Your content is helping the engineers of our future. Thank You for the extremely well produced video, I was able to grasp the material very well compared to what I hear in lecture.
Straight to the point. I was able to grab the main concept within just the duration of this video. Very helpful. Thanks man keep producing more videos.🙏🔥💯
Had a good sleep (first time in a while), waking up, feeling great. Open RUclips, see this on recommended, try it out. Feels even better cause now i understand everything on how and why they are like that ! 100% Approve from Mechanical Engineering Student. 100% efficiency on the video! Great job!
Hello. I don’t know who you are but thank you for existing and making life easier for a stupid engineering student like myself. I don’t think I’ll pass my statics class without your channel. Thank you, hope you’re doing well.
I am just gonna let you know that if you made it into an engineering course at a university, you're definitely not stupid. Statics will get easier, I promise, as long as you get the fundamentals right. Do as many practice problems as possible, try to solve the problems I solve in these videos without seeing the solution first, and if you get stuck, go through how I solve it. Don't beat yourself up, keep up the hard work and it'll get easier for you. I believe in you! You got this, and I wish you the absolute best with your studies :)
Thank you for your kind words, I really needed it right now. I will definitely remember your advice and this channel. I hope I can help you too someday when I become successful, although I don’t know how. Thank you again. Keep safe.
@@fruitpunch7361 Do your best, you got this! Thank you also for your kind words and let me know if you need clarifications on any part of the videos. I'll do my best to help.
Hello! I just want to thank you again. I’ve just received my second statics assessment results and I got 100%. I really can’t imagine how I’d be able to get that without your videos. Thank you!
@@fruitpunch7361 AWESOME!!! You did really well and I am very happy for you :) Keep up the great work and let me know if you need any clarifications on the videos.
hey! once again to your channel on my new semester. some of your videos are amazingly helpful for understanding the basics. please make more videos about solid mechanics and fluid mechanics.
Thank you very much, really glad to hear these videos helped you out. I have the topics you mentioned on my to do list, though I don't know when I will get to them. I'll do my best!
This is my go to channel whenever i need a refresh on statics and dynamics From solids to structure analysis i come here every few months. Keep up the good work sir. 💜
this legit the best video on this topic. Seen so many videos regarding this topic but some or the other end up not explaining some bit of crutial information in between so I have to find another video to watch. This video itself covered everything I need to know for my assignment. A big thank you for explaining the content thoroughly.
I'm really happy to hear everything you needed was said in the video. Thank you taking the time to write your comment, I appreciate it. I wish you the best with your studies!
This statics playlist is amazing. You did a fantastic job 👏. First, you give the concept and then follow up with solving examples that completely clears up the topic. I just wanna ask that you didn't upload any video on dry friction. Isn't it part of engineering statics?
I think this depends on the curriculum. Some courses cover friction as a first year course, others cover it more in dynamics, and some in second year courses. I didn't cover it because I was going to make more videos in the future, just not yet for statics.
Hey man, this is the most helpful video on this subject I've ever found. You're saving my final exams for real. Could you tell me the logic you use when choosing which direction the M and V point when doing the section method? Your insight would help a lot!
Thank you very much for your kind words! I appreciate it. So to know which direction M and V point, please see this video: ruclips.net/video/LPd4vW8f9Ac/видео.html I explain it within the first minute or so :) If you still have questions, let me know!
For the second example, 6:30, is it good practice to simplify the distributed load by creating a resultant force. couldn’t this lead to inaccurate results for shear force and bending moments
It depends on your requirements. Engineers tend to simplify things to get values, rather than using very specific values and a safety factor is used to account for pretty much most scenarios. So this method is perfectly fine for most applications. If you need exact values, then you can always integrate, which is also shown. I also want to point out that if you are taking a statics course, your professor will require you know how to show a resultant force for a distributed force as this is a key skill you will need.
Thanks it's soo helpful like the way that you used to explain and the linearity of the Examples that you used and you Explain a lot of Examples thanks for your Efforts ❤❤👍👍
4:00 why is the shear force drawn upwards? Is this the same as having it drawn the other way but flipping the sign? If it is wouldnt this still effect calculations?
THANK YOU VERY MUCH SIRRRR!!! AREA MOMENT METHOD IS SO MUCH BETTER since most of us are struggling with the equation methond cause we always assume all the time that x(length) is just equal to distance from the origin to the cut section and we put value in it. and that's why we have wrong results. Thank youuuuuuuuuuuuuuuuuu
At 7:20 when you were finding V, is there a reason you used -40x and didn’t use -320 for the distributed load? Would it be wrong to use -320 instead of -40x?
So we want these equations in terms of length. We don't want fixed values. A fixed value would only give us the shear force at a specific length, but if we write it with respect to "x" then we get it as a function of length.
