Combined axial, torsional, and bending loading

A ship's propeller shaft pushes, twists, and sags all at once: axial thrust from the propeller, torque from the engine, and bending from its own weight and the propeller's overhang. None of these loads act alone, and none of the simple single-load formulas from earlier in the course — plain axial stress, plain torsion, plain bending — is enough by itself to tell you whether the shaft is safe. This node is the capstone of the "Stress Transformation, Combined Loading, and Columns" unit: it asks you to superpose the normal stresses from axial force and bending moment, add the shear stress from torque, assemble the resulting plane-stress element at the one point on the cross-section that matters most, and finally run that element through the principal-stress machinery you already built earlier in the unit. Get the critical point wrong and every later calculation, however careful, answers the wrong question.

The core insight

At the critical point of a circular shaft under axial force, bending moment, and torque together, the normal stress σx=P/A+Mc/I\sigma_x = P/A + Mc/I and the shear stress τxy=Tc/J\tau_{xy} = Tc/J combine into one plane-stress element whose principal stresses and maximum shear you find with the same Mohr's-circle formulas used for any two-way combined loading.

The core idea

Three load types produce stress at a point on a circular shaft cross-section (radius cc), each spreading across the section in its own pattern. Axial force PP produces a normal stress that is uniform everywhere — every point feels the same P/AP/A. Bending moment MM produces a normal stress that varies linearly across the section, zero at the neutral axis and maximum, with opposite signs, at the two extreme fibres. Torque TT produces a shear stress uniform in magnitude around the entire outer circumference, tangent to the circle everywhere. Axial and bending stresses are both normal stresses on the same face, so they add algebraically; torsional shear sits on that same face as the shear component. At the outer fibre where the bending stress is largest and adds to the axial stress, the combined normal stress is

σx=PA+McI\sigma_x = \frac{P}{A} + \frac{Mc}{I}

where: PP is the axial force, A=πc2A = \pi c^2 is the cross-sectional area, MM is the bending moment at the section, cc is the outer radius, and I=πc4/4I = \pi c^4/4 is the second moment of area. At that same point the torsional shear stress is

τxy=TcJ\tau_{xy} = \frac{Tc}{J}

where: TT is the torque and J=πc4/2J = \pi c^4/2 is the polar second moment of area. For a solid circular shaft J=2IJ = 2I, so once II is computed, JJ follows immediately without a second integral — a shortcut worth memorising. With σy=0\sigma_y = 0 (no hoop or pressure loading on a plain shaft), (σx,τxy)(\sigma_x, \tau_{xy}) is exactly the plane-stress element from earlier in the unit, so the familiar transformation formulas apply directly:

σ1,2=σx2±(σx2)2+τxy2,τmax=(σx2)2+τxy2\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: σ1,σ2\sigma_1, \sigma_2 are the principal stresses and τmax\tau_{max} is the maximum in-plane shear stress at that point.

Figure
A circular shaft cross section with a critical point marked on its rim showing an outward normal stress arrow and a tangential shear stress arrow combining into one plane stress element
Critical point on the shaft: normal plus shear stress together critical point (bending max, adds to axial) sigma x tau xy plane stress element sigma x from axial plus bending, tau xy from torque

Locating the critical point is the step unique to three-way problems: since bending stress varies around the circumference while axial stress is uniform and torsional shear is uniform, the point with the largest ∣σx∣|\sigma_x| (axial and bending adding, not cancelling) is normally the critical one — it carries both the largest normal stress and the full torsional shear.

Derivation ACombining axial, bending, and torsional stress into principal stresses at the critical point

The goal is to show, starting from the three separate single-load stress formulas already established earlier in the unit — P/AP/A for axial, Mc/IMc/I for bending, Tc/JTc/J for torsion — exactly how they combine into one plane-stress element and why the ordinary principal-stress formula from the transformation unit then applies unchanged, with no new physics beyond superposition of stresses that act on the same plane.

Step 1. Identify what plane each stress acts on. Axial force and bending moment both produce a normal stress on the cross-sectional face (perpendicular to the shaft axis) — call this the xx-face. Torque produces a shear stress also acting on that same cross-sectional face, tangential to the circle. Because all three loads act on the identical face, they assemble into one stress element without any coordinate rotation first.

Step 2. Superpose the two normal-stress contributions algebraically. Axial stress P/AP/A is the same sign and magnitude everywhere on the section; bending stress ±Mc/I\pm Mc/I depends on which side of the neutral axis you stand. At the fibre where the bending contribution has the same sign as the axial contribution, the two add: σx=P/A+Mc/I\sigma_x = P/A + Mc/I. At the diametrically opposite fibre they instead partially cancel. The larger-magnitude fibre is normally the one carried forward.

Step 3. Attach the torsional shear stress to that point. Torsional shear magnitude Tc/JTc/J is uniform around the whole rim, so it applies unchanged at whichever fibre Step 2 selected, giving the stress pair (σx,τxy)(\sigma_x, \tau_{xy}) with σy=0\sigma_y = 0, since nothing loads the perpendicular face of a plain shaft.

Step 4. Recognise this as an ordinary plane-stress element and transform it. The principal-stress and Mohr's-circle formulas derived earlier in the unit apply exactly as before: σ1,2=σx/2±(σx/2)2+τxy2\sigma_{1,2} = \sigma_x/2 \pm \sqrt{(\sigma_x/2)^2 + \tau_{xy}^2} and τmax=(σx/2)2+τxy2\tau_{max} = \sqrt{(\sigma_x/2)^2 + \tau_{xy}^2}. Combined loading only changes how σx\sigma_x and τxy\tau_{xy} are built up, not how they are subsequently processed.

