3A2 Structural Design Notes PDF

2023 3A2 Structural Design DESIGN OF STEEL STRUCUTRES JOHN HICKEY  Course Content & Assessment 3A2 is composed of two sections, stell design and reinforced concrete design. Half the lecture time is dedicated to each section. Assessment is mainly via the exam, where there will be 3 questions on...

2023 3A2 Structural Design DESIGN OF STEEL STRUCUTRES JOHN HICKEY  Course Content & Assessment 3A2 is composed of two sections, stell design and reinforced concrete design. Half the lecture time is dedicated to each section. Assessment is mainly via the exam, where there will be 3 questions on each section, with students required to answer 4 questions in total, two from each. Additionally, for the steel part, there will be two design assignments and two laboratories. A Note on the Notes For the steel part of the 3A2 course there are three different sets of notes; the notes in this document, slides that are used to guide lectures and related examples that are done by hand in class. All of these will be available to students on Blackboard. All cover the same topics – however the ‘class slides’ are intended to guide the learning in class, rather than being a comprehensive set of notes. The notes in this document are intended as a complete and comprehensive resource that students can for revision. Note however that there are no examples in these notes – so you need to come to class to see the examples! There are also hundreds of videos on the internet that cover the content. Perhaps the most useful is this youtube channel; Application of Eurocode 3: https://www.youtube.com/playlist?list=PLePDbQ0-ThidXpNxttiK31qujlNdmcRdT 1 Introduction This part of 3A2 concerns the design of steel structures. What does this mean? The aim of the structural engineer is to design a structure to perform adequately. This is a vague sentence, so we need to discuss briefly the various terms. Firstly, we need to ask what qualifies as a structure? The Institute of Civil Engineers says that a structure is ‘anything that is constructed or built from different interrelated parts with a fixed location on the ground.' This includes both building structures (which of course range from a garden shed to a skyscraper) and non-building structures, such as bridges, wind turbine towers, transmission towers, dams and tunnels or more obscure structures like a signpost. Personally, I would argue that the ‘on the ground’ term in the definition is too restrictive; an aeroplane wing is also a structure – attached to a fixed point on the aircraft fuselage. Some would argue that thigs like the human skeleton or a tree are structures too – although this aren’t designed by an engineer. Maybe we should call these ‘non-engineered structures’. However, the exact definition of a structure that we use isn’t especially important for this course, where we will look primarily at conventional civil engineering structures like buildings and bridges. One challenge you may face as a young civil/structural engineer is the amount of new vocabulary you will come across, often without proper explanation. During these notes we will try to highlight a few of these, starting with ‘mechanism’. While talking about what constitutes a structure, it is relevant to point out that a structure is different from a ‘mechanism’. A structure provides some resistance against deformation when subjected to a force, whereas a mechanism offers none. For example, in a bicycle the frame is a structure (when you sit on it, it supports your weight without changing shape too much) but the pedals, cranks and chain ring form a mechanism (when you push the pedals they rotate the chain ring, without any resistance except for some friction). A structure fails when it becomes a mechanism – we won’t think about this statement too much now, but it is an idea we will touch on during the course. Secondly, what does it mean to design a structure? This involves choosing things such as: 1. An appropriate layout 2. Appropriate Materials 3. Appropriate element sizes For example, if we are designing a bridge we need to choose whether it will be an arch bridge, a suspension bridge, a cable stay bridge or something else (i.e. the ‘layout’). We also need to choose what materials we use to construct it (steel, concrete, timber, carbon fibre, glass?). The process of selecting the layout and materials is sometimes referred to as ‘initial design’ or ‘structural scheme design’, and there is often no one ‘correct’ answer; for example look at the variety of different bridges that successfully cross the Liffey in Dublin. Many factors, including constructability, sustainability, economic and social factors and aesthetics need to be considered by the designer. As there is no ‘correct’ answer, often when discussing this part of design you will see or hear statements along the lines of ‘this is where engineering differs from science’; there is a creative aspect in this part of structural engineering. Once we have a layout and an idea of what material we wish to use, we need to select the sizes of the various members or elements that form our structure. This later aspect of design is primarily what we will consider in this course - we could maybe call this design for serviceability and safety. Finally, our initial statement mentioned that we want our structures to perform adequately. We will discuss this more as we advance, but for now we can say that we want or structures to have sufficient strength and stiffness so that they 1) don’t fail catastrophically in extreme conditions and 2) are practically useable under everyday conditions. 1.1 Structural Design in 3A2 This part of 3A2 looks at structural design for steel structures. Design we had said involves selecting as appropriate layout, materials and element sizes. For this part of the course at least, our material is steel. The other part of the course looks at design using Reinforced Concrete (RC). In both sections of the course we will generally restrict ourselves to looking at the element sizing aspect; choosing an appropriate layout is considered more in your 3A9 project and 4A6(1) next year. So: 1. An appropriate layout – Mostly consider in 3A9 and 4A6(1) 2. Appropriate Materials – In this part of the course we stick to steel 3. Appropriate element sizes – This is the main focus here In order to judge ‘performance’ we will use the rules set out in the European standards (‘Eurocodes’), which specify how structural design should be conducted within the EU (and UK). We will discuss this in more detail later. 1.2 Some Basics in Structural Engineering The following section discusses some basic concepts for building structures. The aim of this section is to provide context for what follows in the remainder of the course. The purpose of a structure is to transfer applied force, or ‘loads’, to a fixed point, which for a building structure is generally the foundations. We can think of forces as ‘flowing’ through the structure, the sketch in Figure 1-1 gives a nice example. Applied loads include vertical loads due to gravity, including the weight of the structure itself and the weight of its occupants, as well as horizontal or ‘lateral’ loads due to wind, earthquakes or settlement. We will consider loading, and how exactly we quantify it, in more detail later on.  Figure 1.1 Sketch of typical ‘load paths’ in a building structure Most larger, modern building structures are formed from a structural frame. This is a network of vertical columns and horizontal beams and floor slabs, as sketched in Figure 1-2. In the vast majority of modern buildings this frame is constructed from steel or reinforced concrete, although sometimes we might use a different material like timber. The other parts of a building, such as most walls and windows, generally don’t have a structural function, i.e. they are not carrying load to the foundations. Figure 1-3 shows some examples of what reinforced concrete and steel frames look like during construction. In many cases when a building is completed we can’t see the structural frame (although with a little experience you can begin to guess how it is done).  Figure 1.2 Sketch of structural frames (a) (b) Figure 1.3 Examples of (a) reinforced concrete and (b) steel frame structural frames In this course we are of course thinking about steel frames. The pictures below show a couple of examples of steel frame structures that you may be familiar with. The first is One Molesworth Street, which is essentially located on Dawson Street (you may know it as the building where ‘The Ivy’ restaurant is situated) just outside the Arts block in Trinity. Figure 1-4 (a) shows the structural frame reasonably early in the construction program, you can make out the steel beams and columns. In Figure 1-4(b) the construction has progressed and the concrete floor slabs are now in place. The finished building is shown in Figure 1-5. Here we can see that a glass and stone façade covers the structure, to the point we can’t really tell any more that it is a steel frame.  (a) (b) Figure 1.4 One Molesworth Street, a steel frame structure, during construction Figure 1.5 The finished building at One Molesworth Street One Molesworth Street is a reasonably standard development. The second example here is the Main Stand at Anfield, Liverpool’s football ground, which is a little more exciting. In Figure 1-6 we can see the stand under construction. The roof is supported by a massive truss (this is of course because columns would block the view of spectators), which is being installed in the pictures. You can also see that the part of the stand behind the viewing area is a regular steel frame.  (a) (b) Figure 1.6 The Main Stand at Anfield during construction Figure 1-7 (a) shows some more pictures, where you can begin to appreciate how the structure works (have a go a sketching the load path). Figure 1-7(b) shows the finished stand, with the façade. Again it is hard to tell from the outside that it is a steel frame structure. (a) (b) Figure 1.7 The Main Stand at Anfield (a) under construction and (b) completed The idea in this part of 3A2 is that we will learn how to design the individual members (this is another structural engineering word, a member is any structural element – beam, column etc) within such a system. In other words, by the end of this course you should be able to select the correct size and type of steel section for any of the elements circled in Figure 1-8. When we do this we will look at lots of individual elements and simplified diagrams, but it is important to bear in mind that these represent part of larger structures. In 3A4, 3A9 and 4A6 you will think more about how these individual elements piece together. Figure 1.8 Examples of the type of structural steel elements we will consider designing in this course 2 Steel as a Material 2.1 Materials in Construction So far we have given a brief introduction to some basic ideas about structures and some examples of steel building structures. The next thing we need to think a little about is steel itself. Most of this will probably be a simple revision of your materials course from previous years. The vast majority of construction in the UK and Ireland uses either steel or concrete (usually reinforced ‘in-situ’ concrete that is poured on site but sometimes precast concrete that is cast in a factory and transferred to site). Both materials have advantages and disadvantages, some of which are summarised in Table 2-1. Table 2-1 Steel vs Reinforced Concrete Steel Reinforced Concrete Stronger than concrete – so we need to use less More durable that steel to do the same ‘job’. This means 1) smaller foundations, 2) that self-weight isn’t such a big issue and 3) steel is good for longer spans Quicker to erect – comes ‘ready to go’ and no Better performance in fire time required for pouring or setting Better quality control – made in a factory rather Can be made locally – no steel is produced in than on site. We can have more trust that the Ireland, so everything has to be imported. material is actually as we specify. More amenable than steel for alterations later in the design process 2.1.1 Sustainability Of course nowadays we need to consider sustainability and embodied carbon. It is surprisingly tricky to establish whether steel or concrete is better from this viewpoint without detailed calculations. Steel requires a lot of energy to produce initially, but can be recycled. There is a large amount of un-verified, and often contradictory, information on the internet. For example, if you google the question ‘Which is more sustainable steel or concrete?‘ the first search result says: ‘Steel. Of all the metals used in construction, steel is amongst the most environmentally-friendly.’ In contrast, googling ‘Is steel or concrete worse for the environment?’ leads to: ‘Steel. By weight, steel has a much higher embodied carbon footprint than concrete does’. This teaches us two things: 1) it is difficult to say whether steel or concrete is ‘better’ environmentally and 2) more generally, in the broader sustainability area there is a lot of poor quality or biased information (sometimes termed ‘greenwashing’) on the internet. One major advantage of steel is that it is recyclable; it can be melted down and reused again. While undoubtedly a good thing, it is important to remember that this is an energy intensive process (you need to generate lots of heat) so ‘recycling’ isn’t entirely ‘green’. Timber is a much more sustainable building material than either steel or concrete. However it has a number of problems, including performance in fire (which isn’t as bad as you might think, but is certainly worse than concrete), susceptibility to vibration problems and poor durability, especially in the Irish climate. The last point is quite important, as there is not a strong tradition of timber construction and many British and Irish engineers and constructors lack experience or confidence when working with timber (for example timber design is not really taught at Trinity!). Although there is an increasing amount of timber frame housing in Ireland, it is rare to see timber used in larger projects. Building regulations are quite conservative with regards to timber in Ireland, it is difficult to go over 3 storeys using timber because of fire requirements. In less humid climates, for example in Scandinavia or the Alps, there is a much greater tradition of using timber as a structural material and large multi-storey timber developments are not uncommon, in Norway for example there are a number of timber buildings taller than any building in Ireland. There are some people who predict that in future that for sustainability reasons timber will become the dominant structural material, but it is hard to tell exactly at this point. 2.1.2 Some Brief History While timber may or may not be the future, understanding the past is also increasingly important for civil and structural engineers. This is because in Europe and the US much of our infrastructure (rail, road, sewage etc.) relies on structures built over 100 years ago and designed without the aid of modern engineering or computers and for very different types of loads. A continuous challenge for engineers is to work out if these old structures are safe to use today. If they are not, what we do about it can be difficult as many are protected and we can’t simply knock them down and replace them. Historically, most buildings tended to be built using timber and stone, or ‘masonry’ as we call it technically – think of various historic cathedrals around Europe for example. Around the time of the industrial revolution engineers began to use iron in buildings and bridges. Cast Iron (where iron is cast in a mould) was dominant in the 19th century, however it is a brittle material and prone to sudden failure. Therefore, with cast iron structures that are still in use we need to be very cautious. Wrought Iron became the material of choice in the late 19th and early 20th century and generally has better properties than cast iron, except for being more prone to corrosion. By around the time of the first world war, steel had generally replaced iron. While we won’t consider iron directly in this course, you will hopefully emerge with an understanding as to why steel is a better engineering material and why cast iron structures require lots of caution. 2.2 Material Propertied of Steel Looking now specifically at steel, we need to address a number of points. Firstly, let’s examine the material properties. Figure 2-1 shows a typical ‘stress-strain’ curve for Mild steel (another engineering word, mild steel is the regular steel we use in everyday structural engineering, as opposed the special high-strength steel we might use in high-spec applications). We can see that while the stress is below a certain value, 𝜎𝑦 , the material behaves linearly and stress and strain (𝜖) are related by a constant we call Young’s Modulus, 𝐸: 𝜎 = 𝐸𝜖 2.1 (An interesting aside; the idea of a material having an elastic modulus was initially proposed by a Swiss mathematician, Leonhard Euler (pronounced ‘oil-er’), who was probably the most prolific mathematician and engineer of all time, particularly impressive as he was blind for most of his life, and who we will meet again when we talk about buckling and bending. However, his idea of elastic modulus got lost somewhere along the way before reaching Britain. A few years later, it was also proposed by an English scientist, Thomas Young. The dispute over who’s idea it is has been settled by calling it ‘Young’s Modulus’ but denoting it 𝐸, which stands for Euler).  Figure 2.1 Sketch of the general shape of the stress strain curve for mild steel While this relation applies, we say that the material is behaving linearly, and if we unload it (i.e. remove the stress) there will be no permanent deformation. Once the stress is increased beyond the yield point, the behaviour is no longer linear and strain increases quite a lot with little extra stress. Now the material is behaving nonlinearly or plastically. (There is a difference between linear/nonlinear and elastic/plastic – linear refers to the straight-line relation between stress and strain and elastic refers to the fact that it returns to it’s original shape when unloaded, but this distinction is not important for steel, which is either linear and elastic or nonlinear and plastic. Other materials, like some types of rubber can be non-linear elastic). As stress is increased the material reaches it’s ultimate strength and eventually fractures. Figure 2-2 shows typical stress strain plots for different types of mild steel we use in structural engineering. There are called ‘Structural Steels’ and are denoted by ‘S’ followed by their yield stress in 𝑁/𝑚𝑚2.  Figure 2.2 Stress strain plots for typical structural steels Table 2-2 gives the yield strengths of various structural steel. At this point it is important to note that engineers generally use the symbol 𝜎 to refer to stress, but Eurocode uses 𝑓𝑦 to denote yield strength. Table 2-2 Yield strength of common structural steels Steel Grade 𝑓𝑦 𝑁/𝑚𝑚2 𝑓𝑦 𝑁/𝑚𝑚2 𝑡 < 16𝑚𝑚 16𝑚𝑚 < 𝑡 < 40𝑚𝑚 S235 235 235 S275 275 265 S355 355 345 S440 440 430 An important point that we can make about steel from looking at these plots is that it is very ductile. In simple terms, it can stretch a lot after it yields but before it fractures; fracture strain is about 100 times the yield strain. This is very advantageous, as it gives lots of ‘warning’ before it fails. This is in contrast to brittle materials, like concrete, cast iron (there were a series of failures of cast iron railway bridges in the 19th centaury, https://en.wikipedia.org/wiki/Tay_Bridge_disaster ) or glass, which fail suddenly with little warning. The idea of brittle and ductile materials is illustrated in Figure 2-3. Another important point to make is that (as long as it is prevented from buckling) steel behaves similarly in tension and compression. This is not true for all materials, especially concrete which is much weaker in tension than compression.  