ࡱ>   eiwY bjbjWW ==Ob]bbbb`}L6pP(]2\0UCWCWCWCQCLFL@J$OQdJ;"]dJ_:bbSp_:_:_:ZbRUCZ \ (bbbbUC_:._:A:%B&"UC tu/ KC  The relevance of materials and structures Every building has to be made out of real materials. The design process includes selecting appropriate materials. Depending on their location in the building, the materials have to perform certain functions. So we need to understand how materials perform (e.g., durability, strength, ability to be worked and joined, etc). Every building is subjected to loads. Loads come from the materials themselves, the people and things in the building, and from wind, earthquakes etc. Therefore the building needs a structural system that will safely resist all the loads and transfer them to the ground. A structural system is described in terms of its elements (columns, beams, slabs, walls etc), their materials and sizes, the way they are located, and the way they are connected together. Contents 1 INTRODUCTION 3 2 STRUCTURAL SYSTEMS 9 3 LOADS 15 4 BEAMS 22 5 MATERIAL BEHAVIOUR AND SECTIONS 29 6 TRUSSES 38 7 AXIALLY LOADED MEMBERS 42 8 MEMBER SELECTION 46 9 CONCRETE MAKING AND TESTING 47 1 INTRODUCTION Descriptions and administration Outline of Unit-of-Study Assessment Reading Material This unit-of-study introduces the basic principles of building structures and the structural use of materials. The concepts, methods of analysis and techniques that are introduced form the bases for subsequent units-of-study in this area. The unit-of-study introduces the idea of a structural system for the building as a whole to provide the necessary stability, strength and stiffness. It examines the loads, the behaviour of each of the elements of the building, and the contribution of the materials and sections to that behaviour. The unit-of-study gives the opportunity to examine some aspects of the adequacy of the structural system in a simple building designed by the student. The unit-of-study then introduces loads and forces, and studies the equilibrium of elements and freebodies, including moments and the resolution of forces, and the graphical representations of internal actions in shear force and bending moment diagrams. The requirement of structural performance in linear structural systems is introduced through the properties of cross-sections of members, and the selection of sections in relation to the properties of the material. The unit-of-study provides the knowledge to select structural assemblies of linear elements, and to select sizes for some of these elements, for simple configurations and loading conditions. Unit-of-study structure The unit-of-study consists of lectures, studio tutorial sessions, laboratory classes and computer modelling exercises. Many of the exercises will be closely related to the students concurrent work in Design. Assessment and criteria for passing Assessment will be based upon the following activities, spread throughout the semester. Progress will be indicated to students as each part is assessed. The submission dates will be advised as each assignment is issued Attention is drawn to the Facultys policy on late submissions, issued on enrolment. Class quizzes (4-off) Lab reports (3-off) There will also be a number of non-assessed tutorials, laboratory classes and computer tutorials which are intended to form part of the learning process. Originality, copying and plagiarism Attention is drawn to Facultys policy document on copying and plagiarism. Students are encouraged to discuss the issues involved in all aspects of the Unit and assist each other to learn, but all work submitted must be the students own intellectual property unless specifically referenced to another source. Reading material Topics similar to this are taught in Architecture schools throughout the world, and most of the teachers have tried to write textbooks or sets of notes. They all have different starting-points, or different emphases, or a different idea of what the architecture student should be able to do at each stage of study. Therefore there are very many textbooks available. Those listed are only a small sample. You will hardly ever want to read a book from cover to cover. Skim the Contents and the Introduction, try to understand where it is leading, and dip into the parts that you are concerned with at the time. If you get stuck, read the part once more to see if it comes clearer, otherwise try another book or talk to someone about it. Recommended books We recommend that you become familiar with these books in the Library, and suggest that you should own at least one of them. In the notes that follow, references to pages or sections in CGW, Schodek, and Wyatt refer to these three books. Where there are references to other books, the book title is also given. Cowan, H J, Gunaratnam, D, and Wilson F (abbreviated as CGW in these Notes)Structural SystemsDept. Architectural and Design Science, Uni of Sydney, 1995 (In the Arch Library at 624.17/41)An easy-to-read but detailed book with many diagramsWyatt, K JPrinciples of Structure NSWUP, 1979 (reprint 1992)A concise introductory book. Some examples, worked and for practice.Schodek, D LStructuresPrentice-Hall, 1983Useful for several years of Structures, many worked examplesZalewski, W and Allen, EShaping Structures / StaticsJohn Wiley and Sons, 1998Easy to read. Many illustrations of interesting buildings.Both American and SI units.Salvadori, M G, and Heller, RStructure in ArchitecturePrentice-Hall Editions 1975, 1986A good descriptive book, concentrating on the principles. Easy to read. Not much numerical material. It is in American units. Additional references We recommend that you acquaint yourself with a number of books on the subject, so that you have somewhere to go for a particular topic when you need to. There is a vast range the following are just a few.Cowan, H J and Smith, P RThe Science and Technology of Building MaterialsVan Nostrand Reinhold 1988Chapters 1,2,3,8 and 20 may be useful on Materials.Morgan, W, Williams, D T, and Durka, FStructural MechanicsLongman, 1986A useful book covering structural mechanics from the simple concepts through to realistic structures. Contains examples if you want more practice.Cowan, H J Architectural StructuresPitman 1980Easy to read, many worked examplesCowan, H J (Ed)Handbook of Architectural TechnologyVan Nostrand Reinhold 1991A Handbook is used for reference, after you have studied a subject. It has more information and data, with less explanation, than a textbook. Chapter 11 deals with Structural Systems. As you progress in the course, find a handbook that you can use for quick reference to things that you used to know. Codes of Practice You should familiarise yourself with the location of the Australian Standards in the Library (they are also available electronically in the Library). In the lectures on Loads, we will refer to the Loading Code (AS1170). This Code will come up from time to time throughout the structures stream of units-of-study. Some students prefer to have their own copy, but as Codes are amended from time to time, most prefer to use the Library copies.In some of the laboratory experiments, reference will be made to Codes of Practice for the testing of materials.  Concepts and units Units and numbers in the SI systemForce, mass and weightVectors and momentsStress and strainElastic, brittle and ductile behaviourEquilibrium SI system of unitsNote that the building industry deals in m and mm, (not cm).Notation for large and small numbersIt is easier to use powers of 10 for very large or small numbers. The building industry uses engineering notation, i.e. 103, 106, 10-3 etc, instead of scientific notation, which uses every power of 10. By using multiples of 103, it is easy to move between metres and millimetres, kilograms and tonnes, etc.Force (external or internal to the structure)The formal definition of a force involves mass and acceleration. It is easier to think of the intuitive idea of force. Most forces acting on buildings are caused by Gravity acting on the building and its contents. The force (load) is mass times the acceleration of gravity (g = 9.81 m/s2), and is in Newtons (or kiloNewtons). Wind acting on a surface. (The moving air is being decelerated.) Earthquake. (The movement of the ground is trying to accelerate the building.) Forces acting on the building are resisted by internal forces in the structure.Mass and weightIn common language we use these terms interchangeably, but mass is a property of an object (how much stuff there is in it), while weight is a property of the object in a particular force field (generally, in the earths gravity field). Weight is a force (in Newtons etc.).VectorA force is a vector, i.e. it has both magnitude and direction (and position).MomentThe moment of a force about a point is the product of the force and the lever arm. A torque wrench measures moment; the principle of leverage depends on moments.EquilibriumThe state of a body that isnt accelerating (and, if it is a building, not moving at all). The concept allows us to make assumptions about the internal forces that resist the applied forces.StressA force, divided by the area that it acts on. Conceptually the same as a pressure. We usually work in MegaPascals (MPa) = MegaNewtons /m2. We are interested in loads on whole elements, and stresses on parts of the material that the elements are made of.StrainWhen a material is stressed, it can only resist that stress by being deformed. The strain is the deformation, expressed as a ratio of the original length. A dimensionless ratio, and usually very small.Rigid, elastic, brittle and ductile materialsThese descriptions refer to the way the material behaves under load. A single material may exhibit several of these characteristics at different stress levels.StrengthStrength can be a property of a material, an element, or a whole assembly of elements. It refers to its ability to carry load without breaking.StiffnessStiffness can be a property of a material, an element, or a whole assembly of elements. It refers to its ability to carry load without deforming too much. Everything that carries load deforms a bit.SafetyIn building, nothing is exact, nothing is absolutely certain. Good design implies adequate margins of safety so that failure is extremely unlikely, yet the building is not grossly over-designed. The minimum margins of safety required are given in Codes of Practice. SI prefixes, and numbers The SI system uses kilo-, mega-, milli-, and micro- to mean 1000x, 1000 000x, 1/1000, 1/1000 000, etc It is convenient to write big or small numbers as 2x106, 3.5 x 10-3, etc. You can enter these on your calculator, usually like 2 exp 6 The calculator may display them as 2 E 06 but sometimes it will spell them out in full Precision of numbers Dont necessarily use all the numbers on your calculator. A carpenter cant measure to nanometres Weights and forces Weight, given in newtons, is a force We use forces (mainly for the loads on buildings) in kilonewtons (= 1000 N) 1kg produces a force of 9.81 N on earth but lets call it 10 in most cases 1kN = 1000 N, etc, using the standard SI prefixes Weights and forces Weight, given in newtons, is a force We use forces (mainly for the loads on buildings) in kilonewtons (= 1000 N) 1kg produces a force of 9.81 N on earth but lets call it 10 in most cases 1kN = 1000 N, etc, using the standard SI prefixes 1kN is about the load of a large footballer standing on top of you Mass and weight In common language, we dont distinguish between them I weigh 70 kg (common usage) Mass is a property of a piece of material, anywhere My mass is 70 kg (Correct, but sounds pompous) Mass and weight Weight is a property of a mass in a force field My weight is 700 Newtons on earth, 5000 N at blastoff, and 120 N on the Moon (An astronaut in this context it is relevant) Vectors A vector has magnitude and direction (like go 500m Northeast) Vectors A force is a vector (like 2kN, downwards) It can be drawn to scale, (using any suitable scale)  Adding vectors Add graphically (go from A to B, then from B to C) Add algebraically (use sines and cosines) Resolve into vertical and horizontal components and add them Resolution of forces is the analytical equivalent of coordinate geometry or orthographic projection.  The vertical and horizontal components can be visualised as drawing the inclined vector in elevation and plan, respectively. Wyatt p 11; Schodek p 35-6 The resolution of forces is the replacement of one force by two or more forces which act in specified directions. We choose two orthogonal directions, e.g. up and horizontal, as in coordinate geometry. The component of a force in one direction is an imaginary force which has the same effect as the force in that direction. The two perpendicular components can be shown to be Fx = F cos q Fy = F sin q These steps can be used to find the resultant of a number of concurrent forces  just add all the x components, and all the y components. If you need the actual resultant, it is the vector sum of the two resultant components R2 = S Fx2 + SFy2 The resultant of two concurrent forces can be found graphically by the parallelogram of forces. The triangle of forces is a shorthand form of the parallelogram.   It can also be found analytically R2 = F12 + F22 +2 F1 F2 cos q (from the cosine rule for solving triangles - note the + sign because q is the angle between the original forces).  The resultant of a number of co-planar concurrent forces can be determined by the method of polygon of forces, which is a series of superimposed triangles of forces. (The closing line of the polygon is the equilibrant , i.e. the force that would put the set of forces into equilibrium. The resultant is the reverse direction of the equilibrant.)  Moments A moment is a turning effect, i.e. a force multiplied by a distance  In the diagram below, the resultant of the two forces F1 and F2 is both a force (F1 - F2) acting to the right, and a moment about any particular point in the plane.  Wyatt p 11; Schodek p 35-6; CGW p36-39 The moment of a force is its turning effect about a given point. If the line of action of the force passes through the chosen point, the moment is zero. The perpendicular distance is referred to as the lever arm of the force. Moments have the units of Nm, kNm, Nmm, etc. Sign convention - clockwise moments are treated as positive and anti clockwise moments as negative. Moment of a couple  Whatever point you take as a pivot point, the moment of the couple is Fa. A couple is a force system consisting of two parallel forces having equal magnitude and opposite in sense (and not in the same line). As the forces are equal in magnitude and opposite in sense, there is no resultant translational effect. The rotational effect of these forces can be determined by taking moments about any point. The rotational effect is defined as the moment of the couple and is the same at all points. moment of couple at o1 = F x1 + F (a - x1) = F a moment of couple at o2 = F x2 + F (a - x2) = F a a is referred to as the lever arm of the couple. Stress Stress is force per unit area 1 pascal = 1 newton per square metre Stress tells how a material will behave  To discuss stress and strain, the following terms need to be defined: Stress = force / area f = P/A It can be tensile or compressive, or shear. Usual units for Stress We use stresses in megapascals (MPa) for most materials 1 megapascal = 106 N/m2 = 1 N / mm2 (remember 1 m2 = 106 mm2) We use stresses in kilopascals for floor loads, and foundation pressures Stress has the dimension of Newtons/metre2, = Pascals or more commonly N/mm2, or MN/m2 (=MPa), or sometimes kN/mm2 (=GPa) or kN/m2 (=kPa). Strain Strain is the change in length, per unit of original length, caused by a stress It is a ratio (dimensionless) Except for rubber bands, strains are very small (often not visible) The more a material strains under load, the more the structure will deflect Strain = change in length / orig. length e = "L/L It is a nondimensional quantity and usually results from the application of stress to a material. Elastic strain is reversible strain that disappears when the stress is removed. For most materials, strain is proportional to stress within this range. Plastic strain is defined as permanent strain that is not recoverable on removal of stress. This occurs at and beyond a stress level referred to as the yield stress, which is a characteristic of the particular material. Plastic strain can continue to increase while the material is at the yield stress. How materials react to stress and strain Buildings are made out of real materials We learned something about materials in MAFIB In this unit, we look further at structural properties of materials We can plot the variation of stress against strain on a diagram, like this one.  