The beam that is a different beam upside down

The beam that fails sideways gave the critical moment for a doubly symmetric beam, in a form with two mechanisms under a root, and at the end it stated what the expression assumes: the shear centre is at the centroid, and for a section with unequal flanges the capacity depends on which flange is in compression. The flange that swings is in tension and the restraint that beats the gradient repeated that caveat on the way past, and neither put a number on it.

So how much does it depend? Most engineers, asked, would guess at a modest correction, a few tens of per cent perhaps, on the reasoning that the same steel is present in both cases and a moment’s sign can only matter a little. A symmetric beam can’t tell whether it’s hogging or sagging, and it seems unlikely that a mild asymmetry would change this much. The answer is that it changes everything, that it does so at very ordinary proportions, and that a whole class of composite bridge girders is built the unfavourable way up during the one stage when it’s unbraced.

Three terms under a root, and one in front #

Kitipornchai and Trahair’s form for a section with one axis of symmetry, under a uniform moment over a span $L$, is

$$M_{cr} = \frac{\pi^2 EI_y}{L^2}\left[\pm\frac{\beta_x}{2} + \sqrt{\left(\frac{\beta_x}{2}\right)^2 + \frac{I_w}{I_y} + \frac{GJ\,L^2}{\pi^2 EI_y}}\,\right].$$

Set $\beta_x = 0$ and it is the earlier essay’s formula. The new term, $\beta_x$, has the dimension of a length. It is

$$\beta_x = 2y_0 - \frac{1}{I_x}\int_A y\,(x^2 + y^2)\,dA,$$

where $y$ is measured from the centroid and $y_0$ is the distance from the centroid to the shear centre. Taken as positive for a larger top flange, it goes with the plus sign when the top is in compression and the minus when it’s in tension.

Take a real section: a 600 × 10 mm web, with 450 mm of flange at 20 mm thick to be divided between top and bottom, so every section examined has the same steel. Equally divided, 225 mm each, that is $I_y = 38 \times 10^6$ mm⁴, $J = 140 \times 10^4$ mm⁴, and $I_w = 3.65 \times 10^{12}$ mm⁶. At a span of 8 m and with $\beta_x = 0$, it has $M_{cr} = 534$ kN·m either way. Now take 300 mm on top and 150 mm on the bottom.

The working is short. $I_y$ rises to $50.7 \times 10^6$, as a wide flange counts as its cube, so the Euler term is $P_e = \pi^2 E I_y / L^2 = 1{,}642$ kN. The warping constant falls to $1.92 \times 10^{12}$, so $I_w/I_y = 37{,}870$ mm². $GJ/P_e = 69{,}060$ mm². And $\beta_x = 448$ mm, so $(\beta_x/2)^2 = 50{,}180$ mm². The root is $\sqrt{157{,}110} = 396$ mm. Then

$$M_{cr}^{+} = 1{,}642 \times (396 + 224) = 1{,}018 \text{ kN·m}, \qquad M_{cr}^{-} = 1{,}642 \times (396 - 224) = 283 \text{ kN·m}.$$

The same beam, and a factor of 3.6.

The two lines cross at equal flanges, as they must, and they separate rapidly. At a 250 mm top flange, with only a slight difference of 250 to 200, the two are 668 and 427, already a factor of 1.57. A designer asked to choose between those two flanges for the sake of a small fabrication convenience probably wouldn’t expect the choice to be worth more than the whole of a normal safety margin. And at the far right of the figure, with almost all the flange on top, the section is close to a tee, and the factor passes 10.

The fibres lean, and which ones matters #

The term looks like a piece of integration that has no business mattering, and it has a clear physical reading once the section is seen in the act of twisting.

When a section twists about its shear centre by a small angle $\phi$, a fibre at distance $r$ from that centre swings sideways by $r\phi$. Over the length of the beam, the twist varies, so the fibre’s path along the beam is not straight. It leans, with a slope of $r\,d\phi/dz$. Each fibre is carrying a stress, and a force along a leaning line has a sideways component, which has a lever arm $r$ about the centre of twist. So a stressed fibre puts a torque on the section, in proportion to its stress and to $r^2$. For a compressed fibre, that torque pushes the twist further: this is how the column that twists instead of bending buckles. A fibre in tension pulls the twist back. That’s the effect Herbert Wagner identified in 1929, and it’s also what a stretched wire does: a tensioned string has a torsional stiffness it does not have when slack.

