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/m^{3}

Thickness of slab = 150mm

Dimension of columns = 230 x 230mm

Supporting beams = 300 x 230mm

Weight of water = 10 KN/m^{3}

Size of tank = 2000 litres

Modulus of elasticity of concrete = 21.7 × 10^{6} KN/m^{2}

**Load Analysis**

Self weight of slab = 24 KN/m^{3} × 0.15m × 2.5m × 2.5m = 22.5 KN = 2293.75 kg

Weight of supporting beams on the four sides = 4 × 24 KN/m^{3} × 2.27m × 0.23m × 0.3m = 15.036 KN = 1532.77kg

Weight of water = 10 KN/m^{3} × 2m^{3} = 20 KN = 2038.73 kg

Weight of tank (assume) = 75kg

* At the 1st level *= 2293.75 + 1532.77 + 2038.73 + 75 = 5940.25 kg

*= 2293.75 + 1532.77 + 2(2038.73) + 2(75) = 8053.98 kg*

**At the 2nd level**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 × 0.23^{3})/12 = 2.332 × 10^{-4} m^{4}

EI_{C} = (2.332 × 10^{-4} m4) × (2.1 × 10^{7} KN/m^{2}) = 4897.2 KN.m^{2}

The mass matrix of the structure is given below;

**Determination of the stiffness matrix**

(12EI_{C})/L^{3} = (12 × 4897.2)/3^{3} = 2176.533 KN/m

K_{11} = (12EI_{C})/h^{3} + (12EI_{C})/h^{3}

K_{11} = 2176.533 + 2176.533 = 4353.067

K_{21} = -(2176.533 + 2176.533) = -4353.067

K_{22} = 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 × 10^{3} – 5940.25ω^{2}) × (8706.132 × 10^{3} – 8053.98ω^{2}) – (4353.067 × 10^{3} × 4353.067 × 10^{3}) = 0

On simplifying;

(4.783 × 10^{7})ω^{4} – (8.677 × 10^{10})ω^{2} + (1.8949 × 10^{13}) = 0

On solving;

ω_{1}^{2} = 253.9146 rad^{2}/sec^{2}

ω_{2}^{2} = 1559.9 rad^{2}/sec^{2}

ω_{1} = 15.9347 rad/sec

ω_{2} = 39.4951 rad/sec

**The natural frequencies;**

F_{1} = ω_{1}/2π = 2.5361 Hz

F_{2} = ω_{2}/2π = 6.2858 Hz

**Determination of the mode shapes**

For ω_{1}^{2} = 253.9146 rad^{2}/sec^{2}

Considering the first row of the matrix;

(2.8448 × 10^{6})A_{1} – (4353.067 × 10^{3})A_{2} = 0

*Setting A _{1} = 1.0*

A

_{2}= (2.8448 × 10

^{6})/(4353.067 × 10

^{3}) = 0.6780

Therefore;ϕ

_{1}= [1, 0.6535]

^{T}

For ω_{2}^{2} = 1559.9 rad^{2}/sec^{2}

Considering the first row of the matrix;

(-4.9129 × 10^{6})A_{1} – (4353.067 × 10^{3})A_{2} = 0

*Setting A _{1} = 1.0*

A

_{2}= (-4.9129 × 10

^{6})/(4353.067 × 10

^{3}) = -1.128

Therefore;ϕ

_{2}= [1, -1.128]

^{T}

As a verification;

K_{1}/M_{1} = 2381668.751 / 9379.8 = ω_{1}^{2} = 253.9146 rad^{2}/sec^{2}

^{ }K_{2}/M_{2} = 25251112,921 / 9379.8 = ω_{2}^{2} = 1559.9 rad^{2}/sec^{2}

**Eigenvector Mass Orthonormalization**

The idea is to obtain;

ϕ_{1}^{T} Mϕ_{1} = 1.0

The conversion factor;

Therefore, C_{1} = 1/√(9379.8) = 0.0103

Hence;

Similarly;

Therefore, C_{2} = 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_{g}x = 0

This leads to the following differential equations;

ẍ_{1}(t) + 253.9146x_{1}(t) = 0 ——————– (a)

ẍ_{2}(t) + 1559.9x_{2}(t) = 0 ——————– (b)

Let the initial condition be a unit velocity;

x_{1}(0) = 0; ẋ_{1}(0) = 0

Solving the differential equation by Laplace Transform;

**For mode 1;**

ẍ_{1}(t) + 253.9146x_{1}(t) = 0 x_{1}(0) = 0; ẋ_{1}(0) = 1.0

(s^{2}x ̅ – sx_{0} – x_{1}) + 253.9146x ̅ = 0

s^{2}x ̅ – 1 + 253.9146x ̅ = 0

x ̅ (s^{2 }+ 253.9146) = 1

x ̅ = 1 / (s^{2 }+ 253.9146)

On solving by Laplace Transform;

x_{1} = (1/√253.9146) sin(√253.9146)

x_{1} = 0.0627 sin(15.935t)

**For mode 2;**

ẍ_{2}(t) + 1559.96x_{2}(t) = 0 x_{1}(0) = 0; ẋ_{1}(0) = 1.0

(s^{2}x ̅ – sx_{0} – x_{1}) + 1559.9x ̅ = 0

s^{2}x ̅ – 1 + 1559.9x ̅ = 0

x ̅ (s^{2 }+ 1559.9) = 1

x ̅ = 1 / (s^{2 }+ 1559.9)

On solving by Laplace Transform;

x_{2} = (1/√1559.9) sin(√1559.9)

x_{2} = 0.0253 sin(39.495t)

Knowing that

u(t) = Φχ(t)

The displacement time history is therefore;

u_{1}(t) = 0.00064581 sin(15.935t) + 0.0001986 sin(39.495t)

u_{2}(t) = 0.0004219 sin(15.935t) – 0.000224 sin(39.495t)

The displacement time history graph is shown below;

Hello Bro,

I think there are a few typo's in your post, specifically Load Analysis.

What would you consider to be your effective span of slab and beam?

Beam DIm (230mm x 300mm)

Wouldn't your effective span be 3 metres ?

Fc

Thank you for the observation. I used a former picture and the correction has been effected.

This may be a trivial question, I am assuming that the tanks are some how connected to the slabs hence the inclusion of the mass?

what would be the dynamic effect if only in contact with slab due to gravity and you introducing a vibration?

Fc

Good post as usual

FC

Could you let us download this file. Thanks.

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