Bending

A bending moment of 40 kN m has been sitting in the diagrams of the third lesson without a verdict attached to it, because dividing a moment by an area gives nothing meaningful. Torsion showed how to handle a case where the stress varies over the section; bending is the same idea applied in the direction beams actually work, and it is the single most used result in structural engineering.

The geometry of a bent beam

Consider a length of beam under pure bending, meaning a constant moment $M$ and no shear. Draw a grid on its side before loading. After loading, the lines that ran across the beam are still straight and still perpendicular to the curved axis; they have simply rotated relative to one another. The lines that ran along the beam have become arcs, longer near the convex face and shorter near the concave one.

The observation that the cross sections stay plane is the Bernoulli assumption, and like the corresponding assumption in torsion it is an experimental fact for slender members rather than a theorem. Jacob Bernoulli reasoned about the bent elastica in 1694, and Leonhard Euler put the theory in its modern form in 1744, though the correct location of the neutral axis waited until Navier's lectures in 1826.

Somewhere between the stretched face and the compressed one is a surface whose length does not change, the neutral surface, and its trace on any cross section is the neutral axis. Measure $y$ from that axis, and let $R$ be the radius of curvature of the neutral surface.

A fibre at distance $y$ originally had length $R d φ$ along with the neutral surface and now has length $(R + y) d φ$ . So its strain is

ε = \frac{(R + y) d φ - R d φ}{R d φ} = \frac{y}{R}

Strain varies linearly with distance from the neutral axis, and while the material is elastic so does stress:

σ = E ε = \frac{E}{R} y

That is the entire physical content. Everything else is bookkeeping over the section.

Where the neutral axis sits

Nothing so far said where the neutral axis is, and equilibrium settles it. Under pure bending there is no net axial force on the section, so

\int σ d A = \frac{E}{R} \int y d A = 0

Since $E / R$ is not zero, $\int y d A = 0$ , and that integral vanishing is precisely the definition of the centroid. So the neutral axis passes through the centroid of the cross section, whatever its shape.

For a rectangle or an I-section symmetric about its horizontal axis, that is at mid-depth and the maximum tensile and compressive stresses are equal. For a T-section, a channel, or a rail, the centroid is nearer one face, and the two extreme stresses differ, sometimes by a factor of two or more. That asymmetry is exploited on purpose in cast iron beams and in reinforced concrete, where the material is much better in compression than in tension and the section is arranged so the small stress lands on the weak face.

The flexure formula

Now take moments about the neutral axis. Each element of area contributes a force $σ d A$ at a lever arm $y$ , so

M = \int σ y d A = \frac{E}{R} \int y^{2} d A = \frac{E I}{R}

where

I = \int y^{2} d A

is the second moment of area about the neutral axis, with units of length to the fourth power. Eliminating $E / R$ between this and the stress relation gives the flexure formula, written as the same chain of ratios as in torsion:

\frac{M}{I} = \frac{σ}{y} = \frac{E}{R}

Three readings, all used constantly. The first gives the stress at any depth from the moment. The third says the curvature is $1 / R = M / E I$ , so $E I$ is the flexural rigidity and it is what the deflection lesson integrates. And the quantity $I$ is again a purely geometric property of the section, with $y$ squared, so material far from the neutral axis matters far more than material near it.

The largest stress occurs at the extreme fibre, at $y = y_{\max}$ , so it is convenient to combine the two geometric quantities into the section modulus

Z = \frac{I}{y_{\max}} σ_{\max} = \frac{M}{Z}

$Z$ has units of length cubed and is the single number a beam catalogue quotes for strength in bending. Steel section tables give it in cubic centimetres, which is why a design calculation shuttles between millimetres and centimetres more than it should.

Second moments, and the parallel axis theorem

For a rectangle of width $b$ and depth $d$ bent about its horizontal centroidal axis, the integral runs from $- d / 2$ to $d / 2$ over strips of area $b d y$ :

I = \int_{- d / 2}^{d / 2} b y^{2} d y = \frac{b d^{3}}{12}

For a solid circle of diameter $d$ it comes out as $π d^{4} / 64$ , exactly half the polar value from the torsion lesson, which is what you would expect since $J = I_{x} + I_{y}$ and the two are equal by symmetry.

Real sections are built up from rectangles that are not all centred on the same axis, and the parallel axis theorem handles that. For an area $A$ whose own centroidal second moment is $I_{c}$ , about an axis a distance $h$ away,

I = I_{c} + A h^{2}

The proof is one line: expand $\int (y + h)^{2} d A$ into $\int y^{2} d A + 2 h \int y d A + h^{2} \int d A$ , and the middle term vanishes because $y$ is measured from the centroid.

Example. A T-section has a flange $150$ mm wide and $20$ mm thick sitting on top of a web $20$ mm wide and $130$ mm tall, so the overall depth is $150$ mm. Find the position of the neutral axis and the two section moduli.

Flange: area $3000$ mm², centroid $140$ mm above the bottom. Web: area $2600$ mm², centroid $65$ mm above the bottom. The centroid of the whole is

\overline{y} = \frac{3000 \times 140 + 2600 \times 65}{5600} = \frac{589 000}{5600} = 105.2 mm

Applying the parallel axis theorem to each piece about that level:

I = [\frac{150 \times 20^{3}}{12} + 3000 (34.8)^{2}] + [\frac{20 \times 130^{3}}{12} + 2600 (40.2)^{2}] = 1.160 \times 10^{7} {mm}^{4}

The extreme fibres are $105.2$ mm below and $44.8$ mm above the neutral axis, so $Z_{bottom} = 1.160 \times 10^{7} / 105.2 = 110 300$ mm³ and $Z_{top} = 1.160 \times 10^{7} / 44.8 = 258 700$ mm³. The smaller one governs, so this section is limited by the stress in its bottom fibre, which under sagging is tension.

