Truss analysis by method of joints: worked example #1

[@kevinnarine1106]

Very helpful. This chapter has been the hardest for me yet and I believe I finally have this stuff figured out. I appreciate you explaining the "why" aspect of what you're doing rather than assuming that the audience knows as much in Statics as you do. I think some professors need to remind themselves that this stuff is going to be obvious to them when they've been doing this stuff for 20+ years, and not so obvious for the students learning this for the first time.

[@Engineer4Free]

Yessss glad that my explanation worked. What you described is exactly what I try to accomplish in the videos. Cheers 🤜🤛

[@justind5742]

What am I even going to college for? lol. You explained analysis of trusses better in 15 minutes than my prof did in 2x 2 hour lectures

[@Engineer4Free]

That's a good question.

[@papess690]

College? Is for us to know that there are many things we don't know...

[@the_candid_mechanic]

That's true

[@anuragexploration212]

what if do not change directions of truss at any specific joint at all(in other words leave all forces at every joint going outwards(tensile) no matter what and just while substituting the force while solving the next joint put a minus sign i.e if got a force as compressive the ans is -ve so instead of changing the direction of force just substitute the given -ve force directly while solving next joint.

[@lingeswwarannedunsezian1697]

How to calcute the reaction at support(im not sure whether its positive or negative)

[@veniaminnossov4929]

This is the one topic I have struggled to grasp and you sir have explained it and presented it flawlessly. I cannot thank you for how much you have improved my chances of getting a better grade.

[@Engineer4Free]

Thanks for letting me know Venya.. really happy to hear that I could help!! I made several other truss and statics videos here: engineer4free.com/statics =)

[@shanehew2373]

Best "method of joints" tutorial I could find on youtube. it really helps thanks for sharing. I hope u r doing great wherever you are now.

[@Engineer4Free]

Thanks for the kind words Max! Hope you are doing great too :)

[@Rino10493]

+Engineer4Free I also agree! Thank you so much for helping me

[@muhammadazrul3055]

True

[@starboii568]

Is the same true if i get -350 for my ay?

[@himanshukumarsingh8234]

this is the best video on KZbin for the analysis of truss, short, simple, straightforward, clean, exact and has got the best explanation. Thank you so much for the amazing video.

[@Engineer4Free]

Thx bro. Please share my vids with some other classmates!

[@himanshukumarsingh8234]

@@Engineer4Free sure, will do the same

[@SnapToothbrush2]

I’m in the middle of taking my final and I think it’s safe to say you’ve saved my grade i wish you were my professor

[@Engineer4Free]

Thanks!! Glad you like the video. Also.... you are in the middle of your final and on YT? Zoom University lol

[@ralff4272]

I am so doing this tomorrow lol

[@Freakybananayo]

Very clean presentation, straight to the point, no time wasted drawing all it out in real time. very useful video

[@Engineer4Free]

Yo Freaky Banana, thanks for the feedback...glad you like it =)

[@HazzCraft]

my mans a literal life saver, amazing work diving into topics specific to an engineering course. you explain it heaps better than my lecturers with thick accents and the long tangents they go in mid class

[@RAZZ_ARCH]

You explain better than my uni Lecturer will do in 1hr without even hearing him not to talk of understanding how he's going about it Thanks so much

[@Jesssuuusss_.]

This is the best KZbin I've ever watched.. Wow...you got good stuff

[@brightappiah3581]

This video has really explained things to my understanding. In fact I'm speechless. This video has just made me pass my elementary structures tomorrow... I'm really grateful

[@Engineer4Free]

Amazing!!! Thanks for letting me know!!! If you have more left in the course, you might want to check out the other examples @ engineer4free.com/statics =)

[@painteur8169]

i have to say this is the best lesson out of the south sahara , you explained a whole 4 hour lecture in two simple videos , ps ive subscribed to the channel and ready for more engineering videos

[@Engineer4Free]

Awesome, thanks Kelvin!!! =) =)

[@aaronconnolly12]

This is hands down the best video I have ever seen . Thank you!!

[@Engineer4Free]

Wow, thanks Aaron! =)

[@jamesevans1105]

Brilliant video, learnt more in this than I did in my lectures and attempting worksheets

[@gauteantonsen3907]

Best "method of joints" tutorial I could find on youtube!

[@Engineer4Free]

Thanks man! That's awesome to hear!!

