With knowledge of elasticity, one can follow the following procedure to estimate the maximum height of a mountain on Earth. 1. Understanding Elasticity: The term elasticity has a very clear definition: Elasticity is defined as the capability of a material to deform when a stress is applied and recovRead more
With knowledge of elasticity, one can follow the following procedure to estimate the maximum height of a mountain on Earth.
1. Understanding Elasticity: The term elasticity has a very clear definition: Elasticity is defined as the capability of a material to deform when a stress is applied and recover its original form after the removal of the applied stress. The crust of Earth acts elastically for some stress ranges.
2. Estimating Stress: The maximum height of a mountain is influenced by the stress exerted by the weight of the mountain itself. This stress can be estimated using the formula:
σ = F / A
where σ is the stress, F is the force (weight of the mountain), and A is the area over which the force is distributed.
3. Material Property: The yield strength of the rock material constituting the mountain is to be considered. If the stress caused by the mountain’s weight exceeds the yield strength, the deformation of the rock is permanent; therefore, the mountain’s height cannot increase beyond that limit.
4. Finding Maximum Height: The maximum height can be found using the formula for stress, strain, and Young’s modulus E (a measure of stiffness). The formula can be written as:
σ = E * ε
where ε is the strain. This can be rearranged as:
ε = σ / E
The change in height (strain) can be approximated by:
Δh = ε * h
where h is the original height before deformation.
5. Setting Up the Calculation: Substituting the values for stress based on the mountain’s mass and area, along with the elastic properties of the rock, you can estimate the maximum height before reaching the elastic limit.
6. Comparative Analysis: Compare the calculated maximum height with known mountain heights, such as Mount Everest, to validate the estimation. Erosion, tectonic activity, and geological history also play a role in actual mountain heights.
In conclusion, through the knowledge of elasticity, a mountain’s height can be approximated by simply calculating the amount of stress which the mountain imposes, the type of material the rocks are made up of, and the strain as a result of this stress.
Elastic hysteresis represents the process in which, in loading and unloading cycles, the stress-strain curve of the material under study shows different paths. This can be considered as an indication of energy dissipation within the material due to internal friction or microstructural change. This iRead more
Elastic hysteresis represents the process in which, in loading and unloading cycles, the stress-strain curve of the material under study shows different paths. This can be considered as an indication of energy dissipation within the material due to internal friction or microstructural change. This is to say that when a material is stretched or compressed, it doesn’t revert back to its original state in the same way that it was deformed, which produces a loop-shaped graph when plotting stress versus strain. The area within this loop represents the energy lost as heat or other forms during the cyclic loading and unloading process.
Applications of Elastic Hysteresis
1. Rubber and Elastomers: For rubber materials, elastic hysteresis is very important, especially in applications such as tires, where the energy loss through hysteresis impacts fuel efficiency and wear resistance.
2. Damping Systems: In mechanical engineering, hysteresis is used in damping systems, such as shock absorbers, where energy dissipation helps to control vibrations and improve stability.
3. Seismic Isolation: In civil engineering, hysteretic materials are used in seismic isolation systems to absorb energy from ground motion during earthquakes, thus reducing the impact on structures.
4. Soft Robotics: Elastic hysteresis is utilized in soft robotics to design compliant actuators that can store and release energy, allowing for smoother and more adaptive movements.
5. Biomedical Devices In the design of some biomedical implants and devices, the understanding of hysteresis improves the fatigue life and performance of the material subjected to cyclic loading.
Elastic hysteresis is one of the crucial concepts in material science and engineering, affecting design and functionality across multiple fields in different applications.
Calculate thickness of metallic ropes in cranes. The thickness of metallic ropes used in a crane is given to ensure that heavy loads can be safely picked up without exceeding the elastic limit of the material. The design process involves the following steps: 1. Elastic Limit: It is the maximum stresRead more
Calculate thickness of metallic ropes in cranes.
