A 3.0 m deep compacted fill is to be placed over the soil profile shown in Figure 11.19. A consolidation test on a sample from point A produced the following results:
$C_{c} = 0.40$

$C_{r} = 0.08$

$e_{0} = 1.10$

$σ^{\prime }_{c} = 70.0 kPa$
This sample is representative of the entire soft clay stratum. Compute the ultimate consolidation settlement due to the weight of this fill.

Question

A 3.0 m deep compacted fill is to be placed over the soil profile shown in Figure 11.19. A consolidation test on a sample from point A produced the following results:

[latex]C_{c} = 0.40[/latex]

[latex]C_{r} = 0.08[/latex]

[latex]e_{0} = 1.10[/latex]

[latex]σ^{\prime }_{c} = 70.0 kPa[/latex]

This sample is representative of the entire soft clay stratum. Compute the ultimate consolidation settlement due to the weight of this fill.

Accepted Answer

Using Equation 11.2:

[latex]\sigma ^{\prime }_{zf}=\sigma ^{\prime }_{z0}+\gamma _{fill}H_{fill}=\sigma ^{\prime }_{z0}+\left(19.2 kN/m^{3}\right) \left(3.0 m\right) =\sigma ^{\prime }_{z0}+57.6 kPa[/latex]

Compute the initial vertical stress at sample location, using Equation 10.34:

[latex]\sigma ^{\prime }_{z0}=\sum{\gamma H - u} =\left(18.5 kN/m^{3}\right) \left(1.5 m\right) +\left(19.5 kN/m^{3}\right) \left(2.0 m\right) +\left(16.0 kN/m^{3}\right) \left(4.0 m\right) - \left(9.8 kN/m^{3}\right) \left(6.0 m\right) =72.0 kPa[/latex]

[latex]\frac{C_{c}}{1 + e_{0}} =\frac{0.40}{1 + 1.10} =0.190[/latex]

At the sample [latex]\sigma ^{\prime }_{c}≈\sigma ^{\prime }_{z0} → ∴[/latex] clay is normally consolidated

If the soil at the sample depth is normally consolidated, and the sample is truly representative, then the entire stratum is normally consolidated.

Assume the sand also is normally consolidated, which is conservative. For the sand strata, use [latex]C_{c} /(1 + e_{0}) = 0.008[/latex], per Table 11.3.

Soil Type	[latex]C_{c}/(1 + e_{0})[/latex]
Soil Type	[latex]D_{r}=0[/latex]%	[latex]D_{r}=20[/latex]%	[latex]D_{r}=40[/latex]%	[latex]D_{r}=60[/latex]%	[latex]D_{r}=80[/latex]%	[latex]D_{r}=100[/latex]%
Medium to coarse sand, some fine gravel (SW)	-	-	0.005	-	-	-
Medium to coarse sand (SW/SP)	0.010	0.008	0.006	0.005	0.003	0.002
Fine to coarse sand (SW)	0.011	0.009	0.007	0.005	0.003	0.002
Fine to medium sand (SW/SP)	0.013	0.010	0.008	0.006	0.004	0.003
Fine sand (SP)	0.015	0.013	0.010	0.008	0.005	0.003
Fine sand with trace fine to coarse silt (SP-SM)	-	-	0.011	-	-	-
Find sand with little fine to coarse silt (SM)	0.017	0.014	0.012	0.009	0.006	0.003
Fine sand with some fine to coarse silt (SM)	-	-	0.014	-	-	-

At midpoint of layer
Layer	H (m)	[latex]\sigma _{z0}^{\prime }[/latex] (kPa) Eqn 10.35 [latex]\sigma ^{\prime }_{z}=\sum{\gamma H} +\sum{\left(\sigma _{z}\right) } _{induced} - u[/latex]	[latex]\sigma _{zf}^{\prime }[/latex] (kPa)	[latex]\frac{C_{c}}{1 + e_{0}} [/latex]	Eqn.	[latex]\left(\delta _{c}\right) _{ult}[/latex] (mm)
1	1.5	13.9	71.5	0.008	11.23 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{c}}{1 + e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	8
2	2.0	37.4	95.0	0.008	11.23 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{c}}{1 + e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	6
3	3.0	56.4	114.0	0.19	11.23 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{c}}{1 + e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	174
4	3.0	75.0	132.6	0.19	11.23 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{c}}{1 + e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	141
5	4.0	96.7	154.3	0.19	11.23 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{c}}{1 + e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	154
					[latex]\left(\delta _{c}\right) _{ult}=[/latex]	483

Round off to:

[latex]\left(\delta _{c}\right) _{ult}=[/latex] 480 mm

Notice how we have used the same analysis for soils above and below the groundwater table, and both are based on saturated [latex]C_{c} /(1 + e_{0})[/latex] values. This is conservative (although in this case, very slightly so) because the soils above the groundwater table are probably less compressible. Section 11.9 discusses unsaturated soils in more detail.

Question 11.5: A 3.0 m deep compacted fill is to be placed over the soil pr...

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