Ahhhh good luck for real Matt!!! Check out engineer4free.com/statics for all my statics videos and then also the first section of engineer4free.com/structural-analysis if you ned more practice with SFD/BMD!
Looking up old notes for a job interview and left my notes at my parents house in California. I now live in Washington. This was a lot better than digging through notes and more straightforward and easy to understand. Thanks for the help!
I have mechanics of materials finals tomorrow and I want to say my teacher tried to explain this for 2 weeks and couldn’t do what you did in 10 mins! Thank you
UR A LIFE SAVING GENIUS!!! I have watched so many videos to try to figure out how to determine the location the sheer diagram crosses zero and all of them were so messy and complicated! - this was so easy and straight forward! May God bless ur soul!
Thanks Sarah 😊. You might find the rest of the examples helpful too, they are here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams
Thanks Elias. I definitely recommend checkout out the other videos I did on SFD and BMD for more practice too: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams =)
(10kN/m)*(6m)=60kN is the magnitude of the entire distributed load. It’s centroid (location of resultant) is in its middle, which is 3m from either side of it. That means the resultant is 3m away from point A. So (60kN)*(3m)=180kNm is the magnitude of the moment that the distributed load causes about A. Note that the units check out too for the units of a moment!
@@lemitowfik5881 It's physics, the moment is the force times the moment arm(distance). So what happens is you have an integral F(force)x(distance)dx, from in the case of the video from 0 to 6. That integral's solution is (Fx^2)/2 = (10(kN)6^2)/2 = (10x360)/2= 180kN. It could be -180kN, but you have to look if that makes sense in the specific problem. I have this as a one time course in my computer-related physics major (can't translate it) I've had only 2 lessons, don't know anything.
Where the BMD is positive, the curvature will be concave up, where the BMD is negative, the deflected beam will be concave down. At points where the BMD switches from positive to negative, you have an inflection point in the curvature. This is a good example that highlights it: engineer4free.com/4/shear-force-and-bending-moment-diagram-practice-problem-8 but I also recommend just watching videos 1-9 here: engineer4free.com/structual-analysis for more examples and practice to get the hang of these
This was great! You explained it in a very good and understanding way. Thank you. I will become a civil engineer in a few years and this has helped me a lot.
The line where I write sum of moments about A has a single unknown, which is B (Rb). Just rearrange to solve for B. For A (Ra), just take the sum of forces in the vertical direction. Knowing that Rb is 30kN up from the previous step, the sum of forces in y is (-10kN/m)(6m) - 20kN + 30kN + A = 0 ... just simplify and rearrange for A ... A = 50kN up. Hope that helps.
Wow, thank you, sir, It's hard for me to understand the mechanics of deformable bodies in this online class. I'm glad that I found your playlist sir thank you very much 💚💚
Glad I can help!! I have 3 playlists related to structures. Depending on the topic, the videos may be in one or another: engineer4free.com/statics engineer4free.com/mechanics-of-materials engineer4free.com/structural-analysis =)
The load on the beam is 10kn/m covering a distance of 6m at the left. You needed to replace the 10kn/m with 60kn (10kn/m*6m) before solving this problem. Your new results should be A=35kn and B=-45kn.
The moment is F(force)*x(distance) - to solve for a force applied at a large area you use the mathematical tool of integrals. So the moments integral is F*xdx - the solution is (10(kN)x^2)/2 = (10*6^2)/2 = 10*36/2 = 180kN, if you do the math it's correct, 50 on A and 30 on B.
I thought it was wrong at first too, that 60kn would be the answer for that part, until I realised how the distributed load works. One way to look at it is converting the distributed load to a point load at its center, then multiplying that value by the 3m distance from A (center of W from A).
you missed out in the last part of the SFD part where you are supposed to connect the line back to 0 in the x axis as the last support gives a force of 30 kN upwards so from -30 + 30 = 0 so just draw a perp line from the end point and stop at the y-axis = 0. i hope you get what i meant.
Yeah it's true. You will sometimes see them with or without that vertical line. As we take the virtual cut a distance dx away from the right hand side support, and as dx approaches zero, the shear remains unchanged. This value of shear that is infinitesimally close to but not touching the reaction is actually we're concerned with. Like if someone asked you what the shear at B is in this case, we would tell them it's -30kN, not 0kN. But yeah should have drawn in the vertical line anyways I suppose!
Hey, the A comes from the sum of force equation in y direction. I skipped the work. Sum of forces in y = A + B - 10kN*6m - 20kN = 0 .... A + 50kN - 10kN*6m - 20kN = 0 ... A = 60kN + 20kN - 50kN ... A = 30kN. And for why 60 is multiplied by 3: that is happening in the sum of moments equation about A. The resultant of the distributed load (overall magnitude) is (10kN/m)*6m =60kN. You need to also know how far away from A this is acting, and the location of a resultant for a uniformly distributed load is in the centre of it, so 3m from either side if it's 6m long. So the moment caused by the udl is 60kN*3m =180kN. Hope that clears it up!
