# Statics Two Force Member Problem

I was given a challenge question from my teacher that I need some assistance in solving. The problem is to merely identify the two-force members. I understand the ideas of how to determine a two-force member (colinear forces, equal magnitude, etc.), but in this problem, they just aren't clicking for me. Thanks in advance for the help!

• At a glance, I want to say... none of them? Like, the double-box at the top is useless and won't have any internal forces (it'll simply allow for rigid body rotations), but I wouldn't consider those to be two-force members. And the rest will clearly suffer bending moments, and therefore won't be two-force members either. I'm low on time now, but will convert this to an answer later.
– Wasabi
Oct 9, 2019 at 1:28
• Yeah, that's what I was thinking about the box in the top right. But I was thinking that A-F might possibly be one. Or possibly some member with one of the pins as an end might be a two-force member. Oct 9, 2019 at 1:36

## None of these elements are two-force members

Before anything, it's worth remembering what two-force members are: bars which only have two equal-and-opposite colinear forces applied, one at each extremity. Given these conditions, we also know the forces must be purely axial, with no shear component. Given they have no shear component, they'll also suffer no bending moment.

Now, it's worth mentioning that there are actually two ways of looking at that definition. One is strictly based on the structure itself, while the other considers the applied loading. The simple case which exemplifies the difference is a cantilever beam with only an axial force applied.

Is that beam a two-force member? If you only look at the structure, then no. It's obvious that under basically any other loading condition, this beam would present shear and bending moment. But if you only think of the currently applied forces, then yes: it only suffers equal-and-opposite colinear axial forces.

I'm of the opinion that the best way to think about it is by only looking at the structure: if you can come up with a classic loading pattern which generates bending moment, it's not a two-force member1.

So let's look at the structure ignoring its loads:

The bottom beam AF is basically a simply-supported beam with a cantilever. If you put a vertical load anywhere along that span, the whole thing'll have shear and bending moments, so not a two-force member.

The bend which connects AF to the rest of the structure is obviously not a two-force member: axial loads aren't collinear. And even if you consider that as two separate members, their shared node is fixed. Therefore, any deflections will cause relative rotations and therefore bending moments.

As for the "box" at the top, well, that's also a bunch of fixed beams. Any loading in that region will lead to bending moments all over the place.

For the record, for those who prefer to consider the given loading, the same conclusion applies:

That's the bending moment diagram (ignore the actual values, I got lazy with the loading inputs). As we can see, all relevant members have bending moment and the box is unloaded (which in my book doesn't count as being a two-force member).

1 For the pedants thinking of truss members with uniform loads applied, just remember the "classic loading pattern" for truss members is for uniform loads to be replaced by nodal forces. And nobody likes a pedant.

• Awesome! Thanks for the answer, makes sense. From what I gather, there can be two ways of looking at two-force members. If we take into account the loading then there will be no two-force members. But you're saying that if you only look at the applied forces there would be a/some two-force member(s)? Oct 10, 2019 at 1:40
• @ThatOneNerdyBoy no. I'm saying that if you look at the structure in isolation, without the specific loading applied, you'll see that all the beams are liable to suffer bending moment (and therefore wouldn't be two-force members) under trivial loading conditions. And, that even if you look at the specific loading given, the above still applies: all the beams suffer bending.
– Wasabi
Oct 10, 2019 at 2:11
• Ok, I understand now! Thanks for the help! Oct 10, 2019 at 11:55