[@mateusz4611]

My man just explained this better in 15 min than my lecturer managed to in 5 hours.

[@Engineer4Free]

😏

[@miliamokhtar4506]

Thank u so much for teaching this for free! I watched most of ur videos whenever i hv trouble understanding something and in need of more examples to solve a problem. Theyve helped me a lot and u make it so easy to understand 😁

[@Engineer4Free]

Awesome!! Thanks for the great feedback 😁😁

[@lilJazza360]

You just saved my exam for tomorrow, god bless you

[@Engineer4Free]

Awesome, glad to hear it! How did the exam end up going?

[@cosmicdust5436]

and here i am ..saved for my exam for tomorrow..LMAO

[@youssefwalid5163]

u saved me for tomorrow's exam too lol

[@danieldelgado5235]

@@Engineer4Free I guess we’ll never know

[@human2373]

got a test tomorrow too lol

[@aafaqkhan9565]

It feels like i have wasted my whole semester classes by listening to boring professor. Many Thanks for saving my paper. Cheers!

[@Engineer4Free]

Sorry to hear, but you’re not alone. Glad I could help 🤜🤛

[@Engineer4Free]

Sorry to hear, but you’re not alone. Glad I could help 🤜🤛

[@darrylros681]

this tutorial explains everything I cant learn in school, I dont know why some teachers in universities dont have this kind of talent when it comes to teaching.

[@Engineer4Free]

Thanks Darryl! In case you haven't seen the rest of the playlist too, its here: engineer4free.com/statics There are more truss tutorials in videos 42-51👌

[@darrylros681]

@@Engineer4Free Thank you! 💙

[@CreatorFlex]

watched this legit 20 minutes before my exam and it helped so much

[@Engineer4Free]

Glad to hear it! Now you know where to come before your next exam too

[@omarfaruk121]

I don't understand where do you find 2 m and 1.5m

[@OrangCerdas-sc7cw]

you dont find it, it's just an example if it is

[@anishahasnat2837]

omg. finally a video that actually makes sense. youve made this so easyyyy. oh god thank you so much

[@sussy_61]

this guy explained it better than my lecturer, thanks man

[@Engineer4Free]

You're welcome, thanks for watching! Check out the full playlist at engineer4free.com/statics 👌

[@kvpgamer3168]

for joint c, why is the horizontal force of AC not -28.85 cos60?

[@kvthe2nd903]

You literally saved my day, thankyou! I've been told this 3 times before but didnt understand it.

[@Engineer4Free]

Really glad I could help. Check out all the videos I did at engineer4free.com/statics , I have a few more on this method there too

[@nigel8221]

Grear refresher for me since we've been in quarantine here in the Philippines for about 2 months now, I forgot my statics😂, I'm now in my Deformable, I'm trying to remember how to solve indeterminates using stress/strain. GOOD WORK SIR THANK YOU🙏

[@Engineer4Free]

Glad I can help!!! Keep it up my friend 🤜🤛

[@Engineer4Free]

PS I’ve got tutorials on indeterminates in these other courses: engineer4free.com/mechanics-of-materials and engineer4free.com/structural-analysis ... if you haven’t seen them already!

[@sterioboy676767]

Thank you so much, literally saved me from hours of frustration.

[@mrmoeguy4664]

This video is always a life saver the night before exams

[@christianfekete]

Very good demonstration, thank you d’or sharing. With your nodes analysis this covers it all. So helpful!

[@s.amirmohammadhassanli9890]

Hi :-) Thank you for your time to making this video! this is really helpful!!! I hope u r doing great wherever you are now ;-)

[@ahmedothman6642]

this guy is always saving me in the day before the exam, god bless you brother

[@dram906]

when I did sin60 my ti-84 gives me -.3048 is there a reason our numbers are different?

[@ahmedothman6642]

@@dram906 maybe ur calculator, reset it because sometimes its on radian system change it to degree

[@dram906]

@@ahmedothman6642 thankyou got it now!

[@ahmedothman6642]

@@dram906 no need, good luck

[@dellpi3911]

kzbin.info/www/bejne/nIPchpljfL5qa5Y

[@luuhax]

I missed the exam last year for this topic, and now when I'm about to retake it I knew i could just come back here to refresh my memory. Excellent video

[@gauravnerkar2982]

Thanks sir for this great tutorial on 'The analysis of Truss'.... because of you I got the simple method to solve problems without any confusion....

