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Matrix Analysis of Structures,. Second Edition. Aslam Kassimali. Publisher, Global Engineering: Christopher M. Shortt. Acquisitions Editor: Randall Adams. PDF | On Jan 26, , Muhammed A. Husain and others published Matrix Structural Analysis: Lecture Notes (Handwritten). Structural analysis (Engineering) 2. Matrices. I. Gere, James M. II. Title. T A W36 'dc20 C1P Preface Matrix analysis of structures.
Divided into 12 chapters, Matrix Methods for Advanced Structural Analysis begins with an introduction to the analysis of structures fundamental concepts and basic steps of structural analysis, primary structural members and their modeling, brief historical overview of methods of static analysis, programming principles, and suggestions for the rational use of computer programs. This is followed by the principal steps of the Direct Stiffness Method including plane trusses, plane framed structures, space trusses, and space framed structures. The case of plane or space framed structure, including possible rigid elements at their beam ends rigid joints is discussed in detail. Other topics discussed in this reference include the procedure for analyzing beams with internal releases partial connection of beam elements and elastic hinges, as well as the alternative handling of internal releases by modifying the element stiffness matrix. Furthermore, the Method of Substructures is demonstrated for the solution of large-scale models in terms of the associated number of degrees of freedom.
The Matrix Analysis of Vibration. Structural analysis of complex networks. Global Structural Analysis of Buildings. Introduction to matrix analysis. Matrix analysis for statistics. Topics in matrix analysis. Recommend Documents. Theory of Matrix Structural Analysis Matrix Structural Analysis, Second Edition Dold and B.
Matrix Analysis of Structures Your name. Recommend Documents. Bank R. Graham J. Stoer R. Varga H. Yserentant Theory of Matrix Structural Analysis Introduction to Matrix Analysis, Second Edition! P Introduction to '1! Due to electronic rights restrictions, some third party content ma Matrix Analysis 2nd edition Matrix Analysis Second Edition Linear algebra and matrix theory are fundamental tools in mathematical and physical scie Your name.
In dynamic analysis it is normally acceptable to avoid this complexity by lumping masses at selected points and considering only a finite number of degrees of freedom. By an obvious extension. Figure 2.
In principle. In static analysis. In a frame loaded normal to its plane. As shown. In rigid frames. Each displacement component illustrated in Figure 2. But as long as we restrict our consideration to linear elastic action. Except where otherwise noted. Under this principle. Positive values of the force and displacement components correspond to the positive sense of the coordinate axes.
That is. Under the action of this force the particle displaces to the point h. A force vector with components Fxg. These axes remain fixed throughout the deformation of the structure.
Consider a particle located at point g when the structure is in the unloaded. When it is necessary to consider moments and rotational displacements. We'll use it. As thus described in the coordinate system of the figure. Direct forces and moments may of course be present at both ends.
There are cases. A measure of displacement at the joints of such structures is the rotation. In frame analysis the simplifying assumption is generally made that a line drawn normal to the elastic line of the beam or frame element remains normal to that line as the structure deforms under load. They are generally the physical joints of the structure. We shall use it consistently. The distinction should always be clear from the context. The term release is often used in framework computer analysis programs.
This convention is carried over into space structures. Given the elastic properties of an element or structure in terms of local coordinates. The first step in the formation of the force and displacement vectors is the definition of the nodal points and their location with respect to a coordinate system. The joint at or closest to the origin of local coordinates is designated as joint 1 and the other node as joint 2. This is illustrated in Figure 2. We use local axes in almost all of our formulations of element equations.
The local x' axis is directed along the axis of the member. Global axes figure prominently in the development of the global equations the equations of the complete structure in Sections 3.
For the case of direct forces and moments at both ends. An individual entry. When different types of axes are being compared or when they appear in the same portion of the text. For the element in Figure 2. They will be discussed further and more precise definitions will be given in Section The transformation procedures could be introduced at this point. The individual truss elements are isolated. For the purpose of forming a mathematical model the state of stress in the members is represented by forces at the element ends.
