Question 8.2: A cylindrical pressure vessel is constructed from a long, na...
A cylindrical pressure vessel is constructed from a long, narrow steel plate by wrapping the plate around a mandrel and then welding along the edges of the plate to make a helical joint (Fig.8-9). The helical weld makes an angle α = 55° with the longitudinal axis. The vessel has inner radius r = 1.8 m and wall thickness t = 20 mm. The material is steel with modulus E = 200 GPa and Poisson’s ratio ν = 0.30. The internal pressure p is 800 kPa.
Calculate the following quantities for the cylindrical part of the vessel: (a) the circumferential and longitudinal stresses \sigma_{1} and \sigma_{2}, respectively; (b) the maximum in-plane and out-of-plane shear stresses; (c) the circumferential and longitudinal strains \varepsilon_{1} and \varepsilon_{2}, respectively; and (d) the normal stress \sigma_{w} and shear stress \tau_{w} acting perpendicular and parallel, respectively, to the welded seam.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
1. Conceptualize: The circular cylindrical tank is a thin-walled pressure vessel. Evaluate membrane stresses at the outer surface of the tank. Evaluate circumferential stresses in the wall using Eq. (8-11) and longitudinal stresses using Eq. (8-12). Computed stresses will be closer to theoretically exact values if you use the inner radius of the shell rather than the mean radius
\sigma_{1}=\frac{p r}{t} (8-11)
\sigma_{2}=\frac{p r}{2 t} (8-12)
2, 3. Categorize, Analyze:
Part (a): Circumferential and longitudinal stresses.
The circumferential and longitudinal stresses \sigma_{1} and \sigma_{2}, respectively, are pictured in Fig. 8-10a, where they are shown acting on a stress element at point A on the wall of the vessel. Calculate the magnitudes of the stresses from Eqs.(8-11) and (8-12):
The stress element at point A is shown again in Fig. 8-10b, where the x axis is in the longitudinal direction of the cylinder and the y axis is in the circumferential direction. Since there is no stress in the z direction (\sigma_{3} = 0), the element is in biaxial stress.
Note that the ratio of the internal pressure (800 kPa) to the smaller in-plane principal stress (36 MPa) is 0.022. Therefore, the assumption that any stresses in the z direction can be disregarded and all elements in the cylindrical shell, even those at the inner surface, are in biaxial stress is justified.
Part (b): Maximum shear stresses.
The largest in-plane shear stress is obtained from Eq.(8-14):
\left(\tau_{\max }\right)_{z}=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{\sigma_{1}}{4}=\frac{p r}{4 t}=18 MPaBecause the normal stress in the z direction is disregarded the largest out-of-plane shear stress is obtained from Eq. (8-15a):
\tau_{\max }=\frac{\sigma_{1}}{2}=\frac{p r}{2 t}=36 MPaThis last stress is the absolute maximum shear stress in the wall of the vessel.
Part (c): Circumferential and longitudinal strains.
Since the largest stresses are well below the yield stress of steel (see Table I-3, Appendix I), assume that Hooke’s law applies to the wall of the vessel. Then obtain the strains in the x and y directions (Fig. 8-10b) from Eqs. (7-40a and b) for biaxial stress:
\varepsilon_{x}=\frac{1}{E}\left(\sigma_{x}-\nu \sigma_{y}\right) \quad \varepsilon_{y}=\frac{1}{E}\left(\sigma_{y}-\nu \sigma_{x}\right) (a,b)
Note that the strain \varepsilon_{x} is the same as the principal strain \varepsilon_{2} in the longitudinal direction and that the strain \varepsilon_{y} is the same as the principal strain \varepsilon_{1} in the circumferential direction. Also, the stress \sigma_{x} is the same as the stress \sigma_{2}, and the stress \sigma_{y} is the same as the stress \sigma_{1} . Therefore, the preceding two equations can be written in the forms:
\varepsilon_{2}=\frac{\sigma_{2}}{E}(1-2 \nu)=\frac{p r}{2 t E}(1-2 \nu) (8-19a)
\varepsilon_{1}=\frac{\sigma_{1}}{2 E}(2-\nu)=\frac{p r}{2 t E}(2-\nu) (8-19b)
Substituting numerical values:
\varepsilon_{2}=\frac{\sigma_{2}}{E}(1-2 \nu)=\frac{(36 MPa )[1-2(0.30)]}{200 GPa }=72 \times 10^{-6}\varepsilon_{1}=\frac{\sigma_{1}}{2 E}(2-\nu)=\frac{(72 MPa )(2-0.30)}{2(200 GPa )}=306 \times 10^{-6}
These are the longitudinal and circumferential strains in the cylinder.