I WISH you were around when I took Statics (Mechanics of Solids) hahaha Much better than my professor. I'm almost graduated now but if I may suggest something, please make videos on the stress tensor, equivalent loads, yield criterion (tresca and von mises) and beam deflection/slope/moment/shear/load for the future generation of students. That way you'd have more or less the whole mechanics of solids class for them 😊
You'are just amazing.i hope that you talk about mechanics of materials for the next people who will want to know about it. this course needs your explanation and I know what I'm talking about haha!. thank you again
So the 1200 N•m is an external moment applied to the beam. It's like someone randomly trying to turn it about that point. That needs to be accounted for in our graph. Clockwise means it needs to be added, counter-clockwise means it needs to be subtracted.
I’m talking about -M-40x(x/2)+133.75 why is the moment of -40x(x/2) negative when initially it was positive when we were trying to find the reaction force at B(y) 7:49
@@luyandochisampala At 7:49, you're looking at the moment created by the distributed load about point B (that's where we made the cut). If the beam was freely rotating about point B, then the distributed load would create a negative moment since it's a counter-clockwise spin. That's because I picked clockwise to be positive. At 8:33, we are again, looking at the moments created by forces about point B. Here, the 20kN force creates a clockwise moment, so it's positive.
So when you're doing the segments, you're looking at it from the perspective of the cut (where the cut was made). So here, the 133.75 kN force pushes up, causing it to rotate clockwise but the 40x force pushes down, causing a counter-clockwise moment.
@@JessicaColin-vc1uf For the first question, please see this video first: ruclips.net/video/LPd4vW8f9Ac/видео.html For the second question, we need the values with respect to distances. If we use 320, we will get one single point of data. If we use x, we can write an equation that allows us to plot a whole set of data points.
At 15.02, you have a shear value of -14.3, that extends for 2 m, so you get -14.3 x 2 = -28.6. Sorry but I don't understand what you mean by why it's different from 11:55? Is it because it's a length of 1m vs 2 m? Please elaborate on your question. For 15:46, you have (-22.3 x 2)+(-22.9) because remember, you're continuing off from the place you remained at. So at this point, you have an additional -22.3 x 2 value added to the starting point of -22.9.
For the example 2, i am trying out method 2 but stuck at 9.23 for finding moment -210. How do we find -210 moment at length 8m using shear graph diagram alone and not an equation. or is it that some problems need to be solved using segment method? Please help
So the first section, 0 to 4 m, we get a moment of (133.75 x 4) = 535. The second segment, so from 4 to 8 m, we have (535 - (186.25 x 4)) = -210. I hope that helps!
If you picked clockwise as positive, and an external moment applied is counter-clockwise, you would subtract it. If you picked counter-clockwise as positive, and an external moment was counter-clockwise, you would add it.
@@QuestionSolutions ok but is this gonna change the rest of the solution because normally when i see left a point i take clockwise as the positive but when i see it from the right i take counterclockwise as the positive
@@S_gfq I am not sure I follow what you're saying. How do you write your moment equations? You have to assume a positive direction first. So it doesn't matter how you assume that direction, but if an external moment is applied in the opposite direction to your assumed positive direction, then you subtract it. If it's in the same direction, then you add it. I hope that helps!
Hi there! I appreciate the videos a lot they are truly a life saver and blessing. I do request that you make a similar video to this on method of integration. Although not hard, I personally don’t understand how to get the constants in the integrals, if this could be explained for others, it would be greatly appreciated. Thanks
Thank you for the feedback! I will add that topic to my list of things to do in the future. I can't say when I can get around to it, but I will do my best :)
It depends on which direction you assume to be positive. Regardless of the direction you pick to be positive, you will still get the same answer. So for example, look at 3:17 moment equation. We assumed clockwise was positive. You can do the same and assume counter-clockwise is positive. You will still get the same diagram.
Please see: www.physicsforums.com/attachments/ys8brde-png.189019/ So what we're doing is looking to see how a beam bends. You can use an opposite sign convention if you'd like, but you have to do it for everything, not just a moment applied at a point.
Thank you very much for the videos! Can I ask, for the first method used, how do you know when you need to solve shear and moment forces forth both pieces cut? I have seen examples in my notes where only the shear force and moment for one half of the cut member is solved, and used for the whole Shear force and moment diagrams.
If it's just 2 pieces, or easy to figure out in your mind, you only need to solve for one half, since you can figure out what comes next, they all have to come back to their return points on the graph. If you do enough questions, you can see what comes next without going through the steps. If you're new to this though, I think its better to solve for all the pieces :)
Hi, I understand everything clearly. My only issue is: how do you know what sign convention to use when assigning internal loading directions during sectioning? If you use a positive shear (V), the equation for internal shear turns into a positive slope. The equation would be: V=40x-133.75 which is incorrect, since at x=0, V= -133.75. Yet, when using the method of sections, a positive shear is the correct sign convention. Im at a loss, any clarification would be appreciated, thank you.
The first method works for all of them, just as how the 2nd method also works for all of them. The only difference is, it's a bit easier to use the 2nd method when you have multiple "breaks" on the beam.