How these idealizing decisions are made. The corresponding displacements of these nodes-the degrees of freedom-are employed in the characterization of the displaced state of the element. The concepts of idealization and analysis can be described with respect to the simple ttuss in Figure 2. By summing up the truss element forces in each direction at each joint in each coordinate direction and equating the result to the corresponding applied loads. In the interest of focusing on the techniques of analysis.
The truss is next reconstructed analytically by examining the equilibrium of member forces at each joint. The joining of truss members at the joints also ensures that the truss displaces as a structural entity without any discontinuities in the pattern.
I Framed structures are almost limitless in variety. All can be broken down into line elements. The solution is exact within the confines of linear elastic analysis. They will apply to the most complex framed structure.
In beams. Because of its limited use. Force-displacement equations. An individual term of the [k] matrix. Venant torsion.
In Chapters we treat only cases in which the component elements are prismatic. Nonprismatic elements are covered in Chapter 7. Consider a simple truss member directed along a local axis x antj having a known 5This element is "complete" in the sense that it is capable of accounting for axial force. Other reasons for neglecting certain effects have been cited in Section 2.
The complete framework element has 12 nodal degrees of freedom and 12 nodal force components. If a displacement! See Section 7. To write the stiffness equations for this member. Thus the correct direction of any force can be determined from the sign of the analytical result. In Figure 2. In analysis. Since it is assumed to be pin-ended and therefore capable of resisting only an axial force. Force-Displacement Relationships The same interpretation can be placed on the second column of the element stiffness matrix see Figure 2.
In Figures 2. The column of j oint forces for this case is equal to the column of stiffness coefficients corresponding to di. It is used to emphasize the fact that these vectors exclude components related to the support conditions. An element flexibility coefficient.
This is a consequence of the necessary reciprocity of displacements. The matrix is also symmetrical. The suppressed displacements associated with any such support conditions are sets of displacements associated with rigid body motion. Flexibility relationships can also be derived for elements supported in a statically indeterminate manner but they would not be of general use since.
They can be applied in as many ways as there are stable and statically determinate support conditions. A complete set of stiffness equations for an element.
The reason is that. Such degrees of freedom. The subscript f on the force and displacement vectors refers to the degrees of freedom that are free to displ. In contrast to the stiffness matrix of Equation 2. The difference is that. If it had not been. It has known properties and lies in an x-y plane. Consider the axial force member in Figure 2. Mathematical transformations are not essential. Element stiffness equations in arbitrary coordinates-either local or global-can be developed from basic principles.
Equation 2. It yields the element stiffness equations in arbitrary coordinates directly. The element stiffness matrix must now be of order 4 x 4. Although not in themselves examples of matrix structural analysis. To determine the column of stiffness coefficients relating the force components to u.
Example 2. Following our assumption that the components of the force vector remain unchanged as node 2 moves frcm 2 to 2'. Example What is the displacement of a? In Example 2.
X lQ-. What are the bar forces and the displacement of a? L aq thus.. From the first part of Equation 2. What is the magnitude and direction of the force P at point a required to displace that point vertically downward 5 mm without any horizontal displacement?
E Assume the pin at b is disconnected and calculate the gap due to free thermal contraction of bar be due to a temperature change Determine the displacement of b and the force in the bars. EA c Case I: If the point a is moved move up the slope? First write the element stiffness equations using Equation 2.
Boundary condition: From a. From b. Each supporting link has the same A. Consider the beam itself undeformable. Equilibrium requirement at node b: Using stiffness equations for the support elements. Eis constant. What would be the corresponding va?
If so. L2 2. Is this a function of the magnitude of the applied load? Compare elements of [K with values obtained in Example 2. Calculate its magnitude. Write the stiffness equations for the assembled system: Demonstrate that there is a mathematical analogy between force.
As in Problem 2. Demonstrate that there is a mathematical analogy between force and current. Cross-sectional areas mm2 x are indicated on each bar. Calculate the displacement at b in part b. Assume that the structural arrangement is such that the two parallel bars must elongate equally. Make the same assumption regarding the elongation of the parallel bars as in Problem 2. Acylincter mm.
