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.
Tejas Rahane said:
6 years ago
Central dip h=(wl^2)/8P.
h= dip.
l = span.
P=pull.
w=load.
h= dip.
l = span.
P=pull.
w=load.
Zeba khan said:
6 years ago
What is the maximum pull in the cable?
Mala said:
6 years ago
Support reactions V = (0.5x 400)/2 = 100 t.
For horizontal trust, equate Mc =0 , moment at centre.
100*200-20*H-0.5*200*100= 0,
=> H = 500.
Tmax = √ (V^2+ H^2).
Tmin = H.
At center.
Hence the answer is 500t at centre.
For horizontal trust, equate Mc =0 , moment at centre.
100*200-20*H-0.5*200*100= 0,
=> H = 500.
Tmax = √ (V^2+ H^2).
Tmin = H.
At center.
Hence the answer is 500t at centre.
Shobhit pal said:
5 years ago
Center of dip = (wl'2/8p).
Susheel kumar said:
4 years ago
Tmin = H,
Tmax = (H^2+V^2)^1/2.
H=wl2/8h.
Minimum tension will be at dip or center.
Maximum tension will be at support.
Tmax = (H^2+V^2)^1/2.
H=wl2/8h.
Minimum tension will be at dip or center.
Maximum tension will be at support.
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