Part 6 - U-frames

Something from the railways

There must be someone out there reading these posts because I had a request for a subject to cover. So, thanks to Jeyanth, I am now adding a bit on U-frames. This will lead on to some bits about bearing stiffeners. It has taken a long time because it is very complicated.

Unfortunately, these are some of the most complex bits in steelwork and I have marked the particularly complex bits “Advanced” in the header so that you can ignore them initially and then come back to them later once you have glued your exploding heads back together.

U- frames

Typically, a metal bridge will either have continuous restraint of the top flange (provided by the slab in a steel/concrete composite bridge) or cross bracing at intervals (the over-the-top part of a through truss bridge) to control the slenderness of the compression part. With a continuous restraint, the effective length i.e. the equivalent buckling length, is zero, and with intermittent cross bracing the effective length will typically equal the spacing (although you do have to worry about the entire bridge behaving as one wider structure with no bracing).

The basic alternative is for the top flange to be unbraced, with an effective length equal to the length between supports. But unless you have a very short beam then your maximum stress will be so limited that the beam will not be at all efficient. Railway engineers can rarely use cross bracing or a top slab (see Railway oddities below), so U-frames were developed.

Figure 43 - U-frame on simple deck

Aside: Engineers are not subtle (I may have mentioned this before). Since we are used to “I” girders being so named because they look like capital I’s (this obviously looks better with the old Courier font i.e. I) it should not be too much of a stretch that U-frames are shaped like a square capital “U”. Strictly speaking, we probably ought to refer to three girder U-frame bridges as “W” frames. We do get more direct when talking about one-sided U-frames, which are referred to as “L” frames – unfortunately it is hard to find a font where the vertical is shorter than the horizontal is wide.

U-frame bracing is sometimes referred to as “Partially effective” bracing because it reduces the effect of slenderness by providing a series of springs, rather than providing a series of hard restraints. If the springs are like small rubber bands, then they will have no effect, whereas if the springs are massively stiff then they will eventually approach the behaviour of rigid bracing: anywhere in between they will give partial restraint.

On an unbraced compression flange, it is the transverse stiffness of the flange itself that controls slenderness. Initial imperfections (out of straightness, usually curvature), combined with the axial force, create a transverse moment (from the eccentric forces), which may then cause further curvature, which then etc. etc. until either the system becomes stable or it buckles.

U-frames springs add to the transverse stiffness, reducing the further eccentricities and effective length. The two effects are different. The stiffness of the top flange causes bending moments due to curvature, which is a local effect of one part of the flange relative to the neighbouring sections. The springs, however, provide restraint relative to a local datum and try to fix the flange still “In space”.

Railway oddities

Unless you happen to work in the railway field, there is a good chance that you will not have any idea what a U-frame is. I worked on a wide variety of bridges for 20 years before I moved over to railways and never came across one.

Railways work to very different rules to highways. Unless you are looking at the French Train à Grande Vitesse (TGV – high speed train) where the maximum longitudinal track gradient is 3.5% (powerful engines, and trains are reputed to use momentum to go up hills) then a 1% gradient is quite steep. This should be compared to a highway where 5% is not unusual and the UK record is probably Hard Knott Pass at around 30%. Railway lines cannot therefore go up and down very fast to get over an obstacle and railway engineers therefore had to be inventive to minimise the amount of rise and fall. If you have to take one highway over another highway, then the first highway has to be (say) 5.5m plus the thickness of the bridge construction. Having a 2m deep highway bridge is fine because going up 7.5m will take 150m using highway gradients but almost a kilometre using railway gradients. But if you can cut that down to 0.5m construction depth then you can save several hundred metres of approach gradient.

Because of this, a typical highway bridge might be steel girders supporting a composite concrete slab, which then supports the surfacing and finally the highway. The clearance depth is the overall depth.

A railway is more likely to use big beams either side of the track, which can be as deep as you like to span the distance, with shallow cross girders spanning transversely between the main girders. This reduces the clearance depth to just the thickness of the cross girders plus ballast etc.

This is all well and good if you only have a very short span to cross, but typically a steel beam will require some sort of bracing on the top flange. If your bridge is deep enough then the train can pass under the high level cross bracing, but that only works with a long truss. It would be very silly to design a bridge with cross bracing two metres above the track because the 4m high train would find it very difficult to get through, and a 6m high beam on a 10m span would be ridiculously inefficient. Railway engineers therefore came up with the idea of U-frames.

U-frame formulae

BS 5400 bases the stiffness on the following formula:

Figure 44 - definition of terms in equation

The equation has three components which are generally easy to take apart – “Generally” because one component is not obvious.

