2. Displacement method of analysis (also known as stiffness matrix method).

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FLEXIBILITY MATRIX METHODS

Since twentieth century, indeterminate structures are being widely used for its obvious merits. It may be recalled that, in the case of indeterminate structures either the reactions or the internal forces cannot be determined from equations of statics alone. In such structures, the number of reactions or the number of internal forces exceeds the number of static equilibrium equations. In addition to equilibrium equations, compatibility equations are used to evaluate the unknown reactions and internal forces in statically indeterminate structure. In the analysis of indeterminate structure it is necessary to satisfy the equilibrium equations (implying that the structure is in equilibrium) compatibility equations (requirement if for assuring the continuity of the structure without any breaks) and force displacement equations (the way in which displacement are related to forces). We have two distinct method of analysis for statically indeterminate structure depending upon how the above equations are satisfied:

1. Force method of analysis (also known as flexibility method of analysis, method of consistent deformation, flexibility matrix method)

2. Displacement method of analysis (also known as stiffness matrix method).

In the force method of analysis, primary unknown are forces. In this method compatibility equations are written for displacement and rotations (which are calculated by force displacement equations). Solving these equations, redundant forces are calculated. Once the redundant forces are calculated, the remaining reactions are evaluated by equations of equilibrium.

In the displacement method of analysis, the primary unknowns are the displacements. In this

method, first force -displacement relations are computed and subsequently equations are written

satisfying the equilibrium conditions of the structure. After determining the unknown

displacements, the other forces are calculated satisfying the compatibility conditions and force

displacement relations. The displacement-based method is amenable to computer programming

and hence the method is being widely used in the modern day structural analysis. In general, the

maximum deflection and the maximum stresses are small as compared to statically determinate

structure. For example, consider two beams of identical cross section and span carrying

uniformly distributed load as shown in Fig. 7.1a and Fig. 7.1b.

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The loads are also the same in both cases. In the first case, the beam is fixed at both ends and thus is statically indeterminate. The simply supported beam in Fig. 7.1b is a statically

determinate structure. The maximum bending moment in case of fixed- fixed beam is wL

²

/12 (which occurs at the supports) as compared to wl

²

/8 (at the centre) in case of simply supported

beam. Also in the present case, the deflection in the case of fixed- fixed beam wl

⁴

/384EI is five times smaller than that of simply supported beam 5wl

⁴

/384EI . Also, there is

redistribution of stresses in the case of redundant structure. Hence if one member fails, structure does not collapse suddenly. The remaining members carry the load. The determinate structural system collapses if one member fails. However, there are disadvantages in using indeterminate structures. Due to support settlement, there will be additional stresses in the case of redundant structures where as determinate structures are not affected by support settlement.

The analysis of indeterminate structure differs mainly in two aspects as compared to determinate structure.

a) To evaluate stresses in indeterminate structures, apart from sectional properties (area of cross section and moment of inertia), elastic properties are also required.

b) Stresses are developed in indeterminate structure due to support settlements, temperature change

and fabrication errors etc.

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After reading this chapter the student will be

1. Able to analyse statically indeterminate structure of degree one.

2. Able to solve the problem by either treating reaction or moment as redundant.

3. Able to draw shear force and bending moment diagram for statically indeterminate beams.

4. Able to state advantages and limitations of force method of analysis.

7.1 Introduction.

In this lesson, a general introduction is given to the force method of analysis of indeterminate structure is given. In the next lesson, this method would be applied to statically indeterminate beams.

Initially the method is introduced with the help of a simple problem and subsequently it is discussed in detail. The flexibility method of analysis or force method of analysis (or method of consistent deformation) was originally developed by J. Maxwell in 1864 and O. C. Mohr in 1874. Since flexibility method requires deflection of statically determinate structure, a table of formulas for deflections for various load cases and boundary conditions is also given in this lesson for ready use.

The force method of analysis is not convenient for computer programming as the choice of redundant is not unique. Further, the bandwidth of the flexibility matrix in the force method is much larger than the stiffness method. However it is very useful for hand computation.

