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Influence Line of Beams And Determinate Structures

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Influence Line of Beams And Determinate Structures

  1. 1. INFLUENCE LINES FOR STATICALLYDETERMINATE STRUCTURES ! Influence Lines for Beams ! Influence Lines for Floor Girders ! Influence Lines for Trusses ! Maximum Influence at a Point Due to a Series of Concentrated Loads ! Absolute Maximum Shear and Moment 1
  2. 2. Influence Line Unit moving load A B 2
  3. 3. Example 6-1Construct the influence line for a) reaction at A and B b) shear at point C c) bending moment at point C d) shear before and after support B e) moment at point Bof the beam in the figure below. C A B 4m 4m 4m 3
  4. 4. SOLUTION • Reaction at A 1 x C A Ay By 8m 4mx Ay0 1 1 + ΣMB = 0: − Ay (8) + 1(8 − x ) = 0, Ay = 1− x4 0.5 88 0 Ay12 -0.5 1 0.5 8m 12 m x 4m -0.5 4
  5. 5. • Reaction at B 1 x C A Ay By 8m 4mx By0 0 14 0.5 + ΣMA = 0: B y (8) − 1x = 0, By = x 88 112 1.5 By 1.5 1 0.5 x 4m 8m 12 m 5
  6. 6. • Shear at C 0≤ x<4 4 < x ≤ 12 1 x C A Ay By 4m 4m 4m 1 x C 1 + ΣFy = 0: 1 − x − 1 − VC = 00≤ x≤4 A MC 8 1 1 Ay = 1− x VC = − x 8 VC 8 4m x C 14 < x ≤ 12 A MC + ΣFy = 0: 1 − x − VC = 0 1 8 Ay = 1− x 1 8 VC VC = 1− x 4m 8 6
  7. 7. 0≤ x<4 4 < x ≤ 12 x VC 1 1 0 0 VC = − x x C 8 4- -0.5A 4+ 0.5 1 VC = 1− x 8 0 8Ay By 4m 4m 4m 12 -0.5 1 VC VC = 1− x 8 0.5 4m 8m 12 m x -0.5 -0.5 1 VC = − x 8 7
  8. 8. • Bending moment at C 0≤ x<4 4 < x ≤ 12 1 x C A Ay By 4m 4m 4m 1 x C 10≤ x≤4 A MC + ΣMC = 0: M C + 1(4 − x) − (1 − x )(4) = 0 1 8 Ay = 1− x 1 8 VC MC = x 4m 2 x C 14 < x ≤ 12 A MC + ΣMC = 0: M C − (1 − x)( 4) = 0 1 8 Ay = 1− x 1 8 VC MC = 4 − x 4m 2 8
  9. 9. 0≤ x<4 4 < x ≤ 12 x MC 1 1 0 0 x MC = x C 2 4 2A 1 8 0 MC = 4 − xAy By 2 12 -2 4m 4m 4m 1 MC = x 1 2 MC = 4 − x MC 2 2 8m 12 m x 4m -2 9
  10. 10. Or using equilibrium conditions: • Reaction at A 1 x C A Ay By 8m 4m 1 + ΣMB = 0: − Ay (8) + 1(8 − x ) = 0, Ay = 1− x 8 Ay 1 0.5 8m 12 m x 4m -0.5 10
  11. 11. • Reaction at B 1 x C A Ay By 8m 4m Ay + ΣFy = 0: Ay + B y − 1 = 0 1 0.5 B y = 1 − Ay 8m 12 m x 4m -0.5 By B y = 1 − Ay 1.5 1 0.5 x 4m 8m 12 m 11
  12. 12. • Shear at C 0≤ x<4 4 < x ≤ 12 1 x C A Ay By 4m 4m 4m 1 x C + ΣFy = 0: Ay − 1 − VC = 00≤ x≤4 A MC 1 VC = Ay − 1 Ay = 1− x VC 8 4m x C4 < x ≤ 12 A MC + ΣFy = 0: Ay − VC = 0 1 Ay = 1− x 8 VC VC = Ay 4m 12
  13. 13. CA B 4m 4m 4m Ay1 0.5 8m 12 m x 4m VC = Ay − 1 VC = Ay -0.5 VC 0.5 4m 8m 12 m x -0.5 -0.5 13
  14. 14. • Bending moment at C 0≤ x<4 4 < x ≤ 12 1 x C A Ay By 4m 4m 4m 1 x C0≤ x≤4 A MC + ΣMC = 0: Ay (4) + 1( 4 − x ) + M C = 0 1 Ay = 1− x VC M C = 4 Ay − (4 − x) 8 4m x C4 < x ≤ 12 A MC + ΣMC = 0: − Ay (4) + M C = 0 1 Ay = 1− x M C = 4 Ay 8 VC 4m 14