The formula calculates the transverse stiffness of the top of the frame i.e.  deflection of the tip under a horizontal unit load: this is 1/spring stiffness.

Figure 45 - contribution from cantilever

The first element is simply the deflection at the tip of a cantilever under a unit load. I₁ is the second moment of area of the vertical part of the frame. The depth is measured to the top of the cross girder since the vertical is assumed to be supported by the cross girder top flange, even if the connection is actually via the rivets/bolts at a slightly lower level.

Figure 46 - contribution from corner flexibility

The third element calculates the moment as the unit force · the lever arm d₂, which then results in a rotation in the corner due to that moment, and that rotation · d₂ gives the resulting deflection. This explains the square of the distance.

Figure 47 - contribution from cross girder deflection (corrected - Sept 2019)

The second element is the one that causes a little confusion. Similar to the third term, the moment in the corner (shown above for ONE FORCE at the top of ONE SIDE) is calculated as the product of the unit force x the lever arm d₂. This moment is then applied to the cross girder, of length B and second moment of area I₂, to calculate the rotation in the corner. But the formula for rotation at the end of the simply supported beam is θ = M.L / 3EI. This rotation is then transformed into another deflection by multiplying by d₂ again. But this would result in a component of 0.33.B.d₂²/EI₂, not 0.5.

The confusion is caused because both of the top flanges can have inwards imperfections, and hence deflect inwards (or both outwards although this is usually less critical since vertical load on the cross girder causes tip movement inwards - but initial imperfections can go either way): this results in TWO restraint forces i.e. ONE FORCE on EACH tip. When the moment is applied to the remote end of the girder, there is an extra rotation at the near end equal to half of the primary rotation. Hence the coefficient ”u” changes from 0.33 to 0.33 + 0.167 = 0.5 for external girders.

The internal girder calculation follows similar principles, although it is a bit more complex. Applying the moment to the middle upright will divide the moment both ways hence 1/6. Applying the two moments to the two ends cause two rotation components of 0.083 (1/12) on the middle girder. Hence the rotation of the middle girder is 1/6 + 2 x 1/12 = 1/3 resulting in the coefficient of 0.333 for internal girders. OH WELL - WE ALL MAKE THINGS TOO COMPLEX - corrrecting after five years, that should have read - Applying the moment to one end would cause less than the standard 1/6 rotation at the middle, due to continuity, but applying the opposite moment at the far end causes exactly the same rotation so there is no conflict and, effectively, no continuity. The total is therefore 1/6 + 1/6 = 1/3.

One other important subtlety

The equation gives the highest flexibility (which gives the lowest restraint/capacity for the compression flange/chord) when there are equal and opposite forces on the two tips of the U-frame: in this situation, the two forces balance out. But all of elements 1 and 3, and 2/3 of element 2 only require one force on one tip. Obviously the relative proportions of one force versus two forces depends on the configuration of the U-frame, but it is reasonable to expect that a single force (i.e. restraining one compression element only) will be a significant proportion of the maximum load.

Fundamental: The horizontal forces not only need to be resisted by the U-frame itself, but the U-frames need to be restrained against global movement. The normal requirement for U-frames to be effective is that the bottoms of the U-frames have to be fixed in space: this is usually achieved by either plan bracing (forming a truss on plan with the beams as “Top and bottom” chords) or with a deck plate (which forms the web of a girder with the beams as the top and bottom flanges). Being engineers, we often cheat a bit and assume that transverse troughing will act as a deck plate: this is reasonable for trapezoidal troughing but may be suspect for rectangular troughing (although normally accepted because ….).

If there is no form of plan bracing then we need to add the transverse flexibility of the main girders to the flexibility of the U-frame. Unfortunately, most beams (or trusses) are an order or two stiffer vertically than transversely and if there is no plan bracing then we assume that the combination will make the U-frame ineffective. Please note that bracing that merely runs perpendicular between the two chords is not counted: this is because both flanges could be initially deformed in the same direction so that they both move together without support.

Aside: Network Rail recently commissioned an investigation into the behaviour of U-frames, mainly on footbridges, but also, to a limited extent, on rail bridges. The work was led by someone for whom I have a great deal of respect, but I have problems with at least one of the results. Unfortunately, I did not get a chance to discuss this at length when he presented the results, and I will put my reservations here.

