Part 4 - why dinosaurs still rule the earth (at least in engineering)

Old fashioned techniques

Time for a little light relief….

One of the biggest problems that I had when I was at university, second only to the occasional Tyrannosaurus Rex wandering past the window, was knowing what was important and what was just included because it filled the lecturers’ spare time.

If you are at university and trying to get a degree, then concentrate on what your lecturers and supervisors tell you, because it will be hard to get a decent job if you do not get a decent degree. One way to look at it is that even if what they are teaching you is totally irrelevant, it is still giving your mind a good workout.

If you are interested in being a practical engineer, however, then most of  what you were taught is totally irrelevant, what is more important is understanding what is going on with whatever you are analysing. It is asking the right questions, and being able to understand what the answer will be before the computer spits it out, that tells you how you are getting on.

There are a lot of ideas in university courses that may have been used early last century but which are now not used with the widespread availability of structural analysis software. That does not, however, make them useless, it just means that they can be useful in other more indirect ways.

Influence lines

I always got confused by influence lines at University (another one of my recurring themes ...). In my defence, this was possibly because the influence line for bending on a simply supported beam is very similar to the bending moment diagrams that result from applying point loads on the same beam. It is also drawn in the opposite direction which might have been a help or a hindrance to my poor overloaded mind. As usual, let us go back to the absolute basic definition.

An influence line for an effect is a plot of how much of a particular effect there is, at a point under consideration on a structure, from applying a unit point load anywhere on that structure.

…. but what does this actually mean ? The clue is in the term “Influence”. The plot shows how much influence a unit load (or action for those whippersnappers who like Eurocodes) anywhere on the structure has on an effect (such as bending moment, shear or reaction) at a particular point on the structure.

Fundamental: I cannot stress this enough but an influence line is a plot of how much effect a series of unit loads applied individually across the entire structure has at one point only.

The effect of applying a point load of x kN to a part of the structure that has a value of y on the influence line for bending at a particular location is to create a moment of x,y kNm at that location. Similarly, applying a UDL over a specific length would result in a bending moment equal to the relevant area under the graph multiplied by UDL. These are demonstrated in Figure 25 below.

Figure 25 - general use of influence lines p.s. sorry about the sign conventions but influence lines are usually drawn up for some reason – go figure

Aside: although it is not a standard, by any means, I usually teach my grads to differentiate between bending moment diagrams and influence line diagrams by the addition of a simple vertical line at the point under consideration, with a cryptic notation like IL BM.

The family of influence lines for bending moment on a simply supported beam are a bit boring, if you are working with generally uniform loads, as can be seen from Figure 26 below. Adding a load anywhere on the span will increase the bending moment at all points on the span.

Figure 26 – family of  bending moment influence lines for a simply supported span

 But if you have a loading pattern that is not uniform, then the form of the bending moment influence line is not quite so boring and can make it a bit easier to apply loading in a sensible manner, without having to think too hard. In my current field, of railway bridge assessment, we have to apply a series of axle loads to a bridge, with two different axle loads and lots of different spacings between the 16 axles (plus a uniform bit at the end). Choosing exactly where to apply the load can be a bit of a trial, but you can rationalise your thoughts by considering how the moving load works with an influence line.

Figure 27 - influence line for bending with train of axles

 Figure 27 shows the railway loading axles (you actually get four groups of four axles but only worry about the ones on the bridge) applied to the bending moment influence line at two locations. The top is at mid span and the second is at a quarter span. If you were to analyse this in a computer programme (or in a spreadsheet because frame analysis is over the top for simple structures) and to run the train across the bridge, then you would find that these are the two critical points for this particular span and these proportions. Why ?

Duh, (says you) because the programme told me so – obvious innit ?

Aside: the experienced engineer is either a) rolling his eyes at you and making patronising noises about the yoof of today (and how they were lucky to live in a cardboard box, and eat gravel three meals a day) or b) preparing to slap you on the back of the head or c) sobbing quietly in a corner.

This is actually where these simple line diagrams can help you to be an engineer rather than a panderer to some electronic black box. Moving the train in either direction will affect the result and this can help you to visualise what is going on.

