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
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.
Example 6 etermine the support reactions on the frame shown. EI is constant. 8 kn/m 5 m 4 m ctual Frame 3
Principle of Superposition 8 kn/m 5 m 4 m ctual Frame y inspection the frame is indeterminate to the first degree. 4
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
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
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
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
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
Δ L Mm 5 dx EI 166.7 166.7 EI EI x 4x.8x dx 1 x 4 1 1 1 1 EI dx EI dx x 1 M 8 kn/m x 4x 1 1 1 1 kn x 1 m1. 8x 1 kn.8 kn M m 1x x x 1 kn kn 1.8 kn
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
f L mm dx EI 6.7 1.3 EI EI.8x dx 1 x 4 5 1 1 EI 48. EI EI dx dx 1 kn.8 kn x 1 m1. 8x 1 m 1x x 1 kn.8 kn 1
ompatibility Equation Δ x f (1) Substituting the data in Eq. (1) 166.7 EI x 48. EI x 3.47 kn NS 13
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
Equilibrium ondition x 4 kn y.5 m.5 m 4 m 3.47 kn Fx ; x 3. 47 x 3. 47kN y M. 5 5-3. 474. kn ; 4 y y 8 Fy ; y 4. 8 y 17. kn 15
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
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
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
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
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.
ompatibility Equation Reference to point 5 lb 3 5 4 M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M1 9. 17x 1 m1 1. 833x 1 x 1 9.17 lb.833 lb x 1 1 lb.ft 3 lb 1
For θ we require application of real loads and a virtual unit couple moment at 5 lb 3 5 4 M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M1 9. 17x 1 m1 1. 833x 1 x 1 9.17 lb.833 lb x 1 1 lb.ft 3 lb
L Mm 8 dx EI 518.5 33. 81.8 EI EI EI 9.17x 1.833x dx 5 96.7x 5x.67x 1 1 1 EI EI dx 5 lb 3 5 4 M x.5 lb 37.8 lb 96.7 lb.7x 5 96 x x.5 lb m. 667x.833 lb.667 lb M1 9. 17x 1 m1 1. 833x 1 x 1 9.17 lb.833 lb x 1 3 lb 1 lb.ft 3
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 m1 1. 833x 1.833 lb x 1 1 lb.ft 4
L m m 8 dx EI 3.85.185 4.4 EI EI EI 1.833x dx 5.67x 1 1 EI.5 lb EI dx x m. 667x.833 lb.667 lb m1 1. 833x 1.833 lb x 1 1 lb.ft 5
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
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
Principle of Superposition 1 k k/ft 15 ft ctual Frame 3 ft egree of indeterminacy = 8
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
Principle of Superposition Primary structure is obtained by removing the hinged support at point. 1 k k/ft Primary Structure 3
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
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
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
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
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) -15-15+1x 1-3 -x -15 35
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) -15-15+1x 1 -x 1-3 -x -15-15 -x 3 36
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) -15-15+1x 1 -x 1 3-3 -x -15 x -15 -x 3 37
Δ x Δ x 15 1x x 3 x 15 L Mm 15 x 1 1 dx dx EI EI 15 x3 dx3 EI 41875 3 16875 135 k ft EI 1 EI dx Δ y Δ y L Mm EI y dx 15 15 1x 3 3 x x EI 1 dx 6415 43875 5 k ft EI 1 3 EI dx 38
f xx L m m EI x x dx 15 x 3 15 15 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 15 3 3 x EI dx 1 EI dx f yy 5 ft 3 EI 39
f xy f yx L m x m EI y dx 15 x 3 3 15x 1 EI dx 1 EI dx f xy f yx 115 ft 3 EI 4
Δ x 41875 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
Now put these values in the Equations (1) and () 41875 9x 115y 6415 115x 5y (1) () y solving (1) and () simultaneously we get x y 1.53 k 33.6 k 4
pplying equations of equilibrium, we have the other support reactions as 1 k k/ft.53 k 53. k-ft 1.53 k 6.774 k 33.6 k 43
Shear diagram 6.774 1.53 -.53-33.6 44
Moment diagram 118.447 6.765 157.545 6.765 157.545 53. 45
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
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
The truss is statically indeterminate to the first degree. b + r = j 6 + 3 = (4) 9 > 8 9 8 = 1 st degree 4 lb 6 ft 8 ft ctual Truss 48
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
4 lb ctual Truss = 4 lb F Δ + F F f Primary Structure Redundant F applied 5
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
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
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
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. +4 4 lb -.8-5 +3 -.6 1 lb 1 lb -.6 4 lb +4 x +1 +1 -.8 3 lb 3 lb 54
nnl E E 11 E.8 4 8.6 6.6 3 6 1 5 1 1 1 E E E E +4 4 lb -.8-5 +3 -.6 1 lb 1 lb -.6 4 lb +4 x +1 +1 -.8 3 lb 3 lb 55
