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Applied Mechanics - Engineering

Q1:

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

A 200 tonnes at the centre

B 500 tonnes at the centre

C 200 tonnes at the right support

D 200 tonnes at the left support.

ANS:B - 500 tonnes at the centre

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. 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. 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.