Method of Consistent Deformation

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1 Method of onsistent eformation Structural nalysis y R.. Hibbeler Theory of Structures-II M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering and Technology, Wah antt

2 FRMES Method of consistent deformation is very useful for solving problems involving statically indeterminate frames for single story and unusual geometry. Problems involving multistory frames, or with high indeterminacy are best solved using the slope deflection or moment distribution or the stiffness methods.

3 Example 6 etermine the support reactions on the frame shown. EI is constant. 8 kn/m 5 m 4 m ctual Frame 3

4 Principle of Superposition 8 kn/m 5 m 4 m ctual Frame y inspection the frame is indeterminate to the first degree. 4

5 Principle of Superposition We will choose the horizontal reaction at support as the redundant. The pin at is replaced by the roller, since a roller will not constraint in the horizontal direction. 8 kn/m Primary Structure Δ 5

6 Principle of Superposition 8 kn/m 8 kn/m Primary Structure = + Δ 5 m ctual Frame 4 m redundant x applied x 6 x f

7 ompatibility Equation 8 kn/m 8 kn/m Δ x f Primary Structure = + Δ 5 m ctual Frame 4 m redundant x applied x 7 x f

8 ompatibility Equation Δ x f The terms Δ and f will be computed using the method of virtual work. The frame s x coordinates and internal moments are shown in figure. It is important that in each case the selected coordinate x 1 or x be the same for both the real and virtual loadings. lso the positive directions for M and m must be same. 8

9 ompatibility Equation For Δ we require application of real loads and a virtual unit load at x 1 M 8 kn/m 1 x1 4 x 1 1 kn x 1 m1. 8x 1 kn.8 kn M m 1x x x 1 kn kn.8 kn 9

10 Δ L Mm 5 dx EI EI EI x 4x.8x dx 1 x EI dx EI dx x 1 M 8 kn/m x 4x kn x 1 m1. 8x 1 kn.8 kn M m 1x x x 1 kn kn 1.8 kn

11 ompatibility Equation For f we require application of real unit load acting at and a virtual unit load acting at 1 kn.8 kn x 1 m1. 8x 1 m 1x x 1 kn.8 kn 11

12 f L mm dx EI EI EI.8x dx 1 x EI 48. EI EI dx dx 1 kn.8 kn x 1 m1. 8x 1 m 1x x 1 kn.8 kn 1

13 ompatibility Equation Δ x f (1) Substituting the data in Eq. (1) EI x 48. EI x 3.47 kn NS 13

14 Equilibrium ondition Showing x on the free body diagram of the frame in the correct direction, and applying the equations of equilibrium, we have 4 kn x y.5 m.5 m 4 m y 3.47 kn 14

15 Equilibrium ondition x 4 kn y.5 m.5 m 4 m 3.47 kn Fx ; x x 3. 47kN y M kn ; 4 y y 8 Fy ; y 4. 8 y 17. kn 15

16 Example 7 etermine the moment at fixed support for the frame shown. EI is constant. 1 lb/ft 5 ft 4 ft 8 ft 3 ft ctual Frame 16

17 Principle of Superposition 1 lb/ft 5 ft 4 ft 8 ft ctual Frame 3 ft y inspection the frame is indeterminate to the first degree. 17

18 Principle of Superposition M can be directly obtained by choosing as the redundant. 1 lb/ft The capacity of the frame to support a moment at is removed and therefore a pin is used at for support. ctual Frame 18

19 ompatibility Equation Reference to point M (1) 1 lb/ft 1 lb/ft = + θ M α M actual frame primary structure Redundant M applied 19

20 ompatibility Equation M (1) The terms θ and α will be computed using the method of virtual work. The frame s x coordinates and internal moments are shown in figure.

21 ompatibility Equation Reference to point 5 lb M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M x 1 m x 1 x lb.833 lb x 1 1 lb.ft 3 lb 1

22 For θ we require application of real loads and a virtual unit couple moment at 5 lb M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M x 1 m x 1 x lb.833 lb x 1 1 lb.ft 3 lb

23 L Mm 8 dx EI EI EI EI 9.17x 1.833x dx x 5x.67x EI EI dx 5 lb M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M x 1 m x 1 x lb.833 lb x 1 3 lb 1 lb.ft 3

24 For α we require application of real unit couple moment and a virtual unit couple moment at.5 lb x m. 667x.833 lb.667 lb m x lb x 1 1 lb.ft 4