EMBED Word.Picture.6  Equilibrium Forces on the whole structure, and each part, just balance to zero For simplicity, it doesnt move up-and-down, or sideways, or spin around The three conditions of equilibrium say that the sum of all the vertical forces is zero; the same applies to the sum of horizontal forces, and to the sum of the moments about any point S V = 0 S H = 0 S M = 0 If we know the external forces acting on the structure, we use these equations to find the unknown reactions and internal forces. To express the conditions in general terms, it is useful to resolve all forces in two directions (vertical and horizontal, or x and y). Any other pair of directions at right angles to each other is equally valid, but most times vertical and horizontal are more convenient. These conditions give us a maximum of three simultaneous equations, which can solve up to three unknowns. Components are defined later. Briefly, a component is the effect of a force in a particular direction. A horizontal force has a vertical component of zero. 2 STRUCTURAL SYSTEMS Purpose and function of structural systems The purpose of a structural system is to resist all the loads acting on the building, and transfer them to the ground. The loads originate either within the materials themselves (the self-weight), or on the surfaces of the floors, walls and roof due to occupancy and wind loads.CGW Chap 7 There is seldom a single right answer for the structural system of any particular building. However, there are some constraints that will point toward suitable systems. EMBED Word.Picture.6 The building must be stable as a whole There must be enough structural elements They must be in suitable places They must be strong and stiff enough, and suitably connected together How and where can we carry vertical loads down to the ground? Anywhere that there are solid walls, we can use either a loadbearing wall or a system of columns. Anywhere there are windows or doors, or in the middle of most rooms, we cant have supports. In a multistorey building, the vertical supports should line up all the way down. It is very expensive otherwise.How and where can we carry horizontal loads down to the ground? Any solid wall can be made stiff in its own plane (either in brick or concrete, or a framed wall with bracing). Windows and doors interrupt this stiffness. The wall has to extend to the ground or to another stiff structure. Alternatively, a set of columns and beams with rigid connections can form a rigid frame. We need horizontal resistance in two directions at right angles. Brick walls meeting at a corner support each other in all directions. Horizontal resistance should be roughly symmetrical about the centre of the building. Qualitative and quantitative understanding First we will try to understand the structural system qualitatively If we dont have the right kind of supports in the right kind of places, it wont work When we have a credible system, we can use precedent and simple rules to get credible sizes We need to understand the quantitative basis of a final structural designA simple post-and-beam system needs to be made stable  EMBED Word.Picture.6 Unless something is done to stabilise it this frame can fall over (in any direction) Bracing one panel makes it stiff in that plane  EMBED Word.Picture.6 Diagonal or cross bracing limits the locations where doors and windows can be put. You can only use it where it wont be in the way. A braced frame does not rely directly on the stiffness of the individual members. It is very stiff against wind load.A masonry wall is automatically stiff in its own plane (but needs support at right angles) This can be a masonry infill wall between columns, or the wall itself may be loadbearing.Columns built into the ground provide some resistance to wind load  EMBED Word.Picture.6 This system relies on the stiffness of the poles. It is ok for a shed, which can accept some swaying. For other buildings, the poles must be very stiff to avoid cracking windows, sticking doors and noticeable swaying in the wind. For pole-frame houses, bracing is often added as well. If the poles are timber, there are moisture and termite problems.Rigid Frame systems suit larger buildings  EMBED Word.Picture.6 As an alternative to diagonal bracing, a set of columns and beams can be made stiff by making the joints rigid. This happens automatically when reinforced concrete is cast in one piece. It can be done in steel using welding or high-strength bolts. It is very difficult in timber. A rigid frame relies on the stiffness of the individual members. Knee-bracing is an alternative to rigid joints  EMBED Word.Picture.6 Knee-bracing is somewhere between cross-bracing and making rigid joints. The short braces across the corners have the effect of making the joints rigid. Like a rigid frame, a knee-braced frame relies on the stiffness of the individual members. Knee-bracing is common in timber frames because it is the most practical way of making a rigid joint. The knee-braces may intrude into the room.Visualising the deflected shape If we can visualise how a structure would bend under load (exaggerating the deflections so we can see them), we can appreciate which members would be stretched or compressed, and which would bend, and which way they would bend. In the computer lab, we will learn to use a structural analysis program that calculates and draws the deflected shapes. That is helpful to understand how the members are working. A rigid right-angle joint remains a right angle after deflection, although it may rotate as a whole. A non-rigid joint might become an acute or obtuse angle. Equilibrium (of the whole building and its parts)  EMBED Word.Picture.6 The building supports its loads All loads are finally resisted by the ground Parts of the building intervene between loads and foundation The building must not overturn as a whole The building must not sink into the ground Each part of the building must resist its own share of the loadsThe equilibrium argument SV = 0 SH = 0 SM = 0 This is a shorthand way of saying that a building doesn t move up and down, or sideways, or turn over. The Greek Sigma (S) means "the sum of .. The supports to a planar structure must provide at least three reaction components, if the structure is to remain stable and in equilibrium. If there are only three reaction components, these allow us to write three equations of equilibrium. Such structures are referred to as (externally) statically determinate. If there are more than three reaction components then the structures are referred to as statically indeterminate (they cannot be worked out by statics alone). Overall equilibrium what do we need to know about the building as a whole? All the possible loads  EMBED Word.Picture.6  How big do the reactions have to be, for equilibrium? Where can the building be supported?What are the problems? Vertical loads Downward loads just need big enough footings  EMBED Word.Picture.6 Equilibrium (forces in line) Every force is resisted by an equal and opposite one, exactly in line  EMBED Word.Picture.6 Horizontal loads might overturn a tall building  EMBED Word.Picture.6 Wider base and heavy building are more stable Equilibrium (forces out of line) Tendency to overturn can be resisted by other out-of-line forces The turning effect is a moment  EMBED Word.Picture.6   Equilibrium of each part what do we need to know? Equilibrium of the building elements If it is stable overall, what happens to the elements inside the building structure? Compression and tension elements (such as columns, and bracing) Bending elements (such as beams and slabs) Elements in shear (such as bracing panels) Trusses Connections between the elementsCompression elements Vertical elements (columns, walls, piers) Resist vertical loads by compression Transferring a load straight down a vertical stick is very simple. There are two main modes of failure to be avoided. If the column is slender, it will buckle long before the material actually crushes. This is the case with most columns in small to medium buildings. At domestic scale (say 2.4m high), a steel or timber column is usually at least 75mm in each direction. Tubular steel columns are very efficient because they resist buckling well. Brick piers are likely to be at least 230mm wide for this height. Only in very short, fat columns is the material in danger of crushing before it buckles. Compression and tension elements Diagonal bracing can be in tension or compression (Takes horizontal loads down to ground) Steel strapping wont work in compression, so only one of the pair is working at any time Bending elements Horizontal elements (beams and slabs) resist vertical loads by bending, and transfer loads from midspan to supports The loads try to bend the beam. To resist bending, the beam depends on its material, and its cross-section. The most important factor is its DEPTH. We often use span-to-depth ratios to estimate the required depth. All beams deflect, but if it is too shallow, deflection will be unacceptably large. If we have beams in two directions, do we sit one on top of the other, or between? That affects the total depth of the structure. A slab works like a wide beam. A reinforced concrete slab can work in both directions if it is supported all around.Elements in shear Mainly applies to plywood and hardboard bracing panels Beams are also subject to shear  Equilibrium of elements Total downward load = total upward reaction If the load is symmetrical, so are the reactions The Tributary Area is the area of floor, etc, supported by the element we are considering. Generally, the Tributary Area is the Span x the Spacing of a series of beams. The total load on a beam is the load per square metre x the tributary area (plus its own weight, plus any other beams or columns supported by it).Total downward load is carried down by columns The reactions supporting this beam become the loads on whatever is below it - usually another beam, or a column. EMBED Word.Picture.6 If the load is off-centre, so are the reactions Most building loads are symmetrical, but a load concentrated near one end puts more load on the support at that end, and less on the other one. EMBED Word.Picture.6  Cantilevers A cantilever on one end of a beam puts more load on the support at that end, and less on the other one. A cantilever generally shouldnt be more than one-third of the main span (to avoid any risk of tipping over, and to avoid having the beam too large) Trusses A truss is like a deep beam with bits missing It should make a series of triangles Members are in tension or compression only It can be almost any shape, but the two shown are most common A beam tends to be stronger and stiffer, when it is deeper. However, a long-span beam with a light load may waste a lot of material just carrying itself. A truss achieves depth without too much extra material. Schodek: p112-115; Wyatt Ch 5; CGW p79-81 There are a lot of connections between members in a truss. Materials and details must be chosen so that they can be made effectively, cheaply (and neatly). It may be triangular or rectangular in elevation (or any other shape that fits the job). Bracing systems are deep trusses. If exposed to view, the character of a truss is different from a beam (lighter, but busier).When to use a truss Useful when there is plenty of depth, and a relatively light loading. (A truss acts like a deep beam with holes in the web.) Loads preferably only occur at the panel points (otherwise the chords have bending as well as axial force). A pitched roof with flat ceiling provides a good shape for a triangular truss.Disadvantages of trusses High labour costs may outweigh the economy of materials. Painting and cleaning may be easier on a smooth beam than a complicated truss.Trusses - lateral stability  Part of a three-dimensional truss - triangular cross-section - can be made from steel tubes. Sometimes three-dimensional trusses are used, to provide stiffness in all directions within the one structure.Since trusses use material economically, they are likely to be made of very slender pieces. It is necessary to stop the truss from falling over; to stop the whole thing from buckling; and to stop the compression members from buckling in their own length. This can all be done quite neatly in a domestic roof, where there is an enclosing material top and bottom to cover up all the necessary bracing pieces. If the trusses are exposed you need to think about the appearance of the bracing.  Other systems Tension members Tension members are very efficient, and they can be very slender and elegant. They mostly cant cope with stress reversal. If the wind can get under the awning in the diagram, there may be a problem. 3 LOADS Loads The building materials impose dead loads (fixed, vertical) The occupants and contents impose live loads (variable, mostly vertical) Wind and earthquake impose loads (variable, mostly horizontal)The loads exert forces on the building. We have to work with these forces and analyse them. Forces have magnitude, direction and position. We can visualise which portion of the buildings loads is transmitted to which structural element, by following the load paths.  