Page 1 {}[] 1 72.855 1 ' 0.707 0.707 1.688 kN 55.97 7.07 AE p AE ?? =- - =- ?? - ?? (16) Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 4 3 8 7 1 07 . 7 ' u u u u m l m l AE p {} [] {} 1 1 ' 0.707 53.825 5.38 kN 7.07 AE p AE == (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is zero. . .e i displacement along " y is zero in the present case. Page 2 {}[] 1 72.855 1 ' 0.707 0.707 1.688 kN 55.97 7.07 AE p AE ?? =- - =- ?? - ?? (16) Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 4 3 8 7 1 07 . 7 ' u u u u m l m l AE p {} [] {} 1 1 ' 0.707 53.825 5.38 kN 7.07 AE p AE == (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is zero. . .e i displacement along " y is zero in the present case. If the stiffness matrix of the entire truss is formulated in global co-ordinate system then the displacements along y are not zero at the oblique support. So, a special procedure has to be adopted for incorporating the inclined support in the analysis of truss just described. One way to handle inclined support is to replace the inclined support by a member having large cross sectional area as shown in Fig. 25.3b but having the length comparable with other members meeting at that joint. The inclined member is so placed that its centroidal axis is perpendicular to the inclined plane. Since the area of cross section of this new member is very high, it does not allow any displacement along its centroidal axis of the jointA. Another method of incorporating inclined support in the analysis is to suitably modify the member stiffness matrix of all the members meeting at the inclined support. Page 3 {}[] 1 72.855 1 ' 0.707 0.707 1.688 kN 55.97 7.07 AE p AE ?? =- - =- ?? - ?? (16) Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 4 3 8 7 1 07 . 7 ' u u u u m l m l AE p {} [] {} 1 1 ' 0.707 53.825 5.38 kN 7.07 AE p AE == (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is zero. . .e i displacement along " y is zero in the present case. If the stiffness matrix of the entire truss is formulated in global co-ordinate system then the displacements along y are not zero at the oblique support. So, a special procedure has to be adopted for incorporating the inclined support in the analysis of truss just described. One way to handle inclined support is to replace the inclined support by a member having large cross sectional area as shown in Fig. 25.3b but having the length comparable with other members meeting at that joint. The inclined member is so placed that its centroidal axis is perpendicular to the inclined plane. Since the area of cross section of this new member is very high, it does not allow any displacement along its centroidal axis of the jointA. Another method of incorporating inclined support in the analysis is to suitably modify the member stiffness matrix of all the members meeting at the inclined support. Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate axes of the member. At node 1, the global co-ordinate system xy is also shown. At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular to oblique support. Let 1 ' u and 2 ' u be the displacements of nodes 1 and 2 in the local co-ordinate system. Let 1 1 ,v u be the nodal displacements of node 1 in global co-ordinate systemxy. Let 22 ", " uv be the nodal displacements along " x - and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, x x v u u ? ? sin cos ' 1 1 1 + = " 2 " 2 2 sin " cos " ' x x v u u ? ? + = (25.1) This may be written as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 " " 2 1 " " sin cos 0 0 0 0 sin cos ' ' v u v u u u x x x x ? ? ? ? (25.2) Denoting " " sin " ; cos " ; sin ; cos x x x x m l m l ? ? ? ? = = = = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 2 1 " " " " 0 0 0 0 ' ' v u v u m l m l u u (25.3a) or {} []{} u T u ' ' = where [] ' T is the displacement transformation matrix. Page 4 {}[] 1 72.855 1 ' 0.707 0.707 1.688 kN 55.97 7.07 AE p AE ?? =- - =- ?? - ?? (16) Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 4 3 8 7 1 07 . 7 ' u u u u m l m l AE p {} [] {} 1 1 ' 0.707 53.825 5.38 kN 7.07 AE p AE == (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is zero. . .e i displacement along " y is zero in the present case. If the stiffness matrix of the entire truss is formulated in global co-ordinate system then the displacements along y are not zero at the oblique support. So, a special procedure has to be adopted for incorporating the inclined support in the analysis of truss just described. One way to handle inclined support is to replace the inclined support by a member having large cross sectional area as shown in Fig. 25.3b but having the length comparable with other members meeting at that joint. The inclined member is so placed that its centroidal axis is perpendicular to the inclined plane. Since the area of cross section of this new member is very high, it does not allow any displacement along its centroidal axis of the jointA. Another method of incorporating inclined support in the analysis is to suitably modify the member stiffness matrix of all the members meeting at the inclined support. Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate axes of the member. At node 1, the global co-ordinate system xy is also shown. At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular to oblique support. Let 1 ' u and 2 ' u be the displacements of nodes 1 and 2 in the local co-ordinate system. Let 1 1 ,v u be the nodal displacements of node 1 in global co-ordinate systemxy. Let 22 ", " uv be the nodal displacements along " x - and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, x x v u u ? ? sin cos ' 1 1 1 + = " 2 " 2 2 sin " cos " ' x x v u u ? ? + = (25.1) This may be written as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 " " 2 1 " " sin cos 0 0 0 0 sin cos ' ' v u v u u u x x x x ? ? ? ? (25.2) Denoting " " sin " ; cos " ; sin ; cos x x x x m l m l ? ? ? ? = = = = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 2 1 " " " " 0 0 0 0 ' ' v u v u m l m l u u (25.3a) or {} []{} u T u ' ' = where [] ' T is the displacement transformation matrix. Similarly referring to Fig. 25.5, the force 1 ' p has components along x and y axes. Hence x p p ? cos ' 1 1 = (25.4a) x p p ? sin ' 1 2 = (25.4b) Page 5 {}[] 1 72.855 1 ' 0.707 0.707 1.688 kN 55.97 7.07 AE p AE ?? =- - =- ?? - ?? (16) Element 6: . 07 . 7 , 135 m L = ° = ? Nodal points 4-2 {}[] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = 4 3 8 7 1 07 . 7 ' u u u u m l m l AE p {} [] {} 1 1 ' 0.707 53.825 5.38 kN 7.07 AE p AE == (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide Fig. 25.3a). At a roller support, the displacement perpendicular to roller support is zero. . .e i displacement along " y is zero in the present case. If the stiffness matrix of the entire truss is formulated in global co-ordinate system then the displacements along y are not zero at the oblique support. So, a special procedure has to be adopted for incorporating the inclined support in the analysis of truss just described. One way to handle inclined support is to replace the inclined support by a member having large cross sectional area as shown in Fig. 25.3b but having the length comparable with other members meeting at that joint. The inclined member is so placed that its centroidal axis is perpendicular to the inclined plane. Since the area of cross section of this new member is very high, it does not allow any displacement along its centroidal axis of the jointA. Another method of incorporating inclined support in the analysis is to suitably modify the member stiffness matrix of all the members meeting at the inclined support. Consider a truss member as shown in Fig. 25.4. The nodes are numbered as 1 and 2. At 2, it is connected to a inclined support. Let ' 'y x be the local co-ordinate axes of the member. At node 1, the global co-ordinate system xy is also shown. At node 2, consider nodal co-ordinate system as " "y x , where " y is perpendicular to oblique support. Let 1 ' u and 2 ' u be the displacements of nodes 1 and 2 in the local co-ordinate system. Let 1 1 ,v u be the nodal displacements of node 1 in global co-ordinate systemxy. Let 22 ", " uv be the nodal displacements along " x - and " y - are in the local co-ordinate system " "y x at node 2. Then from Fig. 25.4, x x v u u ? ? sin cos ' 1 1 1 + = " 2 " 2 2 sin " cos " ' x x v u u ? ? + = (25.1) This may be written as ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 " " 2 1 " " sin cos 0 0 0 0 sin cos ' ' v u v u u u x x x x ? ? ? ? (25.2) Denoting " " sin " ; cos " ; sin ; cos x x x x m l m l ? ? ? ? = = = = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? 2 2 1 1 2 1 " " " " 0 0 0 0 ' ' v u v u m l m l u u (25.3a) or {} []{} u T u ' ' = where [] ' T is the displacement transformation matrix. Similarly referring to Fig. 25.5, the force 1 ' p has components along x and y axes. Hence x p p ? cos ' 1 1 = (25.4a) x p p ? sin ' 1 2 = (25.4b) Similarly, at node 2, the force 2 ' p has components along " x and " y axes. 32 "'cos x pp ? ' ' = (25.5a) 42 "'sin x pp ? ' ' = (25.5b) The relation between forces in the global and local co-ordinate system may be written as, 1 21 32 4 0 cos 0 ' sin cos "' 0 sin " 0 x x x x p p p p p p ? ? ? ? ?? ?? ?? ?? ? ? ?? ?? = ?? ? ? '' ?? ?? ?? ?? ?? '' ?? ?? ? ? (25.6) {} [] {} ' ' p T p T = (25.7) Using displacement and force transformation matrices, the stiffness matrix for member having inclined support is obtained. [] [ ] [ ][ ] ' ' ' T k T k T = [] ? ? ? ? ? ? ? ? ? ? ? ? - - ? ? ? ? ? ? ? ? ? ? ? ? = " " 0 0 0 0 1 1 1 1 " 0 " 0 0 0 m l m l L AE m l m l k (25.8) Simplifying, [] ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - - - - - - - = 2 2 2 2 " " " " " " " " " " " " " " m m l mm lm m l l ml ll mm ml m lm lm ll lm l L EA k (25.9) If we use this stiffness matrix, then it is easy to incorporate the condition of zero displacement perpendicular to the inclined support in the stiffness matrix. This is shown by a simple example.Read More

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