## Fundamentals: Beam Analysis

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

That’s why beam analysis is so critical. Mastering thesetechniques allows us to adequately design beams to withstand stresses andforces while at the same time balancing costs, building codes and designrequirements. To appropriately analyse a beam, you’ll need to understandstructural mechanics, design principles and material properties.

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

### Shear Force & Bending moment

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

**The Beam**

To illustrate a complete beam analysis to determine its behaviourunder different loading patterns, we’ll be looking at straight beams withsimple rectangular cross sections.

**Loads**

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 some different ways beams can be subjected to external loading.

**Supports**

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 a support, thenthere is a corresponding reaction at that location. For example, pinnedsupports allow rotation, so there will be no reaction moment. But since pinnedsupports restrict displacement in both the x and y direction, there will be areaction in both of those directions.

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

**Forces **

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

Theinternal forces in a beam have two components: shear forces in the verticaldirection and normal force in the axis of the beam. If you apply a concentratedload 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 theforces at the top of the beam will be compressive. Simultaneously the bottomside 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 abeam runs continuously over supports, the beam deforms upwards; this is knownas **hogging action**. The bottom sideof 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 isstretching.

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. Oneis a shear force, which is the resultant force of all internal vertical forces,and one is 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 thebeam is supported. As a result, shearforce and bending moment vary in magnitude along the length of the beam. Might need to combine this into oneimage to work in webflow - Hopefully, Jigar can sort that out.

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

The data you need to design an RC beam properly is often found ina Shear Force Diagram (SFD) and Bending Moment Diagram (BMD). Plotting theseenables you to figure out the shear and bending moment profile along the lengthof the beam, and therefore the forces you need to design for at any point alongthe beam. We’ll go into detail on how to interpret and draw SFDs and BMDs next.

**Sign Convention**

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

Whenever you need to analyse a structural member by hand, you willneed to make a ‘cut’ somewhere along the member to examine the internal forcesat that location. Ensuring you use the appropriate sign convention means youcan 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.

### Shear Force and Bending Moment Diagrams

By drawing an SFD and a BMD, we can graphically identify themaximum and minimum design forces in our member, and account for them in ourdesign.

There are generally four steps to shear force and bending momentdiagrams:

**1. Draw a Free Body Diagram**

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

What an FBD typically looks like:

**2. Calculate any un-solved forces **

Applied loads are usually known;however, the reaction forces and moments at thesupport 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 withreaction forces. That is to say, if the beam isn’t moving, the forces acting onit are in equilibrium. In other words, the sum of theforces in the horizontal direction, the sum of the forces in the verticaldirection and the sum of the moments taken at any point should equal zero:

ΣF_x=0

ΣF_y=0

ΣF_M=0

If you can solve all equilibrium equations than yourbeam is statically determinate.

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

**3. Follow the forces or determine the internal shear and bending moment forces at every location.**

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

Remember, whenever you make an imaginary cut at anylocation 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 tomake the cut. You can therefore use the concept of equilibrium and the formulasgiven in step 2 to calculate the shear and bending moment forces along the beam.

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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 making the cut, 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 forShear Force Diagrams and Bending Moment Diagrams:

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

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

Where there are supports, applied external loads, andapplied 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, orapplied moment at that position.

More things to note about shear force and bending moment diagrams:

SFD’s and BMD’s give you a graphical perspective ofthe moment and the internal shear forces that a beamexperiences when subjected to certain loads. A cool relationship to remember isthat you can draw your BMD directly from your SFD or vice versa. The area undera shear force diagram between distances along the beam gives you the bendingmoment values, and the slope of distances inthe bending moment diagram gives you the shear force value.

In otherwords, the shear force is the first derivative of bending moment or bendingmoment is the integral of shear force Again, when itcomes to things like reinforcing a concrete beam, being able to look at abending moment diagram helps. Since a BMD will tell you which part of a beam issagging or hogging, it gives an engineer important information when decidingwhere exactly to put reinforcement.

These steps are best understood through an example.

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

### 1. Draw a Free-Body Diagram

The first step is to draw a free-bodydiagram of the system of the structural system shown above. The FBD mustinclude all externally applied forces and allunknown reactions and moments. The pinned support at point A restrictshorizontal and vertical translation so there will be reactions in thosedirections. The roller support restricts vertical translation only so therewill be a reaction in the y-direction. The FBD becomes:

### 2. Calculate any missing forces

From our FBD diagram, we can see 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:

ΣF_x=0

ΣH_A=0

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

The suzm 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:

ΣF_y=0

R_A+R_B-20-5=0

R_A+R_B=25

In order 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 that the forces try to rotate the beam relative to where you are taking moments from (in this case, support B). You can adopt 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:

ΣF_M=0

(-R_A×10)+(20×7)+(5×2)=0

R_A=15.

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

15+R_B=25

R_B=10

You can now complete the FBD as follows:

### 3. Follow forces and draw SFD and BMD

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

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

V_x=15 kN

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

〖-15 × x+M〗_x=0

M_x=15x

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

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

The SFD and BMD are then updated as follows. Note that if youchoose to draw your SFD by simply following the forces, an applied force ontothe beam is shown by a step down by the magnitude of the force for any pointforces 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 pointforces going up. For any UDLS, you will represent this as a linear slope wherethe magnitude of the distributed force is the slope of the line, bearing inmind your sign convention.

The same procedure is followed again, this time, you complete the wholelength of the beam

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

### Bending Stress

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

As previously mentioned, internal beam forces consist of shearforces 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 thetop (I.e., the underside of the beam is in tension), and the amount it isstretched changes as you go through the cross-section of the beam. There’s anarea in the middle of the beam’s cross-section that is neither stretched norsquared, and this is known as the neutral axis. This scenario is shown below:

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

Stress is given by the following formula:

σ=My/I

Where,

M = Internal bending moment

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

I = Second moment of Inertia (We’ve got an article all about this here!)

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

To find maximum bending stress, you first need to establish themaximum moment (M*) experienced. Go back to your bending moment diagram and seewhat your maximum bending moment is.

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

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

### Deflection

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 quickly approximate the deflected shape of the beam. 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 inflexion 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., the exact location 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 utilising the properties of the area under the bending moment diagram. In other words, the curvature of the beam is proportional to the associated bending moment.

Constructing SFDs and BMDs are an important first step to beam analysis. Through equilibrium equations, you can determine the magnitude and the nature of internal forces 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 second moment of area (I), determines a section’s resistance to bending and deflection and has many uses in other engineering applications (calculating

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

Knowing where to reinforce a beam and how much reinforcement isrequired to resist bending moments and stresses induced ensures that structuralmembers will not crack or fail under loading. This knowledge is acquired aftermastering BMD’s. Whether a beam requires stirrups and how far apart they shouldbe spaced along a beam to resist shear stresses can only be done if you knowabout the internal forces in a beam and how to visualise them through an SFD.This process can be done manually, as we have demonstrated here, but theprocess is time-consuming and gets more complicated when you have to analysemembers in a complex structure. This is often done with the help of software.Nevertheless, understanding the basic techniques is essential knowledge, and itwill allow you to better understand structural mechanics, material properties,design principles and failure mechanisms in structural design.

I think this point should goelsewhere. Perhaps in the section above?

I think it's good here. Thesections above and the example all referred torectangular beams, whereas this part is more of a segway intoreal-life examples and scenarios, so I thought it would be a good way tointroduce an I beam. Let me know if you agree