How this instrument works
The polar moment of inertia, J, measures how a cross-section's material is arranged around its own centre axis, and that arrangement is what resists twisting. Formally it is J = ∫r² dA, the sum of every small patch of area dA weighted by the square of its distance r from the centre. Material far from the axis contributes far more than material near it, because the r² term grows faster than the area shrinks — which is exactly why a solid shaft's stiffness is so sensitive to diameter.
For a solid circular shaft that integral collapses to the closed form J = πd⁴⁄32, with d the full diameter. The fourth power is not a rounding of convenience; it falls straight out of integrating r² over a disc, where both the r² weighting and the area element scale with radius. Double the diameter and J rises by a factor of sixteen, not two or four — the reason a modest increase in shaft size buys a disproportionate gain in torsional capacity, and conversely why a worn or undersized shaft loses strength so fast.
The formula only holds for solid, or with a subtracted inner term hollow, circular cross-sections, because only a circle keeps every point at a fixed distance r from the axis without the section warping out of its own plane under twist. A square or I-beam shaft twists non-uniformly and needs a torsion constant taken from tables or numerical methods, not this closed form — a distinction that trips up anyone reaching for J on a non-round part.
- Enter the Shaft diameter — the full diameter of the solid round shaft, not the radius. The unit menu accepts mm, cm, or in.
- Leave the section solid; this instrument assumes a solid circular shaft, not a hollow tube.
- Read the Polar moment of inertia, m⁴ field — the engine converts your diameter to metres internally before applying J = πd⁴⁄32.
- Use the result with τ = Tr/J to find shear stress at a given radius, or rearrange it to size a shaft for a known torque.
Worked example — a 50 mm steel drive shaft
Take a solid steel shaft 50 mm in diameter, a common size for a small machine-tool drive shaft. Convert to metres first: d = 0.05 m. Then apply the formula: J = π(0.05)⁴ ⁄ 32 = π × 6.25×10⁻⁶ ⁄ 32 = 6.13592315154×10⁻⁷ m⁴. That figure, not the diameter itself, is what a torsion calculation actually needs.
With J in hand, an engineer checking this shaft under a 200 N·m torque finds the maximum shear stress at the outer surface, r = 0.025 m, from τ = Tr/J = 200 × 0.025 ⁄ 6.13592315154×10⁻⁷ ≈ 8.15 MPa — well inside the roughly 80–100 MPa allowable for common shaft steels, confirming the 50 mm size has ample margin before diameter needs to grow.
Questions
What is the difference between polar moment of inertia and moment of inertia?
Polar moment of inertia (J) measures resistance to torsion, twisting about the shaft's own axis, while the ordinary area moment of inertia (I) measures resistance to bending. For a solid circular shaft, J = πd⁴⁄32 is exactly twice I = πd⁴⁄64, because J sums r² over the whole disc while a bending axis only captures half that distribution. Confusing the two gives a torsion answer that is off by a factor of two.
Does this formula work for a hollow shaft or tube?
No, this instrument computes J for a solid circular cross-section only. A hollow shaft with outer diameter d_o and inner diameter d_i uses J = π(d_o⁴ − d_i⁴) ⁄ 32, subtracting the missing inner material. Because the inner region sits closest to the axis, removing it costs relatively little stiffness, which is why tubes are common where torsional stiffness per unit weight matters.
Why does polar moment of inertia use the fourth power of diameter?
Because J is the integral of r² over the cross-sectional area, and for a disc both that weighting and the area element scale with radius, compounding to a fourth-power relationship. Practically it means diameter is the single most powerful design lever: doubling it multiplies J, and torsional stiffness, by sixteen, while halving it divides J by the same factor.
How is J used to find shear stress in a twisted shaft?
Through τ = Tr/J, where T is the applied torque and r is the distance from the axis to the point of interest, usually the outer surface, where shear stress peaks. A 50 mm shaft with J = 6.136×10⁻⁷ m⁴ carrying 200 N·m sees about 8.15 MPa at its surface; the same torque on a slimmer shaft concentrates into a far higher stress because J has fallen so much faster than the diameter.
Can I use this for a square or rectangular shaft?
Not directly. The closed-form J = πd⁴⁄32 depends on every point of the cross-section sitting at a well-defined radius from the axis, which only a circle or annulus does. Non-circular sections warp out of plane under torque and need a separate torsion constant, usually taken from handbook tables such as Roark's Formulas for Stress and Strain rather than a single formula.
What units does the result come out in?
Metres to the fourth power, m⁴, regardless of which unit you typed the diameter in. Enter millimetres, centimetres, or inches and the instrument converts to metres before applying the formula, so a 50 mm shaft and a 1.9685 in shaft return the identical J. Handbooks sometimes quote J in mm⁴ or in⁴ instead; multiply m⁴ by 10¹² or divide by 0.0254⁴ to convert.