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Statics Question regarding Joint type & Reactions

  1. Jun 16, 2010 #1
    When solving for reactions in a 2D diagram as seen below, how does each type of joint affect the summation of forces in the x & y components?

    My problem asks to find the summation of forces in the x & y components, and has a smooth pin connection in the left of the beam and roller on the right of the beam. (This is a conceptual question and not particularly about my HW problem)

    Here's an example of what I was referring to ( the two left figures).

    [URL]http://www.engin.brown.edu/courses/en3/notes/Statics/Constraints/Constraints_files/image030.gif[/URL]

    Thanks.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jun 20, 2010 #2
    Statics can often be an exercise in bookkeeping.

    Your question: "how does each type of joint affect the summation of forces in the x & y components?" I suggest may be rewritten as:

    "What reaction forces are present for each kind of support?"

    The reason I say this is that your summation of forces will follow directly once you discern which forces are present and which are not. That is to say, your equilibrium equations follow directly from free body diagrams (FBDs) you may draw. In a course like statics, you should *always* be drawing FBDs, they are informative and necessary for a complete solution (as well as allowing you to keep track of your signs).

    When thinking of which reaction forces are produced as a result of some support, you need only consider "which motion does said support prevent?" If a support prevents motion in some direction, a reaction force is responsible. It is convenient to resolve these into both x- and y-directions, where at times you may find it useful to incline your axis with respect to a horizontal/vertical system for cleaner equations (e.g. this may inherit the quality that less cosine/sine terms needed to isolate component forces along both directions). Once you decide on a coordinate axis, and draw your FBDs, you may write down component equations (in x and y if you like). Think about what each support is physically to gauge which reaction forces are present, there is nothing to memorize here.

    (1) smooth pin support (2D): this affixes whatever it is attached to so it cannot translate. In general, it is not possible to determine the reaction force's actual direction by inspection. Luckily, you may always resolve forces into components. The reaction force"s" here are then two unknowns, one along x, one along y. The proviso here is that a system is subjected to a variety of loads, and the reaction force actually does point in a definite direction, but that direction is intractable by inspection. Modeling the forces as components in x and y allows you to pick the directions along these known directions easily. Should you wish, you could figure out which way the reaction really points (i.e. the "resultant" force of adding the x and y components together), although this is not usually of concern. Note though, that if you actually knew the direction of the force, you could model the reaction as only one unknown. However, in most every case this information is not known, so we are forced to represent this reaction by two unknowns (components that we choose to lie along the x and y directions).

    Thus, in 2D, a pin support means one force each in the x and y directions. Bear in mind, the 3D analog is a ball-and-socket (in which case you would have no translation in any direction, so 3 reactions along x, y, and z).

    (2) Roller: A roller is a wheel or just a ball that impinges on a surface of an object. The roller pushes on the object, and the object pushes back on it (static equilibrium). In other words, should the object move against the roller, the roller *prevents* motion by pushing back on it. As discussed, this is the definition of a reaction force. The roller can only push, so you know its definite direction (even the sign of it). A reaction force is then easily modeled as a force perpendicular to the object's surface where it contacts the roller.

    My bad about being so loquacious. Make sense?

    You may find the following worked problem useful: http://www.itcanbeshown.com/For_Students/Solutions/EMA_201_Static_Determinacy_and_Fixity_Soln.pdf [Broken] . Whatever is written on that is not relevant for this question. I just wanted to draw your attention to the FBDs which involve all the types of reactions discussed above. In the figures, the top left is a pin support, the ball is a roller, and in the later parts a link support is shown.
     
    Last edited by a moderator: May 4, 2017
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