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Name: Date:Subject: CEE/CNE 210 StaticsArizona State UniversityProblem 1 The frame consists of four rigid bars pinned together and it is externally supported by pins at Aand F. The self-weights of all bars are small compared to the external loads and can be neglected in this case. There is a uniformly distributed load acting along the length of bar BCthat has a magnitude of force per unit length, w.Determine all of the forces on each of the bars in the frame.Diagram is not to scale.Standard SolutionJune 19, 2015Assessment 05Frame AnalysisA. Modeling & Techniques•Since there are no forces in the e3direction and no couple moments in the e1and e2directions, this problem can be analyzed as 2-D. Therefore, force vector equilibrium provides two equivalent scalar equations (in the e1and e2directions) and moment vector equilibrium provides one equivalent scalar equation (in the e3direction). •The external supports at Aand Fprovide only force reactions that each have unknown magnitude and direction.•By inspection we recognize bar CDas a two-force-member. This means that the forces on pins Cand Dmust be equal in magnitude and along the line CD. Therefore, the directions of those forces are known. This also means that an FBD of bar BChas only 3 unknowns, which can be solved directly.•The distributed load must be integrated along the length of bar BCto determine the net force and moment it causes.We will separate the frame into its individual members. We will break it up at the pins because, at these locations, we know there are no internal moments. To solve this problem, (1)We’ll start with the FBD of bar BC. Summing moments about Bwill allow us to solve for the forces on the two-force member CDand then force equilibrium give us the force at B. (2)Next, for an FBD of bar DEF, we’ll sum moments about Fand use the triple scalar product to determine the e1component of the force on pin E.(3)At this point bar ABEwill have only three unknowns and we’ll sum moments about Ato determine the e2component of the force at pin E.