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Home / Technical Articles / Electromagnetic stresses on a busbar system and methods of reducing

Electromagnetic Stresses

When a conductor carries a current it creates a magnetic field which interacts with any other magnetic field present to produce a force. When the currents flowing in two adjacent conductors are in the same direction the force is one of attraction, and when the currents are in opposite directions a repulsive force is produced.

Electromagnetic Stresses On Busbar System
Electromagnetic Stresses On Busbar System (photo credit:

In most busbar systems the current-carrying conductors are usually straight and parallel to one another.

The force produced by the two conductors is proportional to the products of their currents.

Normally in most busbar systems the forces are very small and can be neglected, but under short-circuit conditions, they become large and must be taken into account together with the conductor material fibre stresses when designing the conductor insulator and its associated supports to ensure adequate safety factors.

The factors to be taken into account may be summarised as follows:

  1. Stresses due to direct lateral attractive and repulsive forces.
  2. Vibrational stresses.
  3. Longitudinal stresses resulting from lateral deflection.
  4. Twisting moments due to lateral deflection.

In most cases the forces due to short-circuits are applied very suddenly. Direct currents give rise to unidirectional forces while alternating currents produce vibrational forces.

Maximum stresses

When a busbar system is running normally the interphase forces are normally very small with the static weight of the busbars being the dominant component.

Under short-circuit conditions this is very often not the case as the current rises to a peak of some thirty times its normal value, falling after a few cycles to ten times its initial value.

These high transitory currents create large mechanical forces not only in the busbars themselves but also in their supporting system.

This means that the support insulators and their associated steelwork must be designed to withstand these high loads as well as their normal structural requirements such as wind, ice, seismic and static loads.

The peak or fully asymmetrical short circuit current is dependent on the power factor (cos φ) of the busbar system and its associated connected electrical plant. The value is obtained by multiplying the r.m.s. symmetrical current by the appropriate factor given in Balanced three-phase short-circuit stresses.

If the power factor of the system is not known then a factor of 2.55 will normally be close to the actual system value especially where generation is concerned. Note that the theoretical maximum for this factor is 2√2 or 2.828 where cos φ=0.

These peak values reduce exponentially and after approximately 10 cycles the factor falls to 1.0, i.e., the symmetrical r.m.s. short circuit current.

The peak forces therefore normally occur in the first two cycles (0.04 s) as shown in Figure 1 below.

Short-circuit current waveform
Figure 1 – Short-circuit current waveform

In the case of a completely asymmetrical current wave, the forces will be applied with a frequency equal to that of the supply frequency and with a double frequency as the wave becomes symmetrical. Therefore in the case of a 50 Hz supply these forces have frequencies of 50 or 100 Hz.

The maximum stresses to which a bus structure is likely to be subjected would occur during a short-circuit on a single-phase busbar system in which the line short-circuit currents are displaced by 180°.

In a three-phase system a short-circuit between two phases is almost identical to the single- phase case and although the phase currents are normally displaced by 120°, under short-circuit conditions the phase currents of the two phases are almost 180° out of phase. The effect of the third phase can be neglected.

In a balanced three-phase short-circuit, the resultant forces on any one of the three phases is less than in the single-phase case and is dependent on the relative physical positions of the three phases.

In the case of a single-phase short-circuit, the forces produced are unidirectional and are therefore more severe than those due to a three-phase short-circuit, which alternate in direction. The short-circuit forces have to be absorbed first by the conductor. The conductor therefore must have an adequate proof strength to carry these forces without permanent distortion.

Copper satisfies this requirement as it has high strength compared with other conductor materials (Table 2 below).

Table 2 – Typical relative properties of copper and aluminium

Properties of Cu and AlCopper(CW004A)Aluminium (1350)Units
Electrical conductivity (annealed)10161% IACS
Electrical resistivity (annealed)1.722.83μΩ cm
Temperature coefficient of resistance(annealed)0.00390.004/° C
Thermal conductivity at 20°C397230W/mK
Coefficient of expansion17 x 10–623 x 10–6/° C
Tensile strength (annealed)200 – 25050 – 60N/mm2
Tensile strength (half–hard)260 – 300 85 – 100N/mm2
0.2% proof stress (annealed)50 – 5520 – 30N/mm2
0.2% proof stress (half–hard)170 – 20060 – 65N/mm2
Elastic modulu116 – 13070kN/mm2
Specific heat385900J/kg K
Melting point1083660°C

Because of the high strength of copper, the insulators can be more widely spaced than is possible with lower-strength materials.

Methods of reducing conductor stresses

In cases where there is a likelihood of vibration at normal currents or when subjected to short- circuit forces causing damage to the conductor, the following can he used to reduce or eliminate the effect:

a) Reduce the span between insulator supports.

This method can be used to reduce the effects of both continuous vibration and that due to short-circuit forces.

b) Increase the span between insulator supports.

This method can only be used to reduce the effects of vibration resulting from a continuous current. It will increase the stresses due to a short-circuit current.

c) Increase or decrease the flexibility of the conductor supports.

This method will reduce the effects of vibration due to continuous current but has very little effect on that due to short-circuit forces.

d) Increase the conductor flexibility.

This can only be used to reduce the effects of vibration due to a continuous current. The short-circuit effect is increased.

e) Decrease the conductor flexibility.

This method will reduce the effects of vibration due to either a continuous current or a short-circuit.

It will be noted that in carrying out the various suggestions above, changes can only be made within the overall design requirements of the busbar system.

Reference: Fundamentals of Power System Protection

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More Information

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.


  1. Ashok menghani
    May 27, 2017

    I wish to have some knowledge on thermal imaging of the switchboard.

  2. Fred
    Mar 05, 2017

    Sir,I want to know if the busbars are already under mechanical stresses ,is it affect the conductivity of the busbar

  3. Hussain
    Jun 30, 2014

    Hi Sir.
    Im still waiting for your reply please.

  4. Hussain
    Jun 30, 2014

    Hi Sir Engineer.
    Thanks for your lovely topics…Its really interesting….I just want your opinion on what sofware you recomend to use in desining the installation wiring which can save the time and effort of engineers and installation designers….provided that the software can be used or comply with AS/NZ STANDARED employed in Australia and Newsland.

    Many thanks with best regards.


    • Edvard
      Jun 30, 2014

      What do you consider by desining the installation? Voltage level? Electrical design, or control or protection wiring schemes… or maybe just the design of outlets and lighting?

      Make your question a little bit more clearer ;) There are many aspects of electrical design.

  5. Qadir
    Apr 08, 2014

    one 500kva 11kv/430v transformer got damaged earlier at site. After check delta connection on hv side was found open which was reconnected and after putting load on it, it stood ok for some 15 hours and again delta connection got open with burning of ht fuses. Again it was repaired and same thing happened. what could be the problem? load side ok.

  6. Diganta mech
    Apr 02, 2014

    Sir,i want to know about details of lighting transformer.please

    • Edvard
      Apr 05, 2014

      Lighting transformer is used for lighting bulbs that require reduced voltage. If you ask me, there is nothing special about them that differs from other transformers. However, you can read more about this topic in category ‘Lighting’ here:

      Kind regrds

    • Disant
      May 02, 2014

      Lighting transformers are basically the same as of power transformer however this type of transformers provides the power for normal illumination circuit.This type of transformer is used in the power system as the lighting loads are normally inductive in nature that may affect the Voltage current current curve of a power system,hence in order to correct this a lighting transformer is used in circuit that basically act as an Isolation Transformer in circuit.

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