The thickness of metallic ropes used in a crane is given to ensure that heavy loads can be safely picked up without exceeding the elastic limit of the material. The design process involves the following steps:
1. Elastic Limit: It is the maximum stress that can be applied to a material such that it comes back to its original shape after the removal of load. For stresses beyond the elastic limit, the material can suffer permanent deformation.
2. Computation of the Load: First, the load (W) that the rope needs to raise should be known. This refers not only to the weight but also to additional forces acting on the rope like dynamic loads and safety factors.
3. Factor of Safety: This is a measure of safety where uncertainties in both the load and material properties are accounted for. It is described as:
FOS = (Maximum Load) / (Allowable Load)
The allowable load can be determined from the following relationship:
P_allow = σₑ / FOS
4. Loading in Terms of Rope Thickness: The maximum tensile load in a rope may be expressed using the following expression:
P = A * σₑ
where P is the load, A is the cross-sectional area of the rope, and σₑ is the elastic limit of the material.
5. Determining the Cross-Sectional Area: The cross-sectional area, A, for a cylindrical rope can be found in terms of its diameter d as follows:
A = (π / 4) * d²
6. Combining Equations: By combining the equations, we can express the load in terms of the diameter:
P_allow = (π / 4) * d² * σₑ / FOS
Rearranging gives:
d² = (4 * P_allow * FOS) / (π * σₑ)
Thus, the required diameter (d) can be calculated as:
d = √[(4 * P_allow * FOS) / (π * σₑ)]
Conclusion: With these principles, engineers can determine the appropriate thickness of metallic ropes for cranes, ensuring safety and reliability while lifting heavy loads.
Elastic after effect, which is also called elastic hysteresis or viscoelastic behavior, refers to the time-dependent deformation phenomenon of a material after the removal of a load. A material deforms elastically when subjected to stress and returns to its original shape upon unloading. In viscoelaRead more
Elastic after effect, which is also called elastic hysteresis or viscoelastic behavior, refers to the time-dependent deformation phenomenon of a material after the removal of a load. A material deforms elastically when subjected to stress and returns to its original shape upon unloading. In viscoelastic materials, however, part of the deformation may persist even after the removal of load. This occurs when the original shape is regained over time and hence returns to equilibrium with a delay.
Elastic After Effect Significance
1. Material Selection Material selection for a given application requires the knowledge of elastic after effect as well as specific performance characteristics of an application. Materials with a higher degree of elastic after effect would be suitable for applications such as shock absorbers where energy dissipation is desired.
2. Structural Design: In engineering and construction, accounting for the elastic aftereffect helps ensure that structures can withstand dynamic loads and vibrations. This is important, especially in regions prone to seismic activities, where materials are stressed for a longer period of time.
3. Durability and Fatigue Analysis: The elastic after effect is important for the durability and fatigue life of materials. Engineers can predict how materials will behave under repeated loading and unloading cycles, informing design decisions and maintenance schedules.
4. Biomedical Applications: The elastic after effect in biomedical engineering plays an important role in the design of implants and prosthetics. Materials with properties that resemble viscoelastic properties of biological tissues could better suit applications.
5. Consumer Products: The application of elastic after effect is used to understand and optimize sports equipment, footwear, and furniture materials in order to maximize comfort and performance by better absorption of energy and resistance to impacts.
In summary, the elastic after effect plays a critical role in material science, engineering, and product design, influencing how materials respond to stress and ensuring safety, performance, and longevity in various applications.
Definition of Elastic Potential Energy: Elastic potential energy is the energy stored in an elastic material when it is deformed, such as when it is stretched or compressed. This energy is released when the material returns to its original shape. Derivation of Elastic Potential Energy: 1. Consider aRead more
Definition of Elastic Potential Energy:
Elastic potential energy is the energy stored in an elastic material when it is deformed, such as when it is stretched or compressed. This energy is released when the material returns to its original shape.
