Over the last six steps, we've learned to read the stress anywhere in a member and check it against yield. It's tempting to think that's the whole job: if every fibre is below yield, the member works.
Unfortunately, it isn't quite that simple. In this step, we'll look at two members that both read comfortably below yield and would still fail. They would fail by buckling, a failure mode that a stress check simply can't detect.
Two members below yield that would still fail. Orbit, read the results and hover the members; nothing you do here changes the model.
Members can copy this model into their own projects and work the step in it. Join or sign in.
1.0 Buckling, and why a stress check can't see it
A stress check asks one question: is the material overstressed at this point? That's the right question for a stocky member, where yielding of the material is what limits its capacity. But slender steel members usually fail by buckling which is a failure due to instability of the whole member rather than a local material failure.
Buckling depends on the stiffness of the whole member, its length and how it's restrained along that length, not on the stress at any one point. In other words, two members with the same cross-section and the same stress can have very different buckling capacities, simply because one is longer, or less well restrained, than the other. Nothing in the stress results will identify this.
The two classic cases are:
Flexural buckling of a strut. A slender member in compression bows sideways, about its weaker axis. Euler gave us the elastic critical load (often just called the Euler load) for a pin-ended strut,
Pcrโ=L2ฯ2EIโ
Lateral-torsional buckling of a beam. The compression flange of a beam behaves like a strut. If nothing restrains it laterally, it tries to buckle sideways. The tension flange is being pulled straight and resists moving with it, so instead the whole section moves sideways and twists at the same time. For a simply supported beam with no restraint between its supports, the elastic critical moment is,
Mcrโ=C1โL2ฯ2EIzโโIzโIwโโ+ฯ
where Itโ is the torsion constant (the J of step 6), Iwโ is the warping constant and C1โ allows for the shape of the moment diagram (1.13 for a uniform load). This version is for a load applied at the shear centre. A load sitting on the top flange lowers Mcrโ further: as the section twists, a load above the shear centre moves sideways with it and adds to the twist rather than resisting it.
Notice that the length appears squared in the denominator of both formulas. This is why length matters so much: double the unrestrained length of a strut and its Euler load falls to a quarter, with the cross-section and its stresses unchanged. If you'd like a refresher on Euler buckling, take a look at this EngineeringSkills tutorial on column buckling.
Don't get caught out by the notation here. This formula uses Eurocode 3's axes, where z is the minor axis, so Izโ here is the second moment of area about the weak axis. That's the opposite of the member axes we used in step 2, where z was the strong axis. In this step, Izโ always means the weak axis.
2.0 The model
Our 2D model has two members side by side, both in S355 steel, each loaded so that its stresses sit comfortably below yield. On the left is a 10 m UB 406 x 178 x 60 beam, simply supported and carrying 15 kN/m. It's modelled as two members, beam-1 and beam-2, either side of a node at its midspan. There's no lateral restraint anywhere between its supports. In a real building, this would correspond to a beam with no slab, secondary beams or bracing holding its compression flange in place along its length.
On the right is an 8 m UC 152 x 152 x 37 strut (the member named strut), pinned at both ends and carrying 400 kN of compression. Each member has its own load case, and the combination Both members loaded puts both loads on at once.
Things to try
Press Load the model and pick Both members loaded as the result set. (The model opens on the beam's load case alone, so you'll only see the beam loaded until you switch.)
Switch on Member stress and select the ฯ chip. Hover over the beam at midspan and then over the strut, and note the peak stress in each. Then select the U chip and do the same.
Switch to the U* chip. Nothing turns magenta. Remember from step 4 that U* mutes everything below yield and leaves only the fibres beyond it in magenta. So, as far as the material is concerned, nothing in either member is overstressed.
Switch off Member stress, switch on the Bending moment layer and read the midspan moment in the beam. Then switch on Axial force and read the force in the strut.
Now work out Mcrโ and Pcrโ in the hand checks below, and compare them with what you just read. This comparison is the buckling check that the stress results leave out.
