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Homework Help: Pinned Jointed Frames

  1. Aug 11, 2011 #1
    The problem is attached to this post, basically i need the exterior reactions and how to work them out.

    i know i need to take moments im just not sure how to work it out.


    Basically what i worked out

    is

    i assumed horizontal members are equally to 1

    sum of the horizontal = 0 5Kn- HA - HE=0

    Sum of the vertical =0 10KN- VA =0

    But that is where im stuck i know the numbers should be roughly

    HA=20
    VA= 10
    HE = 15

    But i know it is more accurate than that im not sure if i need to include sine or cosine in the solution

    Thanx in advance
     

    Attached Files:

  2. jcsd
  3. Aug 11, 2011 #2
    To add to it i have just tried

    sum MA= 0

    Therefore he x 2 - 5x1 - 10x2 =0

    he =12.5

    sum of horizontal 5+12.5 =17.5

    therefore ha = -17.5

    sum of the vertical

    Va -10 x 1 =0

    so Va = 10

    Is that right?
     
  4. Aug 12, 2011 #3
    Im desperate and really need someones help urgently.
     
  5. Aug 12, 2011 #4

    Q_Goest

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    Hi matthew,
    Since you're taking moments around A and since the 5 kn and 10 kn forces produce opposite moments around A, one of your terms needs to be positive and one needs to be negative. You have a negative moment around A for BOTH of these forces.

    Once you get the horizontal force from E you can sum forces and get the horizontal and verticle forces on A.
     
  6. Aug 12, 2011 #5
    ok so the new calculated resultant forces would be

    He= 15 ( (-10 x 2) - (+5x1) + He=0)

    Ha= -20 (+15 + 5 - ha= 0 therfore Ha= -20)

    Va= -10 (+10 -Va=0)


    But now checking my lecturers answers he gives

    14.82 for He

    And a resultant force for a 22.2 acting 26.46 degrees from the horizontal


    But with my results the resultant for a is 22.36 and 26.56 from the horizontal.

    who is right, if he is how can i account for the discrepencies
     
  7. Aug 12, 2011 #6

    Q_Goest

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    Those answers are all pretty close, so I would guess the differences are in rounding error. Regardless, I can't check the math because the figure you provide has no dimensions on it.
     
  8. Aug 13, 2011 #7
    All forces are in KN
     
  9. Aug 13, 2011 #8

    Q_Goest

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    I can see the forces. There are no linear dimensions on your picture.
     
  10. Aug 13, 2011 #9

    nvn

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    matthew_hanco: The dimensions are already fully defined by the diagram given in post 1. The resultant reaction force at point A from your lecturer, 22.20 kN, is correct. However, the orientation angles given by both you and your lecturer are currently incorrect. The resultant reaction force at point A is 22.20 kN acting 26.77 deg from the horizontal. Keep trying. Your discrepancies are not due to round-off error. First, figure out the correct dimensions of your structure, and label them on your free-body diagram. You currently appear to be using wrong dimensions.
     
  11. Aug 14, 2011 #10
    Got it, the length should be root 3 instead of 1.

    so it should be

    (-10 x (root3x2) - +5x1)/2=Ha
     
  12. Aug 14, 2011 #11

    nvn

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    Excellent, matthew_hanco.
     
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