SA05: Truss Analysis: Zero-force Members

[@sketzartist]

what are the answers for the 3 problems at the end? So I can reaffirm if I actually know how to do this

[@DrStructure]

(A): None of the interior members (BF, BE, CE, DF, IH, HJ, GJ and GD) carries any force. The three members forming the outer triangle (AD, KD and AK) are the only ones that carry any force. (B): BD, DE, CD, CA and CF are zero-force members. (C): GH, FG, EF, DE, CD and BC are zero-force members.

[@oscarpaez8679]

Dr. Structure In truss B, why member AB and AD are zfm if they are on a pin connection ? If we take the sum of moments about point I, we find that there are reactions at point A. So there are forces that should be transferred by members connected to point A.

[@basketball12125]

Dr. Structure For (B) why is CA a zero member, you said member who are supported by a pin cant be zero members?

[@DrStructure]

basketball12125 1. According to Rule 3, BD and ED are zero-force members. For clarity, redraw the structure without them. 2. Since BD and ED are no longer shown in the truss, by examining joint D (using Rule 3) we can conclude that DC is also a zero-force member. Now redraw the truss without DC, BD and ED (redraw the truss without the zero-force members). 3. Joint C connects two members: CA and CF. Using Rule 1, we can conclude that both CA and CF are zero-force members.

[@DrStructure]

Oscar paez AD is not a zero-force member. See my comment above for why AB (same as AC) is a zero-force member.

[@joeschey]

Such a simple concept that some professors seem to have trouble explaining, but you covered it perfectly Dr. Structure! Much thanks for the awesome video, looking forward to checking out the rest of your videos!

[@DrStructure]

Rasika Thorat In the first example how is rule 3 applied to G and E? In the first example which starts at 4:17, G connects two members (GF and AG). Further, there is a vertical roller at that node. The roller provides a horizontal reaction force. The force in member GF is also horizontal. The two forces therefore are collinear. There is another (non collinear) force at G, the force in member AG. So, three forces act at G, two of them are collinear, therefore, the third one must be a zero force (Rule 3). We can apply Rule 3 to E only after Rule 1 is applied to joint D. That is, after we have determined that CD and ED are zero-force members, then only three forces remain at E: the vertical applied force, the force in member CE and the force in member EF. But since the applied force and Fce are collinear, then the other force (Fef) must be zero.

[@noraaismyname]

does it not matter that joint G is supported by a roller or that joint E has a load "P"?

[@DrStructure]

Nora Castro No, it does not matter. What do a truss member, a roller support reaction and an applied load have in common? Each can be represented by a force vector. In a structure that is in static equilibrium, when these forces intersect each other, we can draw certain conclusions, hence the rules. It does not make any difference whether a force is a support reaction, an applied load or a member force. In the eyes of nature, they are all the same!

[@vigneshkannan1094]

the rules are starting with the conditions of that joint should not have pin or roller but at joint G there is roller then how rule is applicable ????

[@lusoguta]

Bom dia, At 3:15 you should correct AB and AC are zero force members. Congratulations for permitting us to have a glimpse of the future in education.

[@venkatesh2285]

Dont underestimate the power of teachers... infact most of the educational institutions do not provide such facilities neither they are not ready to invest money to buy the software and tools required to prepare such videos. There are lot of teachers in world wide like dr.structure..

[@DrStructure]

***** What do a truss member, a roller support reaction and an applied load have in common? Each can be represented by a force vector. In a structure that is in static equilibrium, when these forces intersect each other, we can draw certain conclusions, hence the rules. It does not make any difference whether a force is a support reaction, an applied load or a member force. In the eyes of nature, they are all the same!

[@skyraider996]

Thank you so much for the "how to identify zero-force members." I had a question for an online quiz that I had no idea where even to start. That section made it so I got it correct.

[@paxlepham]

the problem (A), the zero-force members are : BF,BE, EF, EC, ED AND IH,HJ,JG,HG,DH the problem (B), BF,BE, EF, EC, ED AND IH,HJ,JG,HG,DH THE PROBLEM (c) HG,GF,FE,DE,DC,BC DR. THAT VIDEO IS AWESOME, i HAVE READ MY BOOK A LOT A LOT OF TIMES "HOW TO DETERMINE THE ZERO FORCE MEMBERS", BUT I STILL NOT BE ABLE TO INDICATE THE ZERO-FORCE MEMBERS. ENGLISH IS MY SECOND LANGUAGE BUT JUST AFTER THREE TIME HEAR YOUR EXPLANATION. I FEEL MORE CONFIDENT TO SOLVE THAT PROBLEM.

[@harishtiwari9218]

whats the ans of the questions asked at the last ! ?'

[@rasikathorat9146]

In the first example how is rule 3 applied to G and E?

[@kingcjokay]

Very clear explanation better than my courseware! Thank you!

[@w0w..325]

probably one of the best explanation...very very helpful...thnk u sir...

[@midadrod]

Really a nice tutorial package , That having awsome presentation, relavent example. I have downloaded all of these videos. Thanks Dr. Structure for these tutorials .

[@podger9999]

been looking for clear concise vids on this topic for a while, these are great thank you so much

[@craigevangelista8239]

so im thinking number one is only left with triangle ADK, number two os left with the right side plus beam ADF, and number 3 is left with angle AIB. am i right?

[@DrStructure]

Yes, yes and yes!

[@quangtran7229]

Example B, right side plus ACF, right? Not ADF.

