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Question 13.EP.2: A bronze bushing is to be installed into a steel sleeve as i...

A bronze bushing is to be installed into a steel sleeve as indicated in Figure 13_6. The bushing has an inside diameter of 2.000 in and a nominal outside diameter of 2.500 in. The steel sleeve has a nominal inside diameter of 2.500 in and an outside diameter of 3.500 in.

1. Specify the limits of size for the outside diameter of the bushing and the inside diameter of the sleeve in order to obtain a heavy drive fit, FN3. Determine the limits of interference that would result.

2. For the maximum interference from 1, compute the pressure that would be developed between the bushing and the sleeve, the .stress in the bushing and the sleeve, and the deformation of the bushing and the sleeve. Use E = 30 × 10^{6} psi for the steel and E = 17 × 10^{6} for the bronze. Use v= 0.27 for both materials.

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For 1, from Table 13_4,

Nominal
size range
(in)		 Class FN1 Class FN2 Class FN3 Class FN4 Class FN5
Limits of interterence Slundand
limits		 Limits of interterence Slundand
limits		 Limits of interterence Slundand
limits		 Limits of interterence Slundand
limits		 Limits of interterence Slundand
limits
Over To Hole Shaft Hole Shaft Hole Shaft Hole Shaft Hole Shaft
0-0.12 0.05
0.5		 +0.25
– 0		 + 0.5
+ 0.3		 0.2
0.85		 +0.4
– 0		 +0.85
+ 0.6		 0.3
0.95		 +0.4
– 0		 +0.95
+0.7		 0.3
1.3		 +0.6
– 0		 + 1.3
+0.9
0.12-0.24 0.1
0.6		 + 0.3
– 0		 + 0.6
+ 0.4		 0.2
1.0		 + 0.5
– 0		 + 1.0
+ 0.7		 0.4
1.2		 +0.5
– 0		 + 1.2
+ 0.9		 0.5
1.7		 +0.7
– 0		 + 1.7
+ 1.2
0.24-0.40 0.1
0.75		 + 0.4
– 0		 +0.75
+ 0.5		 0.4
1.4		 +0.6
– 0		 + 1.4
+ 1.0		 0.6
1.6		 +0.6
– 0		 + 1.6
+ 1.2		 0,5
2,0		 +0.9
– 0		 +2.3
+ 1.6
0.40-0.56 0.1
0.8		 + 0.4
– 0		 + 0.8
+ 0.5		 0.5
1.6		 + 0.7
– 0		 + 1.6
+ 1.2		 0.7
1.8		 +0.7
– 0		 + 1.8
+ 1.4		 0.6
2.3		 + 1.0
– 0		 +2.3
+ 1.6
0.56-0.71 0.2
0.9		 + 0.4
– 0		 + 0.9
+ 0.6		 0.5
1.6		 + 0.7
– 0		 + 1.6
+ 1.2		 0.7
1.8		 +0.7
– 0		 + 1.8
+ 1.4		 0.8
2.5		 + 1.0
– 0		 + 2.5
+ 1.8
0.71-0.95 0.2
1.1		 + 0.5
– 0		 + 1.1
+ 0.7		 0.6
1.9		 +0.8
– 0		 + 1.9
+ 1.4		 0.8
2.1		 +0.8
– 0		 + 2.1
+ 1.6		 1.0
3.0		 + 1.2
– 0		 + 3.0
+ 2.2
0.95-1.19 0.3
1.2		 + 0.5
– 0		 + 1.2
+ 0.8		 0.6
1.9		 +0.8
– 0		 + 1.9
+ 1.4		 0.8
2.1		 +0.8
– 0		 + 2.1
+ 1.6		 1.0
2.3		 +0.8
– 0		 + 2.3
+ 1.8		 1.3
3.3		 + 1.2
– 0		 +3.3
+ 2.5
1.19-1.58 0.3
1.3		 + 0.6
– 0		 + 1.3
+ 0.9		 0.8
2.4		 + 1.0
– 0		 + 2.4
+ 1.8		 1.0
2.6		 + 1.0
– 0		 + 2.6
+ 2.0		 1.5
3.1		 + 1.0
– 0		 + 3.1
+ 2.5		 1.4
4.0		 + 1.6
– 0		 +4.0
+ 3.0
1.58-1.97 0.4
1.4		 + 0.6
– 0		 + 1.4
+ 1.0		 0.8
2.4		 + 1.0
– 0		 + 2.4
+ 1.8		 1.2
2.8		 + 1.0
– 0		 + 2.8
+ 2.2		 1.8
3.4		 + 1.0
– 0		 + 34
+ 2.8		 2.4
5.0		 + 1.6
– 0		 +5.0
+4.0
1.97-2.56 0.6
1.8		 +0.7
– 0		 + 1.8
+ 1.3		 0.8
2.7		 + 1.2
– 0		 + 2.7
+ 2.0		 1.3
3.2		 + 1.2
– 0		 + 3.2
+ 2.5		 2.3
4.2		 + 1.2
– 0		 + 4.2
+ 3.5		 3.2
6.2		 + 1.8
– 0		 +6.2
+ 5.0
2.56-3.15 0.7
1.9		 +0.7
– 0		 + 1.9
+ 1.4		 1.0
2.9		 + 1.2
– 0		 + 2.9
+ 2.2		 1.8
3.7		 + 1.2
– 0		 + 3.7
+ 3.0		 2.8
4.7		 + 1.2
– 0		 + 4.7
+ 4.0		 4.2
7.2		 + 1.8
– 0		 +7.2
+6.0
3.15-3.94 0.9
2.4		 + 0.9
– 0		 + 2.4
+ 1.8		 1.4
3.7		 + 1.4
– 0		 + 3.7
+ 2.8		 2.1
4.4		 + 1.4
– 0		 +4.4
+ 3.5		 3.6
5.9		 + 1.4
– 0		 + 5.9
+ 5.0		 4.8
8.4		 +2.2
– 0		 + 84
+7.0
3.94-4.73 1.1
2.6		 + 0.9
– 0		 + 2.6
+ 2.0		 1.6
3.9		 + 1.4
– 0		 + 3.9
+ 3.0		 2.6
4.9		 + 1.4
– 0		 +4.9
+ 4.0		 4.6
6.9		 + 1.4
– 0		 + 6.9
+ 6.0		 5.8
9.4		 +2.2
– 0		 +94
+ 8.0
4.73-5.52 1.2
2.9		 + 1.0
– 0		 + 2.9
+ 2.2		 1.9
4.5		 + 1.6
– 0		 + 4.5
+ 3.5		 3.4
6.0		 + 1.6
– 0		 +6.0
+ 5.0		 5.4
8.0		 + 1.6
– 0		 + 8.0
+ 7.0		 7.5
11.6		 + 2.5
– 0		 + 11.6
+10.0
5.52-6.30 1.5
3.2		 + 1.0
– 0		 + 3.2
+ 2.5		 2.4
5.0		 + 1.6
– 0		 + 5.0
+ 4.0		 3.4
6.0		 + 1.6
– 0		 + 6.0
+ 5.0		 5.4
8.0		 + 1.6
– 0		 + 8.0
+ 7.0		 9.5
13.6		 + 2.5
– 0		 + 13.6
+12.0