Derivation of Elastic Potential Energy:
1. Consider a wire of length L and cross-sectional area A with Young’s modulus Y.
2. The tensile force F stretches the wire by an amount x.
3. Stress in the wire is given by
Stress (σ) = F / A
4. Strain in the wire is given by
Strain (ε) = x / L
5. Young’s modulus is defined as
Y = σ / ε = (F/A) / (x/L) => F = (Y * A * x) / L
6. The work done (W) in stretching the wire is given by the area under the stress-strain curve:
W = ∫(from 0 to x) F dx = ∫(from 0 to x) (Y * A * (x/L)) dx
7. Integrating:
W = (Y * A / L) * ∫(from 0 to x) x dx
= (Y * A / L) * [x²/2] (from 0 to x) = (Y * A / L) * (x²/2)
8. Hence, the elastic potential energy (U) stored in the wire is:
U = (Y * A * x²) / (2L)
Proof of Elastic Energy Density is 1/2 Stress x Strain:
1. The density of elastic energy:
is energy per volume
u = U / (A * L)
= [(Y * A * x²) / (2L)] / (A * L)
= (Y * x²) / (2L²)
2. If one starts by assuming the presence of Young’s modulus:
Y = (F / A) / (x / L)
3. Thus, the stress is:
Stress (σ) = F / A = (Y * x) / L
4. And strain is:
Strain (ε) = x / L
5. Substituting for stress and strain:
u = (1/2) * σ * ε
Hence, the elastic energy density is equal to 1/2 stress x strain.
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fRead more
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fatigue is very important in engineering and in material selection, especially when it comes to components in bridges, vehicles, and machinery that are subject to fluctuating loads.
Poisson's ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as folloRead more
Poisson’s ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as follows:
u = – (transverse strain)/(axial strain) = – (Δd/d)/(ΔL/L)
Where:
– Δd is the change in diameter (transverse deformation),
– d is the original diameter,
– ΔL is the change in length (axial deformation),
– L is the original length.
Important Points
Poisson’s ratio is a dimensionless quantity.
It generally lies between 0 and 0.5 for most materials with values near to 0.5 showing that the material is almost incompressible.
A value of 0 means that there is no transverse deformation when the material is stretched or compressed.
To find the isothermal bulk modulus K of an ideal gas we can use the formula: K = - V (∂P/∂V)_T For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle's Law: PV = nRT Differentiating this equation while keeping the temperature constant giRead more
To find the isothermal bulk modulus K of an ideal gas we can use the formula:
K = – V (∂P/∂V)_T
For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle’s Law:
PV = nRT
Differentiating this equation while keeping the temperature constant gives us:
∂P/∂V = -nRT/V²
Thus the isothermal bulk modulus becomes:
K = -V (-nRT/V²) = nRT/V
Since nRT = PV we can substitute that in the equation too;
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young's modulus: The stress in the wire, σ, can be found by the relation as follows: σ = F/A where F represents the appliedRead more
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young’s modulus:
The stress in the wire, σ, can be found by the relation as follows:
σ = F/A
where F represents the applied force. The strain ε is described by:
ε = x/L
As related by Young’s modulus Y,
Y = σ/ε = (F/A)/(x/L)
From this we can write the force F:
F = (YAx)/L
The work done W when the wire is stretched by an amount x is given by the area under the stress-strain curve which is the integral of force over displacement:
W = ∫ F dx = ∫ (YAx/L) dx
Evaluating the integral we get:
W = (Y A/L) ∫ x dx = (Y A/L) * [x²/2] from 0 to x = (Y A/L) * (x²/2)
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution. Data given are as follows: - Mass of ball 1 is m₁ = m Velocity of ball 1 is u₁ = 2 m/s Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)Read more
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution.
