The cylindrical pressure vessel from Example 8-2 (see photo) is now placed on simple supports and is acted on by uniformly distributed load q = 150 kN/m, which includes the weight of the tank and its contents. The 6-m-long tank has an inner radius r = 1.2 m of and a wall thickness of t = 19 mm .

Question

The cylindrical pressure vessel from Example 8-2 (see photo) is now placed on simple supports and is acted on by uniformly distributed load q = 150 kN/m, which includes the weight of the tank and its contents. The 6-m-long tank has an inner radius r = 1.2 m of and a wall thickness of t = 19 mm . The material is steel with a modulus of E = 200 GPa and the internal pressure P = 270 kPa .

In Example 8-2, we investigated the longitudinal and circumferential stresses and strains, as well as the maximum in-plane and out-of-plane shear stresses. Now, we will investigate the effect of distributed load q to find states of stress at element locations A and B (see Fig. 8-25) due to the combined effects of internal pressure and transverse shear and bending moment (shear-force and bending-moment diagrams are given in Figs. 8-25c and d). Element A is on the outer surface of the vessel, just to the right of the left-hand support; element B is located on the bottom surface of the tank at the midspan.

Accepted Answer

The stresses in the wall of the pressure vessel are caused by the combined action of internal pressure and transverse shear and bending.
At point A, we isolate a stress element similar to that shown in Fig. 8-26a. The x axis is parallel to the longitudinal axis of the pressure vessel, and the y axis is circumferential. There are shear stresses acting on element A due to load q (we assume we are a sufficient distance from the support so that any stress concentration effects are negligible). The stresses are computed as

[latex]\begin{aligned}&\sigma_{x}=\sigma_{L}=\frac{p r}{2 t}=\frac{720 \mathrm{~kPa}(1.2 \mathrm{~m})}{2(19 \mathrm{~mm})}=22.7 \mathrm{~MPa} \&\sigma_{y}=\sigma_{r}=\frac{p r}{t}=\frac{720 \mathrm{~kPa}(1.2 \mathrm{~m})}{(19 \mathrm{~mm})}=45.5 \mathrm{~MPa}\end{aligned}[/latex]

where [latex]\sigma_{L}[/latex] is the longitudinal stress and [latex]\sigma_{r}[/latex] is the circumferential (or radial) stress due to internal pressure p. There are no normal stresses due to bend-ing moment because the longitudinal axis of the vessel lies in the neutral plane for bending. Next we compute shear stress [latex] au_{xy}[/latex] using Eq. (5-48) [latex] au_{\max }=\frac{V Q}{I b}=\frac{4 V}{3 A}\left(\frac{r_{2}^{2}+r_{2} r_{1}+r_{1}^{2}}{r_{2}^{2}+r_{1}^{2}} ight)[/latex], where, from the shear diagram, V = 3qL/10 . We have

[latex]\begin{aligned} au_{x y} &=\frac{-4}{3} \frac{V}{A}\left(\frac{r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}}{r_{1}^{2}+r_{2}^{2}} ight) \ au_{x y} &=\frac{-4}{3} \frac{\left[\frac{3}{10}\left(150 \frac{\mathrm{~kN}}{\mathrm{~m}} ight)(6 \mathrm{~m}) ight]}{\pi\left[(1.219\mathrm{~m})^{2}-(1.2 \mathrm{~m})^{2} ight]}\left[\frac{(1.2 \mathrm{~m})^{2}+1.2 \mathrm{~m}(1.219 \mathrm{~m})+(1.219 \mathrm{~m})^{2}}{(1.2 \mathrm{~m})^{2}+(1.219 \mathrm{~m})^{2}} ight] \&=-3.74 \mathrm{~MPa}\end{aligned}[/latex]

Shear stress [latex] au_{xy}[/latex] is negative (downward on the positive face of the element) in accordance with the sign convention established in Section 1.7.
Principal stresses and maximum shear stresses at point A. The principal stresses are obtained from Eq. (7-17), which is repeated here:

[latex]\sigma_{1,2}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2} ight)^{2}+ au_{x y}^{2}}[/latex]

so

[latex]\begin{aligned}&\sigma_{1}=34.1 \mathrm{~MPa}+11.96 \mathrm{~MPa}=46.1 \mathrm{~MPa} \&\sigma_{2}=34.1 \mathrm{~MPa}-11.96 \mathrm{~MPa}=22.1 \mathrm{~MPa}\end{aligned}[/latex]

The principal stresses are shown on an element rotated through [latex] heta_{p}=9.11^{\circ}[/latex] in Fig. 8-26b.
The maximum in-plane shear stress is computed using Eq. (7-28c):

[latex] au_{\max }=\frac{\sigma_{1}-\sigma_{2}}{2}=12 \mathrm{~MPa}[/latex]

but the maximum out-of-plane shear stress controls using Eqs. (7-28b):

[latex] au_{\max }=\frac{\sigma_{1}}{2}=23.1 \mathrm{~MPa}[/latex]

Because the principal stresses have the same signs, we knew in advance that one of the out-of-plane shear stresses would be the largest shear stress (see the discussion following Eqs. (7-28a, b, and c).

At point B, the stress element is located on the bottom surface of the vessel and (as we look up at it from the bottom of the tank) is oriented as shown in Fig. 8-26c. The x axis is parallel to the longitudinal axis of the pres-sure vessel, and the y axis is circumferential. There are no shear stresses act-ing on element B due to load q, because element B is on the bottom free surface, but normal tensile stress is maximum due to bending. The stresses are computed as

[latex]\sigma_{x}=\sigma_{L}+\frac{M r}{I_{z}}[/latex]

where [latex]I_{z}[/latex] for the vessel is

[latex]I_{z}=\frac{\pi}{4}\left[(r+t)^{4}-r^{4} ight]=0.10562 \mathrm{~m}^{4}[/latex]

so

[latex]\begin{aligned}\sigma_{x} &=\frac{p r}{2 t}+\frac{\left(\frac{q L^{2}}{40} ight)(r+t)}{I_{z}} \&=\frac{720 \mathrm{~kPa}(1.2 \mathrm{~m})}{2(19 \mathrm{~mm})}+\frac{\left[150 \frac{\mathrm{~kN}}{\mathrm{~m}} \frac{(6 \mathrm{~m})^{2}}{40} ight](1.219\mathrm{~m})}{0.10562 \mathrm{~m}^{4}} \\sigma_{x} &=22.74 \mathrm{~MPa}+1.558 \mathrm{~MPa}=24.3 \mathrm{~MPa} \\sigma_{y} &=\sigma_{r}=\frac{p r}{t}=\frac{720 \mathrm{~kPa}(1.2 \mathrm{~m})}{(19 \mathrm{~mm})}=45.5 \mathrm{~MPa}\end{aligned}[/latex]

Because there are no shear stresses acting at B, normal stresses [latex]\sigma_{x}[/latex] and [latex]\sigma_{y}[/latex] are the principal normal stresses, (i.e.[latex]\sigma_{x}=\sigma_{2} ext { and } \sigma_{y}=\sigma_{1}[/latex]). The maximum in-plane and out-of-plane shear stresses can be found from Eqs. (7-28a, b and c).
The maximum in-plane shear stress is computed using Eq. (7-28c) as

[latex] au_{\max }=\frac{\sigma_{1}-\sigma_{2}}{2}=10.6 \mathrm{~MPa}[/latex]

but the maximum out-of-plane shear stress controls using Eq. (7-28b) for

[latex] au_{\max }=\frac{\sigma_{1}}{2}=23 \mathrm{~MPa}[/latex]

Question 8.5: The cylindrical pressure vessel from Example 8-2 (see photo)...

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