Mechanical & Structural Engineering
Bredt-Batho Closed Thin-Walled Box Torsion and Shear Flow Formula
Calculate closed-tube shear flow q = T / (2*A_m), thinnest-wall peak shear stress, and Bredt’s torsion constant J_closed vs open-slit stiffness collapse.
See by hand why slitting a closed box girder destroys 99.5% of torsional stiffness
In a closed thin-walled box section of median centerline width b_m and height h_m (enclosed median area A_m = b_m * h_m), torque T circulates as a constant Bredt-Batho (1896) shear flow q_flow = T / (2 * A_m) around the entire perimeter. Peak shear stress occurs in the THINNEST wall t_min via tau_max = T / (2 * A_m * t_min), and Bredt’s closed torsion constant is J_closed = 4 * A_m^2 / SUM(s_i / t_i).
Verify exact numbers on a 200 x 100 mm median rectangular steel box: b_m = 0.20 m (B2), h_m = 0.10 m (B3, so A_m = 0.020 m^2), flange thickness t_f = 0.008 m (B4, 8 mm), web thickness t_w = 0.005 m (B5, 5 mm), and torque T = 4,000 N*m (B6). Shear flow is 4,000 / 0.040 = 100,000 N/m, peak web shear stress is 100,000 / 0.005 = 20.00 MPa, line integral SUM(s/t) = 2*0.20/0.008 + 2*0.10/0.005 = 50 + 40 = 90, and J_closed = 1,777.7778 cm^4 — 232.09x stiffer than slitting one corner open (J_open = 7.6600 cm^4)!
| Box Torsion Metric | Bredt-Batho Formula | Exact Output |
|---|---|---|
| Shear Flow q_flow (N/m) | T / (2 * b_m * h_m) | 100,000 N/m (100 N/mm) |
| Peak Wall Shear Stress (MPa) | 10^-6 * q_flow / MIN(t_f, t_w) | 20.0000 MPa (in 5 mm web) |
| Closed Torsion Constant J_closed | 10^8 * 4 * A_m^2 / (2*b_m/t_f + 2*h_m/t_w) | 1,777.7778 cm^4 |
| Open-Slit Torsion Constant J_open | 10^8 * (2*b_m*t_f^3 + 2*h_m*t_w^3) / 3 | 7.6600 cm^4 (232.09x drop!) |
Peak Closed-Box Wall Shear Stress (MPa) Formula
Use MIN(t_f, t_w) automatically so whichever plate gauge is thinner governs peak Bredt shear stress.
=LET(b_med, B2, h_med, B3, t_min, MIN(B4, B5), t_nm, B6, IF(OR(b_med <= 0, h_med <= 0, t_min <= 0, t_nm < 0), "Invalid Tube Input", ROUND((10^-6) * t_nm / (2 * b_med * h_med * t_min), 4)))Bredt Closed Torsion Constant J_closed (cm^4) Formula
Evaluate 4 * A_m^2 / (2*b_m/t_f + 2*h_m/t_w) and multiply m^4 by 10^8 to report cm^4 alongside open-slit J_open.
=LET(b_med, B2, h_med, B3, t_fl, B4, t_wb, B5, IF(OR(b_med <= 0, h_med <= 0, t_fl <= 0, t_wb <= 0), "Invalid Tube Input", LET(a_enc, b_med * h_med, s_int, (2 * b_med) / t_fl + (2 * h_med) / t_wb, ROUND((10^8) * 4 * (a_enc^2) / s_int, 4))))Try the example
Use the original values or change them to test your understanding.