Polar Moment of Inertia Calculator

What Is the Polar Moment of Inertia Calculator?

How to use the polar moment of inertia calculator

1. Choose the unknown. The Solve for menu opens on Polar moment of inertia, and the other two choices are Outer diameter and Bore diameter. The applied torque is never the unknown.
2. Type the outer diameter. The full width across the shaft, in mm, cm, m or in, opening on 30 mm with presets at 17, 30 and 400 mm. The box accepts 1e-4 to 2 m.
3. Type the bore diameter. The hole up the middle, as a diameter — not a radius and not a wall thickness. A 30 mm shaft bored to 21 mm has a 4.5 mm wall. Zero is a solid shaft.
4. Type the applied torque. It takes N.m, kN.m or N.mm and opens on 300 N·m. It moves the stress reading and never touches J.
5. Read the answer. The headline carries the answer to six significant figures, trailing zeros dropped, in the unit that box was showing, and it switches to scientific notation below 0.0001 and at or above 10 million. Set the unit first, point Solve for at the target second and type your figures last — switching the target wipes every box back to its default and only the unit menus survive it.
6. Read the surface stress. T · (D/2) / J at the torque you gave, with what fraction of 250 MPa that is beside it. The general form is used, never the solid-only shortcut.
7. Read the stiffness per unit of material. Exactly 1 + k² with k = d/D: 1.0000x solid, 1.6400x at an 80 per cent bore. The surface stress goes up at the same time.
8. Read the equivalent solid shaft. The plain bar that is exactly as hard to twist as the section you typed, D(1 − k4) to the quarter power.
9. Open Show working. The steps give the relation, the boxes in SI base units, the substitution, the wall your bore leaves, where the stress is measured from, the 1 + k² trade, the equivalent solid diameter, the fourth-power result, the assumptions and every unit conversion.
10. Try the inverse modes. Solve for the outer diameter to size a shaft from a stiffness, or for the bore to see how much of the middle you can take out. The bore mode needs care: solve for J first, then switch.

This page exists for the one step that turns a cross-section into a torsional stiffness. The polar moment of inertia guide is the prose half of it, where the fourth-power argument is made and the problems are worked. The quantity that only shares the name — the mass moment of inertia, I in kg·m², which needs a mass and says how hard a body is to spin up rather than to twist — belongs to the moment of inertia guide and to the moment of inertia calculator instead.

Two mistakes account for most wrong answers here. The first is entering a bore radius, or a wall thickness, where the bore diameter belongs. The second is expecting a material box: the shape fixes J and the surface stress outright, and only the twist angle needs a shear modulus.

Worked example: change one thing at a time

The table starts at the defaults, runs the unknown through all three modes on one state, bores the same shaft out twice, loads the real shafts, shows the fourth power as an exact doubling, walks every unit menu, and finally reaches the entries the calculator declines. Every answer and reading cell was taken from the running widget rather than worked out by hand. Where a cell and the tool ever part company, believe the tool.

Rows 1 to 3 are one state — a solid 30 mm shaft at 300 N·m — with the Solve for menu in each of its three positions. All three reading columns are identical down all three rows, because the readings describe the section rather than the mode. Row 3 is the one worth trying: the bore that gives a solid shaft its own J is 0 mm, not an error, and getting that answer rather than a failure is the whole reason this page computes the bore the way it does.

Rows 4 to 6 bore the same 30 mm shaft out and read the trade off the tool. At a 21 mm bore J falls from 7.95216 cm^4 to 6.04284 cm^4 — it keeps 0.7599 of the solid value — while the metal in the face falls to 0.5100 of it, so the stiffness per unit of material reads 1.4900x. At 27 mm those become 0.3439 and 0.1900, and the reading is 1.8100x. And the surface stress rises both times, from 56.588 MPa to 74.468 MPa to 164.549 MPa, because the same torque is carried on a smaller J. The gain is per unit of material, never per shaft.

Rows 7 to 9 are the honest rows. The same 300 N·m on a 17 mm spindle reads 310.989 MPa and on a 6 mm shank 7.074 GPa, which the verdict column marks OVER by 1.24x and OVER by 28.29x. The arithmetic is linear elastic and it never stops, so it goes on printing numbers for shafts that are no longer shafts. The 400 mm turbine row is the source’s own example, and the torque in it is what 1000 megawatts makes at 50 hertz — a step the power guide covers, since power is torque times angular speed.

Rows 10 and 11 are the fourth power with nothing hidden in it. The 50 mm shaft reads 61.3592 cm^4 and the 25 mm one 3.83495 cm^4: exactly 16 times, for every torque and every material, because doubling D multiplies D4 by 16. The surface stress goes the other way by exactly 8, from 97.785 MPa to 12.223 MPa. Row 13 is the cheaper version of the same lesson: 10 per cent more diameter is 1.1⁴ = 1.4641 times the J, about 46 per cent more stiffness for 21 per cent more metal.

