When you twist a non-circular member, the cross-section doesn't stay flat. It warps out of plane, producing a complex, non-uniform shear stress distribution that you can't analyze with the simple $\tau = \frac{Tr}{J}$ formula from circular shafts. Understanding how and why this happens is essential for designing any structural member with a rectangular, elliptical, or other non-circular cross-section.

Cross-sectional behavior during torsion

Circular shafts keep their cross-sections perfectly plane during torsion. Every radial line stays straight and in-plane as the section rotates. Non-circular members behave differently: the cross-section warps, meaning points move out of the original plane. A rectangular bar, for instance, develops noticeable out-of-plane bulging at its corners when twisted. This warping is the root cause of the more complicated stress patterns you'll see throughout this topic.

Shear stress distribution

For circular members, shear stress varies linearly from zero at the center to a maximum at the outer surface:

$\tau = \frac{Tr}{J}$

For non-circular members, the distribution is non-linear and varies across the entire cross-section. The maximum shear stress does not necessarily occur at the point farthest from the center. In an elliptical cross-section, for example, the maximum shear stress occurs at the ends of the minor axis (the points closest to the center), not the major axis. This is counterintuitive and a common exam mistake.

Torsional stiffness

For circular shafts, torsional stiffness depends on the polar moment of inertia $J$ , which is purely a function of the radius.
For non-circular members, you use a torsional constant $K$ (sometimes written $C$ or $J_t$ in different texts) instead of $J$ . The value of $K$ depends on the specific shape and its dimensions, and it's always less than $J$ for a circular section of the same area.
Rectangular cross-sections, for instance, have significantly lower torsional stiffness than circular ones of equal area because energy is "lost" to warping deformation.

Shear stress along the perimeter

On a circular shaft, shear stress is the same magnitude at every point on the outer perimeter. On non-circular members, shear stress varies around the boundary. A key result: shear stress at sharp corners is zero, not maximum. For a triangular cross-section, the maximum shear stress occurs at the midpoints of the sides, while the corners carry no shear stress at all. This is another place where intuition can mislead you.

Membrane analogy for torsion

Membrane analogy concept

The membrane (or soap-film) analogy, introduced by Prandtl, gives you a physical way to visualize torsion of non-circular sections. Imagine stretching a thin elastic membrane over an opening shaped like the cross-section, then applying a small uniform pressure from one side.

The key correspondences are:

The height of the deflected membrane at any point corresponds to the value of the Prandtl stress function $\Phi$ at that point.
The slope (gradient) of the membrane surface at any point is proportional to the shear stress at that point.
The volume enclosed between the membrane and the flat base is proportional to the torque carried by the section.

This means steep slopes on the membrane indicate high shear stress, and flat regions (like near corners) indicate low shear stress. The membrane is fixed (zero deflection) along the boundary of the cross-section.

Governing equation and boundary conditions

The Prandtl stress function $\Phi$ satisfies the Poisson equation:

$\nabla^2 \Phi = \frac{\partial^2 \Phi}{\partial x^2} + \frac{\partial^2 \Phi}{\partial y^2} = -2G\theta'$

where $G$ is the shear modulus and $\theta'$ is the rate of twist (angle of twist per unit length, $\theta' = d\theta/dz$ ). Note that the right-hand side involves the rate of twist, not the torque directly.

Boundary conditions:

$\Phi = 0$ along the perimeter of the cross-section (the membrane is clamped at the edges).
The stress function is single-valued throughout the cross-section.

Solving the membrane analogy

Once you solve the Poisson equation for $\Phi$ , you can extract everything you need:

Shear stress components from the partial derivatives of $\Phi$ :
- $\tau_{xz} = \frac{\partial \Phi}{\partial y}$
- $\tau_{yz} = -\frac{\partial \Phi}{\partial x}$
Resultant shear stress at any point:
- $\tau = \sqrt{\tau_{xz}^2 + \tau_{yz}^2}$
Torque from the volume under the stress function:
- $T = 2 \iint_{A} \Phi \, dA$

Analytical solutions exist for simple shapes (ellipses, equilateral triangles). For a rectangle, series solutions or tabulated coefficients are used. For complex shapes, numerical methods (finite elements, finite differences) are typically required.

