‘Understanding structure’ considers the science of how structures work. How do structures resist all of the demands made on them by forces of self-weight, people moving about, or wind that may blow a hurricane? These forces have to ‘flow-though’ the components of the structure. Some simple structures are statically determinate so the internal forces can be calculated by just making sure all of the forces are in equilibrium. More complex structures have too many unknowns to be solved this way—they are statically indeterminate. The internal forces in these structures are found by satisfying the three conditions of equilibrium, constitutive relations, and compatibility of displacements when the total potential is a minimum.
Most of us, I suspect, have tried our hands at structural engineering by building a tower of children’s play bricks. If you have then you will know that there is a limit to the height you can achieve. Intuitively you realize when you are getting close. At low heights a slight disturbance, say by just touching the tower, has little effect—it is quite stable. As the tower gets taller it gets more and more wobbly until it eventually topples. The height limit depends on a number of factors. For example the surface you are building on makes a difference—you can build a tower taller on a hard flat surface than on a soft fluffy carpet. In other words the foundations have to be suitable. Another factor is the size and shape of the bricks and, in particular, their cross section—the bigger the cross section the taller the tower. Yet another is whether each layer of bricks simply sits on the one below or is connected to it like a Lego brick.
In this chapter we will build on these familiar experiences to understand better how structures work. You’ll remember that in Chapter 2 we showed (in Figure 9) the possible six degrees of freedom or independent directions (in line or rotating) in which a structure or any part of a structure can move or deform. We said that the key to structural art was to create pathways for forces to flow logically, and that the pathways are defined using these local p. 56↵or global directions when free movement is prevented. The restraint may be external, such as a foundation; or internal, such as a connection to another part of the structure. We found that a rope hanging in a catenary has only one degree of freedom and hence one force pathway, because it can only stretch as it resists tension. It cannot resist compression and can only resist a lateral force by changing its shape until it resists that force through direct tension (Figure 4).
Our toy brick tower also only has one degree of freedom and hence one force pathway—but now the force is compressive and the behaviour is more complicated. If we push down onto the top of a very low tower (say only one or two bricks) the bricks will simply squash under the load. You may be able to detect the amount of squash if you have soft foam or rubber bricks but you won’t notice the very small squashing of wooden bricks—you would need a special instrument to measure it. Eventually, under a very heavy weight, the bricks will either squash flat or crush into fragments.
If we push down on a much taller tower then the behaviour is different. A tall tower will topple sideways rather suddenly. The bricks just want to get out from under the downward force or load so they take the path of least resistance and move sideways. As the towers get taller the load at which this happens will get smaller, until they reach the height limit and the tower falls without any load at all. Also the higher the tower the more sensitive it becomes to any slight sideways push or any tendency for the downward load being off-centre.
We can understand this better perhaps through another thought experiment—this time we’ll push down at an angle so there is a downward and a sideways force. Imagine gently pushing the top brick of a tower sideways. Four things can happen depending on several different factors. First, if the bricks are smooth and the downward push load is small compared to the sideways push load, p. 57↵then the top one might just slide over the one underneath until its centre of gravity is just past the edge—and it falls. Second, if the contact surfaces between the bricks are rough then the friction between them will resist the sideways push—F, in Figure 11a. The net result is that a reaction force R1 is generated that is a combination of the weight of the brick and the downward push (together labelled W) and the sideways force F. Furthermore, we can calculate R1 using the triangle of forces that we introduced in Chapter 2—see Figure 11b. However it is important to realize that R1 is only developed because the brick underneath stands firm. We can check that it does stand firm by going through exactly the same logic on the second brick down in Figure 11c. We can check the size of its reaction force R2 using the triangle of forces but using 2W for the two bricks, as in Figure 11d. Then we can continue this process one by one for all of the bricks underneath until we get to the last one (in Figure 11e this is the 4th one) that sits on the ground with a reaction R4 calculated in Figure 11f. If the ground can resist this reaction required from the bottom brick then the tower will stand.
A third possible behaviour of the tower is that the friction between the bricks is so great that the top brick doesn’t slide but eventually starts to rotate about an edge, as shown in Figure 12a. Then as soon as its centre of gravity is poised just over the edge of the lower brick it will topple over, as shown in Figure 12b. Alternatively, if the bond between the bricks is strong enough then the whole tower might rotate about the bottom edge, as shown in Figure 12c. The degree of rotation (and hence the sideways force required) is much smaller for a taller rigid wall, as you can see from the diagram. A fourth possible group of behaviours for the tower is shown in Figure 12d, where the very bottom brick is prevented from rotating and the tower may fall, depending on whether the top one, two, three, four, or more bricks rotate together.
