Formulas - Search term: Shear str...

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Angle α_- Between Local Axis x (or y) and Principal Axis 1 (or 2) for Stresses on Negative Surface Side

α_{-} = \frac{1}{2} \arctan 2 (\frac{2 τ_{x y, -}}{σ_{x, -} - σ_{y, -}}) \in (- 90 °, 90 °]

Angle α₊ Between Local Axis x (or y) and Principal Axis 1 (or 2) for Stresses on Positive Surface Side

α_{+} = \frac{1}{2} \arctan 2 (\frac{2 τ_{x y, +}}{σ_{x, +} - σ_{y, +}}) \in (- 90 °, 90 °]

Angle α_m Between Local Axis x and Principal Axis 1 for Membrane Stresses

α_{m} = \frac{1}{2} \arctan 2 (\frac{2 τ_{x y, m}}{σ_{x, m} - σ_{y, m}}) \in (- 90 °, 90 °]

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

Bending Moment m_x (Creating Stresses in Direction of Local x-Axis)

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

Bending Moment m_y (Creating Stresses in Direction of Local y-Axis)

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

Bending Normal Stress According to Reuleaux

σ_{b} = \frac{δ}{D} \cdot E

σ_{x} (x, z) = - κ (x) Ez

Chord prestress coefficient kₚ EN 1993-1-8

k_{p} = 1 + 0.3 \cdot n_{p} \cdot (1 - n_{p})

n_{p} = \frac{f_{p}}{f_{y}}

f_{p} = \frac{N_{p}}{A_{0}} - \frac{|M_{0}|}{W_{0}}

Contact Stress σ_z ("Soil Pressure") in Direction of Surface Axis z

σ_{z} = \frac{F_{z}}{A}

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

Design of Concrete Compressive Stress and Concrete Compression Strut, Method of Damage Equivalent Stress Range According to 6.8.7

E_{c d, m a x, e q u} + 0.43 \sqrt{1 - R_{e q u}} \leq 1

R_{e q u} = E_{c d, m i n, e q u} / E_{c d, m a x, e q u}

E_{c d, m i n, e q u} = σ_{c d, m i n, e q u} / f_{c d, f a t}

E_{c d, m a x, e q u} = σ_{c d, m a x, e q u} / f_{c d, f a t}

Design of Concrete Compressive Stress and Concrete Compression Struts, Simplified Analysis Method According to 6.8.7 (2)

\frac{σ_{c, m a x}}{f_{c d, f a t}} \leq 0.5 + 0.45 \frac{σ_{c, m i n}}{f_{c d, f a t}} \begin{array}{rcl} \leq & 0.9 & for f_{ck} \leq 50 N / mm ² \\ \leq & 0.8 & for f_{ck} \geq 50 N / mm ² \end{array}

Design Shear Resistance at Interface 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{a_{s}}{0.4} \cdot 435 \cdot (1.2 \cdot 0.9 \cdot 1 + 0)

\to 0.51 + \frac{a_{s}}{0.4} \cdot 469.56 = 1.736 N / {mm}^{2}

\to a_{s} = 10.44 {cm}^{2} / 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}

Design Value of Shear Resistance According to DIN EN 1992-1-1, 6.2.2 (1)

V_{Rd, c, out} = C_{Rd, c} \cdot k \cdot {(100 \cdot ρ_{l} \cdot f_{ck})}^{1 / 3} + k_{1} \cdot σ_{cp} \geq (v_{\min} + k_{1} \cdot σ_{cp}) V_{Rd, c} = 0.10 \cdot 2.0 \cdot {(100 \cdot 0.0166 \cdot 35)}^{1 / 3} + 0.10 \cdot 0 = 0.774 MN / m^{2} \geq 0.586 MN / m^{2}

where

C_{Rd, c} = 0.15 / γ_{c} = 0.15 / 1.5 = 0.10

Design Value of Shear Resistance According to DIN EN 1992-1-1, 6.2.2 (1)

v_{Rd, c} = C_{Rd, c} \cdot k \cdot {(100 \cdot ρ_{l} \cdot f_{ck})}^{1 / 3} + k \cdot σ_{cp} = 0.10 \cdot 1.3953 \cdot {(100 \cdot 0.00703 \cdot 25)}^{1 / 3} + 0.10 \cdot 0 = 0.37323 N / {mm}^{2}