The analysis used geometrically (I totally agree) and materially (I am dubious) non-linear analysis to investigate the behaviour of footbridges, with initial imperfections. One of the major findings was that it made no difference whether or not there was plan bracing, all that was required was for the two bottom chords to be tied together. The suggested reason for this was that local loading on the cross girders would cause both girders to rotate inwards, over riding the sense of the initial imperfections. The two restraint forces would then act in opposing directions and balance out, removing the global net force on plan. I have two major reservations with this, one for footbridges and one for rail bridges.

Footbridges floors may span longitudinally onto the cross girders (which will cause the relevant bending) but floors that span transversely onto the bottom chords will not. I do not see how we can guarantee that the effect will occur in all cases. Also, we cannot guarantee that the load induced deflection will outweigh the effects of the initial imperfection and the forces may not then balance.

In rail bridges, with a single deck and two edge girders, the loading will often come from at least two tracks. Loading one track will usually apply around ¾ of the load to one girder and only ¼ to the other. This will give a large imbalance to the two sets of restraint forces, preventing the forces from balancing. To some extent this could also happen on the footbridge case and I am always concerned about relying on the two forces balancing so precisely. If the flexibility of the plan bracing is ten times higher than the U-frame flexibility, then you only need an imbalance of 10% for the U-frame flexibility to double.

Lack of plan bracing (advanced)

As I mentioned above, the normal assumption, where there is a lack of plan bracing, is that U-frame behaviour is not effective and the top flange/chord (generally referred to here as top flange because I get tired of referring to both every time) is assumed to act without any restraint. There are, however, a couple of alternative methods that can be attempted, even if there is little likelihood of success.

The first method is to model the structure as a frame on plan (or possibly as a full three-dimensional model) and investigate the global effect of horizontal loads. Each U-frame is assumed to have a pair of loads at the tops, with unit loads at midspan, dropping off parabolically towards the supports. The deflection at midspan is added to the U-frame deflection. The forces on the U-frames should be applied in the same direction, which would allow the coefficient of the second local U-frame deflection to be reduced from 0.5 to 0.167 since the transverse bending parts would counteract each other to some extent.

The second method can use the rotational stiffness of the beams/cross girders to reduce the slenderness. The following is based on intuition, rather than being able to cite evidence. You will find that I say this a lot.

Aside: you can either look at this as a damning indictment on my lack of knowledge or take solace in the fact that you are not as ignorant as you might have originally thought. I regularly ask other experienced engineers for the background behind large swathes of the code and I have rarely been given any good answers.

Let us imagine that there is some magical device that managed to keep the top and bottom chords moving exactly together. The compression force in the top flange will almost exactly equal the tension force in the bottom flange. If the top flange has an imperfection then there is a series of forces that push outwards. But if the bottom flange is following the same profile (because of the magical device) then there will be a series of equal inward forces due to the tension forces in the identically shaped bottom chord. In the real world, the magical device does not exist and the torsional stiffness provides by the U-frame will provide something between zero and infinite stiffness. I presume that the formulae in the code calculate the effect of this partial balancing of forces. If anyone can explain the background, I would be very interested to know.

Aside: be very careful, however, not to make assumptions that are too generous. I recently applied the torsional stiffness calculation to a series of cross girders that had no deck plate to keep the top flange stable. I used the bending stiffness of the four lines of rail bearers (short members, one under each rail, spanning between the cross girders) to provide torsional restraint of the cross girder. The stiffness of the cross girders, assuming that they had moment connections to the cross girders, reduced the effective length to 1.25 times the rail bearer spacing, which would have been excellent. BUT (and you knew it was coming) the connections between the ends of the rail bearers and the cross girder consisted of pairs of angle cleats. When I modelled this connection using solid modelling (with a fine mesh of elements ½” cubed) the flexibility of the angles, which acted as plates in torsion, was an order of magnitude higher than that of the beams, so the torsional restraint from the rail bearers was significantly reduced.

Back to the standard U-frames

The behaviour of the top chord is analysed using the principles of a “Beam on an Elastic Foundation”. These principles are normally used for vertical loads applied to a relatively flexible sitting on the ground. Although, again, I cannot describe exactly how these principles are converted into the formulae in the codes (cleverer minds than mine …) the basic behaviour can help you to understand the trends in your calculations.

Unfortunately, according to Dr. J. Tubman (I am quoting from Correspondence on “U-frame restraint against instability of steel beams in bridges – Mr E. Jeffers” published in The Structural Engineer/Volume 69/No. 23/3 December 1991) “… the deliberations of the original (BS5400 Part 3 - 1982) drafting committee were not recorded or archived to any great extent.” A lot of my commentary is therefore based on the “Commentary on BS 5400-3: 2000 Code of practice for the design of steel bridges” published by The Steel Construction Institute in 1991 and revised in 2000. This was edited by C W Brown MA CEng FICE and D C Iles MSc ACGI DIC CEng MICE. This has proved to be my best source of information and I can thoroughly recommend it even though some of it goes over my head.

