A transmission line has a span of 214 m between level supports. The conductos have a cross-sectional area of 3.225 cm². Calculate the factor of safety under the following

conditions;

Vertical sag 2.35m

Breaking stress = 2540 kg/cm²

Weight of conductor = 1.125 kg/m

Wind pressure = 1.50 kg/m

Cross-sectional area of conductor

=3.225 cm²

Vartical Sag = 2.35 m

Weight of conductor per meter meter run = 1.125 kg/m

Wind pressure per meter run, $W_w$ = 1.5kg / m

Breeaking strss=2540kg/cm²

L = 214m

$W_t=\sqrt{W^2+W_W^2}$

$=\sqrt{1.{125}^2+1.5^2}$ = 1.875kg

যদি f সেফটি ফ্যাক্টর হয়, তাহলে, ওয়ার্কিং টেনশন

$T=\frac{Breaking stress \times conductor area}{safety factor}$

$=\frac{2540 \times 3.225}{f}$

$Cos\theta =\frac{1.125}{1.875}=0.6$

Slant Sag, $S=\frac{Vertical Sag}{Cos \theta }=\frac{2.35}{0.6}=3.92 m$

We know,

$Sag=\frac{W_t{I}^2}{8T}$

$=\frac{1.875\times {214}^2}{8\times 3.92}$

= 2738 kg

Safety factor, $f=\frac{Breaking stress}{Working stress}=\frac{8191}{2738}$ = 3 Ans.