastic limit, yielding sequence, failure
- This sequence is incorrect because the proportional limit should be followed by the elastic limit, not the other way around.
B: Elastic limit, proportional limit, yielding, failure
- This sequence is incorrect because the elastic limit should be followed by the proportional limit, not the other way around.
C: Yielding, Proportional limit, elastic limit, failure
- This sequence is correct because it follows the correct order of mechanical properties.
D: None of the above
- This option is incorrect because option C is the correct sequence.
Therefore, the proper sequence of mechanical properties is option C: Yielding, Proportional limit, elastic limit, failure.
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Test: Strength of Materials Level - 1 - Question 3
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Hooke's law holds good up to

Detailed Solution for Test: Strength of Materials Level - 1 - Question 3
Hooke's law holds good up to the elastic limit.
Hooke's law is a principle in physics that states that the force needed to extend or compress a spring by some distance is proportional to that distance. It can be mathematically represented as F = kx, where F is the force applied, k is the spring constant, and x is the displacement.
To understand why Hooke's law holds good up to the elastic limit, let's consider the following points:
1. Hooke's law and elasticity: Hooke's law is specifically applicable to elastic materials, which are those that can return to their original shape and size after being deformed by an external force. These materials exhibit a linear relationship between the applied force and the resulting displacement.
2. Proportional limit: The proportional limit is the point at which the relationship between the applied force and the displacement ceases to be linear. Beyond this point, the material begins to exhibit non-linear behavior, and Hooke's law is no longer valid.
3. Elastic limit: The elastic limit is the maximum stress or force that a material can withstand without permanently deforming. It is the point at which the material's behavior transitions from elastic to plastic (permanent) deformation. Hooke's law holds good up to this limit, as long as the material remains in its elastic range.
4. Yield point and breaking point: Beyond the elastic limit, the material enters the plastic deformation region, where it undergoes permanent changes in shape and size. The yield point is the stress at which significant plastic deformation occurs, while the breaking point is the stress at which the material fractures or breaks. Hooke's law is no longer applicable beyond these points.
In conclusion, Hooke's law holds good up to the elastic limit of a material. Once the material exceeds this limit and enters the plastic deformation region, the linear relationship between force and displacement is no longer valid.
Test: Strength of Materials Level - 1 - Question 4
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The deformation of a bar under its own weight is __________ the deformation, if the same body is subjected to a direct load equal to weight of the body.
Detailed Solution for Test: Strength of Materials Level - 1 - Question 4
Deformation of a bar under its own weight compared to a direct load equal to the weight of the body:
When a bar is subjected to its own weight, it experiences a certain amount of deformation or strain. The question asks us to compare this deformation with the deformation that would occur if the bar is subjected to a direct load equal to its weight.
To understand the relationship between these two scenarios, let's analyze the situation:
1. Deformation under its own weight:
- When a bar is subjected to its own weight, it experiences a compressive force along its length.
- This compressive force causes the bar to deform, resulting in a change in its shape or length.
- The amount of deformation depends on the material properties of the bar, such as its Young's modulus and cross-sectional area.

2. Deformation under a direct load equal to its weight:
- If the same bar is subjected to a direct load equal to its weight, it experiences a compressive force of greater magnitude compared to its own weight.
- This increased compressive force will cause a greater deformation compared to the deformation under its own weight.