Figure 2.3 Sketch of stress-strain plots for a brittle and ductile material For clarity, we also need to discuss briefly some of the terms we see associated with steel. As you learned in materials last year, steel is an alloy of iron, carbon and some small amounts of other elements. ‘Mild Steel’, also known as ‘Low carbon Steel’, which is what we typically use in structural engineering, has quite a low amount of carbon, maybe 0.05% to 0.25% by weight. Other types of steel have more carbon, generally this gives increased yield strength but poorer ductility. Stainless steel contains chromium, which prevents rusting, but makes the yield point less well defined, as illustrated in Figure 2-4. The method of manufacture also influences material properties. The majority of steel sections are manufactured using ‘hot rolling’. This, as the name suggests, involves rolling the section into the desired shape at a very high temperature, before being allowing it to cool. As it cools, the steel shrinks slightly meaning the manufacturer does not have complete control over the final shape, making it less suitable for high precision applications. Therefore hot rolled steel is generally used in applications where minutely specific dimensions aren’t crucial. ‘Cold Formed’ steel is essentially hot rolled steel that is further processed after cooling. This gives a better finish and better control and production accuracy meaning it is used for precision applications or thinner sections. The process of cold working also increases material strength through work hardening (see Figure 2-5). Due to the extra processing, cold formed sections are more expensive than their hot rolled equivalents. In civil engineering applications, cold formed steel is generally used for very thin members (as a brief example, if you look at the standard sections available in the UK and Ireland you can get a 25x25x2mm cold formed steel tube, but the smallest available hot rolled tube is around 40x40x3mm) or steel sheeting, but most common sections are hot rolled. For the purpose of this course, we will consider hot rolled sections only – this means that at later stages when you are looking at various codes and standards you should look at the sections for ‘Hot Rolled’ steel. Figure 2-4 shows the typical stress-strain curves for different type of steel.  Figure 2.4 Comparative tensile stress-strain plots for different structural steels Figure 2.5 An example of cold-formed thin steel sheeting A final implication of the fact that steel is manufactured in a factory environment, is that it is generally comes in standard section shapes and sizes. This is in contrast to reinforced concrete, which is poured into forms (moulds) on site meaning the engineer has a little more freedom to choose section dimensions (although various constraints still exist, you will discuss these in the other part of this course). Typical shapes for steel sections include hollow square (‘Square Hollow Sections’, or ‘SHS’), rectangular (RHS) or circular (CHS), universal beams and columns with ‘I’ or ‘H’ cross sections and various channel (‘U’ shaped) and angle (‘L’ shaped) sections. The standard sections available in the UK and Ireland are detailed in a manual that engineers call the ‘Blue Book’, which is freely available online (https://www.steelforlifebluebook.co.uk) and on blackboard. The Blue Book has a series of tables giving the various properties (e.g. mass, area, second moment of area and many more) of the standard available sections. Of course, non-standard section shapes can be manufactured, but this is generally more expensive than using standard sections. 3 Limit State Design & Eurocodes We have introduced some rather philosophical ideas about design and briefly discussed some material properties of steel. The last part of our introduction before we can begin to do some useful calculations involves discussing the design standards we employ, called ‘Eurocodes’, and the design philosophy underpinning them, which we call ‘Limit State Design’. The Eurocodes are the European structural design standards and specify how structural design should be performed in the European Union. There are many countries outside the EU that also use them, including the UK. There are 10 different parts to the Eurocode, which we call ‘Eurocode 1’, ‘Eurocode 2’ and so on. ‘Eurocode 1’, (sometimes called EC1, or to give it its full title: IS EN 1991 Eurocode 1: Actions on Structures) deals with how to calculate loads on structures. ‘Eurocode 3’ (EC3 or in full IS EN 1993 Eurocode 3: Design of Steel Structures) goes through how to design steel structures. Eurocode adopts what is called a Limit State Design approach. A ‘Limit State’ is a condition beyond which a structure no longer fulfils relevant design criteria. The condition generally refers to the loading on the structure, while the criteria refer to structural integrity, fitness for use, durability or other design requirements. In very simplistic terms, a limit state is generally reached when the load on the structure becomes too great, and this is not good! Eurocode requires the designer to check that any proposed design for two Limit States, called Ultimate Limit State (ULS) and Serviceability Limit State (SLS). ULS is concerned with life safety of people and the structure; clearly as an engineer we don’t want catastrophic failure like a bridge collapsing, even under a unusually large load. SLS is concerned with the day-to-day function of the structure, the comfort of occupants and appearance. As a designer we want to avoid excessive deformations, vibrations or damage to the structure under everyday loading. For example, we don’t want a tall building moving in the wind to the extent that some occupants start to feel sea-sick. 3.1 Examples of Different Failures 3.1.1 ULS Failures There are many examples of ULS failures of engineering structures throughout history; you will study some of these in detail in 4A6(1). A quite recent example you may remember is that of the Morandi Bridge in Genoa in Italy. This bridge collapsed during a rainstorm in 2018, killing 43 people. The exact cause is still debated (some people claim it was a fundamentally bad design while other say that the maintenance was inadequate) and the official court trial has started recently, but collapse was due to failure of pre-stressed concrete cables, probably from corrosion.  Figure 3.1 ULS failure of the Morandi Bridge, Genoa, Italy Figure 3.2 ULS failure of the Morandi Bridge, Genoa, Italy Another example of ULS failure closer to home is the Malahide Railway Viaduct, which partially collapsed due to damage to a bridge pier (resulting from a phenomenon called ‘scour’, where water washes sand and gravel out from around the pier) in 2009. Very fortunately, nobody was hurt.  Figure 3.3 ULS failure of the Malahide railway viaduct An example of a relatively new steel structure failing at ULS is the collapse of the roof of the AZ Stadium in Alkmaar in the Netherlands. This stadium was completed in 2007 but the roof collapsed under strong wind in 2018. Poor quality welding was blamed for the collapse. Figure 3.4 ULS failure of the roof of the AZ Alkmaar stadium in the Netherlands 3.1.2 SLS Failures These type of ULS failures are obvious, and clearly as structural engineers we want to avoid them. SLS failures are often more subtle. Two interesting examples are the Millennium Bridge and the Strata SE1 building, both in London. The Millennium Bridge was opened in the year 2000, but when large crowds of people began to walk across it vibrated excessively from side to side. The bridge was closed almost instantly, and remained out of service for almost two years while a large experimental program was undertaken to investigate the problem. Eventually, the bridge was retrofitted with a series of damping devices, like shock-absorbers in a car, to fix the issue. The bridge didn’t collapse, or was never in any danger of collapse, but the initial structure still failed as it wasn’t comfortable for users. This is a classic example of SLS failure. Figure 3.5 Millennium Bridge, London, which failed in SLS The Strata SE1 building is a slightly less famous but equally interesting example of SLS failure. The building contains three wind turbines at the upper levels, which were intended to meet some of the building’s energy needs (in fact the energy production ability would have been quite small, the turbines were more a marketing idea, what we sometimes call ‘greenwashing’, than a particularly effective energy producer). However, when the turbines were switched on the top third of the building became uninhabitable due to noise and vibration. The most economical solution here has proved to be to fix the turbines, so they don’t turn or produce any electricity. Again, the building was never in any danger of collapse, but it has failed to perform as initially envisaged, so again the design has failed in SLS.  Figure 3.6 Strata SE1 building in London, another example of an SLS failure When we carry out designs to Eurocode, we are required to check that the designs fail in neither of these two limit states. For regular structural members, this typically involves checking that the neither forces or deflections are too great. 