Part 0 - the simple bits

This should be the first section in the blog, but I cannot work out how to rearrange things. I suggest that you read the first part of the Introduction to see where I am coming from and then carry on with this section
Dinosaurs rool OK!

IF YOU ARE LOOKING AT THIS THROUGH A COMPANY COMPUTER THEN YOUR IT PEOPLE MAY HAVE BLOCKED DOWNLOADING OF IMAGES FROM GOOGLE DRIVE, WHICH IS HOW BLOGGER INSISTED THAT I INSERTED THEM. IF YOU HAVE THIS PROBLEM, LOOK AT IT ON YOUR HOME COMPUTER. APOLOGIES BUT THIS IS OUTSIDE OF MY CONTROL.

The whinge

Recently, to my surprise, I have been let loose interviewing graduates. My manager is not totally daft and, rather than doing the professional serious bit, I got to give the poor little dears what we thought was a simple test and then to put them through a short coaching session to see how they react when we discuss the bits that they got wrong. Finding out how someone reacts WHEN they get something wrong is an important thing to find out: it also allows a bit of coaching, and it is useful for prospective employees to find out how we would treat them in a working environment.
I was surprised to find that a large proportion of candidates could not do calculations that they should have been able to solve whilst at school. As part of the coaching sessions, however, I found out that this was often because they had been taught methods that, while rigorously correct, were counter-intuitive and mathematically difficult.
One of the problems is that there are often a number of different ways that a problem can be solved but some of the graduates had been taught methods that caused great amusement/concern to working engineers.

Simple or complex ?

As usual, academics insist on using complicated terms that make little sense if you were looking out of the window during the one lecture that they briefly explained it.
Aside: perhaps lecturers need flash cards, like Sheldon Cooper in The Big Bang Theory, but rather than flashing up “Sarcasm” they need to have a rarely used “Fundamental” flash card, as well as a more widely used “Useful example” as well as the inevitable “This is my speciality and I am”, “ going to teach this obscure bit of research”,  “as though the world revolves around it” (three flash cards probably required for that last one).
Further aside: I am quite worried about the number of people who suggested that I watch Big Bang Theory, because they thought that I had similar character traits to Sheldon.
The terms that are commonly used are “Statically determinate” and “Statically indeterminate” – partial translation - “Solvable using statics” and “Not solvable using statics”.

And in English ?

On a Plane structure (i.e. one where you can draw the entire structure and plot all of the deflections in a single view on a single piece of paper) solving a problem using statics just means that to find the unknowns (usually the support reactions) the only techniques that you need are resolving forces in two directions (usually horizontally and vertically) and taking moments ONCE. A shorter term to use, and one that I will use from now on, would be “Simple”.
Aside: I have never managed to work out why, but taking moments TWICE will just give you another equation that could be found by rearranging the other three equations: this would not give you any extra information.
Statically indeterminate just means that you cannot work out the unknowns using statics alone and need to use a more complicated method. I will just refer to these structures as “Complex”.
You will be pleased to know that a large proportion of structures are simple, or can be treated as simple through a bit of fast talking. After a bit more explaining I will take you through easy methods that include everything that you need to know to solve every Simple structure. These methods will also demonstrate some principles that will allow you to check the results for more complex structures, although that is all I will cover here for complex structures.

Simple ??????

So how do we decide whether something is simple ?
I seem to recall being given some very complicated rules about degrees of freedom (how many joints there are combined with how many ways they can move or rotate) numbers of members and numbers of supports: but it did not make any sense to me at the time and I have completely forgotten it. In thirty plus years I have never needed to use the rule and it does not seem to be necessary. The rule was written to cover all sorts of eventualities, and was probably rigorously correct, but common sense can cover most situations. My method is a combination of simple ideas, but requires you to picture the behaviour of the structure. To keep things simple, although the principles apply to all linear structures, I will mainly talk about structures where all of the loads, and all of the movements of the structure, are in a single vertical Plane: as I mentioned above (but it never hurts to mention the important things twice – flash card : fundamental) these structures are called Plane structures.
The first thing to look at is the structure as an overall rigid body: by this I mean that a structure, even if it has many elements like a truss, is treated as a single body. Using statics, there are only three equations that can be written so you can only solve for three unknowns. Looking at this the other way round, if there are more than three unknowns then you cannot solve using statics and it is Complex.
Looking at the simplest structure, i.e. a single beam, the most common form is to have two supports, distantly followed by the form where one end is clamped to form something like a diving board. For a plane structure, supports can typically be fixed against either horizontal or vertical movement in the plane, or against rotation about an axis perpendicular to the plane. These fixities are the unknowns that we need to calculate using statics.
fixities.png

