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- 1. Structural Analysis - II Indeterminate structures;Indeterminate structures; Force method of analysis 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. Module IModule I Statically and kinematically indeterminate structures • Degree of static indeterminacy, Degree of kinematic indeterminacy, Force and displacement method of analysis Force method of analysis • Method of consistent deformation-Analysis of fixed and i b Force method of analysis continuous beams • Clapeyron’s theorem of three moments-Analysis of fixed and continuous beams • Principle of minimum strain energy-Castigliano’s second theorem- Dept. of CE, GCE Kannur Dr.RajeshKN 2 p gy g Analysis of beams, plane trusses and plane frames.
- 3. Types of Framed Structures • a. Beams: may support bending moment, shear force and axial force yp • b. Plane trusses: hinge joints; In addition to axial forces, a member CAN have bending moments and shear forces if it has Dept. of CE, GCE Kannur Dr.RajeshKN 3 loads directly acting on them, in addition to joint loads
- 4. • c. 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 supporting a twisting momentg • 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 4
- 5. • e Grids: all forces normal to the plane of the grid all couples• 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 torsion Dept. of CE, GCE Kannur Dr.RajeshKN 5 g , ,
- 6. 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 axial Space trusses axial Plane frames flexural and axial Grids flexural and torsional Space frames axial, flexural and torsional Dept. of CE, GCE Kannur Dr.RajeshKN 6
- 7. Types of deformations in framed structures b) axial c) shearing d) flexural e) torsional Dept. of CE, GCE Kannur Dr.RajeshKN 7 b) axial c) shearing d) flexural e) torsional
- 8. Equilibrium •Resultant of all actions (a force, a couple or both) must vanish for static equilibrium q 0F∑ 0F∑ ∑ •Resultant force vector must be zero; resultant moment vector must be zero 0xF =∑ 0yF =∑ 0zF =∑ 0xM =∑ 0yM =∑ 0zM =∑∑ y∑ ∑ •For 2 dimensional problems (forces are in one plane and 0F =∑ 0F =∑ 0M =∑ •For 2-dimensional problems (forces are in one plane and couples have vectors normal to the plane), 0xF =∑ 0yF =∑ 0zM =∑ Dept. of CE, GCE Kannur Dr.RajeshKN 8
- 9. Compatibility •Compatibility conditions: Conditions of continuity ofp y y displacements throughout the structure •Eg: at a rigid connection between two members, theg g , displacements (translations and rotations) of both members must be the same Dept. of CE, GCE Kannur Dr.RajeshKN 9
- 10. Indeterminate Structures Force method and Displacement method • Force method (Flexibility method) • Actions are the primary unknowns St ti i d t i f k ti th• Static indeterminacy: excess of unknown actions than the available number of equations of static equilibrium • Displacement method (Stiffness method) • Displacements of the joints are the primary unknowns • Kinematic indeterminacy: number of independent Dept. of CE, GCE Kannur Dr.RajeshKN Kinematic indeterminacy: number of independent translations and rotations
- 11. Static indeterminacy • Beam: • Static indeterminacy = Reaction components number of eqns• Static indeterminacy = Reaction components - number of eqns available 3E R= − • Examples:• Examples: • Single span beam with both ends hinged with inclined l dloads • Continuous beam • Propped cantilever • Fixed beam Dept. of CE, GCE Kannur Dr.RajeshKN
- 12. Degree of StaticStructure Degree of Static Indeterminacy Structure 0 1 3 Dept. of CE, GCE Kannur Dr.RajeshKN 12
- 13. Degree of Static I d t i Structure 0 Indeterminacy 0 1 55 2 Dept. of CE, GCE Kannur Dr.RajeshKN