25 L m m 8 dx EI EI EI EI 1.833x dx 5.67x 1 1 EI.5 lb EI dx x m. 667x.833 lb.667 lb m x lb x 1 1 lb.ft 5

26 Substituting these results into Eq. (1), and solving yields M 81.8 M EI 4 lb. ft 4.4 EI NS 1 lb/ft The negative sign indicates M acts in opposite direction to that shown in figure. M =4 lb.ft 6

27 Example 8 etermine the reactions and draw the shear and bending moment diagrams. EI is constant. 1 k k/ft 15 ft ctual Frame 3 ft 7

28 Principle of Superposition 1 k k/ft 15 ft ctual Frame 3 ft egree of indeterminacy = 8

29 Principle of Superposition We will choose the horizontal reaction x and vertical reaction y at point as the redundants. ctual Frame 1 k k/ft x y 9

30 Principle of Superposition Primary structure is obtained by removing the hinged support at point. 1 k k/ft Primary Structure 3

31 Principle of Superposition Primary structure is subjected separately to the external loading and redundants x and y as shown. 1 k k/ft Δ y Primary Structure Δ x 31

32 Principle of Superposition Primary structure is subjected separately to the external loading and redundants x and y as shown. x Δ yx = x f yx Redundant x applied Δ xx = x f xx 3

33 Principle of Superposition Primary structure is subjected separately to the external loading and redundants x and y as shown. Δ yy = y f yy y Redundant y applied Δ xy = y f xy 33

34 ompatibility Equation ' ' Δ Δ Δ x xx xy Δ x x f xx y f xy (1) Δ y Δ ' yx Δ ' yy Δ y x f yx y f yy () The equations for bending moments for the members of the frame due to external loading and unit values of the redundants are tabulated in the table. y applying the virtual work method, we will find Δ x, Δ y, f xx, f yx, f xy, f yy, 34

35 ompatibility Equation 1 k k/ft M x M 1 M 3 1 k x 1 x 3 15 k-ft 6 k Member Origin Limits M (k-ft) x 1-3 -x

36 ompatibility Equation m x x m x1 m x3 1 k x 1 x 3 1 k Member Origin Limits M (k-ft) m x (k-ft/k) x 1 -x 1-3 -x x 3 36

37 ompatibility Equation m y x m y1 m y3 3 k-ft 1 k x 1 x 3 1 k Member Origin Limits M (k-ft) m x (k-ft/k) m y (k-ft/k) x 1 -x x -15 x -15 -x 3 37

38 Δ x Δ x 15 1x x 3 x 15 L Mm 15 x 1 1 dx dx EI EI 15 x3 dx3 EI k ft EI 1 EI dx Δ y Δ y L Mm EI y dx x 3 3 x x EI 1 dx k ft EI 1 3 EI dx 38

39 f xx L m m EI x x dx 15 x x 1 EI dx 1 EI dx 3 EI dx 3 f xx 9 ft 3 EI f yy L m y m EI y dx x EI dx 1 EI dx f yy 5 ft 3 EI 39

40 f xy f yx L m x m EI y dx 15 x x 1 EI dx 1 EI dx f xy f yx 115 ft 3 EI 4

41 Δ x EI k ft 3 Δ y 6415 EI k ft 3 f xx f yy 9 ft 3 EI 5 ft 3 EI f xy f yx 115 ft 3 EI 41

42 Now put these values in the Equations (1) and () x 115y x 5y (1) () y solving (1) and () simultaneously we get x y 1.53 k 33.6 k 4

43 pplying equations of equilibrium, we have the other support reactions as 1 k k/ft.53 k 53. k-ft 1.53 k k 33.6 k 43

44 Shear diagram

45 Moment diagram

46 TRUSSES The degree of indeterminacy of a truss can be find using Equation b+r > j. where b = unknown bar forces, r = support reactions, j = equations of equilibrium This method is quite suitable for analyzing trusses that are statically indeterminate to the first or second degree. 46

47 Example 9 etermine the force in member of the truss shown. E is same for all members. 4 lb 6 ft 8 ft ctual Truss 47

48 The truss is statically indeterminate to the first degree. b + r = j = (4) 9 > = 1 st degree 4 lb 6 ft 8 ft ctual Truss 48

49 Principle of Superposition The force in member is to be determined, so member is chosen as redundant. This requires cutting this member, so that it cannot sustain a force, making the truss S.. and stable. 4 lb 6 ft ctual Truss 8 ft 49

50 4 lb ctual Truss = 4 lb F Δ + F F f Primary Structure Redundant F applied 5

51 ompatibility Equation With reference to member, we require the relative displacement Δ, which occurs at the ends of cut member due to the 4-lb load, plus the relative displacement F f caused by the redundant force acting alone, be equal to zero, that is F f 4 lb F Δ F F f Primary Structure Redundant F applied 51

