1. 1 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 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES

2. 2 Influence Line Unit moving load A B

3. 3 Example 6-1 Construct 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 B of the beam in the figure below. B A C 4 m

4. 4 SOLUTION Reaction at A +  M B  = 0: 1 4 m 8 m12 m -0.5 0.5 AyAy x A C 4 m8 m AyAy ByBy 1 x 0 4 8 12 x 1 0.5 0 -0.5 AyAy

5. 5 A C 4 m AyAy ByBy 8 m Reaction at B +  M A  = 0: 1 x 4 m8 m12 m ByBy x 1.5 1 0.5 0 4 8 12 x 0 0.5 1 1.5 ByBy

6. 6 Shear at C A C 4 m AyAy ByBy 1 x  F y = 0: + + VCVC MCMC VCVC MCMC A C x 4 m A C 1 x

7. 7 0 4-4- 4+4+ 8 12 xVCVC 0 -0.5 0.5 0 -0.5 4 m 8 m12 m VCVC x -0.5 0.5 -0.5 A C 4 m AyAy ByBy 1 x

8. 8 Bending moment at C A C 4 m AyAy ByBy 1 x VCVC MCMC VCVC MCMC +  M C  = 0: + A C x 4 m A C 1 x

9. 9 0 4 8 12 xMCMC 0 2 0 -2 4 m 8 m12 m MCMC x 2 -2 A C 4 m AyAy ByBy 1 x

10. 10 Or using equilibrium conditions: Reaction at A +  M B  = 0: 1 4 m 8 m12 m -0.5 0.5 AyAy x A C 4 m8 m AyAy ByBy 1 x 0 4 8 12 x 1 0.5 0 -0.5 AyAy

11. 11 Shear at C A C 4 m AyAy ByBy 1 x  F y = 0: + + VCVC MCMC A C x 4 m VCVC MCMC A C 1 x

12. 12 1 4 m 8 m12 m -0.5 0.5 AyAy x -0.5 VCVC 8 m12 m x 4 m 0.5 -0.5 B A C 4 m

13. 13 Bending moment at C A C 4 m AyAy ByBy 1 x +  M C  = 0: VCVC MCMC A C x 4 m VCVC MCMC A C 1 x +  M C  = 0:

14. 14 2 1 4 m 8 m12 m -0.5 0.5 AyAy x 4 m 8 m12 m MCMC x -2 B A C 4 m

15. 15 Shear before support B A C 4 m AyAy ByBy 1 x 1 8 m12 m -0.5 0.5 AyAy x -0.5 V B - = A y V B - = A y -1 1 AyAy 8 m VB-VB- MBMB x AyAy VB-VB- MBMB V B- x -0.5

16. 16 Shear after support B A C 4 m AyAy ByBy 1 x 1 8 m12 m -0.5 0.5 AyAy x V B+ x 1 V B + = 0 4 m VB+VB+ MBMB 1 V B + = 1 4 m VB+VB+ MBMB

17. 17 Moment at support B A C 4 m AyAy ByBy 1 x 1 8 m12 m -0.5 0.5 AyAy x -4 MBMB x 1 M B = 8A y -(8-x) M B = 8A y 1 AyAy 8 m VB-VB- MBMB x AyAy VB-VB- MBMB

18. 18 P = 1 Reaction Influence Line for Beam C A B P = 1 AyAy ByBy  y = 1 C A B L

19. 19  y = 1 C A B P = 1 AyAy ByBy C A B L

20. 20 L A B a - Pinned Support RARA RARA x 1 b C AB

21. 21 AB AB ab L RARA RARA x 1 1 - Fixed Support

22. 22 Shear C AB P = 1 a b L  y =1  yL  yR A B VCVC VCVC P = 1 AyAy ByBy  y =1

23. 23 L A B a VCVC VCVC VCVC x 1 b L -a L 1 Slope s A = 1 L Slope s B = 1 L Slope at A = Slope at B - Pinned Support b C AB

24. 24 AB ab L A B VBVB VBVB VBVB x 11 - Fixed Support

25. 25 Bending Moment a b L 1 A B MCMC MCMC P = 1 AyAy ByBy h C AB

26. 26 a L A B Hinge MCMC MCMC ba  A = b L B=B= a L  C =  A +  B = 1 MCMC x ab L - Pinned Support b C AB

