1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

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

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    Excellent, matthew_hanco.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pinned Jointed Frames
  1. Pin Jointed Frame (Replies: 5)

Loading...