Tributary areas A beam picks up the load halfway to its neighbours Each member also carries its own weight Loads applied to different parts of a building will eventually have to be transferred to the foundations. The structure should provide a safe path for this load transfer. In order to design a structural element, it is necessary to determine the loads transferred to it by other building or structural elements. We often use a set of parallel beams. It is easy to visualise the area carried by each beam it is a rectangle as long as the span of the beams, and extending halfway toward each adjacent beam. Dont forget the horizontal wind loads carried by each piece of the walls. Turn this diagram around so that the beams are vertical columns.A column generally picks up load from halfway to its neighbours It also carries the load that comes from the floors above In the common case of a series of columns in both directions, each internal column carries the load on a rectangle halfway to each of the adjoining rows. Wyatt p 36 and following; Schodek p 213 Nonstructural building elements may also be required to transfer loads , if the loads do not act directly on the structure. This is particularly the case with the walls and windows. Dead loads on elements Code values are given per cubic metre or square metre Multiply by the volume or area supported Live loads on elements Code values per square metre Multiply by the area supportedDistributed loads and point loads Floors, walls and roofs are generally distributed loads (kN per m) Other beams are point loads (kN) Effect of one member on another The forces at the supports are the reactions For equilibrium, the reactions just balance the loads How does the ground know how hard to push up on my feet? The forces that each element exerts on its supports depend on the loads on that element. These forces are reactions to the upper element, and loads on the lower one. It must be possible to follow the path of every load, all the way to the foundations.How many reactions? Any 2-D element needs 3 reactions Usually 2 vertical and one horizontal The horizontal reaction is often zero No horizontal reaction = standing on rollerskates The nature of the reactive forces that are possible will depend on the type of support provided. If a particular direction of force, or a moment reaction, is required but cannot be provided, then the structure is not satisfactory. If the forces are all vertical, there should be the possibility of a horizontal reaction, but it will be zero until a horizontal force is introduced. (Otherwise it is like standing still on roller-skates).  EMBED Word.Picture.6  Beam with only two reactions If a body is subject to only two forces, the forces must be collinear for equilibrium Hence for a bar subject to forces at the ends only, the resultant force must be along the bar for equilibrium. If a body is subject to only three forces, the forces must be either concurrent or parallel. A minimum of three restraints are necessary to produce a stable structure. The restraints must not, however, be equivalent to either a parallel or concurrent force system.Types of reactions simple support Provides vertical support only Allows rotation - no moment developed Types of reactions cable support Provides vertical support only Allows rotation - no moment developedTypes of reactions roller support Provides vertical support Deliberately avoids horizontal restraint (allows expansion) Types of reactions hinge support Provides vertical support and horizontal restraint Allows rotation - no moment developedTypes of reactions rigid support Provides V, H, and moment restraint Suits cantilever beams or posts, and rigid frames Make sure you can physically achieve it!  How big are the loads? Dead loads How much does the stuff weigh? How much of each material is there? (We have to guestimate how big the elements are to start with) Appendix A of SAA loading code AS 1170.1 tabulates the unit weights of a number of materials and the weight per unit surface area of construction. Typical values are given in the table below (note that they are already in gravitational units) DEAD LOADSBulk MaterialWeight/unit volume Concrete, dense aggregate, unreinforced23.5 kN/m3Timber, Australian hardwood11.0 kN/m3Steel76.9 kN/m3Glass25.5 kN/m3Walls, brickwork (0.19 kN per 10mm thickness) =19.0 kN/m3Sheet or surface MaterialWeight/unit area Ceilings, Gypsum plaster, 13 mm thick 0.22 kN/m2Floors, Terrazo paving, 16 mm thick0.43 kN/m2Roofs, fibre cement, corrugated sheeting 6mm thick0.11 kN/m2Walls, concrete hollow block masonry 150 mm thick1.73 kN/m2The Code deliberately gives these values to 2 or 3 significant figures. Any more precision is not warranted, since they are estimates of the weights of generic materials. Live Loads Building Codes give minimum values Domestic live loads range from 1.5 kPa Corridors and balconies are generally 4kPa, to allow for crowding Most stores and workshops are 5 kPa upwards AS 1170.1 specifies the minimum floor live loads to be considered for occupancy and floor use categories. Typical values of uniformly distributed loads specified are given below (note that these loads are given in kPa, = kN/m2 , that is, per unit area of floor) LIVE LOADSOccupancy or useMinimum live loadInstitutional buildingsAssembly areas (class rooms, lecture rooms etc) 3.0 kPaLibraries (reading areas)2.5 kPaLibraries (stacking areas)3.3 kPa per m of usable heightOffice buildings (Offices, banks etc)Corridors, foyers etc (subject to crowd loading)4.0 kPaoffices3.0 kPaDomestic buildings houses1.5 kPamotels etc, rooms2.0 kPacorridors4.0 kPa Factors in wind speeds General wind speed in the region (pressure varies with square of the speed) Local topography affects wind patterns Wind speed increases with altitude Wind speed decreases with terrain roughness  When the wind hits any obstruction, the pressure it produces will depend on: the local wind speed, and the shape of the object. Wind Speed At any location, longterm weather records will indicate the maximum wind speed expected at the standard 10m height used by meteorologists. (For Sydney, 41m/s maximum). For the particular building, this will be modified by: The height of the building. Windspeed is less near the ground. The topography. Windspeed is greater at the top of a hill than in a valley. The terrain roughness. Wind near the ground is stronger over flat country or the ocean, than over trees or houses. The Wind Code has four different Terrain Categories. The basic pressure (i.e. if the air comes to a complete stop against an obstruction) goes up with the square of the wind speed. It is 0.6 V2 x 10-3 kPa (where V is the speed in m/s). Factors in wind loads Shelter from anything permanent will reduce loads Shape of building (mostly boxy rather than streamlined) affects loads The shape of the building Most buildings are not streamlined like cars. Modern car makers boast about their low drag coefficients, the best being below 0.3 for smooth, curved body shapes. The total drag on the car (the sum of the pressure on the front face and the suction on the back) is the drag coefficient multiplied by the basic wind pressure, above. The drag coefficient for rectangular buildings is between 1.0 and 1.3. A house wouldnt do very well on the race track. Wind loads on buildings Pressure on the windward face Suction on other faces Suction on lowpitched roofs Buildings need bracing and tying-down Wind can come from any direction Wind load is the main horizontal load on most buildings. The load can be pressure or suction on the walls, and on the roof. AS 1170.2 sets out two procedures for determining design wind speeds and wind loads  EMBED Word.Picture.6  To design elements in each wall and the roof, we need to know how the wind forces are distributed between the various surfaces. Remember that the windward side can be any side, depending on the wind direction. The average wind pressure coefficients vary a bit with the proportions, but are about: Windward face: pressure +0.8 times basic wind pressure Leeward face: suction, from -0.2 to -0.5 times basic wind pressure Side walls, suction -0.6 times basic wind pressure Roof, mostly suction, about -0.9 times basic wind pressure. In the case of a pitched roof facing the wind, pitches over about 30 develop pressure on the windward side (about +0.5 for steep pitches). Roofs tend to blow off rather than blowing inward. There will also be an internal pressure or suction, depending whether there are open doors and windows facing toward or away from the wind; the coefficient ranges from -0.3 for mainly leeward openings, to +0.8 for large windward openings. Normally you would close everything in a strong windstorm, but ... (see Cyclone section below). Each wall and the roof of the building are subject to a design wind force, F, which is the product of its area A and the nett wind pressure (pe - pi) on it. Fsurface = A (pe - pi). If A is the area of the surface in m2, and pe and pi are the external pressures in kPa, then F will be the force on the surface in kN. For a small building in a sheltered location, wind loads are mostly in the range 0.5 to 1.0 kPa. For a taller building or an exposed location, wind loads are likely to be 1.0 to 2.0 kPa, and higher in cyclone areas. Wind loads on buildings Wind tends to overturn a tall building  Wind loads on elements Many variables involved in using Code values Multiply by the area exposed to wind In non-cyclone areas, wind loads in the 1kPa range Timber Framing Code has a procedure for finding maximum wind speeds Timber Framing Code also has simplified rules for bracing single-storey houses Localised extreme pressures The figures above are the average values for each wall, etc. Right near the corners, there can be localised pressures and suctions up to double these values. When you see photos of buildings damaged by the wind, it is often the ridge cappings and the corners worst affected. Fixings in these areas need to be especially well considered. Cyclone areas Coastal Australia north of Lat. 27 is subject to tropical cyclones. These produce very high winds, along with heavy rainfall. It can be assumed that most windows will be open (broken by flying debris). The ground will be softened by rain, so posts built in to the ground may not work as lateral load-resisting elements in these conditions. The wind changes direction rapidly, and tends to loosen nails. Special holding-down details are used in cyclone areas. Seismic loads Earthquakes cause damage by horizontal acceleration Heavy buildings attract more load Brittle materials fail, ductile materials may survive Positive connections are essential If a tall building is likely to oscillate under earthquake conditions, then a thorough vibration analysis is necessary. Otherwise, an equivalent static load can be used as described below. Seismic loads occur when the ground moves horizontally and the building tries to stay still. In the simplest analysis, the forces involved are the product of the mass of the building and the acceleration of the ground under it. The horizontal forces due to earthquake are a function of the mass of the building. Lightweight buildings attract lower forces. Ductile materials, with everything strongly connected together, are most likely to survive earthquake. Buildings with low and tall sections joined together, or with an open ground floor and rigid walls above, are most likely to be damaged. Masonry parapets and chimneys are easily dislodged, causing damage to the building and danger to people in the street. Looking inside the elements Freebody diagrams Elements as freebodies  We can isolate any member (or part of it) to study it We must put back artificial forces to replace whatever supports we cut away This is a useful tool to see inside a structure Schodek p 45 We can consider a system of forces and the necessary reactions to produce equilibrium in a whole building, or in a discrete element (such as a beam), or in any part of the structure we care to consider. We do this by making imaginary cuts in the structure wherever we want to. All that is necessary is to carefully include all the external loads and reactions that act on that part of the structure, and also all the internal actions that are provided by the parts that we have cut away. The part we have cut out is sometimes called a freebody because it behaves as if it were a separate element. Usually, some of these internal actions are the unknowns that we want to find. By making the cuts at suitable places, we can find out most things we need to know. If a structure is in equilibrium under the action of external forces and reactions, we must check that the different components and parts of the elements of the structure are also in equilibrium.Internal forces and moments are developed to hold the different parts of the structure together. (The members and connections must be able to withstand these internal forces).In writing equilibrium conditions for a part of the structure or structural element, these internal forces become external for that part of the structure or element. By first drawing the freebody diagram for the part, the equilibrium conditions can be derived.Finding the reactions of a beam "V = 0 tells us the total of the Vs "H = 0 finds the horizontal reaction (if any) "M = 0 lets us distribute the total between two ends Work smarter, not harder  EMBED Word.Picture.6 Most beams are symmetrical If the reactions are equal, don t make hard work of it The left-hand beam is symmetrically loaded. The reactions are each half of the total load. End of story. For the right-hand one, a simple calculation is needed.The "M condition says the sum of moments about any point is zero That means you can pick any point on the page. Pick a point that eliminates one of the unknowns, to make it easyFinding the reactions of a cantilever  EMBED Word.Picture.8 A cantilever has one V, one H, and one M reaction Vertically: R = W1 + W2 Horizontally: H=0 unless there is a horizontal load Moments: M = W1.x1 + W2.x2 M is a negative moment see BEAMS section 4 BEAMS What beams have to do Be strong enough for the loads Not deflect too much Suit the building for size, material, finish, fixing etc Schodek p53-62 and p210; CGW p41-43 and 5.1-5.4; Wyatt Ch 7 Columns and beams are the fundamental units of most linear structural systems. To carry its load safely between its supports, a beam has to resist bending and shear, which are structural requirements. We are also interested in limiting its deflection, which is a serviceability requirement. Loads on beams Point loads, from concentrated loads or other beams Distributed loads, from anything continuous    Loads coming from other beams, or columns, or concentrated pieces of equipment, are point loads. Loads coming directly from floors, or continuous walls, are distributed loads. Many small and equal loads, such as joists supported on a bearer, can be considered as equivalent to a distributed load. Linear structural elements, such as beams, usually support loads that are distributed along the length of the member. These loads are specified in terms of their intensity, which is defined as the magnitude of the load distributed over a unit length. These loads can be treated as parallel forces and it can be shown that the resultant of these loads pass through the centroid of the load diagram. The magnitude of the resultant load is given by the area of the load diagram. The reaction pressure on a foundation may also be uniformly or non-uniformly distributed. The position of the resultant of the loads on the footing will dictate the position of the resultant of the reaction pressure from the foundation. The most common distributed loads are uniform, but they may be triangular or other shapes. In describing distributed loads, we must be clear whether we mean the total load in kN, or the load per unit length in kN/m. What the loads do The loads bend the beam, and try to shear through it   Wyatt p73-7; Schodek p 53-57; CGW p40-43 If we made a beam with a hinge in the middle of it, it would collapse. Why? Because the loads and reactions are combining to try to bend it. The beam is fighting back and trying not to bend. A bending moment is the moment resulting from external forces acting on a beam and causing it to bend. The bending moment is resisted by internal actions within the beam. Shear is the effect of sliding