If we consider the beam with two supports, then one will typically provide both vertical and horizontal support and the other will only provide vertical support. There are three unknowns and they can be calculated using the three equations.
If there were an extra fixity, such as fixing the second support horizontally, then you could get one support working against the other and you cannot simply resolve forces horizontally to split how much of any horizontal load is resisted by either end. This is a statically indeterminate structure, i.e. one that I refer to as Complex.
Aside: if you remove one of the fixities then the structure will not be stable: removing the horizontal fixity would allow the beam to slide freely and removing one vertical support would allow the structure to rotate freely: both of these are referred to as mechanisms and cannot be analysed as they are not stable.
If we look at the springboard then one end is fixed to prevent both vertical and horizontal movement and is also fixed against rotation. Again, there are three unknowns and three equations allowing you to solve the problem, meaning that it is Simple. Adding another simple support at the far end (propped cantilever) makes the structure complex, requiring special treatment, and removing any of the fixities would create a mechanism.
If we extend the form of structure by adding an extra span, with the two spans acting continuously (i.e. the shared ends of the two spans, over the support, are rigidly linked horizontally, vertically and against rotation) then you will need at least four supports to keep it stable. You will always need one horizontal fixity to stop the structure from sliding off, and adding three vertical supports would make four in total. Since this is more than the three equations it is a Complex structure. That is the numerical way of deciding that it is Complex, but a better way is to decide whether or not you can move, or rotate, any one of the supports without any of the structure being stressed in any way.
The structure below (Example the First) is a simply supported structure, with one horizontal fixity.
ssbeam 3 fix moving
The original location is shown behind in grey for reference.
If you move the left support to the right, the right support is free to slide so there is no problem.
If you lift either the left or right support then the structure is free to move as a rigid body. There are no axial forces, shears or bending moments under any of these movements.
Conclusion: the structure is Simple.

The structure below (Example the Second) is a continuous structure with four pinned vertical supports and one horizontal fixity.
3 span moving
If you move the left support to the right the other supports are free to slide so there is no problem.
If you lift any of the supports then the structure bends. There are no axial forces, but there are bending moments and shear forces.
Conclusion: the structure is Complex.
Note that in this problem there are 5 fixities so we would not expect the problem to be Simple.
I will no longer labour the point about which forces etc are occurring since it will (hopefully) be clear by now what I am talking about.

Aside: Being a mean little Adrian, who loves winding up his grads, one of my favourite recent course developments has been to get them all standing by the projection screen, showing a series of structures, and then getting them to stand in front of markers saying Simple, Complex or Something funny. After the initial problem where they all waited, until one of my brighter grads moved first, and then followed like sheep (I made him stand still until they had all chosen for later problems) it was an amusing (and mildly sadistic) way of checking what they were learning.
I will come back later to some more interesting structures, and discuss whether they are Simple or Complex, but the next stage is to work out how to analyse a Simple structure.

Stage the First

The first operation is to calculate the forces and moments in the structure supports. This is simply done by taking moments about one of the supports. Taking moments about that one point allows you to calculate the reaction in the other support. There are some little variations that sometimes come up, but I will confuse you with those at the very end of this part of the guide.
Once you have calculated one reaction then the sum of the two vertical reactions will equal the total applied vertical load: this is resolving vertically and allows you to calculate the second reaction.
Finally, if there are any horizontal forces then these will be resisted by the horizontal fixity: this is resolving horizontally.
OK, you have done the hard bit since you have calculated all of the unknowns.

Stage the Second:

What are bending moments and axial and shear forces ?
A bending moment is similar to a normal moment (i.e. force x lever arm) but it is an internal moment that is acting on a cross section as a result of all of the loads (including reactions) acting on that structure.
Axial force is just the force acting along the member, stretching or compressing it, with shear force being the force acting to try to shear the member.


