After the settlement due to the fill described in Example 11.5 is completed, a 20 m diameter, 10 m tall cylindrical steel water tank is to be built. The bottom of the tank will be at the top of the fill, and it will have an empty mass of 300,000 kg. Ultimately, the water inside will be 9.5 m deep.

Question

After the settlement due to the fill described in Example 11.5 is completed, a 20 m diameter, 10 m tall cylindrical steel water tank is to be built. The bottom of the tank will be at the top of the fill, and it will have an empty mass of 300,000 kg. Ultimately, the water inside will be 9.5 m deep. Compute the ultimate consolidation settlement beneath the center of this tank due to the weight of the tank and its contents. Assume the new fill is overconsolidated with [latex]C_{r}/( 1 +e _{o}) = 0.002[/latex].

Accepted Answer

Strategy

The settlement due to the fill is now complete, so the values of [latex]\sigma ^{\prime }_{zf}[/latex] from the solution of Example 11.5 are now the initial stresses, [latex]\sigma ^{\prime }_{z0}[/latex]. We will compute new [latex]\sigma ^{\prime }_{zf}[/latex] values using [latex]Δσ_{z}[/latex], from Equation 10.25 [latex]\sigma _{z}=q\left[1 - \left\lgroup\frac{1}{1+\left\lgroup\frac{B}{2z_{f}} \right\rgroup^{2} } \right\rgroup ^{1.50}\right][/latex]. Note how the increase in total stress, [latex]Δσ_{z}[/latex], due to the weight of the tank diminishes with depth. This is because the loads from the tank are distributed over a much smaller area compared to the wide fills of previous examples.

Solution

Compute weights:

[latex]W_{tank}=Mg=\left(300,000 kg\right) \left(9.8 m/s^{2}\right) \left\lgroup\frac{1 kN}{1000 N} \right\rgroup =2900 kN[/latex]

[latex]W_{water}=V_{tank}\gamma _{w}=\frac{\pi B^{2}H}{4} \gamma _{w}=\frac{\pi \left(20.0 m\right) ^{2}\left(9.5 m\right) }{4} \left(9.8 kN/m^{3}\right) =29,200 kN[/latex]

The weight of the water is much greater than that of the empty tank, so it is reasonable for us to assume the bearing pressure [latex]q[/latex] is constant across the bottom of the tank.

[latex]q=\frac{W}{A} =\frac{2900 kN+29,200 kN}{\pi \left(20.0 m\right) ^{2}/4} =102 kPa[/latex]

Layer	H (m)	At midpoint of layer						Eqn.	[latex]\left(\delta _{c}\right) _{ult}[/latex] (mm)
Layer	H (m)	[latex]\sigma ^{\prime }_{z0}[/latex] (kPa) Eqn 10.34 [latex]\sigma ^{\prime }_{z}=\sum{\gamma H - u}[/latex]	[latex]z_{f}[/latex] (m)	[latex](\sigma _{z})_{induced}[/latex] (kPa) Eqn. 10.25 [latex]\sigma _{z}=q\left[1 - \left\lgroup\frac{1}{1+\left\lgroup\frac{B}{2z_{f}} \right\rgroup^{2} } \right\rgroup ^{1.50}\right] [/latex]	[latex]\sigma ^{\prime }_{zf}[/latex] (kPa) Eqn. 11.3 [latex]\sigma ^{\prime }_{zf}=\sigma ^{\prime }_{z0}+\left(\sigma _{z}\right) _{induced}[/latex]	[latex]\frac{C_{c}}{1+e_{0}} [/latex]	[latex]\frac{C_{r}}{1+e_{0}} [/latex]	Eqn.	[latex]\left(\delta _{c}\right) _{ult}[/latex] (mm)
1	3.0	28.8	1.5	101.7	130.5		0.002	11.24 [latex]\left(\delta _{c}\right) _{ult}=\sum{\frac{C_{r}}{1+e_{0}} } H\log \left\lgroup\frac{\sigma ^{\prime }_{zf}}{\sigma ^{\prime }_{z0}} \right\rgroup [/latex]	4
2	1.5	71.5	3.7	97.7	169.2	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]	4
3	2.0	95.0	5.5	90.6	185.6	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]	5
4	3.0	114.0	8.0	77.1	191.1	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]	128
5	3.0	132.6	11.0	60.7	193.3	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]	93
6	4.0	154.3	14.5	45.1	199.4	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]	85
[latex]\left(\delta _{c}\right) _{ult}=[/latex]									319

Round off to:

[latex]\left(\delta _{c}\right) _{ult}=[/latex] 320 mm

Commentary

1. If the tank were built immediately after the fill was placed, then [latex]\sigma ^{\prime }_{z0}[/latex] would be the same as in Example 11.5, and everything else would remain unchanged. Such a solution would illustrate the use of superposition of stresses.
2. The values of [latex](σ_{z})_{induced}[/latex] beneath the edge of the tank are less than those beneath the center (see Figure 10.10). Thus, the consolidation settlement also will be less and the bottom of the tank will settle into a dish shape. The difference between these two settlements is called differential settlement. We will discuss differential settlements in more detail in Chapter 17.

Question 11.8: After the settlement due to the fill described in Example 11...

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