- 14. • Rigid frame (Plane):g ( ) • External indeterminacy = Reaction components - number of eqns available 3E R= −available • Internal indeterminacy = 3 × closed frames 3E R 3I a=• 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 14
- 15. Example 1 Example 2p p ( )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= × − + × =
- 16. Degree of Static I d t i Structure Indeterminacy 3 63 Dept. of CE, GCE Kannur Dr.RajeshKN 16
- 17. Degree of Static I d t i Structure Indeterminacy 2 1 Dept. of CE, GCE Kannur Dr.RajeshKN
- 18. • Rigid frame (Space):g ( p ) • External indeterminacy = Reaction components - number of eqns available 6E R= −available • Internal indeterminacy = 6 × closed frames 6E R Example 1 ( ) ( ) 6 6T E I R a= + = − + ( )4 6 6 6 1 24= × − + × = If i l d f ti l t d t ti i d t i i tIf axial deformations are neglected, static indeterminacy is not affected since the same number of actions still exist in the structure Dept. of CE, GCE Kannur Dr.RajeshKN
- 19. 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 ( )3 2 3 2 3m j j= + − = −Hence, number of members for stability, Any additional joint requires 2 additional members. ( )2 3I m j= − −Hence, internal indeterminacy, Total (Internal and external) indeterminacyTotal (Internal and external) indeterminacy ( )3 2 3 2 T E I R m j m R j = + = − + − − = + − 2m R j+ • m: number of members • R: number of reaction components Dept. of CE, GCE Kannur Dr.RajeshKN • j : number of joints • Note: Internal hinge will provide additional eqn
- 20. E l 1Example 1 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 R p 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= − =
- 21. Example 4 2 7 3 2 5 0T m R j= + − = + − × =Example 4 2 7 3 2 5 0T m R j+ + 3 3 3 0E R= − = − = 0I T E 0I T E= − = 2 6 4 2 4 2T m R j= + − = + − × = 3 4 3 1E R= = = Example 5 3 4 3 1E R= − = − = 1I T E= − = 2 11 3 2 6 2T m R j= + − = + − × =Example 6 3 3 3 0E R= − = − = 2I T E= − = Dept. of CE, GCE Kannur Dr.RajeshKN 2I T E= =
- 22. • Wall or roof attached pin jointed plane truss (Exception to the• 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) Example 1 Example 3Example 2Example 1 Example 3p 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 22 7 2 3 1= − × =
- 23. • Space Truss: External indeterminacy = Reaction components - number of eqns available Minimum 6 members and 4 joints 6E R= − ( )6 3 4 3 6m j j= + − = −Hence, number of members for stability, Minimum 6 members and 4 joints. Any additional joint requires 3 additional members. ( )6 3 4 3 6m j j+Hence, number of members for stability, ( )3 6I m j= − −Hence, internal indeterminacy, Total (Internal and external) indeterminacy ( )6 3 6T E I R m j= + = − + − − 3m R j= + − Dept. of CE, GCE Kannur Dr.RajeshKN
- 24. Example 3T m R j= +Total (Internal and external) indeterminacy 3T m R j= + −Total (Internal and external) indeterminacy 12 9 3 6 3T∴ = + − × = 6 9 6 3E R= − = − = Dept. of CE, GCE Kannur Dr.RajeshKN
- 25. Kinematic indeterminacyKinematic indeterminacy • joints: where members meet, supports, free ends • joints undergo translations or rotations •in some cases joint displacements will be known, from the restraint conditions •the unknown joint displacements are the kinematically•the unknown joint displacements are the kinematically indeterminate quantities odegree of kinematic indeterminacy: number of degrees of f dfreedom Two types of DOF • Nodal type DOF J i DOF Two types of DOF Dept. of CE, GCE Kannur Dr.RajeshKN 25 • Joint type DOF
- 26. • degree of kinematic indeterminacy (degrees of freedom) is d fi ddefined as: • the number of independent translations and rotations in a structure. DOF = 1 DOF = 2 Dept. of CE, GCE Kannur Dr.RajeshKN DOF 2
- 27. • in a truss the joint rotation is not regarded as a degree of• in a truss, the joint rotation is not regarded as a degree of freedom. joint rotations do not have any physical significance as they have no effects in the members of the truss • in a frame, degrees of freedom due to axial deformations can be neglected Dept. of CE, GCE Kannur Dr.RajeshKN
