Question

Calculate the sag correction for a 30m steel under a pull of 100N in a three span of 10m each. Weight of steel tape per meter run = 0.624 N

Answer 1

123×234×234+345÷134÷2234÷233+334-2343344= 0

Answer 2

Answer:

The sag correction of the steel tape, $C_{sag}$, calculated is $4.8\times 10^{-5} m$.

Step-by-step explanation:

Given data,

The length of the steel tape, L = $30m$

The pull applied on the steel tape, P = $100N$

The weight of the steel tape per meter run, W = $0.624N$

The sag correction of the steel tape, $C_{sag}$ =?

From the formula of sag correction given below, we can find out the sag correction of the steel tape:

• $C_{sag} = \frac{W^{2}L }{24P^{2} }$

After putting the given values of length, weight, and pull in the equation, we get:

• $C_{sag} = \frac{(0.624)^{2}\times 30 }{24(100)^{2} }$
• $C_{sag} = \frac{0.389\times 30 }{24\times 10000 }$
• $C_{sag} = \frac{0.389\times 3 }{24\times 1000 }$
• $C_{sag} = 0.048\times 10^{-3}$ or $4.8\times 10^{-5} m$

Hence, the sag correction of the steel tape, $C_{sag}$ = $4.8\times 10^{-5} m$.