Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

y
σ hoop
z
x
σ long
P = pπr 2
σ long
FIGURE 14.5a Cylindrical pressure vessel.
FIGURE 14.5b Free-body diagram exposing σ long .
The free-body diagram used to determine the longitudinal
stress (Figure 14.5b) is similar to the free-body diagram of Figure
14.1b, which was used for the sphere, and the results are the
same. Specifically,
pr pd
σ long = =
(14.4)
2t
4t
To determine the stresses acting in the circumferential direction
of the cylindrical pressure vessel, the free-body diagram
shown in Figure 14.5c is considered. This free-body diagram exposes
a longitudinal section of the cylinder wall.
There are two resultant forces P x acting in the x direction,
which are created by pressure acting on the semicircular ends of
the free-body diagram. These forces are equal in magnitude, but
opposite in direction; therefore, they cancel each other out.
In the lateral direction (i.e., the z direction), the resultant
force P z due to the pressure p acting on an internal area 2r∆x is
y σ hoop
z x P x
P x
P z = p2r∆x
σ hoop
FIGURE 14.5c Free-body diagram exposing σ hoop .
∆x
2r
t
t
P
z
= p2r∆x
where ∆x is the length of the segment arbitrarily chosen for the free-body diagram.
The area of the cylinder wall exposed by the longitudinal section (i.e., the exposed z
surfaces) is 2t∆x. The internal pressure in the cylinder is resisted by normal stress that acts
in the circumferential direction on these exposed surfaces. The total resultant force in the z
direction from these circumferential stresses is
Rz = σ hoop(2 t∆x)
The summation of forces in the z direction gives
∑ Fz = Rz − Pz = σ hoop(2 t∆x) − p2r∆ x = 0
From this equilibrium equation, an expression for the circumferential stress in the cylinder
wall can be derived in terms of the inside radius r or the inside diameter d:
pr pd
σ hoop = =
t 2t
(14.5)
In a cylindrical pressure vessel, the hoop stress σ hoop is twice as large as the longitudinal
stress σ long .
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