In this article (and video above), we guide you through the Combined Stress Formula for a cantilevered beam, a valuable skill that will enhance your FE Exam prep and strengthen your understanding of essential engineering concepts.
Question:
In this scenario, we have a 300 mm cantilevered beam with a 20 mm x 20 mm square cross-section. One end of the beam is securely fixed, while the other end carries two applied loads: an axial force Fx = 200 N and a perpendicular force Fy = 150 N. Our goal is to determine the maximum combined stress produced by these forces at the base of the beam, focusing specifically on point A.
Problem Context: Combined Stresses
Before we jump into the question, let’s set the stage by exploring the concept of combined stress. This combined stress state occurs when an object experiences different kinds of loads at the same time. For instance, a beam subject to both axial and bending forces will develop a combination of these induced stresses throughout its structure. In general, there are five key stress types you should be familiar with to be able to solve these kinds of problems, and they are listed here as:
- Axial stress
- Bending stress
- Shear stress
- Torsional stress
- Thermal stress
We’ll start off with a quick explanation of each stress case, starting off with axial stress.
Axial Stress
This occurs when a force is applied along the axis of an object, placing it in tension or compression. If a beam is stretched along its length, it creates positive or tensile stress, while compression results in negative stress. This applied force creates a uniform normal stress distribution across the entire cross-sectional area of the beam, because every part of the cross-section experiences the same amount of force per unit area. The FE Handbook also supplies us with a formula to calculate this normally distributed stress, which is simply and intuitively defined as the applied axial force divided by the object’s cross-sectional area.
Bending Stress
Now, let’s explore how bending stress can affect an object. When we apply two bending moments to either side of a beam as shown here, the moments cause one side of the beam to expand while compressing the opposite side. In this specific graphic, the top side of the beam compresses, resulting in negative stress in this area, while the bottom is in tension, translating to positive stress. And since we have this negative stress on the top side with positive stress on the other end, we can intuitively deduce that the stress should be zero at the centreline of the beam, which is referred to as the beam’s neutral axis. As we move further away from the neutral axis, the magnitude of stress increases, which results in the linear bending stress distribution shown here. The FE Handbook once again supplies us with a formula to determine the normal stress created by this bending, at any location along the beam’s cross section. However, we will be looking at this in more detail once we get to our problem.
Shear Stress
Now, let’s talk about transverse shear stress. This type of stress happens when opposing forces act along the beam’s cross-section, like the forces shown here. These forces try to slide the layers of material past each other, causing the beam to deform. The stress isn’t the same across the section—it follows a parabolic pattern, with zero stress at the outer edges and maximum stress near the centre. This happens because the outer layers move less, while the inner layers experience more sliding. The greatest resistance to this sliding occurs at the neutral axis, which is why the stress is highest there. For reference, we’ve also included the FE handbook formula used to calculate shear stress.
Torsional Stress
Now, let’s take a look at torsional shear stress. This stress happens when a twisting force acts on a beam, as shown here. The stress at the centre of the cross-section is zero and increases toward the edges. This is because the fibres near the centre barely twist, while the outer fibres experience the most rotation. As you move outward from the centre, the shear stress increases linearly, reaching its maximum at the outer surface, where the material resists twisting the most. For reference, we’ve included the FE handbook formula here again.
Thermal Stress
Lastly, let’s talk about thermal stress. This type of stress happens when a material expands or contracts due to a change in temperature, which is especially important in structures exposed to different environmental conditions. The FE handbook includes a formula to calculate how much a beam will expand or shrink with temperature changes, and this displacement can then be used to determine the resulting stress if needed. But now that we’ve covered the main types of stress, let’s move on and figure out which of these apply to our example.
Step 1 – Evaluate Stresses Acting at A
Let’s start by taking a closer look at our system, focusing specifically on point A. We can isolate this location by examining an imaginary cross-section of the beam right at that point. Given the axial force Fx and the perpendicular force Fy applied to the beam, we can identify that the beam will experience both axial stress and bending stress. But what about shear stress? To answer this, let’s refer to the transverse shear stress distribution. As shown, shear stress is always zero at the outer edge of a cross-section—exactly where point A is located.
This means point A will not experience any contribution from shear stress, leaving us with only the axial and bending stresses to consider in our analysis.