So, if I'm needing to find a generalized equation to demonstrate forces at any point along a beam, or cable, or some other rigid body, the moment and force diagrams are how you get there?
Hmm, I guess you can sort of say that, but simply put, shear force and moment diagrams are there to help us understand the effects happening on an object at any given point.
Is the shear force diagram continuous or there is jump discontinuity at the point loads ( like we have placed the cut either immediate left or immediate right of the point load but not exactly at the point load) so does that mean the sf diagram is discontinuous at the point load
Shear force diagrams are drawn continuously. They show a big jump whenever an external force is applied but it's still continuous. You can google "shear force diagrams" to get a better example too.
Thank you for the reply 🙏 I have one last doubt Timestamp 4.45 at the point where 30 KN load is applied in the shear force diagram there is a vertical jump Doubt - what is the value of shear force exactly at the point where 30 KN load is applied because on the immediate left of 30 KN point load I am having a shear force of 20 KN positive. But on the immediate right of 30 KN point load i am having a shear force -10 KN. But what is the value of shear force exactly at the point where 30 KN point load is applied Thank you and great work 👍
amazing content. I do have a concern, though. When you say clockwise moments are positive, this directly contradicts what my prof told me. He said to think of which way the beam bends due to the moment. If the beam makes a happy face then the moment is positive. So given a standard beam, if the moment is on the left side, a clock-wise moment is positive. If the moment is on the right side, a counter clock wise moment is postive. Can you explain your moment sign convention plz
Moments aren't positive or negative, they are either "clockwise", or "counter-clockwise" in 2D space, and in 3D space, you would use the right hand rule to determine the direction of the vector. Even in 2D space, the moment vector is still determined using the right hand rule. If it's a counterclockwise moment, then the moment vector would be straight of the screen towards you, and vice versa. This is why people usually pick counterclockwise to be positive. I don't like counterclockwise being positive, it's just a personal preference. It makes no difference to the answer. In fact, I encourage you to try it both ways, you will still get the same answer. This is what is important: -If you pick counterclockwise to be positive, and your answer is positive, then your moment is counterclockwise. -If you pick counterclockwise to be positive and your answer is negative, then your moment is clockwise. -If you pick clockwise to be positive and your answer is positive, then your moment is clockwise. -If you pick clockwise to be positive and your answer is negative, then your moment is counterclockwise. You can pick whatever side you want to be positive, like when you pick up to be positive, or down to be positive. It's just an assumption. As a convention, people generally consider positive moments as counterclockwise since they are directed along the positive z axis (out of the screen/page). It's completely up to you.
Thanks a bunch! It is helpful! However, I don't know how to use the second method if a distributed load gas a triangular shape... It is still ambiguous.
Yes, I agree, I will probably cover a example like that in the future, but the general idea is the same, your textbook/ course material should have an example with a triangular distributed load.
I have a question: When exactly is it absolutely necessary to split the beam into sections (i.e: using x for length, etc) when making shear/moment diagrams? Is there a way to tell which method to use when given a problem?
If you're referring to the 2 methods I show in this video, then you can do either one to get the same answer. Its faster to not split the beam and use method 2, especially when you become comfortable with the relationships between shear and moment diagrams.
Is it possible you could make a video using x as a length in your problems and having to solve in relation to x to find the shear force and moment diagrams?
I am not entirely sure of your question. Is there an example you can give me so I can take a look to see what type of problem you're talking about? Many thanks!
@@arthurbosch9041 I don't know which question that is, but is it something like the length represented as a variable instead of a fixed length? If so, the process is the same, you'd just have your graphs as a function of x.
Is the 40x(x/2), thats just the distributed load but you put it as a point load in the centroid of that load? is that how you get the integral of the shear load? sorry my brain is tired
Thank you for this wonderful video, I was able to grasp the topic. I was wondering if you plan on creating videos about strengths of materials or mechanics of deformable bodies, it will be a great help.
my statics course uses counterclockwise moment as positive, my equations result in the same magnitude just opposite signs. when drawing my shear force and moment diagrams would i draw it the opposite way?
I have some questions about the last class. In the first example, when x=2 shows two separate regions, it is obvious that v is in the same direction. However, why are the regions of v different when drawing, one is 20 (the upper half axis of x) and the other is -10 (the lower half axis of x)? Is the final shape of the graph different for everyone (because everyone's initial assumptions are different)
we don't consider x = 2, we consider 0 =< x < 2, and 2 < x = 2, we have a shear force of -10. The difference between the two parts, so 20 - (-10) = 30 kN, that is the applied force of the beam. The diagram will look the same for every student, there shouldn't be any difference. There are no assumptions when drawing these diagrams.
this guy explains things better in 16 min than my professor does in 1.5h
Thank you very much, glad to hear the explanations are good 👍 Best wishes with your studies!
No doubt you said the bitter truth.
True that
Better than 99.9% of SFD and BMD diagram tutorial videos out there, you explained the method very clearly.
