The mechanical effects of short-circuit currents in - Montefiore

In circular profiles, every axis through the center of
gravity can be axis of inertia, also the drawn axis
O-O in which the vector 1 2 M M M
r r r
= + lies and is
simultaneously the neutral axis. The absolute value
of M r is given by the equation (2.63):
(2.63)
r
M = M =
r
M
2
1
r
+ M 2
With σ1max = M1/Z, σ2max = M2/Z and σmax = M/Z the
maximum stress becomes:
(2.64)
σ
max
= ±
σ
2
1max
+ σ
Equation (2.64) is also valid for tubes.
a)
b)
2
2
2max
Figure 2.16 Stresses caused by orthogonal moments
a) rectangular profile
b) circular profile
2.3. SPECIAL CONFIGURATIONS
2.3.1. Associated phase structures busbar on a
common support
In the associated phase structures, the busbar is
often set horizontally on a common support as
shown on the Figure 2.17.
28
connection
post insulator
Bus support
ground
a
y
z
x
foundation block
Figure 2.17 Common support of busbar
The dynamic response of the tubes obeys the
following law :
(2.65)
∂ u
∂u
∂ u
i
i
i y
( ρS)
+ c + ( EI)
= F () t
tube
2
∂t
tube
2
tube
tube
∂t
tube
4
∂x
tube
where ui is the displacement function of a tube,
E modulus of elasticity,
I moment of inertia,
(ρS)tube own weight per unit length.
In the case of two- or three-phase faults, instant
LAPLACE’s forces meet the following equations :
(2.66) ∑
i=
r
y
Fi
1..
3
r
= 0
If, in fact, we call i1, i2, i3 the instant currents in
phases 1, 2, 3, these forces can be written as
follows :
0 ⎛ i3
⎞
(2.67) F1 = i1.
⎜i2
+ ⎟
2 a ⎝ 2 ⎠
y µ
π
0
(2.68) F i . ( i i )
y
−
2
µ
=
2πa
0 ⎛ i1
⎞
(2.69) F3 = − i3.
⎜i2
+ ⎟
2 a ⎝ 2 ⎠
y µ
π
By using the differential operator’s linearity
properties and posing :
(2.70) U ∑u
2
tube
=
i
3
tube
i
one could easily check the following differential
equation independent of the shape of the current
chosen (phase to earth, earthed or insulated phase to
phase, three-phase):
1
4
i