Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Torsional Stiffness Calculator

Compute the torsional stiffness k = GJ/L of a uniform shaft or bar from shear modulus, torsion constant J and length, in newton-metres per radian.

Torsional Stiffness Calculator

Pa
m⁴
m
Torsional stiffness
1,316.6667
The torque needed to twist one end of the member one radian relative to the other, in newton-metres per radian.
Model scope
Uniform linear-elastic Saint-Venant torsion model; J must be the correct torsion constant for the section and is not generally the polar moment for noncircular sections.

Background.

Twist a steel bar and it behaves like a spring, winding up an amount of rotation in proportion to the torque on it. Torsional stiffness, k_t, is the constant of that spring: the torque needed to turn one end of a member one radian relative to the other, in newton-metres per radian. It decides how far a drive shaft winds up under load, how accurately a machine spindle holds its angle, whether a car body feels taut or loose, and at what frequency a rotating system will vibrate.

For a uniform bar three things set it. The shear modulus G measures how strongly the material resists being sheared; steel is around 79 GPa, roughly three times typical aluminium alloys. The torsion constant J captures the size and shape of the cross-section. Length works against both, since k_t = GJ/L: a shaft twice as long twists twice as far under the same torque.

Geometry is usually the strongest lever. For a solid round shaft J = πd⁴/32, so stiffness climbs with the fourth power of diameter, and a shaft 25% thicker is about 2.44 times as stiff. Tubes exploit this by placing material far from the axis, where it resists twisting most effectively.

The classic mistake is using the wrong J. For solid circles and round tubes the torsion constant equals the polar moment of inertia; for every other shape it is smaller, sometimes drastically so. Open sections such as channels, angles or a tube slit along its length keep only a small fraction of the stiffness of the closed tube they resemble.

The model behind the result is Saint-Venant torsion: a straight member of constant cross-section, linear-elastic material, free to warp, and twisted through small angles.

What is torsional stiffness calculator?

Torsional stiffness is the ratio of applied torque to the resulting angle of twist, k_t = T/φ, for a member loaded in pure torsion. Its SI unit is the newton-metre per radian. For a straight bar of constant cross-section made of a linear-elastic material, Saint-Venant theory gives k_t = GJ/L, where G is the shear modulus, J the torsion constant of the section and L the length between the two sections whose relative rotation is of interest. A high value means a member that barely rotates under load.

How to use this calculator.

  1. Enter the Shear modulus in pascals: 79,000,000,000 for typical steel. If you only know Young's modulus E and Poisson's ratio ν for an isotropic material, use G = E ÷ (2(1 + ν)).
  2. Enter the Torsion constant J in m⁴. For a solid round shaft J = πd⁴/32 and for a round tube J = π(D⁴ − d⁴)/32, with diameters in metres; a diameter typed in millimetres makes J a trillion times too large. For other shapes take J from a section table, not the polar moment.
  3. Enter the Member length in metres, measured between the two sections whose relative twist you care about.
  4. Read the stiffness in N·m/rad. Divide a torque by it for the twist in radians, or multiply it by π/180 for newton-metres per degree.

The formula.

k_t = GJ / L

k_t = G × J ÷ L. G, the shear modulus in pascals, is a material property: shear stress divided by shear strain in the elastic range. J, the torsion constant in m⁴, depends only on the shape and size of the cross-section. L is the length in metres over which twist accumulates. Pascals times metres to the fourth, divided by metres, leaves newton-metres, and because the angle is in radians the stiffness is quoted per radian. The equation is the angle-of-twist formula for a prismatic bar, φ = TL/(GJ), rearranged as T/φ. The result is proportional to G and J and inversely proportional to L, and since J grows with the fourth power of a shaft's diameter, size is usually the dominant term. It holds only while the material stays elastic and the section is free to warp; tapered members, restrained warping in open sections and yielding all fall outside it.

A worked example.

Example

Take the default steel member: G = 79,000,000,000 Pa, J = 2 × 10⁻⁸ m⁴ and L = 1.2 m. A torsion constant of that size belongs to a solid round shaft about 21.2 mm across, since d = (32J/π)^¼. First GJ = 79,000,000,000 × 0.00000002 = 1,580 N·m². Dividing by the length, 1,580 ÷ 1.2 = 1,316.6667 N·m/rad, which is 1,316.67 × π/180 = 22.98 N·m for each degree of twist. Under a 100 N·m torque the shaft therefore turns 100 ÷ 1,316.67 = 0.0759 rad, or 4.35°, end to end. Made from aluminium alloy with G of about 26 GPa, the same shaft would manage only 433.33 N·m/rad and twist roughly three times as far.

torsion Constant M40
shear Modulus Pa79,000,000,000
member Length M1.2

Frequently asked questions.

Why can't I use the polar moment of inertia for a square or rectangular bar?
Because non-circular sections warp out of plane when twisted, which makes them less stiff than the polar moment implies. For a solid square of side a the torsion constant is about 0.141a⁴, while the polar moment is a⁴/6, roughly 0.167a⁴, so using it overstates stiffness by about 18%. For thin flat strips the error is far larger.
How much torsional stiffness does a tube lose if it is slit along its length?
Almost all of it. A closed thin tube of mean radius r and wall t has J ≈ 2πr³t; slit open it becomes J ≈ (1/3)(2πr)t³. The ratio is 3(r/t)², so a 25 mm radius tube with a 2 mm wall drops from about 1.96 × 10⁻⁷ to 4.19 × 10⁻¹⁰ m⁴, a factor of roughly 469.
How do I combine sections of a stepped shaft?
Lengths carrying the same torque one after another act like springs in series, so their twists add: 1/k = 1/k₁ + 1/k₂ + …. A segment of 2,000 N·m/rad followed by one of 1,000 N·m/rad gives about 666.67 N·m/rad overall, less than either. Members sharing a torque side by side add directly instead.
How do I find the shear modulus if a datasheet only gives Young's modulus?
For an isotropic material use G = E ÷ (2(1 + ν)), with ν the Poisson's ratio. Steel with E = 200 GPa and ν = 0.3 gives about 76.9 GPa, close to the 79 GPa default. The relation does not hold for wood, fibre composites or other materials whose stiffness depends on direction.
How is torsional stiffness used in vibration problems?
A shaft carrying a flywheel, rotor or gear behaves as a torsional spring–mass system. With k_t in N·m/rad and the attached mass moment of inertia I in kg·m², the natural frequency is √(k_t/I) in radians per second; divide by 2π for hertz. Stiffer or shorter shafts push that frequency up.
Does the result account for I-beams whose ends are prevented from warping?
No. Saint-Venant theory assumes every section is free to warp. When the ends of an open section such as an I-beam or channel are held, warping torsion adds stiffness, which matters most in short members. Capturing it needs the section's warping constant and a more detailed analysis than GJ/L.

References& sources.

  1. [1]Gere and Goodno, Mechanics of Materials, 9th ed. (PRINT), chapter 3, angle of twist for prismatic bars.
  2. [2]MIT OpenCourseWare, Unified Engineering, torsion of shafts.
  3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.

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k_t = GJ / L
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