Thank you very much :)
Life saving. I haven't been able to understand these in years.
I am glad this helped you. Best wishes with your studies :)
Your content is helping the engineers of our future. Thank You for the extremely well produced video, I was able to grasp the material very well compared to what I hear in lecture.
Really glad to hear that :) Thank you for your kind comment! Best wishes with your studies.
exactly! we need more educators like you
Honestly clearest explanation I’ve heard all month
Thanks! I hope it was helpful.
Single handedly pulling me through my statics class🙌🏽
Thank you 🙏🏽
That's awesome to hear! Keep up the great work and best wishes with your studies.
Thank you so much, you took my exam anxiety and taught me the basics! Make more mechanics and elastostatics videos please :)
Really happy to hear that. Keep up the awesome work and best wishes with your exams! :)
Literally my paper is in 2 hours and this is the only topic left to cover. You're the best man!
I hope you did well on your exam and everything went smoothly!
Statics becomes more easy with you. Animations are perfect to understand the concept. I wish you success.
Thank you, glad to hear it helps :) I also wish you much success in everything you do!
I can’t thank you enough, quick, accurate, detailed and sharp 😍♥️.
Thank you very much for your kind comment! :)
wow!! two days of lecture in 16 minutes! Thank U!!
Glad it was helpful! Keep up the good work.
This explains everything I had been seeking for a long time ago. This is marvelous!
Thank you and I am really happy this is what you were looking for. Keep up the great work and best wishes with your studies.
Straight to the point. I was able to grab the main concept within just the duration of this video. Very helpful. Thanks man keep producing more videos.🙏🔥💯
Thank you very much! Glad to hear you understood the concept for a short video, keep up the great work. Best wishes with your studies. 🔥
Had a good sleep (first time in a while), waking up, feeling great. Open RUclips, see this on recommended, try it out. Feels even better cause now i understand everything on how and why they are like that !
100% Approve from Mechanical Engineering Student. 100% efficiency on the video! Great job!
Well, I hope you get many good sleeps in your future! And thank you, glad this video was helpful :)
Hello. I don’t know who you are but thank you for existing and making life easier for a stupid engineering student like myself. I don’t think I’ll pass my statics class without your channel. Thank you, hope you’re doing well.
I am just gonna let you know that if you made it into an engineering course at a university, you're definitely not stupid. Statics will get easier, I promise, as long as you get the fundamentals right. Do as many practice problems as possible, try to solve the problems I solve in these videos without seeing the solution first, and if you get stuck, go through how I solve it. Don't beat yourself up, keep up the hard work and it'll get easier for you. I believe in you! You got this, and I wish you the absolute best with your studies :)
Thank you for your kind words, I really needed it right now. I will definitely remember your advice and this channel. I hope I can help you too someday when I become successful, although I don’t know how. Thank you again. Keep safe.
@@fruitpunch7361 Do your best, you got this! Thank you also for your kind words and let me know if you need clarifications on any part of the videos. I'll do my best to help.
Hello! I just want to thank you again. I’ve just received my second statics assessment results and I got 100%. I really can’t imagine how I’d be able to get that without your videos. Thank you!
@@fruitpunch7361 AWESOME!!! You did really well and I am very happy for you :) Keep up the great work and let me know if you need any clarifications on the videos.
Thank you for helping a foreign student studying in Germany. Keep on posting engineering stuffs.
You're very welcome. I wish you the best with your studies!
hey! once again to your channel on my new semester. some of your videos are amazingly helpful for understanding the basics. please make more videos about solid mechanics and fluid mechanics.
Thank you very much, really glad to hear these videos helped you out. I have the topics you mentioned on my to do list, though I don't know when I will get to them. I'll do my best!
This is my go to channel whenever i need a refresh on statics and dynamics
From solids to structure analysis i come here every few months.
Keep up the good work sir. 💜
I am glad these videos help you out :) ❤❤
thanks for this! I just understood a month-worth of lectures in minutes
I am really glad to hear that. Keep up the awesome work! :)
The shear quality of all your videos is mind-boggling!
These puns are too much!
@@QuestionSolutions Just let me know if you ever want to hear a pun on a particular topic; I don't want to Force things though...
One of the Most Crystal Clear Video Regarding SFD & BMD. ❤❤❤
Thank you very much ❤❤
this legit the best video on this topic. Seen so many videos regarding this topic but some or the other end up not explaining some bit of crutial information in between so I have to find another video to watch. This video itself covered everything I need to know for my assignment. A big thank you for explaining the content thoroughly.
I'm really happy to hear everything you needed was said in the video. Thank you taking the time to write your comment, I appreciate it. I wish you the best with your studies!
Best channel for Engineers.
Thank you.
Go ahead.
👍
YOU ARE BECOMING SO IMPORTANT TO ME.NO NEED OF ATTENDING MY FAKE LECTURERS.
😅 Glad to hear these videos are helping.
