For some types of structures, the equilibrium equations are not sufficient to determine
the reactions or the internal stresses. These structures are called statically indeterminate.
For the analysis of such structures, additional equations must be written on the basis of a knowledge of the elastic deformations. Hence methods of analysis that enable deformations to be evaluated in terms of unknown forces or stresses are important for the solution of problems involving statically indeterminate structures.
Some of these methods, like the method of virtual work, are also useful in solving complicated problems involving statically determinate systems.
A virtual displacement is an imaginary small displacement of a particle consistent with the constraints upon it. Thus, at one support of a simply supported beam, the virtual displacement could be an infinitesimal rotation d of that end but not a vertical movement. However, if the support is replaced by a force, then a vertical virtual displacement may be applied to the beam at that end.
Virtual work is the product of the distance a particle moves during a virtual displacement by the component in the direction of the displacement of a force acting on the particle. If the displacement and the force are in opposite directions, the virtual work is negative. When the displacement is normal to the force, no work is done.
Suppose a rigid body is acted upon by a system of forces with a resultant R.
Given a virtual displacement ds at an angle with R, the body will have virtual work done on it equal to R cos ds. (No work is done by internal forces. They act in pairs of equal magnitude but opposite direction, and the virtual work done by one force of a pair is equal but opposite in sign to the work done by the other force.) If the body is in equilibrium under the action of the forces, then R 0 and the virtual work also is zero.
Thus, the principle of virtual work may be stated: If a rigid body in equilibrium is given a virtual displacement, the sum of the virtual work of the forces acting on it must be zero.
When an elastic body is deformed, the virtual work done by the internal forces is equal to the corresponding increment of the strain energy dU, in accordance with the principle of virtual work.
which states that the partial derivative of the strain energy with respect to any specific deformation gives the corresponding force.
Suppose, for example, the stress in the vertical bar in Fig. 5.50 is to be determined.
All bars are made of the same material and have the same cross section.
If the vertical bar stretches an amount e under the load P, the inclined bars will each stretch an amount e cos . The strain energy in the system is [from Eq. (5.30)]
This is known as Castiglianos first theorem. (His second theorem is the principle of least work.)
Method of Least Work
If displacement of a structure is prevented, as at a support, the partial derivative of the strain energy with respect to that supporting force must be zero, according to Castiglianos first theorem. This establishes his second theorem:
The strain energy in a statically indeterminate structure is the minimum consistent with equilibrium.
As an example of the use of the method of least work, we shall solve again for the stress in the vertical bar in Fig. 5.50. Calling this stress X, we note that the stress in each of the inclined bars must be (P - X)/2 cos . With the aid of Eq. (5.30), we can express the strain energy in the system in terms of X as
Dummy Unit-Load Method
In Art. 5.2.7, the strain energy for pure bending was given as U = M^2 L/2EI in Eq. (5.33). To find the strain energy due to bending stress in a beam, we can apply this equation to a differential length dx of the beam and integrate over the entire span.
To find the vertical deflection of a beam, we apply a vertical dummy unit load at the point where the deflection is to be measured and substitute the bending moments due to this load and the actual loading in Eq. (5.96). Similarly, to compute a rotation, we apply a dummy unit moment.
Beam Deflections. As a simple example, let us apply the dummy unit-load method to the determination of the deflection at the center of a simply supported, uniformly loaded beam of constant moment of inertia (Fig. 5.51a). As indicated in Fig. 5.51b, the bending moment at a distance x from one end is (wL/2)x – (w/ 2)x2. If we apply a dummy unit load vertically at the center of the beam (Fig.
Beam End Rotations. As another example, let us apply the method to finding the end rotation at one end of a simply supported, prismatic beam produced by a moment applied at the other end. In other words, the problem is to find the end rotation at B, B, in Fig. 5.52a, due to MA. As indicated in Fig. 5.52b, the bending moment at a distance x from B caused by MA is MAx/L. If we applied a dummy unit moment at B (Fig. 5.52c), it would produce a moment at x of (L - x) /L (Fig. 5.52d). Substituting in Eq. (5.96) gives
The partial derivative in this equation is the rate of change of axial stress with the load P. It is equal to the axial stress u in each bar of the truss produced by a unit load applied at the point where the deformation is to be measured and in the direction of the deformation. Consequently, Eq. (5.98) can also be written
To find the deflection of a truss, apply a vertical dummy unit load at the panel point where the deflection is to be measured and substitute in Eq. (5.99) the stresses in each member of the truss due to this load and the actual loading. Similarly, to find the rotation of any joint, apply a dummy unit moment at the joint, compute the stresses in each member of the truss, and substitute in Eq. (5.99). When it is necessary to determine the relative movement of two panel points, apply dummy unit loads in opposite directions at those points.
