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

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

🙂 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.

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

Thanks!

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

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

Yup, don't over think it!

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

Thanks LARKZ 🙌

Good luck friend!

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 😊😊

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

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

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

Thanks!! Hope it went well 🙂🙂

...still waiting

did you pass

Thanks Tomasi! 🙂

Awesome!!! hope it went well =)

Awesome, glad to hear it John!! =)

Thanks and you're welcome!!

You’re welcome Jackie!! 🙂

You’re welcome Jackie!! 🙂

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.

@@patricktho6546 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.

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

hahaha how'd it go???🤘

You're welcome!! =)

Glad you liked it! 😊

Thanks for letting me know!

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

Thanks!! 😁

Thanks!!

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. ....

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?

Thanks for commenting!

You're welcome! 😊

Thanks for commenting. glad you liked it!

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.

Same question. Probably dynamic analysis including integrals.

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!

tx bae

Are you referring to the vertical member that goes from the pin to the roller? If so, you can't infer that it is a zero force member by inspecting the pin support. It's not possible to determine if the vertical force from the reaction will be directed into the vertical member, or the member on an angle. The only way to determine if that vertical member is a ZFM at a glance, is by observing that the roller support can't provide a vertical reaction, and therefor the joint at the roller would be out of equilibrium if there was any axial force in the vertical member. Rollers can only provide a reaction that is normal to the surface, and in this case, that reaction force would be horizontal. In order for that joint at the roller to be in equilibrium, the vertical member must then carry no force under the current loading conditions. The reaction at the roller will cancel out with the force in the horizontal member there, and that's it. Because the vertical member carries no force, that means at the pin, 100% of the vertical reaction will be carried by the vertical component of the internal force in the angled member (so that the magnitude internal force of the angled member will be greater than the vertical reaction, but the magnitude of the vertical component of the internal force of the angled member is equal to the vertical reaction).

You're welcome, thanks for the message

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

Your welcome, glad it helps!

Hey Billaros, you can find a full list of hardware and software that I use at engineer4free.com/tools

Thanks Nilesh!! 🙂

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

Thanks for the feedback!

You’re welcome!! Thanks for watching 😁

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

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

ZFMs provide lateral bracing to the two inline members that they join. If the ZFM was not there, that joint would need the be solid, and ultimately the two short members would be replaced by one long member. Long slender members in compression are susceptible to buckling. ZFMs also are not always ZFMs. In this problem, the load does not move, and we have just identified that the member is a ZFM when the load is in its current position. If the load moves (imagine a car moving across a bridge) then some members will sometimes be ZFMs and other times not as the load moves. I've got some videos on buckling here engineer4free.com/mechanics-of-materials (videos 50-56), and at the moment I don't have any on moving loads, but you can look up "influence lines" to get an idea of how that works.

Thanks for asking! You are ignoring that there is a 4th non-ZFM member at that joint that goes diagonally up to the left. The other end of that member is in contact with the joint that had the applied force acting on it, so it will be carrying some axial load. That axial load has a vertical component to it because it's angled, so it will be putting some non-zero vertical load on that bottom middle joint. The fully vertical member that your asking about will have to provide and equal and opposite non-zero vertical force, therefor making it not a ZFM. Hope that clears it up!

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

I’m not actually removing the members. They are there always. I’m only erasing them to help visualize which other members are ZFMs. By erasing known ZFMs, it makes it easier to spot the 3 cases highlighted on the left of the screen. But all members that I erased are still there, they just carry no internal force. The ZFMs need to be there for other reasons, like bracing or if the applied load changes location the members which are ZFMs will change.

Zero force members are not always zero force members. It depends on where the load is as to whether or not a member can be considered to have zero force. The thing you are looking for is called "influence lines" give it a search and you will see how to analyze a truss or beam for a load that moves across it.

This is for static load only..

Awesome, glad to hear it Rigel =)

Haha, good luck!!

=)

The upper pin DOES provide a vertical reaction. BUT when we do these problems, sometimes we can only find information about a member from one of the two joints. According to the bottom joint, this pin must absolutely 100% be a ZFM as described in the video. Because the member is absolutely 100% a ZFM, then the entire vertical force applied by the upper reaction force must be countered by the vertical component of the diagonal member that is in contact with it. Hope that makes sense.

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

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!

Yes exactly. These members that are identified as ZFMs will only be ZFMs with this exact loading. If the loading changes, we have to assess again. If you have a vehicle moving across a bridge, then it is much more complicated that this example. The level of instruction in this video is very basic and I just to introduce the concept of a ZFM. For vehicles moving across a bridge, you should look into some tutorials on influence lines. Unfortunately I don’t have any of my own at the moment, but there are plenty on KZfaq.

Yes. The internal forces here are aligned along the axis of the members. Imagine each joint as a 2D particle problem, where the joint is the particle, and the members are force vectors. The sum of forces on the particle must be zero because the joint is not accelerating. It should be easy to see that if for example two two members are co-axial, and the third is not, then that member must carry no force, because otherwise the joint would accelerate in some direction that is not the axis of the two other members.

th8nks weaseldude

Great thanks for the feedback Adam!

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.

Yeah, basically for redundancy and rigidity. When you change the loading, the zfms will change. In real life, the loading will change all the time, and won't be as simplified as these examples.

@@Engineer4Free I’m amazed by how fast you replied. Thanks for that explanation I appreciate it! 😁

I gochu! 🤜🤛

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