On the midspan case, the slopes on either side have the same gradient, ignoring a little thing like signs (or sines – mathematical humour is an acquired taste). Moving the axles to the right would move the first two axles up as much as the second two axles would move downwards, the effect is therefore neutral. The first smaller axle would move down (and the second, and irrelevant small axle would drop off) reducing the overall effect. Moving the axles to the left would move the first three axles (total 600 kN) down and the second three axles (total 500 kN) up. Moving the full set to the left would therefore also cause a reduction. Since moving either left or right causes a reduction, then the current position is a maximum.

On the three-quarter span case, the slope to the left is three times as steep (no poor excuses for a joke this time). Moving the loads to the right would drop 900 kN down the gentle slope, but would move 200 kN up a slope three times steeper: this would reduce the effect overall. Moving to the left would move 700 kN up the gentle slope but drop 600 kN down the steep slope. Again, any movement causes a drop and we are at a maximum.

Since the effect of a load at a point is equal to that load multiplied by the local height of the influence line, maximising the overall height of the loads on the influence line will maximise the effect and allows you to directly visualise what effect any decision on positioning your load will have.

Aside: there is a well-known special case for maximising moment on a simply supported beam when there are two equal point loads. If the loads are positioned anywhere astride mid span then moving the pair left or right would move one load up and one load down, in equal amounts. Hence the positioning of the load is not critical for moment at mid span. But mid span is not actually the critical position for the beam. The height of the influence line at one quarter of the axle spacing away from mid span is only slightly lower than the one at mid span. If the axles are then set with one axle on the peak and the other one three quarters of the spacing the other side of  mid span, then the moment at the peak will be slightly higher than the moment at mid span.

The family of influence lines for shear are more interesting. I have highlighted the influence line for quarter span, by shading it in a tasteful mauve, but the influence line at any point is simply drawn by following the lower diagonal line from the left hand end down to the relevant location, going vertically up to meet the upper diagonal line, and then following the upper line down to the right hand end. Unlike the bending moment influence line, the line for shear runs both above and below the line, so for any point, except at the supports, applying load over the full span would include mainly adverse loading, but also some relieving load. Hence to maximise shear at a point, the load should only be applied up to that point and no further: you can maximise shear and calculate coincident moment, or vice versa, but you cannot have a load case that maximises shear and bending at the same time.

Figure 28 - influence line for shear with loads

And that is where influence lines are so useful: they allow you to intelligently target your load cases to achieve the maximum effect. Of course, if you actually want to use them for doing calculations then that works too.

Text books, like the Steel Designer’s Manual, include influence lines for a number of standard cases, and these can be very illuminating as a guide. Unfortunately the continuous bridge cases are all for equal spans, and it is almost unknown to have a bridge with equal spans. This would result in sagging moments under full length UDL that are over three times higher in the side spans than in the middle of the main span. Normally, the side spans will be around 80% of the main span to minimise that effect, making the standard diagrams useful as a guide, but useless for actual calculation.

Influence lines can be very useful as a short cut for spreadsheet calcs on simply supported beams. But if you want to calculate influence lines for something more interesting then there are two methods that can be used, one of which is very simple but labour intensive, and the other is technically more challenging but only involves one operation.

Figure 29 - WHAT THE HECK ????

 Method 1 is shown in Figure 29 and is the basic method. DO NOT PANIC – if you ever have to create one then a simple knowledge of spreadsheets will help you out. Split the structure into a significant number of members. Apply load case 1 with a unit load applied to point 1. This is repeated with load case 2 with a unit load applied to point 2, and repeated with a load case with a unit load for every position on the structure. Figure 29 actually shows 7 different load cases, at various different locations, all to plot one influence line, although the same loadcases could then be used to plot any number of other influence lines at other locations.

The influence line plotted here is for bending at 28% of the span (and why not ?). Forget the jumble, just concentrate on, say, the lovely purple set, with the load on the right hand side. The aim of the influence line is to see how much effect (influence) this has at the relevant point and to then plot the result at the place where the load is applied. Hence the value that is plotted at the right hand side is the point where the (solid purple) bending moment diagram cross the black chain dotted line. The purple dotted line runs horizontally back to where the load is applied to show the point on the influence line at the load. This is repeated in a tasteful selection of colours another 6 times, to give a spread of results. There are also another two load cases being inferred, with zero moment when the unit loads are applied to the supports.

The actual influence line, shown in solid thick red, just joins up the dots and in this case forms a pair of straight lines.