one part of the material over the other. Schodek p231-6; Wyatt p71-3, 103-7   In a beam, shear can occur as vertical and horizontal shear. Vertical shear can be visualised in the same way as shearing through a sheet of paper with scissors (or through a sheeps wool with comb and cutter, if you prefer). Horizontal shear can be imagined by taking the pages of a book, holding them together to form a thick slab, and bending it. If you release the grip a little, the pages will slide over each other, and will bend much more easily. Vertical and horizontal shear always co-exist at the same point. Failure may occur in one of these directions, or in materials like concrete, it occurs as tension at 45 to these directions. Timber, being easy to split with the grain, has a low resistance to horizontal shear. Steel is equally strong in all directions. It has no preference between horizontal and vertical shear, and because it does not have the directional weakness of timber, it can be used in sections with very thin webs. How we quantify the effects First, find ALL the forces (loads and reactions) Make the beam into a freebody (cut it out and artificially support it) Find the reactions, using the conditions of equilibrium  Localised effects along the length The shearing effect is greatest near the ends We can plot its variation along the beam The bending effect is generally greatest near the middle We can plot its variation along the beam  Local effects of shear Use the freebody idea to isolate a little bit of the beam See what vertical force is needed at the cut, for equilibrium  Local effects of bending Use the freebody idea to isolate a little bit of the beam See what moment is needed at the cut, for equilibrium  The Bending Moment is an internal action acting at all points along the beam. It is resisted by internal stresses in the beam, forming the moment of resistance. If we take the whole beam as a freebody, it is in equilibrium and the sum of the moments about any point is zero. To find the bending moment at a point within the beam, we have to look at part of the beam. It is simple to start with a cantilever (see diagram). The beam shown has a single point load of 2kN at the end. The reaction at the left end consists of a vertical force of 2kN, and a moment of (2 x 5) = 10 kNm. A cantilever always has a moment reaction - otherwise it would fall down.  Calculating bending moments along a cantilever beam Suppose we now make an imaginary cut in the beam 1m from the end, and consider the part to the right as a freebody. What are the actions exerted on that part by the remainder of the beam, that cause the end part to be in equilibrium? The shear force diagram for this beam is a constant 2 kN for the whole length. The shear force means that each part is exerting a vertical force of 2 kN on the part next to it. The load and the shear force form a couple acting on the freebody. To restore equilibrium, there must also be a moment of (2kN x 1m) = 2 kNm acting at the cut section. If we move the cut section back a distance x from the right-hand end, the internal moment is Wx at each point. That gives us the Bending Moment Diagram as shown. Details of shear force diagrams To plot a diagram, we need a sign convention The opposite convention is equally valid, but this one is common There is no difference in effect between positive and negative shear forces   Wyatt p 73 Shear force in a beam is caused by the imposed loads and the reactions. The simplest example is a cantilever beam with a point load at the end. There is a constant shear force throughout the beam, equal to the load. A shear force diagram is a graphical way of showing how the shear force varies along the length of the beam. (In this first example, it shows how the shear force doesnt vary!) In a beam with a support at each end, the shear force near the support is equal to the reaction. As we move along the beam, the shear force reduces as each load is encountered. At the other end it is equal to the other reaction. Sign convention. In a normal beam, the shear force is called positive on the left and negative on the right. It doesnt really matter, except for consistency. There is no practical difference between positive and negative shear (unlike tension and compression). In a beam with a uniformly distributed load, the shear force diagram reduces constantly (it slopes down, rather than having a series of steps). The rate of decrease is the same as the intensity of load per unit length. The shape of the shear force diagram is always one order greater than the shape of the loading diagram. It can be shown that Shear = (loading) dx with appropriate constants of integration. Plotting the shear force diagram Starting at the left hand end, imitate each force you meet (up or down)  EMBED Word.Picture.6  Shape of the shear force diagram Point loads produce a block diagram Distributed loads produce triangular diagrams  What shear force does to the beam Although the shear forces are vertical, shear stresses are both horizontal and vertical Timber may split horizontally along the grain Shear is seldom critical for steel Concrete needs special shear reinforcement  How to calculate the shear stress It depends on the beam cross-section for rectangular sections, 1.5 x (shear force) / (area of section) For I-beams, (shear force) / (area of web)  Schodek p 235 The average shear stress in a rectangular beam is simply the shear force divided by the cross-sectional area. However, the stress is not uniform. It is greater at the centre of the section, and reduces in a parabolic shape, to zero at the top and bottom surfaces. The maximum shear stress occurs at mid-depth and is given by fv = 1.5 V/bd, where fv = shear stress V= shear force b and d = dimensions of section.  In a steel I section, the shear is resisted mostly by the thin web. Although it also varies a little, we usually work with the average value, which is given by fv = V / bw d, where bw = thickness of the web. In a structural steel UB section, bw is typically about 6% of the total width of the section. Therefore the shear stress is very much higher than it would be in a rectangular section. This is achievable because the shear strength of steel is only a little less than its strength in bending. In timber and concrete, the shear strength is much less than the strength in bending. What to do with the shear stress Codes give maximum allowable stresses Timber, depending on grade, can take around 1 MPa Steel can take around 90 MPa Using Codes comes later in the course Details of bending moment diagrams To plot a diagram, we need a sign convention This convention is almost universally agreed  A positive moment is also called a sagging moment because of the deflected shape. A negative moment is also called a hogging moment.  Sign convention for moments: A moment causing compression at the top of a beam is positive. The bending moment in a cantilever is negative. In a normal simply supported beam, it is positive. Where to draw the bending moment diagram Positive moments are drawn downwards (textbooks are divided about this)  Sign convention for diagrams: In these notes I have drawn positive bending moments below the horizontal line. Many textbooks draw them this way, to imitate the shape of the deflected beam. Other textbooks draw them the other way, as for normal graphs. Both these conventions are fairly common, and most people will understand you if you follow either of them. Of course you could do your own calculations using whatever convention you like, if you stick to it. You could take your own lecture notes in Esperanto, too, if you wanted to. Positive and negative moments Cantilevers produce negative moments Simple beams produce positive moments Continuous beams have both, with negative over the supports  Plotting the bending moment diagram Using the freebody idea, find the bending moment at significant points Join them up, knowing the shape you expect   Consider a simple beam with a central point load. Each reaction is half the load. Make an imaginary cut just alongside the load and consider the part to one side as a freebody. The shear force in the left-hand half is 5 kN. The actions acting on our imaginary cut are the shear force, plus a bending moment to counteract the moment of the vertical forces: M = 5 kN x 2m = 10 kNm. At any other point along the left-hand half, distant x from the left support, the moment is R.x where R is the reaction force. This gives rise to the bending moment diagram as shown. (It is symmetrical on the right hand side.) The maximum value occurs in the centre, under the load. Since R = W/2, and the distance to the support = L/2, the maximum bending moment in this type of beam is M = WL/4. Note that this formula (M = WL/4) is only true for a point load in the centre of the span. Shape of the bending moment diagram Point loads produce triangular diagrams Distributed loads produce parabolic diagrams  Consider a beam with a uniformly distributed load (UDL) totalling W. (UDLs are sometimes given as total, sometimes per unit length). Each reaction is equal to half the total load (=W/2). Take an imaginary cut at the centre. The shear force at the centre is zero (see Shear Force section). The vertical forces are the reaction (W/2 upwards) and the UDL acting on the freebody (W/2 downwards), which are in equilibrium vertically. The moment required to restore equilibrium is (Reaction x L/2) - (W/2 x L/4) = WL/8. Note that we must consider only the loads acting on the part that is our freebody. Only the left-hand half of the UDL is considered. It is equivalent to a single load acting at its centre. If we make cuts at other points along the beam, the reaction remains W/2, but the amount of UDL on the freebody reduces as we get closer to the left end. The diagram does not vary uniformly, but parabolically.  The bending moment for a UDL is WL/8 or wL2/8, where W is the total load or w is the load per unit length. Note the difference between big W and little w. The shape of the bending moment diagram is always one order higher than the shear force diagram. It can be shown that Bending moment = (shear) dx = (loading)dxdx with appropriate constants of integration. It follows that the shear force is the differential coefficient of the bending moment. Therefore the maximum bending moment occurs where the shear force is zero. Finding Bending Moments from the Shear Force Diagram  This is an alternative method of finding bending moments. Some people have already learnt it at school, and for some people it is just easier. Because the bending moment diagram is the integral of the shear force diagram, the value of the bending moment at any point can be found by calculating the area under the shear force diagram on one side of the point under consideration. The diagram at the left shows part of the shear force diagram for the beam above, with a UDL of 2kN/m over a span of 4m. To find the BM at the midpoint, find the area of the part of the Shear Force diagram to one side of it (shaded): Area of triangle = 4kN x 2m /2 = 4 kNm. Notice that we multiply the units as well as the values. The units for a moment are always (Force) x (distance). Shape of the bending moment diagram Usually no re-entrant angles We are mainly concerned with the maximum values  Can we reduce the maximum values? We have seen that a cantilevered end reduces the positive bending moment Built-in and continuous beams also have lower maximum BMs and less deflection  Beams with cantilevers  Bending moment diagram for a beam with cantilevers at both ends. The cantilevers help the midspan beam to carry its load. Schodek p65; CGW Fig 4.6.7 Cowan: Handbook of Architectural Technology, p153-5 A cantilever has a negative bending moment at its support. If it forms part of a beam across an adjoining span, that negative moment is resisted by (and applied to) the adjoining beam. This reduces the maximum positive bending moment in the beam. In the example above, the maximum positive bending moment is reduced from 4 to 3 kNm because of the negative moments of -1 kNm at the ends. To take advantage of this effect, we can rely on only the dead load component on the cantilevers. Note that the total height of the parabolic diagram for the 4m span beam is the same as for a simply-supported beam. Using the Shear Force Diagram for a beam with cantilevers  Part shear force diagram for the beam above As before, we calculate the area of the shaded part of the shear force diagram. The part below the line is regarded as opposite sign to the part above the line. The area of the triangles is: (1/2) x [(4kN x 2m) - (2kN x 1m)] = 3 kNm. Continuous and built-in beams  Schodek p305-318; Cowan: Handbook of Architectural Technology, p154-5 When a beam is built-in at the supports, or continuous over several spans, the effect is the same as applying a restraining moment at the ends. The analysis of statically indeterminate beams is not included in this course, but the principles are important. A continuous beam always has reduced deflections, and usually has reduced bending moments, compared to a series of simple beams over the same spans. The largest values of bending moments are usually the negative ones at the supports. Note that the total height of the bending moment diagram is WL/8 which is the same as for a simply supported beam. The negative moments at the supports merely move the baseline. Reinforced concrete is usually poured in place and is therefore automatically continuous. If you want to make it simply-supported you have to specially detail non-rigid joints. Timber and steel are made in straight pieces. The joints are not rigid unless specifically designed to transfer moment. Standard BM coefficients Use the standard formulas where you can  EMBED Word.Picture.6  Tables of BM coefficients can be found: Wyatt p117; CGW p 45-49; Cowan: Handbook of Architectural Technology , p.156-159 Schodek p65 shows some diagrams Coefficients for their bending moments are listed in textbooks and handbooks. Standard BM coefficients for cantilevers  What bending moment does to the beam Causes compression on one face and tension on the other Causes the beam to deflect  How to calculate the bending stress It depends on the beam cross-section We need some particular properties of the section  EMBED Word.Picture.6  What to do with the bending stress Codes give maximum allowable stresses Timber, depending on grade, can take 5 to 20 MPa Steel can take around 165 MPa Use of Codes comes later in the course So now we need to find the properties of the section 5 MATERIAL BEHAVIOUR AND SECTIONS Physical and mechanical properties of materials Crystal structure There are 14 different arrangements of crystal lattice, each giving rise to a distinctive crystal shape, and also different densities and properties (such as coal and diamond). Some of the technical terms in this Section are described in the following Section Cowan & Smith, Science & Tech of B.Materials, p 11-13 In crystalline materials, repeating groups of atoms or molecules are built up into a crystal lattice. In some materials, particularly gemstones, individual crystals are macroscopic. Crystals are strong but the boundary planes are weaker. The intrusion of impurities into a crystal structure may weaken it, or they may act as anchors and strengthen it. Strong metal alloys and ceramics make use of the irregular size of the different atoms, to make it harder for the material to fail. Heat