Shear force and bending moment diagram practice problem #1

[@maryn392]

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

[@matta7829]

Statics final tomorrow, wish me luck!

[@Engineer4Free]

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!

[@argenisperez3467]

Same here friend, best of luck to you

[@eddiemartin2268]

Hope it went well! Got statics in a few days!

[@azamkhan1526]

how was ur result?

[@venchanpiseth1321]

@@Engineer4Free what is the program that u used to draw this?

[@edwinakwemoh9547]

4 years later and the video still helps. God bless you!

[@Engineer4Free]

Can't believe it's already been 4 years! Thanks for still watching in 2021! =)

[@0skulllittle]

you taught me what all my teachers failed to teach me! thanks

[@Engineer4Free]

Glad I can help

[@cpaulcervantes]

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!

[@Engineer4Free]

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!!

[@sarahhassan4953]

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!

[@Engineer4Free]

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

[@tkc056]

Four years of undergrad ME, and this is the cleanest statics beam bending video I've seen. Good stuff

[@Engineer4Free]

I got 8 more right where that came form: engineer4free.com/structural-analysis 😂😂. But thanks tho

[@yatogami5598]

That's the most humane explanation (simplest) ever. Thank you so much!

[@Engineer4Free]

Thanks Yato! 😊

[@sammiemay138]

I was so confused about this and found it hard until i watched the video. Fantastic explanation clear and easy to understand.

[@yadrzgar9980]

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

[@Engineer4Free]

That's unfortunate that you got that professor, but great feedback about my videos.. Thanks for sharing and glad I could help =)

[@aroom995]

You are helping generations man keep it up

[@Iriee242]

You sir, are a hero. Thank you for this labor of love.

[@Engineer4Free]

Thanks friend 🙂

[@abisber]

idk how many times i go back to see this explanation, much helpful than the teachers slides

[@Engineer4Free]

Glad to hear its helping!! =)

[@sserwaddajoseph5869]

This explanation has saved my life. Big thank you sir. God bless you with whatever you want

[@gladys5753]

Hey cousemate,we a test today hopefully u still remember

[@nadiaditta4647]

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

[@Engineer4Free]

Hope it went well!

[@eliasmakanyire3939]

you are the best ,my lecturer never explained where the area came from

[@Engineer4Free]

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 =)

[@tahiratarr9647]

You sir are a whole legend. A method I can finally understand completely. Many blessings!

[@Engineer4Free]

Thanks Tahira!!!

[@deepakbriglall5845]

Explain so much better than my professor. Thank you for all that you do

[@tyleryatco6591]

Thank you! Saved me. Fell asleep in my statics lecture and was so lost in this

[@Engineer4Free]

😂😴

[@michaelhunt2222]

As an architect student, this is really useful (im learning how to do this before its taught in a lecture lol)

[@Engineer4Free]

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!!

[@ahmedyehia9560]

Where do u use structural analysis as an architect ?

[@michaelhunt2222]

@@ahmedyehia9560 nah lol

[@gadieloros6280]

@@ahmedyehia9560 you dont haha its more for the arch. engineers not much architects

[@ancy200]

It's more useful to Civil engineers not Architecture

[@mcho1094]

You a hero who save my assignment

[@Engineer4Free]

Awesome!!! =) =)

[@book9687]

You saved me! Thanks a lot. wish me luck for the exam in 6days 😭

[@Engineer4Free]

Yeah good luck!! Check out 8 more examples in section one of engineer4free.com/structural-analysis 🙌🙌

[@snazzytiktok4223]

just so you know you're still saving lives

[@somtoobienu5081]

Yep

[@Engineer4Free]

Glad to hear it =) =)

[@bars1734]

I have a statics quiz tomorrow. Thank you for the information. You, sir, are a hero.

[@Engineer4Free]

Good luck!!! =)

[@sourblve]

Duuuude you are the OG of statics!

[@Engineer4Free]

Hahaha thanks Menyota 🤜🤛

[@rmrafsanzanilabib6881]

Your BM calculation and BMD was super helpful. Thank you

[@BlessedGPlays]

I have CVE test in like 2 hours 😂😂 this was a huge help

[@shreyashshankar9819]

Bro you are a real Engineer!!

[@Engineer4Free]

Hahaha, thanks!! 💪

[@NoOne-cg9im]

Sir you have just saved my life

[@Engineer4Free]

Glad to hear it! You can check out more examples here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams =)

[@-surah4505]

You are a mazing thank you so much Good luck From Egypt 🇪🇬

[@Engineer4Free]

Thanks and welcome =) =)

[@mathsconceptspashto8959]

Thank you sir. Your teaching method is easy to understand for a common person like me🤗🤗

[@Engineer4Free]

Awesome, that's the goal!

[@gabbyproeh7158]

how did you get 10(3)(6) where is the 3m coming from if the point load is at 6m

[@Engineer4Free]

(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]

Engineer4Free ....so we only multiply centroids for distributed systems cases??

[@WetPig]

@@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.

[@alupha_216]

Just as lost as you are mate, I’m still looking a video that’ll explain everything from the start ☹️

[@drofeng]

Check out this video on force resultants kzbin.info/www/bejne/l33ZfWRrl8qtqa8

[@alantam4579]

thank you, I understand shear force and bending moment much better.

[@Engineer4Free]

Glad to hear it!! 🙂

[@Raniaska0306]

Sir please know that you are a great teacher, keep it up, I love your videos!!!

[@gamersworld6591]

thank you very much My Teacher , your student from Syria

[@Engineer4Free]

You're welcome, I hope you can find more of my videos helpful!

[@farasabyan45]

This video alone worths more than my three hours class.

[@Engineer4Free]

Thanks Faras!!!! =)

[@christinesison5638]

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 😊

[@Engineer4Free]

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

[@hazy7964]

Bro saved me from solid mechanics 🙏

[@Engineer4Free]

I gochu 🙌

[@oseikuami5430]

THANKS FOR SUCH A NICE LESSON.

[@Engineer4Free]

UR WELCOME

[@AbdullahKhan-mv8rq]

Statics final today. Pray for me.

[@Engineer4Free]

How did it go??

[@adhi9621]

Me today

[@raulmishiyev]

Thank you so much! Studying for the FE exam after years away from school and struggling with this section

[@Engineer4Free]

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

[@gowthamthotapalli8867]

@@Engineer4Free can u elaborate support reaction A.

[@joeyjaboli1641]

You earn my subscription. Thanks a lot man!

[@Engineer4Free]

Honoured to have it, thanks Joey!! 🤜🤛

[@a7medabdelsalam775]

thanks for the great explanation, can you please explain to me how you get A=50KN and why you multiply the 60 by 3m

[@Engineer4Free]

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!

[@adoomyforever3673]

@@Engineer4Free thank you so much

[@mhammadsinjer9313]

thanks from all of my heart

[@josipvlahovic326]

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.

[@Engineer4Free]

Awesome! Th aks for commenting, and good luck in your studies! I've got a lot of videos that can help you 🙂🙂

[@Pikapchuu]

What a helpful video! Thanks for helping me review for the FE :)

[@Engineer4Free]

Thanks for the feedback Priscilla, good luck on your exam!!

[@faizanalam855]

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.

[@Engineer4Free]

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!

[@alexnelson8]

Good PE exam review. Thanks bud.

[@Engineer4Free]

Thanks for the feedback, hope the exam goes well!

[@Jjared-ff9km]

i'm having trouble getting Ra and Rb i wish you do it step by step

[@Engineer4Free]

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.

[@tehshowtime]

You Sir are a legend

[@Engineer4Free]

Thanks Nadeem =)

[@pine4pple808]

cant thank you enough, truly appreciated

[@opokuprince253]

God bless you for your nice explaination.

[@augustamoulton2380]

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.

[@WetPig]

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.

[@emily8878]

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).

[@Holyacrobat1]

Could you please do a video on bending & shear force diagrams including equations please

[@Engineer4Free]

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 👌👌

[@Holyacrobat1]

@@Engineer4Free thank you ! Love your videos!

[@greatpharoh0303]

God bless you my brother, Thank you so much!

[@Engineer4Free]

Your welcome Phil!! 😊

[@harshsaini9537]

Quality stuff. just lovin' these lectures

[@Engineer4Free]

Thanks!! =)

[@Valantis_Stylianou]

Sir you are the best! Thank you very much for sharing this video! You've helped me a lot

[@Engineer4Free]

Awesome! Glad to hear it, hope you find my other vids helpful too 🙂🙂

[@rogerzheng2814]

just want to say thank you.

[@Engineer4Free]

Thanks for taking time to leave the comment, you're welcome!

[@karimalsaid1951]

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 ?

[@calvoh9715]

american using metric units,wow.i love your videos.

[@Engineer4Free]

Thanks bud, glad you like them! 🇨🇦

[@babakiani]

Thank you very much Sir for the great content!

[@muhammadzaidhasan1426]

thanks a lot sir.........keep up the good work.....may you propsper more and more

[@Engineer4Free]

Thanks friend!!!

[@boyka712]

thank you my friend for this simplify and hope you the best

[@Engineer4Free]

You're welcome bud!

[@xxxiaoty]

TYSM I HAVE AN EXAM THIS MORNING

[@Engineer4Free]

Good luck!!!! =)

[@spencerpearson168]

THANK YOU FOR THIS VIDEO

[@hannahnkunika1976]

The video is really helpful

[@arconiz]

x can also be solved as 50KN /10KN/m

[@moviesfreak154]

It is helpful. Thank you sir.

[@jamesschroeder2040]

Good job man! helped me a lot :)

[@Engineer4Free]

Awesome glad to hear it!!

