Internal forces in beams – calculation example

Let’s return to the topic of “Internal forces in a beam”.

Understanding the distribution of internal forces in a beam is essential for assessing its strength. In this article, I will present an example calculation of internal forces, such as axial, shear, and bending moments. I will focus on the effects of different types of loads, so you can understand how internal force diagrams are drawn in practice. We will analyze both concentrated loads and moments as well as distributed (continuous) loads.

Our agenda:

• Internal forces analysis – simply supported beam
• Calculating internal forces
• Equation method
  • Calculations for the first section
  • Calculations for the second section
  • Calculations for the third section
  • Calculations for the third section from the right side
• Internal force diagram
• Graphical method
• Determining internal forces – summary

This example and exercises with EquiBeam will help you accurately calculate the values of internal forces and draw their diagrams.

Internal forces analysis – simply supported beam

In this article, we will analyze internal forces in a beam with a static scheme shown in the diagram:

If you’ve read the posts related to reaction calculation examples, you’ve likely already seen this beam and know how to conduct static and kinematic analysis and determine the support reactions. If you haven’t read those, I encourage you to check it out – the article can be found here.

Below you will see the scheme after releasing the supports (with determined reactions) and the previously calculated reaction values.

H_B=-qL,\hspace{12px}V_D=2.5qL,\hspace{12px}V_B=2.5qL

Calculating internal forces

Once we know all the loads and reaction values, we can fully determine the values of internal forces and draw their diagrams.

There are three methods for this:

• the equation method – uses the static scheme to write equations that describe the change in internal forces
• the second is based on general “patterns” of these changes, allowing diagrams to be drawn and values calculated without creating equations and substituting values,
• the superposition method – allows for quickly drawing diagrams for each load separately and then combining the resulting diagrams (I won’t discuss this last method in this article).

Equation method

In this method, we must first identify the areas of the beam where the internal force equations will not change form. Any such change occurs due to the application of a new load, reaction, or, for example, the end of a continuous load. The areas where the equations don’t change are called sections.

In our beam, we have 3 sections:

• the first section starts at point A and ends with the reaction applied at point B. The length of this section is L,
• the second section begins at point B and ends where the continuous load q ends and force P₁ is applied at point C. The length of this section is 2L,
• the last section starts at point C and ends at the beam’s end. The length of the last section is L.

As you may have noticed, in EquiBeam, the basic unit for creating a beam is sections where you can input length and section and material data.

Calculations for the first section

Let’s analyze the first section – we will describe the points in this section using the position variable x. In this case, due to the section length, x ∈ [0, L].

We write the internal force equations according to the principles outlined in theoretical articles on internal forces. Thus, for section 1, they will be in the form:

N_1=0,\hspace{10px}T_1 (x)=-qx,\hspace{10px} M_{g1} (x)=K_1-\frac {qx^2} 2

Normal forces are equal to 0 for an obvious reason – there are no horizontal forces at the beginning of the beam. Here’s a primary note: we only consider the forces and loads before the location of the analyzed point. The shear force T₁ changes with the length of the continuous load according to x. The bending moment from the load is the product of force qx and the moment arm with respect to the point at location x. The moment arm is x/2, giving qx²/2.

Based on the above equations, internal force values can be calculated by substituting location x, e.g., for the section ends:

N_1=0, \hspace{12px} T_1 (0)=0,   \hspace{12px}T_1 (L)=-qL,  \hspace{12px} M_{g1} (0)=2qL^2, \hspace{12px}  M_{g1} (L)=1.5qL^2

Calculations for the second section

The points in this section will still be described by the position variable x. If we interpret x from the start of the beam (though there are methods that differ for ease of calculating certain integrals), the location change will be described as x∈ [L3L].

For the second section, the internal force equations will be:

N_2=-H_B,\hspace{10px}T_2 (x)=-qx+V_B,\hspace{10px}M_{g2} (x)=K_1-\frac {qx^2} 2+V_B (x-L)

Normal force N₂ depends on reaction H_B. The shear force T₂ continues to change with the length of the continuous load (since it hasn’t ended), but concentrated force V_B is added.

The bending moment also includes reaction V_B, but it’s important to consider the moment arm of this force from the point at location x. The arm is not equal to x because the reaction’s point of application is a distance L from the beam’s start – thus, the arm is x-L.

Based on these equations, internal force values can again be calculated by substituting x, e.g., for the section ends:

N_2=qL,  \hspace{10px} T_2 (L)=1.5qL,   \hspace{10px}T_2 (3L)=-0.5qL,   \hspace{10px}M_{g2} (L)=1.5qL^2,  \hspace{10px} M_{g2} (3L)=2.5qL^2  

Now we should note an important point – shear forces at the section ends have opposite signs. This means that somewhere the shear force passes through zero, leading to a bending moment extremum. This extremum should be calculated as it could be the most loaded and potentially most dangerous point of the beam from a strength perspective.

To calculate the extremum, it is enough to find the zero of the shear force function:

T_2 (x)=-qx+V_B=0→x_{ekstr}=\frac{V_B} q=2.5L→M_{g2} (x_{ekstr} )=2.625 qL^2

As we can see, the extremum of the bending moment is indeed greater than the values obtained at the section ends – we were right ????. Let’s move on to the third section.

Calculations for the third section

In the third section, x ∈ [3L, 4L]. The internal force equations will be:

N_3=-H_B,T_3=-q⋅3L+V_B-P_1,

M_{g3} (x)=K_1-q⋅3L⋅(x-1.5L)+V_B (x-L)-P_1 (x-3L)

Normal force N₃ still depends only on reaction H_B since no other horizontal force has appeared.

