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The question is: Which force causes the reaction moment at a fixed joint ?

Lets consider a very simple example, a beam with one end connected by a fixed support to a wall. enter image description here

I understand that at A, there will be a vertical reaction force, a horizontal reaction force and a reaction moment. My understanding is that a moment about an axis is always caused by a force. Which force F causes the reaction moment?

Focusing on this example, suppose F is the horizontal reaction force. From the force equilibrium condition along the x direction, we get that this force has zero magnitude and thus cannot generate the reaction moment.

Suppose F is the vertical reaction force. Apparently, when we cut the beam at some C and apply the moment equilibrium condition at C, we are supposed to include both the reaction moment and the moment generated by the vertical reaction force, which indicates that the moment generated by the vertical reaction force and the reaction moment at A are two different moments caused by different forces.

So, to conclude, if the moment at A is caused neither by the vertical nor the horizontal reaction force, by which force is it caused?

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  • $\begingroup$ Can you identify and explain what type of loading "w = 10 N/m" represents, and in what direction it acts? $\endgroup$
    – r13
    Commented Oct 30, 2021 at 17:56
  • $\begingroup$ @r13 represents the load of the beam. It is not clearly stated in the problem, but propably is the weight of the homogeneous beam $\endgroup$
    – Stamatis
    Commented Oct 30, 2021 at 17:59
  • $\begingroup$ It represents a "uniformly distributed load" acting in the direction shown by the arrows, so F = 10N/m * 2m = 20 N. Do you think this (F) is a horizontal load? $\endgroup$
    – r13
    Commented Oct 30, 2021 at 21:40
  • $\begingroup$ @r13 I do not, I think it is vertical $\endgroup$
    – Stamatis
    Commented Oct 31, 2021 at 12:50

3 Answers 3

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I interpreted your question as 2 separate requests.

  1. What fundamental physics causes a fixed support to create a moment around the central axis of the beam?

  2. How can I calculate the internal forces in a section 'C' of a beam

Q1:

Beam-no-deflection

enter image description here

enter image description here

enter image description here You can imagine the beam is fixed to the wall with glue. As the beam tries to rotate around, the beam is going to push into the wall at the bottom, and get pulled back to the wall from the top by the glue. A similar process happens if the beam is screwed, welded or bolted to the wall. The part of the wall at the bottom is in 'compression' and a material under compression tries to expand back to its original size and hence helps to exert the force required to rotate the beam around its axis (the moment).

Keep in mind; The >net< force at the reaction point in the x direction is still 0, because the top sucking and the bottom pushing sums up to be Zero (0) newtons in the x direction.

Q2: Here is the final solution to a section cut 'C' at 1 meter into the beam. PLEASE keep in mind you should make sure you fully understand how I calculated all the forces and moments here. Essentially, you want to cut the beam, and make sure the sum of the forces on (both of) the beam(s) and the sum of moments is 0, just like you would do for other statics problems. Again, the new pivot point is 'fixed'

The internal moment of the beam at 1 meter from A is 5N-m because through static equilibrium:

$\Sigma Fx = 0; \Sigma F_y = 0 = F_{c_{y}} - 10(N)$

$\Sigma M = 0 = -10(N) * .5(m) + M_c$

$M_c = 5Nm$

enter image description here

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  • $\begingroup$ First, thanks for the reply! Second, what would happen if we tried to find the total moments on the point we made the cut, but at the left beam instead of the right? From what I understand we would have, Sum_M at C = 0 => F_wall*1m - M_reaction - 10*.5m + M_C = 0. Why do we include M_reaction? Is M_reaction not about the perpendicular axis at the wall and not at C? Many thanks! $\endgroup$
    – Stamatis
    Commented Oct 30, 2021 at 17:57
  • $\begingroup$ @Stamatis You can do the analysis at point A or point C, for the left beam. As long as you keep your plus and minus signs correct, you should find it sums to 0, if the problem is static. If it doesn't sum to zero, then you either did the check or the problem wrong and should recalculate. Your equation looks correct to me. ||20N * 1m - 20Nm - 10N * .5m + 5Nm = 0|| is a true statement. Mc (20Nm) occurs around the 'z' axis of this problem. It is useful to know that a moment can be drawn anywhere on a beam; its location on the drawing matters for intuition, not for calculation, unlike forces. $\endgroup$
    – WhisperingShiba
    Commented Oct 30, 2021 at 18:27
  • $\begingroup$ I see what you are saying, but why does the location of a moment does not matter for calculation? For example: The moment that a force creates at x_1 is different from the moment it creates at x_2 by M_2 - M_1 = Fx_2 - Fx_1 = F*(x_2-x_1). When using the moment about z at A, why do we not adjust it by a factor similar to (x_A - x_C) when we try to see the effect of the force that caused M_reaction_wall at C (our cross cut)? $\endgroup$
    – Stamatis
    Commented Oct 30, 2021 at 18:33
  • $\begingroup$ @Stamatis If you ask too many of these why questions you will eventually have to go to the theology or philosophy stack exchanges. I will not be able to explain 'Why' but I can help build some intuition behind this. Imagine a propeller blade with 2 symmetric, 1m blades. If you push with a constant 20 N at 1 m, you have 20Nm torque. If you then removed one of the propeller blades, and extended the remaining blade by 1 meter, but pushed with the same 20 N and 1 m from the pivot axis, you would still have the same torque on the system... Your moment of inertia may be different; same torque. $\endgroup$
    – WhisperingShiba
    Commented Oct 30, 2021 at 19:06
  • $\begingroup$ @stamatis Regarding why your example is correct, and my statement is correct. You are confusing moment, as an abstraction of force acting at a distance; and the actual Newtonian force, acting at a distance. The location of forces is important, but if you skim your statics textbook or look into your next semester, at 'rigid body dynamics' you will see the moment will be drawn anywhere on rotating members. $\endgroup$
    – WhisperingShiba
    Commented Oct 30, 2021 at 19:13
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the moment at pont A is the sum of contribution of all small defferential dx moments along the beam's length.

