Combined axial and torsional loading

A propeller shaft on a ship is pushed by the thrust bearing while it is also twisted by the engine — a bolt tightened down grips its shank in tension while the wrench simultaneously twists it. Neither load acts alone, and treating them as separate one-dimensional problems misses the real danger point on the shaft's surface. Combined axial and torsional loading is the technique of writing both stresses on the same tiny stress element and then using the principal-stress formulas to find what the material actually feels. This topic sits directly after stress transformation and Mohr's circle in the ME10001 sequence, and it is the first place those tools earn their keep on a genuine engineering member. You already know $\sigma = P/A$ for axial load and $\tau = Tc/J$ for torsion separately; here the two combine on one element, and a shaft that looks safe by either formula alone can still fail once the principal stresses are computed correctly.

The core insight

On a shaft carrying both axial force and torque, the axial normal stress and the torsional shear stress act on the same stress element at the outer surface, so they must be combined through the principal-stress formulas — never added directly.

The core idea

Take a circular shaft of outer radius $c$ carrying an axial force $P$ and a torque $T$ . At any point on the outer surface, the axial load produces a uniform normal stress along the shaft axis, and the torque produces a shear stress that is maximum at that same surface:

\sigma_x = \frac{P}{A}, \qquad \tau_{xy} = \frac{Tc}{J}, \qquad \sigma_y = 0

where: $\sigma_x$ is the axial normal stress (Pa), $A$ is the cross-sectional area, $\tau_{xy}$ is the torsional shear stress on the cross-sectional plane, $J$ is the polar second moment of area, and $\sigma_y = 0$ because a plain, unpressurized shaft has no stress in the circumferential direction. These two stresses act on the same small element at the surface — $\sigma_x$ pulling or pushing along the axis, $\tau_{xy}$ shearing the cross-section — so the element is exactly the plane-stress state already studied in the transformation and Mohr's-circle topics. Once the element is set up, the ordinary principal-stress and maximum-shear-stress formulas apply with $\sigma_y$ set to zero:

\sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2+\tau_{xy}^2}, \qquad \tau_{max} = \sqrt{\left(\frac{\sigma_x}{2}\right)^2+\tau_{xy}^2}

where: $\sigma_1$ is the larger (more tensile) principal stress, $\sigma_2$ the smaller (more compressive), and $\tau_{max}$ the maximum in-plane shear stress at that point. Because $\sigma_x$ is uniform over the whole cross-section while $\tau_{xy} = T\rho/J$ grows linearly with radius $\rho$ and peaks at $\rho = c$ , both contributions are simultaneously largest at the outer surface — which is why the outer fibre, not the centre, is the location that governs design. A useful way to picture the two stresses fighting it out on the same element is the pair of arrows below: a normal stress pulling along the shaft axis and a shear stress sliding the cross-section, both acting on the identical surface point, exactly what Mohr's circle then resolves into principal directions.

Figure

A small square stress element on a shaft surface showing a normal stress arrow along the axis from the axial load and a shear stress arrow along the edges from the torque, both acting on the same point

If the axial load is applied through the centroid with no eccentricity, this two-stress analysis is the complete picture. An eccentric axial load additionally bends the shaft, adding a bending stress that must be superposed too — that extension belongs to the next topics in the unit and is not needed here.

Derivation APrincipal stresses on a shaft under combined axial load and torque

The goal is to show, starting only from the two separate stress formulas for axial loading and for torsion, exactly how they combine into the principal stresses and maximum shear stress felt at the outer surface of a shaft carrying both a force $P$ and a torque $T$ at once, since simply adding $\sigma_x$ and $\tau_{xy}$ as if they were the same kind of quantity is not physically meaningful and gives a wrong answer.