27. 27 AB ab L A B MCMC MCMC MBMB x 1 -b - Fixed Support

28. 28 General Shear x V BL x VCVC -1/4 3/4 1 x VDVD -2/4 2/4 1 -3/4 x VEVE 1/4 1 A CDEBFGH L L/4

29. 29 x V BL x VGVG 1 x VFVF 1 x V BR 1 A CDEBFGH L L/4

30. 30 General Bending Moment s A = 3/4 s B = 1/4  = s A + s B = 1 s A = 1/2 s B = 1/2  = s A + s B = 1 s A = 1/4 s B = 3/4  = s A + s B = 1 x MCMC 3/16 x MDMD 4/1 6 x MEME 3/1 6 A CDEBFGH L L/4

31. 31 x MBMB 3L/4 1 x MGMG L/4 1 x MFMF 2L/4 1 A CDEBFGH L L/4

32. 32 Example 6-2 Construct 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 C D E 2 m 4 m Hinge

33. 33 A C D E 2 m 4 m B RARA x 1 RARA SOLUTION

34. 34 A C D E 2 m 4 m B RCRC RCRC x 1 8/6 4/6

35. 35 A B C D E 2 m 4 m RERE RERE x -2/6 2/6 1

36. 36 A B C D E 2 m 4 m VDVD VDVD VDVD x 1 2/6 1 s E = s C s E = 1/6 s C = 1/6 = -2/6 = 4/6

37. 37 Or using equilibrium conditions: V D = 1 -R E 1 VDVD x 2/6 -2/6 4/6 RERE VDVD MDMD 4 m 1 x V D = -R E RERE VDVD MDMD 4 m RERE x -2/6 1 2/6 A B C D E 2 m 4 m Hinge

38. 38 A B C D E 2 m 4 m 4 -1.33 MDMD x (2)(4)/6 = 1.33  D =  C +  E = 1  C = 4/6 2 2/6 =  E MDMD MDMD

39. 39 Or using equilibrium conditions: M D = -(4-x)+4R E 1 RERE VDVD MDMD 4 m 1 x M D = 4R E RERE VDVD MDMD 4 m RERE x -2/6 1 2/6 MDMD x -8/6 8/6 A B C D E 2 m 4 m Hinge

40. 40 A B C D E 2 m 4 m V CL x

41. 41 A B C D E 2 m 4 m Or using equilibrium conditions: 1 RARA x 1 V CL = R A - 1 V CL x V CL = R A RARA 1 V CL MBMB RARA MBMB

42. 42 A B C D E 2 m 4 m V CR x 1 0.333 0.667

43. 43 A B C D E 2 m 4 m Or using equilibrium conditions: 1 V CR x 0.333 1 0.667 RERE x -2/6 = -0.333 1 2/6=0.33 V CR = -R E RERE V CR MCMC V CR = 1 -R E RERE V CR MCMC 1

44. 44 A B C D E 2 m 4 m MCMC MCMC MCMC x 1 -2

45. 45 A B C D E 2 m 4 m MCMC x 1 -2 Or using equilibrium conditions: 1 RERE x -2/6 = -0.333 1 2/6=0.33 M C = 6R E RERE V CR MCMC 6 m RERE V CR MCMC 1

46. 46 Example 6-3 Construct the influence line for - the reaction at A and C - shear at D, E and F - the moment at D, E and F Hinge AB CD EF 2 m

47. 47 SOLUTION RARA RARA x 1 1 0.5 -0.5 2 m A B CD E F

48. 48 2 m AB C D E F RCRC x RCRC 1 0.5 1.5 2

49. 49 2 m AB C D E F VDVD VDVD VDVD x 0.5 -0.5 1 1 = =

50. 50 AB CD E 2 m F VEVE VEVE VEVE x 1 -0.5 0.5 -0.5 = =

51. 51 2 m AB CD E F VFVF VFVF VFVF x 1 = =

52. 52 2 m AB CD E F MDMD MDMD MDMD x -2-2  D = 1 1 2

53. 53 2 m AB CD E F  E = 1 MEME MEME MEME x  C = 0.5  B = 0.5 (2)(2)/4 = 1 -2

54. 54 2 m AB CD E F MEME MEME MFMF x  F = 1 -2

55. 55 Example 6-4 Determine the maximum reaction at support B, the maximum shear at point C and the maximum positive moment that can be developed at 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 4 m A B C

