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 =)

[@Jesssuuusss_.]

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

[@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

[@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! =)

[@gauteantonsen3907]

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

[@Engineer4Free]

Thanks man! That's awesome to hear!!

[@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 😁😁

[@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 👌

[@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

[@anishahasnat2837]

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

[@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 🤜🤛

[@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

[@mateusz4611]

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

[@Engineer4Free]

😏

[@mrmoeguy4664]

This video is always a life saver the night before exams

[@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! 💙

[@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!

[@UltraBallers]

Hello please how did you get sin60 and the cos60 here 5:17 ....I'm confused 😕 I'm not seeing and angle of 60⁰ so why sin60 and cos60

[@davidmuniz5307]

Your explanation made these problems much easier to solve. Thanks

[@Engineer4Free]

Thanks for letting me know David, cheers!

[@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

[@joelpinnock8131]

@3:05. i got inverse tan(0.5/1) = 63.44, where did i go wrong please. why did you then go on to put sin60 and not sin 63.44? thank you

[@sterioboy676767]

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

[@christianfekete]

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

[@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🤜🤛

[@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

[@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

[@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

[@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

[@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 :)

[@kvpgamer3168]

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

[@andrenjiomou4342]

Thanks

[@Engineer4Free]

Your welcome, and thanks so much!! 🙌🙌

[@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

[@iloveketchup20]

5:35 Why isn't FABy negative like when you calculated it for the A joint? Shouldn't it also be -57.7 N like before? And shouldn't the sum be FY = FAB*Sin(60) + FBC*sin(60)? If I solve for FBC this way I get a posistive 57.7N,

[@rousrudolfraemona.3898]

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

[@Engineer4Free]

Thanks for the sub!! =)

[@cellerysandwich]

clear and straight to the point. amazing explanation.

[@Engineer4Free]

Thanks for watching!!! =)

[@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 ;-)

[@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 👌

[@adegboyepromise4683]

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

[@marcinbenedek2089]

One part @5:50 confuses me: why BC*sin60 and not BC*cos30... ? When looking for the y component of BC, I am just imagining a vertical line that divides the angle into two and creates the y component of BC...

[@marcinbenedek2089]

Well now I get it's the same, it's just a bit of a jump from, me, but I guess I should just practice :)

[@Engineer4Free]

Yeah for any force F on an angle theta, Fsin(theta) = Fcos(90-theta). And Fcos(theta)=Fsin(90-theta). Depending on how it's drawn will determine whether you are using sin or cos functions for the x and y components. Glad you were able to figure it out. Keep practicing!!

[@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!

[@ahmedkere-ahmed9594]

Thank you very much for the in-depth explanation

[@Engineer4Free]

Glad it was helpful! thanks for watching =)

[@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!!! 🤜🤛

[@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!

[@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

[@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,

[@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.

[@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

[@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!

[@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?

[@ateate650]

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

[@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

[@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.

[@mitchellberends103]

You are the best saved me alot of time

[@Engineer4Free]

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

[@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!

[@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.

[@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

[@harisjasic7353]

11:00 how does 200N go to 100? I don’t understand. Can someone please help me?

[@Engineer4Free]

57.7 * sin60 = 57.7 * 0.866 = 50. That 50 gets added to the -200 beside it, becoming -150. The expressions goes like this: (57.7)(sin60) - 200 - DE(sin60) = 0 (57.7)(0.866) - 200 - DE(0.866) = 0 50 - 200 - DE(0.866) = 0 -150 - DE(0.866) = 0 -DE(0.866) = 150 DE(0.866) = -150 DE = -150/0.866 DE = -173.2 kN DE = 173.2 kN (compression)

[@banrtv]

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

[@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?

[@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 ?

[@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.

[@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!!

[@cbr16327488]

if the member DE were vertical and the applied force is also vertical would you be able to start at the applied force?

[@Engineer4Free]

Hmmm not sure exactly what you're imagining, but either way in these problems, you should always figure out the reaction forces first, by drawing the fbd of the overall structure. Once you have those, then you can assess where you want to go next, and usually it's starting from one of the joints at a reaction because it will have the least number of unknowns

[@foofoogaming551]

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

[@SHIZZLE71]

When doing the FBD for each joint, the problem im doing, asks me to find member forces for more of the right side of the truss, so can i start doing it from the right side???

[@Engineer4Free]

Yep, same either way. I just tend to start on the left out of habit. If you only need to find the force in one single member, use the method of sections. See videos 49 - 51 here: engineer4free.com/statics 👍

[@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

[@christophercraughwell485]

How is the truss height factored in tho? As a higher section should reduce the forces on the members

[@thomasrichardson504]

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

[@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.

[@baiklahtu]

Hi sir, at minute 10:21 value of CE is positive it mean Compression right? not a tension if im not mistaken

[@ARMY_RUGS]

in choosing the joing do younhave to follow some pattern or any joint can be chosen as much as is has max of two forces to be determined

[@Engineer4Free]

Generally you want to start at a support with known reaction force, but as long as you have enough info to solve at the joint, feel free to start wherever

[@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 🙌

[@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

[@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 ✌️

[@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

[@thinalnim]

so neatly explained :) thank you sir!

