Question 12.4: Compute the magnitude of the maximum longitudinal, hoop, and...

Objective Compute the maximum stresses and specify a material.

Given Pressure = p = 69 MPa; D_{o} = 200 mm; D_{i} = 160 mm

Analysis Use Procedure C from this section.

Results  Step 1. D_{m} = (D_{o} + D_{i} )/2 = (200 + 160)/2 = 180 mm

Step 2. t = (D_{o} + D_{i} )/2 = (200 – 160)/2 = 20 mm

D_{m}/t = 180/20 = 9.0 Because D_{m}/t < 20, cylinder is thick.

Step 3. This step does not apply. Cylinder is thick.

Step 4. Use equations from Table 12–1.

a = D_{i}/2 = 160 mm/2 = 80 mm

b = D_{o}/2 = 200 mm/2 = 100 mm

Hoop stress = \sigma_{1} = \frac{p(b^{2} + a^{2})}{b^{2} – a^{2}} = \frac{69  MPa(100^{2} + 80^{2})mm^{2}}{(100^{2} – 80^{2})mm^{2}} = 314.3   MPa

Longitudinal stress = \sigma_{2} = \frac{p a^{2}}{b^{2} – a^{2}} = \frac{69  MPa(80^{2}  mm)^{2}}{(100^{2} – 80^{2})mm^{2}} = 122.7 MPa

Radial stress = \sigma_{3} = -p = -69 MPa

All three stresses are a maximum at the inner surface of the cylinder.

Step 5. Let the design stress = σ_{d} = s_{y} /4.

Step 6. The maximum stress is the hoop stress, \sigma_{1} = 314.3 MPa. Then the required yield strength for the material is

s_{y} = N( \sigma_{2}) = 4(314.3 MPa) = 1257 MPa

Step 7. From Appendix A–10, we can specify SAE 4140 OQT 700 steel that has a yield strength of 1462 MPa.