All questions of Springs for Mechanical Engineering Exam

To calculate the maximum force acting on each spring, we can use the principle of conservation of energy. When the wagon comes to rest, the kinetic energy of the wagon is converted into potential energy stored in the compressed springs.

Given:
Speed of the wagon (v) = 1.5 m/s
Mass of the wagon (m) = 100 kg
Compression of the springs (x) = 125 mm = 0.125 m

1. Calculate the initial kinetic energy of the wagon:
Initial kinetic energy (KEi) = 1/2 * m * v^2
= 1/2 * 100 kg * (1.5 m/s)^2
= 112.5 J

2. Calculate the potential energy stored in the compressed springs:
Potential energy (PE) = 1/2 * k * x^2
= 1/2 * k * (0.125 m)^2
= 1/2 * k * 0.015625 m^2

3. Equating the initial kinetic energy to the potential energy stored in the springs:
112.5 J = 1/2 * k * 0.015625 m^2

4. Solve for the spring constant (k):
k = (2 * 112.5 J) / (0.015625 m^2)
= 14400 N/m

5. Calculate the maximum force acting on each spring:
Maximum force (F) = k * x
= 14400 N/m * 0.125 m
= 1800 N

Therefore, the maximum force acting on each spring is 1800 N, which is option 'C'.

Understanding Maximum Shear Stress in Helical Springs
In a close coiled helical spring, the distribution of shear stress is influenced by the geometry and loading conditions. Here’s a detailed explanation of why the maximum shear stress occurs on the innermost fibers.
Shear Stress Distribution
- In a helical spring, the coils are subjected to torsion when a load is applied.
- The shear stress is not uniform across the spring's cross-section. It varies from the outermost to the innermost fibers.
Location of Maximum Shear Stress
- The maximum shear stress occurs where the radius is smallest, which is at the innermost fibers of the spring.
- As the load creates torsion, the outer coils experience some degree of twisting, but the innermost fibers bear the brunt of this torsional stress.
Comparison with Other Fibers
- Outermost Fibers: While they experience shear stress, it is lower due to their larger radius.
- Mean Diameter Fibers: These fibers experience intermediate shear stress but still not at the maximum level.
- End Coils: They may have different stress distributions but do not reach the maximum shear stress level.
Conclusion
- The geometry of the helical spring and the nature of torsional loading means that the innermost fibers endure the highest shear stress.
- This phenomenon is critical for the design and analysis of springs to ensure they can withstand operational loads without failure.
Understanding these principles is essential for mechanical engineers when designing spring systems for various applications.

To solve this problem, we can use the principle of conservation of energy.

Given:
Mass of the body (m) = 1000 kg
Fall height (h) = 8 cm = 0.08 m
Stiffness of the spring (k) = 500 kg/cm = 500,000 N/m

1. Calculate the potential energy:
The potential energy (PE) of the body when it is at a height h is given by:
PE = mgh

Substituting the values:
PE = 1000 kg × 9.8 m/s² × 0.08 m
PE = 784 N·m

2. Calculate the spring potential energy:
When the body hits the spring, some of the potential energy is transferred to the spring as elastic potential energy. The elastic potential energy (PEs) of a spring is given by:
PEs = 0.5 kx²

Where x is the deformation of the spring.

3. Equate the potential energy and spring potential energy:
Since energy is conserved, the potential energy of the body must be equal to the elastic potential energy of the spring:
PE = PEs

Substituting the values:
784 N·m = 0.5 × 500,000 N/m × x²

4. Solve for x:
x² = (784 N·m) / (0.5 × 500,000 N/m)
x² = 0.003136
x = √(0.003136)
x ≈ 0.056 m
x ≈ 5.6 cm

Therefore, the deformation of the spring will be approximately 5.6 cm. However, none of the given options match this result. It seems that there might be an error in the question or the options provided.

Martin is a common given name derived from the Latin name Martinus, which means "warrior" or "dedicated to Mars," the Roman god of war. It is a popular name in many different cultures and has variations in different languages, including Martín in Spanish and Martinus in Dutch.

Famous people named Martin include:

1. Martin Luther King Jr. - American civil rights leader and activist.
2. Martin Scorsese - American film director and producer known for movies such as "Taxi Driver" and "Goodfellas."
3. Martin Freeman - British actor known for his roles in "The Office" and as Bilbo Baggins in "The Hobbit" trilogy.
4. Martin Sheen - American actor known for his roles in "Apocalypse Now" and as President Josiah Bartlet in "The West Wing."
5. Martin Garrix - Dutch DJ and record producer known for his hit songs "Animals" and "Scared to be Lonely."

