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
200 tonnes at the centre
500 tonnes at the centre
200 tonnes at the right support
200 tonnes at the left support.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
15 comments Page 1 of 2.

Farah Fateh said:   3 years ago
The formula for horizontal reaction at supports is wl^2/8h.

Where w=load = 0.5.
l = length between supports.= 400m.
h = length of dip = 20m,
0.5 * 400^2/8 * 20 = 500 tonnes.
(5)

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.

Karthik said:   4 years ago
The range of dip of the cable generally lies? Can anyone answer this?
(1)

Shobhit pal said:   5 years ago
Center of dip = (wl'2/8p).

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.

Pushkar jha said:   6 years ago
Central dip h = (wl^2)/8P.
(1)

Zeba khan said:   6 years ago
What is the maximum pull in the cable?

Tejas Rahane said:   6 years ago
Central dip h=(wl^2)/8P.
h= dip.
l = span.
P=pull.
w=load.

Shobha said:   7 years ago
Minimum pull in a suspended cable with supports at two ends is equal to HORIZONTAL THRUST.

AMOL BHALERAO said:   7 years ago
(Wl^2)/(8*span).

(0.5*400^2)/(8*20) = 500.


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