Containment · 6 July 2026, updated 3 August 2026 · The force goes as the square of the peak
BS 7671 requires every conductor to withstand the electromechanical forces a fault puts on it. It says so in four separate places. It gives you no method whatsoever for working out what those forces are — and the current they depend on is not the one your fault study hands you.
Two long parallel conductors carrying currents i1 and i2, separated by a distance s, push on each other:
F / L = μ0 · i1 · i2 / (2π · s) = 2×10−7 · i1 i2 / s N/m
Currents in the same direction attract; currents in opposite directions repel. A short circuit between two conductors is exactly the opposite-direction case, so the conductors are thrown apart. That is Ampère's force law — the relationship the ampere itself was once defined by — and it needs no standard, no manufacturer's data and no software.
Under normal load this is a curiosity: a few newtons per metre, comfortably carried by whatever is holding the cable up. Under fault it is the whole story, and it arrives within the first few milliseconds — long before any protective device has finished thinking about it.
Look again at the numerator: i1 · i2. In a fault the same current flows in both conductors, so the force goes as the square of the current. A 10 % error in the current is a 21 % error in the force. Double the fault level and the force quadruples.
That squaring matters twice, and the second time is the one that catches people out: it applies to the peak factor as well as to the current. Which is the next section, and the reason this article exists.
A fault does not start at a convenient point on the voltage waveform. Depending on when it strikes, the current can be fully offset: instead of swinging symmetrically about zero it sits on a decaying DC component, and the first loop reaches far higher than the steady symmetrical value.
ip = κ · √2 · Ik
κ depends on the X/R ratio of the circuit and is bounded between 1 and 2 — 1 for a purely resistive circuit, approaching 2 for a very inductive one, which is why the peak can reach 2√2 ≈ 2.83 times the RMS for a fully offset fault close to a transformer. Our fault level calculator now reports it alongside the RMS value, and the fault level explainer covers where the RMS figure itself comes from.
The worked case: an 800 kVA transformer at 5 % impedance, single-core cables in a touching trefoil of 60 mm overall diameter, cleats at 600 mm centres. The RMS symmetrical fault is 25.4 kA, giving 1,863 N/m — a number you could reasonably look at and think manageable. The peak is 58.0 kA, giving 9,711 N/m, and at 600 mm centres that is 5,826 N on every cleat. Call it 594 kgf: roughly a grand piano, on each fixing, for the duration of the first loop.
You will see the force written as 0.17 × î² / s, with the peak current î in kA and the spacing s in metres. That is not a different formula — it is the same one with the units folded in. Putting î in kA multiplies by 106, so 2×10−7 becomes 0.2; the √3/2 factor for the worst conductor of a three-phase group turns 0.2 into 0.173. Worth knowing, but worth knowing why: the coefficient already contains the three-phase assumption, so it is wrong for a two-conductor fault, where the figure is 0.2.
For a two-conductor fault — single-phase, or line-to-line — Ampère's law applies directly and exactly. For a three-phase fault the three currents peak at different instants, so the worst conductor does not see the naive two-conductor value: the standard result is √3/2, about 0.87, of it.
Trefoil and flat share that worst-case magnitude. What differs is the direction, and therefore what fails:
You cannot change the fault current — that belongs to the supply, and reducing it is a transformer decision, not a containment one. Two things are yours:
The distinction matters the moment someone asks for a relaxation on site. “Can we go to 900 mm centres?” is a 50 % increase in the load on every cleat, and is answerable straight from the calculation. “Can we push the cables apart a bit?” changes the force, the grouping factor and possibly the cable size — and is not.
This is the part that needs no calculation to make its point. The video below is a short-circuit test on a trefoil of single-core cables: the cable ties break and the cleats hold.
The engineering point underneath it is not that ties are weak. It is that a cable tie has a static loop tensile strength and no declared short-circuit withstand at all. Those are different quantities. You cannot compare a calculated force against a number that does not exist, so a tie cannot satisfy Regulation 521.5.201 — not because it fails the comparison, but because there is no comparison to make. It is also why “we have always used ties on this size” is not evidence of anything.
Four places in BS 7671:2018+A4:2026 require this to be dealt with. Not one of them tells you how:
| Where | What it requires | Method? |
|---|---|---|
| 521.5.201 Electromechanical stresses |
Every conductor shall have adequate strength and be so installed as to withstand the electromechanical forces of any current it may have to carry, including fault current. | No method. |
| 434.5.3(a) Busbar trunking and powertrack |
Icw not lower than the RMS prospective fault current, and the rated peak withstand current not lower than the prospective fault peak. | Names the comparison, not the peak. |
| Part 2 The symbol I<sub>pk</sub> |
Defines Ipk as the rated peak withstand current — the quantity you are required to compare against. | No method. |
| Chapter 13 Conductor selection |
Lists the electromechanical stresses due to short-circuit and earth fault currents as a selection factor, alongside temperature and volt drop. | No method. |
Regulation 521.5.201 is worth reading twice for the phrase “and be so installed as to”. The fixing is inside the requirement, not an installation detail bolted on after the design is finished.
