Buckling

Every check so far has compared a stress against a material limit, and for members in tension that is the whole story. Push on a metre rule instead of pulling it and something else happens entirely: it bows out sideways and loses all resistance, long before anything in it has come close to yielding. That failure has nothing to do with strength, and it is the reason compression members look nothing like tension members.

Why compression is not tension with a sign flipped

A tie in tension is self-straightening. If it starts slightly bent, the pull tends to pull it straight, and any small sideways disturbance is corrected by the load itself. A strut in compression is the opposite: a small bow gives the axial load an eccentricity, the eccentricity produces a bending moment, the bending moment increases the bow. Whether that loop settles or runs away depends on how stiff the member is in bending against how hard it is being pushed.

So a compression member has two entirely separate failure loads. It can be squashed, when the stress reaches the yield strength, at a load $σ_{y} A$ . Or it can buckle, at a load that turns out to depend on $E$ , on $I$ , and on the length, and not at all on the material's strength. The lower of the two governs, and for anything slender the second is far lower.

This is the failure mode that has to be found by calculation rather than by inspection, because a buckled member is not overstressed anywhere at the moment it goes.

Euler's derivation

Leonhard Euler solved the problem in 1744, and the derivation uses the beam equation of the previous lesson with one change of sign.

Take a strut of length $L$ , pinned at both ends so that neither end can move sideways and neither resists rotation, carrying an axial compressive load $P$ . Suppose it has adopted a bowed shape $v (x)$ . At a section $x$ , the free body between the end and the cut has the load $P$ acting at the end, offset from the cut by the deflection $v$ , so the internal bending moment is $M = - P v$ , the sign because the moment acts to increase the bow. Substituting into $E I v'' = M$ ,

E I \frac{d^{2} v}{d x^{2}} + P v = 0

This is the equation of simple harmonic motion with $x$ in the role of time. Writing $k^{2} = P / E I$ , the general solution is

v = A \sin k x + B \cos k x

The boundary conditions are $v = 0$ at both ends. The first gives $B = 0$ . The second gives $A \sin k L = 0$ , and here is the whole subtlety: either $A = 0$ , meaning the strut stays perfectly straight, which is a valid solution at every load, or $\sin k L = 0$ , which requires

k L = n π \Rightarrow P = \frac{n^{2} π^{2} E I}{L^{2}}

At those particular loads, and only at those, a bowed shape is possible alongside the straight one. The smallest, with $n = 1$ , is the Euler critical load

P_{c r} = \frac{π^{2} E I}{L^{2}}

and the shape that goes with it is a single half sine wave.

The equation never determines $A$ . That is not a failure of the algebra; it is the physics. At the critical load the strut is indifferent to being bowed by any amount, which is exactly what neutral equilibrium means, and it is why buckling is an eigenvalue problem rather than a stress problem. A small increase past $P_{c r}$ and the deflection is limited only by the large-deflection terms that were dropped, which is to say the member folds.

Reading the result

Three things are in the formula and one large thing is not.

The load falls as the square of the length, so doubling a strut's length quarters what it can carry. Nothing else in this course punishes length so heavily.

It rises with $E I$ , so it is stiffness that resists buckling. In particular, $I$ must be taken about the axis the strut will actually bend about, which is the one with the smallest second moment. A strut buckles the easy way, and a rectangular strut $50$ mm by $100$ mm will always go about the weak axis unless something restrains it there.

And the strength of the material is absent entirely. A strut in S275 steel and a geometrically identical one in S355 buckle at the same load, because $E$ is the same for both. Paying for higher grade steel in a slender compression member buys nothing at all, which is one of the more expensive lessons in the subject to learn on site.

Example. A solid steel rod of $50$ mm diameter is used as a pin-ended strut $2.5$ m long. Find its Euler load, and compare it with the load that would squash it, taking $σ_{y} = 275$ MPa and $E = 210$ GPa.

$I = π \times 50^{4} / 64 = 3.068 \times 10^{5}$ mm⁴, and $A = 1963$ mm². Then

P_{c r} = \frac{π^{2} \times 210 000 \times 3.068 \times 10^{5}}{2500^{2}} = 101.7 kN

The squash load is $275 \times 1963 = 540$ kN. Buckling governs at under a fifth of it, so four fifths of this rod's strength is unusable at this length.

Now you. The same rod is used over $1.5$ m instead. Find the Euler load.

Answer

$P_{c r} = π^{2} \times 210 000 \times 3.068 \times 10^{5} / 1500^{2} = 282.6$ kN. Shortening the strut by $40$ per cent has nearly tripled it, and it is still only half the squash load.

Effective length

The derivation used pinned ends because they give the simplest boundary conditions. Other end conditions change the shape of the buckled curve, and the change can be absorbed entirely into an effective length $L_{e}$ , the distance between the points of inflection in the buckled shape, so that

P_{c r} = \frac{π^{2} E I}{L_{e}^{2}}

Four cases cover almost everything. Pinned at both ends gives $L_{e} = L$ , by definition. Fixed at both ends gives $L_{e} = 0.5 L$ , so four times the load. Fixed at one end and pinned at the other gives $L_{e} = 0.7 L$ , roughly twice the load. Fixed at one end and entirely free at the other, a flagpole, gives $L_{e} = 2 L$ , so a quarter of the load.

The spread between the best and worst of those is a factor of sixteen, which makes end restraint the cheapest thing an engineer can buy in a compression member.

Example. Take the same $2.5$ m, $50$ mm rod. What is its critical load if both ends are fully fixed, and what if it is a free-standing post fixed only at its base?