52 ompatibility Equation Here the flexibility coefficient f represents the relative displacement of the cut ends of member caused by a real unit load acting at the cut ends of member. F f 4 lb F Δ F F f Primary Structure Redundant F applied 5

53 ompatibility Equation This term, f, and Δ will be computed using the method of virtual work. F f 4 lb F Δ + F F f Primary Structure Redundant F applied 53

54 ompatibility Equation For Δ we require application of the real load of 4 lb, and a virtual unit force acting at the cut ends of member lb lb 1 lb lb +4 x lb 3 lb 54

55 nnl E E 11 E E E E E +4 4 lb lb 1 lb lb +4 x lb 3 lb 55

56 ompatibility Equation For f we require application of the real unit forces acting on the cut ends of member, and virtual unit forces acting on the cut ends of member lb 1 lb 1 lb 1 lb x

57 f n L E E E -.8 E 1 E lb 1 lb lb 1 lb x

58 Substituting the data into Eq. (1) and solving yields E E F 34 lb (T) F NS Since the numerical result is positive, is subjected to tension as assumed. Using this result, the forces in other members can be found by equilibrium, using the method of joint. 58

59 Example 1 etermine the force in member of the truss shown. 6 ft 8 ft ctual Truss 59

60 Example 11 etermine the reactions and the force in each member of the truss shown in Fig. shown. E = 9, ksi E (6 in. ) F 8 k (6 in. ) (4 in. ) (4 in. ) (4 in. ) (6 in. ) 15 ft (6 in. ) (6 in. ) (6 in. ) 5 k 5 k 3 panels at ft = 6 ft ctual Truss 6

61 The truss is statically indeterminate to the first degree. b + r = j = (6) 13 > = 1 st degree E (6 in. ) F 8 k (6 in. ) (4 in. ) (4 in. ) (4 in. ) (6 in. ) 15 ft (6 in. ) (6 in. ) (6 in. ) 5 k 5 k 3 panels at ft = 6 ft ctual Truss 61

62 E F 8 k ctual Truss x x y 5 k 5 k y x at hinged support is selected as Redundant. Primary structure is obtained by removing the effect of x and replacing hinge by roller support there. Primary structure is subjected separately to external loading and redundant Force x. 6

63 E F 8 k Primary Structure x Δ y 5 k 5 k y E 4 F 8 k k Δ 5 k 5 k 18 k 3 k 63

64 Δ is horizontal deflection at point of primary structure due to external loading. 3 E F 8 k Primary Structure 8 k Δ 5 k 5 k 18 k 3 k 64

65 E F Redundant x is applied x Δ = x f E F 1 k f 1 k 65

66 E F Redundant x is applied x Δ = x f Δ is horizontal deflection at point due to redundant force x. 66

67 f is horizontal deflection at point due to unit force. E F 1 k f 1 k 67

68 ompatibility Equation x f 8 k x x y 5 k 5 k = y x 8 k Primary Structure Δ + Redundant x is applied x y 5 k 5 k y Δ = x f 68

69 We will use virtual work method to find Δ and f. eflection of truss is calculated by where nnl E n = N = axial force in truss members due to virtual unit load acting at joint and in the direction of Δ axial force in truss members due to real load acting that causes Δ 69

70 We will use virtual work method to find Δ and f. eflection of truss is calculated by where n L E f n = n = axial force in truss members due to real unit load acting at joint and in the direction of Δ axial force in truss members due to virtual unit load acting at joint and in the direction of Δ 7

71 TLE Member L (in.) (in. ) N (k) n (k) nnl/ (k/in.) n L/ F=N+n x , , , EF F F E F F f nnl 5,493.6 k/in. E E n L 1 (1/in.) E E 5,

72 nnl E E 6E 6E 8 6E E k/in E E 7

73 n L E f f E 6E 6 E f 1 (1/in) E Now put these results into Equation (1) E x 1 E x k 73

74 F F F F N n x (T) () () Equation of Equilibrium F x x y y k 18 k 3 k x = k y =18 5 k 5 k y =3 74 x =45.78

75 E F 8 k Primary Structure x =17.78 x = k 5 k y =18 y = 3 E 4 F 8 k x = x = y =18 5 k 5 k y = 3 75

76 Example 1 etermine the reactions and the force in each member of the truss shown in Fig. shown. E = constant. E = GPa., = 4 mm F G H 7 kn 1m E 8 kn 8 kn 4 panels at 1m= 4m ctual Truss 76

77 Principle of Superposition egree of Indeterminacy = b + r > j > 8 18 > 16 F G H 7 kn x E y 8 kn 8 kn y E y ctual Truss 77

