How to identify zero force members in trusses

[@kylehaase7703]

I got a final that counts for 50 percent of my grade tomorrow, and these vids are definitely helping me grasp the concepts much better.

[@Engineer4Free]

Great to hear it... good luck!!

[@isisyasmim639]

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

[@luisanunez6654]

@@isisyasmim639 omg same, but 3 years later

[@isisyasmim639]

@@luisanunez6654 the cycle continues

[@turboenterprise790]

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

[@CC-jy4gr]

I feel like I can truss this guy.

[@Engineer4Free]

Ayoooooo

[@shivankitss8396]

@C C 😂

[@capuccinolight]

Get out!

[@rockerpat1085]

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!!!

[@amanrajananda7070]

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

[@himanshukumarsingh8234]

I covered the whole truss topic watching your videos and learnt the whole thing in just an hour whereas my professor covered the whole topic in 2 weeks and even after that I had no idea what was going on in the classes. Thank you so much sir.

[@Engineer4Free]

🙂 Glad to be of service!

[@MrTechBossawesome]

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.

[@soliewx433]

Just wanted to say, even after 7 years your video is helping people! I am currently studying for my FE exam and structural analysis was always hard for me but this video makes it so easy to understand. THANK YOU

[@Engineer4Free]

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

[@voice4voicelessKrzysiek]

I'm reading the book "Structures" by J.E. Gordon (I am an electrical engineer) (I think it was on Musk's important books list) and it made me wonder about trusses so that's how I got here. This video and others on the trusses subject by Engineer4Free I found very well presented and easy to understand. I appreciate these free tutorials a lot. Thank you very much, Engineer4Free!

[@Engineer4Free]

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

[@patrickgullia102]

you're description was phenomenal. the other videos don't explain why...good job

[@Matrix-Glitchs]

Finally I understand this. Lost so many grades in quizzes but will certainly answer the final exam tomorrow. Thanks bro.

[@M4tthew._.]

I NODE I could TRUSS this guy from the MOMENT I heard his voice, I'm glad to be a MEMBER of this page. I know these puns are SHEAR idiocy but don't worry I can DEFLECT any hate as I'm quite the STABLE person and I'm DETERMINATE to be funny ;) for real tho please don't PILE anymore STRESS on me as I'm already feeling STRAINED, and I'm too YOUNG's (modulus) to be feeling like this. Anyway imma stop HOGGING the attention and do my assignment, after all we are all EQUILIBRIUM

[@barnabymackay-howse3047]

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

[@Engineer4Free]

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

[@nlo114]

This is great in terms of pure mathematical vector addition. The 'zero-force' members are there because in the real world, the minimum net is not physically perfect. Heavy wind-blown drifted snow on one side upsets the equilibrium of the 'perfect' net. Remove the zero-load trusses, the roof collapses.

[@clivejacobz2090]

Thanks your comment makes a lot of sense

[@Engineer4Free]

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. 🍻

[@khiareozmakhiar3722]

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!

[@emmata98]

@@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.

[@khiareozmakhiar3722]

@@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.

[@csyoung82]

I'm studying for the Structural Systems component of my ARE and this is by far the most succinct explanation for identifying zero force members. Thank you so much for this video.

[@Kyavata]

I'm a truck driver, why do I need to know this? What has KZbin done to me? Is it technology or civilization that has jumped the shark? I don't even... Also: excellent video.

[@abrahkadabra9501]

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.

[@nateschmitz9827]

@@abrahkadabra9501 lol paranoid much

[@lochtlmao]

sitting in my halloween costume trying to learn this before my exam tomorrow

[@Engineer4Free]

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

[@billtaylor2167]

@@Engineer4Free been 4 years and u still replying?

[@Engineer4Free]

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! 🤜🤛

[@lochtlmao]

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

[@Engineer4Free]

Awesome!! Glad to hear it 😊😊

[@alsubaihawiful]

Thank you! Very clear and easy to follow. I'm taking the Civil FE exam in one week from today. I will update you on the zero member question after the exam.

