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02 influenceline inde_3

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  • 1. 1 ! Comparison Between Indeterminate and Determinate ! Influence line for Statically Indeterminate Beams ! Qualitative Influence Lines for Frames INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS
  • 2. 2 A B CED RA A B CED RA Indeterminate Determinate Comparison between Indeterminate and Determinate 11
  • 3. 3 A B CED RA A B CED RA A B CED ME A B CED ME A B CED VD A B CED VD 1 1 1 Indeterminate Determinate 11 1
  • 4. 4 f1j fjj ∆´1 = f1j + 1 1 2 3j4 Redundant R1 applied 1 1 = f11 fj1 ×R1×R1 f11 Influence Lines for Reaction Compatibility equation: 011111 =∆=+ Rff j ) 1 ( 11 11 f fR j−= )( 11 1 1 f f R j =
  • 5. 5 1 1 2 3j4 1 =+ ×R2 Redundant R2 applied fjj f2j 1 fj2 f22 Compatibility equation. 0' 22222 =∆=+∆ Rf 022222 =∆=+ Rff j ) 1 ( 22 22 f fR j−= )( 22 2 2 f f R j =
  • 6. 6 1 2 3j4 fj4 1 1 Influence Lines for Shear 4 44 ) 1 ( jE f f V = f44 1 1
  • 7. 7 4 44 ) 1 ( jE fM α = 1 2 3j4 Influence Lines for Bending Moment 1 1 α44 fj4 1 1
  • 8. 8 R3 )( 33 3 3 f f R j = R2 )( 22 2 2 f f R j = R1 )( 11 1 1 f f R j = 1 2 3j4 1 1 • Influence line of Reaction 1 Using Equilibrium Condition for Shear and Bending Moment 1 11 11 = f f 11 41 f f 11 1 f f j 1 22 22 = f f 22 2 f f j 22 42 f f 1 33 33 = f f 33 3 f f j 33 43 f f
  • 9. 9 1 2 3j4 4 V4 V4 = R1 R1 V4 M4 1 • Unit load to the right of 4 • Influence line of Shear V4 = R1 V4 = R1 - 1 1 R2 R3R1 1 1 1 R1 1x V4 M4 1 • Unit load to the left of 4 V4 = R1 - 1 01;0 41 =−−=Σ↑+ VRFy 1 R1 0;0 41 =−=Σ↑+ VRFy
  • 10. 10 1 2 3j4 • Influence line of Bending moment M4 = - l + x + l R1 M4 - 1 (l-x) - l R1 = 0+ Σ M4 = 0: R1 1x V4 M4 l 1 • Unit load to the left of 4 M4 = l R1 R1 V4 M4 • Unit load to right of l 4 1 1 R2 R3R1 1 l1 l2 l M4 4 1 R1 1 1 M4 = - l + x + l R1 M4 - l R1 = 0+ Σ M4 = 0: M4 = lR1
  • 11. 11 Influence Line of VI Maximum positive shear Maximum negative shear Qualitative Influence Lines for Frames I 1 1
  • 12. 12 Influence Line of MI Maximum positive moment Maximum negative moment I 11
  • 13. 13 A D G 15 m15 m Ay GyDy MA Dy 1.0 Ay 1.0 Gy 1.0 Influence Line for MOF
  • 14. 14 MH MA 1 1 A H G 15 m15 m D
  • 15. 15 RA 1.0 RB 1.0 MB 1 MG 1 A E G B C D
  • 16. 16 VG 1 VF 1 VH 1 A E G B C D
  • 17. 17 Example 1 Draw the influence line for - the vertical reaction at A and B - shear at C - bending moment at A and C EI is constant . Plot numerical values every 2 m. A BC D 2 m 2 m 2 m
