Internal forces in beams

In this post:

1. What are internal forces and types of internal forces

2. Determining internal forces for a beam

3. Internal force diagrams.

Internal forces in beams and their types

To explain what the internal forces in beams are, I will use the mental intersection method. In Fig.1. below there is a bridge supported at both ends by supports. We introduce reactions at the supports and the entire system is in balance.

What if we mentally cut the bridge into two parts now?

Without introducing internal forces at the intersection, the bridge will collapse and our deer will fall into the water😊

Therefore, in order for both parts of the bridge to be in balance, we must replace the interaction of the bridge fragments by applying internal forces.

Fig.1. Internal forces.

We have the following internal (cross-sectional) forces:

• N - normal force (axial, longitudinal) - in the beam it is the force acting in the horizontal x direction, parallel to the beam axis

• T - shear force (transverse) - in the beam it is the force acting in the vertical y direction, perpendicular to the beam axis

• M - bending moment

Determining internal forces for a beam

Let's move on to determining the internal forces in the beam. Let's do this using the example of a simply supported beam in Fig. 2.

Fig.2. Simply supported beam - determination of internal forces

The ranges are marked with numbers from 1 to 3.

We add a range when a new load or support appears.

We will determine the internal forces for each compartment. Before we start determining the forces, we must first discuss the labeling of internal forces

Fig.3. Marking of internal forces

In Fig.3. The method of marking the normal force, shear force and bending moment is presented above. Vectors directed this way have a positive value and you need to remember this. Vectors directed oppositely will have a negative sign. As you can see, it is different on the left and right side of the beam.

Remember that the sign of the bending moment used to determine the reaction may differ from that used for internal forces. I advise separating these two stages of taking the bending moment signs.

Fig.4. Bending moment marking

Additionally, as an association of a positive sign of the bending moment with a smiling face (the bent ends of the beam create a smile). And the negative sign creates a sad face.

First of all, before we start calculating the internal forces, we need to check the static determinacy and calculate the support reactions .

Fig.5. Checking static determinacy

The system is statically determinate - we can proceed to the reaction.

Using the equilibrium equations, we calculate the values of support reactions for a simply supported beam.

Fig.6. Calculation of support reactions

Having correctly determined the support reactions in the simply supported beam , we can begin to calculate the internal forces for individual compartments .

The beam diagram and all calculations are generated in my beam calculator . You can use it to obtain a detailed solution for each statically determinate beam.

Compartment 1

For the first interval, x is from 0 to 2m. I marked the marking system of shearing and normal forces for the left-sided cross-section in blue .

Fig. 7. First compartment - internal forces in the beam

Normal force N:

As for the force N1(x) in the first interval, it is -HA (HA is opposite to our marking system, hence the minus sign), which after substituting the value gives us 10 [kN]. The value is positive, so we have a stretching of the cross-section. As you can see, I use the notation N1(x), which means that N1 is a function of x. We can insert any x from 0 to 2 and we will get the result of the normal force for this x coordinate.

For the purposes of drawing internal force diagrams, we will calculate characteristic points, i.e. the beginning and end of the interval.

Shear force force T:

The shear force T1(x) is VA (positive sign, meaning consistent with our shear force labeling). We have a constant value of the shear force over the entire interval. After substituting Va we have T1=-5.3 [N]

Bending moment Mg:

The most important stage in solving beams is determining bending moments. This is also the most difficult part of solving beam problems.

The bending moment is a function of Va*x. As we know, moment is force multiplied by the arm. The force is the shear force - the arm is our x. The further we are from the support A, the greater the moment from the VA reaction. After substituting x for the beginning of the interval M1(0) = 0 and the end of the interval M1(2) = -10.6 [Nm].

If there is no concentrated moment applied at the beginning or end of the beam , the bending moment value will always be zero

We have designated the first compartment. Let's move on to the next one.

Compartment 2

People often ask whether I should also include forces from the first range or omit it? The answer is:

When it comes to the equations describing the forces in each subsequent interval, we take into account everything that happens from the beginning of the beam, i.e. the forces from each preceding interval are also taken into account.

In the second interval, x is from 2 to 6 m.

I marked the marking system of shear and normal forces for the left-sided cross-section in blue.

Fig. 8. Second compartment - internal forces in the beam

Fig.9. Equations of internal forces for the second part of the beam

Normal force N:

As for the force N2(x) in the second interval, we subtract the force F2 from the value of the previous one, we get -HA-F2. After substituting the values, it gives us 0 [kN], which means there is no normal force in this range.

Shear force T:

The shear force T2(x) added to VA becomes F1. We have a constant value of the shear force over the entire interval, after substituting Va we have T2= 2.7 [N]

Bending moment Mg:

To Va*x we get F1*(x-2) the sign of the moment from F1 is positive. X is reduced by 2m, i.e. by the amount of force F1 that is distant from the beginning of our beam. The actual arm of force F1 is (x-2). After substituting x for the beginning of the interval M2(2) = -10.6 [Nm] and the end of the interval M2(6) = 0.2 [Nm].

Compartment 3

In the last interval, x is from 6 to 10 m.

I marked the marking system of shearing and normal forces for the left-sided cross-section in blue.

Fig. 10. Third compartment - internal forces in the beam

Fig.11. Equations of internal forces for the third part of the beam

Normal force N:

As for the force N3(x), it is equal to the force N2, nothing changes.

Shear force T:

In the formula for the shear force T3(x), we have the continuous load q multiplied by the length over which it occurs, i.e. the section (x-6). After substituting the beginning of the interval for x, we have T3(6)= 2.7 [N] and the end of the interval T3(10)= -5.3[N]

Bending moment Mg:

In the equation for the moment we have what is in the range 2. Additionally, we add M and subtract the value of the moment from the continuous load q. This value is g multiplied by (x-6), which gives us the force, and multiplied by 0.5*(x-6), which is the arm of the force. The sign is negative because the moment from g will act opposite to our assumed positive sign of the moment. After substituting x for the beginning of the interval M3(6) = 5.2 [Nm] and the end of the interval M3(10) = 0 [Nm].

And so we finished determining the internal forces in our beam. Based on the results obtained, charts are prepared as in Fig. 12 below. But more about that in the next entry.

Fig.12. Diagrams of internal forces of a simply supported beam.

Thank you, see you next post .