Great to hear it... good luck!!

What a coincidence! I also have a final that counts for 50% of my grade tomorrow. But it's one year later.

@@isisyasmim639 omg same, but 3 years later

@@luisanunez6654 the cycle continues

@@isisyasmim639 I have my exam on monday :SSS im not that exited about that :D

🙂 Glad to be of service!

Well, that’s coz you probably weren’t focusing in class and gave your 100% in the video. Not spreading hate but explaining why it happens with most people, incl me.

Ayoooooo

@C C 😂

Get out!

I wouldn't trust this guy to design a thing!!! It's this kind of thinking that causes bridges to fail!!! Not to mention all the other things we buy that fail in short amount of time!!!

@@rockerpat1085 Have you ever worked in structural projects? Just asking

Cool, thanks for letting me know!!! =)

Glad to hear it! After all these years the principles still apply haha 😋

Ahahahaha nice one!!! :P made me laufgh i like pun sorry for bad English

😂 😂 wow I would have been impressed with just the first few.. Well done!!! hahaha

Thanks your comment makes a lot of sense

ZFMs play an important role in stability, bracing,m etc. If the load changes, the ZFMs change, so in real life, don’t expect to have a single pint load acting on your structure! Also in this video when I’m erasing the ZFMs, I’m not actually removing them, they are still very much there on the structure. I just erase them to make it easier to spot the other ZFMs in this loading scenario. 🍻

and also in reality the members themselves have weight which would act perpendicular to the horizontal plane. this simplifications are made based on the assumption that the members are weightless so that the understanding of static forces in truss members become easier on engineers. cheers!

@@khiareozmakhiar3722 another assumtion is, that the members/struts are non-elastic and the one you mentioned, that the members are only able to apply/transport force in its tangential direction.

@@emmata98 exactly. elasticity of members could change the statics problem into a strength of materials one where you should use strains and deflections to calculate the correct forces.

Good luck! Make sure to check the rest of the videos at engineer4free.com/statics too 🎃🦇🕸️

@@Engineer4Free been 4 years and u still replying?

Yesss. I'm still trying to respond to every question on every video. 500+ videos and the oldest ones are nearly 9 years ago by now. The notification system is bad for comment replies tho so I still miss a lot, but I try! 🤜🤛

@@Engineer4Free i love your stuff! really easy to grasp and overall just rly helpful. thank you!!

Awesome!! Glad to hear it 😊😊

You’re not alone, glad I can help!!!

@@Engineer4Free I never understood anything in school. I went to KZfaq Univesity and paid an institution for a paper that says I have a degree.

​@@laurap.607 Yes the whole system is so messed up. I also paid an institution for a degree because that's just what you do, but in reality I learned almost everything online too or just from reading old used textbooks.

Agreed

Google has probably been using your smartphone's camera (without your permission) and took some pictures of trusses or maybe some trusses appeared in one of your Web searches. Either way you have NO PRIVACY when you use Google.

@@abrahkadabra9501 lol paranoid much

Thanks LARKZ 🙌

Good luck friend!

Glad I can help!! It shouldn't be complicates, but sometimes it seems that way when it's first taught in a lecture!!

I also found kzfaq.info/get/bejne/qJiHgppn2pm6l2Q.html video useful to clear concept on zero forces

Thanks!! Hope it went well 🙂🙂

...still waiting

did you pass

Yup, don't over think it!

Yep. Also, if the loading changes, the internal forces in each member will most likely change, so a ZFM will not always be a ZFM. Also in real life, the loading is not so simplified, nor is the structure. But good to learn them this way to train the brain!

hahaha how'd it go???🤘

Hi George save yourself a lot of time and money and build your roof trusses with best practice building standards related to specific areas conditions ie Cyclone area etc

This video is for theoretical use. Loads on a trusses are never as simple as one load in one point as shown. Use your local building code to design your trusses.

@@NathanNostaw AHHHHH!!!!! the good old Theory .... Well that is exactly what it is THEORY like religion ... Theory

Thanks!

