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 1 of 2.
Divyesh patel said:
1 decade ago
I want to know how? Please reply.
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.
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.
Salman said:
1 decade ago
The question is about minimum tension not maximum. If it would ask about maximum then 500 is correct.
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.
AMOL BHALERAO said:
7 years ago
(Wl^2)/(8*span).
(0.5*400^2)/(8*20) = 500.
(0.5*400^2)/(8*20) = 500.
Shobha said:
7 years ago
Minimum pull in a suspended cable with supports at two ends is equal to HORIZONTAL THRUST.
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?
Pushkar jha said:
6 years ago
Central dip h = (wl^2)/8P.
(1)
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