Structural Analysis. Method of Joints. Method of Pins

Transcription

1 Structural nalysis / Method of Pins I m reading a book about an1gravity. It s impossible to put down. Pop Quiz What is today s date? 2 1

2 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 3 ridge Truss 4 2

3 Roof Truss 5 Load Transfer 6 3

4 Tools Equations of equilibrium Free ody iagrams Trig and algebra Visualization 7 Trusses simple truss analysis, which we will do in this class, does not consider the weight of the members which make up the truss simple truss analysis considers that all loading are made at connections simple truss analysis considers that all truss members are two-force members 8 4

5 Trusses This means that we can break a truss down into a collection of elements for analysis 9 Trusses n example truss 10 5

6 Trusses The red arrows are the loads (applied externally) 11 Trusses The dark black lines are the members 12 6

7 Trusses Each member goes from connection to connection 13 Trusses The black line along the bottom of the truss represents three separate members 14 7

8 Trusses The support conditions in this example are a pin at the left and a roller at the right 15 Trusses t each intersection of two or more members, we consider that the members are pinned together 16 8

9 Trusses To simplify our analysis, we assign each connection point with a letter 17 Trusses I avoid using F and I as connection point assignments but your text does not 18 9

10 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 E 19 Trusses t the point labeled on the diagram, we have four members coming to a point connected by a single pin E 20 10

11 Trusses Each of the members can now be labeled using the points of connection E 21 Trusses In this truss we have the following members:,,, E, E,,,, and E E E 22 11

12 Trusses The order of the letters is not important, I choose to put them in the order they appear in the alphabet E E E 23 To begin the analysis of the truss using this method, we start by identifying all external forces acting on the truss E E E 24 12

13 To do the analysis, we don t always have to solve for the external reactions but it is important that we at least identify that they are there. E E E 25 In this case it would be the forces represented by the red lines The forces (loads) must always be applied at the connections (joints) E E E 26 13

14 This is usually given in the problem E E E 27 Then we remove any supports and replace them with reactions provided by the supports E E E 28 14

15 We have a pin at the left so we will have an x and a y reaction y E x E E 29 nd we have a roller at the right so we only have a y reaction y E x E E y 30 15

16 We are also usually given the dimensions and geometry of the truss in the given y E x E y 31 Now we might solve for the reactions at the supports using the methods developed previously y E x E y 32 16

17 If and which reactions we need to solve for will be dictated by the problem. y E x E y 33 We will use the entire truss as the F and ignore the internal members (for the moment) y x E y 34 17

18 Summing forces in the x-direction x F x x = = 0 = 0 0 x y E y 35 Summing moments about M = 0 ( m)( ) ( m)( N) ( m)( N) = 0 y y ( 4m)( ) + ( 2m)( ) = = 6m y x E y 36 18

19 Summing forces in the y-direction y = 0 + = 0 y + = 0 y F = 700N y y x y E y 37 Redrawing our system with the known reactions, we have E E 700N 38 19

20 Now for the problems that we had done earlier, we would be finished E E 700N 39 In a truss analysis, we want to know what force is being carried in each of the members E E 700N 40 20

21 Each member of the truss can be in one of three conditions: l It can be in OMPRESSION, that means that it will be pushing on each end where it is connected l It can be in TENSION, that means that it will be pulling on each end where it is connected l It can have no force in the member E E E 700N 41 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 E E E 700N 42 21

22 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 F of the pin itself and showing the forces from the members onto the pin itself E E E 700N 43 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 E E E 700N 44 22

23 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 E E E 700N 45 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 E E E 700N 46 23

24 If a member is pushing on a pin at one end, it is always pushing on the pin at its other end E E 700N 47 onversely, if a member is pulling at a pin on one end, it is pulling on the pin at its other end Never push pull! E E 700N 48 24

25 t any joint, you can only solve for two unknowns!!!! E E 700N 49 Now to decide where to start the analysis, we need to find a joint where there are only two unknowns E E 700N 50 25

26 In our case, the pins at and at both only have two unknowns, so we can start at either point E E 700N 51 If you thought that there were three unknowns at and because of x and y, remember that the force in the member has a line of action along the member 52 E E Wednesday, October 700N 17,

27 We choose to start our analysis We now draw a F of the pin at E 53 E Wednesday, October 700N 17, 2012 I always assume that the members are in compression so our F would look like 1 1 E E E 54 Wednesday, October 700N 17,

28 Since I assume compression, the forces in the members are pushing on the connection (pin) E E Wednesday, October 700N 17, 2012 Now we can use our two equilibrium conditions to solve for the forces in the members F y = 0 1 = 0 2 = 500 2N 1 1 E 700N E 56 28

29 We report all the magnitudes of the forces as positive numbers and the forces themselves as either compression or tension 57 F y = 0 1 = 0 2 = N 1 1 E E 700N Solving for F x = 0 1 = = N = 500 N 2 = 500 N T 1 1 E 700N E 58 29

30 Now that and are known forces, we can revisit the F of the complete truss and find another joint/pin that has only two unknowns 59 E E 700N Usually this will be adjoining the pin that we just solved for For this truss it could be either pin or 60 E E 700N 30

31 t, we have,, and as unknowns Three unknowns, so this pin won t work 61 E E 700N t, we have two unknowns,, and We can proceed to work at 62 E E 700N 31

32 We draw a F of the pin at 63 E E 700N We know that is in compression, so we are not making an assumption about 64 E E 700N 32

33 We are assuming that and are in compression for this F 65 E E 700N E 700N E Now we use our conditions of equilibrium to solve for and 1 2 F x = 0 = 0 1 = N = 500 N 2 = 500 N 66 33

34 E 700N E Now we use our conditions of equilibrium to solve for and 1 2 F y = 0 + = = 2 N + N = 0 67 E 700N E You might be asking yourself why 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 68 34

35 E 700N E Now we could go to either joint or joint t, we have, E, and as unknowns t, we have and Our next point is 69 E 700N E The F at Notice that we have drawn as a tension member, we know that from our earlier solution, it is no longer an assumption E 70 35

36 E 700N E We also have no longer included the force from, we know that it carries no force under there loading conditions E 71 E 700N E Using our equilibrium conditions F y = 0 1 = 0 2 = 0 E 72 36

37 E 700N E Using our equilibrium conditions F x = 0 = = = 500 N T E 73 E 700N E 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 heck the sum of the forces in the x and y direction at that point to see if the truss closes 74 37

38 Homework Problem 6-2 Problem 6-5 Problem