Combined axial and bending loading

Push down on a wooden ruler exactly through its centre and it just gets a little shorter — pure compression, spread evenly across the section. Push down a few millimetres off-centre and something surprising happens: the ruler also bends, one edge squeezed much harder than the other, while the far edge may even go into tension. That off-centre push is an eccentric axial load, really two loads in disguise — a centroidal force plus a bending moment. Combined axial and bending loading is the technique for adding their two stress fields back together at every point on the section, so you find the true peak stress instead of trusting a misleadingly comfortable average. It matters wherever a load cannot be guaranteed to land dead-centre: brackets, offset columns, crane hooks, punch-press frames. Get the superposition wrong and a member that "looks" safely compressed can silently be cracking in tension on its far face.

The core insight

An eccentric axial load $P$ at eccentricity $e$ is statically equivalent to a centroidal force $P$ plus a moment $M = Pe$ , so the total normal stress at any point is the simple sum $\sigma = P/A \pm My/I$ , evaluated at the extreme fibres as $\sigma = P/A \pm Mc/I$ .

The core idea

Start from the two stress formulas you already know separately: the uniform axial stress $\sigma = P/A$ from a centroidal force, and the flexure stress $\sigma = My/I$ from a bending moment. When an axial force $P$ is applied off the centroidal axis, at eccentricity $e$ , it is statically identical to moving that same force $P$ to the centroid and adding a couple $M = Pe$ — the same force–couple equivalence used to relocate forces in statics. Because both resulting stresses act perpendicular to the same cross-section, they are both normal stresses on the same plane, so combining them is a direct algebraic sum rather than a vector or Mohr's-circle problem. The total normal stress at a point a distance $y$ from the neutral axis is

\sigma = \frac{P}{A} + \frac{My}{I} = \frac{P}{A} + \frac{Pe\,y}{I}

where: $P$ is the axial force (tensile positive), $A$ is the cross-sectional area, $M = Pe$ is the equivalent bending moment, $y$ is the signed distance from the centroidal (neutral) axis to the point of interest, and $I$ is the second moment of area about that axis. The sign of $My/I$ must track which side of the neutral axis the point sits on, so that it correctly adds to or subtracts from the axial term. Because bending stress is largest at the extreme fibres, that is where you usually need the answer. Setting $y = \pm c$ , the outer fibre distances, gives the two extreme stresses directly:

\sigma_{max,min} = \frac{P}{A} \pm \frac{Pec}{I} = \frac{P}{A} \pm \frac{Mc}{I}

where: $c$ is the distance from the centroid to the extreme fibre on either side. One extreme is always more compressive (or more tensile) than the axial value alone; the other is relieved by the bending term, and can flip sign entirely if bending dominates. That sign-flip possibility motivates the kern (core) of a section — the zone within which an eccentric compressive load produces no tension anywhere on the cross-section. Setting the relieved extreme stress to exactly zero, $P/A = Mc/I$ , gives the boundary eccentricity. For a rectangular section of width $b$ and depth $h$ , loaded along the depth with $c = h/2$ and $I = bh^3/12$ , this works out to the well-known middle-third rule: $e_{kern} = h/6$ . Load anywhere within the middle third of the depth and the whole section stays in compression; load outside it and part of the section goes into tension.

Figure

A rectangular column cross section with an off centre compressive load, showing a linear stress distribution that is more compressive on the near side and tensile on the far side

Derivation of the combined-stress formula

Derivation ACombined normal stress from an eccentric axial load

The goal is to show, starting only from the two separately-known stress formulas $\sigma = P/A$ and $\sigma = My/I$ and the force–couple equivalence for moving a force, that an eccentrically applied axial load $P$ at eccentricity $e$ produces a total normal stress $\sigma = P/A \pm Mc/I$ at the extreme fibres of a cross-section, with the bending term written using the equivalent moment $M = Pe$ .

Step 1. Replace the eccentric load with an equivalent system at the centroid. A force $P$ applied at a point offset by $e$ from the centroidal axis is statically equivalent to the same force $P$ acting through the centroid plus a couple of moment $M = Pe$ — the standard "move a force to a parallel line of action" equivalence: the net force and moment are unchanged, only the computation is made convenient.

Step 2. Write the stress each part produces on its own. The centroidal force $P$ produces the uniform axial stress $\sigma_{axial} = P/A$ at every point, independent of position. The couple $M$ produces the familiar flexure stress $\sigma_{bend} = My/I$ , which varies linearly with the signed distance $y$ from the neutral axis and is zero at the centroid.

Step 3. Superpose the two fields. Both stresses act normal to the same cross-sectional plane (neither is a shear stress), so they add algebraically at any point without any transformation: $\sigma(y) = P/A + My/I$ . Substituting $M = Pe$ gives $\sigma(y) = P/A + Pe\,y/I$ , in terms of the original eccentric load alone.