7.2 Simple Example

Consider a propped cantilever beam (of constant flexural rigidity EI, and span L), which is carrying uniformly distributed load of as shown in Fig. 7.2a. The beam is statically indeterminate i.e. its reaction cannot be evaluated from equations of statics alone. To solve the above problem by force method proceeds as follows.

1) Determine the degree of statically indeterminacy. In the present case it is one. Identify the reaction, which can be treated as redundant in the analysis. In the present case or can be treated as redundant. Selecting R

B

as the redundant, the procedure is illustrated.

Subsequently, it will be shown how to attack the problem by treating M

A

as redundant.

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1. Introduction 1.1 Background

Up to now we have concentrated on the elastic analysis of structures. In these analyses we used superposition often, knowing that for a linearly elastic structure it was valid. However, an elastic analysis does not give information about the loads that will actually collapse a structure. An indeterminate structure may sustain loads greater than the load that first causes a yield to occur at any point in the structure. In fact, a structure will stand as long as it is able to find redundancies to yield. It is only when a structure has exhausted all of its redundancies will extra load causes it to fail. Plastic analysis is the method through which the actual failure load of a structure is calculated, and as will be seen, this failure load can be significantly greater than the elastic load capacity.

To summarize this, Prof. Sean de Courcy (UCD) used to say:

“a structure only collapses when it has exhausted all means of standing”.

Before analyzing complete structures, we review material and cross section behaviour beyond the elastic limit.

Basis of Plastic Design 2.1 Material Behaviour

A uniaxial tensile stress on a ductile material such as mild steel typically provides the

following graph of stress versus strain:

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As can be seen, the material can sustain strains far in excess of the strain at which yield occurs before failure. This property of the material is called its ductility.

Though complex models do exist to accurately reflect the above real behaviour of the material, the most common, and simplest, model is the idealized stress-strain curve. This is the curve for an ideal elastic-plastic material (which doesn’t exist), and the graph is:

As can be seen, once the yield has been reached it is taken that an indefinite amount of strain can occur. Since so much post-yield strain is modeled, the actual material (or cross section) must also be capable of allowing such strains. That is, it must be sufficiently ductile for the idealized stress-strain curve to be valid. Next we consider the behaviour of a cross section of an ideal elastic-plastic material subject to bending. In doing so, we seek the relationship between applied moment and the rotation (or more accurately, the curvature) of a cross section.

Cross Section Behaviour

Moment-Rotation Characteristics of General Cross Section

We consider an arbitrary cross-section with a vertical plane of symmetry, which is also the plane

of loading. We consider the cross section subject to an increasing bending moment, and assess

the stresses at each stage.

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Moment-Rotation Curve Stage 1 – Elastic Behaviour

The applied moment causes stresses over the cross-section that are all less than the yield stress of the material.

Stage 2 – Yield Moment

The applied moment is just sufficient that the yield stress of the material is reached at the outermost fibre(s) of the cross-section. All other stresses in the cross section are less than the yield stress. This is limit of applicability of an elastic analysis and of elastic design. Since all fibres are elastic, the ratio of the depth of the elastic to plastic regions,

Stage 3 – Elasto-Plastic Bending

The moment applied to the cross section has been increased beyond the yield moment. Since by the idealized stress-strain curve the material cannot sustain a stress greater than yield stress, the fibres at the yield stress have progressed inwards towards the centre of the beam. Thus over the cross section there is an elastic core and a plastic region. The ratio of the depth of the elastic core to the plastic region is . Since extra moment is being applied and no stress is bigger than the yield stress, extra rotation of the section occurs: the moment-rotation curve losses its linearity and curves, giving more rotation per unit moment (i.e. looses stiffness).

Stage 4 – Plastic Bending

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The applied moment to the cross section is such that all fibres in the cross section are at yield stress. This is termed the Plastic Moment Capacity of the section since there are no fibres at an elastic stress, Also note that the full plastic moment requires an infinite strain at the neutral axis and so is physically impossible to achieve. However, it is closely approximated in practice. Any attempt at increasing the moment at this point simply results in more rotation, once the cross- section has sufficient ductility. Therefore in steel members the cross section classification must be plastic and in concrete members the section must be under-reinforced.

Stage 5 – Strain Hardening

Due to strain hardening of the material, a small amount of extra moment can be sustained.