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 -.8 -.8 1 lb 1 lb 1 lb 1 lb -.6 -.6 -.6 -.6 +1 +1 x +1 +1 -.8 -.8 56
f n L E.8 8.6 6 1 E 34.56 E -.8 E 1 E -.8 -.6 1 lb 1 lb -.6 -.6 1 lb 1 lb -.6 +1 +1 x +1 +1 -.8 -.8 57
Substituting the data into Eq. (1) and solving yields 11 34.56 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
Example 1 etermine the force in member of the truss shown. 6 ft 8 ft ctual Truss 59
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
The truss is statically indeterminate to the first degree. b + r = j 9 + 4 = (6) 13 > 1 13 1 = 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
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
E F 8 k Primary Structure x Δ y 5 k 5 k y E 4 F 8 k 3 11.67 18 5 53.33 8 k 5 4.67 4.67 Δ 5 k 5 k 18 k 3 k 63
Δ is horizontal deflection at point of primary structure due to external loading. 3 E 4 11.67 18 5 F 8 k 53.33 Primary Structure 8 k 5 4.67 4.67 Δ 5 k 5 k 18 k 3 k 64
E F Redundant x is applied x Δ = x f E F 1 k 1 1 1 f 1 k 65
E F Redundant x is applied x Δ = x f Δ is horizontal deflection at point due to redundant force x. 66
f is horizontal deflection at point due to unit force. E F 1 k 1 1 1 f 1 k 67
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
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
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
TLE Member L (in.) (in. ) N (k) n (k) nnl/ (k/in.) n L/ F=N+n x 4 6 5 1,8 4 6. 4 6 4.67 1 1,76.8 4-3.11 4 6 4.67 1 1,76.8 4-3.11 EF 4 6-4 -4 F 18 4 18 18 F 18 4 5 5 E 3 6-3 -3 F 3 4 11.67 11.67 F 3 6-53.33-53.33 f nnl 5,493.6 k/in. E E n L 1 (1/in.) E E 5,493.6 1 71
nnl E 51 1 4.671 1 4.671 1 6E 6E 6E 8 6E 176.8 6E 5493.6 k/in E 176.8 6E 7
n L E f f 11 1 11 1 11 1 6E 6E 6 E f 1 (1/in) E Now put these results into Equation (1) 5493.6 E x 1 E x 45.78 k 73
F F F F N n x 5 1 45.78 6. (T) 45.78 3.11 () 4.67 1 3.11 () Equation of Equilibrium F x x y y 8 45.78 17.78 k 18 k 3 k x =17.78 8 k y =18 5 k 5 k y =3 74 x =45.78
E F 8 k Primary Structure x =17.78 x = 45.78 5 k 5 k y =18 y = 3 E 4 F 8 k 3 11.67 18 5 53.33 x =17.78 6. 3.11 3.11 x = 45.78 y =18 5 k 5 k y = 3 75
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
Principle of Superposition egree of Indeterminacy = b + r > j 14 + 4 > 8 18 > 16 F G H 7 kn x E y 8 kn 8 kn y E y ctual Truss 77
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
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
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
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
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
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
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
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
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
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
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
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
N = member forces due to external loading F G H 85 85 7 kn 116.6738 3.536 19.6 19.6 7 15.5 15.5 77.5 77.5 E 8.5 8 kn 8 kn 77.5 9
n = member forces due to unit load at joint F G H.5.5.354.354.354 1 1.61.5.5.75.75 E.5 1 kn.75 91
n G = member forces due to unit force in member G F G H.77.77 1 1 1.77.77 E 9
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) 1 15.5 -.5-381.5.65 18.373 1 15.5 -.5 -.77-381.5-178.175.65 5 1.768 14.65 1 77.5 -.75-581.5 5.65 5.1 E 1 77.5 -.75-581.5 5.65 5.1 FG 1-85.5 -.77-45 6.95.5 5-3.535-6.855 GH 1-85.5-45.5-36.747 F 1 8 -.77-565.6 5 55.891 G 1 -.77 5-4.19 H 1-1 1-96.57 F 14.14-116.673.354-584.96 1.77-8.51 G 14.14 1 14.14 34.1 F 14.14 3.536 -.354 1-17.7 5.6 1.77 14.14-5.6 3.473 H 14.14 19.6.354 548.697 1.77 143.765 EH 14.14-19.6 1.61-1644.541 15.9-7.8-447.64-99.819 48.736 48.84-6.773 93
ompatibility Equation G 4,47.64 kn. m E 99.819 kn. m E 48.736 m f E G y f FG f, G y f G, FG f G, G 48.84 m E y substituting these values into the above equations f G, G f G, f, G 6.773 m E 94
ompatibility Equation 4,47.64 48.736 y 6.773F G 99.819 6.773 y 48.84F G Solving these equations simultaneously for y and F G y F G 96.57 kn 34.1kN 95
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
7 8.5.5 116.673 F 85 G 85 H 8 3.536 19.6 7 kn 19.6 15.5 15.5 77.5 77.5 8 kn 8 kn.354 F.5 G.5 H.354 +.354 1 1.61.5.5.75.75 1 kn + E 77.5 E.75.77 F.77 G H 1 1 1.77.77 E = x =7 8.51 F 6.855 G 36.747 H 55.891 3.473 7 kn 7.8 E 18.373 14.65 5.1 5.1 8 kn 8 kn y =58.373 y =96.57 E y =5.1 34.1 4.19 143.765 96.57 ctual Truss 97
ctual Truss F G H 6.855 36.747 7 kn x =7 8.51 55.891 3.473 34.1 4.19 96.57 143.765 18.373 14.65 5.1 5.1 7.8 E 8 kn 8 kn y =58.373 y =96.57 E y =5.1 98