- 28. Kinematic Degree f F d Structure of Freedom 2 11 22 Dept. of CE, GCE Kannur Dr.RajeshKN
- 29. Kinematic Degree f F d Structure of Freedom 4 If the effect of the cantilever portion is considered as a joint load atIf the effect of the cantilever portion is considered as a joint load at the roller support on the far right, kinematic indeterminacy can be taken as 2. Dept. of CE, GCE Kannur Dr.RajeshKN
- 30. Kinematic Degree f F d Structure of Freedom 2 3 Dept. of CE, GCE Kannur Dr.RajeshKN
- 31. Kinematic Degree f F d Structure of Freedom 6 2 5 Dept. of CE, GCE Kannur Dr.RajeshKN
- 32. Method of Consistent Deformation Illustration of the method Problem Released structure Dept. of CE, GCE Kannur Dr.RajeshKN 32 Released structure
- 33. D fl ti f l d 4 5wL Δ Deflection of released structure due to actual loads 384 B EI Δ = C Deflection of released structure due to redundant applied as app load Dept. of CE, GCE Kannur Dr.RajeshKN 33
- 34. 3 BR L Deflection due o 48 t B B R L I R E = 3 BR L Δ 48 B B EI Δ = 4 3 5 BwL R L Compatibility condition (or equation of superposition or equation of geometry) 5 384 48 BwL R L EI EI = 5 B wL R∴ = 8 B Dept. of CE, GCE Kannur Dr.RajeshKN 34
- 35. •A general approach (applying consistent sign convention for loads and displacements): RA l it l d di t BR•Apply unit load corresponding to 3 48 B L EI δ =Let the displacement due to unit load be BR B BR δDisplacement due to is Dept. of CE, GCE Kannur Dr.RajeshKN 35
- 36. 44 5 384 B wL EI Δ = − (Negative, since deflection is downward) 0B B BR δΔ + = (Compatibility condition) B B B R δ Δ = − 5 8 B wL R∴ = Bδ 8 Dept. of CE, GCE Kannur Dr.RajeshKN 36
- 37. Example 1: Propped cantilever AM PP A 2l 2l2l 2l A B AV BV Choose VB as the redundant P AM 2l 2l AV Dept. of CE, GCE Kannur Dr.RajeshKN 37 Released structure
- 38. Find deflection at B of the released structure 3 1 2 5PL l l l Pl− −⎛ ⎞ Δ = + =⎜ ⎟Pl− . . . 2 2 2 2 3 2 48 B EI EI Δ = + =⎜ ⎟ ⎝ ⎠2 Pl EI M diagram EIEI Apply unit load on released structure corresponding to VApply unit load on released structure corresponding to VB and find deflection at B 3 3 B l EI δ = 1 l 3EI Dept. of CE, GCE Kannur Dr.RajeshKN 38
- 39. 0R δΔ 3 5 3 5B Pl EI P R −Δ = = =0B B BR δΔ + = 3 . 48 16 B B R EI lδ = = = +ve sign indicates that VB is in the same direction of the unit load. i.e., in the upward direction. P 2l 2l 5 3 2 16 16 A Pl Pl Pl M = − =Other reactions 2l 2l 5 16 P11 16 P 16 5 32 Pl Bending moment diagram 3Pl Dept. of CE, GCE Kannur Dr.RajeshKN 39 16 −
- 40. Example 2: Continuous beam 10kN A 2m 2m A B 4m C 10kN A B C V VVAV CVBV Choose VB as the redundant Dept. of CE, GCE Kannur Dr.RajeshKN 40
- 41. 10kN 2 6 A BΔ C 2m 6m B Released structure 2.5 kN7.5 kN 15 EI 10 EI 1 251 358 6+⎛ ⎞ A B C EI EI 4m Conjugate beam of the released structure (to find ∆ ) ( ) 1 25 0.5 15 8 35 EI EI =× × − 1 358 6 80.5 15 8 3EI EI +⎛ ⎞ ÷ =× × ×⎜ ⎟ ⎝ ⎠ Conjugate beam of the released structure (to find ∆B) 1 73.3334 25 4 0 5 10 4 ⎛ ⎞∴Δ ⎜ ⎟ (numerically) Dept. of CE, GCE Kannur Dr.RajeshKN 41 25 4 0.5 10 4 3 B EI EI ⎛ ⎞∴Δ = =× − × × ×⎜ ⎟ ⎝ ⎠ (numerically)
- 42. 1kN1kN 4m A B 4m C 4m 4m Released structure with unit load corresponding to V (to find δ )Released structure with unit load corresponding to VB (to find δB) 3 3 8 10.667 B l δ = = = (numerically) (di i i ∆ i d d )48 48 B EI EI EI δ 0R δΔ + 73.333 6 875B EI R kN −Δ − (direction is same as ∆B i.e., downwards) 0B B BR δΔ + = . 6.875 10.667 B B B R kNm EIδ = = = − i i di t th t V i i th it di ti f th it l d-ve sign indicates that VB is in the opposite direction of the unit load. i.e., in the upward direction. Dept. of CE, GCE Kannur Dr.RajeshKN 42
