Free Vibration of Water Tank Stand When Filled With Water

It is desired to find the eigenfrequencies and eigenvectors (modal parameters) of concrete water tank support when filled with water. This is a case of undamped free vibration, treating the stored water as a lumped mass on the tank stand. The tank will be treated as a system with 2 degrees of freedom, considering lateral displacement only.

Data and Load Analysis;
Density of concrete = 24 KN/m³
Thickness of slab = 150mm
Dimension of columns = 230 x 230mm
Supporting beams = 300 x 230mm
Weight of water = 10 KN/m³
Size of tank = 2000 litres
Modulus of elasticity of concrete = 21.7 × 10⁶ KN/m²

Load Analysis
Self weight of slab = 24 KN/m³ × 0.15m × 2.5m × 2.5m = 22.5 KN = 2293.75 kg
Weight of supporting beams on the four sides = 4 × 24 KN/m³ × 2.27m × 0.23m × 0.3m = 15.036 KN = 1532.77kg
Weight of water = 10 KN/m³ × 2m³ = 20 KN = 2038.73 kg
Weight of tank (assume) = 75kg

At the 1st level = 2293.75 + 1532.77 + 2038.73 + 75 = 5940.25 kg
At the 2nd level = 2293.75 + 1532.77 + 2(2038.73) + 2(75) = 8053.98 kg

We are assuming that the floor slab is very stiff compared to the columns. So the flexural rigidity is taken as infinity.

Geometrical properties of the sections
Columns = 23cm x 23cm = 0.23m x 0.23m
Moment of inertia of column I_C = (0.23 × 0.23³)/12 = 2.332 × 10^-4 m⁴
EI_C = (2.332 × 10^-4 m4) × (2.1 × 10⁷ KN/m²) = 4897.2 KN.m²

The mass matrix of the structure is given below;

(12EI_C)/L³ = (12 × 4897.2)/3³ = 2176.533 KN/m
K₁₁ = (12EI_C)/h³ + (12EI_C)/h³
K₁₁ = 2176.533 + 2176.533 = 4353.067
K₂₁ = -(2176.533 + 2176.533) = -4353.067
K₂₂ = 2(2176.533) + 2(2176.533) = 8706.132

The stiffness matrix is therefore;

For an undamped free vibration, the equation of motion is;

Mü + ku = 0 —————– (1)

Arranging this in matrix form, we obtain;

On solving;

((4353.067 × 10³ – 5940.25ω²) × (8706.132 × 10³ – 8053.98ω²) – (4353.067 × 10³ × 4353.067 × 10³) = 0

On simplifying;

(4.783 × 10⁷)ω⁴ – (8.677 × 10¹⁰)ω² + (1.8949 × 10¹³) = 0

On solving;

ω₁² = 253.9146 rad²/sec²
ω₂² = 1559.9 rad²/sec²

ω₁ = 15.9347 rad/sec
ω₂ = 39.4951 rad/sec

The natural frequencies;
F₁ = ω₁/2π = 2.5361 Hz
F₂ = ω₂/2π = 6.2858 Hz

Determination of the mode shapes

For ω₁² = 253.9146 rad²/sec²
Considering the first row of the matrix;
(2.8448 × 10⁶)A₁ – (4353.067 × 10³)A₂ = 0

Setting A₁ = 1.0
A₂ = (2.8448 × 10⁶)/(4353.067 × 10³) = 0.6780
Therefore;ϕ₁ = [1, 0.6535]^T

For ω₂² = 1559.9 rad²/sec²
Considering the first row of the matrix;
(-4.9129 × 10⁶)A₁ – (4353.067 × 10³)A₂ = 0

Setting A₁ = 1.0
A₂ = (-4.9129 × 10⁶)/(4353.067 × 10³) = -1.128
Therefore;ϕ₂ = [1, -1.128]^T

As a verification;
K₁/M₁ = 2381668.751 / 9379.8 = ω₁² = 253.9146 rad²/sec²
 K₂/M₂ = 25251112,921 / 9379.8 = ω₂² = 1559.9 rad²/sec²

Eigenvector Mass Orthonormalization
The idea is to obtain;

ϕ₁^T Mϕ₁ = 1.0

The conversion factor;

Therefore, C₁ = 1/√(9379.8) = 0.0103

Hence;

Similarly;

Therefore, C₂ = 1/√(16188.005) = 0.00785

Hence;

We can therefore obtain;

Let χ_i(t) be the generalised coordinate which represent the amplitude of the orthonormalised mode shape.

The displacement time history response is therefore; u(t) = Φχ(t)

But the modal equation of motion

Mü + ku = 0 —————– (1)

Realise that;

Substituting ü and u into equation 1 and multiplying by Φ^T , we obtain;

Φ^T MΦẍ + Φ^T KΦx = 0
Which finally leads to;

ẍ + K_gx = 0

This leads to the following differential equations;

ẍ₁(t) + 253.9146x₁(t)  = 0 ——————– (a)
ẍ₂(t) + 1559.9x₂(t)  = 0 ——————– (b)

Let the initial condition be a unit velocity;
x₁(0) = 0;  ẋ₁(0)  = 0

Solving the differential equation by Laplace Transform;

For mode 1;
ẍ₁(t) + 253.9146x₁(t)  = 0    x₁(0) = 0; ẋ₁(0)  = 1.0

(s²x ̅  – sx₀ – x₁) + 253.9146x ̅ = 0
s²x ̅  – 1 + 253.9146x ̅ = 0
x ̅ (s²   + 253.9146) = 1
x ̅ = 1 / (s²   + 253.9146)

On solving by Laplace Transform;

x₁ = (1/√253.9146) sin(√253.9146)
x₁ = 0.0627 sin(15.935t)

For mode 2;
ẍ₂(t) + 1559.96x₂(t)  = 0    x₁(0) = 0; ẋ₁(0)  = 1.0

(s²x ̅  – sx₀ – x₁) + 1559.9x ̅ = 0
s²x ̅  – 1 + 1559.9x ̅ = 0
x ̅ (s²   + 1559.9) = 1
x ̅ = 1 / (s²   + 1559.9)

On solving by Laplace Transform;

x₂ = (1/√1559.9) sin(√1559.9)
x₂ = 0.0253 sin(39.495t)

Knowing that
u(t) = Φχ(t)

The displacement time history is therefore;

u₁(t) = 0.00064581 sin(15.935t) + 0.0001986 sin(39.495t)
u₂(t) = 0.0004219 sin(15.935t) – 0.000224 sin(39.495t)

The displacement time history graph is shown below;