It is worth noting that members that are not stressed by the actual loads or the dummy loads do not enter into the calculation of a deformation.
As an example of the application of Eq. (5.99), let us compute the deflection of the truss in Fig. 5.53. The stresses due to the 20-kip load at each panel point are shown in Fig. 5.53a, and the ratio of length of members in inches to their crosssectional area in square inches is given in Table 5.5. We apply a vertical dummy unit load at L2, where the deflection is required. Stresses u due to this load are shown in Fig. 5.53b and Table 5.5.
The computations for the deflection are given in Table 5.5. Members not stressed by the 20-kip loads or the dummy unit load are not included. Taking advantage of the symmetry of the truss, we tabulate the values for only half the truss and double the sum.
Also, to reduce the amount of calculation, we do not include the modulus of elasticity E, which is equal to 30,000,000, until the very last step, since it is the same for all members.
Reciprocal Theorem and Influence Lines
Consider a structure loaded by a group of independent forces A, and suppose that a second group of forces B are added. The work done by the forces A acting over the displacements due to B will be WAB.
Now, suppose the forces B had been on the structure first, and then load A had been applied. The work done by the forces B acting over the displacements due to A will be WBA.
The reciprocal theorem states that WAB WBA.
Some very useful conclusions can be drawn from this equation. For example, there is the reciprocal deflection relationship: The deflection at a point A due to a load at B is equal to the deflection at B due to the same load applied at A.
Also, the rotation at A due to a load (or moment) at B is equal to the rotation at B due to the same load (or moment) applied at A.
Another consequence is that deflection curves may also be influence lines to some scale for reactions, shears, moments, or deflections (Muller-Breslau principles).
(Influence lines are defined in Art. 5.5.8.) For example, suppose the influence line for a reaction is to be found; that is, we wish to plot the reaction R as a unit load moves over the structure, which may be statically indeterminate. For the loading condition A, we analyze the structure with a unit load on it at a distance x from some reference point. For loading condition B, we apply a dummy unit vertical load upward at the place where the reaction is to be determined, deflecting the structure off the support. At a distance x from the reference point, the displacement in dxR and over the support the displacement is dRR. Hence WAB = 1 (DxR) RdRR. On the other hand, WBA is zero, since loading condition A provides no displacement for the dummy unit load at the support in condition B. Consequently, from the reciprocal theorem,
Since dRR is a constant, R is proportional to dxR. Hence the influence line for a reaction can be obtained from the deflection curve resulting from a displacement of the support (Fig. 5.54). The magnitude of the reaction is obtained by dividing each ordinate of the deflection curve by the displacement of the support.
Similarly, the influence line for shear can be obtained from the deflection curve produced by cutting the structure and shifting the cut ends vertically at the point for which the influence line is desired (Fig. 5.55).
The influence line for bending moment can be obtained from the deflection curve produced by cutting the structure and rotating the cut ends at the point for which the influence line is desired (Fig. 5.56).
And finally, it may be noted that the deflection curve for a load of unity at some point of a structure is also the influence line for deflection at that point (Fig. 5.57).
The principle of superposition applies when the displacement (deflection or rotation) of every point of a structure is directly proportional to the applied loads. The principle states that the displacement at each point caused by several loads equals the sum of the displacements at the point when the loads are applied to the structure individually in any sequence. Also, the bending moment (or shear) at every point induced by applied loads equals the sum of the bending moments (or shears) induced at the point by the loads applied individually in any sequence.
induced at the point by the loads applied individually in any sequence.
The principle holds for linearly elastic structures, for which unit stresses are proportional to unit strains, when displacements are very small and calculations can be based on the underformed configuration of the structure without significant error.
As a simple example, consider a bar with length L and cross-sectional area A loaded with n axial loads P1, P2 . . . Pn. Let F equal the sum of the loads. From Eq. (5.23), F causes an elongation S = FL/AE, where E is the modulus of elasticity of the bar. According to the principle of superposition, if e1 is the elongation caused by P1 alone, e2 by P2 alone, . . and en by Pn alone, then regardless of the sequence in which the loads are applied, when all the loads are on the bar,
In the preceding equations, L/AE represents the elongation induced by a unit load and is called the flexibility of the bar.