- 43. 10kN Other reactions 10kN A B CB C 4.063AV kN= 0.938CV kN=6.875 kNA Bending moment diagramg g A B C 8.122 kNm A C 3.752 kNm− Dept. of CE, GCE Kannur Dr.RajeshKN
- 44. Example 3A: 2m 3m CD 30kN B EI is constant 2m D MD A 2m 6m C D HD 30kN B VD 2m B HA Choose H as the redundant Dept. of CE, GCE Kannur Dr.RajeshKN 44 A Choose HA as the redundant
- 45. D CD To find ∆HA and δHA (using unit load method) 10kN 2m3m CD 2m6m B C 2m A B 2m A B 1kNE=2x105N/mm2 I= 2x108mm4 Portion Origin Limits M m Mm m2 AB A 0-2 0 x 0 x2AB A 0 2 0 x 0 x BC A 2-4 -10(x-2) x -10x2+20x x2 CD A 0 3 20 4 80 16 ( ) ( ) 4 3 21 1 10 20 80 Mmdx x x dx dxΔ = = + +∫ ∫ ∫ CD A 0-3 -20 4 -80 16 ( ) ( ) 2 0 10 20 80HA x x dx dx EI EI EI Δ = = − + + −∫ ∫ ∫ [ ] 43 321 1x⎡ ⎤ 306 67− Dept. of CE, GCE Kannur Dr.RajeshKN [ ] 32 0 2 1 1 10 10 80 3 HA x x x EI EI ⎡ ⎤ Δ = − + + −⎢ ⎥ ⎣ ⎦ 306.67 EI − =
- 46. 4 32 2 4 33 69 3316⎡ ⎤ ⎡ ⎤ 4 32 2 0 0 16 HA m dx x dx dx EI EI EI δ = = +∫ ∫ ∫ 33 00 69.3316 3 xx EIEIEI ⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ 0HA A HAH δΔ + = 306.67 4.4. 69 3 3 3 2B A EI H EI kN δ −Δ = = = HA A HA 69.33B EIδ Dept. of CE, GCE Kannur Dr.RajeshKN 46
- 47. Example 3B: 2I CD 2m 6m CD 30kN 2m I B M 2m A 2 6m 2I C D HD MD A 30kN 2m 6m IVD Choose H as the redundant 2m B H Dept. of CE, GCE Kannur Dr.RajeshKN Choose HA as the redundant A HA
- 48. 2I D 2I CD To find ∆HA and δHA (using unit load method) 30kN 2m6m I CD 2m6m I B C 2m A B 2m A B 1kNE=2x105N/mm2 I= 2x108mm4 Portion Origin Limits M m Mm m2 AB A 0-2 0 x 0 x2 BC A 2-4 -30(x-2) x -30x2+60x x2 CD A 0-6 -60 4 -240 16 ( ) ( ) 4 6 21 1 30 60 240 Mmdx x x dx dxΔ = = + +∫ ∫ ∫ CD A 0 6 60 4 240 16 ( ) ( ) 2 0 30 60 240 2 HA x x dx dx EI EI EI Δ = = − + + −∫ ∫ ∫ [ ] 4 63 21 1 10 30 240⎡ ⎤Δ 920− Dept. of CE, GCE Kannur Dr.RajeshKN [ ] 63 2 02 1 1 10 30 240 2 HA x x x EI EI ⎡ ⎤Δ = − + + −⎣ ⎦ 920 EI − =
- 49. 4 62 2 16d 4 63 69 3338x⎡ ⎤ ⎡ ⎤2 2 0 0 16 2 HA m dx x dx dx EI EI EI δ = = +∫ ∫ ∫ 3 00 69.3338 3 xx EIEIEI ⎡ ⎤ ⎡ ⎤= + =⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦ 0H δΔ + = 920 . 13.269 69 333 HA A EI H kN EIδ −Δ = = =0HA A HAH δΔ + = 69.333 A HA EIδ Dept. of CE, GCE Kannur Dr.RajeshKN
- 50. Example 4: Two or more redundants 60kN A B C 80kN 50kN D 3m 3m A B C 3m 3m 3m 3m 60kN 80kN 50kN 3m 3m A B C D 3m 3m 3m 3m Choose VB and Vc as the redundant Dept. of CE, GCE Kannur Dr.RajeshKN 50
- 51. 60kN 80kN 50kN A B C D BΔ CΔB C 1kN A B D δ C BBδ CBδ 1kN A B D δ C BCδ CCδ Dept. of CE, GCE Kannur Dr.RajeshKN 0B B BB C BCV Vδ δΔ + + = 0C B CB C CCV Vδ δΔ + + =
- 52. 60kN A B C 80kN 50kN DA B 3m 3m 3m 3m 3m 3m A B C D 150kNm 15 120kNm 60kNm 3m 15m 525kN825kN 240kNm 240kNm A B C D 360kNm 9 9 240kNm 240kNm 9m 9m 1620kN1620kN A B C D 125kNm 100kNm 50kNm Dept. of CE, GCE Kannur Dr.RajeshKN 15m 3m 687.5kN437.5kN
- 53. 3420 8280 2325 14025BEIΔ = + + = 2790 8280 2850 13920CEIΔ = + + = D2kNm 4kNm A B C D 12m 6m 16kN20kN 2kNm 16kN20kN 96BBEIδ = 84CBEIδ = A D2kNm 4kNm A B C D 12m 6m 20kN16kN 2kNm Dept. of CE, GCE Kannur Dr.RajeshKN 84BCEIδ = 96CCEIδ =
- 54. 0V Vδ δΔ + + = 14025 96 84 0B CV V⇒ + + =0B B BB C BCV Vδ δΔ + + = 0C B CB C CCV Vδ δΔ + + = 14025 96 84 0B CV V⇒ + + 13920 84 96 0B CV V⇒ + + =C B CB C CC B C 82BV kN= − 73.25CV kN= − ( )82BV kN= ↑ ( )73.25CV kN= ↑ ( )19.25AV kN= ↑ ( )15.5DV kN= ↑ ( )B ( )C Dept. of CE, GCE Kannur Dr.RajeshKN
- 55. Clapeyron’s theorem of three moments 1. Uniform loading w kN m w kN m1w kN m 1 1l I A B C 2w kN m 2 2l I M 1 1,l I 2 2,l I 1 AM EI B C 1 BM EI 2 BM EI 2 CM EI A B C M/EI diagram for joint moments + B C 2 1 1 18 w l EI 2 2 2 28 w l EI Dept. of CE, GCE Kannur Dr.RajeshKN A B C M/EI diagram for simple beam moments