The above moment-rotation curve represents the behaviour of a cross section of a regular elastic- plastic material. However, it is usually further simplified as follows:

With this idealized moment-rotation curve, the cross section linearly sustains moment up to the plastic moment capacity of the section and then yields in rotation an indeterminate amount.

Again, to use this idealization, the actual section must be capable of sustaining large rotations – that is it must be ductile.

Plastic Hinge

Note that once the plastic moment capacity is reached, the section can rotate freely – that is, it

behaves like a hinge, except with moment of Mp at the hinge. This is termed a plastic hinge, and

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hence rotations can increase.

Analysis of Rectangular Cross Section

Since we now know that a cross section can sustain more load than just the yield moment, we are

interested in how much more. In other words we want to find the yield moment and plastic

moment, and we do so for a rectangular section. Taking the stress diagrams from those of the

moment-rotation curve examined previously, we have:

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Shape Factor

Thus the ratio of elastic to plastic moment capacity is:

This ratio is termed the shape factor, f, and is a property of a cross section alone. For a

rectangular cross-section, we have:

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And so a rectangular section can sustain 50% more moment than the yield moment, before a plastic hinge is formed. Therefore the shape factor is a good measure of the efficiency of a cross section in bending. Shape factors for some other cross sections are

Methods of Plastic Analysis 3.1 Introduction

There are three main approaches for performing a plastic analysis:

The Incremental Method

This is probably the most obvious approach: the loads on the structure are incremented until the first plastic hinge forms. This continues until sufficient hinges have formed to collapse the structure. This is a labour-intensive, ‘brute-force’, approach, but one that is most readily suited for computer implementation.

The Equilibrium (or Statical) Method

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overlaid to identify the likely locations of plastic hinges. This method therefore satisfies the equilibrium criterion first leaving the two remaining criterion to derived there from.

The Kinematic (or Mechanism) Method

In this method, a collapse mechanism is first postulated. Virtual work equations are then written for this collapse state, allowing the calculations of the collapse bending moment diagram. This method satisfies the mechanism condition first, leaving the remaining two criteria to be derived there from.

We will concentrate mainly on the Kinematic Method, but introduce now the Incremental Method to illustrate the main concepts.

Incremental Analysis

Illustrative Example – Propped Cantilever

We now assess the behaviour of a simple statically indeterminate structure under

increasing load. We consider a propped cantilever with mid-span point load:

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Since the peak moments are less than the yield moments, we know that yield stress has not been

reached at any point in the beam. Also, the maximum moment occurs at A and so this point will

first reach the yield moment.

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Equilibrium Method

Introduction

To perform this analysis we generally follow the following steps:

1. Find a primary structure by removing redundant until the structure is statically determinate;

2. Draw the primary (or free) bending moment diagram;

3. Draw the reactant BMD for each redundant, as applied to the primary structure;

4. Construct a composite BMD by combing the primary and reactant BMDs;

5. Determine the equilibrium equations from the composite BMD;

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6. Choose the points where plastic hinges are likely to form and introduce into the equilibrium equations;

7. Calculate the collapse load factor, or plastic moment capacity as required.

For different possible collapse mechanisms, repeat steps 6 and 7, varying the hinge locations.

We now apply this method to the Illustrative Example previously analyzed.

Steps 1 to 3 of the Equilibrium Method are illustrated in the following diagram:

For Step 4, in constructing the Composite BMD, we arbitrarily choose tension on the underside

of the beam as positive. By convention in the Equilibrium Method, instead of drawing the two

BMDs on opposite sides (as is actually the case), the reactant BMD is drawn ‘flipped’ over the

line and subtracted from the primary BMD: the net remaining area is the final BMD. This is best

explained by illustration below:

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For Step 7, we solve this equation for the collapse load:

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Kinematic Method Using Virtual Work

Introduction

Probably the easiest way to carry out a plastic analysis is through the Kinematic Method using virtual work. To do this we allow the presumed shape at collapse to be the compatible displacement set, and the external loading and internal bending moments to be the equilibrium set. We can then equate external and internal virtual work, and solve for the collapse load factor for that supposed mechanism.