Rows 14 to 18 change nothing but the units and the torque. Taking the torque off suppresses the stress reading rather than printing 0 Pa beside a yield verdict, because an unloaded shaft still has a J. Typing the diameter as 1.1811023622 in, or the same J as 79521.564044 mm^4 or 0.19105132182 in^4, reproduces the figures they came from; the inch units are offered because the source names inches to the fourth power as the US customary unit, and 1 in is 0.0254 m exactly. If the torque you were handed is in pound-feet, the Nm to ft-lb converter puts it in the unit this box wants.

Rows 19 to 26 are the refusals, and they are refusals on the input rather than tidied-up output. A bore equal to the shaft gives exactly J = 0 and then an infinite stress; a larger bore gives a negative J; a J of 7.96 cm^4 is above the solid J of a 30 mm shaft, so no bore produces it. In every one of them the answer is words rather than an infinity or a NaN.

Formula and symbol reference

One relation does all the work: J = π(D^4 − d^4) / 32, for a circular section that is solid or bored concentrically, with d = 0 giving the solid πD^4 / 32. For a solid shaft of radius R the same quantity is πR^4 / 2, which is exactly twice the planar second moment of the same circle. Everything else on this page is that relation rearranged, or a quantity derived from the section it describes.

The stress that goes with it is tau = T · r / J at a distance r from the shaft’s own axis, so the surface value is T · (D/2) / J. The familiar 16T / (πD³) is that expression with the solid J already substituted in, which makes it wrong for a bored section; this page never uses it. Reading the relation backwards gives D = (32J / π + d^4) to the quarter power and d = (D^4 − 32J / π) to the quarter power. Where the T itself comes from — a force and a lever arm — is the torque guide.

There is deliberately no box for the material, and that is the result rather than a gap. The source states that the rigidity comes from the cross-sectional area only and does not depend on material composition or shear modulus, so J and the surface stress are identical in steel and in aluminium. A shear modulus is needed only for the twist angle, and it is a different modulus from the stretching one that the Young’s modulus calculator works in.

The physics: why the fourth power decides everything

Twist a round shaft and each ring of metal resists in proportion to how far out it sits, twice over: once because it is being sheared further, and once because its lever arm about the axis is longer. That is why the quantity is an integral of r² over the area rather than the area itself. Integrate r² over a disc and the fourth power of the diameter falls out.

Every consequence worth knowing is in that exponent. Doubling the diameter multiplies J by exactly 16, so the twist falls by 16 and the surface stress by 8 — whatever the torque, whatever the length, whatever the metal. Ten per cent more diameter is 1.1⁴, about 46 per cent more stiffness. Nothing on this page needs a material to say any of that.

The same exponent is why hollow wins. Boring a concentric hole of ratio k = d/D leaves J at exactly 1 − k4 of the solid value while the area falls to 1 − k², and the ratio of those two is exactly 1 + k². At k = 0.8 the section keeps 59.04 per cent of the stiffness on 36.00 per cent of the metal, a gain of 1.6400 — pure geometry, with no material in it anywhere.

The reason is the stress distribution, and it is worth seeing rather than taking on trust. tau = T · r / J is exactly linear in r, so it is exactly zero on the axis, exactly half the surface value at mid-radius, and exactly k times the surface value at the bore wall. The metal in the middle was barely carrying anything, which is why a driveshaft, a bicycle spindle and an aircraft control rod are tubes rather than bars — and why the bored shaft’s surface stress is nonetheless higher, because the same torque now runs through a smaller J.

One bridge is worth stating because it settles the name. The mass moment of inertia of a uniform shaft is exactly J times the density times the length, which is why one is in m⁴ and the other in kg·m²: the two differ by exactly a mass per unit of cross-sectional area. That is an identity, not a figure — no density is quoted anywhere in this cluster, so no kilogram metre squared number is either.

Where this arithmetic breaks down

The arithmetic here is exact to the last digit a double can hold. What goes wrong is physical, and one item on this list matters more than all the others.

The arithmetic is linear elastic and it never stops

This is the one to read. Stress is proportional to torque for ever in this model, so the page keeps printing numbers long after a real shaft would have broken. A 6 mm shank at 300 N·m prints 7.074 GPa, which is 28.29 times the only yield figure this cluster has a source for. The reading beside the answer says so in words; the headline J never will, because J is a property of the shape and cares nothing about whether the shaft survived.

The 250 MPa is one shaft’s assumed yield, not steel’s

It is the figure the source states for the steel of its own example shaft, inside its own sample calculation, and it is quoted here as a reference to measure against rather than as a specification for anything. A real part has its own material certificate, its own heat treatment and its own factor of safety. Nothing on this page is a design allowable.

Circular sections only — and the source says why

The polar second moment of area can be insufficient for use to analyze beams and shafts with non-circular cross-sections, due their tendency to warp when twisted, and a torsion constant is substituted instead. That is the source’s own sentence, including its own grammatical slip, and it is the whole of what this page says about square, splined or keyed sections. No figure for any of them is published here.

