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!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

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

Have something to add?
Draft saved Draft deleted