1. CE 201 - Statics Chapter 6 – Lecture 20

2. THE METHOD OF JOINTS All joints are in equilibrium since the truss is in equilibrium. The method of joints is applied using equilibrium equations at each joint of the truss. 30  F B C A F BA F BC F 60  B

3. To do this, the free-body diagram of the joints has to be dawn keeping the geometry of the truss in mind. Always start with a joint that has at least on known and at most two unknowns. Since all forces passes the joint, then  M B is automatically zero.  Fx = 0 and  Fy = 0 need to be solved to determine the unknowns. Assume the direction of the force acting on the joint. If the results produced positive scalar, then your assumption is correct, otherwise the force is acting in the opposite direction if the results were negative.

4. Procedure for Analysis Draw free-body diagram of the joint Stat with the joint that has at least one known and at most two unknowns External reactions at truss supports must be known Assume direction of unknowns Apply equilibrium equations (  Fx = 0 and  Fy = 0) Solve for unknowns and verify their assumed directions Continue the analysis for each joint. Always choose joints with at most two unknowns and at least one known A force member found from one joint can be used at the other end of the member for analysis of forces acting on that member

14. Zero-Force Members Zero-force members are used to: Increase stability of the truss during construction Provide support if applied loading is changed As a general rule: 1. Zero-force members are formed when only two member form a truss joint and no external load or support reaction is applied to that joint

15. A B C D E F  F AF F AB Joint A x  F DE F DC x y Joint D  Fx = 0 ; F AB = 0  Fy = 0 ; F AF = 0  Fy = 0 ; F DC sin  = 0 » F DC = 0  Fx = 0 ; F DE = 0

16. 2. If three members form a truss joint for which two of the members are co-linear, the third member is a zero-force member provided that no external force or support reaction is applied to the joint. A C D E 

17. x y F DA F DE F DC  Fx = 0 ; F DA = 0  Fy = 0 ; F DC = F DE x y F CA F CD F CB  Fx = 0 ; F CA sin  = 0 » F CA = 0  Fy = 0 ; F CB = F CD Joint C