# CEE210 Assess05-Frames-15Su-SOLN.pdf - CEE/CNE 210 Statics...

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Name: Date: Subject: CEE/CNE 210 Statics Arizona State University Problem 1 The frame consists of four rigid bars pinned together and it is externally supported by pins at A and 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 BC that 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 Solution June 19, 2015 Assessment 05 Frame Analysis A. Modeling & Techniques Since there are no forces in the e 3 direction and no couple moments in the e 1 and e 2 directions, this problem can be analyzed as 2-D. Therefore, force vector equilibrium provides two equivalent scalar equations (in the e 1 and e 2 directions) and moment vector equilibrium provides one equivalent scalar equation (in the e 3 direction). The external supports at A and F provide only force reactions that each have unknown magnitude and direction. By inspection we recognize bar CD as a two-force-member. This means that the forces on pins C and D must 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 BC has only 3 unknowns, which can be solved directly. The distributed load must be integrated along the length of bar BC to 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 B will allow us to solve for the forces on the two-force member CD and then force equilibrium give us the force at B . (2) Next, for an FBD of bar DEF , we’ll sum moments about F and use the triple scalar product to determine the e 1 component of the force on pin E . (3) At this point bar ABE will have only three unknowns and we’ll sum moments about A to determine the e 2 component of the force at pin E .