treatment Some metals can be made hard and brittle, or soft and ductile, by heat treatment. Some processes, such as welding, may cause unwanted heat treatment and embrittlement. In most materials, the crystal structure changes at very high or low temperatures, and when the material melts it becomes amorphous (non-crystalline). In the ferrous metals, the iron-carbon compounds go into or out of solid solution at different temperatures, and give rise to different strength characteristics. Higher-carbon steels can be hardened or annealed by heat treatment. Work-hardening In most metals, straining beyond the yield point will rearrange the crystal structure and make the material harder and more brittle. Metal sheet is made stronger by cold-rolling, and wire is made stronger by cold die-drawing. As the strength increases, the maximum elongation (ductility) reduces. See the diagram for steel below. Reinforcing bars should not be bent too tightly, or straightened and re-bent, for fear of brittle cracking. Work-hardened material can usually be annealed by suitable heat treatment. Elastic behaviour All materials deform under load Until you damage the molecular structure, the strain is proportional to stress, and it recovers when you remove the load The MODULUS OF ELASTICITY is stress divided by strain. It is a property of the material  The material property that influences stiffness is the Modulus of Elasticity, also called Young's modulus, E. E = f / e = P.L / A. "L Modulus of Elasticity is the ratio of applied stress to the resulting elastic strain. It has the same units as stress. It measures the stiffness of the material. the relationship between stress and strain can be displayed as a graph. the slope of the line represents the modulus of elasticity of the material. In the diagram it is represented by tan . Typical values of E for common structural materials are: E steel = 200 kN/mm2 = 200 000 MPa E concrete = 25 kN/mm2 = 25 000 MPa (varies) E timber = 10 kN/mm2 = 10 000 MPa (varies a lot) Brittle behaviour All materials deform under load In a brittle material, all the molecular bonds break suddenly at a certain stress level The material fails suddenly (like glass or brick) Brittle materials have only a small amount of elongation at fracture. When brittle materials fail, they do it suddenly and catastrophically. the strength of ductile material is approximately the same in tension and compression. the strength of brittle material is much higher in compression than it is in tension. In very cold conditions (Antarctic, or high mountains in winter), many materials tend to become brittle. Special care may have to be taken in choosing metal alloys for these conditions. High-carbon tool steels can be brittle. Most masonry materials (concrete, brick, stone) are brittle, and so is glass. Ductile (plastic) behaviour All materials deform under load In a ductile material, the molecular bonds gradually break and re-form The material can be greatly bent and reshaped without breaking (like soft metals, or plasticene) Ductility - is the ability to deform permanently prior to fracture. It is measured in terms of percentage elongation at fracture. Most metals are ductile. Mild steel is very ductile, and higher-strength steels are less so. Ductile materials can accommodate local stress concentrations, and they tend to hang together and survive after earthquakes and similar damage. Yield stress is the stress at which a material begins to strain rapidly with little increase in stress. Only very ductile materials show it clearly. Proof stress is the stress at which a given strain is observed (usually 0.2%). It proves that the material can support that stress without yield or brittle failure, and it saves having to do a complete fracture test. Real elasto-plastic materials Most materials are elastic at low stresses Materials like steel become ductile at a high stress level (called the yield stress) We can use them safely below the yield stress If we overstress them, they deform dramatically, but dont immediately break  Typical stress-strain diagram for structural steel The stress-strain diagram for a structural steel specimen in tension shows the following trends. there is an initial elastic range followed by plastic flow at constant stress (yield stress). beyond a particular strain, further deformation occurs under increased stress. This is referred to as strain hardening. failure occurs when the strain attains a large value. (Up to 25% of the original length). the engineering stress-strain graphs are based on stresses defined with original area. this capacity for large plastic deformation prior to failure is referred to as ductility.  Typical stress/strain diagrams for various grades of steel What is the section? The size and shape of the cross-section of the piece of material used For timber, usually a rectangle For steel, various formed sections are more efficient For concrete, either rectangular, or often a Tee Why different shapes and sizes What shapes are possible in the material? What shapes are efficient for the purpose? Obviously, bigger is stronger, but less economical The strength and stiffness of each member is determined by both the shape and size of its cross-section, and the properties of its material. We can make some simple rules about the kinds of sections that are most suitable as beams, columns etc. , and the kinds of section shapes that are possible in the different materials. More on this later. We can calculate the properties of a particular section, to determine how much stress the material will have to bear, or how much the member will deflect. The section properties are easily determined by geometry. Wyatt Ch 8 Which way around? Beams are oriented one way The X-axis is the strong way Some lateral stiffness is also needed Columns need to be stiff both ways (X and Y)  What happens when a beam bends? Draw two parallel lines across the beam They stay straight, but no longer parallel The central layer neither stretches nor squashes  Take a beam and draw two parallel lines down its face. Now bend the beam into a shallow arc. Theory and experiment both indicate that these two lines remain straight, but they now arent parallel. The fibres on the top (the compression side) have got shorter, and those on the bottom (the tension side) have got longer, as we would expect. There is a layer in the middle that stays the same length. We will need to find the layer which is in neither compression nor tension. It is called the neutral surface, and the line where it occurs on any cross-section is called the neutral axis of that section. It happens that the neutral axis of any section passes through its geometric centroid. For a symmetrical section, that is in the middle.  The T-section is unsymmetrical one way; the - L-section is unsymmetrical both ways. Schodek p520-2; Wyatt p85,87 There can be any number of neutral axes, depending on which way we bend the beam. Practical beams are usually upright in section and we bend them in the vertical plane, i.e. about a horizontal axis. There is often a need to know the properties about the vertical axis as well. The horizontal and vertical axes usually represent the strongest and weakest orientations of the beam (but not with the L-section). Where elasticity comes in Stress is proportional to strain Parts further from the centre strain more The outer layers receive greatest stress  The section fights back The stresses developed resist bending Equilibrium happens when the resistance equals the applied bending moment How much curvature exists at this stage?   The internal moment of resistance of the beam is caused by the loads bending the beam and therefore stressing its material. Schodek p212-218; CGW p57 If the material is elastic, the stresses in each of these fibres will be related to the strain in each of them. So there will be some tension and some compression acting on the cross-section, with a lever-arm between these forces. This forms a couple. If the total amount of compressive force is C and the total tension is T , (for horizontal equilibrium C=T) and the lever arm between them is a , then the moment of the couple is M = Ca = Ta. This is the internal moment of resistance of the section. A measure of stiffness I Simple solutions for rectangular sections Doing the maths (in the Notes) gives the Moment of Inertia  Below, we show one of the intuitive approaches to the analysis. Following that is a slightly more rigorous approach. It is important to see the assumptions that go toward these solutions, and to understand how to use the properties that are derived. In a later section we use these properties in the design of beams to demonstrate the uses. Schodek p522-3; Wyatt p87,89 When a beam bends, one face is stretched and the other face is compressed. When this happens, the fibres of the beam resist these deformations and the beam fights back against the attempt to bend it. The beam will bend just enough so that its moment of resistance balances the externally-applied bending moment. Obviously the shape of the cross-section of the beam has something to do with both strength and stiffness - and also the properties of the material do. Just try to bend a scale on its flat, and on its edge. It is stiffer one way than the other. (It is also stronger one way than the other, but dont break it just to find out!) In following paragraphs we look at ways of analysing a rectangular cross-section. Similar principles apply to other shapes, but they are more complicated. A measure of stiffness I The bigger the Moment of Inertia, the stiffer the section It is also called Second Moment of Area Contains d3, so depth is important The bigger the Modulus of Elasticity of the material, the stiffer the section A stiffer section develops its Moment of Resistance with less curvature Simple approach to finding the Moment of Inertia  Think of the beam as being made up of many fibres, all laid together as a bundle. The stress in any of these fibres is proportional to its strain. From the way the beam bends (see above), we can see that the further from the neutral axis, the more is the lengthening or shortening of the fibres. Each fibre contributes a little to the total compressive or tensile force. Each of these little forces contributes to the moment of resistance about the neutral axis. For a fibre distant y from the neutral axis, (a) the stress is proportional to y, and (b) the lever arm is proportional to y. so that its contribution to the moment of resistance is proportional to y2. The total moment of resistance is made up by summing the effects of all these fibres. This property is called the second moment of area. It is given the symbol I in English-speaking books. In the diagram, consider the group of fibres distant y from the neutral axis. The force contributed by this strip is proportional to b "y y, and the moment is proportional to b"yy2. The total contribution is I = "b"yy2, summed across the whole section. The second moment of area, which is a measure of bending stiffness, is a function of b and d3. Depth of section contributes greatly to both strength and stiffness. I = bd3/12 (mm4). Expressed as an integral, this is I = by2dy, between the limits of y =d/2. For the rectangular section, this gives I = bd3/12. (If b and d are both in mm, I is in mm4.) A more rigorous approach: This diagram shows part of a beam, bent into a circular arc, with the top face stretched and the bottom face compressed.   When the beam bends into a circular arc, one face gets shorter while the other gets longer. There will be a layer in the middle which stays the same length, called the neutral axis. In a rectangular section, it is halfway down. How much strain at the top face? In the diagram, we have drawn two vertical lines distant x apart. Since x is the length of the piece of the neutral axis we are looking at, it represents the  original length before this piece of the beam was bent. Strain is change in length / original length. The diagram shows "x as the change in length of the top face due to bending. So the strain at the top face e = "x / x But the two shaded triangles shaped  are similar triangles. So the ratio of their short sides = the ratio of their long sides, i.e. "x / x = y / R (= e) How much stress at the top face? The stress at the top face = strain x Modulus of Elasticity (f = e E) f = E. y / R but since y = d/2, f = E d / 2R How does the section resist the bending moment? Looking at the lower diagram, all the tensile stresses in the top half combine into a tensile force T, and all the compressive stresses in the bottom half combine into a compressice force C. For horizontal equilibrium, C = T. Now we have to find the value of T (or C). The tensile stresses act on half the total cross-section, i.e an area of bd/2. But the stresses vary from a maximum of Ey / R at the outside face, to zero at the neutral axis. So the total force is average stress x area, T =  EQ \F(Ed, 4R) .  EQ \F(bd,2)  What is the lever arm between T and C? In the lower diagram, the stress distribution is triangular. The centroid of a triangle is two-thirds of the way in from the sharp end, so the lever arm is (2/3 x d/2) twice, which is 2d/3. Now we can put it all together and say the Moment of Resistance M (which equals the Bending Moment) M = T x  EQ \F(2d,3)  M =  EQ \F(Ed b d,8R) \F(2d,3) =  EQ \F(E,R) .  