Shear stress in non-circular members

Shear stress components

The shear stress at any point in a non-circular cross-section under torsion comes from the gradient of the Prandtl stress function:

$\tau_{xz} = \frac{\partial \Phi}{\partial y}$
$\tau_{yz} = -\frac{\partial \Phi}{\partial x}$

The resultant shear stress magnitude is:

$\tau = \sqrt{\left(\frac{\partial \Phi}{\partial y}\right)^2 + \left(\frac{\partial \Phi}{\partial x}\right)^2}$

Wherever the stress function surface is steepest, the shear stress is highest. Wherever it's flat (at corners, for example), the shear stress drops to zero.

Maximum shear stress

Where the maximum shear stress occurs depends on the shape:

Cross-section	Location of $\tau_{max}$
Ellipse	Ends of the minor axis
Rectangle	Midpoints of the longer sides
Equilateral triangle	Midpoints of each side

For a general non-circular section, the maximum shear stress can be written as:

$\tau_{max} = \frac{T}{\alpha b t^2}$

where $b$ and $t$ are characteristic dimensions and $\alpha$ is a shape-dependent coefficient found from tables or solutions.

Closed-form solutions for common cross-sections

Elliptical cross-section (semi-axes $a$ and $b$ , with $a > b$ ):

$\tau_{max} = \frac{2T}{\pi a b^2}$

This maximum occurs at the ends of the minor axis ( $b$ ).

Rectangular cross-section (width $a$ , thickness $b$ , with $a \geq b$ ):

$\tau_{max} = \frac{T}{\alpha a b^2}$

where $\alpha$ is a tabulated coefficient that depends on the ratio $a/b$ . For a square ( $a/b = 1$ ), $\alpha \approx 0.208$ . As $a/b$ increases toward a thin rectangle, $\alpha$ approaches $1/3$ .

Torsional resistance and stiffness

Torsional constant

The torsional constant $K$ (sometimes called $J_t$ ) replaces the polar moment of inertia $J$ for non-circular sections. It's defined through the torque-twist relationship, not as a simple geometric integral of $r^2 \, dA$ .

$K$ is computed from the stress function:

$T = 2 \iint_{A} \Phi \, dA$

Closed-form results for common shapes:

Ellipse (semi-axes $a$ , $b$ ): $K = \frac{\pi a^3 b^3}{a^2 + b^2}$
Equilateral triangle (side length $s$ ): $K = \frac{s^4 \sqrt{3}}{80}$
Rectangle (dimensions $a \times b$ , $a \geq b$ ): $K = \beta a b^3$ , where $\beta$ is a tabulated coefficient depending on $a/b$

Torsional stiffness

The torque-twist relationship for any prismatic member is:

$\theta = \frac{TL}{GK}$

where $\theta$ is the total angle of twist, $T$ is the applied torque, $L$ is the member length, $G$ is the shear modulus, and $K$ is the torsional constant. This has the same form as the circular shaft equation $\theta = TL/(GJ)$ , but with $K$ in place of $J$ .

The torsional stiffness (torque per unit twist) is:

$\frac{T}{\theta} = \frac{GK}{L}$

For a rectangular section with $a/b = 1$ (square), $\beta \approx 0.141$ , giving $K = 0.141 \, b^4$ . Compare this to a circle of the same area: the square has roughly 84% of the circular section's torsional stiffness.

Comparison with circular members

For the same cross-sectional area, circular sections are always the most torsionally efficient shape. Non-circular sections pay a penalty because:

Part of the deformation goes into warping rather than pure rotation.
The non-uniform stress distribution means material near corners is underutilized (carrying little or no shear stress).

As a practical rule, the more elongated or irregular the cross-section, the greater the stiffness penalty. This is why shafts designed primarily to transmit torque are almost always circular, while non-circular sections appear where other loading or geometric constraints govern the design.