Data given are as follows:
– Mass of ball 1 is m₁ = m
Velocity of ball 1 is u₁ = 2 m/s
Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)
Velocity of ball 2 is u₂ = 0 m/s (ball is at rest)
Coefficient of restitution is e = 0.5
Step 1: Apply the principle of conservation of momentum
The total momentum before the collision is equal to the total momentum after the collision:
m₁ * u₁ + m₂ * u₂ = m₁ * v₁ + m₂ * v₂
Substitute the given values:
m * 2 + 2m * 0 = m * v₁ + 2m * v₂
2m = m * v₁ + 2m * v₂
Divide both sides by m:
2 = v₁ + 2v₂ (Equation 1)
Step 2: Utilization of the coefficient of restitution
Coefficient of restitution is defined by the following equation:
e = (relative speed after collision) / (relative speed before collision)
In this case, it can be derived as
e = (v₂ – v₁) / (u₁ – u₂)
Using the values:
0.5 = (v₂ – v₁) / (2 – 0)
0.5 = (v₂ – v₁) / 2
Multiply both sides by 2
1 = v₂ – v₁ Equation 2
Step 3: Solution of equations
We have two equations
1. v₁ + 2v₂ = 2
2. v₂ – v₁ = 1
From the last equation we may write
v₂ = v₁ + 1
This must be put in the first:
v₁ + 2v₁ + 2 = 2
3v₁ + 2 = 2
3v₁ = 0
v₁ = 0
Inserting the latter into the last of our initial equations:
v₂ = 0 + 1
v₂ = 1
Final velocities after collision:
Velocity of ball 1 (v₁) = 0 m/s
Velocity of ball 2 (v₂) = 1 m/s
Final Answer:
The final velocities after the collision will be 0 m/s, 1 m/s.
How can the knowledge of elasticity be used to estimate the maximum height of a mountain on earth?
With knowledge of elasticity, one can follow the following procedure to estimate the maximum height of a mountain on Earth. 1. Understanding Elasticity: The term elasticity has a very clear definition: Elasticity is defined as the capability of a material to deform when a stress is applied and recovRead more
With knowledge of elasticity, one can follow the following procedure to estimate the maximum height of a mountain on Earth.
1. Understanding Elasticity: The term elasticity has a very clear definition: Elasticity is defined as the capability of a material to deform when a stress is applied and recover its original form after the removal of the applied stress. The crust of Earth acts elastically for some stress ranges.
2. Estimating Stress: The maximum height of a mountain is influenced by the stress exerted by the weight of the mountain itself. This stress can be estimated using the formula:
σ = F / A
where σ is the stress, F is the force (weight of the mountain), and A is the area over which the force is distributed.
3. Material Property: The yield strength of the rock material constituting the mountain is to be considered. If the stress caused by the mountain’s weight exceeds the yield strength, the deformation of the rock is permanent; therefore, the mountain’s height cannot increase beyond that limit.
4. Finding Maximum Height: The maximum height can be found using the formula for stress, strain, and Young’s modulus E (a measure of stiffness). The formula can be written as:
σ = E * ε
where ε is the strain. This can be rearranged as:
ε = σ / E
The change in height (strain) can be approximated by:
Δh = ε * h
where h is the original height before deformation.
5. Setting Up the Calculation: Substituting the values for stress based on the mountain’s mass and area, along with the elastic properties of the rock, you can estimate the maximum height before reaching the elastic limit.
6. Comparative Analysis: Compare the calculated maximum height with known mountain heights, such as Mount Everest, to validate the estimation. Erosion, tectonic activity, and geological history also play a role in actual mountain heights.
In conclusion, through the knowledge of elasticity, a mountain’s height can be approximated by simply calculating the amount of stress which the mountain imposes, the type of material the rocks are made up of, and the strain as a result of this stress.
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Describe elastic hysteresis. Mention its few applications.
Elastic hysteresis represents the process in which, in loading and unloading cycles, the stress-strain curve of the material under study shows different paths. This can be considered as an indication of energy dissipation within the material due to internal friction or microstructural change. This iRead more
Elastic hysteresis represents the process in which, in loading and unloading cycles, the stress-strain curve of the material under study shows different paths. This can be considered as an indication of energy dissipation within the material due to internal friction or microstructural change. This is to say that when a material is stretched or compressed, it doesn’t revert back to its original state in the same way that it was deformed, which produces a loop-shaped graph when plotting stress versus strain. The area within this loop represents the energy lost as heat or other forms during the cyclic loading and unloading process.
Applications of Elastic Hysteresis
1. Rubber and Elastomers: For rubber materials, elastic hysteresis is very important, especially in applications such as tires, where the energy loss through hysteresis impacts fuel efficiency and wear resistance.