Awesome!!! hope it went well =)

Thanks Ryan, glad my video is getting the job done 🙂. And no, it is not a ZFM. If it were the only member with a vertical component at the bottom joint, then it would be ZFM, but the non-ZFM diagonal member there as well gives an equal/opposite vertical load to it, at that bottom joint. Hope that makes sense.

Engineer4Free ah yeah it does, cheers

+Érika Paulus glad that I was able to help! Make sure you check out the rest of my free courses at engineer4free.com :)

Thank you my friend!! 😁😁

Glad you liked it! 😊

Thanks Tomasi! 🙂

+Taufiqmusic thanks for the kind words, and hope your finals went well!

Yes very true. When I'm erasing the members, I'm just doing it to visualize which members actually need to be analyzed for the current loading. You're right, we can't just remove it for realz. Sometimes people thing ZFMs might as well not be there, but they serve a purpose like you mentioned, of another example would be if the loading changed, the ZFMs would most likely change too!

Yes. Also adding gravity; member section and material properties induce displacement displacement cause ZFM to be non zero and recieve load depending on fixity of joints etc.

How'd it go?

Mine in...1Hr... Just clearing some last minute doubts

Rutik Rojekar How’d it go?

anthoty looks like it probably wasn’t good.

How did it go?

ChickenSmackBoy remember trusses are modeled as pinned connections, so although there may be an external force applied at the top of that vertical member, at the bottom there are no members to resist that force. forces can't act through 90° and so the vertical member must have 0 forces because of equal and opposite reactions. sidenote: if the bottom chord of the truss were to strain to an excessively large amount due to tension forces, that zero force member may actually start taking on a little bit of compression. also, remember that in the real world, beams and chords can only span a certain distance before they start to sag due to their own weight, so that zero force member may be carrying the dead load of the bottom chord.

Change ur view buddy......just forget about the loads , members .... Just concentrate on the joints.......if any external force is not there on that joint then do as shown.

@@JjoshD in a practical application, house roof trusses made from 2x4 lumber seldom have spans between connections that are more than 8'. The video addressed point load condition. Dead load and live load values lead to the familiar truss

Its hard to belief, but the system is perfect so to speak. Just 0,0001 + or (-) degree the rollers would rush up or down in real life 😊. That’s why we need zero force members. Out there. ....

Same question. Probably dynamic analysis including integrals.

Thanks!! 😁

Glad it helped!!! More at engineer4free.com/statics =)

🤠

Totally. Thanks for contributing. The truss in this video and the rest of the examples of the trusses section of engineer4free.com/statics are simplified. We don't consider them to have any self weight, they can only support axial loads (no lateral), and are purely pin jointed. Real trusses are jot like this, but this is how we introduce them at the introductory level I statics. Buckling, self weight, racing, and all the things you mentioned come up later in studies of structural engineering, and are very real considerations in real life! 👌

aayyyyy thanks for the compliment haha. Welcome aboard!

Haha, good luck!!

You’re welcome Jackie!! 🙂

You’re welcome Jackie!! 🙂

True that if you were to remove that member from the truss in real life the truss would no longer be rigid. But in this idealised case, where the truss members are assumed not to be exerting any force themselves, in this configuration that particular member can't be exerting any force, because the only force exerted on the bottom end is at right angles to that member (because of the roller). So for the purposes of eliminating members which don't matter in this particular configuration in order to simplify the calculations for determining the forces in the remaining members, you can ignore it. If the load on the truss were different, and a vertical force were applied to the bottom end, then it would become important and in that situation you would be obliged to leave it in. Which members are zero force is highly dependant on the load applied to the truss, when you 'remove' them the equivalent truss you get is only equivalent in that specific situation. I hope that makes sense!

Yeah thanks for that Tom! And just to clarify, when I am erasing members, I’m not “removing” them from the structure. They are still there for stability etc. I just erase them to visualize easier which other members will also be ZFMs

Yes I have, please see videos 42 - 51 here: engineer4free.com/statics

@@Engineer4Free Okay!! Thanks a lot!

🤜🤛

Thanks and you're welcome!!