  • 18. 18 BBf BBf BBf 1= BBf • Influence line of RB 1 A BC D 2 m 2 m 2 m ABf CBf DBf BBf
  • 19. 19 Real Beam A BC D 2 m 2 m 2 m Conjugate Beam • Find fxB by conjugate beam 11 6 kN•m x =−+ EI x EIEI x 1872 6 3 EI 6 EI 18 EI 18 EI 72 3 x 3 2xV´x M´x x EI 72 EI 18 EI x 2 2 EI x
  • 20. 20 x (m) 0 2 4 6 Point B D C A x fxB / fBB 1 0.518 0.148 0 A BC D 2 m 2 m 2 m 1 fBB fxB 0 72/EI 37.33/EI10.67/EI EI x EIEI x Mf xxB 1872 6 ' 3 −+== EI 72 EI 33.37 EI 67.10 0 /72 = 0.518 /72 = 0.148 1 37.33 10.67 0 Influence line of RB 1 72 72 == BB BB f f fxB
  • 21. 21 1 kN Influence line of RA A BC D 2 m 2 m 2 m AA CA f f AA DA f f 1= AA AA f f
  • 22. 22 Conjugate Beam xV´x M´x EI B y 18 ' = EI x 2 2 EI x 3 x 3 2x A BC D 2 m 2 m 2 m Real Beam 1 kN • Find fxA by conjugate beam 1 kN 6 kN•m EI M A 72 ' = x EI B y 18 ' = =− EI x EI x 6 18 3 EI 18 EI 6
  • 23. 23 Point B D C A fxA / fAA 0 0.482 0.852 1.0 EI x EI x Mf xxA 6 18 ' 3 −== x (m) 0 2 4 6 x A BC D 2 m 2 m 2 m 1 kN fAA fxA 72/EI 61.33 /EI 34.67 /EI fxA 0 EI 67.34 EI 33.61 EI 72 34.67 61.33 72 1 kN Influence line of RA /72 = 1.0 /72=0.852 /72 = 0.482 1= AA AA f f
  • 24. 24 AB BA y RR RR F −= =−+ =Σ↑+ 1 01 ;0 A BC D 2 m 2 m 2 m 1x RA MA RB 0.148 .518 1 RB 1 RB 0.482 0.852 1.0 1 kN RA Alternate Method: Use equilibrium conditions for the influence line of RA RA = 1- RB
  • 25. 25 VC 1 RB 0.148 0.518 1 A BC D 2 m 2 m 2 mRA MA RB 0.852 0.482 -0.148 1 x 1 Using equilibrium conditions for the influence line of VC VC = 1 - RB • Unit load to the left of C RB 1 x VC MC 01 ;0 =+−+ =Σ↑+ BC y RV F RB VC MC VC = - RB 0 ;0 =++ =Σ↑+ CB y VR F • Unit load to the left of C
  • 26. 26 BA BA A RxM RxM M 66 06)6(1 ;0 ++−= =+−−− =Σ A BC D 2 m 2 m 2 mRA MA RB 1 x 1 MA 1 RB 0.148 0.518 1 -1.112 -0.892 Using equilibrium conditions for the influence line of MA +
  • 27. 27 MC C 1 RB 0.148 0.518 1 A BC D 2 m 2 m 2 m 1 x RA MA RB 0.074 1 Using equilibrium conditions for the influence line of MC 0.592 RB 1 x VC MC • Unit load to the left of C RB VC MC 4 m MC = 4RB + 04 ;0 =+− =Σ BC C RM M • Unit load to the left of C 4 m MC = -4 + x + 4RB 04)4(1 ;0 =+−−− =Σ BC C RxM M+
  • 28. 28 Example 2 Draw the influence line and plot numerical values every 2 m for - the vertical reaction at supports A, B and C - Shear at G and E - Bending moment at G and E EI is constant. A B CD E F 2@2=4 m 4@2 = 8 m G
  • 29. 29 Influence line of RA 1 A B CD E F 2@2=4 m 4@2 = 8 m G 1= AA AA f f AA DA f f AA EA f f AA FA f f
