Understanding Stresses in Beams

If we apply a load to a beam, it will deform by bending. This generates internal stresses, which can be represented by a shear force acting in the vertical direction, and a bending moment. The shear force is the resultant of vertical shear stresses, which act parallel to the cross-section, and the bending moment is the resultant of normal stresses, called bending stresses, which act perpendicular to the cross-section. It’s important to have a good understanding of these stresses because any design or analysis of a beam will involve calculating them.

 

Let’s look at bending stresses first. To keep things simple we’ll consider a case of pure bending. A section of a beam is said to be in a state of pure bending when the shear force along it is equal to zero, and so there is a constant bending moment along its length, like there is for this beam loaded by two moments. We also have a case of pure bending over the middle section of this beam, where the bending moment is constant. Let’s look at how a beam deflects when it has a constant bending moment along its length.

 

If we imagine the beam as a collection of very small fibres, as the beam deflects the fibres at the top of the beam get shorter, meaning that they are in compression. And those at the bottom of the beam get longer, so they are in tension. Somewhere between the top and the bottom of the cross-section there will be a surface containing fibres which stay the exact same length. This is called the neutral surface.

 

It passes through the centroid of the cross-section. When looking at the beam in two dimensions, we refer to it as the neutral axis. Let’s try and quantify the bending stresses that develop within the beam to resist these applied moments. First let’s calculate the strains in the beam. This can be done quite easily just by considering the geometry of the deformation.

 

Let’s watch how a fibre at the neutral axis between points A and B and a fibre between points C and D located at a distance Y from the neutral axis deform. Since this is a case of pure bending, we can see that the fibres bend into a perfectly circular arc. We’ll call the centre of the circle O. Before any deformation, the fibres are all the same length. After the deformation, the length of the neutral axis has stayed the same, but the length of the fibre between points C and D has increased. If theta is the angle of the arc, and R is the radius of the arc to the neutral axis, we can calculate the length of the arc between A and B, like this. And we can calculate the length of the arc between C and D in the same way. Strain is defined as the change in length divided by the original length, and so we can derive an equation for bending strain at any distance Y from the neutral axis.

 

We defined the distance Y as being positive downwards, and so this equation will give us a positive strain for the bottom of the cross-section, which is in tension. Sometimes you’ll see this equation written with a minus sign, but that’s because Y was defined as being positive upwards. If we assume that stresses remain within the elastic region of the stress-strain curve, we can then apply Hooke’s law for uniaxial stress to calculate the bending stresses. This gives us the equation for bending stress as a function of the radius of curvature R of the deformation. But what we’re really interested in is how the bending moment M affects the bending stress.

  Understanding Plane Stress

 

If we make an imaginary cut through the beam we can expose the internal bending stresses, represented here as a few discrete forces. The resultant moment of these internal forces must be equal to the bending moment M, and so we can calculate M by integration, like this. Now we can plug in the equation for bending stress we just derived. When rearranged into this form, we can notice that the integral on the right is the definition of the area moment of inertia.

 

This parameter, which I’ve covered in detail in a separate video, defines the resistance of a cross-section to bending due to its shape, and is denoted using the letter I. We can combine this equation for the bending moment with the bending stress equation to obtain what is known as the flexure formula. So what does it tell us? Bending stress increases linearly as the bending moment and the distance from the neutral axis increase. And it decreases as the area moment of inertia increases. The maximum stress occurs at the fibres furthest from the neutral axis.

 

The term I over Y-max depends only on the geometry of the cross-section, and so it is called the section modulus, and is denoted using the letter S. You will often see the section modulus listed for a range of common beam cross-sections in reference texts. The I-beam is a commonly used cross-section because it has a large area moment of inertia, which results in lower stresses.

 

Here’s how the bending stresses are distributed over an I-beam cross-section. They are zero at the neutral axis, and reach a maximum at the outside surfaces of the flanges. For a T-section the neutral axis is shifted upwards, and so the bending stress distribution looks like this. So we’ve established how to calculate the bending stresses, which are normal stresses, for a case of pure bending.

 

Most of the time we won’t have pure bending as there will also be a shear force acting on the beam cross-section, like the beam we saw at the start of the video. It turns out that the presence of a shear force doesn’t normally significantly affect the bending stresses, and so luckily we can consider the flexure formula we derived earlier for pure bending to be valid for a more general case of bending. The shear force V is the resultant of shear stresses which act vertically, parallel to the cross-section. We denote the shear stresses using the Greek letter Tau.

 

To maintain equilibrium, these vertical shear stresses have complementary horizontal shear stresses, which act between horizontal layers of the beam. One way to visualise these horizontal stresses is to consider a beam made up of several planks of wood. When a load is applied, there is a tendency for the planks to slide relative to one another. Now let’s glue the planks together.