Fundamental: Bending moments are the internal moments that would be required to keep the two sides of a structure in equilibrium if you were to take a saw and cut through that structure at the point in question. The oddity is that there are actually two moments required at any section, one for each cut face, and these moments are of opposite sign, as shown above, because one will be acting in a clockwise direction and the other will be anti-clockwise.  Axial force is the internal in-line force required to keep the structure in equilibrium and the shear force is the internal normal force required to do the same: again, both of these will be of opposite sign on the two cut faces.
Fundamental: if the term “Shearing” is unfamiliar, try the following: hold a ruler horizontally, but with the wide face set vertically. Grasp the ruler in both hands, with your hands touching each other. Try to move your left hand upwards, and your right hand down, but using your grip to clamp the ruler so that the ends stay level. You will not be able to do much to the ruler, but if you were to cut the ruler vertically, between your hands (I am not sure how many hands you would need for that in total, and I am sure that the Elf ‘n Safety guys would complain about a blade so close to flesh) then the left side of the cut would move up, past the other side moving down. Shear force is the internal force resisting this movement in the uncut beam.
Aside: before I go on any further, I have an important thing to say …..
I am sorry. I am really very, very sorry. Please accept my most heartfelt apologies. I am really, really, really, incredibly, really, massively sorry.
I am not apologising for something that I have done, but am apologising on behalf of the engineering community over the past 150 years. The sign conventions for moments and shears are rigorous but very confusing. It does not actually matter what sign convention you use, provided that you are consistent. But I will use the convention used by my favourite analysis programme, STAAD (as well as Superstress, which is widely used for simpler analyses in good old Blighty).

Fundamental: in the diagram above, the sign convention is that sagging moments are positive and are plotted DOWNWARDS. Although this is very odd, it means that the sign of the moments for simply supported beams, which used to make up the vast majority of structures, did not require the engineer to put a negative sign in front of every number, but the moment diagram was drawn on the tension side, and was drawn downwards, which seemed more logical when the beam was deflecting down (but please do not get confused that the moment diagram will always follow deflection because it does not do so over internal supports on multi-span continuous structures, as you can see in the diagram above).
The shear force diagram is defined positive upwards, when looking at the left hand end of a cut structure, and that is plotted UPWARDS. Grrrrrrhhh !! This is consistent with what I will discuss later, about gradients, but only because the first red section on the left is going downwards which is actually upwards when you consider the sign convention. Confused much ?
There is a very simple way to calculate the values. Any structure is subject to a series of applied moments and forces, and also to a series of reaction forces and moments. I will demonstrate this later for simple structures (but all that more complex methods do is calculate reactions on Complex structures). Once you know all of these reactions, applying them back onto the structure will result in a structure that is in equilibrium with the applied loads.
In the earlier diagram, the bending moment at mid span is (as we all know) wL2/8. The reactions are both wL/2. The method for working out the bending moment at mid span is to cut the structure in half and remove the left hand side.
Taking moments about the cut, M = wL/2 . L/2 – wL/2 . L/4 = wL2/8. In a similar manner, the shear force is the vertical force required to keep the cut structure in equilibrium and in his case it is wL/2 – wL/2 = 0. Finally, the axial force P is simply F since there is no horizontal reaction at the sliding support.
And that is all there is to it.
Aside: recently, whilst interviewing, I was surprised to find that most of the graduates tried to solve a series of examples by calculating the shear force diagram and then integrating the shear force to get the moment. Although this is a perfectly legitimate method it is ridiculously complex and relies on getting the directions of the reactions correct, which is not always easy to do: it also carries an error made in one part of the calculation through to the end (although at least it allows you to check that everything sums to zero for a pin end). As part of the interview, I spend time coaching them on the bits that they got wrong, partially as part of my relentless war to educate, but partially to test their character when they get something wrong (as we all do). I am often flattered/amazed/worried by the number of grads who say “But your method is really easy. I wish that we had you lecturing us instead”. Much as I love being mean to lecturers, I have to add that coaching one to one is a lot easier than lecturing a roomful of hung over students, but teaching a more simple method has got to be better.

Example the First:

This is a typical example that covers the majority of simply supported beams that you can analyse using statics.

The Method:

First off you need to calculate the reactions, either as detailed earlier or by using some more advanced method for a complex structure.

SSBeam1
The beam has three fixities and can therefore be analysed using statics.
Taking moments about A:
10.RB = 10.2+20.5 = 120
Hence RB = 120/10 = 12kN
Resolving vertically: RA+RB = 10+20 = 30
Hence RA = 30-RB = 30-12 = 18kN
To show that taking a second set of moments does not produce extra info (although it does provide a check:
Taking moments about B:
10.RA = 10.8+20.5 = 180
Hence RA = 180/10 = 18kN (amazing)
So now we have both unknowns that we are interested in (there are no horizontal forces).

