Mechanical Design of
Transmission Lines
Farhan Mahmood
EED, UET Lahore.
1
Main Considerations in the
Mechanical Design

The main considerations in the mechanical design
of an overhead transmission line are:



Adequate clearance between conductor and ground
High mechanical strength of the conductors
Tension or working stress of the conductor < ultimate
tensile strength
Ultimate tensile strength = F.O.S Χ working stress
2
Basic Design Considerations

While erecting an overhead line, it is
important that conductors are under safe
tension.


If the conductors are too much stretched between
supports to save the conductor material, the
stress in the conductor may reach unsafe value
and in certain cases, the conductor may break due
to excessive tension.
In order to permit safe tension in the conductors,
they are not fully stretched but are allowed to
have a dip.
3
Few Important Terms in the
Mechanical Design


TENSION, a force tending to stretch or
elongate a conductor.
ULTIMATE
TENSILE
STRENGTH,
maximum stress, which a conductor can
withstand without failure.
4
Few Important Terms in the
Mechanical Design

SAG, the vertical distance (d) between
the mid-point of a conductor to the line
joining the two supports level.
5
Few Important Terms in the
Mechanical Design

CATENARY‘S CURVE, When the conductor is
suspended between two supports at the same level,
it takes the shape of catenary's curve. However, if the
sag is very small as compared with the span, then
sag-span curve is parabola.
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Few Important Terms in the
Mechanical Design

SPAN, the horizontal distance (L)
between the two adjacent supports.
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Points to Remember

The following points are to be noted,



The tension at any point on the conductor
is tangent to that point
The horizontal component of the tension is
constant throughout the length of wire.
The tension will be maximum at the
supports and minimum at the lowest point
of the curve.
8
Factors affecting Sag

SAG plays a very important role in the mechanical design of an
overhead line. It is not a good practice to provide either too
high or too low sag.
Sag (Too Low)
Sag (Too High)
1. Tension in the conductor is too high
1. Tension in the conductor is too low
2. Less conductor length is required
2. More conductor length is required
3. Lower supports are required
3. Higher supports are required

It is always desired that tension and sag should be as low as
possible, which is not possible simultaneously.
Low Sag ------> tight wire & high tension
High Sag ------> loose wire and low tension
Therefore, a COMPROMISE is made between the two.
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Factors affecting Sag

The factors affecting the sag of a
conductor strung between supports are




Weight of conductor
Distance between the
length)
Working tensile strength
Temperature
supports
(span
10
Sag Calculations

A conductor AOB of length
l’ is suspended at two
towers A and B and are A
spaced L unit apart. Let O is
the lowest point of the wire.
Consider a length OP of the
curve length s.
 w = weight/unit length,
 H = tension at point O
 T = tension at point P,
H
x
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Sag Calculations

Three forces are acting on it



Horizontal tension H at the lowest point
Weight ws of OP acting through its center of gravity
Tension T at point P along tangent to the curve at P.
For equilibrium, horizontal forces in one
direction must be balanced by horizontal
forces in the other direction. Same is true for
vertical forces.
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Sag Calculations
Let θ be the angle which the tangent at
P makes with the horizontal.
T sin θ
T cos θ
ws
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Components of Tension
………(1)
…….(2)
Dividing equation 1 by 2 we get
…….(3)
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Length
…..(4)
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Length
From equation 3 we have
So eq. 4 becomes
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Length
Integrating both sides , we have
Where A is the integration constant
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Length
Using intial values as
x=0
s=0
we get
A = 0,
So we have,
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Length
…….(5)
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Calculation of Sag
As we know that
So,
Since
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Calculation of Sag
or
Integrating both sides we have
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Sag Calculations
Where B is the Integration constant
 Using initial values

x=0

y=0
We get

B = -H/w

So,
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Sag Calculations
This equation is called the equation of catenary
On Expanding we get
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Calculation of Tension
Neglecting high powered terms we get
The tension at point P is,
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Calculation of Tension
When x=L/2 , y is equal to the sag or deflection ‘d’
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Calculation of Length
Using the equations,
After putting the value of H from expression of d into l,
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Solve Example
The weight of a overhead conductor of a
line is 4.0 N/m. The ultimate strength is
8000 N. If safety factor is 4 and span
length is 160 m, find (a) sag and (b)
total length of the line between spans.
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Solution
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Continued…
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Supports at different levels
(unsymmetrical span)
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Continued…

Let P1 and P2 are two points at heights h1 and h2 from the
ground respectively,
where h is difference between the elevations of two supports.
If the span length is L, x1 + x2 = L. Therefore,
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Continued…
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Continued…
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Solve Example
An overhead transmission line conductor has the
following data
 Weight = 0.35 kg/m;
 Maximum allowable strength = 800 kg;
 Safety factor = 2;
 Span length = 160 m.
Supports are at different levels where one support is
at 70 m from the ground. Find the minimum
clearance from the ground and the minimum point of
the catenary from the supports when the second
support is at (a) 40 m and (b) 65 m.
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Solution
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Continued…
This shows that the minimum point lies
outside of the span is 134.29 m from
the lower span. Therefore, the
minimum ground clearance is 40 m that
is the height of lower support.
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Continued…
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EFFECT OF ICE- AND WINDLOADING
1.
2.
3.
The sag and tension of lines are different in
normal weather conditions. Since lines are
designed for all the conditions, it is
important to calculate the sag and tension
during the ice- and wind-loading conditions.
Conductor weight
Ice Loading
Wind Loading
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Conductor Weight
The weight of the conductor acts
vertically downwards and depends upon
the type of the conductor used. The
weight of the conductor per unit length
is available from the table giving the
mechanical characteristics of the
conductor.
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Ice Loading

In snowy areas, ice is deposited on the
conductors and its accumulation on the
conductor affects the design of line,



By increasing the weight per unit length.
By increasing the projected surface area
subjected to wind pressure
In calculations, it will be assumed
that ice is uniformly on the surface of
the conductor.
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Ice Loading
Cross-section area of conductor = π d2 / 4
Cross-section area of ice coated conductor = π (d + 2t)2 / 4
Cross-section area of ice
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Ice Loading
If density of ice = ρi
Weight of ice Wi = Ai х ρi
Total weight of conductor per unit length WT = W + Wi
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Combined effect of wind & ice
Projected area of the conductor = (D+ 2t) x 1
ww = (D + 2t)ρ kg/m
Where ρ is the wind pressure per unit area acting in a
direction normal to the direction of span
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Effective Loading
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Assignment



An ACSR conductor has the following data: normal
copper area = 120 mm2 size = (30 + 7)16.30 mm;
weight = 0.4 kg/rn, tensile strength = 1250 kg,
safety factor = 5. If span length is 200 m, find
Sag in still air
Sag, if the conductor is covered with 0.5-cm thick ice
(ice density of 915 kg/rn3)
Sag (total and vertical), if the conductor is covered
with ice of 0.5-cm thickness and a wind pressure of
10 kg/rn2 is acting on the projected area.
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Solution…
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Continued…
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Continued…
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Continued…
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