# Is an axial force applied at the centre of a beam considered a compressive force, tensile force, or both?

I was looking through some course notes and realized that I never fully understood this concept. So, looking at the diagram below we can see that there is a 20kN axial force being applied in-between sections (1) and (2).

Looking at the force from section (1)'s point of view, this would be a compressive force, as it it acting inwards towards the section. However, from section (2)'s point of view it is tensile, as it is acting away from the center of the section.

From what I can think, in this case I believe that when drawing the axial force diagram we simply go up (+'ve compressive axial force) when drawing from left to right. Meanwhile, if we were to draw this axial force diagram from right to left we would go down (-'ve tensile force).

Am I interpreting this correctly? Any help/insight would be really appreciated!

Edit: Another way I was interpreting this was, if I were to scan, starting from the leftmost side towards the right. I would initially see a compressive force onto the beam from R, then, I would see compressive forces (by using the section method at point 1). However once I am coming up to the 20kN force, I would, from my perspective, see a 20kN force being applied towards me, and from my section's perspective this would be compressive (-'ve force), so shouldn't I go down in the Force/distance chart? Since -'ve forces are compressive and +'ve ones are tensile.

Edit 2: Looking at some more Youtube videos: https://www.youtube.com/watch?v=Q7kYdDNKE8E, it looks like what happens at the point of force application isn't exactly necessary. Mostly we just calculate the force in-between everything, and just simply connect the lines.

• Hi @twowayspeedofLight, given that there are already three answers with totally different perspectives, it would be helpful (more for yourself than anyone else), if you provided some feedback with respect to how well they answer your question. IMHO, your question is stated in such a way that there is a lot of room to interpret it differently (which doesn't necessarily help you at the end of the day).
– NMech
Commented Jun 22, 2021 at 8:27
• I don't know the original intentions behind the figure, but your view of this loading situation depends greatly on whether the beam is perfectly rigid or not. If perfectly rigid, there is only one axial force applied, equal to the difference of those displayed; i.e., 10 kN and there is no particular location at which it's applied. If the beam is elastic, the exact location of the applied forces is important, including the area over which they are applied. In the latter case, the figure is ill defined, leading me to conclude it's a perfectly rigid beam. Commented Jun 25, 2021 at 3:05

I think your big misunderstanding is in this paragraph:

[...] if I were to scan, starting from the leftmost side towards the right. I would initially see a compressive force onto the beam from R, then, I would see compressive forces (by using the section method at point 1). However once I am coming up to the 20kN force, I would, from my perspective, see a 20kN force being applied towards me, and from my section's perspective this would be compressive (-'ve force) [...]

The big problem here is that you are changing your perspective between these two moments. You start by "looking to the left" and then end by "looking to the right".

It's important to remember that the sign convention for internal stresses depends on the face you're looking at: a force is tensile if it's to the right on the right-hand face or if it's to the left on the left-hand face.

When you are scanning along a beam, you need to choose which face you're going to look at. Or, well, you'll choose the direction in which you'll move and that'll define the face you're going to think about.

If you choose to go from left to right (as one usually does), then you obviously must only look at the left-hand face of the element. After all, you don't know what's going on to the right of the element, only to its left.

So, we start at the support, where we see the left-hand face receiving a rightward force. Looking at our sign convention, that's negative (compressive). So far, so good.

We then keep moving to the right, until we reach the point of application of those 20 kN. Where is that force applied? Well, we're only looking at the left-hand face, so we'll only notice that force once we've "moved past it" such that the force is applied on the left-hand face*. At which point we'll have a leftward force on the left-hand face, which in our sign convention is positive (tensile). So now we'll have a net 10 kN leftward force on the left-hand face, which means we're now under tension.

We then keep moving right until we reach the free end with the 10 kN force applied. Once again, we'll only notice that force once it's reached the left-hand face, at which point we'll once again have a rightward force on the left-hand face, which means it's negative (compression). Adding that to what we'd already accumulated, we get the expected zero force at the free end.

Going from right to left will give the same result, so long as you remember to always look at the right-hand face instead.

* The term "moved past it" is merely metaphorical: remember, the element of the sign convention is infinitesimal!

• Thanks for your help! Yea, this seems to be my mistake. I was not being consistent with my reference frame. Commented Jun 22, 2021 at 15:50

Technically a force on a structure is neither compressive nor tensile.

The effect a force has on a structure or the internal reactions can be characterised as tensile or compressive.

So in the following example in all cases there is a F force on the right. Then only difference between case A and case B is that in case A the support is on the left, while on Case B is on the right.

Notice that the force has not changed direction in the global Coordinate system. The only thing that changes is the support, and (even in this simple example) suddenly you have compressive forces, instead of tensile.

In the same manner, if the force was placed in the center of the beam, then the equivalent diagrams would be as follows.

Again as you can see, the direction of the external force F is horizontal to the right, however, the internal reaction is in one case tensile, and in the other (D) compressive. Additionally different parts of the beam are affected.

So, IMHO this is the most important thing that you need to take away from this.

Disclaimer: This is my interpretation based on how you expressed your question. I felt it necessary to point this out, given that Wasabi answered from a totally different perspective (which is entirely valid), I realize that I might have misinterpreted your question (your feedback will be very useful to improve my/our answer).

• Hey. Thanks for the reply. Though not what I was confused about I did get a better perspective on how different parts of the bar can be in compression/tension. Commented Jul 13, 2021 at 5:22

Conventionally, structural diagrams (V, M, T) are drawn from left to right. In the case of the axial force diagram, we need to set up/state the sign convention then draw the diagram accordingly. (Note that it does not matter where to start drawing, the diagram remains the same, as it must conform to the stated sign convention.)

Sign convention: (+) Compression, (-) Tension

Your question in the subject title can be answered by the method of superposition as shown on the diagram below - a "20 kN compressive force" was added to the rod that was subjected to 10 kips tensile force...

The diagram below shows the boundary forces at several internal points along the rod over the compression zone and tension zone. Note the displacement of the rod and the balance of the load in each stage. For the entire system, $$\sum F = 0$$ and $$\sum \Delta = 0$$, that leads to the conclusion - The stress is a function of the boundary forces and the crossectional area; the deflection is a function of the boundary forces, the crossectional area, and length. For an axially loaded member, with a constant crossectional area, the stress remains constant on any segment within the boundary points, but the magnitude of deflection changes due to variation in segment length.

• I see. Yea this makes sense. Thanks for your reply. I haven't thought of looking at things from this perspective! Commented Jul 13, 2021 at 5:23