Which ones are zero force members?

[Kristofeles]

Any help please.
Picture in attachments

 

[Brian_C]

Are we supposed to do the work for you?

 

[Kristofeles]

Haha. No. I have to find the load in each member. I know some of those members are probably zero force members but which ones?

 

[Kristofeles]

LN is a zero force member?

 

[Studiot]

Look again at joint L

If LN is redundant then so is LC.

What makes you think this framework is redundant?
What do you know about the conditions for static indeterminancy?

 

[Gannet]

The members that are perpendicular to the slope are zero force members, because the pin joints must be in equilibrium and zero moments

 

[Studiot]

The members that are perpendicular to the slope are zero force members

Do you mean like LC?

Look again at joint L

 

[Gannet]

Do you mean like LC?

Look again at joint L

No member LC is not a zero force member

The members that are zero force members are: BM, DN, FO, and HJ; because at pin joints B, D, F, & H these members cannot have load because there would be no way of obtaining force equilibrium at these pin joints.

 

[Kristofeles]

Okay. I will double check with my teacher. Thanks

 

[PhanthomJay]

No member LC is not a zero force member

The members that are zero force members are: BM, DN, FO, and HJ; because at pin joints B, D, F, & H these members cannot have load because there would be no way of obtaining force equilibrium at these pin joints.

That would be the case only if there were no load applied at B, D, F, & H. Since there is a load applied at these joints, these are not zero force members.

 

[Gannet]

That would be the case only if there were no load applied at B, D, F, & H. Since there is a load applied at these joints, these are not zero force members.

You are correct Jay I didn't notice the external loads at Joints B & D; Then, the only zero force members are FO and HJ. Thanks Jay

 

[Studiot]

Then, the only zero force members are FO and HJ. Thanks Jay

I did pass comment on your last effort, which you seem to have ignored. The same comment applies again.

If the force in HJ is zero then, by vertical equilibrium of joint J, GJ is also zero.

I ask again why do you think there are any redundant members.

and again

What do you know about the conditions for static determinancy/ indeterminancy?

 

[Kristofeles]

I did asked my teacher and he said that there is no zero forces on this structure

 

[Studiot]

Kris I was referring to Gannet.

I hope your teacher also told you the formula for a perfect frame (statically determinate, no redundant members) - which this is

m = 2j -3

Where m = number of members
j = number of joints.

You can work out for yourself that this frame conforms.

I leave the task of calcualting the member forces to you - thre are many methods available.

 

[nvn]

It currently appears the zero-force members are EO, FO, GJ, GK, GO, HJ, and KO.

 

[Studiot]

It currently appears the zero-force members are EO, FO, GJ, GK, GO, HJ, and KO.

This false statement helps the student in precisely what way?

 

[Kristofeles]

Probably he just don't want us to include the zero force members.

 

[Gannet]

Kris I was referring to Gannet.

I hope your teacher also told you the formula for a perfect frame (statically determinate, no redundant members) - which this is

m = 2j -3

Where m = number of members
j = number of joints.

You can work out for yourself that this frame conforms.

I leave the task of calcualting the member forces to you - thre are many methods available.

Because of the question, I was assuming that Kris was just starting "Theory of Structures" and was using Method of Joints for analyzing Statically Determinant Truss Structures. If Kris is analyzing Statically Indeterminant Truss Structures then he would be using Deflection Methods.

Kris, I am sorry that I mislead you because I missed the two external loads at Joints B and D, the sketch is really light.

 

[Studiot]

Certainly an idea of what you are studying and at what level would be useful to pitch responses and explanations to.

 

[PhanthomJay]

Kris, I am sorry that I mislead you because I missed the two external loads at Joints B and D, the sketch is really light.

So light, in fact, that the loadings on the right side of the frame appear to be missing. If the frame is symmetrically loaded, there are no zero force members; if the frame is loaded only on the side left of center, then there are many zero force members (see nvm's response). It appears there is a 24 unit force reaction at the left support, implying that the frame is symmetrically loaded,and that the applied joint loads on the right side of the frame are missing.