[@Engineer4Free]

Glad to hear it man! Thanks for watching, and if you haven't seen the rest of my statics tutorials, then do check them out at engineer4free.com/statics :)

[@davidmuniz5307]

Your explanation made these problems much easier to solve. Thanks

[@Engineer4Free]

Thanks for letting me know David, cheers!

[@krystallineheart]

Well hats off to Engineer4Free for still replying to comments and keeping people updated even after 4 years. You're amazing

[@Engineer4Free]

Thanks Atharva! Still trying🤜🤛

[@saketpandey18]

Thanks a lot this video helped me a lot in understanding joint method. Love from India. You are doing a great job.

[@Engineer4Free]

Thanks Saket!! I have more tutorials at engineer4free.com/statics see #42-51 for more truss examples👍

[@saketpandey18]

@@Engineer4Free You are welcome. I will surely watch those videos. I have issue with section method

[@JohnDiggle21]

I have a solid mechanics exam today and this just helped me understand this topic so quickly! Thanks!

[@Engineer4Free]

Good luck!!! Plenty more vids btw at engineer4free.com/statics and engineer4free.com/mechanics-of-materials 👌

[@anask4052]

i got one in 3 days lol

[@JohnDiggle21]

@@anask4052 good luck

[@adegboyepromise4683]

You just explained in 15mins what engineer hamzat couldn't make me understand in 2years ...hamzat

[@elijahcholayiik1794]

The best methods of joint you real elaborate whoever coming to cross these tutorial, the easy way of getting all forces acting at X-direction and Y-direction just you need to dismantle the Truss by FBD

[@themansans1929]

This... this is amazing. You just earned another sub!!

[@Engineer4Free]

Awesome!! Welcome 😊😊

[@superslayer7040]

Well this helps can’t wait to learn more when I’m done middle school

[@Engineer4Free]

Keep on it!!! 🤜🤛

[@freserlyn1]

Wow. This is much understandable than my 2-hour class. I hope you’d do videos in all my engineering subjects too. Hahaha Thanks!

[@Engineer4Free]

I'm working on it! Check out engineer4free.com for all the courses, new vids and courses are always being added!!

[@Babiesworld-mm]

Ah the problem is that when i tried to draw it as exactly as you say, all members 1m long and the 60 degree using AutoCad. The height of truss isn't 1m.

[@Engineer4Free]

Yeah, that 1m vertical measurement is a typo, the height of the truss is 0.866m. But it never impacts or is used in the calculation. Sorry if that confused you. Every member is 1m long, and the angled ones are all at 60 degrees from the horizontal.

[@darwinlim4837]

I really try to see this comment relating to this since all my calculations are different bec at start it uses the wrong dimension. But gladly the lesson didnt affect the essence of knowing the method.

[@andrenjiomou4342]

Thanks

[@Engineer4Free]

Your welcome, and thanks so much!! 🙌🙌

[@ateate650]

One question brother! #7:31, 57.7cos60° +57.7cos60° is going to be 57.7N why not 115.4 with negative sign

[@rousrudolfraemona.3898]

Auto subscribe, this tutorial so far is the best for me. Thank you sir!

[@Engineer4Free]

Thanks for the sub!! =)

[@ahmedkere-ahmed9594]

Thank you very much for the in-depth explanation

[@Engineer4Free]

Glad it was helpful! thanks for watching =)

[@cellerysandwich]

clear and straight to the point. amazing explanation.

[@Engineer4Free]

Thanks for watching!!! =)

[@danargumelarinc9660]

This is totally a breakthrough .... Good job.

[@Engineer4Free]

Awesome!! Make sure you check out the rest of my statics vids at engineer4free.com/statics , the truss videos are numbers 42-51

[@someguy351]

statics test in an hour, life saver lmao

[@Engineer4Free]

haha how did it go?

[@jellyman-]

@@Engineer4Free rip

[@Engineer4Free]

😐

[@faridb40]

@@jellyman- hahaha, how do you know? lol My finals in 8 hours...wish me luck

[@victormike7696]

Great content! quick question though, when solving for joint B on the Y axis, you used AB as 57.7 instead of -57.7. Why?

