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T/F: maximum sag in stringing condition

Short answer

FALSE. Stringing sag is smaller than the sag in the easiest condition because the conductor is at a lower temperature (around 25°C) than the highest ambient, so tension is higher and sag is lower. It lies between the toughest and easiest extremes.

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: Maximum sag in stringing condition is greater than in toughest and easiest conditions.

FALSE.

The sag at stringing lies between the sags of the two extreme design conditions. It is never larger than both of them.

  • Toughest condition (lowest temperature, maximum wind, ice if any): tension is at its maximum, T=UTS/FST = \text{UTS}/\text{FS}. The effective weight W1=(wc+wi)2+ww2W_1=\sqrt{(w_c+w_i)^2+w_w^2} is also at its maximum. The resulting sag is inclined and is used for checking strength.
  • Easiest condition (highest temperature, still air, no ice): the conductor is at its longest and tension is lowest. This gives the maximum vertical sag, which fixes the ground clearance and the tower height.
  • Stringing condition (normal erection temperature, about 20–30°C, no wind, no ice): the temperature is below the easiest-condition temperature. The conductor is therefore shorter and tighter, so
Dstringing=wcL28Tstr<Deasiest=wcL28TeasiestD_{\text{stringing}} = \frac{w_c L^2}{8T_{\text{str}}} < D_{\text{easiest}} = \frac{w_c L^2}{8T_{\text{easiest}}}

because Tstr>TeasiestT_{\text{str}} > T_{\text{easiest}} for the same weight wcw_c.

Example: for a 300 m span of a 462.6 mm² ACSR conductor, the toughest-condition sag is about 9.9 m and the stringing sag at 25°C is about 8.8 m. So the stringing sag is the smaller one.

The stringing sag is found from the change-of-state (stringing) equation, starting from the toughest condition. The conductor is pulled to this sag during erection so that the line meets its limits at the two extremes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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