Identification of zero-force members

[ChloeYip]

https://drive.google.com/file/d/0B39n6QfDU8f3SE9nUHhHckRIRFE/view?usp=sharing https://drive.google.com/file/d/0B39n6QfDU8f3SE9nUHhHckRIRFE/view?usp=sharing

1. Homework Statement
I am not sure about the concept of zero-forced members.
There are forces applied on the members, why are they zero-forced members? Or it seems no force in implied on them why are they non-zero-forced members

Homework Equations

the method of joints

The Attempt at a Solution

If two non-collinear members are connected to a joint that has no external loads or reactions applied to it or three members, two of which are collinear, are connected to a joint that has no external loads or reactions applied to it, it is zero-forced members.

Thank you.

 

[haruspex]

If two non-collinear members are connected to a joint that has no external loads or reactions applied to it or three members, two of which are collinear, are connected to a joint that has no external loads or reactions applied to it, it is zero-forced members.

Is that a rule you have been given, or one you have thought up?
If you look at the second diagram, the first part of that rule appears to say that the two diagonal members are unforced, which is not true.
Anyway, such a rule will not help with the third diagram since it depends critically on the two F forces being equal.

In the first two diagrams, imagine removing the vertical strut. What to you think would happen in each case?
(Note: it appears that the midpoint at the bottom of the triangle is a joint, so there are 5 struts altogether.)

 

[ChloeYip]

Is that a rule you have been given, or one you have thought up?

It is given by the powerpoint given by my instructor, who have given no explanation for the part...
After watching ,
Do you agree with this definition rather than the one given by my instructor?
Is that all the zero forced members in first 3 diagram & member 6+7 are because there are truss joint not support by pin or roller and not carry loads?

it appears that the midpoint at the bottom of the triangle is a joint, so there are 5 struts altogether

Thanks for reminding me on this. For 2nd diagram, after removing the vertical strut, the 2 bottom struts shouldn't have component to support y-axis force, thus the structure will fail, am I right?

However, I still don't understand the third diagram... (member 3,4,5,8,9)
Thank you for helping me.

 

[haruspex]

Ok, that video helps.
In the first part of the rule in your initial post, you need to clarify that it refers to exactly two members meeting at a joint.
In the second part, the two collinear members may be under tension. It is only the third member that is sure to be zero force.

In the third diagram, you can use the rules (from the video) to eliminate three struts. Which three? The other two can only be solved by analysing the forces. That is because if the applied forces were a bit different then those two would be carrying forces.

 

[ChloeYip]

In the first part of the rule in your initial post, you need to clarify that it refers to exactly two members meeting at a joint.
In the second part, the two collinear members may be under tension. It is only the third member that is sure to be zero force.

I understand this part now. Thanks.

In the third diagram, you can use the rules (from the video) to eliminate three struts. Which three? The other two can only be solved by analysing the forces.

However, I am sorry but I don't really understand what you mean by here. Which three and which two?

If you look at the second diagram, the first part of that rule appears to say that the two diagonal members are unforced, which is not true.

That is exactly what I feel like for member 3 and 4... Which I think it is not correct. Why is there a pin support exert on member 3&4 and force applied on 4&5 but still they are zero force member?

Thanks.

 

[haruspex]

Why is there a pin support exert on member 3&4

Quite so, which is why the rules you have do not eliminate 3 and 4. As I posted, there are two members which will turn out to have zero force, but you cannot eliminate them by the rules. Those are the two. If you were to change one of the Fs to 2F, say, they would not be zero force.

force applied on 4&5

Watch the video again. There is a similar situation in one of the diagrams analysed there. You can invent an extra strut above the joint where 4 and 5 meet that carries the force F. That leaves strut 5, meeting the two horizontal struts at the bottom, satisfying the the three strut rule.
It might help you if you understood the logic behind these rules. If there were a force in strut 5 then at its joint with the two horizontal struts there would be no force capable of balancing it, so the joint would move. The rules assume the system is in equilibrium.

 

[ChloeYip]

Do you mean the strut within the rectangle 5,6,7,8,9 are already balanced, so strut 3+4 are zero-forced?
Strut 5 is because the bottom is connect to horizontal joint which have no support or load, therefore it is zero forced too?
But why is strut 8 is still non-zero forced? Isn't strut 9 already can balanced the y-component of "the strut of diagonal of force and roller support"?
Thanks.

 

[haruspex]

the strut within the rectangle 5,6,7,8,9 are already balanced, so strut 3+4 are zero-forced?

Yes.

Strut 5 is because the bottom is connect to horizontal joint which have no support or load, therefore it is zero forced too

Yes.

why is strut 8 is still non-zero forced?

You are right, 8 also carries no force.

 

[ChloeYip]

Thank you very much~