Why is this a two-force member?

1. Oct 16, 2011

eurekameh

http://imageshack.us/photo/my-images/855/unledvny.png/

It is saying that member AC and member DC are both two force members. However, isn't there a downward applied force of 600 N acting on both of these two members as well?

2. Oct 16, 2011

lewando

Considering the definition of a two-force member, AC and DC are both pinned at each end and there are no forces in between the pins, so they are both two-force members.

3. Oct 16, 2011

eurekameh

The 600 N force is acting on both members, no? I don't intuitively see it just acting on just the pin.

4. Oct 17, 2011

lewando

You could say the 600N force is acting on both members if you want. It results in tension in AC and compression in DC, but it is acting only at the pin as indicated in the diagram (which sums everything up quite nicely). If you don't see it acting at just the pin, where else do you see it acting? Are you taking exception with some aspect of the diagram?

5. Oct 17, 2011

eurekameh

Without the FBD's, I wouldn't think it would be a two force member if I was only given the diagram of just the setup and if I was asked to draw the FBD's myself, I would draw the 600 N force acting on both members, not just on the pin.

6. Oct 17, 2011

lewando

Could you then please draw your take on the FBDs to enable further exploration of your reasoning?

7. Oct 17, 2011

eurekameh

http://imageshack.us/photo/my-images/197/unledzvh.png/
Drawing it, I can now see why it is a two-force member, because it still has two points where forces are applied. If the 600 N force was acting somewhere other than the pin on each of the two members, then it wouldn't be a two-force member.
However, I'm still confused if the 600 N force is directly acting on member AC and DC or just the pin.

8. Oct 18, 2011

lewando

Don't forget to draw a FBD for the pin, it is an essential element of the system. When you say you are applying the 600N force to C, you need to be exactly clear on how you are doing this. You can attach the 600N force to one of the three elements, the "C-end" of member DC or to the "C-end" of member AC or the pin at C. These are 3 unique locations the the resulting FBDs should reflect it. I would think attaching the force to the pin would simplify things.

9. Oct 20, 2011

eurekameh

Are you saying that we have a choice in attaching the 600 N force to which ever element we want? If so, then wouldn't the "two-force member" definition be relative and dependent on whereever you attach the forces?
Let's say I attach the force to member DC. Would this not be a two-force member then?

10. Oct 20, 2011

lewando

I think you are having some confusion about the definition of a 2-force member. This is my informal working definition--pins at both ends: check, no (external) forces in between the pins: check. Those two conditions are all you need to identify the member as a 2-force member. You can have 1, 2, 3 or N forces applied at the pins, but the neat thing is, they all resolve into 2 co-linear forces (acting along the line between the pins) putting the member into compression or tension.

Last edited: Oct 20, 2011