In this entry:

Steps for calculating support reactions in a beam
Freely supported beam-reaction calculation
Cantilever beam-reaction calculation

The process of calculating the reaction in a beam

We start by introducing the appropriate support reactions at the support site. Read more about this in the publication Reaction forces.
Then we check whether the ray is statically determinate. Read more about this in the publication statically determinate.
In the next step, we write down the equilibrium equations. Read more about this in the publication Equilibrium equations.

Freely supported beam-calculation of support reactions for beams

We start by taking a coordinate system and assuming a positive counterclockwise torque.

Below I have included an example of a freely supported beam diagram. This is what we call a beam, supported by articulated supports at both ends. We will assign support reactions for this beam.

Reactions have already been added to the beams drawing. So, at point A, we have a pivot bearing support, so we add a horizontal reaction HA and a vertical va. At point B at the end of the beam, we have a hinged sliding support, so we add one vertical reaction VB. Then let's check the static notability.

N=R-J-3=3-0-3=0 - the beam is statically determinable
where:
N - degree of static indeterminacy
R =3 - number of support reactions
J =0 - number of internal joints
3 - the number of equilibrium equations. In static systems it is 3

It's time for the equilibrium equations. Remember that for a planar system of forces, we have three equations:

$\F_{ix} = 0$ - sum of the projections of forces on the x-axis

$\F_{iy} = 0.$ - sum of the projections of forces on the y-axis

$\■Sigma M_{i} = 0$ - sum of moments at a point

Let's start with the first and simplest equation. Sum of force projections on the x-axis.

Since in our example a freely supported beam does not have any component of the force acting in the x-axis direction, the reaction HA=0.

Then we move on to the third equation, the sum of moments at a point.

The choice of point is up to you. I chose point A.

When determining the equilibrium equations for the sum of moments, it is best to choose the point at which one of the supports is located.

In our example, we have the choice of punk A or B. By choosing one of the supports, we make the reaction for that support not appear in our equation for the moment, because the moment is the force multiplied by the hand. If the arm is zero (the force passes through our point a), then the moment of this force will also be zero, so we can omit it in the equation.

In the equation, we have:

Reaction VB multiplied by the distance 12, that is, as much as between points A and B.
Force F multiplied by 2, i.e. the distance of force F from point A
We do not multiply the bending moment M of the moment by the distance.
The continuous load q is multiplied by the length 4 on which it acts and 6, that is, the distance of the center of q to point A.
We pay attention to the signs of moments in accordance with what we assumed at the beginning in Fig. 1

After the transformations, we get the force value VB, so we get the calculated reaction.

Finally, we will write an equilibrium equation for forces towards the y-axis.

In the equation, we have:

The VA response is positive, since the return of the VA force corresponds to the return of the y-axis
Reaction VB with a positive sign, since the return of force VA corresponds to the return of the y-axis
Continuous load q multiplied by 4, i.e. the length at which it operates
Force F with the sign ujemy, because the return of force F is opposite to the y-axis

After converting and substituting VB values, we get the strength value VA. So we calculated all the reactions.

I have posted the entire solution below. This solution comes from my beam calculator. In this application you can calculate reactions, shear forces and bending moments for any statically determinable beam.

Cantilever beam, fixed-calculation of support reactions for beams

I've included an example of a cantilever beam diagram below. This is what we call a beam fixed at one end. We will assign support reactions for this beam.

Reactions have already been added to the beams drawing. So, at point A, we have a fixation, so we add the horizontal reaction HA and vertical VA and the moment of fixation Ma. Then let's check the static notability.

It's time for the equilibrium equations. Remember that for a planar system of forces, we have three equations:

$\F_{ix} = 0$ - sum of the projections of forces on the x-axis

$\F_{iy} = 0.$ - sum of the projections of forces on the y-axis

$\■Sigma M_{i} = 0$ - sum of moments at a point

As before, let's start with the first equation. Sum of force projections on the x-axis.

In the equation, we have:

HA reaction with a positive sign, since the return of the HA force corresponds to the return of the x-axis
The horizontal component of the force F with the sign ujemy, since the return of the force F is opposite to the x-axis

After the transformations, we get the strength value HA. We calculated the first reaction.

Next, we will write an equilibrium equation for forces in the direction of the y-axis.

In the equation, we have:

The VA response is positive, since the return of the VA force corresponds to the return of the y-axis
Continuous load q multiplied by 5, i.e. the length at which it operates
The vertical component of the force F with a positive sign, since the return of the force F corresponds to the Y-axis

After the transformations, we get the force value VA. We already have two reactions counted😊

Finally, we turn to the third equation, the sum of moments at a point.

The choice of point is up to you. I chose point A. As in a freely supported beam, it is good to choose a point where we have reactions.

We obtain the following equation:

In the equation, we have:

The moment of approval has as a reaction
Force Fsin45 multiplied by 5, i.e. the distance of force F from point A
We do not multiply the bending moment M of the moment by the distance. With a minus, because it returns the opposite of our positive return
The continuous load q is multiplied by the length 5 at which it operates and 12.5, that is, the distance of the center of q to point A.

After the transformations, we get the value of the moment Ma. We have all the reactions scheduled. Cool!!

Below I put the whole solution with Beam Calculator