Remember:

Equilibrium set: the internal bending moments at collapse;

Compatible set: the virtual collapsed configuration (see below).

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Note that in the actual collapse configuration the members will have elastic deformation in between the plastic hinges. However, since a virtual displacement does not have to be real, only compatible, we will choose to ignore the elastic deformations between plastic hinges, and take the members to be straight between them.

Actual Collapse Mechanism

So for our previous beam, we know that we require two hinges for collapse (one more than its

degree of redundancy), and we think that the hinges will occur under the points of peak moment,

A and C. Therefore impose a unit virtual displacement at C and relate the corresponding virtual

rotations of the hinges using ,

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Other Collapse Mechanisms

For the collapse mechanism looked at previously, it seemed obvious that the plastic hinge in the span should be beneath the load. But why? Using virtual work we can examine any possible collapse mechanism. So let’s consider the following collapse mechanism and see why the plastic hinge has to be located beneath the load.

Plastic Hinge between A and C:

Imposing a unit virtual deflection at B, we get the following collapse mechanism:

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And so we see that the collapse load factor for this mechanism depends on the position of the plastic hinge in the span.

Theorems of Plastic Analysis 4.1 Criteria

In Plastic Analysis to identify the correct load factor, there are three criteria of importance:

1. Equilibrium: the internal bending moments must be in equilibrium with the external loading.

2. Mechanism: at collapse the structure, or a part of, can deform as a mechanism.

3. Yield: no point in the structure can have a moment greater than the plastic moment capacity of the section it is applied to.

Based on these criteria, we have the following theorems.

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The Upper bound (Unsafe) Theorem This can be stated as:

If a bending moment diagram is found which satisfies the conditions of equilibrium and mechanism (but not necessarily yield), then the corresponding load factor is either greater than or equal to the true load factor at collapse.

This is called the unsafe theorem because for an arbitrarily assumed mechanism the load factor is either exactly right (when the yield criterion is met) or is wrong and is too large, leading a designer to think that the frame can carry more load than is actually possible.

The Lower bound (Safe) Theorem

If a bending moment diagram is found which satisfies the conditions of equilibrium and yield (but not necessarily that of mechanism), then the corresponding load factor is either less than or equal to the true load factor at collapse.

Plastic Analysis of Beams

Example 1 – Fixed-Fixed Beam with Point Load

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To start the problem, we examine the usual elastic BMD to see where the plastic hinges are likely to form:

We also need to know how many hinges are required. This structure is 3˚ statically indeterminate and so we might expect the number of plastic hinges required to be 4. However, since one of the indeterminacies is horizontal restraint, removing it would not change the bending behaviour of the beam. Thus for a bending collapse only 2 indeterminacies apply and so it will only take 3 plastic hinges to cause collapse.

So looking at the elastic BMD, we’ll assume a collapse mechanism with the 3 plastic hinges at

the peak moment locations: A, B, and C.

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And so the applied load is in equilibrium with the free BMD of the collapse BMD.

2. Mechanism:

From the proposed collapse mechanism it is apparent that the beam is a mechanism.

3. Yield:

From the collapse BMD it can be seen that nowhere is exceeded. P M

Thus the solution meets the three conditions and so, by the Uniqueness Theorem, is the correct solution.

Example 2 – Propped Cantilever with Two Point Loads

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For the following beam, for a load factor of 2.0, find the required plastic moment capacity:

Allowing for the load factor, we need to design the beam for the following loads:

Once again we try to picture possible failure mechanisms. Since maximum

moments occur underneath point loads, there are two real possibilities:

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Mechanism-1

Mechanism-2

Therefore, we analyse both and apply the Upperbound Theorem to find the design plastic moment capacity.

Mechanism 1: Plastic Hinge at C:

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Mechanism 2: Plastic Hinge at D:

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1. Equilibrium:

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Using the BMD at collapse, we’ll check that the height of the free BMD is that of the equivalent simply-supported beam. Firstly the collapse BMD from Mechanism 1 is:

Hence, the total heights of the free BMD are:

Checking these using a simply-supported beam analysis:

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Thus, using appropriate free body diagrams of AC and DB:

And so the applied load is in equilibrium with the free BMD of the collapse BMD.