Overall, Martin is a versatile and widely used name with a long history.

Understanding Belleville Springs in Series
When two Belleville sprigs are arranged in series, their deflection characteristics change significantly compared to a single sprig. Here’s a detailed explanation of how this configuration affects deflection.
Deflection in Series Arrangement
- When two springs are placed in series, the total deflection for a given load is the sum of the individual deflections of each spring.
- If we denote the deflection of a single Belleville sprig under a certain force as 'd', then the total deflection 'D' when two are in series can be expressed as:
D = d1 + d2
- For identical Belleville sprigs, the deflection of each is the same under the same load, hence:
D = d + d = 2d
Force Distribution
- In a series arrangement, the same load (force) is applied to both springs.
- However, the effective spring constant (stiffness) of the system reduces because the springs share the load.
Resulting Deflection
- The effective spring constant (k) for two identical springs in series is halved:
k_eff = k/2
- Consequently, for the same force, the deflection is effectively doubled, meaning that the total deflection is:
D = 2d
Conclusion
- Thus, under the same force, the deflection obtained when two Belleville sprigs are arranged in series is indeed double that of a single sprig.
- Therefore, the correct answer is option 'B': Double deflection.

A) Yes

Bending stress in graduated length leaves is generally more than that in full length leaves. This is because the shorter leaves experience a greater deflection and bending moment compared to the longer leaves, resulting in higher bending stress.

Bending stresses are the main cause of fatigue failure in many engineering applications. When a component is subjected to cyclic loading, such as repeated bending or flexing, the material experiences alternating tensile and compressive stresses. Fatigue failure occurs when the material is unable to withstand these cyclic stresses, leading to crack initiation and propagation.

Here's a detailed explanation of why fatigue failure is primarily due to tensile stresses:

1. Introduction to Fatigue Failure:
- Fatigue failure is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading.
- It is characterized by the initiation and growth of cracks, which eventually lead to catastrophic failure.

2. Fatigue Crack Initiation:
- When a component is subjected to cyclic loading, small cracks can initiate at locations of high stress concentrations, such as notches or surface defects.
- These cracks initially form due to the tensile stresses experienced during the cyclic loading.

3. Formation of Fatigue Crack:
- Once a crack initiates, it begins to grow under the influence of cyclic loading.
- The crack propagates by the repetitive process of crack tip blunting, crack opening, and crack closure.
- During each loading cycle, the crack tip experiences tensile stresses that cause crack growth.

4. Effect of Compressive Stresses:
- While compressive stresses can help to retard crack growth by closing the crack faces, they do not directly cause crack initiation or propagation.
- Compressive stresses can only influence crack growth when they are combined with tensile stresses.
- In bending conditions, the outer fibers of a component experience tensile stresses, while the inner fibers experience compressive stresses.
- The tensile stresses dominate and lead to fatigue crack initiation, propagation, and ultimate failure.

5. Factors Affecting Fatigue Failure:
- Several factors affect the fatigue life of a component, such as stress amplitude, mean stress, surface finish, material properties, and environmental conditions.
- However, the primary cause of fatigue failure is the cyclic tensile stresses induced during bending or flexing.

In conclusion, fatigue failure is primarily due to tensile stresses induced during cyclic loading. While compressive stresses can affect crack growth, they do not directly cause fatigue crack initiation or propagation. Understanding the mechanisms of fatigue failure is crucial for designing and analyzing components to ensure their reliability and durability.

Belleville springs, also known as conical disc springs or Belleville washers, are commonly used in various mechanical applications to provide a predetermined amount of load or tension. These springs are designed to exhibit a specific load-deflection characteristic, which describes the relationship between the applied load and the resulting deflection or compression of the spring.

Linear Load-Deflection Characteristics:
Belleville springs are known for their ability to produce linear load-deflection characteristics. This means that the relationship between the applied load and the resulting deflection is linear, following Hooke's Law. Hooke's Law states that the deformation of an elastic material is directly proportional to the applied load, as long as the material remains within its elastic limit.

Explanation:
Belleville springs are typically made from a disc-shaped piece of material that is formed into a conical shape. When a load is applied to the spring, it compresses and deflects in a linear manner, with the deflection directly proportional to the applied load. This linear behavior is a result of the material's elastic properties, which allow it to deform and return to its original shape when the load is removed.

Non-Linear Load-Deflection Characteristics:
While Belleville springs are primarily known for their linear load-deflection characteristics, it is also possible to obtain non-linear characteristics by altering the design or using specific materials. By modifying the shape or thickness of the spring, or by using materials with non-linear stress-strain curves, the load-deflection relationship can be made non-linear.