Worked example

A solid circular shaft, d=60 mmd = 60\text{ mm}, transmits a torque T=800N ⋅ mT = 800\text{N}\,\cdot\,\text{m}, carries an axial tensile force P=15 kNP = 15\text{ kN}, and at the section of interest the bending moment is M=500N ⋅ mM = 500\text{N}\,\cdot\,\text{m}. Find the maximum principal stress and the maximum shear stress at the critical point.

Step 1. Section properties. c=0.03 mc = 0.03\text{ m}, A=πc2=π(0.03)2=2.827×10−3 m2A = \pi c^2 = \pi(0.03)^2 = 2.827\times10^{-3}\text{ m}^2, I=πc44=π(0.03)44=6.362×10−7 m4I = \dfrac{\pi c^4}{4} = \dfrac{\pi(0.03)^4}{4} = 6.362\times10^{-7}\text{ m}^4, and by the J=2IJ = 2I shortcut, J=1.272×10−6 m4J = 1.272\times10^{-6}\text{ m}^4.

Step 2. Axial and bending stress separately.

σaxial=15×1032.827×10−3=5.31 MPa,σbending=500(0.03)6.362×10−7=23.58 MPa\sigma_{axial} = \frac{15\times10^3}{2.827\times10^{-3}} = 5.31\text{ MPa}, \qquad \sigma_{bending} = \frac{500(0.03)}{6.362\times10^{-7}} = 23.58\text{ MPa}

Step 3. Combine at the critical point (the fibre where both add).

σx=5.31+23.58=28.89 MPa\sigma_x = 5.31+23.58 = 28.89\text{ MPa}

Step 4. Torsional shear stress at the same point.

τxy=800(0.03)1.272×10−6=18.87 MPa\tau_{xy} = \frac{800(0.03)}{1.272\times10^{-6}} = 18.87\text{ MPa}

Step 5. Principal stresses and maximum shear.

R=(28.89/2)2+18.872=14.452+18.872=564.9=23.77 MPaR = \sqrt{(28.89/2)^2+18.87^2} = \sqrt{14.45^2+18.87^2} = \sqrt{564.9} = 23.77\text{ MPa}
σ1=14.45+23.77=38.2 MPa,σ2=14.45−23.77=−9.3 MPa,τmax=23.8 MPa\sigma_1 = 14.45+23.77 = 38.2\text{ MPa}, \qquad \sigma_2 = 14.45-23.77 = -9.3\text{ MPa}, \qquad \tau_{max} = 23.8\text{ MPa}

The maximum principal stress at the critical point is 38.2 MPa — larger than any one of the three individual stress contributions taken alone, which is exactly why all three loads must be checked together rather than in isolation.

Competitive stretch

If instead the axial load were compressive (P=−15 kNP = -15\text{ kN}, everything else unchanged), the two candidate fibres become σx=23.58−5.31=18.27 MPa\sigma_x = 23.58 - 5.31 = 18.27\text{ MPa} versus σx=−23.58−5.31=−28.89 MPa\sigma_x = -23.58 - 5.31 = -28.89\text{ MPa} on the opposite fibre. The second fibre now has the larger ∣σx∣|\sigma_x| and becomes critical instead — a reminder that "critical point" depends on the sign of the axial load, not just its magnitude, so both extreme fibres should always be checked when signs could combine either way.

Common traps

  • Not checking both extreme fibres: bending stress reverses sign across the neutral axis, so the fibre where it adds in magnitude to the axial stress — not simply the geometric "top" — is the one to analyse.
  • Using II where JJ is needed, or vice versa: axial and bending stress use II, torsional shear uses JJ; for a solid circular shaft J=2IJ = 2I, and mixing them up is a frequent slip.
  • Skipping the "locate the critical point" step: running principal-stress formulas correctly at the wrong point still gives a wrong answer.
  • Forgetting transverse shear is usually zero at the extreme fibre (since Q=0Q = 0 there for beam shear τ=VQ/Ib\tau = VQ/Ib), so it is normally, but not automatically, safe to ignore at the critical point — verify rather than assume.

Ask the tutor: if you want practice, ask for a shaft-design problem where you must find the minimum safe diameter given an allowable stress and all three loads combined.

Why do axial and bending stresses simply add, but torsion needs the full principal-stress formula?

Axial stress and bending stress are both normal stresses acting on the identical face of the same point, so they are literally the same kind of quantity and add algebraically like ordinary numbers, respecting sign. Torsional shear stress is a different kind of quantity, a shear stress, acting on that same face but not simply addable to a normal stress. To combine a normal stress with a shear stress correctly you need the plane-stress transformation formulas, which is why the last step of any combined-loading problem always ends in a principal-stress or Mohr's-circle calculation rather than a plain sum.

Do I always need to check both extreme fibres, or is there a shortcut?

There is no reliable shortcut without checking sign. When the axial load and the bending stress at a fibre have the same sign, that fibre gives the larger normal stress and is usually critical; when they have opposite signs, it is the other fibre, where the axial sign instead matches the bending sign, that becomes critical. Since axial force can combine with either fibre depending on load direction, the safe habit is to compute sigma x at both extreme fibres and keep the larger magnitude, exactly as done in the Competitive stretch above.

Why is transverse shear from bending left out of this analysis?

Beam (transverse) shear stress, τ=VQ/Ib\tau = VQ/Ib, depends on the first moment of area QQ of the material beyond the point in question. At the extreme outer fibre of a cross-section, there is no material further out, so Q=0Q = 0 there and transverse shear vanishes exactly at that point — even though it is nonzero elsewhere on the section, such as at the neutral axis. Since the critical point for combined loading is normally the extreme fibre, where bending normal stress is largest, transverse shear is usually zero right where you need it, which is why it is dropped from the plane-stress element. It should not be dropped automatically at any other point on the section.

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Where this leads