# How to calculate indeterminacy of pin-jointed frame?

I understand that to calculate the indeterminacy of a pin-jointed frame, the unknowns have to be considered i.e. the number of elements corresponding to the number of internal forces, plus the number of reaction forces.

The equations that can be written to solve for these unknowns are vertical and horizontal equilibrium at each node i.e. 2 equations per node.

So, the number of unknowns minus the number of equations results in the indeterminacy of the pin-jointed frame.

However, why do we not include the possibility of being able to write moment equilibrium as one of our equations? For the analysis of beams, the moment equilibrium equation is considered for calculating the indeterminacy of the beam.

• For moment equilibriums: it depends whether the point around which you write them is fixed or allows rotation. – Karlo Mar 20 '17 at 10:39

The entire point of a "pin" connection is that it does not resist a moment. If you move two parts connected via a pin connection, the parts are supposed to move freely, but rotate relative to each other at the pin. As such, the moments are always considered 0 at a pin connection.

Let's consider classical static analysis frames from second year. These always have the loads applied directly to the pins, never on the members. As such, the moment arm is zero, and therefore the moment equation is irrelevant. Now let's consider real world analysis, when a load is applied to the middle of a member. Since the member is considered to have pinned connections, where the moment at each pin is zero, we do not need to count on other members transmitting that load. Instead, that single member that is loaded could be treated as a pinned connection beam. Once the load is resolved, the pins still have - no moments. Again, no additional equations to resolve the indeterminacy.

Note this applies only for static frames. When frames move dynamically, the members can rotate, and this rotation is caused by moments from dynamic movements in the members.

• I fail to see the relevance of the fact that pin connections don't resist moment, and that the result there is therefore zero. As the OP mentions, simply-supported beams (i.e. beams with pin connections) can only be solved by including a moment equation precisely because the moment at the pins is known to be zero. – Wasabi Aug 8 '18 at 18:53
• I see what you're getting at - I'll update. – Mark Aug 8 '18 at 19:01
• @Wasabi - I think this may be a better explanation. I wonder if a more formal analysis was done by someone else. TO me this seems trivial but I think you're emphasis helped convey the discrepancy. – Mark Aug 8 '18 at 19:13

To be clear, just as you consider global bending moment for beams, you also consider it for trusses.

However, note that this is the global bending moment. Once you've calculated that once, there's no reason to calculate it again. Sure, you choose some arbitrary point $A$ from which to calculate $\sum M_A=0$, but the point chosen is completely immaterial: no matter which point you select, the result will be the same. Doing the work again for another point $B$ will grant you absolutely no new information, since both $\sum M_A=0$ and $\sum M_B=0$ are representing the exact same thing.

This is why, when calculating beams, we don't calculate the sum of moment around each support: if $\sum M_A=0$ is satisfied, and so are the global force equations $\left(\sum F_x=\sum F_y=0\right)$, then the global bending moment from the other support is guaranteed to be satisfied.

And so, just as with beams, you do need that global bending moment calculation. But you only need it once.

You also don't need to bother calculating the internal bending moment at any of the truss' pins because, well, pins are defined as having zero bending moment. And, fundamentally, that's why each joint's bending moment isn't counted when thinking about a truss' determinacy: determinacy is about the balance between the number of global equilibrium equations we have available and the number of unknowns in the system. The bending moment on a truss' joint isn't an unknown; it is defined as zero.

That's why the static determinacy equation for trusses is

\begin{alignat}{4} \text{if 2D: }& 2j − b &&= 3 \\ \text{if 3D: }& 3j − b &&= 6 \\ \end{alignat}

where $j$ is the number of joints and $b$ is the number of bars.

If the equation is satisfied, the structure is statically determinate. If $2j-b<3$ (or the 3D equivalent), then it is indeterminate. If $2j-b>3$ (or the 3D equivalent), then it is unstable.

The constant ahead of $j$ represents the number of internal forces existing in the structure: each joint has $F_x$ and $F_y$ (and $F_z$ in 3D). The more joints you have, the more internal forces you need to solve for. Sure, each joint also has $M_z$ (and $M_x$ and $M_y$ in 3D), but we know it's zero so we don't need to solve for it.

And then why is the right-hand side equal to 3 (or 6 in 3D)? Well, that's precisely because that's how many global equilibrium equations we have: $\sum F_x=\sum F_y=\sum M_z=0$ (and $\sum F_z=\sum M_x=\sum M_y=0$ in 3D). So there's the global moment equation you were looking for.

For a stable and determinant truss we can say that:

(# External Reactions) + (# Members) = 2*(# Joints)

Why we only have two equations per joint and not three (i.e. why we ignore moment) is largely due to the assumptions we made when dealing with a truss to begin with.

The typical assumptions are:

1. Members are straight and carry only axial load
2. Joints are frictionless pins (cannot carry moment)
3. Loads are applied only at the joints

As a result of these assumptions, each joint represents a concurrent force system. That is, all forces pass through a common point. If we consider our handy statics potato, we can see that this mean moment equilibrium is automatically satisfied. We can only write the trivial equation ‘sum of moments = 0’. None of our unknowns of interest participate in this equation and therefore the ‘sum of the moments’ equation cannot help us solve for our member forces. Therefore we only have two equations available per joint.

Going back to that initial equation then. When we say “determinant” all we mean is the structure under consideration can be solved with the equations of statics.

For this, we must have:

Number of things we want to solve for = Number of equations we can write

We’ve established that the number of equations we can write = 2*No. Joints. The things we want to solve for are the external reactions and the internal member forces (axial only because of our initial assumption).

• +1 for "Handy statics potato". That's made my day. – Mark Aug 9 '18 at 14:46