3.2 Design to Eurocode The basic steps in Design are as follows: 1. Work out the load on the structure 2. Work out the forces and moments in each member (we call this ‘Structural Analysis’) 3. Check whether the member is adequate to carry these forces and moments We will talk through these briefly now. 3.2.1 Loading to Eurocode The Limit State design process in Eurocode requires us to firstly estimate the external loads that will act on a structure. The process for doing this is detailed in Eurocode 1. In structural engineering we traditionally divide loads into ‘dead’ and ‘live’ loads. Dead loads refer to loads that are always present on the structure, while live loads are those that change over time. The terms used in Eurocode are slightly different to these traditional terms, but essentially have the same meaning. Firstly, Eurocode uses the work ‘Action’ instead of ‘Load’ or ‘Force’. This is because ‘Actions’ include both forces and deformations (for example due to settlement or temperature). Secondly, Eurocode uses the terms ‘Permanent’ and ‘Imposed’ instead of the more traditional ‘Dead’ and ‘Live’. Therefore, when we use Eurocode, we say we are designing for ‘Permanent Actions’ and ‘Imposed Actions’. Eurocode uses the notation 𝐺 to refer to permanent actions and 𝑄 to refer to imposed actions. Here is the relevant paragraph from Eurocode 0: Some other types of actions, such as accidental actions (for example an explosion, or a vehicle driving into a bridge pier) or seismic actions from earthquakes (none in Ireland) are also considered, but we won’t think about them in this course. Permanent loads are calculated from the self-weight of the structure and the weight of any components that won’t change much over time, for example ceilings or air-conditioning or ventilation systems (some engineers term these permanent loads that aren’t from the structure itself as ‘superimposed dead loads’ or ‘super-dead’ loads). Here is what Eurocode 1 says about permanent loads, the list in paragraph (3) and paragraph (4) gives you an idea of the things we need to include in addition to the weight of the structure itself. Imposed loads account for the weight of people, furniture and other objects that can be moved or carried, like furniture in a building or vehicles on a bridge. Here is what Eurocode 1 says about Imposed Loads: Eurocode 1 gives typical values per m2 of live loads for almost every use of building structure imaginable. For example, a classroom needs to be designed for a live load of 3 𝑘𝑁/𝑚2 , while a hotel bedroom needs is assigned a live load of 2 𝑘𝑁/𝑚2. Figure 3.7 Imposed Loads, as specified by Eurocode 1, for different floor uses Finally, while we are discussing loads, it is good to take a minute to consider the question ‘how heavy is a kilo-Newton?’ 1 N is approximately 0.1kg (precisely 9.81 N = 1kg), so 1000 N (or 1 kN) is approximately 100 kg. Personally, I find a nice way to give some context is to remember that a big man, or an average male international rugby player (for those interested, a large back or a smaller forward, Robbie Henshaw or Josh van der Flier for example), weighs around 100 kg, or 1kN. That means when we are designing for an imposed load of 3𝑘𝑁/𝑚2 we are essentially designing for the scenario where 3 large men occupy every square metre.  Figure 3.8 1kN 3.2.1.1 Partial Factors Eurocode defines combinations of actions that we use in order to check that a design works at both Limit States. For Ultimate Limit State the main combination of actions we need to consider is: 𝐸𝑑 = ∑ 𝛾𝐺 𝐺𝑘 + 𝛾𝑄 𝑄𝑘 3.1 𝐸𝑑 is the termed Design Value of Actions. 𝐺𝑘 and 𝑄𝑘 are the values of permanent and imposed actions that we need to consider. 𝛾𝐺 and 𝛾𝑄 are termed partial factors (traditionally we might have called these factors of safety). For ultimate limit state these are equal to 1.35 and 1.5 respectively. Therefore, for ULS or design action can basically be thought of as ‘1.35 x Dead Load +1.5 x Live Load’. At SLS we generally take the partial factors, 𝛾𝐺 and 𝛾𝑄 to equal 1. This corresponds to the ‘everyday’ scenario. It is important to state that the code is quite complicated here, beyond the scope of this course, and in practice we are required to check a variety of different combinations of actions. For example, the code allows us to apply various reduction factors to imposed loads that account for the fact that there is a very low probability of experiencing peak values of all imposed loads at the same time. There are also additional combinations we need to consider for accidental and seismic actions. In a real building we may be required to consider over 100 different combinations of actions. However, to introduce the basic idea of partial factors, for this course we will just consider the following two simplified load combinations: At ULS: 𝐸𝑑 = ∑ 1.35𝐺𝑘 + 1.5𝑄𝑘 3.2 At SLS: 𝐸𝑑 = ∑ 𝐺𝑘 + 𝑄𝑘 3.3 See the examples in the lecture notes for application. 3.2.2 Structural Analysis In the design any structure, once we evaluate the design actions, we need to analyse the structure to figure out what the internal forces and moments are. This is a fundamental part of structural engineering and can be summarised as follows: we input the actions and the outputs are a set of forces, moments and deflections in every part of the structure. Once we have the forces, moments and deflections we are in a position to check that the structural members we use are adequate, i.e. we can ‘design’ our structure. Figure 3.9 Summary of structural analysis. We use the outputs in design. In this course, we will only consider structures that are relatively easy to analyse. You will learn more about analysis in 3A4 Analysis can be quite a complicated process; you started to learn how to do this in 1E7 for very simple structures, then advanced to slightly more complicated structures in 2E4. Next semester, you have an entire course devoted structural analysis in 3A4. In practice, structural analysis is nowadays generally carried out by a computer program, however setting up such programs correctly and checking the output is a skill in itself. In this course we won’t focus so much on structural analysis. Instead, for this course we will generally look at scenarios where (given what you learned in 1E7 and 2E4) you should be easily able to calculate the forces and moments anywhere in the structure. Some examples of the kind of structures you are expected to be able to analyse include: An axially loaded (tension or compression) bar A simply supported beam A simple truss structure Also, for this course, we will restrict ourselves to looking at ‘statically determinant’ structures. These are a special case where the forces and moments don’t depend on the structural members we select, so we only need to perform analysis once. For ‘statically indeterminate’ structures, we need to re-analyse every time we change the size of a structural member, but you will not be asked to analyse such structures in this course. However, having said that, it is worth revising one key concept before we proceed: namely the free body diagram. You saw lots of these in 1E7, and a few more in 2E4, and it is a concept that is central to structural engineering. As you will remember, the idea is that if we make an imaginary cut in a structure that is not moving, the part of the structure that we are interested in (we’ll call it the substructure) must be in equilibrium. Take the diagram in Figure 310. The loads in blue are called external forces. The force in red is sometimes called an internal forces or sometimes member force, and the key point is that this internal force is such that equilibrium of our substructure is maintained. Internal forces are more than a mathematical construct. The are very much real forces – the various molecules inside the structure will feel these forces. Or put another way, if you could somehow replace the part of the structure with your hand or finger or some other body part, you would definitely feel the forces! Applying the free body diagram technique allows us to work out what the internal forces are at any point in a structure. We can then use these forces to perform our structural design. Figure 3.10 Internal force in a bar tension 4 Design of Steel Members for Tension We have now completed our introduction to steel design. We have revised some basic ideas about strain, stress and material properties and introduced the Eurocodes and the ideas of SLS and ULS failure. We are now ready to begin designing! There are four main types of stress a structural member can be subjected to: Axial stress Bending stress Shear stress Torsional stress Axial stresses arise from tension or compression forces – i.e. when an element is pulled or pushed. Although steel behaves similarly in tension and compression, steel sections are very often prone to buckling in compression. This means that we consider design for tension and compression separately. This section considers design of steel members for subjected only to tension forces. On a very general level, we want the tension force in our steel member to be less than the design resistance. In fact this is the basic idea of structural design to Eurocode: we want resistance or capacity to be greater than the applied action. In Eurocode, we denote the design value of the applied force 𝑁𝐸𝑑 and the design tension resistance 𝑁𝑡,𝑅𝑑. We want our steel section to satisfy: 𝑁𝐸𝑑 ≤ 1.0 4.1 𝑁𝑡,𝑅𝑑 The applied tension force, 𝑁𝐸𝑑 , is an output from structural analysis. The design tension resistance is very closely related to the stress-strain curve we saw in Figure 2.1. Remember that axial stress is force divided by area. For a simple bar in tension, which we call a tie (another structural engineering word), we say that the stress should not exceed the yield stress. The design plastic resistance, 𝑁𝑝𝑙,𝑅𝑑 is the product of the area and the yield stress, 𝑓𝑦 (the notation in Eurocode can be a little off-putting, stress, not force, is denoted 𝑓). We also include a material partial factor, 𝛾𝑀0 to account for uncertainty in the material properties. We will touch on these in a minute. 