Torsion in Steel Members – Theory and Worked Examples
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 SectionsBecause 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 CHSSection: 168.3 × 6.4 CHS (Grade C350L0)Applied torsion, T* = 20 kNmLength, L = 3.0 mMaterial and Section PropertiesYield strength, fy = 350 MPa Shear modulus, G ≈ 80,000 MPa Outer radius, ro = 84.15 mm Inner radius, ri = 77.75 mmTorsion Capacity CheckTorsional section modulus: C = π (ro⁴ − ri⁴) / (2 ro) C = 254 × 10³ mm³ Design torsion capacity: ϕTu = 0.9 × 0.6 × fy × C ϕTu ≈ 48 kNmTwist CheckAngle of twist: θ = T* L / (K G) Torsion constant: K = 0.5 π (ro⁴ − ri⁴) = 21.4 × 10⁶ mm⁴ θ ≈ 0.035 radians ≈ 2°------------------------------------------------------------------------ Example 2 – Warping Torsion Using the Twin Beam MethodSection: 150UC37 (Grade 300)Applied torsion, T* = 2.5 kNmBeam length, L = 1.0 mBoundary condition: Fixed at both endsSection PropertiesFlange 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 mmEquivalent Flange ForceFf* = T* / e = 16.6 kNFlange Bending DemandFor torsion applied at mid-span:Mf* = Ff* × a × (L − a) / LMf* ≈ 4.2 kNmFlange Bending CapacityϕMf = 0.9 × fyf × tf × bf² / 6ϕMf ≈ 4.6 kNm
