# Module 1. Energy Methods in Structural Analysis. Version 2 CE IIT, Kharagpur

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1 Module 1 Energy Methods in Structural Analysis

2 Lesson 1 General Introduction

4 1.2 Classification of Structures All structural forms used for load transfer from one point to another are 3- dimensional in nature. In principle one could model them as 3-dimensional elastic structure and obtain solutions (response of structures to loads) by solving the associated partial differential equations. In due course of time, you will appreciate the difficulty associated with the 3-dimensional analysis. Also, in many of the structures, one or two dimensions are smaller than other dimensions. This geometrical feature can be exploited from the analysis point of view. The dimensional reduction will greatly reduce the complexity of associated governing equations from 3 to 2 or even to one dimension. This is indeed at a cost. This reduction is achieved by making certain assumptions (like Bernoulli-Euler kinematic assumption in the case of beam theory) based on its observed behaviour under loads. Structures may be classified as 3-, 2- and 1-dimensional (see Fig. 1.1(a) and (b)). This simplification will yield results of reasonable and acceptable accuracy. Most commonly used structural forms for load transfer are: beams, plane truss, space truss, plane frame, space frame, arches, cables, plates and shells. Each one of these structural arrangement supports load in a specific way.

5 Beams are the simplest structural elements that are used extensively to support loads. They may be straight or curved ones. For example, the one shown in Fig. 1.2 (a) is hinged at the left support and is supported on roller at the right end. Usually, the loads are assumed to act on the beam in a plane containing the axis of symmetry of the cross section and the beam axis. The beams may be supported on two or more supports as shown in Fig. 1.2(b). The beams may be curved in plan as shown in Fig. 1.2(c). Beams carry loads by deflecting in the

6 same plane and it does not twist. It is possible for the beam to have no axis of symmetry. In such cases, one needs to consider unsymmetrical bending of beams. In general, the internal stresses at any cross section of the beam are: bending moment, shear force and axial force. In India, one could see plane trusses (vide Fig. 1.3 (a),(b),(c)) commonly in Railway bridges, at railway stations, and factories. Plane trusses are made of short thin members interconnected at hinges into triangulated patterns. For the purpose of analysis statically equivalent loads are applied at joints. From the above definition of truss, it is clear that the members are subjected to only axial forces and they are constant along their length. Also, the truss can have only hinged and roller supports. In field, usually joints are constructed as rigid by

7 welding. However, analyses were carried out as though they were pinned. This is justified as the bending moments introduced due to joint rigidity in trusses are negligible. Truss joint could move either horizontally or vertically or combination of them. In space truss (Fig. 1.3 (d)), members may be oriented in any direction. However, members are subjected to only tensile or compressive stresses. Crane is an example of space truss.

8 Plane frames are also made up of beams and columns, the only difference being they are rigidly connected at the joints as shown in the Fig. 1.4 (a). Major portion of this course is devoted to evaluation of forces in frames for variety of loading conditions. Internal forces at any cross section of the plane frame member are: bending moment, shear force and axial force. As against plane frame, space frames (vide Fig. 1.4 (b)) members may be oriented in any direction. In this case, there is no restriction of how loads are applied on the space frame.

9 1.3 Equations of Static Equilibrium Consider a case where a book is lying on a frictionless table surface. Now, if we apply a force F 1 horizontally as shown in the Fig.1.5 (a), then it starts moving in the direction of the force. However, if we apply the force perpendicular to the book as in Fig. 1.5 (b), then book stays in the same position, as in this case the vector sum of all the forces acting on the book is zero. When does an object

10 move and when does it not? This question was answered by Newton when he formulated his famous second law of motion. In a simple vector equation it may be stated as follows: n i=1 F i = ma (1.1)