Some of my explanations will be simplifications of the original text and other bits will be my own arithmetic based on the principles that they described. Please note that I will usually refer to struts rather than beams since the top flange of a beam acts very similarly to a strut in compression and the explanations are a lot easier. If you want more information, then you should probably be researching your PhD and not reading this J.

Calculating effective lengths based on U-frame restraint

I am afraid that this is going to be very short and it will basically tell you to follow your code of choice.

1                    calculate the stiffness of your U-frames and bearing stiffeners against unit transverse loads at the tips

2                    calculate the properties of your compression flange (please note that Network Rail include the top third of the web and above)

3                    enter these, and a few other easy to find constants, into the equations to calculate your effective length

4                    if your code requires it, divide the actual length by the effective length and then round down to the integer below to get the number of half wavelengths of buckling (???? Please see rants later if your mind has not exploded so far)

5                    carry on like any other beam but accounting for the ratio of l_e/l_w if your code requires it.

Principles of U-frame restraint calculation (fairly advanced)

If you want a bit more (OK one heck of a lot more) background using BS5400 as an example…

Figure 48 - buckling of strut on an elastic foundation

Beam on an elastic foundation theory predicts that a strut that is continuously supported by a foundation, with a stiffness of k per unit length along the strut, will buckle at a load of :

P_cr = 2(kEI_C)^0.5

Fundamental: An awful lot of the code is based on the assumption that two elements that buckle at the same load will behave in the same manner at lower levels of loading. Although I can see the rationale behind this, I have some grave reservations about this when it is stretched to cover more and more complex situations: unfortunately, this is the fundamental (but hidden) basis of BS 5400 and is explicitly the basis for the Eurocodes. As an example, I would point out that the strut above has a constant force along its entire length, which is very different to the forces in the top flange of a simply supported beam, where the force varies parabolically under a uniform load, and more dramatically under discrete loads.

Standard Euler buckling, of an elastic strut under axial load, predicts a critical load:

P_cr = π²EI_C/l_e²

Setting these two as equal gives:

2(kEI_C)^0.5 = π²EI_C/l_e²

Which can be rearranged to give:

l_e = π/√2.(EI_C/k)^0.25

Figure 49 - equivalence with discrete springs

Of course, U-frames do not provide a perfectly uniform continuous elastic support and, remembering that δ_R is a flexibility i.e. the reciprocal of a stiffness, the discrete U-frames can be smeared so that k = 1/(δ_R.l_R) this can be rearranged to give:

l_e = (π/√2).(E.I_C.δ_R.l_R)^0.25

where π/√2 = 2.22.

This is represented in BS5400 9.6.4.1.1.2 by equating l₁ = (E.I_C.δ_R.l_R)^0.25 .

The effective length l_e is then calculated with a couple of coefficients, k₂ and k₃, which are typically 1.00 and another coefficient  k₅, which accounts for end stiffness, and which can vary between 2.22 for infinitely stiff (the assumption in the derivation - surprise !) and 3.60 for completely floppy bearing stiffness.

Rant: You may be able to see some of my reservations by now. The design of beams, in particular, is being controlled by equivalence to an elastic buckling analysis under a uniform axial load. In practice, design should really be controlled by the additional stresses that build up on the way to buckling and a comparison to the yield stress. Do not get me wrong, unless you are going to do a geometrically non-linear P.δ analysis, you are going to have to apply these rules, which then link back into a series of curves based on a best fit line to represent a series of tests to failure in a lab. I am concerned that the entire method is a structure that is sitting on some very shaky foundations, even though there do not appear to be failures when using the methods. The methods do, however, get a lot more convoluted later on to account for some of the known problems.

The obvious question is how the bearing stiffener can affect the effective length. To investigate this, I created a series of models of an imperfect strut. To make the point, the plan imperfection is exaggerated at length/100 (ten times the normal assumption). Six variations are shown below with the deflected shape shown in red. Up, down, left and right, below, refer to the orientation of the diagram.