[@abhinavnair3962]

Cuz the vector is headed up and to the right. Remember that up right and an anti-clockwis moment are all the 3 positive directions that he is using. If he was using down left and clockwise as positive directions then that would be -57.7

[@dsaun777]

AB is in compression meaning it is pushing out internally a force of 57.7 towards B

[@Tony_Leigh_Gaming]

Where did you get the 2 meters from on the free body diagram?

[@Engineer4Free]

Each member is 1m long (pink text, top left of screen at beginning), so the distance from A to E is 2 members, or 2m. Please note that the 1m vertical measure is a typo, it should be 0.866m, but doesn't actually affect the problem. There's a note in the description mentioning it.

[@Tony_Leigh_Gaming]

@@Engineer4Free Thank you so much!

[@Engineer4Free]

You're welcome! Plz share my statics videos with some other students too 🙌

[@meetlucas]

I think I've forgotten how to do basic trig lol. Can anyone explain to me why sin is used when it is, and why cos is used when it is?

[@Engineer4Free]

Haha all good. Take an angle that you know. Construct a right angle triangle such that the two members on each side of the angle are two of the sides of said triangle. Draw the third side in a way that makes it a right angle triangle. Identify which side is the hyp, opposite, and adjacent from the perspective of the known angle, then just plug and chug with SOH CAH TOA to find any unkniwn sides that you need. Example drawing is here: imgur.com/a/YCT6I1a 👌

[@mitchellberends103]

You are the best saved me alot of time

[@Engineer4Free]

Awesome!! Check out engineer4free.com/statics for the rest of the statics videos =) =)

[@jcnotnot8120]

Why does A have two components in Ax & Ay while E only gets a vertical component?

[@christianmedina74]

If you look closely at E(the figure). The "triangle" has two circles underneath, that means it will only have the y-component

[@jcnotnot8120]

Christian Medina thanks. Another question: if the 200N force was applied at E instead of D, would there be a reaction at D and B?

[@engahmedou2024]

JC NotNot because the pin provides two reaction but the roller provides one reaction force

[@Engineer4Free]

Yeah thanks Christian, you’re right, A is a pin, and E is a roller. Some people draw these slightly differently

[@Engineer4Free]

You’re getting your definitions mixed up. Reaction force is a force exerted on the structure from its connection points to sold ground. The sum of the reactions forces will be equal and opposite to the applied force(s). B and D don’t have reaction forces acting on them, because they are not the points where this structure is connected to solid ground. The 200N force is the applied force. If it were moved to E rather than D, the reaction forces at A and D would be different than they currently are, and the internal forces in all members, including those that touch D and B would be different,

[@aliyubabamohammed7532]

yo thanks a lot at first I wasn't understanding then later I actually did understand keep doing your thang god bless

[@Engineer4Free]

Word glad to hear it. Check out all the truss videos at engineer4free.com/statics if you haven’t already, they are numbers 42-51 ✌️

[@GHOST84559]

Thank you very much.....God will surely bless you for helping us understand My problem was the direction of the forces in the members at first and now you've helped me understand it I'm Soo happy ....thanks!

[@victoromenye6807]

If BC from joint B is in tension, wouldn't it make BC from joint C negative? ....why was positive BC used when calculating in CD?

[@edwardchavez5621]

Yeah I've gone through this video twice and this exact part confused me as well. I believe he simply made a mistake, because for all other joints he followed his positive sign convention of: right for x-directions and up for y-directions. Just be sure to follow the positive sign convention to determine if the value should be negative or positive. The negative magnitude only means the beam is in compression, and thus shifts directions.

[@Engineer4Free]

Hey so I'm not sure what you mean by "make BC from joint C negative." When drawing the FBD of joint B, we determine that member BC is in tension, and that internal axial (tensile) force is 57.7N. When a member is in tension, it is in tension everwhere. A two force member in tension will "pull" on the pins that support it. So when we look at FBD of joint C, member BC must "pull" on it, so the arrow that represents that force on the FBD points away from joint C. Now, when we solve for CD, we use the FBD of joint C. Just like any other basic FBD, we consider left to be the positive x direction and up to be the positive y direction. BC points up and to the left, so its y component is positive, and its x component is negative. The equation that is solving for CD is the sum of forces in y expression, so when we write the y component of BC, it gets a positive value. I mention it at 7:52

[@Engineer4Free]

@@edwardchavez5621 Hey I replied, to the parent comment... just letting you know incase you don't get a notification for it.

