Force on angle tower: 2T sin(α/2)

At an angle tower the line changes direction by an angle α. The conductor tensions on the two sides no longer cancel, and their resultant acts along the bisector of the angle.

Proof

Let the original line be along the x-axis. Span 1 pulls the tower backward along the original direction. Span 2 pulls forward along a direction turned by α.

       R (along bisector)
        ^          T (span 2)
         \        /
          \      /
           \    /  alpha
            \  /
  T <--------O - - - - - - -> original direction
  (span 1)

Resolve the two pulls, each of magnitude $T$:

• Span 1: $(-T,\ 0)$
• Span 2: $(T\cos\alpha,\ T\sin\alpha)$

The resultant components are:

$$\begin{aligned} R_x &= T\cos\alpha - T = -T(1-\cos\alpha) = -2T\sin^2\frac{\alpha}{2}\\ R_y &= T\sin\alpha = 2T\sin\frac{\alpha}{2}\cos\frac{\alpha}{2} \end{aligned}$$

Magnitude:

$$\begin{aligned} R &= \sqrt{R_x^2 + R_y^2} = 2T\sin\frac{\alpha}{2}\sqrt{\sin^2\frac{\alpha}{2} + \cos^2\frac{\alpha}{2}}\\ &= 2T\sin\frac{\alpha}{2} \end{aligned}$$

Direction: $\tan\phi = R_y/|R_x| = \cot(\alpha/2)$. So $R$ acts along the bisector of the inner angle between the two spans, pulling the tower towards the inside of the bend.

$$\boxed{F = 2T\sin\frac{\alpha}{2}}$$

Use in design

• This force acts on every conductor and earth wire at its point of attachment. The bending moment at the base is $\sum 2T_i\sin(\alpha/2)\,h_i$, plus the wind load.
• Example: $T = 6000$ kg and $\alpha = 30^\circ$ give $F = 12000\sin15^\circ = 3106$ kg per conductor. This is about 2.6 times the wind load on a 300 m span of a 4 cm conductor at 100 kg/m² ($100\times0.04\times300 = 1200$ kg). This is why angle towers (B, C, D types) are much heavier than tangent towers, and why tower types are classified by deviation angle (0–2°, 2–15°, 15–30°, 30–60°).