Figure 50 - transverse deflections of struts with different bearing stiffener spring stiffnesses

In every case, the right-hand end is free to move left to right, and the same horizontal force is applied at that end. The left-hand ends are fixed and give reactions that are equal and opposite. All of the internal springs are the same and all of the members have the same section. The only variation between each strut is that the bottom strut has a totally rigid supports up and down, the second strut supports have 100 times the standard internal springs, dropping down by a factor of 10 with each successive strut with the top strut having no vertical fixity at the ends. The fourth strut has the same end stiffness as the internal U-frames, so you could argue that it has no bearing stiffener, and the last two even more so.

As you can see, all of the struts deflect upwards in the middle, as would be expected for an upwards imperfection. In the top example, the tips of all of the internal U-frames are deflecting upwards. These U-frames are resisting the upwards movement and apply a series of forces downwards on the strut. Obviously, for all of the forces to balance, since there is no external force being applied, there must be an overall equal and opposite reaction to balance these and that is provided by the reaction of the rigid bearing stiffeners, which will push upwards on the strut.

We will never have an infinitely rigid bearing stiffener, but the second strut, with bearing stiffeners 100 times stiffer than the U-frames, comes close. The deflection of the support is very small and the deflected shape of the strut is practically identical to that with a rigid bearing stiffener. The deflections in the diagram below are the same as those above, but they have been superimposed.

Figure 51 - previous deflections superimposed on one plot

There are, however, subtle differences between the first two cases. The midspan deflection is almost identical, but the small downwards deflection at the support gives a slightly larger relative deflection between the ends and the middle, giving a slightly larger curvature and hence moment.

As we move onto the third and fourth struts, the displacements at midspan, relative to the ends, increase, again increasing curvature and moment. However, as we then continue onto the fifth and sixth struts, where the bearing stiffener is less stiff than a normal U-frame, or even zero, we tend to a maximum value of relative displacement and an amplified moment I should point out that these analyses are not related to any particular load but it should give an indication of how the end restraint, or lack of it, can affect the behaviour of the strut and hence the moments along it.

The last two cases are particularly spurious since it is unlikely that the bearing stiffeners will be less stiff than the internal U-frame stiffeners but they do indicate that the behaviour is not sensitive to the stiffness of the bearing stiffeners, once you go below reasonable limits.

Confused Author: although rather wordy (it is not called the chatty guide for nothing) nothing in the above is actually very complicated and no unusual effects came out. It even goes to show why the code can take into account some of the U-frame stiffeners to beef up the bearing stiffeners. But now we get on to the bit that I cannot get my head around, and would love it if someone more knowledgeable could comment, and that is how the strength of the stiffeners comes into the calculation. In design codes, the bearing stiffener has to be strong enough, but in the Network Rail assessment code, clause 9.6.1A modifies the effective length using an equation that I cannot even begin to back analyse. This equation, which can increase the effective length by up to 63% in extreme cases, combines the basic effective length (based on stiffnesses) with a multiplier based on the ratio of load/strength. Taken very simplistically, the bearing stiffener will act normally up until its yield strength, and then become a weaker spring as it starts to go plastic. But once it has become fully plastic, it will not have any stiffness and its deflection will shoot up, swamping any stiffening it had up until that point. As far as I can see, the old method, where the effective length was increased by 25% if the bearing stiffener failed (regardless of extent) seems more reasonable although the 25% does appear a little arbitrary. Comments anybody ?

Half wavelength of buckling (for advanced masochists only)

I really wish that someone had written down the reasoning when they wrote BS5400 since it is almost impossible to tell the difference between someone who is writing their opinion of what the code is based on and someone who actually knows why things were decided (except for me who clearly is playing fast and loose with evidence). One of the biggest sources of my confusion is the idea of half wavelengths with U-frames, where the actual length is divided by the effective length and the number of half wavelengths taken as the integer below that ratio.

I have given a direct example where multiple half wavelengths of buckling might occur in my section on “Compact versus non compact and various other states”: this was where opposite face of a square hollow section buckle in and out, like a sine wave (with adjacent faces going the other way) because the face is very stiff transversely. If you have not seen this then please look since it took me a long time to do the sketch. I can therefore completely understand how very stiff U-frames might cause a strut (or top flange of a beam) to buckle into multiple half wavelengths. But that is the situation where you have so much transverse stiffness that the longitudinal stiffness of the “Strut”, in this case the plate that forms one side of the section, is so much lower than the transverse stiffness, that you get a dramatic shortening in the effective length. In my experience of bridge assessment, you are doing well if your effective length is half of the overall span, and one third is almost unheard of. We are therefore working in the grey area where U-frames help but do not act as anything close to a rigid brace.

