Im looking for clarification on how the solution identified the moment to be hogging, as well as the axial stress to be in compression?

Also I'm confused as to how the solution came to a conclusion that the bottom of the cross section is compressive stress, whilst the top of the cross section is tensile stress.

Please reference the photo below with the red box's

enter image description here


  • 1
    $\begingroup$ The questions you’re asking seem to center around trying to understand bending stresses (this post and other recent ones). It might help if you edited your question to specify what concepts make sense to you and where you become uncertain. For example, is the confusion arising over the idea of an eccentric axial load producing a moment? Is there confusion over how to calculate that moment? What about understanding what the distribution of axial and bending stress will look like along the cross section (and calculating the specific values)? $\endgroup$
    – CableStay
    Jun 22, 2019 at 1:27
  • $\begingroup$ @CableStay I have my assumption that the force being applied at the lower half of the beam is what causes the beam to be hogging, as the beam is forced into a upward arrow shape. Whilst the force on the lower half of the beam is causing compression as its directed towards the beam. Leaving the top half of the beam to be in tension. Im not sure if my theory is correct, I'm looking for confirmation. $\endgroup$ Jun 22, 2019 at 1:31
  • 1
    $\begingroup$ What you are forgetting is the axial forces are actually calculated according to 1st moments of area and then 1st moment of Inertia. $\endgroup$
    – Rhodie
    Jun 27, 2019 at 14:56

2 Answers 2


Once again, the superposition principle is our friend. It basically means we can look at each load in isolation, find the individual results, and then add them all together to get the final result.

In this case, we are actually dealing with two loads:

  • There's obviously the force $P$;
  • And then there's the bending moment $P\cdot e$, where $e$ is the eccentricity of the load $P$ in relation to the beam's neutral axis. Since the beam is rectangular with height 200 mm, we trivially know the neutral axis is at 100 mm. And since the load is only 30 mm from the bottom, we know $e=100-30=70\text{ mm}$.

Now let's look at these loads in isolation:

The force $P$

This one is pretty obvious. The force alone (remember, we're not yet considering the bending moment effect, so you can think of the force as acting at the neutral axis) obviously causes uniform compression throughout the beam. This means the beam remains perfectly straight, just a bit shorter. Not much to discuss here, I believe.

The bending moment $P\cdot e$

Herein lies the rub.

The bending moment is counter-clockwise in this case. This seems pretty intuitive to me, so I won't spend any time demonstrating this.

The issue is how the beam deflects given this bending moment.

Looking at the free end, it can rotate in one of two ways: clockwise or counter-clockwise. It once again seems pretty intuitive to me that the beam will rotate counter-clockwise, but here's a demonstration, just in case it isn't to you:

Let's transform the bending moment into a force couple: two opposing forces which cancel each other out other than generating a bending moment. A classic example is a crank valve.

enter image description here

So, if you had a crank valve at the end of a beam, and you applied the forces shown above, how would you expect it to rotate? It seems clear the beam would rotate counter-clockwise. By this I mean that the free end (and indeed any other slice along the beam) would rotate counter-clockwise, such that the top fiber moves a bit to the left and the bottom fiber moves to the right.

And that tells us all we need to know: if the top fiber moves to the left, then the total length of that fiber has increased (from 700 mm to... 701 mm or whatever). And we know that if an element is lengthened, it must be under tension1. Likewise, if the bottom fiber moved to the right, that fiber's length shortened and it is therefore under compression.

Total result

So, summing both loads' results, we get that the top fiber is under compression due to the force and under tension due to the bending moment. Whether the final result is compression or tension will depend on which stress is greater.

The bottom fiber is under compression due to both loads, so that's simple enough.


As for why the moment is considered hogging, well, that's just the technical definition of hogging moment.

When the beam bends in such a way that it forms concavity downwards (cup-shaped) it is called as sagging. Whereas the bending which results in convexity upwards (like a hump); is known as hogging. [source]

Basically, positive moment is sagging, negative moment is hogging.

And if you remember your sign conventions:

enter image description here

You can see that counter-clockwise moments on the left side of the beam are negative, and therefore hogging.

1 Or possibly under a positive thermal load, if you want to be a pedant.

  • $\begingroup$ I thought engineers tended make great pedants! In reference to your second bullet point "Since the beam is rectangular ..." might benefit from the addition of "Since the beam is rectangular and homogeneous...". I only mention this as one of OP's recent questions have dealt with a steel and aluminum section. $\endgroup$
    – Forward Ed
    Jun 24, 2019 at 14:24

Your beam will bend down as opposed to what you think, under the moment of 100kN*((200/2)-30 =70 mm)and will at the same time compress under the stress of P/A= 100kN/ (20*200).

Because the force of 100kN is applied below the neutral axis of the beam it causes counterclockwise moment hence compression below the neutral axis.

As a rule of thumb you always get compression on the part of the beam's section which is under axial load.

  • $\begingroup$ can you clarify that last sentence abut under axial load always being compression? As a stand alone sentence it strikes me as odd since a part or all of a beam can be in tension and that is also an axial load and the complete opposite of compression. $\endgroup$
    – Forward Ed
    Jun 22, 2019 at 13:58

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