2. Mechanism:

From the proposed collapse mechanism it is apparent that the beam is a mechanism. Also, since it is a propped cantilever and thus one degree indeterminate, we require two plastic hinges for collapse, and these we have.

3. Yield:

From the collapse BMD it can be seen that nowhere is the design exceeded. 144 kNm Thus by the Uniqueness Theorem we have the correct solution.

Lastly, we’ll examine why the Mechanism 2 collapse is not the correct solution. Since the virtual work method provides an upperbound, then, by the Uniqueness Theorem, it must not be the correct solution because it must violate the yield condition.

Using the collapse Mechanism 2 to determine reactions, we can draw the following BMD for

collapse Mechanism 2:

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From this it is apparent that Mechanism 2 is not the unique solution, and so the design plastic moment capacity must be 144 kNm as implied previously from the Upperbound Theorem.

Basic Collapse Mechanisms

In frames, the basic mechanisms of collapse are:

Beam-type collapse

Sway Collapse

Combination Collapse

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One of the most powerful tools in plastic analysis is Combination of Mechanisms. This allows us to work out the virtual work equations for the beam and sway collapses separately and then combine them to find the collapse load factor for a combination collapse mechanism.

Combination of mechanisms is based on the idea that there are only a certain number of independent equilibrium equations for a structure. Any further equations are obtained from a combination of these independent equations. Since equilibrium equations can be obtained using virtual work applied to a possible collapse mechanism, it follows that there are independent collapse mechanisms, and other collapse mechanisms that may be obtained form a combination of the independent collapse mechanisms.

Simple Portal Frame

In this example we will consider a basic prismatic (so all members have the same plastic moment capacity) rectangular portal frame with pinned feet:

We will consider this general case so that we can infer the properties and behaviour of all such frames. We will consider each of the possible mechanisms outlined above.

Beam collapse:

The possible beam collapse looks as follows:

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Sway Collapse

The virtual deflection for the sway collapse is:

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Combined Collapse

The virtual deflection for this form of collapse is:

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Collapse Mode

Since we don’t know the relative values of H and V, we cannot determine the

correct collapse mode. However, we can identify these collapse modes if we plot

the three load factor equations derived above on the following interaction chart:

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Notice that each mechanism defines a boundary and that it is only the region inside all of these boundaries that is safe. Now, for a given ration of V to H, we will be able to determine the critical collapse mechanism. Note also that the beam collapse mechanism is only critical for this frame at point P on the chart – this point is also included in the Combined mechanism.

The bending moment diagrams corresponding to each of the mechanisms are approximately:

Beam Sway Combined

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An interesting phenomenon is observed at point Q on the chart, where the Sway and Combined mechanisms give the same result. Looking at the bending moment diagrams, we can see that this occurs as the moment at the top of the left column becomes equal to the mid-span moment of the beam:

Important Problems

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UNIT-3 STIFFNESS MATRIX METHODS

The given indeterminate structure is first made kinematically determinate by introducing constraints at the nodes. The required number of constraints is equal to degrees of freedom at the nodes that is kinematic indeterminacy ∝ k. The kinematically determinate structure comprises of fixed ended members, hence, all nodal displacements are zero. These results in stress resultant discontinuities at these nodes under the action of applied loads or in other words the clamped joints are not in equilibrium. In order to restore the equilibrium of stress resultants at the nodes the nodes are imparted suitable unknown displacements. The number of simultaneous equations representing joint equilibrium of forces is equal to kinematic indeterminacy ∝ k. Solution of these equations gives unknown nodal displacements. Using stiffness properties of members the member end forces are computed and hence the internal forces throughout the structure. Since nodal displacements are unknowns, the method is also called displacement method. Since equilibrium conditions are applied at the joints the method is also called equilibrium method.

Since stiffness properties of members are used the method is also called stiffness method.

Introduction

All known methods of structural analysis are classified into two distinct groups:- (i) force method of analysis and

(ii) displacement method of analysis.

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In module 3, the force method of analysis or the method of consistent deformation is discussed. An introduction to the displacement method of analysis is given in module 3, where in slope-deflection method and moment- distribution method are discussed. In this module the direct stiffness method is discussed. In the displacement method of analysis the equilibrium equations are written by expressing the unknown joint displacements in terms of loads by using load-displacement relations.