I was always troubled by this idea of half wavelengths, especially since there is a step change in behaviour between l_e = 0.499L and l_e = 0.501L. In the first case, l_w = 0.500L and l_e/l_w = 2. In the second case, l_w = 1.000L and l_e/l_w = 1. This gave rise to some very interesting discussions between assessor and checker since the difference in strength on Figure 11 could suddenly jump by 25% just because you squeezed the calculation just past the limit – I always wondered if the bridge beam knew about the calculations and how much the beams cared anyway (in case you are wondering, the censored answers are “Nothing” and “Not at all”). I was therefore heartened to read Section 9.8 of the SCI commentary on BS5400:3 (go and buy a copy) which states “If the simply supported beam is partially restrained … the effective length may reduce slightly but the beam will still buckle over the length between supports. Since the effective length depends on the full span, this needs to be recognised when determining the limiting resistance based on the effective length. This is achieved by increasing the effective imperfection in the ratio of l_w/l_e.”.

I can understand the need for some correction since (and this is one of the places where the tyranny of effective length concerns me) all of the effective length calculation is about the equivalence of an almost perfect Euler strut to an almost perfect strut on an elastic foundation. But we then start comparing the behaviour of a beam with effective length l_e to a shorter beam which is physically “l_e” long and then plug this all into a set of curves that are based on real beam tests. To me, the important words in the commentary are “…but the beam will still buckle over the length between supports.” The imperfections should, obviously, be linked to the full span and be larger than that for a physically shorter strut, but why should there be any sudden step as the effective length reaches half the span since the beam will not suddenly snap through just because some arbitrary ratio has been reached. We should also remember that our structures are a lot stockier than most people imagine, and it is the combination of axial load plus induced bending that matters, which usually happens at relatively low load compared to elastic buckling, and we are therefore unlikely to reach loads that will cause sufficient eccentricity to make snap through likely (just a personal view and not based on anything rigorous).

Unfortunately, we are required to apply the code as is, because we have no basis to decide if it is wrong. It would, however, seem more sensible to me to have curves that correct for l_e/L but these do not exist. If anybody wants a PhD subject then investigating this using normal English would be a useful research subject, although it might be difficult to write it in a way that will pass the review and still be intelligible to someone without that special translation chip inserted in their brain J. As an example, I found a paper that refers to a strut resting on a Winkler foundation, and it took a ten-minute search to find that this referred to a series of elastic springs: the first requires one to learn a new language to join the club whilst the second means the same but is easier to understand.

INTERESTINGLY, the Highways Agency assessment code for steel bridges, BD 56/10, has completely removed the references to the half wavelength in Figure 11. I feel that this is great, since it removes the sudden jump mentioned above, but it ignores the larger imperfections that occur in a 20m beam (with an effective length of 10m) compared to a real 10m beam. Hmmmmh.

Forces on U-frame stiffeners

The first thing to remember about U-frame stiffeners is that they are also going to be web stiffeners.

Forces on transverse web stiffeners

When the Eurocodes were launched, the ICE held a two-day conference to discuss the background and implications. One of the major thrusts that there should not be any great difference between the results of designs to BS 5400:3 and to the Eurocodes, and there was some discussion about trying to adopt Part 3 as Non Contradictory Complementary Information so that we could just ignore the Eurocodes: one big fly in the ointment was forces on web stiffeners. The following comments follow on from me reading “Transverse web stiffeners and shear moment interaction for steel plate girder bridges” by Chris Hendy and Francesco Presta (the Structural Engineer, November 2008).

The Eurocodes follows Höglund’s rotated stress field theory and require the stiffeners to be stiff, but not strong. BS 5400 follows Rockey’s tension field theory, where the stiffeners act as the verticals in a truss, reacting against the tension forces that happen across the diagonals of the webs. The Eurocode was bastardised (the only thing worse than a committee is a committee made up of representatives from lots of other committees, especially when national pride is at stake L) so that the higher BS 5400 forces were included leading to some inconsistencies. Because of this, I will not go into too much detail about how the forces are derived, since even the great and the good seem to have a problem with them. Since I work mainly in assessing old bridges, using codes based on BS 5400:3, I will unapologetically refer to the British clauses and leave it up to you to compare to the Eurocodes et al.

Fundamental: one problem that I regularly see when looking at other people’s assessments is that the loads on transverse web stiffeners, which are “… to be considered” according to the title of clause 9.13.3.1 are just applied regardless. Some of these forces will compress the stiffener, some will stretch the stiffener, and some can go either way. Merely adding the effects together, when they will often work in opposite direction, relieving each other, is ultra conservative. Including tensile forces and checking them as a buckling effect is ultra censored. I will give some pointers about how I apply them, but it is up to you to consider whether the stiffener is acting as a stiffener or just forming part of the connection between the cross girder and the main beam.

