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Engineering Statics Equilibrium

  1. Dec 6, 2012 #1
    hi, for question two on the attached pdf, how would the equilibrium in the x work? From what i understand there must be an x and y component for the pin support at A because it acts at an angle but the support at B only has a y component because it is a rocker.
     

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  3. Dec 6, 2012 #2

    PhanthomJay

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    If there was an x (horizontal) component at the pin support A, then the beam could not be in equilibrium, which you appear to have concluded. So what does that imply about your statement that "there must be an x and y component for the pin support at A"? Is that true?
     
  4. Dec 6, 2012 #3
    But if the support at A is acting on an angle, doesn't there mathematically have to be an x and y component to that force?
     
  5. Dec 6, 2012 #4

    PhanthomJay

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    From equilibrium considerations, the reaction at A must be vertical. You can resolve that resultant reaction force into its vector components perpendicular and parallel to the incline if you want, but the resultant must be vertical.
     
  6. Dec 6, 2012 #5
    Okay so basically rewriting it and just showing A as an upward force and ignoring the x-component altogether.
     
  7. Dec 6, 2012 #6

    PhanthomJay

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    Well yes, A is upward, but it is not correct to say that you are ''ignoring'' the x-component, because no such x component exists. What you are ignoring is the slope when you solve for your shears and moments in the beam.
     
  8. Dec 6, 2012 #7
    okay so the horizontal and vertical are not related to each other through the angle of the pin
     
  9. Dec 6, 2012 #8

    PhanthomJay

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    Since there is no horizontal, there is no such relation. Whether the slope is 3:4 or 1:1 or vertical, or horixontal, or anything, the reaction is still upward of the same magnitude. The slope comes into play when you are considering the design of the support itself, not the beam.
     
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