- 56. To find slopes at B using Conjugate Beam Method: 2 1 1 2 2M l l M l l w l l⎡ ⎤⎡ ⎤ From span AB: 1 1 1 1 1 1 1 11 1 1 1 1 1 1 2 2 . . . . . . . 2 3 2 3 3 8 2 A B B M l l M l l w l l lV l EI EI EI ⎡ ⎤⎡ ⎤ += + ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 3 1 1 1 1 1 1 1 16 3 24 A B B BA M l M l w l V EI EI EI θ⇒ = + + = From span BC: 2 2 2 2 2 2 2 2 22 2 2 2 2 1 1 2 2 . . . . . . . 2 3 2 3 3 8 2 B C B M l l M l l w l l lV l EI EI EI ⎡ ⎤⎡ ⎤ += + ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦ 3 2 2 2 2 2 3 6 24 B C B BC M l M l w l V EI EI EI θ⇒ = + + = Dept. of CE, GCE Kannur Dr.RajeshKN 2 2 23 6 24EI EI EI
- 57. BAθ A B C BA BCθ Deflected shape 0BA BC BA BCθ θ θ θ+ = ⇒ = − 33 2 2 2 21 1 1 1 3 6 246 3 24 B CA B M l M l w lM l M l w l EI EI EIEI EI EI ⎛ ⎞ + +⇒ + + = −⎜ ⎟ ⎝ ⎠2 2 21 1 1 3 6 246 3 24 EI EI EIEI EI EI ⎝ ⎠ 3 3 1 2l lM l M l w l w l⎛ ⎞1 21 2 1 1 2 2 1 21 2 1 2 2 4 4 A C B l lM l M l w l w l M EI EIEI EI EI EI ⎛ ⎞ +⇒ + + = − −⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN
- 58. 2. General loadingg 1a 2a A B C x x1x 2x 1 1 1 1 1 1 1 16 3 A B BA M l M l a x EI EI EI l θ = + + 2 2 2 2 2 2 2 23 6 B C BC M l M l a x EI EI EI l θ = + + 2 2 2 2 1 21 2 1 1 2 26 6 2A C l lM l M l a x a x M ⎛ ⎞ +⇒ + + = − −⎜ ⎟BA BCθ θ= − Dept. of CE, GCE Kannur Dr.RajeshKN 1 21 2 1 1 2 2 2 BM EI EIEI EI EI l EI l +⇒ + + = − −⎜ ⎟ ⎝ ⎠ BA BCθ θ
- 59. 2. General loading with support settlement 1a 2a ;A B B C δ δ δ δ > < A B C x x1x 2x M l M l δ δ1 1 1 1 1 1 1 1 16 3 A B A B BA M l M l a x EI EI EI l l δ δ θ − = + + + 2 2 2 2 2 2 2 2 23 6 B C C B BC M l M l a x EI EI EI l l δ δ θ − = + + + BA BCθ θ= − ( ) ( )1 21 2 1 1 2 2 1 21 2 1 1 2 2 1 2 6 66 6 2 A B C BA C B l lM l M l a x a x M EI EIEI EI EI l EI l l l δ δ δ δ− −⎛ ⎞ +⇒ + + = − − − −⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 1 21 2 1 1 2 2 1 2l l l l⎝ ⎠
- 60. Example 1: 20kN m 20kN m 4m A B C 4m EI is constant 3 3 l lM l M l l l⎛ ⎞ 1 2l l= 3 3 1 21 2 1 1 2 2 1 21 2 1 2 2 4 4 A C B l lM l M l w l w l M EI EIEI EI EI EI ⎛ ⎞ ++ + = − −⎜ ⎟ ⎝ ⎠ 1 2w w=2 2 4 4 B wl M = − 0A CM M= = 0A CM M 2 8 B wl M∴ = − 2 20 4 40 8 kNm × = − = − Dept. of CE, GCE Kannur Dr.RajeshKN 8 8
- 61. Example 2: W l A B Cl l D EI is constant EI EI Span ABC 0M = EI is constant 3 3 l lM l M l l l⎛ ⎞ 1 2l l= 1 2EI EI= 0AM = 3 3 1 21 2 1 1 2 2 1 21 2 1 2 2 4 4 A C B l lM l M l w l w l M EI EIEI EI EI EI ⎛ ⎞ ++ + = − −⎜ ⎟ ⎝ ⎠ 4 0A B CM M M⇒ + + = 4 0M M⇒ + = ( )1 Dept. of CE, GCE Kannur Dr.RajeshKN 4 0B CM M⇒ + = ( )1
- 62. Wl 1 Wl a l= l A B Cl l D Span BCD 4 Wl EI 2 2 . . 2 4 2 a l l x = = 0M ( ) 1 1 2 26 6 4 a x a x l M M M+ + 2 20DM = ( )2 1 4 6 lWl l M M l ⎛ ⎞ ⎜ ⎟ ( ) 1 1 2 2 4B C Dl M M M l l + + = − − ( )2 4 6 . . 22 4 B Cl M M l ⎛ ⎞+ = − ⎜ ⎟ ⎝ ⎠ 3 4 Wl M M ( )24 8 B CM M+ = − ( )2 Wl− Wl M Dept. of CE, GCE Kannur Dr.RajeshKN 10 C Wl M = 40 BM =( ) ( )&1 2
- 63. 40 Wl 5 Wl A B C 40 10 Wl− D 5 10 Bending moment diagram Dept. of CE, GCE Kannur Dr.RajeshKN 63
- 64. Example 3: 20kN m 20kN mp 6mA B C 8mEI is constant 20kN m 20kN m 6mA B C 8mImaginary span A’Aspan A A Span A’AB 1 2 0 20 w w kN m = = ' 0AM = 1 2EI EI= Span A AB 3 3 1 2' 1 2 1 1 2 2 2A B A l lM l M l w l w l M EI EI ⎛ ⎞ ++ + = − −⎜ ⎟ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 1 21 2 1 24 4 A EI EIEI EI EI EI⎜ ⎟ ⎝ ⎠
- 65. 3 20 6×20 6 2 6 6 4 A BM M × × + × = − 2 180M M+ ( )12 180A BM M+ = − Span ABC ( )1 1 2 20w w kN m= = 3 3 1 21 2 1 1 2 2 1 21 2 1 2 2 4 4 A C B l lM l M l w l w l M EI EIEI EI EI EI ⎛ ⎞ ++ + = − −⎜ ⎟ ⎝ ⎠ 0CM = 1 21 2 1 2⎝ ⎠ 3 3 20 6 20 8 6 2 14 4 4 A BM M × × + × = − − 4 4 3 14 1820A BM M+ = − ( )2 ( ) ( )&1 2 28AM kNm= − 124BM kNm= − Dept. of CE, GCE Kannur Dr.RajeshKN
