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![Lesson 24: Direct Stiffness Method: Truss Analysis.pdf ->->->-> http://tinyurl.com/y9jz9gbd
If the axial deformations are neglected then each node of the beam will have two degrees of freedom: a vertical displacement (corresponding to shear) and a rotation (corresponding to bending moment)Version 2 CE IIT, Kharagpur = 0 0 0 0 0 271.0071.000071.0071.020.00 071.0271.0020.0071.0071.000 00271.0071.020.00071.0071.0 020.0071.0271.000071.0071.0 071.0071.020.0071.0271.0071.000 071.0071.000071.0271.0020.0 20.00071.0071.000271.0071.0 00071.0071.002.0071.0271.0 0 10 5 3 2 1 8 7 6 5 4 u u u EA p p p p p (9) Thus, = 3 2 1 271.0020.0 0271.0071.0 20.0071.0271.0 0 10 5 u u u (10) Solving which, yields AE u AE u AE u 825.53;97.55;855.72 321 = == Now reactions are evaluated from the equation, = 3 2 1 8 7 6 5 4 071.020.00 071.000 0071.0071.0 0071.0071.0 071.000 u u u p p p p p (11) 4 5 6 7 83.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kNp p p p kN p= = = = = In the next step evaluate forces in membersWrite the global load-displacement relation for the beamThis is preferred while solving the problem on a computer4Element 1: .00.5,0 mL == Nodal points 2-1 Version 2 CE IIT, Kharagpur { } [ ] = 2 1 4 3 1 0.5 ' u u u u mlmlAEp { } [ ]1 53.8251' 1 1 3.80 kN 72.8555.0 AEp AE = = (12) Element 2: .00.5,90 mL == Nodal points 4-1 { } [ ] = 2 1 8 7 1 5 ' u u u u mlmlAEp { } [ ]1 01' 1 1 11.19kN 55.975 AEp AE = = (13) Element 3: .00.5,0 mL == Nodal points 3-4 { } [ ]{ } 000 5 '1 == AEp (14) Element 4: .00.5,90 mL == Nodal points 3-2 { } [ ] = 4 3 6 5 1 5 ' u u u u mlmlAEp { } [ ] { }1 1' 0 53.825 0 5 AEp AE = = (15) Element 5: .07.7,45 mL == Nodal points 3-1 { } [ ] = 2 1 6 5 1 07.7 ' u u u u mlmlAEp Version 2 CE IIT, Kharagpur { } [ ]1 72.8551' 0.707 0.707 1.688 kN 55.977.07 AEp AE = = (16) Element 6: .07.7,135 mL == Nodal points 4-2 { } [ ] = 4 3 8 7 1 07.7 ' u u u u mlmlAEp { } [ ] { }1 1' 0.707 53.825 5.38 kN 7.07 AEp AE = = (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide FigIn fact the load displacement relation for the entire structure was derived from fundamentalsThus each span is treated as an individual beamIt is noticed that, in this case, nodes are located at the supportsThe beam is divided into three beam members27.1 Introduction27.1 IntroductionVersion 2 CE IIT, Kharagpur 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 supportHowever sometimes it is required to consider a node between support points2{ } [ ] = 6 5 4 3 1 31.2 ' u u u u mlmlAEp { } [ ]1 6.6671' 1 1 2.88 kN 13.3342.31 AEp AE = = (16) Example 25.2 Determine the forces in the truss shown in FigHere lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedomElement 1: .00.5,0 mL == Nodal points 2-1 [ ] 2 1 4 3 0000 0101 0000 0101 0.5 2143 1 = EAk (1) Element 2: .00.5,90 mL == Nodal points 4-1 Version 2 CE IIT, Kharagpur [ ] 2 1 8 7 1010 0000 1010 0000 0.5 2187 2 = EAk (2) Element 3: .00.5,0 mL == Nodal points 3-4 [ ] 8 7 6 5 0000 0101 0000 0101 0.5 8765 3 = EAk (3) Element 4: .00.5,90 mL == Nodal points 3-2 [ ] 4 3 6 5 1010 0000 1010 0000 0.5 4365 4 = EAk (4) Element 5: .07.7,45 mL == Nodal points 3-1 [ ] 2 1 6 5 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 07.7 2165 5 = EAk (5) Element 6: .07.7,135 mL == Nodal points 4-2 Version 2 CE IIT, Kharagpur [ ] 4 3 8 7 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 07.7 4387 6 = EAk (6](http://cdn08.dayviews.com/500/_u3/_u8/_u7/_u9/_u9/_u1/u3879915/19471_1500440580/Lesson_24_Direct_Stiffness_Method_Truss_Analysis_pdf_http_tinyurl_com_y9jz9gbd_If_the_axial.jpg)
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Lesson 24: Direct Stiffness Method: Truss Analysis.pdf ->->->-> http://tinyurl.com/y9jz9gbd
