# Influence Lines of Determinate Structures

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### Influence Lines of Determinate Structures

1. 1. INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 1
2. 2. INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES - AN OVERVIEW • • • • • • • Introduction - What is an influence line? Influence lines for beams Qualitative influence lines - Muller-Breslau Principle Influence lines for floor girders Influence lines for trusses Live loads for bridges Maximum influence at a point due to a series of concentrated loads • Absolute maximum shear and moment 2
3. 3. Definition of Influence Lines QUESTION Where should a structural engineer place the live load on the structure so that it creates the greates influence at a specified point? INFLUENCE LINES Variation of Reaction, Shear, Moment or Deflection at a SPECIFIC POINT due to a concentrated force moving on member
4. 4. LIVE LOADS Common Structures Bridges Industrial Crane Rails Conveyors Floor Girders any other structure where loads move across its span
5. 5. ANALYSIS OF FORCES DUE TO LIVE LOADS LIVE or MOVING LOADS • • Variable point of Application Temporary Loads P V1 P Shear Diagrams V1  V2 V2
6. 6. Influence Lines for Beams 6
7. 7. B
8. 8. Example
9. 9. Example
11. 11. 3.3 MOVING CONCENTRATED LOAD 3.3.1 Variation of Reactions RA and RB as functions of load position 1 x B A C 10 ft 3 ft MA =0 (RB)(10) – (1)(x) = 0 RB = x/10 RA = 1-RB = 1-x/10 1 x A C RA=1-x/10 B RB = x/10 x A RA=1-x/10 C RB = x/10 31
12. 12. RA occurs only at A; RB occurs only at B 1 Influence line for RA 1-x/10 x Influence line for RB 10-x 1.0 x/10 x 10-x 32
13. 13. 3.3.2 Variation of Shear Force at C as a function of load position 0 < x < 3 ft (unit load to the left of C) x 1.0 C A B 3 ft RB = x/10 RA = 1-x/10 10 ft C x/10 Shear force at C is –ve, VC =-x/10 33
14. 14. 3 < x < 10 ft (unit load to the right of C) x B C A 3 ft RB = x/10 RA = 1-x/10 C RA = 1-x/10 Shear force at C is +ve = 1-x/10 1 0.7 -ve Influence line for shear at C +ve 0.3 1 34
15. 15. 3.3.3 Variation of Bending Moment at C as a function of load position 0 < x < 3.0 ft (Unit load to the left of C) x C A B 3 ft RA = x/10 RA = 1-x/10 10 ft x/10 (x/10)(7) x/10 (x/10)(7) C x/10 x/10 Bending moment is +ve at C 35
16. 16. 3 < x < 10 ft (Unit load to the right of C) 1 x ft B C A 3 ft RA = x/10 RA = 1-x/10 10 ft 1-x/10 (1-x/10)(3) 1-x/10 (1-x/10)(3) 1-x/10 C (1-x/10)(3) Moment at C is +ve (1-7/10)(3)=2.1 kip-ft Influence line for bending Moment at C +ve 36
17. 17. PROBLEM • Beam mass=24kg/m • Dolly can travel entire length of beam CB • A is pin and B is roller • Determine Ra and Rb and MD Max • Ra=4.63, Rb=3.31 and MD Max=2.46 37
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19. 19. QUALITATIVE INFLUENCED LINES - MULLER-BRESLAU’S PRINCIPLE • The principle gives only a procedure to determine of the influence line of a parameter for a determinate or an indeterminate structure • But using the basic understanding of the influence lines, the magnitudes of the influence lines also can be computed • In order to draw the shape of the influence lines properly, the capacity of the beam to resist the parameter investigated (reaction, bending moment, shear force, etc.), at that point, must be removed • The principle states that:The influence line for a parameter (say, reaction, shear or bending moment), at a point, is to the same scale as the deflected shape of the beam, when the beam is acted upon by that parameter. – The capacity of the beam to resist that parameter, at that point, must be removed. – Then allow the beam to deflect under that parameter – Positive directions of the forces are the same as before 40
20. 20. PROBLEM • Find MD Max when • Conc moving load=4000 lbs • UD Moving load=300 lbs/ft • Beam wt=200 lbs/ft • MD Max=22.5 k-ft 49
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28. 28. • For example, to develop the shear and moment envelopes for the beam shown, first sketch the influence lines for the shear and moment at various locations. The influence lines for Va-R, Vb-L, Vb-R, Mb, Vs, and Ms are shown 58