Step 2 – Evaluate Axial Stress
Let’s start by taking a closer look at the axial stress in our beam. With the axial force Fx pulling on one end, the beam responds with an equal but opposite reaction force at the fixed support, putting the entire beam in tension. This tension creates a uniform stress distribution across the beam’s cross-section, as shown in the diagram. And because the beam is in tension, we expect the axial stress to be positive. Now, to calculate the stress at point A, we can use the simple formula from the FE handbook. All we have to do is plug in the given values, and we get an axial stress of 0.5 MPa. And with that done, we’re ready to move on to the next part of the question—evaluating the bending stress.
Let’s get a clearer picture of bending stress by looking at the beam from the side. When the force Fy is applied, the beam responds with an equal but opposite reaction force at the fixed end, which could induce shear stress. However, as we already know, shear stress doesn’t contribute to the stress at point A, so we can set that aside for now. What’s important here is that this force also creates a reaction moment at the fixed end, M, which is responsible for the bending stress in our section. Now, according to the FE Handbook, we can classify bending as either positive or negative. A positive bending moment curves the beam upward, with the top fibres in compression and the bottom in tension. But in our case, the reaction moment bends the beam downward, meaning we’re dealing with negative bending. This gives us a useful framework for how the stresses behave in this scenario.
Now, let’s use the equation from the FE Handbook to calculate the bending stress at any vertical point along the beam’s cross-section. To do this, we’ll need three key values: the bending moment M, acting on the section, the moment of inertia I, of the cross-section, and the distance y, from the neutral axis to the point of interest. It’s important to note that positive y refers to a point above the neutral axis, while negative y corresponds to a point below it. For point A, since it’s located at the top of the beam, y is positive and equals half the beam’s thickness—which gives us 10 mm or 0.01 m. Next, let’s calculate the bending moment M. Remember, we’ve already established that because the beam is experiencing negative bending, the moment will be negative. But to find the moment’s magnitude, we simply multiply the force Fy by the distance from the fixed end, which is 0.3 m. This gives us a moment of negative 45 Nm.
The last variable we need to account for is the inertia of the beam. The FE Handbook defines this as the cross-sectional area’s base multiplied by its height cubed, all divided by 12. These variables are both given as 20 mm in our case, so we find the inertia to be 13.3 × 10-9 m4. And remember since we’re working with a square, you could’ve used either of those formulas, since the beam’s base is equal to its height. With all the values in place, we can now calculate the bending stress at point A. Just a quick heads-up: make sure to substitute the reaction moment as negative—those double negatives in the equation will cancel each other out, giving us a positive stress value. The bending stress at A is then found to be positive 33.75 MPa, which matches our earlier expectations. Since the beam is being stretched and expanded on its top surface, it is also experiencing positive tensile stress.
Step 3 – Evaluate Combined Stress
Finally, all that’s left to do is look at the combined effect of both these stresses. In this case, we got lucky because both the axial and bending stresses act in the same direction, allowing us to simply add the two stress values from each loading case. When we sum them up, the total combined stress at point A comes out to be 34.25 MPa.
The answer is C!
And that completes our problem! Comparing our solution to the given options, we find that the correct answer is C.
Final Remarks:
Before we wrap up, let’s touch on a few key considerations that will help you tackle similar problems. While today’s problem focused on axial and bending stresses, it’s equally important to understand shear, torsional, and thermal stresses. These forces often appear in combination, so recognizing and knowing how to analyze them is essential for solving more complex problems. So, on that note, let us know if you’d like us to tackle these topics in a future video.
You’ll also want to become familiar with the Von Mises stress formula, which helps assess whether a material will yield under multi-directional loading. In this question, we had two stress cases acting in the same direction, so a simple addition of these stresses was sufficient, but if you ever come across stresses acting at different angles, von Mises will help you calculate the combined stress.
Lastly, remember to stay consistent with your sign conventions, especially for bending moments. A small mistake with positive and negative signs can throw off your entire analysis, so double-checking your work is crucial. Mastering these concepts will not only prepare you for the FE Exam but also strengthen your engineering skills moving forward. Let us know in the comments if there are other topics you’d like us to cover next!
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Anthony Fasano, P.E.
Engineering Management Institute
Author of Engineer Your Own Success
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