2. Damping Systems: In mechanical engineering, hysteresis is used in damping systems, such as shock absorbers, where energy dissipation helps to control vibrations and improve stability.
3. Seismic Isolation: In civil engineering, hysteretic materials are used in seismic isolation systems to absorb energy from ground motion during earthquakes, thus reducing the impact on structures.
4. Soft Robotics: Elastic hysteresis is utilized in soft robotics to design compliant actuators that can store and release energy, allowing for smoother and more adaptive movements.
5. Biomedical Devices In the design of some biomedical implants and devices, the understanding of hysteresis improves the fatigue life and performance of the material subjected to cyclic loading.
Elastic hysteresis is one of the crucial concepts in material science and engineering, affecting design and functionality across multiple fields in different applications.
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The thickness if metallic ropes used in cranes to lift heavy loads is decided from the knowledge of the elastic limit of the material and the factor of safety.
Calculate thickness of metallic ropes in cranes. The thickness of metallic ropes used in a crane is given to ensure that heavy loads can be safely picked up without exceeding the elastic limit of the material. The design process involves the following steps: 1. Elastic Limit: It is the maximum stresRead more
Calculate thickness of metallic ropes in cranes.
The thickness of metallic ropes used in a crane is given to ensure that heavy loads can be safely picked up without exceeding the elastic limit of the material. The design process involves the following steps:
1. Elastic Limit: It is the maximum stress that can be applied to a material such that it comes back to its original shape after the removal of load. For stresses beyond the elastic limit, the material can suffer permanent deformation.
2. Computation of the Load: First, the load (W) that the rope needs to raise should be known. This refers not only to the weight but also to additional forces acting on the rope like dynamic loads and safety factors.
3. Factor of Safety: This is a measure of safety where uncertainties in both the load and material properties are accounted for. It is described as:
FOS = (Maximum Load) / (Allowable Load)
The allowable load can be determined from the following relationship:
P_allow = σₑ / FOS
4. Loading in Terms of Rope Thickness: The maximum tensile load in a rope may be expressed using the following expression:
P = A * σₑ
where P is the load, A is the cross-sectional area of the rope, and σₑ is the elastic limit of the material.
5. Determining the Cross-Sectional Area: The cross-sectional area, A, for a cylindrical rope can be found in terms of its diameter d as follows:
A = (π / 4) * d²
6. Combining Equations: By combining the equations, we can express the load in terms of the diameter:
P_allow = (π / 4) * d² * σₑ / FOS
Rearranging gives:
d² = (4 * P_allow * FOS) / (π * σₑ)
Thus, the required diameter (d) can be calculated as:
d = √[(4 * P_allow * FOS) / (π * σₑ)]
Conclusion: With these principles, engineers can determine the appropriate thickness of metallic ropes for cranes, ensuring safety and reliability while lifting heavy loads.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
What is elastic after effect? What is its importance?
Elastic after effect, which is also called elastic hysteresis or viscoelastic behavior, refers to the time-dependent deformation phenomenon of a material after the removal of a load. A material deforms elastically when subjected to stress and returns to its original shape upon unloading. In viscoelaRead more
Elastic after effect, which is also called elastic hysteresis or viscoelastic behavior, refers to the time-dependent deformation phenomenon of a material after the removal of a load. A material deforms elastically when subjected to stress and returns to its original shape upon unloading. In viscoelastic materials, however, part of the deformation may persist even after the removal of load. This occurs when the original shape is regained over time and hence returns to equilibrium with a delay.
Elastic After Effect Significance
1. Material Selection Material selection for a given application requires the knowledge of elastic after effect as well as specific performance characteristics of an application. Materials with a higher degree of elastic after effect would be suitable for applications such as shock absorbers where energy dissipation is desired.
2. Structural Design: In engineering and construction, accounting for the elastic aftereffect helps ensure that structures can withstand dynamic loads and vibrations. This is important, especially in regions prone to seismic activities, where materials are stressed for a longer period of time.