Therefore, the deformation of the bar under its own weight is less than the deformation that would occur if the same body is subjected to a direct load equal to its weight.
In conclusion, the answer is option B: half.
Test: Strength of Materials Level - 1 - Question 5
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The elongation of a conical bar under its own weight is ......that of prismatic bar of the same length.
Detailed Solution for Test: Strength of Materials Level - 1 - Question 5
Explanation:
To determine the elongation of a conical bar under its own weight, we need to compare it to a prismatic bar of the same length. Let's consider the following:
Conical Bar:
- A conical bar is in the shape of a cone, with a varying cross-sectional area along its length.
- When the conical bar is subjected to its own weight, it experiences a distributed load, with higher stress at the narrow end and lower stress at the wide end.
- The elongation of the conical bar would depend on its geometry, material properties, and the distribution of stress along its length.
Prismatic Bar:
- A prismatic bar has a constant cross-sectional area along its length, which means the stress distribution is uniform.
- When the prismatic bar is subjected to its own weight, it experiences a distributed load with a constant stress along its length.
- The elongation of the prismatic bar would depend on its material properties and the magnitude of the stress.
Comparison:
- The elongation of a conical bar under its own weight is one-third that of a prismatic bar of the same length.
- This is because the varying stress distribution in a conical bar results in a lower overall elongation compared to a prismatic bar with a uniform stress distribution.
Therefore, the correct answer is C: one-third.
Test: Strength of Materials Level - 1 - Question 6
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Modular ratio of the two materials is the ratio of
Detailed Solution for Test: Strength of Materials Level - 1 - Question 6
Modular Ratio:
The modular ratio is a property used to compare the behavior of two different materials under the same conditions. It is defined as the ratio of their modulus of elasticity.
Modulus of Elasticity:
The modulus of elasticity is a measure of a material's ability to deform elastically under applied stress. It represents the stiffness of the material and is defined as the ratio of stress to strain.
Linear Stress and Linear Strain:
When a material is subjected to a tensile or compressive force, it experiences linear stress and linear strain. Linear stress is the ratio of the applied force to the cross-sectional area of the material, while linear strain is the ratio of the change in length to the original length of the material.
Shear Stress and Shear Strain:
Shear stress and shear strain occur when a material is subjected to forces that cause it to deform in a parallel manner. Shear stress is the ratio of the force applied parallel to the cross-sectional area of the material, while shear strain is the ratio of the displacement parallel to the original length of the material.
Explanation:
The modular ratio compares the modulus of elasticity of two different materials. It does not consider the linear stress and linear strain or the shear stress and shear strain. Therefore, options A and B are incorrect.
The modulus of elasticity represents the stiffness of a material and is a fundamental property that characterizes its behavior under stress and strain. Therefore, option C is correct.
The modulus of rigidity is a measure of a material's resistance to shear deformation. It is defined as the ratio of shear stress to shear strain. However, the question asks for the modular ratio, which is the ratio of the modulus of elasticity. Therefore, option D is incorrect.
In conclusion, the correct answer is option C: the modular ratio of the two materials is the ratio of their modulus of elasticities.
Test: Strength of Materials Level - 1 - Question 7
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A bolt is made to pass through a tube and both of them are tightly fitted with the help of washers and nuts. If the nut is tightened, then
Detailed Solution for Test: Strength of Materials Level - 1 - Question 7

