Formulas - Search term: Shear Forc...

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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

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 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})

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}

Ideal Shear Area in y-Direction

A_{y} = \frac{{I_{z}}^{2}}{G_{ref} \cdot \int_{A *} (^{\frac{S_{z}}{t *}) 2} \cdot \frac{1}{G} d A *}

Ideal Shear Area in z-Direction

A_{z} = \frac{{I_{y}}^{2}}{G_{ref} \cdot \int_{A *} (^{\frac{S_{y}}{t *}) 2} \cdot \frac{1}{G} d A *}

Ideal Shear Stiffness of Stiffening Bracing

s_{id} = E_{D} \cdot A_{D} \cdot \sin (^{α) 2} \cdot \cos (α)

Ideal Shear Stiffness of Stiffening Bracing

s_{id} = E_{D} \cdot A_{D}' \cdot \sin (^{α) 2} \cdot \cos (α)

Limit Shear Stress τ_max

τ_{\max} = \frac{f_{y}}{\sqrt{3}}

Longitudinal Shear Stress in Flank Fillet Welds from Torsional Moment

τ_{∥, M_{x}} = \frac{M_{x}}{2 \cdot A_{m} \cdot a_{w}}

Longitudinal Shear Stress in Neck Seams from Shear Force

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

Maximum Resistance Check of the Concrete Compression Strut of the Shear Truss According to ACI 318-19, 22.5.1.2

Vu ≤ Vc + 8 · fc' · bw · d

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}

Maximum Shear Stress

τ_{\max} = \frac{1}{2} (σ_{1} - σ_{3})

Maximum Shear Stress According to DIN EN 1992-1-1, 6.4.3 (3)

v_{Ed} = \frac{β \cdot V_{Ed}}{u_{1} \cdot d} = \frac{1.10 \cdot 0.7873 MN}{4.19 m \cdot 0.19 m} = 1.088 MN / m

Maximum Shear Stress τ_max Perpendicular to Surface

τ_{\max} = \sqrt{{τ_{x z}}^{2} + {τ_{y z}}^{2}}

Maximum Value of Punching Shear Rresistance According to DIN EN 1992-1-1/NA, 6.4.5 (3)

v_{Rd, \max} = 1.4 \cdot v_{Rd, c} = 1.4 \cdot 0.929 = 1.300 MN / m^{2} \geq v_{Ed} = 1.088 MN / m^{2}

Maximum Value of Shear Resistance of Concrete Vc, max According to ACI 318-19, 22.5.5.1.1

Vc,max = 5 · λ · fc' · bw · d

Maximum Value of Shear Resistance of Concrete Vc, max According to ACI 318-19, 22.5.5.1.1

V_{c, \max} = 5 \cdot λ \cdot \sqrt{f_{c}^{'}} \cdot b_{w} \cdot d

Minimum Shear Reinforcement av,min According to ACI 318-19, 9.6.3.4

a_{V, \min} = \frac{A_{v}}{s} = \max [0.75 \cdot \sqrt{f_{c}^{'}} \cdot \frac{b_{w}}{f_{yt}}; 50 \cdot \frac{b_{w}}{f_{yt}})]

Modulus of Displacement per Shear Plane

K_{ser} = ρ_{m}^{1.5} \cdot \frac{d}{23}

Modulus of Elasticity, Shear Modulus, and Poisson's Ratio

E = 2 G (1 + ν)

Mximum Value of Design Shear Resistance at Interface acc. to DIN EN 1992-1-1, 6.2.5 (1)

v_{Rd, \max} = 0.5 \cdot ν \cdot f_{cd} = 0.5 \cdot 0.70 \cdot 14.167 = 4.958 N / {mm}^{2}

where

ν = 0.70 for the indented interface