Now let’s think about a quite different kind of tower—like a lamp post, a mast of a ship, or, more simply, a slender wooden cane p. 58↵p. 59↵pushed into the ground. Here we don’t have separate bricks but just one continuous piece of wood which doesn’t just sit on the ground but is firmly held by it, as in Figure 13a. Indeed, we will assume for the moment that the cane can’t move or rotate at the bottom but, of course, the rest of the cane above the ground can deflect and bend. It’s rather like a diving springboard in a swimming pool, except there the board is horizontal—other much more complex but real examples are an aeroplane wing and a tall building or skyscraper, which we look at in a bit more detail in Chapter 5. It is quite hard to visualize what is going on in all three dimensions so we’ll simplify a bit. Figure 13a shows the cane bending due to a pushing force F only in the x direction—there are no forces in the y or z directions. The small element of the cane shown in Figure 13b has moved laterally in the x direction and rotated about the y axis, perpendicular to the paper. The turning force created by F (i.e. the moment that bends the element) will be F multiplied by the vertical p. 60↵distance from the horizontal line of action of F to the element. Consequently the maximum turning force or bending moment in the cane caused by F, will be at the base of the cane and equal to the force F multiplied by the total height of the cane. The ground must be able to resist that moment or the cane will rotate into the ground.
Unfortunately, real wooden cane towers don’t just have one degree of freedom like our brick tower—they have all of the six we introduced in Chapter 2, in Figures 9 and 10. The forces and movements in the first three directions are perhaps the easiest to imagine. The first is simply pushing or pulling, a side to side movement in the x direction creating a tension or a compression in the x direction, as in Figure 10a. Then we may have a back to front movement in the y direction and an axial, up (pulling) and down (pushing or squashing) of the cane along its length in the z direction.
p. 61The movements in the second set of three directions are more difficult to visualize. They are the action forces on each face of the element of the cane that cause that element to rotate. These action forces are resisted by the reaction forces on the opposite face of the element and there is a pair for each axis. Together they cause the element to bend or twist as shown in Figures 10d–h.
We have now identified all of the forces acting along our six degrees of freedom on any piece of real structure we care to isolate. Whilst the elements in Figures 9 and 10 are rectangular blocks, in general our pieces of structure can be any size or shape, as seen in Figure 2. They can be anything from the small element of cane we have been examining so far, to any large piece of structure like the deck of a large bridge or the hull of a ship.
Some structures are relatively easy to determine
The next question is ‘How do we find out how big the forces and movements are?’ It turns out that there is a whole class of structures where this is reasonably straightforward and these are the structures covered in elementary textbooks. Engineers call them ‘statically determinate’—we began to look at them in Chapter 2. For these structures we can find the sizes of the forces just by balancing the internal and external forces to establish equilibrium. In other words equilibrium between external and internal forces is both necessary and sufficient to determine the magnitude of them all.
Let’s investigate this further by turning the cane on its side so we now have a cantilever beam. The cane is now like the diving springboard that we mentioned earlier. Imagine that F now represents a diver standing on the end of the horizontal springboard. We can find all of the internal forces in the springboard just by establishing equilibrium, because the cane tower and the springboard beam are statically determinate. In Figure 14a we have exposed the internal forces in the beam by p. 62↵yet another thought experiment. This time we mentally cut out an element of the beam and show the internal forces that balance the external force of F. The beam is of span length l and the element of length is δx. The element can be positioned anywhere along the length of the beam. In Figure 14a it is shown as a distance x from F. Let’s now focus on the piece of beam (length x) between F and the first cut. We can see that there has to be a vertically upwards shear force to balance the weight of the diver F. At the cut there must be an equal and opposite shear force reaction on the right hand end of element δx. This is also true of the force downwards on the right hand end of the stub of beam (l – x) connected to the support. Finally the support must also be able to supply a reaction F.
In just the same way, at the first cut there has to be a turning force or bending moment M1. This occurs because the forces F on the outer piece of beam (length x) create a couple of F times x. Without M1 this piece of beam would rotate freely. With M1 in p. 63↵place it will still rotate but the amount of rotation is constrained because of the way it is connected to the rest of the beam. The equal and opposite moment on the left hand cut is M2, bigger than M1 because the lever arm of the couple has increased from x to (x + δx) and so M2 = F x (x + δx). At the fixed left hand support the bending moment is M3 and equal to F times the whole span length of l so M3 = F x l.
Unfortunately many real structures can’t be fully explained in this way—they are ‘statically indeterminate’. This is because whilst establishing equilibrium between internal and external forces is necessary it is not sufficient for finding all of the internal forces.
