Page 156 - Marks Calculation for Machine Design

P1: Rakesh
January 4, 2005
Brown˙C03
Brown.cls
138
U.S. Customary 14:16 STRENGTH OF MACHINES SI/Metric
Example 7. Given a set of standard fit dimen- Example 7. Given a set of standard fit dimen-
sions, calculate the maximum and minimum sions, calculate the maximum and minimum
radial interferences, (δ max ) and (δ min ), and the radial interferences, (δ max ) and (δ min ), and the
associated interface pressures, (p max ) and associated interface pressures, (p max ) and
(p min ), for a solid shaft and collar assembly, (p min ), for a solid shaft and collar assembly,
with both parts aluminum, where with both parts aluminum, where
Hole: D max = 1.5010 in Hole: D max = 4.0025 cm
D min = 1.5000 in D min = 4.0000 cm
Shaft: d max = 1.5016 in Shaft: d max = 4.0042 cm
d min = 1.5010 in d min = 4.0026 cm
R = 1.5 in R = 4.0 cm = 0.04 m
r o = 3in r o = 8.0 cm = 0.08 m
9
2
2
6
E = 11 × 10 lb/in (aluminum) E = 77 × 10 N/m (aluminum)
solution solution
Step 1. Calculate the maximum radial inter- Step 1. Calculate the maximum radial inter-
ference (δ max ) from Eq. (3.17) as ference (δ max ) from Eq. (3.17) as
1 max min 1 max min
δ max = d shaft − D hole δ max = d shaft − D hole
2 2
1 1
= (1.5016 in − 1.5000 in) = (4.0042 cm − 4.0000 cm)
2 2
1 1
= (0.0016 in) = (0.0042 cm)
2 2
= 0.0008 in = 0.0021 cm = 0.000021 m
Step 2. Calculate the minimum radial interfer- Step 2. Calculate the maximum radial inter-
ence (δ min ) from Eq. (3.18) as ference (δ min ) from Eq. (3.18) as
1 min max 1 min max
δ min = d shaft − D hole δ min = d shaft − D hole
2 2
1 1
= (1.5010 in − 1.5010 in) = (4.0026 cm − 4.0025 cm)
2 2
1 1
= (0.0000 in) = (0.0001 cm)
2 2
= 0in = 0.00005 cm = 0.0000005 m
Step 3. Using the maximum radial interface Step 3. Using the maximum radial interface
(δ max ) found in Step 1, calculate the maximum (δ max ) found in Step 1, calculate the maximum
interface pressure (p max ) from Eq. (3.15) as interface pressure (p max ) from Eq. (3.15) as
2 2

Eδ max R Eδ max R
p max = 1 − p max = 1 −
2R r o 2R r o
9
6
2
2
(11 × 10 lb/in )(0.0008 in) (77 × 10 N/m )(0.000021 m)
= =
2 (1.5in) 2 (0.04 m)

2 2

1.5in 0.04 m
× 1 − × 1 −
3in 0.08 m
8,800 lb/in 1,617,000 N/m
= (1 − 0.25) = (1 − 0.25)
3in 0.08 m
2
2
= (2,933 lb/in )(0.75) = (20,212,500 N/m )(0.75)
2
2
= 2,200 lb/in = 2.2 kpsi = 15,160,000 N/m = 15.2MPa