Theory of Structures- A Must Read for Be's

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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.

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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:

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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.

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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:

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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.

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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.

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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.

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Examples

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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.

Loads

Loads are defined as forces (or moments) acting on a given structure. Structural analysis must generally consider loads originating from a variety of causes, including:

1. Dead load. This load is caused by the weight of the structure itself as well as all fixed nonstructural objects that the structure carries. Dead load of nonstructural items is sometimes referred to as superimposed dead load. In a bridge, for

example, superimposed dead load would include the weight of railings and asphalt paving. Designers calculate dead load of a given structure based on dimensions of structural members and unit weights of standard building materials, normally specified in design standards. Dead loads are applied at the centroid of structural components.

2. Live load. This load is caused by the use of a given structure. For buildings, live load includes the weight of occupants as well as any movable items of machinery. For bridges, live load consists of the weight of heavy trucks using the bridge. Live load values are generally specified in design standards. Designers must arrange live load on structural models to produce the most severe effect.

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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.

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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.

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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:

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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

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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

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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.

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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.

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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.

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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

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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.

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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.

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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.

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Example

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

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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

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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

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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.

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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.

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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.

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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.

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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. What are the reactions at the supports?

2. What is the shape of the cable? 3. 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.

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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 wL}2_{/(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.

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This calculation yields several important observations: 1. The cable has a parabolic shape.

2. The horizontal component of force in the cable, H(x), is constant over the length of the cable and is equal to wL2_/(8f).

3. 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.

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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.

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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),

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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,

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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.

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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.

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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

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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:

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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).

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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.

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

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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.

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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.

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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

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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 right-hand 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.

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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.

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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 right-hand 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.

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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.

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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:

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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 M in the figure