m4l24

m4l24 - Module 4 Analysis of Statically Indeterminate...

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Module 4 Analysis of Statically Indeterminate Structures by the Direct Stiffness Method Version 2 CE IIT, Kharagpur
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Lesson 24 The Direct Stiffness Method: Truss Analysis Version 2 CE IIT, Kharagpur
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Instructional 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. 24.1 Introduction 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. 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. Version 2 CE IIT, Kharagpur
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Now consider a truss member having cross sectional area A , Young’s modulus of material E , and length of the member L . Let the member be subjected to axial tensile force as shown in Fig. 24.2. Under the action of constant axial force , applied at each end, the member gets elongated by as shown in Fig. 24.2. F F u The elongation may be calculated by (vide lesson 2, module 1). u AE FL u = (24.1) Now the force-displacement relation for the truss member may be written as, u L AE F = (24.2) Version 2 CE IIT, Kharagpur
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F ku = (24.3) where L AE k = is the stiffness of the truss member and is defined as the force required for unit deformation of the structure. The above relation (24.3) is true along the centroidal axis of the truss member. But in reality there are many members in a truss. For example consider a planer truss shown in Fig. 24.3. For each member of the truss we could write one equation of the type along its axial direction (which is called as local co-ordinate system). Each member has different local co ordinate system. To analyse the planer truss shown in Fig. 24.3,
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m4l24 - Module 4 Analysis of Statically Indeterminate...

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