EQ \F(bd3,12)  This property of the section,  EQ \F(bd3,12) , is called the Moment of Inertia (or sometimes the Second Moment of Area), and in English is usually given the symbol I; so M = E I / R. But where does the stress come back into it? Combining the equation f = E y / R with M = E I / R, we get f =  EQ \F(M y,I) . The property of the section, I / y, is called the Section Modulus, and is given the symbol Z. So f =  EQ \F(M,Z) . For a rectangular section when the maximum value of y = d/2, Z =  EQ \F(I,y) =  EQ \F(bd2,6) . Naviers Theory of Bending  relates the stress, the applied bending moment, the section properties, and the radius of curvature caused by the bending. A measure of strength Z Simple solutions for rectangular sections Doing the maths (in the Notes) gives the Section Modulus   For unsymmetrical beams, there is a different Z top and bottom. The greatest stress occurs at the thin end. Schodek p220-2 and 526-7; Wyatt p93 and 99-101 We can use the f = M y / I part of Naviers Theorem to find the stress at any point on the cross-section. But we are mostly interested in the maximum stress if that is o.k., it will be o.k. everywhere else. The maximum stress occurs when y is a maximum, i.e. at the extreme top or bottom fibre of the beam. The property I/y is called the Section Modulus, and is given the symbol Z . For a rectangular section, ymax = d/2, and therefore Z = I/y = (bd2 / 12) / (d / 2) = bd2 / 6. (If b and d are both in mm, Z is in mm3). For a T or L shaped section, the neutral axis is not halfway down. There is a different ymax for top and bottom, and therefore a different Z, and a different stress at the top and bottom extreme fibres. The thin end is further from the neutral axis. The highest stress occurs at the thin end. Structural steel manufacturers provide handbooks listing the properties of their sections. A measure of strength Z The bigger the Section Modulus, the stronger the section Contains d2, so depth is important The bigger the allowable stresses on the material, the stronger the section Other properties The area tells how much stuff there is Used for columns and ties Directly affects weight and cost The radius of gyration is a derivative of I, used in slenderness ratio  The radius of gyration, r If all the material in the section were concentrated at a distance r from the neutral axis, it would have the same stiffness as the actual section. It is a useful parameter for comparing the buckling resistance of different-shaped sections. The radius of gyration is useful in determining resistance to buckling. It is given by . (If I is in mm4 and A is in mm2, r is in mm.) What about non-rectangles? Can be calculated, with a little extra work Manufacturers publish tables of properties How do we use these properties Go back to the bending moment diagrams Stress = (Bending moment) / (Section modulus) Maximum stress occurs where bending moment is a maximum  Using Z to check the strength of a beam Given the beam size and material Given the maximum bending moment Use Stress = Moment/Section Modulus Compare this stress to the Code allowable stress  Designing a member implies that you know the bending moment, shear force and other requirements, and want to select a suitable section. It is assumed that you also know the maximum allowable stresses. Checking a member implies that you have been given the requirements and the section, and you want to check whether the stresses are within the allowable limits. The allowable stress in bending for normal structural steel adequately braced against lateral buckling is 165 MPa. (Light-gauge roll-formed steel sections are normally made of higher-strength steel, but we will use this value.) Allowable bending stresses for softwoods are in the range 5 to 8 MPa, and for hardwoods 8 to 11 MPa (with specially selected species much higher). There are many modifying factors, especially for timber. Designing to Codes of Practice is covered in more detail in later years. Checking Example: Take the beam with a UDL that we analysed previously. Maximum bending moment = 4 kNm. For this beam, check whether a softwood section 250 x 50 with an allowable stress of 8 MPa, is satisfactory. The maximum stress f = M / Z . Section modulus Z = bd2 / 6 = 50 x 2502 / 6 = 521000 mm3 = (0.521 x 106) mm3. Therefore f = 4 / (0.521) = 7.68 MPa. This is less that the allowable value of 8, so the section is o.k. Using Z to design a beam for strength Given the maximum bending moment Given the Code allowable stress for the material Use Section Modulus = Moment/Stress Look up a table to find a suitable section  Design Example. For the same beam, find a suitable steel purlin section or hardwood timber section. Steel using an allowable stress of 165 MPa Using f = M/Z, we have M/Z < 165 MPa, or Z > M/165 in appropriate units. Now M is usually in kNm, which is the same as (MN)(mm), and f is in MPa, which is (MN)(mm)-2. Using the numbers for M and f directly, we get Z in 106 mm3. Steel handbooks list values of Z in 103 mm3. Required Z > 4/165 = 0.024 x 106 mm3 = 24 x 103 mm3. Looking through a catalogue of steel purlin sections we find C15020 (C-section, 150 deep, 2.0mm thickness) has a Z of 27.89 x 103 mm3. This is the smallest section that equals or exceeds the required strength. Timber using an allowable stress of 11 MPa Z > M / 11 gives Z > 4 / 11 = 0.3636 x 106 mm3. We have a choice of values for two variables (b and d) and only one equation. Therefore there is no single correct answer. Small timber beams are commonly 50 or 75 mm wide, so let s try b=50 to start. Z = bd2/6 >363600. All dimensions are in mm, so Z is in mm3. If b = 50, d > "(363600 x 6 / 50) = 209 mm If b = 75, d > "(363600 x 6 / 75) = 171 mm Timber comes in multiples of 25mm, so suitable sizes are 225 x 50 or 175 x 75. What controls deflection? Both E and I come into the deflection formula (Material and Section properties) The load, and (span)3 Note that I has a d3 factor Span-to-depth ratios are often used as a guide  Schodek p498, 504, 508 Note that the deflection formula has L3 on the top line and I on the bottom line. I is a function of d3. Therefore the ratio L/d (the span-to-depth ratio) is a useful guide to preliminary sizing of beams. Some rule-of-thumb guides were used in MAFIB and will also be discussed next year. Schodek p 241-4; Wyatt Ch 11 A beam subject to a constant bending moment bends into a circular arc. As the bending moment reduces, the radius of curvature increases, until when M = 0, the beam remains straight. If the bending moment changes sign (as in a beam with a cantilever at the end), the curvature changes direction. It changes direction at the point where the bending moment passes through zero, which is called a point of inflexion. Visualising and sketching the (exaggerated) deflected shape of a structure is useful in understanding the behaviour of more complicated structures, such as rigid frames. Computer programs (such as Multiframe, which will be used in an assignment in this unit-of-study) can plot deflected shapes as well as bending moment and shear diagrams. Deflection formulas simple beams  Deflection formulas cantilevers  Deflection formula built-in beam The deflection is only one-fifth of a simply supported beam Continuous beams are generally stiffer than simple beams  Schodek p241-3; Wyatt p113-7 The maximum deflection of a beam is given by , where k is a coefficient that depends on the type of loading and the support conditions, as listed in the slides above. Tables showing the deflection coefficients for a range of loading conditions are given in the textbooks, e.g. Wyatt, p117 Note particularly the difference between the deflection of a simply supported beam and a built-in beam. Making beams continuous over supports or connecting them rigidly to stiff columns will greatly reduce deflections. However, there are often constructional problems in achieving this. The deflected shape is always two orders higher than the shape of the bending moment diagram. Expressed as an integral, it is Deflection = (slope) dx = (moment) dx dx = (loading) dx dx dx dx. Using I to check the stiffness of a beam Given the beam size and material Given the loading conditions Use formula for maximum deflection Compare this stress to the Code allowable deflection  Example. Check the beam designed above (steel channel solution) for deflection. The maximum deflection should not exceed span/500. The deflection for a simply-supported beam with a UDL is " = (5/384) x WL3/EI. Because of the various units in use, let s work entirely in N and mm (and MPa, which are N/mm2).  Section = C15020 E = 200 000 MPa I = 2.119 x 106 mm4 " = (5/384) x 8000 x 40003 / (200000 x 2.119 x 106) mm = 16 mm. Now the maximum allowable deflection is (span/500) = 4000/500 = 8 mm. So the section we have chosen deflects too much. We would need to choose a stiffer one.  Sections under consideration By proportion, we can see that we need just twice as much I. We could use two of the same section back-to-back (using 100% more material), or a deeper section. A C20020 (200 deep, 2.0 thick) has more than twice the I, yet it uses only 27% more material. Most texts suggest that beams be designed for strength and checked for deflection. Experience indicates that this is a good strategy for heavily-loaded beams, but with lightly-loaded ones deflection is more likely to be critical. Always allow enough depth for the structural system. Skimping on headroom may produce an impossible structural solution, or at best an expensive one. Using I to design a beam for stiffness Given the loading conditions Given the Code allowable deflection Use deflection formula to find I Look up a table to find a suitable section  At the end of the previous example, we re-designed the beam to provide the required stiffness. What section to use?  Certain shaped sections are useful for beams, or columns, or both. But the shapes that are available depend on the material and method of manufacture. Schodek p 247, Fig 6-24 Beams need a large value of I and Z in the direction of bending. They need some stiffness the other way to resist lateral buckling. Columns usually need a large value of r in both directions. In deriving the Moment of Inertia (I), we noted that the contribution of any part of the section was a function of y2, where y is the distance from the neutral axis. Therefore the I -beam uses material efficiently, because most of the material is a long way from the neutral axis in one plane. Hollow sections have the material spaced away from the neutral axis in both planes. I - and H -sections can be hot-rolled in steel (relatively heavy sections). C - and Z - sections can be roll-formed from light-gauge steel, fairly flexible sideways and in torsion, need corrosion protection because of thinness. Round and square tubes can be formed and welded from steel plate. Because of its high shear strength, steel only needs a thin web. Thicker T - and inverted L -sections can be formed in concrete. When concrete is used for a floor, a flat top surface is necessary. Concrete is weak in tension and shear, but steel reinforcement is added to resist these stresses. Timber is most easily sawn into rectangular sections. The low strength in horizontal shear means that a thin web is not a good idea anyway, unless it is made of plywood. What else can go wrong? Deep beams are economical but subject to lateral buckling Beams have to be connected to the rest of the structure The worst condition may happen under partial load or wind uplift, rather than maximum downward load 6 TRUSSES Why trusses? A truss provides depth with less material than a beam It can use small pieces Light open appearance (if seen) Many shapes possible Schodek p115-120, 145-149; Lateral stability p159-161 CGW p79-80; 3-dimensional p145-148 Consider the alternatives. A beam will do the same job as a parallel-chord truss. If there is plenty of height available, an arch or suspension cable might also be possible. A beam is likely to be shallower, but heavier than a truss. Materials. Most trusses are of steel or timber. Reinforced concrete is possible but unusual. Loading. Determine the applied loads. Are they applied to top or bottom chord? Is the loading continuous or discrete? Member layout. If possible, arrange for loads to fall on panel points. Compression members are subject to buckling, so the shorter the better, within reason. Why not trusses? Much more labour in the joints More fussy appearance, beams have cleaner lines Less suitable for heavy loads Needs more lateral support Jointing. The material will influence the jointing methods. Steel trusses are normally welded, timber uses bolts, nailplates, or steel gussets (details are covered in Year 2). Lateral stability. Members are restrained in the plane of the truss at each panel point. To resist lateral buckling of individual members or the truss as a whole, the whole truss must be braced against falling out of its plane. This requires lateral support to both chords. Real applications Domestic roofing, where the space is available anyway Longspan flooring, lighter and stiffer than a beam Bracing systems are usually big trusses  Realistic shapes Span-to-depth ratios are commonly between 5 and 10 This is at least twice as deep as a similar beam Depth of roof trusses is to suit roof pitch  Truss shape. This will usually be determined by the available height and the required line of top and bottom chord. Span-to-depth ratios between 5 and 10 are common. Bracing systems often form very deep trusses. Shallow trusses (L/D of 15 or 20) are possible, but unlikely to offer much advantage over beams. The shallower the truss, the greater the force in each chord. Making the joints Gangnail joints in light timber Gusset plates (steel or timber) Nailplate joints (Gangnail is a trade name) are at present the cheapest and simplest way of making joints in a light timber truss. They allow all the members to lie in one plane, which has some advantages over using double members. They are installed in a factory (using a hydraulic press), so the dimensions of the finished truss must be compatible with transport. Gusset plates (of plywood or steel) have long been used. They are effective but bulky and expensive. Older methods of making timber joints included various timber connectors placed between overlapping members, all held together with a bolt. Making the joints Welded joints in steel Various special concealed joints in timber In steel, welded joints are simple and effective. An elegant, but labour-intensive solution for timber trusses is to fabricate steel plates that are set into a slot within the members (or between double members), and bolted through. Only the timber and the bolt-heads are seen. How trusses work The members should form triangles Each member is in tension or compression Loads should be applied at panel points Loads between panel points cause bending also Supports must be at panel points  The chords and the web The top and bottom chord resist the bending moment The web members resist the shear forces In a triangular truss, the top chord also resists shear  What we need to know For detailed design, forces in each member For feasibility design, maximum values only are needed  How to analyse a truss Find all the loads and reactions (like a beam) Then use freebody argument to isolate one piece at a time Isolate a joint, or part of the truss  As with any other element, we first need to find all the external forces acting on the truss (including the reactions). So far, the truss acts exactly like a beam. Now we want to find the forces in the internal members. We know several things that will be useful: each joint of the truss is in equilibrium under all the forces acting on that joint. We know the line of action, because the forces in the members are either direct tension or compression. if we want to cut off part of the truss as a freebody, then that part is in equilibrium under all the forces acting on it. That includes the forces in the members that we have cut through. Generally we will have cut two or three members. By doing this we temporarily turn the internal forces into external forces, so we can get at them. Method of joints isolating a joint Have to start at a reaction Time-consuming for a large truss  This method is good if there arent too many members, or if you only need the ones near the supports. Otherwise it is time-consuming. Schodek p123-6 ; Wyatt p57-59 Find the reactions. Once you have the loading system, consider the whole truss as a single unit (freebody). Find the reactions using SV=0, SH=0, SM=0. Start at one support. Now consider the joint as a freebody, acted on by the reaction, a load (if any), and the unknown member forces that meet there. All these forces are in equilibrium. Resolve them all vertically and horizontally. In most cases you can solve two unknown forces without needing simultaneous equations. Move to the next joint. Consider this joint as a freebody. You already know the force in one of the members, and with luck can get the next two. Move from joint to joint. By choosing the right sequence, you should be able to move right through the truss without having to carry forward simultaneous equations. Keep to a sign convention. The usual convention is that all members are assumed to be in tension. A negative answer means it is really in compression. For a member in tension, the arrow acting on the joint goes away from the joint. (For the left end of a member, the arrow goes to the right. For the right end of the same member, it goes to the left.)  