In this entry:

1. What are internal forces and types of internal forces
2. Determination of internal forces for a beam
3. Internal force diagrams.

Internal forces in beams and their types

To explain what the internal forces in the beams are I will use the method of intersection by thought. In Fig.1 below we have a bridge supported at two ends by hinged supports. At the point of the supports we introduce reactions and the whole system is in equilibrium.

Now if we were to mentally cut the bridge in two?

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

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

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

• N - normal force (axial, longitudinal) - in a beam it is the force acting in the horizontal direction x, parallel to the beam axis
• V - shear force ( transverse) - in a beam is a force acting in the vertical direction y, perpendicular to the beam axis
• M - bending moment

Determination of internal forces for a beam

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

The numbers from 1 to 3 indicate the ranges.

A span is added when a new load or support appears.

We will do the determination of internal forces for each span in turn. Before we start determining the forces, we must first discuss the labeling of internal forces

The figure above shows how to label the normal force, the shear force and the bending moment. Vectors directed in this way have a positive value, and this is what you need to remember. Oppositely directed vectors will have a negative sign. As you can see, it is different on the left and right side of the beam.

Remember that the bending moment sign adopted in determining the reaction may differ from that used for internal forces. I advise you to keep the two steps of adopting bending moment signs separate.

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

In the first place, before we proceed to calculate the internal forces we need to check static determinability and calculate support reactions .

The system is statically determinable - we can proceed to the calculation of reactions.

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

Having correctly determined support reactions in the beam freely supported , we can proceed to calculate the internal forces for each span .

The beam diagram and all calculations are generated in my Beam Calculator. You can use it to get a detailed solution for any statically determinate beam.

Span 1

For the first interval x should be from 0 to 2m. I have marked in blue marking layout shear and normal forces for the left-hand section.

Normal force N:

As for the force N1(x) in the first interval, it is -HA (the return of 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 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 get the result of the normal force for this x coordinate.

For the purpose of drawing internal force diagrams, we will calculate the characteristic points, that is, the beginning and end of the interval.

Shear force V:

The shear force T1(x) is VA (positive sign, return consistent with our marking of the shear force). 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 step in solving beams is determining bending moments. This is also the most difficult part of solving beam tasks.

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

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

We have the first span determined. Let's move on to the next one.

Span 2

You often ask if I should also include the forces from the first span or omit that? The answer is:

As for the equations describing the forces in each successive span are We take into account everything that happens from the beginning of the beam, That is, forces from each preceding interval are also taken into account.

In the second range x should be from 2m to 6m.

In blue I have marked the marking system of shear and normal forces for the left-hand section.

Normal force N:

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

Shear force V:

Shear force V2(x) to VA we arrive at F1 We have a constant value of the cutting force over the whole interval, after substituting Va we have V2= 2.7 [kN].

Bending Moment Mg:

To Va*x we get F1*(x-2) the sign of the moment from F1 is positive. X is minus 2m, which is how far the F1 force is from the beginning of our beam. The actual arm of the F1 force is (x-2). After substituting for x the beginning of the interval M2(2)= -10.6 [kNm] and the end of the interval M2(6) = 0.2 [kNm].

Span 3

In the last range x should be from 6 to 10m.

In blue I have marked the marking system of shear and normal forces for the left-hand section.

Normal force N:

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

Shear force V:

In the formula for the shear force V3(x), we get the continuous load q multiplied by the length over which it occurs, that is, the interval (x-6). After substituting the beginning of the interval for x, we have V3(6)= 2.7 [kN] and the end of the interval V3(10)= -5.3[kN].

Bending Moment Mg:

In the equation for the moment we have all that in interval 2. In addition, 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 for x the beginning of the interval M3(6)= 5.2 [Nm] and the end of the interval M3(10) = 0 [Nm].

And so we finished the determination of internal forces in our beam. Based on the results obtained, graphs are made as in Fig.12 below. But about that in the next entry.

Thank you, until the next entry.