Thank you for this was panicking as about to take a statics final and forgot how to do this and this made it easy to understand thanks
You're very welcome! Best of luck on your final 👍
that's me right now lol
You are the best teacher i ever seen before in my life thank you❤😍
You're very welcome and thank you for your kind compliment :)
This statics playlist is amazing. You did a fantastic job 👏. First, you give the concept and then follow up with solving examples that completely clears up the topic. I just wanna ask that you didn't upload any video on dry friction. Isn't it part of engineering statics?
I think this depends on the curriculum. Some courses cover friction as a first year course, others cover it more in dynamics, and some in second year courses. I didn't cover it because I was going to make more videos in the future, just not yet for statics.
Hey man, this is the most helpful video on this subject I've ever found. You're saving my final exams for real.
Could you tell me the logic you use when choosing which direction the M and V point when doing the section method?
Your insight would help a lot!
Thank you very much for your kind words! I appreciate it.
So to know which direction M and V point, please see this video: ruclips.net/video/LPd4vW8f9Ac/видео.html I explain it within the first minute or so :) If you still have questions, let me know!
i can't believe i've been struggling with this for months and I've only watched this video and understood the concept
I am really glad to hear this video helped you out. Keep up the great work and I hope you do amazingly on your courses.
For the second example, 6:30, is it good practice to simplify the distributed load by creating a resultant force. couldn’t this lead to inaccurate results for shear force and bending moments
It depends on your requirements. Engineers tend to simplify things to get values, rather than using very specific values and a safety factor is used to account for pretty much most scenarios. So this method is perfectly fine for most applications. If you need exact values, then you can always integrate, which is also shown. I also want to point out that if you are taking a statics course, your professor will require you know how to show a resultant force for a distributed force as this is a key skill you will need.
my deepest thanks for your quick and detailed lectures, you made a big impacts for engineer students around the world
You're very welcome! Thank you for the really nice comment. :)
Thanks!
Thank you very much for supporting the channel :)
Thank you so much! You are a blessing to have.
And may God bless you!
You're very welcome! Keep up the good work and best wishes with your studies.
Thanks it's soo helpful like the way that you used to explain and the linearity of the Examples that you used and you Explain a lot of Examples thanks for your Efforts ❤❤👍👍
You're very welcome! I appreciate the comment :) I wish you the best with your studies ❤❤
4:00 why is the shear force drawn upwards? Is this the same as having it drawn the other way but flipping the sign? If it is wouldnt this still effect calculations?
Please see this video first: ruclips.net/video/LPd4vW8f9Ac/видео.html
tysm : )
Thank YOU for supporting the channel. I really appreciate it. :)
Your contents are amazing. Its way better than what the lecturer had taught us in the university.
Thank you very much! I wish you the best with your studies.
Brother, Cant thank you enough. May Allah bless you...
You are most welcome!
THANK YOU VERY MUCH SIRRRR!!! AREA MOMENT METHOD IS SO MUCH BETTER since most of us are struggling with the equation methond cause we always assume all the time that x(length) is just equal to distance from the origin to the cut section and we put value in it. and that's why we have wrong results. Thank youuuuuuuuuuuuuuuuuu
You are very welcome!
Oh my !ur video truly save the all students of mechanical department
I am really glad to hear this video is helpful to you. Keep up the great work!
awesome video for recap just before exams, very clear concepts .
Glad it was helpful! :)
Finally after 3 weeks i found this video and finally got it. thank you dude.
😀
Glad to hear that! Best wishes with your studies :)
Thank you so much, it really did help. Because you made this easy for me, people will also make it easy for you. Thanks a lot.
Glad to hear that and you are very welcome! :)
You're a legend man.Thanks for the video.
Thank you and you're very welcome!
At 7:20 when you were finding V, is there a reason you used -40x and didn’t use -320 for the distributed load? Would it be wrong to use -320 instead of -40x?
So we want these equations in terms of length. We don't want fixed values. A fixed value would only give us the shear force at a specific length, but if we write it with respect to "x" then we get it as a function of length.
I WISH you were around when I took Statics (Mechanics of Solids) hahaha
Much better than my professor.
I'm almost graduated now but if I may suggest something, please make videos on the stress tensor, equivalent loads, yield criterion (tresca and von mises) and beam deflection/slope/moment/shear/load for the future generation of students.
That way you'd have more or less the whole mechanics of solids class for them 😊
Thank you very much, for the kind comment and the recommendations on topics to cover. :)
You'are just amazing.i hope that you talk about mechanics of materials for the next people who will want to know about it. this course needs your explanation and I know what I'm talking about haha!. thank you again
That's on my list of things to do. I will do my best to do a series on that subject as well! :)
At 11:58, why was the 1200 N•m moment added in the diagram? Whats the reasoning behind it and why wasnt it subtracted?
So the 1200 N•m is an external moment applied to the beam. It's like someone randomly trying to turn it about that point. That needs to be accounted for in our graph. Clockwise means it needs to be added, counter-clockwise means it needs to be subtracted.