[@Engineer4Free]

Thanks!! Hope it went well 🙂🙂

[@troemer100]

...still waiting

[@be4913]

did you pass

[@jarrredl]

Got a statics final in 4 hours so this will save my grade, thanks

[@Engineer4Free]

Good luck friend!

[@stormykat9923]

Thank you so much, this really helped me understand the concept of zero force members! I was having a lot of problems analyzing trusses because I couldn't figure out which members were zero force members (my professors lecture slides were rather unclear) so I got lost trying to figure out free body diagrams for the more complicated trusses and this video helps so much!!

[@HarshadPatankar]

I also found kzbin.info/www/bejne/qGO6gGpsqpiJnZI video useful to clear concept on zero forces

[@adexfunkyadex2337]

Finally understand zero force members teach made it so damn hard but your explanation was amazing

[@Engineer4Free]

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

[@g0dsm4ck100]

What I don't understand is why the member underneath the applied load in the first member is a 0 force member. If there is a vertical load on it, shouldn't there be a compression force on it pushing on the two co-linear members perpendicular to it, therefore not making it a 0 force member?

[@JjoshD]

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.

[@AbhishekVerma-so4kr]

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.

[@markdaniel8740]

@@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

[@andreasmerlung]

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. ....

[@guitarmyart2]

Thank you for uploading these videos. I felt like I never understood any of this in school, but you make it so that its understandable.

[@Engineer4Free]

You’re not alone, glad I can help!!!

[@laurap.607]

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

[@Engineer4Free]

​@@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.

[@vodkacid233]

Agreed

[@larkzproduction6495]

You’re a Legend, swear to god I can’t understand my lecturer but your videos clears everything up...🙏🙌

[@Engineer4Free]

Thanks LARKZ 🙌

[@tiffanypersaud3518]

Got a test Monday. You're saving me!

[@Engineer4Free]

Awesome!!! hope it went well =)

[@johndavidlapena3720]

few minutes in the video and i finally understand it, thankyou very much ❤

[@adewalekehinde5057]

Waoh, so easy . Thanks for explaining in a simple way. Teaching shouldn't be difficult.

[@natashasingh1739]

manifesting a slay in my statics exam thanks to you!!

[@Engineer4Free]

hahaha how'd it go???🤘

[@Engineer9736]

I’m not a student of this subject so i may interpretet the definition of “zero force” wrong, but i wouldn’t remove that last removed beam attached to that roller at 9:12. Now the up/down stability is gone of the horizontal member that’s left. But true, it does not carry the mainload.

[@tombackhouse9121]

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!

[@Engineer4Free]

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

[@williamstaib4623]

FE exam in 30 min... thanks

[@Engineer4Free]

How'd it go?

[@rutikrojekar7094]

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

[@latitude.]

Rutik Rojekar How’d it go?

[@Johnnyy832]

anthoty looks like it probably wasn’t good.

[@MyMemesAreTerrible]

How did it go?

[@georgewashington1106]

I was gonna build some trusses for my barn, sorta like the top pic. Now that I know there's no force on all that stuff in the middle, I can take it all out. Sweet. Saved me a bunch of money on 2x4's

[@altaylor3988]

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

[@NathanNostaw]

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.

[@altaylor3988]

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

[@abdoulxc]

Why don't we take into account the reaction forces coming from fixed pin (horizontally and vertically)? isn't there a force aligned with the zero force member that we removed at 8:36

[@Engineer4Free]

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

[@CMRHM]

great explanation for 0 force members in truss. I understand them now. Bravo job and thank you!