Next, you get two pieces of paper. On one piece you draw the structure and plot all of the loads and reactions. Be careful that the loads are plotted in the right direction, with gravity loads acting downwards and reactions generally (BUT NOT ALWAYS, AS WE WILL SEE IN AN EXAMPLE AT THE END) acting upwards.
SSBeam2
The force acting on a support will go in one direction but we are interested in the forces on the structure (i.e. the reaction) which act in the opposite direction. On the second piece of paper, you draw a dot on the left hand side of the page, a moment arrow, going in a clockwise direction and a vertical arrow.
The final stage is to cover all of the structure, except for the right hand end, and then calculate the moment (bending moment) and total vertical force (shear) for all of the forces NOT covered up by the sliding sheet at any point of interest on the structure. What you are effectively doing is to cut the structure and throw one side away: calculating the equilibrium moment and vertical force gives you the bending moment and shear force. Plotting out these results for all points will give you the bending moment and shear force diagrams for the structure under the applied load case.
The final little niggle is the sign conventions. It is so blindingly obvious that it is really hard to describe exactly what to do, but I will try anyway. In the diagram, the curved moment arrow drawn is a positive moment, which is a sagging moment. This will result in a tensile stress in the bottom of the beam (as shown) and that is the side that you plot the bending moment diagram.
The translucent bit is supposed to represent a piece of paper that has been laid over the drawing. The dot represents the point that we are taking moments about, as well as the location where we are resolving forces. In theory we are setting it an infinitesimally small distance either side of supports or load points. This makes no difference for the moments, since a tiny change in location will make no difference in the length of the lever arms and having a force appearing with a zero lever arm will contribute no moment. It makes a massive difference for the shear forces since the shear force is the total of all of the forces that have been revealed and moving to one side of the load or another means that the force gets either included or not. Hence for the figure above:
Moment = 12.0 = 0kNm
Shear force = -12kN

ssbeam3
The 20kN force has not appeared so is not included.
Moment = 12.5 = 60kNm
Shear force = -12kN

 ssbeam4
Now the 20kN force has appeared so it is included.
Moment = 12.5 = 60kNm (no change from previous calculation)
Shear force = -12 + 20 = 8kN (step change from previous calculation)

 ssbeam5
Moment = 12.8 - 20.3 = 96 - 60 = 36kNm
Shear force = -12 + 20 = 8kN

 ssbeam6
Moment = 12.8 - 20.3 = 96 - 60 = 36kNm
Shear force = -12 + 20 + 10 = 18kN

 ssbeam7
Moment = 12.10-20.5-10.2 = 120-100-20 = 0kNm
Shear force = -12 + 20 + 10 = 18kN

 ssbeam8
Simples !
The above method can be used to analyse absolutely any Simple structure.
Fundamental: you will often hear me banging on about us all making mistakes and looking for ways to check things. Having started at the right hand end to uncover the structure, and using the sign convention that I have used, you will notice that the step changes in the shear force diagram, at the concentrated forces, follow the direction and magnitude of the reactions and applied forces. UDL loads actually do the same thing, but you need to visualise the load as a series of small discrete loads and the sloping line as a series of small steps.
This example also shows another way that you can check any complex structure that you analyse as well as showing the overly complicated method that I was so rude about earlier. If you look at the slope of the moment diagram between A and the 10kN load, you can see that it is sloping downwards. Actually, it is sloping upwards since the moment is plotted positive downwards. The slope is (36-0)/2 = 18, which is also the shear force over that length. Between the 10kN and 20kN loads, the slope is (60-36)/3 = 8, which is the shear force between those points. Finally, the change between the 20kN load and B is (0-60)/5 = -12, which is …. You get the idea. It is a fundamental that the shear force is always equal to the slope (differential) of the moment diagram. The complicated method (about which I was so rude earlier) is to calculate the shear force along the entire structure and then to integrate the shear force to calculate the bending moments (which is the reverse operation). This is technically correct but relies on getting the reactions correct and, more importantly, in the right direction. It is also easy to integrate uniform shear patterns, but more complex for uniformly distributed loads, and worse for variable distributed loads.
The shear force diagram for a structure with point loads will always be a series of rectangles, since there are step changes at the load points and no other loads in between. Since the shear force diagram is always the slope of the bending moment diagram, the series of shear force rectangles will result in the diagram being a series of straight lines. Likewise a UDL will give a uniformly changing shear force diagram, which results in the gradient of the BM diagram uniformly changing, resulting in a parabolic moment diagram.
Below is the bending moment diagram for a three span continuous beam with a single point load on the first span and a UDL on the third span.
3 span moment and shear 
The resulting shear force on the first span has a stepped shear force diagram resulting in two straight lines for the moment diagram. The middle span has a uniform shear force resulting in a single sloping moment diagram. The shear force in the final span has a uniform slope, giving a parabolic bending moment diagram. The rule works for every single structure and is a good check of what you work out, no matter what method you used to work it out.
Fundamental: although we have been using statics to analyse simple structures, they are also vital for complex structures, especially if you are analysing something on a computer. Although you will not be able to actually analyse a complex structure by resolving forces, if you think of the structure as a single body, with reactions being just another set of forces, then you can pick up a lot of mistakes just by adding all of the applied loads in a direction (usually downwards but it also works horizontally) and comparing the applied total to the total of all of the reactions. If they are not equal then you have a problem, and this is usually caused by you applying the loads incorrectly (although it can also be because your check calcs are wrong).

