Question 7.4: 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 = 55 kN (Fig. 7-18a). The cylinder has inner radius r = 50 mm and wall thickness t = 4 mm.
Determine the maximum allowable internal pressure p_{allow} based upon an allowable shear stress of 45 MPa 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. 7-18a), we isolate a stress element as shown in Fig. 7-18b. 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. 7-7a and Eq. 7-6) minus the compressive stress produced by the axial force; thus,
\sigma _2=\frac{pr}{2t} ( Eq. 7-6)
\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. 7-7a and Eq. 7-5):
\sigma _y=\frac{pr}{t} (Eq. 7-5) (g)
Note that \sigma_y is algebraically larger than \sigma_x.
Since no shear stresses act on the element (Fig. 7-18), 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(50 mm)}{4 mm} = 12.5 p
\sigma _2=\frac{pr}{2t} -\frac{P}{2\pi rt}=\frac{p(50 mm)}{2(4 mm)} – \frac{55 kN}{2\pi (50 mm)(4 mm)}
= 6.25p – 43.77 MPa
in which \sigma_1, \sigma_2, and p have units of megapascals (MPa).
In-plane shear stresses. The maximum in-plane shear stress (Eq. 6-26) is
\tau _{max}=\frac{\sigma _1-\sigma _2}{2} (Eq. 6-26)
=\frac{1}{2} (12.5p-6.25p+43.77 MPa)=3.125p+21.88 MPa
Since \tau _{max} is limited to 45 MPa, the preceding equation becomes
45 MPa = 3.125p + 21.88 MPa
from which we get
p=\frac{23.12 MPa}{3.125 } = 7.39 MPa or (p_{allow})_1=7.3 MPa
because we round downward.
Out-of-plane shear stresses. The maximum out-of-plane shear stress (see Eqs. 6-28a and 6-28b) is either
(\tau_{max})_{x_1} =\pm \frac{\sigma_2}{2} (\tau_{max})_{y_1} =\pm \frac{\sigma_1}{2} Eqs. 6-28a, b
\tau _{max }=\frac{\sigma _2}{2} or \tau _{max}=\frac{\sigma _1}{2}
From the first of these two equations we get
45 MPa = 3.125p =- 21.88 MPa or (p_{allow})_2= 21.4 MPa
From the second equation we get
45 MPa = 6.25p or (p_{allow})_3= 7.2 MPa
Allowable internal pressure. Comparing the three calculated values for the allowable pressure, we see that (p_{allow})3 governs, and therefore the allowable internal pressure is
p_{allow} = 7.2 MPa
At this pressure the principal stresses are \sigma_1 = 90 MPa and \sigma_2 = 1.23 MPa. 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. 6-28a, b, and c).
(\tau_{max})_{z_1} =\pm \frac{\sigma_1-\sigma_2}{2} (Eq. 6-28c)
Note: In this example, we determined the allowable pressure in the vessel assuming that the axial load was equal to 55 kN. 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.)
