Torsional Analysis of Structural Steel Members — Part 2: Design Procedure, Worked Examples & Code Checks

11. The seven-step design procedure

Step 1 — Establish the load and its eccentricity

• Locate the point of load application on the section.
• Locate the shear center — for a Channel it lies outside the web.
• Compute the eccentricity e, then T = P × e (concentrated) or t = w × e (distributed).

Step 2 — Screening check: is a torsion analysis needed at all?

T_r / T_c ≤ 0.20   →   torsion may be neglected  (AISC 360 §H3.1)

Quick estimate of T_c:
  Open section  : T_c = φ · 0.6 · F_y · J / t_max
  Closed section: T_c = φ · 0.6 · F_y · C   (C from the AISC Manual)

Steps 3 to 5 — Properties, load case, derivatives

Compute a = √(E·Cw / G·J) and λ = L/a, pick the Case from the 12 in DG9, then evaluate four quantities at the critical location.

• θ — angle of twist, used for the serviceability check.
• θ' — gives the St. Venant shear stress τ_t.
• θ'' — gives the warping normal stress σ_w.
• θ''' — gives the warping shear stress τ_w.

Step 6 — Compute the three torsional stresses

τ_t = G · t · θ'              use the THICKEST element
σ_w = E · W_no · θ''          governs at the outer flange tip
τ_w = E · S_w · θ''' / t      usually at the flange–web junction

12. Combining stresses and checking to AISC 360

The total stress at a point is the algebraic sum of every component acting at that same point.

Total normal stress:  f_n = f_bx + f_by + σ_w + f_a
Total shear stress :  f_v = f_vx + f_vy + τ_t + τ_w

f_bx, f_by = major- and minor-axis bending   f_a  = axial
f_vx, f_vy = shear on each axis

Closed sections (HSS) — the H3-6 interaction equation

( P_r/P_c + M_rx/M_cx + M_ry/M_cy ) + ( V_r/V_c + T_r/T_c )² ≤ 1.0

Take the LARGER of the two axes for V_r/V_c.

Nominal torsional strength:  T_n = F_cr · C
  Rectangular HSS:  C ≈ 2·(B−t)·(H−t)·t
  Round HSS      :  C = π·(D−t)²·t / 2

Open sections — check several points

AISC 360 gives no simple interaction equation for open sections. Compute the total stress and compare with the capacity point by point — at least four points on the section.

13. Formulas for the common loading cases

Case 1 — Pinned–Pinned, concentrated T at midspan

The most common case in practice. Boundary conditions: θ = 0 and θ'' = 0 at both ends.

For 0 ≤ z ≤ L/2:

  θ(z)    = T/(G·J) · [ z/2 − (a/2)·sinh(z/a)/cosh(L/2a) ]
  θ'(z)   = T/(2·G·J) · [ 1 − cosh(z/a)/cosh(L/2a) ]
  θ''(z)  = −T/(2·G·J·a) · sinh(z/a)/cosh(L/2a)
  θ'''(z) = −T/(2·G·J·a²) · cosh(z/a)/cosh(L/2a)

Extreme values at midspan:
  θ_max  = T/(2·G·J) · [ L/2 − a·tanh(L/2a) ]
  θ'_max = T/(2·G·J) · [ 1 − 1/cosh(L/2a) ]

Case 3 — Pinned–Pinned, uniform t over the full span

This is the spandrel beam carrying a wall load.

θ'(z)   = t/(G·J) · [ (L−2z)/2 + a·sinh((L−2z)/2a)/cosh(L/2a) ]
θ''(z)  = t/(G·J) · [ −1 + cosh((L−2z)/2a)/cosh(L/2a) ]
θ'''(z) = t/(G·J·a) · sinh((L−2z)/2a)/cosh(L/2a)

Case 9 — Fixed–Free (cantilever), T at the free end

θ'(z)   =  T/(G·J) · [ 1 − cosh((L−z)/a)/cosh(L/a) ]
θ''(z)  =  T/(G·J·a) · sinh((L−z)/a)/cosh(L/a)
θ'''(z) = −T/(G·J·a²) · cosh((L−z)/a)/cosh(L/a)

Extremes:  θ and θ' peak at z = L (free end)
           θ'' and θ''' peak at z = 0 (fixed support)

14. Worked example 1 — W-shape under an eccentric load

Member        : W16×26  (A992, F_y = 50 ksi)
Span          : L = 20 ft = 240 in
Load          : P = 10 kips at midspan, eccentricity e = 6 in
End conditions: Pinned–Pinned (free warping both ends)

Design torque: T = P × e = 10 × 6 = 60 kip-in → use Case 1.

a = √( 29,000 × 456 / (11,200 × 0.262) ) = √4,507 = 67.1 in
L/a = 240 / 67.1 = 3.58
L/2a = 1.789     cosh(1.789) = 3.121     tanh(1.789) = 0.946

θ_max = 60/(2 × 11,200 × 0.262) × [ 120 − 67.1 × 0.946 ]
      = 0.01022 × 56.5 = 0.578 rad