56. 56 0.5(12)(1.5) = 9 SOLUTION RBRB x 1 1.5 0.5 = 48000 N = 48 kN (R B ) max + (8000)(1.5)= (1000)(9)+ (3000)(9) 8000 N 1000 N/m 3000 N/m 4 m A B C

57. 57 0.5(4)(-0.5) = -1 0.5(4)(0.5) = 1 0.5(4)(-0.5) = -1 = (1000)(-2+1) 4 m A B C VCVC x 0.5 -0.5 1000 N/m (V C ) max + (8000)(-0.5) = -11000 N = 11 kN + (3000)(-2) -0.5 3000 N/m 8000 N

58. 58 8000 N 3000 N/m 1000 N/m (1/2)(4)(2) = 4 +(1/2)(8)(2) = 8 4 m A B C 3000 N/m (M C ) max positive = (8000)(2) = 44000 Nm = 44 kNm + (8-4)(1000)+ (3000)(8) MCMC x 2 -2

59. 59 ABCDEF G H Influence Line for Girder Forces Apply to the Girder (b/L)P1(b/L)P1 (a/L)P1(a/L)P1 P2P2 P1P1 ab P2P2 L A B P 1 D P 2 A BC DE F G H

60. 60 A BC DE F G H Reaction 1 0.5 RGRG RHRH RHRH 1

61. 61 A BC DE F G H RGRG 1 0.5 RGRG 11 RHRH

62. 62 A BC DE F G H Shear L abbb 1 0.5 1 1 VCVC VCVC VCVC x 2b/L -(a+b)/L 1

63. 63 A BC DE F G H L V CD 1 0.5 1 V CD x -a/L b/L ab

64. 64 A BC DE F G H Bending Moment L ab ab/L 1 0.5 MCMC x 1 1 MCMC MCMC

65. 65 A BC DE F G H L MFMF x 1 0.5 1 MFMF MFMF ab ab/L

66. 66 Example 6-5 Draw the influence for - the Reaction R G and R F - Shear V CD - the moment M C and M H 1.5 m A BC DE H F 3 m G

67. 67 1 1.5 m A BC DE H F 3 m G RGRG x RGRG 1.333 0.667 0.333

68. 68 1.5 m A BC DE H F 3 m G RFRF x RFRF 1 0.667 0.333 -0.333

69. 69 1.5 m A BC DE H F 3 m G V CD x 1 0.5 4.5/9 -4.5/9 0.333 -0.333 0.333

70. 70 1.5 m A BC DE H F 3 m G MCMC MCMC 1 1 MCMC x (3)(6)/9 = 2 -2 1

71. 71 1.5 m A BC DE H F 3 m G MHMH MHMH 1 0.5 MHMH x (4.5)(4.5)/9 = 2.25 1.5 -1.5

72. 72 Example 6-6 Draw the influence line diagrams of girder for - the reaction at C and G, - shear at E and H, - bending moment at H. A G B 6 @ 4 m = 24 m C DE FH 2 m

73. 73 A G B 6 @ 4 m = 24 m C DE FH2 m SOLUTION RCRC RCRC 1 1 1 0.75 0.50 0.25 1.25

74. 74 A G B 6 @ 4 m = 24 m C DE FH2 m RGRG RGRG 1 1 1 0.25 0.50 0.75 -0.25

75. 75 A G B 6 @ 4 m = 24 m C DE FH2 m VEVE 1 1 1 1 VEVE VEVE 8/16 = 0.5 -8/16 = 0.5 -0.25 0.25 8 m

76. 76 A G B 6 @ 4 m = 24 m C DE FH2 m 6/16 = 0.375 -10/16 = 0.625 VHVH 1 1 VHVH VHVH 1 0.5 -0.25 0.25 -0.50 10 m 6 m

