## Formal Piping Flexibility Analysis Not Required

Section 319 of ASME B31.3 requires piping systems to have adequate flexibility to prevent thermal expansion or contraction or movements of piping supports and terminals from causing failure of piping or supports from overstress or fatigue, leakage at joints, detrimental stresses or distortion in piping and valves or in connected equipment, resulting from excessive thrusts and moments in the piping. The code holds the designer responsible to the owner for assuring that all the engineering design complies with the requirements of the code.

As per para 319.4 of the code, all piping systems require a formal analysis of adequate flexibility with the exception of the following:

1. Those that are duplicates of operating installations with a successful service record,
2. Those that can be judged as adequate by comparison with previously analyzed systems and
3. Systems of uniform size that have no more than two anchor points, no intermediate restraints, and fall within the limitation of the following empirical equation:

$$\frac{Dy}{(L - U)^2}$$ ≤ K1

where:

• D = outside diameter of pipe, mm
• y = resultant total displacement strains, mm, to be absorbed by the piping system
• L = developed length of piping between anchors, m
• U = anchor distance, straight line between anchors, m
• K1 = 208000 Sa/Ea for SI units (mm/m)2
• Ea = reference modulus of elasticity at 21°C

Example

Using the above equation, determine if a flexibility analysis is required for the following installation. A two-anchor routing of a DN 200 (NPS 8) schedule 40 carbon steel pipe is shown in figure below. The design temperature of the system is 93°C and the installation temperature is 21°C.

(e = 0.8 mm/m at 93°C from Table C-1)

Figure - Two-anchor piping system Solution:

D = 219.1 mm

Sa = 20 ksi

Ea = 28.8 x 103 ksi

K1 = (208000)(20)/(28800) = 144.4

y = $$\sqrt[]{ΔX^2 +ΔY^2}$$

Thermal displacement in X-direction ΔX = 8.0 m (0.8 mm/m) = 6.4 mm

Thermal displacement in Y-direction ΔY = 4.0 m (0.8 mm/m) = 3.2 mm

Resultant Thermal displacement y = $$\sqrt[]{(6.4 mm)^2 +(3.2 mm)^2}$$ = 7.16 mm

Developed piping length between anchors L = 8.0 m + 4.0 m = 12.0 m

Straight length between anchors U = $$\sqrt[]{(8.0 m)^2 +(4.0 m)^2}$$ = 8.944 m

then, $$\frac{(219.1 mm) (7.16 mm)}{(12.0 m - 8.944 m)^2}$$ = 168

Thus $$\frac{Dy}{(L - U)^2}$$ ≥ 144.4

This two-anchor problem does not fall within the limits of the above equation and thus requires further formal flexibility analysis.

If the above piping was DN 150 (NPS 6), the replacing OD = 219.1 by OD = 168.3 gives us

ratio as, $$\frac{(168.3 mm) (7.16 mm)}{(12.0 m - 8.944 m)^2}$$ = 129

Thus $$\frac{Dy}{(L - U)^2}$$ ≤ 144.4

For DN 150 pipe size the two-anchor problem will fall within the limits of the above equation and would not require a formal flexibility analysis to be carried out.

Although this simple equation is useful in determining the need for formal stress analysis, the code provides a a warning regarding its limitations. No general proof can be offered to assure that the formula will yield accurate or consistently conservative results. Users are advised to be cautious in applying it to configurations such as unequal leg U-bends or near-straight saw-tooth configurations, or for large diameter thin wall pipe (stress intensification factors of the order ≥ 5), or in situations where extraneous displacements other than in the direction connecting the anchor points constitute a large proportion of the total displacement. Further, the code provides no assurance that the terminal reactions will be within the allowable limits, even if the piping sysem falls within the limitations of the above equation.