  • 30. 30 4 /EI 1 A B CD E F 4 m 6 m2 m • Find fxA by conjugate beam Real beam 0.51.5 0 18.67/EI 64/EI 4/EI 4/EI Conjugate beam EI 33.5 EI 67.10 0 EI 67.10 EI Mf AAA 64 ' ==
  • 31. 31 x1x2 =− EI x EI x 1 3 1 33.5 12 EIEIEI x 67.1864 6 3 2 −+= V´ x1 M´ x1 x1EI x 2 1 EI 33.5 EI x 4 2 1 3 2 1x 3 1x Conjugate beam 4/EI 4 m 8 m 64/EI EI 33.5 EI 67.18 M´ x2 V´ x2 x2 EI x2 EI x 2 2 2 3 2x 3 2 2x EI 67.18 EI 64
  • 32. 32 CtoBfor EI x EI x Mf xxA , 33.5 12 ' 3 1 −== EI 28 EI 64 BtoAfor EIEI x EI x Mf xxA , 6467.18 6 ' 2 3 2 2 +−== fxA 0 0 EI 16 − EI 10 − EI 14 − x (m) 0 2 4 6 Point C F E D B A 8 12 G 10 fxA / fAA 0 -0.1562 -0.25 -0.2188 0 1 0.4375 x1x2 1 fAA fxA A B CD E FG 4 m 6 m2 m EI 10− EI 16− EI 14− EI 28 EI 64 1 Influence line of RA -0.219 -0.25 -0.156 0.438 1= AA AA f f
  • 33. 33 A B CD E FG 4 m 6 m2 m Using equilibrium conditions for the influence line of RB RB RCRA x1 RA 1 -0.219 -0.25 -0.156 1 0.438 RB 1 0.485 0.8751.078 10.59 RB 0 0.485 0.875 1.078 1 0.5939 0 x (m) 0 2 4 6 Point C F E D B A 8 12 G 10 RA 0 -0.1562 -0.25 -0.2188 0 0.4375 1 AB AB C R x R RxR M 8 12 8 0128 ;0 −= =−+− =Σ+
  • 34. 34 A B CD E FG 4 m 6 m2 m RA 1 -0.219 -0.25 -0.156 1 0.438 RB RCRA x1 x (m) 0 2 4 6 Point C F E D B A 8 12 G 10 RA 0 -0.1562 -0.25 -0.2188 0 0.4375 1 RC 1 0.6719 0.375 0.1406 0 -0.0312 0RC 1 1 0.672 0.3750.141 -0.0312 Using equilibrium conditions for the influence line of RC 1 8 5.0 08)8(14 ;0 +−= =−−− =Σ+ x RR RxR M AC CA B
  • 35. 35 A B CD E FG 4 m 6 m2 m • Check ΣFy = 0 RB RCRA x1 RA 1 -0.219 -0.25 -0.156 1 0.438 RB 1 0.49 0.875 1.081 0.59 RC 1 1 0.672 0.3750.141 -0.0312 1 ;0 =++ =Σ↑+ CBA y RRR F RC 1 0.6719 0.375 0.1406 0 -0.0312 0 Point C F E D B A G RA 0 -0.1562 -0.25 -0.2188 0 0.4375 1 RB 0 0.485 0.875 1.078 1 0.5939 0 ΣR 1 1 1 1 1 1 1
  • 36. 36 VG RA 1 -0.219 -0.25 -0.156 1 0.438 1 x VG = RA RA A VG MG • Unit load to the right of G 0.438 -0.219 -0.25 -0.156-0.562 1 A B CD E FG 4 m 6 m2 m Using equilibrium conditions for the influence line of VG VG = RA - 1 RA 1x A VG MG • Unit load to the left of G 01;0 =−−=Σ↑+ GAy VRF
  • 37. 37 VE RC 1 1 0.672 0.3750.141 -0.0312 1 x 0.625 0.328 0.0312 -0.141 -0.375 1 A B CD E FG 4 m 6 m2 m Using equilibrium conditions for the influence line of VE RC VE ME • Unit load to the left of E VE = - RC 0;0 =+=Σ↑+ ECy VRF VE = 1 - RC • Unit load to the right of E RC 1 x VE ME 01;0 =+−=Σ↑+ CEy RVF
  • 38. 38 MG RA 1 -0.219 -0.25 -0.156 1 0.438 1 x -0.438 -0.5 -0.312 1 A B CD E FG 4 m 6 m2 m Using equilibrium conditions for the influence line of MG 0.876 MG = -2 + x + 2RA • Unit load to the left of G RA 1x A VG MG 2 m 02)2(1 ;0 =−−+ =Σ AG G RxM M+ MG = 2RA • Unit load to the right of G RA A VG MG 2 m 02 ;0 =− =Σ AG G RM M+
