Civil Engineering - Surveying - Discussion
Discussion Forum : Surveying - Section 2 (Q.No. 24)
24.
A tape of length l and weight W kg/m is suspended at its ends with a pull of P kg, the sag correction is
.Discussion:
31 comments Page 1 of 4.
Gokul said:
6 years ago
W=udl.
So, W=wl,
= l*(wl)^2/24p^2.
sag correction = l^3w^2/ 24p^2.
So, W=wl,
= l*(wl)^2/24p^2.
sag correction = l^3w^2/ 24p^2.
(3)
Sarvesh Prajapati said:
8 years ago
Here -> (L*w2)÷(24p2).
Raya said:
3 years ago
{L(WxW)}/{24(nxn)(PxP)}.
Madan said:
3 years ago
Agree, A is the correct answer.
GIRIJA PRASAD said:
5 years ago
Option A is the right answer.
As w is given in kg / m then the above formula is correct.
But the formula will change whenever W becomes TOTAL WEIGHT. Then the formula.
= (W^2*L) ÷ 24p^2.
As w is given in kg / m then the above formula is correct.
But the formula will change whenever W becomes TOTAL WEIGHT. Then the formula.
= (W^2*L) ÷ 24p^2.
Lohith GC said:
5 years ago
The Correct answer is W^2L/24P^2.
Orack said:
7 years ago
Answer A is correct because weight is taken as w kg/m length of tape.
So W=wl.
Correction for sag is l/24*(W/P)^2.
l/24*(wl/P)^2.
So W=wl.
Correction for sag is l/24*(W/P)^2.
l/24*(wl/P)^2.
Kibs said:
7 years ago
Sag correction = L(W)^2/24(P)^2.
Chaugule B M said:
7 years ago
Sag correction = W^2l/24p^2.
Prashanth said:
7 years ago
Correction due to sag (same in levels).
C = w^2 * l/24 * n^2 * p^2.
Where,
W- Total weight of the chain,
n- Number of supports.
If 1 support n=2.
If 2 support n=3.
Correction due to sag (Different levels).
C = Csag*cos^2 &teeta;
C = w^2 * l/24 * n^2 * p^2.
Where,
W- Total weight of the chain,
n- Number of supports.
If 1 support n=2.
If 2 support n=3.
Correction due to sag (Different levels).
C = Csag*cos^2 &teeta;
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