11 n F i i= 1 where is the vector sum of all the external forces acting on the body, m is the total mass of the body and a is the acceleration vector. However, if the body is in the state of static equilibrium then the right hand of equation (1.1) must be zero. Also for a body to be in equilibrium, the vector sum of all external moments ( M = 0) about an axis through any point within the body must also vanish. Hence, the book lying on the table subjected to external force as shown in Fig. 1.5 (b) is in static equilibrium. The equations of equilibrium are the direct consequences of Newton s second law of motion. A vector in 3-dimensions can be resolved into three orthogonal directions viz., x, y and z (Cartesian) coordinate axes. Also, if the resultant force vector is zero then its components in three mutually perpendicular directions also vanish. Hence, the above two equations may also be written in three co-ordinate axes directions as follows: x F = 0 ; F = 0 ; F = 0 (1.2a) y z M = 0 ; M = 0; M = 0 (1.2b) x y z Now, consider planar structures lying in xy plane. For such structures we could have forces acting only in x and y directions. Also the only external moment that could act on the structure would be the one about the z -axis. For planar structures, the resultant of all forces may be a force, a couple or both. The static equilibrium condition along x -direction requires that there is no net unbalanced force acting along that direction. For such structures we could express equilibrium equations as follows: x F = 0 ; F = 0 ; M = 0 (1.3) y Using the above three equations we could find out the reactions at the supports in the beam shown in Fig After evaluating reactions, one could evaluate internal stress resultants in the beam. Admissible or correct solution for reaction and internal stresses must satisfy the equations of static equilibrium for the entire structure. They must also satisfy equilibrium equations for any part of the structure taken as a free body. If the number of unknown reactions is more than the number of equilibrium equations (as in the case of the beam shown in Fig. 1.7), then we can not evaluate reactions with only equilibrium equations. Such structures are known as the statically indeterminate structures. In such cases we need to obtain extra equations (compatibility equations) in addition to equilibrium equations. z

12

13 1.4 Static Indeterminacy The aim of structural analysis is to evaluate the external reactions, the deformed shape and internal stresses in the structure. If this can be accomplished by equations of equilibrium, then such structures are known as determinate structures. However, in many structures it is not possible to determine either reactions or internal stresses or both using equilibrium equations alone. Such structures are known as the statically indeterminate structures. The indeterminacy in a structure may be external, internal or both. A structure is said to be externally indeterminate if the number of reactions exceeds the number of equilibrium equations. Beams shown in Fig.1.8(a) and (b) have four reaction components, whereas we have only 3 equations of equilibrium. Hence the beams in Figs. 1.8(a) and (b) are externally indeterminate to the first degree. Similarly, the beam and frame shown in Figs. 1.8(c) and (d) are externally indeterminate to the 3 rd degree.

14 Now, consider trusses shown in Figs. 1.9(a) and (b). In these structures, reactions could be evaluated based on the equations of equilibrium. However, member forces can not be determined based on statics alone. In Fig. 1.9(a), if one of the diagonal members is removed (cut) from the structure then the forces in the members can be calculated based on equations of equilibrium. Thus,

15 structures shown in Figs. 1.9(a) and (b) are internally indeterminate to first degree.the truss and frame shown in Fig. 1.10(a) and (b) are both externally and internally indeterminate.

16 So far, we have determined the degree of indeterminacy by inspection. Such an approach runs into difficulty when the number of members in a structure increases. Hence, let us derive an algebraic expression for calculating degree of static indeterminacy. Consider a planar stable truss structure having m members and j joints. Let the number of unknown reaction components in the structure be r. Now, the total number of unknowns in the structure is m + r. At each joint we could write two equilibrium equations for planar truss structure, viz., F x = 0 and F. y = 0 Hence total number of equations that could be written is 2 j. If 2 j = m + r then the structure is statically determinate as the number of unknowns are equal to the number of equations available to calculate them. The degree of indeterminacy may be calculated as i = ( m + r) 2 j (1.4)

17 We could write similar expressions for space truss, plane frame, space frame and grillage. For example, the plane frame shown in Fig.1.11 (c) has 15 members, 12 joints and 9 reaction components. Hence, the degree of indeterminacy of the structure is i = ( ) 12 3 = 18 Please note that here, at each joint we could write 3 equations of equilibrium for plane frame.

18 1.5 Kinematic Indeterminacy When the structure is loaded, the joints undergo displacements in the form of translations and rotations. In the displacement based analysis, these joint displacements are treated as unknown quantities. Consider a propped cantilever beam shown in Fig (a). Usually, the axial rigidity of the beam is so high that the change in its length along axial direction may be neglected. The displacements at a fixed support are zero. Hence, for a propped cantilever beam we have to evaluate only rotation at B and this is known as the kinematic indeterminacy of the structure. A fixed fixed beam is kinematically determinate but statically indeterminate to 3 rd degree. A simply supported beam and a cantilever beam are kinematically indeterminate to 2 nd degree.

19 The joint displacements in a structure is treated as independent if each displacement (translation and rotation) can be varied arbitrarily and independently of all other displacements. The number of independent joint displacement in a structure is known as the degree of kinematic indeterminacy or the number of degrees of freedom. In the plane frame shown in Fig. 1.13, the joints B and C have 3 degrees of freedom as shown in the figure. However if axial deformations of the members are neglected then u 1 = u 4 and u2 and u4 can be neglected. Hence, we have 3 independent joint displacement as shown in Fig i.e. rotations at B and C and one translation.