where

C_{Rd, c} = 0.15 / γ_{c} = 0.15 / 1.5 = 0.10

k = 1 + \sqrt{\frac{200}{1280}} = 1.3953 \leq 2.0

ρ_{1} = \frac{A_{sl}}{40 \cdot 128} = 0.007703 \leq 0.02

Design Value of Shear Stress acc. to DIN EN 1992-1-1, 6.2.5 (1)

v_{Edi} = β \cdot \frac{V_{Ed}}{z \cdot b_{i}} = 1.0 \cdot \frac{800 \cdot 10^{3}}{1152 \cdot 400} = 1.736 N / {mm}^{2}

where

β = 1.0

z = 0.9 \cdot d = 0.9 \cdot 1280 = 1152 mm

Eccentric Oriented Shear Stiffness G12

G_{12}^{"} = (G_{x z, i} - G_{y z, i}) \sin (φ - ß_{i}) \cos (φ - ß_{i})

Equivalent Stress According to Tresca

σ_{v} = \sqrt{{(σ_{x} - σ_{y})}^{2} + 4 {τ_{x y}}^{2}}

Equivalent Stress from Basic Stresses According to von Mises

σ_{v} = \sqrt{σ_{x}^{2} + σ_{y}^{2} - σ_{x} σ_{y} + 3 (τ_{x y}^{2})}

Equivalent Stress σ_{v, Bach}

σ_{v, Bach} = max [|σ_{1} - ν (σ_{2} + σ_{3})|; |σ_{2} - ν (σ_{3} + σ_{1})|; |σ_{3} - ν (σ_{1} + σ_{2})|]

Equivalent Stress σ_{v, Rankine,m}

σ_{v, Rankine, m} = \frac{1}{2} |σ_{x, m} + σ_{y, m}| + \frac{1}{2} \sqrt{{(σ_{x, m} - σ_{y, m})}^{2} + 4 {τ_{xy, m}}^{2}}

Equivalent Stress σ_v,Bach,- on Negative Surface Side

σ_{v, Bach, -} = \max (\frac{1 - ν}{2} |σ_{x, -} + σ_{y, -}| + \frac{1 + ν}{2} \sqrt{{(σ_{x, -} - σ_{y, -})}^{2} + 4 {τ_{xy, -}}^{2}}; ν |σ_{x, -} + σ_{y, -}|)

Equivalent Stress σ_v,Bach,+ on Positive Surface Side

σ_{v, Bach, +} = \max (\frac{1 - ν}{2} |σ_{x, +} + σ_{y, +}| + \frac{1 + ν}{2} \sqrt{{(σ_{x, +} - σ_{y, +})}^{2} + 4 {τ_{xy, +}}^{2}}; ν |σ_{x, +} + σ_{y, +}|)

Equivalent Stress σ_v,Mises from Basic Stresses

σ_{v, Mises} = \sqrt{σ_{x}^{2} + σ_{y}^{2} + σ_{z}^{2} - σ_{x} σ_{y} - σ_{x} σ_{z} - σ_{y} σ_{z} + 3 (τ_{x y}^{2} + τ_{x z}^{2} + τ_{y z}^{2})}

Equivalent Stress σ_v,Mises from Principal Stresses

σ_{v, Mises} = \sqrt{\frac{1}{2} [(σ_{1} - σ_{2})^{2} + (σ_{1} - σ_{3})^{2} + (σ_{2} - σ_{3})^{2}]}

Equivalent Stress σ_v,Mises,- on Negative Surface Side

σ_{v, Mises, -} = \sqrt{{σ_{x, -}}^{2} + {σ_{y, -}}^{2} - σ_{x, -} σ_{y, -} + 3 {τ_{x y, -}}^{2}}

Equivalent Stress σ_v,Mises,+ on Positive Surface Side

σ_{v, Mises, +} = \sqrt{{σ_{x, +}}^{2} + {σ_{y, +}}^{2} - σ_{x, +} σ_{y, +} + 3 {τ_{x y, +}}^{2}}