Try a heavier section: W16×50

W16×50:  J = 1.52 in⁴    C_w = 2,340 in⁶

a = √( 29,000 × 2,340 / (11,200 × 1.52) ) = √3,986 = 63.1 in
L/2a = 1.902     tanh(1.902) = 0.956

θ_max = 60/(2 × 11,200 × 1.52) × [ 120 − 63.1 × 0.956 ]
      = 0.001762 × 59.7 = 0.105 rad  ≈ 6°

15. Worked example 2 — HSS, a far simpler route

Member  : HSS 8×6×3/8  (A500 Gr.C, F_y = 50 ksi)
Span    : L = 20 ft
Load    : P = 10 kips at midspan, eccentricity e = 6 in
Also    : M_x = 50 kip-ft, V = 10 kips

T_r = P × e = 60 kip-in

h/t = 20.0    versus    2.45·√(E/F_y) = 2.45·√(29,000/50) = 59.0
→ 20.0 is below 59.0  →  limit state: YIELDING

F_cr = 0.6 × 50 = 30 ksi
T_n  = F_cr × C = 30 × 72.7 = 2,181 kip-in
T_c  = φ × T_n = 0.90 × 2,181 = 1,963 kip-in   (LRFD)

T_r / T_c = 60 / 1,963 = 0.031  ≤  0.20   →  TORSION NEGLECTED

For reference: running H3-6 anyway

M_rx = 600 kip-in     M_cx = 0.90 × 50 × 21.3 = 958.5 kip-in
V_r  = 10 kips        V_c  ≈ 150.8 kips

( 0 + 600/958.5 + 0 ) + ( 10/150.8 + 60/1,963 )²
= 0.626 + (0.066 + 0.031)² = 0.626 + 0.0094 = 0.635  ≤  1.0   ✓

16. AISC versus Eurocode (SCI P385)

The bimoment concept in SCI P385

B   = E · C_w · θ''            (bimoment)
σ_w = B · W_no / C_w

Bending–torsion interaction (simplified, open sections):
  M_y,Ed / M_y,Rd + σ_w,Ed / (f_y/γ_M0) ≤ 1.0

17. Design strategy and six traps

Priority order when torsion appears

• 1. Eliminate it — put the load through the shear center, add lateral bracing, detail connections so reactions pass through the shear center. Always the best option.
• 2. Minimise it — cut the eccentricity, add bracing points, use a composite slab (the concrete restrains warping of the top flange).
• 3. Switch to HSS — if torsion is significant and unavoidable. The calculation is far simpler because warping drops out.
• 4. Full open-section analysis — only when a W-shape or Channel is mandatory.

Six traps that catch engineers out

Which W-shape if you must use one?

• Wide, thick flanges raise both J and Cw — clearly better.
• A thick web raises J.
• Greater depth raises Cw but lowers J — a trade-off.
• Prefer wide-flange series (W14, W12); avoid slender sections such as W24×55 or W21×44 — their J is tiny.

18. Torsion design checklist

• Does the load pass through the shear center? If so, there is no torsion.
• Compute T = P × e using a correctly located shear center.
• Check T_r/T_c ≤ 0.20 — if satisfied, stop; torsion is negligible.
• Choose the section type: HSS if torsion is significant.
• Look up J, Cw, W_no, S_w from the AISC Manual and DG9 Appendix A.
• Compute a = √(E·Cw / G·J) and L/a.
• Determine the end conditions: Fixed, Pinned or Free.
• Select the load Case from the 12 in DG9 Appendix B.
• Evaluate θ, θ', θ'', θ''' at the critical location.
• Compute τ_t, σ_w and τ_w.
• Combine f_n and f_v at no fewer than four points on the section.
• Check f_n ≤ φ·F_y and f_v ≤ φ·0.6·F_y.
• Check the twist θ against the serviceability limit.
• For HSS: check equation H3-6 ≤ 1.0.

19. Quick reference tables

MATERIAL CONSTANTS (steel)
  E = 29,000 ksi = 200,000 MPa
  G = 11,200 ksi =  77,200 MPa
  ν = 0.30

LRFD LIMITS
  Normal stress : f_n ≤ 0.90 · F_y
  Shear stress  : f_v ≤ 1.00 · 0.6·F_y
  Von Mises     : √(f_n² + 3f_v²) ≤ 0.90 · F_y

J, Cw and a for common W-shapes

Channels and HSS

20. Seven key recommendations

• Try to detail the torsion away before you start calculating.
• If torsion is significant, use a closed section (HSS) — simpler and far more effective.
• Locate the shear center precisely, especially for Channels.
• Get the end conditions right — Fixed versus Pinned changes the answer dramatically.
• Always check the twist θ alongside the stresses.
• Combine stresses at the same point — never add extremes from different locations.
• Check the DG9 errata before trusting any formula.

Part of the “Industrial structural design guide” series by Roberto Structural. The content is technical guidance only; the engineer remains responsible for verifying and adapting it to each project and the governing code. Codes change — always check the current edition, including the AISC Design Guide 9 errata.