[@DrStructure]

ADF is correct. BD is a zero-force member, so is ED. Then, looking at joint D, we can conclude that DC is also a zero-force member. Now moving to joint C, we can see that both CE (and EF) and CB (and BA) are zero-force members. Then, the only remaining part of the right side is ADF.

[@DrStructure]

@@danielimm In Exercise Problem B, there is no reason to believe that GK is a zero-force member. None of the rules apply to either joint G or joint K. At joint G, we have a force in member HG, and a force in member GJ. And, there are three other forces acting on G: the forces in members FG, GI and GK. Since none of these forces have been determined to be zero, we cannot remove them from the truss. Therefore, we have a total of five forces acting at G. None of the rules that we have allow us to infer one of the five forces is zero. So, joint G cannot be used to conclude that GK is a zero-force member. Examining joint K, we find three acting member forces: the force in HK, JK, and GK. If HK and JK were collinear (if the angle between members HK and JK was 180 degrees), Rule 3 would have applied and GK would have been a zfm. But since the angle between the two members is 90 degrees, GK cannot be a zfm. For joint K to remain in equilibrium the algebraic sum of the three forces must be zero, which means neither of them can be zero.

[@capoman1]

Dang, you put alot of work into the production here!

[@nerd2544]

5:20 is genius. no other video and neither my professors mentioned that you can treat external forces at a joint as a pseudo-member i.e. the same way as internal truss forces at a joint so you can extend the use of the 3-member rule. this is the key to so many zero force member questions where I always missed one or two members that seemingly fit none of the rules. thank you so much bro. EDIT: wait, doesn't this violate the rule where there should be no external forces at a joint for zero force members to be connected to it??

[@DrStructure]

You may want to take a look at a different approach for identifying zero-force members here: kzbin.info/www/bejne/ZnKpoKhtar2ZqpI

[@alialrumhi5408]

Thank you doctor ..That was a great explanation keep it up

[@sameersinghania100]

nice session rules is explained clearly thnx

[@rahulr.a34]

Thank you very much Dr. Structure

[@illilichannel]

I have a couple of questions: 1) when it comes to rule 3, is it supposing there is no load or support on that joint? 2) as for the use of ZFMs, the first reason given said "to prevent a force carrying compression member from buckling", so does that mean the ZFM can only be in compressing when subjected to a time variant load? Can you explain the first reason further? I don't really understand the difference between the 2 reasons given for the necessity of ZFMs either, so maybe I'm misunderstanding something. In all, thank you for putting this up. I can't find anything that really explains ZFMs to my satisfaction.

[@DrStructure]

1) Correct, Rule 3 applies only to joints that carry no load and are not supported by a pin or a roller. 2) No, the zero-force member is not the compression member. Rather, it is the member that prevents the compression member from buckling. So, let's refer to the compression member as A and the member that braces it as B. If the compressive force in A is not large enough to cause buckling, then B carries no force. If, however, the compressive force in A is large enough to cause buckling, if it were braced by B, then as a result of preventing A from buckling, a secondary force would develop in B. We call this a secondary force because it is not due to the applied load but due to the tendency of A to buckle (that is due to the fact that A is either pushing B down or pulling it up). This secondary force, could be either tensile or compressive. For example, in the truss example shown at the beginning of the video, if the horizontal compression member tends to buckle downward, then the bracing *the vertical member) will go under compression. If however, the compression member (member A) wants to buckle upward, then the vertical bracing (member B) will go under tension. But, keep in mind that B carries no force (it remains a zero-force member) unless the compression force in A is large enough to cause buckling.

[@illilichannel]

Thank you for the explanation!

[@alaahesham250]

Answers are :number one is only left with triangle ADK, number two os left with the right side plus beam ADF, and number 3 is left with angle AIB. credits to craig evangelista

[@Twitter.345]

But saying AIB will create a doubt here bcs AB is not here as a link so I think we can say that only AI and BI members will have non zero forces... Isn't it?

[@AmanAriyaSarina]

Dr. Structure, for example (A) were all the internal members for triangle ADK zero force members? First, I applied rule 1 to the following angles: BEC; FBE; IHJ; GJH. Then applying rule 3 to lines: AFG & KGF resulted in DF & DG to be zero force members, respectively. Similar methodology was applied to example (C), is this a correct analysis?

[@DrStructure]

+Aman Neha Yes, your analysis is correct.

[@ravigautam8685]

Dear Dr. Structure, Thank you for your response. As I understand, zero force member concept holds only for pin joints and not for welded joints. In many research papers people apply method of joints for free body diagram of bicycle frame in which all the joints are welded joints. They also apply for chairs and other furniture. It so confusing. In one blog they said for chair having four identical legs with each leg being bent rod then all the legs are zero force members. I don;t know if there exist an authentic comprehensive book which can explain basics. I also need to understand if in a frame I have some elements as bent bar then how do I find stress on each member. How does bending moments cancels out to achieve equilibrium. Is it that we do not have theory for frames with welded joints. If there is where do I find them. Regards, Ravi

[@ambarshrivastava1131]

very good explaination of zero force member. please upload the answer to the questions

[@farrukhaqil24]

Beautifully explained! :)

[@chungken8496]

Excuse me, at 03:19, it mentioned a truss joint do not carry any loading, what is that mean? Does the loading means external loading? At 03:19, does joint C also do not carry any loading?

[@DrStructure]

That is correct. The rule works if there is no external load applied to the joint. And yes, there is no external load applied to joint C.

[@rosy7849]

At 4:15, for rule 3, it says that "AD is a zero force member" however in the video you said that CD was the zero force member. Is CD the zero force member? Also can you explain why EF is a zero force member according rule 3 at 5:42? Since E does not connect members that are collinear.

