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DC EC 07: reinforcement in beam shear and torsion (ULS design)

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
T=6\text{kNm} at end of span
Internal forces M_{Ed.ULS}=162\text{kNm}
V_{Ed.ULS}=40.5\text{kN}
T_{Ed.ULS}=8.10\text{kNm}
Standard EN 1992-1-1 [- -]

Independent reference results

Open handcalculations

The beam is subjected to a bending moment M, shear force V and torsion T.

  • The bending moment M will cause longitudinal reinforcement.
  • The shear force V will cause transverse reinforcement,
    And, if the shear force is so high that stirrups under 45° are required, also longitudinal web reinforcement (see DC EC 09).
  • The torsion T will cause transverse and longitudinal reinforcement.

We calculate the required reinforcement for each of these internal forces and superpose them.

Reinforcement for M_{Ed.ULS}

The result will be the same as in example DC EC 11.

  • At the begin point: A_{y.M.top}=1304mm^2 and A_{y.M.bottom}=144mm^2
  • At the end point: A_{y.M.top}=A_{y.M.bottom}=107mm^2.

Reinforcement for V_{Ed.ULS}

The result will be the same as in example DC EC 01.

  • At the begin point: A_{wz.V}=160mm^2/m
  • At the end point: A_{wz.V}=173mm^2/m.

Reinforcement for T_{Ed.ULS}

Is torsional reinforcement needed?

The torsional resistance of a concrete beam cannot be separated from the resistance to shear force. Because the lowest shear resistance was found at the end of the beam (see DC EC 01), we will also determine the torsional reinforcement there.

    \[A = b \cdot h = 200mm \cdot 400mm=80000mm^2\]

    \[u=2 \cdot \left ( b+h \right )=2 \cdot \left ( 200mm+400mm \right )=1200mm\]

    \[t_{ef}=max\left (\frac{A}{u},2 \cdot c \right )= max\left (\frac{80000mm^2}{1200mm},2 \cdot 40mm \right )=80mm\]

    \[b_{ef}=b- t_{ef}= 200mm-80mm=120mm\]

    \[h_{ef}=h- t_{ef}= 400mm-80mm=320mm\]

    \[A_k=b_{ef} \cdot h_{ef}= 120mm \cdot 320mm = 38400mm^2\]

    \[T_{Rd.c}=f_{ctd} \cdot t_{ef} \cdot 2 \cdot A_k = 1.71\text{MPa} \cdot 80mm \cdot 2 \cdot 38400mm^2=10.51 \text{kNm}\]

This beam is loaded with shear and torsion, so we should consider the combination of both:

    \[\frac{T_{Ed.ULS}}{T_{Rd.c}} +\frac{ V_{Ed.ULS} }{V_{Rd.c}} > 1  \Rightarrow \text{calculated reinforcement is required}\]

V_{Rd.c} was calculated in DC EC 01 and equals 29.05\text{kN}.

Verification of the compression strut

    \[\nu =0.6 \cdot \left ( 1-\frac{f_{ck}}{250\text{MPa}} \right )=0.6 \cdot \left ( 1-\frac{25\text{MPa}}{250\text{MPa}} \right )=0.54\]

(6.30)   \begin{align*} T_{Rd.max} &=2 \cdot \nu \cdot f_{cd} \cdot A_k \cdot t_{ef} \cdot sin(\theta) \cdot cos(\theta) \\ &=2 \cdot 0.54 \cdot 16.6\text{MPa} \cdot 38400mm^2 \cdot 80mm \cdot sin(31^\circ) \cdot cos(31^\circ)\\ &=24.4\text{kNm} \end{align*}

This beam is loaded with shear and torsion, so we should consider the combination of both:

    \[\frac{T_{Ed.ULS}}{T_{Rd.max}} + \frac{V_{Ed.ULS}}{V_{Rd.max}} < 1  \Rightarrow \text{compression strut is OK}\]

V_{Rd.max} was calculated in DC EC 01 and equals 257.47\text{kN}

Note: if \frac{T_{Ed.ULS}}{T_{Rd.max}} + \frac{V_{Ed.ULS}}{V_{Rd.max}} >1, then the compression strut would be insufficient and you’d see a skull in Diamonds.