77. 77 A G B 6 @ 4 m = 24 m C DE FH2 m MHMH 1 1 1 0.5 MHMH MHMH (10)(6)/16 = 3.75 1.50 2.50 -1.50 3 10 m 6 m

78. 78 Draw the influence for - the Reaction R G and R H - Shear V C and V CD - the moment M C, M D, and M F and determine the maximum for - the Reaction (R G ) max and (R H ) max - Shear (V CD ) max due to - a uniform dead load 2 kN/m - a uniform live load 5 kN/m - a concentrated live load 50 kN Example 6-7 2 m 4 m 2 m A BC DE F G H

79. 79 2 m 4 m 2 m A BC DE F G H RGRG 1 0.5 10/14 6/14 2/14 1/14 RGRG 11 RHRH

80. 80 2 m 4 m 2 m A BC DE F G H 1 0.5 RGRG RHRH RHRH 1 12/14 8/14 4/14 6/14

81. 81 2 m 4 m 2 m A BC DE F G H VCVC VCVC 1 0.5 VCVC x 8/14 -6/14 1 1 1 -2/14 4/14 -1/14

82. 82 -8/14 6/14 2 m 4 m 2 m A BC DE F G H VFVF VFVF 1 0.5 1 V CD x -1/14 -2/14 -6/14 4/14

83. 83 2 m 4 m 2 m A BC DE F G H (6)(8)/14 = 3.43 1 0.5 MCMC x (4/8)(3.43) (2/6)(3.43) 1 1 MCMC MCMC

84. 84 2 m 4 m 2 m A BC DE F G H 1 0.5 1 1 MDMD x (10)(4)/14 = 2.86 (6/10)(2.86) (2/10)(2.86) MDMD MDMD

85. 85 (8)(6)/14=3.428 2 m 4 m 2 m A BC DE F G H MFMF x 1 0.5 1 2.285 2.571 0.857 0.429 MFMF MFMF

86. 86 [0.5×4 (2/14)] + [(0.5)(2/14 + 1)(12) = 7.143 RGRG 1 10/14 6/14 2/14 1/14 Maximum Reaction 2 kN/m 5 kN/m 50 kN (R G ) max = (2)(7.143) + (5)(7.143)+ (50)(1) = 100 kN 2 m 4 m 2 m A BC DE F G H

87. 87 2 m 4 m 2 m A BC DE F G H (0.5)(16)(12/14) = 6.857 RHRH 12/14 8/14 4/14 6/14 Maximum Reaction 2 kN/m 5 kN/m 50 kN (R H ) max = (2)(6.857)+ (5)(6.857)+ (50)(12/14) = 90.86 kN

88. 88 2 m 4 m 2 m A BC DE F G H + (0.5)(5.6)(4/14) = + 0.8 -[0.5(4)(2/14) + 0.5(2/14) + (6/14)(4) + 0.5(2.4)(6/14)] = -1.943 2.4 m V CD -1/14 -2/14 -6/14 4/14 Maximum Shear 2 kN/m 5 kN/m 50 kN (V CD ) max =(2)(-1.943 + 0.8)(5)(-1.943)+ (50)(-6/14) = -33.43 kN

89. 89 Influence Line for Trusses RARA x 1 RERE x 1 (3/4) (2/4) (1/4) (2/4) (3/4) (1/4) A BC E G H D F 4 @ x m h

90. 90 A BC E G H D F 4 @ x m h 1 kN Consider the right hand F CB RERE F GB F GH F BC x (2x/h)(1/4) +  M G  = 0: RERE x 1 (2/4) (3/4) (1/4)

91. 91 A BC E G H D F 4 @ x m h 1 kN Consider the left hand F HG F BC RARA F BG +  M G  = 0: F BC x (2x/h)(1/4) RARA x 1 (3/4) (2/4) (1/4) (2x/h)(1/4) (2x/h)(2/4)

92. 92 Determine the maximum force developed in member BC, BG,and CG of the truss due to the wheel loads of the car. Assume the loads are applied directly to the truss. the wheel loads of the car 4 kN1.5 kN 2 m Example 6-8 A BC E G H 4 m D F 3 m