[@DrStructure]

@4:15-- The writing "AD is a zero force member" pertains to Rule 3 and refers to the drawing to the left of the simple truss. Using Rule 3, we can conclude that CD (the vertical member in the truss) indeed is a zero force member. @5:42-- Truss members carry axial loads only. Put it differently, if you see an axial force (like the applied vertical force P at E, you may assume an imaginary member there that contains/carries that force). So, for the sake of this analysis, we can replace P with a vertical (downward) truss member at joint E. Doing so gives us three members at E, the imaginary member carrying P, member CE and member EF. Two of these members are collinear (CE and the imaginary member), therefore EF has to carry a zero force.

[@dashnduke]

Why is AG in the first example (at 5:50) zero force member if there is a force from the pin going upwards?

[@DrStructure]

There are three forces acting at joint G: the force in member GF, the force in member AG and the horizontal reaction at the roller support. Two of these forces are acting in the horizontal direction (along the x axis). The other force (Fag) is in the vertical direction (along the y axis). Sum of the forces at G must be zero. This means sum of the forces in the x direction must be zero, and sum of the forces in the y direction must be zero. But, since there is only one force acting in the y direction, for the sum to be zero, the force (Fag) must be zero. Therefore, AG is a zero-force member.

[@hrishikeshdas9754]

sir in the video 1:35 min in the video the vertical member at joint 3 remains a zero force member for any point of the moving load, then why that member is provided????

[@DrStructure]

The member could be used to prevent buckling (due to a large compressive force) of the top horizontal member. So, the zero-force member serves as a stabilizer for the compression member.

[@hrishikeshdas9754]

Thank u sir... can u please make a video on plastic analysis and matrix method to find deflection..

[@sportingaesenal428]

why member AF is not eliminated using the third rule at 6:13?we have two forces Ax and AB colinear

[@DrStructure]

A is a pin support, there are two reaction forces acting at A: Ax and Ay. Rule 3 would have been applicable if Ay was not there.

[@ryandonex9452]

Thnkyou my lect is using this to teach us :)

[@chungken8496]

Excuse me, at 05:40, how could joint E apply rule 3? First, it need a truss joint is not carrying any loads & not supported by a pin or roller. Does joint E is that king of truss joint?

[@DrStructure]

Truss members are axially loaded only. For analysis purposes, we can represent the member as a force vector. In a way, we can think of a truss member and a force vector as one and the same. When we see a member, we can think of a force. And when we see a force, we can think of a member. In this sense, the force and the member are interchangeable. At Joint E, I see a force (Gx), so I can think of a member. For our analysis purposes, in my mind, I can replace that force with a member. If I do that, then I end up with three members at G, two of which are collinear, so the third one must be a zero-force member.

[@chungken8496]

Thank you for your response. At 03:19, the principle of the rule must satisfy the truss joint is not carrying any external load and not supported by a pin or roller. At joint E, there have an external loading, can it apply the rules under this condition?

[@DrStructure]

@3:19, joint E connects three members, with no pair being collinear, none of the rules apply to the joint. However, once we apply Rule 1 to F concluding the EF is a zero-force member, we can then revisit joint E (which now has one less member), and apply Rule 1 to conclude that the other two members also carry a zero force.

[@chungken8496]

Sorry, I made some mistake on the last reply. I present again at follow: At 03:19, the principle of the rule must satisfy the truss joint is not carrying any external load and not supported by a pin or roller. However, at 05:40, joint E have an external loading, can it apply the rules under this condition?

[@DrStructure]

Yes, the same line of reasoning applies here too. Forces and truss members are interchangeable. Except that at E, the two forces are collinear and one of them is not zero. So, the other one cannot be a zero force member either.

[@deeman.n6064]

thanks This vedio made me understand the zero force member clearly thanks Dr. :)

[@nandinidas9265]

Thank you, this is very helpful. On rule #1, (at 3:12), you mean AB and AC are zero force members, not AB and BC, right?

[@DrStructure]

Yes, I've added a note correcting the error using the annotation tool.

[@fathereshark8421]

For rule 2, what is meant by it doesn't carry any internal forces? As in there is no point load on the joints the member is connected to? Are the red arrows on the first 3 examples point loads?

[@DrStructure]

Truss members carry axial force. If the magnitude of such a force is zero-if the member carries no axial force- we call it a zero-force member. It is not necessary for a joint attached to the member to be loaded for the member to carry a force. Axial forces propagate through the structure without a load being present at every joint of the truss. Yes, we use arrows to represent forces and loads.

[@latinmusic1328]

what is the reasons of zero force if there is no force to act ore react then why do we need to use and spend the iron for those ????

[@DrStructure]

ali akbari Good question. It was asked, and answered, starting at @0:24 in the video. Let me know if you have any additional questions on this.

[@KhabisangLekena]

Well presented!! Thanks, I now understand zero force member

[@Chikokishi]

Dr. Structure, I am attempting to learn about how to solve for forces in the Zero members on a real life structure. For example, your truss at 6:40, when under the load provided, would actually deflect to the right a little bit. Also, assuming the truss is something weak like aluminum, would deflect a lot. So, during this time the distance perpendicular through segment AG through point F would get smaller, causing a compression force inside all of the zero force members as they try to hold the structure up. Do you know where I could search, or what I should search for, to find information on this situation? Everything I try leads me back to the "they have no force... ever" answer. Thank you.

[@DrStructure]

Yes, you want to look into second order structural analysis, which involves determining internal forces induced by deformation/displacement (P-Delta effects). The topic is rather advanced and beyond the scope of most undergraduate structural analysis courses as it generally requires solving systems of nonlinear equations.