[@semisikaufusi2467]

Very concise walkthrough of the problem...thank you TTK

[@Engineer4Free]

Thanks Tomasi! 🙂

[@عبداللهالبسام-ت6ب]

From Saudi Arabia you are the best teacher🇸🇦🇸🇦🙏👍👍

[@Engineer4Free]

Thank you my friend!! 😁😁

[@user-yz5qy7fi9n]

At 6:00, there is a force directly acting on vertical member from upside, though there is no force from downside, but it has to balance or resist that upward force. HOW can it be a zero force member

[@Holsink7]

This is a really good video, you saved my exam, thank you

[@danialalkhoury5136]

Very nice video and explanations, have to add something 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... so yes you can solve the problem, but don't draw it like that because I won't make sense.

[@Engineer4Free]

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!

[@benjaminlavigne2272]

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.

[@SazikimiJaeger]

What a simple explanation...Awesome tutorial.Really helped a lot..Thanks :)

[@afiqsafwan4714]

OMG its that simple.....Thank you so much kind sir. You just made my day a whole lot easier.

[@Engineer4Free]

Yup, don't over think it!

[@usmanfarooq745]

Such a nice explanation, thumbs up

[@Engineer4Free]

Glad you liked it! 😊

[@aSpyIntheHaus]

Aw Yis! I'm going to re-install Poly Bridge and get cracking on optimising my designs

[@acapjumpen]

One of the best videos on structures out here on KZbin! Thank you Sir

[@Engineer4Free]

Thanks!

[@netron66]

Zero force member does not mean it is useless, there are something called bucking when comes to design and those extra member is really to prevent this from happening

[@Engineer4Free]

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!

[@delaniecornwell6578]

Ridiculously helpful! Thank you!

[@Engineer4Free]

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

[@ryanm.191]

16 hours of tuition at university, each hour costing £135. Useless. 9 minute video on KZbin, absolutely free. Understood in 4 minutes Thank you very much In the first truss diagram wouldn’t the centre vertical member be a zero force as well? If not why?

[@Engineer4Free]

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.

[@ryanm.191]

Engineer4Free ah yeah it does, cheers

[@saarojrawal3331]

In problem (b) at the last the inclined member should also be zero, cause there is a horizontal reaction on the hinge making the inclined member zero with the two colinera members?

[@erikapaulus7961]

wow, save my life last minute for my homework and prep for my final exam, thanks a lot

[@Engineer4Free]

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

[@dixonchan0621]

although the second example's leftmost member is a zero member, I think it cannot be erased, otherwise, it becomes externally non-static. what do you think?

[@Engineer4Free]

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

[@sidmarker8739]

While discussing the last zero force member, there is a reaction force acting at the hinge support. So shouldn't a vertical force be considered in the last zero force member you considered?

[@mohammedbadawi2361]

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.

[@AbhishekVerma-so4kr]

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

[@danialalkhoury5136]

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.

[@VenturiClasses]

You can watch solved example using this concept in the attached video. kzbin.info/www/bejne/a3iyc2ipqrt2fck

[@ravigautam8685]

Dear Engineer4Free, Thank you for your response. It was helpful for me know that "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." Only thing could not understand axis of both arms of V-shaped rod or L-shaped rod (basically bent rod) are different. How the internal force in the bent rod look like compression, tension, moment. Suppose I have fixed one arm of vertically aligned V-shaped rod to the ground and hanged a weight (10kN) on its upper arm. what will be the stress on the rod. If possible kindly provide some reference. Statics as subject is confusing for newcomer like me because many authors call structures as truss which do not have pinned joints at all. Also its difficult to imagine any real life structure which have pinned joints. For example river Bridges, bicycles etc can not afford to have pinned joints. joints are either riveted or welded. I one blog (Quora), some body posted a figure of chair with identical four legs with legs being non-straight rods (they were bent rods) and he said all four legs are zero force member. In many places all V-shaped members (unloaded at its vertex) are said to be zero force members with same analysis using static equilibrium condition at its vertex. Should static equilibrium condition be applied at its vertex of V-shaped member?

[@Engineer4Free]

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.

[@anwarulbashirshuaib5673]

College teachers: Blaeh Bluh Bleh bluu TrUSs done. Test tomorrow. Engineer4Free: I'm about to end their whole career

[@Engineer4Free]

🤠

[@itzzsaroj5647]

Great explanation !!!

