3: Forces in Statically Determinate Trusses

3.2: Determinacy and Stability

3.2.1: Determinacy

If there are $m$ members in a truss and $r$ independent reaction components, then the total number of unknowns to to be determined is $m+r$. We will see later that the number of independent equations of equilibrium available to solve for this is $2j$, where $j$ is the number of joints in the truss.

One requirement for stability is that

\[m + r \ge 2j\]

As usual, that is a necessary, but not sufficient, for stability.

When $m + r > 2j$, then the degree of statical indeterminacy is

\[m + r - 2j\]

3.2.2: Stability

In the worst case, it can be very difficult to tell by observation whether a truss is stable or not. We will start by learning to recognize some special configurations, and at the same time introduce a classification system that can help.

Simple Trusses

A simple truss is one that can be built up starting from a triangular arrangement of 3 members and 3 joints, such as that shown in Fig. 3.2.2-1.

Figure

Figure 3.2.2-1: Triangle (base truss unit)

The truss is to be built up in steps by adding 2 members and a joint at each step, as shown in Fig. 3.2.2-2. The original triangle is stable, and if, at each step, the two new members are not co-linear, the result will be stable as well.

Figure

Figure 3.2.2-2: Simple truss built from base unit

The base triangle has $m=3$ and $j=3$. The minimum stable support set consists of a pin and a roller, so $r=3$ for the case of a simply supported base. Overall, we see that $m+r=2j$ and thus the triangular base unit is statically determinate when simply supported (by a pin and roller). As we add 2 to $m$ and 1 to $j$ at each, step the equality holds and the resulting truss remains statically determinate.

Simple trusses are stable internally and statically determinate. If the reactions are stable, they are stable externally as well.

Recognizing simple trusses is a good first step in evaluating stability. For example, see Fig. 3.2.2-3 showing a simple truss, such as might be used for a small bridge. We note that it is supported by a pin and a roller, thus ensuring that it is externally stable.

Figure

Figure 3.2.2-3: Simple truss

To determine that it is a simple truss, and therefore stable, we imagine the construction of that truss, in steps, starting from a base triangle. One possible set of construction steps is show in Fig. 3.2.2-4. That figure shows the construction of the truss adding 2 members and a joint at each step.

Figure

Figure 3.2.2-4: Truss “construction”

Because the result is a simple truss, and because it is simply supported, it is both stable and statically determinate.

Consider the truss of Fig. 3.2.2-5. Is it stable and statically determinate? Because we can locate a starting triangle and can see how the truss is built up 2 members and a joint at a time, the answer to both questions is “yes”.

Figure

Figure 3.2.2-5: Simple truss.

Compound Trusses

Compound trusses are two or more simple trusses joined by linking members. It can sometimes be quite difficult to tell by inspection whether the resulting truss is stable or not (you can always judge determinacy by counting the members, joints and reactions).

Consider the compound truss shown in Fig. 3.2.2-6, part a). For this truss, we see that $m=16$, $r=4$ and $j=10$. Therefore $m+r=2j$ and the truss is statically determinate, if it is stable.

Figure

Figure 3.2.2-6: Compound Truss

To judge stability, one tactic is to identify the simple parts; we know that those parts are each stable (i.e., rigid). These are identified in part b) of the figure. We then try to imagine consistent small geometry changes; if we can find one, then we know the truss is unstable.

On this truss, we note that the left rigid unit is supported on a pin and a roller, so therefore the left portion of the structure is stable; no geometry changes are possible in that portion.

We can see that the two horizontal links between the two parts prevent sideways displacement of the right portion. The roller at c prevents vertical displacement. In addition, the top horizontal link and the roller at c prevent rotation about c (the top link would have to change in length to allow that). The bottom link prevents rotation about the right end of the top link. So, under these support conditions, the truss is stable.

What would happen if the roller at c were was moved to one of the other pins on the right rigid unit, say to d, as seen in Fig. 3.2.2-7?

Figure

Figure 3.2.2-7: Inconsistent Displacements

In this case, the horizontal links will still prevent a lateral displacement of the right portion. Now the pin at c may move vertically. For that to happen, both links must rotate counter clockwise about their left ends, as shown in the figure. If that happens, the vertical member above c will move straight up, remaining vertical. The only way that is possible is if the whole right portion also moves straight up, with no rotation, also as shown in the figure(it is rigid, after all). But that is inconsistent with the prevention of vertical displacement by the roller, now at d, and is therefore not possible.

The truss is stable. In fact, a constraint against vertical displacement anywhere on the right portion will render the structure stable.

Compound Trusses - Common Forms

We finish this section by illustrating a few common forms of compound trusses.

Figure

Figure 3.2.2-8: Compound Truss

Figure

Figure 3.2.2-9: Compound Truss

The truss shown in Fig. 3.2.2-9 is particularly attractive for use in large roof trusses.
Each portion can be assemblied in the shop, trucked to the site, and placed by making a field conenction at the top and inserting the tie. And yes, it is really a simple truss rather than a compound truss by our criteria; we mention it in this section because of its normal fabrication practice.