93. 93 RARA x 1 0.75 0.5 0.25 RERE x 0.5 0.75 0.25 1 SOLUTION A BC E G H 4 m D F 3 m

94. 94 A BC E G H 4 m D F 3 m Influence Line for F BC 1 kN Consider the right hand F BC = (6/4) R E = 1.5 R E F CB RERE F GB F GH RERE x 0.5 0.75 0.25 1 F BC x 1.5(0.25) = 0.375 +  M G  = 0:

95. 95 A BC E G H 4 m D F 3 m 0.375 1.5(0.5) = 0.75 1 kN Consider the left hand F HG F BC RARA F BG RARA x 1 0.75 0.5 0.25 F BC = 1.5 R A F BC x 0.375 +  M G  = 0:

96. 96 A BC E G H 4 m D F 3 m RERE x 0.5 0.75 0.25 1 Influence Line for F BG 1 kN Consider the right hand  F y = 0 ; F BG cos  = R E F GH F GB F CB RERE  F BG = 1.25 R E F BG (4/5) = R E 1.25(0.25) = 0.3125 F BG x

97. 97 A BC E G H 4 m D F 3 m 1 kN Consider the left hand  F y = 0 ; F BG cos  + R A = 0 RARA x 1 0.75 0.5 0.25 F HG F BC RARA F BG  0.3125 F BG x F BG = -1.25 R A F BG (4/5) = -R A -0.3125 -1.25(0.5) = -0.625

98. 98 A BC E G H 4 m D F 3 m F CG x Influence Line for F CG 1 kN 0 0

99. 99 A BC E G H 4 m D F 3 m F CG x 1 kN 1 1

100. 100 A BC E G H 4 m D F 3 m (F BC ) max by Loads (F BC ) max = (4)(0.75)+ (1.5)(0.5) = 3.75 kN # 0.375 0.75 F BC x 0.375 4 kN1.5 kN 2 m 0.5

101. 101 A BC E G H 4 m D F 3 m (F BG ) max by Loads (F BG ) max = (4)(-0.625)+ (1.5)(-0.417) = -3.126 kN # F BG x -0.3125 0.3125 -0.625 4 kN1.5 kN 2 m -0.417

102. 102 A BC E G H 4 m D F 3 m (F CG ) max by Loads F CG x 1 0.333 (F CG ) max = (4)(1)+ (1.5)(0.333) = 4.50 kN # 4 kN1.5 kN 2 m

103. 103 Maximum Influence at a Point Due to a Series of Concentrated Loads F2F2 x1x1 x2x2 F1F1 F3F3 x MCMC ab/L VCVC x (b/L) (-a/L) A B C ab

104. 104 A B C ab x MCMC ab/L F2F2 x1x1 x2x2 F1F1 F3F3 VCVC x (b/L) (-a/L)

105. 105 A B C ab VCVC x (b/L) (-a/L) x MCMC ab/L F2F2 x1x1 x2x2 F1F1 F3F3

106. 106 A B C ab VCVC x (b/L) (-a/L) x MCMC ab/L F2F2 x1x1 x2x2 F1F1 F3F3

107. 107 VCVC x 0.75 -0.25 4 kN 1 m3 m 1 kN 6 kN Shear A B C 2 m6 m

108. 108 Case 1 VCVC x 0.75 -0.25 4 kN 1 m3 m 1 kN 6 kN 0.625 0.25 (V C ) 1 = 1(0.75) + 4(0.625) + 6(0.25) = 4.75 kN A B C 2 m6 m

109. 109 Case 2 VCVC x 0.75 -0.25 4 kN 1 m3 m 1 kN 6 kN (V C ) 2 = 1(-0.125) + 4(0.75) + 6(0.375) = 5.125 kN -0.125 0.375 A B C 2 m6 m

110. 110 Case 3 VCVC x 0.75 -0.25 4 kN 1 m3 m 1 kN 6 kN (V C ) 3 = 6(0.75) = 4.5 kN (V C ) max = (V C ) 2 = 5.125 kN A B C 2 m6 m

