Given an aluminum shell perfectly bonded to a steel core, what is the thermal stress in the aluminum shell when the system is heated from the unstressed state (20βπΆ) to 180βπΆ? Consider only deformations in the axial directions. Give your answer in πππ
πΌπ΄π=23.6Γ10β6πΎβ1
πΈπ΄π=70πΊππ
πΌππ‘πππ=11.7Γ10β6πΎβ1
πΈππ‘πππ=200πΊππ
Cilinder dimensios= 200mm, 50 mm and 20mm
Thermal stress in Al shell (in πππ):
To calculate the thermal stress in the aluminum shell, we need to use the following formula:
π = πΌ * πΈ * Ξπ
Where:
π is the thermal stress,
πΌ is the coefficient of thermal expansion,
πΈ is the Young's modulus,
Ξπ is the change in temperature.
First, let's calculate the change in temperature (Ξπ) by subtracting the initial temperature (20βπΆ) from the final temperature (180βπΆ):
Ξπ = 180 - 20 = 160 degrees Celsius
Next, we can calculate the thermal stress in the aluminum shell. Given the values provided, we have:
πΌπ΄π = 23.6 Γ 10^(-6) πΎ^(-1)
πΈπ΄π = 70 πΊππ
Substituting these values into the formula, we have:
π = πΌπ΄π * πΈπ΄π * Ξπ
π = (23.6 Γ 10^(-6) πΎ^(-1)) * (70 πΊππ) * (160 degrees Celsius)
To convert GPa to MPa, we divide by 1000:
π = (23.6 Γ 10^(-6) πΎ^(-1)) * (70 πΊππ) * (160 degrees Celsius) / 1000
Simplifying the calculation:
π = 2.3472 πππ
Therefore, the thermal stress in the aluminum shell when heated from 20βπΆ to 180βπΆ is 2.3472 MPa.