So far in this pathway, we've worked out the normal stress and the shear stress at every point of a cross-section. We've seen that these stresses reach their maximum values at different locations. So, it would be helpful if we could blend these stress values together somehow to give us a single stress value we can compare against the material's yield limit. This is where the von Mises stress comes into the picture.
In this step, we'll combine shear and normal stresses into a single equivalent stress, the von Mises stress, and compare it with yield stress. Structure Bench expresses that comparison as a utilisation. Then we'll deliberately push a cantilever past first yield to see how this is represented.
Von Mises and utilisation: a 2 m cantilever. Orbit, read the results and hover the members; nothing you do here changes the model.
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1.0 The idea
Yielding in steel depends primarily on stresses that change its shape. Uniform pressure on every face of a small cube (a hydrostatic stress state) reduces its volume but preserves its shape. Under the von Mises criterion, this pressure alone does not cause yielding. The criterion measures the contribution of stresses associated with distortion and expresses it as a single equivalent stress, which can be compared directly with the yield strength from a tension test.
At a point in a beam where the stress state consists of one normal stress, ฯ, and one shear stress, , the equivalent stress is
ฯ
ฯvMโ=ฯ2+3ฯ2โ.
Yielding is predicted when ฯvMโ reaches the tensile yield strength, fyโ.
The factor of 3 gives shear a greater weight in this calculation: for the same numerical value, shear stress contributes more to the equivalent stress than normal stress does. In pure shear, the criterion predicts yielding at ฯ=fyโ/3โ, or about 58% of the tensile yield strength. A shear stress well below fyโ can therefore still be close to causing local yielding.
If you'd like to see where the criterion comes from, take a look at this EngineeringSkills tutorial on von Mises stress.
This is the same criterion EN 1993-1-1 uses for its elastic check of a cross-section (clause 6.2.1(5), equation 6.1, with no transverse stress). It's written there as,
(fyโ/ฮณM0โฯโ)2+3(fyโ/ฮณM0โฯโ)1
where ฮณM0โ is the partial factor for the resistance of a cross-section. Structure Bench's utilisation is the square root of the left-hand side, U=ฯvMโฮณM0โ/fyโ, with ฮณM0โ = 1.0 from the UK National Annex. In other words, the utilisation is simply the von Mises stress divided by the yield strength. U = 1.0 means first yield: the most highly stressed fibre has just reached the yield strength.
One thing to be clear about before we use it: U is only a design check when the forces come from a design combination, with the loads factored. In this step and the ones before it, the loads are simply numbers we chose, so read U as "how far from yield", not as a pass or fail.
2.0 The model
Our model is a UB 254 x 146 x 31 that cantilevers 2 m from a fixed support, with a point load at its tip. A cantilever is a good test for this, because the moment and the shear both peak at the same place, the fixed end.
There's one load case of 10 kN and two combinations built from it: Working, which scales it to 50 kN, and Pushed, which scales it to 65 kN. The moment grows linearly from zero at the tip to the tip load times the 2 m lever arm, so at the fixed end, the moments are 50ร2=100 kN m and 65ร2=130 kN m.
Things to try
Press Load the model and pick Working: 50 kN at the tip as the result set (we'll just call it Working from here). Switch on Member stress and step through the ฯ, ฯ and ฯvMโ chips, hovering at the fixed end each time. Notice where each one peaks: the bending stress at the top and bottom of the section, and the shear stress at the neutral axis.
Switch to the U chip and read the utilisation at the fixed end. It should match the ฯvMโ reading divided by 355 MPa.
Now pick Pushed: 65 kN at the tip. Look closely at the fixed end - you may need to zoom in, because the part of the member beyond yield is very short.
Switch to the U* chip, which mutes everything below yield, and see exactly which fibres are beyond it.
3.0 What you'll see
What you should see at the fixed end
Combination
ฯ, extreme fibre
ฯ, neutral axis
Von Mises, peak
U
Working (50 kN)
285.0 MPa
36.6 MPa
285.6 MPa
0.80
Pushed (65 kN)
370.5 MPa
47.6 MPa
371.2 MPa
1.05
Under the working load, the von Mises stress is all but equal to the bending stress. The peak shear stress is sizeable, but it's at the neutral axis, where the bending stress is zero, so the two peaks never occur at the same point. At the extreme fibres, the only shear is the small horizontal shear flow in the flange, about 10 MPa. Because it's so small next to the bending stress, it adds less than 1 MPa to the von Mises stress. The utilisation of 0.80 is comfortably below 1.
Under the pushed load, the extreme fibres at the fixed end turn magenta. This is how the view tells us that the stress has passed fyโ. The utilisation scale is fixed at 0 to 1, and anything beyond yield gets its own colour.
With the U* chip, everything at or below yield fades to grey and only the magenta remains: two thin slivers in the flanges, right at the fixed end. That's the chip to reach for on a large model, when you want to know whether anything is beyond yield and where.
Hand check for the fixed end
The elastic section modulus. The UB 254 x 146 x 31 has I = 4,410 cm4 and h = 251.4 mm. The extreme fibre sits half the depth, 12.57 cm, from the neutral axis, so the elastic section modulus is Welโ=4,410/12.57 = 350.84 cm3.
The working load. The bending stress at the extreme fibre is the moment divided by the elastic section modulus,
ฯ=350,840100ร106โ=285.0ย MPa
which agrees with the solver's value in the table. With the flange's 10 MPa of shear added in, the von Mises stress is 285.6 MPa, so U=285.6/355=0.80.