Part (d): Normal and shear stresses acting on the welded seam.
The stress element at point B in the wall of the cylinder (Fig. 8-10a) is oriented so that its sides are parallel and perpendicular to the weld. The angle θ for the element is
\theta=90^{\circ}-\alpha=35^{\circ}as shown in Fig. 8-10c. Either the stress-transformation equations or Mohr’s circle may be used to obtain the normal and shear stresses acting on the side faces of this element.
Stress-transformation equations: The normal stress \sigma_{x1} and the shear stress \tau_{x1y1} acting on the x_{1} face of the element (Fig. 8-10c) are obtained from Eqs.(7-4a and b), which are repeated here
\sigma_{x 1}=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{\sigma_{x}-\sigma_{y}}{2} \cos 2 \theta+\tau_{x y} \sin 2 \theta (8-20a)
\tau_{x 1 y 1}=-\frac{\sigma_{x}-\sigma_{y}}{2} \sin 2 \theta+\tau_{x y} \cos 2 \theta (8-20b)
Substitute \sigma_{x} = \sigma_{2} = pr/2t , \sigma_{y} = \sigma_{1} = pr/t , and \tau_{xy} = 0, to obtain
\sigma_{x 1}=\frac{p r}{4 t}(3-\cos 2 \theta) \quad \tau_{x 1 y 1}=\frac{p r}{4 t} \sin 2 \theta (8-21 a,b)
These equations give the normal and shear stresses acting on an inclined plane oriented at an angle θ with the longitudinal axis of the cylinder.
Substitute pr/4t = 18 MPa and θ = 35° into Eqs. (8-21a and b) to obtain
\sigma_{x 1}=47.8 MPa \tau_{x 1 y 1}=16.9 MPaThese stresses are shown on the stress element of Fig. 8-10c.
To complete the stress element, calculate the normal stress \sigma_{y1} acting on the y_{1} face of the element from the sum of the normal stresses on perpendicular faces [Eq. (7-6)] \sigma_{x 1}+\sigma_{y 1}=\sigma_{x}+\sigma_{y}:
\sigma_{1}+\sigma_{2}=\sigma_{x 1}+\sigma_{y 1} (8-22)
Substitute numerical values to get
\sigma_{y 1}=\sigma_{1}+\sigma_{2}-\sigma_{x 1 }=72 MPa +36 MPa -47.8 MPa =60.2 MPaas shown in Fig. 8-10c.
From the figure, the normal and shear stresses acting perpendicular and parallel, respectively, to the welded seam are
\sigma_{ w }=47.8 MPa \quad \tau_{ w }=16.9 MPa3. Finalize:
Mohr’s circle: The Mohr’s circle construction for the biaxial stress element of Fig. 8-10b is shown in Fig. 8-11. Point A represents the stress \sigma_{2} = 36 MPa on the x face (θ = 0) of the element, and point B represents the stress \sigma_{1} = 72 MPa on the y face (θ = 90°). The center C of the circle is at a stress of 54 MPa, and the radius of the circle is
R=\frac{72 MPa -36 MPa }{2}=18 MPaA counterclockwise angle 2θ = 70° (measured on the circle from point A) locates point D, which corresponds to the stresses on the x_{1} face (θ = 35°) of the element. The coordinates of point D (from the geometry of the circle) are
\sigma_{x 1}=54 MPa -R \cos 70^{\circ}=54 MPa -(18 MPa )\left(\cos 70^{\circ}\right)=47.8 MPa\tau_{x 1 y 1}=R \sin 70^{\circ}=(18 MPa )\left(\sin 70^{\circ}\right)=16.9 MPa
These results are the same as those found earlier from the stress-transformation equations.