3. Durability and Fatigue Analysis: The elastic after effect is important for the durability and fatigue life of materials. Engineers can predict how materials will behave under repeated loading and unloading cycles, informing design decisions and maintenance schedules.
4. Biomedical Applications: The elastic after effect in biomedical engineering plays an important role in the design of implants and prosthetics. Materials with properties that resemble viscoelastic properties of biological tissues could better suit applications.
5. Consumer Products: The application of elastic after effect is used to understand and optimize sports equipment, footwear, and furniture materials in order to maximize comfort and performance by better absorption of energy and resistance to impacts.
In summary, the elastic after effect plays a critical role in material science, engineering, and product design, influencing how materials respond to stress and ensuring safety, performance, and longevity in various applications.
Click here for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
What is meant by elastic potential energy? Derive an expression for the elastic potential energy of stretched wire. Prove that its elastic energy density is equal to 1/2 stress x strain.
Definition of Elastic Potential Energy: Elastic potential energy is the energy stored in an elastic material when it is deformed, such as when it is stretched or compressed. This energy is released when the material returns to its original shape. Derivation of Elastic Potential Energy: 1. Consider aRead more
Definition of Elastic Potential Energy:
Elastic potential energy is the energy stored in an elastic material when it is deformed, such as when it is stretched or compressed. This energy is released when the material returns to its original shape.
Derivation of Elastic Potential Energy:
1. Consider a wire of length L and cross-sectional area A with Young’s modulus Y.
2. The tensile force F stretches the wire by an amount x.
3. Stress in the wire is given by
Stress (σ) = F / A
4. Strain in the wire is given by
Strain (ε) = x / L
5. Young’s modulus is defined as
Y = σ / ε = (F/A) / (x/L) => F = (Y * A * x) / L
6. The work done (W) in stretching the wire is given by the area under the stress-strain curve:
W = ∫(from 0 to x) F dx = ∫(from 0 to x) (Y * A * (x/L)) dx
7. Integrating:
W = (Y * A / L) * ∫(from 0 to x) x dx
= (Y * A / L) * [x²/2] (from 0 to x) = (Y * A / L) * (x²/2)
8. Hence, the elastic potential energy (U) stored in the wire is:
U = (Y * A * x²) / (2L)
Proof of Elastic Energy Density is 1/2 Stress x Strain:
1. The density of elastic energy:
is energy per volume
u = U / (A * L)
= [(Y * A * x²) / (2L)] / (A * L)
= (Y * x²) / (2L²)
2. If one starts by assuming the presence of Young’s modulus:
Y = (F / A) / (x / L)
3. Thus, the stress is:
Stress (σ) = F / A = (Y * x) / L
4. And strain is:
Strain (ε) = x / L
5. Substituting for stress and strain:
u = (1/2) * σ * ε
Hence, the elastic energy density is equal to 1/2 stress x strain.
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What is elastic fatigue? What is its importance?
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fRead more
Elastic fatigue is a condition in which a material loses its elastic properties slowly over time as it experiences repeated or cyclic loading and unloading. Continuous cycles of stress cause the material to lose its elastic recovery properties, thus failing to return to its original shape. Elastic fatigue is very important in engineering and in material selection, especially when it comes to components in bridges, vehicles, and machinery that are subject to fluctuating loads.
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Define Poisson’s ratio.
Poisson's ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as folloRead more
Poisson’s ratio (ν) is defined as the ratio of the transverse strain to the axial strain when a material is subjected to uniaxial stress. It denotes how much a material deforms in the lateral direction when it is stretched or compressed along its length. Mathematically it can be represented as follows:
u = – (transverse strain)/(axial strain) = – (Δd/d)/(ΔL/L)
Where:
– Δd is the change in diameter (transverse deformation),
– d is the original diameter,
– ΔL is the change in length (axial deformation),
– L is the original length.
Important Points
Poisson’s ratio is a dimensionless quantity.
It generally lies between 0 and 0.5 for most materials with values near to 0.5 showing that the material is almost incompressible.
A value of 0 means that there is no transverse deformation when the material is stretched or compressed.
