Beam torsion design per ACI 318-19

Q: Can I neglect torsion entirely?

Yes, if Tu <= phi*Tth per Cl 22.7.4.1, where Tth = lambda*sqrt(f'c)*Acp^2/pcp. Below this threshold the torque is small enough that it does not need to be considered in design.

Q: Why does calcnote use a 45 degree truss angle for torsion?

ACI 318-19 permits the strut angle theta to be taken as 45 degrees for nonprestressed members, which gives a simple, conservative design with cot(theta) = 1. The code allows other angles between 30 and 60 degrees, but a 45 degree angle is the standard simplification and is what calcnote uses.

Q: How do I combine torsion and shear stirrups?

Add the demands per leg: the required area per leg is Av/(2s) for shear plus At/s for torsion, where At is the area of one leg. The stirrup must be a closed stirrup per Cl 9.7.5, because torsion needs a closed loop around the section.

Torsion twists a beam about its longitudinal axis and is resisted by closed stirrups plus longitudinal bars distributed around the perimeter. ACI 318-19 first asks whether torsion can be neglected, then whether it is equilibrium or compatibility torsion, and finally sizes the transverse and longitudinal steel with a space-truss model. Torsion is combined with shear through the stirrup area and a cross-section adequacy check. For the shear part, see beam shear design.

Applicable clauses

ACI 318-19 reference	Topic
Cl 22.7.1.1	Torsion design applicability
Cl 22.7.4.1	Threshold torsion T_th (can neglect)
Cl 22.7.5.1	Cracking torque T_cr
Cl 9.5.4.1	Equilibrium torsion: design for full T_u
Cl 9.5.4.2	Compatibility torsion: reduce T_u to φT_cr
Table 22.7.6.1	T_n (nominal torsional strength)
Cl 9.6.4.2	Minimum transverse reinforcement (combined shear + torsion, and torsion alone)
Cl 9.6.4.3	Minimum longitudinal torsional reinforcement
Cl 9.7.5	Torsion detailing (closed stirrups, spacing)
Cl 22.7.7.1	Cross-section adequacy (RSS interaction)

When to design for torsion (Cl 22.7.4.1)

Torsion may be neglected when the factored torque is below the threshold torque:

T_u \leq \phi\,T_{th}, \qquad T_{th} = \lambda\,\sqrt{f'_c}\;\frac{A_{cp}^2}{p_{cp}} \quad (\phi = 0.75)

Here A_cp is the area enclosed by the outside perimeter of the concrete section, p_cp is that perimeter, and λ is the lightweight concrete factor (1.0 for normal-weight concrete). √f′_c is in psi, giving T_th in lb-in. If T_u exceeds φT_th, torsional reinforcement is required.

Equilibrium vs compatibility torsion (Cl 9.5.4.1, 9.5.4.2)

Equilibrium torsion is required to keep the structure in static equilibrium. Examples are a cantilevered slab framing into a beam, or a spandrel with no alternative load path. The beam must be designed for the full T_u.

Compatibility torsion arises only because adjoining members must deform together, for example an interior spandrel that can shed load to the slab once it cracks. After cracking the beam twists freely and the torque redistributes, so T_u may be reduced to the cracking torque:

T_u = \phi\,T_{cr}, \qquad T_{cr} = 4\,\lambda\,\sqrt{f'_c}\;\frac{A_{cp}^2}{p_{cp}}

The member and its supports must then be checked for the redistributed forces. When in doubt about the load path, treat the torsion as equilibrium torsion.

Torsional section properties

For a rectangular beam of width b and height h:

A_{cp} = b\,h, \qquad p_{cp} = 2\,(b + h)

The reinforcement is measured on the centerline of the outermost closed stirrup. With clear cover c to the stirrup and stirrup bar diameter d_s:

x_1 = b - 2c - d_s, \qquad y_1 = h - 2c - d_s

A_{oh} = x_1\,y_1, \qquad p_h = 2\,(x_1 + y_1)

A_oh is the area enclosed by the stirrup centerline and p_h is its perimeter. The gross-section properties (A_cp, p_cp) drive the threshold and cracking torque; the core properties (A_oh, p_h) drive the reinforcement.

Nominal torsional strength (Table 22.7.6.1)

Torsional strength comes from the closed stirrups acting as the tension ties of a space truss:

T_n = \frac{2\,A_o\,A_t\,f_{yt}}{s}\,\cot\theta, \qquad A_o = 0.85\,A_{oh}, \quad \theta = 45^\circ

where A_t is the area of one stirrup leg and s is the stirrup spacing. Setting φT_n = T_u and rearranging gives the required transverse reinforcement:

\frac{A_t}{s} = \frac{T_u}{\phi\,2\,A_o\,f_{yt}\,\cot\theta} \quad (\phi = 0.75)

With θ = 45°, cotθ = 1 and the expression reduces to A_t/s = T_u / (φ · 2 · 0.85A_oh · f_yt).

Longitudinal torsional reinforcement (Cl 9.6.4.3)

The same truss pushes the diagonal struts outward along the beam, and additional longitudinal steel is needed to balance that axial component:

A_l = \frac{A_t}{s}\,p_h\,\frac{f_{yt}}{f_y}\,\cot^2\theta

The minimum longitudinal torsional reinforcement is:

A_{l,min} = \frac{5\,\sqrt{f'_c}\;A_{cp}}{f_y} - \frac{A_t}{s}\,p_h\,\frac{f_{yt}}{f_y}

In the A_l,min expression, A_t/s is not taken less than 25b_w/f_yt. The bars are distributed around the perimeter of the closed stirrups, with at least one bar in each corner. They are added to the flexural steel; they do not replace it.

Combined shear and torsion (Cl 9.6.4.2)

Shear stirrups have two legs sharing the shear demand, while torsion is resisted by the single outside leg on each side. The area required per leg and per unit length is the sum:

\frac{A_{v+t}}{s} = \frac{A_v}{2\,s} + \frac{A_t}{s} \quad \text{(per leg)}

The total must also satisfy the combined minimum of Cl 9.6.4.2: (A_v + 2A_t)/s ≥ max(0.75√f′_c, 50) b_w / f_yt (f′_c, f_yt in psi). The stirrups must be closed per Cl 9.7.5, and the stirrup spacing limits still apply; torsion adds a spacing cap of the lesser of p_h/8 and 12 in.

Cross-section adequacy (Cl 22.7.7.1)

Shear and torsion stresses add on the same section. For solid sections the root-sum-square (RSS) interaction must satisfy:

\sqrt{\left(\frac{V_u}{b_w\,d}\right)^2 + \left(\frac{T_u\,p_h}{1.7\,A_{oh}^2}\right)^2} \;\leq\; \phi\left(\frac{V_c}{b_w\,d} + 8\sqrt{f'_c}\right)

This limits diagonal crushing of the concrete. If the check fails, no amount of stirrups can fix it and the section must be enlarged (wider, deeper, or higher f′_c).

What calcnote computes

The beam torsion check performs the following steps:

Compute A_cp, p_cp, A_oh, p_h from the section and stirrup geometry.
Compare T_u against φT_th (Cl 22.7.4.1); neglect torsion when below the threshold.
Apply equilibrium (full T_u) or compatibility (φT_cr) torque as selected.
Check the RSS cross-section adequacy (Cl 22.7.7.1) and reject sections that fail.
Compute the required A_t/s and A_l, and the minimums of Cl 9.6.4.2 and 9.6.4.3.
Combine A_t/s with the shear stirrup demand and report the governing spacing.

Check your beam torsion design: run a beam calculation or see a torsion demo result.

Sources

ACI 318-19: Cl 22.7.1.1 (applicability), Cl 22.7.4.1 (T_th), Cl 22.7.5.1 (T_cr), Cl 9.5.4.1 and 9.5.4.2 (equilibrium and compatibility torsion), Table 22.7.6.1 (T_n), Cl 9.6.4.2 and 9.6.4.3 (minimum reinforcement), Cl 9.7.5 (detailing), Cl 22.7.7.1 (cross-section adequacy).

FAQ

What is the difference between equilibrium and compatibility torsion?

Equilibrium torsion is required for static equilibrium: if the member does not resist the torque, the structure cannot carry the load (for example a cantilevered slab framing into a beam). Compatibility torsion arises only from deformation compatibility between members and can be redistributed after the beam cracks, so T_u may be reduced to φT_cr.

Can I neglect torsion entirely?

Yes, if T_u ≤ φT_th per Cl 22.7.4.1, where T_th = λ√f′_c A_cp²/p_cp. Below this threshold the torque is small enough that it does not need to be considered in design.

Why does calcnote use a 45 degree truss angle for torsion?

ACI 318-19 permits the strut angle θ to be taken as 45° for nonprestressed members, which gives a simple, conservative design with cotθ = 1. The code allows other angles between 30 and 60 degrees, but a 45° angle is the standard simplification and is what calcnote uses.

How do I combine torsion and shear stirrups?

Add the demands per leg: the required area per leg is A_v/(2s) for shear plus A_t/s for torsion, where A_t is the area of one leg. The stirrup must be a closed stirrup per Cl 9.7.5, because torsion needs a closed loop around the section.