Attention should be paid to what happened on the side of shear forces and bending moments. Shear force T₃ no longer depends on variable x! This is naturally due to the end of the continuous load, which now acts as a concentrated force (with a value of q⋅3L) at its center of gravity.

Additionally, negative force P₁ appears in the equation. In the bending moment, we also see interesting changes – with the end of the continuous load, the moment arm description must change.

The equivalent force of the continuous load is located at a distance of 1.5L, so the moment arm is x-1.5L. Based on the above equations, internal force values can again be calculated by substituting x, e.g., for the section ends:

N_3=qL,   \hspace{10px}T_3=-2.5qL,  \hspace{10px} M_{g3} (3L)=2.5qL^2,  \hspace{10px} M_{g3} (4L)=0

Calculations for the third section from the right side

As you’ve noticed, with each section, the internal force equations become more complex, requiring more mathematical operations to calculate the values.

This is, of course, a significant downside to this method, which can be quickly mitigated by analyzing the beam from the other side. Remember that when analyzing the system from the right side, the internal force sign conventions change.

For the right side, the equations will be:

N_3=P_2,\hspace{10px}T_3=-V_D,\hspace{10px}M_{g3} (x_p )=V_D⋅x_p

At first glance, these are much simpler, allowing for faster value calculations. However, note that the description and interpretation of the position variable X_p also change. In this case, X_p belongs to the section [0, L], where 0 is the right end of the beam. Substituting the data:

N_3=qL,\hspace{10px}T_3=-2.5qL,\hspace{10px}M_{g3} (0)=0,\hspace{10px}M_{g3} (L)=2.5 qL^2, 

Naturally, regardless of the method chosen, the result must be the same. This is also an interesting way to verify whether the previously obtained values are correct.

Internal force diagram

In this way, we obtained a set of functions and values describing the internal forces in the analyzed beam. Based on them, we can easily draw the internal force diagrams, as we have the complete data set for it.

The values at the section edges are recorded directly from previous calculations.

If the function is constant (as with the entire normal force diagram) or changes linearly (like almost the entire shear force diagram), we simply connect the section end values with a line segment.

If the diagram is described by a quadratic function (the first two sections on the bending moment diagrams), it is easiest to draw the diagram by using the basic relationship between shear force and bending moment: the shear force is the derivative of the bending moment. Thus, if the shear force is positive, the bending moment increases; if it is negative, it decreases. When the shear force passes through zero, we have an extremum in the bending moment. The greater the shear force, the faster the moment diagram rises, and so on.

Graphical method

At the end of the article, I’ll briefly present how the values of internal forces can be calculated using the graphical method, which, to be honest, is much faster and can allow internal force diagrams to be drawn in less than a minute (sometimes required of students in early mechanics classes… seriously!).

Like before, the method requires analyzing the load changes on the beam and understanding how the diagrams behave under certain loads (since we don’t have equations to determine the function curves). The diagram is drawn at each stage of the calculation. Below is the “analysis” of changes for our beam

Normal forces:

• we start (always) from a value of 0,
• at point A, there are no horizontal concentrated forces, so we stay at 0 until the next horizontal load,
• the concentrated force H_B causes a jump on the diagram by its value (note the sign – the force compresses the second section, so it should be negative – but it itself is negative).
• we continue with the value qL until the next load P₂,
• the concentrated force causes a jump by its value (compressing the nonexistent fourth section), so theoretically, we return to 0 (though this isn’t shown on the diagram).

Shear forces:

• we start (always) from a value of 0
• at point A, there are no vertical concentrated forces (so no jump), but there is a continuous load, which causes a decrease of qL per unit distance L. At the section’s end, we thus have a value of –qL,
• At point B, there is force V_B, which causes a jump of 2.5qL, giving a value of 1.5qL,
• the continuous load continues to act, so over the next distance 2L, we decrease to -0.5qL,
• we automatically detect the crossing through zero – its location is obtained by dividing the initial section value (1.5qL) by the load q.
• The extremum is 1.5L from point B.
• At point C, we have a jump of -2qL (force P₁).
• The continuous load no longer acts, so we remain at -2.5qL until the beam’s end, where we again have a jump of 2.5qL (V_D), theoretically returning to 0 (though again, this isn’t shown on the diagram).

Bending moment diagram – we create this by analyzing concentrated moments (jumps) and calculating areas under the shear force diagram (since the bending moment is the integral of the shear force, and the area under the diagram is a graphical interpretation of the integral).

• we start at 0 (ALWAYS) and immediately have a jump at point A of 2qL² caused by concentrated moment K,
• to proceed to point B, we calculate the triangle area (note – negative). The change from the area is -0.5qL², reaching 1.5qL². The diagram shape is determined by the shear force trend,
• no concentrated moments appear in the rest of the beam – everything is calculated from areas under the shear force diagram,
• in the second section, we have two triangles – the first has a base of 1.5L, the second 0.5L – as calculated in the shear force stage,
• the first triangle area gives a change of 1.125qL², reaching an extremum of 2.625qL²,
• the second triangle changes by -0.125qL², ending the section at 2.5qL²,
• the last section is a change by the rectangle area – the area is -2.5qL², finishing at a value of 0!

As you can see, simply checking whether the diagram “zeros out correctly” allows for a quick check of its accuracy ????.

Determining internal forces – summary

In the article, we analyzed how to determine the values of internal forces such as normal, shear, and bending moments, which are crucial for assessing the strength of beams.

Understanding how these forces act and creating accurate internal force diagrams provides a complete picture of how a beam responds to various types of loads.

Thanks to the presented calculations and examples available in the EquiBeam app, you now have the tools to effectively analyze the distribution of internal forces in beams and prepare for more complex strength topics.