Mechanical Strength Requirements in Busbars

All busbar systems have to be designed to withstand the mechanical forces to which they may be subjected, whether these be due to their own weight, wind and ice loads, or short-circuit forces. This force becomes more onerous with increasing voltage and decreasing current due to respectively longer insulators and smaller conductors.
The conductor itself should have sufficient material strength under all operational conditions. It must be able to support itself without undue deflection under normal working conditions, and not suffer permanent damage under abnormal conditions. The following section enables the mechanical strength requirement of a conductor to be calculated using the short-circuit forces obtained from the formulae given previously. 

The maximum deflection of a beam carrying a uniformly distributed load and freely supported at each end is given by the formula:
where D = maximum deflection, mm
w= weight per unit length of loaded beam, N/mm
L = beam length between supports, mm
E = modulus of elasticity (124 x 103N/mm2)
I = moment of inertia of beam section, mm
If one end of a beam is rigidly fixed in a horizontal position the deflection is 0.415 times that given by the above formula and it follows that if a freely supported beam is also supported at its mid-point then its maximum deflection is reduced to 0.025 of its former value. If both ends of a beam are rigidly fixed in a horizontal position the deflection is 0.2 times that given by the above formula.
Thus with a continuous beam freely supported at four or more points the maximum deflection in the centre spans may be assumed to be 0.2 times that given by the formula, while the deflection in the end spans is 0.415 times. The deflection in the end spans, therefore, may be assumed to be twice that in the centre spans, assuming equal span distances.
Moments of inertia
In the above formula the moment of inertia I for the section of the beam has to be calculated about the neutral axis which runs parallel to the beam where the beam has zero tensile forces. In most cases this is the same axis of the centre of cross-section.
For a rectangular section of depth D and breadth B
For a circular section of diameter D
For a tubular section of internal diameter d and external diameter D
It should be noted that the value of I for a given cross-section is dependent on the direction in which each individual force is applied. Moments of inertia for a range of copper rods, bars, sections and tubes are given in Tables 12 – 16 (Appendix 2).
The natural frequency of a beam simply supported at its end is
and for a beam with both ends fixed horizontally it is
where fn = natural frequency, Hz
D = deflection, mm
As the deflection with fixed ends is 0.2 times the value with freely supported ends it follows that the natural frequency is increased by 2.275 times by end-fixing; fixing one end only increases the natural frequency by about 50%. Where equipment is to be mounted outside, natural frequencies of less than 2.75 Hz should be avoided to prevent vibration due to wind eddies.
In considering the loading of a conductor for outdoor service not only must the weight of the conductor itself be taken into account but also the weight of a coating of ice which it may carry, together with the wind pressure on the ice loaded conductor.
The maximum thickness of the ice and the maximum wind speed are normally specified by the purchaser of the busbars but where these are not specified they are usually available from national standards bodies within the country where the equipment is to be installed.
The wind and ice loading can be calculated using the following formulae:
Wind loading:
Ww  = p(D+2t) x 105
Ice loading:
where ww = wind loading, N/m
wi = ice loading, N/m
p = wind pressure, N/mm2
D = diameter, mm
t = ice thickness, mm
It is assumed that the wind load is at right angles to the vertical load of the conductor weight, and that its ice load and hence the resultant load on the conductor has to be added vertically. The resultant load is given by:
where R = resultant load, N/m
w = conductor weight per unit length, N/m
and where R acts at an angle q° to the vertical given by the formula
The vertical sag or deflection in the conductor span is given by
where Di is the sag in mm in a plane inclined at an angle q to the vertical.
The maximum permissible stress in a conductor is the resultant of its own natural weight (w) and the additional forces of wind (ww) and ice (wi) loadings (see above) and the magnetic forces resulting from a short circuit. It should be noted that the direction of a short-circuit force (Ws) depends on the position of adjacent phases and the direction of the currents in them.
In a general case the following method should be used for calculating the resultant force and its direction:
and
The maximum skin stress in the conductor can then be calculated using the following formula:
where f = maximum skin stress, N/mm2
M = maximum bending moment, N mm
Z = section modulus, mm3
For a single beam of length L (mm) uniformly loaded and freely supported at both ends or freely supported at one end and fixed at the other,
where W = load, N/mm
L = span, mm
For a circular section of external diameter D or for a rectangular section of depth D,
where I = moment of inertia, mm
D = diameter, mm
Then the maximum stress
The maximum permissible stress is dependent on the conductor material, temper, etc., but must not exceed the material proof stress or permanent deformation will occur. For a conductor manufactured from hard drawn copper the value is approximately 245 N/mm2.
For a beam which is horizontally fixed at both ends the bending moment at the centre is reduced to one third and that at its ends to two-thirds of those for a simple supported beam.
If the changes in length that occur in a conductor as it expands and contracts with temperature variations are not allowed for, undue forces will be set up in the conductor support system or in the equipment to which the busbar is connected.
The coefficient of linear expansion for copper may be taken as 17 x 10–6 /°C (for temperatures from ambient up to 200°C) compared with 23 x 10–6 /°C for aluminium. The lower value for copper is of great importance when allowing for thermal expansion under both normal and transitory conditions, as up to 25% less expansion need be accommodated for a particular length of busbar. If a length of copper bar were to be kept from expanding or contracting, a force of nearly 2 N per mm2 of cross-sectional area would be developed for a temperature change of 1°C.
In most cases the supports expand far less due to much smaller temperature changes and lower thermal expansion coefficients. It is therefore normal practice to allow for the full expansion using flexible conductor connections at suitable points.
Types of expansion joints
In the case of short bars it is usually not necessary to make any special provision to accommodate expansion. There will normally be one or two reasonably flexible bends capable of relieving any undue stresses which may be set up.
Figure 17 Types of expansion joints in copper conductors
To relieve intermediate supports of stress, clamps which allow the conductor to move freely in the longitudinal direction should be provided. These clamps must be designed and arranged with care to avoid the danger of stresses building up at any point at which the bar may become wedged or prevented from moving freely.
In the case of long straight runs it is advisable that expansion joints should be introduced. The joints may use laminated thin copper strips or leaves and have the same total current rating as the busbar itself.
As an alternative to laminated flexible joints, copper braid may be used. This type of joint is usually more expensive to manufacture but has the advantage that it can accommodate expansion in more than one direction (in most cases three directions) and also tends to eliminate vibration forces being passed from one piece of equipment to another. It is important that the ferrule into which the copper braid is clamped is of sufficient thickness to ensure consistent high conductivity after manufacture and during its service life. Where high resistances develop in the joint after manufacture, overheating and ultimately braid failure due to oxidation of the braid material may result.

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