Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 6 (Q.No. 15)
15.
The sag of 50 m tape weighing 4 kg under 5 kg tension is roughly
Discussion:
24 comments Page 1 of 3.
Ahmed Mohamed Adan said:
7 years ago
Given:
(W)^2=(4kg )^2 = 16kg.
L=50.
p=5^2=25.
Sag correction = 16*50/24*25.
800/600 = -1.33ft.
Note: Sag correction is Always negative.
(W)^2=(4kg )^2 = 16kg.
L=50.
p=5^2=25.
Sag correction = 16*50/24*25.
800/600 = -1.33ft.
Note: Sag correction is Always negative.
(3)
Suhasish Munshi said:
7 years ago
The correct steps:
(W^2 * L)÷(24 *P^2).
W=weight of tape in kg/m,
L=length of tape in m,
P= pull applied or tension in kg,
So, W= 5kg/50m= 0.1kg/m.
L=50m and P=5kg.
(W^2 * L)÷(24 *P^2).
W=weight of tape in kg/m,
L=length of tape in m,
P= pull applied or tension in kg,
So, W= 5kg/50m= 0.1kg/m.
L=50m and P=5kg.
(2)
Prazwol said:
10 months ago
(W^2×l)/24 P^2.
W = weight of tape
P = tension
So, sag correction = (4^2× 50)/ (24 × 5^20).
= 1.333.
W = weight of tape
P = tension
So, sag correction = (4^2× 50)/ (24 × 5^20).
= 1.333.
(2)
Johny Chauhan said:
6 years ago
.083 m is not the correct answer for given data.
Here, Given data will be same except W= 1kg
So, ( 1kg)^2 * (50 m) /[ 24 * ( 5kg)^2]= .083 m.
Here, Given data will be same except W= 1kg
So, ( 1kg)^2 * (50 m) /[ 24 * ( 5kg)^2]= .083 m.
(2)
Rishabh purohit said:
6 years ago
Correction for sag = W^2*L / 24*P^2 , where w is in kg,N,kN etc.
Correction for sag = w^2*L^3 / 24*P^2 , where w is in kg/m, N/m etc.
Correction for sag = w^2*L^3 / 24*P^2 , where w is in kg/m, N/m etc.
(2)
Vintage said:
3 years ago
50/ 24*5^2 = 0.0833.
(1)
Manish said:
5 years ago
You are correct, Thanks @Johny.
(1)
Vaish said:
6 years ago
((4/50)^2 * 50)/(24*5^2) = 0.0833m.
Y (4/50)----->w is the weight per meter length.
Y (4/50)----->w is the weight per meter length.
(1)
Raval chaitanya said:
8 years ago
@Adi
Formula = l3w2/24p2.
Formula = l3w2/24p2.
Sandeep said:
9 years ago
Please give the formula, to solve the question.
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