Now you. A T-section has a flange $200$ mm by $25$ mm on a web $25$ mm by $175$ mm, overall depth $200$ mm. Find the neutral axis position and $I$ .

Answer

The flange has area $5000$ mm² at $187.5$ mm and the web $4375$ mm² at $87.5$ mm, so $\overline{y} = (937 500 + 382 813) / 9375 = 140.8$ mm from the bottom. Then $I = [200 \times 25^{3} / 12 + 5000 (46.7)^{2}] + [25 \times 175^{3} / 12 + 4375 (53.3)^{2}] = 3.476 \times 10^{7}$ mm⁴, and the governing section modulus is the bottom one, $3.476 \times 10^{7} / 140.8 = 246 800$ mm³.

Why beams are deep

The cube on the depth in $b d^{3} / 12$ is the most consequential exponent in structural engineering, and it is worth meeting it on a real member.

Example. A timber floor joist $50$ mm wide and $200$ mm deep spans $4$ m and carries $3$ kN m⁻¹. Find the maximum bending stress, and compare it with the same joist laid flat.

The maximum moment is $w L^{2} / 8 = 3 \times 16 / 8 = 6$ kN m. On edge, $I = 50 \times 200^{3} / 12 = 3.333 \times 10^{7}$ mm⁴ and $Z = I / 100 = 333 300$ mm³, so $σ = 6 \times 10^{6} / 333 300 = 18.0$ MPa. Grade C24 softwood has a characteristic bending strength of $24$ MPa, so this joist is working hard but plausibly.

Now you. The same joist is laid flat, $200$ mm wide and $50$ mm deep. Find the stress.

Answer

$I = 200 \times 50^{3} / 12 = 2.083 \times 10^{6}$ mm⁴ and $Z = I / 25 = 83 330$ mm³, so $σ = 6 \times 10^{6} / 83 330 = 72.0$ MPa, four times as much, from exactly the same piece of wood. It would break.

The factor of four is not an accident of these numbers. Swapping $b$ and $d$ in a rectangle changes $Z = b d^{2} / 6$ by the ratio $(d / b)$ , so a joist four times deeper than it is wide is four times stronger in bending on edge. In stiffness the penalty is worse still, since $I$ changes by $(d / b)^{2}$ , a factor of sixteen here.

The I-beam

If material far from the neutral axis earns its keep and material near it does not, the logical section puts almost everything at the extremes and leaves just enough in the middle to hold the two halves together. That is the I-beam, and it is why steel is rolled into that shape rather than any other.

Example. An I-section has two flanges $200$ mm wide and $15$ mm thick with a web $8$ mm thick, giving an overall depth of $300$ mm. Compare it with a solid square of the same cross-sectional area.

Compute $I$ as a full rectangle minus the two voids beside the web: $I = (200 \times 300^{3} - 192 \times 270^{3}) / 12 = 1.351 \times 10^{8}$ mm⁴. The area is $2 \times 200 \times 15 + 270 \times 8 = 8160$ mm², and $Z = I / 150 = 900 500$ mm³.

A solid square of $8160$ mm² has a side of $90.3$ mm, so $I = {90.3}^{4} / 12 = 5.55 \times 10^{6}$ mm⁴ and $Z = 5.55 \times 10^{6} / 45.2 = 122 900$ mm³.

The I-section has $7.3$ times the bending strength and $24$ times the bending stiffness of the same weight of steel arranged as a square. That is the entire argument for the shape, and it is why rolled sections are quoted by depth and mass per metre rather than by area.

Now you. A welded plate girder has flanges $300$ mm wide and $20$ mm thick with a web $10$ mm thick, giving an overall depth of $800$ mm. Find $I$ and $Z$ .

Answer

$I = (300 \times 800^{3} - 290 \times 760^{3}) / 12 = 2.191 \times 10^{9}$ mm⁴, and $Z = I / 400 = 5.479 \times 10^{6}$ mm³, six times the section modulus of the $300$ mm deep section above for only $2.4$ times the area.

The limit on the argument is that flanges cannot be made indefinitely wide and thin, nor webs indefinitely slender, because both then buckle locally under the compression they are carrying. Steel design codes classify sections into four classes on exactly that basis, and a section too slender to reach its yield stress before a flange buckles is designed to a reduced capacity.

What the flexure formula assumes

Five assumptions were made, and each has a range.

The material is linear and elastic, so the formula stops at first yield. Steel design routinely goes beyond, using a plastic section modulus that assumes the whole section has yielded, which for a rectangle gives $b d^{2} / 4$ against the elastic $b d^{2} / 6$ , a reserve of fifty per cent.

The beam is initially straight and bent about a principal axis of the section. Bend a channel about a random axis and it twists as well as bends, which is why unsymmetric bending is a separate topic.

The section is constant and the beam is slender: the plane sections assumption gets steadily worse as the span-to-depth ratio falls below about ten, and a deep beam or a corbel is designed by other methods entirely.

The bending is pure. Real beams carry shear at the same time as moment, and shear makes the sections warp slightly out of plane, contradicting the derivation. The effect on bending stress is small for slender beams, which is why the formula survives, but the shear itself is not small and it has its own stress distribution to work out. That is next.

Finally, the beam is free to bend in the plane of the load. A tall narrow beam loaded on its strong axis can fail instead by buckling sideways and twisting, at a moment well below the one this lesson would predict. That is lateral torsional buckling, and it belongs with the buckling lesson.