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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 &#x2206;&#xB4;1 = f1j + 1 1 2 3j4 Redundant R1 applied 1 1 = f11 fj1 &#xD7;R1&#xD7;R1 f11 Influence Lines for Reaction Compatibility equation: 011111 =&#x2206;=+ Rff j ) 1 ( 11 11 f fR j&#x2212;= )( 11 1 1 f f R j =
• 5. 5 1 1 2 3j4 1 =+ &#xD7;R2 Redundant R2 applied fjj f2j 1 fj2 f22 Compatibility equation. 0' 22222 =&#x2206;=+&#x2206; Rf 022222 =&#x2206;=+ Rff j ) 1 ( 22 22 f fR j&#x2212;= )( 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 &#x3B1; = 1 2 3j4 Influence Lines for Bending Moment 1 1 &#x3B1;44 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 &#x2022; 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 &#x2022; Unit load to the right of 4 &#x2022; Influence line of Shear V4 = R1 V4 = R1 - 1 1 R2 R3R1 1 1 1 R1 1x V4 M4 1 &#x2022; Unit load to the left of 4 V4 = R1 - 1 01;0 41 =&#x2212;&#x2212;=&#x3A3;&#x2191;+ VRFy 1 R1 0;0 41 =&#x2212;=&#x3A3;&#x2191;+ VRFy
• 10. 10 1 2 3j4 &#x2022; Influence line of Bending moment M4 = - l + x + l R1 M4 - 1 (l-x) - l R1 = 0+ &#x3A3; M4 = 0: R1 1x V4 M4 l 1 &#x2022; Unit load to the left of 4 M4 = l R1 R1 V4 M4 &#x2022; 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+ &#x3A3; 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 &#x2022; 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 &#x2022; Find fxB by conjugate beam 11 6 kN&#x2022;m x =&#x2212;+ EI x EIEI x 1872 6 3 EI 6 EI 18 EI 18 EI 72 3 x 3 2xV&#xB4;x M&#xB4;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 &#x2212;+== 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&#xB4;x M&#xB4;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 &#x2022; Find fxA by conjugate beam 1 kN 6 kN&#x2022;m EI M A 72 ' = x EI B y 18 ' = =&#x2212; 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 &#x2212;== 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 &#x2212;= =&#x2212;+ =&#x3A3;&#x2191;+ 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 &#x2022; Unit load to the left of C RB 1 x VC MC 01 ;0 =+&#x2212;+ =&#x3A3;&#x2191;+ BC y RV F RB VC MC VC = - RB 0 ;0 =++ =&#x3A3;&#x2191;+ CB y VR F &#x2022; Unit load to the left of C
• 26. 26 BA BA A RxM RxM M 66 06)6(1 ;0 ++&#x2212;= =+&#x2212;&#x2212;&#x2212; =&#x3A3; 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 &#x2022; Unit load to the left of C RB VC MC 4 m MC = 4RB + 04 ;0 =+&#x2212; =&#x3A3; BC C RM M &#x2022; Unit load to the left of C 4 m MC = -4 + x + 4RB 04)4(1 ;0 =+&#x2212;&#x2212;&#x2212; =&#x3A3; 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 &#x2022; 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 =&#x2212; EI x EI x 1 3 1 33.5 12 EIEIEI x 67.1864 6 3 2 &#x2212;+= V&#xB4; x1 M&#xB4; 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&#xB4; x2 V&#xB4; 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 &#x2212;== EI 28 EI 64 BtoAfor EIEI x EI x Mf xxA , 6467.18 6 ' 2 3 2 2 +&#x2212;== fxA 0 0 EI 16 &#x2212; EI 10 &#x2212; EI 14 &#x2212; 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&#x2212; EI 16&#x2212; EI 14&#x2212; 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 &#x2212;= =&#x2212;+&#x2212; =&#x3A3;+
• 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 +&#x2212;= =&#x2212;&#x2212;&#x2212; =&#x3A3;+ x RR RxR M AC CA B
• 35. 35 A B CD E FG 4 m 6 m2 m &#x2022; Check &#x3A3;Fy = 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 =++ =&#x3A3;&#x2191;+ 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 &#x3A3;R 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 &#x2022; 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 &#x2022; Unit load to the left of G 01;0 =&#x2212;&#x2212;=&#x3A3;&#x2191;+ 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 &#x2022; Unit load to the left of E VE = - RC 0;0 =+=&#x3A3;&#x2191;+ ECy VRF VE = 1 - RC &#x2022; Unit load to the right of E RC 1 x VE ME 01;0 =+&#x2212;=&#x3A3;&#x2191;+ 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 &#x2022; Unit load to the left of G RA 1x A VG MG 2 m 02)2(1 ;0 =&#x2212;&#x2212;+ =&#x3A3; AG G RxM M+ MG = 2RA &#x2022; Unit load to the right of G RA A VG MG 2 m 02 ;0 =&#x2212; =&#x3A3; 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 &#x2022; Unit load to the left of E ME = 4RC + &#x3A3;ME = 0; ME = - 4 + x+ 4RC &#x2022; Unit load to the right of E RC 1 x VE ME 4 m 04)4(1 ;0 =+&#x2212;&#x2212;&#x2212; =&#x3A3;+ 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 &#x2022; Find fxA by conjugate beam 1 kN 8 kN&#x2022;m x (m)V (kN) 1 1 + x (m)M (kN&#x2022;m) 8 4 + 2EI fAA = M&#xB4;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 &#x2022; 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&#x2022;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&#x2022;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&#x2022;m) + - -46.48 26.76 0.14 RA 0.614 0.294 0.081 1 0 RB=23.31 kN MB=46.48 kN&#x2022;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 &#x2022;Muller-Breslau for the influence line of reaction, shear and moment &#x2022;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&#xB4;x M&#xB4;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 &#x2022; Find fxA by conjugate beam 1 kN 6 kN&#x2022;m EI M A 72 ' = x EI B y 18 ' = 6 /EI EI 18 =&#x2212; 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 &#x2212;== 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 &#x2022; 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 &#x2022; Find fxB by conjugate beam 11 6 kN&#x2022;m x =&#x2212;+ EI x EIEI x 1872 6 3 EI 6 EI 18 EI 18 EI 72 3 x 3 2xV&#xB4;x M&#xB4;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 &#x2212;+== 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&#x2022;m 1 kN 1 kN2 k 1 kN 1 kN 2 kN&#x2022;m 1 kN2 kN1 kN &#x2022; 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 &#x2022;m) x (m) 4 Conjugate beam 4/EI M&#xB4;D 1 kN 1 kN 2kN
• 60. 60 2 m 2 m 2 m 2 m M&#xB4;D Conjugate beam 4/EI A B C D E 4/EI 0 8/EI m 3 8 &#x2022; Determine M&#xB4;D at D EI3 16 EI3 8 128/3EI 4/EI 8/EI m 3 8 EI3 16 EI3 40
• 61. 61 =&#x2212;=&#x2212; )2)( 3 8 () 3 2 )( 2 ( 4 EIEIEI EIEIEIEI 3 52 )2)( 3 40 ( 3 128 ) 3 2 )( 2 ( =&#x2212;+= EIEIEI 3 76 )2)( 3 40 () 3 2 )( 2 ( &#x2212;=&#x2212;= 128/3EI 2 m 2 m 2 m 2 m Conjugate beam 4/EI A B C D E EI3 8 EI3 40 V&#xB4;DL M&#xB4;DL 2/EI 2/EI 3 2 EI3 40 EI3 40 V&#xB4;DR M&#xB4;DR 2/EI 3 2 128/3EI 2/EI 2/EI V&#xB4;E M&#xB4;E 3 2 EI3 8
• 62. 62 128/3EI = M&#xB4;D = fDD V &#xB4; x (m) &#x3B8; fxD = M &#xB4; x (m) &#x2206; EI3 76 &#x2212; EI3 40 &#x2212; EI3 34 &#x2212; EI3 2 EI3 16 &#x2212; EI3 8 EI3 52 EI 4 &#x2212; 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 &#x2022;m1 kN &#x2022;m &#x3B1;DD MD DD EDf &#x3B1; DD DDf &#x3B1;