[@Chikokishi]

Excellent, thank you for that. I am an undergrad in mechanical engineering, but the way I solve problems and understand is through my reasoning. Which leads me into predicaments such as this where I see more than my class is allowing me to understand. Which, in the end, makes it nearly impossible for me to comprehend the method of solving my problems. So, I will look into this second order stuff until the point where I feel confident enough to justify the simplifications for myself. Thanks again, we need more professors like you.

[@arijitsaha1472]

How can we take the point E where it is experiencing an external force?And also point G where there is a support reaction??

[@jacott1382]

I have a question. I understand rule 1 and 3 in your explanation, but doesn't rule 2 mean that I would first have to calculate the reaction forces in point A to know if AB is a zero force member?

[@DrStructure]

Rule 2 usually works in conjunction with the other rules. For example using Rule 3, we may determine that a member (say, AB) carries a zero force. If AB makes a 180 degree angle with another member (say, BC), then using Rule 2 we can conclude that BC also carries a zero force. So, by examining joint A we’ve concluded that AB is a zero-force member via Rule 3, and by examining joint B (via Rule 2), we have concluded that BC is also a zero-force member. You may want to see an alternative version of this lecture in our free online course. See the video description field for the link.

[@rahulanand9851]

@5:23 p is consider as axial force , what mates it axial force ?

[@DrStructure]

Any force that causes tension or compression in a member is considered an axial force. The fact that P is pulling down on the vertical member @5:23, makes it an axial force for that member.

[@kadejourney9152]

Thanks for the video, it is quite helpful. For the question (B), why are AC and CF zero force members, is joint A not supported by a reaction won't it affect the value of member AC.

[@DrStructure]

With regard to Problem B, if we remove the obvious zero-force members (ED and BD), we can see that DC is also a zero-force member. If we remove DC from the structure, then examine joint C (there are only two members connected to the joint at this time, CF and AC), we can conclude that AC abd CF are indeed zero-force members. The support at A has two reactions, one in the x direction and one in the y direction. These are the x and y components of the axial force in member AF. That is, the reaction at A is a force equal in magnitude but opposite in direction to the axial force in AF.

[@kadejourney9152]

Thanks so much

[@olanrewajuoludare8396]

@@DrStructure BUT THERE IS A AXIAL FORCE FROM THE VERTICAL REACTION AT A PASSING THROUGH AC. i guess that would not allow rule 1 to hold at joint C

[@DrStructure]

@@olanrewajuoludare8396 True, there is vertical reaction at A. But how did you arrive at the conclusion that is must pass through member AC? Cannot it not pass through member AD (or AF)? We draw the conclusion that AC is a zero-force member not by examining joint A, but by examining joint C, after DB, DE, and DC are shown to be zero-force member and removed from the structure. For an alternative discussion on zero-force member in trusses see: kzbin.info/www/bejne/ZnKpoKhtar2ZqpI Better yet, login to our (free) online course and access the updated version of our lectures there. The link is given in the video description field.

[@fontanillacarlamariellie.7261]

Thank you so much. This is so helpful.

[@ozgunsen4354]

In ex 1 why FB is not a zero force member cuz there is 180 degree line GE or same as FC and FA "Rule number 3" ?! 4:22

[@DrStructure]

In Example 1, FB cannot be a zero-force member. Rule 3 does not apply to either of its end joints. Here is why: FB has two end joints: F and B. Joint F connects five members (GF, FE, FC, FH and FB). One of these members (FE) turns out to carry no load. That leaves four other members at the joint. Rule 3 requires three members (or forces) to work, so it cannot be applied to Joint F. Joint B connects four members (BH, AB, BC and FB). One of them (BH) is a zero force member. That leaves three members at the joint (AB, BC and FB). For Rule 3 to apply, AB and BC have to be separated by 180 degrees, but they are not. The angle between them is less than 180 degrees. So no Rule 3 here either.

[@hvg71928]

Great explanation....so are the answers to the last three problems anywhere?

[@DrStructure]

Check comments for answers.

[@farrasdaffa1418]

hi Dr!! i want to ask, so the external load effected the joint when it's collinier to the frame on that joint? is it right?? sorry for late comment hehe. i can't understand why DG inn the second example (triagle with pin support on f and rol in a) not a zero force member?? ugh so hard...

[@farrasdaffa1418]

so if we say that DG on second example has a force, so why DF and DG were not an zero force member?

[@DrStructure]

DG is a zero-force member in that example.

[@netbilly8487]

Muito bom!! Excellent videos

[@husseinhussein-vi2ip]

In 6:48 how about force FD It apears that it can be zero with 3rd rule

[@DrStructure]

There is a pin support at F, the rule does not apply.

[@husseinhussein-vi2ip]

Dr. Structure sorry.in5:55 at G there is a roller and rule 3 is applied

[@DrStructure]

Yes, the rule applies at G.

[@ildemcnar8449]

@Dr. Structure Sir, I don't understand why CA and CF are zero members? they are on a pin connection?

[@DrStructure]

I am assuming your question is in reference to Exercise Problem (B). It is true that there is a pin connection at end A, but we arrive at the conclusion that CA is a zero-force member by examining joint C as explained below. Before we can draw that conclusion, we need to take three steps. Step 1: By examining joint B, since there is no load or support at the joint, we can conclude that BD is a zero-force member. Step 2: By examining joint E, we can conclude that ED is a zero-force member. Step 3: After removing BD and ED from the truss, only three members remain at joint D. Therefore, we can easily conclude that DC is a also zero-force member. After removing the above zero-force members from the truss, only two members remain at joint C: CA and CF. Since there is no support or load at C, CA and CF have to be zero-force members.