[@deezynar]

All that's true if you don't count gravity pulling on the length of every member that's not perfectly vertical. If the truss is small, it's not significant. But if the truss is large, long members need to be reinforced against gravity tugging them out of straight. You can make those members beefier, but there are times when adding 1 or 2 light members keeps the critical members straight while adding negligible weight. Also, buckling is unavoidable in a member under compression when its cross section to length ratio is below what's needed. So there are conditions when one or more members will need additional members added to keep them from bending, and or buckling. Those members may appear to be zero force, but they serve a critical purpose. If you look at trusses in the real world, they often have many members that would be counted as zero thrust from this assessment.

[@Engineer4Free]

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! 👌

[@Taufiqmusic]

I've never seen anything more beautiful. Thanks, this helps a lot and my finals is in 2 days. You're a life saver!

[@Engineer4Free]

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

[@Fluy91]

Awesome tutorial. Thank you!

[@Engineer4Free]

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

[@Fluy91]

@@Engineer4Free nothing like 2 weeks before exam refresher

[@jackieelder4290]

Thank you so much for posting this. This is so very helpful.

[@Engineer4Free]

You’re welcome Jackie!! 🙂

[@Engineer4Free]

You’re welcome Jackie!! 🙂

[@aelhkmm6009]

man you should be the dean at where ever you're teaching! love from egypt

[@utkarshsaxena3511]

very nice explanation

[@Engineer4Free]

Thanks!! 😁

[@bonvivant9726]

hey in the above diagram , i mean the first one , the top vertex can also be a joint hence central line is a zero force member ?? explain please why is it not

[@Engineer4Free]

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!

[@HuyPham-bw2fd]

Wouldn't the member that your mouse hovered over at 9:19 also be 0 force? by the same logic you used to identify your last 0 force member?

[@CCNproductions1324]

If you're talking about the last two "internal lines", the upper part of it is not zero force bc the pinned support has a reaction in the x and y direction. The lower part is also not a zero force because the two lines it connects with are not co-linear.

[@ravigautam8685]

In you video titled "Two force members explained (statics)" you said L-shaped rod is in equilibrium and without being zero-force member. Though there was some difference from 3rd zero force member that you describe here. But I am learner. I need to understand from you.

[@Engineer4Free]

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.

[@dogukan733]

Perfect explanation! Thanks a lot!

[@Engineer4Free]

Thanks and you're welcome!!

[@mr_slidey]

This is a great recap video thanks so much

[@EJ-yz8lx]

If I may ask what was the point of zero force members? I heard that due to weather or like snow, they may be required to act as support right?

[@Engineer4Free]

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.

[@EJ-yz8lx]

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

[@Engineer4Free]

I gochu! 🤜🤛

[@nikunjarya9641]

Please make a video over the Graphical method of analysis of truss.

[@Engineer4Free]

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

[@nikunjarya9641]

@@Engineer4Free Okay!! Thanks a lot!

[@Engineer4Free]

🤜🤛

[@akhilsuresh7971]

In the 2nd problem u worked out there is a horizontal reaction in the hinged joint so the member you made a zero force member at the end of the video should be there to counteract that reaction right..

[@Engineer4Free]

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

[@VenturiClasses]

You can watch solved example using this concept in the attached video. kzbin.info/www/bejne/a3iyc2ipqrt2fck

[@NomadUrpagi]

This lesson is true only for static objects right? If we consider a car moving through a bridge what would our analysis be like? Would we have to integrate sum of forces over the distance?

[@Engineer4Free]

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 KZbin.

[@bradley9856]

Very useful video, thank you

[@khinlapyaewonn2011]

Thanks alot for this video, worth watching it.

[@maudejuly]

8:50 what about the upper pin? Doesn't it need to affect as reaction force to the x-axis just as the roller pin?

[@Engineer4Free]

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.

[@shellyashleymagat8330]

Wow. Thank you so much sir! You're way better than our prof. Lol. I'm subrcribing in your channel now.

