The case where there is no net effect of the forces acting on a rigid body

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1 The case where there is no net effect of the forces acting on a rigid body Outline: Introduction and Definition of Equilibrium Equilibrium in Two-Dimensions Special cases Equilibrium in Three-Dimensions Constraints and Determinacy 1

2 We will now use concepts from previous chapters to solve static equilibrium problems Requirements for equilibrium: For a RB in equilibrium, the external forces impart no translation or rotation motion So, the RB (Rigid Body) is not moving, or is moving at a constant velocity (translational and rotational) 2

2D, is relatively straightforward since F z = 0 so, M x = 0, M y = 0 therefore M z = M o We use: since the location of the origin is arbitrary, we can also say M o = M A = M B = M C = etc = 0 we have

3 2D, is relatively straightforward since F z = 0 so, M x = 0, M y = 0 therefore M z = M o We use: since the location of the origin is arbitrary, we can also say M o = M A = M B = M C = etc = 0 we have multiple M equations that we can use in general, we have 3 equations in total, and we can solve for up to 3 unknowns 1 fixed support, or 2 rollers and a cable, or 1 roller and 1 pin in a fitted hole 3

4 For example: Truss loaded as shown Given P, Q, S, find reactions at A and B Draw FBD, include known forces, weight and reaction forces at A and B y x 4

5 we know P, Q and S, so pick the equations that will isolate the unknowns at A and B Case 1 (always works) use M A = 0 (to get B y ) use F x = 0 (to get A x ) use F y = 0 (to get A y ) y x You can get the same answers by using different equations in different orders, but sometimes the equations will be more complicated to solve. For example - in this case solving ΣF y = 0 first give 2 equations and 2 unknowns 5

6 you can use 3 other equations if you wanted Case 2: F x = 0 M A = 0 M B = 0 - valid so long as A and B aren t on the same line parallel to the y-axis Case 3: F y = 0 M A = 0 M B = 0 - valid so long as A and B aren t on the same line parallel to the x-axis note: since A and B are often where ground rxn forces apply, these 3 eqns are generally not a good choice Case 4: M A = 0 M B = 0 M C = 0 - valid so long as A, B and C are not in a line i.e. collinear Using an inappropriate equation will not give you the wrong answer, but you will not be able to solve (0 = 0). 6

In general, choose equations of equilibrium that have 1 unknown each (where possible) by: i] Summing moments at points of intersection of unknown forces ii] Summing components in a direction

7 In general, choose equations of equilibrium that have 1 unknown each (where possible) by: i] Summing moments at points of intersection of unknown forces ii] Summing components in a direction perpendicular to their common direction, if they re parallel Figure at right - use ( to F A and F B, to get F Dx ) - use (intersection for F A and F D, to get F By ) - use (intersection for F B and F D, to get F ay ) 7

8 A fixed crane has a mass of 1000 kg and it is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. 8

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11 The uniform 100 kg I-beam is supported initially by its end roller and pin on the horizontal surface at A and B. By means of the cable at C it is desired to elevate end B to a position 3 m above end A. Determine the required tension P and the reaction at A, when end B is 3 m higher than A and C is directly below the pulley. P A C B 6 m 2 m 11

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13 TWO-FORCE ELEMENT a special case where forces are only applied at 2 points on a Rigid Body (RB) if a two-force element is in equilibrium then the two forces have e.g. to satisfy: F x = 0 F y = 0 M = 0 F A F B 13

14 if they didn t have the same line of action (seen by joining the two points of application for the forces) then sum of moment equal zero. F A F B we can treat small (or thin) RBs such as links, where their weight is small compared to external forces, as 2 force elements in general, if we can reduce forces acting on an RB to F s at only 2 points, we can create a convenient twoforce element problem (this helps in Chap 9) 14

15 THREE-FORCE ELEMENT another special case is where the RB is subjected to forces acting at 3 points if a RB is in equilibrium, then the lines of action will be either concurrent (act through the same point) or parallel F 2 F 1 D F 3 acts at dot D is the point of intersection of F 1 & F 2 To satisfy M D =0, the LOA of F 3 must act through D. this situation can make finding the directions of all unknown forces easier the only exception is when all forces are parallel (no intersection) 15

16 Four types important elements i) two-force RB is subject to forces at 2 points ~ Resultants of forces acting at each point have the same magnitude, line of action and opposite sense ~ The direction is along the 2 points of application ii) three-force RB is subject to forces at 3 points ~ Resultants of forces acting at each point must have lines of action that are either concurrent or parallel ~ Use intersection of LOAs of 2 force to get the 3 rd iii) Particle is subjected to forces at only 1 point ~ All forces are concurrent ~ Moment condition is automatically satisfied iv) Frictionless pulleys only change the direction of a cable 16

17 The lever ABC is pin supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A. Pins at A & D 4 unknowns Look at BD and AB separately ENGR 1205 Chapter 3 17

18 START WITH BD: Pins at B & D direction and magnitude of force is unknown But only 2 forces so for equilibrium, LOA must run through B & D and forces must be equal yet opposite ENGR 1205 Chapter 3 18

19 NOW LOOK AT AB: Pins at A & B unknowns But F B acts at 45 from BD AB is a 3-force member so all the forces meet at 1 point We can find (for direction of F A ) ENGR 1205 Chapter 3 19