Under an axial load, all fibres have the same sign, so the effect is purely destabilising. Under a bending moment, half are in compression and half in tension, which suggests a cancellation, and in a symmetric section that’s exactly what happens. In a monosymmetric section it doesn’t, since the shear centre has moved. It sits close to the larger flange: in the section above, 179 mm above the centroid, which itself is 62 mm above the web’s mid-height. So:

• With the larger flange in compression, the compressed fibres have a short distance to the centre of twist and little destabilising torque. The tensioned fibres of the small flange and the web’s lower half stand far from the centre, with a large stabilising torque. On balance, bending stress stiffens the section in torsion.
• With the larger flange in tension, the same distances apply with the signs reversed. The small flange’s compressed fibres are far out, and they destabilise, and the moment softens the section.

The moment has, in effect, added to the torsional stiffness by an amount $M\beta_x$. The critical moment is where a demand and a stiffness meet:

$$\frac{M^2}{P_e} = GJ + \frac{\pi^2 EI_w}{L^2} \pm M\beta_x.$$

The parabola is the demand and each straight line is a stiffness. A symmetric beam would have a flat line at one. The picture shows why the effect is so large: the moment itself is what is being solved for, and it enters on both sides. Where the two lines cross the parabola depends strongly on the slope, and that slope is $\beta_x$ against a modest constant torsional stiffness. A section that has a small $GJ$ will have a large response, because the Wagner torque is then large by comparison with what it’s being added to.

The fit in common use comes from Kitipornchai and Trahair, $\beta_x \approx 0.9h(2\rho - 1)(1 - (I_y/I_x)^2)$, where $\rho = I_{y,c}/I_y$ is the compression flange’s share of the lateral stiffness.

The integral and the fit agree to within 4 per cent. This is a good sign that the integral was set up correctly, and it also shows how little depends on the detail. The term is of the order of the depth of the section, and it goes as twice the flange share less one. This means a section with a share of 0.9 has a Wagner term of about 0.7 depths. That’s a long length compared with the other terms in the bracket, $\sqrt{I_w/I_y}$ at 195 mm for this section. So once a beam has a noticeably unequal pair of flanges, $\beta_x$ is the largest term in the formula, and it’s the one left out of the symmetric expression.

With the span and with a tee #

The factor between the two falls as the span rises, from 5.5 at 4 m to 2.7 at 12 m. At long spans, the torsional term $GJL^2/\pi^2 EI_y$ under the root has grown large and it swamps $\beta_x/2$. Short spans are where the term in front dominates, so the effect is greatest, though at those spans a section might be limited by yield rather than by buckling, and the favourable curve is well above any section’s strength. The unfavourable curve is what stays of interest. At 4 m, the section with its smaller flange in compression has only 660 kN·m, well below its yield moment, on a span at which a designer would hardly expect buckling to govern.

The limit of the whole argument is a tee, where the small flange has gone altogether and only one flange and a stem are left. A thin-walled tee has no warping constant, so all it has are $I_y$, $J$ and $\beta_x$.

The American steel specification treats tees separately for this reason, with its closed form $M_{cr} = \pi\sqrt{EI_yGJ}\,(B + \sqrt{1 + B^2})/L$, $B = \pm 2.3(d/L)\sqrt{I_y/J}$, where the sign says which part is compressed. The formula’s constant of 2.3 is a simplification of $\beta_x$ for a typical tee. The two approaches agree in shape and they differ by 10 or 15 per cent here. It is notable that the formula’s sign convention is what gets a tee with its stem in compression its low value, and that this one small sign is the entire difference between a tee cantilever and a tee span. A cantilevered tee with its flange on top has its stem in compression at the root and is at the bad end of this curve, and this is the ordinary roof-canopy detail that the flange that swings is in tension dealt with for symmetric sections.

What the beam can actually reach #

A critical moment isn’t a capacity. The compression-flange-large case is limited by the first yield of the small flange, since that is the furthest fibre. So the two limits need to be read together.

This is the designer’s figure. If the moment is always of one sign and the span can’t be braced, shifting steel from the tension flange to the compression flange raises the beam’s capacity from 534 kN·m to 918, 72 per cent more with no extra steel. Past that point, first yield of the small flange takes over. This is why crane girders have a cap channel on their top flange, where the wheels are. It has been done for a century for local reasons as well, as the wheel’s sideways surge is resisted by the widened flange, but the effect on the buckling moment is real, and it is usually much larger than the designer has credited.

