--- title: "Force on angle tower: 2T sin(α/2)" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:angle-tower-force-2t-sin" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/angle-tower-force-2t-sin site: BCT Engineering --- # Force on angle tower: 2T sin(α/2) Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2067 Chaitra (old course), 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/angle-tower-force-2t-sin - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Show that the force acting on a angle tower due to horizontal turning of the conductor by an angle of α with conductor tension T could be expressed as 2T sin α/2. ## Short answer Resolving the two conductor tensions T into components, the resultant force magnitude is √[(Tcosα – T)² + (Tsinα)²] = 2T sin(α/2). The force acts along the bisector of the turn angle, pulling the tower inward. ## Answer 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. ```text 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°). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. At an angle tower, why is there a net force? A. Conductor weight differs B. The tensions of the two spans act in different directions and do not cancel C. Wind acts on the tower D. The tower tilts Answer: B. The spans pull at an angle alpha, so their vector sum is non-zero. (Key point: Setup) 2. With span 1 as (-T, 0), what is span 2? A. (T cos(alpha), T sin(alpha)) B. (T sin(alpha), T cos(alpha)) C. (T, T) D. (T cos(alpha/2), 0) Answer: A. Span 2 is turned by alpha from the original direction. (Key point: Components) 3. Which identity reduces Rx and Ry to R = 2T sin(alpha/2)? A. tan(alpha) = sin/cos only B. 1 - cos(alpha) = 2 sin^2(alpha/2) and sin(alpha) = 2 sin(alpha/2)cos(alpha/2) C. cos(alpha) = 1 - sin(alpha) D. sin(2 alpha) = 2 sin(alpha) Answer: B. Rx = -2T sin^2(alpha/2), Ry = 2T sin(alpha/2)cos(alpha/2); the sum of squares leaves 2T sin(alpha/2). (Key point: Simplification) 4. Where does R act? A. Along the original line B. Perpendicular to the original line C. Along the bisector of the inner angle, pulling towards the inside of the bend D. Outward from the bend Answer: C. tan(phi) = cot(alpha/2); the resultant lies on the bisector towards the inside. 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