Beams are all around us, and they’re used in many structural and mechanical engineering applications. Not only that but all structures can be approximated as beams or a collection of beam-like members. Beams are also integral for stability, whether used as internal supports to a building’s roof, floor, or walls, or in moment-resisting frames or to transfer loads to a building’s foundations.

That’s why beam analysis is so critical. Mastering these techniques allows us to adequately design beams to withstand stresses and forces while simultaneously balancing costs, building codes, and design requirements. To appropriately analyze a beam, you’ll need to understand structural mechanics, design principles, and material properties.

In this article, we’ll go through all the significant components of beam analysis in detail. You’ll gain insight into fundamental concepts that underlie the beam design decisions engineers make daily.

Before we explain what exactly shear force and bending moments are and how to find them, let's lay down some ground rules first.

To illustrate a complete beam analysis to determine its behavior under different loading patterns, we’ll be looking at straight beams with simple rectangular cross-sections.

Beams can be loaded in several ways. A beam can be loaded typically through:

- A concentrated force (also known as a point load)
- A uniformly distributed force
- An applied moment, or
- A combination of these.

The free-body diagrams below show different ways beams can be subjected to external loading.

For a beam to remain in equilibrium (i.e., for the beam to remain standing), they need supports to constrain them when they are subjected to external loads. In theory, supports can be pinned, rollers (releasing the beam axially), or fixed. At any point of support, there are a maximum of 6 degrees of freedom, which essentially describe the number of possible movements (i.e., translations, displacements, and rotations) that can occur at that specific node.

For a 2D beam, the directions these movements can happen in are the axial direction (x-axis), the transverse direction (y-axis), and rotation (z-axis). Check out the local coordinate system for a beam below.

If a certain degree of freedom is restrained at supports, then there is a corresponding reaction at that location. For example, pinned supports allow rotation, so there will be no reaction moment. But since pinned supports restrict displacement in both the x and y direction, there will be a reaction in both of those directions.

A summary of all support and their boundary conditions is shown below.

When a beam is loaded, internal forces develop within it to resist the imposed load, thereby maintaining equilibrium. Let’s say I gave you a wooden ruler and told you to break it in half. I can say with almost absolute certainty that you would not try to grab onto it to pull it apart or squash it together. Those are examples of applying axial forces. Most likely, you’ll rest the ruler on the edge of a surface or a table and snap it in half, or, more likely than that; you would bend the ruler until it breaks. In those two examples, you’ve essentially applied a shear force (snapping the ruler over a surface) and a bending moment (bending the ruler until it breaks). Shear force and bending moment are vital because they affect beams' structural integrity.

The internal forces in a beam have two components: shear forces in the vertical direction and normal force in the axis of the beam. If you apply a concentrated load on the top side of a beam, the beam will be curving downwards, also known as sagging action.

When a beam is sagging, the top of the beam will shorten, and the forces at the top of the beam will be compressive. Simultaneously the bottom side of the beam will stretch, and the fibres in this part will be in tension. When the load is applied from the underside of the beam or when a beam runs continuously over supports, the beam deforms upwards, known as **hogging action**. The bottom side of the beam is now shortening, the normal forces in the bottom are compressive, and the forces on the top of the beam are in tension because the top half is stretching.

Each normal tensile force has a corresponding compressive force equal in magnitude but opposite in direction; therefore, a net normal force is not produced. Instead, these internal forces produce a moment.

[1] Consequently, we can represent all internal forces in a beam with two resultant forces. One is a shear force, which is the resultant force of all internal vertical forces, and one is a bending moment which is the resultant of the normal internal forces. The resultant forces will depend on the loads acting on the beam and how the beam is supported. As a result, shear force and bending moment vary in magnitude along the length of the beam.

A method to calculate these forces and locate them analytically and graphically is essential. Why? Well, think about some of the materials you may use in design. We know that concrete performs well under compression but is quite terrible in tension, so often, we need to reinforce it with steel. But how do we know where exactly to reinforce it? Well, that’s where this knowledge and ability comes in.

The data you need to design an RC beam properly is often found in a Shear Force Diagram (SFD) and Bending Moment Diagram (BMD). Plotting these enables you to figure out the shear and bending moment profile along the length of the beam, and, therefore the forces you need to design for at any point along the beam. We’ll detail how to interpret and draw SFDs and BMDs next.

Before we outline the steps to making killer Shear Force and Bending Moment diagrams, we’ll outline a sign convention. While it seems nefarious, missing this step can trip you up later.

Whenever you need to analyze a structural member by hand, you will need to make a ‘cut’ somewhere along the member to examine the internal forces at that location. Ensuring you use the appropriate sign convention means you can obtain accurate results every time. A common sign convention is as follows:

- Applied forces pointing downwards are negative (-)
- Reaction forces pointing upwards are positive (+)
- After making an imaginary ‘cut’, shear forces pointing downwards on the left side of the cut are positive because they will try and make the beam concave upwards.
- On the right side of the cut, shear forces pointing upwards are positive because not only would it make the beam concave down, but it has to have an equal and opposite direction to the left side of the cut in order to keep the beam in equilibrium.
- If the bending moment causes the beam to sag, then it is positive; if it causes the beam to hog, then the bending moment is negative.

By drawing an SFD and a BMD, we can graphically identify the maximum and minimum design forces in our members and account for them in our design.

There are generally four steps to shear force and bending moment diagrams:

A free-body diagram (FBD) is a simplified sketch of the structural system. In simplified 1D and 2D elements, it shows all applied loads and reaction forces acting on a member, and it’s critical in determining shear force and bending moment.

What an FBD typically looks like:

Applied loads are usually known; however, the reaction forces and moments at the support are often unknown. Before drawing SFDsand BMDs, you must establish these reaction forces.

Supposing you start with a beam in static equilibrium, calculating the reaction forces is done by balancing out applied forces with reaction forces. That is to say, if the beam isn’t moving, the forces acting on it are in equilibrium. In other words, the sum of the forces in the horizontal direction, the sum of the forces in the vertical direction, and the sum of the moments taken at any point should equal zero:

If you can solve all equilibrium equations then your beam is statically determinate.

**Remember**: if you have uniformly distributed loads (UDL), you need to convert them into equivalent point loads. This is done by multiplying the value of the UDL by the loading length. The equivalent point load will then act at the mid-point of the UDL’s loading length and position.

After determining all unknown forces, you can draw shear force and bending moment diagrams.

Remember, whenever you make an imaginary cut at any location in the beam, the internal forces must be in equilibrium with all external forces. This includes any reactions or applied loads present just before where you choose to cut. You can therefore use the concept of equilibrium and the formulas given in step 2 to calculate the shear and bending moment forces along the beam.

The tendency is to start from the left side of the beam and move along, but you can make a cut and start from anywhere you like. After cutting, solve for internal shear (Vx) and bending moment forces (Mx)using equilibrium equations and bearing in mind the sign convention.

Continue doing this along the beam, mindful of any additional external forces as you go along, and this will give you the SFD and BMD.

Underneath your FBD, have the following templates for Shear Force Diagrams and Bending Moment Diagrams:

These axes should line up below your free body diagram with the x-axis representing the location along the beam, and the y-axis representing the magnitude of force at that location.

As you track the forces along the beam, add these values to your SFD and BMD as points. The sign convention we’ve chosen is that positive forces are above the line and negative forces are below the line.

Where there are supports, applied external loads, and applied moments, these are known as discontinuities in the diagram. In your SFDor BMD, the values of the discontinuities (vertical movement in the diagram) should equal the value of the reactions, point loads, or applied moment at that position.

- Point loads cause a vertical jump in the shear diagram, in the same direction as the26/12/2022 sign of the point load (i.e., (+) reaction forces cause an upward jump on an SFD and (-) point loads cause a downward jump on an SFD)
- Uniformly distributed loads (UDLs) are shown graphically on an SFD as a straight, sloped line with a gradient equal to the value of the distributed load. This is because when solving for a shearing force under a UDL, the equation will be that of a straight line,
- Where there is no applied load, the SFD is constant and horizontal until the next applied force is reached.

- At points in the beam where there are no applied loads, the moment diagram is a straight, sloped line, where the gradient is equal to the value of the shear force

- UDLs are shown graphically on a BMD as a parabolic curve because when solving for bending moment, the equation is that of a parabola.

- Maximum/minimum values or peaks in a BMD occur at points where the shear force is zero. View your finished BMD to find the absolute minimum or maximum bending moment values.

SFDs and BMDs give you a graphical perspective of the moment and the internal shear forces that a beam experiences when subjected to certain loads. A cool relationship to remember is that you can draw your BMD directly from your SFD or vice versa. The area under a shear force diagram between distances along the beam gives you the bending moment values, and the slope of distances in the bending moment diagram gives you the shear force value.

In other words, the shear force is the first derivative of the bending moment or the bending moment is the integral of the shear force. Again, when it comes to things like reinforcing a concrete beam, being able to look at a bending moment diagram helps. Since a BMD will tell you which part of a beam is sagging or hogging, it gives an engineer important information when deciding where exactly to put reinforcement.

These steps are best understood through an example.

Let’s take a simply supported beam carrying two loads:

The first step is to draw a free-body diagram of the system of the structural system shown above. The FBD must include all externally applied forces and all unknown reactions and moments. The pinned support at point A restricts horizontal and vertical translation, so there will be reactions in those directions. The roller support restricts vertical translation only so there will be a reaction in the y-direction. The FBD becomes:

Our FBD diagram shows that the forces we need to solve for are HA, RA, and RB. We can use equilibrium equations to solve for these reaction forces.

The sum of forces in the horizontal ‘x’ direction has to equal zero, so:

Since HA is the only force in the horizontal direction, it has to equal zero.

The sum of all the forces in the vertical ‘y’ direction has to equal zero. Remember the sign convention that all externally applied forces pointing downwards are negative and all applied forces pointing up are positive:

To solve the above, we can take moments at any point of the beam which has to sum up to zero. Let’s take moments at Support B. Remember that moment is the product of a force and the distance from the point of the force. Also, take note of the direction in which the forces try to rotate the beam relative to where you are taking moments from (in this case, support B). You can adapt the sign convention that if a force will try to rotate the beam clockwise, then the clockwise moment will be negative; conversely, if a force tries to rotate the beam anti-clockwise, then the anti-clockwise moment is positive. In this case, RA is trying to rotate the beam in a clockwise direction, so it is negative, and the applied forces will try to push the beam in an anti-clockwise direction, so they cause a positive moment. It does not matter which sign convention you adopt, as long as you are consistent:

Now, plug the value for RA into the second equilibrium equation:

You can now complete the FBD as follows:

Now the BMD and SFD can be drawn. Starting from the left side of the Beam, from Support A, make a cut ‘x’ metres to the right of the 15 kN reaction force at Point A but before the applied force of 20 kN:

To maintain equilibrium, the shear force (Vx) has to balance the 15 kN reaction force, so:

The internal moment (Mx) must also balance the moment that is generated by the 15 kN force:

The shear force will remain constant as you go along the beam at 15kN until the next applied force is reached. You can start drawing the SFD. The BMD is shown graphically as a straight line:

Then repeat the process, and keep moving your imaginary cut along the beam. This time your cut will be immediately after the 20kN point load. The Free Body Diagram of this is shown below. The same equilibrium equations shown in Step 2 are used, and the corresponding shear force and moment formula is shown.

The SFD and BMD are then updated as follows. Note that if you choose to draw your SFD by simply following the forces, an applied force onto the beam is shown by a step down by the magnitude of the force for any point forces going down in the SFD. For forces going up (typically reactions forces), the line jumps upwards in your SFD by the magnitude of the force for any point forces going up. For any UDLS, you will represent this as a linear slope where the magnitude of the distributed force is the slope of the line, bearing in mind your sign convention.

The same procedure is followed again. This time, you complete the whole length of the beam.

The simplest way to check whether you’ve done your SFDs and BMDs correctly is that you should always wind up coming back to zero on your diagrams once you’ve reached the end of your beam. If you haven’t, go back and check your work.

Beam Flexure is often the governing factor in beam design, so it is crucial to understand bending stress and how to calculate it.

As previously mentioned, internal beam forces consist of shear forces in the vertical direction and normal forces along the axis of the beam. Say you bend a beam, the bottom part of the beam is stretched compared to the top (I.e., the underside of the beam is in tension), and the amount it is stretched changes as you go through the cross-section of the beam. There’s an area in the middle of the beam’s cross-section that is neither stretched nor squared, and this is known as the neutral axis. This scenario is shown below:

We’ve learned that shear force and Bending moment change as you go along the beam, and so do bending stresses. Combining these gives us the information we need to determine where and how much the maximum stress will be.

The following formula gives stress:

Where,

M = Internal bending moment

y = Perpendicular distance from the Neutral Axis to the point of interest in the beam

It’s clear to see the larger the value for ‘y’ (i.e., the further the distance away from the N.A), the bigger the stress generated for a given cross-section. This aligns with the fact that bending stresses are the biggest on the outer edges of a beam (this is why I-beams, which have thick flanges at maximum distance from the N.A., are so efficient!).

You first need to establish the maximum moment (M*) experienced to find maximum bending stress. Go back to your bending moment diagram and see what your maximum bending moment is.

Then, find your N.A. Usually, RC beams will have regular cross-sections like squares or rectangles, in which case the N.A. is central to the beam. However, steel members are typically more complex shapes, like i-beams, square hollow sections, and T-beams. For these, a few extra steps are required to find the N.A.

Once the N.A. is established, finding the max bending stress is simply a case of plugging in the maximum bending moment and the furthest distance (y) from the N.A. into the formula above.

When addressing a beam for serviceability conditions, you must understand how it will deflect under various loading conditions. While there are specific formulas and methods to calculate deflection. Your bending moment diagram allows you to approximate the deflected shape of the beam quickly. This can tell you some quick but essential information regarding where to limit the beam’s deflection to ensure the safety and functionality of your design.

To approximate and sketch the deflected shape of a beam from the bending moment diagram, here are a **few general rules:**

● Beam supports act as clamps for deflection

● If the beam experiences a positive moment over a certain distance, it will concave upwards.

● If the beam experiences a negative moment over a certain distance, it will concave downwards

● Points of inflection are located and points where the moment is equal to zero. Graphically speaking, a point of inflection is where the curvature changes its sign, i.e., where the beam goes from hogging to sagging or vice versa.

In cases where it is **assumed that deformation** is within the elastic range and that displacements are small, the quickest way to compute deflection at a specific location is by utilizing the area's properties under the bending moment diagram. In other words, the curvature of the beam is proportional to the associated bending moment.

Constructing SFDs and BMDs is an important first step to beam analysis. You can determine the magnitude and nature of internal forces through equilibrium equations and use them to determine the stresses, curvature, and deflection at any point in a beam.

So far, we’ve mostly discussed straight, simply-supported beams with rectangular cross-sections. But in reality, beams come in various shapes and sizes and are made from diverse materials. Also, beam loading is rarely consistent, and support arrangements can be highly varied. From cantilevered to fixed supports or simply supported with overhangs and continuous beam runs, real-world supports never perfectly match theory. So it’s important to understand that the process described in this article, and most analysis software are approximations of reality.

This makes the application of beam analysis in the field more complex. But when you reduce these more complicated problems down to their basic principles, you will still follow the same general steps outlined in this article.

Some beam cross-sections are more efficient than others regarding resisting bending and contending with biaxial moments and flexural and shear forces. Commonly, you’ll see beams with an I-shaped cross-section (shown below) used in construction. Cross-sectional properties, like the second moment of area (I), determine a section’s resistance to bending and deflection and has many uses in other engineering applications.

As it turns out, an I-beam is one of the most efficient beam cross-sections because it is stiffer and has a larger second moment of area, so it is more resistant to bending and deflection than a simpler beam cross-section like a rectangle or square. We’ve written an article explaining this theory in detail - read it here!

Knowing where to reinforce a beam and how much reinforcement is required to resist bending moments and stresses induced ensures that structural members will not crack or fail under loading. This knowledge is acquired after mastering BMDs. Whether a beam requires stirrups and how far apart they should be spaced along a beam to resist shear stresses can only be done if you know about the internal forces in a beam and how to visualize them through an SFD.

This process can be done manually, as we have demonstrated here, but the process is time-consuming and gets more complicated when you have to analyze members in a complex structure. This is often done with the help of software. Nevertheless, understanding the basic techniques is essential knowledge, and it will allow you to understand better structural mechanics, material properties, design principles, and failure mechanisms in structural design.