Nozzle Reinforcement per UG-37: A Worked Example

When you cut a hole in a pressure vessel shell, the metal you removed used to be carrying hoop stress. That metal is gone now, but the pressure is not, and the load it was carrying has to go somewhere. UG-37's area-replacement method asks one straightforward question: is there enough excess metal near the opening to carry what the removed metal used to carry?

The logic is bookkeeping, not magic. You compute the area the opening must replace, called the required area. Then you count up every area available to replace it: excess shell thickness, excess nozzle wall thickness, weld metal, and, if those fall short, a reinforcing pad. All of this counting only happens within defined limits around the opening, because metal far from the hole is not doing any real work reinforcing it. If the available areas add up to at least the required area, the nozzle passes. If they don't, you add a pad, a thicker nozzle, or both, and check again.

Below is the method laid out step by step, followed by a complete numeric example for a 12-inch nozzle in a 60-inch vessel, with every substitution shown so you can follow the arithmetic from start to finish.

What area does the opening have to replace?

$A_{req} = d\,t_r\,F + 2\,t_n\,t_r\,F\,(1 - f_{r1})$

Each symbol in that equation carries a specific meaning, and getting any one of them wrong throws off the whole calculation:

• $d$: the finished opening diameter, measured in the corroded condition. For hillside nozzles, this is the chord length at the mid-surface, not the simple diameter.
• $t_r$: the required shell thickness per UG-27, computed using a joint efficiency of E = 1, regardless of the actual joint efficiency of the shell seam.
• $F$: a correction factor, equal to 1.0 in almost every case, except for integrally reinforced nozzles in specific orientations where a lower value applies.
• $t_n$: the nozzle wall thickness, also in the corroded condition.
• $f_{r1}$: the strength ratio of the nozzle material to the shell material, $S_n/S_v$, capped at 1.0. When the nozzle and shell are the same material, this ratio is exactly 1.0 and the second term of the equation drops out entirely, which is why so many worked examples use matching materials to keep the arithmetic simple.

The first term of the equation is doing most of the work in a typical case: it is essentially the diameter of the hole multiplied by the thickness the shell would need anyway, which is a reasonable way to size how much metal a hole of that size actually removed. The second term only matters when the nozzle material is weaker than the shell material, in which case the required area grows to compensate for the weaker nozzle's reduced ability to help carry the load.

What areas count toward replacement?

Once you know how much area is required, the next question is where you are allowed to find it. The Code recognizes five categories of available area, summarized here:

Area	What it is
$A_1$	Excess thickness in the shell beyond what UG-27 requires
$A_2$	Excess thickness in the nozzle wall beyond what UG-45/UG-27 requires
$A_3$	Nozzle wall projecting inward, if any
$A_{41}, A_{42}, A_{43}$	Weld metal: outward nozzle weld, pad weld, inward weld
$A_5$	Reinforcing pad area

The intuition behind $A_1$ and $A_2$ is straightforward once you see it: if the shell or the nozzle wall is thicker than the bare minimum the pressure calculation demands, that extra thickness is metal doing no work under normal conditions, and the Code lets you count it toward reinforcement. $A_3$ shows up on set-through nozzles where the nozzle projects into the vessel interior. The weld areas, $A_{41}$ through $A_{43}$, credit the actual cross-sectional area of the welds themselves, since weld metal is structural too. $A_5$ only exists if you add a reinforcing pad, and it is often the area that closes the gap when the shell and nozzle alone come up short.

None of this area counts unless it falls inside the UG-40 limits of reinforcement. Parallel to the shell, the limit is the larger of $d$ or $R_n + t_n + t$, measured each side of the nozzle centerline. Normal to the shell, the limit is 2.5 times the smaller of the shell or nozzle thickness, with the pad thickness added in for the pad-side limit. Metal outside those boundaries might be real and might be thick, but it is too far from the opening to be doing meaningful reinforcing work, so the Code simply does not let you claim it.

Worked example, 12-inch nozzle, 60-inch vessel

Given. Vessel 60 in ID, P = 150 psi, shell SA-516-70 (S = 20,000 psi), shell nominal t = 0.500 in, corrosion allowance 0.125 in, E = 1.0. Nozzle NPS 12 Sch 40 (12.75 in OD, 0.406 in wall), same material, set through, full-penetration weld with 0.375-in cover fillet.

Every calculation from here forward uses the corroded dimensions, not the as-ordered nominal dimensions, because that is the condition the vessel has to survive over its service life. It is worth keeping the given values in view as you work through the steps, since it is easy to lose track of which number came from where.

Step 1, corroded dimensions. Shell available thickness $t = 0.500 - 0.125 = 0.375$ in. Nozzle wall $t_n = 0.406 - 0.125 = 0.281$ in. Opening $d = (12.75 - 2 \times 0.406) + 2 \times 0.125 = 12.188$ in.

Notice that corrosion moves the opening diameter in the opposite direction from the wall thicknesses. As the shell and nozzle wall corrode, they get thinner, but the opening itself gets larger, because corrosion eats away at the nozzle bore too. Both effects work against you, which is exactly why the Code insists on evaluating everything corroded.

Step 2, required thicknesses. Shell: $R = 30.125$ in (corroded). $t_r = \dfrac{150 \times 30.125}{20{,}000 \times 1.0 - 0.6 \times 150} = \dfrac{4{,}519}{19{,}910} = 0.227$ in. Nozzle: $R_n = 6.094$ in. $t_{rn} = \dfrac{150 \times 6.094}{19{,}910} = 0.046$ in.