Shear = slope of moment diagram

In case you are interested, it is very easy to show why this is. The diagram below (which may look familiar from above) shows a very short length (cut out of beam) of length δL.
 demonstrate differentiation
At one end there is Bending Moment M and Shear Force V: at the other end these have changed to M+δM and V+δV. Taking moments about the right end M – (M+δM) + V.δL = 0, hence V = δM/δL which is the first differential of the bending moment diagram and also the slope.
By the way, if you took moments about the other end then there would be a term δV.δL which would tend to zero.
Fundamental: please bear in mind that, when taking moments about a point, the moment due to a force is the product of force x perpendicular distance, but the “Moment” due to an applied moment is just the moment (i.e. it does not matter where that moment is actually applied).

Time to test yourself

Before you can try to apply statics, you need to be able to determine if statics will work. My method is to check whether or not you could move all of the restraints, one at a time, without causing any stress in the structure e.g. by trying to bend or shorten it. If you can then it is Simple, if you cannot then it is Complex.
To check your logic, and to test the results, move down to the end of the examples. Try hard not to move too far because it will be hard not to see the answers unintentionally.

Example the Third – two fixed pins

Example the Fourth – two sliding pins

Example the Fifth – encastré cantilever

Example the Sixth – pinned cantilever

Example the Seventh – propped cantilever

Example the Eighth – three span with drop in beam

















The Answers


Example the Third – two fixed pins
ssbem 4 fix moving
If you move the left support to the right, the right support is fixed so the beam is squashed (bad).
If you lift either the left or right support then the structure is free to move as a rigid body (fine).
Conclusion: the structure is Complex.
If the structure gets hot or cold then it will be stressed because it will not be able to expand or contract as it wants. Also, since the structure has depth, the bottom fibre will get longer under bending which will also be resisted.
HOWEVER a lot of the old structures that I work with are beams sitting on padstones (stone bearings) so they are not free to move. We do, however, cheat and assume that the walls will flex enough to allow for the relatively small movements required under bending or temperature effects.









Example the Fourth – two sliding pins
ssbeam 2 fix moving
If you move the left support to the right, the right support is free to slide so there is no problem with the beam. Unfortunately the structure has no way to restrain any horizontal load and is a skate board – good to let off some stream – bad as a structure.
If you lift either the left or right support then the structure is free to move as a rigid body.
Conclusion: the structure is a mechanism and cannot be analysed.







Example the Fifth – encastré cantilever
cantilever 3 fix moving
If you lift the support then the structure follows as a rigid body.
If you move the support to the right then the structure follows as a rigid body.
If you rotate the support then the structure rotates without restraint.
Conclusion: the structure is Simple.








Example the Sixth – pinned cantilever
canti 2 fix moving
Bad case of brewer’s droop (if English is not your first language, then good luck translating that J). There is nothing to stop the structure from just dropping.
Conclusion: the structure is a mechanism.







Example the Seventh – propped cantilever
canti 4 fix moving
Lifting the encastré left support will bend the structure.
If you move the left support to the right, the right support is free to slide so there is no problem.
Rotating the support will bend the structure.
Dropping the pinned support will bend the structure.
Conclusion: the structure is Complex.