  • 39. 39 RC 1 1 0.672 0.375 0.141 -0.0312 ME 1 x 1.5 0.688 1 4 m 6 m2 m A B CD E FG Using equilibrium conditions for the influence line of ME -0.125 0.564 RC VE ME 4 m • Unit load to the left of E ME = 4RC + ΣME = 0; ME = - 4 + x+ 4RC • Unit load to the right of E RC 1 x VE ME 4 m 04)4(1 ;0 =+−−− =Σ+ CE E RxM M
  • 40. 40 Example 3 For the beam shown (a) Draw quantitative influence lines for the reaction at supports A and B, and bending moment at B. (b) Determine all the reactions at supports, and also draw its quantitative shear, bending moment diagrams, and qualitative deflected curve for - Only 10 kN downward at 6 m from A - Both 10 kN downward at 6 m from A and 20 kN downward at 4 m from A 10 kN A C B 4 m 2EI 3EI 2 m 2 m 20 kN
  • 41. 41 1 fCA fAA fEA fDA 1 /fAA /fAA /fAA /fAA fCA fAA fEA fDA A C B 4 m 2EI 3EI 2 m 2 m Influence line of RA
  • 42. 42Conjugate Beam A C B 4 m 2EI 3EI 2 m 2 m 1 Real Beam • Find fxA by conjugate beam 1 kN 8 kN•m x (m)V (kN) 1 1 + x (m)M (kN•m) 8 4 + 2EI fAA = M´A = 60.44/EI 60.44/EI 12/EI 1.33EI EI 67.2
  • 43. 43 EI 67.2 2EI 60.44/EI Conjugate Beam12/EI 1.33EI 37.11/EI 17.77/EI 4.88/EI 60.44/EI 0 RA = fxA/fAA 0.614 0.294 0.081 1 0 fxA A C B • Quantitative influence line of RA
  • 44. 44 A B 4 m 2 m 2 m Using equilibrium conditions for the influence line of RB and MB RB = 1 - RA 1 0.386 0.706 0.919 0 MB = 8RA - (8-x)(1) 0 -1.352 -1.088 -1.648 0 RA RB MB RA 0.614 0.294 0.081 1 0 1x
  • 45. 45 A B 4 m 2 m 2 m Using equilibrium conditions for the influence line of VB RA 0.614 0.294 0.081 1 0 -1 -0.386 -0.706 -0.919 0 VB = RA -1 RA 1x VB = RA - 1
  • 46. 46 A B 4 m 2 m 2 m C Using equilibrium conditions for the influence line of VC and MC RA RA 0.614 0.294 0.081 1 0 1x RB MB MC = 4RAMC = 4RA - (4-x)(1) MC VC = RAVC = RA - 1 VC 1 -0.386 -0.706 0.3240.456 1.176 0.294 0.081 0
  • 47. 47 A B 4 m 2 m 2 m 10 kN RA 0.614 0.294 0.081 1 0 10(0.081)=0.81 kN MA (kN•m) + - -13.53 4.86 V (kN) -9.19 0.81 0.81 - The quantitative shear and bending moment diagram and qualitative deflected curve RB=9.19 kN MB= 13.53 kN•m
  • 48. 48 A B 4 m 2 m 2 m 10 kN The quantitative shear and bending moment diagram and qualitative deflected curve 20 kN 20(.294) +1(0.081) = 6.69 kN V (kN) -23.31 6.69 6.69 - -13.31 -23.31 MA (kN•m) + - -46.48 26.76 0.14 RA 0.614 0.294 0.081 1 0 RB=23.31 kN MB=46.48 kN•m
