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DC EC 01: shear reinforcement in beam

Description

Geometry Cross-section: b=200mm
h=400mm
Material: C25/30
Concrete cover: c=40mm
Load Self-weight:
Dead loads:		 neglected
P=30\text{kN} at end of span
Internal forces V_{Ed.ULS}=40.5\text{kN}
Standard EN 1992-1-1 [- -]
Reinforcement At begin point A_{s1}=1304mm²
At end point A_{s1}=107mm²

Independent reference results

Open handcalculations

EN 1992-1-1 [–] allows the shear reinforcement to be calculated using the variable strut inclination method. In this method, you can choose

  • the inclination of the struts \theta between 1 < \cot (\theta) < 2.5.
    Diamonds starts its calculations with \theta=31° (more info).
  • and the inclination of the stirrups \alpha. In practice there are 2 options: either you have vertical stirrup (\alpha=90°) or inclined stirrups (\alpha=45°).
    Diamonds starts with vertical stirrups (\alpha=90°). If that is insufficient, it will try with inclined stirrups (\alpha=45°).

The variable strut inclination method in EN 1992-1-1 is often mixed up with the standard method used in ENV 1992-1 because of their similarity. But while the variable strut inclination method allows you to choose \theta, while the standard method always assumes \theta=45°.

For this beam we’ll calculate the shear reinforcement at two locations: at the end en begin point of the bar. We start with the end poit, because that’s the most interseting one.

At end point of the bar

Is shear reinforcement needed?

    \[d=h-c=400mm-40mm=360mm\]

    \[k=min\left ( 1+\sqrt{\frac{200mm}{d}},2 \right )=min\left ( 1+0.75,2 \right )=1.75\]

(6.2a)   \begin{align*} V_{Rd.c1} &=b \cdot d \cdot \left \frac{0.18}{\gamma _c} \cdot k \cdot \left ( \frac{100 \cdot A_{s1} \cdot f_{ck} }{b \cdot d} \right )^{\frac{1}{3}} + k \cdot \sigma \right \\ &=200mm \cdot 360mm \cdot \left \frac{0.18}{1.5} \cdot 1.75 \cdot \left ( \frac{100 \cdot 107mm^2 \cdot 25\text{MPa} }{200mm \cdot 360mm} \right )^{\frac{1}{3}} + 0 \right\\ &=23.36\text{kN} \end{align*}

\sigma is the compression (or tensile) stress due to the axial force. If the beam would be loaded with axial compression, V_{Rd.c1} would increase. If the beam with be loaded with axial tension, V_{Rd.c1} would decrease. Since no axial force is present in this example, \sigma equals zero.

(6.2b)   \begin{align*} V_{Rd.c2} &=b \cdot d \cdot 0.035 \cdot k^{\frac{3}{2}} \cdot \sqrt{f_{ck}} \\ &=200mm \cdot 360mm \cdot 0.035 \cdot 2^{\frac{3}{2}} \cdot \sqrt{25 \text{MPa}}\\ &=29.05\text{kN} \end{align*}

    \[V_{Rd.c}=max\left ( V_{Rd.c1},V_{Rd.c2} \right )=29.05\text{kN}\]

    \[V_{Ed.ULS} > V_{Rd.c} \Rightarrow \text{shear reinforcement is required}\]

Verification of the compression strut

(6.6N)   \begin{align*} \nu &=0.6 \cdot \left ( 1-\frac{f_{ck}}{250\text{MPa}} \right )=0.6 \cdot \left ( 1-\frac{25\text{MPa}}{250\text{MPa}} \right )\\ &=0.54 \\ \end{align*}

(6.9N)   \begin{align*} V_{Rd.max} &=0.9 \cdot d \cdot b\cdot \nu \cdot f_{cd}\cdot \frac{\cot (\theta )+\cot (\alpha )}{1+\cot (\theta )^2} \\ &=0.9 \cdot 360mm \cdot 200mm \cdot 0.54 \cdot 16.67\text{MPa}\cdot \frac{\cot (31^{\circ})+\cot (90^{\circ})}{1+\cot (90^{\circ})^2}\\ &=257.47\text{kN} \end{align*}

    \[V_{Ed.ULS}< V_{Rd.max} \Rightarrow \text{compression strut is OK}\]

Note: if V_{Ed.ULS} > V_{Rd.max}, then the compression strut would be insufficient and you’d see a skull in Diamonds.

Calculate required amount of shear reinforcement

(6.8)   \begin{align*} A_{sw} &=\frac{V_{Ed.ULS}}{0.9\cdot d\cdot f_{ywd}\cdot (\cot (\theta)+\cot (\alpha))\cdot \sin (\alpha)} \\ &=\frac{40.5 \text{kN}}{0.9\cdot 360mm\cdot 434.8 \text{MPa}\cdot (\cot (31^{\circ})+\cot (90^{\circ}))\cdot \sin (90^{\circ} )}\\ &=173mm^2/m \end{align*}

At begin point of the bar

Is shear reinforcement needed?

    \[d=h-c=400mm-40mm=360mm\]

    \[k=min\left ( 1+\sqrt{\frac{200mm}{d}},2 \right )=min\left ( 1+0.75,2 \right )=1.75\]

(6.2a)   \begin{align*} V_{Rd.c1} &=b \cdot d \cdot \left \frac{0.18}{\gamma _c} \cdot k \cdot \left ( \frac{100 \cdot A_{s1} \cdot f_{ck} }{b \cdot d} \right )^{\frac{1}{3}} + k \cdot \sigma \right \\ &=200mm \cdot 360mm \cdot \left \frac{0.18}{1.5} \cdot 1.75 \cdot \left ( \frac{100 \cdot 1304mm^2 \cdot 25\text{MPa} }{200mm \cdot 360mm} \right )^{\frac{1}{3}} + 0 \right\\ &=53.75\text{kN} \end{align*}

\sigma is the compression (or tensile) stress due to the axial force. If the beam would be loaded with axial compression, V_{Rd.c1} would increase. If the beam with be loaded with axial tension, V_{Rd.c1} would decrease. Since no axial force is present in this example, \sigma equals zero.

(6.2b)   \begin{align*} V_{Rd.c2} &=b \cdot d \cdot 0.035 \cdot k^{\frac{3}{2}} \cdot \sqrt{f_{ck}} \\ &=200mm \cdot 360mm \cdot 0.035 \cdot 2^{\frac{3}{2}} \cdot \sqrt{25 \text{MPa}}\\ &=29.05\text{kN} \end{align*}

    \[V_{Rd.c}=max\left ( V_{Rd.c1},V_{Rd.c2} \right )=53.75\text{kN}\]

    \[V_{Ed.ULS} < V_{Rd.c} \Rightarrow \text{only minimum shear reinforcement is required}\]

Calculate required amount of shear reinforcement

(9.5)   \begin{align*} A_{sw.min}&=\frac{0.08\cdot \sqrt{f_{ck}}}{f_{yk}}\cdot b\cdot 10^3 \\ &=\frac{0.08\cdot \sqrt{25\text{MPa}}}{500\text{MPa}}\cdot 200mm\cdot 10^3=160mm^2/m \end{align*}

Diamonds results and comparison

Shear reinforcement calculated by Diamonds

Results Independent reference Diamonds Difference
Shear reinforcement at begin point 173 mm²/m 173 mm²/m 0%
Shear reinforcement at end point 160 mm²/m 160 mm²/m 0%

References

  • Tested in Diamonds 2025.

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