Diagrams for example (The actual dimensions dont matter, if we know the angles. If units are given for the loads, they should be included in the answers.) Example (see diagram above). Find all the member forces. Reactions. By moments, or by symmetry, each reaction is half the total load. At A - (vertically). 4+AB = 0. AB = -4 (i.e., compression). (horizontally) AD = 0. At B. (vertically) -1 -(-4) -BD cos 45 = 0. BD = + 4.24 (horizontally) BC + 4.24 sin 45 = 0. BC = -3 At D (vertically) 4.24 cos 45 + DC = 0. DC = -3 (horizontally) -0 -4.24sin 45 + DF = 0. DF = +3. At C (vertically) -2 -(-3) -CF cos 45 = 0. CF = +1.41 (horizontally) -(-3) + CE + 1.41 sin45 = 0. CE = -4. At E (vertically) -2 -EF = 0. EF = -2. That completes half the truss. The other half is the same by symmetry. In this notation it doesnt matter whether we say AB or BA, etc. Note that the biggest chord forces are near the middle, and the biggest web forces are near the ends. This is always true for parallel chord trusses with fairly uniform loading. Method of joints dealing with inclined forces Resolve each force into horizontal and vertical components  Method of sections cutting through members Quick for just a few members  This method is good if we only want to know a few member forces - say the end diagonal and the middle chords.  Diagrams for example Schodek p130-8; Wyatt p59-61 Find the reactions. As before, find the reactions using SV=0, SH=0, SM=0. Make an imaginary cut in the truss, passing through the member you want to find. Keep to a sign convention as before. Example (see diagram at left). Find the forces in BD, CE, DF. Reactions. By moments, or by symmetry, each reaction is half the total load. Make a cut to pass through one of the members we want (in this case BD). Consider the part on the left as a freebody. Mark all the forces acting on it, including the members we have cut off. The freebody is in equilibrium. Using SV = 0, 4 -1 -BD cos 45 = 0. BD = -4.24. Make another cut to pass through CE and DF. Again, mark all the forces acting on the piece on the left. We can use any of the equations of equilibrium. It is more elegant (and easier) to pick one that gives the answer simply. If we take moments about a point through which several unknowns pass (therefore they have no moment about that point), it is usually possible to get the wanted unknown in one go. Using SM = 0 about F, considering all forces acting on the bit to the left, 4 x 6 - 1 x 6 - 2 x 3 + CE x 3 = 0. CE = -4. Using SM = 0 about C, 4 x 3 - 1 x 3 - DF x 3 = 0. DF = +3. We have found only the members we wanted. We could have gone on to find any or all the others. This method is very useful to find the maximum chord forces in a truss like this:  Graphical method drawing conclusions* Uses drafting skills Quick for a complete truss  This method was once popular with both architects and engineers, because it could be done on the drawing board instead of using long calculations. With the use of calculators and computers, it is less popular today. Schodek p127; Wyatt p53-7 When several forces meet at a point, the resultant or equilibrant can be found graphically, by drawing a polygon of forces. The equilibrant is the line that closes the polygon. Imagine solving a truss by starting at one end like the Method of Resolution at Joints, but doing it graphically instead. The unknown at one joint becomes one of the known forces at the next. If we overlay the diagrams, we dont need to do too much drawing to solve the whole thing. The diagram is called the Maxwell Diagram. For a fuller description, see one of the references. A key to the use of the Maxwell diagram is the notation, called Bows Notation, in which the spaces between the members, and between the loads, are given names instead of numbering the joints. Begin by selecting a scale that will fit on the page, and draw a line representing the loads and the reactions. (If all the loads are vertical, so is the line). This is line a b c d e f g in the diagram above. ab is 1 unit down, bc is 2 units down etc, because these are the forces between the zones a, b, c etc. fg is 4 units upward, as is ga, because these are the reactions. By equilibrium, the sum of the loads is equal and opposite to the sum of the reactions, so this line is self-closing. (There is also a graphical way of finding the reactions, but it is much easier by calculation.) Now try to find one of the zones near a reaction - say h. We know that the member separating a and h is vertical, and hg is horizontal. Draw lines in these directions through a and g. They meet at g!. The length of hg, and therefore the force in the member between h and g is zero. Try to find the next zone - i. We know that bi is horizontal and ih is at 45. Draw these two lines, and they meet in i. Now ij is vertical and jg is horizontal. This locates j. Proceed to k in the same way, and half the truss is solved. Complete the other half, which is symmetrical if the loading and layout of the truss is symmetrical. Measure the length of each line, using the scale you started with. This gives the force in each member. There is a convention for determining tension or compression, given in the references. * (That was a play on words, not the official title of the method. ) Quick assessment parallel trusses The chords form a couple to resist bending moment This is a good approximation for long trusses  This method gives the correct answer for one of the chords, and errs on the safe side for the other one. The more panels in the truss, the less the error. The depth, d, is the centre-to-centre distance of the chords, not the overall depth. It only finds the chord forces, not the webs, but the chords are usually critical to see whether the truss is feasible. Quick assessment pitched trusses The maximum forces occur at the support  This finds the maximum chord forces, which are critical for the feasibility of the truss. As the truss gets shallower, the forces go up rapidly. 7 AXIALLY LOADED MEMBERS Axial tension members Tension members occur in trusses, and in some special structures Load is usually self-aligning Efficient use of material Stress = Force / Area The connections are the hardest part Tension members are a very economical way to use material. A member in tension will always tend to pull straight, so slender members, or flexible cables, can be used. Schodek p278-287; CGW p53 The members of a pin-jointed truss are subject to axial tension or compression. There is practically no bending, unless loads are located between the panel points. The stress is force/area. The whole cross-section of the member is used, and therefore a truss is fairly economical of material. In a tension member we have to deduct any holes used for connections (bolt holes etc.), since the member will fail through the thinnest part. Axially loaded piers For short piers, Stress = Force / Area For long columns, buckling becomes a problem Load is seldom exactly axial Compression members are more common than tension ones. (All columns are included.) They are also more complicated and rather less economical, because of the buckling problem. Columns, and the compression members in trusses, carry axial compression. A compression member will only fail in true compression (by squashing) if it is fairly short; it is called a short column. Otherwise it will buckle before its full compressive strength is reached; then it is called a long column. The overturning moment (OTM) Horizontal load x height Load x eccentricity  The overturning effect is the same, whether caused by the load itself being off-centre, or by a separate horizontal force pushing the pier over. Eccentrically loaded piers The average compressive stress = Force / Area But it isnt uniform across the section Stresses can be superimposed  Stress diagrams. The shaded stress diagrams shown here are simply a graphical representation of how the magnitude of the stress varies as we move across the base of the pier. The bigger the vertical dimension, the bigger the stress. I have drawn compression downwards (the pier is pushing down on the foundation), and tension upwards (the pier is trying to lift off the foundation). In an elastic material, where stress is proportional to strain, it is acceptable to split up the internal actions into several distinct parts (such as a bending moment and an axial force), determine the stresses resulting from each of them, and add up the stresses. Does tension develop? Stress due to vertical load is P / A, all compression Stress due to OTM is M / Z, tension one side and compression on the other Is the tension part big enough to overcome the compression? What happens if it is? Schodek p282-3 When a pier or column is loaded concentrically, the stress is a uniform P/A across the whole section. (In the following discussion, let us assume that the force P is the resultant of the weight of the pier and an applied load.) If the load P is eccentric by a distance e, the effect is the same as a concentric load P and a bending moment equal to Pe. The stress is then the sum of P/A and Pe/Z . The P/A component is compressive across the whole section, and the Pe/Z varies from compressive on one side to tensile on the other. For a rectangular pier, of dimensions d x b, the section modulus Z = bd2/6, and the area A = bd. Therefore the stresses at the two extreme edges of the pier are given by f = (P/bd) 6Pe/bd2. If the pier is asymmetrical in plan, there will be a different value of Z for each face. Does tension develop? If eccentricity is small, P/A is bigger than Pe/Z If eccentricity is larger, Pe/Z increases Concrete doesnt stick to dirt tension cant develop!  Middle-third rule the limiting case For a rectangular pier Reaction within middle third, no tension Reaction outside middle third, tension tries to develop  From the diagrams above, it is obvious that a slightly eccentric load will leave the whole pier in compression, while a more eccentric load will cause tension on one side. The value of the eccentricity e that corresponds to the boundary between these two conditions occurs when the lower value of f equals zero: 0 = (P/bd) - 6Pe/bd2, which gives e = d/6. This is the middle-third rule (since the resultant can fall anywhere within d/6 either side of centre without causing tension). The middle-third rule always applies when the material cannot develop tension - e.g. masonry laid in lime mortar, or masonry separated by a damp-proof course, or a footing sitting on the foundation material. In these cases, if the resultant falls within the base but outside the middle third, part of the base ceases to carry any stress, and the stress under the remaining part increases rapidly. Horizontal loads on piers The overturning effect is similar to eccentric loading We treat them similarly There is only the weight of the pier itself to provide compression  When the overturning is caused by a horizontal load, there is only the weight of the pier to counteract it. When it is caused by an eccentric vertical load, that load is added to the weight of the pier itself. Extra weight helps Extra load helps to increase the compression effect, and counteract tension  Medieval cathedrals make great use of extra decorative pinnacles, to add weight to the walls and piers. They have to counteract an outward thrust from the roof (roof trusses generally werent used). How safe is my pier? Will it sink? (Can the material stand the maximum compressive stress?) Will it overturn? Reaction within the middle third factor of safety against overturning usually between 2 and 3 Reaction outside middle third factor of safety inadequate Reaction outside base no factor of safety In small-scale buildings, the strength of the foundation is less likely to be a problem (but we cant ignore it!). When the pier or wall gets beyond the safe situation towards overturning, a very high stress is put on the outside edge of the footing. The middle-third rule is a good safety rule against the pier letting go on the tension side. It is useful, but not definitive, against sinking on the compression side. Slender columns A slender column buckles before it squashes A slender column looks slender We can quantify slenderness by a ratio The mimimum breadth, B, or the radius of gyration, r The effective length, L The slenderness ratio is L/B or L/r CGW p54-56; Schodek p283; Wyatt p30. Buckling is an elastic phenomenon. A slender elastic member may become unserviceable by excessive buckling without suffering any permanent damage to itself. The buckling load depends on the Modulus of Elasticity of the material, and a ratio taking into account the length and stiffness of the actual cross-section used. This is usually expressed in terms of the moment of inertia, I or the radius of gyration, r of the section. For a rectangular section, a ratio using the actual width of the material can also be used. The slenderness ratio For timber and concrete limit for L/B is about 20 to 30 For steel, limit of L/r is about 180 At these limits, the capacity is very low: efficient use of material, the ratios should be lower The ratio L/r (length divided by radius of gyration) is called the slenderness ratio. It is dimensionless. The effect of buckling is calculated by the Euler Buckling Formula, which gives either the critical buckling load, PCr, or the critical buckling stress, FCr. PCr =  EQ \F(2 EI,L2) , or FCr =  EQ \F(2E,(L/r)2) . This occurs in a different form in Wyatt (p30) where the radius of gyration for a rectangular section, D/"12,is used directly. The stress FCr can be multiplied by the cross-sectional area to give the load PCr. The Euler formula assumes that the column is pin-jointed at both ends. A real column might have the ends built-in, in which case the effective length is less than the real length. For a flagpole (not restrained at the top at all), the effective length is twice the real length. For end conditions and effective lengths, see Schodek p 289 The diagram below shows the column load as calculated by the long column formula, (where buckling is the criterion), and by the short column formula, where compression is the criterion. Design Codes adopt a compromise formula , which is safe in both long and short regions.  Allowable stress in columns depends on the slenderness ratio Which value of I or r do we use? Unless the strut is a round or square section, it is stiffer in one plane than the other. It will always tend to buckle in its weakest direction, so we use the smallest r. Sometimes a strut is restrained at closer intervals in one plane than in the other. In that case we use the combination that gives the largest value of L/r. The implication of this is that struts are more efficient if they are stiff in both directions. Hollow tubes (round or square) are the most efficient shapes, because they use a minimum of material to create reasonable stiffness in both planes. The buckling stress The buckling stress increases with E (so steel is better than aluminium) The buckling stress reduces with (L/r)2 (so a section with a bigger r is better) This just tells us that some combinations of material, section, and layout of connecting members, might give a more efficient use of material for columns. A final decision must take into account many aspects of the building, apart from the efficiency of one particular element. How do we improve performance? L/r may be different in each direction Can we support the column to reduce L? Can we use a section with a bigger r in both directions? A common case is the studs in a stud-frame wall. Their weaker dimension gets extra support from the noggings, so that the L/B ratio ends up roughly equal in both directions. (for 100 x 50 in a 2400 storey-height, with one row of noggings, it is 24 each way). Good sections for columns Tubular sections are stiff all ways Wide-flange (H) beams better than I-beams Squarish timber posts rather than rectangular  EMBED Word.Picture.6  Hollow sections ()can be made in metal, but they are not easy to connect together. Large structural steel columns are usually of broad H rather than the more slender I section beams. Timber columns are usually square, or sometimes spaced sections II joined together at intervals. 