Engineering Mechanics NPTEL 2017 Analysis of plane trusses Analysis of trusses is the central to designing of structural elements in any system. For example, the chassis of a car is made of many structural elements. One can consider it as a 3D space truss and find the forces acting on each element of the structure. We will now look at methods of analysing plane trusses. Plane trusses are 2D trusses. We will look at how to analyse plane trusses using an example. Method of joints Basic steps to follow The following are the basic steps to be followed while analysing trusses: 1. Determine the nature of the supports. 2. Draw the free body diagram of the truss system. 3. Determine the reactions offered by the supports. 4. Draw free body diagram of each joint in the truss with the convention that forces due to the members act away from the joints. 5. Apply force balance on each joint along x and y directions. We will end up with a set of linear equations with the forces in the members as variables. 6. Solve these equations to determine the forces in each member. 7. If the force in a member is positive, it is in tension. If it is negative, it is in compression. Example Step 1: The support at joint A is a pinned support while that at joint B is a roller support. A pinned support provides x and y reactions while a roller support provides only y reaction. Step 2: The free body diagram of the above truss system is : 1 Engineering Mechanics NPTEL 2017 Step 3: x and y balance of forces give us: Rx − 10 = 0 (1) Ry + N − 20 = 0 (2) Moment balance about point A gives us: N ∗ (2) + 10 ∗ √ 3 1 − 20 ∗ =0 2 2 (3) Solving equations (1), (2) and (3), we get: Rx = 10kN (4) Ry = 19.330127kN (5) N = 0.66987kN (6) Step 4: Free body diagram of all the pins: 2 Engineering Mechanics NPTEL 2017 Step 5: x and y balance of forces on each pin gives: TAB + TAE + Rx = 0 2 √ 3TAB + Ry = 0 2 TBE TAB TBC + − =0 2 2 √ √ 3TAB 3TBE 20 + + =0 2 2 TCD TCE TBC + − + 10 = 0 2 2 √ √ 3TCE 3TCD + =0 2 2 √ 3TCD =0 N+ 2 TCD TED + =0 2 TCE TBE − − TAE = 0 TED + 2 2 √ √ 3TCE 3TBE + =0 2 2 (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) Step 6: Solving the equations (4)- (16), we get the forces in the members as: TAB = −22.3205kN (17) TBC = −10.7735kN (18) TCD = −0.7735kN (19) TED = 0.38675kN (20) TAE = 1.16025kN (21) TBE = −0.7735kN (22) TCE = 0.7735kN (23) Step 7: Members CD and AE are in tension while the rest are in compression. Easier and more quicker way of using the above method Instead of balancing forces on each joint and then trying to solve all the equations, we can simultaneously balance forces and solve for forces in members. Start with a joint which is acted on by forces among which less than two are unknown. Since we’ll have two equations, one for x balance and one for y balance, we can obtain those unknown forces. Now, again look for a joint which is acted on by forces among which less than two are unknown. Keep following the same procedure untill all forces are determined. In the above example, if we look at the free body diagram of all joints, we can see that joint A and joint D are acted on by forces among which only two are unknown in each. Choose any one joint. Let’s say joint D. If we √ 3TCD = 0 and N + = 0, since N has already been look at the force balance equations i.e. TED + TCD 2 2 determined, only TCD and TED are unknown. Using these two equations, we can obtain TCD and TED . Now, since TCD has been determined, even joint C is acted on by forces among which only two are unknown. Similarly, we can solve for the two unknown forces with the two force balance equations. By marching from one joint to another using this technique, we can be quicker in determining the forces in the memebers. 3 Engineering Mechanics NPTEL 2017 Method of sections Method of sections is another method to determine forces in members of a truss structure. In order to find unknown forces in members using method of sections, one must section the truss structure into two halves such that not more than 3 members are being sectioned. The section must contain the member who’s force is to be determined. The net force on one part of the section must be zero since the isolated section does not move. This method is often faster because instead of moving from joint to joint with the method of joints, the members of interest can be immediately isolated. Example Let’s consider the same example we used to explain method of joints. Let’s say, we are interested in finding the force in member BE. We need to section the truss into two halves such that BE is sectioned and there are not more than 3 members are being sectioned. One way of doing that is to section AE, BE and BC. Now, the sectioned truss looks like: Now, consider any one of the halves and apply equillibrium equations. Let use do this for the right half. The force balance and momemt balance about D give: TBC + TAE + TBE + 10 = 0 2 √ 3TBE +N =0 2 √ √ √ 3 3 3TBE 10 ∗ ( ) + TBC ∗ ( )− ∗ (1) = 0 2 2 2 We have three unknowns and three variables. Solving them, we get: (24) (25) (26) TBE = −0.7735kN (27) TAE = 1.16025kN (28) TBC = −10.7735kN (29) These values are consistent with the values obtained from method of joints. 4