Applications of Non-Linear Belleville Springs:
Non-linear Belleville springs find applications in situations where a non-linear load-deflection characteristic is desired. For example, in some engineering applications, it may be necessary to have a spring that provides different levels of resistance at different levels of deflection. Non-linear Belleville springs can be used in such scenarios to achieve the desired load-deflection relationship.

Conclusion:
In summary, Belleville springs are capable of producing both linear and non-linear load-deflection characteristics. While their default behavior tends to be linear, it is possible to design Belleville springs with non-linear characteristics by altering the shape or thickness of the spring, or by using materials with non-linear stress-strain curves. Therefore, the correct answer to the given question is option 'B' - Belleville springs can exhibit both linear and non-linear load-deflection characteristics.

A) Half force is obtained for a given deflection.

To find the maximum force that can act on the spring, we need to calculate the spring constant (k) first.

The spring constant (k) can be calculated using Hooke's Law:

k = (F_max * d) / x

where:
k is the spring constant
F_max is the maximum force that can act on the spring (70 kN = 70,000 N)
d is the distance between the eyes of the spring (1.2 m)
x is the deflection of the spring (unknown)

Since the spring constant is related to the modulus of elasticity (E) and the dimensions of the spring, we can rearrange the equation to solve for x:

k = (E * A) / L

where:
E is the modulus of elasticity (207,000 N/mm^2 = 207,000,000 N/m^2)
A is the cross-sectional area of the spring (width * thickness)
L is the length of the spring (number of leaves * leaf length)

The cross-sectional area (A) can be calculated as follows:

A = width * thickness

Substituting the values, we have:

A = 100 mm * 12 mm = 1200 mm^2 = 0.0012 m^2

The length of the spring (L) can be calculated as follows:

L = (3 extra leaves + 14 graduated leaves) * leaf length

The leaf length is unknown, so let's represent it as "L_leaf".

L = (3 + 14) * L_leaf
L = 17 * L_leaf

Now we can substitute the values into the equation for the spring constant:

k = (E * A) / L
k = (207,000,000 N/m^2 * 0.0012 m^2) / (17 * L_leaf)

We can simplify this equation further by canceling out the units:

k = (207,000,000 * 0.0012) / (17 * L_leaf)
k = 2,484,000 / (17 * L_leaf)
k = 146,117.65 / L_leaf

Now we can substitute the spring constant (k) and the distance between the eyes of the spring (d) into the equation for the maximum force:

k = (F_max * d) / x
146,117.65 / L_leaf = (70,000 N * 1.2 m) / x

To find x, we can rearrange the equation:

x = (70,000 N * 1.2 m) / (146,117.65 / L_leaf)

This gives us the deflection of the spring (x) when the maximum force (F_max) is applied.

Note: The value of L_leaf is not given, so we cannot calculate the exact deflection of the spring without that information.

Explanation:

When a spring is cut into two equal parts, each part will have half the number of coils compared to the original spring. This is because the number of coils in a spring is directly proportional to its length.

Stiffness constant (k):
The stiffness constant of a spring is a measure of its ability to resist deformation under an applied force. It is given by the equation k = (Gd^4)/(8ND^3), where G is the modulus of rigidity, d is the wire diameter, N is the number of coils, and D is the mean coil diameter.

Effect of cutting the spring:
When a spring is cut into two equal parts, the number of coils (N) in each part is reduced by half. However, the wire diameter (d) and mean coil diameter (D) remain the same for both parts.

Stiffness constant of the new spring:
Using the equation for the stiffness constant (k), we can see that cutting the spring into two equal parts will not change the stiffness constant.

- The modulus of rigidity (G) is a material property and remains the same for both parts.
- The wire diameter (d) and mean coil diameter (D) are also the same for both parts.

Therefore, the stiffness constant of each part of the spring will remain the same as the original spring, which is k.

Conclusion:
The statement that the stiffness constant of the new spring will be k/2 is false. The stiffness constant of each part of the spring will be equal to the stiffness constant of the original spring, which is k.

To calculate the bending stress induced in the strip of the helical spring, we can use the formula for bending stress:

Bending stress = (M * y) / I

where:
M = applied moment = 1250 N-mm
y = distance from the neutral axis to the outermost fiber = thickness/2 = 1.5mm / 2 = 0.75mm = 0.75e-3 m
I = moment of inertia of the strip = (1/12) * b * h^3 = (1/12) * 11mm * (1.5mm)^3 = 0.0006234375 m^4

Plugging the values into the formula:

Bending stress = (1250 N-mm * 0.75e-3 m) / 0.0006234375 m^4
Bending stress = 508.8 N/mm

Therefore, the bending stress induced in the strip of the helical spring is 508.8 N/mm.

