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!

Determine forces in beam with hinges

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data


    Is it possible to find the VA , VB and VC without ' breaking ' the beam into 2 parts ?



    2. Relevant equations


    3. The attempt at a solution
    Here's my working , i gt

    Moment about A = 10(1) -VB(2) -VC(4) +5(3)(4 + (3/2) ) = 0 Hence , 92.5-2VB -4VC= 0

    moment about B = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) ) = 0 Hence , 42.5 + 2VA -2VC = 0

    Moment about C = VA(4) -10(3) +VB(2) + (5)(3)(3/2) =0, Hence , -7.5 +4VA +2VB = 0

    and sum of vertical force = VA +VB +VC = 25

    Then , i solve the 4 equations using online calculator , i found that some of the values couldnt be found .

    Or there's something wrong with my working ?
     

    Attached Files:

    • 446.PNG
      446.PNG
      File size:
      60.5 KB
      Views:
      37
    • 447.PNG
      447.PNG
      File size:
      39.7 KB
      Views:
      29
  2. jcsd
  3. Nov 29, 2016 #2
    Bump
     
  4. Nov 29, 2016 #3

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    you are making a few errors here, particularly when including the hinge force at B in your equations when looking at the whole beam, because it is not an external support. But anyway, looking at the entire beam, you can sum moments = 0 about countless points and get an infinite number of equations , but you won't get anywhere, because you only have --?-- equilibrium equations to work with. Break the beam apart.
     
  5. Nov 29, 2016 #4
    Do you mean for Moment about A = 10(1) -VC(4) +5(3)(4 + (3/2) ) = 0 Hence , 92.5 -4VC= 0 , it should be like this ?

    For Moment about C = VA(4) -10(3) + (5)(3)(3/2) =0, Hence , -7.5 +4VA = 0 , is should be like this ?

    For moment about B , = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) ) = 0 Hence , 42.5 + 2VA -2VC = 0 ?

    sollving 3 equations , i have VA = 1.875 , VC = 23.125 , VB =0 ?
    bUT , my ans is still different from the author ans
     
  6. Nov 29, 2016 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You are not paying attention , you forgot to include the unknown moment , M, at the fixed end. The problem cannot be solved unless you break up the beam at B. Follow the book solution.
     
  7. Nov 30, 2016 #6
    after making correction ,

    Moment about A = 10(1) -VC(4)+M +5(3)(4 + (3/2) ) = 0 Hence , 92.5 -4VC +M= 0


    For moment about B , = VA(2) -10-VC(2) +(5)(3)(2 + (3/2) )+M = 0 Hence , 42.5 + 2VA -2VC +M= 0 ?


    For Moment about C = VA(4) -10(3) +M+ (5)(3)(3/2) =0, Hence , -7.5 +4VA+M = 0 ,

    here's my answer , M = 0 , VA = 1.875 , VC = 23.125 , which is still not same as the author's working
     
    Last edited: Nov 30, 2016
  8. Nov 30, 2016 #7

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    well, you didn't do it right. You do not get any more useful equations when summing moments about multiple different points. Suppose you had 2 equations, x + y = 0 and 2x +2y = 0. 2 equations, 2 unknowns, but I bet you can't solve for x and y.
    If you do not wish to believe that you must break up the beam to solve, then I can't help you any further.
     
  9. Nov 30, 2016 #8
    do you mean my equation is correct ? just my concept is incorrect ?
     
  10. Nov 30, 2016 #9

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, your equations are ok but if you try to solve them correctly you get 0 = 0. The concept is wrong. When you look at the entire beam, you have 4 unknowns....the moment at A, VA, HA, and VC. Four unknowns, with only 3 useful equilibrium equations: sum of forces in y direction = 0, sum of forces in x direction =0, and sum of moments about a point = 0. So you need to resort to another method to solve, which is the one described in the solution.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Determine forces in beam with hinges
  1. Forces on beam (Replies: 9)

  2. Beam with 2 hinges (Replies: 6)

Loading...