9.13.3.2 Axial force due to tension field action

This one is the bone of contention above, but unless you have a Departure from Standard agreed with your Technical Approval Authority, then you will need to apply it in the UK.

Consider a beam that is being sheared, whilst ignoring any bending. The primary effect is that the rectangular web panel lozenges, as shown below.

Figure 52 - tension field action (in British Standards)

The red diagonal gets shorter, so the panel is compressed in that direction. The black diagonal gets longer and is therefore stretched. Web plate panels are generally relatively thin and having a diagonal with length/thickness that might be 100 is going to lead to a plate that is prone to buckling. Once it buckles, which it will do with a series of ripples, it will not be able to take any more compressive load. But the tension diagonal, which will generally be running (sort of) parallel to the buckles, can take load. This tension tie acts like the tension member in a truss and gives a vertical force in the vertical, i.e. the web stiffener. This force will always be compressive (if it actually exists as noted above) and I always include it if it occurs. Please note that this is a non-linear calculation since there is no force if the web can take all of the load in basic shear, but an increasing load beyond that point.

9.13.3.3 Axial force representing the destabilising influence of the web

According to the commentary, this is a totally imaginary axial force that allows a bit of extra axial capacity instead of applying the moments caused by the of plane deformations of the imperfectly flat webs. Just apply it and do not worry about where it comes from because I have no idea.

9.13.3.1 c) axial force due to transfer of load through a cross frame or beam

This one should be applied only if you are very sure. Personally, I almost never apply it to a stiffener. In my field, these cross girders will only be attached to the stiffener if it is part of a U-frame (oh yes – them). In this situation, the cross girder is connected onto the bottom part of the stiffener, or the stiffener might even terminate on the top of the cross girder. The force on the end of the cross girder will always be downwards (unless someone has managed to switch gravity) which will, at most, put the stiffener into tension, opposing the tension field force and certainly not contributing to buckling. But the major effect of any end connection is to spread load into the web of the main girder so even if the girder is indirectly connected, the load will quickly spread out into the web, and the load will not go into the web stiffener.

9.13.3.1 d) axial force due to load applied at flange level

This one should be applied, but it is only really relevant where there is a deck slab (with the load transferred as a compression through the top flange) and you should try to spread the load out as far as possible before it reaches the stiffener. Again it has no reason to go into the stiffener but will primarily spread into the web of the main girder.

9.13.3.4 axial force due to curvature of flange

This one is a bit rare since most beams have parallel flanges. When the top flange is set on a curve (when viewed from the side) the compression force acts like an arch and tends to push upwards, pulling away from the stiffener and putting it into tension: but again, the stiffener is just a hard point that will divert a little bit of the force from the web, before distributing the load out into the web again as it stretches. If the bottom flange is curved downwards (a fish belly beam) then that will have the opposite effect and will put the web/stiffener into a bit of compression.

9.13.3.1 f) axial force due to change of slope of flange

This is similar to the effect above, but results in a localised single force that is calculated by resolving forces parallel and perpendicular to the flange on one side of the kink.

9.13.3.1 g) bending moment about the centroidal axis …

Not exactly sure where this one would come from or how we would apply the first bit on a symmetric stiffener. It says to take into account the eccentricity of the previous loads to the centroid of the stiffener, but when the cross girder spans, it spans between the main girder shear centres and hence has no eccentricity, and most of the other effects come from the web which is also concentric. I suppose that you could do something if there were only a stiffener on one side but I do not come across that very often.

It does also slip in “…or from flexure of a cross beam, U-frame or deck” at the end, which is a bit naughty.

If there is a moment connection at the end of the cross girder, which would be required for a U-frame connection, then you can get some resistance to rotation due to the torsional stiffness of the main girder. This is, however, usually quite small and it is normal to ignore the effect and assume that the cross girder is simply supported applying no moment to the stiffener.

And then we get on to the U-frame forces …

Forces on U-frame stiffeners (finally)

There are a number of these and, despite many days trying, I have not managed to justify the forces or moments specified in the codes.

9.12.2 Elements providing discrete intermediate restraints

The fundamental idea of U-frames is that they provide some restraint to the top flange of a beam by acting as springs. But when these springs act to restrain the flange, there are a series of equal and opposite reactions from the flange onto the stiffeners.