29. 29. These influence lines are used to determine where to place the uniform live load to yield the maximum positive and negative values for the different functions. For example; Support removed, unit load applied, and resulting influence line for support reaction at A · The maximum value for the positive reaction at A, assuming no partial loading, will occur when the uniform load is applied on the beam from A to B (load case 1) Load case 1 59
30. 30. · The maximum negative value for the reaction at A will occur if a uniform load is placed on the beam from B to C (load case 2) · Load case 1 is also used for: maximum positive value of the shear at the right of support A maximum positive moment Ms · Load case 2 is also used for: maximum positive value of the shear at the right of support B maximum negative moments at support B and Ms · Load case 3 is required for: maximum positive reaction at B 60 maximum negative shear on the left side of B
31. 31. Load case 3 · Load case 4 is required for the maximum positive shear force at section s Load case 4 · Load case 5 is required for the maximum negative shear force at section s 61
32. 32. Load case 5 To develop the shear and moment envelopes, construct the shear and moment diagrams for each load case. The envelope is the area that is enclosed by superimposing all of these diagrams. The maximum positive and negative values can then be determined by looking at the maximum and minimum values of the envelope at each point. 62
33. 33. Individual shear diagrams for each load case; 63
34. 34. Superimpose all of these diagrams together to determine the final shear envelope. Resulting superimposed shear envelope 64
35. 35. Individual moment diagrams for each load case; 65
36. 36. Superimpose all of these diagrams together to determine the final moment envelope. 66
37. 37. Influence Lines for Floor Girders 67
38. 38. INFLUENCE LINE FOR FLOOR GIRDERS Floor systems are constructed as shown in figure below, 68
39. 39. INFLUENCE LINES FOR FLOOR GIRDERS (Cont’d) 69
40. 40. IL FOR SHEAR IN PANEL CD OF FLOOR GIRDER 73
41. 41. 74
42. 42. INFLUENCE LINES FOR FLOOR GIRDERS Force Equilibrium Method: Draw the Influence Lines for: (a) Shear in panel CD of the girder; and (b) the moment at E. x A´ B´ C´ D´ E´ F´ A B C D E 5 spaces @ 10´ each = 50 ft F 75
43. 43. Place load over region A´B´ (0 < x < 10 ft) Find the shear over panel CD VCD= - x/50 At x=0, VCD = 0 At x=10, VCD = -0.2 C F D Shear is -ve Find moment at E = +(x/50)(10)=+x/5 At x=0, ME=0 At x=10, ME=+2.0 E +ve moment RF=x/50 F RF=x/50 76
44. 44. Continuation of the Problem x -ve 0.2 I. L. for VCD +ve 2.0 I. L. for ME 77
45. 45. Place load over region B´C´ (10 ft < x < 20ft) VCD = -x/50 kip At x = 10 ft VCD = -0.2 At x = 20 ft VCD = -0.4 ME = +(x/50)(10) = +x/5 kip.ft At x = 10 ft, ME = +2.0 kip.ft At x = 20 ft, ME = +4.0 kip.ft C D Shear is -ve E D Moment is +ve F RF = x/50 F RF = x/50 78
46. 46. x B´ C´ -ve 0.2 0.4 I. L. for VCD +ve 4.0 2.0 I. L. for ME 79
47. 47. Place load over region C´D´ (20 ft < x < 30 ft) When the load is at C’ (x = 20 ft) C D Shear is -ve RF=20/50 =0.4 VCD = -0.4 kip When the load is at D´ (x = 30 ft) A B C RA= (50 - x)/50 D Shear is +ve VCD= + 20/50 = + 0.4 kip 80
48. 48. ME = + (x/50)(10) = + x/5 x Load P E +ve moment A´ B´ C´ A B D´ C 0.2 RF= x/50 +ve -ve 0.4 D I. L. for VCD +ve 2.0 4.0 6.0 I. L. for ME 81
49. 49. Place load over region D´E´ (30 ft < x < 40 ft) VCD= + (1-x/50) kip A B RA= (1-x/50) C D E Shear is +ve ME= +(x/50)(10) = + x/5 kip.ft E Moment is +ve RF= x/50 At x = 30 ft, ME = +6.0 At x = 40 ft, ME = +8.0 82
50. 50. Problem continued x A´ B´ C´ D´ 0.4 E´ +ve 0.2 I. L. for VCD 2.0 4.0 +ve 8.0 6.0 I. L. for ME 83
51. 51. Place load over region E´F´ (40 ft < x < 50 ft) VCD = + 1-x/50 At x = 40 ft, VCD= + 0.2 At x = 50 ft, VCD = 0.0 x 1.0 A B C D E RA= 1-x/50 Shear is +ve ME= + (1-x/50)(40) = (50-x)*40/50 = +(4/5)(50-x) x A B RA=1-x/50 C D E F At x = 40 ft, ME= + 8.0 kip.ft At x = 50 ft, ME = 0.0 Moment is +ve 84
52. 52. x A´ B´ 1.0 C´ D´ E´ 0.4 0.2 -ve +ve F´ 0.2 0.4 I. L. for VCD 2.0 4.0 6.0 +ve 8.0 I. L. for ME 85
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54. 54. Draw IL for moment at point F for the floor girder 88
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58. 58. 92
59. 59. Influence Lines for Trusses 93
60. 60. 99
61. 61. INFLUENCE LINES FOR TRUSSES Draw the influence lines for: (a) Force in Member GF; and (b) Force in member FC of the truss shown below in Figure below G F E 20 ft 10(3)1/3 600 A B 20 ft D C 20 ft 20 ft 100
62. 62. Place unit load over AB (i) To compute GF, cut section (1) - (1) (1) G 1-x/20 1 F E x/20 x 600 A B RA= 1- x/60 At x = 0, FGF = 0 At x = 20 ft FGF = - 0.77 C (1) D RD=x/60 Taking moment about B to its right, (RD)(40) - (FGF)(103) = 0 FGF = (x/60)(40)(1/ 103) = x/(15 3) (-ve) 101
63. 63. (ii) To compute FFC, cut section (2) - (2) G x (2) F E 1 x/20 1-x/20 reactions at nodes A 300 600 B RA =1-x/60 D C (2) RD=x/60 Resolving vertically over the right hand section FFC cos300 - RD = 0 FFC = RD/cos30 = (x/60)(2/3) = x/(30 3) (-ve) 102
64. 64. At x = 0, FFC = 0.0 At x = 20 ft, FFC = -0.385 20 ft I. L. for FGF -ve 0.77 -ve 0.385 I. L. for FFC 103
65. 65. Place unit load over BC (20 ft < x <40 ft) [Section (1) - (1) is valid for 20 < x < 40 ft] (i) To compute FGF use section (1) -(1) G (1) F E (40-x)/20 x (x-20)/20 1 reactions at nodes A 20 ft RA=1-x/60 B C (1) (x-20) (40-x) D RD=x/60 Taking moment about B, to its left, (RA)(20) - (FGF)(103) = 0 FGF = (20RA)/(103) = (1-x/60)(2 /3) At x = 20 ft, FFG = 0.77 (-ve) At x = 40 ft, FFG = 0.385 (-ve) 104
66. 66. To compute FFC, use section (2) - (2) Section (2) - (2) is valid for 20 < x < 40 ft G (2) F (40-x)/20 (x-20)/20 1 x E FFC 300 600 A B RA =1-x/60 C (2) D RD=x/60 Resolving force vertically, over the right hand section, FFC cos30 - (x/60) +(x-20)/20 = 0 FFC cos30 = x/60 - x/20 +1= (1-2x)/60 (-ve) FFC = ((60 - 2x)/60)(2/3) -ve 105
67. 67. At x = 20 ft, FFC = (20/60)(2/ 3) = 0.385 (-ve) At x = 40 ft, FFC = ((60-80)/60)(2/ 3) = 0.385 (+ve) -ve 0.385 0.77 I. L. for FGF 0.385 -ve I. L. for FFC 106
68. 68. Place unit load over CD (40 ft < x <60 ft) (i) To compute FGF, use section (1) - (1) G (1) F E (x-40) 1 x (60-x)/20 A 20 ft RA=1-x/60 B (1) C reactions at nodes (60-x) (x-40)/20 D RD=x/60 Take moment about B, to its left, (FFG)(103) - (RA)(20) = 0 FFG = (1-x/60)(20/103) = (1-x/60)(2/3) -ve At x = 40 ft, FFG = 0.385 kip (-ve) At x = 60 ft, FFG = 0.0 107
69. 69. To compute FFG, use section (2) - (2) reactions at nodes G F E (60-x)/20 (x-40)/20 FFC x 1 300 600 A B RA =1-x/60 C x-40 60-x (2) D RD=x/60 Resolving forces vertically, to the left of C, (RA) - FFC cos 30 = 0 FFC = RA/cos 30 = (1-x/10) (2/3) +ve 108
70. 70. At x = 40 ft, FFC = 0.385 (+ve) At x = 60 ft, FFC = 0.0 -ve 0.385 0.770 I. L. for FGF +ve -ve 0.385 I. L. for FFC 109
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75. 75. Influence lines for wheel loads
76. 76. Live Loads for Highway BRIDGES •Caused by Traffic •Heaviest: Series of Trucks Specifications: AASHTO American Association of State and Highway Transportation Officials Load Desigantions 15 Tons (10-20) H 15-44 2 Axle Truck 20 Tons HS 20-44 Year of Specifications 2 Axle Truck & 1 Axle Semitrailer Year of Specifications Simplification: Uniform Load plus a Concnetrated force placed at critical locations
78. 78. Maximum Influence due to a Series of Concentrated Loads - Shear
79. 79. Maximum Influence due to a Series of Concentrated Loads - Shear
80. 80. Maximum Influence due to a Series of Concentrated Loads - Shear
81. 81. Maximum Influence due to a Series of Concentrated Loads - Shear V12  5.375  5.25  0.125 V23  2.5  5.375  2.875
82. 82. Maximum Influence due to a Series of Concentrated Loads - Shear
83. 83. Maximum Influence due to a Series of Concentrated Loads - Moment
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