Es T T Problem 2. Calculate the changes in lengths and the internal forces in bar and cylinder. The bolt is loaded to an initial tension of kN by tightening the nut. Plot a diagram of the force in the bolt versus T. In extending the scope and power of structural analysis. The same skill. And it merely involves the application of the conditions of equilibrium and continuity of displacement at the joints of the analytical model.
This is accomplished either by writing the equations directly in global coordinates as in Figure 2. Chapter 3 Formation of the Global Analysis Equations In this chapter we discuss the fundamentals of the direct stiffness approach to the formation of the equations of analysis. This approach requires only the notion and the algebraic form of the element stiffness matrix.
Only axial force members are considered. The element does not have a defined support condition. There are no inherent restrictions on the ways of designating global joints but.
For the axial fqrce member the transformation matrix is rectangular see Section 5. The truss bars shown all lie in the x-y plane. Even without the benefit of the mathematics that precedes it. When the global analysis equations are formed using the direct stiffness method. The quantities in the x direction at this point are designated by the subscript i. To illustrate the procedure.
The global numbering system is completely independent of the local element numbering system. From the condition of x-direction equilibrium. Appropriate subscripts or superscripts are assigned in each case. The capitalized terms K Ff in terms of the corresponding element degrees of freedom A1. In Equation 3. Note that each of the bars meeting at the indicated junction points possesses stiffness coefficients with common subscripts.
Substitution of such expressions into Equation 3. The force-displacement equations for the elements. Consider now what happens when an additional member.
When the subscripts of the coefficients of two or more different elements are identical. For junction point equilibrium. Thus it is a simple matter to revise the stiffness equations to include another member.!.
These will be taken up in Section But there are simple procedures for obtaining approximate solutions when structural changes are made.
When a new member is added. There are no other contributions of member H to joint equilibrium equations. Figure 3. Write the member force-displacement relationships in global I coordinates In the determination of the displacements. Example for structural changes. Assemble the global stiffness equations. Equation The ways in which various stiffness equations are used to determine displacements.. Member force-displacement relationships see Equation 2. This 3. Adding rows 1 and 3 of the global stiffness 0.
Rigid body motion. Global stiffness equations in matrix form see Equation 3. Use equations of Example 3. Calcuiate the bar forces. This is a signal that. Calculate the reactions. Calculate the displacements at a and b. Pi P2 6. Bar forces. The bar forces may now be obtained from the member stiffness equations: The upper three global stiffness equations can be written as follows: The lower three stiffness equations now yield the reactions: F"" Member force-displacement relationships at node a see Equation 2.
No nonzero degrees of freedom have been added. The result is a complete set of coefficient:. A search is then made through the member stiffness coefficients. Each time a coefficient is placed in a location where a value has already been placed.
When a coefficient is reached whose designating first subscript is 1. At the completion of this step all terms in row 1 have achieved their final value. Parts of its mathematical basis that have not been covered already are best explained after outlining the procedure: The first subscript row pertains to the force equation.
Each element stiffness coefficient. The array is illustrated in Figure 3. The process of steps 2 and 3 is r e peat e d for all other rows in order. Reactions may be obtained by adding an equation for Rxrl. Each term in this array is identified by two subscripts. The procedure of step 2 is continued until all elements have been searched.
The first subscript. Provision is made for a square matrix whose size is equal to the number of degrees of freedom in the complete system. This approach was implicit in Example 3.
After reordering rows and columns to separate quantities pertaining to the s11p- 'The treatment of nonzero supports that follow a prescribed behavioral rule or are otherwise constrained is covered in Section It is assumed that the rationale for steps is clear from the previous discussions of Equations 3. The surplus equations are those that pertain to the external forces at the support points.
Thus it will be found that an equal number of rows and columns have been extracted from the array. The set of equations remaining after step 5 is solved for the remaining degrees of freedom. Steps then lead to the reduced stiffness matrix and step 5 is eliminated. These quantities may require a further transformation from global to local coordinates and finally a transformation into stresses.
The immediate result. This process. The logic of the foregoing process will now be reviewed using a matrix formulation. Expanding Equation 3. These transformations are often made part of the basic analysis. Such a numbering scheme may be advantageous.
