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Angle α_m Between Axis x and Principal Axis 1 (for Axial Force n₁)

α_{m} = \frac{1}{2} [\arctan (\frac{2 n_{x y}}{n_{x} - n_{y}}))]

Angle β_b Between Principal Shear Force v-max,b and Local Axis x

β_{b} = \arctan \frac{v_{y}}{v_{x}}

Applicable Shear Force

η = 0.8

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

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

F_{s, k} = 46.32 k N / m

Axial force - calculated by standard SRSS rule

N_{SRSS} = \sqrt{{(1.361 kN)}^{2} + {(- 0.246 kN)}^{2} + {(0.815 kN)}^{2} + {(- 2.322 kN)}^{2}} = 2.823 kN

Axial Force n₁ in Direction of Principal Axis 1

n_{1} = \frac{1}{2} [n_{x} + n_{y} + \sqrt{(^{n_{x} - n_{y}) 2} + 4 {n_{x y}}^{2}})]

Axial Force n₂ in Direction of Principal Axis 2

n_{2} = \frac{1}{2} [n_{x} + n_{y} - \sqrt{(^{n_{x} - n_{y}) 2} + 4 {n_{x y}}^{2}})]

Axial Force n_x in Direction of Local x-Axis

n_{x} = \int_{- \frac{d}{2}}^{+ \frac{d}{2}} σ_{x} d z

Axial Force n_y in Direction of Local y-Axis

n_{y} = \int_{- \frac{d}{2}}^{+ \frac{d}{2}} σ_{y} d z

Calculation of Unreinforced Foundation

0.85 \cdot \frac{d}{a} \geq \sqrt{\frac{3 σ_{gd}}{f_{ctd, pl}}}

Coefficients of equivalent linear combination for maximum axial force

f_{maxN, LF 1} = \frac{1.361 kN}{2.823 kN} = 0.482

f_{maxN, LF 2} = \frac{- 0.246 kN}{2.823 kN} = - 0.087

f_{maxN, LF 3} = \frac{0.815 kN}{2.823 kN} = 0.289

f_{maxN, LF 4} = \frac{- 2.322 kN}{2.823 kN} = - 0.822

Correlation Between Support Force and Friction Force

p_{support} = μ \cdot p_{axial force}

Correlation of Hinge Friction Force and Internal Force

P_{hinge} = μ \cdot P_{axial force / shear force}

Corresponding shear force - calculated with equivalent linear combination

V_{z, maxN} = 0.482 \cdot 0.480 kN - 0.087 \cdot (- 1.635 kN) + 0.289 \cdot (- 0.556 kN) - 0.822 \cdot 1.546 kN = - 1.058 kN

Critical Compressive Force

N_{cr} = \frac{π^{2} \cdot E I}{l_{cr}^{2}}

Dead 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

Design Shear Resistance at Interface Without Shear Reinforcement acc. to DIN EN 1992-1-1, (NDP) 6.2.5 (1)

v_{Rdi} = c \cdot f_{ctd} \cdot μ \cdot σ_{n} + ρ \cdot f_{yd} \cdot (1.2 \cdot μ \cdot \sin (α) + \cos (α)) = 0.5 \cdot 1.02 + 0.9 \cdot 0 + \frac{0}{0.4} \cdot 435 \cdot (1.2 \cdot 0.9 \cdot 1 + 0) = 0.510 N / {mm}^{2}

Determining Material Properties of Steel-Fiber-Reinforced Concrete and Their Application in RFEM

{f^{f}}_{cflk, Li} = e^{({Lf}^{f}_{cflm, Li} - k_{s} \cdot L_{s})} \leq 0, 51 \cdot {f^{f}}_{cflm, Li}

Equation 6.2: Linear Interaction of Axial Force and Moment Loading

\frac{N_{E d}}{N_{R d}} + \frac{M_{y, E d}}{M_{y, R d}} + \frac{M_{z, E d}}{M_{z, R d}} \leq 1

Force Components

N_{⊥, Ed} = \frac{F_{1 ⊥, Ed}}{l_{w}} \pm \frac{M_{Ed}}{(}

V_{⊥, Ed} = \frac{F_{2 ⊥, Ed}}{l_{w}}

V_{∥, Ed} = \frac{F_{∥, Ed}}{l_{w}}

Force Resultant

F_{w, Ed} = \sqrt{({N^{2}}_{⊥, Ed} + {V^{2}}_{⊥, Ed} + {V^{2}}_{∥, Ed})}

Formula 1: Forces Due to Bending

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Formula 2: Forces Due to Torsion

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Formula 5: Shear Forces of Partial Cross-Section

\begin{array}{l} V_{y, 1} = \frac{100 \cdot (1.084 \cdot 12.268)}{12.268 \cdot 1.093} = 99.18 kN \\ V_{y, 2} = \frac{100 \cdot (0.009 \cdot 12.268)}{12.268 \cdot 1.093} = 0.823 kN \end{array}

Longitudinal Reinforcement Design, Method of Damage Equivalent Stress Range According to 6.8.5 (3)

γ_{F, f a t} \times ∆ σ_{s, e q u} (N *) \leq ∆ σ_{R s k} (N *) / γ_{s, f a t}

Longitudinal Reinforcement Design, Simplified Design Method According to 6.8.6

η = \frac{∆ σ_{S, m a x}}{k_{1}}

Longitudinal Shear Stress in Neck Seams from Shear Force

τ_{∥, Vz} = - \frac{V_{z} \cdot S_{y}}{I_{y} \cdot \sum a}

Maximum axial force - calculated by equivalent linear combination

N_{maxN} = 0.482 \cdot 1.361 kN - 0.87 \cdot (- 0.246 kN) + 0.289 \cdot 0.815 kN - 0.822 \cdot (- 2.322 kN) = 2.823 kN

Maximum Resulting Shear Force v_max,b from Bending Components

v_{\max, b} = \sqrt{{v_{x}}^{2} + {v_{y}}^{2}}

Maximum Shear Force v_max,m from Membrane Components

v_{\max, m} = \frac{\sqrt{(^{n_{x} - n_{y}) 2} + 4 {n_{x y}}^{2}}}{2}

Mechanical Percentage of Reinforcement

ρ = \frac{A_{s} \cdot f_{yd}}{A_{c} \cdot f_{cd}}