# Confused about internal hinges for calcualting reactions

[]

I'm really confused about internal moments.

Regarding image 1,

When taking moments about D (the hinge), why does the book only consider the right side? Could you technically consider the left side (-Va x (L1+L2) - (Vb x L2) = 0) as well? By technically i mean unrealistically because i understand that considering the left side would introduce unwanted unknowns into the 4th equation (first 3 are sum of x,y,moment about A) as the equation would become:

VcL3 - P(L3+L4) - Va x (L1+L2) - (Vb x L2) = 0 am i right???

Regarding image 2,

Why must the triangular distributed load be split as shown to calcualte Ma? I got the question wrong when i assumed the UDL was concentrated at its centroid (right of hinge) as a single force. I feel this contradicts with image 1 (3) because Vc is also a single force on the right hand side of the hinge? My understanding is that normally (without hinge) the whole triangular load would act at the centroid of the whole triangle?

I do understand that internal hinges don't restrict moments and that only shear forces are transferred across but this doesn't make sense.

Thank you and sorry for the long quesiton.

## An extended discussion on internal vs. external forces

We usually like to describe hinges as "places where the moment is always zero."

But, wait a minute, the moment is always zero anywhere in a stable structure. Don't believe me? Let's take a look at the most trivial structure ever, a simply-supported beam with a uniform load:

So, it has a span of 8 m and a load of 1 kN/m. Obviously, the reactions are 4 kN each.

We also know that the bending moment at the midspan is given by $$M = \dfrac{qL^2}{8}$$, which in this case gives us 8 kNm, as shown in the diagram above.

But let's calculate the moment at midspan by hand, using our trusty sum of bending moments approach:

\begin{align} \sum M_{\text{midspan}} &= M_{\text{left reaction}} + M_{\text{right reaction}} + M_{\text{load}} \\ &= -(4\cdot4) + (4\cdot4) + (1\cdot8\cdot0) \\ &= 0\text{ kNm} \end{align}

So... what's going on here? What's that 8 kNm at midspan if we actually obviously have zero moment there?

Well, what's going on is that internal moment is one thing, external is another, and we can't get them mixed up.

For a more obvious distinction between the two, let's instead look at axial loads: imagine you have a sugar cube between your fingers and you start squeezing it. What'll happen? Well, if you look at the applied forces generally, you'll conclude that nothing will happen: you are applying equal forces in opposite directions, so the net force on the sugar cube is zero! It doesn't matter whether you are squeezing the cube with barely any force at all or if you're putting all your strength into it: you'll always be applying two forces in opposite directions and the net force will always be zero.

But we know that's not how this works; if you squeeze the cube enough, it'll crumble between your fingers. Because the net external force may be zero, but the cube is under extreme internal forces. And the value of this internal force is equal to the force applied by one of your fingers (if you squeeze the cube with a force of 1 kN from each side, the internal force in the cube will be of 1 kN).

The definition of a stable structure is that the external forces are all balanced in every point in space (it doesn't even need to be on the beam. You could calculate the moment at $$x = 1000\text{ m}$$ and you'll still get zero moment). And when we calculate moment as we did above, what we're calculating is the external moment. Hinges are in no way special with regards to external moment (indeed, nowhere is).

So external moment (and other forces) is useful to know whether the structure is stable: if the external moment were non-zero, that'd mean we're actually dealing with a mechanism that will accelerate over time.

Internal moment (and other forces), however, is useful to know whether the structure can withstand the applied load. And just as the internal force in the cube is equal to the load applied to one side of the sugar cube, so is the internal bending moment equal to the bending moment on one side of the point of interest.

So, if we recalculate the bending moment at midspan looking just at the load to its right, we get:

\begin{align} \sum M_{\text{midspan}}^+ &= M_{\text{right reaction}} + M_{\text{load}} \\ &= (4\cdot4) - (1\cdot4\cdot\dfrac{4}{2}) \\ &= 8\text{ kNm} \end{align}

Notice that since we are calculating the moment to the right of the midspan, I had to "break up" the uniform load so that I'm considering the effect of its right half. After all, that uniform load is arbitrary. We drew it as one uniform load along the entire span, but should we expect a different result if we instead had two uniform loads of equal value, one on each side of the midspan? Of course not!

So just because we drew the uniform load as covering the entire span (and with a centroid at the midspan), that doesn't mean we can ignore its effect on the internal bending moment calculated when looking at only one side of the beam.