To see why let’s change the problem yet again. This time we decide to prop up the end of the cantilever as shown in Figure 14b. An example is a balcony with a column supporting the outer edge. A person stands somewhere near the middle of the span. Unfortunately this seemingly small change makes finding the internal forces much more difficult. We cannot now calculate the internal forces in the beam just by establishing equilibrium. You will recall that in Chapter 2 we said that Gaudí focused on equilibrium and did not consider the two other requirements of modern structural analysis—the material properties (or so-called constitutive equations that are part of the subject of ‘strength of materials’) and the geometry of deformation (the so-called ‘compatibility equations’). These second two factors become important when structures are ‘statically indeterminate’.
Let’s think about this further using yet another example—this time the difference between a three-legged and a four-legged stool. When you sit on a three-legged stool, what are the forces in each leg? The stool is statically determinate, and the external and internal forces must be in equilibrium. The sum of the forces in each leg is equal to your weight. If everything is symmetrical and you are sitting centrally then each leg must take one-third of your weight, whether the floor is flat or not, and whatever the length p. 64↵of each leg. Of course if you are not sitting centrally then one of the legs may take a higher force than the others—indeed, if you were to sit exactly over one of the legs then it would have to take your whole weight (but this would be difficult to do in practice since you would have to balance yourself by constantly adjusting your position).
Now let’s contrast our three-legged stool with a four-legged one. If you are again sitting centrally does each leg take one-quarter of your weight? The answer is, no. The four-legged stool is statically indeterminate. You will begin to understand this if you have ever sat at a four-legged wobbly table, say in a restaurant, which has one leg shorter than the other three legs. There can be no force in that leg because there is no reaction from the ground. What is more, the opposite leg will have no internal force either because otherwise there would be a net turning moment about the line joining the other two legs. Thus the table is balanced on two legs—which is why it wobbles back and forth. If you, like most people, insert some packing to stabilize the table then each leg will have an internal force. Alternatively, if the floor is not too uneven or stepped, you can often find a stable position by turning the table around its central vertical axis. It is then more difficult to find the magnitude of each force because each leg has one degree of freedom but we have only three ways of balancing them in the (x, y, z) directions. In mathematical terms, we have four unknown variables (the internal forces) but only three equations (balancing equilibrium in three directions). It follows that there isn’t just one set of forces in equilibrium—indeed, there are many such sets. However, only one set will satisfy the other two requirements we mentioned earlier—the constitutive equations and the compatibility conditions.
Going back to our propped cantilever in Figure 14, we can now see why it is statically indeterminate. First of all, note that the local and global axes are the same and that there are no horizontal forces and hence no axial forces in the beam. Consequently there p. 65↵are only two degrees of freedom or only two ways of transmitting force within the beam—bending and shear. However, simply balancing the shear forces and the bending moments isn’t enough because there are three unknown constraining forces—the prop reaction, the shear force, and the bending moment at the fixed end. So we have more unknown variables than we have equilibrium equations.
Another example is shown in Figure 15a, a two span beam resting on two supports with a person standing in the middle of each span. Many highway bridges are two span beams. The beam is statically indeterminate because we have (again, no horizontal loads) two degrees of freedom (moment and shear) and three unknown reactions RL, RC, and RR. We can easily see that equilibrium is not enough by finding a set of forces in equilibrium that have incompatible displacements. For example in the middle diagram we obtain a set of forces in equilibrium just by thinking of each span as a simple beam with reactions at each end of W/2. Thus the total reactions are RL = W/2, RC = W/2 + W/2 = W and RR = W/2 as shown. So we have established equilibrium but, as the diagram shows, the corresponding displaced shapes can only happen if the beam cracks over the central support. If we enforce the condition that the beam doesn’t crack, then the slopes either side of the central prop will be compatible but the reaction forces and internal forces in the beam will change.
Another example of statical indeterminacy is shown in Figure 15b. This is a pin-jointed square truss with one diagonal, no external loads, and a second diagonal that is too short to fit by an amount u. The structure without the ill-fitting diagonal is statically determinate. However, if we stretch the second diagonal to make it fit then internal forces are set up in the whole truss. An unknown force R is created in the second diagonal as we stretch it. We can use the triangle of forces (Chapter 2) to calculate all of the other internal self-balancing tensions and compressions in terms of R. But we cannot easily calculate the magnitude of R, since the p. 66↵stretched second diagonal and the rest of the truss will deform until all of the forces are in equilibrium and all of the deformations (stretching of tension members and shortening of members in compression) are compatible.
