Question 3.119: The table gives data concerning the shrink fit of two cylind...

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
Material did _{ i } d0d _{ 0} Material DiD _{ i } D0D _{ 0}
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
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From Table A-5, Ei=10.4 Mpsi,Eo=30 Mpsi, vi=0.333,vo=0.292E_{i}=10.4  Mpsi , E_{o}=30 \text { Mpsi, } v_{i}=0.333, v_{o}=0.292 .

ri=0,R=1 in, ro=1.5 in r_{i}=0, R=1 \text { in, } r_{o}=1.5 \text { in }

The minimum and maximum radial interferences are

δmin=12[2.0032.002]=0.0005 inδmax=12[2.0062.000]=0.003 in\begin{aligned}&\delta_{\min }=\frac{1}{2}[2.003-2.002]=0.0005  in \\&\delta_{\max }=\frac{1}{2}[2.006-2.000]=0.003  in\end{aligned}

Eq. (3-56),

p=δR[1Eo(ro2+R2ro2R2+vo)+1Ei(R2+ri2R2ri2vi)]p=\frac{\delta}{R\left[\frac{1}{E_{o}}\left(\frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}+v_{o}\right)+\frac{1}{E_{i}}\left(\frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}-v_{i}\right)\right]}

 

p=δ1[130(106)(1.52+121.5212+0.292)+110.4(106)(12+0212020.333)]p=\frac{\delta}{1\left[\frac{1}{30\left(10^{6}\right)}\left(\frac{1.5^{2}+1^{2}}{1.5^{2}-1^{2}}+0.292\right)+\frac{1}{10.4\left(10^{6}\right)}\left(\frac{1^{2}+0^{2}}{1^{2}-0^{2}}-0.333\right)\right]}

 

p=6.229(106)δ psi pmin=6.229(106)δmin=6.229(106)(0.0005)=3114.6 psi=3.11 kpsipmax=6.229(106)δmax=6.229(106)(0.003)=18687 psi=18.7 kpsi\begin{aligned}&p=6.229\left(10^{6}\right) \delta  psi  \\&p_{\min }=6.229\left(10^{6}\right) \delta_{\min }=6.229\left(10^{6}\right)(0.0005)=3114.6  psi =3.11  kpsi \\&p_{\max }=6.229\left(10^{6}\right) \delta_{\max }=6.229\left(10^{6}\right)(0.003)=18687  psi =18.7  kpsi\end{aligned}

 

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

(σt)imin=pminR2+ri2R2ri2=(3.11)12+021202=3.11 kpsi\left.\left(\sigma_{t}\right)_{i}\right|_{\min }=-p_{\min } \frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}=-(3.11) \frac{1^{2}+0^{2}}{1^{2}-0^{2}}=-3.11  kpsi

 

(σt)omin=pminro2+R2ro2R2=(3.11)1.52+121.5212=8.09 kpsi\left.\left(\sigma_{t}\right)_{o}\right|_{\min }=p_{\min } \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}=(3.11) \frac{1.5^{2}+1^{2}}{1.5^{2}-1^{2}}=8.09  kpsi

 

Maximum interference:

(σt)imax=pmaxR2+ri2R2ri2=(18.7)12+021202=18.7 kpsi(σt)omax=pmaxro2+R2ro2R2=(18.7)1.52+121.5212=48.6 kpsi\begin{aligned}&\left.\left(\sigma_{t}\right)_{i}\right|_{\max }=-p_{\max } \frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}=-(18.7) \frac{1^{2}+0^{2}}{1^{2}-0^{2}}=-18.7  kpsi \\&\left.\left(\sigma_{t}\right)_{o}\right|_{\max }=p_{\max } \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}=(18.7) \frac{1.5^{2}+1^{2}}{1.5^{2}-1^{2}}=48.6  kpsi\end{aligned}

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Eq. (3-56) : p=δR[1Eo(ro2+R2ro2R2+vo)+1Ei(R2+ri2R2ri2vi)]p=\frac{\delta}{R\left[\frac{1}{E_{o}}\left(\frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}+v_{o}\right)+\frac{1}{E_{i}}\left(\frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}-v_{i}\right)\right]}

Eq. (3-58) : (σt)ir=R=pR2+ri2R2ri2\left.\left(\sigma_{t}\right)_{i}\right|_{r=R}=-p \frac{R^{2}+r_{i}^{2}}{R^{2}-r_{i}^{2}}

Eq. (3-59) : (σt)or=R=pro2+R2ro2R2\left.\left(\sigma_{t}\right)_{o}\right|_{r=R}=p \frac{r_{o}^{2}+R^{2}}{r_{o}^{2}-R^{2}}

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