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A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is
To find the isothermal bulk modulus K of an ideal gas we can use the formula: K = - V (∂P/∂V)_T For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle's Law: PV = nRT Differentiating this equation while keeping the temperature constant giRead more
To find the isothermal bulk modulus K of an ideal gas we can use the formula:
K = – V (∂P/∂V)_T
For an ideal gas at constant temperature (isothermal) the relation between pressure P and volume V is given by Boyle’s Law:
PV = nRT
Differentiating this equation while keeping the temperature constant gives us:
∂P/∂V = -nRT/V²
Thus the isothermal bulk modulus becomes:
K = -V (-nRT/V²) = nRT/V
Since nRT = PV we can substitute that in the equation too;
K = PV/V = P
Thus the isothermal bulk modulus of the gas is:
K = P
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A wire of length L, and cross-sectional area A is made of a material of Young’s modulus Y. If the wire is stretched by an amount x, the work done is
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young's modulus: The stress in the wire, σ, can be found by the relation as follows: σ = F/A where F represents the appliedRead more
To calculate the work done W by stretching a wire of length L and cross-sectional area A through an amount x, we could use the following relationship between stress strain and Young’s modulus:
The stress in the wire, σ, can be found by the relation as follows:
σ = F/A
where F represents the applied force. The strain ε is described by:
ε = x/L
As related by Young’s modulus Y,
Y = σ/ε = (F/A)/(x/L)
From this we can write the force F:
F = (YAx)/L
The work done W when the wire is stretched by an amount x is given by the area under the stress-strain curve which is the integral of force over displacement:
W = ∫ F dx = ∫ (YAx/L) dx
Evaluating the integral we get:
W = (Y A/L) ∫ x dx = (Y A/L) * [x²/2] from 0 to x = (Y A/L) * (x²/2)
Work done is hence,
W = (Y A x²)/(2 L)
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A ball moving with velocity 2 m/s collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5, then their velocities (in m/s) after collision will be
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution. Data given are as follows: - Mass of ball 1 is m₁ = m Velocity of ball 1 is u₁ = 2 m/s Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)Read more
To solve for the velocities of the balls after collision, we apply the principles of conservation of momentum and the coefficient of restitution.
Data given are as follows:
– Mass of ball 1 is m₁ = m
Velocity of ball 1 is u₁ = 2 m/s
Mass of ball 2 is m₂ = 2m (it is double in mass compared to ball 1)
Velocity of ball 2 is u₂ = 0 m/s (ball is at rest)
Coefficient of restitution is e = 0.5
Step 1: Apply the principle of conservation of momentum
The total momentum before the collision is equal to the total momentum after the collision:
m₁ * u₁ + m₂ * u₂ = m₁ * v₁ + m₂ * v₂
Substitute the given values:
m * 2 + 2m * 0 = m * v₁ + 2m * v₂
2m = m * v₁ + 2m * v₂
Divide both sides by m:
2 = v₁ + 2v₂ (Equation 1)
Step 2: Utilization of the coefficient of restitution
Coefficient of restitution is defined by the following equation:
e = (relative speed after collision) / (relative speed before collision)
In this case, it can be derived as
e = (v₂ – v₁) / (u₁ – u₂)
Using the values:
0.5 = (v₂ – v₁) / (2 – 0)
0.5 = (v₂ – v₁) / 2
Multiply both sides by 2
1 = v₂ – v₁ Equation 2
Step 3: Solution of equations
We have two equations
1. v₁ + 2v₂ = 2
2. v₂ – v₁ = 1
From the last equation we may write
v₂ = v₁ + 1
This must be put in the first:
v₁ + 2v₁ + 2 = 2
3v₁ + 2 = 2
3v₁ = 0
v₁ = 0
Inserting the latter into the last of our initial equations:
v₂ = 0 + 1
v₂ = 1
Final velocities after collision:
Velocity of ball 1 (v₁) = 0 m/s
Velocity of ball 2 (v₂) = 1 m/s
Final Answer:
The final velocities after the collision will be 0 m/s, 1 m/s.
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