Complex Trusses

Trusses that cannot be classified as either simple or compound are called complex. A couple of examples are:

Figure

Figure 3.2.2-10: Complex Truss

Figure

Figure 3.2.2-11: Complex Truss

It can be very difficult to tell by inspection whether a complex truss is stable or not. For example, in Fig. 3.2.2-10, $m=11$, $j=7$ and $r=3$ and in Fig. 3.2.2-11, $m=9$, $r=3$, and $j=6$ and so $m+r=2j$ for both. While the supports and connectivity are the same in both, the truss in Fig. 3.2.2-10 is stable and the one in Fig. 3.2.2-11 is not.

A truss that is not stable must undergo small geometry changes in order to equilibrate with external forces, and is said to possess a “critical form”.

A "critical form" is a truss arrangement that is unstable.

Critical Form of a Hexagon Truss

To demonstrate a procedure that can be used to determine if a truss has a critical form, consider the regular hexagon truss shown in Figure 3.2.2-12. Our job is to determine if small geometry changes can be made to this arrangement. In other words, can the joints move while retaining the member lengths of all the members and also maintaining all joint and support constraints?

Start by observing that joint a is prevented from moving in any direction. Similarly, joint f is prevented from moving vertically by the support and horizontally by member af, so it is fixed in location as well.

Figure

Figure 3.2.2-12: Hexagon Truss

We then examine the ends of the members attached to those fixed support joints - these are members ab, ad, fc and fe. We sketch paths along which the joints at the other ends can move (joints b, c, d and e). This is shown in Figure 3.2.2-13. The joints at the free ends can move along paths that are perpendicular to each of the members; the path through b perpendicular to ab, etc.

Note that in the truss, all internal angles are 60 degrees, so members ab and fc are parallel, as are members ad and fe. That means the paths at joints b and c are parallel, as are the paths at joints d and f.

Figure

Figure 3.2.2-13: Paths of “Free” Nodes

The first step in imagining the displacements is to choose one of the free joints, say joint b, and draw it moving along its path. Also show the member attached to both b and the fixed joint a. This is shown in Figure 3.2.2-14. Plainly, ab does not change in length.

Figure

Figure 3.2.2-14: Step 1

Now, in step 2 shown in Figure 3.2.2-15, investigate some of the other members attached to joint b. If member be is not to change in length, then joint e will move leftward the same amount as joint b. If joint e moves to the left the same amount as joint b, it will move up the same amount as b moves downward and member be will not change in length. Member fe will retain its length as well as it follows joint e. So far, all displacements are possible.

Figure

Figure 3.2.2-15: Step 2

In step 3, we move joint c along its path the same amount as joint b, as we see in Figure 3.2.2-16. Member bc remains parallel to its original location, so neither it nor member cf change in length. Joint d and members de and ad are similar.

Figure

Figure 3.2.2-16: Step 3

Finally, the last step is shown in Figure 3.2.2-17. We have one remaining member to fit: cd. Joints c and d plainly displace horizontally the same amount. As these are very small displacements, the distance c’d’ is the same as cd and so member cd does not change in length.

Figure

Figure 3.2.2-17: Step 4

We have determined that we can make a small, consistent set of displacements in the truss. Therefore, it is unstable, and the original regular hexagon is a critical form.

In the example above, the three long members cross in the middle and are not joined. If, instead of crossing, the members were joined with a normal pin joint at the centre, as in Figure 3.2.2-18, the truss would be stable because it would be a simple truss. But this would add 3 to $m$ and 1 to $j$ meaning that $m+r > 2 j$ and the truss would be statically indeterminate to 1 degree.

Figure

Figure 3.2.2-18: Stable version of hexagon truss

Example of Non-critical Form

We will now illustrate the above procedure of manually attempting to find a critical form for a stable truss. Consider the complex truss show below in Figure 3.2.2-19 (a repeat of a truss shown earlier).

Figure

Figure 3.2.2-19: Complex Truss

The first step is to sketch possible paths of joint displacements as in Figure 3.2.2-20. Here we note that there are two triangles: a-b-g and e-f-g, each attached to a support. The only possible displacements for joints b and e are horizontal and for joint g vertical, corresponding to rigid-body rotations of the triangles about supports a and f.

Also shown are possible paths for joints c and d, perpendicular to members fc and ad respectively (which are constrained to rotate about non-movable supports a and f)

Figure

Figure 3.2.2-20: Step 1 - Possible Paths

Figure 3.2.2-21 shows step 2, imagining the rotations of the two triangles.

Figure

Figure 3.2.2-21: Step 2

Step 3 in Figure 3.2.2-22 is to imagine the consistent displacement of members bc and de. c and d must follow their paths, with the result that both c and d must move upwards and inwards toward the centre-line of the truss.

Figure

Figure 3.2.2-22: Step 3

Figure 3.2.2-23 shows what would have to happen at step 4, showing that member cd would have to shorten, as the paths at the ends through joints c and d are not parallel.

Figure

Figure 3.2.2-23: Step 4

That plainly is not possible and so the displacement pattern we attempted to show is not valid.

As the only prossible displacements must involve the rotations of triangles a-b-g and e-f-g, we see that it is impossible to find a consistent displacement pattern.

Therefore the truss is stable.

If no small displacement pattern is possible, a truss is stable.