50kN comes from the sum of forces in the y direction. The sum of forces in y is: (-10kN/m)(6m) - 20kN + A + B = 0 -----> (-10kN/m)(6m) - 20kN + A + 30kN = 0 -----> -60kN - 20kN + A + 30kN = 0 -----> A = 60kN + 20kN - 30kN -----> A = 50kN. Check out videos 66 -72 here: engineer4free.com/statics
Hey Adrian, sorry I did that with not much explanation. Look at the diagonal line on th eSFD that goes from x=0m to x=6m. At x=0, v=50kN and at x=6m, v=-10kN. Use these as the coordinates of the end points of a hypotenuse for a right angle triangle. The third point would be located at (x=0, v=-10kN). Se we are just superimposing a triangle over the diagram. The base of that triangle is 6m and the height is 60kN (50kN + 10kN). That is our first triangle. There is a smaller similar triangle of base=xm and height=50kN nested inside. To find the unknown base of the smaller triangle, we wet up a similar triangle equation which is like this: (rise1/run1)=(rise2/run2) so I wrote the smaller one as triangle1 and the bigger one as triangle2 to get (50kN/xm)=(60kN/6m) and rearrange to solve for x which turns out to be 5m. You could also identify that because v=+50kN and the the uniformly distributed load (udl) is 10kN/m down, then the shear should drop by 10kN/m as we move to the right, and would have to hit v=0kN at x=5m, but not all problems have such nice numbers to work with. Hopefully that helps to clear it up, I recommend working through the examples in videos 1-9 here: engineer4free.com/structural-analaysis =)
I have a question, we were taught (long time ago, already forgot alot) that for all structural analysis on civil engineering problems, we are to put the positive moments at the bottom side of the axis, so that the bmd correlates with tension, and whenever there is a distributed load, bmd has to be parabolic, with the arc bending with the direction of the dis. load, now looking at other people's examples, haven't seen anyone mention the inverted signs on bmd; so now im trying to wrap my head around if that holds true (specifically the arc bend going with the direction of udl) in all examples ?
To Whom It May Concern, When we are looking at the BMD (body moment diagram) on the third break in line 9 meters across we multiply 10KN * 3m. I can see that this works just by looking at the graph but I don't know why. Can anyone tell me why we multiply 10kN * 3m instead of what I would expect to see 20kN * 3m? Thank you for the video it's a huge help.
He multiplied 10kN*3m to get the area of the rectangle between 6m to 9m. 10*3 = 30m for the area (l*w) so the BMD from 6m to 9m is going to end 30m less than the previous height. (Less than since the area is negative bc under x axis and in a linear function since the SFD is a line) This is the same concept for the first area that was found, it being 2 triangles from 0m to 6m. T1 = 1/2bh = .5*5*50=125 T2 = 1/2bh = .5*1*10 = 5 Since T1 is in the positive side, the BMD is going to be a parabolic function to a height of 125, and since T2 is negative the BMD will be a parabolic decrease in height to 125-5 or 120. The reason it’s parabolic in the BMD 0m to 6m is because it is linear in the SFD from 0m-6m And it’s linear in the BMD 6m to 9m because it’s a horizontal line in the SFD on that interval. I hope this helps!
Hey Christian I have several that include the equations, they’re just in a different playlist. Please see videos 66-72 here: engineer4free.com/statics 👌👌
The quickest way to do it is as follows. Pretend that you're going to do a numerical integration, and slice up the triangle shape on the SFD into many skinny rectangles. The area under the curve on the SFD represents change in magnitude across that same section of BMD. So where the rectangles are tall, the change in magnitude will be greater across their skinny width than where they are shorter. If all slices are the same width of dx, then that means a greater change in magnitude corresponds to a greater slope in that region of the BMD. So the tall side of the triangle on the SFD will correspond to the side of the parabola on BMD with steeper slope, and the short side of the triangle will correspond to the more gentle slope on parabola. Also knowing that a positive area on SFD will cause a positive change in magnitude on BMD and a negative area on SFD will cause a negative change in magnitude on BMD, then it always only ever leaves one possible option for the concavity of the parabola. That's a pretty quick and dirty way to do it, but once you get used to it, it's extremely fast. I recommend checking out videos 1-9 here: engineer4free.com/structural-analysis for more examples. I think with more practice it gets easier to notice the pattern.
Good explanation dear. I have questions . Doesn't the curve of parabola will be the opposite? I mean starts at 125 and curves untill it reaches zero, with its inside part points upward. Because the slope at the left is very much than that at the right where it tends to zero actually. I wonder
We are basically just integrating to go from the SFD to the BMD. The area of the SFD translates into the slope or "rate of change" to the corresponding part of the BMD. If you solved this numerically, you would slice up that first triangle into many tall and thin sections (lets say we slice it into 100 sections, each 0.05m wide). We would calculate the area of each of those sections in kNm one-at-a-time, and then each answer would translate to the change in magnitude of the BMD from the beginning of that section to the end of that section (covering a distance of 0.05m). Because this triangle has it's tall side on the left, the areas on the left side will be larger than the areas on the right side. This means we will get larger changes in magnitude per section on the left side than the right side where the areas are smaller. This is why the parabola is oriented as it is, with larger rates of change (or slopes) on the left hand side, and then the rate of change decreasing as we go to the right. Does that clear it up?
