Question 3.9: Thick cylinder with pistons A cylinder with 50 mm bore and 1...
Thick cylinder with pistons
A cylinder with 50 mm bore and 100 mm OD is subjected to an internal pressure of 400 bar. The end loads are supported by pistons which seal without restraint. Determine the distributions of stress across the cylinder wall.
p = 400 bar = 400 × 100 kPa
= 40 × 1000 kPa
= 40 N/mm²
Since there is no axial load on the cylinder, then σ_{z} = 0.
For a thick cylinder,
and
\sigma _{\theta } = A + \frac{B}{r^{2}}At r = 25 mm, σ_{r} = – 40 N/mm²
∴ – 40 = A – \frac{B}{625} (3.54)
At r = 50 mm, σ_{r} = 0
0 = A – \frac{B}{2500} (3.55)Eliminating A from equations (3.54) and (3.55) gives:
40 = B \left(\frac{1}{625} – \frac{1}{2500} \right)= B\left(\frac{4 – 1}{2500} \right)
∴ B = \frac{40 \times 2500}{3}
Substituting for B into equation (3.55) gives:
0 = A – 40 \times \frac{2500}{3 \times 2500}∴ A = \frac{40}{3}
Hence,
\sigma _{\theta } = \frac{40}{3} + \frac{40 \times 2500}{3r^{2}} = \frac{40}{3} \left(1 + \frac{2500}{r^{2}} \right)and
\sigma _{r } = \frac{40}{3} – \frac{40 \times 2500}{3r^{2}} = \frac{40}{3} \left(1 – \frac{2500}{r^{2}} \right)At r = 25 mm,
\sigma _{\theta } = \frac{40}{3} \times 5 N/mm^{2} = 66.7 N/mm^{2}and
\sigma _{r } = \frac{40}{3} \times (-3) N/mm^{2} = – 40 N/mm^{2}At r = 50 mm,
\sigma _{\theta } = \frac{40}{3} \times 2 N/mm^{2} = 26.7 N/mm^{2}and
\sigma _{r } = 0
