Module 13 · Lesson 13.4

Euler–Bernoulli and Timoshenko

Every deflection calculation so far has used a theory nobody named, resting on an assumption nobody stated. This lesson states it, drops it, and finds out what changes.

Why this matters

Two lessons ago you computed shear stress in a beam. One lesson ago you computed its deflection. Nothing connected them, and that should be uncomfortable — because the deflection calculation assumed the beam has no shear strain at all.

It did that quietly, through a sentence that sounds like a description rather than an assumption: plane sections remain plane and normal to the deformed axis. The word doing the work is normal. If the section must stay at right angles to the axis, then the material cannot distort in shear, and a beam that cannot distort in shear is stiffer than a real one.

That theory is Euler–Bernoulli. The theory that drops the word normal is Timoshenko. This lesson derives both, and finds where the difference stops being academic.

By the end of this lesson you should be able to

State the kinematic assumption behind every deflection formula you have used
Write the shear strain as the gap between the axis slope and the section rotation
Derive both governing equations from the same starting point
Show that the shear share of deflection goes with (d/L)²
Say from what slenderness Euler–Bernoulli is adequate, and why
Recognise the inconsistency Euler–Bernoulli lives with

What you should already know

Bending stress and the moment–curvature relation (Module 9)
Shear stress in beams, τ = VQ/It (Module 10)
Deriving the beam deflection equation — this module's first lesson, which derives EI v″ = M
Deflection due to shear — this module's lesson on unsymmetrical bending, which computes it

One assumption, in two halves

Both theories start by saying that a plane cross-section stays plane — it does not warp into a curved surface. That is a real assumption and it is not exactly true, but it is very nearly true for a beam, and both theories keep it.

Euler–Bernoulli adds a second half: the plane section stays normal to the deformed axis. Timoshenko does not.

That is the entire difference. Everything else follows.

To see what it costs, give the two things separate names:

dw/dx — the slope of the deformed beam axis at a point.
φ — the rotation of the cross-section at the same point.

Euler–Bernoulli says these are the same number. Timoshenko lets them differ, and the difference is the shear strain:

γ = dw/dx − φ

Read that as a sentence. The shear strain is the amount by which the section has failed to keep up with the axis. Setting γ = 0 is exactly the statement that the section stays normal.

Note

Where this sits against the other two lessons in this module

This module has already done two things that belong to this subject, and it is worth being explicit about the join rather than leaving a reader to wonder.

"Deriving the beam deflection equation" derives EI v″ = M from exact curvature and the small-slope approximation. That derivation is Euler–Bernoulli — it just did not use the name, because there was nothing yet to distinguish it from.

"Deflection due to shear" computes the term this lesson is about, and gets the same 1/L² behaviour. What it does not do is say which theory that term belongs to, or why the theory that produced the deflection formula had no room for it.

One notational warning, because the two lessons use opposite conventions and they look like a contradiction. That lesson uses a form factor k = 6/5, writing the shear deflection as k·VL/(GA). This one uses a shear coefficient κ = 5/6, writing it as VL/(κGA).

They are reciprocals of each other, and the two expressions are identical. Both say a rectangle stores shear strain energy as though only five sixths of its area were working.

Both conventions are in use in the literature. Read which one a source means before substituting a number into it.

Try it

What the two theories actually assume

One subtraction separates them. The section rotates by φ, the axis has slope dw/dx, and the gap between them is the shear strain. Euler–Bernoulli sets that gap to zero.

Theory

Shear strain, magnified12×

Slope of the axis, dw/dx: 0.00400 rad
Rotation of the section, φ: 0.00397 rad
Shear strain, γ = dw/dx − φ: 3.302e-5 rad
Departure from normal: 0.0019 °
Shear force carried: 400 kN
Shear rigidity κGA: 12.12 × 10⁶ kN

Things worth trying

Start on Timoshenko. The solid line is the cross-section; the dashed line is where it would be if it stayed at right angles to the axis. They differ, and the difference is the shear strain.
Switch to Euler–Bernoulli. The two lines coincide exactly — that is the assumption, drawn. The section is DEFINED to be normal to the axis.
Now read the departure figure on Timoshenko. It is around two thousandths of a degree — for a 300 × 600 steel beam carrying 400 kN of shear, which is a lot. That is the real angle in an ordinary beam, and it is why the assumption survived two centuries without anyone minding.
Drag the magnification to zero. The two theories become visually identical while the numbers stay different — a reminder that the drawing is magnified and the physics is not.
The subtraction γ = dw/dx − φ is the whole of the distinction. Everything else in both theories follows from whether that quantity is allowed to be non-zero.
Notice what Euler–Bernoulli does next, and why it is strange: having assumed zero shear STRAIN, it then computes a non-zero shear STRESS from equilibrium — the τ = VQ/It of Module 10. The two statements are inconsistent, and the theory gets away with it because the energy involved is tiny in a slender beam.