Yea no worries it's a good question. Moments are in units of kNm. If you multiply 10kN/m * 6m as you suggested, you get 60 kN ... the m's cancel out and your left with just kN. kN is the units of a force, not a moment. 60 kN is important tho, it is the total force exerted on the beam by the whole distributed load. Ten kilonewtons per meter times 6 meters is 60 kilonewtons. Moments basically need a force, and a distance to a point that the force is perpendicular to. For the purpose of calculating the moment about a point caused by a distributed load, we consider the resultant force (the total force exerted by the whole distributed load) to be located in the center of the distributed load (for uniformly distributed ones like this). So the resultant force is in the middle, which is 3m away from either side of it. That mens it's also 3m away from point A. Now we have the whole force (60 kN) and the perpendicular distance to A from the line of action of the force (3m) and if we multiply them together we get 180 kNm. We now have the right units of kNm for a moment. I just did all of that in one step by doing 10kN/m * 6m * 3m = 180 kNm. Hope that helps clear it up, if you need extra help on this stuff, check out engineer4free.com/statics videos 25-34 for moment, videos 58-65 for distributed loads, an 66-72 for SFD & BMD basics. There are more examples to work through on SFD/BMD too in videos 1-9 at engineer4free.com/structural-analaysis . Cheers hope that helps!
@@Engineer4Free Damn! Honestly wish I found your channel earlier, gotta blame my own procrastination though. Honestly you've helped A LOT though, A massive thank you from me!!
Trying to explain to my son. Indeterminate Beam with even distributed load and 3 support, and different length between supports. Can´t find any explanation how to divide into separate cases and superposition for that?
Superposition + indeterminate problem is referred to as the force method. You can find some tutorials that I did on force method here: engineer4free.com/structural-analysis On that page you'll also find a section dedicated to superposition for statically determinate problems that would be a good review/intro. Also, you could solve your problem with a different method, such as the slope deflection method which there is also a section on that page describing it. Cheers =)
Do you have a tutorial on how to make shear force and bending moment diagram? I'm an incoming 3rd yr college student, and I have no idea what that is since my class in mechanics of deformable bodies was cut off because of the pandemic, we were basically half way in deformable (I think) our last topic was stresses like axial, and shearing. I really like your videos I hope you have a tutorial for that. Much love from the Philippines. Keep safe!
Hey yeah. In your case, I recommend watching videos 66-72 here: engineer4free.com/statics and then videos 1-9 here: engineer4free.com/structural-analysis. AFTER that, I'd recommend poking through any videos in engineer4free.com/mechanics-of-materials as needed. Good luck!!
Very informative - only complaint was that it was hard to hear you at times - you talk quietly and even at my pc's max volume I had to rewind some parts to try to hear what you said. Other than that - fantastic!
Thank for the comment Ryan and glad it helped! I have a constant struggle with microphones and sometimes videos come out with the wrong audio level, hopefully one day I just find a mic that works every time =)
The maxi shear and bending moment can be read off the graphs. The max shear is +50 kN, and the max bending moment is +125 kNm. If you do SFDs and BMDs with this method, you' will always be able to read the max values on the diagram. If you study structures further, the sign (max positive or max negative) will become important too, but still you will be able to just read it off the diagram.
Hey there. Your video is really helpful. Which software you are using? I would like to write important notes on this and probably, one example to remind me your teaching.
If the entire BMD looks flipped about the axis to you then that's okay, I draw BMDs on the compression side, some people draw them on the tension side. Follow the convention that is mostly used in your country. If it looks like it's on the right side, but you were expecting the first part to be "concave up" then you need to look a little closer. The slope of the BMD is basically the value of the SFD. Where SFD has a larger value, the slope on the BMD will be greater. Thats why the left side of the first section has a steeper slope than the right side. For that to be true, and get the proper change in magnitude, it can only be concave down as drawn (when drawing BMD on compression side). Hope that helps clear it up :)
Hi sir. You really have a great lecture and really appreciate your tutorial. But somehow I think the BMD is slightly incorrect as it should have been drawn as downward curve to represent the tension face of the beam.
Hey thanks for the feedback. The convention for drawing BMDs around the world is not the same everywhere. Some countries draw it the way I do, and others draw it inverted, with tension on the positive side like you are referring to. Either way is fine, but you should follow the convention in your country, whichever it may be. It's good to know though that many people in other countries will be drawing an inverted diagram too. Keep a close look out for the positive and negative labels on the vertical axis of the BMD!
You know the area of the SFD is positive, so the change in magnitude on the BMD will be positive from left to right. Next, do a quick and dirty numerical integration in your head of the triangular area on the SFD. slice it up into many tall skinny rectangles. The rectangles will be taller on the left, and shorter on the right. Area on SFD = change in magnitude on BMD across that same section, so the taller rectangles have more area than shorter ones. This means that the change in magnitude across each dx will be more on the left side than the right side. Greater change in magnitude per dx means greater slope. So the parabola is steeper on the left and less steep on the right. It also has positive change in magnitude from left to right. This leaves only one option for the parabola to be drawn (which is concave down in this case).
Thanks so much for the tutorial! Please may you elaborate on the "similar triangles" solution to find x on the SF diagram? Also, are there other ways of finding x? Thank you.
So the 50/x is the smaller triangle on the SFD. x is the distance (run) you are looking for. So you divide the 50/x (rise/run). Now equal it to the bigger triangle which has a rise of 10(6) = 60 load and divide it with 3+3 = 6 m. You solve for x by cross multiply it so 50/x = 60/6 would become 60(x)=50(6) and just solve for x. Hope this helps.
