Shaft in Torsion (Solid, Circular)

stresstorque-powermass-cost

Verified build 6 relations · 5 identities proven · 1 modeling step · 6 parity samples

Every machine that spins — every motor, gearbox, drill, driveshaft, wind turbine — pushes its power through a shaft in torsion. Twist a solid circular bar and the cross-sections rotate rigidly, shearing the material between them; the strain (and so the stress) grows linearly from zero at the axis to a maximum at the surface:

$\tau_{max} = \frac{16\,T}{\pi d^3} \qquad \theta = \frac{T L}{G J} \qquad P = T\,\omega$

Three things to notice, one per formula:

The stress contains no material property. $\tau$ is set by torque and diameter alone — swap steel for aluminum in the widget and watch $\tau_{max}$ refuse to move while everything else changes. Material choice never changes the stress in a statically determinate part; it changes what the part can tolerate ( $\sigma_y$ → SF) and how much it distorts ( $G$ → θ).
Diameter is cubed. A 26% larger shaft halves the stress; doubling it cuts the stress eightfold. No alloy upgrade competes with a few millimeters of radius — and since the middle of a solid shaft barely works (the stress cone is zero at the axis), hollow shafts keep most of the strength at a fraction of the mass.
Torque, not power, sizes the shaft. $P = T\omega$ means the same 50 kW arrives as gentle torque at high speed or brutal torque at low speed. The power-in configuration makes the widget find $T$ first — drop the speed knob and watch the stress climb at constant power. This is the whole reason gearboxes exist: they trade speed for torque, and the slow side always needs the fatter shaft.

The twist matters more often than beginners expect: machine tools lose accuracy, control linkages feel spongy, and long drivelines resonate — all serviceability failures governed by $G$ and $J$ , arriving long before anything yields.

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Configuration

Material

T3, bare flat sheet 0.010-0.128 in. thick, AMS 4037 / AMS-QQ-A-250/4 (MIL-HDBK-5J Table 3.2.3.0(b1), p. 3-71)

Bound properties of 2024-T3 aluminum sheet (bare)
G	4 Msi	typical	mil-hdbk-5j
sigma_y	47 ksi	design min.	mil-hdbk-5j
rho	0.1 lb/inch**3	typical	mil-hdbk-5j

4 materials in the database are not listed here: no published value in our cited sources for every property this THING needs.

Materials modeled here: 2024-T3 aluminum sheet (bare) 304 stainless steel 6061-T6 aluminum 7075-T6 aluminum AISI 1045 medium-carbon steel AISI 4340 low-alloy steel (Ni-Cr-Mo) ASTM A36 structural steel (hot-rolled) C26000 Cartridge Brass (70/30) Ti-6Al-4V

Governing relations

J = \frac{\pi d^4}{32}

Assumes: solid circular cross-section

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: strain kinematics, the torsion formula, angle of twist).

\tau_{max} = \frac{T\,(d/2)}{J} = \frac{16\,T}{\pi d^3}

Assumes: linear elastic, shear stress grows linearly from zero at the axis to a max at the surface; pure torsion — no bending, no axial load, no stress concentrations (keyways, steps) · Valid while: Surface shear stress exceeds the shear yield strength (σ_y/2 by the maximum-shear-stress criterion) — the outer fibers have yielded and every elastic number here stops being the truth.

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: strain kinematics, the torsion formula, angle of twist).

\theta = \frac{T\,L}{G\,J}

Assumes: linear elastic, uniform section and torque along the length; plane cross-sections rotate rigidly (exact for circular sections, wrong for everything else)

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: strain kinematics, the torsion formula, angle of twist).

P = T\,\omega

Assumes: steady rotation — torque and speed constant

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §3-12 (power transmission, P = Tω), §5-4 (maximum-shear-stress theory, S_sy = 0.5 S_y).

\mathrm{SF} = \frac{\sigma_y/2}{\tau_{max}}

Assumes: maximum-shear-stress (Tresca) criterion — shear yield at σ_y/2; conservative next to the distortion-energy value 0.577 σ_y

m = \rho \frac{\pi d^2}{4} L

Assumes: prismatic solid shaft, uniform density

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: strain kinematics, the torsion formula, angle of twist).

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

\gamma = \frac{r_{c} \theta}{L}

1. The modeling step: cross-sections of a circular shaft rotate rigidly without warping, so a straight line scribed along the surface becomes a shallow helix. The shear strain at radius r is the helix angle, γ = rθ/L — zero on the axis, maximum at the surface. — kinematics: plane sections rotate rigidly modeling step

\tau = \frac{G d \theta}{2 L}

2. Hooke's law in shear, τ = Gγ, evaluated at the surface r = d/2 where the strain peaks. The stress distribution is a cone: zero at the center, linear to the rim — which is why hollow shafts are such a good deal; the middle barely works. — Hooke's law in shear

\theta = \frac{L T}{G J}

3. The torque is the resultant of the stress distribution: T = ∫ r·τ(r) dA = (Gθ/L)∫ r² dA = GθJ/L, where J = ∫ r² dA is the polar second moment. Solve for the twist. (The build's test suite re-runs that integral with a computer algebra system as an independent check.) — integrate the stress distribution

\tau = \frac{16 T}{\pi d^{3}}

4. Substitute J = πd⁴/32 and r = d/2. Stress falls with the cube of diameter — upsizing a shaft by 26% halves the stress, which is why diameter, not material, is the first lever a designer reaches for. — substitute the solid circular section

