Module 13 · Lesson 13.6
Deflection under unsymmetrical bending, and deflection due to shear
Beams that move sideways, and the deflection that bending theory leaves out.
Why this matters
Two loose ends from the deflection story. A section without an axis of symmetry deflects in a direction you did not load it in — which matters for the sideways movement of a purlin as much as the vertical. And every deflection so far has come from bending alone, ignoring the shear strain that is undeniably present. Usually that is fine. Knowing when it is not is the point.
By the end of this lesson you should be able to
- Resolve a load onto the principal axes and combine the two deflections
- Explain why the deflection is perpendicular to the neutral axis, not parallel to the load
- Apply the moment–area method to unsymmetrical bending
- Estimate shear deflection and judge when it matters
What you should already know
- Unsymmetrical bending and principal axes (Module 9)
- Standard deflection formulae and moment–area (this module)
- Shear strain and the shear modulus G (Module 7)
A beam bends about its neutral axis and deflects perpendicular to it. When the section is symmetric those two directions line up with the load and nothing surprising happens. When it is not, they do not.
The reliable procedure is to work in the principal axes, where the product of inertia vanishes and ordinary bending theory applies:
- 1.Find the principal axes and the principal second moments I₁ and I₂.
- 2.Resolve the applied load into components along those two axes.
- 3.Calculate the deflection about each axis separately, using any standard method — the load component divided by the relevant EI.
- 4.Add the two deflections as vectors to get the true movement.
The result generally has a component at right angles to the applied load. For an unequal angle used as a purlin, a purely vertical load produces a substantial horizontal deflection — which is why purlins are so often tied back with sag rods, and why the check that matters may be the sideways one.
What it calculates: the true magnitude and direction of the deflection
- δ₁, δ₂
- deflections about the two principal axes (mm)
- β
- angle of the resultant from the axis-1 direction (degrees)
This assumes
- Linearly elastic behaviour, so the two components superpose
- The load acts through the shear centre, so there is no twist to add
In plain terms: Because I₁ and I₂ differ, the two components are divided by different stiffnesses. The resultant therefore leans towards the weak axis — the beam sags preferentially in the direction it finds easiest.
Predict first
An unequal angle purlin carries a purely vertical load through its shear centre. In what direction does it deflect?
The moment–area theorems carry over unchanged, applied once about each principal axis. Draw the M/EI diagram for the component of moment about axis 1 and get δ₁; repeat for axis 2 and get δ₂; combine as vectors. Nothing about the theorems changes — only that you now do the work twice, in a rotated frame.
The same is true of Macaulay's method and of the unit-load integral. Every technique in this course works on an unsymmetrical section, provided you first resolve onto the principal axes.
Now the second loose end. Every deflection formula so far has come from EI v″ = M, which accounts only for bending strain. Shear strain is also present, and it adds a deflection of its own:
δshear = k V L / GA
for a cantilever with an end load, where k is the shear form factor — 6/5 for a rectangle, 10/9 for a circle, and roughly the gross area over the web area for an I-section.
Worked example
Shear deflection of a stubby and a slender cantilever
Given
- Rectangular steel cantilever, 150 mm wide × 400 mm deep, tip load 50.0 kN
- E = 205 000 N/mm², G = 79 000 N/mm², shear form factor k = 6/5
- Two cases: span 2.00 m and span 6.00 m
Find
The bending and shear deflections in each case, and their ratio.
Practice
A rectangular steel cantilever 150 mm wide × 400 mm deep spans 2.00 m and carries a 50.0 kN tip load. With G = 79 000 N/mm² and k = 6/5, what is the shear deflection at the tip, in mm?
Practice
For that same beam and load, what is the bending deflection at the tip, in mm? (E = 205 000 N/mm², I = 8.00 × 10⁸ mm⁴.)
Practice
The same beam is now made 6.00 m long instead of 2.00 m. By what factor does the ratio of shear deflection to bending deflection change?
Worked example
Moment–area on an unsymmetrical section
Given
- Cantilever of span 2.00 m built from an equal angle 100 × 100 × 10 mm
- Principal second moments I₁ = 2.866 × 10⁶ mm⁴ and I₂ = 0.734 × 10⁶ mm⁴, with the principal axes at 45°
- Vertical tip load P = 2.00 kN, E = 205 000 N/mm²
Find
The tip deflection about each principal axis, and the resultant.
Assumptions
- The load acts through the shear centre, so the angle bends without twisting
- Elastic behaviour, so the two components superpose
Practice
An equal angle cantilever has I₁ = 2.866 × 10⁶ mm⁴ and I₂ = 0.734 × 10⁶ mm⁴. Equal load components act along each principal axis. What is the ratio of the two tip deflections, δ₂/δ₁?
Summary
- A beam deflects perpendicular to its neutral axis, not along the load
- Resolve onto the principal axes, deflect about each, then add as vectors
- Every deflection method works unsymmetrically once you are in principal axes
- Shear adds δ = kVL/GA, with k = 6/5 for a rectangle and 10/9 for a circle
- Bending deflection goes as L³ and shear as L, so their ratio goes as 1/L²
- Negligible for slender beams; a few per cent or more below a span-to-depth ratio of about 5
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