I use a rather un-mathematical (but fast) way to do it: Do a fake numerical integration of the triangular shape of the the section on the SFD. Slice it into many vertical bits. The bits that are taller have more area than the bits that are shorter. More area in any given section of SFD means more change in magnitude across same section on BMD. The area of this triangle is all positive, so the change in magnitude from left to right on BMD will always be increasing. This triangle is taller on the left, and shorter on the right. This means the slices on the left will have greater slope than the slices on the right. By knowing the values of BMD on the left side (0kNm) and the right side (125kNm), and that its a parabolic shape, and that its slope is greater on the left than it is on the right, there is only one way to draw a parabola that works (it must be concave down in this case). Writing this explanation out never seems to come across that simple, but think about it, and try it a few times, and you’ll realize that its 100% the fastest and most foolproof way to do it if you only need to draw the general shape and the endpoints. Keep in mind where a SFD crosses the axis (has a sign change, that is a local max or min on the BMD, but it shouldn’t affect anything in your determination of the concavity. I really hope that helps!! This is an easy trick that often gets left out of instruction!!
![Kodak Black - Sharp Vibes [Official Music Video]](http://i.ytimg.com/vi/oEV0OoLJeIs/mqdefault.jpg)
![Uniformly Distributed Load (UDL): Shear Force and Bending Moment Diagram [SFD BMD Problem 4]](/img/1.gif)
Statics final tomorrow, wish me luck!
Ahhhh good luck for real Matt!!! Check out engineer4free.com/statics for all my statics videos and then also the first section of engineer4free.com/structural-analysis if you ned more practice with SFD/BMD!
Same here friend, best of luck to you
Hope it went well! Got statics in a few days!
how was ur result?
@@Engineer4Free what is the program that u used to draw this?
You made it look so easy to understand. I used to struggle really hard drawing this diagram but you made it look simple. Thank you very much
you taught me what all my teachers failed to teach me! thanks
Glad I can help
Looking up old notes for a job interview and left my notes at my parents house in California. I now live in Washington. This was a lot better than digging through notes and more straightforward and easy to understand. Thanks for the help!
Glad I could help! Check out engineer4free.com/structural-analysis for some more examples if you haven’t already, and good luck with the interview!!
Four years of undergrad ME, and this is the cleanest statics beam bending video I've seen. Good stuff
I got 8 more right where that came form: engineer4free.com/structural-analysis 😂😂. But thanks tho
I have mechanics of materials finals tomorrow and I want to say my teacher tried to explain this for 2 weeks and couldn’t do what you did in 10 mins! Thank you
That's unfortunate that you got that professor, but great feedback about my videos.. Thanks for sharing and glad I could help =)
4 years later and the video still helps. God bless you!
Can't believe it's already been 4 years! Thanks for still watching in 2021! =)
That's the most humane explanation (simplest) ever. Thank you so much!
Thanks Yato! 😊
UR A LIFE SAVING GENIUS!!! I have watched so many videos to try to figure out how to determine the location the sheer diagram crosses zero and all of them were so messy and complicated! - this was so easy and straight forward! May God bless ur soul!
Thanks Sarah 😊. You might find the rest of the examples helpful too, they are here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams
You sir, are a hero. Thank you for this labor of love.
Thanks friend 🙂
idk how many times i go back to see this explanation, much helpful than the teachers slides
Glad to hear its helping!! =)
I was so confused about this and found it hard until i watched the video. Fantastic explanation clear and easy to understand.
This explanation has saved my life. Big thank you sir. God bless you with whatever you want
Hey cousemate,we a test today hopefully u still remember
thank you so much for these amazing videos!! my structural mechanics exam tomorrow and i’m terrified but a bit less so now that i’ve found these
Hope it went well!
As an architect student, this is really useful (im learning how to do this before its taught in a lecture lol)
Smart move my friend, it will make your life much easier to be ahead both in the sense of time, and understanding with this stuff!!
Where do u use structural analysis as an architect ?
@@ahmedyehia9560 nah lol
@@ahmedyehia9560 you dont haha its more for the arch. engineers not much architects
It's more useful to Civil engineers not Architecture
You sir are a whole legend. A method I can finally understand completely. Many blessings!
Thanks Tahira!!!
You are helping generations man keep it up
You saved me! Thanks a lot.
wish me luck for the exam in 6days 😭
Yeah good luck!! Check out 8 more examples in section one of engineer4free.com/structural-analysis 🙌🙌
you are the best ,my lecturer never explained where the area came from
Thanks Elias. I definitely recommend checkout out the other videos I did on SFD and BMD for more practice too: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams =)
Thank you sir. Your teaching method is easy to understand for a common person like me🤗🤗
Awesome, that's the goal!
how did you get 10(3)(6) where is the 3m coming from if the point load is at 6m
(10kN/m)*(6m)=60kN is the magnitude of the entire distributed load. It’s centroid (location of resultant) is in its middle, which is 3m from either side of it. That means the resultant is 3m away from point A. So (60kN)*(3m)=180kNm is the magnitude of the moment that the distributed load causes about A. Note that the units check out too for the units of a moment!
Engineer4Free ....so we only multiply centroids for distributed systems cases??
@@lemitowfik5881 It's physics, the moment is the force times the moment arm(distance). So what happens is you have an integral F(force)x(distance)dx, from in the case of the video from 0 to 6. That integral's solution is (Fx^2)/2 = (10(kN)6^2)/2 = (10x360)/2= 180kN. It could be -180kN, but you have to look if that makes sense in the specific problem. I have this as a one time course in my computer-related physics major (can't translate it) I've had only 2 lessons, don't know anything.
Just as lost as you are mate, I’m still looking a video that’ll explain everything from the start ☹️
Check out this video on force resultants ruclips.net/video/aLvL15bezsM/видео.html
Thank you! Saved me. Fell asleep in my statics lecture and was so lost in this
😂😴
just so you know you're still saving lives
Yep
Glad to hear it =) =)
I have a statics quiz tomorrow. Thank you for the information. You, sir, are a hero.