The unknown joint displacements (the degrees of freedom of the structure) are calculated by solving equilibrium equations. The slope-deflection and moment-distribution methods were extensively used before the high speed computing era. After the revolution in computer industry, only direct stiffness method is used.

The displacement method follows essentially the same steps for both statically determinate and indeterminate structures. In displacement /stiffness method of analysis, once the structural model is defined, the unknowns (joint rotations and translations) are automatically chosen unlike the force method of analysis. Hence, displacement method of analysis is preferred to computer implementation. The method follows a rather a set procedure. The direct stiffness method is closely related to slope-deflection equations.

The general method of analyzing indeterminate structures by displacement method may be traced to Navier (1785-1836). For example consider a four member truss as shown in Fig.23.1.The given truss is statically indeterminate to second degree as there are four bar forces but we have only two equations of equilibrium. Denote each member by a number, for example (1), (2), (3) and (4). Let

i

α be the angle, the i-th member makes with the horizontal. Under the action of external loads and, the joint E displaces to E’. Let u and v be its vertical and horizontal displacements. Navier solved this problem as follows.

In the displacement method of analysis u and v are the only two unknowns for this structure. The elongation of individual truss members can be expressed in terms of these two unknown joint displacements. Next, calculate bar forces in the members by using force–displacement relation.

Now at E, two equilibrium equations can be written viz., 0=ΣxFand 0=ΣyF by summing all

forces in x and y directions. The unknown displacements may be calculated by solving the

equilibrium equations. In displacement method of analysis, there will be exactly as many

equilibrium equations as there are unknowns.

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of force. In module 1, we have discussed force- displacement relationship. The force (F) is related to the displacement (u) for the linear elastic material by the relation

where the constant of proportionality k is defined as the stiffness of the structure and it has units

of force per unit elongation. The above equation may also be written as

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Chapter-2 Objectives

After reading this chapter the student will be able to 1. Derive member stiffness matrix of a truss member.

2. Define local and global co-ordinate system.

3. Transform displacements from local co-ordinate system to global co-ordinate system.

4. Transform forces from local to global co-ordinate system.

5. Transform member stiffness matrix from local to global co-ordinate system.

6. Assemble member stiffness matrices to obtain the global stiffness matrix.

7. Analyse plane truss by the direct stiffness matrix.

An introduction to the stiffness method was given in the previous chapter. The basic principles

involved in the analysis of beams, trusses were discussed. The problems were solved with hand

computation by the direct application of the basic principles. The procedure discussed in the previous

chapter though enlightening are not suitable for computer programming. It is necessary to keep hand

computation to a minimum while implementing this procedure on the computer. In this chapter a

formal approach has been discussed which may be readily programmed on a computer. In this lesson

the direct stiffness method as applied to planar truss structure is discussed.

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Plane trusses are made up of short thin members interconnected at hinges to form triangulated

patterns. A hinge connection can only transmit forces from one member to another member but not

the moment. For analysis purpose, the truss is loaded at the joints. Hence, a truss member is

subjected to only axial forces and the forces remain constant along the length of the member. The

forces in the member at its two ends must be of the same magnitude but act in the opposite directions

for equilibrium as shown in Fig. 24.1.

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Problem-1

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UNIT-4- CABLES AND SUSPENSION BRIDGES

Instructional Objectives:

After reading this chapter the student will be able to 1. Differentiate between rigid and deformable structures.

2. Define funicular structure.

3. State the type stress in a cable.

4. Analyse cables subjected to uniformly distributed load.

5. Analyse cables subjected to concentrated loads.

31.1 Introduction

Cables and arches are closely related to each other and hence they are grouped in this course in the same module. For long span structures (for e.g. in case bridges) engineers commonly use cable or arch construction due to their efficiency. In the first lesson of this module, cables subjected to uniform and concentrated loads are discussed. In the second lesson, arches in general and three hinged arches in particular along with illustrative examples are explained. In the last two lessons of this module, two hinged arch and hingeless arches are considered.