Fixed at both ends, $L_{e} = 1250$ mm and $P_{c r} = 407$ kN, four times the pinned value and now approaching the squash load. As a free-standing post, $L_{e} = 5000$ mm and $P_{c r} = 25.4$ kN, one quarter of the pinned value.

Now you. The same rod with one end fixed and the other pinned. What is $P_{c r}$ ?

Answer

$L_{e} = 0.7 \times 2500 = 1750$ mm, so $P_{c r} = 101.7 \times (2500 / 1750)^{2} = 207$ kN, or equivalently $101.7 / 0.49$ .

In practice, fully fixed ends do not exist. A real connection has some rotational stiffness and codes require effective length factors that are deliberately less generous than the theoretical ones, typically $0.7$ where theory says $0.5$ , precisely because assuming restraint you have not actually built is unsafe in the direction that matters.

Slenderness

Comparing struts of different sections needs the result in terms of stress rather than load. Divide by the area and define the radius of gyration $r = \sqrt{I / A}$ , the distance at which all the area could be concentrated without changing $I$ . Then

σ_{c r} = \frac{P_{c r}}{A} = \frac{π^{2} E}{(L_{e} / r)^{2}} = \frac{π^{2} E}{λ^{2}}

where $λ = L_{e} / r$ is the slenderness ratio, a dimensionless number that is the single most useful descriptor of a compression member. The critical stress depends on nothing but $E$ and $λ$ , so one curve serves every steel section ever rolled.

Example. A square hollow section $100$ mm across with a $6$ mm wall is used as a pin-ended column $4$ m long. Find its slenderness, critical stress and critical load.

Ignoring corner radii, $A = 100^{2} - 88^{2} = 2256$ mm² and $I = (100^{4} - 88^{4}) / 12 = 3.336 \times 10^{6}$ mm⁴, so $r = \sqrt{3.336 \times 10^{6} / 2256} = 38.5$ mm. Then $λ = 4000 / 38.5 = 104$ , and

σ_{c r} = \frac{π^{2} \times 210 000}{104^{2}} = 192 MPa

which is below the yield of $275$ MPa, so buckling governs, and $P_{c r} = 192 \times 2256 = 432$ kN.

Now you. The same section over a $3$ m length. What is the slenderness and the critical stress, and what does the answer mean?

Answer

$λ = 3000 / 38.5 = 78$ , giving $σ_{c r} = π^{2} \times 210 000 / 78^{2} = 341$ MPa. That is above the yield stress, so the Euler formula is predicting a buckling stress the material can never reach. It is invalid here, and the real capacity is set by yielding and inelastic buckling, somewhat below the squash load of $275 \times 2256 = 620$ kN.

Where the Euler curve stops being true

The last answer exposes the boundary. Setting $σ_{c r} = σ_{y}$ gives the slenderness at which the two failure modes coincide:

λ_{transition} = π \sqrt{\frac{E}{σ_{y}}}

For S275 steel that is $π \sqrt{210 000 / 275} = 87$ , and for S355 it is $76$ . Above those values a strut is genuinely slender and Euler is roughly right. Below them, Euler predicts a stress the material cannot deliver, and the real capacity is governed by yielding.

Real columns do not follow either line. Tests on struts scatter well below the lower envelope of the two, and worst of all near the transition, where a column is neither clearly slender nor clearly stocky. Three effects account for the shortfall.

Initial crookedness. No member is straight. An initial bow $e_{0}$ is amplified by the axial load by a factor $1 / (1 - P / P_{c r})$ , so a strut loaded to half its Euler load has double its original bow, and the resulting bending stress adds to the direct stress. There is no sudden buckling at all in this picture: deflection grows from the first newton and the member fails when the combined stress reaches yield.

Accidental eccentricity. Loads are never applied exactly on the centroid, and connection details usually guarantee an offset.

Residual stresses. A rolled or welded section cools unevenly, leaving parts of the section already in compression before any load is applied. Those parts yield early, so the effective stiffness of the section drops before the nominal stress reaches yield.

Design codes handle all three together with a column curve, an empirical reduction factor applied to the squash load as a function of slenderness. The Perry-Robertson formulation, which underlies the British and European curves, models the whole thing as an equivalent initial bow calibrated to test data, and codes publish several curves because a hot-rolled I-section, a welded box and a hollow section have different residual stress patterns. The Euler load remains the upper bound they all approach at high slenderness, and it stays worth computing as the sanity check.

Buckling is not only a strut problem

The same instability appears whenever a compressed piece of a structure can move sideways.

A deep, narrow beam bent about its strong axis has a compression flange that is effectively a strut held only by the web. It can buckle sideways while the beam twists, at a moment well below the one the flexure formula allows. That is lateral torsional buckling, and it is why beams are given lateral restraint at intervals and why a joist is fixed against rotation at its supports.

A thin flange or web in compression can buckle locally in a short wavelength without the member as a whole moving at all, which is local buckling and is why steel codes classify sections into slenderness classes. A thin cylinder under axial compression is the extreme case: it is so sensitive to imperfections that real cylinders fail at a fraction of the theoretical load, and shell design is done almost entirely from test-based knockdown factors.

None of these is covered by comparing a stress against a strength. A structure has to be checked for stability separately, and stability is the thing that fails without warning.

The remaining gap is that members rarely carry one action alone. A column carries compression and bending together, and a shaft carries torque and moment together, so the next lesson has to combine what the last several have kept apart.