 

[Studiot]

I had assumed the following

1) The truss was symmetric and symmetrically loaded. Because of symmetry there was no need to replicate on the right hand half so it was 'greyed out'. this is common practice.

2) The loads were purlin loads and perhaps snow loads.

3) No wind loads were considered.

It should be pointed out that some inaccurate statements have been made about the conditions for recognising redundant or 'zero force' members. In particular the fact that a member joins at right angles is no guarantee of redundancy.
In a plane truss if both diagonals in a quadrilateral are present and crossing one will be redundant.
But the only guaranteed method is to apply the formula or work it out from first principles.

 

[PhanthomJay]

I had assumed the following

1) The truss was symmetric and symmetrically loaded. Because of symmetry there was no need to replicate on the right hand half so it was 'greyed out'. this is common practice.

Might be common practice, but that intrinsically is not made clear.

2) The loads were purlin loads and perhaps snow loads.

I thought they were tiny birds perched on the joints. Is there a difference in the analysis if the loads come from birds or purlins?

3) No wind loads were considered.

I didn't see any lateral loads either, but the print was so light, maybe we missed 'em...

It should be pointed out that some inaccurate statements have been made about the conditions for recognising redundant or 'zero force' members. In particular the fact that a member joins at right angles is no guarantee of redundancy.
In a plane truss if both diagonals in a quadrilateral are present and crossing one will be redundant. But the only guaranteed method is to apply the formula or work it out from first principles.

but a truss can be statically determinate and still have zero force redundant members

 

[Studiot]

Please help me here Jay.

Why the sarcastic comments?

Since what is obviously some form of coursework is posted in the general engineering section I assume the originator is a budding engineer or architect.

Having responded in a way intended to help the OP, and others viewing this thread, a better understanding of his subject in general and this problem in particular we resolved the issue at hand between all of us.

There were some incorrect statements made by some and I did not want to let these lie to confuse future viewers.
It is helpful to learners to experience and assimilate real action so I posted my outset assumptions, provided a few pointers that a keen student might follow up, and corrected the incorrect.

It is open for any to ask for more explanation or seek it out themselves.

And yes a manifold redundant truss may be statically determinate or indeterminate. I never said otherwise.

If you are short of threads about structural engineering how about adding some help in this one?

https://www.physicsforums.com/showthread.php?t=382150

 

[Redbelly98]

Moderator's note:

Thread moved from "General Engineering". In the future, please help out by reporting misplaced homework threads.

Thank you,

Redbelly98
Physics Forums Mentor

 

[pongo38]

It is quite common for beginners to mistakenly call zero-force members 'redundant', not realizing that there is a special and different meaning of that term in relation to statically indeterminate structures. However, Kristofoles needs the encouragement to look at the equilibrium of joints F and H to start with, not using horizontal x and vertical y axes, but u and v inclined axes. I'm afraid I disagree with studiot on one count. One is that m = 2j-3 is a dangerous 'rule' because there are mechanistic (deficient) structures that satisfy this expression.

 

[Studiot]

Good heavens this is a bog standard question which is discussed at great length on pages 55 - 57 of Whitlow.

If you come across an unfamiliar configuration ( = a mechanism) it pays to understand how the formula arises so it can be dealt with from first principles.

Ask yourself why are these zero force members included in a design? They are certainly redundant when they carry nothing.

The answer is that in the real world the loading diagram provided by Kris would only be part of the story. Real roof trusses have to suport load variation, point loads for maintenance, loads for services and ceilings etc etc. And of course the joints are not actually pin joints.
So how often is zero actually achieved?

But Pongo you are right it is possible to construct academic counter examples that break the rules.

 

[Redbelly98]

Moderator's note, read this before posting further:

The usual rules on homework help apply in this thread, and the OP has not shown any of his/her work at this point.

Kristofeles, just how far have you gotten with solving this problem?

Anybody not familiar with PF guidelines should read the 3-paragraph section titled Homework Help at

https://www.physicsforums.com/showthread.php?t=5374​