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.
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.
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.
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.
Where w=load = 0.5.
l = length between supports.= 400m.
h = length of dip = 20m,
0.5 * 400^2/8 * 20 = 500 tonnes.
(5)
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.
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.
Salman said:
1 decade ago
The question is about minimum tension not maximum. If it would ask about maximum then 500 is correct.
Shobha said:
7 years ago
Minimum pull in a suspended cable with supports at two ends is equal to HORIZONTAL THRUST.
Karthik said:
4 years ago
The range of dip of the cable generally lies? Can anyone answer this?
(1)
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.
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