In this post you will learn:

What are internal force diagrams
What internal forces do we draw on the graph
Internal force diagram - a simply supported beam
Internal force diagram - restrained beam

Internal force diagrams is a graphical representation of the cross-sectional forces arising in our system (for example, in a beam, truss or frame). These forces arise under the influence of an applied load.

Internal force diagrams - types

We distinguish the following internal force diagrams:

normal force diagram - N
shear force diagram - V
diagram of bending moments - M

The rules for drawing internal force diagrams for all structures are the same. It doesn't matter if it's a beam, frame or truss.

Simply supported beam

To be able to draw a force diagram, we must first calculate these forces. With an understanding of the method of determining internal forces in beams, this post will help: Internal forces in beams.

We will draw force diagrams for the beam in the figure below

As you remember we start by checking static determinability and determination of reactions for the beam.

N=R-J-3
where:
N - degree of static indeterminacy
R - the number of support reactions. That is, the sum of all reactions for our supports
J - number of internal joints - if not present P=0
3 - the number of equilibrium equations. In static systems it is 3

When we have the support reactions counted, we can move on to determining the internal forces In compartments. We will have 3 compartments in our example beam. For each compartment, we write an equation for the normal force N(x), shear force T(x) and bending moment M(x). The values of each force are a function of x, which means that we can substitute any value of x from our interval and we will get the result. To draw the graphs, we only need the beginning and end of the interval. In this way we will calculate the so-called. landmarks.

Below you will find the written equations and the calculated beginnings and ends of the compartments for our beam.

The beam diagram, all calculations and internal force diagrams are generated in my Beam calculator. You can create beam diagrams online and get a detailed solution for each statically determinate beam.

Span 1

Having calculated forces for the first interval, let's attempt to draw graphs. Below in Figure 5 I have plotted the characteristic points for N, T and M in the first interval. We then connect these points with lines.

If the function N(x), T(x) or M(x) is:
1. a constant value independent of x is the graph is horizontal line
2. a linear function from x is graph is straight line tilted at an angle
3. a quadratic function from x is the graph is parabola

In compartment 1 we have local maximum bending moment. This is the highest point of the parabola. How to determine it? Very simply. This is the place where the value of the shear force is zero.

After determining the xmax coordinate, we calculate the bending moment value for this coordinate M1(xmax) = 34.613 [kNm].

Span 2

In compartment two we do exactly the same as in the first compartment. We apply points and connect them with lines. We get the following parts of the graphs, which I have shown in Fig.

Note that the bending moment function in the second interval is no longer a quadratic function, so the graph is linear.

Span 3

In the third last compartment, we have the following values of internal forces:

Thus, we have drawn complete diagrams of the cross-sectional forces normal force N(x), shear force T(x) and bending moment M(x).

Sometimes one encounters bending moment diagrams with inverted. That is, positive values are below and negative values are above the horizontal axis.

Below I have included charts obtained from beam calculator. As you can see, all the values are identical. This is the kind of graphs my beam calculation application generates.

Cantilever beam (restrained)

Below is an example that we will address.

As in a simply supported beam, we need to calculate the internal forces to draw the diagram.

we start by checking static determinability and determination of reactions for the beam.

When we have the support reactions counted, we can move on to determining the internal forces In compartments. We will have 2 compartments in our example restrained beam. For each compartment, we write an equation for the shear force V(x) and bending moment M(x). This time we skip calculating and drawing the graph of normal forces N(x).

In some beams, there is no need to calculate normal forces, because the value is zero for the entire beam. These are beams in which there is no horizontal component of the load.

Below you will find the written equations and the calculated beginnings and ends of the compartments for our beam.

Span 1

Having calculated forces for the first interval, we draw graphs.

Span 2

In compartment two we do exactly the same as in the first compartment. We apply points and connect them with lines. We get the following parts of the graphs, which I have shown in Fig.