Example the Eighth – three span with drop in beam
3 span drop in moving               
If you move the left support to the right, the other supports are free to slide so there is no problem.
If you lift any of the supports then the structure just goes with it and none of the beams are stressed.
Conclusion: the structure is Simple (Fairly).
Surprise ! There are 5 fixities so we would not expect the problem to be Simple but because of the releases this actually is (Fairly) Simple. In order to analyse this structure (which is a quite common construction form from the 70’s, but rarely used now due to problems of maintaining the bearings in the hinges as well as hidden corrosion) you have to work from the simple bit and work out. The drop in span is a simply supported structure and can be easily solved. Then just take the span out whilst applying the end reactions from the drop in span as loads on the remaining two structures.

Principle of superposition

This is one of those lovely phrases that does not leap out as either a) English, or b) useful: it is however both of these.
All it means is that for certain types of structures you can analyse a series of load cases and get a series of results. If you add the series of results together then you will get the same answer as if you analysed the structure for one load case with all of the loads applied at once.
This means that if you have a structure with lots of complicated loads then you can break it down into little bits and solve them one at a time.
The relevant type of structure is called linear and covers probably 99% of all structures that we have to deal with. Oh, and to be honest, if you are reading this then you probably are not going to be working on non-linear structures J.
Linear structures are those structures that do not change their shape very much under load, so suspension bridges and structures that are on the way to buckling are GEOMETRICALLY non-linear, whilst most cable stayed structures, as well as practically everything else, are geometrically linear. There is also another form of analysis where the structures that are loaded so heavily that they go beyond elastic behaviour and these are MATERIALLY non-linear. We generally try to avoid pushing structures that far, so material non-linearity is rare.
Hence linear structures are anything that are neither geometrically or materially non-linear, and it is for these structures that superposition works.

Simple and Complex Truss structures

The structures that I was discussing before were all beam structures but there are similar rules for truss structures. The first thing to do is to treat the truss as a rigid body and to think of it in the same way as a beam. Hence a truss with two fixed supports is likely to be Complex whilst one with one fixed support and one sliding one is likely to be Simple.
But trusses also have a further set of guidelines that determine whether or not they are Simple, and which can break the 3 fixity guidance. If the structure is made out of a series of triangulated pin jointed truss members (i.e. ones where the members can take tension or compression but are free to rotate at the joints) then it is very likely to be Simple. If, however, any of the members are continuous and can carry bending moments then the structure will be Complex. Most real world truss structures are a hybrid of pin ended truss members and bending members. The primary support mechanism for most trusses is the pin ended behaviour, but most trusses will also have parts that are continuous that will attract bending moments whether you want them to or not.
Aside: there is a form of “Truss” where there are no triangles, and everything is a rectangular grid of members that resist bending as well as axial forces, and these are called Vierendeel Trusses. They are much beloved by Architects (and I hope that you are impressed by the fact that I did not stick a rude adjective before the A word) because they are open structures that look pretty. Unfortunately they are massively inefficient and one small area of corrosion at a joint can cause progressive collapse of the entire structure. Bless ! Winning awards and keeping structures up are two different things, obviously.
Back to the real world. Typically the top and bottom chords (as well as the end diagonal if sloped) will be continuous with each other and relatively stiff in bending, whilst the internal diagonals and verticals will typically be more slender, be connected more lightly and may often be treated as pin ended. These structures are Complex and require complex analysis.
Aside: If you are looking into the primary support mechanism, then the above truss can be treated as pinned, allowing Simple analysis, but I have concerns about this method since it is fine as a plastic analysis (i.e. if the local sections are so chunky that they can form plastic hinges and rotate if necessary) but a lot of trusses have cross sections that are slender locally, and I would only consider this if all relevant sections were compact i.e. chunky enough that they could form a full plastic hinge without local buckling.