The method can then be repeated for shear forces and for reactions. A very simple method, that can be applied to any form of structure, but a very time consuming one which cannot avoid a lot of calculation.

 I will cheat massively and skimp on the calculations for method 1. Consider the simplest influence line diagram, i.e. for bending moment at mid span of a simply supported beam of length L. If we apply a unit load at the end of the beam, i.e. to a support, then there is no moment at mid span, so the influence line at the end is zero. If we apply the unit load at quarter span then the moment at mid span is L/8, and applying it at mid span moment causes a moment of L/4. Plotting these out, and mirror imaging about the centre, gives a triangular influence line peaking at L/4 at mid span.
BUT WHY DO THINGS THE HARD WAY  ??

Method 2, which can either be done accurately, or just used to sketch out general behaviour, only involves one operation for each effect at each point. I can never remember what the name of the particular theory is, but the method involves modelling a distortion of a particular type at the relevant location and then plotting out the deflected shape. I will say this now, and probably say it again later, but the important thing to remember is that it always involved plotting deflections and does not have anything to do with plotting out bending moments, shear forces or reactions.

Fundamental: there is one little wrinkle that you need to get your head around before you try this method, and that is that most computer analysis uses something called small deflection theory, runs counter to common sense. Imagine that you are rotating the end of a rigid beam, which is fixed to a wall at the other end with a hinge, through a very small angle: it will cause a small vertical deflection at the end but will not cause any deflection sideways. Now common sense tells us that if we apply a large rotation to the hinged end of the beam, not only will we get a large vertical deflection, we will also get a small horizontal deflection as the tip starts to swing around. But in the computer analysis if you apply 100 times as much load you will get 100 times the small deflection, since it is a linear analysis. But with no horizontal deflection under a small load, 100 times no deflection is still no deflection. This is, of course totally fascinating, but what does it mean in practice? That is where Figure 30 comes in.

Figure 30 - real world geometry versus small deflection theory

Let us imagine that we have a straight member, of length L, that is horizontal. If we rotate one end of the member by 0.5 radians then we know that the other end will move down by L.sin(0.5 radians) and move sideways by L.*[1-cos(0.5 radians)]. But under small deflection theory, the above situation would be 100 times a rotation of 0.005 radians. Since 1 - cos(0.005 radians) is as near as dammit to zero then  under small deflection theory a rotation of 0.5 radians will give a horizontal deflection of 100 x (as near as dammit to zero) = zero. But the vertical deflection will be 100 x L.sin(0.005 radians), or 0.5.L.

And so on to Method 2 (starting with the influence line for bending moments) ….

Figure 31 - single span influence line for bending

Method 2 involves applying a 1 radian distortion at the point under consideration so it forms a shallow inverted Vee. This is where we have to remember the small deflection theory and not worry that the tips of the vee would sweep inwards, and imagine that they just move downwards and are still a distance L apart. The next stage is to fit the structure to the supports since we know that the deflection will be zero at the supports. There is only one way that it can fit, and that is to form a shallow vee dropping from mid span to the supports. Due to symmetry the angles at the end will be the same and both be 0.5 radians (notice the sneaky coincidence there …). Under small deflection theory, the deflection (not the moment or the shear force, or whatever) will be 0.5 times the length, or 0.5 x L/2 = L/4. The arms of the vee have not had to bend to fit the supports, so they will be straight so we have a deflected shape that is a triangle peaking at L/4 at mid span. Does that sound at all familiar ?

Method 2 is generally applicable to any structure formed from beams. If you use an analysis package that allows for concentrated distortions, then you can model it directly. But you can also eye it in to get a qualitative idea of where you need to put the loads, which is how I cut down on the amount of work that I have to do (I am soooo lazy, although at interviews I refer to this as efficient). The way that I look at these things is to imagine my structure floating with gravity turned off (turns on reverb - bridges in space – turns reverb off) and then fit it to the supports, starting with two simple ones and then cranking the structure one bit at a time, until it all fits the supports. The deflected shape of the structure is the influence line for bending at that point.

The method for a three span structure is shown in Figure 32. The structure is distorted (think taking the structure, cutting it and welding it back together with a kink in it) and it fits easily to two supports. Then it gets bent so that the remaining two support points also fit. Note that the peak in the middle has dropped, that the lines are now curved and that adding load on the adjacent spans actually reduces the moment at the point under consideration.

