Analysis of Statically Determinate Structures

Transcription

1 nalysis of Statically Determinate Structures! Idealized Structure! Principle of Superposition! Equations of Equilibrium! Determinacy and Stability! eams! Frames! Gable Frames! pplication of the Equations of Equilibrium! nalysis of Simple Diaphragm and Shear Wall Systems Problems 1

2 lassification of Structures Support onnections weld stiffeners typical pin-supported connection (metal) weld typical fixed-supported connection (metal) typical roller-supported connection (concrete) typical fixed-supported connection (concrete) 2

3 pin support pin-connected joint fixed support fixed-connected joint torsional spring support torsional spring joint P P L/2 L/2 L/2 L/2 actual beam idealized beam 3

4 Table 2-1 Supports for oplanar Structures Type of onnection Idealized Symbol Reaction Number of Unknowns (1) θ Light cable θ θ One unknown. The reaction is a force that acts in the direction of the cable or link. (2) rockers rollers F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. (3) F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. (4) F One unknown. The reaction is a force that acts perpendicular to the surface at the point of contact. 4

5 Type of onnection Idealized Symbol Reaction Number of Unknowns (5) Smooth pin or hinge F y F x Two unknowns. The reactions are two force components. (6) slider F M Two unknowns. The reactions are a force and moment. fixed-connected collar (7) fixed support F x M F y Three unknowns. The reactions are the moment and the two force components. 5

6 Idealized Structure. 4 m F 4 m F actual structure idealized structure 6

7 D joist slab column girder idealized framing plan fixed-connected beam idealize beam fixed-connected overhanging beam Idealized beam 7

8 idealized framing plan idealized framing plan 8

9 Tributary Loadings. stringer slab girder veihicle slab stringer girder floor beam floor beam deck girder pier 9

10 spandrel beam beam 2 nd floor joist slab joist supported slab beam foundation wall 1 st floor stairs landing slab on grade column wall footing basement spread footing 10

11 One-Way System. 4 m E D 0.5 kn/m 2 D 1 m 1 m 1 m 4 m F 2 m E F 1 m 2 m idealized framing plan 1 kn/m 1 kn 2 kn 1 kn D F 2 kn 2 kn 4 m idealized beam 2 m 2 m idealized girder 11

12 column girder beam concrete slab is reinforced in two directions, poured on plane forms L 1 /2 L 1 /2 E L 2 D F L 1 L 1 Idealized framing plan for one-way slab action requires L L 2 2 / 1 12

13 Two-Way System. 4 m 4 m D 4 m 0.5 kn/m 2 45 o 45 o 2 m L 2 /L 1 = 1 D 4 m 1 kn/m 2 m 2 m idealized beam, all idealized framing plan L 2 /L 1 = 1.0 < 2 2 m 45o 6 m 45 o 4 m 1kN/m 1 kn/m D idealized framing plan 2 m 2 m 2 m 2 m 2 m idealized beam 13

14 Principle of Superposition P = P 1 + P 2 d = d P 1 Two requirements must be imposed for the principle of superposition to apply : 1. The material must behave in a linear-elastic manner, so that Hooke s law is valid, and therefore the load will be proportional to displacement. σ = P/ δ = PL/E + d P 2 2. The geometry of the structure must not undergo significant change when the loads are applied, i.e., small displacement theory applies. Large displacements will significantly change and orientation of the loads. n example would be a cantilevered thin rod subjected to a force at its end. 14

15 Equations of Equilibrium ΣF x = 0 ΣF y = 0 ΣF z = 0 ΣM x = 0 ΣM y = 0 ΣM z = 0 N M M N V V internal loadings 15

16 Determinacy and Stability Determinacy r = 3n, statically determinate r > 3n, statically indeterminate n = the total parts of structure members. r = the total number of unknown reactive force and moment components 16

