Structural Analysis. Method of Joints. Method of Pins

Transcription

1 Structural Analysis / Method of Pins "This professor would make a good parking lot a6endant. Tries to tell you where to go, but you can never understand him." Trusses Trusses are common means of transferring loads from their points of application to the supports They vary in design but their analysis always follows the same pattern 2 1

2 Tools Equations of equilibrium Free Body Diagrams Trig and algebra Visualization 3 Trusses A simple truss analysis, which we will do in this class, does not consider the weight of the members which make up the truss A simple truss analysis considers that all loading is made at connections A simple truss analysis considers that all truss members are two-force members 4 2

3 Trusses This means that we can break a truss down into a collection of elements for analysis 5 Trusses An example truss 6 3

4 Trusses The red arrows are the loads (applied externally) 7 Trusses The dark black lines are the members 8 4

5 Trusses Each member goes from connection to connection 9 Trusses The black line along the bottom of the truss represents three separate members 10 5

6 Trusses The support conditions are the pin at the left and the roller at the right 11 Trusses At each intersection of two or more members, we consider that the members are pinned together 12 6

7 Trusses For example, at the point labeled G on the diagram, we have four members coming to a point connected by a single pin 13 Trusses To simplify our analysis, we assign each connection point with a letter 14 7

8 Trusses I avoid using F and I as connection point assignments but your text does not 15 Trusses Pick a starting point and label each connection, I chose to start at the lower left and move across the top and then across the bottom 16 8

9 Trusses Each of the members can now be labeled using the points of connection 17 Trusses In this truss we have the following members: AB, BC, CD, DE, CE, CG, EG, BG, and AG 18 9

10 Trusses The order of the letters is not important, I choose to put them in the order they appear in the alphabet 19 To begin the analysis of the truss using this method, we start by identifying all external forces acting on the truss 20 10

11 In this case it would be the forces represented by the red lines The forces (loads) must always be applied at the connections (joints) 21 This is usually given in the problem 22 11

12 Then we remove any supports and replace them with reactions provided by the supports 23 We have a pin at the left so we will have an x and a y reaction 24 12

13 And we have a roller at the right so we only have a y reaction 25 We are also usually given the dimensions and geometry of the truss in the given 26 13

14 Now we can solve for the reactions at the supports using the methods developed previously 27 We will use the entire truss as the FBD and ignore the internal members (for the moment) 28 14

15 Summing forces in the x-direction 29 Summing moments about D 30 15

16 Summing forces in the y-direction 31 Redrawing our system with the known reactions, we have 32 16

17 Now for the problems that we had done earlier, we would be finished 33 In a truss analysis, we want to know what force is being carried in each of the members 34 17

18 Each member of the truss can be in one of three conditions: It can be in COMPRESSION, that means that it will be pushing on each end where it is connected It can be in TENSION, that means that it will be pulling on each end where it is connected It can have no force in the member 35 In the method of joints, we are going to assume the condition of each of the members, draw a free body diagram of the connection points, and apply two of our three equilibrium conditions to solve for the force actually in the members 36 18

19 When we used a pin connection previously, we used an x and a y reaction to show what was being provided by the pin Here we are actually drawing a FBD of the pin itself and showing the forces from the members onto the pin itself 37 You can assume any condition that you want for the members, I usually choose to assume that they are all in either tension or compression and then let the signs of the solutions dictate their actual condition 38 19

20 Since we can only use two of the equations of equilibrium, the sum of the forces in the x and the sum of the forces in the y direction, our analysis pattern will be dictated by the makeup of the truss itself 39 First, we can look at all the forces acting on each pin of the truss It is critical to the analysis that you do not exclude either external forces or support reactions at each pin 40 20

21 If a member is pushing on a pin at one end, it is always pushing on the pin at its other end 41 Conversely, if a member is pulling at a pin on one end, it is pulling on the pin at its other end Never push pull! 42 21

22 At any joint, you can only solve for two unknowns!!!! 43 Now to decide where to start the analysis, we need to find a joint where there are only two unknowns 44 22

23 In our case, the pins at A and at D both only have two unknowns, so we can start at either point 45 If you thought that there were three unknowns at A and D because of AB x and AB y, remember that the force in the member has a line of action along the member 46 23

24 We choose A to start our analysis We now draw a FBD of the pin at A 47 I always assume that the members are in compression so our FBD would look like 48 24

25 Since I assume compression, the forces in the members are pushing on the connection (pin) 49 Now we can use our two equilibrium conditions to solve for the forces in the members 50 25

26 We report all the magnitudes of the forces as positive numbers and the forces themselves as either compression or tension 51 Solving for AG 52 26

27 Now that AB and AB are known forces, we can revisit the FBD of the complete truss and find another joint/pin that has only two unknowns 53 Usually this will be adjoining the pin that we just solved for For this truss it could be either pin B or G 54 27

28 At G, we have BG, CG, and EG as unknowns Three unknowns, so this pin won t work 55 At B, we have two unknowns, BC, and BG We can proceed to work at BG 56 28

29 We draw a FBD of the pin at B 57 We know that AB is in compression, so we are not making an assumption about AB 58 29

30 We are assuming that BC and BG are in compression for this FBD 59 Now we use our conditions of equilibrium to solve for BC and BG 60 30

31 Now we use our conditions of equilibrium to solve for BC and BG 61 You might be asking yourself why BG is even included in the truss This is a very preliminary analysis of one loading condition, there is actually much more to truss design than this 62 31

32 Now we could go to either joint C or joint G At C, we have CG, CE, and CD as unknowns At G, we have CG and EG Our next point is G 63 The FBD at G Notice that we have drawn AG as a tension member, we know that from our earlier solution, it is no longer an assumption 64 32

33 We also have no longer included the force from BG, we know that it carries no force under there loading conditions 65 Using our equilibrium conditions 66 33

34 Using our equilibrium conditions 67 You can continue on with the pattern until all the members in the truss are solved for You will always have one extra pin/joint available after all the members in the truss have been solved Check the sum of the forces in the x and y direction at that point to see if the truss closes 68 34