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=1 ext { in, } r_{o}=1.5 ext { in }[/latex]

The radial interference is [latex]\delta=\frac{1}{2}(2.002-2.000)=0.001 in[/latex]

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]=\frac{30\left(10^{6} ight) 0.001}{2\left(1^{3} ight)}\left[\frac{\left(1.5^{2}-1^{2} ight)\left(1^{2}-0 ight)}{\left(1.5^{2}-0 ight)} ight] \&=8333 psi \doteq 83.3 ext { kpsi } \end{aligned}[/latex]

The 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}}=-(8333) \frac{1^{2}+0^{2}}{1^{2}-0^{2}}=-8333 psi \doteq-8.33 kpsi[/latex]

[latex]\left.\left(\sigma_{t} ight)_{o} ight|_{r=R}=p \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}=(8333) \frac{1.5^{2}+1^{2}}{1.5^{2}-1^{2}}=21670 psi \doteq 21.7 kpsi[/latex]

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

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