Determining the indeterminate
Detailed analyses of these statically indeterminate problems are beyond the scope of this book, but we will look at some of the basic principles. As we have said, we have to apply the three conditions: equilibrium, constitutive relations, and compatibility of displacements.
To understand this, we need now to look at equilibrium slightly differently. You will recall from Chapter 2 that when a structure is in equilibrium it has minimum strain energy. First, it is p. 67↵important to note that this has nothing to do with the first law of thermodynamics and the principle of the conservation of energy. These laws require that the total energy is constant but do not refer to the way it changes. By contrast the various theorems that are used to analyse statically indeterminate structures refer not to total energy but only to changes in energy. Specifically they refer to maximum and minimum values. One way of thinking about this is to imagine an undulating surface, like a mountain range, with peaks and troughs. The surface represents our structure under various forces and displacements. The height of the surface represents the energy at a point on the surface, and the peaks and troughs represent points of equilibrium—points where the energy is a local maximum or minimum. Now imagine a ball rolling on this surface and perhaps balancing on a peak or settling in a trough. Wherever the ball is balanced we are going to change its position by perturbing it slightly. If it is balanced on an energy peak, rather like a ball balanced on a seal’s nose, then it will roll away from its equilibrium position (the seal can only maintain balance by moving its nose). This is called ‘unstable equilibrium’. It is the sort of equilibrium you experience when you stand on one leg or balance on two legs of a three-legged stool. We will look at it further in Chapter 5. For now imagine that the ball has settled in one of the troughs—perturb it and it rolls back. This is called stable equilibrium and is the most desirable form. However, in both cases, stable or unstable, the equilibrium is local minimum or maximum because a very large movement or perturbation might push the ball over a local peak and into a new and different part of the energy surface.
There is a further complication too. Strictly speaking, minimum strain energy as a criterion for equilibrium is true only in specific circumstances. To understand this we need to look at the constitutive relations between forces and deformations or displacements. Strain energy is stored potential energy and that energy is the capacity to do work. The strain energy in a body is there because work has been done on it—a force moved through a p. 68↵distance. Hence in order to know the energy we must know how much displacement is caused by a given force. This is called a ‘constitutive relation’ and has the form ‘force equals a constitutive factor times a displacement’. The most common of these relationships is called ‘linear elastic’ where the force equals a simple numerical factor—called the stiffness—times the displacement (see Figure 16a where the graph is a straight line). A material is elastic if, on decreasing the force (unloading), the graph follows the same path back. The inverse of the stiffness is called flexibility and if the graph is a curve then the elasticity is p. 69↵non-linear. The strain energy can be calculated for a given force because it is simply the area under the graph as shown. The area above the curve also turns out to be important in structural analysis and is called the complementary energy. For a linear elastic material the strain energy equals the complementary energy because the areas are the same. A last complication is that many real materials do not follow such simple curves (see Figure 16c) or their unloading path is a loop as Figure 16b—a phenomenon called hysteresis. In this case some of the energy is converted into other forms such as heat and sound (forms of kinetic energy) as permanent deformations are created.
Solutions using virtual work
There are two approaches to satisfying the three conditions, and both are very general and very powerful although they only apply to elastic materials as Figure 16a and not as in Figure 16b. They again depend on thought experiments. The first approach relies on imaginary or virtual displacements. We impose these virtual displacements on our structure. In other words we imagine every node moving an amount dictated by the chosen set of virtual displacements. We then see what happens when we vary them whilst keeping the forces unchanged. We do this to find the equilibrium state using a principle we’ll come to in a moment. The virtual displacements are chosen so that the external and internal displacements are compatible but they may result in forces that are not in equilibrium.
For the second approach we choose to apply imaginary or virtual forces. We impose these virtual forces on our structure. We then see what happens when we vary them whilst keeping the displacements unchanged. We do this to find the equilibrium state, again using a principle we’ll come to in a moment. The virtual forces are chosen so that the external and internal forces are in equilibrium but they may result in incompatible displacements, as we saw earlier for the two span beam of Figure 15a.
p. 70The two principles we use to find a state of equilibrium when we vary the virtual displacements or virtual forces are: first, ‘the principle of the minimum total potential’; and, second, ‘the principle of the minimum of total complementary potential’. As these statements imply, when the total potential or the total complementary potential is a minimum our structure is in equilibrium. Total potential consists of the internal strain energy of the structure plus the potential energy of the external action forces (loads) on the structure. Complementary potential is the internal complementary energy of the structure plus the potential energy of the external reactions. (In passing, note that the latter is often zero if the support reactions do not move).