[@banrtv]

I hve an excellent teacher, but on trusses i missed the class, so i m watching u thanks

[@thinalnim]

so neatly explained :) thank you sir!

[@Engineer4Free]

Most welcome!!! More truss vids @ engineer4free.com/statics =)

[@thomasrichardson504]

Honestly mate, you're a legend!!!!!!

[@chandiyadevi3040]

at 6:42 why have you taken the value of AB as positive, since the actual value of AB you have got is -57.7

[@Engineer4Free]

Yeah so initially AB is drawn as an unknown in tension. It's found to have a negative magnitude at 3:27 which indicates it's actually in compression. At 4:47 I update the FBD of joint A and B by switching the direction of AB on each one. This switch goes from the forces pulling on each joint (tension) to the forces pushing on each joint (compression). Negative tension = positive compression and vice versa. This is why I write (C) and (T) beside each answer as we go as well, because it's easy to get lost in here with what is in tension and what is in compression, but by writing the letter beside it, it hopefully reduces the chance for confusion or error. What I could have done to be more clear would be to write the magnitude and sense on the main drawing to the left each time I solve one. This would have included switching the direction of the AB arrows at 4:47 like I did, but then also writing "57.7N (C)" in between. Without doing that, I see how it's easy to get confused about whether it is considered positive or negative. I hope that helps to clear it up!

[@jessefaden3379]

0:37 i don't understand why the hight is 1m

[@Engineer4Free]

Oh yeah that's a typo, but doesn't actually affect the problem. All members are 1m, so the height is actually 0.866m sorry for the confusion!

[@jessefaden3379]

@@Engineer4Free Thank you so much Your video helped me a lot

[@BODYBUILDERS_AGAINST_FEMINISM]

7:30 it didn't click until you said this. Now I understand. If it's pushing on the joint it's compression, if it's pulling the join its tension.

[@Engineer4Free]

Yessss glad to hear it makes sense now 🙌

[@crispin3475]

I had a similar problem but the truss members were 1.5 meters instead. I got the exact same answers (forces) in the vid tho which makes me kinda suspicious of my own answers.

[@Engineer4Free]

It actually doesn't matter how long the members are, just that they all are the same length. That creates equilateral triangles, which give the same proportions and 60° internal angles no matter the length 👌

[@crispin3475]

@@Engineer4Free awesome! Thanks a lot

[@emmanmismanos6694]

man, you're a life saver!

[@Engineer4Free]

Happy to help! You should check out the rest of the statics vids I did here too: engineer4free.com/statics =) =)

[@foofoogaming551]

I have wasted 8hrs in classroom but you made it worth in 15mins........

[@chryssemansmilanes7304]

Hi, great vid! However i'm a bit confused as to why you removed the negative sign in 5:32 and 10:30 (-57.7N, C)?

[@fubrian2945]

Same, did you figure it out?

[@chryssemansmilanes7304]

@@fubrian2945 Hi, yes, apparently you have to assume that all forces in joints are in tension, even though you know that a force member is not. The sign just varies on the direction of the force. You can construct an XY plane to make your FBD easy :)

[@chryssemansmilanes7304]

@@fubrian2945 kzbin.info/www/bejne/rqO4ZJ9nh9usn6c here is a wonderful tutorial too

[@Engineer4Free]

Yeah, it's best to always assume every unknown to be in tension, and always draw them pulling on the joints. Then if you get a negative number, you know it's actually in compression. The next time that compressive member is used in calculation at a joint, actually draw it in compression, pushing on the joint, because it is now actually known, but then draw the other unknown forces in tension and repeat.

[@fubrian2945]

@@Engineer4Free tension is when the forces point towards each other in the truss member right?

[@parkerfoutch1739]

How did you get the formula ( EFx=50N+AB SIn(60)=0) at about three minutes in for the ABC joint. specifically the sin (60) part

[@Engineer4Free]

All angles in this problem are 60 degrees. That is because all members are the same length, so every triangle is equilateral, so every angle is 180/3=60 degrees. So AB is 60 degrees off the horizontal. We use the SOH CAH TOA trig functions to find its x and y components. Draw a right angle triangle such that AB is the hypotenuse, and the 60 degree angle is included. ABx = ABcos(60) and ABy = ABsin(60). See this diagram for help, it's a slightly different orientation, but the same idea: i.imgur.com/Dvdqf2t.jpg

[@parkerfoutch1739]

@@Engineer4Free thank you so much. I know this video is old and my question was probably below your ability, so I appreciate your reply.