Pure torsion, with no features

No bending, no axial load, no keyway, no shoulder, no press fit and no step in the diameter. Those are exactly where a real shaft fails, and it fails at the feature rather than in the plain section. This page will not quantify that, because nothing it is built on sources a stress concentration factor.

The bore mode is the one ill-conditioned direction

Going from J and D to a bore takes a fourth root of the difference between two nearly equal numbers, and the fourth root magnifies. Near a solid shaft a J quoted to six significant figures pins the bore only to about 2.7 per cent of the diameter, which is 0.8 mm on a 30 mm shaft. That is why the J box opens on eleven figures, why a J even a whisker above the solid value is declined outright, and why the safe route is to solve for J first and then switch.

A concentric bore, and nothing else

The hole is assumed to be on the axis. An eccentric bore, a cross-drilled hole or a hollow section with a seam is a different calculation that this page does not do, and offering a figure for one would be inventing it.

This tau is not viscosity’s tau

The symbol is shared and the quantity is not: here it is the elastic shear stress in a twisted solid, set by a geometry, and in a fluid it is set by a velocity gradient and a viscosity. The viscosity guide is where that other one lives, and nothing on this page applies to it.

A clean answer can be a coincidence of the numbers

The 25 mm and 50 mm rows differ by exactly 16, and the 80 per cent bore reads exactly 1.6400x. Those are exact because the ratios are exact powers, not because the arithmetic is tidy — change the diameter to 33 mm and every figure turns ragged while every identity still holds to the last digit.

The calculator measures nothing — you supply the numbers

Two diameters and a torque go in, one relation is used, and the answer describes the section you described. The case names on the buttons are shorthand for the four figures behind them and are not claims about any particular shaft, spindle or machine.

Where the polar moment is actually used

Deciding whether to bore a shaft out

Load the same shaft, 70 per cent bored and the trade is on screen in one line: 6.04284 cm^4 against 7.95216 cm^4, a reading of 1.4900x per unit of material, and a surface stress up from 56.588 MPa to 74.468 MPa. That is the whole decision — stiffness per gram against stress at the skin.

Sizing a shaft from the stiffness you need

Point Solve for at the outer diameter and the question becomes the one a design actually asks. Because J goes as the fourth power, doubling the stiffness needs only about 19 per cent more diameter, and sixteen times the stiffness is exactly twice. The working prints that factor for your own figures.

Finding how much of the middle you can take out

Point Solve for at the bore, give the J you must keep and the diameter you have, and the answer is the largest hole that still leaves it. Ask for the J of a 21 mm bore on a 30 mm shaft and you get 21 mm back; ask for the solid J and you get 0 mm, which is the honest answer rather than a failure.

Checking a stress against a yield figure

The reading beside the answer turns a J into something you can argue about. 310.989 MPa on the 17 mm spindle comes with OVER by 1.24x attached, which is a far more useful sentence than the stress on its own. The yield it is measured against is the source’s assumption about one shaft, so treat the percentage as a comparison and not a verdict on your part.

Reproducing a published sizing calculation

Load a 1000 MW turbine shaft: 400 mm solid at the torque 1000 megawatts makes at 50 hertz, which the source rounds its own radius to. The page prints 253.303 MPa for it — above the 250 MPa it was sized against, because stress goes as one over the radius cubed and the published radius was rounded down. That is why the source then applies a factor of safety.

Telling the two quantities apart

If what you actually need is how hard a wheel, a disc or a rotor is to spin up, that is the mass moment of inertia in kg·m², it needs a mass, and the moment of inertia calculator is the page for it. This one needs no mass at all. The two differ by exactly a mass per unit of cross-sectional area, which is also why their units differ.

Working in inches to the fourth power

Section tables in US customary units quote this quantity in in⁴, which the source names as the customary unit for it. The J menu offers it, and because 1 in is 0.0254 m exactly nothing is rounded on the way across. A 30 mm solid shaft is 0.19105132182 in^4.

Where to go next

For the argument this arithmetic supports — the fourth power, what hollow really buys, and why this quantity is not the one it is named after — read the polar moment of inertia guide, which this tool is the calculating half of. The moment of inertia guide is the other quantity: I in kg·m², for nine common shapes, about how hard a body is to spin up.

Four more posts pick up the pieces this page uses but does not explain. The torque guide is where the T comes from, the power guide is how a megawatt rating becomes a torque, the physics symbols guide is where the clash between this J and the other I is catalogued, and the mechanics formula sheet is the one-page reference.

Four tools sit next to this one. The moment of inertia calculator is the mass quantity, the torque calculator gives the torque from a force and a lever arm, the Young’s modulus calculator is the stretching modulus rather than the twisting one, and the Nm to ft-lb converter changes the torque unit. The moment of inertia simulator is the interactive version of the quantity this one is confused with.

The full physics lab library and the calculator index are open too.