The reciprocal, AE/L, represents the force that causes a unit elongation and is called the stiffness of the bar.
Analogous properties of beams, columns, and other structural members and the principle of superposition are useful in analysis of many types of structures. Calculation of stresses and displacements of statically indeterminate structures, for example, often can be simplified by resolution of bending moments, shears, and displacements into components chosen to supply sufficient equations for the solution from requirements for equilibrium of forces and compatibility of displacements.
Consider the continuous beam ALRBC shown in Fig. 5.58a. Under the loads shown, member LR is subjected to end moments ML and MR (Fig. 5.58b) that are initially unknown. The bending-moment diagram for LR for these end moments is shown at the left in Fig. 5.58c. If these end moments were known, LR would be statically determinate; that is LR could be treated as a simply supported beam subjected to known end moments ML and MR. The analysis can be further simplified by resolution of the bending-moment diagram into the three components shown to the right of the equal sign in Fig. 5.58c. This example leads to the following conclusion:
The bending moment at any section of a span LR of a continuous beam or frame equals the simple-beam moment due to the applied loads, plus the simple- beam moment due to the end moment at L, plus the simple-beam moment due to the end moment at R.
When the moment diagrams for all the spans of ALRBC in Fig. 5.58 have been resolved into components so that the spans may be treated as simple beams, all the end moments (moments at supports) can be determined from two basic requirements:
1. The sum of the moments at every support equals zero.
2. The end rotation (angular change at the support) of each member rigidly connected at the support is the same.
A matrix is a rectangular array of numbers in rows and columns that obeys certain mathematical rules known generally as matrix algebra and matrix calculus. A matrix consisting of only a single column is called a vector. In this book, matrices and vectors are represented by boldfaced letters and their elements by lightface symbols, with appropriate subscripts. It often is convenient to use numbers for the subscripts to indicate the position of an element in the matrix. Generally, the first digit indicates the row and the second digit the column. Thus, in matrix A, A23 represents the element in the second row and third column:
Methods based on matrix representations often are advantageous for structural analysis and design of complex structures. One reason is that matrices provide a compact means of representing and manipulating large quantities of numbers. Another reason is that computers can perform matrix operations automatically and speedily. Computer programs are widely available for this purpose.
Matrix Equations. Matrix notation is especially convenient in representing the solution of simultaneous liner equations, which arise frequently in structural analysis.
For example, suppose a set of equations is represented in matrix notation by
Application to a Beam. A general method for determining the forces and moments in a continuous beam is as follows: Remove as many redundant supports or members as necessary to make the structure statically determinant. Compute for the actual loads the deflections or rotations of the statically determinate structure in the direction of the unknown forces and couples exerted by the removed supports and members. Then, in terms of these forces and couples, treated as variables, compute the corresponding deflections or rotations the forces and couples produce in the statically determinate structure (see Arts. 5.5.16 and 5.10.4). Finally, for each redundant support or member write equations that give the known rotations or deflections of the original structure in terms of the deformations of the statically determinate structure.
For example, one method of finding the reactions of the continuous beam AC in Fig. 5.59a is to remove supports 1, 2, and 3 temporarily. The beam is now simply supported between A and C, and the reactions and moments can be computed from the laws of equilibrium. Beam AC deflects at points 1, 2, and 3, whereas we know that the continuous beam is prevented from deflecting at these points by the supports there. This information enables us to write three equations in terms of the three unknown reactions that were eliminated to make the beam statically determinate.
To determine the equations, assume that nodes exist at the location of the supports 1, 2, and 3. Then, for the actual loads, compute the vertical deflections d1, d2, and d3 of simple beam AC at nodes 1, 2, and 3, respectively (Fig. 5.59b). Next, form two vectors, d with element d1, d2 and R with the unknown reactions R1 at node 1, R2 at node 2, and R3 at node 3 as elements. Since the beam may be assumed to be linearly elastic, set d = FR, where F is the flexibility matrix for simple beam AC. The elements yij of F are influence coefficients. To determine them, calculate column 1 of F as the deflections y11, y21, and y31 at nodes 1, 2, and 3, respectively, when a unit force is applied at node 1 (Fig. 5.59c). Similarly, compute column 2 of F for a unit force at node 2 (Fig. 5.59d) and column 3 for a unit force at node 3 (Fig. 5.59e). The three equations then are given by