Step 4. Evaluate at the extreme fibres, where design usually cares most. Setting $y = \pm c$ for the two outer surfaces gives $\sigma_{max,min} = P/A \pm Pec/I = P/A \pm Mc/I$ . The plus sign applies on the side the moment compresses further (same sense as the eccentric load), the minus sign on the side it relieves — and if the relief term exceeds the axial term, that side flips sign entirely.

Worked example

A short rectangular post, $b = 100\text{ mm}$ wide and $h = 200\text{ mm}$ deep, carries a compressive load $P = 120\text{ kN}$ applied at eccentricity $e = 40\text{ mm}$ from the centroidal axis, along the depth direction. Find the maximum and minimum normal stress on the cross-section, and check the result against the kern.

Step 1. Section properties. $A = bh = 0.1\times0.2 = 0.02\text{ m}^2$ . $I = \dfrac{bh^3}{12} = \dfrac{0.1(0.2)^3}{12} = 6.667\times10^{-5}\text{ m}^4$ . Extreme fibre distance $c = h/2 = 0.1\text{ m}$ .

Step 2. Axial stress. $\sigma_{axial} = P/A = \dfrac{120\times10^3}{0.02} = 6.0\times10^6\text{ Pa} = 6.0\text{ MPa}$ (compressive, uniform across the section).

Step 3. Equivalent moment and bending stress. $M = Pe = 120\times10^3\times0.04 = 4800\text{N}\,\cdot\,\text{m}$ . $\sigma_{bending} = \dfrac{Mc}{I} = \dfrac{4800(0.1)}{6.667\times10^{-5}} = 7.2\times10^6\text{ Pa} = 7.2\text{ MPa}$ .

Step 4. Superpose at the two extreme fibres. $\sigma_{max(comp)} = -6.0 - 7.2 = -13.2\text{ MPa}$ , and $\sigma_{min} = -6.0 + 7.2 = +1.2\text{ MPa}$ — a small tension on the far side, even though the post is nominally under an overall compressive load.

Step 5. Check against the kern. $e_{kern} = h/6 = 200/6 = 33.3\text{ mm}$ . Since the actual eccentricity $e = 40\text{ mm}$ exceeds $33.3\text{ mm}$ , the load falls outside the middle third, so tension on the far face is expected — exactly what Step 4 found. The section sees a maximum compressive stress of 13.2 MPa and a maximum tensile stress of 1.2 MPa.

Competitive stretch

If instead the same post were loaded at the kern boundary exactly, $e = 33.3\text{ mm}$ , the minimum stress would come out to zero: $\sigma_{bending} = Mc/I = (120\times10^3 \times 0.0333)(0.1)/6.667\times10^{-5} = 6.0\text{ MPa}$ , so $\sigma_{min} = -6.0 + 6.0 = 0$ . This confirms the kern definition directly — load at $e_{kern}$ and the far fibre is exactly on the verge of tension, with the whole section still at or below zero (compression only).

Common traps

Skipping the force–couple step. Trying to write a combined-stress formula without first converting the eccentric load into a centroidal $P$ plus a moment $M = Pe$ leads to a wrong or incomplete moment term.
Sign confusion between the two terms. The axial term is the same sign everywhere on the section; the bending term changes sign across the neutral axis. Track both separately at each extreme fibre rather than guessing.
Treating the extreme fibre as automatically the point of maximum stress. That is only true for pure axial-plus-bending. If shear or torsion also act on the member, the true maximum principal stress may occur elsewhere and needs a full stress-transformation check.
Mixing this up with column buckling. This is a cross-sectional stress superposition with no instability involved. Euler buckling and beam-column interaction are a separate topic later in the unit.

Ask the tutor: if you would like to work through locating the kern for a circular or T-shaped section, or try a case with eccentricity in two directions at once, just ask.

Why can a "compressive" load ever cause tension anywhere on the section?

Because the bending term $Mc/I$ from the eccentricity can be larger than the axial term $P/A$ on the relieved side. The two stresses are added algebraically: on the far side from the load, the bending stress opposes the axial compression, and if it is big enough it overwhelms the axial term and flips the net stress to tension. This is exactly what the worked example shows — a nominally compressive $120$ kN load still leaves $+1.2$ MPa of tension on the far face because the eccentricity exceeds the kern limit.

Why does the middle-third rule use h over 6 and not h over 3?

The "middle third" describes the width of the safe zone measured across the whole depth $h$ , centred on the centroidal axis — it spans from $-h/6$ to $+h/6$ , a total width of $h/3$ . The eccentricity itself, measured from the centroid to one side, is only half that span, which is why the boundary value used in the stress formula is $e_{kern} = h/6$ , even though the safe zone as a whole is one-third of the depth.

Do I need Mohr's circle to solve a combined axial and bending problem?

Not for this topic on its own. Both $P/A$ and $Mc/I$ are normal stresses acting on the same plane (the cross-section), so they add directly by ordinary algebra — no transformation is needed to combine them. Mohr's circle or stress transformation only becomes necessary if you also have a shear or torsional stress at the same point and need the true maximum principal stress, which is a separate combined-loading case (axial plus torsion, or axial-bending plus shear).