Structure may be classified into rigid and deformable structures depending on change in

geometry of the structure while supporting the load. Rigid structures support externally applied

loads without appreciable change in their shape (geometry). Beams trusses and frames are

examples of rigid structures. Unlike rigid structures, deformable structures undergo changes in

their shape according to externally applied loads. However, it should be noted that deformations

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support suspension roofs, bridges and cable car system. They are also used in electrical transmission lines and for structures supporting radio antennas. In the following sections, cables subjected to concentrated load and cables subjected to uniform loads are considered.

The shape assumed by a rope or a chain (with no stiffness) under the action of external

loads when hung from two supports is known as a funicular shape. Cable is a funicular

structure. It is easy to visualize that a cable hung from two supports subjected to external

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load must be in tension (vide Fig. 31.2a and 31.2b). Now let us modify our definition of cable. A cable may be defined as the structure in pure tension having the funicular shape of the load.

As stated earlier, the cables are considered to be perfectly flexible (no flexural stiffness) and inextensible. As they are flexible they do not resist shear force and bending moment. It is subjected to axial tension only and it is always acting tangential to the cable at any point along the length. If the weight of the cable is negligible as compared with the externally applied loads then its self weight is neglected in the analysis. In the present analysis self weight is not considered.

Consider a cable as loaded in Fig. 31.2. Let us assume that the cable lengths and sag at () are known.

The four reaction components at ACDEB andB, cable tensions in each of the four segments and three sag values: a total of eleven unknown quantities are to be determined. From the geometry, one could write two force equilibrium equations (0,0== ΣΣ

^yx

FF) at each of the point and DCBA,,,E i.e. a total of ten equations and the required one more equation may be written from the geometry of the cable.

For example, if one of the sag is given then the problem can be solved easily. Otherwise if the total length of the cable is given then the required equation may be written as

Cable subjected to uniform load.

Cables are used to support the dead weight and live loads of the bridge decks having long spans.

The bridge decks are suspended from the cable using the hangers. The stiffened deck prevents

the supporting cable from changing its shape by distributing the live load moving over it, for a

longer length of cable. In such cases cable is assumed to be uniformly loaded.

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Consider a cable which is uniformly loaded as shown in Fig 31.3a

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Due to uniformly distributed load, the cable takes a parabolic shape. However due to its own dead weight it takes a shape of a catenary. However dead weight of the cable is neglected in the present analysis.

Example 31.1

Determine reaction components at A and B, tension in the cable and the sag of the cable

shown in Fig. 31.4a. Neglect the self weight of the cable in the analysis

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Taking moment about E, yields

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UNIT-5 SPACE TRUSSES

Tetrahedron: simplest element of stable space truss (six members, four joints) expand by adding 3 members and 1 joint each time

Determinacy and Stability b + r < 3j unstable

b + r = 3j statically determinate (check stability)

b + r > 3j statically indeterminate (check stability)

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Method of Sections Method of Joints

Numerical example

In the following example we shall construct the internal forces diagrams for the given in Fig. 8 space frame structure. The introduced global coordinate system is shown in the same figure.

The introduced local coordinate systems of the different elements of the space frame are

presented in Fig. 9. The typical sections where the internal forces must be calculated, in order to

construct the relevant diagrams, are numbered from 1 to 8 in the same figure. The typical

sections are placed at least at the beginning and at the end of each element (segment) of the

frame. The internal forces diagrams, in the limits of each element, could be derived by using the

corresponding reference and base diagrams.

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Internal Forces

In order to obtain the internal forces at a specified point, we should make section cut perpendicular to the axis of the member at this point. This section cut divides the structure in two parts. The portion of the structure removed from the part into consideration should be replaced by the internal forces. The internal forces ensure the equilibrium of the isolated part subjected to the action of external loads and support reactions. A free body diagram of either segment of the cut member is isolated and the internal loads could be derived by the six equations of equilibrium applied to the segment into consideration.

We shall skip the derivation of internal forces in section 1 from Fig. 8, because they can be derived

without any troubles. Let us go direct to the internal forces in section 2. If we pass a section cut at

point 2 the space frame will be separated as shown in Fig. 10. The positive directions of internal

forces, in accordance with the introduced local coordinate system for the members

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Equilibrium of the joints

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Numerical Example-2

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