Page 192 - Marks Calculation for Machine Design

P1: Shibu
January 4, 2005
Brown˙C04
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174
U.S. Customary 14:25 STRENGTH OF MACHINES SI/Metric
Example 7. Determine the maximum biaxial Example 7. Determine the maximum biaxial
stresses (σ xx ) and (σ yy ) due to a combination of stresses (σ xx ) and (σ yy ) due to a combination of
axial and pressure loads like those for the piping axial and pressure loads like those for the piping
installation in Fig. 4.20, where installation in Fig. 4.20, where
L o = 12 ft (1/16 of an inch too short) L o = 4m
L installed = 12.00521 ft L installed = 1.0016 m
2
6
9
2
E = 30 × 10 lb/in (steel) E = 207 × 10 N/m (steel)
p i = 200 psi p i = 1.4 MPa = 1,400,000 N/m 2
r m = 1.5 in r m = 4cm = 0.04 m
t = 0.3 in t = 0.8 cm = 0.008 m
solution solution
Step 1. Calculate the axial strain (ε axial ) due Step 1. Calculate the axial strain (ε axial ) due
to improper installation using Eq. (4.14) as to improper installation using Eq. (4.14) as
L L installed − L o L L installed − L o
ε axial = = ε axial = =
L L o L L o
(12.00521 ft) − (12 ft) (4.0016 m) − (4m)
= =
(12 ft) (4m)
0.00521 ft 0.0016 m
= = 0.000434 = = 0.0004
12 ft 4m
Step 2. Calculate the axial stress (σ axial ) Step 2. Calculate the axial stress (σ axial ) due
because of improper installation, again using to improper installation, again using Eq. (4.14)
Eq. (4.14) as as
σ axial = Eε axial σ axial = Eε axial
6
2
2
9
= (30 × 10 lb/in )(0.000434) = (207 × 10 N/m )(0.0004)
2
2
= 13,021 lb/in = 13.0 kpsi = 82,800,000 N/m = 82.8MPa
Step 3. Calculate the axial stress (σ axial ) due Step 3. Calculate the axial stress (σ axial ) due
to the internal pressure from Eq. (4.15) as to the internal pressure from Eq. (4.15) as
p i r m p i r m
σ axial = σ axial =
2 t 2 t
2
2
(200 lb/in )(1.5in) (1,400,000 N/m )(0.04 m)
= =
2 (0.3in) 2 (0.008 m)
300 lb/in 56,000 N/m
= =
0.6in 0.016 m
2
2
= 500 lb/in = 0.5 kpsi = 3,500,000 N/m = 3.5MPa
Step 4. Calculate the hoop stress (σ hoop ) due Step 4. Calculate the hoop stress (σ hoop ) due
to the internal pressure from Eq. (4.16) as to the internal pressure from Eq. (4.16) as
p i r m p i r m
σ hoop = σ hoop =
t t
2
2
(200 lb/in )(1.5in) (1,400,000 N/m )(0.04 m)
= =
(0.3in) (0.008 m)
300 lb/in 56,000 N/m
= =
0.3in 0.008 m
2
2
= 1,000 lb/in = 1.0 kpsi = 7,000,000 N/m = 7.0MPa