There should also be a couple of extra words in my description of a simple truss: the words “Non-overlapping” should be added before “Triangulated”. If you have a football (soccer to our colonial cousins) goal with hinged supports and hinges in the corner then it can fall over sideways and is a mechanism.
mechanism

If you add one diagonal joining one support to the opposite corner then it becomes a triangulated structure and is therefore stable.
Simple

But if you add a second diagonal across the opposite direction then you could tension up one diagonal then that would also tension up the other one.
Complex

This means that one member can work against the other with no external load being applied: this form of overlapping structure is Complex. This could, for instance, occur if someone were to stick ice packs around both of the diagonals, so that they are both trying to contract, then these two diagonals would be working against each other and the rest of the structure. If you want to picture this behaviour, sit down in a chair, hold on to the sides of the seat and pull hard. You will be able to pull as hard as you like but you will not move. You will be feeling an internal force in your arms but no external force is being applied. Apart from looking at your white knuckles an outside observer would not be able to tell how much force there was in your arms: the same thing could be arranged in the two diagonals and they could both be stressed without the structure noticeably deflecting.
In a more practical arrangement, in the example above, the two diagonals would take the same compression and tension, IF they were the same section. But if one were a lot stiffer than the other then it would take a lot more load (this would require complex analysis and is not covered here).
Back to my simple rule, in a Simple truss you can change the length of any of the members by some amount without the rest of the structure being stressed.
Lengthened truss
In the truss above, one top chord member has been lengthened and the structure has rearranged itself to suit. The structure is therefore Simple. Notice that the sliding bearing has slid sideways: if the bearing had been fixed then the structure would not have been free to rearrange itself in an unstressed way and the structure would have been complex.
Back in the good old days (more memories from when dinosaurs roamed the Earth) trusses were analysed by calculating the reactions in the normal way, followed by a large series of resolutions of forces, horizontally and vertically, at each joint in turn. This allowed the tension and compression forces to be calculated in every member. This was only possible if the truss were treated as Simple, but would have given reasonable results if there were some continuity because the pin jointed truss was the primary support mechanism.

Members at an angle

sliding ladder.png

If the free end allows movement towards the other end then it does not matter which end you take moments about to calculate reactions. If the free movement does line up with the fixed end then you need to take moments about the fixed end. If you take moments about the top reaction, in the problem above, then you have two unknowns, which makes life a little too interesting. But if you take moments about the lower point then two of the unknown reactions drop out of the equation and the equations are much simpler.
This is the sort of problem that I learnt at school when I was fourteen and we had to show whether or not the ladder would slip, by comparing the friction required to the maximum available of μ.RV. The only thing that we never worked out was whether or not the ladder would snap under the load, by comparing the bending moment to the bending strength: the principles for calculating reactions were, however, exactly the same as we use in structural analysis.
Aside: at school we, of course, became obsessed by the impossibility of the frictionless wall. The teachers told us that it could not be solved if there was friction on both surfaces. If only they had told us, as someone with an engineering PhD would, that it was statically indeterminate then all would have become clear. For those who do not know me, the last bit was written with an ironic tone in my head.
Further aside: just to clear up one point – I have met several PhD’s for whom I have a lot of respect as engineers. They just tend to be the ones who talk a recognisable language rather than dropping “But of course it is just a negative second order differential” to describe a hog curve.

Final example

I will now demonstrate a solution to one problem to demonstrate most of what we have been looking at, as well as to point out a common mistake that people make.
The beam is simply supported, with a fixed pin at one end, a sliding pin inset from the other end, self-weight applied over the entire length and two point loads, one in the middle of the main span and one at the end of the overhanging section.
last example 1
1) Taking moments about A:
8.RB = 6.5.12 + 10.4 + 20.12 = 640
Hence RB = 640/8 = 80kN
Resolving vertically: RA + RB = 5.12 + 10 + 20 = 90
Hence RA = 90 - RB = 90 - 80 = 10kN

At right hand end:
Moment = 0kNm
Shear force = 20kN

At B (RHS):
Moment = -4.5.2 - 20.4 = -120kNm
Shear force = 4.5 + 20 = 40kN

At B (LHS):
Moment = -4.5.2 - 20.4 = -120kNm
Shear force = 4.5 + 20 - 80 = -40kN

At midspan (RHS):
Moment = -8.5.4 - 20.8 + 4.80 = 0kNm
Shear force = 8.5 +20 - 80 = -20kN

At midspan (LHS):
Moment = -8.5.4 - 20.8 + 4.80 = 0kNm
Shear force = 8.5 +20 +10 - 80 = -10kN

At A:
Moment = -5.12.6 – 10.4 – 20.12 + 80.8 = 0kNm
Shear force = 5.12 + 10 + 20 - 80 = 10kN

