Problem 10.15. Elastic foundation -water tank

The cross section of a water tank is given in the Figure. The top of the walls is connected by a floor which hinders displacement and allows rotation of the edge of the wall. 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. The tank is fully filled with water.
a) Determine the moments, M_x in the walls and in the slab, calculate the tensile force in the floor. (Consider only a strip of unit width far from the end walls. The strip is long, λL > 4.)
b) Determine the internal forces caused by 30 degrees heating of the floor. (Elastic compressibility of the floor is neglected.) Thermal expansion coefficients of steel and concrete are the same: α = 1.2×10^-5 1/°C.
c) How the moments due to the water pressure will change if the connection of the floor and the wall fails?

Solve Problem

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Problem a)

Moment at the bottom of the wall, M[kNm/m]=

Tensile force in the floor, N[kN/m]=

Problem b)

Tensile force in the floor, N[kN/m]=

Problem c)

Moment at the bottom of the wall, M[kNm/m]=

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Steps

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Problem a)

See Example 10.10.

Solve the statically indetermined structure with the force method.

Section 9.1.

Step 1. Define the primary structure. Draw moment diagram of the primary structure from the load. Calculate the displacement at the removed support.

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The primary structure is obtained by introducing a hinge at the intersection of the wall and the slab. The corresponding deflected shape and bending moment curve are shown in the Figure, the relative rotation in the hinge (simply supported beam subjected to a triangular load) is

$a_{0} = φ_{0} = \frac{(γ H) H^{3}}{45 D_{w}} = \frac{10 \times 4 . 0^{3}}{45 \times 9300} = 0.00612$

See Problem 6.8.

Step 2. Draw moment diagram of the primary structure from the redundant. Calculate the displacement at the removed support.

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The redundant is a moment couple, the deflections and moments are shown in Figure. The relative rotation contains two terms, the first one is the end rotation of the wall (simply supported beam subjected to an end moment), while the second one is the end rotation of the slab on elastic foundation is given below.

Since $λ = \sqrt[4]{\frac{c}{4 D_{s}}} = \sqrt[4]{\frac{100000}{4 \times 11600}} = 1.211 \frac{1}{m}$ and $\frac{4}{λ} = 3.303 < L = 15 m$ , the slab is considered to be long, and we write

$a_{1} = φ ‘_{1} + φ ‘_{2} = \frac{H}{3 D_{w}} + \frac{1}{λ D_{s}} = \frac{4.0}{3 \times 9300} + \frac{1}{1.21 \times 11600} = 0.000214 \frac{1}{kN}$

Figure 10.45c)

Step 3. Write compatibility condition. Determine the redundant. Give the internal forces of the statically indetermined structure.

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The compatibility condition is:

Eq.(9-1)

a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = - \frac{0.00611}{0.000214} = - 28.52 kNm / m

The bending moment is obtained by superposition:

$M = M_{0} + M_{1} X M_{bottom} = X = - 28.52 kNm / m$

Tensile force in the top support (floor) is

$N = \frac{γ H^{2}}{6} + \frac{X}{H} = \frac{10 \times 4 . 0^{2}}{6} - \frac{28.52}{4.0} = 19.54 kN / m$

Problem b)

Solve the statically indetermined structure with the force method.

Section 9.1.

Step 1. Define the primary structure. Calculate the displacement at the removed support from the temperature load.

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The primary structure is a cantilever.

Top displacement is the half of the elongation of the heated top floor:

Eq.(5-2)

$a_{0} = ∆ L = \frac{1}{2} L α ∆ T = \frac{1}{2} L α ∆ T = \frac{1}{2} 15.0 \times 1.2 \times 10^{- 5} \times 30 = 0.0027 m$

(From the water load there is no displacement on top, see Problem a) )

Step 2. Give the top displacement from the redundant.