Step 1. Identify the two stresses separately, exactly as in the earlier single-load topics. Axial loading through the centroid gives a uniform normal stress $\sigma_x = P/A$ acting along the shaft's longitudinal axis. Torsion of a circular shaft gives a shear stress that varies linearly with radius, $\tau = T\rho/J$ , reaching its largest value $\tau_{xy} = Tc/J$ at the outer surface $\rho = c$ . Both formulas come from the earlier topics and are not re-derived here.

Step 2. Recognize that both stresses act on the same material point and the same pair of orthogonal faces of a tiny stress element cut from the outer surface: the face perpendicular to the shaft axis carries the normal stress $\sigma_x$ , and that same cross-sectional face carries the shear stress $\tau_{xy}$ from the torque. Because there is no external pressure or hoop load on a plain shaft, the transverse normal stress $\sigma_y$ is zero. This is precisely the general plane-stress element $(\sigma_x, \sigma_y=0, \tau_{xy})$ already handled by stress transformation.

Step 3. Apply the standard principal-stress result for plane stress, substituting $\sigma_y = 0$ . The general formula $\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$ reduces, with $\sigma_y=0$ , to $\sigma_{1,2} = \frac{\sigma_x}{2}\pm\sqrt{\left(\frac{\sigma_x}{2}\right)^2+\tau_{xy}^2}$ . The same substitution in the maximum in-plane shear formula gives $\tau_{max} = \sqrt{\left(\frac{\sigma_x}{2}\right)^2+\tau_{xy}^2}$ , identical in form to the radius of Mohr's circle centred at $\sigma_x/2$ .

Step 4. Read off the physical meaning. The centre of Mohr's circle sits at $\sigma_x/2$ (half the axial stress alone), and the radius is enlarged beyond a pure-torsion circle by the presence of $\sigma_x$ . Because $\sigma_x$ is constant over the cross-section but $\tau_{xy}$ is largest at the outer surface, the combined circle — and hence the worst-case principal stresses — is always evaluated at the outer fibre, confirming that this is the critical location for design against combined loading. Verified: setting $\sigma_y=0$ in the general plane-stress principal-stress formula reproduces the combined-loading result exactly, and it collapses correctly to $\sigma_{1,2}=\pm\tau_{xy}$ for $\sigma_x=0$ (pure torsion) and to $\sigma_1=\sigma_x,\ \sigma_2=0$ for $\tau_{xy}=0$ (pure axial load).

Worked example

A solid circular shaft of diameter $d = 50\text{ mm}$ carries an axial tensile force $P = 40\text{ kN}$ and a torque $T = 900\text{N}\,\cdot\,\text{m}$ . Find the principal stresses and the maximum shear stress at a point on the outer surface.

Step 1. Section properties. The outer radius is $c = d/2 = 0.025\text{ m}$ . The area is $A = \pi c^2 = \pi(0.025)^2 = 1.963\times10^{-3}\text{ m}^2$ , and the polar moment is $J = \dfrac{\pi c^4}{2} = \dfrac{\pi(0.025)^4}{2} = 6.136\times10^{-7}\text{ m}^4$ .

Step 2. Axial stress. $\sigma_x = \dfrac{P}{A} = \dfrac{40\times10^3}{1.963\times10^{-3}} = 20.37\times10^6\text{ Pa} = 20.4\text{ MPa}$ , a uniform tensile stress along the shaft axis.

Step 3. Torsional shear stress at the outer surface. $\tau_{xy} = \dfrac{Tc}{J} = \dfrac{900(0.025)}{6.136\times10^{-7}} = 36.67\times10^6\text{ Pa} = 36.7\text{ MPa}$ .

Step 4. Principal stresses. The Mohr's-circle radius is $R = \sqrt{(20.4/2)^2+36.7^2} = \sqrt{10.2^2+36.7^2} = \sqrt{104.0+1346.9} = \sqrt{1450.9} = 38.1\text{ MPa}$ , so $\sigma_1 = 10.2+38.1 = 48.3\text{ MPa}$ and $\sigma_2 = 10.2-38.1 = -27.9\text{ MPa}$ .

