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# Module1 1 introduction-tomatrixms - rajesh sir

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### Module1 1 introduction-tomatrixms - rajesh sir

1. 1. Structural Analysis - III Introduction to M t i M th dMatrix Methods Dr. Rajesh K. N. Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering Govt. College of Engineering, Kannur Dept. of CE, GCE Kannur Dr.RajeshKN 1
2. 2. Module I Matrix analysis of structures Module I • Definition of flexibility and stiffness influence coefficients – d l t f fl ibilit t i b h i l h & Matrix analysis of structures development of flexibility matrices by physical approach & energy principle. Flexibility method • Flexibility matrices for truss beam and frame elements –• Flexibility matrices for truss, beam and frame elements – load transformation matrix-development of total flexibility matrix of the structure –analysis of simple structures – l t ti b d l f d l l dplane truss, continuous beam and plane frame- nodal loads and element loads – lack of fit and temperature effects. Dept. of CE, GCE Kannur Dr.RajeshKN 2
3. 3. Force method and Displacement methodp •These methods are applicable to discretized structures•These methods are applicable to discretized structures of all types • Force method (Flexibility method) • Actions are the primary unknowns • Static indeterminacy: excess of unknown actions than the available number of equationsq of static equilibrium Dept. of CE, GCE Kannur Dr.RajeshKN
4. 4. • Displacement method (Stiffness method) • Displacements of the joints are the primary unknowns • Kinematic indeterminacy: number of d d l d ( hindependent translations and rotations (the unknown joint displacements) • More suitable for computer programming Dept. of CE, GCE Kannur Dr.RajeshKN
5. 5. Types of Framed Structures • a. Beams: may support bending moment, shear force and Types of Framed Structures axial force • b. Plane trusses: hinge joints; In addition to axial forces, a member CAN have bending moments and shear forces if it Dept. of CE, GCE Kannur Dr.RajeshKN 5 g has loads directly acting on them, in addition to joint loads
6. 6. • c. Space trusses: hinge joints; any couple acting on ac. Space trusses: hinge joints; any couple acting on a member should have moment vector perpendicular to the axis of the member, since a truss member is incapable of ti t i ti tsupporting a twisting moment • d. Plane frames: Joints are rigid; all forces in the plane of the frame all couples normal to the plane of the frame Dept. of CE, GCE Kannur Dr.RajeshKN 6 the frame, all couples normal to the plane of the frame
7. 7. G id ll f l t th l f th id ll• e. Grids: all forces normal to the plane of the grid, all couples in the plane of the grid (includes bending and torsion)) • f. Space frames: most general framed structure; may support bending moment shear force axial force and Dept. of CE, GCE Kannur Dr.RajeshKN 7 support bending moment, shear force, axial force and torsion
8. 8. Deformations in Framed Structures , ,x y zT M MThree couples:, ,x y zN V VThree forces: •Significant deformations in framed structures:•Significant deformations in framed structures: Structure Significant deformationsStructure Significant deformations Beams flexural Plane trusses axialPlane trusses axial Space trusses axial Plane frames flexural and axialPlane frames flexural and axial Grids flexural and torsional Space frames axial flexural and torsional Dept. of CE, GCE Kannur Dr.RajeshKN 8 Space frames axial, flexural and torsional
9. 9. Types of deformations in framed structures Dept. of CE, GCE Kannur Dr.RajeshKN 9 b) axial c) shearing d) flexural e) torsional
10. 10. Static indeterminacy B Static indeterminacy • Beam: • Static indeterminacy = Reaction components - number ofy p eqns available 3E R= − • Examples: • Single span beam with both ends hinged with inclined loads C ti b• Continuous beam • Propped cantilever • Fixed beam Dept. of CE, GCE Kannur Dr.RajeshKN
11. 11. Dept. of CE, GCE Kannur Dr.RajeshKN 11
12. 12. • Rigid frame (Plane):g d a e ( a e): • External indeterminacy = Reaction components - number of il bleqns available 3E R= − • Internal indeterminacy = 3 × closed frames 3I a= • Total indeterminacy = External indeterminacy + Internal indeterminacy ( )3 3T E I R a= + = − + • Note: An internal hinge will provide an additional eqn Dept. of CE, GCE Kannur Dr.RajeshKN 12
13. 13. Example 1 Example 2Example 1 Example 2 ( )3 3T E I R a= + = − +( )3 3T E I R a= + = − + ( ) ( )3 3 3 3 2 12= × − + × =( )2 2 3 3 0 1= × − + × = Example 3 Example 4 ( ) ( ) 3 3T E I R a= + = − + ( ) ( ) 3 3T E I R a= + = − + Dept. of CE, GCE Kannur Dr.RajeshKN ( )3 2 3 3 3 12= × − + × = ( )4 3 3 3 4 21= × − + × =
14. 14. Dept. of CE, GCE Kannur Dr.RajeshKN 14
15. 15. • Rigid frame (Space):g d a e (Space): • External indeterminacy = Reaction components - number of il bleqns available 6E R= − • Internal indeterminacy = 6 × closed frames Example 1 ( )6 6T E I R a= + = − + Example 1 ( )4 6 6 6 1 24= × − + × = If axial deformations are neglected, static indeterminacy is not affected since the same number of actions still exist in Dept. of CE, GCE Kannur Dr.RajeshKN not affected since the same number of actions still exist in the structure
16. 16. Plane truss (general): External indeterminacy = Reaction components - number of eqns available 3E R= − Minimum 3 members and 3 joints. A dditi l j i t i 2 dditi l b Hence, number of members for stability, Any additional joint requires 2 additional members. ( )3 2 3 2 3m j j= + − = − Dept. of CE, GCE Kannur Dr.RajeshKN
17. 17. ( )2 3I m j= − −Hence, internal indeterminacy, l ( l d l) dTotal (Internal and external) indeterminacy ( )3 2 3T E I R m j= + = − + − − 2m R j= + − b f b• m : number of members • R : number of reaction components • j : number of jointsj j • Note: Internal hinge will provide additional eqn Dept. of CE, GCE Kannur Dr.RajeshKN 17
18. 18. Example 1p 2 9 3 2 6 0T m R j= + − = + − × = 3 3 3 0E R= − = − = 0I T E= − = Example 2 2 15 4 2 8 3T m R j+ + ×2 15 4 2 8 3T m R j= + − = + − × = 3 4 3 1E R= − = − = 2I T E= − = Example 3 2 6 4 2 5 0T m R j= + − = + − × = ( )3 1 4 4 Hinge at0 AE R= − + = − = Dept. of CE, GCE Kannur Dr.RajeshKN 0I T E= − =
19. 19. Example 4Example 4 2 7 3 2 5 0T m R j= + − = + − × = 3 3 3 0E R= − = − =3 3 3 0E R= = = 0I T E= − = 2 6 4 2 4 2T m R j= + − = + − × = Example 5 2 6 4 2 4 2T m R j+ + 3 4 3 1E R= − = − = 1I T E 1I T E= − = Example 6 2 11 3 2 6 2T m R j= + − = + − × = 3 3 3 0E R= − = − = p Dept. of CE, GCE Kannur Dr.RajeshKN 2I T E= − =
20. 20. • Wall or roof attached pin jointed plane truss (Exception• Wall or roof attached pin jointed plane truss (Exception to the above general case): • Internal indeterminacy 2I m j= − • External indeterminacy = 0 (Since, once the member forces are determined, reactions are determinable)forces are determined, reactions are determinable) Example 1 Example 3Example 2Example 1 Example 3Example 2 2 6 2 3 0 T I m j= = − = − × = 2 5 2 1 3 T I m j= = − = − × = 2 7 2 3 1 T I m j= = − Dept. of CE, GCE Kannur Dr.RajeshKN 20 6 3 0 7 2 3 1= − × =
21. 21. • Space Truss: • External indeterminacy = Reaction components - number of equations available 6E R= − E l 1 3T m R j= + −Total (Internal and external) indeterminacy Example 1 Total (Internal and external) indeterminacy 3T m R j= + − ( ) y 12 9 3 6 3T∴ = + − × = Dept. of CE, GCE Kannur Dr.RajeshKN 6 9 6 3E R= − = − =
22. 22. Actions and displacements •Actions: p • External actions (Force or couple or combinations) and • Internal actions (Internal stress resultants – BM, SF, axial forces, twisting moments) Dept. of CE, GCE Kannur Dr.RajeshKN 22
23. 23. •Displacements: A translation or rotation at some pointsp ace e ts: t a s at o o otat o at so e po t •Displacement corresponding to an action: Need not be caused b th t tiby that action Dept. of CE, GCE Kannur Dr.RajeshKN
24. 24. •Notations for actions and displacements:•Notations for actions and displacements: D32 D33 Dept. of CE, GCE Kannur Dr.RajeshKN 24
25. 25. Equilibrium •Resultant of all actions (a force, a couple or both) must vanish for static equilibrium q q •Resultant force vector must be zero; resultant moment vector must be zero 0xF =∑ 0yF =∑ 0zF =∑ ∑ ∑ ∑ must be zero 0xM =∑ 0yM =∑ 0zM =∑ F 2 di i l bl (f i l•For 2-dimensional problems (forces are in one plane and couples have vectors normal to the plane), 0xF =∑ 0yF =∑ 0zM =∑ •In stiffness method, the basic equations to be solved Dept. of CE, GCE Kannur Dr.RajeshKN 25 In stiffness method, the basic equations to be solved are the equilibrium conditions at the joints
26. 26. Compatibility •Compatibility conditions: Conditions of continuity of displacements throughout the structure •Eg: at a rigid connection between two members, the displacements (translations and rotations) of both membersdisplacements (translations and rotations) of both members must be the same I fl ibili h d h b i i b l d•In flexibility method, the basic equations to be solved are the compatibility conditions Dept. of CE, GCE Kannur Dr.RajeshKN 26
27. 27. Action and displacement equations A SD=D FA=•Spring: 1 S F− =•Stiffness 1 F S− =•Flexibility: •The above equations apply to any linearly elastic structure Dept. of CE, GCE Kannur Dr.RajeshKN 27
28. 28. •Example 1: Flexibility and stiffness of a beam subjected to a single load 3 48 L F EI =Flexibility 3 48EI S L =Stiffness Dept. of CE, GCE Kannur Dr.RajeshKN 28 48EI L
29. 29. •Example 2: Flexibility coefficients of a beam subjected to several loads A ti th bActions on the beam Deformations corresponding to actions Dept. of CE, GCE Kannur Dr.RajeshKN 29 Deformations corresponding to actions
30. 30. Dept. of CE, GCE Kannur Dr.RajeshKN 30 Unit load applied corresponding to each action, separately
31. 31. 1 11 12 13D D D D= + + 2 21 22 23D D D D= + + 3 31 32 33D D D D= + + 1 11 1 12 2 13 3D F A F A F A= + + 2 21 1 22 2 23 3D F A F A F A= + + 3 31 1 32 2 33 3D F A F A F A= + + 11 12 13, ,F F F Flexibility coefficientsetc. • Flexibility coefficient F12: Displacement corresponding to A1 caused by a unit value of A2.caused by a unit value of A2. • In general, flexibility coefficient Fij is the displacement Dept. of CE, GCE Kannur Dr.RajeshKN 31 In general, flexibility coefficient Fij is the displacement corresponding to Ai caused by a unit value of Aj.
32. 32. •Example 3: Stiffness coefficients of a beam subjected to several loads Actions on the beam Deformations corresponding to actions Dept. of CE, GCE Kannur Dr.RajeshKN 32 Deformations corresponding to actions
33. 33. Unit displacement applied corresponding to each DOF separately Dept. of CE, GCE Kannur Dr.RajeshKN 33 Unit displacement applied corresponding to each DOF, separately, keeping all other displacements zero
34. 34. A A A A= + +1 11 12 13A A A A= + + A S D S D S D+ +1 11 1 12 2 13 3A S D S D S D= + + 2 21 1 22 2 23 3A S D S D S D= + + 3 31 1 32 2 33 3A S D S D S D= + + 11 12 13, ,S S S Stiffness coefficients:etc. • Stiffness coefficient S12: Action corresponding to D1 caused by a unit value of D2.a unit value of D2. • In general, stiffness coefficient Sij is the action corresponding Dept. of CE, GCE Kannur Dr.RajeshKN 34 In general, stiffness coefficient Sij is the action corresponding to Di caused by a unit value of Dj.
35. 35. •Example 4: Flexibility and stiffness coefficients of a beamExample 4: Flexibility and stiffness coefficients of a beam Flexibility coefficients:y Dept. of CE, GCE Kannur Dr.RajeshKN 35
36. 36. Stiffness coefficients: Dept. of CE, GCE Kannur Dr.RajeshKN 36
37. 37. •Example 5: Flexibility and stiffness coefficients of a trussExample 5: Flexibility and stiffness coefficients of a truss Dept. of CE, GCE Kannur Dr.RajeshKN 37
38. 38. Flexibility and stiffness matrices 1. Flexibility matrix Th tibilit ti 1 11 1 12 2 13 3 1... n nD F A F A F A F A= + + + + oThe compatibility equations are: 2 21 1 22 2 23 3 2... ............................................................ n nD F A F A F A F A= + + + + 1 1 2 2 3 3 ...n n n n nn nD F A F A F A F A= + + + + oIn matrix form, 1 11 12 1 1 2 21 22 2 2 ... ... n n D F F F A D F F F A ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬ 1 2 ... ... ... ... ...... ...n n nn nn F F F AD ⎢ ⎥=⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭ Dept. of CE, GCE Kannur Dr.RajeshKN 38 1 2 11 n n nn nn n n nn × ×× ⎣ ⎦ ⎩ ⎭⎩ ⎭
39. 39. D = FA {D} Displacement matrix (vector),{D} Displacement matrix (vector), [F] Flexibility matrix, {A} A ti t i ( t ){ } [ ]{ }D F A= {A} Action matrix (vector){ } [ ]{ }D F A F are the flexibility coefficientsijF are the flexibility coefficients Dept. of CE, GCE Kannur Dr.RajeshKN 39
40. 40. 2. Stiffness matrix2. Stiffness matrix oThe equilibrium equations are: 1 11 1 12 2 13 3 1... n nA S D S D S D S D A S D S D S D S D = + + + + = + + + +2 21 1 22 2 23 3 2... ............................................................ n nA S D S D S D S D A S D S D S D S D = + + + + + + + +1 1 2 2 3 3 ...n n n n nn nA S D S D S D S D= + + + + oIn matrix form, 1 11 12 1 1 2 21 22 2 2 ... ... n n A S S S D A S S S D ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎨ ⎬ ⎨ ⎬ 1 2 ... ... ... ... ...... ...n n nn nn S S S DA ⎢ ⎥=⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎣ ⎦ ⎩ ⎭⎩ ⎭ Dept. of CE, GCE Kannur Dr.RajeshKN 40 1 2 11 n n nn nn n n nn × ×× ⎣ ⎦ ⎩ ⎭⎩ ⎭
41. 41. A = SD {A} Action matrix (vector) , { } [ ]{ }A S D= [S] Stiffness matrix, {D} Displacement matrix (vector)p ( ) ijS are the stiffness coefficientsijS are the stiffness coefficients • Relationship between flexibility and stiffness matrices { } [ ]{ } [ ][ ]{ }A S D S F A Relationship between flexibility and stiffness matrices [ ] [ ] 1 F S − ∴ { } [ ]{ } [ ][ ]{ }A S D S F A= = Dept. of CE, GCE Kannur Dr.RajeshKN 41 [ ] [ ]F S∴ =
42. 42. •Example: Cantilever element 3 L 2 L L 11 ; 3 L F EI = 21 12 ; 2 L F F EI = = 22 L F EI = 3 2 L L3 2 1 1 2 3 2 L L D A A EI EI = + 2 2 1 2 2 L L D A A EI EI = + 3 2 L L⎡ ⎤3 2 1 1 2 3 2 L L D AEI EI D AL L ⎡ ⎤ ⎢ ⎥⎧ ⎫ ⎧ ⎫ ⎢ ⎥=⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎩ ⎭ ⎩ ⎭2 2 2 D AL L EI EI ⎢ ⎥⎩ ⎭ ⎩ ⎭ ⎢ ⎥⎣ ⎦ Dept. of CE, GCE Kannur Dr.RajeshKN 42[ ]F
43. 43. 12EI 6EI− 4EI 11 3 12 ; EI S L = 21 12 2 6 ; EI S S L = = 22 4EI S L = 1 1 23 2 12 6EI EI A D D L L = − 2 1 22 6 4EI EI A D D L L − = + 3 2 12 6EI EI A DL L −⎡ ⎤ ⎢ ⎥⎧ ⎫ ⎧ ⎫3 2 1 1 2 2 2 6 4 A DL L A DEI EI L L ⎢ ⎥⎧ ⎫ ⎧ ⎫ = ⎢ ⎥⎨ ⎬ ⎨ ⎬ −⎩ ⎭ ⎩ ⎭⎢ ⎥ ⎢ ⎥⎣ ⎦L L⎢ ⎥⎣ ⎦ [ ][ ] [ ][ ] [ ] 1 0⎡ ⎤ [ ]S Dept. of CE, GCE Kannur Dr.RajeshKN 43 [ ][ ] [ ][ ] [ ] 1 0 0 1 F S S F I ⎡ ⎤ = = =⎢ ⎥ ⎣ ⎦ [ ]S
44. 44. •Flexibility matrix and stiffness matrix are relating actions and corresponding displacements The flexibility matrix obtained for a structure analysed by[ ]Fy y y flexibility method may not be the inverse of the stiffness matrix obtained for the same structure analysed by stiffness method [ ]S obtained for the same structure analysed by stiffness method because different sets of actions and corresponding displacements may be utilized in the two methods. Dept. of CE, GCE Kannur Dr.RajeshKN 44
