Page 1 Instructional Objectives After reading this chapter the student will be able to 1. Transform member stiffness matrix from local to global co-ordinate system. 2. Assemble member stiffness matrices to obtain the global stiffness matrix. 3. Analyse plane truss by the direct stiffness matrix. 4. Analyse plane truss supported on inclined roller supports. 25.1 Introduction In the previous lesson, the direct stiffness method as applied to trusses was discussed. The transformation of force and displacement from local co-ordinate system to global co-ordinate system were accomplished by single transformation matrix. Also assembly of the member stiffness matrices was discussed. In this lesson few plane trusses are analysed using the direct stiffness method. Also the problem of inclined support will be discussed. Example 25.1 Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be constant for all the members. Page 2 Instructional Objectives After reading this chapter the student will be able to 1. Transform member stiffness matrix from local to global co-ordinate system. 2. Assemble member stiffness matrices to obtain the global stiffness matrix. 3. Analyse plane truss by the direct stiffness matrix. 4. Analyse plane truss supported on inclined roller supports. 25.1 Introduction In the previous lesson, the direct stiffness method as applied to trusses was discussed. The transformation of force and displacement from local co-ordinate system to global co-ordinate system were accomplished by single transformation matrix. Also assembly of the member stiffness matrices was discussed. In this lesson few plane trusses are analysed using the direct stiffness method. Also the problem of inclined support will be discussed. Example 25.1 Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be constant for all the members. The numbering of joints and members are shown in Fig. 25.1b. Also, the possible displacements (degrees of freedom) at each node are indicated. Here lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedom. Thus displacements 6,7 and 8 are zero due to boundary conditions. First write down stiffness matrix of each member in global co-ordinate system and assemble them to obtain global stiffness matrix. Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 1 EA k (1) Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 Page 3 Instructional Objectives After reading this chapter the student will be able to 1. Transform member stiffness matrix from local to global co-ordinate system. 2. Assemble member stiffness matrices to obtain the global stiffness matrix. 3. Analyse plane truss by the direct stiffness matrix. 4. Analyse plane truss supported on inclined roller supports. 25.1 Introduction In the previous lesson, the direct stiffness method as applied to trusses was discussed. The transformation of force and displacement from local co-ordinate system to global co-ordinate system were accomplished by single transformation matrix. Also assembly of the member stiffness matrices was discussed. In this lesson few plane trusses are analysed using the direct stiffness method. Also the problem of inclined support will be discussed. Example 25.1 Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be constant for all the members. The numbering of joints and members are shown in Fig. 25.1b. Also, the possible displacements (degrees of freedom) at each node are indicated. Here lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedom. Thus displacements 6,7 and 8 are zero due to boundary conditions. First write down stiffness matrix of each member in global co-ordinate system and assemble them to obtain global stiffness matrix. Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 1 EA k (1) Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 . 4 2 EA k (2) Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 3 EA k (3) Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 4 EA k (4) Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 5 EA k (5) The assembled global stiffness matrix of the truss is of the order 8 8 × . Now assemble the global stiffness matrix. Note that the element 1 11 k of the member stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. On the member stiffness matrix the corresponding global degrees of freedom are indicated to facilitate assembling. Thus, Page 4 Instructional Objectives After reading this chapter the student will be able to 1. Transform member stiffness matrix from local to global co-ordinate system. 2. Assemble member stiffness matrices to obtain the global stiffness matrix. 3. Analyse plane truss by the direct stiffness matrix. 4. Analyse plane truss supported on inclined roller supports. 25.1 Introduction In the previous lesson, the direct stiffness method as applied to trusses was discussed. The transformation of force and displacement from local co-ordinate system to global co-ordinate system were accomplished by single transformation matrix. Also assembly of the member stiffness matrices was discussed. In this lesson few plane trusses are analysed using the direct stiffness method. Also the problem of inclined support will be discussed. Example 25.1 Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be constant for all the members. The numbering of joints and members are shown in Fig. 25.1b. Also, the possible displacements (degrees of freedom) at each node are indicated. Here lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedom. Thus displacements 6,7 and 8 are zero due to boundary conditions. First write down stiffness matrix of each member in global co-ordinate system and assemble them to obtain global stiffness matrix. Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 1 EA k (1) Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 . 4 2 EA k (2) Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 3 EA k (3) Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 4 EA k (4) Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 5 EA k (5) The assembled global stiffness matrix of the truss is of the order 8 8 × . Now assemble the global stiffness matrix. Note that the element 1 11 k of the member stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. On the member stiffness matrix the corresponding global degrees of freedom are indicated to facilitate assembling. Thus, [] 8 7 6 5 4 3 2 1 162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0 0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0 0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0 0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0 0 0 0 0 25 . 