Given:
A bolt is made to pass through a tube and both of them are tightly fitted with the help of washers and nuts.
Explanation:
When the nut is tightened, it applies a force on the bolt and tube assembly. This force can be either compression or tension depending on the arrangement and application. Let's analyze the possibilities:
A. Bolt and tube are under tension:
If the nut is tightened, it will pull the bolt and tube towards each other. This will create tension in both the bolt and the tube. However, this is not a common arrangement as it may not provide a secure fit.
B. Bolt and tube are under compression:
If the nut is tightened, it will push the bolt and tube away from each other. This will create compression in both the bolt and the tube. However, this is also an unlikely arrangement as it may cause the bolt and tube to loosen over time.
C. Bolt is under compression and tube is under tension:
If the nut is tightened, it will push the bolt away from the tube, creating compression in the bolt. At the same time, it will pull the tube towards the bolt, creating tension in the tube. This is a more common arrangement as it ensures a secure fit between the bolt and tube.
D. Bolt is under tension and tube is under compression:
If the nut is tightened, it will pull the bolt towards the tube, creating tension in the bolt. At the same time, it will push the tube away from the bolt, creating compression in the tube. This arrangement is less common as it may cause the bolt and tube to loosen over time.
Conclusion:
Based on the given options, it can be concluded that the correct answer is D. When the nut is tightened, the bolt is under tension and the tube is under compression.
Test: Strength of Materials Level - 1 - Question 8
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The Poisson's ratio for steel varies from
Detailed Solution for Test: Strength of Materials Level - 1 - Question 8
Poisson's Ratio for Steel
The Poisson's ratio is a material property that relates the lateral strain to the axial strain when a material is subjected to tensile or compressive loading. The Poisson's ratio for steel can vary depending on the specific type of steel and its composition.
Range of Poisson's Ratio for Steel
The Poisson's ratio for steel typically falls within a specific range. Based on the given options, the correct answer is A: 0.27 to 0.30. This means that the Poisson's ratio for steel can range from 0.27 to 0.30.
Importance of Poisson's Ratio
The Poisson's ratio is an important material property as it characterizes the deformation behavior of a material under different loading conditions. It is particularly relevant in structural engineering and design, as it affects the stability, strength, and performance of structures.
Significance of the Given Range
The given range of 0.27 to 0.30 for the Poisson's ratio of steel is based on the typical behavior observed in various types of steel. Different types of steel, such as carbon steel or stainless steel, may have slightly different Poisson's ratios within this range. However, it is important to note that the exact value of the Poisson's ratio can vary depending on factors such as temperature, strain rate, and manufacturing process.
Conclusion
In conclusion, the Poisson's ratio for steel can vary, but the most accurate range based on the given options is A: 0.27 to 0.30. It is crucial to consider this property when designing and analyzing structures made from steel to ensure their stability and performance.
Test: Strength of Materials Level - 1 - Question 9
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The Poisson's ratio for cast iron varies from
Detailed Solution for Test: Strength of Materials Level - 1 - Question 9
Question: The Poisson's ratio for cast iron varies from?
Answer:
The correct answer to the question is B: 0.23 to 0.27.
Explanation:
To understand why the correct answer is B, let's break down the Poisson's ratio and its range for cast iron.
What is Poisson's ratio?
Poisson's ratio is a measure of the ratio of lateral strain to axial strain for a material under an applied load. It quantifies the material's ability to deform in one direction when being stretched or compressed in another direction.
Poisson's ratio range for cast iron:
- Cast iron is a group of iron-carbon alloys with a high carbon content. It is known for its excellent castability and high strength.
- The Poisson's ratio for cast iron typically falls within a specific range.
- The correct range for the Poisson's ratio of cast iron is 0.23 to 0.27.
- This means that when cast iron is subjected to an axial strain, it will exhibit a lateral strain that is approximately 23% to 27% of the axial strain.
Importance of Poisson's ratio:
- Poisson's ratio is an important material property that affects the behavior of materials under loading conditions.
- It is used in engineering calculations to predict the deformation and behavior of materials.
- Knowing the Poisson's ratio for a material like cast iron is crucial for designing and analyzing structures that use this material.
In conclusion, the correct range for the Poisson's ratio of cast iron is 0.23 to 0.27 (Option B).
Test: Strength of Materials Level - 1 - Question 10
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When a body is subjected to three mutually perpendicular stresses, of equal intensity, the ratio of direct stress to the corresponding volumetric strain is known as
Detailed Solution for Test: Strength of Materials Level - 1 - Question 10
Explanation:
To find the ratio of direct stress to the corresponding volumetric strain, we need to consider the concept of bulk modulus.
Bulk Modulus:
The bulk modulus (K) is a measure of a material's resistance to uniform compression. It is defined as the ratio of direct stress (σ) to the corresponding volumetric strain (εv).
Given:
- The body is subjected to three mutually perpendicular stresses.
- The stresses are of equal intensity.
Key Points:
- When three mutually perpendicular stresses are applied to a body, each stress component acts independently and contributes to the overall deformation of the body.
- The direct stress is the average of the three stress components.
- The volumetric strain is the sum of the three strain components.

- Since the three stresses are of equal intensity, the direct stress is the average of the three stress components. Therefore, the direct stress is equal to one-third of the intensity of each stress component.
- The volumetric strain is the sum of the three strain components. Therefore, the volumetric strain is equal to the sum of the strain components.
Ratio of Direct Stress to Volumetric Strain:
- The ratio of direct stress to volumetric strain is equal to the ratio of the average stress component to the sum of the strain components.
- This ratio is known as the bulk modulus (K).
Conclusion:
Therefore, the correct answer is option C: bulk modulus.
Test: Strength of Materials Level - 1 - Question 11
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The relation between Young's modulus (E) and bulk modulus (K) is given by
Detailed Solution for Test: Strength of Materials Level - 1 - Question 11