[@Engineer4Free]

aayyyyy thanks for the compliment haha. Welcome aboard!

[@jmmifsud1]

great tutorial I picked up a ton - thanks

[@Engineer4Free]

Awesome, glad to hear it John!! =)

[@MostlyLoveOfMusic]

How much force in a Liz Truss?

[@Abohmedjt]

That was really helpful 🌟

[@IshaShrestha]

omg tomorrow is my boards exam thank you so much

[@Engineer4Free]

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

[@oklord6415]

Legit studying for a quiz that starts today lol

[@Engineer4Free]

Haha, good luck!!

[@NileshRoamingNITIAN]

Awesome man 🔥

[@Engineer4Free]

Thanks Nilesh!! 🙂

[@somebuddy8104]

Thank you sir, your videos helped a lot.

[@Engineer4Free]

You're welcome! 😊

[@Mr0iddont0caRe]

Super helpful, book does a horrible job explaining this concept!

[@Engineer4Free]

Glad I could help =)

[@natthekiwi7074]

I'm in school to become an engineer and from my very beginner understanding, more triangles means more stability right? This concept is a bit confusing, are we analyzing from the perspective of each specific node?

[@Engineer4Free]

Yup generally triangles are a good thing especially when these nodes are "pin connected". I really recommend taking an hour or so and watching videos 42 - 51 here: engineer4free.com/statics it will really help to clear up trusses!

[@ellan7286]

this was so incredibly helpful!!! thank you

[@khangdo4959]

I think we have a downward force so the horizontal forces should be 0, and we only have vertical reaction from the above support. So the last zero member you consumed should not zero. Is that right?

[@rajkanttripathi]

at 8:53, as per rule 1, why didn't you take two collinear member as zero force member? Iam talking about the diagonal members....plz make me clear

[@Engineer4Free]

If you draw FBD of the whole thing and do you sum of moments about the upper reaction, you will see that there must be a horizontal force at the lower reaction to cancel out the moment caused by the applied force. The diagonal members that I put a red circle on in that region are ZFMs, and that's why I erase them off the "equivalent" diagram in the bottom left corner of the screen. After we determined that they are ZFMs, we know that all of the horizontal reaction force must be transferred into the horizontal member there at the lower reaction. IF some force was transferred into the diagonal members, they would then also exert a vertical component of force on the joint. BUT because the have zero internal force, they exert zero vertical force on that roller. And if they are not providing and vertical component of force, then as per rule #2, then the fully vertical member that touches the roller must also carry no force, therefor is a ZFM too. Right at 9:00 when I say "two co-linear forces," and draw them in red, one force is the compression in the horizontal member, and the other force is the reaction. They must be equal and opposite to achieve static equilibrium in the horizontal direction. If a third force was introduced with a vertical component, the joint would have the tenancy to accelerate in the direction of that applied force, making it not in static equilibrium. It IS in static equilibrium, so no such force exists at that joint, and therefore the internal force in the vertical member is 0. I'm really going off of rule 2 here. If you replace the "joint and members" in rule 2 with a "particle and forces," it might be more obvious. Does that help?

[@rajkanttripathi]

Indeed its useful....thank you so much sir.

[@CC-jy4gr]

@@Engineer4Free MIC DROP SON!!!

[@advillobotomite1196]

I think it makes more sense to say that if it isn't in equilibrium, it would accelerate, rather than translate or something else.

[@Engineer4Free]

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 🙂

[@enrique2914]

For the second truss if you do sum of the forces at Ax, wouldn't that be 200 k making AB 200? Thanks.

[@Engineer4Free]

Hey sorry I'm not sure what you're asking. I didn't use any letters to identify any points in this video, or numbers to discuss potential magnitudes of internal forces

[@enrique2914]

@@Engineer4Freemy bad I had a similar problem in front of me called AB I was working on, but I meant the last zero force of the horizontal of the second truss you found to be a zero force member. Wouldn't the X direction be a -200 on the pin? Is it zero because the diagonal takes the X force as well? Thanks

[@VndNvwYvvSvv]

We haven't covered this yet, but it looks like this can be figured using vectors. Is that true? To what degree?

