Question

Two towers of height 40 m and 30 m respectively support a transmission line conductor at water crossing. The horizontal distance between the towers is 300 m. If the tension in the conductor is 1590 kg, find the clearance of the conductor at a point mid-way between the supports. Weight of conductor is 0·8 kg/m. Bases of the towers can be considered to be at the water level.

Answer 1

The sag in the tensioned cable depends on the weight of the cable and

the applied tension.

• The clearance midway between the support is approximately 29.34 meters

Reasons:

The given parameters are;

Height of the towers are; h₁ = 40 m and h₂ = 30 m

Horizontal distance between the towers, l = 300 m

The tension in the conductor, T = 1,590 kg

Weight of the conductor, w = 0.8 kg/m

Required:

The clearance space of the conductor mid-way between the support

Solution:

The distance, x₁, to the lowest point on the conductor, from the tower, h₁, is given by the formula;

$\displaystyle x_1 = \mathbf{ \frac{l}{2} -\frac{T \cdot h}{w \cdot l}}$

Where;

h = h₂ - h₁ = 40 m - 30 m = 10 m

Which gives;

$\displaystyle x_1 = \frac{300}{2} -\frac{1590 \times 10}{0.8 \times 300} = \mathbf{83.75}$

The sag at the lowest point is give by the formula;

$\displaystyle d_1 = \mathbf{ \frac{W \cdot x_1^2}{2 \cdot T}}$

Therefore;

$\displaystyle d_1 = \frac{0.8 \times 83.75^2}{2 \times 1590} = \frac{4489}{2544} \approx 1.765$

The lowest point sag, s₁ = 30 m - 1.765 m = 28.235 m

The distance of the midpoint from the lowest point, x, is therefore;

$\displaystyle x = \frac{l}{2} - x_1 = \frac{300}{2} - 83.75 = \mathbf{ 66.25}$

Which gives;

$\displaystyle d_x = \frac{0.8 \times 66.25^2}{2 \times 1590} = \frac{53}{48} \approx 1.104$

The clearance midway between the support ≈ 28.235 + 1.104 ≈ 29.34

The clearance midway between the support ≈ 29.34 m

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