Question

A spherical pressure vessel is formed of 16-gauge (0.0625-in) cold-drawn AISI 1020 sheet steel. If the vessel has a diameter of 15 in, use the distortion-energy theory to estimate the pressure necessary to initiate yielding. What is the estimated bursting pressure

Answer 1

Step-by-step explanation:

FromTable A - 20; the values of yield strength and tensile strength are derived by using the "Deterministic ASTM Minimum Tensile and yield strength for some Hot-Rolled (HR) and Cold - Drawn (CD) steels for cold drawn AISI 1020 sheet steel:

`Yield \ Strength (S_y) = 57 \ kpsi`

`Tensile \ Strength (S_(ut) ) = 68 \ kpsi`

We calculate the ratio of the radius in relation to the thickness of the spherical vessel by using the formula:

`(r)/(t) = (7.5 \ in)/(0.0625 \ in) = 120`

Since the fraction from the ration is higher than 1°, then the shell can be regarded to be a thin spherical shell.

From here; we estimate the tensile stress induced by using the formula:

`\sigma_t = (Pd)/(4t)`

`\sigma_t = (P(15))/(4(0.0625))`

`\sigma_t =60 P`

Therefore; the tensile stress is equal to the stress-induced in the transverse direction; i.e.

`\sigma_2 = 60 P`

Thus, since
`\sigma _1 = \sigma _2 = 60P;`

Then the radial stress
`\sigma_r = \sigma _3 = -P`

By the application of Von Mises Stress; the resultant stress can be estimated as follows:

`(\sigma ') = \sqrt{(1)/(2) ( \sigma_1 -\sigma_2)^2+ ( \sigma_2 - \sigma_3)^2 + ( \sigma_3-\sigma_1)^2}`

`(\sigma ') = \sqrt{(1)/(2) ( 60P -60P)^2+ ( 60P -(-P))^2 + ( (-P) -(60P))^2}`

`(\sigma ') =61P`

Then: by relating the Von Mises stress at yield condition:

`S_y = \sigma '`

`(57)= 61P`

`P = (57)/(61)`

P = 0.934 kpsi

P = 934 psi

Hence, the pressure at the yield condition is 934 psi

Similarly, relating Von Mises stress at the rupture condition

`(S_(ut) )= \sigma'`

68 = 61 P

`P = (68)/(61)`

P = 1.11 kpsi

Hence, the pressure at rupture condition is 1.11 kpsi