It can be simplified as the moment of the total load acting at its CG.

$$M_A=10N*2*1m=20Nm$$

I think your are confusing the free body moment on a random point with the moment at A.

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  • $\begingroup$ I must indeed be confusing it since I am not familiar with the term. It also appears I cannot find it online. By free body moment, are you referring to the moment that would be assigned to A at a free body diagram? $\endgroup$
    – Stamatis
    Commented Oct 30, 2021 at 18:02
  • $\begingroup$ i don't know how much you know. free body moment at A in your beam for the entire beam is achieved by removing the support and replacing it with its reactions. Which gives the same moment at A, 20Nm. But if you need to study forces at a random point, x along the beam, you cut the beam at that point and show all the forces and moments working on that point. For example, moment at X=1m is by cutting the beam at 1m and verifying M at that point $\endgroup$
    – kamran
    Commented Oct 30, 2021 at 18:11
  • $\begingroup$ perfect, that exactly the way I have understood it. So then my question becomes: If we cut at a random point x, and try to evaluate the total moment at the edge of the left beam (towards the wall), why do we include M_reaction_wall? Is that moment not assosiated with the axis perpendicular to the paper at A and not at the edge of the cut? Is every moment by definition only assosiated with respect to a specific axis? Thus if M_reaction wall is with respect to z_axis_at_A, does it not need to be adjusted to be included at C? Do you think I should perhaps ask this as a separate question? $\endgroup$
    – Stamatis
    Commented Oct 30, 2021 at 18:19
  • $\begingroup$ if we cut at random X and need the moment at A we have to add the moment on the right hand of X + the moment caused by w between X and A, we end up with 20Nm again. the moment at A lies in XY plane., the axis of rotationis in z plane. $\endgroup$
    – kamran
    Commented Oct 30, 2021 at 20:14
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IMHO you are confusing the "moment of a force about a point A", with the the "reaction moment" that develops at the support of a structure, and more specifically how they are calculated. Although they are related there are not the same thing.

The first one is a very fundamental concept in physics. Moment of a force is the product of force times the distance of the point from the force carrier (colinear line to the the force).

The reaction moment that develops is developed by the structure. Although in statics we assume rigid structures (i.e. small or nonexistent displacements), in real life all structures deform. The deformation of the structures may have a significant effect on the structure.

What I am trying to arrive at is that "the carrier structure as well as the load determines the support reaction".

Additionally, regarding the part of your question

Apparently, when we cut the beam at some C and apply the moment equilibrium condition at C, we are supposed to include both the reaction moment and the moment generated by the vertical reaction force, which indicates that the moment generated by the vertical reaction force and the reaction moment at A are two different moments caused by different forces.

This sentence is wrong on many levels, so it would be best if you actually did the math so that we could tell you where you got it wrong. I'll try to explain it with a small example: Assume C is right at the middle of the beam in your post.

In that case the CB section has a length of 1m, and the total transverse force on it is 10N. The reaction that develops on C is based on those values, so on C there is a Transverse force of 10 N and a 5 Nm bending.

On the AC section, again the length is 1m, and the total transverse force on AC section is 10N, however on C, there are also present the reactions from the section CB. Those forces are a Transverse force of 10 N and a 5 Nm bending (with opposite direction that above). So although there is a cut at C, the reaction forces that develop at C are such that the entire structure is in static equilibrium.

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