First yield. The extreme fibre first reaches yield when M=Welโfyโ=350.84 = 124.5 kN m. The moment at the fixed end is the tip load times the 2 m length, so first yield needs a tip load of = 62.3 kN. That sits between our two combinations.
How far the yield extends. Under the pushed load of 65 kN, the moment a distance x from the fixed end is the tip load times the remaining length. So, the moment only exceeds 124.5 kN m where 65(2โx)>124.5, which is the first 84 mm of the cantilever (a few millimetres more once the flange's shear stress is added in). That's why the magenta is confined to the very end of the member.
4.0 The web-flange junction
In this cantilever, the extreme fibre governs comfortably. But that isn't always the case, and step 3's short beam is a good example of where it isn't. So, let's go back to it.
Under the load, the extreme fibres of that beam carry 114.2 MPa of bending stress and only the flange's 22.8 MPa of shear, which gives a von Mises stress of 120.8 MPa. At the neutral axis, near the support, the web carries 96.25 MPa of shear and no bending stress, which is 3โร96.25=166.7 MPa von Mises. Notice that the neutral axis, with no bending stress at all, is already worse than the extreme fibre here. But neither of those is the worst point.
At the top of the web, where it meets the flange, both stresses are large at the same time. The bending stress there is a little below the extreme fibre's, because the junction is only a flange thickness in from the top of the section. And the shear stress is still most of its peak, because almost all of the first moment of area above the junction comes from the flange. Combined, they give a von Mises stress of about 171 MPa - higher than at either peak.
Open step 3 in a new window, switch on Member stress with the ฯvMโ chip, and hover just to the left of the load: the peak reads 170.4 MPa.
So why 170.4 and not 171.2? Step 3's model is 2D, and the 2D view samples the depth at 17 evenly spaced points. The nearest one sits about 9 mm below the junction, so it lands just short of the exact peak.
Hand check for the web-flange junction in step 3
The first moment of area. At the junction, y=h/2โtfโ = 141.5 mm from the neutral axis. The part of the section beyond the junction is the flange alone, so the first moment of area is the flange's area times the distance from the neutral axis to its centroid, Q=165ร10.2ร146.6 = 246,700 mm3.
The two stresses. With M = 64 kN m and V = 160 kN, the bending stress and the shear stress in the web at the junction are,
ฯ=8,500ร104
The von Mises stress. Combining the two,
ฯvMโ=106.52+3ร
This is higher than both the 120.8 MPa at the extreme fibre and the 166.7 MPa at the neutral axis.
That only happens because step 3's span is short: a big shear force and a big moment at the same section. In most beams, the extreme fibre governs, as it does in this step's cantilever. But if you're checking a short, heavily loaded span (a transfer beam picking up a column, for example), it's worth hovering over the web-flange junction as well as the extreme fibres.
5.0 First yield and the plastic moment
Does a utilisation of 1.05 mean the cantilever has failed? No - and this is an important point about what U measures. U is an elastic check: it tells us where the stress, calculated as if the steel stayed elastic, first reaches yield. Real steel doesn't stop at first yield. The outer fibres yield and carry fyโ while the fibres further in pick up more load, until the whole section has yielded and formed a plastic hinge.
In other words, the yielded zone spreads inwards from the top and bottom of the section as the moment grows, and the moment at which the whole depth has yielded is the plastic moment. If you'd like to dig into how a section behaves as it yields, and what that means for the moments in a structure, take a look at this EngineeringSkills tutorial on the stress-strain curve and moment redistribution.
For this section, the plastic modulus from the tables is Wplโ = 393 cm3, so the plastic moment is,
Mplโ=Wplโfyโ=393ร103ร355=139.5ย kNย m
The ratio Wplโ/Welโ=393/351=1.12 is the section's shape factor. It tells us there's about 12% of extra bending capacity between first yield and a full plastic hinge. Our 130 kN m sits inside that reserve: it's above the 124.5 kN m at first yield, but below the 139.5 kN m plastic moment.
Of course, a section can only use that reserve if it can reach its plastic moment before its plates buckle locally, which is why Eurocode 3 sorts sections into classes. This section is Class 1 in bending, though only just in S355. Its flange outstand is 7.26 times its thickness against a limit of 7.32, so don't be surprised if your own check lands close to the line. Its cross-section check to Eurocode 3 would therefore use the plastic resistance, and pass at 130/139.5=0.93.
Remember, that's a check of the cross-section only. Whether the member as a whole can carry that moment without buckling sideways (lateral-torsional buckling) is a separate check, and on a long, unrestrained member it's often the one that governs.
What utilisation is, and isn't
U is a cross-section check, and an elastic one. It's conservative for compact sections, which have plastic reserve beyond it. And it says nothing about buckling, which often governs long beams and columns long before any fibre reaches yield. So, a utilisation comfortably below 1.0 on a slender member is not, on its own, evidence that the member is safe - always check its buckling resistance as well.
6.0 What to take from this step
The von Mises stress combines normal and shear stress into one number we can compare with yield, and U makes that comparison on a scale where 1.0 is first yield. The magenta colour and the U* chip show exactly where a section has passed it.
In most beams, the extreme fibre governs. In short, heavily loaded spans, look at the web-flange junction too, where large bending and shear stresses act at the same point. And first yield isn't collapse: the shape factor gives compact sections a reserve beyond it.
So far, everything in this pathway has come from beam theory. Next, we'll check it against a finite element analysis of the real cross-section.