Calculating Thermal Stress on Steel Train Rails: Winter to Summer Transition

In summary, steel train rails are laid in 10m long segments end-to-end and on a winter day with a temperature of -10 degrees Celsius. On a summer day with a temperature of 35 degrees Celsius, the thermal stress on the rails is -1.08x10^8. To prevent thermal stress during summer, a minimum gap of 5.4x10^-3 m should be left between the rails when they are originally laid. These calculations were based on the Young's modulus for steel of 20x10^10 and its coefficient of linear expansion of 1.2x10^-5.
  • #1
jsalapide
40
0
Steel train rails are laid in 10m long segments placed end-to-end. The rails are laid on a winter day when the temperature is -10 degrees Celsius.

a. if the rails are originally laid in contact, what is the thermal stress on them on a summer day when their temperature is 35 degrees Celsius?

b. To prevent thermal stress during summer, what minimum gap between rails must be left when they are originally laid?

my answer in (a) is -3.6x10^7

my answer in (b) is 1.8x10^-3

am i correct?
 
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  • #2
Do you have any work to back those answers up?
 
  • #3
sorry it was the wrong one..

i resolved it and:

a. the young's modulus for steel is 20x10^10 and its coefficient of linear expansion is 1.2x10^-5..
i used the formula (F/A)=-(coefficient of linear expansion)(young's modulus)(delta T)
F/A is the thermal stress..
my answer is -1.08x10^8..

b. i just used the formula delta L=(coefficient of linear expansion)(original length)(delta T)

my answer is 5.4x10^-3 m

is that correct?
 

Related to Calculating Thermal Stress on Steel Train Rails: Winter to Summer Transition

1. What is thermal stress?

Thermal stress is the stress or strain that occurs in a material due to changes in temperature. When a material is heated or cooled, it expands or contracts, causing internal forces that can lead to deformation or failure.

2. What factors contribute to thermal stress?

The main factors that contribute to thermal stress are the temperature change, the material's coefficient of thermal expansion, and the material's ability to conduct heat. Other factors that can influence thermal stress include the shape and geometry of the material, as well as external forces that may be applied.

3. How does thermal stress affect materials?

Thermal stress can cause materials to deform, crack, or fail if the stress exceeds the material's strength. It can also lead to changes in the material's physical and mechanical properties, such as its dimensions, strength, and stiffness. In extreme cases, thermal stress can cause catastrophic failure.

4. How can thermal stress be reduced or controlled?

There are several methods for reducing or controlling thermal stress, including selecting materials with low coefficients of thermal expansion, designing structures with sufficient thermal expansion joints, and using insulation to minimize temperature changes. Additionally, thermal stress can be managed through proper maintenance and monitoring of temperature changes.

5. What are some common applications of thermal stress analysis?

Thermal stress analysis is commonly used in engineering and design to predict the effects of temperature changes on materials and structures. It is used in various industries, including aerospace, automotive, and construction, to ensure the safety and reliability of materials and structures under different thermal conditions. It is also used in the development of new materials and products, as well as in troubleshooting and failure analysis.

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