The 3rd case in this video is not an L shaped member, it is two straight members connected with a pin. A curved or irregular shaped member can be a two force member as long as only 2 forces are acting on it and with their lines of action being the same. That would be for any shaped member with 2 pin connections. Trusses are a special case of two forces members in that they are straight ad considered to only have axial forces. So if there are two truss members that are connected like in the third case, then for the structure to be in equilibrium at that joint, neither may have an internal force making them ZFMs. If they did have a non zero axial force, they would not net to zero because the line of actions in one member is not in line with the other, and the joint woukd accelerate in the net force direction. Hope that clears it up.

You're welcome Zachary!! Check out engineer4free.com/statics for the full playlist if you haven't already!

@@Engineer4Free nothing like 2 weeks before exam refresher

That member is determined to be ZFM exclusively by what's going on at its top right joint. That joint must be in equilibrium, and because 2 of the nonZFM the members are colinear, the third member in question must be ZFM, otherwise the joint would accelerate inline with that axis. Once we determine this member is ZFM we can realize it will not put any force on the other joint, so we can erase it temporarily to continue the analysis. Hopefully that clears it up, I'm not 100% if I answered what you were asking

Awesome, glad to hear it John!! =)

Glad I could help =)

Yes that would have been far more accurate to say, I slipped up in the word there. Thank you for pointing it out and contributing to the discussion! Hopefully others can benefit from your comment 🙂

Thanks Nilesh!! 🙂

Each truss has one pin and one roller. This is the general case for simply supported statically determinate structures. For the first one, the pin prevents movement in the x direction, and the roller does not. Although, there is no horizontally applied force, so technically the pin doesn't even provide a reaction force in the x direction in this case (its Rx = 0), though it exists for stability. In the second case, the pin prevents movement in the y direction, and the pin does not. If there were two pinned connections, the elongation/contraction of individual members would cause unequal reactions in each support due to the eccentric loading, and would require more advanced methods of structural analysis to calculate the reactions. These are simplified problems though. In real life, of course the roller in the second example would fall down, but it is used here as a generalized support that only provides a reaction in the direction normal to the surface, and not parallel to it. You'll see beams, trusses, and frames always supported by one pin and one roller in first year statics classes and textbooks. Once you learn the basics of these statically determinate structures (solveable by the 3 equations of statics alone - ΣFx=0, ΣFy=0, ΣMa=0) then you can move into the world of statically indeterminate structures, which would involve situations with two pins, or other more complicated connections that some more advanced structural analysis would be required to solve the reaction forces at each support. Hope that helps 🙌

You're welcome, hope the exam goes well!!

Yes. I erase only to illustrate that it has no impact on the joints it touches. In reality it’s still there, but you can solve the remaining non-ZFMs as if it’s not there

Great thanks for the feedback Adam!

You're welcome!! =)

You're welcome! 😊

Great!!

Thanks!!

When I use the word truss, I’m referring to the overly simplified version for entry level statics, which means straight, two force members, connected only with pinned joints. IRL trusses are typically made with fixed connections like you’ve suggested, but in statics, we use the simplification just to get the basic principles across. These conditions make it so that each member of a truss can only possibly experience axial internal force (tension and compression) and there is no way for them to experience internal shear or bending moment. For a bent rod like you’re asking about that’s supporting a hanging weight, that member is going to have axial force that is tangent to it’s curve at any given point, internal shear that is perpendicular to the tangent, and an internal bending moment too. That type of problem is probably beyond you’re typical statics level course, so don’t worry about it. Just focus on simplified problems with straight, two force, pinned only members. Now to address the V shaped member and it’s vertex. In a basic statics problem, every truss member is straight. If something is forming a V shape, then it is two straight members connected with a pin. Because they are two different simple members, they both only carry axial force, if any. If it was a solid V shaped member, then we would maybe be getting the internal shear and bending moment, but no statics truss problem will ever have a V shaped member. The reason an isolated V (two members only at one joint that are not aligned, and are not subject to an applied load at the joint) are ZFMs, is that if one had a non zero force acting on the joint, the other could not resist it because it’s not inline, and the joint would not be in equilibrium. All statics problems are assumed to be in equilibrium, and as it should be, otherwise that part of the bridge would accelerate, which it’s not. You will see bent members and rigid connections in “frames and machines” problems in statics, and these are closer to structures that you would encounter IRL. But its very important to realize when you are given a truss problem and when you are given a frame/machine problem, because frame/machines are not entirely made of straight, two force, pin connected members. On engineer4free.com/statics you’ll see a section just for trusses, and a section just for frames and machines. I recommend watching all of them in each section and realizing that “trusses” (in the entry level statics lingo) are more simplified problems. I hope that helps clear it up. For now, enjoy the simplifications, as when you get into mechanics of materials you’ll start dropping the simplifications and getting closer to what happens IRL.

