Linear expansion problem

Question

a wheel is 0.915m in circumference. an iron tire measures

0.912m around its inner face. how much must the temperature of the tire be raised in order that it may just slip onto the wheel?

Answer 1

change to diameter

0.2912535 m

0.2902986 m

difference is 0.0009549 m

linear thermal expansion coef Iron 11.8  x 10^-6/K

0.0009549 = 11.8  x 10^-6/K x T

T = 80.9 degrees C

.

Answer 2

*PI*D=0.915m

D=0.2912535 m

*PI*D=0.912m

D=0.2902986 m

*Length change = Original length x alpha x delta T

alpha

=Fractional expansion in m/m.K x 10^-6 =11.8 x10^-6 /deg K at 20 C

0.0009549=0.2902986*11.8*10^ -6 *delta T

delta T=0.0009549 / 0.2902986*11.8*10^ -6

delta T = 278.76031373750793849470271329158 degree K

delta T = 278.76 degree K

delta T = [278.76 - 273.15] degree C

delta T =5.61  degree C

[Coefficient of Linear Expansion=11.0 For Iron.]

Edit:

*PI*D=0.915m

D=0.2912535 m

*PI*D=0.912m

D=0.2902986 m

*Length change = Original length x alpha x delta T

alpha

=Fractional expansion in m/m.C x 10^-6 =11.8 x10^-6 /deg C at 20 Deg C

0.0009549=0.2902986*11.8*10^ -6 *delta T

delta T=0.0009549 / 0.2902986*11.8*10^ -6

delta T = 278.76031373750793849470271329158 deg C

delta T = 278.76 degree C

[Coefficient of Linear Expansion/deg C=11.8 For Iron.]

delta T =Final temperature - Initial Temperature

delta T  + Initial Temperature=Final temperature

Final temperature=delta T  + Initial Temperature

Final temperature=278.76 + 20 degree C

Final temperature=298.76 degree C

Answer 3

Unrestrained thermal expansion is given by:

delta = alpha * deltaT * length

where delta is the thermal displacement, alpha is the coefficient of thermal expansion, deltaT is the temperature change, and length is the characteristic length of the part (for a wheel it would be the diameter).

Wheel Diameter = 0.29125 m

Tire Diameter = 0.29030 m

delta = Wheel Diameter - Tire Diameter = 0.00095 m

length = 0.29030 m

alpha = 11.486E-6 m/m/K

9.5E-4  m = 11.486E-6 m/m/K * deltaT * 0.29030 m

deltaK = 284.9 K

This is 284.9 K above the temperature at which the wheel & tire circumference is measured.  So, if the temperature was measured at room temperature 21.1 degC  (294.3 K) the required temperature to have a slip fit is 294.3 K + 284.9K = 579.2 K ( 315.4 degC).