Module 10 · Lesson 10.3
Shear in unsymmetrical sections and the shear centre
Why a channel twists when you load it through its centroid.
Why this matters
Load a channel vertically through its centroid and it twists. Nothing is wrong with the channel, and nothing is wrong with your load — the shear stresses inside it have a resultant that does not pass through the centroid, and the difference is a torque. The point you must load through is the shear centre, and for a channel it is not even inside the section.
By the end of this lesson you should be able to
- Apply τ = VQ/It to a thin-walled open section
- Trace the shear flow round a channel and find its resultant
- Define the shear centre and locate it for a channel
- Explain the effect of shear on the assumptions of simple bending theory
What you should already know
- The shear formula τ = VQ/It (this module)
- Shear flow q = VQ/I (this module)
- Unsymmetrical bending and principal axes (Module 9)
The shear formula does not care whether the section is symmetric. What changes is where you have to look.
In a thin-walled open section it is easier to work with shear flow, q = VQ/I, in newtons per millimetre of wall, than with the stress itself. Trace it round the section from a free edge, where Q = 0 and therefore q = 0, accumulating first moment of area as you go.
For a channel bent about its strong axis, with the web vertical:
- Start at the tip of the top flange. q = 0.
- Move along the flange towards the web. Q grows linearly, so q grows linearly, reaching a maximum where the flange meets the web.
- Turn down the web. Q now grows parabolically, so q is parabolic, peaking at the neutral axis.
- Down into the bottom flange, where by symmetry it mirrors the top.
The web carries almost all the vertical shear, as you would expect. But the flanges carry substantial horizontal shear flows — and those are the ones that cause the trouble.
Add up what those flows do.
The two flange flows are horizontal, equal in magnitude, and opposite in direction — one runs inwards along the top flange, the other outwards along the bottom. Equal and opposite forces on parallel lines a distance h apart form a couple, of magnitude Fflange × h.
The web flow gives a vertical resultant equal to the applied shear V.
So the internal shear stresses are statically equivalent to a vertical force V plus a torque. For the section to carry the load without twisting, the applied load must produce exactly that same torque about the same point — which means it cannot act through the centroid. It must act through a point offset from the web, on the opposite side from the flanges.
That point is the shear centre.
Predict first
A channel section is used as a beam and loaded vertically through its centroid. What happens?
One last point completes the theory of bending. Simple bending theory assumes plane sections remain plane. Shear stress violates that assumption: it produces shear strain, and shear strain means the cross-section warps out of plane, more near the neutral axis where shear stress peaks than at the extreme fibres where it is zero.
So the two theories are, strictly, incompatible. In practice this matters very little:
- For a slender beam the warping is tiny and nearly identical at neighbouring sections, so the bending stresses are essentially unaffected.
- The error grows as the span-to-depth ratio falls. For a deep beam — a transfer beam, a deep spandrel — neither simple bending theory nor the shear formula is reliable, and the stress distribution stops being linear.
The conventional dividing line is a span-to-depth ratio of about 4. Below that, treat the member as a deep beam and use a method that does not assume plane sections.
Worked example
Shear flow and the shear centre of a channel
Given
- Thin-walled channel: web 300 mm deep, flanges 100 mm wide, uniform thickness t = 8 mm
- Dimensions taken to the wall centrelines
- Vertical shear force V = 120 kN through the section
Find
The flange shear flow, the flange force, and the distance from the web to the shear centre.
Practice
A thin-walled channel has I = 54.0 × 10⁶ mm⁴. At the flange–web junction the first moment of the flange about the neutral axis is Q = 120 × 10³ mm³. Under a shear force of 120 kN, what is the shear flow there, in N/mm?
Practice
The flange is 100 mm wide and the shear flow varies linearly from zero at the tip to 266.7 N/mm at the web. What is the total horizontal force in one flange, in kN?
Practice
For that channel under 120 kN of shear, how far from the web centreline is the shear centre, in mm?
Summary
- Shear flow q = VQ/I is easier than stress for thin-walled sections; start from a free edge where q = 0
- In a channel the flange flows are horizontal and opposite, forming a couple
- The shear centre is where the resultant of the internal shear flows acts
- For a doubly symmetric section it coincides with the centroid; for a channel it lies outside the section
- For an angle it is at the intersection of the leg centrelines
- Loading away from the shear centre applies a torque to a section that is poor in torsion
- Shear strain warps the cross-section, so plane sections do not strictly remain plane — this matters below a span-to-depth ratio of about 4
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint