Shear and torsion¶

The two are treated together because they compete for the same concrete strut — and it is this competition that the interaction check captures.

Shear¶

NBR 6118 calculation Model I: struts at 45°, with a constant \(V_c\) term.

\[ \tau_{wd} = \frac{V_d}{b_w\,d} \qquad\qquad V_d = \gamma_f\,V_k \]

Strut crushing¶

\[ \tau_{wu} = 0.27\,\alpha_v\,f_{cd} \qquad\qquad \alpha_v = 1 - \frac{f_{ck}}{250} \]

If \(\tau_{wd} > \tau_{wu}\), the section is insufficient and the calculation is stopped. No reinforcement can fix strut crushing.

Transverse reinforcement¶

\[ \tau_c = \begin{cases} \dfrac{0.126\,f_{ck}^{2/3}}{\gamma_c} & f_{ck} \le 50\ \text{MPa} \\[10pt] \dfrac{0.8904\,\ln(1+0.11\,f_{ck})}{\gamma_c} & f_{ck} > 50\ \text{MPa} \end{cases} \]

\[ \tau_d = \max\left[1.11\left(\tau_{wd} - \tau_c\right);\ 0\right] \qquad\qquad \frac{A_{sw}}{s} = \frac{100\,b_w\,\tau_d}{f_{ywd}} \]

Stirrup steel is limited to 435 MPa

\(f_{ywd} = \min\left(f_{yk}/\gamma_s;\ 435\ \text{MPa}\right)\).

The code limits the stress in the transverse reinforcement precisely to control the width of diagonal cracks in service — there is no point using high-strength steel in the stirrup if it is going to crack before it yields.

Minimum reinforcement¶

\[ \rho_{sw,min} = \frac{0.2\,f_{ct,m}}{f_{ywk}} \qquad\qquad f_{ct,m} = \begin{cases} 0.3\,f_{ck}^{2/3} & f_{ck} \le 50 \\[4pt] 2.12\,\ln(1+0.11 f_{ck}) & f_{ck} > 50 \end{cases} \]

with \(f_{ywk} \le 500\) MPa and \(A_{sw,min} = \rho_{sw,min}\cdot 100\, b_w\).

Torsion¶

The equivalent hollow section¶

Torsion is resisted by a closed shear flow near the faces — the core contributes little. That is why the solid section is replaced by an equivalent hollow section of thickness \(t\).

Starting from \(t_0 = \dfrac{b\,h}{2(b+h)}\) and \(c_1 = d'\):

t₀ ≥ 2c₁t₀ < 2c₁

\[ t = t_0 \qquad A_e = (b - t)(h - t) \qquad u_e = 2(b + h - 2t) \]

\[ t = \min(t_0;\ b - 2c_1) \qquad A_e = (b - 2c_1)(h - 2c_1) \qquad u_e = 2(b + h - 4c_1) \]

\(A_e\) is the area enclosed by the centreline of the wall, and \(u_e\) the perimeter of that line.

Stress and limit¶

\[ \tau_{td} = \frac{T_d}{2\,A_e\,t} \qquad\qquad \tau_{tu} = 0.25\,\alpha_v\,f_{cd} \]

Reinforcement¶

\[ \frac{A_{sw,t}}{s} = \frac{100\,T_d}{2\,A_e\,f_{yd}} \qquad\text{(per leg)} \qquad\qquad A_{sl,t} = \frac{T_d\,u_e}{2\,A_e\,f_{yd}} \]

The longitudinal torsion reinforcement is distributed over the faces in proportion to the outer perimeter — which gives shares for the bottom face, the top face and the two sides.

Minimum, when there is torsion:

\[ A_{sl,min} = 0.5\,\rho_{sw,min}\,u_e\,b \]

The interaction check¶

This is the central point of this chapter:

\[ \frac{\tau_{td}}{\tau_{tu}} + \frac{\tau_{wd}}{\tau_{wu}} \;\le\; 1 \]

Shear and torsion add compression in the same strut. Checking each one on its own against its own limit passes sections that the combination fails — which is why the code requires the sum of the ratios.

If the sum exceeds 1, the calculation is stopped: it is crushing, and calls for a larger section.

Maximum stirrup spacing¶

The same indicator governs the spacing:

Sum of the ratios	Maximum spacing
\(\le 0.67\)	\(\min(0.6\,d;\ 30\ \text{cm})\)
\(> 0.67\)	\(\min(0.3\,d;\ 20\ \text{cm})\)

The closer to crushing, the closer the stirrups — they start stitching cracks that have already formed.

Adding it all up¶

The total transverse reinforcement combines both terms, bearing in mind that the torsion stirrup counts two legs:

\[ \frac{A_{sw,tot}}{s} = \frac{A_{sw,v}}{s} + 2\,\frac{A_{sw,t}}{s} \qquad\ge\ \rho_{sw,min}\cdot 100\,b \]

And the longitudinal reinforcement also receives the increase due to shear — the tensile share that the truss model transfers to the tension chord:

\[ \Delta A_{s,V} = \frac{0.5\,V_d}{f_{yd}} \]

Limits of validity¶

Range of application

The torsion implemented is that of a rectangular section. T sections under torsion use the web as the reference section, which is an approximation.
Shear Model I, with struts at 45°. Model II, with variable inclination, is not implemented — it usually gives less reinforcement in heavily loaded members.
The torsion considered is equilibrium torsion. Compatibility torsion, which can be neglected when there is redistribution, is the designer's decision: if it is not entered, the program does not invent it.
There is no check of crack width in service, which in members under torsion is usually what actually governs the detailing.
The limits of \(f_{ywd}\) at 435 MPa and \(f_{ywk}\) at 500 MPa are those of the code for transverse reinforcement.