- 66. Example 4 (Support settlement): 2 26320EI kN 87δ δEI is constant 20kN m 20kN m20kN m 2 26320EI kNm= 87B C mmδ δ= =EI is constant 6mA D C 3m B3m EI EI EI= =0M Span ABC 1 2EI EI EI= =0AM = ( ) ( )3 3 1 21 2 1 1 2 2 66 C BA BA C l lM l M l w l w l δ δδ δ −−⎛ ⎞ ( ) ( )1 21 2 1 1 2 2 1 21 2 1 2 1 2 66 2 4 4 C BA BA C B l lM l M l w l w l M EI EIEI EI EI EI l l δ δδ δ⎛ ⎞ ++ + = − − − −⎜ ⎟ ⎝ ⎠ ( ) ( )3 3 1 20 3 20 6 6 6 00.087 18 6 26320 4 26320 4 26320 3 6 B CM M × × ×− + = − − − − × × Dept. of CE, GCE Kannur Dr.RajeshKN 3 560.78B CM M+ = ( )1
- 67. Span BCD 0DM = ( ) ( )3 3 1 21 2 1 1 2 2 6 6 2 4 4 B C D CB D C l lM l M l w l w l M EI EIEI EI EI EI l l δ δ δ δ− −⎛ ⎞ ++ + = − − − −⎜ ⎟ ⎝ ⎠ D 1 21 2 1 2 1 24 4EI EIEI EI EI EI l l⎜ ⎟ ⎝ ⎠ ( ) ( )3 3 1 20 6 20 3 6 0 6 0.087 6 18M M × × × − +( )6 18 26320 4 26320 4 26320 6 3 B CM M+ = − − − − × × 3 560.78B CM M+ = ( )2 140.195BM kNm= 140.195CM kNm= Alternatively, from symmetry, B CM M= Dept. of CE, GCE Kannur Dr.RajeshKN 67 3 560.78B CM M+ = 4 560.78 140.195B BM M kNm⇒ = ⇒ =
- 68. Examples 5 (Fixed Beam)xa p es 5 ( xed ea ) P A B C l A P A B C A′ B′ l A A′ B Imaginary span A’A Imaginary span BB’span A A span BB 1 21 2 1 1 2 26 6 2A C l lM l M l a x a x M ⎛ ⎞ ++ +⎜ ⎟ Dept. of CE, GCE Kannur Dr.RajeshKN 68 1 21 2 1 1 2 2 1 21 2 1 1 2 2 2A C BM EI EIEI EI EI l EI l ++ + = − −⎜ ⎟ ⎝ ⎠
- 69. Span A’AB 1 l ' 0AM = 4 Pl EI 2 1 . . 2 4 Pl a l l = 1 2EI EI= 4EI 2 2 l x =A B 6Pl6 2 16 A B Pl M M⇒ + = − S ABB’ ( )1 ' 0BM = 1 2EI EI= Span ABB’ 2 1 1, 8 2 Pl l a x= = 8 2 6 2 16 A B Pl M M⇒ + = − ( )2 16 ( ) ( )&1 2 A B Pl M M= = − Dept. of CE, GCE Kannur Dr.RajeshKN 69 ( ) ( )&1 2 8 A BM M
- 70. Examples 6 (Fixed Beam) w xa p es 6 ( xed ea ) lA B CC Dept. of CE, GCE Kannur Dr.RajeshKN 70
- 71. ENERGY PRINCIPLES BASED ON DISPLACEMENT FIELDENERGY PRINCIPLES BASED ON DISPLACEMENT FIELD Principle of Minimum Total Potential Castigliano’s Theorem (Part I)Total Potential Energy (PMTPE) Theorem (Part I) alternative forms ofalternative forms of Principle of Stationary TotalPrinciple of Stationary Total Potential Energy (PSTPE) Dept. of CE, GCE Kannur Dr.RajeshKN 71
- 72. Principle of Stationary Total Potential Energy (PSTPE)p y gy ( ) When the displacement field in a loaded elastic structure is given a ll d bit t b ti i t i i tibilit dsmall and arbitrary perturbation, maintaining compatibility and without disturbing the associated force field, then the first variation of the total potential energy is equal to zero, if the forces are in a state of static equilibrium. Dept. of CE, GCE Kannur Dr.RajeshKN 72
- 73. Alternative form of Principle of Stationary Total Potential EnergyAlternative form of Principle of Stationary Total Potential Energy (PSTPE) The total potential energy π in a loaded elastic structure expressed as a function of n independent displacements D1, D2, … Dn in a compatible displacement field must be rendered stationary with thecompatible displacement field must be rendered stationary, with the partial derivative of π with respect to every Dj being equal to zero, if the associated force field is to be in a state of static equilibrium. Dept. of CE, GCE Kannur Dr.RajeshKN
- 74. Principle of Minimum Total Potential Energy (PMTPE)Principle of Minimum Total Potential Energy (PMTPE) When the displacement field in a loaded linear elastic structure is given a small and arbitrary perturbation, maintaining compatibility and without disturbing the associated force field, then the first variation of the total potential energy is equal to zero, if the forces are in ap gy q , state of static equilibrium. Castigliano’s Theorem (Part I) If the strain energy, U, in an elastic structure, subject to a system of external forces in static equilibrium, can be expressed as a functionq , p of n independent displacements D1, D2, … Dn satisfying compatibility, then the partial derivative of U with respect to every Dj will be equal to the value of the conjugate force, Fj. Dept. of CE, GCE Kannur Dr.RajeshKN 74 Dj will be equal to the value of the conjugate force, Fj.
