--- title: "Bending moment ratio, 132 kV vs 220 kV towers" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:bending-moment-ratio-132-220kv" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/bending-moment-ratio-132-220kv site: BCT Engineering --- # Bending moment ratio, 132 kV vs 220 kV towers Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2068 Bhadra (old course), 20 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/bending-moment-ratio-132-220kv - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Compute the ratio of bending moment acting on transmission tower of a 132 kV and 220 kV single circuit transmission lines of span 300 m with DEER conductor. Use data from Appendix and assume any suitable data if necessary. [The appendix is not reproduced with this paper.] ## Short answer With DEER conductor, 300 m span, and appropriate tower geometries, base bending moment is 44,903 kg·m for 132 kV and 53,833 kg·m for 220 kV. The ratio (132/220 kV) is 0.834. ## Answer The bending moment at the tower base is the sum of each transverse load times its height. Both lines use the same conductor and span, so they have the same sag and the same wind load per wire. The ratio therefore comes from the **tower geometry**: ground clearance, conductor spacing and earth wire height, all of which grow with voltage. #### Data and assumptions | Item | Value | |---|---| | DEER (ACSR 30/7/4.27 mm) | d = 29.89 mm, A = 529.8 mm², w = 1.9714 kg/m, UTS = 179 kN = 18,253 kg | | E, α (30/7 ACSR) | $0.789\times10^6$ kg/cm², $17.73\times10^{-6}$ /°C | | Toughest condition | 0°C, wind 100 kg/m², no ice, $T_1 = \text{UTS}/2$ | | Easiest condition | 60°C, no wind | | Earth wire | 7/3.66 mm GS wire, d = 10.98 mm, one per tower | | Air clearance | 6.5 inch per 10 kV (phase rms of max voltage) + 8 inch | | Insulator string | 254 × 146 mm discs (9 for 132 kV, 14 for 220 kV) + 0.3 m hardware | | String swing | 45°; tower width at cross arm 1.5 m | | Ground clearance | $H_g = 17 + (V-33)/33$ ft | | Towers | Suspension (straight-line); wind span = 300 m | | Conductors | Three levels, one conductor per level | #### Step 1 – maximum sag (same for both lines) $$ \begin{aligned} T_1 &= 18253/2 = 9126.5\ \text{kg}\\ W_1 &= \sqrt{1.9714^2 + (100\times0.02989)^2} = 3.581\ \text{kg/m}\\ AE &= 5.298\times0.789\times10^6 = 4.180\times10^6\ \text{kg}\\ K_2 &= -9126.5 + 17.73\times10^{-6}\times60\times AE + \frac{W_1^2L^2AE}{24T_1^2}\\ &= -9126.5 + 4446.8 + 2412.8 = -2266.8\\ K_1 &= \frac{1.9714^2\times300^2\times AE}{24} = 6.092\times10^{10} \end{aligned} $$ Solving $T_2^2(T_2 - 2266.8) = 6.092\times10^{10}$ gives $T_2 = 4853$ kg. Then $$ D_{max} = \frac{1.9714\times300^2}{8\times4853} = 4.57\ \text{m} $$ #### Step 2 – clearances and tower geometry | Quantity | Formula | 132 kV | 220 kV | |---|---|---|---| | Max phase voltage | $V_m/\sqrt3$ | 83.72 kV | 141.45 kV | | Air clearance a | $6.5V_{ph}/10 + 8$ in | 1.585 m | 2.539 m | | String length l | $n\times0.146 + 0.3$ | 1.614 m | 2.344 m | | Vertical spacing Y | $l + a$ | 3.199 m | 4.883 m | | Cross arm X | $a + l\sin45^\circ$ | 2.727 m | 4.196 m | | Earth wire height $h_e$ above top conductor | $(X + 0.75)/\tan30^\circ$ | 6.022 m | 8.567 m | | Ground clearance $H_g$ | $17 + (V-33)/33$ ft | 20 ft = 6.096 m | 22.67 ft = 6.909 m | Conductor heights at the tower: | Height | 132 kV (m) | 220 kV (m) | |---|---|---| | $h_1 = H_g + D_{max}$ | 10.666 | 11.479 | | $h_2 = h_1 + Y$ | 13.865 | 16.361 | | $h_3 = h_1 + 2Y$ | 17.064 | 21.244 | | $H_t = h_3 + h_e$ | 23.086 | 29.810 | #### Step 3 – transverse loads (wind span 300 m) $$ \begin{aligned} F_c &= p\,d\,L = 100\times0.02989\times300 = 896.7\ \text{kg per conductor}\\ F_e &= 100\times0.01098\times300 = 329.4\ \text{kg (earth wire)} \end{aligned} $$ #### Step 4 – bending moment at the base $$ M = F_c(h_1+h_2+h_3) + F_e H_t $$ $$ \begin{aligned} M_{132} &= 896.7(10.666+13.865+17.064) + 329.4(23.086)\\ &= 37{,}298 + 7{,}605 = 44{,}903\ \text{kg·m}\\ M_{220} &= 896.7(11.479+16.361+21.244) + 329.4(29.810)\\ &= 44{,}013 + 9{,}820 = 53{,}833\ \text{kg·m} \end{aligned} $$ #### Step 5 – ratio $$ \frac{M_{132}}{M_{220}} = \frac{44{,}903}{53{,}833} = 0.834 $$ ```text 220 kV 132 kV E 29.8 m E 23.1 m | | --* 21.2 m --* 17.1 m *-- 16.4 m *-- 13.9 m --* 11.5 m --* 10.7 m ===== ground ===== ground ``` **Answer: the bending moment is about 44.9 t·m for the 132 kV tower and 53.8 t·m for the 220 kV tower, a ratio of M(132) : M(220) ≈ 0.83 : 1. The loads per wire are the same, so the 220 kV tower carries about 20% more moment only because it is taller, through larger clearances and spacing.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Both lines use DEER conductor and 300 m span. Why does the moment differ? A. The wind load per wire is greater at 220 kV B. The tower geometry (clearances, spacing, heights) grows with voltage C. The sag is larger at 220 kV D. The conductor tension is higher at 220 kV Answer: B. Sag and wind load per wire are the same; only tower heights differ through larger clearances and spacing. (Key point: Source of the ratio) 2. How is the base bending moment found here? A. Sum of each transverse load times its height above the base B. Total load times the tower width C. Conductor weight times the span D. Sum of loads divided by the heights Answer: A. M = Fc(h1+h2+h3) + Fe*Ht, the sum of load times height for conductors and earth wire. (Key point: Bending moment formula) 3. For the sag, which condition gives T1 and the loading used in the change-of-state equation? A. Easiest, 60°C no wind, T1 = UTS/2 B. Toughest, 0°C with 100 kg/m² wind, T1 = UTS/2 C. Toughest, 0°C no wind, T1 = UTS D. Easiest, 0°C with ice Answer: B. The toughest condition (0°C, wind 100 kg/m², T1 = UTS/2) is the start state; the maximum sag is then found at 60°C. (Key point: Maximum sag basis) 4. What is the resulting ratio M(132):M(220), and why is it below 1? A. About 1.20, because 132 kV has more load B. About 1.00, because loads are equal C. About 0.83, because the 220 kV tower is taller with the same wire loads D. About 0.50, because 220 kV has twice the wires Answer: C. 44,903 / 53,833 = 0.834; the 220 kV tower is taller, so same loads act at greater heights. 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