I will now split the above into three separate loads to demonstrate superposition.
 last example 2
2) Taking moments about A:
8.RB = 5.12.6 = 360
Hence RB = 360/8 = 45kN
Resolving vertically: RA + RB = 5.12 = 60
Hence RA = 60 - RB = 60 - 45 = 15kN

At right hand end:
Moment = 0kNm
Shear force = 0kN

At B (RHS):
Moment = -4.5.2 = -40kNm
Shear force = 4.5 = 20kN

At B (LHS):
Moment = -4.5.2 = -40kNm
Shear force = 4.5 - 45 = -25kN

At midspan (LHS and RHS):
Moment = -8.5.4 + 4.45 = 20kNm
Shear force = 8.5 - 45 = -5kN

At A:
Moment = -5.12.6 + 45.8 = 0kNm
Shear force = 5.12 - 45 = 15kN


 last example 3
3) Taking moments about A:
8.RB = 10.4 = 40
Hence RB = 40/8 = 5kN
Resolving vertically: RA + RB = 10
Hence RA = 10 - RB = 10 - 5 = 5kN

At right hand end:
Moment = 0kNm
Shear force = 0kN

At B (RHS):
Moment = 0kNm
Shear force = 0kN

At B (LHS):
Moment = 0kNm
Shear force = -5kN

At midspan (RHS):
Moment = 5.4 = 20kNm
Shear force = -5kN

At midspan (LHS):
Moment = 5.4 = 20kNm
Shear force = 10 - 5 = 5kN

At A:
Moment = 5.8 – 10.4 = 0kNm
Shear force = 10 - 5 = 5kN


 last example 4
4) Taking moments about A:
8.RB = 20.12 = 240
Hence RB = 240/8 = 30kN
Resolving vertically: RA + RB = 20
Hence RA = 20 - RB = 20 - 30 = -10kN
Please notice the sign of the reaction. If there were nothing else holding down Bearing A then the beam would lift off and the flip over. In this case, the loads from the other two parts of the overall load case are there at the same time and will hold the beam down overall.

At right hand end:
Moment = 0kNm
Shear force = 20kN

At B (RHS):
Moment = -20.4 = -80kNm
Shear force = 20kN

At B (LHS):
Moment = -20.4 = -80kNm
Shear force = -30 + 20 = -10kN

At midspan (LHS and RHS):
Moment = 30.4 - 20.8 = -40kNm
Shear force = -30 + 20 = -10kN

At A:
Moment = 30.8 - 20.12 = 0kNm
Shear force = -30 + 20 = -10kN

If you add all of the results together, you get 2) + 3) + 4). If you look you can see that these match the result when you apply all cases at once. This may seem trivial, and if you can spare the time then you can split up all of the reactions into their individual components in case 1 and it should be fairly obvious that the results will add together.


RA
A
LHS MS
RHS MS
LHS B
RB
RHS B
R End
1) Reaction
10




80


    Bending mom.

0
0
0
-120

-120
0
    Shear force

10
-10
-20
-40

40
20
2) Reaction
15




45


    Bending mom.

0
20
20
-40

-40
0
    Shear force

15
-5
-5
-25

20
0
3) Reaction
5




5


    Bending mom.

0
20
20
0

0
0
    Shear force

5
5
-5
-5

0
0
4) Reaction
-10




30


    Bending mom.

0
-40
-40
-80

-80
0
    Shear force

-10
-10
-10
-10

20
20
2+3+4) Reaction
10




80


    Bending mom.

0
0
0
-120

-120
0
    Shear force

10
-10
-20
-40

40
20

The other point of interest is the uplift reaction at A for case 4. One of the major drawbacks with calculating reactions, then shears and integrating to get moments (apart from the fact that it is too complex and prone to arithmetic error) is that some people assume that all reactions are upwards and taking the force in the wrong direction will totally mess up the numbers.

Conclusion

Well I could go on and on for a few more hours, but I hope that you have the idea by now. These methods are used for a large proportion of analyses and are very simple when you understand the basics. If you have a structure that is more complex than the simple structures then you will most likely be using a computer analysis programme.
The problem is that these simple methods are some of the first things that you will have been taught at University. Unfortunately, these methods are not interesting for researcher/lecturer types and you will probably have been taught a lot of special cases that you will never use.

There will be some of you who wonder why I have spent so much time producing something so basic. But if you are one of the majority who has learnt something then please be aware that you are not alone.

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