Shear force and bending moment diagram practice problem #1
Vložit
- čas přidán 15. 11. 2017
- Check out www.engineer4free.com/structur... for more free structural analysis tutorials. The course covers shear force and bending moment diagram review, method of superposition, moment area method, force method, displacement method, slope deflection method, and 3 moment equation.
This tutorial goes over how to draw the shear force diagram, bending moment diagram, and deflected shape of a simply supported beam with a distributed load and a point load.
If you found this video helpful, please consider supporting my work on Patreon: / engineer4free
Looking for software? I highly recommend checking out SkyCiv. They make a full suite of online structural analysis software and tools that are useful for both students and professionals, including calculators for beams, trusses, frames, moment of inertia and more. It’s great for checking your work: bit.ly/skyciv-e4f
Don’t forget to subscribe on CZcams and join the Engineer4Free mail list:
Subscribe: czcams.com/users/subscription_...
Mail List: www.engineer4free.com/newsletter
Social Media:
Facebook: / engineer4free
Instagram: / engineer4free
Twitter: / engineer4free
Thanks for watching, The Struggle Is REAL!
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
I was so confused about this and found it hard until i watched the video. Fantastic explanation clear and easy to understand.
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!!
You sir, are a hero. Thank you for this labor of love.
Thanks friend 🙂
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! 😊
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!
You sir are a whole legend. A method I can finally understand completely. Many blessings!
Thanks Tahira!!!
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
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
Your BM calculation and BMD was super helpful. Thank you
idk how many times i go back to see this explanation, much helpful than the teachers slides
Glad to hear its helping!! =)
Sir please know that you are a great teacher, keep it up, I love your videos!!!
What a helpful video! Thanks for helping me review for the FE :)
Thanks for the feedback Priscilla, good luck on your exam!!
Quality stuff. just lovin' these lectures
Thanks!! =)
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
You are helping generations man keep it up
Explain so much better than my professor. Thank you for all that you do
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 =)
Thank you! Saved me. Fell asleep in my statics lecture and was so lost in this
😂😴
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 🙂🙂
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 🙂🙂
Good job man! helped me a lot :)
Awesome glad to hear it!!
thank you my friend for this simplify and hope you the best
You're welcome bud!
You’re amazing thank you so much 😊
Thankssssss =) =)
Thank you for this!
You earn my subscription. Thanks a lot man!
Honoured to have it, thanks Joey!! 🤜🤛
I have a statics quiz tomorrow. Thank you for the information. You, sir, are a hero.
Good luck!!! =)
Thank you very much Sir for the great content!
God bless you my brother, Thank you so much!
Your welcome Phil!! 😊
Absolutely grateful, thank you.
Awesome Ahmed!! 🙂
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 =)
cant thank you enough, truly appreciated
this is so helpful thank you so much!
Thanks for letting me know, glad to hear it!! I have more examples like it at engineer4free.com/structural-analysis check them out!!
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 🙌🙌
Good PE exam review. Thanks bud.
Thanks for the feedback, hope the exam goes 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
Thank you sir. Your teaching method is easy to understand for a common person like me🤗🤗
Awesome, that's the goal!
It is helpful.
Thank you sir.
thank you very much, it helps a lot
Hey thanks for the comment =)
The video is really helpful
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 ?
God bless you for your nice explaination.
THANK YOU FOR THIS VIDEO
Duuuude you are the OG of statics!
Hahaha thanks Menyota 🤜🤛
Thanks for the useful video!!!
You're welcome, glad I can help!!
I have CVE test in like 2 hours 😂😂 this was a huge help
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 a hero who save my assignment
Awesome!!! =) =)
thank you so much
thank you, I understand shear force and bending moment much better.
Glad to hear it!! 🙂
thanks a lot sir.........keep up the good work.....may you propsper more and more
Thanks friend!!!
THANKS FOR SUCH A NICE LESSON.
UR WELCOME
Sorry,, clarify me how did you get that value of x?
This was helpful asf...thank you 😭❤️
Glad it helped! Full playlist is here engineer4free.com/structural-analysis 👍
thanks from all of my heart
Appreciate your work thanks
Thanks Lim!! =)
thank you. clear and informative.
Glad it was helpful! =)
You are a mazing thank you so much
Good luck
From Egypt 🇪🇬
Thanks and welcome =) =)
just so you know you're still saving lives
Yep
Glad to hear it =) =)
Great video
Thanks Devin!! 🙃
thank you very much My Teacher , your student from Syria
You're welcome, I hope you can find more of my videos helpful!
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
You Sir are a legend
Thanks Nadeem =)
nice video, thanks!
Thanks!
thank u bro!
ur welcome marco👌👌
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.
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 czcams.com/video/aLvL15bezsM/video.html
Great videos. Do you know anywhere one could find shear and moment diagram practice problem videos like this for distributed loads of varying shapes, such as parabolas, triangles, etc?
I've got some videos for finding reaction forces with triangular/parabolic etc distributed loads under the "centroids and distributed loads" section of engineer4free.com/statics that would be a good starting point, as that is the first step in drawing the SFDs and BMDs. I should include some actual examples through on doing full problems. Thanks for the suggestion, hopefully I can get around to making some soon!
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 =)
helped alot thx a ton
Glad to hear it! There are some more examples here too: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams =)
Bro you are a real Engineer!!
Hahaha, thanks!! 💪
Thank you very much! 👍
You’re welcome!! 🙂
Nice presentation. May i ask what software are you using?
So much helpful
=) =)
Sir i realy thank for you
You're welcome =) =)
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%)
czcams.com/video/BT0LLe4_M3A/video.html
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 =)
Thank you!!
You're welcome!!! =)
And how to find the bending reinforcement from there ? Would be very pleased, if someone will explain
Thank you!!!
You're welcome Alex!! =)
just want to say thank you.
Thanks for taking time to leave the comment, you're welcome!
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 :)
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.
Hello sir thank you very much. I would like to ask 8:13 is the point of zero shear (crossing zero part in shear) always where you get the maximum moment? Im doing an exercise right now and the maximum moment im getting is not on the point of zero shear.
Hey Nigel, yes, where the SFD crosses the x axis you will always have a local max (or min) on the BMD. Also good to know, where the BMD crosses the x axis, you will have an inflection point on the deflected shape. So it sounds like you just have an error somewhere in your work. Double check that you have solved for the reactions correctly.
It was good Lecture !
Thanks Samuel!! 🙂
This video alone worths more than my three hours class.
Thanks Faras!!!! =)
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
Thanks to you. Question is can we apply this method to the frame problems for finding diagrams?
Yes, see my channel for example frame problems.
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 ✌️
there is another way to solve this question by using Equilibrium equation can you do videos about it plz
Thanks so much sincerly
Any time!
Sir how do u find point of inflection and point of bending moment of this kind of question
THANK YOUUUUU
YOU'RE WELCOMEEEE
Can point loads in the y direction have an effect on the Axial Force Diagram in beams?
For a simply supported beam like this, no. Point loads will only impact shear and bending moment diagrams
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 =)
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!!
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?