45. 45. Equivalent joint loads •Analysis by flexibility and stiffness methods requires that loads must act only at joints Equivalent joint loads must act only at joints. •Thus, loads acting on the members (i.e., loads that are not acting at the joints) must be replaced by equivalent loads acting at the joints. •The loads that are determined from loads on the members are called equivalent joint loads. Dept. of CE, GCE Kannur Dr.RajeshKN 45
46. 46. •Equivalent joint loads are added to the actual joint loads to get combined joint loads. •Analysis carried out for combined joint loads •Combined joint loads can be evaluated in such a manner that the resulting displacements of the structure are same as the di l t d d b th t l l ddisplacements produced by the actual loads •This is achieved thru the use of fixed end actions to getg equivalent joint loads Dept. of CE, GCE Kannur Dr.RajeshKN 46
47. 47. •Example 1•Example 1 Beam with actual applied loads Applied joint loads Dept. of CE, GCE Kannur Dr.RajeshKN 47 Applied joint loads
48. 48. Applied loads other than joint loads (To be converted to equivalent joint loads)( o be co e ted to equ a e t jo t oads) Dept. of CE, GCE Kannur Dr.RajeshKN 48
49. 49. Member fixed end actions (Due to applied loads other than joint loads ) Dept. of CE, GCE Kannur Dr.RajeshKN 49 Fixed end actions for the entire beam
50. 50. Equivalent joint loads (Negative of fixed end actions)Equivalent joint loads (Negative of fixed end actions) Combined joint loads (A li d j i t l d + E i l t j i t l d ) Dept. of CE, GCE Kannur Dr.RajeshKN 50 (Applied joint loads + Equivalent joint loads)
51. 51. Fixed end actions M M ( )2 2 Mb a b l − a b ( )2 2 Ma b a l − 3 6Mab l 3 6Mab l M M 4 M 2 l 4 M 2 l 3 2 M l 3 2 M l Dept. of CE, GCE Kannur Dr.RajeshKN 2l2l
52. 52. 120 kN40 kN/m 20 kN/m •Example 2 A B C D / 20 kN/m 4 m 12 m 12 m 12 m 2 l2 480 12 wl = 480 240 240 wl 240 120 120 240 2 wl = 240 120 120 213 33 106 67213.33 106.67 Fi d d ti Dept. of CE, GCE Kannur Dr.RajeshKN 5288.89 31.11 Fixed end actions
53. 53. 480 266 67 133.33 240 A B C D 480 266.67 B C 240 240 88.89 328 89 + = 120 31.11 151 11 + 120 328.89= 151.11= Equivalent joint loads (Opposite of fixed end actions) Combined joint loads are same as equivalent joint loads here, i th l d li d t j i t di tlsince there are no loads applied to joints directly Dept. of CE, GCE Kannur Dr.RajeshKN 53
54. 54. •Superposition of combined joint loads and restraint actions gives the actual loads.the actual loads. •Superposition of joint displacements due to the combined j i l d d i i i h di ljoint loads and restraint actions gives the displacements produced by the actual loads. •But joint displacements due to restraint actions are zero. Thus, joint displacements due to the combined joint loads give th di l t d d b th t l l dthe displacements produced by the actual loads •But member end actions due to actual loads are obtained byy superimposing member end actions due to restraint actions and combined joint loads Dept. of CE, GCE Kannur Dr.RajeshKN 54
55. 55. SummarySummary Matrix analysis of structures • Definition of flexibility and stiffness influence coefficients – d l t f fl ibilit t i b h i l h & Matrix analysis of structures development of flexibility matrices by physical approach & energy principle. Dept. of CE, GCE Kannur Dr.RajeshKN 55