Thermal Expansion in an aluminum plate

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SUMMARY

The discussion focuses on calculating thermal expansion in aluminum and brass components. For the aluminum plate, the diameter of a hole at 199.0 degrees Celsius is determined using the formula for change in area, with a coefficient of thermal expansion (alpha) of 24e-6. The calculations yield a change in diameter of approximately 5.414e-4 meters. Additionally, the forum addresses the need for similar calculations for a brass sleeve and steel shaft to achieve a shrink-fit, emphasizing the application of thermal expansion equations in both scenarios.

PREREQUISITES
  • Understanding of thermal expansion principles
  • Familiarity with the coefficient of thermal expansion (alpha)
  • Basic knowledge of geometry and area calculations
  • Proficiency in using formulas for temperature-related changes in dimensions
NEXT STEPS
  • Study the derivation and application of the thermal expansion formula
  • Learn about the coefficients of thermal expansion for different materials
  • Explore practical applications of shrink-fitting in engineering
  • Investigate the effects of temperature on material properties in mechanical design
USEFUL FOR

Mechanical engineers, materials scientists, and students studying thermodynamics or material properties will benefit from this discussion, particularly those involved in design and manufacturing processes requiring precise dimensional tolerances.

SoccaCrazy24
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Question 1:
A hole in an aluminum plate has a diameter or 1.178 cm at 23.00 degrees Celsius.
(A) What is the diameter of the hole at 199.0 degrees Celsius?
(B) What temperature is the diameter of the hole equal to 1.176 cm?

For (A) I used the equation Change in Area = 2 * alpha * A * Change in Temp.
alpha = 24e-6
A= pi * r^2 = 1.09e-4
Change in Temp = 176
So Change in Area = 2 * 24e-6 * 1.09e-4 * 176 = 9.207e-7
so change in Diameter = sqroot (A/pi) = 5.414e-4m
This doesn't seem right... am i using the right equation? and then i would use the same equation for (B)

Question 2:
At 12.25 degrees Celsius a brass sleeve has an inside diameter of 2.196 cm and a steel shaft has a diameter of 2.199 cm. It is desired to shrink-fit the sleeve over the steel shaft.
(A) To what temperature must the sleeve be heated in order for it to slip over the shaft?
(B) Alternatively, to what temperature must the shaft be cooled before it is able to slip through the sleeve?

If I use the same equation from question 1 shall i receive the same answer?
 
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not to be repetitve or a forum whore? but does anyone have a clue?
 

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