[@Engineer4Free]

No worries! It's a very common question, hope it makes sense now! There are a few more examples at engineer4free.com/statics in the trusses section that would be wirth watching for more practice!

[@pesachnestlebaum]

One thing I will suggest is not to use the same angles for every angle. It wasn't always clear which angle you were using to make each calculation and I couldn't differentiate because they are all the same. Other than that, great video and thank you!

[@Engineer4Free]

Oh yeah fair enough! I did some examples with 45 and 90 degree members in the truss section of engineer4free.com/statics but thanks for watching and the suggestion!

[@marshalleriksen3164]

PLS POST MORE CONTENT, RELYING ON U TO SURVIVE THROUGH MY COLLEGE !!! MUCH THANKS! Able to post content such as Autocad/Solidworks tutorials, Python and other contents like engineering materials pls. Much thanks! U're a lifesaver

[@Engineer4Free]

Working on it!! Check out all the courses I've completed so far at engineer4free.com 👌

[@nitrospidergaming7394]

How did you get BC to be a negative? When I tried it I got this. 57.7sin(60) - BCsin(60) = 0 57.7sin(60) = BCsin(60) 57.7sin(60)/sin(60) = BCsin(60)/sin(60) 57.7N = BC

[@LC-gd6qj]

What kind of program are you using for these questions? How do you draw and cut the truss etc?

[@Engineer4Free]

This video was drawn in an app called sketchbook, using a MacBook and a bamboo Wacom pen tablet. What you see is a combination of drawing program techniques, and video editing tricks. You can see the full list of hardware and software that I use here: engineer4free.com/tools 🙂

[@the_candid_mechanic]

Superb video Sir....Hats off..

[@Engineer4Free]

Thanks!!! =)

[@1crida1]

First off, thank you for these amazing videos!! The only part I'm confused on is picking between sine and cosine because these can change depending on how you imagine the x and y parts of the forces are drawn. For example, at 11:07, you used sin(60) when solving for the forces in the y direction. Why is that? Looking at CD, I imagine the x-part going to the right and the y-part going upwards, which would mean it's cos(60). Instead, you used the y-part of CD to go upwards, then the x-part to the right, making it sin(60). How do I know which direction to imagine the vectors in?

[@1crida1]

I figured it out y'all! You have to be aware of the angle you are making. When I chose to imagine the vectors of CD going to the right and upwards, I made a 30deg angle instead of 60deg. If I update my equation to be cos(30), it's the equivalent of sin(60), as was shown in this video. cos(30)=sin(60)=sqrt(3)/2. Be careful when choosing your vectors!

[@khoiphan7967]

this is really help, thanks for the lession sir

[@amolmadhale35]

I've got it clearly, thank you very much.

[@Engineer4Free]

Great!!! Thanks for letting me know =)

[@fred4800]

your sign convention is not consistent, please check it

[@unknown-gy8jz]

yeah realised it as well, sometimes he use tension as positive and sometimes negative

[@JohnSmith-je7ip]

why do you add in some cases and subtract in other cases?

[@Engineer4Free]

I should have drawn a coordinate axis that shows positive x is to the right and positive y is up. Because when I'm doing the sum of forces in x direction, I consider an force (or component of) on any FBD that points to the right to be positive, and an force that points to the left to be negative. Same with the sum of force equations in the y direction. Any force (or component of) that points up is considered to be positive and and force that points down is considered to be negative. That's why there are positive and negative terms when summing to zero. Hope that makes sense. If we then get a positive value for the magnitude of an internal force, it means that we correctly assumed its direction (and we always assume unknowns to be in tension, so it confirms that we guess tension correctly). If we get a negative value for the magnitude of an internal force, it means that we incorrectly assumed its direction, and it is actually in compression. That was the case with AB, where on the FBD of A, we had drawn AB in tension. But ultimately we find that AB is actually in compression, so when I draw the FBD for B, AB is now a KNOWN force, so I draw it in it's known direction, which is compression, which means it's pushing on the joint (ie, pointing at it). Hopefully that all makes sense! It's all just vector addition at the end of the day. I have some more tutorials on vector math and also trusses in videos 1-4 and 42-51 respectively, here: engineer4free.com/statics

[@alexcampbell3327]

why when drawing the whole trusses free body diagram do u not include Ax anymore, while including it in the second drawing?

