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.
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.
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.
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 =)
this is the best video on TH-cam for the analysis of truss, short, simple, straightforward, clean, exact and has got the best explanation. Thank you so much for the amazing video.
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
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
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 =)
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
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 😁
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.
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👌
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🙏
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!
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!
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 :)
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
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
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,
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
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
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!
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...
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!!
@@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 :)
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.
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.
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
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?
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!
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
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
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.
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!
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.
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!
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 👌
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 🙂
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,
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
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.
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.
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.
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
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
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.
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.
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
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.
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?
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!
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
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.
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 👌
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
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?
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)
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
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!
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
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
BC is at a 60 degree angle down from the horizontal so the cos60 is there to reveal only its horizontal component. BD is perfectly horizontal, so no cos or sin function is required to reveal its horizontal component, because the whole force is horizontal already.
Ax is determined to be zero in the overall sum of force equation for the whole structure because it's the only external force with an x component so it must be zero. In the fbd of joint A, there's no point in labelling a force with zero magnitude, as it won't have any impact on the force balance, so that's why it's not included.
Always draw unkniwns in tension. If you find the magnitude to be positive, tension is the correct sense. If you find it to be negative, then it is in fact in compression. See videos 42, 43, 49 here: engineer4free.com/statics
The cos60 is there to represent the x component of DE. The equation for sum of forces in X at joint E is like this: DEcos60 + CE = 0. CE is negative because it points to the left, so the equation can be written with the magnitudes as: 173.2cos60 - 86.55 = 0. CE (-86.55N) is wholly in the x direction, so all of its magnitude counts toward the sum of force in x direction expression.
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!
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
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???
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 👍
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.
Yessss glad that my explanation worked. What you described is exactly what I try to accomplish in the videos. Cheers 🤜🤛
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
That's a good question.
College? Is for us to know that there are many things we don't know...
That's true
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.
How to calcute the reaction at support(im not sure whether its positive or negative)
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.
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 =)
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.
Thanks for the kind words Max! Hope you are doing great too :)
+Engineer4Free I also agree! Thank you so much for helping me
True
Is the same true if i get -350 for my ay?
this is the best video on TH-cam for the analysis of truss, short, simple, straightforward, clean, exact and has got the best explanation.
Thank you so much for the amazing video.
Thx bro. Please share my vids with some other classmates!
@@Engineer4Free sure, will do the same
Very clean presentation, straight to the point, no time wasted drawing all it out in real time. very useful video
Yo Freaky Banana, thanks for the feedback...glad you like it =)
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
Thanks!! Glad you like the video.
Also.... you are in the middle of your final and on YT? Zoom University lol
I am so doing this tomorrow lol
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
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
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
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 =)
This is the best TH-cam I've ever watched..
Wow...you got good stuff
Brilliant video, learnt more in this than I did in my lectures and attempting worksheets
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
Awesome, thanks Kelvin!!! =) =)
My man just explained this better in 15 min than my lecturer managed to in 5 hours.
😏
This is hands down the best video I have ever seen . Thank you!!
Wow, thanks Aaron! =)
Best "method of joints" tutorial I could find on youtube!
Thanks man! That's awesome to hear!!
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 😁
Awesome!! Thanks for the great feedback 😁😁
It feels like i have wasted my whole semester classes by listening to boring professor. Many Thanks for saving my paper. Cheers!
Sorry to hear, but you’re not alone. Glad I could help 🤜🤛
Sorry to hear, but you’re not alone. Glad I could help 🤜🤛
this guy explained it better than my lecturer, thanks man
You're welcome, thanks for watching! Check out the full playlist at engineer4free.com/statics 👌
You just saved my exam for tomorrow, god bless you
Awesome, glad to hear it! How did the exam end up going?
and here i am ..saved for my exam for tomorrow..LMAO
u saved me for tomorrow's exam too lol
@@Engineer4Free I guess we’ll never know
got a test tomorrow too lol
This video is always a life saver the night before exams
One question brother! #7:31, 57.7cos60° +57.7cos60° is going to be 57.7N why not 115.4 with negative sign
omg. finally a video that actually makes sense. youve made this so easyyyy. oh god thank you so much
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.
