Chassis Suspension & Motorsport Anti-Roll Bars, Hollow Torsion-Bar d^4 Scaling
Hollow Torsion-Bar Spring Rate, Twist & Shear Formula
Calculate Polar Area Moment J (cm⁴), Torsional Angular Stiffness k_theta (N·m/deg), Suspension Windup Twist Angle (deg), and Outer-Surface Shear Stress tau_max (MPa).
Gun-drilling a 50% ID bore down a 32 mm torsion bar cuts unsprung steel mass by 25%—while losing ONLY 6.25% (0.5^4) of torsional spring rate!
**Automotive Suspension Torsion Bars and Tubular Motorsport Anti-Roll (`Sway`) Bars** (`Outer Diameter d_o` mm, `Inner Bore d_i` mm, `Active Effective Length L` mm, `Shear Modulus G = 78.5` GPa for shot-peened `5160` / `300M` spring steel) exploit **Fourth-Power Polar Moment of Area `J = (pi / 32) * (d_o^4 - d_i^4)`** (`mm⁴`, `/ 10,000` for `cm⁴`)! Because core steel near the neutral axis carries almost zero shear strain (`tau(r) ~ r`), boring out `d_i / d_o = 0.50` removes `(0.50)^2 = 25.0%` of dead weight while subtracting just **`(0.50)^4 = 6.25%` of Angular Spring Rate `k_{theta} = (((G * 1,000) * J) / L) * (pi / 180) / 1,000`** (`N·m/deg`)! Under Control-Arm Bump Torque `T` (`N·m`), **Windup Twist Angle equals `theta = T / k_{theta}`** (`deg`) and **Outer-Fiber Maximum Torsional Shear Stress equals `tau_{max} = (16,000 * T * d_o) / (pi * (d_o^4 - d_i^4))`** (`MPa`)!
Verify on a `32` mm OD x `16` mm ID (`6` mm wall) Motorsport Tubular Anti-Roll / Torsion Bar—Outer_Diameter `d_o` = 32.0 mm (B2), Inner_Bore `d_i` = 16.0 mm (B3), Active_Length `L` = 1,050 mm (B4), Steel_Shear_Modulus `G` = 78.5 GPa (B5), and Full-Bump_Arm_Torque `T` = 1,850 N·m (B6): Polar Moment `J` equals `9.651` cm⁴ (E2), Torsional Spring Rate `k_{theta}` equals **`125.93` N·m/deg** (E3), Windup Angle equals `14.69°` (E4), and Outer-Surface Shear Stress equals **`306.7` MPa** (E5)!
| Hollow vs Solid Torsion-Bar KPI (do=32 mm, L=1,050 mm, G=78.5 GPa, T=1,850 N·m) | Formula | Verified Output |
|---|---|---|
| Polar Moment of Area J (cm⁴) | ((PI() / 32) * (do_mm^4 - di_mm^4)) / 10000 | 9.651 cm⁴ hollow (10.294 cm⁴ solid) |
| Torsional Stiffness (N·m/deg) | (((G_GPa*1000 * J_mm4) / L_mm) * PI/180) / 1000 | 125.93 N·m/deg (134.33 solid = +6.7%) |
| Windup Twist Deflection (deg) | Applied_Torque_Nm / Stiffness_Nm_deg | 14.69° hollow (13.77° solid) |
| Outer-Surface Shear Stress (MPa) | (16000 * T_Nm * do_mm) / (PI * (do^4 - di^4)) | 306.7 MPa hollow (287.5 MPa solid) |
Saint-Venant / Coulomb Hollow-Shaft Torsional Stiffness & Shear Formula
Validate `Outer_Diameter d_o` in `[8, 150]` mm, `Inner_Bore d_i` in `[0, 0.85 * d_o)` (`0` = solid bar!), `Active_Length L` in `[100, 4,000]` mm, `Shear_Modulus G` in `[20, 100]` GPa (`~26` Al, `~44` Ti, `~78.5` steel), and `Torque T > 0` N·m!
=LET(do_m, B2, di_m, B3, lb_m, B4, gs_g, B5, tq_n, B6, IF(OR(do_m < 8, do_m > 150, di_m < 0, di_m >= 0.85 * do_m, lb_m < 100, lb_m > 4000, gs_g < 20, gs_g > 100, tq_n <= 0), "Invalid Torsion Input", ROUND((16000 * tq_n * do_m) / (PI() * ((do_m ^ 4) - (di_m ^ 4))), 1)))Solid-Bar Comparison (`d_i = 0` mm) 25%-Mass-for-6.25%-Stiffness Proof Check
Set `B3 = 0` mm to compare an identical `32` mm solid bar (`33%` heavier than the hollow tube!): Polar Moment E2 nudges up from `9.651` to `10.294` cm⁴, Stiffness E3 rises just **`+6.67%` to `134.33` N·m/deg** (`13.77°` twist in E4), and Peak Surface Shear E5 drops only `6.25%` from `306.7` to `287.5` MPa—confirming why GT race cars and aircraft landing struts always run hollow torsion tubes!
=IF(ISTEXT(E5), E5, LET(jm_q, (PI() / 32) * ((B2 ^ 4) - (B3 ^ 4)), ROUND((((B5 * 1000) * jm_q) / B4) * (PI() / 180) / 1000, 2)))Try the example
Use the original values or change them to test your understanding.