Here goes my efforts to explain (hope isn’t too long)-
A little more info on the “span charts”:
Looking at Navco, p150, the Navco suggested support spacing is based on 2300 psi bending stress. MSS-SP-69, p8, is based on 1,500 psi, although I can’t seem to quickly find it in the text. There are many “company” standard span charts also. Anyway, most tables have notes and qualifications although I’ve seen some “company” charts copy other sources and leave this critical information out. These notes may include items like the following:
*The pipe is assumed to have standard wall thickness with insulation,
*No concentrated loads present such as valves,
*There are no changes of direction in the spans,
*Spans are assumed to run in the horizontal plane,
*The maximum deflection of the span under load is limited to 0.1”, and
*The stress intensification factors of components are not considered.
I am almost certain that SP-69 was generated using allowable stress values from the power piping code. The following is an attempt to show how the span tables might have been generated:
Select a low allowable stress value, e.g., 1,500 psi (Sall(weight)/i), for the combined bending and shear value, such that a sufficient factor of safety is provided, and allowing the standard span chart to be applied to a wide range of piping systems:
Beginning with the generalized primary stress equation,
PD/4t + iM/Z <= kSh
where iM/Z values are code dependent, and
k values are code dependent and occurrence of load (assumed = 1.0 here), and
Sh is code dependent
Using beam formulas for span:
M = 1/2[(wl^2)/8 + (wl^2)/12] is the average of a uniformly loaded simple beam and a uniformly loaded beam fixed at both ends.
This reduces to: M = (wl^2)/10
Substituting this expression of M into the generalized stress equation and rearranging results:
L<=SQRT[(Sh-PD/4t)(10Z/iw)]
So then, an example of a non-critical type system might look like:
Assuming B31.1 and A106 Gr B material
P = 150 psig
T = 350F
Sh = 15,000 psi
Let PD/4t = 3000 psi (approx)
This leaves 12,000 psi. A system design of this type will utilize higher SIFs, say in the range of 6-8, so,
12,000/8 = 1,500 psi
An example of a critical type system might look like:
Assuming B31.1 and A181 material
P = 2600 psig
T = 1060F
Sh = 6,000 psi
Let PD/4t = 3000 psi (approx)
This leaves 3,000 psi. A system design of this type will utilize lower SIFs, say in the range of 1-2, so,
3,000/2 = 1,500 psi
Note that the wall thicknesses will be greater and the fabrication control improved resulting in the lower SIFs.
I’ve left out several steps to save some space, because I think that the most important thing to know when using a standard span table is generally how it was constructed, and what simplifying assumptions were made.
Also related is a calculation (ref. Hick’s) to determine the minimum slope of a pipe to ensure complete draining. On the one hand it is a little conservative since emptying is not considered, but there are other factors that can negate its conservatism. If interest, I can post that relation also. . .