  • 49. 49 Example 1 Draw the influence line for - the vertical reaction at B A BC D 2 m 2 m 2 m APPENDIX •Muller-Breslau for the influence line of reaction, shear and moment •Influence lines for MDOF beams
  • 50. 50 1 kN Influence line of RA A BC D 2 m 2 m 2 m 1= AA AA f f AA CA f f AA DA f f
  • 51. 51 Conjugate Beam xV´x M´x EI B y 18 ' = EI x 2 2 EI x 3 x 3 2x A BC D 2 m 2 m 2 m Real Beam 1 kN • Find fxA by conjugate beam 1 kN 6 kN•m EI M A 72 ' = x EI B y 18 ' = 6 /EI EI 18 =− EI x EI x 6 18 3
  • 52. 52 Point B D C A fxA / fAA 0 0.482 0.852 1.0 x (m) 0 2 4 6 x A BC D 2 m 2 m 2 m 1 kN fAA fxA 72/EI 61.33 /EI 34.67 /EI EI x EI x Mf xxA 6 18 ' 3 −== fxA 0 EI 67.34 EI 33.61 EI 72 34.67 61.33 72 1 kN Influence line of RA /72 = 1.0 /72=0.852 /72 = 0.482 1= AA AA f f
  • 53. 53 • Influence line of RB 1 A BC D 2 m 2 m 2 m 1= BB BB f f BB AB f f BB CB f f BB DB f f
  • 54. 54 Real Beam A BC D 2 m 2 m 2 m Conjugate Beam • Find fxB by conjugate beam 11 6 kN•m x =−+ EI x EIEI x 1872 6 3 EI 6 EI 18 EI 18 EI 72 3 x 3 2xV´x M´x x EI 72 EI 18 EI x 2 2 EI x
  • 55. 55 x (m) 0 2 4 6 Point B D C A x fxB / fBB 1 0.518 0.148 0 A BC D 2 m 2 m 2 m 1 fBB fxB 0 72/EI 37.33/EI10.67/EI /72 = 0.518 /72 = 0.148 /72 = 1 1 37.33 10.67 0 fBB Influence line of RB fBB = 1 72 EI x EIEI x Mf xxB 1872 6 ' 3 −+== fxB 0 EI 72 EI 33.37 EI 67.10
  • 56. 56 Example 2 For the beam shown (a) Draw the influence line for the shear at D for the beam (b) Draw the influence line for the bending moment at D for the beam EI is constant.Plot numerical values every 2 m. A B CD E 2 m 2 m 2 m 2 m
  • 57. 57 A B CD E D 2 m 2 m 2 m 2 m The influence line for the shear at D 1 kN 1 kN D VD 1 kN 1 kN DD ED f f 1= DD DD f f
  • 58. 58 A B CD E 2 m 2 m 2 m 2 m 2 kN•m 1 kN 1 kN2 k 1 kN 1 kN 2 kN•m 1 kN2 kN1 kN • Using conjugate beam for find fxD 1 kN 1 kN
  • 59. 59 A B CD E 2 m 2 m 2 m 2 m 1 kN 1 kN Real beam V( kN) x (m) 1 -1 M (kN •m) x (m) 4 Conjugate beam 4/EI M´D 1 kN 1 kN 2kN
  • 60. 60 2 m 2 m 2 m 2 m M´D Conjugate beam 4/EI A B C D E 4/EI 0 8/EI m 3 8 • Determine M´D at D EI3 16 EI3 8 128/3EI 4/EI 8/EI m 3 8 EI3 16 EI3 40
  • 61. 61 =−=− )2)( 3 8 () 3 2 )( 2 ( 4 EIEIEI EIEIEIEI 3 52 )2)( 3 40 ( 3 128 ) 3 2 )( 2 ( =−+= EIEIEI 3 76 )2)( 3 40 () 3 2 )( 2 ( −=−= 128/3EI 2 m 2 m 2 m 2 m Conjugate beam 4/EI A B C D E EI3 8 EI3 40 V´DL M´DL 2/EI 2/EI 3 2 EI3 40 EI3 40 V´DR M´DR 2/EI 3 2 128/3EI 2/EI 2/EI V´E M´E 3 2 EI3 8
  • 62. 62 128/3EI = M´D = fDD V ´ x (m) θ fxD = M ´ x (m) ∆ EI3 76 − EI3 40 − EI3 34 − EI3 2 EI3 16 − EI3 8 EI3 52 EI 4 − Influence line of VD = fxD/fDD 76 52 4 0.406 = 128 /128 = -0.594 /(128/3) = -0.094 Conjugate beam EI3 8128/3EI 2 m 2 m 2 m 2 m 4/EI A B C D E EI3 40