𝐴𝑓𝑦 𝑁𝑝𝑙,𝑅𝑑 = 4.2 𝛾𝑀0 4.1 Bolted Members Many steel members have some holes in them for bolted connections (Figure 4.1(a)). This means that at certain points along the member, the cross-sectional area is reduced. Therefore, the stress increases (Figure 4.1(b)) near the hole. At these points Eurocode allows us to exploit the ductility of steel and we calculate the ultimate resistance using the ultimate stress, 𝑓𝑢 : 𝐴𝑛𝑒𝑡 𝑓𝑢 𝑁𝑢,𝑅𝑑 = 0.9 4.3 𝛾𝑀2 The area considered is the net area, which is the cross sectional area excluding bolt holes, as opposed to the ‘gross area’ in Equation 4.2. Compared to Equation 4.2, Equation 4.3 also considers additional uncertainty associated with ultimate resistance through the ‘0.9’ and the use of the larger material partial factor 𝛾𝑀2. In a case where we have a bolt hole, we need to make sure that the member won’t fail either near the hole or elsewhere in the section. Therefore, the tension resistance of the member, 𝑁𝑡,𝑅𝑑 , is defined as the smaller of 𝑁𝑝𝑙,𝑅𝑑 and 𝑁𝑢,𝑅𝑑 : 𝑁𝑝𝑙,𝑅𝑑 𝑁𝑡,𝑅𝑑 = min { } 4.4 𝑁𝑢,𝑅𝑑 (a) (b) Figure 4.1 (a) Example of a Square Hollow Section with bolt holes and (b) Concentration of tensile stress next to a hole For square, rectangular or circular cross sections the net area is calculated by simply subtracting the cross sectional area of the bolt(s) from the cross sectional area of the section. Figure 4.2 Example calculation of net and gross area In class we will do an interesting example where we examine the stresses and strains of a member that is allowed to yield in the vicinity of a bolt hole. We also do a laboratory where we look at the ways a bolted connection can fail. 4.2 Partial Factors You will have noticed that the formulae 4.2 and 4.3 contain factors 𝛾𝑀0 and 𝛾𝑀2. These are ‘material partial factors’ that are intended to account for uncertainty in material strength. For steel in Ireland we use the following values: 𝛾𝑀0 = 1.00 𝛾𝑀1 = 1.00 𝛾𝑀2 = 1.25 Note the 𝛾𝑀0 and 𝛾𝑀1 (which arises elsewhere in the code) are specified as 1.0 in Ireland. So in reality they make no impact on our designs. This is because steel is produced in a factory environment and we are generally confident that the steel we use will have a yield strength similar to what we expect it to have. For ultimate strength, we are a little less certain, so we have a slightly higher partial factor, which introduces some more conservatism (or caution) into our design. In my experience some students get a little bothered about the 𝛾𝑀0 = 1.00 term (‘why are we dividing by one?’). The reason we include it is that in other countries the national authorities are less confident about the quality control in steel manufacture, so they set this to something like 𝛾𝑀0 = 1.05, just to be sure. This is partly because it can become important in earthquake resistant design, which we don’t worry about here. All you need to remember is in Ireland 𝛾𝑀0 = 1.00! 5 Design of Steel Member in Compression Design for members in compression is more complicated than tension due to buckling. Most steel elements in compression are shaped and configured in such a way that they buckle in compression. 5.1 Buckling You were introduced to flexural buckling in 2E4 last year (there are other types of buckling we will deal with next year). Buckling is the sudden change in shape of structural component under compression. You may remember the concept of the Euler Buckling Load, 𝑁𝑐𝑟 , which as you can imagine was derived by Euler. Sometimes called the ‘critical load’, this is the (theoretical) load at which a strut (another engineering word, describing a member in compression) in compression will buckle. 𝜋 2 𝐸𝐼 𝑁𝑐𝑟 = 5.1 𝐿2𝑐𝑟 You will notice that the buckling load doesn’t depend on the yield strength of the material. Rather, it just depends on the elastic modulus, the second moment of area (which is essentially a property related to the geometry of the section) and the length and end conditions. Therefore, buckling is a failure mode that relates to stiffness rather than strength. 5.1.1 Why does buckling occur? The why of buckling is complicated and we won’t attack the mathematics of it here; we leave that to Euler. We will limit our discussion to some comments on stability. Buckling is what we call an instability – even if we don’t explore the maths it’s good to know the word and appreciate what it means. We say a system is stable if it remains in it’s original position when it is disturbed slightly. A ball at the bottom of a valley is a nice example of this, if we push it slightly up the hill, it will roll back down to it’s original place. In contrast, a ball on top of a hill is unstable. If we nudge it ever so slightly, it will roll down to the bottom of the valley. A strut in compression is a little similar. We won’t deal with the mathematics of why, but we can show (or Euler showed) that if the compressive force is great enough, the strut is unstable in a straight configuration. Any sort of disturbance causes the strut to jump to a stable position. We call this jump ‘buckling’. Figure 5.1 Example of a stable and unstable system – stuts are similar This is relevant for the modern engineer as it is very important to remember why buckling occurs when using computer analysis software. In the perfect world of computer software, the little disturbances that always exist in the real world and cause buckling don’t exist. This means that regular computer often simulations cannot capture buckling, unless we specifically include some imperfections in compression members. 5.2 Buckling In Design to Eurocode 3 We want to avoid buckling in our designs. Therefore, Eurocode 3 requires us to verify that the design buckling resistance 𝑁𝑏,𝑅𝑑 is greater than the applied compressive force: 𝑁𝐸𝑑 ≤1 5.2 𝑁𝑏,𝑅𝑑 The design buckling resistance is calculated using a ‘buckling reduction factor’, 𝜒: 𝜒𝐴𝑓𝑦 𝑁𝑏,𝑅𝑑 = 5.3 𝛾𝑀1 We can see that this is similar to the expression for plastic tension resistance (4.2), but with the additional factor 𝜒. We don’t need to consider the impact of holes or fastener openings in compression (so long as the hole is filled by a bolt). The questions we need to answer now are ‘what is 𝜒 and how do we calculate it?’ The Equation for critical buckling load can be rewritten so as it depends on something called ‘slenderness’. Slenderness, denoted 𝜆, is essentially a measure of how long and thin a member is, and is given by the formula: 𝐿𝑐𝑟 𝐿𝑐𝑟 𝜆= = 5.5 𝑖 √𝐼 𝐴 𝐼 The denominator, √𝐴, is called the ‘radius of gyration’, or 𝑖, and has various uses in engineering and values are available in the blue book for all standard steel sections. With some simple algebra, we can rewrite the Euler buckling load in terms of 𝜆 as follows: 𝜋 2 𝐸𝐴 𝑁𝑐𝑟 = 5.6 𝜆2 Eurocode presents slenderness in a ‘non-dimensional’ form, defining a ‘non-dimensional slenderness ratio’: 𝐴𝑓𝑦 𝜆̅ = √ 5.7 𝑁𝑐𝑟 We can see that this definition preserves the fact that slenderness squared is inversely proportional to the critical load, but 𝜆̅ accounts for yield stress, which is good as it makes it easy to compare different cross sections. With some manipulation using of Euler’s Equation 5.1, it is easy to show that: 𝐿𝑐𝑟 1 𝜆̅ = 5.8 𝑖 𝜆1 Where:  𝐸 𝜆1 = 𝜋√ 5.9 𝑓𝑦 Now, how does this relate to our buckling reduction factor, 𝜒? Examining Equation 5.7, we can see that effectively, 𝜆̅ is a comparison between yield load with no buckling (𝐴𝑓𝑦 ) and the critical buckling load. This is very similar to our buckling reduction factor, 𝜒 in Equation 5.3. If we equate Euler’s critical buckling load to 𝑁𝑏,𝑅𝑑 (and forget the partial factors for a second), it is easy to show that: 𝑁𝑐𝑟 𝜒= 5.10 𝐴𝑓𝑦 𝑁 Now we have two Equations containing an 𝐴𝑓𝑐𝑟 type term that we can easily compare: Equation 5.7 and 𝑦 Equation 5.10. Doing this allows us to write: 1 𝜒= 5.11 𝜆̅2 We can consider this the ’Euler’ or theoretical buckling reduction factor. If we plot this, we get something like this: Figure 5.2 Theoretical Euler Buckling Reduction Factor We call this kind of plot relating 𝜆̅ and 𝜒 a ‘buckling curve’. However, the Euler buckling curve does not consider the possibility for compressive material failure. For a member with a low slenderness, or what we call a ‘stocky’ member, the material will yield in compression before buckling can occur. We can call this the ‘crushing load’ and it becomes the relevant failure mode when 𝜒 goes above 1 (i.e. when the buckling ‘reduction’ factor actually works to increase the buckling load beyond the plastic resistance of the member), which for an idealized Euler strut occurs when 𝜆̅ < 1. This concept is illustrated in Figure 5-3. Figure 5.3 Combination of material failure in compression and theoretical failure in buckling So Figure 5-3 (b) here gives us a theoretical value for what we would expect the value of 𝜒 to be for a steel member in compression. For 𝜆̅ < 1 it will fail by ‘crushing’ (i.e. material yielding in compression) and for 𝜆̅ > 1 its capacity will be governed by buckling. Now let’s now look at how this theoretical failure curve compares to the values presented in Eurocode. The basis for the buckling curves in the code is a series of experimental tests carried out in the 1970s. In real strucutres, imperfections are unavoiable and experimentally it has been shown that buckling generally occurs at a lower value than the theoreitical buckling load, as can be seen in Figure 5-4. Figure 5.4 Experimental and theoretical buckling loads Eurocode presents a family of different buckling curves based on these tests. The equations that defines the Eurocode buckling curves is obtained by the following two Equations: 𝜙 = 0.5(1 + 𝛼(𝜆̅ − 0.2) + 𝜆̅2 ) 5.12 1 𝜒= 5.13 𝜙 + √𝜙 2 − 𝜆̅2 Looking at these Equations, it can be appreciated that the buckling curve depends on a function of 𝜆̅, as well as a factor 𝛼, which is termed the ‘imperfection factor’. However, Equations 5.12 and 5.13 are quite untidy, and it is difficult to immediately picture how exactly they relate to the theoretical values in Figure 5-3 (b). However, it can be shown that if we set 𝛼 to zero in Equation 5.12, Equation 5.13 equals the theoretical Euler Buckling curve that we derived previously. The 5 different buckling curves given in the code, 𝑎0 , 𝑎, 𝑏, 𝑐 and 𝑑, are shown in Figure 5-5. Each of these is associated with a different value of the imperfection factor, 𝛼. The imperfection factor varies with different types of sections and for different methods of production. The code provides instruction on which buckling curve to employ for various types of sections, as shown in Figure 5-6. Curve 𝑎0 represents cases with the least imperfections, with 𝑑 representing sections likely to have the most imperfections. Generally, for the purpose of this course it will be sufficient to read the appropriate buckling reduction factor off a table. You are not expected to remember and evaluate Equation 5.12 or 5.13, especially in an exam scenario, where you will be provided a table that represents the curves Figure 5-5 if you need it. However, you will need to remember how to evaluate non-dimensional slenderness.  