  15. 15. CA B 4m 4m 4m Ay 1 0.5 8m 12 m x 4mM C = 4 Ay − (4 − x) M C = 4 Ay -0.5 MC 2 8m 12 m x 4m -2 15
  16. 16. • Shear before support B 1 x C A Ay By 4m 4m 4m 1 x MB MB 8m 8mAy VB - Ay VB- VB- = Ay-1 VB- = Ay Ay 1 0.5 8m 12 m x VB- 4m -0.5 x -0.5 -1.0 -0.5 16
  17. 17. • Shear after support B 1 x C A Ay By 4m 4m 4m 1 MB MB 4m 4m VB+ VB+ VB+ = 0 VB+ = 1 Ay 1 0.5 8m 12 m x 4m -0.5 VB+ 1 x 17
  18. 18. • Moment at support B 1 x C A Ay By 4m 4m 4m 1 x MB MB 8m 8mAy VB - Ay VB- MB = 8Ay-(8-x) MB = 8Ay Ay 1 0.5 8m 12 m x MB 4m -0.5 x 1 -4 18
  19. 19. Influence Line for Beam • Reaction P=1 C A B x L P=1 δy 1 δ y sB = = δy = 1 L L C A B Ay By Ay (1) − 1(δ y ) + B y (0) = 0 Ay = δ y 19
  20. 20. P=1 CA B x L P=1 δ y δy = 1 CA B δy 1 sA = = Ay L L By Ay (0) − 1(δ y ) + B y (1) = 0 By = δ y 20
  21. 21. - Pinned Support A C B a b L B A RA RA 1 b L x 21
  22. 22. - Fixed Support A B a b L A B RA RA 1 1 x 22
  23. 23. P=1• Shear A C B a b L VC P=1 1 δ y sB = δy=1 δyR L A B δyL Ay 1 By sA = L VC Ay (0) + VC (δ yL ) + VC (δ yR ) − 1(δ y ) + B y (0) = 0 VC (δ yL + δ yR ) = δ y slopes : s A = sB δy=1 VC = δ y 23
  24. 24. - Pinned Support A C B a b L VC A B VC 1VC 1 Slope s B = L b 1 L x -a 1 L Slope s A = L -1 Slope at A = Slope at B 24
  25. 25. - Fixed Support A B a b L VB A B VB VB 1 1 x 25
  26. 26. • Bending Moment P=1 A C B a b L φ = θ A +θ B = 1 P=1 h MC MC δ y A B h h Ay θA = θB = a b ByAy (0) + M C (θ A ) + M C (θ B ) + 1(δ y ) + B y (0) = 0 h h ( + ) =1 1 a b M C (θ A + θ B ) = δ y h( a + b) ab = 1, h= ab ( a + b) M C = δ y 26
  27. 27. - Pinned Support A C B a b L Hinge A B MC MC MC a b φC = θA + θB = 1 ab a+b x b θA = a L θA = L 27
  28. 28. - Fixed Support A B a b L A B MC MC MB x 1 -b 28
  29. 29. • General Shear C D E B F G H A L/4 L/4 L/4 L/4 L/4 L/4 L/4 L VC 3/4 1 x VD -1/4 2/4 1 x VE -2/4 1/4 x 1 VBL -3/4 x -1 29
  30. 30. C D E B F G HA L/4 L/4 L/4 L/4 L/4 L/4 L/4VBL L x -1VBR 1 xVF 1 xVG 1 x 30
  31. 31. • General Bending Moment C D E B F G H A L/4 L/4 L/4 L/4 L/4 L/4 L/4 L MC 3L/16 φ = sA + sB = 1 x θA = 3/4 θB = 1/4 MD 4L/16 φ = sA + sB = 1 x θA = 1/2 θB = 1/2 ME 3L/16 φ = θA + θB = 1 x θA = 1/4 θB = 3/4 31
  32. 32. C D E B F G HA L/4 L/4 L/4 L/4 L/4 L/4 L/4 LMB x 1 3L/4MF x 1 2L/4MG x 1 L/4 32
  33. 33. Example 6-2Construct the influence line for - the reaction at A, C and E - the shear at D - the moment at D - shear before and after support C - moment at point C A B Hinge D C E 2m 2m 2m 4m 33
  34. 34. SOLUTION B C D E A 2m 2m 2m 4m RARA 1 x 34
  35. 35. B C D EA 2m 2m 2m 4m RC 8/6 1RC 4/6 x 35
  36. 36. A B D C E 2m 2m 2m 4m RE 1RE 2/6 x -2/6 36
  37. 37. VD A B D C E VD 2m 2m 2m 4m 1 4/6 2/6 =VD 1 x sE = 1/6 = sC = 1/6 -1 -2/6 • sE = sC 37