P_{w} = T \omega

5. Power is torque times angular speed. The same 50 kW can be a screaming 500 rad/s at 100 N·m or a grinding 50 rad/s at 1000 N·m — and it is the torque, not the power, that sizes the shaft. This is why gearboxes exist. — definition of mechanical power

SF = \frac{\sigma_{y}}{2 \tau}

6. Yield in shear arrives at σ_y/2 by the maximum-shear-stress criterion. Note what each material property touches: G set the twist (stiffness), σ_y sets this margin (strength), and τ itself contains no material property at all — geometry and load alone. — maximum-shear-stress criterion

How it fails

The widget’s margin is first yield in shear at the surface, under steady pure torsion. Real shafts rarely die that politely:

Fatigue at stress concentrations. Almost every real shaft failure is fatigue, and it starts where the clean $16T/\pi d^3$ field is disturbed: keyway corners, shoulder fillets, circlip grooves, press-fit edges. A sharp keyway can more than double the local stress; millions of rotating cycles do the rest. Shaft design in practice is mostly fatigue design.
Ductile vs brittle torsion fracture — read the break. Torsion puts maximum shear on planes perpendicular to the axis and maximum tension on a 45° helix. A ductile shaft shears off flat, straight across; a brittle one (cast iron, hardened steel, chalk — try it) cracks along the 45° helix, leaving the classic spiral fracture face. The broken end tells you which material regime you were in.
Combined loading. Real shafts also bend — gears and pulleys push sideways between bearings. Bending and torsion combine into a worse principal-stress state than either alone; the pure- torsion safety factor here is an upper bound on the truth.
Torsional vibration and resonance. A long shaft is a torsion spring connecting inertias. Engine pulse frequencies that land on the natural frequency wind the shaft far beyond its static twist — crankshafts carry tuned dampers precisely for this.
Yield ≠ collapse. Past first yield, the cone-shaped stress distribution flattens as the outer fibers saturate; a solid section carries up to 4/3 of its first-yield torque before fully plastic. The margin is real but small, and the shaft keeps a permanent twist as a souvenir.
Non-circular sections warp. Everything here assumed circular symmetry. Square or splined sections warp out of plane, the corner stress is NOT given by any formula on this page, and an open slit (a keyway run wild) can cut torsional stiffness by orders of magnitude.

Fixed-Fixed Torsion Shaft (Interior Torque)
A solid circular shaft built into a wall at BOTH ends, with a torque applied at an interior station. Equilibrium gives one equation for the two wall reaction torques; the missing equation is compatibility — the twist at the load point is single-valued — and the build solves the coupled 2×2 system exactly. The larger reaction lands on the SHORTER segment, and the material cancels out.
- stress
- mass-cost
Rectangular Shaft in Torsion (Saint-Venant)
Twist a solid rectangular bar and the shear stress does something the round shaft never does: it peaks at the MIDDLE of the long side and drops to exactly zero at the corners. Two cited coefficients c1, c2 — functions only of the side ratio a/b — set the peak stress and the twist, and an equal-area round shaft beats it on both counts. Why square shafts are a bad deal.
- stress
- mass-cost
Shaft under Combined Bending + Torsion
Real shafts never get to choose: the belt that twists them also bends them. Bending and torsion land on the same surface element, Mohr's circle finds the worst plane, and two failure criteria — Tresca and von Mises — disagree by up to 15 % about how bad it is.
- stress
- mass-cost
Thin-Walled Tube in Torsion (Bredt)
Why driveshafts, airframes, and bike frames are closed tubes: in torsion, what matters is not how much metal you have but how much AREA the wall encloses. Bredt's shear flow makes any closed section solvable with two knobs — and the isoperimetric inequality polices which sections can exist at all.
- stress
- mass-cost
Cantilever Beam (End Load)
A beam fixed at one end, loaded at the other — the fruit-fly of structures. One widget shows why stiffness (E) and strength (σ_y) are independent axes: swap steel for titanium and deflection goes UP while the safety factor also goes up.
- stress
- mass-cost
Composite Bar (Core + Sleeve)
A solid core inside a concentric sleeve, bonded between rigid end plates and pushed by a centric axial load. The two materials must stretch together, so the load splits in proportion to each member's axial stiffness A·E — and the build solves that coupled 2×2 share exactly. Swap the sleeve's metal and watch the load migrate to the stiffer member.
- stress
- mass-cost

Chains with

Outputs whose SI dimension and quantity kind match another THING's input — the only wires the planner's connectionLegal accepts (invariant 2, computed at build time, not hand-listed). Wire these on the chaining demo.

Impact Loading (Falling Mass, Energy Method)
- m_shaft m
Symmetric Two-Bar Truss
- theta alpha
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
- tau q
- tau sigma_allow
Simply Supported Beam (Center Load + UDL)
- SF SF
Fixed-Fixed Torsion Shaft (Interior Torque)
- T T
Rectangular Shaft in Torsion (Saint-Venant)
- T T
Shaft under Combined Bending + Torsion
- SF SF_t
- T T
Thin-Walled Tube in Torsion (Bredt)
- SF SF
- T T

+ 16 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources

Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — §3.3–3.5 (circular shafts in torsion: strain kinematics, the torsion formula, angle of twist).
Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §3-12 (power transmission, P = Tω), §5-4 (maximum-shear-stress theory, S_sy = 0.5 S_y).

Shaft in Torsion (Solid, Circular)

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Governing relations

Derivation

How it fails

Related THINGs

Chains with

Sources