# Calculating Vertical +Horizontal Forces of Beam in Equilib.

1. Mar 4, 2016

### King_Silver

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

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2. Mar 4, 2016

### SteamKing

Staff Emeritus
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.

3. Mar 4, 2016

### King_Silver

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?

4. Mar 4, 2016

### SteamKing

Staff Emeritus
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.

5. Mar 4, 2016

### King_Silver

Got it sorted! I forgot about the pin and realized where I was going wrong cheers for that!