# Theory of Structures- A Must Read for Be's

August 15, 2017 | Author: mbabu76067 | Category: Euclidean Vector, Structural Load, Force, Truss, Deformation (Mechanics)

#### Short Description

Theory of Structures...

#### Description

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Fundamental Properties of Structural Systems In this section, we discuss two important and related properties of structural systems: stability and statical determinacy. We will focus on external stability and indeterminacy, which is determined primarily by the type of external restraint provided to the structure.

Stability The checks of safety and serviceability that we must perform are based on the assumption that the structures we design (and the structural models we create to analyze them) are stable. Intuitively, we recognize an unstable structure as one that will undergo large deformations under the slightest load, without the creation of restraining forces. This is illustrated by the simple example shown below:

This structure is unstable due to the way it is supported. Given that we assume the roller supports provide no horizontal restraint at all, the slightest horizontal force will be sufficient to make the beam roll horizontally. The magnitude of this displacement cannot be calculated. No restraining forces in the horizontal direction are created, nor can any be created.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

In such cases, it makes no sense to perform calculations of strength and serviceability because the unstable structure is clearly unfit for its purpose. The structure shown in the figure above is two-dimensional. It follows that the equations of equilibrium are: 1. ΣFx = 0 2. ΣFy = 0 3. ΣM = 0 We observe, however, that there are only two unknown forces, the reactions RL and RR. We therefore have too many equations of equilibrium for the unknown forces available. In mathematical terms, we say that the system is under-constrained. As is noted in the figure, the addition of a horizontal restraint creates an unknown reaction at one of the supports, which creates an unknown reaction in the horizontal direction. This restores stability to the system. It is generally not sufficient to count up unknown reactions and compare them to the number of equations of equilibrium. As shown in the figure above, the addition of another roller support increases the number of unknown reactions but does not create a stable system. It can thus be stated that: 1. Systems with fewer reactions than equations of equilibrium are always unstable 2. Systems with number of reactions greater than or equal to the number of equations of equilibrium are not necessarily stable. Such cases must always be investigated by the designer by visualizing the displacement of the structure under the action of forces and moments in all possible directions.

Other Issues Related to Stability The stability issues discussed in the preceding section are referred to as the external stability of structures, since they relate to the number of external reactions relative to the number of equations of equilibrium. Structures that satisfy the conditions described above can have subsystems that are unstable, such as the structure shown in the figure below:

This particular issue, referred to as internal stability, will be dealt with in greater detail in the section on trusses.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Statical Determinacy and Indeterminacy Basic Principles We can extend our comparison of the number of reactions to the number of equations of equilibrium to structures with larger number of reactions. Unless otherwise stated, the remainder of this discussion refers only to stable structures. For the example considered previously, we can create stability by adding horizontal restraint at one end. For this structure, there are three equations of equilibrium and three unknowns. We can thus solve for the unknown reactions for any given load condition, and thus solve for unknown sectional forces for any given free body cut from the structure. This type of structure, for which the number of equations of equilibrium is exactly equal to the number of unknown reactions, is called statically determinate, since we can determine all unknown forces in the structure from the laws of statics alone.

If we add an additional restraint to the structure, such as rotational fixity as shown, we now have four unknown reactions. The number of equations of equilibrium obviously remains unchanged at three. Structures such as this one, for which the number of unknown reactions exceeds the number of equations of equilibrium, are called statically indeterminate structures.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

To solve for the unknown reactions, we need to increase the number of mathematical conditions to equal the number of unknowns. We do this by considering the characteristics of deformation of the structure under load. This topic will be dealt with in detail in the second half of the course. For now, it is sufficient to be able to distinguish between determinate and indeterminate structures. This is important because: 1. The choice of method of analysis is generally determined by whether or not a given structure is statically determinate or indeterminate. The computational effort required for determinate structures is generally manageable. The computational effort required for indeterminate structures, however, can be significant. 2. Determinate structures respond to imposed deformations differently from indeterminate structures. This distinction is a significant issue for structural designers. We will discuss this issue in more detail in the section on actions on structures. We refer to the degree of statical indeterminacy as the difference between the number of unknowns and equations. From a less mathematical point of view, we can regard the degree of statical indeterminacy as the number of restraints we would need to add to the structure to make it statically determinate. The two definitions are equivalent.

Structures with Internal Hinges For structures with internal hinges, we need to extend our definition of statical determinacy. We will explore this issue by means of the example shown below:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This two-dimensional structure has an internal hinge in the right-hand span, which provides no bending restraint but can transmit shear and axial force. We have three equations of equilibrium and four unknown reactions. Based on this consideration alone, it would appear that the structure was statically indeterminate. If we separate the structure at the hinge into two free bodies, however, we observe that the right-hand free body has exactly three unknown forces: the reaction at the support, the unknown shear force at the hinge, and the unknown horizontal force at the hinge. We can therefore solve for these three forces from the equilibrium conditions of the free body. Once the forces at the hinge are known, we can apply them as known quantities to the left-hand free body. This leaves the original three support reactions as unknowns. The system can therefore be solved from equilibrium conditions. We can therefore observe that the presence of the hinge reduces the degree of statical indeterminacy. Whereas a two-span structure without the hinge would have a degree of indeterminacy of 1, the hinge reduces the degree of indeterminacy to zero, i.e., it makes the structure statically determinate.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This observation can be generalized to the following statement: the degree of statical indeterminacy of a given structure is equal to the number of unknown support reactions minus the number of restraints removed by internal hinges, minus the number of equations of equilibrium.

Quality There is no hard rule regarding whether or not determinate systems are better than indeterminate systems, or vice versa. It is up to each individual designer to decide what type of system to use to satisfy the project requirements.

Other Issues Related to Determinacy As with the previous discussion on stability, the discussion of determinacy presented above relates primarily to the conditions of external equilibrium and restraint of a given structure. It is possible for a given structure to be externally statically determinate (all reactions can be calculated from the equations of equilibrium alone) but internally statically indeterminate (equilibrium conditions are not sufficient for the calculation of internal forces). We will examine these issues in greater detail in the section on trusses.

The Designer's Approach Although the mathematical definitions of stability and indeterminacy are useful, it is worthwhile for structural engineers to develop a sense of these two important properties of structural systems. In the case of stability, it is generally useful and practical to visualize how the structure will deflect under all possible directions and points of application of load (not just the loads that have been given). To evaluate the degree of statical indeterminacy, it can be helpful to release restraints in the structure by adding internal hinges or by releasing restraint at supports, working progressively towards a known statically determinate arrangement. The number of released restraints is the degree of indeterminacy. We are generally free in our choice of which restraints to release. In some statically indeterminate systems, however, it is possible to create an unstable structure by releasing fewer restraints than the degree of statical indeterminacy plus one. This can happen, for example, in structures where there is only one reaction providing restraint in a given direction and several reactions providing restraint in the other two. For example, consider the two-span continuous beam shown below. The degree of indeterminacy of the structure is one. We would therefore expect that by removing two restraints, we will make the structure unstable. This would be the case, for example, if R2 and R3 were removed. By removing the restraint in the x-direction H, however, the structure becomes unstable by removing only one restraint.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

If we wish to determine the degree of indeterminacy of a given structure by progressively removing restraints, therefore, we must never completely remove restraint in one of the three possible force directions (Fx, Fy, or M) when there are still several independent restraints in the other two directions. We sometimes encounter indeterminate structures for which reactions and sectional forces can be solved using equilibrium conditions alone, for a specific subset of load arrangements. An example of such a structure is shown in the following figure. It is clear that the load P must travel from one end of the member to the other, creating only axial force in the member, and an equal and opposite reaction at the left end of the beam.

This simple example may appear trivial, but it illustrates a principle that will prove to be of value when we come to the analysis of arches.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Examples Additional examples of stability and indeterminacy are given in the figure below:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Actions on Structures We refer collectively to loads and imposed deformations as actions on structures. We distinguish between the two types of action because they produce a fundamentally different response in statically determinate structures.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

3. Wind load. Structures must be capable of resisting the effects of wind. Wind loads are defined in design standards, usually for a specific geographical location. Mathematically, we model loads as force or moment vectors applied to a given structure. Nontrivial loads always produce nontrivial reactions in a given structure, regardless of of the degree of statical indeterminacy. In all cases, reactions are statically equal and opposite to the applied loads.

Imposed Deformations Imposed deformations may induce stress in structures, but they are not modeled as forces of moments applied to structures. The most common types of imposed deformations include: 1. Strains due to change in temperature. As the temperature of structural members increases, their volume increases. Design values of temperature change are generally specified in design standards. 2. Strains due to creep and shrinkage in concrete. Concrete structures deform progressively over time due to the unique characteristics of the material itself. The most important two mechanisms in this regard are creep and shrinkage. 3. Settlement of supports. Over time, foundations on soil can displace downward due to time-dependent changes in the properties of the soil supporting the foundations. Supports that settle will generally drag the structure along with them. Because there is no applied force or moment vector associated with imposed deformations, these actions induce forces and reactions only when these deformations are restrained. They produce the following response in structures depending on the degree of indeterminacy: 1. In statically determinate structures, imposed deformations induce no sectional forces or reactions. Because the structure is minimally restrained, it can deform freely in a stress-free condition. 2. In statically indeterminate structures, imposed deformations may induce sectional forces and reactions. Because there is no external force or moment vector associated with the imposed deformation, the reactions must be statically equivalent to zero.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example The following example illustrates the differences between imposed deformations in determinate and indeterminate systems:

Forces in the statically indeterminate system due to the support settlement are functions of EI, the stiffness of the beam. This implies that these forces will decrease with decreasing EI. This makes sense, since reducing the stiffness of the beam effectively decreases the degree to which the system can restrain the imposed deformations. In the limiting case of EI = 0, no forces will be produced. This demonstrates another important difference between loads and imposed deformations. For a beam of uniform stiffness EI, the forces and reactions due to load will be independent of EI.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Principles of Graphic Statics Graphic statics is a method of calculating structural response from a pictorial representation of the forces acting on a given structure. The method is based on the principle that forces can be modeled as vectors, which follow well-defined mathematical laws that can be expressed in graphical terms. In this section, we present basic principles of graphical analysis that will be used in subsequent sections to obtain solutions for cables, arches, beams, and trusses.

The Significance of Graphic Statics Graphic statics is a method for calculating the response of structures to actions. The method consists of drawing, to scale, an accurate representation of the loads acting on a given structure and the forces within the structure. If the drawing is correct, the magnitude and direction of specific forces can be obtained directly from simple measurement of the magnitude and direction of the corresponding line segments in the drawing. Graphic statics is based on the principle that forces can be represented as vectors, which are mathematical quantities having both magnitude and direction. The mathematical laws governing the addition of vectors can be expressed in graphical terms. It is thus possible not only to add vectors algebraically, but also to add them graphically by applying the parallelogram rule. The parallelogram rule states that the sum of two coincident vectors is equal (in magnitude and in direction) to the diagonal of the parallelogram formed by the two vectors we wish to add. The equivalence of algebraic and graphical addition of vectors is illustrated in the following example:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We have added two vectors using two methods: algebraic and graphical. The graphical procedure is done completely by drawing and measuring with precision: first we draw to scale the two vectors to be added, we draw the remaining two sides of the parallelogram, and then the diagonal. We then measure the length of the diagonal, which is equivalent to the length of the vector sum of the two original vectors. The implements used to add the vectors graphically were: a sharp pencil, a scale, and two drafting triangles. A calculator was not used in the graphical calculation. As shown in the example, the accuracy of the graphical addition of vectors is good. The relative error in the graphical calculation is less than 1 percent, which is generally acceptable in most structural engineering calculations. Since forces are mathematically equivalent to vectors, it follows that graphic methods for the manipulation of vectors are also valid for the manipulation of forces. This is the mathematical basis of graphic statics. Graphical methods of structural analysis have been used since the nineteenth century. The following figure, for example, which illustrates the graphical analysis of a lattice arch, was taken from the 1906 textbook by Ritter (1906). (Ritter was Professor of Civil Engineering at the Federal Institute of Technology in Zurich, Switzerland, often referred to by its German acronym ETH. The ETH has a strong tradition of education in structural NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

engineering, and its alumni figure among the most illustrious structural designers the world has known.) We can see that the figure presents a carefully drawn diagram of the structure as well as diagrams representing forces (the German word Kräfte (forces) is used in the diagram). The lines in the diagram correspond to the magnitude and direction of the forces in the structure due to the load case shown. For the member linking points 7 and 8 in the diagram of the structure (topmost diagram), for example, we see a corresponding line segment 7-8 in the force diagram (second diagram from the top). The magnitude and direction of this line segment corresponds to the magnitude and direction of the force in Member 7-8 due to the given load case.

With the introduction of digital computers and software for the numerical analysis of structures, the use of graphical methods of analysis gradually declined in engineering practice. This does not imply, however, that graphic statics is defective or inferior to automated, computer-based methods. In fact, graphic statics has many important advantages over other methods of analysis, the most important which are the following: Graphical analysis is fast. The calculation is complete as soon as the structural model and the force diagram corresponding to the given load case have been drawn, especially for arches and cables. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Graphical analysis is accurate. A carefully drawn and measured drawing, suitably sized, yields accuracy that is acceptable for most structural engineering applications. Graphical analysis is easy to interpret and to check. Graphical analysis generally favours calculating the entire response of a given structure rather than the calculation of a single response quantity. Because the entire analysis is contained on one drawing, relations between forces throughout the structure can be readily observed. This greatly increases the ease of understanding the results of the analysis. Because the response of the structure as a whole is made visible on one drawing, it is also much easier to detect values that do not make sense than in a numerical calculation in which only one specific quantity is calculated. Graphical analysis helps to close the gap between analysis and design. The force diagrams obtained from graphic analysis can lead directly to important new ways of arranging structural components to achieve more efficient structural behaviour and aesthetically significant visual forms. This will demonstrated in the section on arches. Finally, graphic statics is an important teaching tool. By representing forces as lines on paper, we are forced to think of them as physical objects with real magnitude and direction, not just as numerical abstractions. This perspective will help you to get a more intuitive feel for how the forces are actually flowing in a structure, which is an important skill for engineers to develop.

References Ritter, W. 1906. Anwendungen der Graphischen Statik. Vierter Teil. Der Bogen. (Applications of Graphic Statics. Part Four. The Arch.) Zurich: Verlag von Albert Raustein.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Principles of Graphic Statics Graphic statics is a method of calculating structural response from a pictorial representation of the forces acting on a given structure. The method is based on the principle that forces can be modeled as vectors, which follow well-defined mathematical laws that can be expressed in graphical terms. In this section, we present basic principles of graphical analysis that will be used in subsequent sections to obtain solutions for cables, arches, beams, and trusses.

Definitions and Conventions Location Plan and Force Plan We will draw forces as vectors in two separate but related representations: the location plan and the force plan. As its name implies, the location plan shows the true location of all forces, i.e., their true lines of action. The force plan is an alternative representation of these same force vectors, in which the magnitude and direction remain unchanged but the vectors have been arranged head to tail. It is important that the vectors always be arranged head to tail in the force plan. Arranging vectors head to head or tail to tail will result in incorrect results. In the force plan, the vectors no longer act at their true location. The force plan is a convenient representation of forces because, as we shall see, it allows vectors to be added and brought into equilibrium by closing the polygon formed by arranging the vectors head to tail.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Notation We will use a notation convention for notation that makes possible a direct correspondence between the location plan and the force plan. In the location plan, the areas bounded by the lines of action of the forces under consideration are given upper case letters A, B, C, etc. In the force plan, the endpoints of the curve created by the force vectors laid out end to end are given lower case letters a, b, c, etc. Forces in the location plan are referred to by a pair of upper case letters AB, BC, CD, etc., according to the two areas separated by a given force. In the force plan, forces are referred to by a pair of lower case letters ab, bc, cd, etc., according to the endpoints of the curve created by the force vectors.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Basic Principles Addition of Two Coincident Forces In the location plan, we add two force vectors using the parallelogram rule, as shown previously for purely abstract vectors. This is illustrated in Part (a) of the figure below. In the force plan, we first arrange the vectors head to tail (the order is not important). We observe that the triangle formed by the free ends of the vectors (tail of the first vector and head of the second vector) is congruent to one half of the parallelogram in the location plan. In the force plan, therefore, the resultant of the two vectors is thus obtained by connecting the two free points of the vectors, from free tail to free head.

