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001816

Dipl.-Ing. (BA) Markus Baumgärtel

2023-09-27

Design of Buoyancy After Comparing Weight Forces and Considering Soil Reactions

This example shows you how to quickly determine the buoyancy or the uplift limit state of a vessel in RFEM.

The present article is about a reinforced concrete tank standing 1.5 m (4.9 ft) submerged in groundwater. Furthermore, we look at the extent to which the application of wall friction affects the design.

The model of this article is adjusted in such a way that all parameters and the geometry can be edited using the "Global Parameters". The model is then updated automatically.

First, we want to see whether the tank shown in the example is floating due to groundwater. There is water in the tank, and the wall friction is neglected. This case reflects the operating state.

Design 1

Self-Weight of Reinforced Concrete Tank

G = 24.0 k N / m^{3} \cdot (2 \cdot 1.5 m \cdot 10 m + 1.5 m \cdot 12 m) = 1, 152 k N / m

Dead Weight of Reinforced Concrete Tank

The uplift pressure at a groundwater level of about 8.5 m (27 ft) is

A_{k} = 10 k N / m^{3} \cdot 8.5 m \cdot 15.0 m = 1, 275 k N / m

Soil Pressure Force

The water in the tank has a stabilizing effect and is added to the tank's self-weight.

G_{W} = 10 k N / m^{3} \cdot 3.5 m \cdot 12.0 m = 420 k N / m

Weight of Water

Next, the limit state design against buoyancy can be summarized as follows:

γ_{G, s t b} \cdot (G + G_{W}) \geq γ_{G, d s t b} \cdot A m i t γ_{G, s t b} = 0, 95 u n d γ_{G, d s t b} = 1, 05

0, 95 \cdot (1152 + 420) = 1493, 4 k N / m

1, 05 \cdot 1275 = 1338, 75 k N / m

1493, 4 \geq 1338, 8

Buoyancy Design

The analysis has shown that the safety against buoyancy of the vessel is sufficient. A residual load-bearing capacity of 1493.4 kN/m - 1338.8 kN/m = 154.65 KN/m remains (CO3).

Security Against Buoyancy

Design 2

The aim is to check whether the reinforced concrete tank has sufficient safety against buoyancy when it is pumped empty. In this design check, the wall friction is to be used.

Layer 1
SU, γ = 17 kN/m³, φ´= 30°

Layer 2
SU, γ`= 9 kN/m³, φ´= 30°

Layer 3
GW, γ´ = 12 kN/m³, φ´= 32,5°

The distribution of the horizontal earth pressure is composed as follows:

α= 0; β= 0; δ= 2/3*φk; k_ah,SU= 0,28; k_ah,GW= 0,25

Concrete Tank

For each of the wall surfaces, the resulting shear force to be applied is:

η = 0.8

F_{s, k} = η \cdot E_{ah, k} \cdot \tan φ_{α}

F_{s, k} = 0.8 \cdot (0.5 \cdot 7.1 kN / m^{2} \cdot 1.5 m) \cdot \tan \frac{2}{3} \cdot 30 ° + 0.5 \cdot (7.1 kN / m^{2} + 20.5 kN / m^{2}) \cdot 5.3 m \cdot \tan \frac{2}{3} \cdot 30 ° + 0.5 \cdot (18.3 kN / m^{2} + 27.9 kN / m^{2}) \cdot 3.2 m \cdot \tan \frac{2}{3} \cdot 32.5 °

F_{s, k} = 46.32 kN / m

Applicable Shear Force

To simplify, the load is applied as a line load at the top edge of the container as a line load in Load Case 4.

Thus, it is possible to determine the uplift limit state of the vessel (CO5):

γ_{G, s t} \cdot G + γ_{G, s t} \cdot F_{s, k} \geq γ_{G, d s t} γ_{G, s t} = 0, 95 γ_{G, d s t} = 1.05

\sum_{}^{} G \cdot 0.95 = 0.95 \cdot (1, 152 + 2 \cdot 46.3) = 1, 182.5 k N / m

A_{k} \cdot 1.05 = 1, 275 \cdot 1.05 = 1, 338.8 k N / m

1, 182.5 \leq 1, 338.8

Uplift Limit State

As expected, Load Combination 5 also becomes unstable in RFEM 6. Consequently, safety against buoyancy is not given in this case.

Knowledge Base

KB 001816 | Design of Buoyancy After Comparing Weight Forces and Considering Soil Reactions

Author

Mr. Baumgärtel provides technical support for Dlubal Software customers.

References

Dörken, W., Dehne, E., & Kliesch, K. (2020). Grundbau in Beispielen Teil 1, (7th ed. Cologne: Reguvis Fachmedien.

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