We can also calculate this from the left-hand side, but then we need to remember the sign convention that, for left-hand internal moments, clockwise is positive:

\begin{align} \sum M_{\text{midspan}}^- &= M_{\text{left reaction}} + M_{\text{load}} \\ &= (4\cdot4) - (1\cdot4\cdot\dfrac{4}{2}) \\ &= 8\text{ kNm} \end{align}

That these two calculations should have equal results is obvious: if the external bending moment is zero at that point, then the internal bending moment must be equal (and opposite, but we dropped the sign for the left-hand side because of the convention) on both sides of that point.

As a refresher, here's the sign convention for internal forces:

## Now to answer your actual question

What we really mean when we say that hinges always have zero moment is that they always have zero internal moment.

As such, when calculating this zero-moment at the hinge, we need to do so by only looking at the loads to one side of the hinge. So for your triangular load in the second example, you need to calculate the moment at the hinge by considering the part of the load that's visible on the side you're calculating.

• Hi! Thank you so much for the thorough response! I'm however still a bit confused about why in the first image (3) equation forces and reactions on the right side of the hinge contributes to the moment about A; but for the triangular load the force cannot be concentrated at the centroid as a point force? I understand everything you explained about internal moments so thank you so much again! Mar 31 at 4:59
• @mdkbear I don't think I understood what you're asking here... if you're looking at the second beam with the triangular load and doing an external moment calculation around A, then you CAN simplify the triangular load to a point force at its centroid. You only need to do the decomposition scribbled over the original image if you're doing moment around the hinge; in that case you need to do a sum of the load to the left of the hinge (a simple triangular load which can itself be simplified into a point force) OR to the right (with two loads, one triangular and one uniform).
– Wasabi
Mar 31 at 15:58

It is important to understand what is an internal hinge, its behavior, and assumptions around it.

An internal hinge represents a discontinuity in the beam, it carries a pair of equal but opposite force couple, one to each side of the hinge, carried over from the beam segment it supports/connect to, so sum Fy is zero. Another important assumption is the hinge can't restrain rotation, therefore, sum M must be zero.

On your questions, the force to the right side of the hinge is the reaction (an unknown) from the right side beam segment, and as stated above, there exists an equal but apposite force to the left, and it becomes an applied load to the left side support system. The sketch below shows the force couple and unknown support reactions (in read), with the available structural equilibrium equations and unknowns indicated below. You should be able to figure out why the solve starts from the right side.

Below is my earlier response, that step by step show you why the solve stars from the right side, rather than the left.

1. After separated at the internal hinge at "D", it leaves 2 unknown (V_C & V_D) at the right side structure, and you have 2 available equations (sum Fy=0 & sum M=0) to find the solution the unknowns. On the other hand, the left side structure has 3 unknowns (V_A, V_B & V_D), but there are only two equations available, so it is not workable.

2. Same as above, you must break up the beam at the internal hinge (point B), and start from the right side with the load directly positioned on it to find the internal force V_B, then reverse the direction, and apply V_B across the hinge as an applied load, then work out the reactions on support A in conjunction with the remaining tringle load on the left side beam segment.

Hope both of the above helps.

• Hi! Thank you for confirming my hypothesis of image 1 it seems like i was right! For 2. What technically is "stopping" me from concentrating the UDL at 1/3 of the beam from the right end and taking moments about A? I fully understand your explanation for why I must split the beam into 2 - because of shear transfers and newtons 3rd law which allows this to happen. But (3) in image 1 allows for Vc to be taken directly directly to A! It's definitely something that doesn't allow me to simplify the UDL into a point force. Mar 31 at 4:55
• I've add additional sketches on the bottom of my response. You should be able to figure it out, and answer your questions. But keep asking, if anything is unclear to you. (Note, without getting V_B first, direct working on the left side beam segment will lead to erroneous results)
– r13
Mar 31 at 7:27
• Add'l notes. For (2), on the right side beam segment, without V_B, the beam is "unstable", so V_B must be pointing up and acting as support. On the left side beam segment, the reactions at A will be wrong, because missing the effect from V_B. The V_B on the left side is in opposite direction to the V_B on the right side, so sum FY about the hinge point B equals zero.
– r13
Mar 31 at 7:55

You answered yourself. No matter how much moment is calculated on the left side of the D it has zero effect on the right side of D.

The moment does not pass through the hinge and in this case, no shear transfers from the left to the right either because the two supports A and B have taken care of the vertical forces.

Regarding image two it's just an easier way of bookkeeping of the areas and their CGs.

And the site policy is one question at a time, please.

• Hi, thank you for answering. Which part of what i said was correct? I'm also confused as to what you mean by "passing through", does this relate to newton's third law? Mar 30 at 6:20
• where you say " I do understand that internal hinges don't restrict moments and that only shear forces are transferred across" you are right. Mar 30 at 6:34