A simple circuit consists of a light bulb connected to the terminals of a battery. A voltmeter shows a potential difference of 1
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Answer:
Boring depth = 6.76m
Explanation:
Given Data:
No. of columns = 30
No. of storeys = 5
Water level =3.5m below ground level
Axial load = 2500kN
Foundation size = 3mx2m
Depth of footing = 1.5m = Df
Solution:
First we need to calculate estimated net stress on footing
As we know that: q₀ = P/A
By putting values we get
q₀ = 2500/(3*2) = 416.677 kN/m³
Now for the depth of bedrock from the bottom surface of footing,
D₃ = total depth of bedrock - Df = 100 -1.5 = 98.5m
Now for the calculation of D1, here D1 is the distance from bottom surface of footing to depth where change in vertical stress = Δσ¹ = (1/10)q₀ ---------equation (1)
Now, second diagram attached below is drawn according to stress distribution of 2V:1H
Δσ¹ = P/A¹ = (2500)/(3+z)(2+z) ---------equation (2)
From equation 1 and 2 we get
0.1*416.677 = 2500/(3+D1)(2+D1)
D1²+5D1 = 54
D1 = 5.2621 m
Now we will calculate D2 where we have Δσ¹ /σ₀¹ = 0.05
Now,
we will check this ratio for bottom most layer first(i.e. clay layer, ρsat=16.9kN/m³), if it is satisfied then O.K other wise we will continue procedure for top layers.
We have, Δσ¹ = σ₀¹(0.05)
First we need to find vertical stress σ₀¹ at depth z
σ₀¹ = (3.5*17) + 1.5*(ρsat-ρw)sand + (ρsat-ρw)clay*(D2-3.5)
By putting values we get
σ₀¹ = 59.5 + 13.035 _ 7.09D2 - 28.815
Putting it in equation 2 , we get
σ₀¹ = 2500/(3+D2)(2+D2) = 0.05 * (7.09D2 + 47.72)
By simplification, we get
7.09D₂³+83.167D₂²282.147D₂-49714 = 0
D₂ = 15.35 m
So from this calculation it is evident that Δσ¹ /σ₀¹ = 0.05 is at depth of 15.35m which is in clay layer, so no need to check in upper layers.
Now, for the calculation of boring depth depth,
Boring depth = D +Df
1st case = 5.26 + 1.5 = 6.76m
2nd case = 15.35 + 1.5 = 16.85m
13rd case = 98.5 + 1.5 = 100m
Boring depth = 6.76m