 

When the load is applied the planks cannot slide, and so horizontal stresses develop between them. If these shear stresses are larger than the shear strength of the glue bond, the glue will fail. These horizontal shear stresses don’t exist if we apply a moment instead of a force, because that gives us a state of pure bending. And so there is no tendency for the planks to slide relative to one another.

  Understanding Material Strength, Ductility and Toughness

 

The presence of these horizontal shear stresses explains why wooden beams sometimes fail by splitting longitudinally. This failure usually occurs close to the neutral axis, for reasons which will soon be obvious. So how can we calculate the shear stresses? We can calculate the average shear stress acting on the cross-section as the shear force V divided by the cross-sectional area.

 

But the shear stresses aren’t distributed uniformly across the beam cross-section. The shear stress has to be zero at the free surfaces at the top and bottom of the beam. So the average shear stress isn’t very useful, since it doesn’t tell us the maximum shear stress. Instead, we can use this equation to calculate the shear stresses acting on the cross-section. I won’t cover the derivation of the equation here, but it’s based on considering equilibrium of stresses acting on a small element within the beam.

 

The equation assumes that the shear stress is constant across the width B of the cross-section, so Tau is a function of the distance along the beam, X, and the distance above the neutral axis, Y. V is the shear force acting on the cross-section, which varies with the distance along the beam. B is the width of the cross-section. It can vary with the distance y from the neutral axis, but in this case the cross-section is rectangular, so b is constant. I is the area moment of inertia, which is a constant value calculated based on the shape of the cross-section. And Q is the first moment of area for the portion of the cross-section above the location we want to calculate the shear stress for.

 

So it varies with the distance Y above or below the neutral axis. It is equal to the product of the area above the location of interest and the distance between the centroid of that area and the neutral axis. If the location of interest is below the neutral axis, we consider the area below the axis, instead of the area above it. To calculate the first moment of area at this line which is at a distance Y from the neutral axis, we multiply the area of the blue rectangle above the line by the distance from the neutral axis to the centroid. Doing this calculation gives us an equation for Q as a function of the distance Y from the neutral axis for a rectangular cross-section.

 

And so we can obtain an equation which describes how the shear stress varies with distance from the neutral axis. The Y term is squared, and so the shear stress varies parabolically over the height of the cross-section, with the maximum shear stress occurring at the neutral axis. This is opposite to the bending stress, which is zero at the neutral axis, and explains why the horizontal shear failure of a wooden beam we saw earlier occurs close to the neutral axis. By setting Y to zero in this equation, we obtain an equation for the maximum shear stress in rectangular cross-sections. It is equal to 1.5 times the average shear stress across the entire cross-section.

  Understanding the Area Moment of Inertia

 

The derivation of this equation for shear stress makes a few assumptions, so we need to be careful with how we apply it. First, it assumes that the shear stresses are constant across the width of the cross-section. For rectangular cross-sections this is a reasonable assumption if the rectangle is thin. But for cross-sections like this one, the shear stresses can vary significantly over the width, and so in these cases the equation can really only give us the average shear stress across the width of the cross-section. It can’t tell us what the maximum shear stress will be. Another assumption this equation makes is that the shear stresses are aligned with the Y axis.

 

Shear stresses act tangentially at a free surface, so for a circular cross-section, for example, we can’t strictly use this equation to get the distribution of shear stresses across the height of the cross-section. But we can still use it to estimate the shear stresses at the neutral axis, because the shear stresses there are aligned with the Y axis. The equation for shear stress at the neutral axis in a circular cross-section is similar to the equation for a rectangular section, where we have the average shear stress V over A, multiplied by a constant. The constant is 4 over 3 for a circular cross-section and 3 over 2 for a rectangular one. We can also use the shear stress equation for thin-walled sections like this I beam, although things are a bit more complicated.

 

Because the vertical shear stresses at the surfaces shown in red must be zero, and because the flanges are very wide, the vertical shear stress in the flanges is very small. This means that the vertical shear stress is distributed like this. The web mostly carries the shear force, and the flanges mostly carry the bending moment, as we saw earlier. You can see that the shear stresses are distributed quite evenly over the height of the web. This is because the flanges contribute significantly to the first moment of area Q when calculating the shear stresses in the web, but they don’t carry much of the vertical shear force. Since the web is thin, the shear stresses are also distributed evenly across its width. Because of this, we can easily calculate the approximate shear stress in the web, like this.

 

More detailed analysis reveals that there are shear stresses in the flanges, but they are acting mainly in the horizontal direction. The horizontal stresses on both sides of the flanges cancel each other out, so the net shear force is still just a vertical force. We can figure out the direction of the horizontal shear stresses based on the direction of the vertical shear stresses by imagining that the stresses are flowing through the cross-section. That’s it for this review of bending and shear stresses in beams. If you enjoyed the video and would like to see more content like this, please consider supporting the channel on Patreon! Thanks for watching!

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