If indexis <3 then stresses are high due to curvature effect.

Concentric Springs and Non-Proportional Force
Concentric springs can indeed be used to obtain a force that is not proportional to its deflection.

Explanation:
- Concentric Springs: Concentric springs consist of two or more springs nested inside each other. When these springs are compressed or extended, they can provide a combined force that is different from what a single spring would provide.
- Non-Proportional Force: By combining different types of springs with varying spring constants in a concentric arrangement, it is possible to create a system where the force exerted is not directly proportional to the deflection. This allows for more complex force-deflection relationships to be achieved.
- Use in Engineering: This feature of concentric springs can be utilized in various engineering applications where a non-linear force-deflection relationship is desired. For example, in shock absorbers or suspension systems, concentric springs can be employed to provide varying levels of resistance based on the amount of deflection.
- Advantages: By using concentric springs to obtain a non-proportional force, engineers can tailor the behavior of the system to meet specific requirements, such as damping characteristics or load-bearing capabilities.
In conclusion, concentric springs can be effectively used to achieve a force that is not directly proportional to its deflection, offering versatility and customization in engineering design.

Helical Torsion Spring and Stress Concentration Factor:

A helical torsion spring is a type of spring that is designed to resist twisting or torsional forces. It is typically used in applications where rotational motion needs to be controlled or resisted, such as in suspension systems or mechanical assemblies.

Stress Concentration Factor:

The stress concentration factor (Kt) is a dimensionless factor that represents the increase in stress at a particular location due to the presence of a geometric feature or a change in geometry. It is used to quantify the effect of stress concentration caused by notches, holes, grooves, or other stress-raising features.

The stress concentration factor at the outer fiber of a helical torsion spring can be determined using the following formula:

Kt = 1 + (4C - 1) / (4C - 4) * (D / d)^(3/2)

Where:
- Kt is the stress concentration factor
- C is the spring index (C = D / d, where D is the mean coil diameter and d is the wire diameter)

Calculation:

Given:
- Spring index (C) = 5

We can calculate the stress concentration factor using the given formula. Substituting the value of C into the formula, we get:

Kt = 1 + (4 * 5 - 1) / (4 * 5 - 4) * (D / d)^(3/2)
Kt = 1 + (19 / 21) * (D / d)^(3/2)

To find the stress concentration factor at the outer fiber, we need to determine the values of D and d. The outer diameter (Do) of the spring can be calculated using the formula:

Do = D + 2d

Since the outer fiber is located at Do, we can substitute the value of Do into the formula:

Kt = 1 + (19 / 21) * (Do / d)^(3/2)

Substituting the given values:

Given:
- Spring index (C) = 5

Using the formula for the outer diameter:

Do = D + 2d
Do = 5d + 2d
Do = 7d

Substituting Do into the stress concentration factor formula:

Kt = 1 + (19 / 21) * ((7d) / d)^(3/2)
Kt = 1 + (19 / 21) * 7^(3/2)

Calculating the value:

Kt = 1 + (19 / 21) * 7^(3/2)
Kt ≈ 1 + (19 / 21) * 7.937
Kt ≈ 1 + 6.952
Kt ≈ 7.952

Therefore, the stress concentration factor at the outer fiber of the helical torsion spring is approximately 7.952.

Conclusion:

The correct answer is option B) 0.87.

It seems that there might be an error in the options provided. The calculated stress concentration factor is 7.952, not 0.87. Please double-check the options or the calculations.

Understanding Surge in Sprigs
Surge is generally considered an undesirable phenomenon in the context of mechanical systems, particularly in sprigs or springs. Here’s why:
Definition of Surge
- Surge refers to an abrupt increase or fluctuation in pressure or flow, often leading to instability in mechanical systems.
Impact on Performance
- Reduced Efficiency: Surge can cause inefficiencies in the operation of mechanical systems, resulting in increased energy consumption.
- Structural Damage: The sudden changes in pressure can lead to fatigue and eventual failure of the structural components of the sprigs.
Mechanical Implications
- Vibration Issues: Surging can lead to excessive vibrations, which not only hinder performance but can also damage surrounding components.
- Inconsistent Behavior: The erratic nature of surge can lead to unpredictable system behavior, making it difficult to maintain control over the mechanical system.
Conclusion
In summary, surge is an undesirable effect in the context of sprigs. Its presence can lead to a variety of negative consequences, including reduced efficiency, potential for structural damage, and operational instability. Thus, the correct answer to the question regarding the desirability of surge in sprigs is indeed option 'B', indicating that surge is not a favorable condition.