Page 168 - Marks Calculation for Machine Design
P. 168
P1: Rakesh
January 4, 2005
Brown˙C03
Brown.cls
150
U.S. Customary 14:16 STRENGTH OF MACHINES SI/Metric
solution solution
Step 1. Convert the units on the speed of Step 1. Convert the units on the speed of
rotation (ω) from rpm to (rad/s) rotation (ω) from rpm to (rad/s)
rev 2πrad 1 min rev 2π rad 1 min
ω = 1,000 × × ω = 1,000 × ×
min rev 60 s min rev 60 s
(1,000)(2 π) rad (1,000)(2 π) rad
= =
60 s 60 s
= 105 (rad/s) = 105 (rad/s)
Step 2. Using Eq. (3.43) calculate the quantity Step 2. Using Eq. (3.43) calculate the quantity
(σ o ) as (σ o ) as
2 2
2 2
σ o = ρω r o σ o = ρω r o
2 2
slug rad 2 kg rad 2
= 15.2 105 (3ft) = 7,850 105 (1m)
ft 3 s 2 m 3 s 2
" # slug · ft ft " # kg · m m
2
2
= (15.2)(105) (3) 2 = (7,850)(105) (1) 2
s 2 ft 3 s 2 m 3
lb N
= 1,508,220 = 9,975 lb/in 2 = 86,546,250
ft 2 m 2
= 10.0 kpsi = 86.5MPa
Step 3. Calculate the maximum tangential Step 3. Calculate the maximum tangential
stress (σ t ) at the inside radius (r i ), where the stress (σ t ) at the inside radius (r i ), where the
radial (σ r ) is zero, using Eq. (3.44). radial (σ r ) is zero, using Eq. (3.44).
2
2
3 + ν 1 − ν r 3 + ν 1 − ν r
σ t max = σ o 1 + i σ t max = σ o 1 + i
4 3 + ν r o 2 4 3 + ν r 2 o
3 + 0.3 3 + 0.3
= (10 kpsi) = (86.5MPa)
4 4
2 2
1 − 0.3 (1in) 1 − 0.3 (0.025 m)
× 1 + × 1 +
3 + 0.3 (36 in) 2 3 + 0.3 (1m) 2
3.3 3.3
= (10 kpsi) = (86.5MPa)
4 4
0.7 1 0.7 0.000625
× 1 + × 1 +
3.3 1296 3.3 1
= (10 kpsi)(0.825) = (86.5MPa)(0.825)
× (1 + 0.00016) × (1 + 0.00013)
= 8.25 kpsi = 71.4MPa