Calculate required amount of torsional reinforcement

  • Stirrups

        \[u_k= 2 \cdot \left ( b_{ef} + h_{ef}\right )= 2 \cdot \left ( 120mm + 320mm\right )=880mm\]

        \begin{align*} A_{w.T}= A_{wy.T}&=\frac{T_{Ed.ULS} \cdot tan(\theta))}{2 \cdot A_k \cdot f_{yd}}\\ &=\frac{8.1\text{kNm} \cdot tan(31^\circ)}{2 \cdot 38400mm^2 \cdot 434.7\text{MPa}}\\ &=146 mm^2/m \end{align*}

    A_{w.T} is the required cross-section of the stirrups on one side of the beam. But A_{wz} in Diamonds represent the cross-section of the stirrups on the left AND right side of the beam. Analogous for A_{wy} which represents the cross-section of the stirrups on the top AND bottom side of the beam. Therefor the result A_{w.T} must be multiplied by 2 to obtain the Diamonds results.

        \[A_{wz.T}=A_{wy.T}=2 \cdot A_{w.T}=2 \cdot 146mm^2/m=292mm^2/m\]

  • Longitudinal

        \[A_{s.T}=\frac{T_{Ed.ULS} \cdot cot(\theta) \cdot u_k}{2 \cdot A_k \cdot f_{yd}}=\frac{8.1\text{kNm} \cdot cot(31^\circ) \cdot 880mm}{2 \cdot 38400mm^2 \cdot 434.7 \text{MPa}}=355mm^2 \]

    This amount needs to be regulary distributed over the cross-section.

        \begin{align*} A_{z.T}&=A_{s.T} \cdot \frac{h-2 \cdot d_1 }{(b-2 \cdot d_1)+(h-2 \cdot d_1)}\\ &=355mm^2 \cdot \frac{400mm -2 \cdot 40mm}{(200mm-2 \cdot 40mm)+(h-2 \cdot 40mm)}\\ &=258mm^2 \end{align*}

        \begin{align*} A_{z.T}&=A_{s.T} \cdot \frac{b-2 \cdot d_1 }{(b-2 \cdot d_1)+(h-2 \cdot d_1)}\\ &=355mm^2 \cdot \frac{200mm -2 \cdot 40mm}{(200mm-2 \cdot 40mm)+(h-2 \cdot 40mm)}\\ &=97mm^2 \end{align*}

Combine reinforcement amounts

At end point
  • Stirrups

        \[A_{wz}=A_{wz.T}+A_{wz.V}=292mm^2/m+173mm^2/m=465mm^2/m\]

        \[A_{wy}=A_{wy.T}+A_{wz.V}=0mm^2/m+173mm^2/m=173mm^2/m\]

  • Longitudinal
    A_{y.T} is neglected because it’s smaller than the minimum reinforcement.

        \[A_{y.top}=A_{y.M.top}+A_{y.T}=107mm^2+0mm^2=107mm^2\]

        \[A_{y.bottom}=A_{y.top}\]

        \[A_{z.left}=A_{z.right}=A_{z.T}=259mm^2\]

At begin point
  • Stirrups
    A_{wz.V} is neglected because the A_{wz.T} is larger than the minimum shear reinforcement.

        \[A_{wz}=A_{wz.T}+A_{wz.V}=292mm^2/m+0mm^2/m=292mm^2\]

        \[A_{wy}=A_{wy.T}=292mm^2/m\]

  • Longitudinal

        \[A_{y.top}=A_{y.M.top}+A_{y.T}=1304mm^2+97mm^2=1401mm^2\]

        \[A_{y.bottom}=A_{y.M.bottom}+A_{y.T}=144mm^2+97mm^2=241mm^2\]

        \[A_{z.left}=A_{z.right}=A_{z.T}=259mm^2\]

Diamonds results and comparison

Longitudonal reinforcement calculated by Diamonds (EN 1992-1-1 [- -])

Results Independent reference Diamonds Difference
At begin point Ay 1401 mm² 1401 mm² 0,00%
Az 259 mm² 259 mm² 0,00%
Awz 292 mm²/m 292 mm²/m 0,00%
Awy 292 mm²/m 292 mm²/m 0,00%
At end point Ay 107 mm² 107 mm² 0,00%
Az 107 mm² 107 mm² 0,00%
Awz 173 mm²/m 173 mm²/m 0,00%
Awy 465 mm²/m 464 mm²/m 0,00%

References

  • Tested in Diamonds 2024r01.

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