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DC EC 09: shear reinforcement (under 45°) in beam

Description

Geometry Cross-section: b=200mm
h=400mm
Material: C25/30
Concrete cover: c=40mm
Load Self-weight:
Dead loads:		 neglected
P=280 \text{kN}
Internal forces V_{Ed.ULS}=330.7\text{kN}
Standard EN 1992-1-1 [- -]
Reinforcement A_{s1}=1327mm² from Diamonds model

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 ‘straight’ stirrup (\alpha=90°) or inclined stirrups (\alpha=45°).
    Diamonds start with straight stirrups (\alpha=90°), if that is insufficient, he 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. The standard method always assumes \theta=45°.

Is shear reinforcement needed?

    \[d=h-c=400mm-40mm=360mm\]

    \[k=min\left ( 1+\sqrt{\frac{200mm}{d}},2 \right )=min\left ( 1+\sqrt{\frac{200mm}{360mm}},2 \right )=1.75\]

(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 1327mm^2 \cdot 25\text{MPa} }{200mm \cdot 360mm} \right )^{\frac{1}{3}} + 0 \right\\ &=54.06\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*}

    \[V_{Rd.c}=max\left ( V_{Rd.c1},V_{Rd.c2} \right )=54.06\text{kN}\]

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

Verification of the compression strut

Check the compression strut using the default angles \theta=31° and \alpha=90°.

(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*}

(6.14)   \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 (31^{\circ})^2}\\ &=257.47\text{kN} \end{align*}

    \[V_{Ed.ULS}> V_{Rd.max} \Rightarrow \text{compression strut is NOT OK}\]

Increase the angle of the compression strut to \theta=45°, while still assuming the stirrups are vertical (\alpha=90°):

(6.14)   \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 (45^{\circ})+\cot (90^{\circ})}{1+\cot (45^{\circ})^2}\\ &=291.6\text{kN} \end{align*}

    \[V_{Ed.ULS}> V_{Rd.max} \Rightarrow \text{compression strut is still NOT OK}\]

Leave the angle of the compression strut on \theta=45°, while assuming the stirrups are inclined (\alpha=45°):

(6.14)   \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 (45^{\circ})+\cot (45^{\circ})}{1+\cot (45^{\circ})^2}\\ &=583.2\text{kN} \end{align*}

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

Calculate required amount of shear reinforcement

(6.13)   \begin{align*} A_{sw.45^{\circ}} &=\frac{V_{Ed.ULS}}{0.9\cdot d\cdot f_{ywd}\cdot (\cot (\theta)+\cot (\alpha))\cdot \sin (\alpha)} \\ &=\frac{330.7\text{kN}}{0.9\cdot 360mm\cdot 434.8 \text{MPa}\cdot (\cot (45^{\circ})+\cot (45^{\circ}))\cdot \sin (45^{\circ} )}\\ &=1660mm^2/m \end{align*}

A contractor will never be keen on placing stirrups inclined. However inclined stirrups can be decomposed into vertical stirrups + web reinforcement, which is easier to place. Diamonds will show those decomposed amounts in its reinforcement results.

    \[A_{sw.Diamonds} =A_{sw.45^{\circ}} \cdot \sin (45^{\circ})=1174mm^2/m\]

    \[A_{z.Diamond.due to shear} =\frac{A_{sw.45^{\circ}} \cdot d}{2} =\frac{=1660mm^2/m \cdot 360mm}{2}=211mm^2/m\]

A_{sw.45^{\circ}} is expressed in mm^2/m. So to convert this amount to mm², we need to multiply it with the useful height d and distribute it between the left and right side of the beam (thus divide it by 2).

Diamonds results and comparison

Shear reinforcement calculated by Diamonds

Web reinforcement due to shear calculated by Diamonds

Results Independent reference Diamonds Difference
Shear reinforcement 1174 mm²/m 1174 mm²/m 0%
Web reinforcement due to shear 211 mm²/m 211 mm²/m 0%

References

  • Tested in Diamonds 2025.

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