Keep up the good work man you are underrated
:) Many thanks!
At 8:33 why is the distributed load negative for the moment if the clockwise moments are positive
Are you talking about: M+20(11-x)+150=0 ? If so, it's not negative, it's positive.
Thank you😌
I’m talking about -M-40x(x/2)+133.75 why is the moment of -40x(x/2) negative when initially it was positive when we were trying to find the reaction force at B(y) 7:49
@@luyandochisampala At 7:49, you're looking at the moment created by the distributed load about point B (that's where we made the cut). If the beam was freely rotating about point B, then the distributed load would create a negative moment since it's a counter-clockwise spin. That's because I picked clockwise to be positive. At 8:33, we are again, looking at the moments created by forces about point B. Here, the 20kN force creates a clockwise moment, so it's positive.
@@QuestionSolutions thank you very much you’ve opened up my mind now
nothing but remarkable.....truly .....😇god bless you
Thank you very much, I appreciate it.
How does sign convention work for moments? I thought at 12:05 that the moment be negative if it was clockwise.
Please see: ruclips.net/user/shortsP029mqnp4XY
Good day! How do you find the vertex of the parabola of the shear moment diagram using the 2nd method mentioned in the video? Thank you!
You can find the x-coordinate using b/2a. The maximum bending moment occurs at the location where the shear force is 0. See 9:40.
4:00 could you please tell me why the shear force is upward in the 2nd section? My basics are bad
Please see this video, especially the first minute: ruclips.net/video/LPd4vW8f9Ac/видео.html
at 7:46, why is it that it's -40x(x/2)? Shouldnt it be +40x(x/2) since the moment that the load causes is clockwise?
So when you're doing the segments, you're looking at it from the perspective of the cut (where the cut was made). So here, the 133.75 kN force pushes up, causing it to rotate clockwise but the 40x force pushes down, causing a counter-clockwise moment.
Hello, I have a question. Do you usually want to make the cut where there is an external load being applied as you did in the first example?
Yes, wherever there is a change, an external force, or a support means a cut has to be made.
thanks you you presentation is clear and well understandable
Glad to hear that :) Thank you!
Good afternoon Professor, on minute 3:55 when drawing the shear forces, how do we know which way they point? Or is this just an estimate? Thank you.
Also another question, in minute 7:07, why wouldn't be contemplate the force 320 KN instead of using 40x?
@@JessicaColin-vc1uf For the first question, please see this video first: ruclips.net/video/LPd4vW8f9Ac/видео.html
For the second question, we need the values with respect to distances. If we use 320, we will get one single point of data. If we use x, we can write an equation that allows us to plot a whole set of data points.
how did you get -28.6 at 15:02? and why is it different from 11:55?
also helptt T.T how did u get 67.5 at 15:46?
At 15.02, you have a shear value of -14.3, that extends for 2 m, so you get -14.3 x 2 = -28.6.
Sorry but I don't understand what you mean by why it's different from 11:55? Is it because it's a length of 1m vs 2 m? Please elaborate on your question.
For 15:46, you have (-22.3 x 2)+(-22.9) because remember, you're continuing off from the place you remained at. So at this point, you have an additional -22.3 x 2 value added to the starting point of -22.9.
For the example 2, i am trying out method 2 but stuck at 9.23 for finding moment -210. How do we find -210 moment at length 8m using shear graph diagram alone and not an equation. or is it that some problems need to be solved using segment method? Please help
So the first section, 0 to 4 m, we get a moment of (133.75 x 4) = 535. The second segment, so from 4 to 8 m, we have (535 - (186.25 x 4)) = -210. I hope that helps!
@@QuestionSolutions Thank you very much! I spent too much time figuring that out. Really appreciate your videos. Very helpful!
This guy is a hero 👏🙌❤.
Many thanks!
Great job. I need to know which programs were used to plot the diagrams. Thanks
The diagrams were drawn on illustrator.
Thanks sir, the concept is very clear.
I am happy to hear that! Keep up the good work.
Thanks for this excellent explanation
But if the concentrated moment is counterclockwise should I subtract it?
If you picked clockwise as positive, and an external moment applied is counter-clockwise, you would subtract it. If you picked counter-clockwise as positive, and an external moment was counter-clockwise, you would add it.
@@QuestionSolutions ok but is this gonna change the rest of the solution because normally when i see left a point i take clockwise as the positive but when i see it from the right i take counterclockwise as the positive
@@S_gfq I am not sure I follow what you're saying. How do you write your moment equations? You have to assume a positive direction first. So it doesn't matter how you assume that direction, but if an external moment is applied in the opposite direction to your assumed positive direction, then you subtract it. If it's in the same direction, then you add it. I hope that helps!