- 75. ENERGY PRINCIPLES BASED ON FORCE FIELDENERGY PRINCIPLES BASED ON FORCE FIELD Principle of Minimum Total Complementary Potential Castigliano’s Theorem (Part II) Complementary Potential Energy (PMTCPE) Theorem of Least Work alternative forms of Principle of StationaryPrinciple of Stationary Complementary Total Potential Energy (PSCTPE) Dept. of CE, GCE Kannur Dr.RajeshKN
- 76. Principle of Stationary Total Complementary Potential EnergyPrinciple of Stationary Total Complementary Potential Energy (PSTCPE) When the force field in a loaded elastic structure is given a small and arbitrary perturbation, maintaining equilibrium compatibility and without disturbing the associated displacement field, then the firstg p , variation of the total complementary potential energy is equal to zero, if the displacements satisfy compatibility. Dept. of CE, GCE Kannur Dr.RajeshKN 76
- 77. Alternative form of Principle of Stationary Total Complementary Potential Energy (PSTCPE)Potential Energy (PSTCPE) The total complementary potential energy π* in a loaded elasticp y p gy structure expressed as a function of n independent forces F1, F2, … Fn in a statically admissible force field must be rendered stationary, with the partial derivative of π* with respect to every Fj being equalw t t e pa t a de vat ve o w t espect to eve y j be g equa to zero, if the associated displacement field is to satisfy compatibility. Dept. of CE, GCE Kannur Dr.RajeshKN
- 78. Principle of Minimum Total Complementary Potential EnergyPrinciple of Minimum Total Complementary Potential Energy (PMTCPE) When the force field in a loaded linear elastic structure is given a small and arbitrary perturbation, maintaining equilibrium compatibility and without disturbing the associated displacement field, then theg p , first variation of the total complementary potential energy is equal to zero, if the displacement satisfy compatibility. Dept. of CE, GCE Kannur Dr.RajeshKN 78
- 79. Castigliano’s Theorem (Part II)Castigliano s Theorem (Part II) If the complementary strain energy, U*, in an elastic structure, with a kinematically admissible displacement field, is expressed as a function of n independent external forces F1, F2, … Fn satisfying equilibrium, then the partial derivative of U* with respect to every Fjq , p p y j will be equal to the value of the conjugate displacement, Dj. If the behaviour is linear elastic, U* can be replaced by U. Thus, Castigliano’s Theorem (Part II) can otherwise be stated as: “If U is the total strain energy in a linear elastic structure due to application of external forces F1 F2 F3 F at points A1 A2 A3 Aapplication of external forces F1, F2, F3, … Fn at points A1, A2, A3,…, An respectively in the directions AX1, AX2, AX3,…, AXn then the displacements at points A1, A2, A3,…, An respectively in the directions AX AX AX AX ∂U/∂F ∂U/∂F ∂U/∂F ∂U/∂F Dept. of CE, GCE Kannur Dr.RajeshKN 79 AX1, AX2, AX3,…, AXn are ∂U/∂F1, ∂U/∂F2, ∂U/∂F3,…, ∂U/∂Fn respectively.”
- 80. Proof for Castigliano’s theorem (Part II)Proof for Castigliano s theorem (Part II) Let x1, x2, x3,… xn be deflections at points A1, A2, A3,… An due to F F F Fdue to F1, F2, F3,… Fn Total strain energy, 1 1 2 2 3 3 1 1 1 1 2 2 2 2 n nU F x F x F x F x= + + + +… (1) Let the load F1 alone be increased by 1Fδ Let 1 2 3, , , nx x x xδ δ δ δ… be the additional deflections at points A1, A2, A3,… An I i t iIncrease in strain energy, 1 1 1 2 2 3 3 1 2 n nU F F x F x F x F xδ δ δ δ δ δ⎛ ⎞ = + + + + +⎜ ⎟ ⎝ ⎠ … 2⎝ ⎠ 1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +… , neglecting small quantities. Dept. of CE, GCE Kannur Dr.RajeshKN 80 (2)
- 81. ( )1 1 2 3, , , nF F F F Fδ+ …Let are acting on the original structure ( )( ) ( ) ( ) ( ) 1 1 1 1 U U F F F F Fδ δ δ δ δ δ Total strain energy, ( )( ) ( ) ( ) ( )1 1 1 1 2 2 2 3 3 3 1 1 1 1 2 2 2 2 n n nU+ U F F x x F x x F x x F x xδ δ δ δ δ δ= + + + + + + + + +… (3) 1 1 1 1 U F F F F+ + + +(1) 1 1 2 2 3 3 2 2 2 2 n nU F x F x F x F x= + + + +…(1) 1 1 1 1 U F F F Fδ δ δ δ δ⎛ ⎞ ⎜ ⎟(3) (1) 1 1 1 1 2 2 2 2 2 2 n nU x F F x F x F xδ δ δ δ δ⎛ ⎞ = + + + +⎜ ⎟ ⎝ ⎠ …(3)-(1) ( )2 U x F F x F x F xδ δ δ δ δ= + + + + (4) 1 1 2 2 3 3 n nU F x F x F x F xδ δ δ δ δ= + + + +…(2) ( )1 1 1 1 2 22 n nU x F F x F x F xδ δ δ δ δ= + + + +… (4) Dept. of CE, GCE Kannur Dr.RajeshKN 81 1 1U x Fδ δ=(4)-(2)
- 82. Uδ U∂ 1 1 U x F δ δ = 1 1 1 0, U When F x F δ ∂ → = ∂ 2 2 2 3 , , n n U U U Similarly, x x x F F F ∂ ∂ ∂ = = = ∂ ∂ ∂ … For example, in the case of bending, 2 2 M U dx EI ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ∫⎝ ⎠ U M M dx F EI F δ ∂ ∂⎛ ⎞ ∴ = = ⎜ ⎟ ∂ ∂⎝ ⎠ ∫F EI F∂ ∂⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 82
- 83. Example 1: Using Castigliano’s Theorem, analyse the continuous w kN/m p g g , y beam shown in figure. A B C / LLL w kN/m A B C / LL RB L 2 BR wL − 2 BR wL − Let RB be the redundant. Dept. of CE, GCE Kannur Dr.RajeshKN 83 B
- 84. 2 wxR⎛ ⎞ 22 B x wxR M xwL ⎛ ⎞= −−⎜ ⎟ ⎝ ⎠ From A to B, 0Bδ = 0 B B U M M dx R EI R ∂ ∂⎛ ⎞ ⇒ = =⎜ ⎟ ∂ ∂⎝ ⎠ ∫ 2 M x R ∂ − = ∂ F th ti AC2BR∂ 2 2M M∂ ⎛ ⎞⎛ ⎞ For the entire span AC 2 2 0 0 22 2 B B M M xR x wx dx dxwLx EI R EI ∂ −⎛ ⎞⎛ ⎞ ∴ = ⇒ =− −⎜ ⎟ ⎜ ⎟ ∂⎝ ⎠ ⎝ ⎠ ∫ ∫ 5 B wL R⇒ = Dept. of CE, GCE Kannur Dr.RajeshKN 4 BR⇒
- 85. Example 2: Using Castigliano’s Theorem analyse the frame shownExample 2: Using Castigliano s Theorem, analyse the frame shown in figure. 3 kN/3 kN/m 2mB C 3 kN/m 2mB C 2m 1m D 2m 1m D A D A VD HD VD V HA VA Let HA be the redundant. Dept. of CE, GCE Kannur Dr.RajeshKN Let HA be the redundant.