Good luck!!! =)
Explain so much better than my professor. Thank you for all that you do
I have CVE test in like 2 hours 😂😂 this was a huge help
You a hero who save my assignment
Awesome!!! =) =)
This video alone worths more than my three hours class.
Thanks Faras!!!! =)
I was little confuse to determine if it is concave upward or concave downward if you sketch in moment diagram but i really appreciate your video 😊
Where the BMD is positive, the curvature will be concave up, where the BMD is negative, the deflected beam will be concave down. At points where the BMD switches from positive to negative, you have an inflection point in the curvature. This is a good example that highlights it: engineer4free.com/4/shear-force-and-bending-moment-diagram-practice-problem-8 but I also recommend just watching videos 1-9 here: engineer4free.com/structual-analysis for more examples and practice to get the hang of these
Bro saved me from solid mechanics 🙏
I gochu 🙌
This was great! You explained it in a very good and understanding way. Thank you.
I will become a civil engineer in a few years and this has helped me a lot.
Awesome! Th aks for commenting, and good luck in your studies! I've got a lot of videos that can help you 🙂🙂
i'm having trouble getting Ra and Rb i wish you do it step by step
The line where I write sum of moments about A has a single unknown, which is B (Rb). Just rearrange to solve for B. For A (Ra), just take the sum of forces in the vertical direction. Knowing that Rb is 30kN up from the previous step, the sum of forces in y is (-10kN/m)(6m) - 20kN + 30kN + A = 0 ... just simplify and rearrange for A ... A = 50kN up. Hope that helps.
Thank you so much! Studying for the FE exam after years away from school and struggling with this section
Hey Raúl, happy to help! Make sure you check out the rest of the vids on engineer4free.com this is from the Structural Analysis playlist
@@Engineer4Free can u elaborate support reaction A.
Your BM calculation and BMD was super helpful. Thank you
Bro you are a real Engineer!!
Hahaha, thanks!! 💪
Wow, thank you, sir, It's hard for me to understand the mechanics of deformable bodies in this online class. I'm glad that I found your playlist sir thank you very much 💚💚
Glad I can help!! I have 3 playlists related to structures. Depending on the topic, the videos may be in one or another:
engineer4free.com/statics
engineer4free.com/mechanics-of-materials
engineer4free.com/structural-analysis
=)
@@Engineer4Free oh thank you very much I appreciated all your effort to help us learn about these lessons thank you very much sir :)
The load on the beam is 10kn/m covering a distance of 6m at the left. You needed to replace the 10kn/m with 60kn (10kn/m*6m) before solving this problem. Your new results should be A=35kn and B=-45kn.
The moment is F(force)*x(distance) - to solve for a force applied at a large area you use the mathematical tool of integrals. So the moments integral is F*xdx - the solution is (10(kN)x^2)/2 = (10*6^2)/2 = 10*36/2 = 180kN, if you do the math it's correct, 50 on A and 30 on B.
I thought it was wrong at first too, that 60kn would be the answer for that part, until I realised how the distributed load works. One way to look at it is converting the distributed load to a point load at its center, then multiplying that value by the 3m distance from A (center of W from A).
Sir please know that you are a great teacher, keep it up, I love your videos!!!
Duuuude you are the OG of statics!
Hahaha thanks Menyota 🤜🤛
thank you very much My Teacher , your student from Syria
You're welcome, I hope you can find more of my videos helpful!
you missed out in the last part of the SFD part where you are supposed to connect the line back to 0 in the x axis as the last support gives a force of 30 kN upwards so from -30 + 30 = 0 so just draw a perp line from the end point and stop at the y-axis = 0. i hope you get what i meant.
Yeah it's true. You will sometimes see them with or without that vertical line. As we take the virtual cut a distance dx away from the right hand side support, and as dx approaches zero, the shear remains unchanged. This value of shear that is infinitesimally close to but not touching the reaction is actually we're concerned with. Like if someone asked you what the shear at B is in this case, we would tell them it's -30kN, not 0kN. But yeah should have drawn in the vertical line anyways I suppose!
TYSM I HAVE AN EXAM THIS MORNING
Good luck!!!! =)
What a helpful video! Thanks for helping me review for the FE :)
Thanks for the feedback Priscilla, good luck on your exam!!
Sir you have just saved my life
Glad to hear it! You can check out more examples here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams =)
You are a mazing thank you so much
Good luck
From Egypt 🇪🇬
Thanks and welcome =) =)
You earn my subscription. Thanks a lot man!
Honoured to have it, thanks Joey!! 🤜🤛
Sorry,, clarify me how did you get that value of x?
Good PE exam review. Thanks bud.
Thanks for the feedback, hope the exam goes well!
thank you, I understand shear force and bending moment much better.
Glad to hear it!! 🙂
Quality stuff. just lovin' these lectures
Thanks!! =)
american using metric units,wow.i love your videos.
Thanks bud, glad you like them! 🇨🇦
Sir you are the best! Thank you very much for sharing this video! You've helped me a lot
Awesome! Glad to hear it, hope you find my other vids helpful too 🙂🙂
how did you get 50 for a ?
Summation of the vertical forces should be zero..