[@Engineer4Free]

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

[@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 🙂

[@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 =) =)

[@katiebarclay3605]

Hey, i have a warren truss question with just a single triangle and a force acting on the top of the truss a size of 15kN to an angle of 60°. Can you help me with this?

[@Engineer4Free]

If a member is oriented at 60 degrees and has an internal force of 15kN, then the x component of that internal force will be 15kN*(cos60)=7.50kN and the y component of that internal force will be 15kN*(sin60)=12.99kN. If the member is in tension, it will pull on each joint, and if it's in compression it will push. You can use those numbers and directions to find the sum of forces in x and y at each joint individually. More tutorials on trusses are also here: engineer4free.com/statics (videos 42-51)

[@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 🙌

[@nuetro1709]

Why did you put negative 57.7 for joint A when u took the sum of x? It's pointing in the positive direction

[@nuetro1709]

Also, do you determine whether things are in compression or tension before the calculations or do you find which one it is DURING calculations? What's the safe approach? Because I was taught to find the tension/compression before u do the problem and i always got confused

[@Engineer4Free]

When I initially write the expression: 50N + ABsin(60) = 0 , both terms on the left hand side are positive, and sum to zero. It only becomes negative once I rearrange the equation by bringing the term to the other side. Essentially, I just subtract 50 from both sides then divide both sides by sin(60) which is 0.866. That's how the right hand side turns from a zero into -57.7N. To answer your other question, look at how I've drawn the FBD of each joint. Every unknown is drawing in tension initially. When I set up the equations then based on each FBD, if I find an unknown value to be positive, I know it is in tension, and if it is negative, I know it is in compression. It's easy that way and makes it hard to mess up. If I determine a member to be in compression, then I will update it on any other fbd that the member connects to, and change it from an unknown value assumed to be in tension, to a known value in compression. For example, after I determine that AB is 57.7N in compression, I update member AB on the FBD for joint B to reflect this BEFORE starting the calculations for that joint. Hope this helps clear it up.

[@asddwwwawd5725]

why use sin60 and cos60 on point b espicially horizontal force i thought the angle should be respect on vertical and horizontal component, why not using cos30? because it has different answer thanks for the answer i really confuse by that

[@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)

[@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!

[@the_candid_mechanic]

Superb video Sir....Hats off..

[@Engineer4Free]

Thanks!!! =)

[@spongebobhang2969]

To determine the (-)or(+) sign to put in the calculation of forces in the x-direction(or y-direction) of a diagonal truss member is that compression member is in (-) and tension member is in (+) or is there any other method we could determine that. I'm struggling with this problem when I'm doing my method of joint calculation for my project. Thanks in advanced

[@Engineer4Free]

Oh man, I just typed up a massive and extremely detailed answer then accidentally deleted it....😭😭😭 Please watch all videos 42-51 here: engineer4free.com/statics and if you're still not sure I can write my essay again

[@spongebobhang2969]

@@Engineer4Free ok thank you so much😄appreciate the quick reply btw

[@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 👌

[@khoiphan7967]

this is really help, thanks for the lession sir

[@gabrielbecklourenco9801]

quick question , sorry if its a stupid one , but how do you know whether to use cosine or sine when finding out the forces? thanks

[@micahrolloque3634]

On the third diagram, the Fx, why is the other one not positive? Well BC is in tension. It has a positive value of 57.7cos60. I am just confused

[@Engineer4Free]

When you’re solving for the sum of forces in x direction for the forces in a free body diagram, any force with a positive x component (that is, pointing to the right, or positive x direction) will have a positive value in the expression. Any force whose x component is pointing toward the negative x direction (to the left), will have a negative value in the expression. The fact that BC is in tension, does not make it’s x component positive, It’s x component points to the left, and is negative. I hope that helps clear it up!

[@micahrolloque3634]

Engineer4Free it did help! Many thanks to you.

[@victorlam2997]

I was wondering why you used sin for the y component when working out BC, because wouldn't using sin work out the x component?

[@Engineer4Free]

Hey, here is an illustration why: imgur.com/a/YCT6I1a You need to re-assess how the right angle triangle is oriented every single time that you use sin or cos in these problems. Hope this clears it up, cheers!

[@victorlam2997]

another question came up, how do you know which angle to use when calculating?

[@Engineer4Free]

It's something that you have to decide every time. Construct a triangle like in the linked image such that one side is known. The known side will always be the hypotenuse or the adjacent side of the right angle triangle, and from there, you should be able to identify the angle that is "opposite" from the opposite side. That is the angle that you use in the sin and cos operations.

[@justinesurabilla2929]

If you are looking for Fy use sin, if you are looking for Fx use cosine.

[@hayfordamissahakesseh6863]

When measuring the angle from x-axis use cos and when measuring from the y-axis use sine