Fundamental Mechanics ENGG 1300 Pin-Jointed Trusses - Method of Joints Ir Dr. Ray Su Faculty of Engineering THE UNIVERSITY OF HONG KONG 1 Structural Analysis of Pin-Jointed Trusses 2 Pin-Jointed Truss A truss is a structure composed of a series of slender members jointed together at their ends. Roof truss Planar trusses lie in a single plane. They are often used for roof structures or bridges. They are typically constructed from triangular (but not quadrilateral) elements to achieve a stable structural form. Truss bridge footbridge 3 Connections The connections of a truss are usually formed by bolting or welding the ends to a common plate, called a gusset plate. bolts Gusset plate Some of them are formed by simply passing a large bolt or pin through each of the members. A pin 4 Assumptions for analysing pin-jointed trusses Analytical model of a bridge truss All loadings are applied at the joints In most situations, such as for bridge and roof trusses, this assumption is true. The external loads are applied at the joints. The self-weight of a member is usually evenly distributed at the end of the member. For simplicity, the self-weight of trusses is neglected in this course, unless otherwise specified. The members are jointed together by smooth (frictionless) pins. The truss members take axial load only, either in tension or in compression. Deformations of the truss are small. 5 Sign convention of pin-jointed truss members The reaction forces at the ends of the truss member must be Why? directed along the axis of the member. If the force tends to elongate the member, it is considered to be in tension. T T If the force tends to shorten the member, it is considered to be in compression. C C Often, compression members must be made thicker to resist buckling effects. 7 For information Buckling of a compression member Buckling is an instability phenomenon in structural engineering, where a small increase in the compression load can lead to a sudden large deformation and catastrophic failure of a member under compression. Before buckling After buckling 8 Method of Joints In previous lecture, we have already learnt how to calculate external reaction forces for beams and frames. In order to fully analyse or solve a truss structure, we need to calculate all the internal forces within every member of the truss. The method of joints simply consists of: 1. Selecting a solvable joint within the truss and isolating it. 2. Drawing a free body of that solvable joint – replacing the truss members with unknown axial forces. 3. Applying the two equilibrium equations to solve for the unknown forces. Why? F H 0 and F V 0 9 Example Determine all the internal forces of the truss by the Method of Joints. Solution The free body diagram of Joint B is shown below. Applying ΣFH = 0, then A pinned support A roller support + + 500 – FBC sin 45º = 0 FBC = 500 / sin 45º = 707.1 N 10 After updating the free body diagram of Joint B, we can now solve for member BA. FBC = 707.1N Applying ΣFV = 0, then + + 707.1 cos 45º – FBA = 0 FBA = 707.1 cos 45º = 500 N 11 Sign convention for tension and compression When using the method of joints, we are considering internal forces of members (in black). Which one is in tension? The forces shown (in red) are the external forces. Remember that when using the method of joints – the joint must be in equilibrium (i.e. external loads balance by internal forces). The best way to remember the sign convention is: All tension forces are drawn AWAY from the joint. 12 Continue from the previous Example, the next joint to analyse is Joint C. FBD of Joint C (C) (T) RC F H 0 + 707.1 cos 45º – FCA = 0 FCA = 500 N F V 0 – 707.1 sin 45º + RC = 0 RC = 500 N You can now transfer your answers to the main FBD. 13 The last joint to analyse is Joint A. FBD of Joint A RAH RAV FAC = 500N (T) RC = 500N F F H 0 RAH = 500 N V 0 RAV = 500 N You can now transfer your answers to the main FBD. 14 RAH = 500N FAC = 500N (T) RAV = 500N RC = 500N Free body diagrams of joints and members Note: You can check your answers by working out the reactions separately and seeing if they match. Example: Take moment at A, -500 N2+Rc2 = 0, Rc = 500 N 15 Procedure of method of joints 1. Draw the FBD for the complete truss. 2. Label all the joints in the truss (preferably A to Z, usually omitting I). 3. Draw and label the external support reactions. 4. Decide whether you need to calculate the external support reactions in order to progress or complete the truss analysis. 5. Determine the forces in each truss member by method of joints: i. Choose an appropriate solvable joint (i.e. a joint with at most two unknown member forces). ii. Draw the FBD of the chosen joint – assuming the unknown member forces to be in tension or compression. iii. Apply the equilibrium equations to solve for the unknown forces. iv. Transfer each member force to the main FBD (of the complete truss), ensuring correct tension or compression signs are expressed. v. Repeat steps i to iv until the entire truss has been solved. 16 Example Determine all the internal forces of the truss by the Method of Joints. pinned support Roller support Here, we need to calculate the reactions first because Joints B and D both have more than 2 unknown member forces. 17 Solution M RCV FBD B RCH AboutC 400 x 3 + 600 x 4 - RA x 6 = 0 RA = 600 N F V D 0 -RCV + RA – 400 = 0, Rcv = 200 N F H RA 0 0 RCH = 600 N Now it is possible to analyse either Joint A or C. 18 Choosing Joint A: 200 N FBD B FBD of Joint A 600 N (C) (T) D 600 N F F V 0 + 600 – (FAB x 4/5) = 0 FAB = 600 x 5/4 = 750 N H 0 – (750 x 3/5) + FAD = 0 FAD = 450 N You can now transfer your answers to the main FBD. 19 Choosing Joint C next: 200 N FBD B FCB 600 N FCD You should be able to see via inspection that: FCB = 600 N (C) FCD = 200 N (C) 450 N (T) D 600 N You can now transfer your answers to the main FBD. 20 Choosing Joint D next: B 600 N (C) 200 N (C) 600 N 450 N (T) FBD of Joint D FDB (C) 200 N 200 N FBD D 600 N F V 0 – 200 – (FDB x 4/5) = 0 FDB = – 200 x 5/4 = – 250 N As FDB is negative, this means that the initial assumption of force sense was incorrect. You can check the results by summing the horizontal forces. FH 0 – 450 + 600 +(–250 x 3/5) = 0 21 You can now transfer your answers to the main FBD. 200 N FBD B 600 N (C) 200 N (C) 600 N 450 N (T) D As FDB is negative, this means that the initial assumption of force sense (C) was incorrect. So we can simply change the sense of the force when transferring it to the main free body diagram as shown. 600 N 22 Method of Joints To sum up, the method of joints consists of: 1. Selecting a solvable joint within the truss and isolating it. 2. Drawing a free body diagram of that solvable joint – replacing the truss members with unknown axial forces. 3. Applying the equilibrium equations to solve for the unknown forces. 23 Ended 24 For information Force equilibrium A joint F H 0 F1 Unknown force F3 Applied forces + The usual sign convention Then F2 - F1 - F2 - F3 = 0 F1 = + F2 + F3 The second equation can actually be interpreted like that the unknown force (F1) is balanced by the applied loads (F2 & F3). The corresponding sign convention is + Unknown force = + Applied loads or + = + 25