  • 63. 63 A B CD E 2 m 2 m 2 m 2 m The influence line for the bending moment at D 1 kN •m1 kN •m αDD MD DD EDf α DD DDf α
  • 64. 64 1 kN•m 0.5 kN 1 kN•m 0.5 kN1 kN0.5 kN 2 m 2 m 2 m 2 m 0.5 kN 0.5 kN 0.5 kN1 k A B CD E 1 kN •m1 kN •m • Using conjugate beam for find fxD
  • 65. 65 2 m 2 m 2 m 2 m Real beam A B CD E 0.5 kN1 kN0.5 kN 1 kN •m1 kN •m V (kN) x (m) 0.5 1 M (kN •m) x (m) 2 2/EI Conjugate beam
  • 66. 66 2/EI 4/EI m 3 8 EI3 8 2/EI 0 4/EI m 3 8 2 m 2 m 2 m 2 m 2/EI Conjugate beam EI3 4 EI3 8 EI3 32 EI3 4
  • 67. 67 EIEIEI 3 26 )2)( 4 () 3 2 )( 1 ( =+= =−=− )2)( 3 4 () 3 2 )( 1 ( 2 EIEIEI 2 m 2 m 2 m 2 m 2/EI Conjugate beam EI3 32 EI3 4 EI 4 M´D V´D 1/EI 1/EI 3 2 EI 4 1/EI 1/EI V´E M´E 3 2 EI3 4
  • 68. 68/(32/3) = -0.188-2 Influence line of MD 813.0 32 26 = xD xDf α αDD = 32/3EI 2 m 2 m 2 m 2 m 2/EI Conjugate beam 4 3EI 4 EI 32 3EI x (m)V ´ θ EI 4 EI3 8 − EI3 1 EI3 4 EI3 17 − EI 5 fxD = M ´ x (m) ∆ -2/EI θD = 0.469 + 0.531 = 1 rad θDL = 5/(32/3) = 0.469 rad. θDR = -17/32 = -0.531 rad. 26/3EI
  • 69. 69 Example 3 Draw the influence line for the reactions at supports for the beam shown in the figure below. EI is constant. A DB C 5 m 5 m 5 m 5 m5 m 5 m GE F
  • 70. 70 Influence line for RD A DB C GE F 5 m 5 m 5 m 5 m5 m 5 m 1 fBD fCD fDD fED fFD fXD 1 fXD/fDD = Influence line for RD DD BD f f 1= DD DD f f DD CD f f DD ED f f DD FD f f
  • 71. 71 Conjugate beam 15 + (2/3)(15) EI 5.112 EI 15 • Use the consistency deformation method 1 + x RG - Use conjugate beam for find ∆´G and fGG ∆´G + fGGRG = 0 ------(1) 1 Real beam 15 m 15 m A G 1 15 1 Real beam 30 m A G 1 30 1 A G 3@5 =15 m 3@5 =15 m fGG ∆´G 1 = 112.5/EI EI M CC 5.2812 '' ==∆ Conjugate beam 20 m EI 15 EI 450 EI Mf GGG 9000 ''' == EI 450
  • 72. 72 x RG = -0.3125 kN 0 90005.2812 =+ GR EIEI ↓−= ,3125.0 kNRG Substitute ∆´G and fGG in (1) : 1 A G 5.625 0.6875 0.3125 11 15 = 1 + 1 30
  • 73. 73 8.182 m 6.818 m EI 16.35 EI 98.15 EI 01.23 ) 3 ( 2 3125.0 2 2 2 xx − 22 ' 13.28 xMx EI =+ ) 2 ( 6875.0625.5 1 2 11 x EI xx − = ) 3 2 ( 2 6875. 1 2 1 x EI x + 1 fBD fCD fDD fED fFD A G 3@5 =15 m 3@5 =15 m Real beam 5.625 0.6875 0.3125 • Use the conjugate beam for find fXD 28.13 EIx1 x2 x1 = 5 m -----> fBD = M´1= 56/EI x1 = 10 m -----> fCD = M´1= 166.7/EI x1 = 15 m -----> fDD = M´1= 246.1/EI x2 = 5 m -----> fFD = M´2= 134.1/EI x2 = 10 m -----> fED = M´2= 229.1/EI x2 = 15 m -----> fDD = M´2= 246.1/EI A G Conjugate beam EI 625.5 EI 688.4− A x1 (5.625-0.6875x1)/EI V´1 M´1 EI 625.5 EI xx 2 11 6875.0625.5 −EI x 2 6875.0 2 1 G x2 0.3125x2 V´2 M´2 EI 13.28 EI x 2 3125.0 2 2