[@princessrollettetolibas4084]

@@DrStructureow i get it!! thankyou po!! ngayon ko lang nabasa toh . this channel is great!! highly recommended :))))

[@yesyes6001]

This is so helpful. Thank you.

[@marufahmednaim2004]

Sir,could you please explain why CD,BC and GH are zero force members in (C)?How does rule 3 apply to them despite having the effects of supports(In case of CD and BC) and load(In case of GH)?

[@DrStructure]

Please take a look at the lecture referenced below. It is a variation of this lecture on zero-force members. If after watching that lecture, you still are not clear, let us know and we will elaborate further. Lecture on zero-force members: kzbin.info/www/bejne/ZnKpoKhtar2ZqpI

[@trmpereira]

The rule 1 say AB and BC, would not be AB and AC?

[@DrStructure]

Thanks for the comment. Yes, it should read AC. This is noted on the video.

[@mamras9238]

thanks for making it so easy...

[@felts_9751]

In question (A), why are DF and DG zero force members when there is a force being applied at D?

[@DrStructure]

+Felts_97 The force applied to D is counteracted/resisted by force in members AD and DK.

[@YouTubeChannel2000]

Hi, Please, what type of program can I use to create such video style ? Kindly advise........

[@DrStructure]

Wajdi Al-Rekaby Various software tools have been used for creating the videos. Check out VideScribe, Camtasia Studio and CrazyTalk.

[@YouTubeChannel2000]

Dr. Structure Thank you very much doctor.

[@vaibhavraj3176]

Hey, Dr.Structure I have got the first two ones correct. However, I am a bit struggling with the question c, Could you plz explain to me how you identify the zero-force member in c? :) Isn't it there support reaction at B?

[@DrStructure]

Yes, there is a pin support at B, and one at A. Start with joint H. Note that GH has to be a zero-force member, since the joints connects three members two of which are colinear. Remove GH from the truss, then examine joint G. Again, the joint connects three members two of which are colinear. That makes the third member (FG) a zero-force member. Remove it from the truss then proceed to joint F. Using the same line of reasoning we can show that EF is a zero-force member, so is DE, CD, and BC.

[@vaibhavraj3176]

@@DrStructure Thank you so much for the explanation :)

[@DrStructure]

@@vaibhavraj3176 You're welcome.

[@syedmuhammadalkherid490]

What about the force FD for (B)? Appreciate your help, thanks!

[@DrStructure]

FD is not a zero-force member.

[@syedmuhammadalkherid490]

Thank you for the response! Why not though, if you don't mind me asking?

[@DrStructure]

Syed Muhammad Al Kherid Simply put, none of the rules applies to FD. To see why that is, remove all the zero-force members from the truss. These would be all the members to the left of DF including BD, DE, CD, CE & EF, and CB & BA. So all the members to the left of AD & DF are removed resulting in a simplified truss with fewer member. Now examine member DF in the simplified truss. At joint D, Rule 2 would have been applicable if AD was a zero-force member, but AD does carry a force, since there is a pin support at A. So Rule 2 does not apply. Examining joint F, Rule 3 would have worked only if FG and FH were co-linear, but they are not. The angle between them is different than 180 degrees. So Rule 3 does not apply either. In summary, neither of the rules can be used to conclude that DF is a zero-force member. Here is a more intuitive explanation. Remove all the zero-force members from the truss. Carefully examine the simplified structure. Since AD and DF are co-linear, let's view them as a single element (say, AF). Clearly AF has to carry a force or else the structure would be unstable. To see why, remove AF. What do we have now? The right part of the structure resting on a roller support only which is going to have a rigid-body motion, it is going to turn counter-clockwise. Therefore, for the truss to remain stable, AF must carry a force which means FD must carry a force.

[@Jigar8320]

at 6:36 how DG became a zero force member ?

[@DrStructure]

If CB, BE an ED are removed from the truss, because they are all zero-force members, then applying Rule 3 to joint D yields DG a zero-force member too. That is, three members are present at the joint, two of them are collinear, therefore the third one (DG) must be a zero-force member.

[@ravigautam8685]

In a video titled "Two force members explained (statics)" is said L-shaped rod is in equilibrium and without being zero-force member. But according to rule 1 it should be zero force member. Kindly help me to understand.

[@DrStructure]

The rules for zero-force member apply only to trusses, slender members connected together with pins. If an L-shaped system does not embody a pin, then it is not a truss configuration. A rigidly built L-shaped member, therefore, does not follow the zero-force member rules for trusses.

[@yanfu3334]

thanks for the video and the explanations on comments!!

[@princessrollettetolibas4084]

@Dr. Structure hi :)) i can't understand why in truss B: member CA is a zfm? there's a force acting at joint A, so why its a zfm? i am sorry it's just confusing, but according to your rule #2 if there's a force acting at joint A, therefore there's a force acting at member AC, joint B is not included because member BD is a zfm.. so why member AC is a Zfm? can you enlighten me pls? tnx in advanced :)))

[@DrStructure]

If you are referring to the exercise problem, please review a different (and more recent) version of this lecture (see link below). There we have actually solved that exercise problem. There is also a pdf version of the lecture that you can download (the download link is given in the description field of that video). kzbin.info/www/bejne/ZnKpoKhtar2ZqpI

[@princessrollettetolibas4084]

@@DrStructure wow thankyou!! godblessyou and i hope you continue what yoou are doing coz its a great help to every students like me

[@anupamkushwaha4840]

AT 5:36 HOW RULE 3 IS APPLIED AT "EF" AND "GA"???...............PLZ EXPLAIN IT

[@DrStructure]

+Anupam Kushwaha @5:36, there are three forces acting at joint E, the applied force (P), the force in member CE and the force in member EF. Two of these forces are co-linear, therefore the third one (the force is member EF) must be zero. @5:50, there are three forces acting at joint G, the horizontal reaction force at G, the force in member GF and the force in member AG. Since two of these forces are co-linear, the third one must be zero.