From first principles

Both governing equations, from the same start

We want to show: To get from one kinematic statement to the two equations that are actually solved — and to see Euler–Bernoulli fall out of Timoshenko as the special case γ = 0.

A beam resists load two ways: by bending, which stores energy in fibres stretching and shortening, and by shearing, which stores energy in the material sliding. Euler–Bernoulli models the first and declares the second impossible. Timoshenko models both, so it needs a second variable to describe the sliding — and that second variable is the section rotation φ, now free of the slope.

How much does it actually matter

The derivation says the theories differ. It does not say by how much, and the answer is more useful than the equations.

Take a simply supported rectangular beam under a uniform load. The two deflection terms are

bending: 5wL⁴/384EI · shear: wL²/8κGA

Divide one by the other and substitute I = bd³/12, A = bd, κ = 5/6 and G = E/2(1+ν). The load cancels. The width cancels. Even the material almost cancels, because only the ratio E/G survives. What is left is:

δ_shear / δ_bending = 0.96 (E/G) (d/L)²

For steel, E/G = 2.6, so the coefficient is about 2.5. For a central point load it is about 3.1.

That is a result worth keeping. Shear deflection is a question about proportions, not about material or loading, and it falls with the square of slenderness.

Try it

How much of the deflection is shear

Euler–Bernoulli reports the bending term only. The curve is the share it leaves out, against span over depth — and the shape of that curve is the whole answer.

Span6.00m

Beam depth600mm

Load case

Span / depth: 10.00
Bending deflection (Euler–Bernoulli): 0.45 mm
Shear deflection (the missing term): 0.01 mm
Total (Timoshenko): 0.46 mm
Shear / bending: 2.5 %
Euler–Bernoulli understates by: 2.4 %
The relationship: 2.50 (d/L)²
Below 5% once L/d exceeds: 7.1

Things worth trying

Start at 6 m on a 600 mm beam — L/d = 10. Shear is 2.5% of the bending deflection, and Euler–Bernoulli understates the total by 2.4%. Nobody would notice.
Now drag the span down to 2 m without touching the depth. L/d falls to 3.3 and the shear share jumps to 22.5%. The beam did not change material or shape — only its proportions.
Watch the curve, not the number. It is an inverse SQUARE: halve the slenderness and the shear share quadruples. That is why the effect appears so suddenly as beams get deep.
Take the span out to 14 m. The share falls to about 0.5% and the two theories are indistinguishable. This is the region every textbook example lives in, which is why the assumption is invisible.
Switch to the point load. Every value rises by exactly 25%, because a central point load carries its full shear right up to mid-span while a uniform load's shear falls to zero there.
Find the crossing of the 5% line. For a uniform load it is L/d ≈ 7.1; for a point load ≈ 7.9. That is the honest version of 'shear deflection can be ignored for ordinary beams'.
Set the depth to 1400 mm and the span to 3 m — a transfer beam. Euler–Bernoulli is now understating the deflection by about a third. A serviceability check on that basis is not conservative.

Worked example

The same beam at three proportions

Given

A 300 × 600 rectangular beam, steel, E = 210 kN/mm², ν = 0.3
Simply supported under a uniform load of 30 kN/m
Spans of 12 m, 6 m and 2 m — the section never changes

Find

How much of the deflection Euler–Bernoulli leaves out at each span

Practice

A steel beam has a span-to-depth ratio of 5 and carries a uniform load. What percentage of the bending deflection does the shear deflection add? Take the relationship as 2.5(d/L)².

Practice

At what span-to-depth ratio does the shear deflection of a uniformly loaded steel beam fall to exactly 5% of the bending deflection? Take the relationship as 2.496(d/L)².

Check yourself

Euler–Bernoulli assumes zero shear strain, and then computes a non-zero shear stress from equilibrium. What should be made of that?

Predict first

Two beams have the same span-to-depth ratio: one is steel (ν = 0.3), the other concrete (ν = 0.2). Which carries the larger shear share of its deflection?

Summary

Every deflection formula you have used solves EI w⁗ = q, which assumes zero shear strain
The assumption is the word 'normal' in 'plane sections remain plane and normal to the axis'
γ = dw/dx − φ: the shear strain is the gap between the axis slope and the section rotation
Timoshenko lets that gap exist and pays with a second unknown and a second equation
Euler–Bernoulli sets it to zero and collapses to one fourth-order equation
The shear share of deflection is about 2.5(d/L)² for a uniform load, 3.1(d/L)² for a point load
So it is a question of proportions: negligible above L/d ≈ 7, a third of the answer at L/d ≈ 2
The load cancels entirely — a heavier load does not make shear deformation more important

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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