In practice, torsion is commonly classified into two distinct behaviours as below:
Uniform (St Venant) Torsion: Uniform torsion occurs primarily in closed sections such as Circular Hollow Sections (CHS) and Rectangular Hollow Sections (RHS). In these members, the torsional shear stresses are distributed around the perimeter of the section, and warping deformations are minimal.
Warping Torsion in Open Sections
Because of this complexity, warping torsion is frequently overlooked or underestimated in design.
Twin Beam Method (Simplified Approach)
Accurate warping torsion calculations can be highly detailed and complex. The twin beam method provides a simplified and conservative approach that is commonly used in practice to assess warping torsion in open sections.
In this method, torsion is resisted by treating the two flanges as individual beams in bending, while the contribution of the web is neglected. This allows engineers to perform a practical hand check without resorting to full numerical analysis.
While FEA is recommended where higher accuracy or greater capacity is required, the twin beam method offers a clear and efficient way to quickly assess whether warping torsion is critical and whether the flange bending capacity is sufficient. More detailed formulations can be found in Roark’s Formulas for Stress and Strain.
The trosion has been calculated for two different sections in the following:
Example 1 – Uniform Torsion in a CHS
Section: 168.3 × 6.4 CHS (Grade C350L0)
Applied torsion, T* = 20 kNm
Length, L = 3.0 m
Material and Section Properties
Yield strength, fy = 350 MPa
Shear modulus, G ≈ 80,000 MPa
Outer radius, ro = 84.15 mm
Inner radius, ri = 77.75 mm
Torsion Capacity Check
Torsional section modulus:
C = π (ro⁴ − ri⁴) / (2 ro)
C = 254 × 10³ mm³
Design torsion capacity:
ϕTu = 0.9 × 0.6 × fy × C
ϕTu ≈ 48 kNm
Twist Check
Angle of twist:
θ = T* L / (K G)
Torsion constant:
K = 0.5 π (ro⁴ − ri⁴) = 21.4 × 10⁶ mm⁴
θ ≈ 0.035 radians ≈ 2°
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Example 2 – Warping Torsion Using the Twin Beam Method
Section: 150UC37 (Grade 300)
Applied torsion, T* = 2.5 kNm
Beam length, L = 1.0 m
Boundary condition: Fixed at both ends
Section Properties
Flange width, bf = 154 mm
Section depth, d = 162 mm
Flange thickness, tf = 11.5 mm
Flange yield strength, fyf = 300 MPa
Distance between flanges, e = 150.5 mm
Equivalent Flange Force
Ff* = T* / e = 16.6 kN
Flange Bending Demand
For torsion applied at mid-span:
Mf* = Ff* × a × (L − a) / L
Mf* ≈ 4.2 kNm
Flange Bending Capacity
ϕMf = 0.9 × fyf × tf × bf² / 6
ϕMf ≈ 4.6 kNm