[@anupamkushwaha4840]

THANK U FOR YOUR RPLY....

[@DrStructure]

+Kosmosis From a practical standpoint, you are correct, we are breaking the rule. From a conceptual standpoint, however, supports and loads can be viewed as forces applied at truss joints. In this sense, we are replacing a roller support with a force to make the rule applicable. The important consideration here is that for Rule 3 to apply, we need to have only three forces acting at the joint regardless of their source.

[@anupamkushwaha4840]

thank you for your explanation

[@rafiqullah7437]

Electricity transmission tower modeling alway consider the legs members are beam and the rest of truss what is the reason and can we consider the also as truss members

[@DrStructure]

"always" is not quite a fitting adverb to use here. The designed configuration of a structure, any kind of structure, depends on many factors. There are no fast rules for a transmission tower to be a hybrid system as you describe it.

[@yogeshpanchal575]

Can you tell me which software you used for this video ??

[@DrStructure]

For this particular video, the notes were traced manually on an iPad, then saved as svg file(s). These files were ran through VideoScribe (an online service) for converting them to video (avi) files. The avi files were compiled into a single file, and the video and audio tracks were synched using Camtasia Studio. The talking head was produced using CrazyTalk.

[@janynnebm]

Thanks! Excellent video!

[@weishentan369]

Why is HK and KJ not a zero force member for question B?

[@DrStructure]

Thank you for the question! Clearly, the stated rules do not permit us to conclude that HK and KJ are zero-force members. But, is it possible to determine if a member carries a zero force any other way? Yes, you can determine the force in a member by actually analyzing the truss. So, when in doubt, analyze the truss. In this case, we can use qualitative reasoning to conclude that neither HK nor KJ is a zero-force member. 1. If either HK or KJ is a zero-force member then GK must also be a zero-force member. This can be determined from force equilibrium at joint K. There are three forces (members) at K, if one of them is zero, the other two must be zero too. 2. Member GH carries an axial compressive force equal in magnitude to the vertical force applied at H. Let's replace member GH by a vertical force at joint G. 3. Member GJ carries an axial compressive force equal in magnitude to the horizontal force applied at J. Let's replace the member with a horizontal force at G. 4. Now, visualize member FI as a single straight truss member with joint G in the middle of it. This becomes an unstable mechanism if any transversal load is applied to G. Since there is a horizontal force and a vertical force at G, the system is considered unstable unless member GK carries a force. But, if GK carries a force, HK and JK cannot be zero-force members (from 1 above).

[@weishentan369]

Dr. Structure Thank you for the detailed explanation.

[@singla6746]

@@DrStructure u are amazing. I love u so much cuz u save my asm!!!

[@felixchikwu9386]

Thank you Dr

[@felixchikwu9386]

Kindly send me the link to the software to enhance my analysis as a fact checker

[@DrStructure]

@@felixchikwu9386 What software are you referring to?

[@DrStructure]

We have developed 2 apps so far, one for truss analysis and one for frame analysis. You can find the download link for each in the description field of the video that introduces each app. iTruss: kzbin.info/www/bejne/i564nGyKdtGle8U iFrame: kzbin.info/www/bejne/l6rCn6uGo7Kfas0 If you experience any difficulty with the download and/or install process, let us know.

[@khondokarabuzafor865]

Sir!! In your lecture you have told that zero force member will only count if joint doesnot carry any support or any load. but in your 1st example u have considered load p as a member and using rule 3 you have declared "EF" as zero force member. why??

[@DrStructure]

We think of a truss member as an axial force. That is what the member carries. Conversely, we can think of any force (i.e., applied load or support reaction as being an axial force in a virtual truss member) for the purpose of identifying zero-force members in trusses. For an alternative presentation on the identification of zero-force members, please see: kzbin.info/www/bejne/ZnKpoKhtar2ZqpI Hopefully that makes the underlying concept and its application more clear.

[@khondokarabuzafor865]

Dr. Structure Thank you sir!! for your help!!

[@uchienhoang3302]

I think in question B example, GK should also be a ZFM right? cause FGI are co linear. pls help

[@uchienhoang3302]

i mean the last problem, question B

[@DrStructure]

No, GK is not a zero-force member. There are five members, not three members, at G. So, the co-linearity of FG and GI does not lead to that conclusion. Neither HG nor JG is a zero-force member, so GK cannot be a zero-force member either.

[@connorwetzel6596]

On problem (B), I am finding that the only members that ARE NOT zero force members are members IG, HG, and JG. Is this correct, or have I inappropriately used the rules somehow?

[@DrStructure]

+Connor Wetzel No, you went too far. Rule 3 at joint B makes BD a zero-force member (zfm). Rule 3 at E makes DE a zfm. Then, Rule 3 at D makes CD a zfm. Then, Rule 1 at C makes ABC and CEF zfms. That is it.

[@mohammedhanii230]

what about GK ?

[@DrStructure]

GK is not a zero-force member.

[@nataliejudkowsky1339]

In rule #1 shouldn't it be AB and AC? (instead of AB and BC?)