Equilibrium of Two Coincident Forces The preceding graphical constructions can be expanded to the calculation of the force required to bring two given forces into equilibrium. This is shown in Part (b) of the figure below. Calling the two original forces F1 and F2, and their resultant FR, we observe that the sum F1 + F2 - FR equals zero. In the force plan, this is equivalent to closing the triangle by connecting the head of the second vector to the tail of the first, such that the path around the triangle is either completely clockwise or completely counter-clockwise. The corresponding set of forces in equilibrium is also drawn in the location plan.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Three or More Coincident Forces We can extend the previous two graphical constructions to a set of three or more coincident forces. As shown in the following figure, the resultant of three or more vectors in the location plan is obtained from two successive applications of the parallelogram rule. In the force plan, we first lay the given forces end to end. We then proceed to close the triangle of a pair of the given forces, in this case ab and bc. The closing segment is the resultant of these two forces, and is called R1. We then close the triangle formed by R1 and cd. The closing segment is the resultant of R1 and cd, and is thus the resultant of the three given forces. We observe that the result could have been obtained directly simply by closing the polygon formed in the force plan. This observation has general validity. For a given set of forces arranged end to end in the force plan, their resultant is given by the segment that closes the polygon. As discussed for the case of two forces, the force required to bring a given set of three or more forces into equilibrium is simply the negative of the resultant of these forces. This is obtained graphically in the force plan by closing the polygon with a vector directed such that all vectors move around the polygon in the same rotation direction, either all clockwise or all counter-clockwise. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The difference between the two types of problem discussed thus far is emphasized: When the task is to calculate the resultant of two or more vectors, the solution is obtained by closing the polygon drawn in the force plan with a vector extending from the tail of the first vector to the head of the last vector. When the task is to calculate the force required to bring a set of two or more vectors into equilibrium, the solution is obtained by closing the polygon drawn in the force plan with a vector extending from the head of the last vector to the tail of the first vector.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Principles of Graphic Statics Graphic statics is a method of calculating structural response from a pictorial representation of the forces acting on a given structure. The method is based on the principle that forces can be modeled as vectors, which follow well-defined mathematical laws that can be expressed in graphical terms. In this section, we present basic principles of graphical analysis that will be used in subsequent sections to obtain solutions for cables, arches, beams, and trusses.

Basic Principles (continued) Non-Coincident Forces We often have to deal with non-coincident forces. This set of forces includes not only parallel forces (which are truly non-coincident), but also forces with a point of intersection at an inconvenient location (such as off the analyst's page). We illustrate the procedure for a pair of parallel forces in the figure below. The task is to calculate the magnitude, direction, and resultant of this pair of forces. In the location plan, we introduce two auxiliary forces that can be freely chosen but which must be equal in magnitude, opposite in direction, and located on the same line of action. We call these forces H and -H. Because these two forces add to zero, introducing them into the solution has no effect on the equilibrium of the system. We add H to AB and -H to BC. This yields resultant forces R1 and R2, the sum of which is the sum of the original forces AB and BC. Alternatively, in the force plan, we draw forces ab and bc head to tail. We know from our the discussions above that vector ac gives the magnitude and direction of the resultant of ab and bc (it closes the "polygon" composed of segments ab and bc). We now lack only the location of this force. To determine this, we select a point O, called the pole. The location of O can be freely chosen. We then draw segments linking O with a, b, and c. Based on our observations of systems of coincident forces, we know that ab is the sum of aO and Ob, and bc is the sum of bO and Oc. Returning to the location plan, we can resolve the given forces AB and BC into components with directions given by aO, Ob, bO, and Oc. We select any point on the line of action of AB as a starting point, and draw the two components parallel to aO and Ob as shown. We intersect the line parallel to Ob with the line of action of BC, and resolve BC into components at this location. The intersection of the components parallel to aO and Oc (the yellow and pink lines in the figure), gives a point on the line of action of the resultant of AB and BC.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example The figure shows an example of a graphic calculation of the resultant of four noncoincident forces.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Uniform Loads Most of the methods of graphic statics work with concentrated loads. For models of real structures that are loaded with uniform loads, it is acceptable to transform the uniform load into a statically equivalent set of concentrated loads. For example, a uniform load of 10 kN/m applied over 50 m could be modeled as ten concentrated loads of 50 kN at 5 m spacing.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Analysis of Cable Systems Cables are perfectly flexible structural members. They establish equilibrium with a given arrangement of load by assuming a shape that allows the load to be carried by the cable in a state of pure axial tension.

Cables are perfectly flexible structural members that carry load in pure tension. Cables establish a state of equilibrium with a given loading by assuming a shape that allows load to be carried with the cable in tension. In this important sense, cables do not have a predefined geometry, as do beam and truss structures. Their geometry is determined by the arrangement of loads they carry.

Examples of Applications of Cables Although cables are used in many types of structure, we are perhaps most familiar with their use as the primary structural members of long-span bridges and roofs. A familiar example of the use of cables in bridge construction is the suspension bridge. The picture below, which shows the Golden Gate Bridge during construction, illustrates the two primary types of cable used in this type of structure. Vertical suspender cables carry load upward the truss that supports the roadway to the main cables. The main cables, which are draped over the entire length of the bridge, carry vertical loads from the suspender cables to the towers and the anchorages at the ends of the bridge. The main cables illustrate how axial tension in a cable can effectively resist loads applied in a direction other than that of the axis of the cable.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The use of cables in buildings can create a bold visual effect, as shown in the figure below, a picture of the main terminal at Dulles International Airport outside of Washington, D. C., U. S. A. In this structure, the cables span between inclined reinforced concrete towers. The surface formed by the draped cables defines the plane of the roof. The roof itself is composed of precast concrete panels attached to the cables. The structural function of these cables is actually similar to that of the main cables of suspension bridges.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Stability and Determinacy of Cables Before we develop procedures for calculating forces in cables due to load, it is important to review issues related to stability and statical indeterminacy.

Degree of Statical Indeterminacy We will examine the issue of statical indeterminacy with the help of the simple cable shown in the figure below. At first glance, it would appear that the cable is statically indeterminate. The number of reactions (four) exceeds the number of equations of equilibrium (three). As shown in the figure, however, we can solve for all of the reactions and forces in the cable using the methods of statics. We would appear to have a paradox.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

To resolve the paradox, we need to find another constraint. One additional constraint plus three equations of equilibrium will give us the four conditions we need to solve for the four unknown reactions. The additional constraint is geometrical. A review of the calculations above shows: 1. The value of RR can be obtained from moment equilibrium, taking moments about the left support. 2. The value of RL can be obtained from vertical force equilibrium and the value of RR. 3. The equation of horizontal equilibrium can be used to express HR as a function of HL. (For the loading shown in the figure above, HR is equal to HL.) So we need to solve for HL. If we call the vertical distance from the left support to the point of application of the leftmost load "a", then breaking apart the left support as a free body and solving for horizontal equilibrium yields HL = RL·x1/a, where x1 in this case equals 20 m. In this case, we know the value of a (it is 15 m), which leads directly to a complete solution of the unknown reactions.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is actually not necessary to know the value of a, the vertical ordinate of the point of application of the left-most load, but merely to know the value of the vertical ordinate of the cable at any one point along its length. This proposition is demonstrated in the following figure. If we cut the structure at a distance x from the left-hand support and treat the structure to the left of the cut as a free body, then we obtain an expression for y, the vertical ordinate of the cable at any given point, that is the product of f(x), a function of known parameters (magnitude and location of loads), and the unknown parameter a.

If a is unknown but the vertical ordinate of the cable is known at any other point (say y(x0) = b for some value of x0), then we can solve for a using the last equation given on the right-hand side of the figure above: a = b/f(x0). It follows, therefore, that knowing the vertical ordinate at one point along the length of the cable provides us with the fourth constraint that is needed to solve for the four unknown equations. The cable is therefore statically determinate. When no vertical ordinate is specified, the structure is under-constrained and can only be solved if we know one of the reactions in advance. This can happen, for example in a case where we are actually jacking a known force into a cable at one end. In such a case, the length of the cable would be adjusted to provide a vertical profile in equilibrium with the loads and the jacking force. From the three equations of equilibrium remaining, we could solve for the vertical ordinate of the cable at a given point.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

When more than one vertical ordinate is specified, the structure is over-constrained and, generally speaking, a state of equilibrium cannot be established.

Stability Are cables stable? Consider the simple cable shown in the figure below. It is apparent that a loose cable is not stable. It can be rearranged into any number of shapes with minimal load. Once load is applied, however, the cable gains an important measure of stability, in the sense that the load required to displace the cable away from its original loaded shape is nontrivial.

Stability of cables thus depends not only on structural arrangement, but also on load. This proposition is illustrated in the figures below. The first figure shows a structure composed of pin-jointed truss members arranged according to the profile of the cable shown in the second figure. The structure is not loaded. It is apparent that the structure is unstable, as is demonstrated by the unstable arrangement of members shown. The structure can assume this arrangement without any work being performed.

When we consider a geometrically identical arrangement, this time a loaded cable, we see that to assume the displaced shape assumed in the previous figure, the loads P1 through P3 are displaced, which means that work has to be performed to displace the cable as assumed. The loaded cable is thus stable.

The relation between load and stability of cables is important. It is no coincidence that cables are most often used for structures for which dead load (loads of fixed magnitude, direction, and location) dominates over live loads (i.e., loads of variable magnitude, direction, and location). Suspension bridges are one example of this type of structure. As shown in the figure below, cables are not well suited for structures that are subject to significant reversals in load. The purple and the green curves are two valid states of

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

equilibrium, for a load P applied in two opposite directions. The change in cable geometry that is required to establish equilibrium with these two forces is major, and would generally not be acceptable in most structural applications.

Cables with Concentrated Loads We will now develop a simple graphic method for calculating cable forces and support reactions. It is based on the methods developed previously for the resolution of two or more non-coincident forces. We will develop the method by means of four examples.

Example 1 Consider the cable shown in the figure below. We must solve for reactions and cable forces, given the geometry, loads, and restraints shown.

We start by drawing the polar diagram of forces in the force plan. Location of the pole O can be freely chosen. In this case, the diagram is very simple because there is only one applied load. This diagram represents one possible state of equilibrium at the point of application of the 20 kN load.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We then transfer the forces aO and Ob back to the location plan. We select the point of intersection of the line of action of the applied load and force aO (called Point S in the figure below) such that the line of action of aO passes through the left-hand support (Point M). The intersection of force Ob with the line of action of the vertical reaction RR will, in general, not be at the support. The second component of reaction at the supports must be on the line passing through M and N, to maintain moment equilibrium. This diagram represents one possible state of equilibrium of the system, but it is not consistent with the support conditions. It is, however, an important step towards a solution.

We augment the force diagram in the force plan to show the state of equilibrium at the two supports (not the actual supports, but the shifted supports as shown in the previous figure). We do this by drawing in line segment WO parallel to MN in the location plan. Triangle aOW thus represents the state of equilibrium at Point M and triangle WOb represents the state of equilibrium at Point N. It thus follows that the vertical leg of each

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

of these two triangles is the vertical reaction at these two points. The magnitude of these reactions can be scaled from the force diagram.

We now correct the force diagram to restore the right-hand support to its correct position. We do this simply by shifting the pole O downward to new pole O', such that segment WO' is parallel to segment MP (line joining the real supports) in the location plan. Points a, W, and b in the force plan do not move. Note that since we maintain the closure of the triangles, the structure remains in equilibrium.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We now redraw the forces in the location plan, parallel to the new set of forces in the force plan. The diagram should close properly. We observe that the sag of the cable, approximately 36 metres, is larger than the required sag, which is 20 m. We thus need to make one more correction to the forces. We compute the ratio of actual sag to required sag, which equals 1.8.

We return to the force plan and scale its horizontal axis by 1.8. This moves the pole to the right, to point O''. This modified force diagram represents the correct solution to the forces in the structure. We can scale all reactions and cable forces off the force diagram.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is instructive to check the results of the graphical calculation with the results of an algebraic analysis. The check shows agreement to within a relative error of 1 percent, which is excellent for structural engineering.

Analysis of Cable Systems Cables are perfectly flexible structural members. They establish equilibrium with a given arrangement of load by assuming a shape that allows the load to be carried by the cable in a state of pure axial tension.

Cables with Concentrated Loads (continued) Example 2: Several Vertical Loads In the second example, we will consider the case of a cable loaded at several locations along its length. We will draw the diagrams in a way that is closer to the way they would be used in an actual calculation, i.e., by superimposing several diagrams on top of each other.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 3: Supports at Unequal Elevations The calculation is similar to the case of a cable with supports at equal elevation, except that we will arrange our support reactions to be vertical and along the line of action of the cable, and not vertical and horizontal as was done previously. Although this simplifies the graphical calculation, it must be remembered that the reaction along the line of the cable has a vertical component, which must be considered when calculating final reactions. The line segment WO' is not horizontal, but rather is drawn parallel to the line joining the two support points. This ensures that the non-vertical component of reaction is also on the line joining the two supports. Once the solution has been completed, we can select another point W", such that W"O" is horizontal. The length of W"O" gives the true horizontal reaction at the supports. aW" and W"b give the true vertical reactions.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 4: Inclined Loads Inclined load will usually create unequal horizontal support reactions. In moving from the force plan to the location plan, we must be careful always to move forces along their line of action, which is not necessarily vertical. This includes not only the inclined loads but also the reaction at the right-hand end. To create the cable profile in the location plan corresponding to the assumed pole O in the force plan, we intersect a line parallel with cO with a line drawn through the right-hand support. This line is not vertical, but rather is parallel to the line ac in the force plan, which is the assumed inclination of the reaction component. The final reactions given by the force plan are horizonal (WO") and inclined (aW and Wc). We can change this to a set of horizontal and vertical reactions at the supports by resolving these vectors graphically. At the left support, the vectors are aW' and W'O". At the right support, the vectors are W"c and W"O".

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4. Cable with Uniform Load As stated in our initial discussion of the principles of graphic statics, uniform loads can always be treated as a group of several equally spaced concentrated loads of equal magnitude and direction. Uniform loads occur sufficiently often, however, to merit a specific discussion. One of the best known instances of uniform load on cables is the case of dead load on suspension bridges. Most of the dead load on suspension bridges comes from the girder or truss that supports the actual roadway. For cables carrying a given uniform load, we wish to answer the following questions: 1. 2. 3.

What are the reactions at the supports? What is the shape of the cable? What are the forces in the cable?

We will answer these questions for the cable shown below. The load, w, is uniform along the projected length of the cable. Because supports are at the same elevation and loading is symmetrical, it follows that slope of the cable at midspan is zero.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Equilibrium of the entire cable requires that RL = RR = wL/2, and HL = HR = H. We now consider equilibrium of a free body cut at midspan of the cable, shown in the figure below. For a given value of sag f, we can solve for the horizontal reaction, H, which equals wL2/(8f). Reactions at the supports are thus wL/2 (vertical) and wL2/(8f) (horizontal).

Next we cut a free body at a given location along the projected length of the cable, x, as shown below. We can solve for the vertical ordinate of the cable, y, as a function of x from the conditions of equilibrium and the known value of H.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This calculation yields several important observations: 1. 2. 3.

The cable has a parabolic shape. The horizontal component of force in the cable, H(x), is constant over the length of the cable and is equal to wL2/(8f). Tension in the cable at any point can be calculated using the Theorem of Pythagoras, from the constant horizontal component H and the vertical component V(x) = wL/2 - wx.

Unequal Support Elevations We will investigate the same issues for cables with unequal support elevations, loaded uniformly along their projected length. Most of the expressions we have defined for the cable with supports at equal elevations are still valid, provided we define sag and vertical ordinates of the cable with respect to the chord linking the two supports.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We observe that, provided we define sag f and vertical ordinate y with respect to the chord between supports, the expressions for H and y derived previously for the cable with supports at equal elevations are valid. Different expressions are required, however, for the vertical reactions.

Analysis of Arches The primary issues related to the analysis of arches are: (1) selection of the shape of the arch to allow the dead load to be carried in pure axial compression, and (2) calculation of bending moments in the arch due to other load cases.

Examples of Applications of Arches Arches are structural systems that carry the dominant permanent load case in pure axial compression and all other load cases in a combination of axial compression and bending. They are highly efficient and visually powerful structural systems. They have been used extensively for bridges since Roman times. An important example of Roman use of the arch is the Pont du Gard in Nîmes, France, shown in the figure below.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Stone masonry, the material used by the Romans, remained the preferred material for the construction of arches up to the twentieth century. One of the most daring achievements in masonry arch design is the Landwasser Viadukt, built in Switzerland near the turn of the twentieth century. This bridge is remarkable for a masonry arch, not only because of the height of it piers (65 m) but also because it is curved in plan (radius 100 m) and it carries heavy railway live load.

In the twentieth century, reinforced concrete, prestressed concrete, and steel have been used to construct arch bridges. Examples of bridges built of each of these materials are the Salginatobel Bridge (reinforced concrete),

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

the Bridge over the Rhine at Reichenau (Switzerland) by Christian Menn (prestressed concrete),

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

and the New River Gorge Bridge in West Virginia, U. S. A. (steel). Arches are used in buildings not only as two-dimensional elements (similar to their use in bridges) but also as domes, where they take on a true three-dimensional character. Examples of domes include the church at Colonia Guell near Barcelona by Antonio Gaudi, a masonry dome,

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

and the reinforced concrete domes of the Swiss engineer Heinz Isler.

