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.
Abhay said:
10 years ago
L = 400 m, δ = 20 m, W = 0.5 t.
Vertical reaction at each end = WL/2 = 0.5*400/2 = 100 t.
Horizontal thrust H = WL^2/8δ = 0.5*400^2/(8*20) = 500 t.
T = Sqrt(V^2+H^2) = sqrt(100^2+500^2) = 509.9 t.
Vertical reaction at each end = WL/2 = 0.5*400/2 = 100 t.
Horizontal thrust H = WL^2/8δ = 0.5*400^2/(8*20) = 500 t.
T = Sqrt(V^2+H^2) = sqrt(100^2+500^2) = 509.9 t.
Salman said:
1 decade ago
The question is about minimum tension not maximum. If it would ask about maximum then 500 is correct.
Sujit said:
1 decade ago
Lets l=400m, yc=20m w= 0.5t.
Vertical reaction v=wl/2.
Horizontal thrust H= WXLXL/8YC.
Tension T = Root over of VXV+HXH.
Vertical reaction v=wl/2.
Horizontal thrust H= WXLXL/8YC.
Tension T = Root over of VXV+HXH.
Jay bermudo said:
1 decade ago
As you can see, there are tonne in the question so the answer must be the greatest tonne at the center ever in the history of applied mechanics.
Divyesh patel said:
1 decade ago
I want to know how? Please reply.
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