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Calculating Vertical +Horizontal Forces of Beam in Equilib.

  1. Mar 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Below are attached 2 beams, A and B. I have to calculate the vertical and horizontal reactions at both supports in both questions. I have done a lot of calculations and believe I have a good understanding of what is needed to be done however my forces are not balancing for some reason. We use a software called LinPro and for Beam A the following answers are given: Ha = -12kN , Va = -39.46kN, Vb = 36.54kN

    2. Relevant equations
    ∑Ma = 0
    ∑Hf = 0
    ∑Vf = 0

    3. The attempt at a solution

    For A I have set it up as follows.
    Va + Vb = 60kN + 16kN (76kN)
    The 16kN comes from 4kN* 4 from the diagonal force given.
    I then took the moments about point A as follows.
    (20kN*3m*1.5m) + 5Vb + (16kN*7) = 0
    5Vb = 202kN
    Vb = 40.4kN which is NOT 36.54kN.

    The horizontal forces are easy to calculate due to Vb being a roller support therefore Hb = 0 which ultimately means Ha = -12kN.

    For the Support B!
    Va+Vb = 50kN
    Ha+Hb = 10kN (once again, Ha = -10kN because of the roller support at Hb = 0kN)

    For this one however the support is at an angle. How to deal with supports at an angle? I figured it would just be Arctan(1/2) but that s giving me some strange answers which I know from just looking at aren't quite right.

    Am I missing something crucial? I feel I must be overlooking something or are the answers I am calculating on Linpro software just incorrect and am I the one who is right? thanks
     

    Attached Files:

  2. jcsd
  3. Mar 4, 2016 #2

    SteamKing

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    You must account for the direction of the individual moments. About point A, the 20 kN dist. load will have a CW rotation as will the 16 kN component force at the end. The reaction Vb will have a CCW rotation.
    For the support on a slope, you must use the correct angle which is normal to the ground. Arctan (1/2) is not the correct angle normal to the ground.
     
  4. Mar 4, 2016 #3
    So you're saying instead of (20kN*3m*1.5m) + 5Vb + (16kN*7) = 0
    (20kN*3m*1.5m) + (16kN*7) = 5Vb
    How would that change my values though? or am I misunderstanding? Also, are the answers Ha = -12kN , Va = -39.46kN, Vb = 36.54kN correct for Beam A?

    Normal to the ground.. a flat surface is 180 degrees. The support is at a diagonal so it will have both a Hf and Vf component no?
     
  5. Mar 4, 2016 #4

    SteamKing

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    I thought you had two separate beams. What you have is a 2-D frame. View A represents a longitudinal elevation of this frame, while view B is looking from say the 2m vertical bar toward the other end. The tiny circle at the end of the 20 kN/m dist. load shows where this 5 m horizontal side beam connects to the 7 m long beam.

    Since this 5 m side beam is loaded, your equilibrium equations are affected, which is why your initial calculations weren't working out.
     
  6. Mar 4, 2016 #5
    Got it sorted! I forgot about the pin and realized where I was going wrong cheers for that!
     
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