The table gives data concerning the shrink fit of two cylinders of differing materials and 3–119 dimensional specification in inches. Elastic constants for different materials may be found in Table A–5. Identify the radial interference δ, then find the interference pressure p, and the tangential

Question

The table gives data concerning the shrink fit of two cylinders of differing materials and 3–119 dimensional specification in inches. Elastic constants for different materials may be found in Table A–5. Identify the radial interference δ, then find the interference pressure p, and the tangential normal stress on both sides of the fit surface. If dimensional tolerances are given at fit surfaces, repeat the problem for the highest and lowest stress levels.

Problem Number	Inner Cylinder			Outer Cylinder
Problem Number	Material	[latex]d _{ i }[/latex]	[latex]d _{ 0}[/latex]	Material	[latex]D _{ i }[/latex]	[latex]D _{ 0}[/latex]
3–116	Steel	0	2.002	Steel	2.000	3.00
3–117	Steel	0	2.002	Cast iron	2.000	3.00
3–118	Steel	0	1.002/1.003	Steel	1.001/1.002	2.00
3–119	Aluminum	0	2.003/2.006	Steel	2.000/2.002	3.00

Accepted Answer

From Table A-5, [latex]E_{i}=E_{o}=30 Mpsi , v_{i}=v_{o}=0.292 . \quad r_{i}=0, R=0.5 ext { in, } r_{o}=1 ext { in }[/latex]

The minimum and maximum radial interferences are

[latex]\begin{aligned}\delta_{\min } &=\frac{1}{2}(1.002-1.002)=0.000 in \\delta_{\max } &=\frac{1}{2}(1.003-1.001)=0.001 in\end{aligned}[/latex]

Since the minimum interference is zero, the minimum pressure and tangential stresses are zero

The maximum pressure is obtained from Eq. (3-57).

[latex]\begin{aligned}&p=\frac{E \delta}{2 R^{3}}\left[\frac{\left(r_{o}^{2}-R^{2} ight)\left(R^{2}-r_{i}^{2} ight)}{r_{o}^{2}-r_{i}^{2}} ight] \&p=\frac{30\left(10^{6} ight) 0.001}{2\left(0.5^{3} ight)}\left[\frac{\left(1^{2}-0.5^{2} ight)\left(0.5^{2}-0 ight)}{\left(1^{2}-0 ight)} ight]=22500 psi\end{aligned}[/latex]

The maximum tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively.

[latex]\left.\left(\sigma_{t} ight)_{i} ight|_{r=R}=-p \frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}=-(22500) \frac{0.5^{2}+0^{2}}{0.5^{2}-0^{2}}=-22500 psi[/latex]

[latex]\left.\left(\sigma_{t} ight)_{o} ight|_{r=R}=p \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}=(22500) \frac{1^{2}+0.5^{2}}{1^{2}-0.5^{2}}=37500 psi[/latex]

_________________________________________________________________________________________________________

Eq. (3-57) : [latex]p=\frac{E \delta}{2 R^{3}}\left[\frac{\left(r_{o}^{2}-R^{2} ight)\left(R^{2}-r_{i}^{2} ight)}{r_{o}^{2}-r_{i}^{2}} ight][/latex]

Eq. (3-58) : [latex]\left.\left(\sigma_{t} ight)_{i} ight|_{r=R}=-p \frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}[/latex]

Eq. (3-59) : [latex]\left.\left(\sigma_{t} ight)_{o} ight|_{r=R}=p \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}[/latex]

Question 3.118: The table gives data concerning the shrink fit of two cylind...

Related Answered Questions