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👌
@@Engineer4Free Thank you! 💙
this guy is always saving me in the day before the exam, god bless you brother
when I did sin60 my ti-84 gives me -.3048 is there a reason our numbers are different?
@@dram906 maybe ur calculator, reset it because sometimes its on radian system change it to degree
@@ahmedothman6642 thankyou got it now!
@@dram906 no need, good luck
th-cam.com/video/fRyUf-GY754/w-d-xo.html
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🙏
Glad I can help!!! Keep it up my friend 🤜🤛
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!
for joint c, why is the horizontal force of AC not -28.85 cos60?
watched this legit 20 minutes before my exam and it helped so much
Glad to hear it! Now you know where to come before your next exam too
You literally saved my day, thankyou!
I've been told this 3 times before but didnt understand it.
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
Well hats off to Engineer4Free for still replying to comments and keeping people updated even after 4 years. You're amazing
Thanks Atharva! Still trying🤜🤛
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
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!
Thank you so much, literally saved me from hours of frustration.
I have a solid mechanics exam today and this just helped me understand this topic so quickly! Thanks!
Good luck!!! Plenty more vids btw at engineer4free.com/statics and engineer4free.com/mechanics-of-materials 👌
i got one in 3 days lol
@@anask4052 good luck
You just explained in 15mins what engineer hamzat couldn't make me understand in 2years ...hamzat
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....
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 :)
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
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
@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
Your explanation made these problems much easier to solve. Thanks
Thanks for letting me know David, cheers!
Well this helps can’t wait to learn more when I’m done middle school
Keep on it!!! 🤜🤛
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
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.
Yessss glad to hear it makes sense now 🙌
I don't understand where do you find 2 m and 1.5m
you dont find it, it's just an example if it is
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,
Very good demonstration, thank you d’or sharing. With your nodes analysis this covers it all. So helpful!
Thanks a lot this video helped me a lot in understanding joint method. Love from India. You are doing a great job.
Thanks Saket!! I have more tutorials at engineer4free.com/statics see #42-51 for more truss examples👍
@@Engineer4Free You are welcome. I will surely watch those videos. I have issue with section method
I have wasted 8hrs in classroom but you made it worth in 15mins........
Thanks
Your welcome, and thanks so much!! 🙌🙌
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
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
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
@@Engineer4Free thank you so much. I know this video is old and my question was probably below your ability, so I appreciate your reply.
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!
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...
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 :)
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!!
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)?
Same, did you figure it out?
@@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 :)
@@fubrian2945 th-cam.com/video/xrU3l1RvyiE/w-d-xo.html here is a wonderful tutorial too
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.
@@Engineer4Free tension is when the forces point towards each other in the truss member right?
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?
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.
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
@@edwardchavez5621 Hey I replied, to the parent comment... just letting you know incase you don't get a notification for it.
Hi :-)
Thank you for your time to making this video!
this is really helpful!!!
I hope u r doing great wherever you are now ;-)
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?
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!
I hve an excellent teacher, but on trusses i missed the class, so i m watching u thanks
With the 200N pointing down from the top, what if the arrow was pointing down but from underneath the truss??
No change whatsoever. Each represents the exact same thing: just a downward force acting on the truss at that joint!
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?
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
AB is in compression meaning it is pushing out internally a force of 57.7 towards B
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
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.
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!
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.
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!
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?
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 👌
What kind of program are you using for these questions? How do you draw and cut the truss etc?
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 🙂
Why does A have two components in Ax & Ay while E only gets a vertical component?
If you look closely at E(the figure). The "triangle" has two circles underneath, that means it will only have the y-component
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?
JC NotNot because the pin provides two reaction but the roller provides one reaction force
Yeah thanks Christian, you’re right, A is a pin, and E is a roller. Some people draw these slightly differently
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,
11:00 how does 200N go to 100? I don’t understand. Can someone please help me?
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)
clear and straight to the point. amazing explanation.