17 Example 2-1 lassify each of the beams shown below as statically determinate or statically indeterminate. If statically indeterminate, report the number of degrees of indeterminacy. The beams are subjected to external loadings that are assumed to be known and can act anywhere on the beams. hinge hinge 17

18 SOLUTION r = 3, n = 1, 3 = 3(1) Statically determinate r = 5, n = 1, 5-3(1) = 2 Statically indeterminate to the second degree hinge r = 6, n = 2, 6 = 3(2) Statically determinate r = 10, n = 3, 10-3(3) = 1 Statically indeterminate to the first degree 18

19 Example 2-2 lassify each of the pin-connected structures shown in figure below as statically determinate or statically indeterminate. If statically are subjected to arbitrary external loadings that are assumed to be known and can act anywhere on the structures. 19

20 SOLUTION r = 7, n = 2, 7-3(2) = 1 Statically indeterminate to the first degree r = 9, n = 3, 9 = 3(3) Statically determinate 20

21 r = 10, n = 2, 10-6 = 4 Statically indeterminate to the fourth degree r = 9, n = 3, 9 = 3(3) Statically determinate 21

22 Example 2-3 lassify each of the frames shown in figure below as statically determinate or statically indeterminate. If statically indeterminate, report the number of degrees of indeterminacy. The frames are subjected to external loadings that are assumed to be known and can act anywhere on the frames. D 22

23 SOLUTION D r = 9, n = 2, 9-6 = 3 Statically indeterminate to the third degree r = 15, n = 3, 15-9 = 6 Statically indeterminate to the sixth degree 23

24 Stability r < 3n, unstable > r 3n, unstable if member reactions are concurrent or parallel or some of the components form a collapsible mechanism Partial onstrains P P M F 24

25 Improper onstraints O O P d d F F F P P P F F F 25

26 Example 2-4 lassify each of the structures in the figure below as stable or unstable. The structures are subjected to arbitrary external loads that are assumed to be known. hinge D 26

27 SOLUTION The member is stable since the reactions are non-concurrent and nonparallel. It is also statically determinate. hinge The compound beam is stable. It is also indeterminate to the second degree. The compound beam is unstable since the three reactions are all parallel. 27

28 The member is unstable since the three reactions are concurrent at. D The structure is unstable since r = 7, n = 3, so that, r < 3n, 7 < 9. lso, this can be seen by inspection, since can move horizontally without restraint. 28

29 pplication of the Equations of Equilibrium D P 1 x D x y x P 2 E y D y Dx P 1 E x y x y E y P 1 x E x P 2 x P 2 x r = 9, n = 3, 9 = 3(3); statically determinate 29

30 P 1 x P 1 y x y P 2 x P 1 y P 2 x x y P 2 x y r = 6, n = 2, 6 = 3(2); statically determinate 30

31 Example 2-5 Determine the reactions on the beam shown. 150 kn o 1 m 2 m 70 kn m 31

32 SOLUTION 265 kn o 1 m 2 m 70 kn m 265 sin 60 o = kn x cos 60 o = kn y 1 m 70 kn m y + ΣF x = 0: x = 0: x = kn, + ΣM = 0: y (4) - (229.5)(3) + (132.5)(0.3) -70 = 0 y = kn, + ΣF y = 0: y = 0 y = kn, 32

33 Example 2-6 Determine the reactions on the beam shown. 15 kn/m 5 kn/m 12 m 33

34 SOLUTION (1/2)(12)(10) = 60 kn 15 kn/m 12 m 5 kn/m 10 kn/m 5 kn/m x M y 4 m 12 m (5)(12) = 60 kn 6 m + ΣF x = 0: x = 0 + ΣF y = 0: y = 0 y = 120 kn, + ΣM = 0: M - (60)(4) - (60)(6) = 0 M = 600 kn m 34

35 Example 2-7 Determine the reactions on the beam shown. ssume is a pin and the support at is a roller (smooth surface). 7 kn/m 4 m 2 m 35