  38. 38. Or using equilibrium conditions: 1 A B Hinge D C E 2m 2m 2m 4m 1 VD x 4m VD MD 4m MD RE RE VD = -RE VD = 1 -RE 1 RE 2/6 x -2/6 4/6 VD 2/6 x -2/6 38
  39. 39. A B MD MD C E D 2m 2m 2m 4m (2)(4)/6 = 1.33 4 2 φD = θC+θE = 1MD θC = 4/6 2/6 = θE x -1.33 39
  40. 40. Or using equilibrium conditions: 1 A B Hinge D C E 2m 2m 2m 4m 1 VD x 4m VD MD 4m MD RE RE MD = 4RE MD = -(4-x)+4RE 1 RE 2/6 x -2/6 8/6 MD x -8/6 40
  41. 41. VCL A C E B D VCL 2m 2m 2m 4mVCL x -1 -1 41
  42. 42. Or using equilibrium conditions: A B 1 D C E 2m 2m 2m 4m 1 MB MB RA VCL RA VCL VCL = RA - 1 VCL = RA RA 1 x VCL x -1 -1 42
  43. 43. VCR A C E B D VCR 2m 2m 2m 4m 1 0.667VCR 0.333 x 43
  44. 44. Or using equilibrium conditions: A B 1 D C E 2m 2m 2m 4m 1 MC MC VCR VCR VCR = -RE RE VCR = 1 -RE RE 1 RE 2/6=0.33 x -2/6 = -0.333 1 0.667 VCR 0.333 x 44
  45. 45. A B MC C MC D E 2m 2m 2m 4mMC x 1 -2 45
  46. 46. Or using equilibrium conditions: A B 1 D C E 2m 2m 2m 4m 1 x MC MC 6m 6m VCR VCR MC = 6RE RE M C = 6 RA − x RE 1 RE 2/6=0.33 x -2/6 = -0.333 MC x 1 -2 46
  47. 47. Example 6-3Construct the influence line for - the reaction at A and C - shear at D, E and F - the moment at D, E and F Hinge A D B E C F 2m 2m 2m 2m 2m 2m 47
  48. 48. SOLUTION D B E C FA RA 2m 2m 2m 2m 2m 2m 1 1RA 0.5 x -0.5 -1 48
  49. 49. A D B E F C RC 2m 2m 2m 2m 2m 2m 2 1.5 1 0.5RC x 49
  50. 50. VD A B E C F D VD 2m 2m 2m 2m 2m 2mVD 1 1 = 0.5 x = -0.5 -1 50
  51. 51. VE A D B E C F VE 2m 2m 2m 2m 2m 2m 0.5VE = 1 x = -0.5 -0.5 -1 51
  52. 52. VF A D B E C F VF 2m 2m 2m 2m 2m 2m 1 =VF x = 52
  53. 53. A D B E C F MD MD 2m 2m 2m 2m 2m 2m 2MD 1 x θD = 1 -1 -2 53
  54. 54. A D B ME ME C E F 2m 2m 2m 2m 2m 2m (2)(2)/4 = 1ME φE = 1 x θB = 0.5 θC = 0.5 -1 -2 54
  55. 55. A D B C ME ME E F 2m 2m 2m 2m 2m 2mMF x θF = 1 -2 55
  56. 56. Example 6-4Determine the maximum reaction at support B, the maximum shear at point C andthe maximum positive moment that can be developedat point C on the beam shown due to - a single concentrate live load of 8000 N - a uniform live load of 3000 N/m - a beam weight (dead load) of 1000 N/m A C B 4m 4m 4m 56
  57. 57. SOLUTION 8000 N 3000 N/m 1000 N/m A C B 4m 4m 4m RB 1.5 1 0.5 0.5(12)(1.5) = 9 x (RB)max = (1000)(9) + (3000)(9) + (8000)(1.5) = 48000 N = 48 kN 57
  58. 58. 8000 N 3000 N/m 3000 N/m 1000 N/m A C B 4m 4m 4mVC 0.5 0.5(4)(0.5) = 1 x 0.5(4)(-0.5) = -1 0.5(4)(-0.5) = -1 -0.5 -0.5 (VC)max = (1000)(-2+1) + (3000)(-2) + (8000)(-0.5) = -11000 N = 11 kN 58
  59. 59. 8000 N 3000 N/m 1000 N/mA C B 4m 4m 4mMC 2 +(1/2)(8)(2) = 8 x (1/2)(4)(2) = 4 -2 (MC)max positive = (8000)(2) + (3000)(8) + (8-4)(1000) = 44000 N•m = 44 kN•m 59

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