Question 12.7: Design a cylinder to be made from aged titanium Ti–6Al–4V to...
Design a cylinder to be made from aged titanium Ti–6Al–4V to carry CNG at 50 MPa. The internal diameter must be 600 mm to provide the necessary volume. The design stress is to be 1/6 of the ultimate strength of the titanium.
Objective Design the cylinder.
Given Pressure = p = 50 MPa; D_{i} = 600 mm
Titanium Ti–6Al–4V; s_{u} = 1170 MPa (Appendix A–11)
Analysis Use Procedure B from this section. = 600 mm
Results Step 1. Let D_{m} = 600 mm
Step 2. Assume thin-walled cylinder.
Step 3. Design stress,
\sigma_{d} = s_{u} /6 = (1170 MPa) /6 = 195 MPa
Step 4. Use Equation (12–20) to compute nominal value for t.
\sigma = \frac{F_{R}}{A_{w}} = \frac{pD_{m}L}{2tL} = \frac{pD_{m}}{2t} (12-20)
t = \frac{pD_{m} }{2\sigma_{d}} = \frac{(50 MPa)(600 mm)}{2(195 MPa)} = 76.9 mm
Step 5. Trial #1: D_{i} = 600 mm; t = 80 mm.; D_{o}= D_{i} + 2t = 760 mm
Step 6. D_{m} = D_{i} + t = 600 + 80 = 680 mm
Step 7. D_{m} /t = 680/80 = 8.5 < 20; thick walled.
Step 8. This step does not apply.
Step 9. Use the equation for σ1 from Table 12–1.
| TABLE 12–1 Stresses in thick-walled cylinders and spheres .^{a} | ||
| Stress at position r | Maximum stress | |
| Thick-walled cylinder | ||
| Hoop (tangential) | \sigma_{1} = \frac{pa^{2}(b^{2}+r^{2})}{r^{2}(b^{2}-a^{2})} | \sigma_{1} = \frac{p(b^{2}+a^{2}}{b^{2}-a^{2} } (at inner surface) |
| Longitudinal | \sigma_{2} = \frac{pa^{2}}{b^{2}-a^{2}} | \sigma_{2} = \frac{pa^{2}}{b^{2}-a^{2}} (uniform throughout wall) |
| Radial | \sigma_{3} = \frac{-pa^{2}(b^{2}-a^{2})}{r^{2}(b^{2}-r^{2})} | \sigma_{3} = -p (at inner surface) |
| Thick-walled sphere | ||
| Tangential | \sigma_{1} = \sigma_{2} = \frac{pa^{3}(b^{3} + 2r^{3})}{2r^{3}(b^{3}-a^{3})} | \sigma_{1} = \sigma_{2} = \frac{p(b^{3} + 2a^{3})}{2(b^{3}-a^{3})} (at inner surface) |
| Radial | \sigma_{3} = \frac{-pa^{3}(b^{3}-r^{3})}{r^{3}(b^{3}-a^{3})} | \sigma_{3} = -p (at inner surface) |
a = \frac{D_{i}}{2} = \frac{600 mm}{2} = 300 mm
b = \frac{D_{o}}{2} = \frac{760 mm}{2} = 380 mm
\sigma_{1} = \frac{p(b^{2}+a^{2})}{b^{2}-a^{2}} = \frac{50 MPa(380^{2} + 300^{2})mm^{2}}{(380^{2} – 300^{2})mm^{2}} = 215 MPa
\sigma_{1} = 215 MPa slightly high. Repeat steps 5 and 9
Step 5. Increase t = 100 mm.; D_{o} = D_{i} + 2t = 800 mm; thick walled.
Step 9. Use the equation for σ_{1} from Table 12-1
a = D_{i}/2 = 600/2 = 300 mm
b = D_{o}/2 = 800/2 = 400 mm
Then σ_{1} = 178.6 MPa. This is less than the design stress. OK.
Comment The wall thickness is quite large, which would result in a heavy cylinder. Consider using a sphere and higher strength material for the vessel. A sphere made from a composite mate-rial may produce a lighter design.