To find the relation between Young's modulus (E) and bulk modulus (K), we can use the definition of Young's modulus and bulk modulus.
Young's modulus (E) is defined as the ratio of stress to strain in the direction perpendicular to the applied force.
Bulk modulus (K) is defined as the ratio of the change in pressure to the fractional change in volume.
Let's derive the relation between E and K step by step:
1. Start with the definition of Young's modulus:
E = stress/strain
2. Express stress in terms of pressure and strain in terms of fractional change in volume:
stress = pressure
strain = -ΔV/V₀
3. Substitute the expressions for stress and strain into the definition of Young's modulus:
E = pressure/(-ΔV/V₀)
E = -V₀Δp/ΔV
4. Rearrange the equation to solve for Δp/ΔV:
Δp/ΔV = -E/V₀
5. Recall that the definition of bulk modulus is the ratio of Δp to ΔV/V₀:
K = -Δp/(ΔV/V₀)
6. Substitute the expression for Δp/ΔV into the equation for K:
K = -(-E/V₀)
K = E/V₀
7. Simplify the equation by multiplying both numerator and denominator by V:
K = EV/EV₀
8. Cancel out the V terms:
K = E/V₀
9. Finally, rearrange the equation to solve for K:
K = E/(V₀)

Hence, the relation between Young's modulus (E) and bulk modulus (K) is given by:
K = E/(V₀)
Answer: D. K = mE/[3(m-2)]
Test: Strength of Materials Level - 1 - Question 12
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The ratio of shear modulus to the modulus of elasticity for a Poisson's ratio of 0.4 will be
Detailed Solution for Test: Strength of Materials Level - 1 - Question 12
Given:

Poisson's ratio (ν) = 0.4


To find:

The ratio of shear modulus (G) to the modulus of elasticity (E)



The ratio of shear modulus (G) to the modulus of elasticity (E) can be calculated using the formula:


G/E = (3 - 2ν)/(2(1 + ν))


Substituting the given value of Poisson's ratio (ν = 0.4) into the formula:


G/E = (3 - 2(0.4))/(2(1 + 0.4))


G/E = (3 - 0.8)/(2(1.4))


G/E = 2.2/2.8


G/E = 11/14


Answer:

The ratio of shear modulus to the modulus of elasticity for a Poisson's ratio of 0.4 is 11/14.

Test: Strength of Materials Level - 1 - Question 13
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The relation between Young's modulus (E), shear modulus (C) and bulk modulus (K) is given by
Detailed Solution for Test: Strength of Materials Level - 1 - Question 13
The relation between Young's modulus (E), shear modulus (C), and bulk modulus (K) is given by:
Answer: C
The correct relation between Young's modulus (E), shear modulus (C), and bulk modulus (K) is given by equation C.
Explanation:
Young's modulus (E), shear modulus (C), and bulk modulus (K) are all measures of a material's elasticity, which describe how it deforms under external forces. The relation between these moduli can be expressed in the following equation:
E = 3K(1 - 2ν)
where:
- E is Young's modulus
- C is shear modulus
- K is bulk modulus
- ν (nu) is Poisson's ratio
Poisson's ratio (ν) is a dimensionless constant that represents the ratio of lateral strain to axial strain when a material is subjected to an external force. It is a property that characterizes the deformation behavior of a material.
In equation C, the relation between Young's modulus (E), shear modulus (C), and bulk modulus (K) is given in terms of Poisson's ratio (ν). The equation shows that Young's modulus (E) is related to the bulk modulus (K) and Poisson's ratio (ν). The shear modulus (C) is not directly related to Young's modulus (E) and bulk modulus (K).
Therefore, the correct relation between Young's modulus (E), shear modulus (C), and bulk modulus (K) is given by equation C.
Test: Strength of Materials Level - 1 - Question 14
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When a body is subjected to a direct tensile stress (sx) in one plane accompanied by a simpl shear stress (txy), the maximum shear stress is
Test: Strength of Materials Level - 1 - Question 15
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The state of stress at a point in a loaded member is shown is fig. The magnitude of maximum shear stress is
Detailed Solution for Test: Strength of Materials Level - 1 - Question 15

Max shear stress = sqrt[{(-40- 40) / 2}^2 + 30^2]. = sqrt[40^2 + 30^2]. = sqrt(2500). = 50 mpa.