111. 111 The critical position of the loads can be determined in a more direct manner by finding the change in shear,  V, which occurs when the load are moved from Case 1 to Case 2, then from Case 2 to Case 3, and so on.If the slope of the influence line is s, then (y 2 - y 1 ) = s(x 2 - x 1 ), and therefore  V = Ps(x 2 - x 1 ) Sloping Line ----------(6-1) If the load moves past a point where there is a discontinuity or “jump” in the influence line, as point C, then the change in shear is simply  V = P(y 2 - y 1 ) Jump ----------(6-2)

112. 112 A B C 2 m6 m 4 kN 1 m3 m 1 kN 6 kN VCVC x 0.75 -0.25  V 1-2 = 1(-0.25 - 0.75) + 1(1/8)(1) + 4(1/8)(1) + 6(1/8)(1) = 0.375 kN (V C ) 2 = (V C ) 1 +  V 1-2 = 4.75 + 0.375 = 5.125 kN s = 1/8

113. 113 4 kN 1 m3 m 1 kN 6 kN VCVC x 0.75 -0.25  V 2-3 = 1(1/8)(1) + 4(-0.25-0.75) + 4(1/8)(2) + 6(1/8)(3) = -0.625 kN (V C ) 3 = (V C ) 2 +  V 2-3 = 5.125 -0.625 = 4.5 kN s = 1/8 A B C 2 m6 m

114. 114 4 kN 1 m3 m 1 kN 6 kN Moment MCMC x (2)(6)/8 = 1.5 A B C 2 m6 m

115. 115 Case 1 MCMC x 1.5 1.25 0.5 (M C ) 1 = 1(1.5) + 4(1.25) + 6(0.5) = 9.5 kNm 4 kN 1 m3 m 1 kN 6 kN A B C 2 m6 m

116. 116 Case 2 MCMC x 1.5 (M C ) 2 = 1(0.75) + 4(1.5) + 6(0.75) = 11.25 kNm 0.75 4 kN 1 m3 m 1 kN 6 kN A B C 2 m6 m

117. 117 Case 3 MCMC x 1.5 (M C ) 3 = 6(1.5) = 9 kNm (M C ) max = (M C ) 2 = 11.25 kNm 4 kN 1 m3 m 1 kN 6 kN A B C 2 m6 m

118. 118 We can also use the foregoing methods to determine the critical position of a series of concentrated forces so that they create the largest internal moment at a specific point in a structure. Of course, it is first necessary to draw the influence line for the moment at the point and determine the slopes s of its line segments. For a horizontal movement (x 2 - x 1 ) of a concentrated force P, the change in moment,  M, is equivalent to the magnitude of the force times the change in the influence-line ordinate under the load, that is  M = Ps(x 2 - x 1 ) Sloping Line ----------(6-3)

119. 119 = 1.75 kNm = 11.25 kNm MCMC x 1.5 4 kN 1 m3 m 1 kN 6 kN A B C 2 m6 m

120. 120 MCMC x 1.5 = -2.25 kNm = 9 kNm 4 kN 1 m3 m 1 kN 6 kN A B C 2 m6 m

121. 121 Example 6-9 Determine the maximum shear created at point B in the beam shown in the figure below due to the wheel loads of the moving truck. 4 kN 9 kN 15 kN10 kN 1 m2 m A C B 4 m

122. 122 A C B 4 m SOLUTION -0.5 0.5 VBVB x 4 kN 9 kN 15 kN10 kN 1 m2 m V BR = 4(0.5) + 9(0.375) + 15(0.125) = 7.25 kN 0.125 0.375 V BL = 4(-0.5) + 9(0.375) + 15(0.125) = 3.25 kN Case 1

123. 123 -0.5 0.5 VBVB x 4 kN 9 kN 15 kN10 kN 1 m2 m -0.375 0.25 V BR = 4(-0.375) + 9(0.5) + 15(0.25) + 10(0) = 6.75 kN V BL = 4(-0.375) + 9(-0.5) + 15(0.25) + 10(0) = -2.25 kN Case 2 A C B 4 m

124. 124 -0.5 0.5 VBVB x 4 kN 9 kN 15 kN10 kN 1 m2 m 0.25 -0.25 -0.125 V BR = 4(-0.125) + 9(-0.25) + 15(0.5) + 10(0.25) = 7.25 kN V BL = 4(-0.125) + 9(-0.25) + 15(-0.5) + 10(0.25) = -7.75 kN Case 3 A C B 4 m

