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Conductor tension in stringing condition

Short answer

Using the stringing equation, the stringing tension at 25°C without wind/ice is found from the toughest condition (UTS/2.5=6364 kg, wind+ice). The required tension is 2211 kg, giving a sag of about 8.78 m.

Taking the factor of safety under toughest condition to be 2.5, compute the tension in the conductor to be given in the conductor for stringing condition for a span of 300 m. Assume the ultimate tensile strength of a conductor is 15910 kg and factor of safety of 2.5. Use the following data: The conductor specifications are: Area = 462.60 mm², Linear expansion coefficient α = 17.73×10⁻⁶/°C, Modulus of elasticity (E) = 0.789×10⁶ kg/cm², Weight (Wc) = 1726 kg/km, Radius = 1.40 cm.

ConditionTemperatureWind pressureIce loading (ice density 950 kg/m³)
Toughest condition−5°C100 kg/m²1 cm
Stringing condition25°C--

[Given stringing equation: T2²[T2 + K2] − K1 = 0, where K2 = −T1 + α(θ2 − θ1)AE + W1²L²AE/(24T1²) and K1 = W2²L²AE/24.]

The stringing tension is found by solving the given stringing equation from the toughest condition (−5°C, wind and ice) to the stringing condition (25°C, still air).

Toughest condition (condition 1)

Allowable tension:

T1=UTSFS=159102.5=6364 kgT_1 = \frac{\text{UTS}}{FS} = \frac{15910}{2.5} = 6364\ \text{kg}

Conductor diameter d=2×1.40=2.8d = 2\times1.40 = 2.8 cm; ice thickness t=1t = 1 cm.

wc=1726 kg/km=1.726 kg/mwi=ρ πt(d+t)=950×π×0.01×(0.028+0.01)=1.134 kg/mww=p(d+2t)=100×(0.028+0.02)=4.80 kg/mW1=(1.726+1.134)2+4.802=5.588 kg/m\begin{aligned} w_c &= 1726\ \text{kg/km} = 1.726\ \text{kg/m}\\ w_i &= \rho\,\pi t(d+t) = 950\times\pi\times0.01\times(0.028+0.01) = 1.134\ \text{kg/m}\\ w_w &= p(d+2t) = 100\times(0.028+0.02) = 4.80\ \text{kg/m}\\ W_1 &= \sqrt{(1.726+1.134)^2 + 4.80^2} = 5.588\ \text{kg/m} \end{aligned}

Stringing condition (condition 2)

W2=wc=1.726W_2 = w_c = 1.726 kg/m, θ2−θ1=25−(−5)=30∘\theta_2 - \theta_1 = 25-(-5) = 30^\circC.

AE=4.626 cm2×0.789×106 kg/cm2=3.650×106 kgAE = 4.626\ \text{cm}^2\times0.789\times10^6\ \text{kg/cm}^2 = 3.650\times10^6\ \text{kg}

Constants

K2=−T1+α(θ2−θ1)AE+W12L2AE24T12=−6364+17.73×10−6×30×3.650×106+5.5882×3002×3.650×10624×63642=−6364+1941.4+10551.0=6128.3 kgK1=W22L2AE24=1.7262×3002×3.650×10624=4.0775×1010 kg3\begin{aligned} K_2 &= -T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2}\\ &= -6364 + 17.73\times10^{-6}\times30\times3.650\times10^6 + \frac{5.588^2\times300^2\times3.650\times10^6}{24\times6364^2}\\ &= -6364 + 1941.4 + 10551.0 = 6128.3\ \text{kg}\\ K_1 &= \frac{W_2^2L^2AE}{24} = \frac{1.726^2\times300^2\times3.650\times10^6}{24} = 4.0775\times10^{10}\ \text{kg}^3 \end{aligned}

Solving T22(T2+6128.3)=4.0775×1010T_2^2(T_2 + 6128.3) = 4.0775\times10^{10}

By trial (or Newton's method):

T2T_2 (kg)T22(T2+K2)T_2^2(T_2+K_2)
20003.25×10103.25\times10^{10} (too low)
22004.03×10104.03\times10^{10} (slightly low)
2211.24.0775×10104.0775\times10^{10} (matches)
T2=2211 kgT_2 = 2211\ \text{kg}

Check: the stringing sag is D=1.726×30028×2211.2=8.78D = \dfrac{1.726\times300^2}{8\times2211.2} = 8.78 m. The sag in the toughest condition is 5.588×30028×6364=9.88\dfrac{5.588\times300^2}{8\times6364} = 9.88 m.

Answer: the conductor should be strung with a tension of about 2211 kg (about 13.9% of UTS) at 25°C, giving a stringing sag of about 8.78 m on the 300 m span.

✓ Answer checked by Claude Opus 5.5 (high effort)

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