Test: Strength of Materials Level - 1 - Question 16
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The strain energy stored in a spring, when subjected to maximum load, without suffering permanent distortion, is known as
Detailed Solution for Test: Strength of Materials Level - 1 - Question 16
Proof Resilience:
- The strain energy stored in a spring, when subjected to maximum load without suffering permanent distortion, is known as proof resilience.
- It is a measure of a material's ability to absorb energy without undergoing permanent deformation.
- Proof resilience is an important property in engineering applications, as it indicates the spring's ability to absorb and release energy repeatedly without failure.
- It is commonly used to evaluate the performance and reliability of springs, which are widely used in various mechanical systems.
- The proof resilience of a spring can be calculated using the formula: Proof Resilience = (1/2) × Stress × Strain × Volume.
- The unit of proof resilience is joules per cubic meter (J/m³).
- Proof resilience is different from impact energy, which refers to the energy absorbed by a material when it is subjected to a sudden impact or shock.
- It is also different from proof stress, which is the maximum stress a material can withstand without permanent deformation.
Conclusion:
The correct answer is B: proof resilience.
Test: Strength of Materials Level - 1 - Question 17
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The bending equation is
Test: Strength of Materials Level - 1 - Question 18
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The neutral axis of the cross-section a beam is that axis at which the bending stress is
Detailed Solution for Test: Strength of Materials Level - 1 - Question 18
The neutral axis of the cross-section of a beam is the axis at which the bending stress is zero.
The bending stress in a beam is caused by the bending moment applied to it, which results in a distribution of stress throughout the cross-section of the beam. The neutral axis is the axis of the beam where the bending stress is zero, meaning that there are no tensile or compressive stresses present at that location.
Explanation:
To understand why the neutral axis is where the bending stress is zero, we need to consider the behavior of the beam under bending.
1. Bending Stress Distribution:
- When a beam is subjected to bending, the top fibers of the beam are under compression, while the bottom fibers are under tension.
- This results in a stress distribution within the beam, with the maximum compressive stress occurring at the top surface and the maximum tensile stress occurring at the bottom surface.
- The stress distribution is linear across the height of the beam, with the highest stress at the extreme fibers and decreasing towards the neutral axis.
2. Neutral Axis Location:
- The neutral axis is the axis within the beam cross-section where the stress is zero. This means that the fibers along the neutral axis do not experience any tensile or compressive stress.
- The neutral axis is located at the centroid of the cross-section, which is the geometric center of the shape.
- For symmetric cross-sections such as rectangles, circles, or I-beams, the neutral axis is located at the center of the shape.
- For asymmetric cross-sections, the neutral axis is shifted towards the side with more material.
3. Implications of Zero Bending Stress:
- The fact that the bending stress is zero at the neutral axis has important implications for the design and analysis of beams.
- It means that the neutral axis is the axis that is least affected by bending, as it experiences no stress.
- The location of the neutral axis determines the distribution of stress within the beam, and it is used to calculate important parameters such as the moment of inertia and section modulus.
- The position of the neutral axis also affects the deflection and stiffness of the beam.
In conclusion, the neutral axis of a beam is the axis at which the bending stress is zero. This axis is located at the centroid of the cross-section and is least affected by bending. Understanding the concept of the neutral axis is crucial for the analysis and design of beams.
Test: Strength of Materials Level - 1 - Question 19
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When a rectangular beam is loaded transversely, the maximum tensile stress is developed on the
Detailed Solution for Test: Strength of Materials Level - 1 - Question 19

Correct Answer :- b

Explanation : Maximum compressive stress in developed at the top layer. And maximum tensile stress is developed at bottom fibre due to elongation of fibre.

Test: Strength of Materials Level - 1 - Question 20
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Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B'. The stiffness of spring 'B' will be
Detailed Solution for Test: Strength of Materials Level - 1 - Question 20

To solve this problem, we need to understand the concept of stiffness or spring constant.
The stiffness or spring constant (k) of a spring is a measure of how much force is needed to stretch or compress the spring by a certain distance. It is directly proportional to the diameter of the wire and inversely proportional to the length of the spring.
Given that the diameter of the wire of spring 'A' is double that of spring 'B', we can assume that the length and the number of coils of both springs are the same.
Now, let's compare the stiffness of spring 'A' and spring 'B':
Spring A:
- Diameter of wire: 2x (double that of spring B)
- Length: same as spring B
- Number of coils: same as spring B
Spring B:
- Diameter of wire: x
- Length: same as spring A
- Number of coils: same as spring A
From the given information, we can conclude that the only difference between spring 'A' and spring 'B' is the diameter of the wire. Since the stiffness of a spring is directly proportional to the diameter of the wire, we can compare the stiffness of spring 'A' and spring 'B' as follows:
Stiffness of spring A = 2x
Stiffness of spring B = x
Therefore, the stiffness of spring 'B' is one-half (1/2) of the stiffness of spring 'A'.
Hence, the correct answer is option D: one-half.
Test: Strength of Materials Level - 1 - Question 21
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Two closely -coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be ....... that of spring 'B'.
Detailed Solution for Test: Strength of Materials Level - 1 - Question 21

