Skip to main content

Bending moment ratio, 132 kV vs 220 kV towers

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.

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.]

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

ItemValue
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)0.789×1060.789\times10^6 kg/cm², 17.73×10−617.73\times10^{-6} /°C
Toughest condition0°C, wind 100 kg/m², no ice, T1=UTS/2T_1 = \text{UTS}/2
Easiest condition60°C, no wind
Earth wire7/3.66 mm GS wire, d = 10.98 mm, one per tower
Air clearance6.5 inch per 10 kV (phase rms of max voltage) + 8 inch
Insulator string254 × 146 mm discs (9 for 132 kV, 14 for 220 kV) + 0.3 m hardware
String swing45°; tower width at cross arm 1.5 m
Ground clearanceHg=17+(V−33)/33H_g = 17 + (V-33)/33 ft
TowersSuspension (straight-line); wind span = 300 m
ConductorsThree levels, one conductor per level

Step 1 – maximum sag (same for both lines)

T1=18253/2=9126.5 kgW1=1.97142+(100×0.02989)2=3.581 kg/mAE=5.298×0.789×106=4.180×106 kgK2=−9126.5+17.73×10−6×60×AE+W12L2AE24T12=−9126.5+4446.8+2412.8=−2266.8K1=1.97142×3002×AE24=6.092×1010\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 T22(T2−2266.8)=6.092×1010T_2^2(T_2 - 2266.8) = 6.092\times10^{10} gives T2=4853T_2 = 4853 kg. Then

Dmax=1.9714×30028×4853=4.57 mD_{max} = \frac{1.9714\times300^2}{8\times4853} = 4.57\ \text{m}

Step 2 – clearances and tower geometry

QuantityFormula132 kV220 kV
Max phase voltageVm/3V_m/\sqrt383.72 kV141.45 kV
Air clearance a6.5Vph/10+86.5V_{ph}/10 + 8 in1.585 m2.539 m
String length ln×0.146+0.3n\times0.146 + 0.31.614 m2.344 m
Vertical spacing Yl+al + a3.199 m4.883 m
Cross arm Xa+lsin⁡45∘a + l\sin45^\circ2.727 m4.196 m
Earth wire height heh_e above top conductor(X+0.75)/tan⁡30∘(X + 0.75)/\tan30^\circ6.022 m8.567 m
Ground clearance HgH_g17+(V−33)/3317 + (V-33)/33 ft20 ft = 6.096 m22.67 ft = 6.909 m

Conductor heights at the tower:

Height132 kV (m)220 kV (m)
h1=Hg+Dmaxh_1 = H_g + D_{max}10.66611.479
h2=h1+Yh_2 = h_1 + Y13.86516.361
h3=h1+2Yh_3 = h_1 + 2Y17.06421.244
Ht=h3+heH_t = h_3 + h_e23.08629.810

Step 3 – transverse loads (wind span 300 m)

Fc=p d L=100×0.02989×300=896.7 kg per conductorFe=100×0.01098×300=329.4 kg (earth wire)\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=Fc(h1+h2+h3)+FeHtM = F_c(h_1+h_2+h_3) + F_e H_t M132=896.7(10.666+13.865+17.064)+329.4(23.086)=37,298+7,605=44,903 kg⋅mM220=896.7(11.479+16.361+21.244)+329.4(29.810)=44,013+9,820=53,833 kg⋅m\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

M132M220=44,90353,833=0.834\frac{M_{132}}{M_{220}} = \frac{44{,}903}{53{,}833} = 0.834
   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.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Transmission and Distribution System Design chapter 4 questions →