Description
| Geometry | Cross-section: |  and  |
|---|---|---|
| Material: | C25/30 | |
| Concrete cover: |  |
|
| Load | Design shear force: |  |
| Standard | EN 1992-1-1 [–] |
Results
Handcalculation
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
between 1 < 
< 2.5.
Diamonds starts its calculations with
=31° (more info). - and the inclination of the stirrups
. In practice there are 2 options: either you have ‘straight’ stirrup (=90°) or inclined stirrups (=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 =45°.
![\[\nu =0.6 \cdot \left ( 1-\frac{f_{ck}}{250MPa} \right )=0.6 \cdot \left ( 1-\frac{25MPa}{250MPa} \right )=0.54\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-9b3f886272dbd319c4081e4347db9f76_l3.png "Rendered by QuickLaTeX.com")
![\[V_{Rd.max}=b\cdot h\cdot \nu \cdot f_{cd}\cdot \frac{\cot (\theta )+\cot (\alpha )}{1+\cot (\theta )^2}\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-a934b12b3f57a5a7cdaf3ebdbb791b45_l3.png "Rendered by QuickLaTeX.com")
![\[V_{Rd.max}=200mm\cdot 400mm\cdot 0.54 \cdot 16.67MPa\cdot \frac{\cot (31^{\circ})+\cot (90^{\circ})}{1+\cot (\theta )^2}=257.47kN\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-4a92ca4a493565e869cacf936ecff5e1_l3.png "Rendered by QuickLaTeX.com")
![\[V_{Ed.ULS}< V_{Rd.max} \Rightarrow OK\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-71a5c7bb03454777ff166d8a48e47abd_l3.png "Rendered by QuickLaTeX.com")
![\[A_{sw}=\frac{V_{Ed.ULS}}{0.9\cdot d\cdot f_{ywd}\cdot (\cot (31^{\circ})+\cot (90^{\circ}))\cdot \sin (90^{\circ} )}=438mm^2/m\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-8077ab541913bc9468a2dd4a17a344c7_l3.png "Rendered by QuickLaTeX.com")
![\[A_{sw.min}=\frac{0.08\cdot \sqrt{f_{ck}}}{f_{yk}}\cdot b\cdot 10^3=\frac{0.08\cdot \sqrt{25MPa}}{500MPa}\cdot 200mm\cdot 10^3=160mm^2/m\]](https://support.buildsoft.eu/wp-content/ql-cache/quicklatex.com-8df847a288daef73fd6cde9cfdb2f94a_l3.png "Rendered by QuickLaTeX.com")
Shear reinforcement calculated by Diamonds
| Results | Independent reference | Diamonds | Difference |
|---|---|---|---|
| Maximum shear reinforcement | 438 mm²/m | 438 mm²/m | 0,00% |
| Minimum shear reinforcement | 160 mm²/m | 160 mm²/m | 0,00% |
References
- EN 1992-1-1: 2005 + AC: 2010 §6.2.3 en §9.2.2
- Van Hooymissen, L., Spegelaere, M., Van Gysel, A., & De Vylder, W. (2002). Gewapend beton. Academia Press.
This reference does not contain this exact example, but it’s a good reference if you want to understand how reinforcement calculations work. However, keep in mind the calculations in this book are made done using the NBN B15-002 which is comparable to ENV 1992-1. The principle is similar to EN 1992-1-1 but some parameters differ slightly.
- Tested in Diamonds 2023r01.
