In this entry:

### The process of calculating the reaction in a beam

We start by introducing appropriate support reactions in place of the supports. Please find more

__here__Then we check whether the beam is statically determinate. You can find more about it in the entry

__here__In the next step, we write the equilibrium equations. You can find more about it in

__Equilibrium equations.__

### Simply supported beam - calculation of reaction force for beam

Let's start with a coordinate system and assuming a positive counterclockwise rotation of the moment.

Below is an example diagram of a simply supported beam. This is the name given to a beam supported by hinged supports at both ends. We will determine the support reactions for this beam.

Reactions have already been added to the beam drawing. And so, at point A we have a non-moving articulated support, so we add the horizontal reaction HA and the vertical reaction VA. At point B at the end of the beam we have a sliding articulated support, so we add one vertical reaction VB. Next, let's check the static determinacy.

N=R-J-3=3-0-3=0 - the beam is statically determinate

Time for equilibrium equations. You remember that for a plane system of forces we have three equations:

Let's start with the first and simplest equation. The sum of the projections of forces on the x axis.

Since in our simply supported beam example there is no force component acting in the x direction, the reaction HA=0.

We will then move on to the third equation, for the sum of moments at a point.

The choice of point is yours. I chose point A.

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

In our example, we can choose point A or B. By selecting one of the supports, the reaction for this support will not appear in our moment equation, because the moment is the force multiplied by arm. If the arm is zero (the force passes through our point A), the moment of this force will also be zero, so we can ignore it in the equation.

In the equation we have:

The VB reaction multiplied by the distance 12, which is the distance between point A and B.

Force F multiplied by 2, i.e. the distance of force F from point A

Bending moment M. The moment is not multiplied by the distance.

Continuous load q multiplied by the length 4 over which it acts and 6, i.e. the distance from the center q to point A.

We pay attention to the signs of the moments in accordance with what we assumed at the beginning in Fig. 1

After the transformations, we obtain the value of the force VB, so we have the next reaction calculated.

Finally, we will write the equilibrium equation for the forces in the y-direction.

In the equation we have:

VA reaction with a positive sign, because the direction of the VA force is consistent with the direction of the y axis

Reaction VB with a positive sign, because the direction of force VA is consistent with the direction of the y axis

Continuous load q multiplied by 4, i.e. the length over which it acts

Force F with a negative sign, because the direction of force F is opposite to the y axis

After transformations and substituting the VB value, we obtain the VA force value. This is how we calculated all reactions.

I have included the entire solution below. This solution comes from __ __ where you can calculate reaction for any statically determinate beam.

### Cantilever beam, restrained - calculation of reaction force for beam

Below is an example diagram of a cantilever beam. This is what we call a beam fixed at one end. We will determine the support reactions for this beam.

Reactions have already been added to the beam drawing. And so, at point A we have restraint, so we add the horizontal reaction HA and the vertical reaction VA and the restraint moment MA. Next, let's check the static determinacy.

N=R-J-3=3-0-3=0 - the beam is statically determinate

Time for equilibrium equations. You remember that for a plane system of forces we have three equations:

As before, let's start with the first equation. The sum of the projections of forces on the x axis.

In the equation we have:

HA reaction with a positive sign, because the direction of the HA force is consistent with the direction of the x axis

The horizontal component of the force F with the sign is negative, because the direction of the force F is opposite to the x axis

After the transformations, we obtain the value of the HA force. We have counted the first reaction.

We will then write the equilibrium equation for the forces in the y direction.

In the equation we have:

VA reaction with a positive sign, because the direction of the VA force is consistent with the direction of the y axis

Continuous load q multiplied by 5, i.e. the length over which it acts

The vertical component of force F has a positive sign because the direction of force F is along the y axis

After the transformations, we obtain the value of the force VA. We've already counted two reactions😊

Finally, we will move on to the third equation, for the sum of moments at a point.

The choice of point is yours. I chose point A. Similarly to a simply supported beam, it is good to choose a point where we have reactions.

We get the following equation:

In the equation we have:

The moment of restraint MA as a reaction

Force Fsin45 multiplied by 5, i.e. the distance of force F from point A

Bending moment M. The moment is not multiplied by the distance. With a minus because it is directed opposite to our positive return

Continuous load q multiplied by the length 5 over which it acts and 12.5, i.e. the distance from the center q to point A.

After the transformations, we obtain the value of the moment MA. We have all the reactions planned. Great!!

I have included the entire solution below. This solution comes from __beam calculator__ where you can calculate reaction for any statically determinate beam.

This concludes the entry for calculating support reactions for beams. Thanks 😊

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