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A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be?
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A rectangular polymer bat of length 40mm fits exactly into a steel mol...
Thermal Expansion and Strain
Thermal expansion is the tendency of a material to change in shape, area, and volume in response to a change in temperature. When a material is heated, its atoms and molecules vibrate more vigorously, causing them to move further apart. This increase in molecular motion leads to an expansion of the material.
Linear Thermal Expansion Coefficient
The linear thermal expansion coefficient (α) is a measure of how much a material expands per unit length when heated by one degree Celsius. It is expressed in units of 1/°C. The formula to calculate the change in length (ΔL) of a material due to thermal expansion is:
ΔL = α * L0 * ΔT
where ΔL is the change in length, α is the linear thermal expansion coefficient, L0 is the initial length, and ΔT is the change in temperature.
Given Data
- Length of the bat (L0) = 40 mm
- Initial temperature (T0) = 20°C
- Final temperature (Tf) = 80°C
- Linear thermal expansion coefficient of the polymer (αp) = 80 * 10^-6/°C
- Linear thermal expansion coefficient of the steel (αs) = 11 * 10^-6/°C
Calculating the Change in Length
Using the formula mentioned earlier, we can calculate the change in length (ΔL) of the bat due to thermal expansion:
ΔL = α * L0 * ΔT
For the polymer bat:
ΔLp = αp * L0 * ΔTp
ΔLp = 80 * 10^-6/°C * 40 mm * (80°C - 20°C)
ΔLp = 80 * 10^-6/°C * 40 mm * 60°C
ΔLp = 0.192 mm
For the steel mold:
ΔLs = αs * L0 * ΔTs
ΔLs = 11 * 10^-6/°C * 40 mm * (80°C - 20°C)
ΔLs = 11 * 10^-6/°C * 40 mm * 60°C
ΔLs = 0.264 mm
Calculating the Strain
Strain is a measure of the deformation of a material in response to an applied stress. It is calculated as the change in length divided by the original length. The formula to calculate strain (ε) is:
ε = ΔL / L0
For the polymer bat:
εp = ΔLp / L0
εp = 0.192 mm / 40 mm
εp = 0.0048 or 0.48%
For the steel mold:
εs = ΔLs / L0
εs = 0.264 mm / 40 mm
εs = 0.0066 or 0.66%
Conclusion
The strain encountered by the polymer sample along its length when heated from 20°C to 80°C is 0.48%. The strain encountered by the steel mold along its length is 0.66%. This means that the polymer sample undergoes a smaller amount of deformation compared
Thermal expansion is the tendency of a material to change in shape, area, and volume in response to a change in temperature. When a material is heated, its atoms and molecules vibrate more vigorously, causing them to move further apart. This increase in molecular motion leads to an expansion of the material.
Linear Thermal Expansion Coefficient
The linear thermal expansion coefficient (α) is a measure of how much a material expands per unit length when heated by one degree Celsius. It is expressed in units of 1/°C. The formula to calculate the change in length (ΔL) of a material due to thermal expansion is:
ΔL = α * L0 * ΔT
where ΔL is the change in length, α is the linear thermal expansion coefficient, L0 is the initial length, and ΔT is the change in temperature.
Given Data
- Length of the bat (L0) = 40 mm
- Initial temperature (T0) = 20°C
- Final temperature (Tf) = 80°C
- Linear thermal expansion coefficient of the polymer (αp) = 80 * 10^-6/°C
- Linear thermal expansion coefficient of the steel (αs) = 11 * 10^-6/°C
Calculating the Change in Length
Using the formula mentioned earlier, we can calculate the change in length (ΔL) of the bat due to thermal expansion:
ΔL = α * L0 * ΔT
For the polymer bat:
ΔLp = αp * L0 * ΔTp
ΔLp = 80 * 10^-6/°C * 40 mm * (80°C - 20°C)
ΔLp = 80 * 10^-6/°C * 40 mm * 60°C
ΔLp = 0.192 mm
For the steel mold:
ΔLs = αs * L0 * ΔTs
ΔLs = 11 * 10^-6/°C * 40 mm * (80°C - 20°C)
ΔLs = 11 * 10^-6/°C * 40 mm * 60°C
ΔLs = 0.264 mm
Calculating the Strain
Strain is a measure of the deformation of a material in response to an applied stress. It is calculated as the change in length divided by the original length. The formula to calculate strain (ε) is:
ε = ΔL / L0
For the polymer bat:
εp = ΔLp / L0
εp = 0.192 mm / 40 mm
εp = 0.0048 or 0.48%
For the steel mold:
εs = ΔLs / L0
εs = 0.264 mm / 40 mm
εs = 0.0066 or 0.66%
Conclusion
The strain encountered by the polymer sample along its length when heated from 20°C to 80°C is 0.48%. The strain encountered by the steel mold along its length is 0.66%. This means that the polymer sample undergoes a smaller amount of deformation compared
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A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be?
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A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be?.
A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be?.
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A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be?, a detailed solution for A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? has been provided alongside types of A rectangular polymer bat of length 40mm fits exactly into a steel mold cavity and the entire assembly was heated from 20to 80c . The linear thermal expansion coefficients of the polymer and steel are 80*10^-6 and 11*10^-6 respectively .The strain encountered bg the polymer sample along the length will be? theory, EduRev gives you an
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