A + 30 = ( 10 × 6 ) + 20
A + 30 = 80
A = 50
thanks for the great explanation, can you please explain to me how you get A=50KN and why you multiply the 60 by 3m
Hey, the A comes from the sum of force equation in y direction. I skipped the work. Sum of forces in y = A + B - 10kN*6m - 20kN = 0 .... A + 50kN - 10kN*6m - 20kN = 0 ... A = 60kN + 20kN - 50kN ... A = 30kN. And for why 60 is multiplied by 3: that is happening in the sum of moments equation about A. The resultant of the distributed load (overall magnitude) is (10kN/m)*6m =60kN. You need to also know how far away from A this is acting, and the location of a resultant for a uniformly distributed load is in the centre of it, so 3m from either side if it's 6m long. So the moment caused by the udl is 60kN*3m =180kN. Hope that clears it up!
@@Engineer4Free thank you so much
I wish you explained how you actually got 50kN. I don't know which equation you actually used to substitute the 30kN to get to 50kN :(
50kN comes from the sum of forces in the y direction. The sum of forces in y is: (-10kN/m)(6m) - 20kN + A + B = 0 -----> (-10kN/m)(6m) - 20kN + A + 30kN = 0 -----> -60kN - 20kN + A + 30kN = 0 -----> A = 60kN + 20kN - 30kN -----> A = 50kN. Check out videos 66 -72 here: engineer4free.com/statics
God bless you my brother, Thank you so much!
Your welcome Phil!! 😊
Good job man! helped me a lot :)
Awesome glad to hear it!!
You Sir are a legend
Thanks Nadeem =)
thank you my friend for this simplify and hope you the best
You're welcome bud!
How did you get 50/x=60/6 to solve for that length?
Hey Adrian, sorry I did that with not much explanation. Look at the diagonal line on th eSFD that goes from x=0m to x=6m. At x=0, v=50kN and at x=6m, v=-10kN. Use these as the coordinates of the end points of a hypotenuse for a right angle triangle. The third point would be located at (x=0, v=-10kN). Se we are just superimposing a triangle over the diagram. The base of that triangle is 6m and the height is 60kN (50kN + 10kN). That is our first triangle. There is a smaller similar triangle of base=xm and height=50kN nested inside. To find the unknown base of the smaller triangle, we wet up a similar triangle equation which is like this: (rise1/run1)=(rise2/run2) so I wrote the smaller one as triangle1 and the bigger one as triangle2 to get (50kN/xm)=(60kN/6m) and rearrange to solve for x which turns out to be 5m. You could also identify that because v=+50kN and the the uniformly distributed load (udl) is 10kN/m down, then the shear should drop by 10kN/m as we move to the right, and would have to hit v=0kN at x=5m, but not all problems have such nice numbers to work with. Hopefully that helps to clear it up, I recommend working through the examples in videos 1-9 here: engineer4free.com/structural-analaysis =)
I think the reaction F from B should be 25kn instead of 30kn. as you see from the moment that 300/12=25
Hey Sipan, the left side of the equation reduces to 360, not 300, so 360/12=30 !! 🙂
THANKS FOR SUCH A NICE LESSON.
UR WELCOME
I feel that no one explains properly on howto actually draw a benidng moment diagram for beginers.
Please watch videos 66-72 here: engineer4free.com/statics for a complete beginners introduction to shear force and bending moment diagrams.
thats true i watched so much vids and its still confusing
This video and the example problems that follow may help ruclips.net/video/c9MESVdTBS4/видео.html
I have a question, we were taught (long time ago, already forgot alot) that for all structural analysis on civil engineering problems, we are to put the positive moments at the bottom side of the axis, so that the bmd correlates with tension, and whenever there is a distributed load, bmd has to be parabolic, with the arc bending with the direction of the dis. load, now looking at other people's examples, haven't seen anyone mention the inverted signs on bmd; so now im trying to wrap my head around if that holds true (specifically the arc bend going with the direction of udl) in all examples ?
To Whom It May Concern,
When we are looking at the BMD (body moment diagram) on the third break in line 9 meters across we multiply 10KN * 3m. I can see that this works just by looking at the graph but I don't know why. Can anyone tell me why we multiply 10kN * 3m instead of what I would expect to see 20kN * 3m? Thank you for the video it's a huge help.
He multiplied 10kN*3m to get the area of the rectangle between 6m to 9m.
10*3 = 30m for the area (l*w) so the BMD from 6m to 9m is going to end 30m less than the previous height. (Less than since the area is negative bc under x axis and in a linear function since the SFD is a line)
This is the same concept for the first area that was found, it being 2 triangles from 0m to 6m.
T1 = 1/2bh = .5*5*50=125
T2 = 1/2bh = .5*1*10 = 5
Since T1 is in the positive side, the BMD is going to be a parabolic function to a height of 125, and since T2 is negative the BMD will be a parabolic decrease in height to 125-5 or 120.
The reason it’s parabolic in the BMD 0m to 6m is because it is linear in the SFD from 0m-6m
And it’s linear in the BMD 6m to 9m because it’s a horizontal line in the SFD on that interval.
I hope this helps!
Yessssss thank you! That is exactly what’s going on here 😁😁
I'm back. This thing never goes away in mechanics :/
Hahaha, welcome back! 😅
You are a lifesaver! had so much trouble with these questions but it all makes so much sense now, thank you :)
Glad I could help!! =)
You’re amazing thank you so much 😊
Thankssssss =) =)
Could you please do a video on bending & shear force diagrams including equations please
Hey Christian I have several that include the equations, they’re just in a different playlist. Please see videos 66-72 here: engineer4free.com/statics 👌👌
@@Engineer4Free thank you ! Love your videos!