These are the same UG-27 thickness equations used to set the shell's own pressure rating, just applied at the shell radius for the shell number and at the nozzle radius for the nozzle number. The nozzle's required thickness comes out much smaller than the shell's, simply because the nozzle has a much smaller radius, and radius drives required thickness directly.

Step 3, required area. Same material so $f_{r1} = 1.0$, which drops the second term of the required area equation entirely: $A_{req} = 12.188 \times 0.227 \times 1.0 = 2.767 \text{ in}^2$

That number, 2.767 square inches, is the target. Everything from here forward is about accumulating available area until it meets or beats that target.

Step 4, available areas. $A_1 = d\,(E_1 t - F t_r) = 12.188 \times (0.375 - 0.227) = 1.804\ \text{in}^2$ $A_2 = \min\big[5t(t_n - t_{rn}),\ 5t_n(t_n - t_{rn})\big] = \min[0.441,\ 0.330] = 0.330\ \text{in}^2$ $A_3 = 0$ (no inward projection) and $A_{41} = (0.375)^2 = 0.141\ \text{in}^2$

$A_1$ is doing the heavy lifting here, drawn from the extra 0.148 in of shell thickness beyond what UG-27 strictly requires (0.375 in available against 0.227 in required), multiplied out over the opening diameter. $A_2$ contributes a smaller amount from the nozzle wall's own excess thickness, capped by the smaller of two limiting expressions. $A_3$ is zero because this is a set-through nozzle with no inward projection to credit. $A_{41}$ picks up the cross-sectional area of the outward nozzle weld, based on the 0.375-in cover fillet leg size given at the start.

Step 5, compare. $1.804 + 0.330 + 0.141 = 2.275 < 2.767$. Short by 0.492 in². Unreinforced, this nozzle fails.

That result is not unusual, and it is not a sign anything went wrong upstream. A 12-inch opening in a 0.500-in shell is a large hole relative to the shell thickness, and it is common for shell and nozzle metal alone to fall short on openings this size. The next step exists for exactly this situation.

Step 6, add a pad. Try 16 in OD by 0.375 in pad, same material ($f_{r4} = 1.0$): $A_5 = (D_p - d - 2t_n)\,t_e = (16 - 12.188 - 0.562) \times 0.375 = 1.219\ \text{in}^2$ Total available $= 2.275 + 1.219 = 3.494 \ge 2.767$. Passes with margin 0.727 in². The pad must also sit inside the UG-40 parallel limit of $2d = 24.4$ in, and a 16-in pad comfortably does.

With the pad added, the nozzle clears the required area with room to spare. The margin, 0.727 in², is not wasted safety factor stacked on top of the Code's built-in margins. It simply reflects that pad sizes come in practical increments, much like plate thickness does, so an exact match to the required area is rarely achievable or even desirable.

Step 7, don't stop here. Passing the area-replacement check is necessary, but it is not the end of the nozzle evaluation. UG-45 still checks the nozzle neck thickness independently, using its own minimum-thickness rules that have nothing to do with area replacement. The weld sizes need their own UW-16 checks to confirm the welds themselves are adequately sized. Weld load paths per UG-41 apply as well, confirming the load actually transfers through the welds the way the area calculation assumes it does. A nozzle that passes UG-37 but fails UG-45 is still a failed nozzle. The two checks answer different questions and both have to clear.

What are the common mistakes?

• Forgetting corrosion in $d$ and $t_n$. The opening grows and the nozzle thins as the vessel corrodes over its life. UG-37 is always evaluated in the corroded condition, never at nominal, as-ordered dimensions.
• Counting pad area outside the UG-40 limits. Only the portion of the pad that falls inside the parallel limit counts toward reinforcement. A pad that overhangs the limit does not get extra credit for the overhanging part.
• Hillside nozzles. $d$ becomes the chord length in the plane being checked rather than the simple bore diameter, and the plane of least reinforcement is the one that governs the evaluation.
• Skipping UG-45 after reinforcement passes. Thin-neck failures are caught at UG-45, not at UG-37. Passing one does not excuse skipping the other.
• Assuming $f_{r1} = 1$ with dissimilar materials. A weaker nozzle material reduces the strength ratio below 1.0, which both reduces the credit available and adds directly to the required area through the second term of the equation.

FAQ

When can reinforcement calculation be skipped entirely? UG-36(c)(3) exempts small openings that fall within stated size limits and spacing rules from the reinforcement calculation altogether. Verify the current limits before relying on the exemption, since the size thresholds are specific and easy to misapply on borderline cases.

Does set-on vs set-through change the calculation? It changes which weld areas exist and where the reinforcement limits land, since the two configurations put metal in different places relative to the shell surface. It does not change the underlying method: you still compute required area and compare it against available area the same way either way.

What about Appendix 1-7 large openings? Openings beyond UG-36's size limits trigger the supplemental rules of Appendix 1-7, which impose additional requirements beyond the standard area-replacement method. Flag any opening larger than half the vessel diameter for that check early, since it can change the reinforcement approach significantly.

Where does this fit in the full design flow? Stage 3 of the spec-sheet-to-U-stamp workflow, tightly coupled with UG-45 and MDMT per UCS-66, since the governing thickness used in the MDMT evaluation often traces back to the same nozzle geometry checked here.

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DeepMechanix runs UG-37, UG-40, and UG-45 together and prints every substitution above automatically. See the report format.