Figure 5.5 Buckling curves in Eurocode 3 Figure 5-6 tells you which buckling curve to use. As you can see, this is depends on the section that you are looking at, and the axis about which buckling occurs (more on that in a second). Also, remember for this course we are sticking to ‘hot rolled’ sections; you’ll need to remember this when we look at SHS, RHS or CHS sections.  Figure 5.6 Instruction on which buckling curve to employ for various cross sections 5.2.1 Summary of Key Equations The previous section gives an overview of the theoretical background underpinning the buckling curves contained in Eurocode 3 and is perhaps a little long-winded. Design of struts to Eurocode 3 can be summarised by the following Equations: 𝑁𝐸𝑑 Check 𝑁 ≤1 𝑏,𝑅𝑑 𝜒𝐴𝑓𝑦 Buckling resistance: 𝑁𝑏,𝑅𝑑 = 𝛾𝑀1 Buckling reduction factor 𝜒 – obtained from buckling curve – easiest to read off curve o Function of 𝜆̅ 𝐿 1 ▪ 𝜆̅ = 𝑐𝑟 𝑖 𝜆1  𝐸 ▪ 𝜆1 = 𝜋√ 𝑓𝑦 Looking at these, to implement the formulae we need 4 values, 𝐸, 𝑓𝑦 , 𝐿𝑐𝑟 and 𝑖. 1. 𝐸 is the elastic modulus of steel – which we always take as 210,000 𝑁/𝑚𝑚2 2. 𝑓𝑦 is the yield stress. This is generally the number that accompanies the ‘S’ in the grade of steel, so for S355 we use 𝑓𝑦 = 355𝑁/𝑚𝑚2. However, note that this value reduces slightly as section thickness increases; see Table 2-2. 3. 𝐿𝑐𝑟 is the critical length of the member. You may remember in 2E4 seeing diagrams like that in Figure 5-7, which give different effective lengths for different end conditions. However, these idealized end conditions are rarely found in real construction. Eurocode 3 is not very clear on what effective length we should use in various scenarios. 𝐿𝑐𝑟 is generally assumed to just equal the length of the member, 𝐿, although there are some cases like a cantilever, when we need to know that it should be greater than 𝐿. For this course, it is probably advisable to assume 𝐿𝑐𝑟 = 𝐿 unless you can find a very good reason not to. Figure 5.7 Sketch showing effective lengths for different idealized end conditions. Eurocode is not clear on how we should choose effective length for realistic end conditions. Generally 𝐿𝑐𝑟 = 𝐿 is the best approach 4. 𝑖 is the radius of gyration. This is where things can get confusing – when you look at your tables there will often be two 𝑖 values. This is because there are two ways a member can buckle. In plain language a strut can buckle forward-back or left-right; in technical language we say it can buckle about the y-y or z-z axis. This concept is illustrate in Figure  In reality, unless it is somehow constrained, a strut will always buckle about the weaker axis. So, for buckling, we generally take the smaller 𝑖 value, unless there is some specific reason not to. Remember of the example of the long thin ruler that we did in class! Finally, there is a nice detailed walk-through of how to apply these Equations available at the following YouTube link: https://www.youtube.com/watch?v=5709a9retXo 5.3 Truss Design We are now at the point where we can design members for tension or compression forces. This allows us to design pin-jointed truss structures. There are structures made up of members, typically arranged in triangular patterns, that are either in tension or compression only. In pin jointed trusses, where loads are only applied at the joints between members, there is no bending or shear forces. Pin -jointed trusses are an effective way to span a long distance using a minimal amount of material. Therefore, they are commonly employed in situations where we require a large span (another structural engineering word, basically meaning length between supports). For example, in Figure 1-7 you saw the truss in the main stand at Anfield, that supports the roof over a span of more than 100m (i.e. with over 100m between supports). Similarly, steel trusses are often employed in bridges or for roofs in locations where we don’t want columns (like a swimming pool or a sports hall, as in Figure 5.8). Figure 5.8 Steel truss supporting the roof in a sports hall We cover a number of examples of truss analysis and design in class and you will design a truss bridge in your first piece of coursework. For analysis, it is good to know how to use both the ‘method of joints’ and the ‘method of sections’. It is also worth noting that many braced frame structures are also behave as trusses – see the 2022 exam for an example. 6 Bending The next section in the course considers bending. Bending, as you should know from 2E4, is key to how beams carry load. If we draw a free body diagram of a beam and equate vertical forces and moments clockwise and anti-clockwise moments, we find there are internal moments and shear forces in the beam that are required to provide equilibrium. This is demonstrated for a simply supported beam in Figure 6.1. Figure 6.1 Shear Force and Bending moment at a distance x from the left hand support of a simply supported beam with a point load at the middle We can repeat this process along the beam to draw a shear force and bending moment diagram, as illustrated in Figure 6.2. You should be very familiar with these ideas from 2E4.  Figure 6.2 Shear Force Diagram and Bending Moment Diagram for a Simply Supported Beam of length L with a point load P applied at mid-span For design, we are generally concerned with the maximum bending moment, i.e. 𝑃𝐿/4 for the simply supported beam with a point load at the middle or 𝑤𝐿2 /8 for a beam with a UDL of 𝑤 𝑁/𝑚. These formulae should be familiar to you from 2E4 (and you should know for the 3A2 exam). 6.1 Bending Stress In order to understand the ideas underpinning design for bending, we need to consider how bending stresses change over the cross section. We use a model called the Euler-Bernoulli Beam Theory or Classical Beam Theory to represent beam behaviour (more work from our friend Euler, along with a few of the Bernoullis; the Bernoullis were a Swiss family who made various contributions to modern engineering and mathematics over a few generations. You can think of them as a bit like engineering’s answer to the Kardashians). Euler-Bernoulli theory is a slight simplification, but it works perfectly well for the vast majority of beams that we encounter in steel structures. Consider the deflected shape of a beam under bending, as in Figure 6.3. We can see that near the top the beam has shortened from it’s original length, which means that the fibres (or molecules or however you wish to think about it) in this region are in compression. Near the bottom the beam has lengthened, which means that these fibres are in tension. There is a point somewhere in the middle, where the fibres have neither lengthened or shortened; we call this the neutral axis. The beam in Figure 6-3 is sketched in sagging, i.e. a ‘u’ shaped deflection. If it was in hogging, an ‘n’ shape, we would of course have the fibres above the neutral axis in tension and below in compression.  Figure 6.3 Deflected shape of a beam in bending Let’s consider a small section of this beam, of length 𝑑𝑠. It is bent as illustrated in Figure 6.4. Figure 6.4 Illustration of relation between strain and distance from the neutral axis, y Figure 6-4 (indirectly) illustrates the main assumption in Euler Bernoulli beam theory; which is that plane sections remain plain. This is a fancy way of saying that lines that the lines/distances we call 𝑦 are always at a right angle to the neutral axis. This might not be the case if the beam is somehow being twisted for example, but we won’t think about that here. It is shown in Figure 6-4 that the variation in strain in the fibres over the depth of the section is linear; that is: 1 𝜖= 𝑦 6.1 𝜌 𝑦 is the distance of any fibre from the neutral axis. We call 𝜌 the radius of curvature, but for now we can just think of it as a proportionality constant, so that we remember that strain is proportional to the distance from the neutral axis. Equation 6.1 is always true, irrespective of whether the beam is behaving linearly or not. If the beam is behaving linearly, we also know that 𝜎 = 𝐸𝜖. This means the stress also varies linearly over the cross section, and we can write: 1 𝜎= 𝐸𝑦 6.2 𝜌 This means that while the material is linear we can draw the variation in stress over the depth of the cross section as shown in Figure 6.5. Figure 6.5 Variation in stress over the depth of a beam in bending Figure 6.6 Stress acting on a small area dA that is a distance y from the neutral axis Now we consider the stress acting on a tiny area in the cross section of size 𝑑𝐴, as in Figure 6.6. The stress leads to a force acting on this tiny area (technically we should call it an ‘infinitesimally small area’), which is equal to the product of the stress, 𝜎, and the area, 𝑑𝐴. The resulting moment about the neutral axis is this force times the distance to the neutral axis, 𝑦. The sum of all the moments from all the little boxes across the entire cross section must equal the bending moment: 𝑀 = ∑ 𝑦𝜎𝑑𝐴 6.3 If we remember that sums over very small areas and integrals are the same thing we can write: 𝑀 = ∫ 𝑦𝜎(𝑦)𝑑𝐴 6.4 This is a really important equation, it ‘defines’ bending moment (and, arguably, if you can understand this the rest of this course is fairly straight-forward). Again, this is always applicable, irrespective of linearity. Note we write stress as 𝜎(𝑦), the ‘y’ in the brackets reminds us that stress changes as ‘y’ changes (i.e. 𝜎 is a function of ‘y’). 1 If behaviour is linear, using Equation 6.2, where had noted that 𝜎 = 𝜌 𝐸𝑦, substituting and rearranging we can write: 1 𝑀= 𝐸 ∫ 𝑦 2 𝑑𝐴 6.5 𝜌 The second moment of area, 𝐼, is defined as ∫ 𝑦 2 𝑑𝐴. This allows us to replace the integral with 𝐼. 1 𝜎 Furthermore, looking back at Equation 6.2, we can see that 𝐸 =. Thus by substituting we can write: 𝜌 𝑦 𝑀𝑦 𝜎= 6.6 𝐼 This is the Equation for bending stress in a beam where the material is behaving linearly, which is also an important Equation. From this, we can see that the maximum bending stress occurs when both 𝑦 and 𝑀 are maximised. 