Hey Kartik, the top vertex is indeed a joint, but it doesn't satisfy one of the three conditions on the left side to immediately assume that the vertical member is a ZFM. The second condition (mid left on the screen) involves 3 members at a joint, but two of the members must be co-linear to assume that the third is a ZFM. When we look at the top vertex of the first truss, it does indeed have 3 members touching it, but none of them are co-linear (in line) with each other. If two of them were, then the third would have to be ZFM because nothing else could react equal and opposite to any of it's magnitude that is perpendicular to the other two's co-linear lines of action. Another way to look at it, is that if the two diagonal members are at the same angle, and if for some reason they had the same magnitude and sense then the the joint would net to zero force in the x direction, but would have a net upward push from them if they were in compression or a net downward pull by them if they were in tension, and either way the vertical member would have to compensate with it's own non-zero force for the joint to remain in equilibrium. So based on what's going on at that joint, the vertical member cannot be a ZFM. You can also confirm that on the bottom middle joint (on the bottom end of that vertical member connected to the top vertex), because the single diagonal member that carries a force has a y component that must be compensated by the vertical member (which only has a y component itself), as the two bottom horizontal members could not do the compensating. So from inspection either joints, that vertical member must carry some internal force and cannot be a ZFM with this exact loading on the overall structure. Hope that helps!

You’re welcome!! Thanks for watching 😁

Glad to hear it Burhan, you're welcome!!

You're welcome! =)

No problem🙂

tx bae

Thanks for letting me know!

Jah welcome

Studying the joint at the bottom, there is no other vertical component to counteract. That means the vertical component of the hinge is zero as well.

That force is being balanced by the external blue load applied on truss

for the last zero force member in the video... it is possible to remove it in order to get the right answer, however it can't be removed from the structure since it is supporting the rolling element that would fall downward (that member has negligible force in it since the rolling element has a negligible weight)... so yes you can solve the problem, but don't draw it like that because I won't make sense.

You can watch solved example using this concept in the attached video. kzfaq.info/get/bejne/a61_dZik2rynd5s.html

Love you too

Thanks 👍

You're welcome, thanks for the message

Sorry, I'm not sure which example you're referring to, they both have rollers. Sounds like you mean the second example tho. If so, we don't know if the line of action of the reaction force is colinear with the diagonal member. If it turned out to be the case then yeah you're right, but we have no grounds to assume that that is the case here. If we were given a magnitude of the blue applied force, then you would determine the magnitude and direction (or just of the x and y components) of the force in the first stage of solving this problem.

Thanks!!

Hmmm sorry you've lost me. In the second problem, the top reaction is a hinge and the bottom reaction is a roller. Both reaction forces have vertical components, but only the top (hinge) reaction has a vertical component. That vertical component is taken entirely by the angled member that touches it. Because the bottom reaction is a roller, it can only have a component that is normal to the surface, which is in the horizontal direction in this case. The horizontal member takes all of that force, and because the roller doesn't provide and force in the vertical direction, the vertical member must have zero internal force (otherwise the joint would accelerate vertically). So the vertical member that touches both joints has to be a zero force member. Hope that helps clarify

You can watch solved example using this concept in the attached video. kzfaq.info/get/bejne/a61_dZik2rynd5s.html

Cole Smith do keep in mind that this is theoretical, and assuming members have no weight. In real live, every member has a force working on it, same as in game, that will introduce a force. And even if there really is a real live zero force member, (which is impossible because gravity) this zero force member will still be useful for stability.

=)

Awesome glad to hear it!

Ur welcome dude

You’re welcome, thanks for watching!! =)