for a part size of 2.50 in at the mating surface, the tolerance limits on the hole in the outer member are  +1.2 and -0 . Applying these limits to the basic size gives the dimension limits for the hole in the steel sleeve: 2.5012 in
2.5000

in For the bronze insert, the tolerance limits are +3.2 and +2.5. Then the size limits for the outside diameter of the bushing are 2.5032 in
2.5025 in

The limits of interference would be 0.0013 to 0.0032 in. For 2, the maximum pressure would be produced by the maximum interference,
0.0032 in. Then, using a = 1.00 in, b= 1.25 in,E_{o}=30 \times 10^{6} \mathrm{psi}, E_{i}=17× 10^{6} psi,and v_{o}=v_{i}=0.27 from Equation (13_3), p=\frac{\delta}{2 b\left[\frac{1}{E_{o}}\left(\frac{c^{2}+b^{2}}{c^{2}-b^{2}}+v_{o}\right)+\frac{1}{E_{i}}\left(\frac{b^{2}+a^{2}}{b^{2}-a^{2}}-v_{i}\right)\right]}

 

p=\frac{0.0032}{(2)(1.25)\left[\frac{1}{30 \times 10^{6}}\left(\frac{1.75^{2}+1.25^{2}}{1.75^{2}-1.25^{2}}+0.27\right)+\frac{1}{17 \times 10^{6}}\left(\frac{1.25^{2}+1.00^{2}}{1.25^{2}-1.00^{2}}-0.27\right)\right]}

p = 3518 psi The tensile stress in the steel sleeve is \sigma_{o}=p\left(\frac{c^{2}+b^{2}}{c^{2}-b^{2}}\right)=3518\left(\frac{1.75^{2}+1.25^{2}}{1.75^{2}-1.25^{2}}\right)=10846 \mathrm{psi} The compressive stress in the bronze bushing is \sigma_{i}=-p\left(\frac{b^{2}+a^{2}}{b^{2}-a^{2}}\right)=-3518\left(\frac{1.25^{2}+1.00^{2}}{1.25^{2}-1.00^{2}}\right)=-16025 \mathrm{psi} The increase in the diameter of the sleeve is \delta_{o}=\frac{2 b p}{E_{o}}\left[\frac{c^{2}+b^{2}}{c^{2}-b^{2}}+v_{o}\right]

 

\delta_{o}=\frac{2(1.25)(3518)}{30 \times 10^{6}}\left[\frac{1.75^{2}+1.25^{2}}{1.75^{2}-1.25^{2}}+0.27\right]=0.00098 \text { in } The decrease in the diameter of the bushing is \delta_{i}=-\frac{2 b p}{E_{i}}\left[\frac{b^{2}+a^{2}}{b^{2}-a^{2}}-v_{1}\right]

 

\delta_{i}=\frac{2(1.25)(3518)}{17 \times 10^{6}}\left[\frac{1.25^{2}+1.00^{2}}{1.25^{2}-1.00^{2}}+0.27\right]=0.00222 \mathrm{in}   Note that the sum of \delta _{o} and \delta _{i} equals 0.0032, the total interference, δ.

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