3.0 What you'll see
What you should see
Beam
Strut
Design effect
187.50 kNยทm
400.00 kN
Peak stress
176.4 MPa
-84.9 MPa
Utilisation U
0.50
0.24
Elastic critical value (from the hand checks)
107.7 kNยทm
228.6 kN
The last row comes from the hand checks below, not from the solve.
By the stress check, both members are fine. The beam is working at half of its yield stress and the strut at a quarter. If we stopped at the stress results, we'd sign both of them off.
Now compare the design effects with the elastic critical values. The beam's applied moment is 1.74 times its elastic critical moment (187.5 kN m against 107.7 kN m), and the strut's applied force is 1.75 times its Euler load (400 kN against 228.6 kN). Both would buckle long before the loads were reached, and nothing in the member stress view hints at it.
The stress view can't warn you, because every fibre really is below yield right up to the point where the member moves sideways. The stresses we read are correct for the straight, untwisted member. However, once the buckling load is exceeded, the members enter a state of unstable equilibrium which for all practical purposes means they will buckle and fail.
Notice that the strut is drawn with its strong axis in the plane of this 2D model, where its Euler load is 716 kN - comfortably above the 400 kN it carries. But it would buckle about its weak axis, out of the plane, at 228.6 kN. A 2D model can only bend in its own plane, so even a buckling check made on this model would look at the wrong axis unless we told it about the weak one. So, whenever you model a compression member in 2D, ask yourself which way it would actually buckle, and remember that it may well be out of the plane you're looking at.
Hand check for the beam
The section properties. From the section tables, the UB 406 x 178 x 60 has Izโ = 1,200 cm4 about its weak axis, Itโ = 33.3 cm4, and hsโ=hโtfโ = 393.6 mm between the flange centres.
The warping constant. For a doubly symmetric I-section, the warping constant is, to a very good approximation,
Iwโ=4
The critical moment. We can now evaluate the formula above one piece at a time, with L = 10 m, E = 210 GPa and G = 80.8 GPa. The term in front of the square root is simply the Euler load of the beam treated as a strut buckling about its weak axis,
L2ฯ2EI
Inside the square root, the first term comes from the section's resistance to warping and the second from its resistance to twisting (St Venant torsion),
Izโ
Putting the pieces together with C1โ = 1.13 for the uniform load,
Mcrโ=1.13ร248.7ร0.0387+0.1082
against an applied moment of wL2/8=15ร102/8=187.5 kN m.
Hand check for the strut
The section properties. The UC 152 x 152 x 37 has I = 706 cm4 about its minor axis and 2,210 cm4 about its major axis.
The Euler loads. Both ends are pinned, so the buckling length is the full length of the member. With L = 8 m,
Pcr,minorโ=
The stress. Dividing the 400 kN load by the section's area of 4,710 mmยฒ, the stress is 400,000/4,710=84.9 MPa, which matches the peak stress in the table.
4.0 What a stress check covers
These elastic critical values are an upper bound, not the answer. A real member is never perfectly straight and is never free of residual stresses, so a design check to Eurocode 3 also allows for the member's initial crookedness, residual stresses and yielding, through the reduction factors ฯ and ฯLTโ, and gives a buckling resistance lower again. For these two members, it's far below the applied loads.
If you want to see how imperfections pull a real strut's capacity down from the Euler load, take a look at this EngineeringSkills tutorial on realistic buckling behaviour.
What a stress check covers
The member stress view and the stress planes check the cross-section: is the material overstressed at this point? They don't check flexural buckling, lateral-torsional buckling, local buckling of thin plates, or anything else that depends on the member's length and restraint. A utilisation below 1.0 means the cross-section is fine, provided its plates are stocky enough not to buckle locally (Class 1 to 3). It doesn't mean the member is fine.
5.0 What to take from this step
A stress check looks at one point at a time, but buckling depends on the whole member - its length and how it's held. We looked at a beam working at half of yield and a strut at a quarter of it, and both would buckle under their applied loads. A member can buckle with every fibre comfortably below yield, and the stress results won't warn you.
So, for any member in compression, and any beam whose compression flange isn't restrained, check buckling separately, using the member's real length and restraints.
Next, the final step: putting everything together on a portal frame.