Page 151 - Marks Calculation for Machine Design

P1: Rakesh
14:16
January 4, 2005
Brown˙C03
Brown.cls
U.S. Customary ADVANCED LOADINGS SI/Metric 133
Step 2. Substitute the pressure (p i ), the inside Step 2. Substitute the pressure (p i ), the inside
radius (r i ), and the outside radius (r o ) into these radius (r i ), and the outside radius (r o ) into these
two expressions to obtain two expressions to obtain
2 2
p i r i 2 r o p i r 2 i r o
σ t = 1 + σ t = 1 +
2
2
r − r i 2 r i r − r 2 i r i
o
o
2
2
(450 lb/in )(2in) 2 (3,150,000 N/m )(0.05 m) 2
= 2 2 = 2 2
(4in) − (2in) (0.1m) − (0.05 m)
2 2
4in 0.1m

× 1 + × 1 +
2in 0.05 m
1,800 lb 7,875 N
= (1 + 4) = (1 + 4)
12 in 2 0.0075 m 2
2
2
= (150 lb/in )(5) = (1,050,000 N/m )(5)
2
2
= 750 lb/in = 750 psi = 5,250,000 N/m = 5.25 MPa
and and
p i r 2 r o 2 p i r 2 r o 2
σ r = i 1 − σ r = i 1 −
2
2
r − r 2 i r i r − r i 2 r i
o
o
2
2
(450 lb/in )(2in) 2 (3,150,000 N/m )(0.05 m) 2
= =
2
2
(4in) − (2in) 2 (0.1m) − (0.05 m) 2
2 2

4in 0.1m
× 1 − × 1 −
2in 0.05 m
1,800 lb 7,875 N
= (1 − 4) = (1 − 4)
12 in 2 0.0075 m 2
2
2
= (150 lb/in )(−3) = (1,050,000 N/m )(−3)
2
2
=−450 lb/in =−450 psi =−3,150,000 N/m =−3.15 MPa
=−p i =−p i
The maximum radial stress (σ max ) occurs at the inside radius, and is the negative of the
r
internal pressure (p i ). The algebraic steps to show this fact are given in Eq. (3.8).
2
2
2
p i r i 2 r o 2 p i r i 2 r − r 2 o p i r i 2 −(r − r )
o
i
i
σ max = 1 − = =
r 2 2 2 2 2 2 2 2 =−p i
r − r r i r − r r r r − r
o i o i i i o i

Eq.(1.55) with r =r i find common denominator rearrange and cancel terms
(3.8)
Axial Stress. If the pressure (p i ) is confined at the ends, an axial stress (σ a ) is also
developed. The geometry and stresses, including the tangential stress (σ t ), are shown in
Fig. 3.6.