Short answer: size the link at half the line conductor — 35 mm² for the 70 mm² feed — unless the adiabatic check proves less. A short 16 mm² link across the gasket stands only when S = I√t/k works out at 16 mm² or less: at 3.3 kA cleared in 0.1 s with copper k = 115, the check demands 9.1 mm², so the link holds. Treat the gasket as the insulator it is. Bond the plate metal-to-metal with serrated washers, or fit a dedicated jumper across the joint, sized by the same equation. Your 300 mm run is judged on cross-section, not on length, and six M10 bolts through clearance holes are not a bond until you make them one.
The default: half the line conductor
The jumper from gland plate to earth terminal is part of the supply circuit's protective conductor. It therefore sizes off the 70 mm² line conductor, not off the terminals that happen to sit on the DIN rail. The wiring rules give the default in a table you can carry in your head: same size as the line conductor up to 35 mm², half the line size above that. BS 7671 states it for UK work; IEC 60364-5-54, the IEC wiring-rules part on earthing arrangements and protective conductors, works the same way.
| Line conductor | Default protective conductor |
|---|---|
| 16 mm² | 16 mm² |
| 25 mm² | 25 mm² |
| 35 mm² | 35 mm² |
| 50 mm² | 25 mm² |
| 70 mm² | 35 mm² |
| 95 mm² | 50 mm² |
For the 70 mm² feed the default is 35 mm², and the 16 mm² DIN-rail earth terminals change none of it. They only tell you where 35 mm² will not land, so you either support the smaller size with the calculation below or give the big link another landing point. The background — what counts as an effective fault path and which metal parts must be bonded — sits in the grounding and bonding guide.
Run the adiabatic check on the short link
Two inputs decide it: the prospective earth-fault current at the board and the time the upstream device takes to clear that current. Measure the current with an earth-fault-loop tester at the board terminals. Read the clearing time from the protective device's time-current curve at that current, taking the worst value the curve shows.
Then run the adiabatic equation: S = I × √t / k, with S in mm², I in amperes and t in seconds. Use k = 115 A√s/mm² for copper with 70 °C thermoplastic insulation, the standard green/yellow PVC, and k = 143 for 90 °C XLPE.
Worked example: a 70 mm² feed into a distribution board, measured prospective fault current 3.3 kA, device clearing in 0.1 s, 16 mm² PVC copper link, 300 mm long.
- √0.1 = 0.316
- 3300 A × 0.316 = 1043
- 1043 / 115 = 9.1 mm²
The check demands 9.1 mm² of copper; the link carries 16 mm², about 1.8× what it needs, so it stands. Length plays no part in that verdict. The 300 mm run sets resistance, resistance sets the fault magnitude you have already measured, and the thermal limit lands on cross-section alone. A 3 m run of the same 16 mm² passes or fails on the same numbers.
What 16 mm² carries, by clearing time
Invert the equation and you get the current the link survives: I = kS/√t. For 16 mm² copper at k = 115:
| Clearing time | Current a 16 mm² link survives |
|---|---|
| 0.05 s | 8.2 kA |
| 0.1 s | 5.8 kA |
| 0.5 s | 2.6 kA |
| 1 s | 1.8 kA |
| 5 s | 823 A |
For the 3.3 kA fault of the example that becomes a time limit: t = (kS/I)² = (1840 / 3300)² = 0.31 s. The 16 mm² link holds while the device clears in 0.31 s or faster. A 35 mm² link stretches the same fault to (4025 / 3300)² = 1.5 s. Past that, copper is the wrong fix: the same 3.3 kA cleared in 5 s demands 3300 × 2.236 / 115 = 64 mm², more copper than the line conductor. Speed the clearing — a faster or cascaded device upstream — then resize.
When the terminals really do cap you at 16 mm², run two equal links in parallel: same length, same size, routed together, each end landed. The 3.3 kA fault then puts 1.65 kA on each link, and each survives (1840 / 1650)² = 1.2 s of it. The pair behaves as one 32 mm² conductor, good for 11.6 kA over 0.1 s.
Make the gasketed gland-plate joint a designed bond
The gasket seals, and it also insulates. Six M10 bolts through clearance holes give contact only where metal bites metal on both faces, and powder coat, paint or anodizing on either side defeats it. NEC 250.12 requires paint, lacquer and enamel to be removed from threads and contact surfaces at grounding and bonding connections. Strip both contact faces to bare metal before you trust any bolted joint.
Then build the bond deliberately, one of two ways:
- Serrated washers at every bolt. Conical or toothed washers under head and nut bite into both faces and hold the contact through thermal movement — the same practice used when bonding steel-wire-armoured cable glands with serrated washers. On a 304 or 316L plate, fit stainless washers: mixed metals at pierced, damp joints invite crevice corrosion, as explained in stainless and galvanized metal contact.
- A dedicated jumper across the gasket. Size it by the same check: the example's 3.3 kA for 0.1 s demands 9.1 mm², so a 16 mm² jumper passes with margin. Land it from a plate bolt to an earth boss or terminal on the enclosure wall, outside the gasket line where you can.
Fit both when outgoing circuits rely on the plate: the washers fold the plate into the bonded envelope, the jumper carries the fault. It mirrors the gasketed door, where the hinge side gets a bonding jumper sized the same way.
The reason a gland needs earthing at all is the fault path itself, and a replacement gland landing in a plain clearance hole needs its own bond exactly as a threaded entry did. If the board has no plate yet, see when a conduit-fed box needs a gland plate. For the joint hardware, the T-joint gasket design for removable seals shows how to detail a seal you can pull and refit.
One line on the hazardous-area board where this question usually lands: on certified Ex e equipment the gland plate, its bolts and its terminal sizes are part of the certificate. Keep the certificate holder's arrangement, and make any change of terminal size through their documentation. We build washdown stainless enclosures, not hazardous-location types.