Step 5. Maximum shear stress equals the circle's radius: $\tau_{max} = 38.1\text{ MPa}$ . The point on the surface sees a maximum tensile principal stress of 48.3 MPa, a compressive principal stress of 27.9 MPa, and a maximum shear stress of 38.1 MPa — noticeably larger than either $\sigma_x$ or $\tau_{xy}$ taken alone, which is exactly why the combined analysis, not simple addition, governs the design check.

Competitive stretch

If the same shaft instead carried a compressive axial load of equal magnitude ( $\sigma_x=-20.4\text{ MPa}$ ) with the torque unchanged, only the sign of $\sigma_x$ flips: $R=38.1\text{ MPa}$ is unchanged (it depends on $\sigma_x/2$ squared), but the circle centre shifts to $-10.2\text{ MPa}$ , giving $\sigma_1=27.9\text{ MPa}$ and $\sigma_2=-48.3\text{ MPa}$ — the same two magnitudes, swapped between tension and compression. Flipping the sign of $P$ mirrors the principal stresses about zero but leaves $\tau_{max}$ unchanged.

Common traps

Adding $\sigma_x$ and $\tau_{xy}$ directly. They are a normal stress and a shear stress on the same element, not two quantities of the same kind — they must be combined through the principal-stress formulas, never summed.
Using the wrong radius for $\tau_{xy}$ . Torsional shear stress varies as $\tau=T\rho/J$ ; always use the outer radius $c$ for the combined analysis, since that is where both stresses are largest together.
Dropping the sign of the axial load. A compressive $P$ gives a negative $\sigma_x$ , which shifts the Mohr's-circle centre in the opposite direction and swaps which principal stress is the larger tensile one.
Inventing a second normal stress. For a plain, unpressurized shaft under axial load and torque only, $\sigma_y=0$ ; there is no hoop or radial normal stress to add.

Ask the tutor: if you'd like to work a design problem where you're given an allowable stress and must find the minimum safe shaft diameter under combined axial and torsional loading, just ask.

Why can't I just add the axial stress and the torsional shear stress to get the worst stress?

Because they are different kinds of stress acting on different faces of the same tiny element — $\sigma_x$ is a normal stress pulling along the shaft axis, while $\tau_{xy}$ is a shear stress sliding the cross-sectional face. Adding them numerically has no physical meaning, the way you would not add a force to a velocity. The correct combination comes from treating them as one plane-stress state and applying the principal-stress formulas (or drawing Mohr's circle), which properly accounts for how normal and shear stress interact when the element is rotated to find its worst orientation.

Why is the outer surface always the critical location, and not somewhere inside the shaft?

Because both stress contributions are largest there at once. The axial stress $\sigma_x=P/A$ is uniform across the entire cross-section, so it does not favour any radius. But the torsional shear stress $\tau=T\rho/J$ grows linearly with radius $\rho$ and is zero at the centre, peaking only at the outer radius $\rho=c$ . Since the combined principal stresses depend on both $\sigma_x$ and $\tau_{xy}$ together, and $\tau_{xy}$ is maximised at the surface while $\sigma_x$ is unchanged there, the surface is where the combined stress state is worst — which is why every worked example evaluates it there.

Does this method change if the shaft is hollow instead of solid?

No, the method is identical — only the section properties change. You still compute $\sigma_x = P/A$ and $\tau_{xy} = Tc_o/J$ at the outer surface, using the hollow shaft's area $A=\pi(c_o^2-c_i^2)$ and polar moment $J=\frac{\pi}{2}(c_o^4-c_i^4)$ , where $c_o$ and $c_i$ are the outer and inner radii. Once $\sigma_x$ and $\tau_{xy}$ are found from the correct hollow-section formulas, the same principal-stress and maximum-shear-stress equations apply exactly as for a solid shaft, and the outer surface is still the critical location.

Combined axial and torsional loading

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads