## Answers

Solution :- Criven that i- Vc 2om Rc re * HA А.

He ac 'AD XCO Н. W WEBORN 6 Criven, Tension in AD = Tension in CD TAD Teo RA Rc Sunt the = Juz+HE Sң со -HA & He=0 HA=He = H (Say) TV+ H² Vu+M? The Left hand side term and Right Hand Side term of above equation will be equal only when VA=Vc = 50 = 25kN t

and we know that VA and No will be canal When load. 'W' is at Cender of Span. that is at ild =lom=H= And (6) Whem, x=5m and Sag, y=im X2=20-5= 15m Evro V-50+ Vero VA TVC = 50KN EM=0 V* 20 * 20 – 50*(20-81)=0 50*120-5) = 0 20 V 50AIS 20 = 37.5KN (^) A from equation ② V=50-VA= 50-37.5=12:5KN Ve=12.5KN EH=0 -HA theo THA= H = H (say)

Moment is too we know that everywhere in the Cable Hence EMD=0 V* Hq - Henyo 12-5*15 = 187.5km He = H = 12:5*15 He=H= 187.5KN () Now, Tension to 12 + H2 112.98+187 Tco=187.916 KN C Whem, Sag, y=1m and H=5m H₂=20-5=15m Total length of Cable = length of As a length of De L E552 + 1152 12 =20.1323m) Ane Any

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