The flat-topped form of each curve shows the limit to the trade. A moment that reverses gets the worse of the two senses, and the best that a section can do against a reversal is to have equal flanges. That’s a surprising use for the symmetric section: it is not generally the strongest, it’s the one that doesn’t need to know which way it will be bent.

There’s a further case, and a costly one. A composite bridge girder has its top flange small, on purpose. The concrete deck will carry the compression in service, so a large steel top flange would be wasted, and the steel is all at the bottom, where the tension is. The girder is then lifted into place and its deck is poured. During the pour, the wet concrete is a load and not a restraint, and the small top flange is the compression flange of a sagging beam that has no deck yet. Of the many ways this structure could be loaded, this is the worst, and it lies at the bad end of the first figure in this essay: the smaller flange in compression, with the Wagner term working against it, on a section whose top flange has been made small because of a stage it isn’t yet in. Girders have fallen sideways at that stage. The most dangerous day gives the general principle, and the box that is a trough until its deck is cast has the same arrangement in a closed section. This essay’s contribution is the size of the number, and the size of the number is why construction-stage bracing for such girders is a design issue and not a detail.

The same section with its moment reversed then flips in the opposite way. A continuous composite girder hogs over its piers, where the bottom flange is in compression. That flange is the large one in that case, and that fact is usually credited to the section’s strength. It also helps with the distortional buckling of the bottom flange, which is held by the web alone.

What the wrong way up costs in bracing #

The usual repair is to brace the compression flange until the beam no longer cares, and the same formula prices that. The beam with its small flange in compression has 283 kN·m on the full 8 m span, and if the wet deck’s moment is around 500 kN·m it needs help. At 6 m between braces its critical moment is 386 kN·m, at 5 m 484, at 4 m 660 and at 3 m 1,036. So a brace is needed about every 4.8 m just to reach 500, and every 4 m to have a working margin over it. The same section the right way up has 1,018 at 8 m and has no need of any.

The equal-flanged beam from the same steel would have 534 kN·m with no bracing at all. Here the cost of the economical division of steel is visible in full: a composite girder saves steel in its service state by making its top flange small, and it pays back at the construction stage with cross-frames or a temporary top chord. In practice, the two are not usually weighed against each other in one calculation, because they are the concerns of different people working at different times. A top flange a little wider than the service state wants, just wide enough to reach the brace spacing the erector would like, may be the better bargain, and only the Wagner term makes that trade worth examining, since the width enters the formula twice, once through $I_y$ and again through $\beta_x$.

What uniform moment and a perfect section leave out #

The moment is uniform. Under a gradient, the sign of the moment may vary along the span, and the Wagner term is not then simply one number: it has to be integrated against the moment along the length. Standards offer a factor, $C_3$ or its like, on this term that depends on the moment diagram. A beam with a sign change along its span is a harder case, and its two parts help and harm one another.

The load is at the shear centre. The load height changes the critical moment in its own way, as the load that moves with the twist has it, and for a monosymmetric section the shear centre has moved. A load on the top flange is much closer to the shear centre of the large-topped beam than the symmetric formula would assume. So the favourable case gets a further benefit, and a wheel on a cap channel is near enough to the shear centre that its usual destabilising term shrinks.

The cross-section keeps its shape. A real plate girder with a thin web can distort, so the flanges rotate differently and the small flange can buckle almost by itself. With unequal flanges and a slender web, as the flange a slender web wants heavier noted, this is an added risk, and this closed form overestimates the moment as a result.

The material is elastic and straight. The numbers are elastic critical values. A real beam’s strength follows a buckling curve below them, and the codes’ curves for monosymmetric beams are based on fewer tests than those for rolled sections. At the low end, the true capacity may be much less than the elastic value, in the way the formula that reads four numbers sets out.

End restraint is simple. A warping-restrained end adds to the bracket through $I_w$, and that term depends on the flange division in the opposite way from $\beta_x$. Thus a heavily unequal section gets little from warping restraint.

Still open: the sign that changes along the span #

A cantilever’s root and a span’s middle give two different problems when they share one beam, and the two regions don’t buckle separately. A span with a hogging end region has a compressed bottom flange near the support and a compressed top flange in the middle. A symmetric section is indifferent to this. For a monosymmetric one, the Wagner torque changes sign along the length and works for the beam in one region and against it in the other. Whether the favourable region can rescue the unfavourable one through the continuity of the twist, and whether a code factor that depends only on the moment diagram captures that, needs an energy solution in which $\beta_x$ is integrated against a moment that changes sign, and the uniform moment used here has no such thing to integrate.