Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 11)
11.
A cable loaded with 0.5 tonne per horizontal metre span is stretched between supports in the same horizontal line 400 m apart. If central dip is 20 m, the minimum tension in the cable, will be
Discussion:
15 comments Page 2 of 2.
AMOL BHALERAO said:
7 years ago
(Wl^2)/(8*span).
(0.5*400^2)/(8*20) = 500.
(0.5*400^2)/(8*20) = 500.
Zeba khan said:
6 years ago
What is the maximum pull in the cable?
Divyesh patel said:
1 decade ago
I want to know how? Please reply.
Pushkar jha said:
6 years ago
Central dip h = (wl^2)/8P.
(1)
Shobhit pal said:
5 years ago
Center of dip = (wl'2/8p).
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