Example 3. The structures used are elementary pin-jointed trusses. Note the separation made in Example 3. The details of a basic system tor forming the global stiffness equations should be clear from this example. Separate operations are required to transform them into: The application of the general procedure is illustrated in Examples 3. We have emphasized that the operation of matrix inversion is symbolic. Member ab F f. K L Sym. Questions of indeterminacy are introduced in Section 3.
Show that the stiffness equations contain rigid-body-motion terms. Write the force-displacement relationships in global coordinates. Global stiffness equations in matrix form see Equations P1 3. Calculate the bar forces. The displacements are indefinite-there may be rigid body motion. Calculate the displacements at b. The truss of Example 3.
Remove columns and rows 1. Adding rows 1. Fae From equilibrium at the force F12 is 2 P2 The remaining stiffness equations rows and are used as follows: See Reference 3. Remove columns and rows 1 to leaving 5 from the stiffness equations. Calculate the displacements at c and d. Whatever the approach employed. In other words. The members shown in Figure 3. All other terms in the row are zero. The structures m Examples 3. Equat10n-solvmg algorithms will be discussed in Chapter When there are many degrees of freedom in the complete structure.
Bars such as E. Even the very elementary problem in Example. This can be done by numbering the degrees of freedom in such a way that the columnar distance of the term most remote from the main diagonal term in each row is minimized. This view. Kinematic indeterminacy refers to the number of displacements that are required to define the response of the structure. In the figure. A flexibility approach to Example 3.
In eval- in the problem 6Support settlement problems in which estimated displacements of the supports are specified are cases in which displacements have predetermined values. From the preceding. The displacement approach appears more automatic since. The unknowns in the analysis are the degrees of freedom. The unknowns in the analysis are the redundant forces at the cut sections or removed supports.
In the early. Since the labor of solution is largely a function of the number of unknowns. The number of such degrees of freedom is the number of equations that are required for analysis. The degree of kinematic redundancy is equal to the number of degrees of freedom that must be conceptually constrained to reduce the system to one in which all joint displacements are zero or have predetermined values.
Both can be clarified by comparing definitions of the two. As an example of the consequence of this. In all cases. This would add additional degrees of freedom to the analysis. What are the approximate values of A and. Cross-sect10nal areas mm 2 x 10 are shown on each bar. Assume the same material is used for all three bars.
E and A of all bars are the same. Problem 3. In its externally unloaded condition part a of the figure it is pretensioned so that the force in ab is 50 kN. What is the significance of your results? A and E are the same for all bars. Sketch the displaced structure. U Three bars of equal A and E are arranged as shown. Consider vertical displacement only. Prentice Hall. SO 8 7m ' e Problem 3. For convenience the unit thickness is shown to reduced scale.
From As mentioned in Chapter 2. Figure 4. Formation of this matrix requires an understanding of the material stress-strain relationships. G The modulus of rigidity shear modulus.
Here we are using the term frames in a generic sense to include structures such as pin-jointed planar and space trusses. Chapter 4 Stiffness Analysis of Frames-I In this and the following chapter we develop the remaining tools needed for the linear elastic stiffness analysis of complete frames. These are reviewed to the extent useful to the development of the stiffness matrix. By the end of Chapter 5 we can demonstrate the solution of complex.
Our main purpose in the present chapter is the development and application of the stiffness matrix of the prismatic. The three properties are: E Young's modulus.
This will be done in local coordinates. By equilibrium. The ultimate compressive strength. At stresses below the yield. The Young's modulus and Poisson's ratio of steel are always close to The most widely used civil engineering structural materials-. For normal weight concrete. In most analyses these shortcomings and the complications they cause are ignored. The mechanical properties of concrete are less predictable than those of steel.
Strain energy and complementary work complementary strain energy densities d Figure 4. Steel doesn't creep at ordinary temperatures. Concrete creeps-that is.
Over a total displacement t Work and But there are many practical problems in which doing this can compromise the validity of the results.. Both materials fall short of the ideals of homogeneity and isotropy.
In this text we are concerned with static behavior. From the development of the equations it is obvious that. Fxu The above considerations.. In this case.
Both the force and the displacement at a point are representable as vectors F and A and. U describes the strain energy of deformation corresponding to the work W.