Arches and Cables Similarities "As hangs a flexible cable, so, inverted, stand the touching pieces of an arch". Robert Hooke (1635-1703). We will take Hooke's observation as our starting point for a discussion of arches. Hooke recognized that there is a fundamental equivalence between the state of equilibrium of a cable in pure tension and the equilibrium of an arch in pure compression. We can express this equivalence in graphical terms as shown in the following figure. We have a cable with loads P1 through P6. The profile of the cable has been chosen such that it equilibrates the given loads.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We then draw an arch with same span, same loads, and inverse profile of the cable, i.e., if the profile of the cable is y(x), then the profile of the arch, called y'(x), satisfies the relation y(x) = -y'(x) for all x. It is a straightforward process to demonstrate that if the cable is in equilibrium, then so is the corresponding arch. We can consider, for example, equilibrium at a given point of application of load, say A for the cable, by drawing the forces acting at this point in the force plan. We do the same for the arch. Since angles θ1 and θ2 are equal (due to the correspondence between the geometries of cable and arch), it follows that the magnitudes of the corresponding forces are equal: T3 = C3 and T4 = C4. We can establish identical relations between corresponding forces at all other points of the arch. It follows therefore that the tensile forces in a cable are equal to the compressive forces in an arch for equal loads and equal geometries.

Differences There are, however, important differences between cables and arches. Cables, being perfectly flexible members, cannot resist compression. Arches must therefore be sufficiently stiff to allow them to carry compressive forces without buckling. Cables deform as required to maintain equilibrium under varying load conditions. When load is increased at a given location, for example by adding load ∆P2 as shown in the figure below, the sag of the cable simply increases as required. The green curve represents the new state of equilibrium.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This mechanism for maintaining equilibrium has two problems for arches. First, the deformations required to maintain equilibrium are often large. Given that arches are stiff members, the deformations required are likely to violate project requirements with regard to safety or serviceability. The second difficulty is illustrated in the figure above. Whereas the cable in tension responds to additional load by sagging more, the arch in compression would actually have to rise up to meet the load, changing its geometry into the green curve. This type of structural response is generally not possible with the structural materials and systems currently in use. Arches are therefore designed to function as follows: 1.

Arches carry the dominant permanent load case (usually full dead load) in pure axial compression. This is accomplished by a suitable selection of the shape of the arch. 2. In general, arches carry additional loads (such as live load) in bending. From the perspective of structural analysis, we therefore have two primary issues to address: 1.

Selection of the shape of the arch to carry the dominant permanent load case in pure axial compression. This shape will be called the pressure line. 2. Calculation of bending moments in the arch due to additional load cases.

Definitions The following terms are used to refer to components of arches: Springing lines: The points at which the arch touches its foundation (the use of "line" here is misleading, but we will use it because it is the common term) Rise: The vertical distance from the midpoint of the chord joining the springing lines to the axis of the arch

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Span: The length of the horizontal projection of the chord joining the springing lines Crown: The point on the arch at midspan.

The Pressure Line As stated previously, our first task in the analysis of an arch is to determine the pressure line. This is an example of a task at the intersection of analysis and design, since the actual shape of the arch does not exist at the outset. We will generally be given the following information: 1. Span length 2. Rise 3. A definition (magnitude and location of loads) of the dominant permanent load case, usually full dead load. (In the preliminary stages of design, the characteristics of the structure will not normally be completely defined. To define dead load for the calculation of the pressure line, designers will normally make initial assumptions of the cross-sections of structural members and produce an initial estimate of dead load. It is necessary to verify the accuracy of this assumption at a subsequent stage in the design process.) Given this information, we need to determine the shape of the arch that allows the load to be carried in pure axial compression. We can proceed graphically, in a similar fashion to the method developed for the analysis of cables. This procedure is demonstrated in the example given in the following figure:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Proceeding as defined for cables, we first draw the loads head to tail in the force plan. It is convenient to select the pole O to the left of the loads. This will yield vectors that are suitable for an arch shape rather than a cable. We then draw the rays Oa through Og in the location plan and close the diagram. These are the green curves in the diagram above. We then draw ray OW, which is parallel to the closing line in the location plan. We select a new pole O' such that O'W is parallel to the chord of the arch. We draw new rays O'a through O'g in the force plan and then transfer them to the location plan. We then make our final adjustment to the pole to give the arch the required rise f. The final diagram in the location plan (drawn in purple above) gives the pressure line of the arch. An arch constructed to follow this profile will carry loads P1 through P6 in pure axial compression. The ordinates of the profile can be scaled directly from the location plan. Reactions and compressive forces in the arch can be scaled off the final diagram in the force plan, exactly as was done for cables. We can observe from the above diagram above that arches are not always curved. Angle breaks in the pressure line are possible when the arch carries concentrated loads. It is also possible to proceed algebraically in a manner similar to that developed for cables subjected to uniform load. We define the ordinates of the pressure line, y(x), as the vertical distance from the chord joining the springing lines to the pressure line. (If the load is known to be uniform, then the expressions developed for H and y(x) for cables subjected to uniform load can be applied directly to arches.) For non-uniform loads, the principles can still be applied. We first solve for the vertical reactions RL and RR from the conditions of equilibrium of the entire structure. If there is no horizontal applied load, then HL = HR = H. We then cut a free body at midspan and solve for H. With the reactions now known, we can solve for any value of the vertical profile of the arch, y(x), for any value of x. This is accomplished by cutting a free body at x and imposing the conditions of equilibrium as functions of the unknown ordinate y(x).

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Provided a given arch is stable, the pressure line can be determined without consideration of the degree of restraint provided at the supports or the arrangement of internal hinges. The pressure line, by definition, carries its defining load case in pure axial compression. Because there is no bending, the degree of bending restraint, which corresponds to the presence or absence of hinges), is of no relevance to the behaviour of the structure for the load case that defines the pressure line. The figure below shows two arches with identical profile. The upper arch has pinned supports; the lower has fully fixed supports. Assuming the arches have been laid out according to the pressure line of the same arrangement of load, there will be no tendency for either structure to bend under the action of that particular load case. The presence or absence of rotational restraint at the springing lines thus has no effect on the calculation of the pressure line.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This situation is similar to the situation observed for a much simpler system during the general discussion of statical indeterminacy in structures.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Analysis of Arches The primary issues related to the analysis of arches are: (1) selection of the shape of the arch to allow the dead load to be carried in pure axial compression, and (2) calculation of bending moments in the arch due to other load cases.

Analysis for Bending Degree of Statical Indeterminacy Although the pressure line can be determined without reference to the degree of statical indeterminacy of the arch, the degree of indeterminacy must be considered when the arch is subjected to loads that do not correspond to the pressure line. As stated previously, loads that do not correspond to the pressure line induce bending in the arch. Calculation of bending moments in a given arch require consideration of the characteristics of bending restraint in the system. We will leave the discussion of statically indeterminate arches for a subsequent section. For now, we will concentrate on the three-hinged arch, which is statically determinate.

The Three-Hinged Arch As the name implies, the three-hinged arch has three hinges, which provide zero rotational restraint. They are generally located at midspan and at or near the springing lines. Maillart's Salginatobel Bridge is a classic example of a three-hinged arch. We will present both a graphical and an analytical method for calculating the bending moments in a three-hinged arch. Unless otherwise stated, it is assumed that the arch geometry has already been correctly defined to match the pressure line of the dominant permanent load case.

Definition of Sectional Forces The following figure gives definitions relating a compressive force in space, which is the conventional representation of the state of equilibrium of an arch under a given loading as determined by a graphical analysis, and sectional forces in the arch itself. The figure shows that the force in space, C, is resolved into components parallel and perpendicular to the axis of the arch, called N (axial force) and V (shear). These forces are then displaced to the axis of the arch. The resulting moment M is the axial force N times the perpendicular distance from force C to the axis of the arch.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The sectional force representation is the format generally most useful for checks of safety and serviceability of the arch member itself. When the solution is obtained graphically, the preceding construction is difficult to perform accurately, since it requires a line perpendicular to the axis of the arch to be drawn. Since it is generally easier to draw vertical lines, we can develop a formulation of sectional forces that allows us to determine M with far greater accuracy when graphical methods of analysis are used. We begin by drawing a vertical line from the axis of the arch at the section in question. We can slide the resultant compressive force C to the point of intersection of the line of action of the force and the vertical without changing the state of equilibrium. We then resolve C into horizontal and vertical components, V* and H. We can slide force V* along its line of action down to the arch. When we displace H down to the axis of the arch, we must add a moment, equal to H times a, the vertical distance from the line of action of C to the axis of the arch. This moment, M, is identical to the moment computed previously.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

This procedure is a more reliable means of computing bending moments in arches using graphical methods. When we use the method, however, we must remember that V* is not the true shear force and H is not the true axial force acting on the arch, since these forces are not truly perpendicular and parallel to the axis of the arch. To obtain the true shear V and axial force N, we must resolve C as shown in the first figure above.

Graphical Solution We will develop the method of graphical solution by examining three examples. Example 1. One Load There is only one possible state of equilibrium for the unloaded half of the arch isolated as a free body: two equal and opposite forces, N, acting on a line of action defined by the chord joining the two hinges. This defines the direction of the reaction at the right-hand support.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The solution is obtained directly by drawing the line of action of the left-hand reaction, which must pass through the point of intersection of the applied load P and the righthand reaction, the line of action of which has already been found. We obtain the magnitude of the reactions from the triangle of forces drawn in the force plan. The triangle can be drawn since we know the magnitude and direction of the applied load, as well as the direction of both reactions. Bending moments in the arch can then be determined using the relations defined in the previous subsection.

Example 2. Several Loads on One Side of the Central Hinge

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The solution can proceed according to the steps defined for a single load, provided we first determine the magnitude, direction, and location of the resultant of all the loads. Although the complete solution can be obtained using one diagram in the location plan, two diagrams have been used for clarity. We first determine the location of the resultant of all the loads using methods developed previously. We then obtain the direction of the reactions following the procedure described under Example 1. We transfer these vectors to the force plan, which gives us a new pole O'. We draw rays from O' to the endpoints of the original load vectors a through d. These rays are then transferred back to the location plan. The resulting diagram in the location plan gives us the location of the resultant compressive force in the arch. The length of the segments issuing from O' in the force plan represent the magnitude of the resultant compressive force along the arch.

Bending moments in the arch are then determined using the relations defined in the previous subsection, based the magnitude of the resultant compressive force in the arch (obtained from the force plan) and the distance of the resultant to the axis of the arch. Example 3. Loads on Both Sides of the Central Hinge

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The method involves drawing a solution for loads applied to the left side of the hinge as presented in Example 2, drawing a separate solution for loads applied to the right side of the hinge as presented in Example 2, and combining the two solutions. We begin by finding the resultant PL of loads P1 and P2 (left half of span), and the resultant PR of loads P3 and P4 (right half of span). For convenience, we draw loads P1 through P4 head to tail in one force plan diagram, but we select two separate poles OL and OR, since we are calculating two separate resultant forces. We then determine the slope of the reactions at the supports for two separate cases. The first case is due to load PL; the second case is due to load PR. The methods of Example 1 are used. We transfer the vectors corresponding to the reactions to the force plan. This yields two new poles, O'L and O'R. We observe that the two components of reaction at the left-hand support are given by O'La and O'Rc in the force plan. The two components of reaction at the right-hand support are given by O'Lc and O'Re. We add O'La and O'Rc in the force plan to obtain the total reaction at the left-hand support, and add O'Lc and O'Re to obtain the total reaction at the righthand support. We accomplish this by translating vectors O'Rc and O'Lc as shown in the force plan. This gives us a new pole for the entire system, O''. From O'', we draw rays to points a through e. This defines the magnitude and direction of the reactions and compressive forces in the arch for the total load case. We transfer these rays to the location plan to show the path of the resultant compressive force required for equilibrium.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

A simple and important check of the graphical calculation is to ensure that the resultant compressive force in the arch intersects the axis of the arch at the three hinges.

Analytical Solution The response of three-hinged arches to load can also be calculated analytically by a variety of methods. The method presented here makes use of a "primary system", obtained by removing the internal hinge and releasing the horizontal restraint at the right-hand support. The primary system is also statically determinate. We calculate the response of the primary system for two separate load cases. In the first case, the primary system is loaded with the given external loads, in this case P. Bending moment at midspan due to this load case is calculated. This moment will be nonzero in the primary system, since we have removed the middle hinge. In the second case, we load the primary system with a horizontal load H at the right-hand support, which is free to displace in the direction of the load. We express the bending moment at midspan as a function of H.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We can replicate the original system by adding the moment diagrams due to the two load cases in the primary system with a suitable choice of H. The value of H required is the value that gives a total midspan moment of zero. This is the condition that is required of the three-hinged arch. Having solved for H, we can establish the total reactions in the original system, and solve for moments and shears in the original system using the expressions given in the figure below.

Example Calculation The task is to calculate bending moments in the three-hinged arch at the point of application of the 100 kN load. Span is 80 m, rise is 16 m. Geometry of the arch is shown in the figure below:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The calculation is done using two separate methods. In the upper half of the figure, a graphical method is used. The lines of action of the forces in the arch are drawn following the methods presented previously in Example 1. The forces are then drawn in the force plan and the horizontal component of force in the arch is determined. Moment at the point of application of load is then determined by scaling the vertical distance between the arch and the line of action of force in the arch (in this case 8 metres), and multiplying this dimension by the horizontal component of force in the arch. The answer is 620 kN·m. In the lower half of the figure, an analytical method is used. The bending moments in the primary system, in which the central hinge has been fixed and the right-hand horizontal reaction has been released, are first calculated for the given load P. These moments are called M0 in the figure above. Moment at midspan for this case is 1250 kN·m. We then calculate moments due to the horizontal reaction H, which has been taken away in the primary system and which we have to restore. These moments are called M1 in the figure above. At midspan, M1 is equal to -16H. Imposing the condition that Mtot (= M0 + M1) must equal zero at midspan, we find that H = 78.1 kN. (Note that this value is very close to the value of H obtained graphically, 77.5 kN.) Finally, we calculate the total moment at the point of application of the load P. We know that M0 at this location is 1720 kN·m. Moment NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

M1 at this location is obtained from the expression M1 = H·y, where the value of y is read from the arch ordinates given. The total moment at this location is thus 1720 - 1090 = 630 kN·m. The relative error of the graphical method is 1.6 percent of the analytically computed result. This is another example of the accuracy of graphical methods when diagrams are drawn with care.

Variations Two important variations on the arches presented thus far are worthy of brief mention.

Tied Arches Instead of resisting the horizontal component of reaction with a foundation, it is possible to resist this force by means of a tie extending from one support to the other. An example of a tied arch bridge, the Fort Duquesne Bridge in Pittsburgh, U. S. A., is shown in the following figure.

Tied arches are used when it is not possible to resist the horizontal component of reaction by means of a foundation. They are often considered when the arch members are laid out above the structure it supports. The fundamental differences and similarities relating classical arches and tied arches are illustrated in the figure below. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Assuming the tie is rigid (i.e., assuming it provides horizontal restraint similar to that of a foundation), we can analyze tied arches using the methods presented in this section.

Frames The principles described for three-hinged arches are also valid for three-hinged structures that do not look like arches, i.e., three-hinged frames. The following figure illustrates the application of the procedures developed in Example 1 to a three-hinged frame.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Analysis of Statically Determinate Beams This section presents graphical and analytical methods for calculating sectional forces (bending moment, shear, and axial force) in beams. Special procedures are presented for several common cases, including indirect loading of beams and loading trains.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The arch is relatively thin. Although it is adequate to carry all dead loads, it has insufficient bending capacity to resist live load on its own, which in this case is a railway loading. By giving the deck girder sufficient strength and stiffness, these loads can be carried by the beam, producing negligible bending stresses in the arch. 3. Beams are well adapted to prefabrication. Because the geometry of beams is not determined by the arrangement of loading, beams can be standardized to allow efficient mass production. The use of prefabricated structural members often results in lower construction costs. Rolled and/or welded steel sections (wide flange members and structural tubes) and precast concrete beams and girders are two examples of the use of prefabricated beams in construction. As a result of these advantages, beams have a vast range of application in construction. The following figures show a few applications of beams as structural members in bridge construction.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Curved concrete box girder constructed from precast segments assembled by posttensioning: left side completed, right side in construction.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Precast concrete I-girders

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Welded steel plate girders

Preliminary Considerations We will first restrict our attention to straight beams loaded in a vertical plane containing the longitudinal axis of the beam. The general loading condition is shown in the diagram in the upper left corner of the following figure:

The general loading condition includes inclined loads, shown here as Q1, Q2, and Q3. These loads can be resolved into horizontal and vertical components. It is convenient to perform separate analyses for the vertical loads and the horizontal loads. The results of

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

these two analyses can then be combined to give the response of the beam to the original loading Q1, Q2, and Q3.

Response of Beams to Vertical Load We will first develop methods of calculating the response of beams to vertical load, followed in a separate section by a consideration of horizontal loads on beams.

Equations of Equilibrium The response of beams to a given load can be expressed mathematically in terms of equations of equilibrium of a differential element at any given location in the beam. The derivation of these equations is straightforward and is shown in the figure below:

These equations are valid for all beams loaded perpendicular to their axis, regardless of the degree of statical indeterminacy. Although the differential equation M'' = -q(x) (where prime denotes derivative with respect to x) can be solved numerically or analytically for a given set of boundary conditions, this is generally not the approach favoured by design engineers. In the following subsections, we will develop suitable methods for calculating M and V for a given loading q(x). The equations do allow us, however, to derive some general insights regarding the behaviour of beams. For a section of beam that is unloaded (such as would occur between the loading points of a beam loaded with a one or more concentrated loads, we have V'(x) = 0 for this interval. Integrating this equation yields the solution that V(x) is constant over this interval, and that M(x) is a linear function of x.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Similarly, if q(x) is equal to a constant C over a given interval within the beam, then integrating equations (1) and (2) yields that V(x) must be a linear function of x and M(x) must be a quadratic function of x, i.e., a parabola, over this interval.

Graphical Methods The analytical methods for calculating bending moments and shear forces in beams were covered in previous courses. In this subsection we will develop graphical methods for performing the same task. In all cases, the beams are loaded with concentrated loads. These methods can also be used for beams with distributed loads by replacing the distributed load with a statically equivalent series of concentrated loads.

Simple Beam The task is to calculate bending moments and shear forces in the beam shown in the figure below. We begin by drawing the given loads in the force plan and calculating the resultant load R. We do this by defining a pole O and drawing rays from the endpoints of the load vectors to O. We then transpose these lines into the location plan. The intersection of lines parallel to aO and Od in the location plan gives one point on the line of action of the resultant load R. We then close the diagram in the location plan and transpose the closing line to the force plan, which gives us point W. We know that the resultant load, vector ad, can be resolved into the sum of vectors aO and Od. In turn, it can be seen from the force plan that aO is the sum of aW and WO, and Od is the sum of OW and Wd. Replacing resultant load R with vectors aW, WO, Od, and OW in the force plan, we find that the vectors parallel to WO and OW cancel out. We are left with a vertical vector aW at the left support and a vertical vector Wd at the right support. It follows that the magnitude of the left reaction is given by the length of aW and the magnitude of the right reaction is given by the magnitude of Wd.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We now proceed to drawing the shear diagram. We consider a free body extending from the left support to a point at a distance x0 from the left support. We know that V(x0) is RL plus the sum of all loads applied between the support and x0. In the force plan, this is expressed graphically as a line of length Wa minus ab, i.e., the reaction minus the applied loads in question. We can therefore draw the shear diagram directly from the force diagram by taking as a datum a horizontal line at the level of W. The diagram is obtained by simply drawing horizontal lines from the endpoints of the load vectors and stepping down at the line of action of the adjacent load.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Equilibrium of the same free body gives us a graphical means of calculating bending moments. The upper of the two free bodies is not in equilibrium. It is used to calculate the resultant of reaction RL and load Q1. We know that this resultant is equal and opposite to the shear force at the section, V(x0). We also know that V(x0) corresponds to vector bW in the force plan. We can resolve bW into components bO and OW in the force plan. By translating bO and OW into the location plan, we can determine the location of the resultant of RL and Q1. We simply extend the corresponding lines in the location plan to their point of intersection, which in this case is point T. The resultant of RL and Q1 is thus a horizontal distance r from the cross-section at x0. To establish equilibrium with the resultant, we add an equal and opposite force (i.e., the shear force V(x0)), and move it to the cross-section. We maintain equilibrium by adding a moment equal to V(x0) times r. Bending moment at section x0 is thus equal to V(x0)·r. We can transform this into a more useful expression observing that triangles TUS in the location plan and ObW in the force plan are similar. It thus follows that y(x0)/r = V(x0)/H, where y(x0) is the vertical ordinate of the cable diagram, i.e., the distance from the closing line to the cable diagram, and H is the horizontal component of force from the force plan. Rearranging, we obtain M(x0) = y(x0)·H. This discussion proves that the bending moment diagram of a simple beam is geometrically similar in shape to a cable subjected to the same arrangement of load. The cable diagram has been left skewed, i.e., the rightmost point has not been displaced upwards to make the closing line level. This has been done to save time. Although this is a valid transformation of the diagram and would make the cable diagram look more like the moment diagrams we are used to, it is not necessary since we compute bending moments from the vertical offset y from the closing line to the cable curve, which will not be changed by such a transformation.

Indirect Loading Graphical methods can be used to develop effective solutions to specialized problems in the analysis of beams. Three of these will be examined, beginning with a system in which load is carried indirectly to the supports. Such systems often consist of relatively short longitudinal beams, often called stringers, which span between transverse members, referred to as floorbeams. These in turn carry load to the main girders, which are supported on piers or other foundations. The figure below shows a graphical analysis of this system. The upper portion of the figure shows a plan (view from above) of the system, showing the two girders running horizontally on the outer edges, five floorbeams running vertically, and short stringers spanning between the floorbeams. It is assumed that all stringers are simply supported at the floorbeams. A load Q of 60 kN is applied to one of the stringers as shown.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We first calculate the bending moments, shear, and reactions produced by the load in the stringer. We consider the stringer as a simply supported beam on rigid supports. This is a valid assumption for the calculation of forces and reactions. We begin by drawing the load vector ab in the force plan, selecting a pole O, drawing vectors aO and Ob, and transposing these lines to the location plan. These lines intersect the lines of action of the supports at points R and S. We then draw the closing line RS and transpose it to the force plan, where it becomes WO. The shear diagram in the stringer can now be drawn following the methods described for simply supported beams. We obtain the maximum bending moment Mmax as the product of ymax, the maximum ordinate between the closing line and the cable diagram, and the horizontal component of force H. We now turn to the calculation of moments, shears, and reactions in the main girder. The stringer reactions are taken by the floorbeams, which distribute these forces to the girders. (The calculations shown here are for the total effects of the load in both girders. To obtain the actual forces in a given girder, the results shown would have to be NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

multiplied by the ratio of the corresponding floorbeam reaction to total reaction. For the example shown here, assuming the stringers are spaced at 5 metre intervals, the given load would be distributed 83 percent to the lower girder in the plan and 17 percent to the upper girder.) We begin by observing that the girder, considered as a free body, is loaded by the reactions from the stringers. Using the same force plan, we therefore can consider the load vectors to be aW and Wb. Transposing rays aO, WO, and bO to the location plan, we can draw a new closing line TU extending from one girder support to the other. This line is transposed onto the force plan as W'O. The reactions at the ends of the girder are thus aW' and W'b. This allows us to draw the shear diagram, taking a horizonal line extending from W' as the datum. The bending moments are obtained in a similar manner to that used for the stringer, i.e., M(x) = y(x)·H.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Analysis of Statically Determinate Beams This section presents graphical and analytical methods for calculating sectional forces (bending moment, shear, and axial force) in beams. Special procedures are presented for several common cases, including indirect loading of beams and loading trains.

Response of Beams to Vertical Load (continued) Graphical Methods (continued)

Overhanging Ends Until now, we have dealt only with situations that produce positive moments. To study the application of graphical methods to beams that resist positive and negative moment, we will consider a beam with overhanging ends. The techniques we have developed thus far are also valid for this situation. The clarity of the results is improved, however, by an additional manipulation of the diagrams, as will be demonstrated in the figure below. We begin by drawing the structure to scale and drawing the loads in the force plan. We select a pole O and draw rays aO, bO, cO, and dO (green lines). We then transpose these lines to the location plan. We draw the closing line RS joining the intersection of line I and the line of action of the left reaction, and line IV and the line of action of the right reaction. We transpose the closing line to the force plan as ray WO. We wish to draw a force diagram that will allow us to determine y-ordinates based on a common datum line. We will use line RS as this datum, and extend it to cover the entire structure. The 15 kN load at the tip of the right cantilever is currently resolved into vectors cO and dO in the force plan (along lines III and IV in the location plan). We now resolve this load into a different pair of vectors, one directed along RS and the other along the line joining points P and T in the location plan. Line PT is transposed as cO'' in the force plan. The triangle cdO'' represents the new state of equilibrium. We make a similar transformation at the left-hand end of the structure, resulting in the state of equilibrium depicted by abO' in the force plan.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is now possible to determine y-ordinates at all locations along the beam from a common datum line. We compute bending moments as usual, i.e., M(x) = y(x)·H. The resulting bending moment diagram is drawn at the bottom of the figure on a horizontal axis. The theoretical values of peak moments are 40 kN·m, 30 kN·m, and 60 kN·m respectively. Also shown in the diagram is the diagram of a cable with loads and vertical support reactions applied. As expected, it has the same shape as the bending moment diagram of the beam.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Gerber Systems Gerber systems are multiple-span girders with internal hinges arranged to make the structure statically determinate. Examples of Gerber systems are shown in the following figure.

(Although Gerber systems were one popular in bridge construction, they are usually not recommended because of problems with durability due to the presence of a large number of expansion joints. They have, however, been used as a temporary structural system to facilitate construction, with a subsequent locking of the hinges and elimination of the expansion joint after the primary structural members have been erected.) The analytical solution of these systems is straightforward, provided one begins with a portion of a beam bounded by two hinges and with no supports between these two hinges. This portion of the beam can be separated as a free body and solved using the equations of equilibrium. In the following figure, solution proceeds from the upper free body downward. The asterisks represent quantities that are unknown at the time the given free body is analyzed.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Alternatively, the graphical techniques developed in the previous example can be used. The use of these techniques will be illustrated for the three-span Gerber system shown in the following figure. The solution is based on the principle that the line of force (i.e., the cable profile) must pass through the hinges. This is the graphic representation of the static condition of zero moment at the hinges. We begin with the leftmost span. We resolve load Q1 into components aO1 and O1b in the force plan. These components are then transferred to the location plan. To satisfy the condition of zero moment at the internal hinge, we draw the closing line RS to pass through the hinge at point S. The y-ordinate of the force diagram in the location plan, measured from the datum line RS, is thus zero at the hinge. We extend RS over the entire length of the structure. This will be our datum for moments at all locations. We now proceed to the force diagram in the rightmost span. We select a pole O3 and draw rays dO3 and O3e. These lines are transposed onto the location plan, making sure to pass the line parallel to dO3 through the internal hinge. The closing line, TU, does not coincide with the desired datum RS, so we shift pole O3 to O'3. The resulting force diagram shares a common datum with the diagram in the leftmost span, and satisfies the statical conditions of zero moment at the hinge and at the end support.

The final step is to draw the force diagram for the middle span. The graphical condition we must satisfy is that, in the location plan, the diagram must pass through points H and G. We select pole O2, draw rays bO2, cO2, and dO2 in the force plan, and transfer these rays to the location plan. The closing line, HI is not coincident with HG, so we must select a different pole O'2. We do so such that W2O'2 is parallel to HG. (In addition, it is convenient for all poles to lie on a common vertical line. In this way, the horizontal component of the rays in the force plan is constant over the entire structure.) Redrawing the rays in the force plan to O'2 and transferring them to the location plan yields a line of force that satisfies the statical conditions of the problem. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The final bending moment diagram is H times the vertical distance between the line of force (green line) and the datum line (line RF). The diagram has been redrawn at the bottom of the figure. Vertical reactions are obtained using reasoning similar to that used on previous problems. at support 2, the reaction is the vertical component of the sum of O1b and bO'2.

Load Trains The live load models used for bridge design generally consist of a set of load vectors at a specified spacing, representing the loads applied to the structure by the wheels of a heavy truck or train. The following figure, taken from the Canadian Highway Bridge Design Code, illustrates one such load model. The arrangement of axle loads and spacings specified in the figure can be applied at any location on the structure. These load models are referred to collectively as load trains.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is the designer's responsibility to ensure that the loads are applied to produce maximum effect. In this subsection, we will develop a procedure to accomplish this. The first task is the following: For a given load train and a given location on the structure, say a distance x0 from the left support, what is the position of the load train that will produce maximum bending moment at x0? We will develop a procedure for this task using the example shown in the following figure. In this case x0 will be taken as 16 m. We define parameters x'i as the distance from the right support to load i, and zi as the distance from x0 to load i. Loads Q1 through Q5 can be placed anywhere on the bridge, but the spacing between the loads remains constant. Because the loading consists of concentrated loads, the bending moment diagram will be piecewise linear. It therefore follows that the maximum bending moment at x0 will occur when one of the loads is positioned at x0. We will assume for now that positioning the train so that Q3 is above x0 produces the maximum moment at x0. The following figure gives a derivation of a method for validating this assumption.

The following numerical example illustrates the use of the inequality derived in the figure above:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The second task of interest is to determine, for a given load in the train Qi, the location in the span x0 such that, when the train is arranged with Qi over x0, the bending moment at x0 is maximized. The method is derived in the figure below.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Moment M(x0) is maximized when load Qi and the resultant of all loads, R, are equidistant from midspan.

Horizontal Load The analysis of beams for horizontal load is generally straightforward. Eccentricity of loads and support reactions must be considered, as shown in the figure below. It is generally a good first step to take all the loads and move them to the centroid of the section, adding moments corresponding to the displacement of the loads. The moments thus calculated can then be added to the moments due to vertical load calculated using the methods presented previously.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Analysis of Statically Determinate Trusses This section presents methods for assessing the stability and degree of statical indeterminacy of plane trusses, as well as graphical and analytical methods for calculating forces in truss members.

Basic Principles Trusses are structural systems composed of straight solid bars connected to each other at their ends only. Although truss members can be connected to each other in many different ways, truss connections are modeled as pinned connections, i.e., connections that cannot resist bending moments. This assumption is valid even for apparently stiff gusset plate connections since the axial stiffness of the truss members is generally sufficiently large to ensure that the bending restraint provided by gussets and other types of right connections is negligible. The points at which truss members are connected to one another are called joints. Trusses are generally designed to be loaded at the joints only. This condition, in addition to the assumption of perfectly pinned connections, implies that truss members work in pure tension or compression only.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Examples of Applications Trusses have been used since the earliest days of post-industrial revolution engineering. Examples of historical trusses include the Quebec Bridge, which still holds the record for the longest-spanning cantilever truss bridge in the world. (A cantilever truss is distinguished from other types of trusses by the need for a downward reaction at the end supports.)

Trusses are also incorporated into arch bridges and cable-supported bridges. An example of the former application is the Bridge over the New River Gorge in West Virginia. In this bridge, the arch is actually a truss curved according to the shape of the pressure line. The use of a truss as opposed to a solid member saves weight and provides considerable stiffness to resist bending due to live load.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The Brooklyn Bridge is an example of a cable-supported bridge that includes a truss as a major structural component. In this case, the truss stiffens the suspension bridge under the action of live load. Although we normally think of steel as the material of choice for trusses, they have been built in wood and even in concrete. In modern bridge construction, we see more and more trusses built of concrete and steel acting together. The Sherbrooke Pedestrian Bridge is a notable example of this type of structural system.

Stability In our previous discussion of stability, we stated the following principles: 1. Systems with fewer reactions than equations of equilibrium are always unstable 2. Systems with number of reactions greater than or equal to the number of equations of equilibrium are not necessarily stable. Such cases must always be investigated by the designer by visualizing the displacement of the structure under the action of forces and moments in all possible directions. The second principle holds particularly true for trusses, in which unstable arrangements of members such as the one shown in the figure below are possible, even though the number of reactions is equal to the the number of equations of equilibrium:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

In general, the best way to check for stability of trusses is by careful inspection. As you gain familiarity with different types of truss arrangements, you will become more adept at assessing stability in this way. A general analytical procedure for assessing stability does exist, but it is generally suitable only for use with a computer. At each joint of the truss, we can formulate two equations of equilibrium (ΣFx = 0 and ΣFy = 0) by cutting the joint away from the rest of the structure as a free body. We can assemble the entire set of equations for each joint of the structure into a matrix. The determinant of this matrix will be zero for unstable structures.

