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

Answer:

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:

$\dfrac{r}{t} = \dfrac{7.5 \ in}{0.0625 \ in} = 120$

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:

$\sigma_t = \dfrac{Pd}{4t}$

$\sigma_t = \dfrac{P(15)}{4(0.0625)}$

$\sigma_t =60 P$

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:

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

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

$(\sigma ') =61P$

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

$S_y = \sigma '$

$(57)= 61P$

$P = \dfrac{57}{61}$

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

$P = \dfrac{68}{61}$

P = 1.11 kpsi

Hence, the pressure at rupture condition is 1.11 kpsi