[@Engineer4Free]

I draw it on the first FBD of the entire structure, and then at 0:40 I acknowledge that Ax=0 because there are no horizontally applied loads acting on the structure. Once you know a force is zero, you don’t need to draw it again, it would just be distracting, so its acceptable to not draw it on the FBD of only joint A.

[@sahilahuja6933]

thank you for this and saving me for my final coming up!

[@ho6756]

Hi sir, here is a quick question of finding CE. At 9:20, x force of CD goes positive direction but u calculated as -57.7cos60. So do I just follow up the results I get from other point, or do I need to figure out everytime whether its going to be plus or minus? Thank you!

[@ho6756]

Nvm, Can u just double check i understood correctly? So CE was found as -57.7N, arrow should be changed towards point C b4 I calculate CE? So it is a negative value for both x and y positions.

[@Engineer4Free]

Yes this is correct. On the fbd it's drawn pulling away from the joint, but is fou d to have a negative value at 8:30. A force with a negative magnitude means it is actually acting the opposite way as indicated in the fbd. So you could redraw the fbd so that CD points toward the joint, but then you would have to change the sign to positive. It's the same as leaving it as is, and having the value be negative. So yeah I think you got it, but hope that helps to clarify!

[@SHIZZLE71]

With the 200N pointing down from the top, what if the arrow was pointing down but from underneath the truss??

[@Engineer4Free]

No change whatsoever. Each represents the exact same thing: just a downward force acting on the truss at that joint!

[@Hasan-nm1rw]

why did you take abSin60 @3:09 i mean how did the sine come here? Please explain?

[@Engineer4Free]

Every angle in this problem is 60 degrees because all the triangles formed by the trusses are equilateral triangles. The sin 60 comes from finding the y component of the vector that is oriented at 60 degrees CCW from the positive x axis. Look at this: i.imgur.com/Dvdqf2t.jpg that is essentially the exact same thing, just flipped vertically. The "opposite" side is the vertical component of the force when theta is taken off the x axis.

[@melvinthomas2128]

@Engineer4Free. Thank You for the video. I was just wondering, how did u get Absin60 at around 3.30 minutes in. I just don't understand that part.

[@Engineer4Free]

All angles in this problem are 60 degrees. That is because all members are the same length, so every triangle is equilateral, so every angle is 180/3=60 degrees. So AB is 60 degrees off the horizontal. We use the SOH CAH TOA trig functions to find its x and y components. Draw a right angle triangle such that AB is the hypotenuse, and the 60 degree angle is included. ABx = ABcos(60) and ABy = ABsin(60). See this diagram for help, it's a slightly different orientation, but the same idea: i.imgur.com/Dvdqf2t.jpg

[@gaberobison680]

The beams you have stated as in tension have compressive forces though? I assume it actually referring to the joints as those are actually experiencing tension

[@Engineer4Free]

It's best to think of only the members in tension or compression. Don't try to contemplate the state of a joint. The joint its self is only a point, and points experience various point loads, not really tension and compression. When a member is in tension, it can only pull on the joint (image it is rope, rope is in tension, and can only pull). Members that are in compression push in the joints. Imagine a column in a building. It is in compression due to the weight above, and it is pressing down on the floor below it. It is also pushing up on the floor above it. Compression members push in their joints. Tensile members pull in their joints. The method of joints has us drawing each joint, not each member, notice that. The diagrams of the joints show the internal force of each push it pull, as felt by the joint. I recommend watching all the truss videos at engineer4free.com/statics, and after a few repetitions it should hopefully be clear

[@ramamaulana4393]

where do you get 1.5 m?

[@Engineer4Free]

Each member is 1m long, and the triangles are equilateral, so the top or bottom point of each triangle is half way along the opposite side, making it 0.5m from the 'base". 1m + 0.5m = 1.5m, that's where it comes from!

[@woodyislit14]

I’m lost because sin(60) doesn’t equal 0.866?

[@Engineer4Free]

There is a typo in this video: The truss is 0.866m tall, not 1m. It's 1m*sin(60)=0.866m. Doesn't actually affect anything in the problem, it was just labelled wrong. There's a note mentioning this in the description, sorry for the confusion!