Hi there! I appreciate the videos a lot they are truly a life saver and blessing. I do request that you make a similar video to this on method of integration. Although not hard, I personally don’t understand how to get the constants in the integrals, if this could be explained for others, it would be greatly appreciated. Thanks
Thank you for the feedback! I will add that topic to my list of things to do in the future. I can't say when I can get around to it, but I will do my best :)
@@QuestionSolutions
Wow this was impressive I don’t really comment this types of videos but well done keep up the good work!
Thank you very much :)
Is it always that a clockwise moment will provide a positive jump on the moment diagram?
It depends on which direction you assume to be positive. Regardless of the direction you pick to be positive, you will still get the same answer. So for example, look at 3:17 moment equation. We assumed clockwise was positive. You can do the same and assume counter-clockwise is positive. You will still get the same diagram.
After watching this finally I understood this. thanks a lot
I am glad to hear that :)
@15:13 what if i use CCW positive would that still work??
Please see: www.physicsforums.com/attachments/ys8brde-png.189019/
So what we're doing is looking to see how a beam bends. You can use an opposite sign convention if you'd like, but you have to do it for everything, not just a moment applied at a point.
Thank you very much for the videos! Can I ask, for the first method used, how do you know when you need to solve shear and moment forces forth both pieces cut? I have seen examples in my notes where only the shear force and moment for one half of the cut member is solved, and used for the whole Shear force and moment diagrams.
for both pieces *
If it's just 2 pieces, or easy to figure out in your mind, you only need to solve for one half, since you can figure out what comes next, they all have to come back to their return points on the graph. If you do enough questions, you can see what comes next without going through the steps. If you're new to this though, I think its better to solve for all the pieces :)
@@QuestionSolutions Thank you for the explanation! :)
I have a question about the second question. When you began drawing the moment diagram, how did you know that 133.75x - 20x^2 ended at the value -210?
So that equation is valid from 0 to 8 m, so all you need to do is plug in 8 m to get the final value.
@@QuestionSolutions Thank you!
@@anellotedesco0 You're very welcome!
How do you determine which directions the shear stress direction in 6:54
Please see this video: ruclips.net/video/LPd4vW8f9Ac/видео.html
You are doing great job man 👏👏👏 Thank you so much
You're very welcome! Keep up the great work and best wishes with your studies.
3:50 how do we know the direction of shear force either it's upwards or downwards?
oh I got it. it's because of the direction conventions.
Please watch this video: ruclips.net/video/LPd4vW8f9Ac/видео.html
Hi, I understand everything clearly. My only issue is: how do you know what sign convention to use when assigning internal loading directions during sectioning? If you use a positive shear (V), the equation for internal shear turns into a positive slope. The equation would be: V=40x-133.75 which is incorrect, since at x=0, V= -133.75. Yet, when using the method of sections, a positive shear is the correct sign convention. Im at a loss, any clarification would be appreciated, thank you.
Please watch this video first, and then if you still have the same concern, send me a comment, thanks! ruclips.net/video/LPd4vW8f9Ac/видео.html
Is the first method not applicable by the other examples? I tried the first method on the examples after 10:12 and it did not work out
The first method works for all of them, just as how the 2nd method also works for all of them. The only difference is, it's a bit easier to use the 2nd method when you have multiple "breaks" on the beam.
@@QuestionSolutions after I tried again, I understood. Thanks a lot. I'll try to use the second method if there are multiple cuts.
So, if I'm needing to find a generalized equation to demonstrate forces at any point along a beam, or cable, or some other rigid body, the moment and force diagrams are how you get there?
Hmm, I guess you can sort of say that, but simply put, shear force and moment diagrams are there to help us understand the effects happening on an object at any given point.
Is the shear force diagram continuous or there is jump discontinuity at the point loads ( like we have placed the cut either immediate left or immediate right of the point load but not exactly at the point load) so does that mean the sf diagram is discontinuous at the point load
Shear force diagrams are drawn continuously. They show a big jump whenever an external force is applied but it's still continuous. You can google "shear force diagrams" to get a better example too.
Thank you for the reply 🙏
I have one last doubt Timestamp 4.45 at the point where 30 KN load is applied in the shear force diagram there is a vertical jump
Doubt - what is the value of shear force exactly at the point where 30 KN load is applied because on the immediate left of 30 KN point load I am having a shear force of 20 KN positive. But on the immediate right of 30 KN point load i am having a shear force -10 KN. But what is the value of shear force exactly at the point where 30 KN point load is applied
Thank you and great work 👍
very helpful, while i was reviewing my course
Glad to hear that :)
Thank you for the video, a very good explanation of the concepts
You're very welcome! :)
Goated youtuber thank you so much I love you
Thank you very much! Keep up the good work :)
You are a tutor indeed.
Thanks!
amazing content. I do have a concern, though. When you say clockwise moments are positive, this directly contradicts what my prof told me. He said to think of which way the beam bends due to the moment. If the beam makes a happy face then the moment is positive. So given a standard beam, if the moment is on the left side, a clock-wise moment is positive. If the moment is on the right side, a counter clock wise moment is postive. Can you explain your moment sign convention plz
Moments aren't positive or negative, they are either "clockwise", or "counter-clockwise" in 2D space, and in 3D space, you would use the right hand rule to determine the direction of the vector. Even in 2D space, the moment vector is still determined using the right hand rule. If it's a counterclockwise moment, then the moment vector would be straight of the screen towards you, and vice versa. This is why people usually pick counterclockwise to be positive. I don't like counterclockwise being positive, it's just a personal preference. It makes no difference to the answer. In fact, I encourage you to try it both ways, you will still get the same answer.
This is what is important:
-If you pick counterclockwise to be positive, and your answer is positive, then your moment is counterclockwise.
-If you pick counterclockwise to be positive and your answer is negative, then your moment is clockwise.
-If you pick clockwise to be positive and your answer is positive, then your moment is clockwise.
-If you pick clockwise to be positive and your answer is negative, then your moment is counterclockwise.
You can pick whatever side you want to be positive, like when you pick up to be positive, or down to be positive. It's just an assumption. As a convention, people generally consider positive moments as counterclockwise since they are directed along the positive z axis (out of the screen/page). It's completely up to you.
Thanks a bunch! It is helpful! However, I don't know how to use the second method if a distributed load gas a triangular shape... It is still ambiguous.
Yes, I agree, I will probably cover a example like that in the future, but the general idea is the same, your textbook/ course material should have an example with a triangular distributed load.
I have a question: When exactly is it absolutely necessary to split the beam into sections (i.e: using x for length, etc) when making shear/moment diagrams? Is there a way to tell which method to use when given a problem?
If you're referring to the 2 methods I show in this video, then you can do either one to get the same answer. Its faster to not split the beam and use method 2, especially when you become comfortable with the relationships between shear and moment diagrams.
@@QuestionSolutions Thanks for the quick response!
@@bagaboiebailey You're very welcome!
Thank you for this video, but I have a question: What program do you use for drawing and explaining?
I use illustrator to draw these diagrams.
8:58 how did you get -186.25 sir ?
So you have to plug in 8m to the first shear force equation. So you get v = 133.75 - 40(8) ===> v = -186.25
how did you start with -28.6 for the last example when drawing the moment diagram
-14.3 x 2 = -28.6 (I show this at 15:04, with a red box on the shear force diagram)
@@QuestionSolutionsI'm confused on why you had to multiply it by 2
You're like the Engineering Plug. Thanks
You're welcome!
Hi, in the last diagram how did you know when drawin the moment diagram, that the area under the 3 and 5 graph was 44,6 KN/M
22.3 x 2 = 44.6. It's a rectangle, so it's base times height. Base is 2, height is 22.3. I hope that helps!
Thank you again for another great explanation ⭐
Glad you liked it!⭐
Is it possible you could make a video using x as a length in your problems and having to solve in relation to x to find the shear force and moment diagrams?
I am not entirely sure of your question. Is there an example you can give me so I can take a look to see what type of problem you're talking about? Many thanks!
@@QuestionSolutions I mean I'm not sure if you use this book or not but for example 7-9 in the engineering mechanics statics by hibbeler in SI units
@@arthurbosch9041 I don't know which question that is, but is it something like the length represented as a variable instead of a fixed length?
If so, the process is the same, you'd just have your graphs as a function of x.
Is the 40x(x/2), thats just the distributed load but you put it as a point load in the centroid of that load? is that how you get the integral of the shear load? sorry my brain is tired
That is correct, please see: ruclips.net/video/lI5klge2GlM/видео.html
@@QuestionSolutions thank you, I've subscribed
You're very welcome :)@@tturi2
You blew away all my doubt for BM SMD
Thank you very much, really glad to hear!
you're the best. thank you
👍 Many thanks!
Thanks a lot for this. Much understood now
Glad to hear :)
Thank you for this wonderful video, I was able to grasp the topic. I was wondering if you plan on creating videos about strengths of materials or mechanics of deformable bodies, it will be a great help.
You're very welcome. I do have plans on creating videos about strengths of materials, but probably not for some time. Best wishes with your studies!
@@QuestionSolutions Looking forward into that. Thank you!
that was insanely good
Thank you very much!
my statics course uses counterclockwise moment as positive, my equations result in the same magnitude just opposite signs. when drawing my shear force and moment diagrams would i draw it the opposite way?
What does your professor want?
I have some questions about the last class. In the first example, when x=2 shows two separate regions, it is obvious that v is in the same direction. However, why are the regions of v different when drawing, one is 20 (the upper half axis of x) and the other is -10 (the lower half axis of x)? Is the final shape of the graph different for everyone (because everyone's initial assumptions are different)
we don't consider x = 2, we consider 0 =< x < 2, and 2 < x = 2, we have a shear force of -10. The difference between the two parts, so 20 - (-10) = 30 kN, that is the applied force of the beam. The diagram will look the same for every student, there shouldn't be any difference. There are no assumptions when drawing these diagrams.