𝑦 is maximised at the extreme fibre of the section, and we can identify where 𝑀 is maximised from the bending moment diagram. This concept is illustrated in Figure 6-6, which shows some results for a simply supported beam with a point load in the middle and linear material behaviour. An important note here, the derivation of Equation 6.6 is ‘in play’ (i.e. it’s a possible part of a question) for the exam this year, even though it hasn’t been asked in previous years.  Figure 6.7 Location of maximum bending stress in a simply supported beam subject to a point load at mid-span 6.1.1 Elastic Modulus We can use Equation 6.6 to perform design. Remember, for design to be adequate, we want the capacity of our member to be greater than the applied moment. If we were to say that the maximum permissible stress is the yield stress, 𝜎𝑦 , Equation 6.6 could be manipulated to give a maximum allowable bending moment, or elastic moment capacity: 𝜎𝑦 𝐼 𝑀𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 6.7 𝑦𝑚𝑎𝑥 In order for a design to conform to this, we would have to choose a beam with a combination of 𝐼 and 𝑦𝑚𝑎𝑥 that is sufficiently large. As these two parameters are connected, we define a property that we call the Elastic Section Modulus, 𝑊𝑒𝑙 : 𝐼 𝑊𝑒𝑙 = 6.8 𝑦𝑚𝑎𝑥 The elastic modulus for standard steel sections is available in the tables in the ‘Blue Book’, meaning we can easily calculate the permissible moment for the section to remain elastic as: 𝑀𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 𝜎𝑦 𝑊𝑒𝑙 6.9  Figure 6.8. Extract from the ‘Blue Book’ showing the Elastic Modulus Note, for bending we generally use the larger value. This is because as engineers, we control which way our beam is orientated, and about which axis bending occurs. 6.1.2 Plastic Design However, as steel is a very ductile material, it may not be optimal to design beams so that the entire cross section remains elastic. Instead, we can allow some yielding to occur. In order to get an idea of what happens after steel begins to yield, we assume what is known as an ‘elastic perfectly plastic’ stress-strain relation, as illustrated in Figure 6-8. Figure 6.9 Idealized ‘Elastic – Perfectly Plastic’ stress strain relation This means that as strain increases beyond the yield strain, stress remains constant at the value of yield stress, 𝜎𝑦. Thus, when we look at the stress and strain actin on a cross section, we get a stress of 𝜎𝑦 when the strain exceeds the yield value, 𝜖𝑦. This is illustrated in Figure 6-9. Figure 6.10 Strain and Stress of a cross section of a beam where strain has exceeded the yield value Calculation of the moment across the section is now a little more challenging. We can see that the formula 𝑀𝑦 𝜎= no longer applies as there is not a linear variation in stress with 𝑦. Instead, in theory we need to 𝐼 do piecewise integration with the various limits defined in Figure 6-10. 𝑦0+ + 𝑦𝑚𝑎𝑥 𝑦0− 𝑀 = ∫ 𝜎(𝑦)𝑦𝑑𝐴 + ∫ −𝜎𝑦 𝑦𝑑𝐴 + ∫ 𝜎𝑦 𝑦𝑑𝐴 6.10 𝑦0− 𝑦0+ − 𝑦𝑚𝑎𝑥 Figure 6.11 Definition of terms in Equation 6.10 This integral can become painfully laborious for complicated section shapes, especially if the section is asymmetric and the ‘+’ terms are not the same as the ‘-‘ terms. However, it is relatively easy to evaluate for a rectangular section of breadth 𝐵 and depth 𝐷. In this case it can be shown (try it yourself) that: 1 𝑀= 𝜎 𝐵𝐷 2 (2 − 𝑛2 + 2𝑛) 6.11 12 𝑦 where, as illustrated in Figure 6-10, 𝑛 is a factor that quantifies how much of the section has yielded. If 𝑛 = 0, Equation 6.11 returns the elastic moment capacity: 1 1 𝑀𝑒𝑙𝑎𝑠𝑡𝑖𝑐 = 𝜎𝑦 𝐵𝐷 2 (2 + 0 + 0) = 𝜎𝑦 𝐵𝐷 2 6.12 12 6 From this, we can easily obtain the elastic modulus we discussed in section 6.1.1 as: 𝑀𝑒𝑙𝑎𝑠𝑡𝑖𝑐 1 𝑊𝑒𝑙 = = 𝐵𝐷 2 6.13 𝜎𝑦 6 If 𝑛 = 1, and Equation 6.14 gives the plastic moment capacity for a rectangular cross section. 1 1 𝑀𝑝𝑙𝑎𝑠𝑡𝑖𝑐 = 𝜎𝑦 𝐵𝐷 2 (2 − 1 + 2) = 𝜎𝑦 𝐵𝐷 2 6.14 12 4 This condition describes the formation of a plastic hinge. In terms of a stress diagram, this is illustrated in Figure 6-12. Once a plastic hinge is formed, the structure behaves as if there was a ‘pin’ connection at that point, i.e. it can no longer resist rotation. This means that unless there is redundancy in the structure, the structure will fail once a plastic hinge forms. If too many plastic hinges for, it will become a mechanism, i.e. start to deform without resistance. However, it is ok to allow plastic behaviour up to the point where the structure becomes a mechanism. Figure 6.12 Stress distribution for plastic hinge, and example of how a simply supported beam becomes a mechanism Given that it is ok to allow plastic deformation up until the beam becomes a mechanism, it makes sense to utilize this capacity in design. If we define a plastic modulus, 𝑊𝑝𝑙 , in a similar way to our elastic modulus, for a rectangular cross section we get: 𝑀𝑝𝑙𝑎𝑠𝑡𝑖𝑐 1 𝑊𝑒𝑙 = = 𝐵𝐷 2 6.15 𝜎𝑦 4 To calculate the plastic modulus of any section shape, we need to evaluate Equation 6.10 for the case where 𝑦0 = 0, or 𝑛 = 1, and divide by 𝜎𝑦. This gives: + 𝑦𝑚𝑎𝑥 0 𝑊𝑒𝑙 = ∫ −𝑦𝑑𝐴 + ∫ 𝑦𝑑𝐴 6.16 0 − 𝑦𝑚𝑎𝑥 Finally, we can compare the values of 𝑊𝑒𝑙 and 𝑊𝑝𝑙 to assess the extra capacity afforded by allowing the section to yield. We call the ratio of these two terms the shape factor for the section, denoted 𝑘. For a rectangular section this is: 𝑊𝑝𝑙 𝑘= = 1.5 6.17 𝑊𝑒𝑙 We can see that for a rectangular section, the shape factor is 1.5, meaning we can get 50% more moment capacity by allowing the beam to yield in bending. This explains why we allow the beam to yield. By doing so we get 50% (in this case) more capacity out of our beam, meaning we can use smaller, cheaper, lighter and less carbon-intensive sections for our design. The blue book gives values for the plastic modulus for standard steel sections. Therefore, if we want to work out the plastic moment capacity of a section we simply need to look up this value and multiply it by the material yield strength. 𝑀𝑝𝑙 = 𝑊𝑝𝑙 𝜎𝑦 6.18 6.1.2.1 Asymmetric Sections In the preceding examples we always knew where the neutral axis was, as the sections were symmetrical. But what if the section we are interested in is not symmetrical, like a ‘T’ section? We need to go back to first principles and remember equilibrium to figure out where the neutral axis lies. We need to work out the point where compression force equals tension force. The tension force is given by: 𝑦𝑛𝑎 𝐹𝑡 = ∫ 𝜎(𝑦)𝑑𝐴 0 While the compression force is given by: 𝑑 𝐹𝑐 = ∫ 𝜎(𝑦)𝑑𝐴 𝑦𝑛𝑎 These must be equal: 𝑦𝑛𝑎 𝑑 ∫ 𝜎(𝑦)𝑑𝐴 = ∫ 𝜎(𝑦)𝑑𝐴 0 𝑦𝑛𝑎 First, let’s assume the section is fully elastic, so our stress profile has a linear variation such that: 𝐸𝑦 𝜎(𝑦) = 𝜌  𝑦𝑛𝑎 𝑑 ∫ 𝑦𝑑𝐴 = ∫ 𝑦𝑑𝐴 0 𝑦𝑛𝑎 This can be rewritten as: ∑ 𝐴𝑖 𝑦𝑖 = 0 𝑖 which defines the ‘centroid’ of a section. Therefore, for elastic behaviour we say the neutral axis is at the centroid of the section. If the beam is fully plastic, the neutral axis is given by: 𝑦𝑛𝑎 𝑑 ∫ 𝜎𝑦 𝑑𝐴 = ∫ 𝜎𝑦 𝑑𝐴 0 𝑦𝑛𝑎 𝑦𝑛𝑎 𝑑 ∫ 𝑑𝐴 = ∫ 𝑑𝐴 0 𝑦𝑛𝑎 This latter equation defines the ‘centre of area’. In other words, the area above the neutral axis is equal to the area below. Practically, we can calculate this as: ∑ 𝐴𝑖 = 0 The centre of area and centroid are co-incident for a symmetric section, but are different for an asymmetric section. 6.1.3 Key Take Aways The preceding section covers a lot of ground. The key points can be summarised as follows: 1. Assuming Euler-Bernoulli beam theory, for any beam a. Strain is proportional to the distance from the neutral axis b. The moment is equal to the integral (i.e. sum) of the stress by area by distance from the neutral axis: 𝑀 = ∫ 𝜎𝑦𝑑𝐴 2. If the material is behaving elastically: 𝑀𝑦 a. Bending stress is given by the formula 𝜎 = 𝐼 b. The blue book gives values for the elastic modulus (𝑀𝑒𝑙 = 𝑊𝑒𝑙 𝜎𝑦 ) for standard steel sections. Therefore, if we want to work out the elastic moment capacity of a section we simply need to look up this value and multiply it by the material yield strength. 3. If the material is behaving inelastically: 𝑀𝑦 a. 𝜎 = 𝐼 no longer applies b. The blue book gives values for the plastic modulus (𝑀𝑝𝑙 = 𝑊𝑝𝑙 𝜎𝑦 ) for standard steel sections. Therefore, if we want to work out the plastic moment capacity of a section we simply need to look up this value and multiply it by the material yield strength. c. The extra capacity we get by allowing the beam to yields is represented by the a term called the shape factor. A quick overview of the general purpose is also worthwhile at this point. The reason why we do all of the above is that we are trying to understand what the maximum stress a beam in bending can resist is. If we have this, we can then select a beam capable of sustaining whatever stress will be applied without failing. Ultimately, this will allow us to design steel beams that are integral to many structures. How we go about figuring out the maximum stress depends on whether the beam is elastic or plastic. But the blue book gives us elastic and plastic moduli for standard steel sections, which make our lives quite simple. And the purpose of Section 6.1 is to try to explain why this is so! 6.2 Design for Bending to Eurocode 3 The preceding section of the notes brought us on a long journey through bending theory, and on to the concepts of plastic and elastic modulus. We devoted some time to thinking about how we might calculate these. Now we need to bring ourselves back from theory towards design. When we considered plastic modulus and shape factors, we have shown that we can increase the moment capacity of a beam by allowing it to yield and behave plastically. The question we need to consider for design is whether it is wise to do so? The answer to this question depends on the beam cross section itself. If parts of the cross section are very slender, local buckling can occur if the compressive stress is too great. In bending, this can happen to parts of the cross section on the compressive side of the neutral axis (e.g. the top flange in sagging – two more structural engineering terms). Some examples of local buckling are given in Figures 6-11 and 6-12. Figure 6.13 Examples of local buckling in a beam flange  Figure 6.14 Example of local buckling in a beam web Local buckling is obviously undesirable. Members with more slender flanges or webs will experience local buckling at lower stresses. However, if we can show local buckling won’t be an issue, it is safe to exploit the plastic capacity of the beam. So the question we need to answer for design becomes ‘how likely is local bucking’? The answer to this lies in ‘section classification’. 