[@DrStructure]

Yes! It should be AB and AC. Thanks for pointing it out.

[@432bismanazir6]

Sir in B anc C ..what are the supports ...are they both hinge supports?

[@DrStructure]

Yes, in both trusses the supports are pin.

[@432bismanazir6]

@@DrStructure sir then the formula ...r+b=2j is not satisfied..if both are pin joints..

[@DrStructure]

For Truss B, there are 11 joints, 18 members, and 4 support reactions. 2*11 = 18 + 4 For Truss C, we have 9 joints, 14 members, and 4 support reactions. 2*9 = 14 + 4

[@432bismanazir6]

@@DrStructure okk.. thanku sir.

[@DrStructure]

You’re welcome.

[@ykirankumar4985]

sir suppose if two 3 membered joint has common non-collinear member and an external force is applied at one joint and at other joint is no force applied ,then what about that member ? as in the sense of no force joint that non-collinear member is zero force member but it is contradict in the sense of force applied joint

[@DrStructure]

If there is a truss joint that connects two (non-collinear) members only, and there is no load or support at the joint, then the members are indeed zero-force members, regardless of the connectivity and load pattern at the other end of the members. Say, Member AB is connected to only one other member at joint A, and the angle between the two members is different than 180 degrees. Further, there is no load or support at joint A. But AB, at joint B, is connected to two other members, and there is an applied load present at B. Can we conclude (by examining joint A only) that AB carries no force? Yes! But what about the applied load at B? That load has to be carried by the other members at B, but not by AB. If the other members at B are unable to effectively carry the load, the truss would collapse. It would be considered unstable.

[@ykirankumar4985]

is it the same in case of non-collinear member connecting two 3 -membered joint if 2 members are collinear at each joint?

[@ykirankumar4985]

what about if applied force is in collinear with non-collinear member at one joint?

[@DrStructure]

Per example above, if at joint A the two members are collinear, then they may not be zero-force members. If the collinearity is at the other joint(s), then AB still is a zero-force member.

[@DrStructure]

Per example above, if a load, collinear with member AB, is applied to joint A, then AB carries that load. But, the other member connected to A would be a zero-force member.

[@AllFictionful]

why is ab not zero for A, if u start at A the pin support should have an X force, and with AF it is 180 degs shouldn't that mean ab is zero member?

[@DrStructure]

I am not exactly sure how you are thinking about this problem. But,... At A, we have Ax and Ay, and there is a force in AF. But why would that make AB is a zero-force member? Why are you assuming that the force in AF equals to the sum of Ax and Ay?

[@AllFictionful]

Understand now, thanks!

[@fanbingbing7390]

I was wondering about rule 3 the third member must perpendicular to colinear ?

[@fanbingbing7390]

oh I just prove it can be not perpendicular to colinear.

[@황태종-o4h]

HMM I have question is Zero-force member existence only structural turss????

[@DrStructure]

The term zero-force member is generally associated with trusses. But, the basic idea behind a structural member carrying a zero force under certain loading conditions is not limited to trusses. When analyzing a frame structure, we could encounter a beam segment with zero bending moment and shear force.

[@benjaminminer8972]

This was really helpful! Thank you!

[@bekis95]

Hi in problem b why aren't DF and GF zero force members? If you are looking at point F

[@DrStructure]

+bekis95 After removing the known zero-force members, three members remain connected to joint F: FD, FG and the horizontal member to the right of F. So, none of the rules applies to joint F. Here is a more intuitive explanation. Except for FD and AD, all the members on the left half of the truss are zero-force members. If FD was also a zero-force member, the applied load had to be supported entirely by the right side of the truss. But that is not possible as there right side rests on one (pin) support only. A stable truss must have at least two support points.

[@bekis95]

+Dr. Structure thanks makes sense now!

[@jaykantariya2403]

after 3 rules in a first example for member FE and AG how 3rd rule is applied ?make me clear..!!

[@DrStructure]

Since truss members carry an axial force only, for the purpose of our analysis, there is no difference between a force/load and a member. So, when needed, we can view a force/load as a truss member. At G, the support reaction (force) can be viewed as a member. The same can be said at E, the applied load can be viewed as a member for the sake of using the rules.

[@jaykantariya2403]

Thank you very much Dr..

[@ahmedkamal123123]

Is it possible for sacked zero member ? and why ?

[@DrStructure]

The design may require the presence of a zero-force member for stability, as was mentioned at the beginning of the video. In such a case, the removal of the zero-force member from the structure could lead to the failure (buckling) of another (compression) member. Further, when the structure is subjected to dynamic (time-variant) loading, a member may not carry the same force at all times. A truss member that carries a compression or tension force at time t1 may become a zero-force member in time t2. Obviously that member cannot be removed from the structure simply because it carries no load at t2 as it does carry loads at some other times. If a truss member does not contribute to the stability of the structure, and carries no load during the life-span of the structure, then from the engineering standpoint it serves no purpose and should not have been placed there to begin with, unless it has some non-engineering (say aesthetic) purpose.

[@nadeemkuskiwala6220]

If truss joint is not carrying any loads and not supported a pin or roller...but according to rule 2 you gave simple example of pin and roller supported joint..so why this members are zero????? means why AB is zero??

[@DrStructure]

+Nadeem Kuskiwala If two members are collinear, as stated in Rule 2, and one of them is a zero force member, then the other one has to be a zero force member as well. Why? because we can think of the two members as one long straight member with a pin in the middle. If one part of the member does not carry any force then the other part cannot carry a force either.

[@mollypete4270]

Why would you not provide the answers at the end?