[@Engineer4Free]

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.

[@JackdeDuCoeur]

Listen at 4:05 and tell me what you hear.

[@muhammadfareeziqmal7494]

"two collinear"

[@aleksandreakhvlediani8034]

why is the last one perpendicular to roller support 0 force member? using same logic I would analyze pin before roller and say that it has reaction in x and y which should be counter-acted, so I don't think the last member was 0 force...

[@Engineer4Free]

Rollers can only apply a force that is normal to the surface they rest on. In this case, the surface is vertical so the force can only be horizontal. This means the reaction force at the roller is horizontal, with absolutely no vertical component. If you draw a fbd of the joint at the roller, you would have a horizontal force from the reaction force, a vertical force from the internal force of the vertical member, an angled force from the internal force of the angled member, and a horizontal force from the internal force of the horizontal member. We know that the angled member is a ZFM so the internal force is 0 and we can ignore or erase it. This leaves us with two horizontal forces and a vertical force. For the joint to be in equilibrium, the horizontal forces must be equal and opposite. The only way for the joint to not accelerate in the vertical direction is if the single vertical force that we drew on the fbd is zero. So that member has zero internal force, it's a zero force member.

[@danialalkhoury5136]

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.

[@keikay2733]

Great video, it helped me alot, thanks so much

[@SophiaTheKittycat]

How about the pin above the roller support it will not effect on the last member you removed?

[@Engineer4Free]

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.

[@jeujopogo3433]

I want to build a steel truss for my own house.. what is the purpose off identifying zero force memder are we going to remove that zero force member out of the truss to reduce the cost?

[@Engineer4Free]

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.

[@kukuyiehalem6391]

on the last example u gave what about the pin? doesnt it have force on y direction?

[@karhukivi]

Some years ago a friend wanted to do an attic conversion himself. He started to cut the various trusses one by one, watching to see if the cut gap opened or closed which he presumed would indicate extension or compression. Nothing happened, so he continued until he had opened the space he required and then used the pieces he had cut out to complete the construction.

[@johnrogers5658]

Are you using coral draw or A great package of the 90s, Struct Plus...

[@mariam-bx8yx]

What is the app you use it to write in this video?

[@adamgiallo340]

Pretty straight forward and easy to understand, thanks mate!

[@Engineer4Free]

Great thanks for the feedback Adam!

[@_chickenhead]

Makes so much sense thank you

[@Engineer4Free]

You’re welcome!! Thanks for watching 😁

[@billkalaitzakis7908]

can you tell me which programme did you use for this tutorial??

[@Engineer4Free]

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

[@haydenmcdow7130]

So I appreciate the video and the explanation, but in the second example, with the pin support at the top, can we assume that it is a zero force member if we don't know the reactions? I'm thinking that if the pin support had a vertical and horizontal reaction component then the member should have a force in it to keep the system in equilibrium.

[@Engineer4Free]

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).

[@lf198]

Why isn’t the left horizontal beam from the top figure truss zero force? My understanding is that if there are only two members and it’s loaded, then if the line of action of the resultant force at a joint is collinear with one of the members, then the other member is zero force. The joint at the top figure I’m referring to in my question is the one with the roller.

[@Engineer4Free]

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.

[@MostlyLoveOfMusic]

I don't know about this... Why are you assuming that the roller doesn't move?

[@Engineer4Free]

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 🙌

[@hunterp913]

What would you do for dynamic loads, say a car driving across the bridge?

[@Engineer4Free]

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.

[@satrioekowicaksono7452]

This is for static load only..

[@Findout1882]

Amazzzing explained thanks man......

[@Engineer4Free]

Thanks!!

[@anton9004]

This was very helpful. Thanks!

[@Engineer4Free]

Thanks for letting me know!