0 0 25 . 0 0 0 433 . 0 0 433 . 0 0 866 . 0 0 0 162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0 094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0 8 7 6 5 4 3 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - - - - - - - - - - - - - = EA K (6) Writing the load-displacement relation for the truss, yields ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - - - - - - - - - - - - - = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0 0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0 0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0 0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0 0 0 0 0 25 . 0 0 25 . 0 0 0 433 . 0 0 433 . 0 0 866 . 0 0 0 162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0 094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0 u u u u u u u u EA p p p p p p p p (7) The displacements 1 u to 5 u are unknown. The displacements 0 8 7 6 = = = u u u . Also 0 5 3 2 1 = = = = p p p p . But 4 10 kN p = - . ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - 5 4 3 2 1 487 . 0 0 433 . 0 094 . 0 054 . 0 0 25 . 0 0 25 . 0 0 433 . 0 0 866 . 0 0 0 094 . 0 25 . 0 0 575 . 0 0 054 . 0 0 0 0 108 . 0 0 10 0 0 0 u u u u u EA (8) Solving which, the unknown displacements are evaluated. Thus, Page 5 Instructional Objectives After reading this chapter the student will be able to 1. Transform member stiffness matrix from local to global co-ordinate system. 2. Assemble member stiffness matrices to obtain the global stiffness matrix. 3. Analyse plane truss by the direct stiffness matrix. 4. Analyse plane truss supported on inclined roller supports. 25.1 Introduction In the previous lesson, the direct stiffness method as applied to trusses was discussed. The transformation of force and displacement from local co-ordinate system to global co-ordinate system were accomplished by single transformation matrix. Also assembly of the member stiffness matrices was discussed. In this lesson few plane trusses are analysed using the direct stiffness method. Also the problem of inclined support will be discussed. Example 25.1 Analyse the truss shown in Fig. 25.1a and evaluate reactions. Assume EA to be constant for all the members. The numbering of joints and members are shown in Fig. 25.1b. Also, the possible displacements (degrees of freedom) at each node are indicated. Here lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedom. Thus displacements 6,7 and 8 are zero due to boundary conditions. First write down stiffness matrix of each member in global co-ordinate system and assemble them to obtain global stiffness matrix. Element 1: . 619 . 4 , 60 m L = ° = ? Nodal points 4-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 1 EA k (1) Element 2: . 00 . 4 , 90 m L = ° = ? Nodal points 2-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 . 4 2 EA k (2) Element 3: . 619 . 4 , 120 m L = ° = ? Nodal points 3-1 [] ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 75 . 0 433 . 0 75 . 0 433 . 0 433 . 0 25 . 0 433 . 0 25 . 0 619 . 4 3 EA k (3) Element 4: . 31 . 2 , 0 m L = ° = ? Nodal points 4-2 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 4 EA k (4) Element 5: . 31 . 2 , 0 m L = ° = ? Nodal points 2-3 [] ? ? ? ? ? ? ? ? ? ? ? ? - - = 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 1 31 . 2 5 EA k (5) The assembled global stiffness matrix of the truss is of the order 8 8 × . Now assemble the global stiffness matrix. Note that the element 1 11 k of the member stiffness matrix of truss member 1 goes to location ( ) 7 , 7 of global stiffness matrix. On the member stiffness matrix the corresponding global degrees of freedom are indicated to facilitate assembling. Thus, [] 8 7 6 5 4 3 2 1 162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0 0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0 0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0 0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0 0 0 0 0 25 . 0 0 25 . 0 0 0 433 . 0 0 433 . 0 0 866 . 0 0 0 162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0 094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0 8 7 6 5 4 3 2 1 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - - - - - - - - - - - - - = EA K (6) Writing the load-displacement relation for the truss, yields ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - - - - - - - - - - - - - = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 162 . 0 0934 . 0 0 0 0 0 162 . 0 094 . 0 0934 . 0 487 . 0 0 0 0 433 . 0 094 . 0 054 . 0 0 0 162 . 0 094 . 0 0 0 162 . 0 094 . 0 0 0 094 . 0 487 . 0 0 433 . 0 094 . 0 054 . 0 0 0 0 0 25 . 0 0 25 . 0 0 0 433 . 0 0 433 . 0 0 866 . 0 0 0 162 . 0 094 . 0 162 . 0 094 . 0 25 . 0 0 575 . 0 0 094 . 0 054 . 0 094 . 0 054 . 0 0 0 0 108 . 0 u u u u u u u u EA p p p p p p p p (7) The displacements 1 u to 5 u are unknown. The displacements 0 8 7 6 = = = u u u . Also 0 5 3 2 1 = = = = p p p p . But 4 10 kN p = - . ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - 5 4 3 2 1 487 . 0 0 433 . 0 094 . 0 054 . 0 0 25 . 0 0 25 . 0 0 433 . 0 0 866 . 0 0 0 094 . 0 25 . 0 0 575 . 0 0 054 . 0 0 0 0 108 . 0 0 10 0 0 0 u u u u u EA (8) Solving which, the unknown displacements are evaluated. Thus, AE u AE u AE u AE u AE u 334 . 13 ; 642 . 74 ; 668 . 6 ; 64 . 34 ; 668 . 6 5 4 3 2 1 = - = = - = = (9) Now reactions are evaluated from equation, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - ? ? ? ? ? ? ? ? ? ? - - - - - - - = ? ? ? ? ? ? ? ? ? ? 334 . 13 642 . 74 668 . 6 64 . 34 668 . 6 1 0 0 0 162 . 0 094 . 0 0 0 433 . 0 094 . 0 054 . 0 094 . 0 0 0 162 . 0 094 . 0 8 7 6 EA EA p p p (10) Thus, 67 8 5.00 kN ; 0 ; 5.00 kN pp p == = . (11) Now calculate individual member forces. Member 1: m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = = . {} [] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 2 1 8 7 1 619 . 4 ' u u u u m l m l AE p {} [] 1 6.667 1 ' 0.5 0.866 5.77 kN 34.64 4.619 AE p AE ?? =- - = ?? - ?? (12) Member 2: m L m l 0 . 4 ; 0 . 1 ; 0 = = = . {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 2 1 4 3 1 0 . 4 ' u u u u m l m l AE p {} [] 1 74.642 1 ' 1 1 10.0 kN 34.64 4.619 AE p AE -?? =- =- ?? - ?? (13) Member 3: m L m l 619 . 4 ; 866 . 0 ; 50 . 0 = = - = .Read More

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