8 MEMBER SELECTION The variables in any given situation are: The spans and arrangement of the building The loads to be carried (which are partly determined by the size and materials of the building) The cross-sections used for the members The stresses that the materials can safely carry. We are not going to be able to come up with a single, mathematically-determined right answer to any problem involving such an array of variables. If we know 1, 2 and 3, we can determine the actual stresses that the materials will be subjected to, and then look up properties of the material to see whether that is safe. This is checking an existing design. In conventional structural design, we usually know 1, 2 and 4. Then we can select cross-sections that will be suitable. In remodelling an existing building, we may know 1, 3 and 4. We can check what loads the building can support. In architectural design, we want to have control over item 1, and to some extent item 3. This unit-of-study will have given us an introduction to the relationships between all these factors. Later units-of-study will add to our experience of working with structural systems and materials. The building and its relation to the site One characteristic that distinguishes a building from most other manufactured artefacts, is its relationship to a particular site. The structural system of the building must transfer all the expected loads to the foundation. The type and size of the footings and the form of the building structure are interdependent. Both are influenced by the magnitude of the loads and the characteristics of the foundation material. The building as a whole Schodek ; stability p12-18; planning & grids p427-436 design issues p445-458 lateral loads p468-475 constructional approaches chapter 15 Wyatt: bracing p63-69 In parts of this unit-of-study we have broken the structure down into its elements in order to understand them separately. It is important first to see the building as a whole. It has functional and aesthetic requirements that influence the form and size, and the choice of materials and of structural system. The structural system must provide stability in all directions. Planar elements are stiff in their own plane but require support against out-of-plane forces. The need for open spaces and glazed or openable walls will determine where support can or cannot be provided. The structural system may consist of the enclosure itself; or an exposed framed system associated with, or separate from, the enclosure; or a framed system concealed within the enclosure. Each of these alternatives will produce a different character in the building, it will put different constraints on the process and sequence of construction, and it will apply different constraints to the initial planning and the future alterability of the building. 9 CONCRETE MAKING AND TESTING Concrete materials Cement Coarse aggregate Fine aggregate Water Additives Concrete is unique among modern building materials in that it is made specially for each job, and its handling on the job affects its quality. Portland Cement The common type of cement is called Type A. There are also rapid-hardening, low-heat, sulphate-resisting, off-white, white and coloured cements. High-alumina cement is chemically different from Portland Cement. It has been found to be unstable and is no longer used structurally. Cowan & Smith, Science & Tech of B.Materials, p 113-117 Cement is made by heating, in a kiln, a mixture of limestone, shale and ironstone. The resulting clinker has compounds of calcium aluminates, silicates, and aluminoferrites. Each component reacts with water to form first a gel, then a rigid mass. The components react at different rates. The process of allowing these reactions to take place is called curing. The chemical reactions are also speeded up by higher temperatures. Concrete gains strength very slowly in the 0-10C range. Below 0C, the water freezes and the concrete is spoiled. Aggregate The grading and shape of the aggregate particles is essential in the design of the mix, but this is mostly done by the pre-mix supplier. Cowan & Smith, Science & Tech of B.Materials, p122-124 The aggregate must be at least as strong as the final concrete. It is there to provide bulk (cheaper than cement), and also to reduce shrinkage and improve durability. Only the cement shrinks. If you want exposed-aggregate finish, then the selection of the aggregate and the cement colour are essential. Structural concrete uses 20mm maximum aggregate. In heavy civil works, larger aggregate allows a more economical mix but it is harder to place and finish, and the spaces between reinforcement must be larger. Water The water is the cheapest, and arguably most critical, of the components. See the effect of water/cement ratio later in this section. Cowan & Smith, Science & Tech of B.Materials, p 128 Impurities may delay or reduce the strength-gaining process, so water must be clean. Additives Many additives are available to improve various aspects of the concretes properties. Most should be treated with caution and expert advice sought. Cowan & Smith, Science & Tech of B.Materials, p 128 Plasticisers and water-reducers. Since more water is needed as a lubricant than is needed for the chemical process of setting, anything that makes the water wetter is an advantage. Commonly used in commercial mixes. Pozzolans. Roman cement used natural volcano byproducts which acted like cement. Modern fly-ash (from coal furnaces) acts similarly, and is routinely added to reduce the cement content of commercial mixes. It has some benefits, (including economy), but requires even more attention to curing than pure cement. Accelerators and retarders. Like the name says. Used in cold or hot weather, or to retard a surface for exposed aggregate. OK if used properly. Waterproofing additives. These are used to seal up the microscopic pores that develop as excess water is evaporated from the concrete. Most users now prefer to use a low-water-ratio concrete, which can be waterproof in itself. Proportions Cement paste coats all surfaces of aggregate Fine fills the spaces between coarse aggregate  The detailed grading of the aggregate is done by the concrete supplier. This is one of the advantages of readymixed over site-mixed concrete. The old 1-2-4 mix (1 cement, 2 sand, 4 gravel) is usually too harsh to finish smoothly, but the ratio of cement to total aggregate is usually about the range 1:4 to 1:7. Weighing and mixing Weigh-batching necessary for accuracy Mixed in factory and transported in agitator truck Site mixing uneconomical, not accurate enough Older textbooks will show various means of site-batching. The quality of the concrete made that way was usually much lower than modern, weigh-batched concrete. Handling on site Chute and barrow Concrete pump Vibrators and vibrating screed Trowelling machine Cowan & Smith, Science & Tech of B.Materials, p 131 Concrete must be placed well before the initial set takes place (which is about 2 hours after adding the water, or less in hot weather), and not disturbed again after that. On site, it is placed either directly from the truck chute, or in barrows on a small site, or by pump or crane-skip. It must never free-fall more than about 1m, to avoid separation of coarse and fine components. Pumping is now commonly used on all but the smallest jobs. To eliminate air pockets and ensure compaction around the reinforcement and into corners of the formwork, it must be worked or vibrated. Immersion vibrators are most common. The surface is roughly levelled with shovels, then screeded to the required levels, and finally trowelled (usually with a trowelling machine). Roads and paths might be finished with a wood float (for a rougher surface), or broomed (for an even better non-slip surface). Stages of setting Initial set Hardening Ageing When cement combines with water, several different chemical compounds are formed, at different rates. They all contribute partly to the strengthening process. The initial set (stiffening of the concrete) begins within about 2 hours of the addition of water. The hardening (the main gaining of strength) begins shortly after this, and the main gain is during the first week. We usually consider the concrete to have gained most of its strength by 28 days, although the process continues at a slower rate after that. Ageing goes on slowly for years. Opinions are divided as to whether it ever stops, although the rate is negligible after a year or so. Curing Keep moisture available throughout the setting process Concrete gains most strength in 28 days, but continues Fast and slow setting cements Temperature affects rate The gain of strength only proceeds as long as water is available to combine with the cement. If the process is interrupted and then resumed, the result is not as good as if moisture is kept up continuously. The test cylinders will be cured under ideal conditions. The results tell you what the concrete would have been like if it was cured under ideal conditions also. Curing is achieved by covering with plastic sheeting or a spray-on wax membrane, or ponding with water, or maintaining a full-time spray system. Concrete will set under water, if you can stop it from washing away in the first few hours. Strength Water/cement ratio affects strength More water = less strength Try to limit water content But it must be workable  The water-cement ratio (ratio by weight) is vital to the strength of the concrete. Too much water leaves minute fissures in the matrix and reduces strength. Less water makes a stronger, more waterproof and durable mix. Too little water prevents proper compaction and leaves air-pockets, ruining the strength. High-strength concrete requires a low water-cement ratio, and therefore the amount of aggregate must be reduced to achieve enough workability (that is the same as saying, add more cement). The chemical reaction requires less water than is needed for workability, but the reaction goes on for months. Moisture must remain available as long as possible. Workability Concrete must be fully compacted to remove air bubbles Water assists compaction Try to have it wet enough Use vibration to improve workability Use additives to improve workability A very dry mix is too difficult to compact. A very wet mix will flow easily into the formwork, but is too sloppy to handle and finish, as well as being too weak. The ideal combination of wet enough and dry enough requires some compromise. Measuring workability The slump test Not perfect, but simple and quick Good guide to uniformity between batches  EMBED Word.Picture.6   The slump test is an internationally recognised measure of workability. Its popularity is its simplicity, not its absolute accuracy. The can is filled with concrete, then lifted, to see how far the wet concrete slumps from its original height. Economy Cement is dearer than aggregate Try to limit cement content The cement content is responsible for the strength, but also for much of the cost, and the shrinkage. Shrinkage Concrete shrinks on setting Cement paste shrinks, aggregate doesnt Try to limit water content Good curing delays shrinkage Reinforcement helps limit cracks  Shrinkage is the dimensional change that occurs in some materials mainly as a result of loss of moisture. Materials like concrete, timber and brick are mainly affected. Shrinkage effects are expressed as strains and are dimensionless quantities. Shrinkage in concrete starts off rapidly, and continues for several years. Maximum values are of the order of 300 x 10-6 to 800 x 10-6. It is caused by the loss of volume when water and cement combine chemically, and by the loss of excess water not required for the chemical process. Good curing reduces the total shrinkage, but also allows the concrete to gain strength before most of the excess water is lost. The stronger concrete can better resist chrinkage cracking. Conflicting requirements The various requirements above are in conflict Good compromise solutions are possible Specifying and measuring strength The cylinder crushing strength (in MPa) Usually measured at 28 days Test cylinders cast on site Cured in lab before testing Strength is specified by the crushing strength expected from standard cylinders, tested at 28 days after making. The actual test results will have some variation, and statistical tests are used to determine whether a group falls within the expected range. Testing cylinders Cylinders crushed in lab at 28 days Capped with sulfur for accurate ends Some early tests at 7 days Drastic consequences if under-strength after 28 days If concrete is found to be under-strength after 28 days, the job has progressed, and serious contractual problems arise. Most suppliers test some samples at 7 days. They know from these results whether there is likely to be a problem at 28 days, and if necessary the job can be delayed. Testing other materials Quality control of other materials is usually done by manufacturers Homogeneous materials like steel are made to close tolerances of strength Natural materials like timber vary greatly, and are classified into several grades The only way to be sure of the strength of a sample of material, is to test it to destruction. Then you dont have it any more. Statistical sampling techniques must be used. They are most effective on large, uniform production runs. That is why a concrete plant, with a large throughput, can get better reliability than a one-off site mixing operation. However, a big civil engineering project in the country would usually establish its own concrete-mixing plant on site. Other materials are also tested, usually by the manufacturers (or, in the case of timber, the processors). Results from registered laboratories are usually accepted. Some typical properties of materials are included in the following table. MaterialTypical yield stress or ultimate stress MPaModulus of Elasticity MPaCoeff. linear expansion K-1Density kg/m3 (approx)Structural steel250 - 300 (high strength wires ~ 1000)20000012 x 10-67800AluminiumVaries, up to 1507000024 x 10-62700Concrete20 ~ 5020000 ~ 3000011 x 10-62400Brick3 ~ 2010000 ~ 250006 ~ 8 x 10-61900Timber30 - 150 (clear timber) 10 ~ 50 as used8000 - 15000Low (moisture movement dominates)600 ~ 1000PlasticsVariable, mostly low<5000High~ 1000GlassSheet glass, ~ 15, (but reduced by defects) Filaments, high500008 x 10-62700The density of water is 1000 kg/m3. Materials lighter than this will float on water. Page  PAGE 8 DESC 1004 Building Principles DESC 1004 Building Principles page  PAGE 9 ,-bqAJ !1#$&78rstu(13hitYZsΨ5CJCJ 56CJ6CJ5CJ6CJOJQJ 5OJQJ65CJOJQJ5OJQJOJQJ56CJ OJQJ56CJ OJQJCJCJ$ 5OJQJ jU?,bqAJK\t$d%d&d'd ! $d%d&d'd ! 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