• 64. 64 1 kN&#x2022;m 0.5 kN 1 kN&#x2022;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 &#x2022;m1 kN &#x2022;m &#x2022; 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 &#x2022;m1 kN &#x2022;m V (kN) x (m) 0.5 1 M (kN &#x2022;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 ( =+= =&#x2212;=&#x2212; )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&#xB4;D V&#xB4;D 1/EI 1/EI 3 2 EI 4 1/EI 1/EI V&#xB4;E M&#xB4;E 3 2 EI3 4
• 68. 68/(32/3) = -0.188-2 Influence line of MD 813.0 32 26 = xD xDf &#x3B1; &#x3B1;DD = 32/3EI 2 m 2 m 2 m 2 m 2/EI Conjugate beam 4 3EI 4 EI 32 3EI x (m)V &#xB4; &#x3B8; EI 4 EI3 8 &#x2212; EI3 1 EI3 4 EI3 17 &#x2212; EI 5 fxD = M &#xB4; x (m) &#x2206; -2/EI &#x3B8;D = 0.469 + 0.531 = 1 rad &#x3B8;DL = 5/(32/3) = 0.469 rad. &#x3B8;DR = -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 &#x2022; Use the consistency deformation method 1 + x RG - Use conjugate beam for find &#x2206;&#xB4;G and fGG &#x2206;&#xB4;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 &#x2206;&#xB4;G 1 = 112.5/EI EI M CC 5.2812 '' ==&#x2206; 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 &#x2193;&#x2212;= ,3125.0 kNRG Substitute &#x2206;&#xB4;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 &#x2212; 22 ' 13.28 xMx EI =+ ) 2 ( 6875.0625.5 1 2 11 x EI xx &#x2212; = ) 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 &#x2022; Use the conjugate beam for find fXD 28.13 EIx1 x2 x1 = 5 m -----&gt; fBD = M&#xB4;1= 56/EI x1 = 10 m -----&gt; fCD = M&#xB4;1= 166.7/EI x1 = 15 m -----&gt; fDD = M&#xB4;1= 246.1/EI x2 = 5 m -----&gt; fFD = M&#xB4;2= 134.1/EI x2 = 10 m -----&gt; fED = M&#xB4;2= 229.1/EI x2 = 15 m -----&gt; fDD = M&#xB4;2= 246.1/EI A G Conjugate beam EI 625.5 EI 688.4&#x2212; A x1 (5.625-0.6875x1)/EI V&#xB4;1 M&#xB4;1 EI 625.5 EI xx 2 11 6875.0625.5 &#x2212;EI x 2 6875.0 2 1 G x2 0.3125x2 V&#xB4;2 M&#xB4;2 EI 13.28 EI x 2 3125.0 2 2
• 74. 74 &#x2022; 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 &#x2022; Use consistency deformations 1 = &#x2206;&#xB4;D + fDDRD = 0 ------(2) - Use conjugate beam for find &#x2206;&#xB4;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 &#x2206;&#xB4;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 '' =&#x2212;+==&#x2206; EIEI MfDD 1125 )15 3 2 ( 5.112 '' =&#xD7;== &#x2193;=&#x2212;==+ ,5.25.2,0 11255.2812 kNkNRR EIEI DDSubstitute &#x2206;&#xB4;D and fDD in (2) : x RD = -2.5 kN = 1 11 30 + 11 15 1.5 7.5 2.5 15 m V&#xB4; M&#xB4; EI 5.112 EI 5.1 450/EI EI 9000 M&#xB4;&#xB4; V&#xB4;&#xB4; EI 15 EI 5.112
• 78. 78 EIEI fBG 5.62 )5 3 2 ( 75.18 &#x2212;=&#xD7;&#x2212;== &#x2022; 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&#xB4;G = 1968.56/EI 168.75/EI EIEI fCG 06.125 )67.6( 75.18 &#x2212;=&#x2212;== A G Conjugate beam EI 5.7&#x2212; 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&#xB4;1 M&#xB4;1 EI 5.7&#x2212; EI 75.18 A 6.67 m V&#xB4;2 M&#xB4;2 EI 75.18 EI 5.7&#x2212; EI 75.18&#x2212;
• 79. 79 =&#x2212;+ 3 3 2 3 75.16856.1968 ) 3 ( 2 x EIEI xx x = 5 m -----&gt; fFG = M&#xB4;= 1145.64/EI x = 10 m -----&gt; fEG = M&#xB4; = 447.73/EI Influence line for RG -0.064-0.032 0.227 0.582 1.0 M&#xB4; G x fGG = M&#xB4;G = 1968.56/EI 168.75/EI x V&#xB4;2 2 2 x 1 fXG EI 5.62&#x2212; EI 125&#x2212; EI 73.447 EI 64.1145 EI 56.1968 56.1968 5.62&#x2212; 56.1968 125&#x2212; 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 &#x2212;&#x2212;==&#x3A3;&#x2191;+ 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 &#x2212;&#x2212;=&#x3A3;+