Fit and land the link
- Measure the prospective earth-fault current at the board with an earth-fault-loop tester and record it (the example: 3.3 kA).
- Read the upstream device's clearing time at that current from its time-current curve, worst value (the example: 0.1 s).
- Compute S = I√t/k with k = 115 for PVC copper or 143 for XLPE; the example gives 9.1 mm².
- Choose the next standard size above the result: 16 mm² covers the example; anything you cannot support with a calculation defaults to 35 mm², half the line.
- Land the 35 mm² where the 16 mm² terminals cannot take it: a 35 mm² compression lug on an M10 earth boss, a terminal block or busbar rated for 35 mm², or two equal 16 mm² links in parallel.
- Build the joint: faces to bare metal, serrated washers at each bolt, and the jumper across the gasket sized by the same equation.
- Route the link short and clear of sharp edges — 300 mm in the example — with a little slack at both ends so vibration works on the loop, not on the lugs.
Checks when done
- Measure plate to main earth terminal through the finished link with a low-resistance ohmmeter. The 300 mm of 16 mm² copper alone reads ρL/A = 1.72 × 10⁻⁸ Ω·m × 0.3 m / 16 × 10⁻⁶ m² = 0.32 mΩ, so expect the whole path, joints included, in single-digit milliohms. A reading near an ohm means paint under a washer or a lug that never crimped.
- Re-run the earth-fault-loop measurement and compare it with your first reading; the finished bond leaves the figure at or below where it started.
- Torque-check the bolted joint after the panel's first heat cycle, and file the fault current, clearing time, calculated size and both meter readings with the board's paperwork.
What breaks the earth path
- Bolts trusted through a gasket: without prepared faces and biting washers there is no designed contact to lean on.
- Paint, powder coat or anodizing left under the washers — the common milliohm-becomes-ohms failure.
- A 16 mm² link fitted because the terminals accept it, with no measured fault current behind the arithmetic.
- Parallel links of unequal length, which share the fault unevenly and overload one of them.
- Plain steel or galvanized washers on a stainless plate, which corrode at the bite marks and raise the joint resistance over time.
- A replacement gland in a clearance hole, bonded the way the old threaded entry was — that is, not at all.
The decision, case by case
- The check passes at your measured current and time — 9.1 mm² demanded, 16 mm² fitted in the example: keep the short link and add serrated washers or the jumper across the gasket.
- The check fails, or you have no measured fault current: fit 35 mm², half the line, on an earth boss, a rated terminal block, or as two 16 mm² links in parallel.
- The fault needs more than 35 mm²: improve the clearing time upstream, then resize — copper alone gets unwieldy fast, 64 mm² for the example's fault at 5 s.
- A certified Ex e board: the certificate governs plate, bolts and terminals, so keep its arrangement. We do not build hazardous-location enclosures.
- Ordering the box with the plate already in it: our custom stainless enclosures take the gland plate, bolt pattern and cutouts from your drawing — the cutout drawing checklist lists what to send — and it is delivered in 10 business days anywhere in the world. For the plate choice itself, see stainless enclosures with a gland plate.
Catalog dimensions to compare
Compare these catalog dimensions with your component layout. Open each enclosure for its rating, material and usable space, then check those specifications against your application.
| Enclosure | Material | Outside H x W x D / display | Rating |
|---|---|---|---|
| 12 × 10 × 6 in | 304 stainless | 12 x 10 x 6 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
| 16 × 14 × 8 in | 304 stainless | 16 x 14 x 8 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
| 20 × 16 × 8 in | 304 stainless | 20 x 16 x 8 in | Rated NEMA 4X / IP66 by the manufacturer; not UL listed |
Questions
How much fault current can two 16 mm² links in parallel carry?
Run the two links the same length, land both at the same pads, and the pair behaves as one 32 mm² conductor: at k = 115 it survives 3680 / 0.316 = 11.6 kA for 0.1 s. For the 3.3 kA example, each link carries 1.65 kA and holds for 1.2 s. Unequal lengths share the fault unevenly, so keep them matched and route them together.
Which k value applies if the earth link is 90 °C XLPE instead of PVC?
Use k = 143 for copper with 90 °C thermosetting insulation, against 115 for 70 °C thermoplastic PVC. A 16 mm² link then survives 2288 / 0.316 = 7.2 kA for 0.1 s, up from 5.8 kA. A bare copper strap bolted straight to the plate carries more again; size it by the same equation with the k value for bare copper conductors.
Will six M10 bolts bond the gland plate through the gasket on their own?
Count on them only when the joint is built for it: contact faces stripped to bare metal on both sides and serrated washers that bite and stay seated. Through paint or gasket squeeze-out the same joint can read whole ohms, while the 300 mm, 16 mm² jumper adds 0.32 mΩ. Fit the jumper as well; it is the path you can measure and trust.
Where do you land a 35 mm² link when the earth terminals only take 16 mm²?
Three landings pass the same check: a 35 mm² compression lug on an M10 earth boss through the plate or wall, a terminal block or busbar rated for 35 mm², or two equal 16 mm² links in parallel, each carrying half the fault. Land the 35 mm² on the main earth terminal and feed the 16 mm² rail from it, and the rail then follows its own check.
Does the enclosure metal carry the fault current so the link can be skipped?
It can when every joint in the path is bonded metal-to-metal: NEC 250.96(A) requires metal parts serving as equipment grounding conductors to be bonded so they safely conduct the fault current likely to be imposed on them. NEC 250.4(A)(5) keeps the earth itself out of that path. On a gasketed gland plate the joints are exactly what the gasket interrupts, so the jumper is the designed path, not a spare.