  • 74. 74 • Influence Line for RD Influence Line for RD 0.2280.677 1.0 0.931 0.545 1 fXD EI 56 EI 7.166 EI 1.246 EI 2.229 EI 1.134 1 fXD/fDD 1.246 56 1.246 7.166 1.246 1.246 1.246 2.229 1.246 1.134
  • 75. 75 Influence line for RG A DB C GE F 5 m 5 m 5 m 5 m5 m 5 m fXG 1 fBG fCG fGG fEG fFG fXG/fGG 1 GG BG f f GG CG f f GG EG f f GG FG f f 1= GG GG f f
  • 76. 76 Conjugate beam 20 m EI 450 EI 30 • Use consistency deformations 1 = ∆´D + fDDRD = 0 ------(2) - Use conjugate beam for find ∆´D and fDD 1 +1 Real beam 30 m A G 1 30 1 Real beam 15 m 15 m A G 1 15 X RD fXG 1 3@5 =15 m 3@5 =15 m ∆´D fDD Conjugate beam 15 + (2/3)(15) EI 15 EI 5.112 450/EI EI 9000
  • 77. 77 EIEIEIEI MD 5.2812 )15( 4509000 ) 3 15 ( 5.112 '' =−+==∆ EIEI MfDD 1125 )15 3 2 ( 5.112 '' =×== ↓=−==+ ,5.25.2,0 11255.2812 kNkNRR EIEI DDSubstitute ∆´D and fDD in (2) : x RD = -2.5 kN = 1 11 30 + 11 15 1.5 7.5 2.5 15 m V´ M´ EI 5.112 EI 5.1 450/EI EI 9000 M´´ V´´ EI 15 EI 5.112
  • 78. 78 EIEI fBG 5.62 )5 3 2 ( 75.18 −=×−== • Use the conjugate beam for find fXG Real beam 1 fBG fCG fGG fEG fFG 3@5 =15 m 3@5 =15 m1.5 7.5 2.5 fGG = M´G = 1968.56/EI 168.75/EI EIEI fCG 06.125 )67.6( 75.18 −=−== A G Conjugate beam EI 5.7− EI 15 10 m 15 + (10/3) = 18.33 m 25 + (2/3)(5) = 28.33 m EI 5.112 EI 75 EI 75.18 A 5 m V´1 M´1 EI 5.7− EI 75.18 A 6.67 m V´2 M´2 EI 75.18 EI 5.7− EI 75.18−
  • 79. 79 =−+ 3 3 2 3 75.16856.1968 ) 3 ( 2 x EIEI xx x = 5 m -----> fFG = M´= 1145.64/EI x = 10 m -----> fEG = M´ = 447.73/EI Influence line for RG -0.064-0.032 0.227 0.582 1.0 M´ G x fGG = M´G = 1968.56/EI 168.75/EI x V´2 2 2 x 1 fXG EI 5.62− EI 125− EI 73.447 EI 64.1145 EI 56.1968 56.1968 5.62− 56.1968 125− 56.1968 73.447 56.1968 64.1145 56.1968 56.1968 1 fXG/fGG
  • 80. 80 Using equilibrium condition for the influence line for Ay A DB C GE F 5 m 5 m 5 m 5 m5 m 5 m 1x MA Ay RD RG Unit load 1 1 Influence Line for RD 0.2280.678 1.0 0.929 0.542 Influence line for RG -0.064-0.032 0.227 0.582 1.0 0.386 Influence line for Ay 0.804 -0.156 -0.124 1.0 CDAy RRRF −−==Σ↑+ 1:0
  • 81. 81 Using equilibrium condition for the influence line for MA A DB C GE F 5 m 5 m 5 m 5 m5 m 5 m 1x MA Ay RD RG x 15 x 30 RD 0.2280.678 1.0 0.929 0.542 1x 5 10 15 20 25 30 RG -0.064-0.032 0.227 0.582 1.0 Influence line for MA 2.54 1.75 -0.745 -0.59 CDA RRxM 30151:0 −−=Σ+

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