Thin Cylindrical Shells

The required thickness of a cylindrical shell due to internal pressure is determined from one of two equa- tions listed in Paragraph UG-27. The equation for the required thickness in the circumferential direction, Fig. 2.1 (a), due to internal pressure is given as

t = PR/(SE – 0.6P), when t < 0.5R or P < 0.385SE (2.1)

where

E = Joint Efficiency Factor P = internal pressure R = internal radius

S = allowable stress in the material t = thickness of the cylinder

This equation can be rewritten to calculate the maximum pressure when the thickness is known. It takes the form

P = SEt/(R + 0.6t) (2.2)

It is of interest to note the similarity between Eq. (2.1) and the classical equation for circumferential mem- brane stress in a thin cylinder (Beer, et al, 2001), given by

The difference is in the additional term of 0.6P in the denominator. This term was added by the ASME to take into consideration the nonlinearity in stress that develops in thick cylinders, i.e., when the thickness of a cylinder exceeds 0.1R. This is demonstrated in Fig. 2.2 for circumferential stress calculated by three different methods. The first is from Eq. (2.3), the theoretical equation for thin cylinders; the second is from Eq. (2.1); and the third is from Lamé’s theoretical equation for thick cylinders and is discussed later as Eq. (2.12).

Similarly, the equation for the required thickness in the longitudinal direction, Fig. 2.1(b), due to internal pressure is given as

t = PR/(2SE + 0.4P), with t < 0.5R or P < 1.25SE (2.4)

or in terms of pressure,

P = 2SEt/(R – 0.4 t) (2.5)

Notice again the similarity between Eq. (2.4) and the classical equation for longitudinal stress in a thin cylin- der given by

t = PR/2SE (2.6)

Equations (2.1) and (2.4) are in terms of the inside radii of cylinders. In some instances, the outside radius of a shell is known instead. In this case, the governing equation for circumferential stress is expressed in terms of the outside radius RO. This equation, which is obtained from Eq. (2.1) by substituting (RO– t) for

R, is given in VIII-1, Appendix 1, Article 1-1, as

t = PRO/(SE + 0.4P), with t < 0.5RO or P < 0.385SE (2.7)

P = SEt/(RO– 0.4t) (2.8)

VIII-1 does not give an equation for the thickness in the longitudinal direction in terms of outside radius

RO. Such an expression can be obtained from Eq. (2.4) as

FIG. 2.2

t = PRO/(2SE + 1.4P) (2.9)

or in terms of P,

P = 2SEt/(RO– 1.4t) (2.10)

Equations (2.1) through (2.10) are applicable to solid wall as well as layered wall construction. Layered vessels consist of thin cylinders wrapped around each other to form a thick cylinder, Fig. 2.3. At any given cross section, a–a, the total thickness consists of individual plate material as well as weld seams. The Joint Efficiency Factor for the overall thickness of a layered vessel is calculated from the ratio

E = (Σ Eiti)/t (2.11)

where

E = overall Joint efficiency Factor for the layered cylinder Ei= Joint Efficiency Factor in a given layer

t = overall thickness of a layered cylinder ti = thickness of one layer

The rules in VIII-1 assume that the longitudinal welds in various layers are staggered in such a way that E in Eq. (2.11) is essentially equal to 1.0.

Example 2.1 Problem

A pressure vessel is constructed of SA 516-70 material and has an inside diameter of 8 ft. The internal design pressure is 100 psi at 450°F. The corrosion allowance is 0.125 in., and the joint efficiency is 0.85. What is the required shell thickness if the allowable stress is 20,000 psi?

Solution

Refer to Paragraph UG-27 of VIII-1. The quantity 0.385SE = 6545 psi is greater than the design pressure of 100 psi. Thus, Eq. (2.1) applies. The inside radius in the corroded condition is equal to

R= 48 + 0.125 = 48.125 in.

t = [PR/(SE – 0.6P)] + corrosion

= [100 × (48.125)/(20,000 × 0.85 – 0.6 × 100)] + 0.125 = 0.41 in.

The calculated thickness is less than 0.5R. Thus, Eq. (2.1) is applicable.

A check of Eq. (2.4) for the required thickness in the longitudinal direction will result in a t = 0.27 in., including corrosion allowance. This is about 60% of the thickness obtained in the circumferential direction. Example 2.2

Problem

A pressure vessel with an internal diameter of 120 in. has a shell thickness of 2.0 in. Determine the maxi- mum permissible pressure if the allowable stress is 20 ksi. Assume E = 0.85.

Solution

For the circumferential direction, the maximum pressure is obtained from Eq. (2.2) as

P = 20,000 × 0.85 × 2.0/(60 + 0.6 × 2.0)

= 556 psi

For the longitudinal direction, the maximum pressure is obtained from Eq. (2.5) as

P = 2 × 20,000 × 0.85 × 2.0/(60 – 0.4 × 2.0)

= 1149 psi

Thus, the maximum pressure permissible in the vessel is 556 psi. Example 2.3

Problem

A vertical boiler is constructed of SA 516-70 material and built in accordance with the requirements of VIII- 1. It has an outside diameter of 8 ft and an internal design pressure of 450 psi at 709°F. The corrosion allowance is 0.125 in., and the joint efficiency is 1.0. Calculate the required thickness of the shell if the allow- able stress is 17,500 psi. Also, calculate the maximum allowable additional tensile force in the axial direc- tion that the shell can withstand at the design pressure.

Solution

From Eq. (2.7), the required thickness is

t = 450 × 48/(17,500 × 1.0 + 0.4 × 450) + 0.125

= 1.222 + 0.125 = 1.35 in.

From Eq. (2.10), the maximum allowable axial pressure is

P = 2 × 17,500 × 1.0 × 1.222/(48 – 1.4 × 1.222)

= 924.0 psi

Subtracting from this value the internal pressure of 450 psi results in the additional equivalent pressure P′, that can be applied to the cylinder during operation.

P′ = 924.0 – 450 = 474.0 psi

Total corroded metal area of cylinder = π(R2 O– R2) = π(482_{– 46.778}2₎ = 363.9 in.2

Hence, total allowable force in cylinder during operation is

F = 474.0 × 363.9

= 172,500 lb

Example 2.4 Problem

What is the required thickness of a layered cylinder subjected to an internal pressure of 1400 psi? Let R = 72 in., S = 18 ksi, ti= 0.25 in. The longitudinal seams of the layers are staggered circumferentially so that

any cross section will have only one longitudinal joint with an efficiency of 0.65. Solution

This problem must be solved by trial and error. Let E = 1.0. Then from Eq. (2.1),

t = 1400 × 72/(18000 × 1.0 – 0.6 × 1400)

= 5.87 in.

Try 24 – 1/4 in. layers with a total thickness of 6.0 in. The joint efficiency from Eq. (2.11) for the total cross section is

E = (23 × 1.00 + 1 × 0.65)/24

= 0.985

Using this Joint Efficiency Factor, recalculate the required thickness:

t = 1400 × 72/(18000 × 0.985 – 0.6 × 1400)

= 5.97 in.

Since this thickness is less than the assumed thickness of 6.0 in., the solution is complete. Hence, 24 – 1/4 in. layers are adequate.