To find the stiffness ratio between two closely-coiled helical springs 'A' and 'B', we need to consider the following:
1. The stiffness of a spring is directly proportional to the square of the number of turns. Therefore, we can write the equation as:
Stiffness of spring 'A' / Stiffness of spring 'B' = (Number of turns of spring 'A')^2 / (Number of turns of spring 'B')^2
2. Given that the number of turns of spring 'A' is double that of spring 'B', we can substitute the values in the equation:
Stiffness of spring 'A' / Stiffness of spring 'B' = (2)^2 / (1)^2
= 4/1
= 4
3. Therefore, the stiffness of spring 'A' is 4 times that of spring 'B'.
4. The options given are:
A. one-half
B. one-eighth
C. one-fourth
D. one-sixteenth
5. From the calculation, we can see that the stiffness ratio is 4, which is equal to one-fourth.
6. Hence, the correct answer is option C: one-fourth.
Test: Strength of Materials Level - 1 - Question 22
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A closely-coiled helical spring is cut into two halves. The stiffness of the resulting spring will be
Detailed Solution for Test: Strength of Materials Level - 1 - Question 22

The stiffness of a spring is determined by its spring constant, which is a measure of how much force is required to stretch or compress the spring by a certain amount.
When a closely-coiled helical spring is cut into two halves, the following changes occur:
1. Length: The length of each half of the spring will be half of the original length.
2. Number of turns: Each half of the spring will have half the number of turns compared to the original spring.
3. Cross-sectional area: The cross-sectional area remains the same as it is not affected by cutting the spring.
4. Material properties: The material properties of the spring remain the same as they are not affected by cutting.
Based on these changes, we can determine the new stiffness of the resulting spring:
- The length of each half of the spring is half of the original length. This means that the spring constant for each half of the spring will be twice the original spring constant. (K' = 2K)
- The number of turns for each half of the spring is half of the original number of turns. This means that the spring constant for each half of the spring will be half the original spring constant. (K'' = K/2)
Therefore, the stiffness of the resulting spring will be double the original stiffness.
The correct answer is option B: double.
Test: Strength of Materials Level - 1 - Question 23
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A composite shaft consisting of two stepped portions having spring constants k1 and k2 is held between two rigid supports at the ends. Its equivalent spring constant is
Detailed Solution for Test: Strength of Materials Level - 1 - Question 23


To find the equivalent spring constant of the composite shaft, we can use the concept of parallel and series combination of springs.
The composite shaft consists of two stepped portions, each with a different spring constant. Let the spring constants of the two portions be k1 and k2.
Step 1: Find the equivalent spring constant of the first portion (k1).
The first portion of the shaft has a spring constant of k1. Therefore, the equivalent spring constant of this portion is k1.
Step 2: Find the equivalent spring constant of the second portion (k2).
The second portion of the shaft has a spring constant of k2. Therefore, the equivalent spring constant of this portion is k2.
Step 3: Find the equivalent spring constant of the composite shaft.
Since the two portions of the shaft are in series, the equivalent spring constant of the composite shaft can be found using the formula:
1/keq = 1/k1 + 1/k2
where keq is the equivalent spring constant of the composite shaft.
Simplifying the equation, we get:
keq = (k1 * k2) / (k1 + k2)
Therefore, the equivalent spring constant of the composite shaft is (k1 * k2) / (k1 + k2).
Answer: (D) (k1 * k2) / (k1 + k2)

Test: Strength of Materials Level - 1 - Question 24
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A composite shaft consisting of two stepped portions having spring constants k1 and k2 is held between two rigid supports at the ends. Its equivalent spring constant is
Detailed Solution for Test: Strength of Materials Level - 1 - Question 24


The equivalent spring constant of the composite shaft can be found by considering the individual spring constants of the two stepped portions.
Let's denote the spring constants of the first and second stepped portions as k1 and k2, respectively.
Using the formula for the equivalent spring constant of springs in series:
1/k = 1/k1 + 1/k2
 the answer is option D.

Test: Strength of Materials Level - 1 - Question 25
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According to Euler's column theory, the crippling load for a column length (l) hinged at both ends, is
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