Thanks for watching!!! =)
if the member DE were vertical and the applied force is also vertical would you be able to start at the applied force?
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
statics test in an hour, life saver lmao
haha how did it go?
@@Engineer4Free rip
😐
@@jellyman- hahaha, how do you know? lol
My finals in 8 hours...wish me luck
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.
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.
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.
Auto subscribe, this tutorial so far is the best for me. Thank you sir!
Thanks for the sub!! =)
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
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
why when drawing the whole trusses free body diagram do u not include Ax anymore, while including it in the second drawing?
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.
Where did you get the 2 meters from on the free body diagram?
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.
@@Engineer4Free Thank you so much!
You're welcome! Plz share my statics videos with some other students too 🙌
This is totally a breakthrough .... Good job.
Awesome!! Make sure you check out the rest of my statics vids at engineer4free.com/statics , the truss videos are numbers 42-51
My problem is similar to yours, however I have another external force pointing down on B, where would I add that into the equations ?
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
Engineer4Free ok thanks very much , I will try it and let you know if it worked , thanks again !
Engineer4Free sorry to bother you again, but where would I include it in the equation for joint B ?
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.
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?
Wow. This is much understandable than my 2-hour class. I hope you’d do videos in all my engineering subjects too. Hahaha Thanks!
I'm working on it! Check out engineer4free.com for all the courses, new vids and courses are always being added!!
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!
Hi sir, at minute 10:21 value of CE is positive it mean Compression right? not a tension if im not mistaken
How is the truss height factored in tho? As a higher section should reduce the forces on the members
at 1:16 why did you multiply by 1.5?
Distance from A to D along the x axis
@@RatedRizk thanks I understand
Thanks!
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
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
Thank you very much for the in-depth explanation
Glad it was helpful! thanks for watching =)
This... this is amazing. You just earned another sub!!
Awesome!! Welcome 😊😊
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.
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 👌
@@Engineer4Free awesome! Thanks a lot
@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.
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
You are the best saved me alot of time
Awesome!! Check out engineer4free.com/statics for the rest of the statics videos =) =)
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
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?
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)
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
I have a question
Why didn't you solve for Ax
And why is 57.7 positive as you're solving for the second one?
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
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!
Engineer4Free it did help! Many thanks to you.
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
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
@@Engineer4Free ok thank you so much😄appreciate the quick reply btw
At 7:05, why didn't you multiply BD with cos60 like you did with BC
BC is at a 60 degree angle down from the horizontal so the cos60 is there to reveal only its horizontal component. BD is perfectly horizontal, so no cos or sin function is required to reveal its horizontal component, because the whole force is horizontal already.
In Joint A, when you Sum all Fx, Ax is not inlcluded? Why?
Ax is determined to be zero in the overall sum of force equation for the whole structure because it's the only external force with an x component so it must be zero. In the fbd of joint A, there's no point in labelling a force with zero magnitude, as it won't have any impact on the force balance, so that's why it's not included.
@@Engineer4Free Thank you :)
How to determine the direction of forces,when they are positive or negative?
Always draw unkniwns in tension. If you find the magnitude to be positive, tension is the correct sense. If you find it to be negative, then it is in fact in compression. See videos 42, 43, 49 here: engineer4free.com/statics
Engineer4Free Thanks a lot!
🤜🤛
I don't seem to understand why at the end. Fx=173.2cos60 -86.55.shouldnt there be a cos 60 too.
The cos60 is there to represent the x component of DE. The equation for sum of forces in X at joint E is like this: DEcos60 + CE = 0. CE is negative because it points to the left, so the equation can be written with the magnitudes as: 173.2cos60 - 86.55 = 0. CE (-86.55N) is wholly in the x direction, so all of its magnitude counts toward the sum of force in x direction expression.
I’m lost because sin(60) doesn’t equal 0.866?
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!
why do you add in some cases and subtract in other cases?
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
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???
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 👍
whys isnt your force at AB for FBD at point B is -57.7 ? when you solved it for that equation? is this where 3rd law comes in?