Thanks sir. But I have a question, is the slope for the first parabolic suppose to curve in the 'n' direction or 'u' direction
The quickest way to do it is as follows. Pretend that you're going to do a numerical integration, and slice up the triangle shape on the SFD into many skinny rectangles. The area under the curve on the SFD represents change in magnitude across that same section of BMD. So where the rectangles are tall, the change in magnitude will be greater across their skinny width than where they are shorter. If all slices are the same width of dx, then that means a greater change in magnitude corresponds to a greater slope in that region of the BMD. So the tall side of the triangle on the SFD will correspond to the side of the parabola on BMD with steeper slope, and the short side of the triangle will correspond to the more gentle slope on parabola. Also knowing that a positive area on SFD will cause a positive change in magnitude on BMD and a negative area on SFD will cause a negative change in magnitude on BMD, then it always only ever leaves one possible option for the concavity of the parabola. That's a pretty quick and dirty way to do it, but once you get used to it, it's extremely fast. I recommend checking out videos 1-9 here: engineer4free.com/structural-analysis for more examples. I think with more practice it gets easier to notice the pattern.
just want to say thank you.
Thanks for taking time to leave the comment, you're welcome!
bro this made me cry ( i just studied method of sections )
Hopefully tears of joy 😂😂
thanks a lot sir.........keep up the good work.....may you propsper more and more
Thanks friend!!!
Good explanation dear. I have questions . Doesn't the curve of parabola will be the opposite? I mean starts at 125 and curves untill it reaches zero, with its inside part points upward. Because the slope at the left is very much than that at the right where it tends to zero actually. I wonder
We are basically just integrating to go from the SFD to the BMD. The area of the SFD translates into the slope or "rate of change" to the corresponding part of the BMD. If you solved this numerically, you would slice up that first triangle into many tall and thin sections (lets say we slice it into 100 sections, each 0.05m wide). We would calculate the area of each of those sections in kNm one-at-a-time, and then each answer would translate to the change in magnitude of the BMD from the beginning of that section to the end of that section (covering a distance of 0.05m). Because this triangle has it's tall side on the left, the areas on the left side will be larger than the areas on the right side. This means we will get larger changes in magnitude per section on the left side than the right side where the areas are smaller. This is why the parabola is oriented as it is, with larger rates of change (or slopes) on the left hand side, and then the rate of change decreasing as we go to the right. Does that clear it up?
Great video
Thanks Devin!! 🙃
thanks from all of my heart
Statics final today. Pray for me.
How did it go??
why is moment about A 10kN/m x 6m x 3m Instead of 10kN/m x 6m? sorry if this is silly
Yea no worries it's a good question. Moments are in units of kNm. If you multiply 10kN/m * 6m as you suggested, you get 60 kN ... the m's cancel out and your left with just kN. kN is the units of a force, not a moment. 60 kN is important tho, it is the total force exerted on the beam by the whole distributed load. Ten kilonewtons per meter times 6 meters is 60 kilonewtons. Moments basically need a force, and a distance to a point that the force is perpendicular to. For the purpose of calculating the moment about a point caused by a distributed load, we consider the resultant force (the total force exerted by the whole distributed load) to be located in the center of the distributed load (for uniformly distributed ones like this). So the resultant force is in the middle, which is 3m away from either side of it. That mens it's also 3m away from point A. Now we have the whole force (60 kN) and the perpendicular distance to A from the line of action of the force (3m) and if we multiply them together we get 180 kNm. We now have the right units of kNm for a moment. I just did all of that in one step by doing 10kN/m * 6m * 3m = 180 kNm. Hope that helps clear it up, if you need extra help on this stuff, check out engineer4free.com/statics videos 25-34 for moment, videos 58-65 for distributed loads, an 66-72 for SFD & BMD basics. There are more examples to work through on SFD/BMD too in videos 1-9 at engineer4free.com/structural-analaysis . Cheers hope that helps!
@@Engineer4Free That's really helpful thanks so much man! I've subscribed and will check out your website:)
Awesome, glad I can help! Please tell some friends about the site too if they could benefit from the videos as well =) =)
@@Engineer4Free Already done it mate!
@@Engineer4Free Damn! Honestly wish I found your channel earlier, gotta blame my own procrastination though. Honestly you've helped A LOT though, A massive thank you from me!!
Trying to explain to my son. Indeterminate Beam with even distributed load and 3 support, and different length between supports. Can´t find any explanation how to divide into separate cases and superposition for that?
Superposition + indeterminate problem is referred to as the force method. You can find some tutorials that I did on force method here: engineer4free.com/structural-analysis On that page you'll also find a section dedicated to superposition for statically determinate problems that would be a good review/intro. Also, you could solve your problem with a different method, such as the slope deflection method which there is also a section on that page describing it. Cheers =)
God bless you for your nice explaination.
Nice presentation. May i ask what software are you using?
Very nice sir and please make video on RCC
Thanks Shekhar, check out videos 32-34 here for RCC beams: engineer4free.com/mechanics-of-matrerials
Thank you for this!
Do you have a tutorial on how to make shear force and bending moment diagram? I'm an incoming 3rd yr college student, and I have no idea what that is since my class in mechanics of deformable bodies was cut off because of the pandemic, we were basically half way in deformable (I think) our last topic was stresses like axial, and shearing. I really like your videos I hope you have a tutorial for that. Much love from the Philippines. Keep safe!