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$a_{1} = a_{11} + a_{12} = \frac{H^{3}}{3 D_{w}} + \frac{H}{λ D_{s}} H = \frac{4 . 0^{3}}{3 \times 9300} + \frac{4 . 0^{2}}{1.21 \times 11600} = 0.00343 \frac{m^{2}}{kN}$

Step 3. Write compatibility condition. Determine the redundant. Give the internal forces of the statically indetermined structure.

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The compatibility condition is:

Eq.(9-1)

a_{0} - a_{1} X = 0 \to X = \frac{a_{0}}{a_{1}} = \frac{0.0027}{0.00343} = 0.787 kN / m N = X = 0.787 kN / m M_{\max} = X H = 4.0 \times 0.787 = 3.15 kNm / m

Problem c)

See Example 10.9b.

Step 1. Calculate the moment of the statically determined structure.

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The moment is determined on a cantilever:
$M_{\max} = \frac{γ H^{3}}{6} = \frac{10 \times 4 . 0^{3}}{6} = 106.7 kNm / m$

Results

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Problem a)

See Example 10.10.

The statically indetermined structure is solved by the force method.

Section 9.1.

$a_{0} = φ_{0} = \frac{(γ H) H^{3}}{45 D_{w}} = \frac{10 \times 4 . 0^{3}}{45 \times 9300} = 0.00612$

See Problem 6.8.

Since $λ = \sqrt[4]{\frac{c}{4 D_{s}}} = \sqrt[4]{\frac{100000}{4 \times 11600}} = 1.211 \frac{1}{m}$ and $\frac{4}{λ} = 3.303 < L = 15 m$ , the slab is considered to be long, and we write

$a_{1} = φ ‘_{1} + φ ‘_{2} = \frac{H}{3 D_{w}} + \frac{1}{λ D_{s}} = \frac{4.0}{3 \times 9300} + \frac{1}{1.21 \times 11600} = 0.000214 \frac{1}{kN}$

Figure 10.45c)

The compatibility condition is

Eq.(9-1)

$a_{0} + a_{1} X = 0 \to X = - \frac{a_{0}}{a_{1}} = - \frac{0.00611}{0.000214} = - 28.52 kNm / m$

The bending moment is obtained by superposition:

$M = M_{0} + M_{1} X M_{bottom} = X = - 28.52 kNm / m$

Tensile force in the top support (floor) is

$N = \frac{γ H^{2}}{6} + \frac{X}{H} = \frac{10 \times 4 . 0^{2}}{6} - \frac{28.52}{4.0} = 19.54 kN / m$

Problem b)

The statically indetermined structure is solved by the force method.

Section 9.1.

The primary structure is a cantilever.

Top displacement is the half of the elongation of the heated top floor:

Eq.(5-2)

$a_{0} = ∆ L = \frac{1}{2} L α ∆ T = \frac{1}{2} L α ∆ T = \frac{1}{2} 15.0 \times 1.2 \times 10^{- 5} \times 30 = 0.0027 m$

(From the water load there is no displacement on top, see Problem a) )

The top displacement from the redundant: $a_{1} = a_{11} + a_{12} = \frac{H^{3}}{3 D_{w}} + \frac{H}{λ D_{s}} H = \frac{4 . 0^{3}}{3 \times 9300} + \frac{4 . 0^{2}}{1.21 \times 11600} = 0.00343 \frac{m^{2}}{kN}$

The compatibility condition is

Eq.(9-1)

a_{0} - a_{1} X = 0 \to X = \frac{a_{0}}{a_{1}} = \frac{0.0027}{0.00343} = 0.787 kN / m N = X = 0.787 kN / m M_{\max} = X H = 4.0 \times 0.0787 = 3.15 kNm / m

Problem c)

See Example 10.9b.

The moment is determined on a cantilever:
$M_{\max} = \frac{γ H^{3}}{6} = \frac{10 \times 4 . 0^{3}}{6} = 106.7 kNm / m$