Note: When seen in a side view, a helix follows the shape of a sine curve (Fig. 8-12). The pitch of the helix is
p=\pi d \tan \theta (8-23)
where d is the diameter of the circular cylinder and θ is the angle between a normal to the helix and a longitudinal line. The width of the flat plate that wraps into the cylindrical shape is
w=\pi d \sin \theta (8-24)
Thus, if the diameter of the cylinder and the angle θ are given, both the pitch and the plate width are established. For practical reasons, the angle θ is usually in the range from 20° to 35°.
| Table I-3 | |||||
| Mechanical Properties | |||||
| Material | Yield Stress \sigma_{Y} | Ultimate Stress \sigma_{U} | percent Elongation (2-in. gage length) |
||
| ksi | Mpa | ksi | Mpa | ||
| Aluminum alloys 2014-T6 6061-T6 7075-T6 |
5–70 60 40 70 |
35–500 410 270 480 |
15–80 70 45 80 |
100–550 480 310 550 |
1–45 13 17 11 |
| Brass | 10–80 | 70–550 | 30–90 | 200–620 | 4–60 |
| Bronze | 12–100 | 82–690 | 30–120 | 200–830 | 5–60 |
| Cast iron (tension) | 17–42 | 120–290 | 10–70 | 69–480 | 0–1 |
| Cast iron (compression) | 50–200 | 340–1400 | |||
| Concrete (compression) | 1.5-10 | 10-70 | |||
| Copper and copper alloys | 8–110 | 55–760 | 33–120 | 230–830 | 4–50 |
| Glass Plate glass Glass fibers |
5–150 10 1000–3000 |
30–1000 70 7000–20,000 |
0
|
||
| Magnesium alloys | 12–40 | 80–280 | 20–50 | 140–340 | 2–20 |
| Monel (67% Ni, 30% Cu) | 25–160 | 170–1100 | 65–170 | 450–1200 | 2–50 |
| Nickel | 15–90 | 100–620 | 45–110 | 310–760 | 2–50 |
| Plastics
Nylon |
6–12 |
40–80 |
20–100 |
||
| Rock (compression)Granite,
marble, quartz Limestone, sandstone |
8–40 3–30 |
50–280 20–200 |
|||
| Rubber | 0.2–1.0 | 1–7 | 1–3 | 7–20 | 100–800 |
| Steel
High-strength |
50–150 |
340–1000 |
80–180 |
550–1200 |
5–25 |
| Steel, structural
ASTM-A36 ASTM-A572 ASTM-A514 |
30–100
36 50 100 |
200–700
250 340 700 |
50–120
60 70 120 |
340–830
400 500 830 |
10–40
30 20 15 |
| Steel wire | 40–150 | 280–1000 | 80–200 | 550–1400 | 5–40 |
| Titanium alloys | 110–150 | 760–1000 | 130–170 | 900–1200 | 10 |
| Tungsten | 200-600 | 1400-4000 | 0-4 | ||
| Wood (bending)
Douglas fir Oak Southern pine |
5–8 6–9 6–9 |
30–50 40–60 40–60 |
8–12 8–14 8–14 |
50–80 50–100 50–100 |
|
| Wood (compression parallel to grain) Douglas fir Oak Southern pine |
4–8 |
30–50 |
6–10 |
40–70 |
|