Determinacy Stable trusses are statically determinate when the following equation is satisfied: M+R=2·J where M is the number of members in the truss, R is the number of reactions, and J is the number of joints. This can be proven by the following reasoning: The unknowns are the member forces and the reactions. The number of unknowns is thus M + R, which is the right-hand side of the equation. The number of equations is equal to two times the number of joints, since there are two equations at each joint (ΣFx = 0 and ΣFy = 0) when we separate a given joint from the rest of the structure as a free body. Generally speaking, trusses with "X" panels will be statically indeterminate. This can be understood by removing one of the diagonals of the X. It is generally the case that the truss will remain stable when the member is removed.

Calculation of Forces in Trusses Using Analytical Methods We will briefly review the methods developed in previous courses for the analytical methods of calculating forces in determinate trusses.

Method of Sections This method consists of cutting the truss in two pieces, treating one of the pieces as a free body, and calculating the forces required to bring the free body into equilibrium.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The method is illustrated in the following figure:

We begin by calculating the reactions at the supports of the truss. We then cut the truss into two pieces, through the members for which we wish to calculate forces. We draw in vectors representing the unknown forces at the section. We then solve for the unknown forces using the three equations of equilibrium. This method works for cuts that produce no more than three unknown forces. Sections with more than three unknown forces exceed the number of available equations of equilibrium.

Method of Joints A more general method is the method of joints, which is based on isolating each joint as a free body and solving the two equations of equilibrium (ΣFx = 0 and ΣFy = 0). (The equation of moment equilibrium is not necessary because all forces pass through a common point at truss joints.) The method is illustrated in the following figure:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We begin by calculating the support reactions. We then isolate a joint that has no more than two unknown forces, isolate it as a free body, and solve for the unknown forces using the equations ΣFx = 0 and ΣFy = 0. We move from joint to joint in this way, being careful only to work with joints that have two unknown forces. When we write forces onto the truss diagram, we can use the letters "C" and "T" for compression and tension, respectively. Another convention is to use positive numbers for tension and negative numbers for compression. Using either convention, though, we must be sure to consider forces correctly in the equations of equilibrium, for which upward forces and forces to the right are positive, regardless of the way they are notated on the truss diagram. It is conceivable, for example, for a tensile force (positive in the diagram) to be directed downward at a joint (negative in the equation of equilibrium).

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Calculation of Forces in Trusses Using Graphical Methods The method of joints can easily be implemented graphically, using the principle that forces in equilibrium acting at a common point form a closed polygon when drawn in the force plan. In the graphical method, we likewise proceed from joint to joint, solving the equations of equilibrium graphically. An example of this method is given in the following figure:

We begin by calculating the support reactions graphically using the methods developed for beams. We then start at a joint with two unknown forces and draw the force diagram representing equilibrium of the joint in the force plan. The geometrical constraints of the truss and the known forces at the joint will lead directly to a graphical solution. In drawing the force diagrams, we must always go around a given joint in the same direction for all joints. In this case, a counter-clockwise direction was chosen. We treat external loads as member forces in this regard. For example, at joint L2, we first draw the force in L1L2, then the 60 kN load, then the force in L2L3, then L2U2, and finally close the figure with L2U1.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The direction of forces is always taken relative to the joint in question. For example at L2, we already know that L1L2 is a tensile force, i.e. a force pulling away from L2. We therefore draw it in the force plan as a force moving from right to left, i.e., away from L2. In a similar manner, the 60 kN load drawn downward. We require a force moving from left to right for L2L3 to close the polygon. This is equivalent to pulling away from L2, so force L2L3 is also tensile. To reduce the possibility of error, it is generally possible to draw contiguous polygons of equilibrium in the force plan. An example of how L0 and U1 could be drawn together is shown in the figure above. Using this method means that it is not necessary to redraw the force in L0U1.

Variations Loads Between Joints Although we generally assume that trusses are loaded at their joints only, there do exist cases in which truss members are loaded between two joints. In such cases, we can calculate stresses as shown in the figure below:

We first isolate loaded member AB as a free body (in this case a simply supported beam) and calculate moments, stresses, and reactions in the members. We then apply the reactions as loads to the truss as a whole, and calculate the axial force in member AB. We

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

can then combine stresses due to axial force in the overall response of the truss to stresses due to bending in the member computed previously.

Redundant Bracing One common case of statically indeterminate trusses that can be analyzed using the methods presented in this section is the case of redundant X bracing. The concepts are illustrated in the figure below:

The original system is a tower with two X-braced panels. The structure is two times statically indeterminate. We can design members AE, DB, DG, and EF to be sufficiently slender so that two of them will buckle when lateral load P exceeds a certain threshold value. Once buckling occurs, say in members AE and DG, the structure remains stable since members DB and FE maintain their integrity in tension. After buckling, we can consider that AE and DG no longer carry load. This implies that we can idealize the behaviour of the system under lateral load as the system shown in the rightmost diagram in the figure above, which is stable and statically determinate. This type of design is not used for structures in which fatigue is an important consideration.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Fatigue Fatigue is the progressive damage in structural members caused by the accumulated effect of repeated cycles of stress. Fatigue can cause members to fail at loads significantly lower than computed ultimate capacity. Designers ensure that structures have sufficient resistance against failure by fatigue by limiting the stress range due to live load to values defined in design standards.

A Simple Example Structural members such as bridge girders, crane rails, and machine parts, all of which undergo a large number of variations in stress due to live load, have been known to fail at loads significantly less than the ultimate capacity calculated on the basis of the section properties of the member and material yield strength. What generally happens is illustrated schematically in the following figure:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We will focus on member AB of the steel truss. By separating AB from the rest of the structure as a free body, we can see that AB is subjected to pure axial tension. The area of the cross-section of AB is As. We assume that we have performed an analysis of the structure for dead and live load and have determined that tension in AB due to dead load is PD and maximum tension in AB due to live load is PL. We say the member is safe when the design load Pf is less than the capacity of the member, AsFy, where Fy is the yield stress of the steel used in member AB. Design load Pf is equal to αDPD + αLPL, where αD and αL are factors of safety. They ensure that the capacity of the structure is adequate to resist not only the most likely combination of loads (PD + PL), but also load combination corresponding to a less likely but possible overload. We plot the stress in member AB as a function of time. Stresses corresponding to capacity (Fy) and design load (σf) are drawn in as constant values. The blue curve is the actual stress in AB, which is equal to dead load stress (σD) when live load is not present on the bridge and is equal to (σD + σL) when live load is present. The failure depicted occurs at a stress that is significantly less than the stress corresponding to the design load of the member. This graph therefore represents a mode of failure that is not adequately prevented by merely proportioning the cross-section to provide a capacity greater than the design load.

The Phenomenon What caused this behaviour? The answer lies in the microscopic structure of metals. (In fact, of all the materials normally used in construction, it is steel that is by far the most prone to fatigue.) Although we generally consider steel and other metals to be homogeneous materials, their crystal structure actually contains discontinuities called microcracks. Investigations into the behaviour of steel and other metals at the tip of microcracks shows that stresses at these locations are generally very high. The effect of repeated cycles of stress on microcracks has been described by Kulak and Grondin (2002): "Fatigue is the initiation and propagation of microscopic cracks into macro cracks by the repeated application of stresses. A discontinuity in the crystal structure of a metal or an initial crack will grow a small amount each time a load is applied to the part. Growth occurs at the crack front, which is initially sharp. Even at relatively low loads, meaning stresses less than the yield strength of the material, there will be a high concentration of stress at the sharp front, and plastic deformation (slip on atomic planes) takes place at the crack front. Continued slip results in a blunted crack tip, and the crack grows a minute amount during this process. Upon unloading, not necessarily to zero, the crack top again becomes sharp. The process, termed fatigue crack growth, is repeated during each load cycle." Fatigue is thus the growth of microcracks into a cracks of dimensions sufficient to impair the effectiveness of a given cross-section to resist load. In the example from the figure above, the cyclic stresses due to live load caused the growth of a crack that eventually

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

reduced the cross-section of member AB from As to A*s. The tensile resistance of the reduced section was thus reduced to A*sFy. Once A*s dropped to a value such that A*sFy was less than tensile demand, PD + PL, the member failed. The quotation by Kulak and Grondin given above makes two important points regarding the nature of fatigue: 1. Fatigue is a process of progressive damage in steel that requires both a discontinuity in the crystal structure (a microcrack) and repeated cycles of loading. Without the discontinuity, the material will be able to undergo repeated cycles of loading without change in its microscopic structure. Without the repeated cycles of loading, a given discontinuity cannot grow. 2. Microcracks do not significantly impair the capacity of structures to resist steadily increasing (i.e. non-cyclic) stresses. As stated in the above quotation, plastic deformation at the tip of the crack blunts the crack, preventing further growth of the crack provided load is not reduced and cycled up again. For this reason, it is perfectly acceptable to calculate the capacity of steel tension members to resist overload on the basis of the gross area of the cross-section As without direct consideration of microcracking.

Design Approach Structural designers do not generally deal with fatigue from a metallurgical point of view, i.e., designers do not calculate the response and grown of microcracks to a given cyclic regime of stress. Rather, we use empirical relations that have been determined relating the severity of microcracks in a given structural component to the expected cyclic stresses that the component will undergo during its service life. Modern design standards formulate these relations in terms of the following parameters: 1. The number of live load stress cycles that a given structural member is expected to undergo during its service life. 2. The stress range (σL,max - σL,min) in the component due to live load. 3. The type of details used in the structural component under consideration. Parameter (3), the type of detail, represents the severity of microcracks in the component in question. For most commonly used structural details, design standards assign a "fatigue category" depending on the presence of holes, welds, discontinuities in geometry, abrupt changes in the direction of stress, and other factors that are known to lead to stress concentrations. The following two figures illustrate how CSA Standard S16.1 Limit States Design of Steel Structures, the Canadian design standard for steel buildings, defines fatigue categories for common types of detail:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The standard describes a detail in words ("General condition" and "Situation") and assigns to it a Detail category (letters A, B, E, etc.). The standard also refers to Figure K2 giving illustrative examples of particular details. This figure assists designers in assigning the correct category to a specific detail. A portion of Figure K2 has been reproduced below.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Parameters (1) and (2) define the expected cyclic stress regime for the component in question. Parameter (2) defines the magnitude of the live load stress range. This value is calculated for live load models defined in design standards using a suitable method of structural analysis. Parameter (1) defines the number of times the stresses are expected to fluctuate through this range during the service life of the structure. For a given detail category and a given number of cycles of stress, design standards define an acceptable stress range that must not be exceeded. The following curves depict the relation between detail category, number of cycles of stress, and acceptable stress range:

For a given detail category, the acceptable stress range is the greater of the heavy solid line and the horizontal dashed line. The acceptable stress range for detail category A, for example, decreases from approximately 450 MPa at 105 cycles down to 165 MPa at 2 · 106

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

cycles. For a number of cycles greater than 2 · 106, the acceptable stress range remains constant at 165 MPa. This reflects an important characteristic of our current fatigue models, namely, that if stress range remains below a threshold value, the structural component in question can withstand an infinite number of stress cycles. Numerical values of this cut-off stress (corresponding to the dashed horizontal line) are given in the following table, under the heading "Constant amplitude threshold stress range":

It is often easier to use the following algebraic expression to calculate the acceptable stress range rather than the graph:

where Fsr is acceptable stress range, γ is the "Fatigue life constant" from Table 4(a) above for the category of detail under consideration, N is the number of cycles of stress, and Fsrt is the constant amplitude threshold stress range, also given in Table 4(a). Using this graph, we can determine the maximum acceptable stress range for a given situation. If we know, for example, that we are dealing with a "category B" detail, and the number of stress cycles is one million, then the acceptable stress range is 158 MPa. We compare this value to the actual live load stress range for the appropriate live load model specified in the applicable design standard. The actual stress range is defined as σL,max - σL,min, where σL,min may be zero. If σL,min is negative (i.e., a compressive stress), we compute the difference algebraically. The actual stress range defined by σL,max = 100 MPa and σL,min = -25 MPa, therefore, would be 125 MPa. It is emphasized that calculation of the actual stress range involves live load only. Dead load stresses are not considered in this calculation.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Design Procedure As designers, we must ensure that our structures are safe under all reasonably foreseeable conditions. For structures such as bridges, crane rails, towers, and others that must resist variable loads that are significant and frequent, we must add a check of fatigue in addition to our usual checks for safety. The following approach is used: 1. Preliminary selection of structural system. Select an initial structural arrangement and member properties. 2. Analysis. For a given structural component, calculate stress due to dead load, σD, and stress due to live load, σL. 3. Check of safety (strength). Calculate the design stress σf = αDσD + αLσL, using factors αD and αL from the applicable design standard. Ensure that σf is less than or equal to Fy. This ensures that the structure will have sufficient capacity to resist an accidental overload. 4. Check of fatigue. Ensure that the actual live load stress range is less than the acceptable live load stress range as defined in the design standard. This ensures that the structure will have sufficient capacity to resist a large number of cycles of a commonly occurring live load. (The approach given above has been simplified somewhat for clarity. You will learn about necessary refinements to this procedure in courses relating to design in specific materials in Third Year. The essence of the approach described here is, however, accurate.) This approach reflects the usual reality of design loadings, namely, that loads that cause fatigue (i.e., the loads that occur most frequently) are not the largest overloads. Two separate checks are therefore required.

Variable Stress Ranges The preceding discussion assumed that there was only one live load stress range to consider in checking fatigue. Although this assumption is often adequate for design, there are cases in which we can clearly identify two important live load models acting on the structure, each with its own stress range and expected number of cycles. We can deal with this situation with an expression called the Palmgren-Miner rule, which is defined by the following equation:

where Ni is the acceptable number of cycles for stress range i and ni is the actual number of cycles for stress range i.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The following example illustrates the use of the Palmgren-Miner rule:

References Kulak, G. L. and G. Y. Grondin. 2002. Limit States Design in Structural Steel. 7th ed. Toronto: Canadian Institute of Steel Construction.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The Method of Virtual Work The principle of virtual work states that a system of external and internal forces is in equilibrium if and only if the virtual work done by the forces as they undergo a given virtual displacement is zero. This principle is applied primarily to the calculation of deflections in structural systems.

Contents 1. The Significance of the Method of Virtual Work 2. Virtual Work 3. Internal Forces, Displacements, and Work 4. The Principle of Virtual Work 5. Application of the Principle of Virtual Work to Rigid Systems 6. Application of the Principle of Virtual Work to Flexible Systems 7. Integration Tables 8. Examples of Calculation of Deflections in Flexible Structures

The Significance of the Method of Virtual Work The remainder of this course will be devoted to the calculation of the response of statically indeterminate structures to loads and imposed deformations. Statically indeterminate structures are those for which the conditions of equilibrium alone are insufficient to allow reactions and/or internal forces to be calculated. There exist several effective methods of calculating the response of statically indeterminate structures. One of the most important of these, called the force method or flexibility method, is based on defining a number of additional conditions that, combined with the equilibrium conditions, will be sufficient to allow the response of a given structure to be calculated. These additional conditions are based on the compatibility of deformations. The following example describes the conceptual steps involved in the force method: The portal frame shown in the upper left-hand corner is statically indeterminate to the first degree, since there are four unknown forces (horizontal and vertical reactions at A, and horizontal and vertical reactions at D) and three equations of equilibrium. The

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

structure is loaded with a vertical load at midspan of the horizontal beam. Calculation of the response of this structure proceeds according to the following steps:

1. Release a sufficient number of restraints in the structure to produce a statically determinate system. In this case we release the horizontal reaction at D. The statically determinate system thus produced is called the primary system. This system is the second of the diagrams in the upper half of the figure. 2. Apply the given loads to the primary system and calculate the conjugate displacement in the direction of the forces that were released to create the primary system. In this case, the displacement is a horizontal deflection at D. This displacement corresponds to the horizontal reaction at D that was released. We call this displacement δ0. 3. Apply a load to the primary system in the direction of the force that was released to create the primary system and calculate the displacement of the structure that is conjugate to this force. In the figure, this force is called H. It corresponds to the horizontal reaction that was released. The displacement to be calculated is called δ1. 4. Calculate the total deflection of the structure in the direction of the released force: δtot = δ0 + δ1. 5. Calculate the value of H that makes δtot = 0. The condition δtot = 0 is called a compatibility condition. By imposing this condition, we effectively restore the original geometrical constraints of the structure, by ensuring that there is no displacement in the

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

direction of the forces that were originally released. This effectively restores the structure to its original statically determinate condition, but with knowledge of the unknown reaction H. We have thus eliminated one of the four unknown forces, leaving three forces that can be solved with the three equations of equilibrium. We see from this procedure that the calculation of deflections is of critical significance to the calculation of the response of statically indeterminate structures using the force method. For obvious reasons, the calculation of deflections is also of significance in and of itself, for structures of any given degree of statical indeterminacy. For example, design standards often define limits on deflections due to specified load cases. Designers must calculate deflections of the structure for these load cases and ensure that they are less than the specified limiting values. The most important method used in the calculation of deflections is the method of virtual work, which will be defined and described in the remainder of this section.