78 Principle of Superposition y at support and force F G in member G are selected as redundants. F G H 7 kn x E y 8 kn 8 kn y E y ctual Truss 78

79 Principle of Superposition The roller support at is removed and member G is cut to make the structure determinate. F G H 7 kn x E y 8 kn 8 kn E y eterminate Truss 79

80 Principle of Superposition This determinate truss is subjected separately to actual loading, redundant y and redundant force in the redundant member G. F G H 7 kn Δ G E 8 kn 8 kn Δ Primary structure subjected to actual loading 8

81 Principle of Superposition This determinate truss is subjected separately to actual loading, redundant y and redundant force in the redundant member G. Δ G, = y f G, F G H Δ = y f E y Redundant y applied 81

82 Principle of Superposition This determinate truss is subjected separately to actual loading, redundant y and redundant force in the redundant member G. Δ G,G =F G f G,G F G H F G F G Δ,G =F G f,g E Redundant F G applied 8

83 F G H 7 kn ctual Truss x y 8 kn 8 kn y E y = F G H 7 kn E Δ G 8 kn 8 kn + Δ E Primary structure Δ G, = y f G, F G H Δ = y f Δ G,G =F G f G,G F G H + y E Redundant y applied F G F G Redundant F G applied Δ,G =F G f,g E 83

84 Δ G, = y f G, F G H Redundant y applied Δ = y f E y F G H f G, Unit load in the direction of y applied f 1 E 84

85 Δ G,G =F G f G,G F G H Redundant F G applied F G F G Δ,G =F G f,g E f G,G F G H 1 1 f,g Unit force in member G applied E 85

86 ompatibility Equation x F G H 7 kn y 8 kn 8 kn y E y = F G H 7 kn E ctual Truss y f FG f, G y f G, FG f G, G G Δ G 8 kn 8 kn + Δ G, = y f G, F G H Δ E Primary structure Δ = y f Δ G,G =F G f G,G F G H + F G F G y Δ,G =F G f,g Redundant y applied E Redundant F G applied E 86

87 ompatibility Equation y f FG f, G G y f G, FG f G, G Δ = vertical deflection at joint of primary truss due to external loading Δ G = relative displacement b/w cutting ends of member G due to external loading f = vertical deflection at joint due to a unit load at joint f G, = relative displacement b/w cutting ends of member G due to unit load at f G,G = relative displacement b/w cutting ends of member G due to unit force f,g = vertical deflection at joint due to a unit force in member G 87

88 ompatibility Equation We will use the method of virtual work to find the deflections f F f G y G, G y f G, FG f G, G Nn L E f nnl E f,g n n E G L G Nn G L E f G n G n E G L f G, n n E G L 88

89 ompatibility Equation Nn L f E G Nn G L E f G n n L E n n L G G E f f,g G, n n ngl E nl E G N = member forces due to external loading n = member forces due to unit load at joint n G = member forces due to unit force in member G The numerical values of the member forces, as computed by the method of joints, are shown in next figures, and are tabulated in the TLE 89

90 N = member forces due to external loading F G H kn E kn 8 kn

91 n = member forces due to unit load at joint F G H E.5 1 kn.75 91

92 n G = member forces due to unit force in member G F G H E 9

93 Member L (m) N (kn) n (kn/kn) n G (kn/kn) Nn L (kn.m) Nn G L (kn.m) n L (m) n G L (m) n n G L (m) F = N + n y + n G F G (kn) E FG GH F G H F G F H EH

94 ompatibility Equation G 4,47.64 kn. m E kn. m E m f E G y f FG f, G y f G, FG f G, G m E y substituting these values into the above equations f G, G f G, f, G m E 94

95 ompatibility Equation 4, y 6.773F G y 48.84F G Solving these equations simultaneously for y and F G y F G kn 34.1kN 95

96 The remaining reactions of the indeterminate truss can now be determined by superposition of reactions of primary truss due to the external loading and due to each of the redundants. The forces in the remaining members of the indeterminate truss can be determined by using the superposition relationship 96

97 F 85 G 85 H kn kn 8 kn.354 F.5 G.5 H kn + E 77.5 E F.77 G H E = x = F G H kn 7.8 E kn 8 kn y = y =96.57 E y = ctual Truss 97

98 ctual Truss F G H kn x = E 8 kn 8 kn y = y =96.57 E y =5.1 98

TYPES OF STRUCUTRES. HD in Civil Engineering Page 1-1

TYPES OF STRUCUTRES. HD in Civil Engineering Page 1-1 E2027 Structural nalysis I TYPES OF STRUUTRES H in ivil Engineering Page 1-1 E2027 Structural nalysis I SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented: