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Beam Analysis and Design Fundamentals

Fundamentals: Second Moment of Inertia

Fundamentals: Newton’s 3 Laws of Motion

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Fundamentals of Engineering 

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 structural 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.

straight beams with simple rectangular cross-sections

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

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.

concentrated load on the top side of a beam

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 

. 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:

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.

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.

: 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.

‍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 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 

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.

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:

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:

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Whenever you see or want to describe the effect of forces on basically any object, 

 likely be using Sir Isaac Newton's Laws of Motion to do it. These laws are the cornerstone of modern physics. They substantiate everything from our knowledge of planetary motion to our understanding of how planes take off and fly. This article will describe Newton's Laws of Motion and how they apply to everyday life.

An object at rest remains at rest, and an object in motion remains in motion at a constant speed and in a straight line unless acted on by an unbalanced force

: Things like to stay where they are; Newton's first law of motion describes an object's tendency to keep doing what it's doing.

 states that all objects will remain at rest or in uniform motion unless an external force changes its state. Changing how something moves requires acceleration, and you need a net force.

An object's tendency to resist change in a state of motion or stationery is known as 

. How is inertia measured? Well, you need to know about mass. Say you throw two balls; one is a bowling ball, and the other is an inflatable ball. Intuition tells you which is harder to move or throw and harder to stop. The bowling ball has more mass; therefore, it has more inertia…more stuff to keep it doing what it's doing before you stop it.  

Newton's first law also applies to bodies at rest. Say you have an object lying on a table. The force of the object downwards onto the table due to its weight and the equivalent upward reaction exerted by the table means that all external forces cancel each other out, and the object remains at rest. In other words, there is no net force acting on the object. 

The same situation would happen if you applied equal but opposing horizontal forces on the object (provided it was on a smooth surface). Say you put two hands on an object from opposite sides and apply the same amount of pressure on each side, the external forces (from your hands) would cancel each other out, and the object will remain stationary. The object will move when the forces are unbalanced (i.e., you remove one hand). Then do the same thing on a rough surface; it might be harder to slide the object across that surface because you've added another force into the mix this time. It's friction.

The concept of inertia and forces connects to Newton's second law.

Newton's second law of motion states that the acceleration of an object depends on the object's mass and the amount of force applied. 

 To overcome inertia, you must apply an unbalanced or net force to an object. A 

 is all the forces left over after adding together all the forces that might otherwise cancel each other out. When forces get unbalanced, that's when exciting stuff happens!

The net force is found using the equation:

You've probably heard 'force equals mass times acceleration' over and over, and for a good reason. It's likely the most used equation in physics. This equation tells us that an object subjected to a net external force will accelerate. The amount that an object will accelerate is directly proportional to the magnitude of the force. Acceleration is also inversely proportional to the mass of the object. This goes back to our example of two balls, one light, and one heavy. If the same force is applied to both, the heavier ball will experience less acceleration than the lighter ball because it has a higher mass.

 the F=ma relation is only suitable for objects with a constant mass!

Essentially, the formula defines a force to be a change in momentum, where momentum represents the mass of an object (m) of an object times its velocity (V). But how has this simple law been derived?

Imagine a car with mass (m) with a starting velocity of (V0) starting at point X0. The car speeds up and reaches another point (X1) at a velocity of (V1). Given that the mass of the car doesn't change and that force is equal to a change in momentum with a change of time, the formula becomes:

mass (m) with a starting velocity of (V0)

What about net force? Say you've slid a ball onto a smooth surface; your action of pushing is a force that isn't being cancelled out by other forces as the ball slides forward. The ball is experiencing acceleration. All the external forces have balanced out when the ball slows down and stops moving.

The gravitational force is the most common net force that makes an object move. If you throw a ball up in the air, the force of gravity pulls down on the ball once the ball loses upward momentum, and it will accelerate downwards. But gravity isn't the only force acting on an object; Newton's third law explains how to calculate net forces.

Newton's third law of motion states that whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first.

 For every action (force), there is an equal but opposite reaction.

 exerts a force on object Y, Y will also exert the same but opposite force on X. If this symmetry did not exist, a chair would collapse every time you sit on it. Your body exerts a force downward onto a chair, and the chair needs to exert an equal force upwards to remain standing.

 A normal force is a contact force that is perpendicular to the surface that an object is on. The normal force is always perpendicular to that object, whether you rest an object on a table or an angled ramp.

A normal force is special because it changes its magnitude. Say you have cling wrap over a deep dish and place one strawberry on it. The strawberry exerts a little force onto the cling wrap; a normal force from the cling wrap pushes right back with equal force. Add another strawberry, and the force due to gravity doubles, and so does the normal force. This will continue until you add enough strawberries that break through the cling wrap, i.e. when the normal force cannot match the force pushing against it.

The term 'for every action there is an equal but opposite reaction,' yet this kind of symmetry needs to explain why we can move things like picking up a cup of coffee. Well, things move because there's more going on other than just the action and reaction forces. There's motion as soon as forces become unbalanced (refer to Law #2).

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Have you ever wondered why many structural elements like beams aren't commonly manufactured as fully-filled simple shapes like circles or squares?  Well, it's all about how efficiently a cross-section resists bending...and that's got a lot to do with a shapes Second Moment of Inertia (I).  

A column or a beam must typically contend with biaxial moments and flexural and shear forces. A member requires a minimum stiffness or, analogously, a minimum cross-sectional moment of inertia to resist these forces and avoid buckling or deflecting.  

So what exactly is the Second Moment of Inertia (I), what determines it, and how can engineers use this to their advantage? Let's take a look at an example to understand it fully.

Suppose you have a rectangular plank of wood and are using it to cross a void. You can orientate it flat, as shown on the right below, or rotate it vertically onto its side, as shown on the left.

Intuitively, you know that the arrangement on the left will perform better,  but the plank cross-sections and the area are the same, and so are its material properties. So what's changed? Well, by rotating the plank onto its side, its geometry and resultant moment of inertia have changed completely.

Some cross-sections are better at resisting bending than others, and the orientation of a shape has a major influence on a cross-section's ability to resist bending. This is precisely what the moment of inertia or the second moment of the area describes. When quantified, it measures how much resistance the cross-section has to bend. 

Let's unpack this example further by looking at the cross-sectional geometry in 2D.

The further the area of a material is concentrated away from its bending axis, also known as the neutral axis (N.A.), the stiffer the section is and the higher its moment of inertia will be, relative to that axis. Just by orientating an identical wooden plank on its side (left), the cross-section has more material located away from the (N.A.), thereby improving its resistance to bending. 

Take any small area in that plank of wood, it will either experience compressive or tension stresses, which increase linearly as you move up or down from the N.A. It will try to resist bending when subjected to a load by creating a moment around the neutral axis. This moment exerted increases linearly as you move away from the N.A. This means that the resistance to bending provided by any point in the cross-section is directly proportional to the distance from the neutral axis squared. Once you integrate the force times the distance from the N.A. on the entire surface, you get the total resistance to bending specific to the axis specified.

This is a critical phenomenon that we can use to our advantage. Let's take it further and add an I beam into the mix:

An I-beam maximizes the amount of material located far away from the NA. This gives it a bigger moment of inertia and greater stiffness for the same mass of material. As a result, it performs most efficiently in bending out of all three cross-sections shown. This is a reason why I-sections are so commonly seen in construction.

While the moment of Inertia (I) is a cross-sectional property that partially determines a section's resistance to bending and deflection, it also has many uses in other engineering applications.

The moment of inertia is used in beam and column analysis, calculating bending stresses and flexural rigidity. It is also important to note that it is not a fixed value that is unique to each cross-section; it changes depending on the location of the bending (or reference) axis. In general, most engineers will take the neutral axis x-x and y-y, located at the centroid of a cross-section profile, as the reference axis used to calculate the second moment of inertia.

How do you calculate the Second Moment of Inertia (I)?

Now that you understand what factors influence a cross-section's Second Moment of Inertia, we will outline the general steps required to calculate the moment of inertia.

1.  Understand the fundamental mathematical expressions

The Second Moment of Inertia is synonymous with the second moment of the area because it is calculated by taking the moment about the area twice: 

Now, you could explore the next steps in calculating the second moment of inertia by hand (which we recommend you do). Still, in case you find yourself pressed for time, we have compiled a summary of second moment of inertia formulas for common shapes and beam sections:

 Second moment of Area formulas for common shapes

2.  Break the cross-section up into segments

When calculating the Moment of Inertia, we must calculate the Moment of Inertia of smaller segments. Try to break them into simple rectangular sections or other sections for which the formula already exists. Here we can break this I-beam into three components:

Or this hollow rectangle into an equivalent outer rectangle minus the inner:

The Neutral Axis (N.A.) or the horizontal x-x axis is located at the centroid or center of mass. Most engineering cross-sectional shapes are symmetrical, so you will only need to find the x-x neutral axis.

Now let's combine the segments using the parallel axis theorem with the formula:

The second moment of inertia is one of the most important material properties of structural elements because it determines an engineer's section selections. Since this value is used to predict the resistance of beams to bending, torsion, and deflection, you'll see it pop up in quite a few engineering concepts and formulas, namely:

 the stress that an object encounters when it is subjected to a large load at a particular point, causing it to bend and become fatigued.

 Deflection is also known as displacement and can occur from externally applied loads or the structure's self-weight. If the deflection is large enough, you can visualize the degree to which an element/structure changes shape or is displaced from its original position- this is something we want to avoid at all costs!

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