--- title: "Derivation of stringing equation" subject: "Transmission and Distribution System Design" chapter: 4 key: "semester_8/transmission_and_distribution_system_design:stringing-equation-derivation" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/stringing-equation-derivation site: BCT Engineering --- # Derivation of stringing equation Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection - Asked in: 2070 Magh, 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/stringing-equation-derivation - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4 ## Question Using the equation of maximum sag & conductor length directly derive the stringing equation. ## Short answer Equating the change in conductor length from sag (L + W^2 L^3/(24 T^2)) to the change from thermal expansion and elastic strain yields the stringing equation: T2^2 [T2 + alpha*(theta2 - theta1)*A*E + (W1^2 L^2 A*E)/(24 T1^2) - T1] = (W2^2 L^2 A*E)/24. Solving gives erection tension T2. ## Answer The stringing equation relates the tension $T_2$ at erection (stringing) to the known tension $T_1$ in the toughest condition. It uses the sag and conductor-length formulas of a parabolic span. **Symbols:** span $L$; area $A$; modulus $E$; coefficient of expansion $\alpha$. Condition 1 (toughest) has $T_1, W_1, \theta_1$. Condition 2 (stringing) has $T_2, W_2, \theta_2$. #### Step 1 – maximum sag and conductor length For a parabolic span with level supports: $$ D = \frac{WL^2}{8T}, \qquad l = L + \frac{8D^2}{3L} $$ Substituting D: $$ l = L + \frac{8}{3L}\cdot\frac{W^2L^4}{64T^2} = L + \frac{W^2L^3}{24T^2} $$ Hence $$ l_1 = L + \frac{W_1^2L^3}{24T_1^2}, \qquad l_2 = L + \frac{W_2^2L^3}{24T_2^2} $$ #### Step 2 – change in length from physics Between the two conditions, the conductor length changes by expansion and by elastic strain: $$ l_2 - l_1 = \alpha(\theta_2-\theta_1)L + \frac{T_2-T_1}{AE}L \qquad (l_1 \approx L) $$ #### Step 3 – equate and simplify $$ \frac{W_2^2L^3}{24T_2^2} - \frac{W_1^2L^3}{24T_1^2} = \alpha(\theta_2-\theta_1)L + \frac{(T_2-T_1)L}{AE} $$ Multiply by $AE/L$: $$ \frac{W_2^2L^2AE}{24T_2^2} = T_2 - T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2} $$ Multiply by $T_2^2$: $$ T_2^2\left[T_2 - T_1 + \alpha(\theta_2-\theta_1)AE + \frac{W_1^2L^2AE}{24T_1^2}\right] = \frac{W_2^2L^2AE}{24} $$ or $$ T_2^2\,(T_2 + K_2) - K_1 = 0 $$ with $K_2 = -T_1 + \alpha(\theta_2-\theta_1)AE + \dfrac{W_1^2L^2AE}{24T_1^2}$ and $K_1 = \dfrac{W_2^2L^2AE}{24}$. This is the **stringing equation**. With $T_1 = \text{UTS}/\text{FS}$ known, the cubic gives $T_2$, and the stringing sag is $D_2 = W_2L^2/(8T_2)$. When $\theta_2$ is the maximum temperature and $W_2$ is the bare conductor weight, the same equation gives the maximum sag in the easiest condition. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What expression for conductor length is used? A. l = L + D^2/L B. l = L + 8D^2/(3L) with D = WL^2/(8T), giving l = L + W^2L^3/(24T^2) C. l = L + WL/(8T) D. l = L + 3D^2/(8L) Answer: B. Parabolic span formula. (Key point: Sag and length) 2. What gives the physical change l2 - l1? A. Only thermal expansion B. Only elastic strain C. Thermal expansion alpha(theta2-theta1)L plus elastic strain (T2-T1)L/(AE) D. Wind and ice weights Answer: C. Both temperature and tension change alter the length. (Key point: Physical change) 3. What is the final stringing equation? A. T2(T2 + K2) = K1 B. T2^2 = K1 K2 C. T2^3 = K1 - K2 D. T2^2(T2 + K2) = K1, with K1 = W2^2L^2AE/24 Answer: D. K2 = -T1 + alpha(theta2-theta1)AE + W1^2L^2AE/(24T1^2). (Key point: Final form) 4. Which tension is known to start with? A. T1 = UTS/FS in the toughest condition B. T2 at stringing C. Tension at maximum temperature D. Average of T1 and T2 Answer: A. Then T2 is solved and D2 = W2L^2/(8T2). 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