Virtual Work To understand the concept of virtual work, we must first define some important terms. Work is a physical quantity defined as the product of a force and a displacement conjugate to that force. Real work is defined as the product of a real force and the conjugate displacement produced by that same force. A real force is a force that corresponds to a state of equilibrium due to actions that correspond to the design requirements of a given structure. For example, real forces include the loads, reactions, and internal forces (moments, shear forces, and axial forces) due to conditions such as dead load, live load, wind load, etc. Real forces are the forces that a given structure will actually experience during its service life, or forces that represent a limit state (such as ultimate limit state) that must be checked to ensure safety and serviceability. A real displacement of a given structure is a displacement that corresponds to the response of the structure to real forces. For example, a beam subjected to dead load will deflect downward. The deflection at a given point along the span is a real displacement because it corresponds to a state of equilibrium defined by real forces. The following figure illustrates the concept of real work. In Example (a), a real horizontal load Q0 is applied to the tip of a cantilever column. The state of equilibrium corresponding to this force consists of the applied load Q0, and reactions H and M. Load Q0 causes the column to deflect. The displacement that is conjugate to Q0 is the horizontal deflection at the tip, δ. Because δ is the deflection caused by real load Q0, it is a real displacement. The real work done by Q0 is thus Q0 · δ.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

In Example (b), a cantilever with horizontal restraint at the tip is loaded with real moment M0 applied at the tip. The reactions required for equilibrium are horizontal forces at the tip and the base, and moment M1 at the base. Load M0 induces the displacements shown in the lower right-hand portion of the figure. The tip of the structure rotates through angles ω. Because this displacement is conjugate to the M0, we conclude that the real work done by M0 is M0 · ω. We note that the reaction H at the tip of the structure performs no work, because there is not displacement of the structure conjugate to H. Force H is horizontal and applied at the tip of the structure, and the structure has zero horizontal displacement at this location. Any force that is not real is called a virtual force. Virtual forces do not correspond to states of equilibrium related to the actions that will be experienced by a given structure or that must otherwise be checked to ensure safety and serviceability. For example, the simply supported bridge shown in the figure below, it is physically impossible to apply a concentrated moment at one end of the span as shown. The forces corresponding to this load case are therefore not real forces, but virtual forces.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

A virtual displacement of a given structure can be any displacement of the structure. Virtual displacements include the displaced shape of structures due to real or virtual loads, as well as displacements caused by the release of internal or external constraints. The three displaced shapes of the beam shown in the figure below all represent valid virtual displacements.

Virtual work is defined as the product of a given force and a given conjugate displacement, regardless of whether force or displacement are real or virtual. In fact, provided force and displacement are conjugate, they can be otherwise completely independent of each other. Real work is a subset of virtual work, corresponding to the case in which the force is real and the conjugate displacement was produced by the real force in question.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Internal Forces, Displacements, and Work The definition of virtual work is general, in the sense that it is valid not only for the displacement of external forces such as loads and reactions, but also for the displacement of internal forces such as bending moments, shear forces, and axial forces. Internal work, i.e., the work performed by internal forces as they undergo a displacement, is given a negative sign when it corresponds to the storage of internal strain energy. The significance of this sign convention will become clear when we consider the principle of virtual work. In order to calculate the work performed by internal forces, it is necessary to have a clear understanding of the displacements that are conjugate to these forces. Conjugate internal displacements are defined for axial force, shear, and bending moment in the figure below:

1. Axial Force Axial force N is applied to a differential element of length dx in the leftmost diagram of the figure above. The internal displacement corresponding to this force is the elongation of the element, which is equal to εdx, where ε is axial strain. The internal work performed by axial force N in differential element dx is given by the following expression: dWi = –N · εdx

2. Shear Force Shear force V is applied to a differential element of length dx in the middle diagram of the figure above. The internal conjugate displacement is the vertical displacement of the force on the right side of the differential element, which is equal to γdx, where γ is shear strain. The internal work performed by shear force V in differential element dx is thus given by the following expression:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

dWi = –V · γdx

3. Bending Moment Bending moment M is applied to a differential element of length dx in the rightmost diagram of the figure above. The internal conjugate displacement is the angle of rotation of one side of the element relative to the other. This is equal to the change in slope of the element, which is given by the expression (dθ/dx) · dx, where θ is slope. But change of slope dθ/dx is equivalent to curvature φ, so we can express the internal conjugate displacement as φ · dx. The internal work performed by moment M in differential element dx is thus given by the following expression: dWi = –M · φdx

The Principle of Virtual Work The principle of virtual work was defined previously in the context of influence lines. It is repeated here for emphasis: A system of forces acting on a given structure is in equilibrium if and only if the virtual work performed by these forces is zero for any compatible virtual state of deformation of the structure. Virtual work is understood to mean the sum of internal work (i.e. work performed by internal forces) Wi and external work (i.e. work performed by loads and reactions) We. In this statement of the principle of virtual work, we consider the term system of forces to include external loads, support reactions, and all internal forces (bending moments, shear forces, and axial forces) required for equilibrium. Virtual displacements are said to be compatible when the external displaced shape of the structure (displacements and rotations) can be determined uniquely from the internal deformations of the structure (strains and curvatures). The principle of virtual work is valid regardless of the relation (or lack thereof) between the system of forces and the state of deformation. In particular, the system of forces and the state of deformation can be completely independent of each other. We express the principle of virtual work mathematically as follows: For a given structure, let Qj be a set of loads, Rj be a set of support reactions, and N(x), V(x), and M(x) be internal forces. Assume all of these forces form a system of forces in equilibrium. Define a compatible state of deformation of the structure such that displacements δj are conjugate to loads Qj, displacements rj are conjugate to the reactions Rj, and ε(x), γ(x), and φ(x) are axial strain, shear strain, and curvature. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We express external work as follows: We = Qj · δj + Rj · rj Internal work is expressed as follows: Wi = –∫N(x) · ε(x)dx – ∫V(x) · γ(x)dx – ∫M(x) · φ(x)dx The principle of virtual work can this be expressed as follows: We + Wi = 0 Substituting in the expressions derived above for We and Wi, we obtain the following expression: Qj · δj + Rj · rj = ∫N(x) · ε(x)dx + ∫V(x) · γ(x)dx + ∫M(x) · φ(x)dx

Application of the Principle of Virtual Work to Rigid Systems The principle of virtual work can be applied to rigid systems. Because these systems by definition do not deform, no internal work is performed. For rigid systems, therefore, the principle of virtual work is used primarily to calculate sectional forces. One of the important applications of the principle of virtual work to rigid systems is in the creation of influence lines. We will not discuss rigid systems further in the current section.

Application of the Principle of Virtual Work to Flexible Systems The primary application of the principle of virtual work to flexible systems is in the calculation of displacements. This application of the principle is often referred to as the moment-area method.

Outline of Method Assume that we must calculate the displacement of a given structure at location x0 in the structure. This displacement will be called δ(x0). This displacement can be a rectilinear deflection or a rotation. The method of virtual work is based on the definition of the following states: 1. A real state of deformation corresponding to a given action of interest. For example, the real state of deformation could be the external and internal displacements of the structure due to dead load, or due to a uniform drop in temperature. Quantities associated with the real state of deformation will be written in green bold type.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

2. A virtual state of equilibrium corresponding to a unit load conjugate to δ(x0), called Q(x0). The state of equilibrium consists of the unit load Q(x0), support reactions, and internal forces. Quantities associated with the virtual state of equilibrium will be written in regular weight type. We first rewrite the mathematical statement of the principle of virtual work derived previously: Qj · δj + Rj · rj = ∫N(x) · ε(x)dx + ∫V(x) · γ(x)dx + ∫M(x) · φ(x)dx All displacement quantities correspond to the real state of deformation; all force quantities correspond to the virtual state of equilibrium. In particular, quantities ε(x)dx, γ(x)dx, and φ(x)dx are the strains corresponding to the given action under consideration, and quantities N(x), V(x), and M(x) are the sectional forces due to the unit load conjugate to the deflection to be calculated. Because we are dealing with a real state of deformation, the displacements of the supports, rj, will all be zero. Because there is only one load, and this load is equal to one, and the conjugate displacement to this load is δ(x), we can transform the equation into the following one:

δ(x0) = ∫N(x) · ε(x)dx + ∫V(x) · γ(x)dx + ∫M(x) · φ(x)dx The deflection of the structure at point x0 can thus be calculated by evaluating the integrals given on the right-hand side of the equation.

Application of Method The two primary applications of the method of virtual work are: (1) deflections of trusses, and (2) deflections of beams and frames.

Deflections of Trusses In trusses, we do not consider bending or shear in members. The previous equation thus simplifies to:

δ(x0) = ∫N(x) · ε(x)dx Furthermore, since force and strain are constant in a given truss member, we can rewrite the equation as:

δ(x0) = Σ(Nj · εj · Lj) where Nj is axial force in member j due to a unit virtual load conjugate to δ(x0), εj is strain in member j corresponding to the real state of deformation, Lj is the length of member j, and the summation is over all members in the truss.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Deflections of Beams and Frames Generally speaking, it is acceptable to neglect axial and shear deformation in beams and frames of conventional proportions. The additional accuracy gained from considering axial and shear deformation is usually negligible, provided the depth of members is small compared to span length. This is generally the case in most situations. If these assumptions are valid, then the equation of virtual work previously defined can be simplified by setting ε(x) and γ(x)dx to zero for all x. We therefore obtain the following equation:

δ(x0) = ∫M(x) · φ(x)dx where M(x) is bending moment due to a unit virtual load conjugate to δ(x0) and φ(x) is curvature corresponding to the real state of deformation. For linear elastic structures, we often make use of the expression φ(x) = M(x)/(EI), where M(x) is bending moment due to the real actions (not the same as the virtual moment M(x)!!!), E is modulus of elasticity, and I is moment of inertia. For linear thermal gradient (temperature varies linearly through the depth of a given member), curvature is given by the following formula:

φ = (Ttop - Tbot) · αT/h where φ is curvature, Ttop and Tbot are temperatures at the top and bottom of the member, respectively, αT is the coefficient of thermal expansion, and h is the depth of the member.

Integration Tables The definite integrals that must be calculated in applications of the method of virtual work generally need not be evaluated by hand from first principles. It is valid to use integration tables, which have been developed for many of the cases commonly encountered in structural analysis. In the examples that will be presented in the remainder of this section, integration tables will be used where appropriate. When the functions that must be integrated cannot be found in the integration tables, a numerical method such as Simpson's rule is recommended.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Examples of Calculation of Deflection in Flexible Structures Example 1: Deflection of a cantilever column due to uniform horizontal load

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 2: Rotation of a cantilever column due to uniform horizontal load

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 3: Deflection of leg of statically determinate portal frame

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 4: Deflection of frame due to thermal gradient

Example 5: Deflection of truss due to load

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 6: Deflection of truss due to temperature Truss is the same as shown in the previous example. (Real) temperature change is given individually for each member in the table below.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 7: Variable moment of inertia When moment of inertia varies within a given member, it is generally not practical to use integration tables. In such cases, it is usually preferable to use numerical integration. The following example shows how to proceed.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is clear from the information given that there is a significant variation in moment of inertia I from the free end to the fixed end of the cantilever. To calculate curvature, we divide the real moment M(x) by EI, which in this case is not a constant, but a function of x. The resulting quotient, M(x)/(EI(x)), does not correspond to any of the functions given in the integration tables. We must therefore proceed numerically. We first generate numerical values for M(x)/(EI(x)) at a number of stations that is large enough to give acceptable accuracy. These values are equal to the real curvature due to the given load. We then apply a unit virtual load conjugate to the deflection to be calculated, i.e. a vertical load at the tip of the cantilever. We the calculate the values of the virtual moment due to this load at the stations under consideration. The numerical values of M(x)/(EI(x)) are multiplied by the values for virtual moment. These products are in turn multiplied by the integration coefficients defined by Simpson's rule. The results are added and the sum is multiplied by the integration sub-interval (in this case 12.5 m) divided by 3. This example illustrates how the virtual moment diagram can be regarded as a weighting function. If we want to change the deflection at the tip of the cantilever by changing the stiffness (EI(x)) of the system, we would do best to change the stiffness where it will have the most effect. Since the virtual moment decreases from left to right, it follows that changing the stiffness of the beam near the tip of the cantilever will have relatively little effect on the tip deflection, since any change in M(x)/(EI(x)) will be dampened out by the virtual moment. For example, if we increase stiffness at Station 3 by 100 percent, we decrease tip deflection by only five percent. In comparison, increasing stiffness at Station 1 by 100 percent decreases tip deflection by 29 percent. This confirms that deflections can be controlled most effectively by changing stiffness at those locations in the structure where the virtual moment is greatest.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The Force Method The force method is used to calculate the response of statically indeterminate structures to loads and/or imposed deformations. The method is based on transforming a given structure into a statically determinate primary system and calculating the magnitude of statically redundant forces required to restore the geometric boundary conditions of the original structure.

Contents 1. Outline of Method 2. A Simple Example 3. Observations 4. Selection of the Primary System 5. Examples

1. Outline of Method The force method (also called the flexibility method) is used to calculate reactions an internal forces in statically indeterminate structures due to loads and imposed deformations. The steps in the force method are as follows: 1. Determine the degree of statical indeterminacy of the structure. Use the parameter n to denote the degree of indeterminacy. 2. Transform the structure into a statically determinate system by releasing a number of statical constraints equal to the degree of statical indeterminacy, n. This is accomplished by releasing external support conditions or by creating internal hinges. The system thus formed is called the primary system. Number the released constraints from 1 to n. 3. For a given released constraint j, introduce an unknown redundant force Xj corresponding to the type and direction of the released constraint. 4. Apply the given loading or imposed deformation to the primary system. Use the method of virtual work to calculate displacements at each of the released constraints in the primary system. These displacements are called δ10, δ20, ..., δn0.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

5. For a given released constraint j, apply a unit load Xj = 1 to the primary system. Use the method of virtual work to calculate displacements at each of the released constraints in the primary system. These displacements are called δ1j, δ2j, ..., δnj. 6. Solve for redundant forces X1 through Xn by imposing the compatibility conditions of the original structure. These conditions transform the primary system back to the original structure by finding the combination of redundant forces that make displacement at each of the released constraints equal to zero. The conditions are expressed mathematically as follows:

δ10 + X1 · δ11 + X2 · δ12 + ... + Xn · δ1n = 0 δ20 + X1 · δ21 + X2 · δ22 + ... + Xn · δ2n = 0 δ30 + X1 · δ31 + X2 · δ32 + ... + Xn · δ3n = 0 ...

δn0 + X1 · δn1 + X2 · δn2 + ... + Xn · δnn = 0 This is a system of n linear equations in n unknowns. The displacements δ are all known. The unknown forces are Xj. (It can thus be seen that the name force method was given to this method because its primary computational task is to calculate unknown forces, i.e., the redundant forces X1 through Xn.) 7. Calculate force S at a given location in the structure using the following combination: S = S0 + X1 · S1 + X2 · S2 + ... + Xn · Sn where quantities Xj have been calculated from the n by n system of equations given in Step 6, S0 is the force due to the given load or imposed deformation in the primary system, and Sj is the force due to Xj = 1 applied to the primary system.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

2. A Simple Example The principles defined in the previous section are illustrated using the following simple example:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

3. Observations Based on the preceding example, the following observations can be made:

Choice of Primary System There is an infinite number of valid primary systems. The following figure shows the original statically indeterminate system, the primary system used in the previous example, and three additional primary systems with redundant forces.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Although there is no limit on the number of feasible primary systems, not all primary systems are equal in terms of the computational effort they require. This issue is discussed in further detail in Section 4 of these notes.

Treatment of M1 and M2 Moments M1(x) are the moments in the statically determinate primary system due to the redundant force X1 = 1 applied as a load to the primary system. For the calculation of the displacements that must be evaluated for the force method, M1(x) must be considered as real moments in certain cases and virtual moments in others. For example, for the calculation of δ10, the displacement in the direction of X1 due to the given loads, moments M1(x) are considered to be virtual moments. The cause of the displacement is the given loads, so moments due to the given loads M0(x) are considered to be real for this calculation. Moments M1(x) are produced by the unit moment X1 = 1 conjugate to the displacement to be calculated, so M1(x) is taken as virtual. For the calculation of δ21, however, moments M1(x) are considered to be real. Displacement δ21 is the displacement in the direction of X2 due to a unit load applied in the direction of X1. Since X1 is the cause of the displacement to be calculated, moments M1(x) are considered to be real moments. In this case, therefore, moments M2(x) are considered to be virtual moments since they correspond to the unit load applied conjugate to the displacement to be calculated, i.e., rotation in the direction of X2. The preceding discussion, of course, applies equally to moments M2(x).