20

21 1.6 Kinematically Unstable Structure A beam which is supported on roller on both ends (vide. Fig. 1.14) on a horizontal surface can be in the state of static equilibrium only if the resultant of the system of applied loads is a vertical force or a couple. Although this beam is stable under special loading conditions, is unstable under a general type of loading conditions. When a system of forces whose resultant has a component in the horizontal direction is applied on this beam, the structure moves as a rigid body. Such structures are known as kinematically unstable structure. One should avoid such support conditions.

22 1.7 Compatibility Equations A structure apart from satisfying equilibrium conditions should also satisfy all the compatibility conditions. These conditions require that the displacements and rotations be continuous throughout the structure and compatible with the nature supports conditions. For example, at a fixed support this requires that displacement and slope should be zero. 1.8 Force-Displacement Relationship

23 Consider linear elastic spring as shown in Fig Let us do a simple experiment. Apply a force P 1 at the end of spring and measure the deformation u1. Now increase the load to P2 and measure the deformation u2. Likewise repeat the experiment for different values of load P, 1 P2,..., P n. Result may be represented in the form of a graph as shown in the above figure where load is shown on y -axis and deformation on abscissa. The slope of this graph is known as the stiffness of the spring and is represented by k and is given by P2 P1 k = = u u 2 1 P u (1.5) P = ku (1.6) The spring stiffness may be defined as the force required for the unit deformation of the spring. The stiffness has a unit of force per unit elongation. The inverse of the stiffness is known as flexibility. It is usually denoted by a and it has a unit of displacement per unit force. the equation (1.6) may be written as 1 a = (1.7) k P = ku u = 1 P = ap (1.8) k The above relations discussed for linearly elastic spring will hold good for linearly elastic structures. As an example consider a simply supported beam subjected to a unit concentrated load at the centre. Now the deflection at the centre is given by 3 PL 48EI u = or P = u 3 (1.9) 48EI L The stiffness of a structure is defined as the force required for the unit deformation of the structure. Hence, the value of stiffness for the beam is equal to 48EI k = 3 L

24 As a second example, consider a cantilever beam subjected to a concentrated load ( P ) at its tip. Under the action of load, the beam deflects and from first principles the deflection below the load (u ) may be calculated as, 3 PL u = (1.10) 3 EI zz For a given beam of constant cross section, length L, Young s modulus E, and moment of inertia I ZZ the deflection is directly proportional to the applied load. The equation (1.10) may be written as u = a P (1.11) 3 L Where a is the flexibility coefficient and is a =. Usually it is denoted by a ij 3EI zz the flexibility coefficient at i due to unit force applied at j. Hence, the stiffness of the beam is 1 3EI k 11 = = (1.12) 3 a L 11 Summary In this lesson the structures are classified as: beams, plane truss, space truss, plane frame, space frame, arches, cables, plates and shell depending on how they support external load. The way in which the load is supported by each of these structural systems are discussed. Equations of static equilibrium have been stated with respect to planar and space and structures. A brief description of static indeterminacy and kinematic indeterminacy is explained with the help simple structural forms. The kinematically unstable structures are discussed in section 1.6. Compatibility equations and force-displacement relationships are discussed. The term stiffness and flexibility coefficients are defined. In section 1.8, the procedure to calculate stiffness of simple structure is discussed. Suggested Text Books for Further Reading Armenakas, A. E. (1988). Classical Structural Analysis A Modern Approach, McGraw-Hill Book Company, NY, ISBN Hibbeler, R. C. (2002). Structural Analysis, Pearson Education (Singapore) Pte. Ltd., Delhi, ISBN

25 Junarkar, S. B. and Shah, H. J. (1999). Mechanics of Structures Vol. II, Charotar Publishing House, Anand. Leet, K. M. and Uang, C-M. (2003). Fundamentals of Structural Analysis, Tata McGraw-Hill Publishing Company Limited, New Delhi, ISBN Negi, L. S. and Jangid, R.S. (2003). Structural Analysis, Tata McGraw- Hill Publishing Company Limited, New Delhi, ISBN Norris, C. H., Wilbur, J. B. and Utku, S. (1991). Elementary Structural Analysis, Tata McGraw-Hill Publishing Company Limited, New Delhi, ISBN MATRIX ANALYSIS of FRAMED STRUCTURES, 3-rd Edition, by Weaver and Gere Publishe, Chapman & Hall, New York, New York, 1990

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