The force F_R is an odd little beast, which the commentary says is conservative because the derivation depends on a uniform force along the entire length of the top chord, equal to the midspan force. It may appear that the force is proportional to this force at midspan, but if you look at the first term σ_fc/(σ_ci - σ_fc) then you can see that this is one stress divided by another, so there is no force, even if there were a term for area included (which there is not). This first term is a multiplier which is zero if there is no load and which increases to infinity when the load approaches the buckling load of the effective beam length. When approaching buckling, the transverse deflection would become massive leading to the infinite load.

The second term has two components. The l_w/667 is the imperfection of the beam l/1000 multiplied by the extra 1.5 factor that BS 5400 builds in. The slight confusion here is that, as mentioned before, the imperfection is based on the full length not the fictional reduced wavelength used here. This initial imperfection is divided by the flexibility of the U-frame spring giving the force that would be induced if the spring were forced over by the initial imperfection amount. Hence a notional value is calculated which is then beefed up as the load increases towards buckling.

Fundamental: this calculation is NON-LINEAR. Most calculations in structural engineering are linear, meaning that the graph of result versus input is a straight line. In practice this means that you can work out the answer for one load case, and then for another, and if you add the answers together then you get the same answer as if you analysed for both loads combined into one. The important, and slightly tricky, term here is σ_fc/(σ_ci - σ_fc).

Imagine that we have σ_ci = 4.0, which is the stress at which the beam will buckle. We will then take σ_fc = 1.0 for dead load and 2.0 for live load.

If we calculate these separately then the terms become

1.0/(4.0 – 1.0) = 0.333 and 2.0/(4.0 – 2.0) = 1.000 giving a total of 1.333.

But if we combine the two stresses to give a total stress of 3.0 then we get

3.0/(4.0 – 3.0) = 3.000, which is significantly higher than merely adding the two results together.

Hence when working out an answer for a non-linear effect like this then you need to apply everything as one load, including all individual loads together with their partial factors. This is one reason why partial factors have to be applied to the correct part of the analysis, load or resistance, rather than just lumping them all together.

9.12.3.3 U-frames with cross members subjected to vertical loading

Simple analyses normally apply loads onto a structure as a UDL, where the intensity depends on length. This is not a real load, just a simplification that will give reasonable results for global analysis. But real loads are lumpier, especially when you have discrete cross girders and highly concentrated loads such as axle loads.

For this load effect we will ignore all of the global effects and consider two situations a) the leading edge of a train where two cross girders are heavily loaded, but subsequent girders have no live loading, and b) the gap between two axles, where the first and third cross girders are loaded but the second is not. In both cases, the only loads considered are live loads since dead loads are likely to be the same causing all beams to deflect the same amount.

Looking at a) local loading on the first two cross girders will cause the cross-girders to bend, resulting in rotations at the ends. If there were no top flange to the main girders, then the tops of the U-frames would be free to deflect inwards. The third cross girder, being unloaded, will have no end rotations and the tips of the U-frames will not move. In reality, however, we have the top flange of the main girder, which will try to stay in a straight line. The force F_C occurs because the various tips end up working against each other, but the amount of force depends on how easily the mismatch can be accommodated. If the top flange were floppy, then the U-frames would merely bend it and the tips would approach their desired movement. Likewise, if the U-frames were flexible then they would not be able to bend the top flange transversely, but then they would not be much good as U-frames so this is a little moot.

Case b) is similar, but it is the first and third tips that move and affect the second stationary tip. Please note that both cases benefit from spread of load along the deck, such as the 25:50:25 split of load between sleepers allowed for ballasted track on the railways. This tends to equalise the loads, minimising the differential movement.

The top term calculates the amount that the second tips are trying to move relative to the ones either side. Obviously, the more this happens, the greater the problem we have. This deflection is divided by the sum of terms related to the stiffness of the U-frame itself and a term relating to the flexibility of the top flange. The frame does not care whether the top flange flexes or if it flexes itself, it is only concerned about how easily the mismatch of movements can be dissipated.

Combining F_R and F_c

This is a little arbitrary and you should discuss this with your Technical Approval Authority. In my case, the U-frame restraint force F_R is highest near midspan, where the load is highest. This high moment depends on load along the entire span and it is not reasonable to combine this with the local loading from the leading edge of the train of loads, case a), since that load configuration would not cause much global loading. The gap load in case b) would, however, load the full span so is compatible with the F_R force. My approach is therefore to take the higher of F_C  and F_R for case a) and the sum of F_R PLUS F_C for case b) since merely combining the highest results would be unrealistic.


Well that is all for now folks. Another post, on bearing stiffeners, will appear very shortly but that is also very complicated and I thought that you needed a rest.