_{} - **the**** equation of the wire state **(the basic reference equation)

This cubic equation _{} or _{} - most often, it is solved by trial method (this what it used to be like, now, with the help of computers, it is solved by bipartitioning and some other methods).

The calculationof this equation could be performed for both mono metal and composite wires. For composite wires the values caracterizing the entire wire (_{}; _{} and others) are inserted into the equation.

Thus, when we have the equation of the wire state, and we know the tension at “m”-conditions, we can find out the tension in the wire material at any atmospheric conditions.

The question is what tension is the source tension. The starting point will the permissible tension. In any case, the tension in the wire material should not exceed the permissible value. We have three values for permissible tension, we should sort out the one we need.

The three values for permissible tension equation – for maximum load, lowest temperature and average annual temperature – were accepted for steel-aluminum wires up until 1975 for (high in case of ice, lower – at lowest temperature), and the same for mono metal wires. In 1975 the values for permissible tension were stated for the lowest temperature as well as for maximum load.

Solving the equation on a slide-rule (Krukov K.P., Novgorodtsev B.P. Constructions and Mechanical Calculation of Power Transmission Line

At the lower scale we mark the supposed value for the sought-for tension _{}and bring either the beginning or the end of the runner scale to this value. The upper movable scale will show the value _{}, which should be multiplied by _{}at the upper scale of the runner; if the value of the product _{} is less than _{}, the value of _{} should be increased. Move the runner and multiply _{} by a new value of _{}until the product equals _{}. After the required product is received, digit position should be checked for possible mistakes.

Example.

A steel-aluminum wire АС120/19 is placed in a span of 300 m with the tension of 13.0 daN/mm2 at the temperature of _{}, ice thickness of _{} and wind pressure of _{}. It is required to define the sag of the wire at _{}. The modulus of elasticity for the wire is _{}and the thermal expansion coefficient is _{}.

_{} _{}

_{}

_{}

_{}

_{}

_{}

_{}

Insert these values into the equation of the wire state:

_{}

_{}

_{} Take the value of_{} on the slide-rule. We 've got the following product:

5,0^{2} (5+10,5)=387>370. Take the lowest value of 4,9, and we’ve got 4,9^{2} (4,9+10,5)=372>370.

At last, in cas _{} we’ll receive the exact value for the product 4,89^{2} (4,89+10,5)=370.

Find out the sag

_{}