8.7 Shear Centre & Unsymmetrical Bending

The Problem the Shear Centre Solves

Ordinary bending theory assumes a beam bends without twisting. For a symmetric section loaded in its plane of symmetry that is true. For a thin-walled open section such as a channel, it is not: load the channel through its centroid and it will bend and twist at the same time.

The explanation lies in shear flow. Transverse shear generates a flow $q = \dfrac{VQ}{I}$ around the section. In a channel, this flow runs horizontally outward along the top flange, vertically down the web, and horizontally inward along the bottom flange. The two flange forces are equal, opposite and separated by the web depth, so they form a couple. Unless the applied load provides an equal and opposite couple, the section twists.

The shear centre is the point through which the resultant of the internal shear flow acts. Apply the external load through it and the two couples cancel:

Load through the shear centre produces bending without twisting. Load elsewhere produces bending plus a torque equal to the load times its eccentricity from the shear centre.

Locating the Shear Centre: General Rules

The angle and T results are worth understanding rather than memorising: in a section made of narrow rectangles all meeting at a point, each rectangle's shear force acts along its own centreline, so the resultant must pass through the common intersection. No couple can arise, so that point is the shear centre.

Shear Centre of a Channel Section

For a channel with flange width $b$, web depth $h$ measured between flange centrelines, and uniform thickness $t$, the shear centre lies at a distance $e$ from the centre of the web:

$e = \frac{b^2h^2t}{4I_{xx}}$

and, substituting the standard approximation $I_{xx} = \dfrac{th^3}{12} + 2\left(bt\right)\left(\dfrac{h}{2}\right)^2$, this reduces for a thin uniform channel to

$e = \frac{3b^2}{6b + h}$

The shear centre lies on the opposite side of the web from the flanges — that is, outside the material of the section entirely. This is initially counter-intuitive and is precisely why it is examined.

Practical consequence

A channel used as a crane runway beam, a conveyor stringer or a purlin will twist if the load hangs from its centroid. Designers respond either by restraining the section against twist at intervals, by using a symmetric section such as an I-beam, or by using a closed box section, whose torsional stiffness is orders of magnitude greater than an open section of the same area.

That last point deserves emphasis. For thin-walled members, torsional constant $J$ for an open section is approximately $\sum \frac{1}{3}bt^3$, which is very small because $t$ is cubed. For a closed section, Bredt's formula gives $J = \dfrac{4A_m^2}{\oint ds/t}$, where $A_m$ is the area enclosed by the median line. Slitting a closed tube longitudinally can reduce its torsional stiffness by a factor of hundreds while barely changing its bending stiffness.

Unsymmetrical Bending

Simple bending theory, $\sigma = My/I$, is valid only when the plane of loading contains a principal axis of the cross-section. When it does not, the beam deflects in a plane different from the plane of loading, and the neutral axis is no longer perpendicular to the load.

Principal axes

Every cross-section has a pair of mutually perpendicular principal axes through the centroid, about which the product of inertia $I_{xy}$ vanishes and the second moments take their maximum and minimum values.

• If the section has any axis of symmetry, that axis is a principal axis, and so is the perpendicular through the centroid.
• For an unequal angle, neither leg is a principal axis; the principal axes are inclined, which is why an unequal angle deflects sideways under a purely vertical load.

The principal axis inclination follows the same transformation mathematics as Mohr's circle for stress:

$\tan 2\alpha = \frac{2I_{xy}}{I_{yy} - I_{xx}}$

and the principal second moments are

$I_{1,2} = \frac{I_{xx} + I_{yy}}{2} \pm \sqrt{\left(\frac{I_{xx} - I_{yy}}{2}\right)^2 + I_{xy}^2}$

The structural identity with the principal-stress equations is exact, and recognising it saves learning a second set of formulae.

The resolution method

For a section whose principal axes are known:

1. Resolve the applied bending moment into components $M_u$ and $M_v$ about the two principal axes $u$ and $v$.
2. Compute the stress from each component independently using simple bending theory.
3. Superpose:

$\sigma = \frac{M_u\,v}{I_{u}} + \frac{M_v\,u}{I_{v}}$

1. The neutral axis is the locus where $\sigma = 0$, giving

$\tan\beta = -\frac{M_v I_u}{M_u I_v}$

which is in general not perpendicular to the plane of loading.

1. Maximum stress occurs at the point of the section farthest from the neutral axis, which for an unsymmetrical section may not be the point farthest from the centroid — a genuine trap.

Worked example in outline

A rectangular section beam $100$ mm deep and $50$ mm wide carries a moment of 5 kN m inclined at 30 degrees to the vertical. Resolving:

$M_x = 5\cos30^\circ = 4.33 \text{ kN m}, \qquad M_y = 5\sin30^\circ = 2.5 \text{ kN m}$

With $I_{xx} = \dfrac{50\times100^3}{12} = 4.167\times10^{6}$ mm$^4$ and $I_{yy} = \dfrac{100\times50^3}{12} = 1.042\times10^{6}$ mm$^4$, the extreme-fibre stress at the corner $(25, 50)$ is

$\sigma = \frac{4.33\times10^{6}\times50}{4.167\times10^{6}} + \frac{2.5\times10^{6}\times25}{1.042\times10^{6}} = 52.0 + 60.0 = 112 \text{ MPa}$

Note that the smaller moment component contributes the larger stress, because it acts about the weaker axis. Ignoring the inclination and computing only $M/Z$ about the strong axis would have underestimated the stress by more than half — the practical reason unsymmetrical bending must be checked.

Summary of the Two Ideas

Both arise from the same underlying point: simple bending theory carries assumptions about symmetry, and thin-walled or asymmetric sections violate them.