Symmetry of Flexibility Coefficients For 1 ≤ i, j ≤ n, quantities δij are called flexibility coefficients. (This term is used because the value of δij increases with increasing values of 1/(EI). Since EI is a measure of the stiffness of a given member, 1/(EI) is a measure of its flexibility. Coefficients δij are thus a NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

measure of the flexibility of a given member.) The matrix of flexibility coefficients δij is called the flexibility matrix, where i and j vary from 1 to n. Coefficient δij is defined as the displacement in the direction of Xi due to a unit load in the direction of Xj. It can be expressed mathematically as:

δij = ∫Mi(x) · (Mj(x)/(EI)) · dx where Mj(x)/(EI) is the real curvature due to a unit load applied in the direction of Xj and Mi(x) is the virtual moment due to a unit load applied in the direction of Xi. This equation can be rearranged as follows:

δij = ∫Mj(x) · (Mi(x)/(EI)) · dx i.e., δij is equivalent to the integral of the product of the curvature due to a unit load applied in the direction of Xi and the moment due to a unit load applied in the direction of Xj. This, however, is the definition of flexibility coefficient δji. It therefore follows that

δij =δji for 1 ≤ i, j ≤ n. In other words, the flexibility matrix is symmetrical about its main diagonal.

Solution The solution of forces X1 through Xn requires inversion of the flexibility matrix. This can be accomplished using any of the familiar methods of linear algebra. It can be seen that, for n > 3, the computational effort becomes significant unless a computer is used.

4. Selection of the Primary System As stated previously, there is no limit to the number of different primary systems that can be generated for a given structure. The choice of primary system, however, must ensure that the primary system is stable. In addition, it is recommended that the primary system be chosen to minimize computational effort and maximize computational accuracy.

Stability of Primary System It is not sufficient merely to release the correct number of statical constraints in generating a primary system. Care must be taken to ensure that the primary system is stable. The following examples illustrate this proposition:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Choice of Primary System to Reduce Computational Effort The computational effort required in calculating the response of a given structure using the force method can vary significantly depending on the choice of primary system. In this regard, there are two issues to consider: 1. Select the primary system to permit the use of integration tables 2. Select the primary system to maximize the number of flexibility coefficients equal to zero These issues are illustrated in the following example:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The structure under consideration is a four-span continuous beam. The degree of indeterminacy, n, is three. Two primary systems are illustrated. On the left-hand side of the figure, the primary system is formed by releasing moment in the beam at the three interior supports. On the right-hand side of the figure, the primary system is formed by releasing the vertical reaction at the three interior supports. For each primary system, bending moments M0, M1, M2, and M3 are drawn. The following observations are made: 1. For the primary system on the left-hand side of the diagram, all integrations required for calculating the coefficients δ can be performed using integration tables. This is not the case for the system on the right-hand side of the diagram, since function M0 does not correspond to a function in the tables. Coefficients δ10, δ20, and δ30 must therefore be calculated using numerical integration or other means. 2. For the primary system on the left-hand side of the diagram, several coefficients are zero. On the right-hand side, however, all coefficients are nonzero. The choice of primary system on the left allowed the influence of a given redundant force Xj to be restricted to a relatively small portion of the structure (two spans in this particular case). For the primary system on the right-hand side, the influence of a given redundant force Xj is felt throughout the structure. The larger the number of zero coefficients, the easier the solution. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We can conclude that the primary system on the right-hand side is preferable because it reduces computational effort by allowing the use of integration tables and by limiting the influence of the given redundant forces, thus increasing the number of zero coefficients δ.

5. Examples Frame with vertical load The task is to calculate bending moments in the frame given below due to the vertical load shown. The example proceeds step by step according to the procedure given previously. Steps 1, 2, and 3:

Step 4. One of the primary tasks in this step is to calculate moment diagrams in the primary system M0 (due to the given loads), M1 (due to X1 = 1), and M2 (due to X2 = 1). Bending moment M0 is shown below:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

M1:

M2:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

These moment diagrams are then used to calculate deflections δ10 and δ20:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Step 5. Calculate the coefficients of the flexibility matrix:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Step 6. Set up the system of linear equations and solve it for X1 and X2:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Step 7. Draw the bending moments M for the original structure using the relation M = M0 + X1 · M1 + X2 · M2

Frame with settlement of a support We will use the same structure as used for the previous example. Steps 1, 2, and 3 are as outlined previously. Step 4: Imposed deformations create no moments in statically determinate structures. A support settlement is an imposed deformation, and the primary system is statically

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

determinate. Moments M0 are therefore zero. We calculate δ10 and δ20 from simple geometrical considerations, as shown below.

Step 5. Because the primary system and redundant forces are identical to the ones selected in the previous example, the flexibility coefficients will also be the same. Step 6. We can also re-use the inverse flexibility matrix computed previously. Step 7. The final moment M is equal to X1 · M1 + X2 · M2, since M0 is zero.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The Displacement Method The displacement method is used to calculate the response of statically indeterminate structures to loads and/or imposed deformations. The method is based on calculating unknown rotations at the joints of frames based on conditions of equilibrium at the joints.

Contents 1. Introduction to the Displacement Method 2. Definitions and Sign Conventions 3. Fixed-End Moments 4. Stiffness Coefficients 5. Member End Moments 6. Equilibrium Conditions (Joint Translation not Considered) 7. Procedure (Joint Translation not Considered) 8. Examples (Joint Translation not Considered) 9. Joint Translation 10. Example (Joint Translation Considered)

1. Introduction to the Displacement Method The force method, derived in a previous section, is a method for calculating the response of statically indeterminate structures by which the unknowns are force quantities (the redundant forces X1, X2, ..., Xn) and the equations used to solve for the unknowns are based on geometrical conditions (compatibility conditions at the location of each redundant force). It is possible to consider an analogous method for calculating the response of statically indeterminate structures in which the unknowns are displacement quantities and the

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

equations used to solve for the unknowns are based on statical conditions (equilibrium conditions). The displacement method, in the form that it will be presented here, is also referred to as the slope-deflection method.

Outline of the Displacement Method A more detailed step-by-step definition of the displacement method will be given later. For now, however, it is helpful to outline the major elements of the method: 1. For a given structure and loading, consider the joints to be fully fixed against translation and rotation. 2. Calculate the moments in each member of the structure due to the given loads, assuming full fixity at the joints. These moments are called fixed-end moments. 3. Calculate moments at the ends of each member due to unit displacements of the joints. 4. Express the total moment at each end of a given member as the sum of the fixed-end moments and the product of unknown joint displacements times the moments produced by unit joint displacements, calculated in Step 3. 5. Generate an equation of moment equilibrium at each joint. 6. Solve the system of equations for the unknown joint displacements. 7. Calculate the member end moments using the expressions derived in Step 4 and the values of joint displacements calculated in Step 6. 8. Calculate all remaining forces in the structure (shear forces and axial forces).

2. Definitions and Sign Conventions One practical difference between the displacement method and the force method is that strict adherence to a unique sign convention is required when using the displacement method.

2.1 Numbering of Joints The displacement method requires that all joints be numbered. Any numbering scheme can be chosen. Internal hinges are not considered as joints and are not numbered.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

2.2 Member End Moments We define member end moment Mik as the moment at joint i in the member ik. The first subscript denotes the joint under consideration, the pair of subscripts denotes the member under consideration. For example, M23 would be the moment at the top of the left column in the frame shown above, i.e., the moment at joint 2 in member 2-3. The moments at the ends of member 2-3 are therefore called M23 (at the top of the member) and M32 (at the bottom of the member). The sign convention for member end moments Mik is counter-clockwise positive when the free body under consideration is a member. It follows from equilibrium considerations that the sign convention for member end moments acting on a joint that has been separated as a free body is clockwise positive.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

A moment externally applied as a load to joint i is called Mi. Moment Mi applied externally to joint i as a load is counter-clockwise positive. This type of moment often occurs when the given structure contains a cantilever member. It is convenient to calculate the cantilever moment, remove the cantilever, and apply the moment as an external load at the joint opposite the free end of the cantilever, as shown in the following figure:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Although the cantilever moment is

2.3 Joint Rotations Two types of rotation angles are considered. The angle of joint rotation, φi, is the angle by which joint i rotates relative to the undeformed structure, assuming that there is no translational displacement of any of the joints in the structure. Joint rotation is defined as counter-clockwise positive. Positive joint rotations φi and φk are shown in the following figure.

2.4 Member Rotations The angle of member rotation, ψik, is created by translational displacements of joints i and k in member i-k, assuming joint rotation is zero at both ends of the member. It is defined as the angle at joint i between the line segment joining the displaced positions of joints i and k and member i-k in the undeformed structure. It follows that ψik = ψki. Member rotation is defined as counter-clockwise positive. Positive member rotations ψik and ψki are shown in the following figure. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

3. Fixed-End Moments The displacement method takes as its starting point a structure in which all joints have been fixed against translation and rotation. We define moment Mik0 as the moment produced by the given loads acting on member i-k, assuming joints i and k are fully fixed against translation and rotation. Moment Mik0 is called a fixed-end moment. In most cases, it is practical to use the force method to calculate fixed-end moments. Solutions have been derived, however, for several frequently occurring cases, which are given below (EI is constant along the given member in all cases):

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Uniform Load (End Span with Pinned Support)

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Concentrated Load (End Span with Pinned Support)

Summary The fixed-end moments derived above are summarized in the following figure. The sign convention is as defined in Section 2. Although the fixed-end moments for these cases can be derived from scratch using the force method, it is preferable to commit them to memory.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4. Stiffness Coefficients Having developed expressions for end moments in members due to the given loads, assuming fixed-end conditions, we must now develop expressions for end moments due to unit end rotations. These relations will then be used to develop a general expression for total end moments, expressed as the sum of the fixed-end moments and moments due to rotations at the ends of the members. The following cases will be considered: 1. Moments due to a unit joint rotation at one end of a beam, with opposite end fixed 2. Moments due to a unit joint rotation at one end of a beam, with opposite end pinned 3. Moments due to a unit joint rotation at one end of a beam with internal hinge, with opposite end fixed 4. Moments due to a unit joint rotation at one end of a beam with internal hinge, with opposite end pinned 5. Moments due to a unit member rotation, with both ends fixed against joint rotation 6. Moments due to a unit member rotation, with one end fixed and one end pinned for joint rotation These results will be derived using the force method. We will use the same primary system used to calculate the expressions derived previously for fixed-end moments. The system, flexibility matrix, and inverse flexibility matrix are shown below. In all cases, we will assume that EI is constant.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We will use the following notation: sik: Quantity s refers to the moment at a given end of the member due to a unit rotation applied at the same end of the member. Quantity sik is thus the moment at end i of member i-k due to unit rotation applied at end i. Quantity ski is the moment at end k of member i-k due to a unit rotation applied at end k. tik: Quantity t refers to the moment at a given end of the member due to a unit rotation applied at the opposite end of the member. Quantity tik is thus the moment at end i of member i-k due to a unit rotation applied at end k. Quantity tki is the moment at end k of member i-k due to a unit rotation applied at end i.

4.1 Moments due to a unit joint rotation at one end of a beam, with opposite end fixed

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4.2 Moments due to a unit joint rotation at one end of a beam, with opposite end pinned

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4.3 Moments due to a unit joint rotation at one end of a beam with internal hinge, with opposite end fixed

4.4 Moments due to a unit joint rotation at one end of a beam with internal hinge, with opposite end pinned

It is clear from the figure above that the member provides no restraint to an imposed rotation applied at end i. Moments sik and tki are thus both zero.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4.5 Moments due to a unit member rotation, with both ends fixed against joint rotation

In the figure above, the quantities sik, tik, ski, and tki refer to the stiffness coefficients developed in Section 4.1.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4.6 Moments due to a unit member rotation, with one end fixed and one end pinned for joint rotation

In the figure above, sik, tik, ski, and tki refer to the stiffness coefficients developed in Section 4.2. Quantities tik, ski, and tki are zero.

5. Member End Moments We combine the expressions developed for fixed-end moments and moments due to joint and member rotation to give total member end moments. The general expressions are as follows for member i-k: Mik = Mik0 + sik · φi + tik · φk – (sik + tik) · ψik, and Mki = Mki0 + tki · φi + ski · φk – (ski + tki) · ψki Using the first equation as an example, the individual terms are defined as follows: Mik0 is the fixed-end moment at end i sik · φi is the moment at end i due to a rotation of φi applied at joint i tik · φk is the moment at end i due to a rotation of φk applied at joint k

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

– (sik + tik) · ψik is the moment at end i due to a member rotation of ψik The equation thus gives moment due to all possible causes: moment due to load assuming fixed ends, plus moments due to all possible rotations of joints and member. In the preceding expressions, fixed-end moments and stiffness coefficients s and t must correspond to the particular conditions at the joints of the member under consideration. In general, member ends are considered to be fixed for all cases except cases of exterior members with a pin or roller at the exterior support.

6. Equilibrium Conditions (Joint Translation not Considered) The preceding expressions for member end forces are general, in the sense that they consider the effect of both joint rotation and member rotation. In developing the remaining steps of the displacement method, we will first proceed under the assumption that the joints do not displace horizontally or vertically, i.e., that joint translation can be neglected. For structures in which axial deformation of members is small compared to flexural deformation, this assumption is often valid.

For example, we can consider joint "j" between beam and column in the two structures shown in the figure above. In both structures, vertical displacement joint "j" is prevented by the column, which we have assumed not to deform axially. In the structure on the left, the left end of the beam is pinned and hence cannot displace horizontally. This condition, together with the axial assumed axial rigidity of the of the beam, ensures that joint "j" does not displace horizontally. We say that the frame on the left is braced against sway. For this frame, ψik will be zero for all members i-k, and we can thus simplify the expressions for member end moments to the following equations: Mik = Mik0 + sik · φi + tik · φk, and Mki = Mki0 + tki · φi + ski · φk In the structure on the right in the figure above, the pin at the left end of the beam is replaced by a roller. As shown by the displaced shape, joint "j" is now free to displace horizontally. We say that this frame is unbraced against sway. It is clear from the figure that the column can undergo a member rotation, and thus the full expressions for member end moments, including ψik, must be used.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Assuming that the joints of the structure cannot translate, we can therefore formulate equations of equilibrium at each joint at which an unknown joint rotation is present. For a given joint "i", the equation of equilibrium is as follows: Mi – ΣMik = 0 where moment Mi is an external moment (considered as a load) applied at joint i and Mik are member end moments incident at joint i. Summation is over the index k. These principles are illustrated in the following figure:

At Joint 2, moment M2 is applied externally. In accordance with the sign convention defined previously, it is shown as counter-clockwise positive. The equation of equilibrium at Joint 2 is thus: M2 – (M21 + M23 + M24) = 0 External supports are defined as supports with only one incident member. Equilibrium conditions are not considered at fixed external supports, since there is no joint rotation at these locations. Equilibrium conditions are not considered at external supports with pins or rollers, since there moment is zero at these locations.

7. Procedure (Joint Translation not Considered) Assuming joints do not translate, the steps of the displacement method are as follows: 1. Number the joints 2. Identify unknown joint rotations φ 3. Compute member end moments for all members using the following expressions: Mik = Mik0 + sik · φi + tik · φk and Mki = Mki0 + tki · φi + ski · φk 4. Compute external moments applied at the joints, Mi 5. For each joint corresponding to an unknown joint rotation, formulate an equation of equilibrium 6. Solve the equations of equilibrium for unknown joint rotations φ

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

7. Calculate member end moments using the equations developed in Step 3 and the solution obtained in Step 6 8. Calculate shear forces using conditions of equilibrium in a given member 9. Calculate axial forces using conditions of equilibrium at a given joint

8. Examples (Joint Translation not Considered) Example 1

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Example 2 The following example illustrates the use of the displacement method for problems with more than one unknown, as well as the treatment of cantilever members.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

9. Joint Translation The method previously developed for structures braced against sway can be expanded to cover the case of structures that are not braced against sway. In such structures, we must consider both joint rotations φ and member rotations, ψ. We will first introduce a procedure that will allow us to determine the number of independent member rotations that must be considered in a given structure. The procedure, illustrated in the figure below, is as follows: 1. Insert a hinge at each joint in the structure 2. Introduce forces to restore stability to the system The number of forces thus determined is the number of independent member rotations that must be considered in the solution.

The independent member rotations are defined by examination of the displacements of the structure with hinges. In the upper structure in the figure above, for example, rotations ψ31 and ψ42 are equal. It is thus necessary only to include one of the two as an unknown. For example, if we consider ψ31 to be the independent unknown member rotation, then we would express the end moments in member 2-4 as functions of ψ31 and not of ψ42.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

When inclined members are present, the relations linking member rotations can become somewhat more complex, as illustrated in the example below. In this case, angle β is a linear function of α, so there is only one independent member rotation.

