Question 8.5: A thin-walled cylindrical pressure vessel with a circular cr...
A thin-walled cylindrical pressure vessel with a circular cross section is subjected to internal gas pressure p and simultaneously compressed by an axial load P = 12 k (Fig. 8-25a). The cylinder has inner radius r = 2.1 in. and wall thickness t = 0.15 in.
Determine the maximum allowable internal pressure p_{allow} based upon an allowable shear stress of 6500 psi in the wall of the vessel.
The stresses in the wall of the pressure vessel are caused by the combined action of the internal pressure and the axial force. Since both actions produce uniform normal stresses throughout the wall, we can select any point on the surface for investigation. At a typical point, such as point A (Fig. 8-25a), we isolate a stress element as shown in Fig. 8-25b. The x axis is parallel to the longitudinal axis of the pressure vessel and the y axis is circumferential. Note that there are no shear stresses acting on the element.
Principal stresses. The longitudinal stress \sigma_{x} is equal to the tensile stress \sigma_{2} produced by the internal pressure (see Fig. 8-7a and Eq. 8-6) minus the compressive stress produced by the axial force; thus,
\sigma_{2}=\frac{pr}{2t} (8-6)
\sigma_{x}=\frac{pr}{2t}-\frac{P}{A}=\frac{pr}{2t}-\frac{P}{2\pi rt} (f)
in which A = 2πrt is the cross-sectional area of the cylinder. (Note that for convenience we are using the inner radius r in all calculations.)
The circumferential stress \sigma_{y} is equal to the tensile stress \sigma_{1} produced by the internal pressure (Fig. 8-7a and Eq. 8-5):
\sigma_{1}=\frac{pr}{t} (8-5)
\sigma_{y}=\frac{pr}{t} (g)
Note that \sigma_{y} is algebraically larger than \sigma_{x}.
Since no shear stresses act on the element (Fig. 8-25), the normal stresses \sigma_{x} and \sigma_{y} are also the principal stresses:
\sigma_{1}=\sigma_{y}=\frac{pr}{t} \sigma_{2}=\sigma_{x}=\frac{pr}{2t}-\frac{P}{2\pi rt} (h,i)
Now substituting numerical values, we obtain
\sigma_{1}=\frac{pr}{t}=\frac{p(2.1_\ in.)}{0.15_\ in.}=14.0p\sigma_{2}=\frac{pr}{2t}-\frac{P}{2\pi rt}=\frac{p(2.1_\ in.)}{2(0.15_\ in.)}-\frac{12_\ k}{}
= 7.0p – 6063 psi
in which \sigma_{1}, \sigma_{2}, and p have units of pounds per square inch (psi).
In-plane shear stresses. The maximum in-plane shear stress (Eq. 7-26) is
Since \tau_{\max} is limited to 6500 psi, the preceding equation becomes
6500 psi = 3.5p + 3032 psi
from which we get
p=\frac{3468_\ psi}{3.5}=990.9_\ psi or \left(p_{allow}\right)_{1}=990_\ psi
because we round downward.
Out-of-plane shear stresses. The maximum out-of-plane shear stress (see Eqs. 7-28a and 7-28b) is either
(\tau_{\max})_{x_{1}}=\pm \frac{\sigma_{2}}{2} (\tau_{\max})_{y_{1}}=\pm \frac{\sigma_{1}}{2} (7-28a,b)
\tau_{\max}=\frac{\sigma_{2}}{2} or \tau_{\max}=\frac{\sigma_{1}}{2}
From the first of these two equations we get
6500 psi = 3.5p – 3032 psi or \left(p_{allow}\right)_{2}=2720_\ psi
From the second equation we get
6500 psi = 7.0p or \left(p_{allow}\right)_{3}=928_\ psi
Allowable internal pressure. Comparing the three calculated values for the allowable pressure, we see that \left(p_{allow}\right)_{3} governs, and therefore the allowable internal pressure is
p_{allow} = 928 psi
At this pressure the principal stresses are \sigma_{1} = 13,000 psi and \sigma_{2} = 430 psi. These stresses have the same signs, thus confirming that one of the out-of-plane shear stresses must be the largest shear stress (see the discussion following Eqs. 7-28a, b, and c).
(\tau_{\max})_{z_{1}}=\pm \frac{\sigma_{1}-\sigma_{2}}{2} (7-28c)
Note: In this example, we determined the allowable pressure in the vessel assuming that the axial load was equal to 12 k. A more complete analysis would include the possibility that the axial force may not be present. (As it turns out, the allowable pressure does not change if the axial force is removed from this example.)