[@DrStructure]

See the comment pinned at the top here

[@ankur29]

could you add the answers to the last three please

[@DrStructure]

You can find the answers in the comments section.

[@mollypete4270]

Where?

[@Muddassir_Ahmed]

can you please explain A problem (you had given in end of video)

[@DrStructure]

Start at Joint C: CE must be a zero-force member since the other two members are co-linear. Remove CE from the structure. Now examine Joint E. BE has to be a zero-force member since the other two members are co-linear. Remove BE, then examine the remaining structure. Using the same line of reasoning, we can conclude that BF is also a zero-force member.' Remove BF, then examine the truss one more time. Joint F now connects three members only, two of them are co-linear. Therefore FD must also be a zero-force member. We can apply the same line of reasoning to the other side of the truss concluding that IH, JH and JG and GD are all zero-force members.

[@Muddassir_Ahmed]

Thanks alot sir you are great

[@krishnagoyal1263]

According to me GK,HK,JK should be zero forces Members but cannot understand why it is not in answers ?? IN TRUSS 2

[@DrStructure]

How do you arrive at that conclusion? Please elaborate.

[@krishnagoyal1263]

@@DrStructure AS we can see in structure that force is acting on H so Fh and HK should be zero,Similarly Force is acting on G so JK should be zero and now if we will draw FBD of point K we will see GK is also zero as Fx and Fy is already zero at K. Thanks

[@DrStructure]

You wrote "... force is acting on H so FH and HK should be zero." Your assertion is not correct. It is true that a force is acting at H, but that does not mean that FH and HK carry no force. All we can conclude from joint H is that (1) the force is GH equals to the applied load, and (2) the force in FH is equal to the force in HK.

[@krishnagoyal1263]

@@DrStructure THANK YOU VERY MUCH FOR CLEARING MY DOUBT .I HAVE NOT YHOUGHT THAT SOMEONW WILL REPLY TO MY QUERY IN SUCH ASHORT TIME. THANKS YOU ARE REALLY DOING GREAT WORK.

[@DrStructure]

You’re welcome.

[@teteyaw4006]

is that Maximillion Pegasus?

[@XanderMande]

How do we apply Rule 3 in FE?

[@DrStructure]

Examine joint E at/around 4:50. How many forces do you see at the joint? You should see three of them. Anytime you have three members, or three forces intersecting at a join, and two of them are colinear, Rule 3 applies.

[@maui_l1675]

how to identify load carrying members?

[@DrStructure]

By actually analyzing the truss.

[@sundaylawal4849]

I have a question but I don't know hw you can help me to solve it

[@shanaesara5550]

Hi +Dr Structure what are the answers in (A)?

[@shanaesara5550]

+Dr. Structure

[@DrStructure]

You should be able to find the answers in this comment section.

[@farrasdaffa1418]

how can in the second exercise, GK have a force?

[@DrStructure]

GJ carries a force and GH carries a force. Their forces are transmitted to joint G, which then needs to be balanced (sum of the forces at the joint must e zero) with the other forces acting at the joint. One of the other forces acting G is GK. So it must play a part in balancing the joint with a non-zero force.

[@rafiurrahman6596]

@@DrStructure But there is IF member for balancing as well.For instance, GF & GI, they will also paly thir role, right?

[@DrStructure]

@@rafiurrahman6596 Correct! They all play a role in transferring the loads.

[@CandidlySubtle]

No solutions to those last 3 problems?

[@DrStructure]

Yes, they are here, in the comments section. They were posted about six months ago.

[@CandidlySubtle]

Dr. Structure The only comment I see from 6 months ago was written by sketzartist. You can also post the solutions in the "About Video" section, right?

[@CandidlySubtle]

Dr. Structure Okay. Now I see it. This? (A): None of the interior members (BF, BE, CE, DF, IH, HJ, GJ and GD) carries any force. The three members forming the outer triangle (AD, KD and AK) are the only ones that carry any force. (B): BD, DE, CD, CA and CF are zero-force members. (C): GH, FG, EF, DE, CD and BC are zero-force members.

[@DrStructure]

Yes.

[@dkarivazhagan]

hi ur explanation is amazing and i need more videos such as SA02 , SA03 , SA08 etc

[@terobau8177]

thanks doc :D for your awesome video.....can you please provide me the anserw for the problem no (B). thanks.

[@DrStructure]

TERO BAU Please review the comment section, You find relevant info about the exercise problems there.

[@RainbowPaint]

how come GK is not a zfm for B?

[@DrStructure]

Since there is a vertical load applied at H, then GH cannot be a zfm. Similarly, since there is a horizontal load applied at J, GJ is not a zfm. Therefore, we cannot conclude GK is a zfm; none of the rules apply to joint G.

[@かっテャン]

thank you very much!!!!!!!!!!!!!!!!!!

[@frankmiguelmenchaca1942]

esta mega cool el video tipo me encanto. cuero el viejito tmb :)

[@emmanuellakegyamensah9536]

Isn't BC a zero force member?

[@DrStructure]

In which truss?

[@anupamkushwaha4840]

IN THE EXAMPLE GIVEN IN FIRST RULE ............ IS "CE" MEMBER ZERO FORCE MEMBER???.........IF NOT THEN PLEASE EXPLAIN IT......

[@DrStructure]

+Anupam Kushwaha Not sure which example you are referring to, please give video location on the timeline.

[@anupamkushwaha4840]

AT 3:19........

[@DrStructure]

+Anupam Kushwaha Yes, CE, and DE, and CB and CD are all zero-force members in the example truss.