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Engineering Statics (Equilibrium of Rigid Bodies)

  1. May 9, 2013 #1
    I have problems understanding the sense of reactive forces generated at pins / connections.
    Book is by Hibbler, I am posting links to images which will better explain my questions.

    The following links are concerning reactive forces.

    1#

    http://i.imgur.com/0x62lp2.png

    2#

    http://i.imgur.com/gedJfXk.png

    3#

    http://i.imgur.com/ytfxbOB.png

    4#

    I don't know completely how the external moment in this case affect the beam / rod and make its pin generate reactive forces.

    http://i.imgur.com/2BOj90e.png

    5#

    The same confusion about the sense and direction of reactive forces at pin

    http://i.imgur.com/GfyBICD.png

    6#

    And have a look at this problem.

    http://i.imgur.com/i9SAgfL.png

    --------------------------------------…

    Kindly explain every image in order? I am not getting a bit from this topic :|
     
  2. jcsd
  3. May 9, 2013 #2

    SteamKing

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    In #6, a couple is generated by the two springs. When the boy is standing on the end of the board, it is horizontal; after he jumps off, the board assumes some angle, which suggests that the springs undergo some extension/compression when someone of the proper weight is standing on the end of the board to make it horizontal.

    In #2, if there were two axial reactions, the beam would be much harder to solve. I agree that with two pins, there should be two axial reactions. IMO, this example is slightly incorrect, but I think it is just for practice in setting up and solving equilibrium equations. With two axial reactions, I think the beam is statically indeterminate, and additional equations of equilibrium derived from strength of materials have to be developed.

    As a general rule, in the other examples, one assumes the positive direction of reactive forces and moments at the start of analyzing for equilibrium. After running thru the equilibrium calculations, if the reactive forces and moments are negative, it means that the assumed directions of action made at the beginning of the calculation were in error.
     
  4. May 9, 2013 #3

    SteamKing

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    In examples #3 and #5, the reactive forces are assumed to act such that they prevent motion of the bodies due to the applied forces. When a body is in equilibrium, there are no net forces and moments acting on the body which would cause either translation or rotation.
     
  5. Jun 16, 2013 #4
    Well I learned the subject by thinking about what would physically happen. For example with a pin, no matter what you do it will not move freely left, right, up or down it is essentially fixed. Therefore since the pin is not able to move up or down (y-direction) freely a reaction force is created, likewise since the pin is not able to move right or left (x-direction) then a reaction force is created. It's the same concept for all supports, a rocker can move side to side freely so no reaction force is created in the x direction, but a rocker cannot move freely vertically so a vertical reaction force is created. Also it doesn't matter what way you draw the arrows on the diagram, once you work out the calculations using the equations of equilibrium if you obtained a negative number than you know you drew your forces in the wrong direction. If you're using Hibbeler's Engineering Mechanics:Statics 13th edition re read the chapter, it discusses all these things and also gives a nice chart of supports and their reactive forces. You can also check YouTube there's a user by the name structurefree and he does an excellent job explaining engineering concepts as well and he has a ton of these problems on his channel. Hope that helps
     
  6. Jun 16, 2013 #5

    nvn

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    jukos: In #1, I think you are correct; I currently think there is no horizontal reaction force at point A. I.e., I currently think Ax = 0 N.

    In #2, the support at point A is shown and clearly stated to be a rocker, not a pin. The rocker at point A is free to rock (translate) horizontally. Therefore, the support at point A generates no horizontal reaction force.

    In #3, they perhaps assumed Ax is directed in the +x direction because the reaction force at point B has a negative horizontal force component (?). But even if they assumed the wrong direction, then the answer for Ax would turn out negative, as stated by SteamKing and caldweab.

    In #4, as stated by others, you do not need to know the direction of the reactions. If you draw any reaction in the wrong direction, then it would turn out negative in your answers. You have three equilibrium equations, and three unknowns. Therefore, you can solve for the reactions using statics (static equilibrium). This is how you will find out how the applied external moment affects the reactions.

    In #6, good catch; you are correct. Assume spring A is connected to the diving board with a pin. Assume spring B is connected to the board with a roller.
     
    Last edited: Jun 17, 2013
  7. Jun 17, 2013 #6
    Number 2 is not statically indeterminate. There's three unknowns and three equations of equilibrium. If we take the moment about B we can solve for Ay, since it eliminates By and Bx from the calculations. We can then sum forces in the Y direction using the value of Ay we found from the moment equation to get By, finally summing forces in the X direction will give you Bx. The support at a is a rocker not a pin, if it were a pin this problem would be a bit more complicated because you'd have two vertical reactive forces and two horizontal reactive forces. 4 unknowns, 3 equations. Although you don't need strength of materials in this case to solve it.
     
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