20 ENGR 1205 Chapter 3 20

21 ENGR 1205 Chapter 3 21

22 F x = 0 F y = 0 F z = 0 M x = 0 M y = 0 M z = 0 these 6 equations can be solved for up to 6 unknowns, usually for rxns at supports or connections generally use vectors to solve 3D problems M = r x F = 0, set each of the components = 0 F = 0 and do the same for its components choose the points around which you calculate moments carefully (can also take moments about axes) 22

23 1. draw the FBD - for each support note the 1-6 rxns (any prevented movements are rxns go through carefully) 2. F = 0 and M = 0 (about any point) will give you 3. seek equations involving as few unknowns as possible e.g. sum moments about ball and socket joints or hinges, or draw an axis through points of application of all but 1 unknown rxn, and then solve for it using mixed triple products 23

24 Rod AB has a weight of 200 N that acts as shown. Determine the reactions at the ball-and-socket joint A and the tension in the cables BD and BE. 24

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A 20-kg ladder used to reach high shelves is supported by two wheels A and B mounted on a rail and by an unflanged wheel (treat like a ball) at C resting against a rail fixed to the wall.

28 A 20-kg ladder used to reach high shelves is supported by two wheels A and B mounted on a rail and by an unflanged wheel (treat like a ball) at C resting against a rail fixed to the wall. An 80-kg man stands on the ladder and leans to the right such that the line of action of the combined weight of the ladder and man intersects the floor at point D. Determine the reactions at A, B, and C. 28

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32 A uniform pipe cover of radius 240 mm and mass 30 kg is held in a horizontal position by cable CD. Assuming that the bearing at B does not exert any axial thrust, determine the tension in the cable and the reactions at A and B. 32

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35 The tension in the cable is 343N, the reaction at A is 49i j k N and the reaction at B is 245i j N 35

36 The bar ABC is supported by ball and socket supports at A and C and the cable BD. The suspended mass is 1800 kg. Determine the magnitude of the tension in the cable. E 36

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38 A 450-lb load hangs from the corner C of a rigid piece of pipe which has been bent as shown. The pipe is supported by the ball-and-socket joints A and D, and by a cable attached at the midpoint BC to the wall at G. Determine where G should be located if the tension in the cable is to be minimum, and the corresponding value of the tension. 38

Completely Constrained means the rigid body can t move under the given loads (or under any loads) Statically Determinate means the values of the unknowns can be

39 Completely Constrained means the rigid body can t move under the given loads (or under any loads) Statically Determinate means the values of the unknowns can be determined under static equilibrium (in 2D 3 unknowns & 3 equations, in 3D 6 unknowns & 6 equations This case is both statically determinate and completely constrained. 39

40 Statically Indeterminate means there are more constraints than needed 4 unknowns (A x, A y, B x, B y ) but only 3 independent equations of equilibrium Can t determine A x and B x separately. M A = 0 gets us B y and M B = 0 gets us A y but F x = 0 only gets us A x and B x 40

Partially Constrained means the rigid body has fewer constraints than necessary (it can move and is unstable) 2 unknowns and 3 equations one equation will usually not be satisfied M A = 0 gets us B

41 Partially Constrained means the rigid body has fewer constraints than necessary (it can move and is unstable) 2 unknowns and 3 equations one equation will usually not be satisfied M A = 0 gets us B y, M B = 0 gets us A y but ΣF x 0 Occasionally in these cases we get lucky with applied forces and F x =0, but we can t ensure that it equals 0 because we can t control the x-direction with a reaction force. 41

42 In order to be statically determinate and completely constrained, we need an equal number of unknowns and equations of equilibrium. If not, the system will be partially constrained or statically indeterminate or both. Having an equal number of unknowns and equations is necessary (but not sufficient ) for static determinacy 42

43 Example 1 3 unknowns and Statically indeterminate can t determine the values of A,B & E Improperly constrained 43

44 Example 2 4 unknowns and ΣM A 0 Statically determinate 4 unknowns and 3 independent equations Improperly constrained ΣM A = 0 doesn t hold since the LOA of the rxn forces all go through the A 44

45 In 3D: if rxns involve > 6 unknowns, some rxns will be statically indeterminate if rxns involve < 6 unknowns, the RB will only be partially constrained (although it may still be in equilibrium depending on specific loads) even with 6+ unknowns, some equations may not be satisfied (such as when rxns are parallel or intersect the same line) in which case the RB is then improperly constrained 45

46 A system is improperly constrained whenever the supports (even if providing enough rxns) are arranged such that the rxns are concurrent (as above in example 2) or in parallel (as in example 1). To get static determinancy, make sure the rxns involve 3 unknowns and that the supports don t require concurrent or parallel rxns supports involving statically indeterminate rxns can be dangerous, so be careful using them when designing things 46

47 Completely Constrained means the rigid body can t move under any loads Statically Determinate means the values of the unknowns can be determined under static equilibrium Statically Indeterminate means there are more constraints than needed Partially Constrained means the rigid body has fewer constraints than necessary In order to be statically determinate and completely constrained, we need an equal number of unknowns and equations of equilibrium and that the supports don t require concurrent or parallel rxns 47

Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd

Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using