- 86. 3 kN/m 2m 1 B C From A to B, x AM H x= − 2m 1m D AH A M x H ∂ = − ∂ A 3 AH V = − AH From B to C, 2 3⎛ ⎞HA 3 2 DV 3 AH V + 2 3 23 22 A x A xH M x H⎛ ⎞= − −+⎜ ⎟ ⎝ ⎠ 3 2 A AV = + 2 2A M x H ∂ = − ∂ F D t C M H x= M x ∂ Dept. of CE, GCE Kannur Dr.RajeshKN From D to C, x AM H x= − A x H = − ∂
- 87. 0 M M dx ∂⎛ ⎞ ∴ ⎜ ⎟∫ 0 A dx EI H ∴ =⎜ ⎟ ∂⎝ ⎠ ∫ 2 2 1 ⎧ ⎫⎛ ⎞⎛ ⎞ ( )( ) ( )( ) 2 2 12 0 0 0 1 3 023 2 22 2 A A AA xH x x dx dx dxH x H xx Hx x EI ⎧ ⎫⎛ ⎞⎛ ⎞⇒ + + =− −−+ − −− −⎨ ⎬⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎩ ⎭ ∫ ∫ ∫ 2 2 13 3 3 2 4 2 3 2 33 03 4A A A A A A H x x H x H x x H x H x x H x x ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ + + =+ − − − − + +⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦0 0 03 2 12 2 16 2 3 A⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 8 8 8 4A AA A H HH H⎡ ⎤8 8 8 4 03 12 2 8 8 3 32 12 2 A AA A A A H HH H H H ⎡ ⎤+ + =+ − − − − + +⎢ ⎥⎣ ⎦ 3 3.19 2 A A H V kN= + =0.39AH kN= H Dept. of CE, GCE Kannur Dr.RajeshKN 3 2.8 2 A D H V kN= − =
- 88. Example 4: Using Castigliano’s Theorem, analyse the truss shownp g g , y in figure. AE is constant. CD CD 10kN CD T10kN CD 3m 3m T 4m A B 4m A B T m . . 0 P L P T AE ∂ = ∂ ∑ 6.25T kN= Dept. of CE, GCE Kannur Dr.RajeshKN
- 89. Example 4: Do the above problem using the method of consistentExample 4: Do the above problem using the method of consistent deformation. P kL′ 10kN 2 P kL AET k L ′ = − ∑ ∑ CD 3m k L AE ∑ A B 4m B P P kT′= + Dept. of CE, GCE Kannur Dr.RajeshKN
- 90. Note 1: If there are two internal redundants in a truss in method ofNote 1: If there are two internal redundants in a truss, in method of consistent deformation, 2 P k L k L k k L′ 2 1 1 1 2 1 2 0 P k L k L k k L T T AE AE AE ′ + + =∑ ∑ ∑ 2 2 1 2 2 1 2 0 P k L k k L k L T T AE AE AE ′ + + =∑ ∑ ∑AE AE AE Note 2: If there are both internal redundants external redundants in a truss, in method of consistent deformation, 2 1 1 1 0BP k L k L k k L T V ′ + + =∑ ∑ ∑1 0BT V AE AE AE + + =∑ ∑ ∑ 2 P k L k k L k L′ Dept. of CE, GCE Kannur Dr.RajeshKN 2 1 1 0B B B B P k L k k L k L T V AE AE AE ′ + + =∑ ∑ ∑
- 91. Example 3: Using Castigliano’s Theorem, analyse the pin-jointed h i fi 3m 4m truss shown in figure. A B C 3m 4m 400A = A 3m 500A = 400A = D 80kN Internally indeterminate to degree 1. Take force in BD as redundant. y g 0 P L P ∂ ∑ Dept. of CE, GCE Kannur Dr.RajeshKN 91 Assume force in BD is T . . 0P T AE = ∂ ∑
- 92. cos45 sin 53.13AD CDF F=cos45 sin 53.13AD CDF F cos45 cos53.13 80AD CDF F T+ + = T 57.14 0.714T− 64.64 0.808T− sin 53.13 cos53.13 80CD CDF F T+ + = 80kN 57.14 0.714CDF T= − i 53 13 ( ) sin 53.13 57.14 0.714 64.64 0.808 cos45 ADF T T= − = − . . 0 P L P T AE ∂ = ∂ ∑ ( ) ( ) ( ) ( ) 4.243 57.14 0.714 5 64.64 0.8083 0.714 3 0.808 0 500 400 400 T TT E E E − −⎛ ⎞ ⎛ ⎞⎛ ⎞ × − + × + × − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ Dept. of CE, GCE Kannur Dr.RajeshKN 9249.09T kN∴ =
- 93. SummarySummary Statically and kinematically indeterminate structures • Degree of static indeterminacy, Degree of kinematic indeterminacy, Force and displacement method of analysis Force method of analysis • Method of consistent deformation-Analysis of fixed and continuous beams • Clapeyron’s theorem of three moments-Analysis of fixed and continuous beams • Principle of minimum strain energy-Castigliano’s second theorem- Analysis of beams, plane trusses and plane frames. Dept. of CE, GCE Kannur Dr.RajeshKN 93

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