Problem 10.16. Elastic foundation -water tank during construction

Determine the internal forces of the slab of the water tank given in the previous problem during construction when the floor is not yet
built. Self-weight of the wall is 5 kN/m². Stiffness of the base slab is: D_s = 11 600 kNm²/m, stiffness of the wall is D_w = 9 300 kNm²/m. The soil is modelled as an elastic (Winkler type) support, the foundation coefficient is: c = 100 000 kN/m³. 

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Steps

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See Example 10.9a.

Step 1.  Give the load on the water tank which cause internal forces.

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Due to the self-weight of the slab the whole tank sinks , which results in no bending moment in the wall. The self-weight of the wall gives a line load:

$F=gH=5.0\times 4.0=20 \mathrm{kN}/\mathrm{m}$

Step 2.  Determine the moment distribution from the load. Give the maximum moment in the slab.

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Figure 10.45a

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{F}{2{\lambda }^3{D}_{\mathrm{s}}}0.64{\lambda }^2{D}_{\mathrm{s}}=\frac{20}{2\times 1.21}0.64\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{28}\mathbf{ }\mathbf{kNm}\mathbf{/}\mathbf{m} \\ \mathrm{where} \\ \lambda =\sqrt[4]{\frac{c}{4D_{\mathrm{s}}}}=\sqrt[4]{\frac{100000}{4\times 11600}}=1.21\frac{1}{\mathrm{m}} \end{gathered}$$

Results

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See Example 10.9a.

Due to the self-weight of the slab the whole tank sinks , which results in no bending moment in the wall. The self-weight of the wall gives a line load:

$F=gH=5.0\times 4.0=20 \mathrm{kN}/\mathrm{m}$

The moment distribution from this load is shown in the Figure below.

Figure 10.45a

The maximum moment is

$$\begin{gathered} {\mathit{M}}_{\mathbf{max}}=\frac{F}{2{\lambda }^3{D}_{\mathrm{s}}}0.64{\lambda }^2{D}_{\mathrm{s}}=\frac{20}{2\times 1.21}0.64\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{28}\mathbf{ }\mathbf{kNm}\mathbf{/}\mathbf{m} \\ \mathrm{where} \\ \lambda =\sqrt[4]{\frac{c}{4D_{\mathrm{s}}}}=\sqrt[4]{\frac{100000}{4\times 11600}}=1.21\frac{1}{\mathrm{m}} \end{gathered}$$