Time to read: 5 min

Products that are subject to a torsional load often require analysis similar to what we use for bending stiffness. The main differences are the specific material property to be used (Shear Modulus of Rigidity) and the polar moment of inertia (which is very similar to the area moment of inertia used in bending). If we define torsion as the act of twisting along a structural component from an applied load, one example of this is the twist induced in a drive shaft when throttle is applied in a race car. This article will walk product designers and engineers through the methods used to analyze and improve torsional rigidity in design.

Problem Statement and Target Objectives

The key problem we’ll solve in this article is that of enhancing the torsional rigidity of our design. The definition of rigid, in this case, is the resistance to twisting as a torque is applied to the component. Going back to our drive shaft example, if we allow too much twist, we may end up with unstable performance. The images below show two shafts that are being torqued by 100 ft-lbs, but they have different torsional rigidities.

shaft that is being torqued by 100 ft-lbsshaft that is being torqued by 100 ft-lbs

There is a small “flag” modeled on the outside of the shaft to give a visual of the feature twisting as the torque is applied. The design on the bottom (Steel) is much more rigid than the design on top (HDPE). The next section will walk through the properties that make this possible.

Key Properties and Formulae

If we want to improve the rigidity of a design, we have a few material properties with which we can play. The list below details the key properties:

  • Shear Modulus, G (modulus of rigidity):

  • Polar Moment of Inertia

Furthermore, we can relate the shear modulus to the modulus of elasticity (E) for isotropic materials using the following formula:

formula for relating shear modulus to modulus of elasticity (E) for isotropic materials


  • E = Modulus of Elasticity

  • v = Poisson’s Ratio

To give a quick example, if we take the modulus of elasticity of steel at approximately 30e6 psi at room temperature, we can convert it to shear modulus with the following math:

if we take the modulus of elasticity of steel at approximately 30e6 psi at room temperature, we can convert it to shear modulus with this equation

Now that we understand the material property of interest, let’s look at the formulas that actually allow us to determine how rigid a component is. We will start with the overall formula for angle of twist (θ) in radians:

overall formula for angle of twist (θ) in radians


  • T = Torque applied to the component

  • L = Length of the Beam

  • J = Polar Moment of Inertia (Torsional Constant)

  • G = Shear Modulus (Modulus of Rigidity)

The image below illustrates what exactly we mean when we say angle of twist:

angle of twist

The polar moment of inertia, J, is the same thing as the area moment of inertia about the long axis. That means the formula for determining J will depend on the shape of your component. If we’re dealing with a solid shaft, the formula is:

shaft equation

It’s important to remember to use the correct value of J for all problems you solve, as it can have a big impact on design decisions. We’ll end on a formula that uses some of the variables above and is known simply as torsional stiffness (k).

torsional stiffness (k) formula

We could use this value to compare the rigidity of components directly, without ever solving for the angle of twist. Now that we know the math, we’ll step into some examples in the next section.

Applied Example

In this section, we’ll solve for polar moment of inertia, shear modulus, stiffness, and angle of twist. We will start with a simple solid circular shaft, for simplicity’s sake. At the end of this section, we’ll look at how to use CAD programs to help us solve for more complex geometry.

First, let’s start with a shaft that has the following geometry:

shaft with specific geometry

To start out, we’ll calculate the polar moment of inertia:

polar moment of intertia

We’ll analyze this part using the properties of aluminum. This means that with an elastic modulus of about 10e6 psi and a Poisson’s ratio of 0.33, we can calculate the shear modulus to be:

shear modulus calculation

Solving for stiffness, we can see that we get a value of:

solving for stiffness value

Finally, let’s look at the angle of twist if we apply 200 in-lb:>

angle of twist

If we go from aluminum to steel, we’ll increase the stiffness three times. This means that we’ll also have one-third of the deflection, or 0.07 rad. If you’re in a situation where the design is very sensitive to positional accuracy, going with a stiff material will be beneficial. If a greater increase in stiffness is needed than can be obtained from a material change alone, increasing the polar moment of inertia will be the best option. As an example, going from 0.5” to 0.75” brings the polar moment of inertia from 0.00614 in4 all the way up to 0.0311 in4. That means we’re making the part five times as rigid, with a 50% increase in diameter.

In many instances, we’re not just dealing with simple geometries. While the equations for angle of twist, stiffness, and shear modulus apply, we can’t use the same formula for polar moment of inertia. As a matter of fact, there may not be a textbook-designed formula for your given geometry. In that event, it’s best to use a CAD system to calculate your polar moment of inertia. This value can be substituted in for J in the remaining equations. The picture below from my previous article on designing for stiffness illustrates how to do this. While the image was captured from CREO, Solidworks can do the same type of analysis.

The highlighted quantities are Area MOI and Polar MOI (J), and as we’ve discussed, the former is the property of choice for this article. Using a CAD system makes this analysis much easier than hand-deriving a value.

Design Guidelines

Based on these formulas and the overall analytical approach, there are a few important guidelines to remember:

  • Increases in stiffness, as a result of increasing diameters in circular or near circular sections, can often greatly exceed a material change.

  • The angle of twist increases linearly with length, so making a member shorter can also reduce the total twist.

  • You can compare Young’s modulus alone for a relative difference in rigidity if all other properties are held constant. There’s no need to calculate shear modulus until you’re looking to solve for stiffness or angle of twist.

  • These formulas don’t solve for stress. If the loads are expected to induce stresses near the material strength, a more thorough stress analysis needs to be completed.

  • CAD programs can be very helpful when trying to gather section properties.

This isn’t an exhaustive list, but it should get you on your way to improving the rigidity of your components.

Main Takeaways

Designing for torsional rigidity may sound daunting, but after breaking down the steps, it hopefully feels more manageable. These principles apply to products spanning drive shafts, quadcopters, 3D printers, and much more. I encourage anyone involved in design to explore the multiple variables here for help figuring out how to accomplish your design objectives.