6.2.1 Section Classification Eurocode 3 divides cross sections into four different classes, as defined in Table. Class 1 cross sections are those which are least prone to local buckling, while Class 4 sections are those which are most likely to experience it. Graphically, this idea is illustrated in Figure 6-14. Table 6-1 Section Classifications according to Eurocode 3 Class 1 Form a plastic hinge with the rotation capacity required from plastic analysis without reduction of the resistance. Class 2 Can develop their plastic moment resistance, but have limited rotation capacity because of local buckling. Class 3 Stress in the extreme compression fibre of the steel member assuming an elastic distribution of stresses can reach the yield strength, but local buckling is liable to prevent development of the plastic moment resistance. Class 4 Local buckling will occur before the attainment of yield stress in one or more parts of the cross-section.  Figure 6.15 Four different section classifications as defined by Eurocode 3 From Figure 6-14, it can be appreciated that design for plastic moment capacity is ok for Class 1 sections, but Class 4 sections might need more consideration, even in cases where we don’t exploit plastic capacity. It’s less clear what we should do with Class 2 or 3 cross sections. The question we need to answer now however is how we do the classification. The answer depends on the width to thickness ratio of the various parts of the section subjected to compression. Various parts of a section – such as the web and flange in a beam – can be different classes. The overall section classification is determined by the least-favourable section; for example a beam with a class 1 web and a class 2 flange would be categorised as a Class 2 cross section. Guidance on the width to thickness ratios for the various classes is given in Table 5.2 of Eurocode 3. We will just focus on the first column – pure bending – in this course. Sheet 1 of the Table deals with ‘internal elements’, like beam webs, which both ends are restrained. Sheet 2 deals with ‘outstand elements’, where there is one free end, like a bema flange, and local buckling is more likely and consequently the ratios for each class are lower. For common steel sections, the ratio of 𝑐/𝑡 are given in the Blue Book. For example for Universal Beams. We do a number of examples in class where we perform section classification and evaluate moment capacity and resistance. This will definitely be a question on the exam in some form, so make sure you know it well! 7 Shear The final action we consider in 3A9 for the design of beams is shear. You should remember from 2E4 how to draw a shear force diagram. Figure 7-1 shows the SFD for a simply supported beam subjected to a point load. A shear force diagram allows us to find the point of maximum shear in a beam, which generally lies near the supports. Figure 7.1 Shear Force Diagram for a simply supported beam subjected to a point load at the centre 7.1 Shear Stress Distribution over Section The distribution of shear stress over the depth of a section is more complicated than bending stress. You might remember the formula from 2E4: 𝑉 ∫ 𝑦𝑑𝐴 𝜏 = 7.1 𝐼𝑡 𝜏 is shear stress, 𝑉 is shear force, 𝐼 is the second moment of area and 𝑡 is thickness. ∫ 𝑦𝑑𝐴 is the first moment of area. Sometimes this equation is presented a little differently, but the general idea stands. Unlike bending stress, we won’t worry about deriving this Equation. It can be hard to visualize how this looks exactly over the depth of a cross section. Figure 7-2 shows the variation over a rectangular and I sections.  Figure 7.2 Distribution of shear stress over the depth of various beam cross secitons Using Equation 7.1 in design can become complicated, particularly for an I section. However, If we look at the shear stress distribution, we can see that: 1) The shear stress is high over only part of the cross section 2) There ‘high’ sear stresses are reasonably uniform This leads to the concept of a ‘shear area’, as sketched in Figure 7-3 for an I section.  Figure 7.3 ‘Shear area’ for an I cross section 𝑓𝑦 At yield, the average shear stress in the shear area is taken to be. This is a simplification of some more √3 complex solid mechanics (Tresca and Von Mises failure criterion if you’re interested) that we won’t go into in this course. Similar to before, for an adequate design, the design action, 𝑉𝐸𝑑 , must be less than the design plastic shear resistance, 𝑉𝑝𝑙,𝑅𝑑 : 𝑉𝐸𝑑 < 𝑉𝑝𝑙,𝑅𝑑 7.2 The design plastic shear resistance is given by: 𝑓𝑦 𝐴𝑣 ( ) 𝑉𝑝𝑙,𝑅𝑑 = √3 7.3 𝛾𝑀0 The shear area of rolled I or H sections is given by: 𝐴𝑣 = 𝐴 − 2𝑏𝑡𝑓 + (𝑡𝑤 + 2𝑟)𝑡𝑓 7.4 Where the various symbols are as in universal beam tables. Conservatively, this can be estimated as: 𝐴𝑣 = 1.04 ℎ 𝑡𝑤 7.5 Similar expressions are given or other cross section shapes, such as RHS or angled sections in Eurocode 3. 7.2 Deflection at SLS The final check we need to make in the design of a beam is for deflection at SLS (serviceability limit state). We need to make sure that under everyday loading the beam won’t deflect too much. This is generally a reasonably easy calculation. 7.2.1 Steps in the Design of a Beam This final example unifies many of the ideas we have encountered in the course. We are looking to design a beam. We must check calculate the design actions and then check the beam has adequate capacity to resist bending and shear at ULS and will also be ok at SLS. The maximum deflection of a simply supported beam under a point load, 𝑃, at the centre is given by:  1 𝑃𝐿3 Δ= 7.6 48 𝐸𝐼 While the maximum deflection of a simply supported beam under a UDL, 𝑤, is given by: 1 𝑤𝐿4 Δ= 7.7 385 𝐸𝐼 These formula allow us to work out what the deflection will be. Note however, that we do this at SLS, so there should be no partial factors applied to the loads for deflection, as per Equation 3.3. Once we have the deflection evaluated, we can compare the value to the allowable deflection specified by Eurocode. Eurocode gives us maximum allowable values for deflection under variable loads, which it terms 𝛿2 , and deflection under combined permanent and variable actions, termed 𝛿𝑚𝑎𝑥 : 𝐿 𝛿2 = 7.8 300 𝐿 𝛿𝑚𝑎𝑥 = 7.9 250 When we design a beam, we need to check that the deflections given by Equations 7.6 and 7.7 are compatible with these limits. 7.2.2 Summary of steps in beam design 1) Calculate the design action 2) Evaluate the maximum bending moment and shear force 3) Choose a cross section we think might work a. Classify the section – can we design plastically? b. Evaluate the bending moment capacity at ULS c. Evaluate the shear capacity at ULS d. Evaluate the deflection at SLS 8 Embodied Carbon The final section of the notes deals with Embodied Carbon. This is clearly a very important topic and is becoming more and more relevant in industry. This is the first year this has been on the course, and, for this year, the examinable information won’t be very taxing, but the information in this section will be needed for your assignments and might also help you in 3A9. This section is based on the Institute of Structural Engineers Guide ‘How to Calculate Embodied Carbon’ (we will call this ‘the guide’ for short) and you will have access to the accompanying calculation tool should you need it. Like many energy calculations, the mathematics behind embodied carbon calculations are simple; you take a figure for embodied carbon per kilo and multiply it by the weight of steel in your design. However, getting a good estimate for the embodied carbon is the tricky part and often depends on a wide variety of different factors. Thankfully, the guide does a lot of this work for us, so the calculation of embodied carbon is actually relatively simple. The guide adopts a life-cycle, or ‘cradle to grave’, approach to calculating embodied carbon, meaning it attempts to quantify the amount t of carbon used at all stages in the life of a structure. As can be seen in Figure 8.1, it splits the life of a structure into four sections, A,B,C and D, which correspond to construction, use, end of life and reuse respectively. For now, we will just focus on section A1-A5, i.e. to practical completion of a structure. Emissions are measured in equivalent carbon, or kgCO2e/kg, which the guide calls a ‘Carbon Factor’. Figure 8.1. IStructE classification of Embodied Carbon The majority of embodied carbon emissions associated with a project are contained in A1-A3. Table 2.3 gives Carbon Factors for steel. We can see that the recommended ‘Embodied Carbon Factor’ (ECF) depends on the type of steel section used, as well as it’s origin. This is because in different countries different amounts of recycled steel are typically used. For open sections, such as a UB, we get somewhere around 1.7kg The discussion of embodied carbon in the guide for steel is reproduced below: Below is an example of the embodied carbon (A1-A3) for 1 ton of steel. Four different CEfs are considered. For context, the estimated embodied carbon produced in a year by a new car on average and 1000 litres of milk are also shown. We can see that a ton of steel requires about a similar amount carbon to produce as these. Emissions from transport (A4) are generally fairly small in comparison to production. Values can be obtained from table 2.4. This is relevant for steel in Ireland, as we have no steel production here, so all the steel we use has to be shipped form the UK (or further afield). However, in the grand scheme of things, transportation actually isn’t a very energy intensive activity. Below is an estimate of the transport emissions for 1 ton of steel manufactured in Port Talbot (the largest steel plant in the UK) and transported to Dublin. We can see the