If the axial deformations are neglected then each node of the beam will have two degrees of freedom: a vertical displacement (corresponding to shear) and a rotation (corresponding to bending moment)Version 2 CE IIT, Kharagpur = 0 0 0 0 0 271.0071.000071.0071.020.00 071.0271.0020.0071.0071.000 00271.0071.020.00071.0071.0 020.0071.0271.000071.0071.0 071.0071.020.0071.0271.0071.000 071.0071.000071.0271.0020.0 20.00071.0071.000271.0071.0 00071.0071.002.0071.0271.0 0 10 5 3 2 1 8 7 6 5 4 u u u EA p p p p p (9) Thus, = 3 2 1 271.0020.0 0271.0071.0 20.0071.0271.0 0 10 5 u u u (10) Solving which, yields AE u AE u AE u 825.53;97.55;855.72 321 = == Now reactions are evaluated from the equation, = 3 2 1 8 7 6 5 4 071.020.00 071.000 0071.0071.0 0071.0071.0 071.000 u u u p p p p p (11) 4 5 6 7 83.80 kN ; 1.19 kN ; 1.19 kN ; 3.8 0 ; 15.00 kNp p p p kN p= = = = = In the next step evaluate forces in membersWrite the global load-displacement relation for the beamThis is preferred while solving the problem on a computer4Element 1: .00.5,0 mL == Nodal points 2-1 Version 2 CE IIT, Kharagpur { } [ ] = 2 1 4 3 1 0.5 ' u u u u mlmlAEp { } [ ]1 53.8251' 1 1 3.80 kN 72.8555.0 AEp AE = = (12) Element 2: .00.5,90 mL == Nodal points 4-1 { } [ ] = 2 1 8 7 1 5 ' u u u u mlmlAEp { } [ ]1 01' 1 1 11.19kN 55.975 AEp AE = = (13) Element 3: .00.5,0 mL == Nodal points 3-4 { } [ ]{ } 000 5 '1 == AEp (14) Element 4: .00.5,90 mL == Nodal points 3-2 { } [ ] = 4 3 6 5 1 5 ' u u u u mlmlAEp { } [ ] { }1 1' 0 53.825 0 5 AEp AE = = (15) Element 5: .07.7,45 mL == Nodal points 3-1 { } [ ] = 2 1 6 5 1 07.7 ' u u u u mlmlAEp Version 2 CE IIT, Kharagpur { } [ ]1 72.8551' 0.707 0.707 1.688 kN 55.977.07 AEp AE = = (16) Element 6: .07.7,135 mL == Nodal points 4-2 { } [ ] = 4 3 8 7 1 07.7 ' u u u u mlmlAEp { } [ ] { }1 1' 0.707 53.825 5.38 kN 7.07 AEp AE = = (17) 25.2 Inclined supports Sometimes the truss is supported on a roller placed on an oblique plane (vide FigIn fact the load displacement relation for the entire structure was derived from fundamentalsThus each span is treated as an individual beamIt is noticed that, in this case, nodes are located at the supportsThe beam is divided into three beam members27.1 Introduction27.1 IntroductionVersion 2 CE IIT, Kharagpur 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 supportHowever sometimes it is required to consider a node between support points2{ } [ ] = 6 5 4 3 1 31.2 ' u u u u mlmlAEp { } [ ]1 6.6671' 1 1 2.88 kN 13.3342.31 AEp AE = = (16) Example 25.2 Determine the forces in the truss shown in FigHere lower numbers are used to indicate unconstrained degrees of freedom and higher numbers are used for constrained degrees of freedomElement 1: .00.5,0 mL == Nodal points 2-1 [ ] 2 1 4 3 0000 0101 0000 0101 0.5 2143 1 = EAk (1) Element 2: .00.5,90 mL == Nodal points 4-1 Version 2 CE IIT, Kharagpur [ ] 2 1 8 7 1010 0000 1010 0000 0.5 2187 2 = EAk (2) Element 3: .00.5,0 mL == Nodal points 3-4 [ ] 8 7 6 5 0000 0101 0000 0101 0.5 8765 3 = EAk (3) Element 4: .00.5,90 mL == Nodal points 3-2 [ ] 4 3 6 5 1010 0000 1010 0000 0.5 4365 4 = EAk (4) Element 5: .07.7,45 mL == Nodal points 3-1 [ ] 2 1 6 5 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 07.7 2165 5 = EAk (5) Element 6: .07.7,135 mL == Nodal points 4-2 Version 2 CE IIT, Kharagpur [ ] 4 3 8 7 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 5.05.05.05.0 07.7 4387 6 = EAk (6) There are eight possible global degrees of freedom for the truss shown in the figure27.2 shows a prismatic beam of a constant cross section that is fully restrained at ends in local orthogonal co-ordinate system I also confirm that I read and I agree with the Privacy policy concerning, among others, how data are used by the websiteVersion 2 CE IIT, Kharagpur { } [ ] = 2 1 6 5 1 619.4 ' u u u u mlmlAEp { } [ ]1 13.334 1' 0.5 0.5 0.866 6.667 5.77 kN 4.619 34.64 AEp AE = = (14) Member 4: mLml 0.31.2;0;0.1 === One way to handle inclined support is to replace the inclined support by a member having large cross sectional area as shown in Fig[ ] [ ] [ ][ ]''' TkTk T= [ ] = "" 00 0011 11 "0 "0 0 0 ml ml L AE m l m l k (25.8) Simplifying, [ ] = 2 2 2 2 """"" """"" "" "" mmlmmlm mllmlll mmmlmlm lmlllml L EAk (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 matrixThis procedure runs into trouble when the structure is large and complex{ } [ ] = 2 1 8 7 1 619.4 ' u u u u mlmlAEp { } [ ]1 6.6671' 0.5 0.866 5.77 kN 34.644.619 AEp AE = = (12) Member 2: mLml 0.4;0.1;0 === c3545f6b32
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