1. Limited time only! Sign up for a free 30min personal 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!

Statics: moments around beam subjected to 3 forces

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    NDCw2.jpg
    The beam is subjected to two oppositve vertical forces and one horizontal force as shown. Determine the reactions at the supports A & C



    2. Relevant equations

    [itex]\sum[/itex]forces in plane = 0
    moment = force x perpendicular distance



    3. The attempt at a solution

    [itex]\sum[/itex]Y = 15 + RyC + RyA - 15 = 0
    [itex]\sum[/itex]x = 20 + RxC = 0
    [itex]\sum[/itex]Moments around A = (-15 * 3) + (15 *12)+(RyC * 9)

    Reaction at C in x = 20 kN
    Not sure about reactions in Y at either, tried to calculate this from the moments but neither of these types of joint provide any reaction force to moments?

    Thanks in advance
     
    Last edited: Nov 3, 2011
  2. jcsd
  3. Nov 3, 2011 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Point A is a roller. Is point C pinned?

    BTW, the moment arm from A to C is 12 m according to the figure, not 15 m as shown in the calculation.
     
  4. Nov 3, 2011 #3
    yep point C is pinned, and thanks for spotting that mistake! I still can't see how to calculate the moment reaction though because A is a roller so has no reaction to the moments?

    Following on from that I get RyC = 135/9 = 15 kN... but that would mean the RyA would have to be -15kN?
     
    Last edited: Nov 3, 2011
  5. Nov 4, 2011 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The roller at A and the pin at C can both exert force on the beam, even though not being able to resist a moment.

    Initially, static reactions at A and C are assumed to exist. Then, by applying the equations of static equilibrium, the magnitudes and directions of the reactions can be determined. Even though neither point A or point C can resist a moment, nevertheless, the sum of the moments about either point due to the applied loads and reactions must sum to zero, just like the forces do.
     
  6. Nov 4, 2011 #5
    Moments around A = (-15 * 3) + (15 *12)+(RyC * 9) = 0

    = -45 + 135 + 9 Ry at C = 0
    = 90 + 9 Ry at C = 0

    90/9 = Ry C = 10

    But that must mean that Ry at A is negative in order for the forces in the y direction to balance?
     
    Last edited: Nov 4, 2011
  7. Nov 4, 2011 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You have an arithmetic mistake in your moment calculations. What is 15 * 12?
     
  8. Nov 4, 2011 #7
    oops thats what I get for working on it so late...180
     
  9. Nov 5, 2011 #8
    Regardless of that error, this still doesn't make sense to me though, there are 2 vertical reactions...and they cannot be equal to 0 because the moments would not be in equilibirum.

    However since the two known vertical forces are equal and opposite, both of the vertical reactions cannot be positive (upwards) or else the beam would not be in equilibrium? But how could the support provide a downwards force?
     
    Last edited: Nov 5, 2011
  10. Nov 5, 2011 #9

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You must correct the arithmetic in your moment equation and calculate the reaction at C.
    The roller at A can only offer an upward reaction, but the pinned connection at C can have a reaction oriented in any direction.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Statics: moments around beam subjected to 3 forces
Loading...