Question

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal is 12.3×10^(-6) /°C, its yield strength is 400 MPa, and its modulus of elasticity is 209 GPa. At room temperature (20°C), the outer and inner diameters of the collar = 95 mm and 60 mm, respectively, and the pin diameter = 60.03 mm. The pin must be reduced in size for assembly into the collar by cooling to a sufficiently low temperature that will give a clearance of 0.06 mm when initially inserted. After insertion, the pin will expand to room temperature to provide a secure fitting with the collar.
Determine (a) the temperature to which the pin must be cooled for assembly, and (b) the radial pressure at room temperature after assembly.
(c) What is the safety factor in the resulting assembly?

Answer 1

Answer:

a)  the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the  safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar$D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$$(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$  $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 =$7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

b)

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c)  the safety factor of the resulting assembly is calculated as:

safety factor =  $\frac{Yield \ strength }{walking \ stress}$

safety factor =  $\frac{400}{62.8}$

safety factor = 6.4

Thus, the  safety factor in the resulting assembly = 6.4