Figure 32 - three span influence line for bending at mid span

 One final bending case is for hogging over an internal support, as shown in Figure 33.

Figure 33 - three span influence line for bending at internal support

 We start with the same 1 radian distortion, but this time it is at the support, since that is where we are creating the influence line. The obvious starting point here is to fix the structure through the first, second and third supports, since that will be symmetric and easy to visualise. The next stage is to pull the structure to fit the fourth support. Please notice that doing this has some interesting effects on the influence line. The first stage effectively gave the influence line for an internal support of a two span structure. The point of maximum effect is not actually at mid span but is actually 55% of the span away from the support. Levering the beam to match the fourth support causes the middle span to rise (showing a reduced influence) and then causing the far span to drop (showing a raised influence): applying loads to the middle of the outer span will have a higher effect than applying the same load to the middle of the inner span.

There is a similar system for shear calculations which uses a slightly odd distortion, but follows the same procedure. The distortion is a 1 metre step this time. The slight twist is that the stepped ends are allowed to rotate, but they must always remain parallel. If you want a physical representation, then picture the two ends connected by a parallelogram linkage. With the simply supported beam shown in Figure 34 the two sloping lines at top and bottom are parallel and 1 metre apart, so every time that a vertical line is drawn it fits the criteria for a shear influence line.

Figure 34 - shear influence line for single span

 Things get a bit more interesting for the three span situation. Again, there is a useful demonstration part way through the process.

Figure 35 - influence line for shear for three spans

 The influence line for shear for a single span is a simple triangle. But pulling down the first part of the beam to meet the first support causes a rotation at the second support. So not only is the first span a positive area for shear, but the rotation causes the end of the second span to rotate up so that the area of the influence line over the second span is greater than just a simple triangle.

Pulling the span to meet the fourth support causes the beam to rotate over the third support which reduces the area over the second span slightly.

Please note that the selection of load patterns depends heavily on the form of the loading. On highways (pre Eurocode anyway) the intensity of the loading depends on the length of the load (the longer the load the lower the intensity). The load pattern chosen depends on a combination of the areas under the graph over the two spans, compared to the intensities. The area under the first span is possibly 15% of the area under the second span, so if the load intensity drops by more than 13% (1/1.15) when the loaded length doubles (likely) then the product of intensity x area will be lower and the loaded length would be taken over the middle span only. If you are working on a railway analysis then the loading does not depend so much on length and the loaded length would probably be the two spans.

FINALLY, there is also an equivalent for reaction influence lines. This is simply the deflected shape of the structure when a support is raised by 1m as shown in Figure 36.

Figure 36 - influence line for reaction

 Again, note how the load on the outer span has more effect on the support reaction than load on the internal span.

These methods can either be applied by eye, if you want to get a rough idea of how to apply loads, or by computer if you want to get an accurate set of numbers to use. Some analysis programmes will allow you to apply local distortions, as a kink (for bending IL) as a step (for shear IL) or as a support displacement (for reaction IL).

Aside: my favourite ever influence line analysis was for a five span structure with four beams throughout, with four supports at each abutment, but only single supports at each internal support. When I applied a 1 radian distortion for the outer beam over an internal support, to determine the influence line for hogging moment, the shape of the deflected structure was similar to the shape in Figure 33 but the point distortion rose up by a small amount. Hence the adverse area of the hogging moment influence line required the area nearest to the support not to be loaded. This seemed a bit strange until I realised that there were two opposing effects, with central loads on the adjacent spans causing a hogging component over the support line, but load to the side of the point support caused the deck to twist causing a local sagging moment. Up to 15% of the span away from the support, the twist was most important and adding load to that area actually reduced the hogging moment.

Rather than me drawing up hundreds of influence lines to prove my point, try looking at standard examples in the Steel Designer’s manual and (if you squeeze the scales a bit to suit) you will always find something that looks very much like a 1 radian distortion at every point under consideration. Please bear in mind that the diagrams are of the DEFLECTED SHAPES and are NOT anything to do with bending moment diagrams.

By the way, if you think of truss structures as being similar to beams, with the chord stresses being the bending stress, and the tensions in the diagonals representing shear on a beam, then you can eye in influence lines for trusses in a similar way.

