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!!!
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.
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.
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!
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
@@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.
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.
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.
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! 🤜🤛
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
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!!
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.
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.
I'm a truck driver, why do I need to know this? What has TH-cam done to me? Is it technology or civilization that has jumped the shark? I don't even... Also: excellent video.
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.
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
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.
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.
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.
16 hours of tuition at university, each hour costing £135. Useless. 9 minute video on TH-cam, 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?
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.
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
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!
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.
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.
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.
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! 👌
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?
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. ....
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.
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 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 🙂
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
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.
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?
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?
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.
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.
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
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
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?
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
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 🙌
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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.
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
I feel like I can truss this guy.
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
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.
🙂 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.
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!
Cool, thanks for letting me know!!! =)
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
Glad to hear it! After all these years the principles still apply haha 😋
you're description was phenomenal. the other videos don't explain why...good job
Waoh, so easy . Thanks for explaining in a simple way. Teaching shouldn't be difficult.
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.
You’re not alone, glad I can help!!!
@@Engineer4Free I never understood anything in school. I went to TH-cam 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
You’re a Legend, swear to god I can’t understand my lecturer but your videos clears everything up...🙏🙌
Thanks LARKZ 🙌
Finally understand zero force members teach made it so damn hard but your explanation was amazing
Glad I can help!! It shouldn't be complicates, but sometimes it seems that way when it's first taught in a lecture!!
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.
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.
Thanks!! Hope it went well 🙂🙂
...still waiting
did you pass
sitting in my halloween costume trying to learn this before my exam tomorrow
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 😊😊
Got a statics final in 4 hours so this will save my grade, thanks
Good luck friend!
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
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
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!!
I also found th-cam.com/video/r2WO76u3Vg0/w-d-xo.html video useful to clear concept on zero forces
great explanation for 0 force members in truss. I understand them now. Bravo job and thank you!
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.
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.
OMG its that simple.....Thank you so much kind sir. You just made my day a whole lot easier.
Yup, don't over think it!
I'm a truck driver, why do I need to know this? What has TH-cam done to me? Is it technology or civilization that has jumped the shark? I don't even...
Also: excellent video.
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
Very concise walkthrough of the problem...thank you TTK
Thanks Tomasi! 🙂
This is a really good video, you saved my exam, thank you
manifesting a slay in my statics exam thanks to you!!
hahaha how'd it go???🤘
What a simple explanation...Awesome tutorial.Really helped a lot..Thanks :)
Got a test Monday. You're saving me!
Awesome!!! hope it went well =)
From Saudi Arabia you are the best teacher🇸🇦🇸🇦🙏👍👍
Thank you my friend!! 😁😁
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
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
Such a nice explanation, thumbs up
Glad you liked it! 😊
Aw Yis! I'm going to re-install Poly Bridge and get cracking on optimising my designs
One of the best videos on structures out here on TH-cam! Thank you Sir
Thanks!
Ridiculously helpful! Thank you!
Glad it helped!!! More at engineer4free.com/statics =)
I've never seen anything more beautiful. Thanks, this helps a lot and my finals is in 2 days. You're a life saver!
+Taufiqmusic thanks for the kind words, and hope your finals went well!
Great explanation !!!
wow, save my life last minute for my homework and prep for my final exam, thanks a lot
+Érika Paulus glad that I was able to help! Make sure you check out the rest of my free courses at engineer4free.com :)
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.
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.
16 hours of tuition at university, each hour costing £135. Useless.
9 minute video on TH-cam, 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?
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
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
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!
FE exam in 30 min... thanks
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?
Wow. Thank you so much sir! You're way better than our prof. Lol. I'm subrcribing in your channel now.
aayyyyy thanks for the compliment haha. Welcome aboard!
very nice explanation
Thanks!! 😁
Thank you so much for posting this. This is so very helpful.
You’re welcome Jackie!! 🙂
You’re welcome Jackie!! 🙂
Perfect explanation! Thanks a lot!
Thanks and you're welcome!!
this was so incredibly helpful!!! thank you
Awesome tutorial. Thank you!
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
man you should be the dean at where ever you're teaching! love from egypt
This is a great recap video thanks so much
great tutorial I picked up a ton - thanks
Awesome, glad to hear it John!! =)
Thanks alot for this video, worth watching it.
Pretty straight forward and easy to understand, thanks mate!
Great thanks for the feedback Adam!
That was really helpful 🌟
Awesome man 🔥
Thanks Nilesh!! 🙂
Thank you sir, your videos helped a lot.
You're welcome! 😊
Very useful video, thank you
Please make a video over the Graphical method of analysis of truss.
Yes I have, please see videos 42 - 51 here: engineer4free.com/statics
@@Engineer4Free Okay!! Thanks a lot!
🤜🤛
Super helpful, book does a horrible job explaining this concept!
Glad I could help =)
Great video, it helped me alot, thanks so much
Thank you for this
is very clear
Great!!
College teachers: Blaeh Bluh Bleh bluu TrUSs done. Test tomorrow.
Engineer4Free: I'm about to end their whole career
🤠
Legit studying for a quiz that starts today lol
Haha, good luck!!
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.
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.
Makes so much sense thank you
You’re welcome!! Thanks for watching 😁
Amazzzing explained thanks man......
Thanks!!
Thanks!
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Very helpful ...thank you
Glad to hear it Burhan, you're welcome!!
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.
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! 👌
Great video
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?
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. ....
thank you dude
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Thankyou sir....from the bottom of my heart , thankuuu
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Thank you for making this!! :)
You're welcome!! =)
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.
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
Thanks a lot
Jah welcome
I think it makes more sense to say that if it isn't in equilibrium, it would accelerate, rather than translate or something else.
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 🙂
This was very helpful. Thanks!
Thanks for letting me know!
one of best explanation
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W video. Very good
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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
very beautifully explained😘😘😘
Not sure why this is being recommended now, but at least I can finally optimize Poly Bridge levels!
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.
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?
Thank you.. It helped
Awesome, glad to hear it Rigel =)
Quite helpful 🌸
=)
nice video.. cleared about it.. thanks
Awesome glad to hear it!
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?
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.
thank you
You're welcome Mahmoud =)
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.
omg tomorrow is my boards exam thank you so much
You're welcome, hope the exam goes well!!
Very helpful, thank you.
Your welcome, glad it helps!
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
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
Thanks for explaination.exam in 3 hour
Good luck, hope it goes well!!
Engineer4Free thank you so much
GOAT
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?
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
I don't know about this... Why are you assuming that the roller doesn't move?
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 🙌
very well expalined!
Thanks for the feedback!