We can use either of these principles with any arbitrary set of action and reaction forces that are in equilibrium—we’ll call that set F. We also use a quite separate, arbitrary, and independent set of external displacements of the loads and supports that are compatible with the internal displacements of the members—we’ll call that D. It is important to note that F and D are totally separate and independent—so we are at liberty to choose them both to our advantage in any particular calculation.
In a proof beyond the scope of this book it can be shown that for any F with any D the virtual work done by the external loads and supports will equal the virtual work done by the internal members as they strain. We call the first the external virtual work and the second the internal virtual work. The only condition on the proof of these theorems is that the displacements are small when compared to the dimensions of the actual structure.
How do we use the first principle of minimizing the total potential? We can choose F to be all of the known actual external forces on the structure. Then we can choose D to be any set of quite arbitrary (but convenient for our purpose) set of virtual displacements. We then calculate the strain energy in all of the p. 71↵internal members of the structure as they undergo the imposed virtual displacements. We calculate the potential of the external loads as the virtual work done by those loads in moving through the virtual displacements. We add the strain energy and the potential of the external loads to get the total potential. Then, as we noted earlier, we keep the forces constant and vary the virtual displacements so that, by using some appropriate mathematics, we can find the set of virtual displacements that minimize the total potential. At that stage the problem is solved because the external forces are the actual forces, the virtual displacements are compatible, we have used the constitutive relationships (in finding the total strain energy) and the structure is in equilibrium. The calculated virtual displacements are therefore the real displacements.
The second principle is minimizing the total complementary potential. Now we choose D to be the actual displacements of the structure. Then we choose F to be a set of quite arbitrary virtual forces and proceed in a similar manner. In the calculation the displacements are kept constant and the virtual forces are varied. Eventually using appropriate mathematics we can find a set of virtual forces that minimize the total complementary potential. So the problem is solved because the displacements are the actual displacements, the virtual external and internal forces are in equilibrium, we have used the constitutive relationships (in finding the total strain energy) and the total structure is in equilibrium. The calculated virtual forces are therefore one possible set of real forces.
The finite element method
Almost all modern structural calculations are performed on a computer. The most well known tool is ‘the finite element (FE) method’. To get a flavour of it here, we will consider only one version, which assumes linear elastic materials and minimizes the total potential. It is commonly called ‘the stiffness method’.
p. 72In brief, the total structure is divided into discrete finite elements or pieces of structure, as we have discussed before. The elements or pieces may be any size but are usually either a whole structural member, such as a beam, or imaginary divisions (of a plate, wall, or floor) in suitable shapes such as triangles and rectangles. The elements are connected only at joints or nodes. These are normally at the ends of a member or the corners of the imaginary triangles or rectangles. External forces are applied at the nodes only. Displacements within each element are calculated using an assumed mathematical relationship with the displacements at the nodes and continuity between adjacent elements. That relationship is called a mathematical shape or displacement function.
Constitutive equations are derived for each separate internal force in turn, along each separate local degree of freedom of the element, at a node. They are called ‘influence stiffness coefficients’ because they are calculated by isolating each node of each member. A virtual displacement is imposed along a degree of freedom at that node with all other forces kept constant. By equating the consequent internal and external virtual work a relationship is developed between a force and a corresponding displacement in a local pathway or degree of freedom. These separate relationships, of force equals a stiffness factor times a displacement, can then be added or assembled together in a large set of simultaneous equations. There is one equation for each global degree of freedom. We can think of each equation as representing the relationship between the external force along one global pathway at a node and the corresponding unknown displacement such that the known force equals the known stiffness times the unknown displacements at that node.
We humans find solving large sets of simultaneous equations very difficult. Computers do it easily. Once we have the unknown displacements the computer uses them to calculate all of the internal forces in the structure.
p. 73It is important to note that all of this can only be done because the matrix equations are linear—they involve no variables that are squared or cubed. We are relying on the principle of superposition—so called because the various parts can be superimposed together. This is not true for non-linear systems, so extended techniques have to be used. For example we can apply loads in small increments, then reset and solve the equations at each step.
There are three key points to take from this chapter. First, some simple structures (as in some of the more elementary textbooks) are statically determinate, so we can calculate the internal forces by just making sure all of the forces are in equilibrium. Unfortunately, more complex structures have too many unknowns to be solved this way—they are statically indeterminate. We find the internal forces in these structures by satisfying the three conditions of equilibrium, constitutive relations, and compatibility of displacements when the total potential is a minimum.
In the next chapter we’ll explore how these ideas are used to design and build very large structures such as jumbo jets and cruise liners.