Introduction:
Prior to the design of underground water retaining structures such as tanks and swimming pools, preliminary sizing of the structural members must be done. However, these sizes must be verified to ensure that they satisfy some adverse effects which the structure would be exposed to in its service life. This verification involves the stabilizing actions that tend to keep the structure stable such as the weight of the structural components of the structure together with the weight of stored water and destabilizing actions that tends to unsettle the structure such as uplift forces from underground water etc.
Generally, verification of stabilizing actions and destabilizing actions should be done in worst case scenario viz:
(1) When the structure is empty, to investigate the adverse effects of earth pressure and groundwater on the base and on the walls of the structure.
(2) When the structure is full, to consider the effects of stored water on the base and walls of the structure without considering the beneficial effect of the near balance of earth pressure and groundwater on external walls.
The aim of the verification is to ensure that the structure is safe in the worst condition. If initial verification fails (i.e. the floatation is critical), maybe due to the fact that the effect of destabilizing forces is more than that of stabilizing forces, the thickness of the wall can be increased or the base of the slab extended (see Figure 1 below) such that the weight of the backfill earth on the extended portion contributes to stabilizing the structure. Another possible means is to lower the water table of the soil.
Verification for Adverse Conditions
Verify the adverse conditions of the swimming pool sections shown below and determine the suitable sizes of the members to satisfy adverse conditions.
A. Stabilizing actions (due to the weight of the pool components when empty)
i. Weight of the base (volume of base x unit weight of concrete)
-Area of base = (4.6 x 2.3) + (4.6 x 4.1) + (4.6 x 3.3) = 10.58 + 18.86 + 15.18 = 44.62 m2
-Volume of base = area x thickness = 44.62 x 0.3 = 13.386 m3
-Weight of base = 13.386 x 25 = 334.65 kN
ii. Weight of perimeter beams
-Depth of perimeter beams excluding part taken by slab = 600 – 300 = 300 mm = 0.3 m. Total length of perimeter beam = 4.6 + 4.6 + 2 + 2 + 4.1 + 4.1 + 3 + 3 = 27.4 m
-Thickness of perimeter beam = 300 mm = 0.3 m
-Volume of concrete due to perimeter beam = 0.3 x 27.4 x 0.3 = 2.466 m3
-Weight of perimeter beams = 2.466 x 25 = 61.65 kN
iii. Weight of longitudinal walls
-Volume of concrete wall at section A = (L x H x T) x 2 = 2 x 1.5 x 0.3 x 2 = 1.8 m3
-Volume of concrete wall at section B = (area of trapezium x thickness of section) x 2 = ½ (1.5 + 2.4)4 x 0.3 x 2 = 4.68 m3
-Volume of concrete wall at section C = 3 x 2.4 x 0.3 x 2 = 4.32 m3
-Total volume = 1.8 + 4.68 + 4.32 = 10.8 m3
Weight of longitudinal walls = 10.8 x 25 = 270 kN
iv. Weight of lateral walls
-Weight of lateral walls = (4.6 x 1.5 x 0.3 x 25) + (4.6 x 2.4 x 0.3 x 25) = 51.75 + 82.8 = 134.55 kN
Total weight, W = 334.65 + 61.65 + 270 + 134.55 = 800.85 kN
The load is for stabilizing action and it is considered favourable load and it is permanent. From Table 1, partial factor, γG = 0.9
Thus, Gstb,d = W x γG = 800.85 x 0.9 = 720.765 kN
B. Destabilizing actions (due to uplift pressure under the base of the pool)
i. Section A (head of groundwater above the base = 1.1 m)
Uplift pressure = 10 x 1.1 = 11 kN/m2
Udst,d = factor for unfavorable permanent action (Table 7) x area of zone A x uplift pressure = 1.1 x 10.58 x 11 = 128.018 kN
ii. Section B (average head of groundwater above the base = 1.55 m)
Uplift pressure = 10 x 1.55 = 15.5 kN/m2
Udst,d = 1.1 x 18.86 x 15.5 = 321.563 kN
iii. Section C (head of groundwater above the base = 2 m)
Uplift pressure = 10 x 2 = 20 kN/m2
Udst,d = 1.1 x 15.18 x 20 = 333.96 kN
Total Udst,d = 128.018 + 321.563 + 333.96 = 783.54 kN
Gstb,d (720.765 kN) < Udst,d (783.54 kN) – this is not satisfactory.
Decision: Let us increase the base thickness to 400 mm and the depth of the perimeter beam to 750 mm
A1. NEW Stabilizing actions
Total weight of the pool when empty comprises
ai. Weight of the base
-Volume of base = area x thickness = 44.62 x 0.4 = 17.848 m3
-Weight of base = 17.848 x 25 = 446.2 kN
aii. Weight of perimeter beams
-Depth of perimeter beams excluding part taken by slab = 750 – 400 = 350 mm = 0.35 m.
-Volume of concrete due to perimeter beam = 0.35 x 27.4 x 0.3 = 2.877 m3
-Weight of perimeter beams = 2.877 x 25 = 71.925 kN
aiii. Weight of longitudinal walls
-Volume of concrete wall at section A (L x H x T) x 2 = 2 x 1.5 x 0.3 x 2 = 1.8 m3
-Volume of concrete wall at section B (area of trapezium x thickness of section) x 2 = ½ (1.5 + 2.4)4 x 0.3 x 2 = 4.68 m3
-Volume of concrete wall at section C = 3 x 2.4 x 0.3 x 2 = 4.32 m3
Total volume = 1.8 + 4.68 + 4.32 = 10.8 m3
Weight of longitudinal walls = 10.8 x 25 = 270 kN
aiv. Weight of lateral walls
-Weight of lateral walls = (4.6 x 1.5 x 0.3 x 25) + (4.6 x 2.4 x 0.3 x 25) = 51.75 + 82.8 = 134.55 kN
Total weight, W = 446.2 + 71.925 + 270 + 134.55 = 922.675 kN
The load is for new stabilizing action and it is considered favourable load and it is permanent. From Table 7, partial factor, γG = 0.9
Thus, Gstb,d = W x γG = 922.675 x 0.9 = 830.4075 kN
Gstb,d (830.4075 kN) > Udst,d (783.54 kN) – this is satisfactory
Thus, the final dimensions of the components of the pool are;
Thickness of walls = 300 mm
Thickness of slab = 400 mm
Depth of external perimeter beams only = 750 mm including thickness of base slab.