36 SOLUTION 7 kn/m 4 m 2 m 28 kn N tan -1 (3/2) = 56.3 o x y 2 m 6 m 90 o o = 33.7 o + ΣM = 0: -28(2) + N sin 33.7(6) + N cos 33.7(3) = 0 N = 9.61 kn + ΣF x = 0: x - N cos 33.7 = 0; x = 9.61cos 33.7 = 8 kn, + ΣF y = 0: y cos33.7 = 0 y = kn, 36

37 Example 2-8 The compound beam in figure below is fixed at. Determine the reactions at,, and. ssume that the connection at pin and is a rooler. 6 kn/m hinge 6 m 4 m 8 kn m 37

38 SOLUTION 36 kn 6 kn/m hinge 6 m 4 m 8 kn m x x x 8 kn m M y 6 m y y y Member Member + ΣM = 0: y (4) - 8 = 0 y = 2 kn, + ΣM = 0: M - 36(3) + 2(6) = 0 M = 96 kn m + ΣF x = 0: x = 0 + ΣF x = 0: x - = 0 ; x = x = 0 + ΣF y = 0: y - y = 0; y = y = 2 kn, + ΣF y = 0: y = 0 y = 34 kn, 38

39 Example 2-9 The side girder shown in the photo supports the boat and deck. n idealized model of this girder is shown in the figure below, where it can be assumed is a roller and is a pin. Using a local code the anticipated deck loading transmitted to the girder is 6 kn/m. Wind exerts a resultant horizontal force of 4 kn as shown, and the mass of the boat that is supported by the girder is 23 Mg. The boat s mass center is at G. Determine the reactions at the supports. 1.6 m 1.8 m 2 m 4 kn 0. G D 6 kn/m roller pin 39

40 SOLUTION 1.6 m 1.8 m 2 m 4 kn 0. G D 6 kn/m roller pin + ΣF x = 0: 4 - x = 0 x = 4 kn, + ΣM = 0: 4 kn 0. D G 6(3.8) = 22.8 kn 1.9 m 2 m x 22.8(1.9) - y (2) (5.4) -4(0.3) = 0 y = kn, + ΣF y = 0: 23(9.81) kn = kn 5.4 m y y y = 0 y = 382 kn, 40

41 Example 2-10 Determine the horizontal and vertical components of reaction at the pins,, and of the two-member frame shown in the figure below. 8 kn 3 kn/m 2 m m 1.5 m 2 m 41

42 SOLUTION Member 2 m 8 kn kn/m 2 m + ΣM = 0: - y (2) +6(1) = 0 y = 3 kn, Member 1.5 m + ΣM = 0: 2 m 6 kn -8(2) - 3(2) + x (1.5) = 0 x = 14.7 kn, + ΣF x = 0: 8 kn (4/5)8 y x 1 m 1 m x y y ΣF y = 0: + x + (3/5) = 0 x = 9.87 kn, (3/5)8 x 1.5 m Member y - (4/5)8-3 = 0 y = 9.4 kn, x y 2 m + + ΣF x = 0: ΣF y = 0: x - x = 0; x = x = 14.7 kn, x = 0 ; y = 3 kn, 42

43 Example From the figure below, determine the horizontal and vertical components of reaction at the pin connections,, and of the supporting gable arch. 15 kn 43

44 SOLUTION 15 kn x x y y Entire Frame + ΣM = 0: y ( 6) 15(3) = 0 ΣF y = 0: + y = 0 y = 7.5 kn, y = -7.5 kn, 44

45 15 kn y x 3.75 kn= x 7.5 kn = y x 7.5 kn x 7.5 kn Member Member + ΣM = 0: 15 (3) (6) + 7.5(3) = 0 + x + ΣF x = 0: 3.75 = 0 x + ΣF x = 0: x = kn, = 0 x x = 3.75 kn x = 3.75 kn, + ΣF y = 0: y = 0 y = 7.5 kn 45