125. 125 -0.5 0.5 VBVB x 4 kN 9 kN 15 kN10 kN 1 m2 m -0.25 V BR = 9(0) + 15(-0.25) + 10(0.5) = 1.25 kN V BL = 9(0) + 15(-0.25) + 10(-0.5) = -8.75 kN The maximum shear created at point B is -8.75 kN Case 4 A C B 4 m

126. 126 Case 1V BR = 7.25 kN Case 2V BR = 6.75 kN Case 3V BR = 7.25 kN Case 4V BR = 1.25 kN 4 kN 9 kN 15 kN10 kN 1 m2 m -0.5 0.5 VBVB x Summary 1-m Movement of 4 kN,  VB = 6.75-7.25 = - 0.5 kN 2-m Movement of 9 kN,  VB = 7.25-6.75 = + 0.5 kN 2-m Movement of 15 kN,  VB = 1.25-7.25 = - 6 kN A C B 4 m

127. 127 Or using the sloping line -0.5 0.5 VBVB x 4 kN 9 kN 15 kN10 kN 1 m2 m A C B 4 m

128. 128 4 kN 9 kN 15 kN 10 kN 1 m2 m -0.5 0.5 VBVB x A C B 4 m 4 kN9 kN 15 kN 10 kN 1 m2 m = +0.5 kN 2-m Movement of 9 kN = -6 kN 2-m Movement of 15 kN = -0.5 kN 1-m Movement of 4 kN The correct position of the loads. A C B 4 m 4 kN 9 kN 15 kN 10 kN 1 m2 m 4 kN 9 kN 15 kN 10 kN 1 m2 m

129. 129 -0.5 0.5 VBVB x V BR = 9(0) + 15(-0.25) + 10(0.5) = 1.25 kN V BL = 9(0) + 15(-0.25) + 10(-0.5) = -8.75 kN The maximum shear created at point B is -8.75 kN (-0.25) 4 kN 9 kN 15 kN 10 kN 1 m2 m A C B 4 m

130. 130 Example 6-10 Determine the maximum positive moment and negative moment created at point B in the beam shown in the figure below due to the wheel loads of the crane. AC B 2 m 3m3m 3 kN 8 kN 4 kN 2 m3 m

131. 131 x MBMB 2(3)/5 = 1.2 -0.8 ACB 2 m 3m3m 3 kN 8 kN 4 kN 2 m3 m SOLUTION M B = 3(1.2) + 8(0) = 3.6 kNm Positive moment CASE I

132. 132 ACB 2 m 3m3m 3 kN 8 kN 4 kN 2 m3 m x MBMB 1.2 -0.8 M B = 8(1.2) + 3(0.4) = 10.8 kNm 0.4 CASE II

133. 133 ACB 2 m 3m3m 3 kN 8 kN 4 kN 2 m3 m x MBMB 1.2 -0.8 M B = 4(1.2) + 8(0) + 3(-0.8) = 2.4 kNm The maximum positive moment created at point B is 10.8 kNm CASE III

134. 134 ACB 2 m 3m3m 3 kN 8 kN 4 kN 2 m3 m x MBMB 1.2 -0.8 M B = 8(-0.8) + 4(0.4) = -4.8 kNm Negative moment 0.4 CASE I

135. 135 3 kN 8 kN 4 kN 2 m3 m x MBMB 1.2 -0.8 M B = 4(-0.8) = -3.2 kNm ACB 2 m 3m3m The maximum negative moment created at point B is 4.8 kNm CASE II

136. 136 Absolute Maximum Shear and Moment L CL L/2 FRFR x F1F1 F2F2 F3F3 d1d1 d2d2 2 AB AyAy ByBy

137. 137 A (L/2 - x) F1F1 d1d1 V2V2 M2M2 For maximum M 2 we require 2

138. 138 L AB CL L/2 F1F1 F2F2 F3F3 x1x1 x2x2 FRFR

139. 139 FRFR F1F1 F2F2 F3F3 x1x1 x2x2 b b FRFR F1F1 F2F2 F3F3 x1x1 x2x2 b b a b M x L AB

140. 140 FRFR F1F1 F2F2 F3F3 x1x1 x2x2 a a FRFR F1F1 F2F2 F3F3 x1x1 x2x2 a a a b M x The absolute maximum moment is comparison by M S1 and M S2. L AB

141. 141 1200 kg 400 kg 8 m Example 6-11 Determine the absolute maximum moment on the simply supported beam cased by the wheel loads. 30 m A B

142. 142 CL 15 m 1200 kg 400 kg 8 m 1600 kg  M @1200 kg = 0 : A B

143. 143 CL M S =(1200)(7.47) + (400)(3.74) = 10460 kgm 14 m 16 m M 1200 x (14)(16)/30 = 7.47 3.74 +  M B  = 0: Global: A y = 746.67 kg 1600 kg 1200 kg 400 kg 6 m 2 m a a 1 m 1 A 14 m 1 A y = 746.67 kg V1V1 M1M1 +  M 1  = 0: M 1 = 10453.38 kgm Or using equilibrium conditions: A B

144. 144 b b 3 m CL M S =(1200)(4)+ (400)(7.2)= 7680 kgm By comparison, M max = 10460 kgm M 400 (18)(12)/30 = 7.2 x 4 18 m 12 m b b 3 m 1600 kg 1200 kg 400 kg 6 m 2 m 1600 kg 1200 kg 400 kg 6 m 2 m A B

145. 145 4.6 T8.2T 4.2 m 1.2 m Example 6-12 Determine the absolute maximum moment on the simply supported beam cased by the wheel loads. 20 m A B

146. 146 4.6 T8.2T 4.2 m 1.2 m  M @ 4.6 T = 0 : F R =21 T 20 m A B

147. 147 A B 4.6 T8.2T F R =21 T 4.2 m 1.2 m 1.875 m 10 m CL M 4.6T x a = 8.125 m b = 11.875 m 4.82 3.12 2.63 M S =(4.6)(4.82)+ (8.2)(3.12) = 69.32 Tm + (8.2)(2.63)

148. 148 4.6 T8.2T F R =21 T 4.2 m 1.2 m 0.225 m CL 10 m M 8.2T x a = 10.225 m b = 9.775 m 5 2.95 4.39 M S =(4.6)(2.95)+ (8.2)(5) = 90.57 Tm + (8.2)(4.39) By comparison, M max = 90.57 Tm +  M B  = 0: Global: A y = 10.74 T Or using equilibrium conditions: 2 +  M 2  = 0: M 2 = 90.50 Tm A 10.225 m 2 A y = 10.74 T V2V2 M2M2 4.6 T 4.2 m A B

149. 149 Example 6-13 Determine the absolute maximum moment on the simply supported beam cased by the wheel loads. 4 kN 9 kN 15 kN10 kN 1 m2 m A B 4 m

150. 150 SOLUTION  M @ 4 kN = 0 : 4 kN 9 kN 15 kN10 kN 1 m2 m F R =38 kN 1.74 m 0.26 m A B 4 m

151. 151 a = 3.13 mb = 4.87 m x M 9 kN (3.13)(4.87)/8 = 1.91 1.13 0.34 1.30 M S = 4(1.30) + 9(1.91) + 15(1.13) + 10(0.34) = 42.74 kNm 4 kN 9 kN 15 kN10 kN 1 m2 m F R =38 kN 0.87 A B 4 m

152. 152 FRFR 0.13 m 4 kN 9 kN 15 kN10 kN 1 m2 m 0.13 m a = 4.13 mb = 3.87 m x M 15 kN (4.13)(3.87)/8 = 2.0 1.03 0.55 0.97 M S = 4(0.55) + 9(1.03) + 15(2.0) + 10(0.97) = 51.17 kNm By comparison, M max = 51.17 kNm 3 +  M A  = 0: Global: B y = 18.38 kN Or using equilibrium conditions: B 2 m 15 kN 10 kN 3 18.38 kN 3.87 m V3V3 M3M3 +  M 3  = 0: -M 3 -10(2) + 18.38(3.87) = 0 M 3 = 51.13 kNm A B 4 m