This question is so fundamentally basic that I am almost embarrassed to ask but it came up at work the other day and and nearly no one in the office could give me a good answer. I was calculating the shear stress in a member using the equation, $\frac{Tr}{J_T}$ and noticed, that for a shaft with a circular cross section, $J_T = I_P$.

Both $I_P$ and $J_T$ are used to describe an object's ability to resist torsion. $I_P$ is defined as, $ \int_{A} \rho^2 dA $ where $\rho$ = the radial distance to the axis about which $I_P$ is being calculated. But $J_T$ has no exact analytical equations and is calculated largely with approximate equations that no reference I looked at really elaborated on.

So my question is, what is the difference between the Polar Moment of Inertia, $ I_P $, and the torsional constant, $ J_T $? Not only mathematically, but practically. What physical or geometric property is each a representation of? Why is $J_T$ so hard to calculate?


3 Answers 3


The torsion constant $J_T$ relates the angle of twist to applied torque via the equation: $$ \phi = \frac{TL}{J_T G} $$ where $T$ is the applied torque, $L$ is the length of the member, $G$ is modulus of elasticity in shear, and $J_T$ is the torsional constant.

The polar moment of inertia on the other hand, is a measure of the resistance of a cross section to torsion with invariant cross section and no significant warping.

The case of a circular rod under torsion is special because of circular symmetry, which means that it does not warp and it's cross section does not change under torsion. Therefore $J_T = I_P$.

When a member does not have circular symmetry then we can expect that it will warp under torsion and therefore $J_T \neq I_P$.

Which leaves the problem of how to calculate $J_T$. Unfortunately this is not straightforward, which is why the values (usually approximate) for common shapes are tabulated.

One way of calculating the torsional constant is by using the Prandtl Stress Function (another is by using warping functions).

Without going into too much detail one must choose a Prandtl stress function $\Phi$ which represents the stress distribution within the member and satisfies the boundary conditions (not easy in general!). It also must satisfy Poisson's equation of compatability: $$ \nabla^2 \Phi = -2 G \theta $$ Where $\theta$ is the angle of twist per unit length.

If we have chosen the stress function so that $\Phi = 0$ on the boundary (traction free boundary condition) we can find the torsional constant by: $$J_T = 2\int_A \frac{\Phi}{G\theta} dA$$

Example: Rod of circular cross section

Because of the symmetry of a circular cross section we can take: $$\Phi = \frac{G\theta}{2} (R^2-r^2) $$ where R is the outer radius. We then get: $$J_T = 2\pi\int_0^R (R^2-r^2)rdr = \frac{\pi R^4}{2} = (I_P)_{circle}$$

Example: Rod of elliptical cross section

$$\Phi = G\theta\frac{a^2 b^2}{a^2+b^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)$$ and $$J_T = \int_A \frac{a^2 b^2}{a^2+b^2}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)dA = \frac{\pi a^3 b^3}{a^2+b^2} $$ which is certainly not equal to the polar moment of inertia of an ellipse: $$ (I_P)_{ellipse} = \frac{1}{4}\pi a b(a^2+b^2) \neq (J_T)_{ellipse}$$

Since in general $J_T < I_P$, if you used the polar moment of inertia instead of the torsional constant you would calculate smaller angles of twist.


This is almost a coincidence, and it is only true for solid or hollow circular cross sections. Of course shafts carrying torsion often are circular, for reasons that are independent of the question!

The torsion of a circular shaft is physically simple because of the symmetry of the circular shape. By symmetry, the stresses and strains at any point can only be a function of the radial distance from the centre line of the shaft. By Pythagoras' theorem, you can take an arbitrary pair of axes and express the radius as $r^2 = x^2 + y^2$.

Using that fact, you can convert the integral over the cross section into the sum of two integrals in the $x$ and $y$ directions, and again by symmetry those two integrals must be equal to each other.

The form of the integrals happen to be exactly the same mathematical form as for the second moments of area of a circular beam, which leads to the result you asked about.

This doesn't work for non-circular sections, because the stress distribution is not radially symmetrical. For example if you compare the torsion constant and polar moment of a solid square section, you will find the "constants" in the two formulas are different. The more the cross section deviates from a circle, the bigger the difference will be.

The torsion constant for an complex shaped section (for example an I-beam) is hard to calculate because the stress distribution over the section is complicated, and there is no simple "formula" for it that you an integrate mathematically. Many of the formulas for torsion in engineering handbooks are based on simplified assumptions rather than "exact" mathematical solutions.

But in real life the "errors" are not too important, because when a torsional load is applied to a non-circular structure, the cross sections "warp", i.e. they no longer remain plane. In real life, the amount of warping is often unknown, because the restraints at the ends of the shaft affect it. If you really need an accurate estimate of the torsional stiffness of a non-circular component, you have to make a full 3-D model of the component itself and how it is fixed to the rest of the structure. If you make a model with that level of detail, there is not much point in reducing the answer to one number just so you can call it "the torsional stiffness".


Polar moment of inertia, Ip, is the resistance of a solid to be torsioned. However, rotational mass moment of inertia, J, is the inertia moment of a rotating solid. See this web.

As I understand, J is the same as normal moment of inertia, but for rotating objects.

  • 1
    $\begingroup$ Don't confuse $I_{zz} = \int r^2 {\rm d}A$ with $I_{zz} = \int r^2 {\rm d}m$. He is asking about the polar moment of area, not the polar moment of inertia. $\endgroup$ Commented Mar 21, 2016 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.