It is required to construct a pile foundation comprised of 20 piles arranged in 5 columns at distances of 90 cm center to center. The diameter and lengths of the piles are 30 cm and 9 m respectively. The bottom of the pile cap is located at a depth of 2.0 m from the ground surface. The details of

Question

It is required to construct a pile foundation comprised of 20 piles arranged in 5 columns at distances of 90 cm center to center. The diameter and lengths of the piles are 30 cm and 9 m respectively. The bottom of the pile cap is located at a depth of 2.0 m from the ground surface. The details of the soil properties etc. are as given below with reference to ground level as the datum. The water table was found at a depth of 4 m from ground level.

Drpth (m)		Soil properties
From	To
0	2	Silt, saturated, [latex]\gamma=16 kN / m ^{3}[/latex]
2	4	Clay, saturated, [latex]\gamma=19.2 kN / m ^{3}[/latex]
4	12	Clay, saturated, [latex]\gamma=19.2 kN / m ^{3}, q_{u}=120 kN / m ^{2}, e_{0}=0.80, C_{c}=0.23[/latex]
12	14	Clay, [latex]\gamma=18.24 kN / m ^{3}, q_{u}=90 kN / m ^{2}, e_{0}=1.08, C_{c}=0.34[/latex].
14	17	Clay [latex]\gamma=20 kN / m ^{3}, q_{u}=180 kN / m ^{2}, e_{0}=0.70, C_{c}=0.2[/latex]
17	-	Rocky stratum

Compute the consolidation settlement of the pile foundation if the total load imposed on the foundation is 2500 kN.

Accepted Answer

Assume that the total load 2500 kN acts at a depth (2/3 )L = (2/3) x 9 = 6 m from the bottom of the pile cap on a fictitious footing as shown in Fig. 15. 31 (a). This fictitious footing is now at a depth of 8 m below ground level. The size of the footing is 3.9 x 3.0 m. Now three layers are assumed to contribute to the settlement of the foundation. They are: Layer 1 — from 8m to 12m (= 4m thick) below ground level; Layer 2 — from 12mtol4m = 2m thick; Layer 3 — from 14 m to 17 m = 3 m thick. The increase in pressure due to the load on the fictitious footing at the centers of each layer is computed on the assumption that the load is spread at an angle of 2 vertical to 1 horizontal [Fig. 15.31(a)] starting from the edges of the fictitious footing. The settlement is computed by making use of the equation

[latex]S_{t}=H_{i} \frac{C_{c}}{1+e_{o}} \log \frac{p_{o}+\Delta p}{p_{o}}[/latex]

where [latex]p_{o}=[/latex] the effective overburden pressure at the middle of each layer,

[latex]\Delta p=[/latex] the increase in pressure at the middle of each layer

Computation of [latex]p_{o}[/latex]

For Layer 1, [latex]p_{o}=2 imes 16+2 imes 19.2+(10-4)(19.2-9.81)=126.74 kN / m ^{2}[/latex]

For Layer 2, [latex]p_{o}=126.74+2(19.2-9.81)+1 imes(18.24-9.81)=153.95 kN / m ^{2}[/latex]

For Layer 3, [latex]p_{o}=153.95+1(18.24-9.81)+1.5 imes(20.0-9.81)=177.67 kN / m ^{2}[/latex]

Computation of [latex]\Delta p=[/latex]

For Layer 1

Area at 2 m depth below fictitious footing [latex]=(3.9+2) imes(3+2)=29.5 m ^{2}[/latex]

[latex]\Delta p=\frac{2500}{29.5}=84.75 kN / m ^{2}[/latex]

For Layer 2

Area at 5 m depth below fictitious footing [latex]=(3.9+5) imes(3+5)=71.2 m ^{2}[/latex]

[latex]\Delta p=\frac{2500}{71.2}=35.1 kN / m ^{2}[/latex]

For Layer 3

Area at 7.5 m below fictitious footing [latex]=(3.9+7.5) imes(3+7.5)=119.7 m ^{2}[/latex]

[latex]\Delta p=\frac{2500}{119.7}=20.9 kN / m ^{2}[/latex]

Settlement computation

Layer 1 [latex]S_{1}=\frac{4 imes 0.23}{1+0.80} \log \frac{126.74+84.75}{126.74}=0.113 m[/latex]

Layer 2 [latex]S_{2}=\frac{2 imes 0.34}{1+1.08} \log \frac{153.95+35.1}{153.95}=0.029 m[/latex]

Layer 3 [latex]S_{3}=\frac{3 imes 0.2}{1+0.7} \log \frac{177.67+20.9}{177.67}=0.017 m[/latex]

Total [latex]=0.159 m \approx 16 cm[/latex].

Question 15.27: It is required to construct a pile foundation comprised of 2...

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