Once the independent member rotations have been identified, it is necessary to express all other member rotations as functions of the independent member rotations. Following this, expressions for member end moments must be established for all members, using the complete expressions developed previously and repeated here: Mik = Mik0 + sik · φi + tik · φk – (sik + tik) · ψik, and Mki = Mki0 + tki · φi + ski · φk – (ski + tki) · ψki In the above expressions, for member i-k, if ψik is not one of the independent member rotations, then it must be replaced by a linear combination of independent rotations.

Equilibrium Conditions Since the number of equilibrium conditions based on joint rotations is equal in number to the number of unknown joint rotations, additional equilibrium conditions are required if member rotations must be considered. We establish these conditions with the help of the principle of virtual work. We proceed as follows: 1. Introduce hinges at the ends of all members and draw in member end moments. 2. Displace the structure thus modified, with external load, according to a given independent member rotation. (Because of the way the hinges were selected, the joints do not translate. The work done by moments at the joints is thus zero.) 3. Calculate total virtual work performed by external loads and member end moments. In accordance with the principle of virtual work, equate the total virtual work to zero. This equation is an additional equilibrium condition. These principles are illustrated in the following figure. NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We can now define the steps for calculations of structures that are not braced against sway.

Procedure 1. Number the joints 2. Identify unknown joint rotations φ 2A. Identify independent member rotations ψ and express all other member rotations as functions of the independent rotations 3. Compute member end moments for all members using the following expressions: Mik = Mik0 + sik · φi + tik · φk – (sik + tik) · ψik, and Mki = Mki0 + tki · φi + ski · φk – (ski + tki) ·

ψki

4. Compute external moments applied at the joints, Mi 5. For each joint corresponding to an unknown joint rotation, formulate an equation of equilibrium

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

5A. For each independent member rotation, formulate an equation of equilibrium based on the principle of virtual work. 6. Solve the equations of equilibrium for unknown joint rotations φ and member rotations

ψ

7. Calculate member end moments using the equations developed in Step 3 and the solution obtained in Step 6 8. Calculate shear forces using conditions of equilibrium in a given member 9. Calculate axial forces using conditions of equilibrium at a given joint

10. Example (Joint Translation Considered)

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

The Method of Moment Distribution The method of moment distribution is an application of the displacement method that uses a method of successive approximations to solve for the unknown joint rotations, rather than matrix methods.

Contents 1. Basic Concepts 2. Procedure (No Joint Translation) 3. Example (No Joint Translation) 4. Simplified Procedure (With Joint Translation) 5. Fixed-End Moments Due to Imposed Translation of Joints 6. General Procedure (With Joint Translation)

1. Basic Concepts The method of moment distribution is a refinement of the displacement method presented previously, by which member end moments are calculated based on a procedure of successive approximations rather than the solution of a matrix equation. The method was developed by Hardy Cross (1936) and was the primary method of analyzing structures that respond primarily in bending until the availability of computer software for structural analysis in the 1970s. To define the procedure to be followed in calculating the response of structures using moment distribution, it is first necessary to examine in greater detail the state of equilibrium and deformation at the joints of a given structure. We will first develop the method of moment distribution assuming the joints do not translate.

Restraint of Joints and Unbalanced Moments In the displacement method, we derived an expression for member end moments, which is repeated here:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Mik = Mik0 + sik · φi + tik · φk – (sik + tik) · ψik Since we have assumed the joints do not translate, it follows that the contribution of member rotation, (sik + tik) · ψik, can be set to zero. We are therefore left with the following expression: Mik = Mik0 + sik · φi + tik · φk If we make the further assumption that the joints do not rotate, then we can set φi and φk to zero. The member end moments can therefore be written as follows: Mik = Mik0 What is left, therefore, are the fixed-end moments. We now investigate the state of equilibrium at joint i. We begin by adding the member end moments Mik for all members incident at joint i. (The sign convention developed for the displacement method, i.e., counter-clockwise positive, must be followed.) The total moment applied to the joint will thus be –ΣMik0, where the summation is over all members incident at joint i. (The minus sign is used because moments applied to the joints are equal and opposite to the member end moments.) In general, this sum will not be equal to zero. To establish equilibrium at joint i, therefore, we have to add a moment equal and opposite to –ΣMik0. We call this moment Mi0: Mi0 = ΣMik0

(Equation 1)

where the summation is over all members incident at joint i. We call Mi0 the unbalanced moment at joint i. It is the moment that we have to apply at joint i to maintain equilibrium with the fixed-end moments, and hence to prevent the joint from rotating.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

These concepts are illustrated in the following figure:

The structure shown is braced against sway, i.e., the joints cannot translate. We can sketch the deflected shape of the structure for the given load, showing clearly that we will get a rotation at joint 2. If we assume that joint 2 is restrained from rotating, however, members 1-2 and 2-3 effectively have fixed ends at joint 2, which means that moments M21 and M23 will be fixed-end moments. The sum of the fixed-end moments applied to joint 2 is -112 kN m clockwise. To establish equilibrium, therefore, we must apply moment M20 at joint 2 equal and opposite to this moment, i.e. -112 kN m counterclockwise.

Restoring Equilibrium at Joints Since equilibrium must be maintained, we must somehow provide moment Mi0 at joint i. This moment cannot be applied externally, i.e., the joint cannot remain "artificially" locked. We can, however, generate Mi0 by allowing joint i to rotate.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

In the development of the displacement method, we developed relations between the angle of rotation at the end of a given member and the moments at the ends of the members. We will use these relations to determine a set of member end moments Mik that produce equilibrium at joint i. When joint i is unlocked, it will rotate through angle φi to establish equilibrium. The relations between rotation and member end moment will then be used to develop expressions for member end moments Mik as functions of the unknown joint rotation φi. The condition of equilibrium at joint i is used to eliminate the parameter φi, which allows us to express Mik as a function of the known unbalanced moment Mi0 alone. These expressions will be derived in the remainder of this subsection. Starting from a system in which all joints have been "locked", i.e., in which rotation has been restrained at all joints, and external load has been applied, we proceed to "unlock" joint i. All other joints remain locked. After joint i has been unlocked, it rotates by angle φi to restore equilibrium. Member end moments Mik produced by rotation φi can be expressed as follows: Mik = sik · φi + tik · φk No fixed-end moment has been included since we are considering only the effect of rotation φi. Since all joints other than joint i remain unblocked, φk must be zero. The expression therefore simplifies to: Mik = sik · φi

(Equation 2)

So the total moment applied at joint i due to rotation φi is equal and opposite to the sum of the member end moments incident at the joint, i.e., –ΣMik. Using Equation 2, this sum can be expressed in terms of φi and the member stiffness coefficients: –ΣMik = –Σsik · φi = – φi · Σsik

(Equation 3)

where the summation is over index k, i.e., over all members incident at joint i, and the minus sign is used again because joint moments are equal and opposite to member end moments. Since the joint must be in equilibrium, the total moment at joint i due to rotation and due to fixed-end moments must be zero: –ΣMik – ΣMik0 = 0,

(Equation 4)

where the first term is moment at joint i due to rotation φi and the second term is moment at joint i due to fixed-end moments. Substituting Equations 1 and 3 into Equation 4, we obtain –φi · Σsik – Mi0 = 0 We can isolate φi in the above equation, yielding

φi = –(Mi0) / Σsik

(Equation 5)

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Substituting this expression for φi into Equation 2 yields the following expression for member end moment Mik: Mik = –Mi0 · (sik / Σsik)

(Equation 6)

We define the quantity sik / Σsik to be the distribution factor, and give it the symbol κik. Equation 6 gives member end moment due to rotation φi. To obtain the total member end moment, we must include the fixed-end moment at the end of the member. Total member end moment is thus given by the following expression: Mik = Mik0 – Mi0 · (sik / Σsik) These concepts are illustrated in the following figure:

Carry-Over In the preceding subsection, we dealt with member end moments at end "i" of member ik, i.e., the end where rotation occurs. From our previous study of stiffness coefficients, we know that, in member i-k, rotation at joint i can produce moments at the opposite end, i.e., at joint k. In the previous subsection, we developed expressions for member moments Mik that did not require knowledge of φi. We will now do the same for member end moments Mki. Assuming joint i has been unlocked, and all other joints remain locked, we are interested in calculating moment Mki, i.e., the moment in member i-k at the end opposite to joint i. We begin with the general expression for this member end moment: NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Mki = Mki0 + tki · φi + ski · φk – (ski + tki) · ψki Since joint translation has been excluded, we can eliminate the terms involving member rotation ψ from the above equation. In addition, since we are dealing only with the effects of rotation φi, we can set the fixed-end moment to zero. Finally, since φk is zero (we have assumed that joint k is still locked), the above equation can be simplified to the following equation: Mki = tki · φi We know, however, that Mik = sik · φi (Equation 2). Substituting for φi, therefore, we obtain: Mki = Mik · (tki / sik) Quantity tki / sik is referred to as the carry-over factor, and is called µik. We thus have an expression for member end moments Mki as a function of known moments only. In the example given above, the carry-over factors would be as follows:

µ21 = t12 / s21 = 0, since t12 = 0 (there is a hinge at joint 1) µ23 = t32 / s23 = 0, since t32 = 0 (there is a hinge at joint 3) µ24 = t42 / s24 = 0.5, since t42 = 2EI/L and s24 = 4EI/L These observations can be generalized to the following statements for members with constant EI: The carry-over factor to a fixed joint or a locked joint is 0.5. The carry-over factor to a joint where moment is zero (pin or roller at the end of a member) is 0.

Outline of a Method The preceding expressions have been developed for one simple situation, namely, a loaded structure in which all joints have been artificially locked and a given joint (say joint i) is then unlocked. The method allows moments Mik and Mki to be calculated for this situation. This is far from a complete solution. Once this procedure has been carried out, however, we can lock joint i up again and repeat the procedure at another joint, say joint j. The unbalanced moment at joint j is defined more generally as the total moment causing disequilibrium at the joint. This may include fixed-end moments ΣMjk0, but can also include any additional moments that have been carried over from joint i in the previous step. Because all joints are locked except joint j, the expressions developed above for moments in members incident at joint i can be used for members incident at joint j.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

We therefore have the elements of an iterative method, in which we unlock a given joint, distribute a moment that equilibrates the unbalanced moment, carry over to the opposite ends of the members framing in that the joint, lock the joint up again, and then repeat these steps at a different joint.

2. Procedure (No Joint Translation) Based on the concepts developed above, we can now define the procedure for calculating the response of structures that are braced against sway: 1. Number the joints 2. Establish that the structure is indeed braced against sway using the considerations presented for the displacement method. 3. Calculate all distribution factors κik = sik / Σsik 4. Calculate all carry-over factors µik = tki / sik 5. Calculate fixed-end moments Mik0. 6. At a given joint i, proceed as follows: (a) Calculate the unbalanced moment Mi0 = ΣMik0 (b) Distribute this moment to establish equilibrium at the joint: Mik = –Mi0 · κik (c) Carry Mik over to the opposite ends of the members framing into joint i: Mki = Mik ·

µik

7. At another joint j, proceed as follows: (a) Calculate the unbalanced moment Mj0. If this is the first time joint j has been unlocked, the unbalanced moment will be the sum of fixed-end moments ΣMjk0. If this is the second or greater time that the joint has been unlocked, the unbalanced moment will be the sum of moments Mjk carried over from adjacent joints since the previous iteration. (b) Distribute this moment to establish equilibrium at the joint: Mjk = –Mj0 · κjk (c) Carry Mjk over to the opposite ends of the members framing into joint j: Mkj = Mjk ·

µjk

8. Repeat Step 6 for other joints until the desired degree of accuracy has been achieved. 9. Calculate shear forces, axial forces, and reactions.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

It is generally preferable to begin the process at the joint with the largest unbalanced moment.

3. Example (No Joint Translation) The following example illustrates the procedure derived above. Distribution factors are shown in boxes attached to members at the corresponding joints. One horizontal line denotes that the joint is in equilibrium. For a given iteration, therefore, the unbalanced moment to consider is the sum of the moments written in after the most recent horizontal line was drawn.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

Comments on the iteration are as follows: Iteration 1, Joint 2: Unbalanced moment = -100 + 267 = 167 kN · m Iteration 2, Joint 3: Unbalanced moment = –267 – 23.9 = –290.0 kN · m, where the moment –23.9 was carried over from Joint 2. Iteration 3, Joint 2: Unbalanced moment = 56.0 kN · m. This moment was carried over from Joint 3. Iteration 4, Joint 3: Unbalanced moment = –8.0 kN · m. This moment was carried over from Joint 2. Carry-over from Joint 3 to Joint 2 after this iteration is 1.5 kN · m. Say stop here. Bending moment, shear, and axial force diagrams have not been drawn in this example. A procedure similar to that used for the displacement method should be followed to draw these diagrams.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

4. Simplified Procedure (With Joint Translation) As with the displacement method, analysis of systems that are not braced against sway using moment distribution requires special consideration. We will develop the major aspects of this method using the simple example shown below.

We first identify the independent member rotations, following a procedure similar to that developed for the displacement method. In this case, there is one independent member rotation, corresponding to horizontal displacement of Joints 2 and 3. So sway must be considered. We will solve the problem in two parts. The given structure to analyze, (a), is expressed as the sum of systems (b) and (c). System (b) is the structure with the given loads and an additional load, F10, which prevents horizontal displacement of joints 2 and 3. System (c) is the structure loaded with a force of magnitude F11, opposite in direction to the corresponding force in System (b). It is clear that (b) + (c) will be equal to (a) provided F10 = F11. We can therefore solve System (b), solve System (c), and combine the results of both analyses to give the correct results for the original system (a).

System (b) Because force F10 effectively braces the structure against sway, we can calculate moments in the structure using the method of moment distribution for structures with no joint translation. The member end moments calculated in this way will be called Mik,0. We then calculate unknown force F10 using the equations of equilibrium or the principle of virtual work (the latter method is usually more convenient).

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

System (c) We wish to use the method of moment distribution to calculate moments due to the load F11 applied in the direction opposite to the force required to restrain sway in System (b). Provided F11 = F10, the combination of Systems (b) and (c) will give the solution to the original structure and loading. There is no direct way, however, to calculate the response of a given structure to this type of load using moment distribution. We can, however, calculate fixed-end moments due to an imposed displacement in the direction of this force. The calculation of these fixed-end moments is described in Section 5 of this article. We therefore impose a displacement δ1 in the direction of F11 and calculate the moments of the structure thus produced. The method of moment distribution for structures with no joint translation can be used for this calculation. The method is valid in this case since the joint displacements are known and fixed. Thus, even though the system undergoes member rotations ψ, all these rotations are known. The member rotations thus do not increase the number of unknowns in the system. Once the moments due to δ1 have been calculated, we can calculate force F*11 required for equilibrium using equations of equilibrium or the principle of virtual work. The magnitude of force F*11 corresponds to the magnitude of the displacement δ1, which was arbitrarily chosen. The required force F11 can therefore be expressed as a constant c1 times F*11: F11 = c1 · F*11 But we know that F10 must equal F11. We can therefore solve for c1 as follows: c1 = F10/F*11 Moments in the original system can therefore be calculated using the following equation: Mtot = MSystem (b) + c1 · MSystem (c) The method described here is valid for systems that have one independent member rotation. For two or more independent joint rotations, it is necessary to calculate moments for each independent displacement of the system and to solve a matrix equation for c1, c2, etc. The required procedure is described in standard texts on structural analysis such as the text by Gere (1963).

5. Fixed-End Moments Due to Imposed Translation of Joints Fixed-end moments due to imposed translation of joints can be obtained directly from the stiffness coefficients derived previously for member rotation. Instead of a unit angle of member rotation ψ = 1, we now have a member rotation equal to imposed translation divided by span, ∆/L.

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

6. Example (With Joint Translation) We will now solve the example presented previously. It is repeated in the figure below for convenience:

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB

ENGINEERS CADD CENTRE (P) LTD

STAAD STUDENTS CAN STUDY THIS ON SUNDAYS OR ANYDAY FROM 7-10AM

References Cross, H. 1936. Analysis of Continuous Frames by Distributing Fixed-End Moments. Transactions of the American Society of Civil Engineers 96 (Paper no. 1793). Gere, J. M. 1963. Moment Distribution. New York: Van Nostrand.

END OF THEORY OF STRUCTURES NOTES

EXCLUSIELY COMPILED FOR OUR BELOVED STUDENTS

NOT FOR DISTRIBUTION AS HARD COPY. ONLY FOR OUR STUDENTS STUDY IN LAB