Regulation 434.5.3 is the sharpest of the four, because it asks for two comparisons on a busbar trunking or powertrack system: the RMS prospective fault current against the system's Icw, and the prospective peak against its rated peak withstand. A system can pass the first and fail the second. An Icw of 30 kA against a 25 kA fault looks comfortable — until you work out that the same fault peaks at 57 kA and the declared peak withstand is 50. That is exactly what the two-part check exists to catch, and it is why a single kA figure is never enough to specify a busbar system.
For the cleat itself, BS EN 61914 is the product standard: it is what a manufacturer's declared short-circuit performance is declared against, and tested data for the exact cleat, cable diameter, formation and spacing is the strongest evidence available. What this page gives you is the other half — the force to compare it with, and some confidence that it was calculated from the right current.
A cleat spacing that cannot be audited is a cleat spacing that will be changed on site. The minimum that makes it checkable:
That last point is where the geometry starts to bite. A cleat is wider than its cable, so on a fully loaded ladder the cleats on adjacent cables clash if they share a rung:
300 mm is a habit, not a design. It can be dangerously wide on a run near a large transformer and needlessly tight on a small final circuit. The spacing follows from the peak fault current at that point, the formation, the centre-to-centre spacing and the cleat's declared short-circuit withstand — four numbers, none of which is 300.
F/L = 2 x 10^-7 x i1 x i2 / s newtons per metre, with the currents in amperes and the spacing in metres. That is Ampere's force law and it needs no standard. For a three-phase group the worst conductor sees root-3/2 of it, which with the peak current in kA and the spacing in metres collapses to the familiar 0.17 x i-peak-squared / s.
The peak. i-peak = kappa x root-2 x Ik, where kappa depends on the X/R ratio and lies between 1 and 2. Because force goes as current squared, the peak factor gets squared too: at X/R 6 the RMS answer understates the force by 5.2 times, and at a high X/R it approaches 8. This is the single most common error in the subject.
It does not. Regulation 521.5.201 requires every conductor to withstand the electromechanical forces of a fault, Regulation 434.5.3 requires a busbar trunking system to withstand the prospective peak, Part 2 defines Ipk, and Chapter 13 lists the stresses as a selection factor. Four requirements, and not one method between them. The magnitude comes from first principles; the cleat comes from BS EN 61914 and the manufacturer's test data.
Both peak at the same magnitude for the worst conductor, root-3/2 of the two-conductor value. They fail differently. Trefoil is radial, so the force tries to burst the bundle apart and a cleat gripping all three resists it directly. In flat formation the middle conductor is pushed alternately both ways as the phases reverse, so it is thrown sideways rather than outward.
Because spacing is measured centre to centre, and for a touching trefoil that distance IS the cable overall diameter. A larger conductor in a thicker sheath sits further from its neighbours and sees a lower force per metre — so the dimension that sets the force is the one on the cable data sheet, not the one on the schedule.
Tested data for the exact cleat, cable diameter, formation and spacing is the strongest evidence there is, and BS EN 61914 is the product standard that performance is declared against. The calculation on this page gives you the force to compare with it. Where your real configuration does not match a tested one, the calculated force against the declared withstand is the route — documented with its inputs, because a checker cannot audit an assumption you did not write down.
Not as the restraint against fault forces. A cable tie has a static loop tensile strength and no declared short-circuit withstand at all, so there is nothing to compare your calculated force against. That is not a marginal call — it is the absence of the evidence the regulation requires. Ties are fine for bundling cable that is already properly restrained.
A cleat is wider than its cable, so on a fully loaded ladder the cleats on adjacent cables would collide if fixed to the same rung. Staggering onto alternate rungs gives every cable its cleat while keeping them clear — but note that it doubles the effective interval for each cable, so it belongs in the calculation rather than being applied afterwards on site.
Verify before you rely on it. The force law is Ampère's and needs no standard. Everything past the magnitude does: conductor stiffness, the dynamic response of the span and the natural frequency of the cable between fixings are what IEC 60865 and the cleat maker's tested data cover. The κ expression used here is attributed to IEC 60909 and its coefficients have not been transcribed from it — what is defensible without it is the shape, that κ lies between 1 and 2. Treat the output as a first-pass magnitude, not a cleat selection.
Specifying containment on a high-fault-level project? We can produce or check the cleat spacing calculations, with the peak current and the assumptions written down — get in touch.