Moment distribution

Nobody in their right minds would consider doing such a calculation nowadays but ….

This is a hand calculation that distributed moments around a statically indeterminate frame (one that it is too complex to be solved by resolution of forces) such as a continuous bridge deck.  It works by assuming that the ends of each span are fixed against rotation. The fixed end moments are worked out for each span from standard solutions and tabulated. If the moments on either side of a joint (a support or an intersection point for a frame) add up to zero then everything balances and there is no tendency for the joint to rotate. If, however, there is an imbalance then the net moment will cause the joint to rotate.  Each side of the joint will rotate by the same amount so the moment will be shared by the elements proportional to their rotational stiffness’s, which are calculated from standard tables. The resulting rotation will transfer half the moment to the other end of each member and these moments are then treated in the same manner, which are then totalled and redistibruted, around and around until all of the out of balance moments, which  halving through each cycle, reduce to nothing.

Aside: what is scary is that I have a recent graduate, who went to the same university as me, and it seems that the stuff that was out of date thirty years ago is still being taught now. But all they teach is the method and do not point out the useful bits

The method is, however, very instructive in understanding structural behaviour. Forget about the actual method and just think about how the ends of the members will rotate and hence how the moments will be distributed.

Consider a three span continuous bridge deck with the outer two spans at about 80% of the middle span. The rotational stiffness’s on either side of the inner piers are roughly equal and the fixed end moments are also fairly equal under uniform loading so the joints will not rotate very much and will act almost as though they are built in. This will attract a significant amount of the moment from the middle of the span to the supports. If the side spans were longer, or the sections smaller, then the ends of the main span would not attract so much moment and the behaviour would be closer to that of simply supported. Conversely, if the side spans were shorter, or of a stiffer cross section, then the ends of the main span would attract more moment and the behaviour would be closer to that of a fixed ended beam.

This is a simple example of how the relative stiffnesses govern how a structure deflects and hence how the moments are distributed around a structure. It is something that could be learnt from the moment distribution method, but which is rarely taught.

Fundamental: this leads on to another basic idea that is often skipped over, the free moment diagram. When a uniform load of w kN/m is applied to a simply supported beam of span L then the moment at the middle will be wL²/8 or 0.125 wL² (which I will refer to as 0.125M in this section). If we apply the same load to all spans of a three even span continuous bridge then the hogging moment at the internal supports will be -0.100M and the midspan moment will be 0.025M. Please note that the difference is 0.125M. If the load is applied to a single span with fixed ends, then the end moments are -M/12 or -0.08333M. The mid span moment will be M/24 or 0.04167M: the difference is 0.125M again. Have we spotted the pattern yet?

Regardless of the form of the loads or the structure, if the same pattern of loads is applied to a simply supported beam of the same length as the span under consideration, then the resulting moment is called the “Free moment diagram”. If the moments at the supports at either end of a span, that forms part of a larger structure, are plotted and a straight line is drawn between, then the full moment diagram can be drawn by superimposing the zero line of the free moment diagram onto the straight line with the rest of the free moment diagram hanging beneath. It does not matter what the load pattern is, the free moment diagram that results can be hung between the moments at the end. Varying the stiffness of the adjacent spans, or applying loads on adjacent spans, or whatever, will change the moments at the supports, which will in turn affect the moment at midspan: it will not, however, change the difference between the moments in the span compared to the straight line drawn between the moments at the ends.

Figure 37 - free moment diagrams

 The principal applies regardless of the pattern of loading and a similar system would apply if the loading were a series of point loads, leading to a series of straight line moment diagrams. 

Aside: the mid span moment on a simply supported beam, under uniform load, is wL²/8, as mentioned above. I had three accountants in a row at one company saying “Structures are easy, all you need to know is wL²/8”. I do not know if they are like American presidents, passing on useful information to their successors in an envelope, or if there is a special course at bean counter school called “How to hack off an engineer”, but it is one of the many annoying tools in an accountant’s arsenal – just remember who actually brings the money in guys …

TO BE CONTINUED …

Apparently, following the blog might tell you when I add new section (dinosaur author please meet the modern world) which happens whenever I feel like it. Comments can also be helpful (preferably polite) since everything that I write is perfectly clear to me, but that does not mean that it is actually clear.