46 Example The side of the building in the figure below is subjected to a wind loading that creates a uniform normal pressure of 1.5 kpa on the windward side and a suction pressure of 0.5 kpa on the leeward side. Determine the horizontal and vertical components of reaction at the pin connections,, and of the supporting gable arch. 2 m 2 m 4 m wind 4 m 46

47 SOLUTION 2 m 2 m 4 m wind 4 m uniform distributed load on the windward side is 6 kn/m 2 kn/m (1.5 kn/m 2 )(4 m) = 6 kn/m uniform distributed load on the leeward side is 6 kn/m 2 kn/m (0.5 kn/m 2 )(4 m) = 2 kn/m 47

48 25.46 cos kn sin sin kn 8.49 kn x 45 o 1.5m 45 o 8.49 cos 45 6 kn x y y Entire Frame + ΣM = 0: -(18+6)(1.5) - ( )cos 45 o (4.5) - (25.46 sin 45 o )(1.5) + (8.49 sin 45 o )(4.5) + y (6) = 0 y = 24.0 kn, + ΣF y = 0: y sin 45 o sin 45 o = 0 y = kn 48

49 25.46 cos sin kn 45 o x 1.5 x 8.49 sin kn 45 o 8.49 cos kn y 3 y 6 kn x x y = 12.0 kn y = 24.0 kn Member + ΣM = 0: + ΣF x = 0: + ΣF y = 0: Member + ΣF x = 0: (25.46 sin 45 o )(1.5) + (25.46cos 45 o )(1.5) + (18)(4.5) + x (6) + 12(3) = 0 x = kn cos 45 o - x = 0 x = 7.5 kn, sin 45 o + y = 0 y = 30.0 kn, cos 45 o x = 0 x = kn, 49

50 nalysis of Simple Diaphragm and shear Wall Systems F F/2 F/8 F/8 roof diaphragm F/8 F/8 F/8 F/8 F/8 F/8 F/8 F/8 F/8 F/8 F/8 F/2 F/8 floor diaphragm F/8 F/8 50

51 roof diaphragm D Wind F D shear walls second floor diaphragm F/2 F/4 F/16 F/16 F/16 F/16 3F/16 3F/16 F/16 roof diaphragm F/16 F/16 2 st floor 3F/16 F/16 F/16 F/16 3F/16 3F/16 3F/16 F/4 3F/16 F/4 3F/16 1 st floor F/4 3F/16 F/4 3F/16 F/4 51

52 Example 2-12 ssume the wind loading acting on one side of a two-story building is as shown in the figure below. If shear walls are located at each of the corners as shown and flanked by columns, determine the shear in each panel located between the floors and the shear along the columns. 30 m 20 m 1.2 kpa D D 4 m 4 m 0.8 kpa 52

53 1.2 kpa F R2 F R1 SOLUTION D 30 m 20 m 0.8 kpa D F R1 = 0.8(10 3 ) N/m 2 (20 m)(4 m) = 64 kn F R1 /2 = 32 F R2 = 1.2(10 3 ) N/m 2 (20 m)(4 m) = 96 kn F R2 /2 = 48 4 m 4 m F R2 /2 = 48 kn kn 12 kn 32 kn F R1 /2 = 32 kn 12 kn 12 kn 32 kn 32 kn 32 kn 40 kn 12 kn roof diaphragm F/8 = 12 kn 12 kn 32 kn 2 st floor 32 kn 12 kn 32 kn 1 st floor 32 kn 40 kn 40 kn 12 kn 12 kn 32 kn 32 kn 40 kn 53

54 12 kn F v F v 4 m + ΣM = 0: 12 kn F v (3) - 12(4) = 0 F v = 16 kn F v 32 kn F v 32 kn 4 m + ΣM = 0: F v(3) - 32(4) = 0 F v = 42.7 kn 54