Hey yeah. In your case, I recommend watching videos 66-72 here: engineer4free.com/statics and then videos 1-9 here: engineer4free.com/structural-analysis. AFTER that, I'd recommend poking through any videos in engineer4free.com/mechanics-of-materials as needed. Good luck!!
Yea currently stressing about stresses
+jimel lalla 😅
cant thank you enough, truly appreciated
Very informative - only complaint was that it was hard to hear you at times - you talk quietly and even at my pc's max volume I had to rewind some parts to try to hear what you said. Other than that - fantastic!
Thank for the comment Ryan and glad it helped! I have a constant struggle with microphones and sometimes videos come out with the wrong audio level, hopefully one day I just find a mic that works every time =)
Finals today, wish me luck
how about the maximum shear force and bending moment? also thnx for the very clear explanation !
The maxi shear and bending moment can be read off the graphs. The max shear is +50 kN, and the max bending moment is +125 kNm. If you do SFDs and BMDs with this method, you' will always be able to read the max values on the diagram. If you study structures further, the sign (max positive or max negative) will become important too, but still you will be able to just read it off the diagram.
Engineer4Free THANKS for the quick and the informative reply 🙌🏻
No worries! Glad I can help =)
Absolutely grateful, thank you.
Awesome Ahmed!! 🙂
Hey there. Your video is really helpful. Which software you are using? I would like to write important notes on this and probably, one example to remind me your teaching.
Hey Nirmal, I've got a full list of the hardware and software that I use to make the videos at engineer4free.com/tools ✌️
Appreciate your work thanks
Thanks Lim!! =)
I think BMD should be concave.
If the entire BMD looks flipped about the axis to you then that's okay, I draw BMDs on the compression side, some people draw them on the tension side. Follow the convention that is mostly used in your country. If it looks like it's on the right side, but you were expecting the first part to be "concave up" then you need to look a little closer. The slope of the BMD is basically the value of the SFD. Where SFD has a larger value, the slope on the BMD will be greater. Thats why the left side of the first section has a steeper slope than the right side. For that to be true, and get the proper change in magnitude, it can only be concave down as drawn (when drawing BMD on compression side). Hope that helps clear it up :)
Hi sir. You really have a great lecture and really appreciate your tutorial. But somehow I think the BMD is slightly incorrect as it should have been drawn as downward curve to represent the tension face of the beam.
Hey thanks for the feedback. The convention for drawing BMDs around the world is not the same everywhere. Some countries draw it the way I do, and others draw it inverted, with tension on the positive side like you are referring to. Either way is fine, but you should follow the convention in your country, whichever it may be. It's good to know though that many people in other countries will be drawing an inverted diagram too. Keep a close look out for the positive and negative labels on the vertical axis of the BMD!
You are right
How to figure out when to bent upward or downward on bending moment diagram of parabolic line?
You know the area of the SFD is positive, so the change in magnitude on the BMD will be positive from left to right. Next, do a quick and dirty numerical integration in your head of the triangular area on the SFD. slice it up into many tall skinny rectangles. The rectangles will be taller on the left, and shorter on the right. Area on SFD = change in magnitude on BMD across that same section, so the taller rectangles have more area than shorter ones. This means that the change in magnitude across each dx will be more on the left side than the right side. Greater change in magnitude per dx means greater slope. So the parabola is steeper on the left and less steep on the right. It also has positive change in magnitude from left to right. This leaves only one option for the parabola to be drawn (which is concave down in this case).
Thank you very much Sir for the great content!
Thanks so much for the tutorial! Please may you elaborate on the "similar triangles" solution to find x on the SF diagram? Also, are there other ways of finding x? Thank you.
So the 50/x is the smaller triangle on the SFD. x is the distance (run) you are looking for. So you divide the 50/x (rise/run). Now equal it to the bigger triangle which has a rise of 10(6) = 60 load and divide it with 3+3 = 6 m. You solve for x by cross multiply it so 50/x = 60/6 would become 60(x)=50(6) and just solve for x.
Hope this helps.
BMD and SFD (100%)
ruclips.net/video/BT0LLe4_M3A/видео.html
How do you know if the parabola in the moment diagram opens up or down? Is there an easier way to know other than the mathematical method?
I use a rather un-mathematical (but fast) way to do it: Do a fake numerical integration of the triangular shape of the the section on the SFD. Slice it into many vertical bits. The bits that are taller have more area than the bits that are shorter. More area in any given section of SFD means more change in magnitude across same section on BMD. The area of this triangle is all positive, so the change in magnitude from left to right on BMD will always be increasing. This triangle is taller on the left, and shorter on the right. This means the slices on the left will have greater slope than the slices on the right. By knowing the values of BMD on the left side (0kNm) and the right side (125kNm), and that its a parabolic shape, and that its slope is greater on the left than it is on the right, there is only one way to draw a parabola that works (it must be concave down in this case). Writing this explanation out never seems to come across that simple, but think about it, and try it a few times, and you’ll realize that its 100% the fastest and most foolproof way to do it if you only need to draw the general shape and the endpoints. Keep in mind where a SFD crosses the axis (has a sign change, that is a local max or min on the BMD, but it shouldn’t affect anything in your determination of the concavity. I really hope that helps!! This is an easy trick that often gets left out of instruction!!
thank you so much
my prof didn't consider my answers using this method on our final exam T^T even though everything was correct, but still thank you for thisss!
Ouch, sorry to hear that! This was the method that my prof taught in my senior structural design course =/