[@allan4775]

Your video helped a lot, thank you!

[@Engineer4Free]

Awesome Allan, glad I could help!! 🙂

[@priviledgetaquezininga9634]

on that part, you are saying on point c the vertical components of the other inclined forces should be bcsin30 and cdsin30.............becox the angle has been halved into two...within the 60

[@Engineer4Free]

Mmm I use 60 degrees in every calculation. You can use 30 degrees if you want but you would switch each instance of sin with cos and vice versa. It would be because you're changing the orientation of the right angle triangle which would half the angle. If's fine to do if you're comfortable with that. Typically for trusses it's better to draw your right angle triangles such that cos functions are finding x components and sin functions are finding y components. It's not mandatory, but usually less confusing.

[@R1ZW4N]

My problem is similar to yours, however I have another external force pointing down on B, where would I add that into the equations ?

[@Engineer4Free]

Include it in the equilibrium equations for the entire structure. It will impact what the reaction forces are. Then also include it in the fbd for joint B when doing them individually

[@R1ZW4N]

Engineer4Free ok thanks very much , I will try it and let you know if it worked , thanks again !

[@R1ZW4N]

Engineer4Free sorry to bother you again, but where would I include it in the equation for joint B ?

[@pikovandropov2086]

Do you not check for solvability first? Or do you only ever use solvable examples? Just wondering. Good, effective video.

[@Engineer4Free]

Thanks Clief, and yes I always do solvable problems, just to demonstrate the given method. I do have a few videos in other topics like differential equations that I go over checking the suitability of certain methods, but in statics I keep it pretty straightforward.

[@yashkapoor5894]

I still hate statics but you did a great job explaining this. Thanks!

[@Engineer4Free]

🤣🤣 glad I can help!

[@mahaelmahmoud1047]

Hi. the video and explanation is great but i keep getting negative values when you get positive ones. I was wondering how you decide on the direction of the components of the force before finding out the direction of the force itself (compression or tension)

[@berryjamesa.6518]

I think the DE is Tension. But sir thank you for your works keep it up.

[@balintboda4970]

I thought so too right?

[@Engineer4Free]

DE must be in compression. You can tell this by inspection of the sum of forces in y direction. The reaction is oriented up, so the vertical component of DE must be oriented down to counteract it. The only way for the vertical component to be oriented down is if DE is in compression.

[@R1ZW4N]

Engineer4Free , did you just forget to put the minuses in for the DE equation?

[@daniellamaedemesa9651]

Can I ask what is the reason why you didn't change the sign of AB? for it to used in getting the AC, I have watched some videos relating to truss and before coming to another joints they are changing the sign already before using it. Im so confuse sorry

[@Engineer4Free]

Instead of changing the sign, I change the direction of arrow on the fbd (at 5:01 ). You either need to change the direction to the known sense (compression) of leave it in the known incorrect sense of tension, and assign it a negative value. Either is fine, but pick only one.

[@nanuar6213]

THANK YOU FOR MAKING THIS VIDEO, IT HELPED ME A TON!

[@Johndavid1451]

On joint A, I'm a little confused as to why you included Ay when solving the forces in the y direction, but didn't include Ax when solving for the x-direction. Is this because AC acts as Ax due to it being perfectly horizontal? (This video is extremely useful to me btw, thank you for making this)

[@Engineer4Free]

Hey John, good question. The reason is that when I did the FBD for the whole structure, I determined Ax to be equal to zero. So when I draw the FBD of joint A, I just don’t bother drawing Ax on, as it doesn’t contribute anything. That means though, that the horizontal component of AB then must be equal and opposite to AC, for the horizontal forces at the joint to balance. I hope that clears it up. Glad you’re like the video, and I also recommend to check out the rest that I did over at engineer4free.com/statics 👍👍

[@Johndavid1451]

Engineer4Free Great! I’ll check it out. Thank you for the help!

[@taskeen2628]

While solving for joint D why was CD taken as positive evne though it is in compression and was found out to be negative? Someone please clear this confusion.

[@Engineer4Free]

CD is found to be in compression so we change its orientation on the fbd of joint D to be pointing towards the joint with RTs known magnitude. Up is considered positive y direction and right is considered positive x direction for the fbd. CD is pointing up and to the right so its y conponent points up and its x component points right, both of which are considered positive for the respective sum of forces equations used. Does that clear it up?