[@Sumankumar-zf4fh]

Very nice sir and please make video on RCC

[@Engineer4Free]

Thanks Shekhar, check out videos 32-34 here for RCC beams: engineer4free.com/mechanics-of-matrerials

[@limliyong9873]

Appreciate your work thanks

[@Engineer4Free]

Thanks Lim!! =)

[@sipanhamo322]

I think the reaction F from B should be 25kn instead of 30kn. as you see from the moment that 300/12=25

[@Engineer4Free]

Hey Sipan, the left side of the equation reduces to 360, not 300, so 360/12=30 !! 🙂

[@norielgarcia9379]

Nice presentation. May i ask what software are you using?

[@nnorugaifex9489]

Thanks sir. But I have a question, is the slope for the first parabolic suppose to curve in the 'n' direction or 'u' direction

[@Engineer4Free]

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.

[@maikokyando4596]

Sorry,, clarify me how did you get that value of x?

[@OGTennyson]

This was helpful asf...thank you 😭❤️

[@Engineer4Free]

Glad it helped! Full playlist is here engineer4free.com/structural-analysis 👍

[@elvinhasanov5285]

Thanks to you. Question is can we apply this method to the frame problems for finding diagrams?

[@drofeng]

Yes, see my channel for example frame problems.

[@nigel8221]

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.

[@Engineer4Free]

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.

[@geraldinepaulinelucas6894]

thank you so much

[@UnOrthodox_Christian]

Great video

[@Engineer4Free]

Thanks Devin!! 🙃

[@mohamadakashah644]

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.

[@Engineer4Free]

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!

[@khodadadvahedi2457]

You are right

[@melissaleffingwell9501]

Thank you for this!

[@lucasgibson9295]

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.

[@latitude.]

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!

[@Engineer4Free]

Yessssss thank you! That is exactly what’s going on here 😁😁

[@hasanegeylmaz4335]

Thanks so much sincerly

[@Engineer4Free]

Any time!

[@mohannadzaaza5162]

thank you. clear and informative.

[@Engineer4Free]

Glad it was helpful! =)

[@zhengyangsong]

Thanks for the useful video!!!

[@Engineer4Free]

You're welcome, glad I can help!!

[@asadgondal3746]

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.

[@sacayan_johnn]

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.

[@olevel_maths1470]

BMD and SFD (100%) kzbin.info/www/bejne/eIWTfX-bacSAaaM

[@amielcolon2382]

Did you use the Area Method for this example?

[@Engineer4Free]

I'm not sure if it's officially called the area method, but I just call it the fast way. But yes, as you can see I'm juts taking the areas of the sections of SFD to calculate change in magnitude of the BMD across the same section. See videos 1 - 9 here engineer4free.com/structural-analysis for a few more examples using this method 👍👍

[@amielcolon2382]

@@Engineer4Free yes I thought so too. Thank u so much for replying :)

[@realismtalks]

Absolutely grateful, thank you.

[@Engineer4Free]

Awesome Ahmed!! 🙂

[@nitikabhattarai8812]

How to figure out when to bent upward or downward on bending moment diagram of parabolic line?

[@Engineer4Free]

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).

[@a7medabdelsalam775]

there is another way to solve this question by using Equilibrium equation can you do videos about it plz

[@kill4vill4]

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?

[@Engineer4Free]

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!!

[@viktoremusic8484]

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?

[@Engineer4Free]

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 =)

[@jessieberro9514]

Sir how do you know what type of parabolic curve in the moment diagram by just looking at the slope in the shear diagram?

[@Engineer4Free]

Hey Jessie. 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.

[@Sh7i__]

how about the maximum shear force and bending moment? also thnx for the very clear explanation !

[@Engineer4Free]

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.

[@Sh7i__]

Engineer4Free THANKS for the quick and the informative reply 🙌🏻

[@Engineer4Free]

No worries! Glad I can help =)

[@brandongarnett1800]

Discovering this a little late but how did you work out the 50 kN at A

[@tomgaboian6279]

sum of forces on the beam = 0, +30-20-60+A=0

[@seanlorenzo5149]

@@tomgaboian6279where you get the 60? Is it 10+20+30?

[@nirmalrana3538]

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.

[@Engineer4Free]

Hey Nirmal, I've got a full list of the hardware and software that I use to make the videos at engineer4free.com/tools ✌️

[@ahmedalmunla6866]

What controls the direction of moment parabola ? , like a if it is , let's say a sad face or smile face ?

[@Engineer4Free]

I like to think of it visually. It’s the integral of the corresponding section of the SFD. If you were to numerically integrate a triangular section of the SFD, you’d cut it up into many thin vertical slices. Where the triangle is taller, those slices have more area, and therefor, the BMD will have a steeper slope. Where those triangular slices are shorter, they will have less area, and therefor the BMD will have a less steep slope. Additionally, where the SFD has a positive are, the change in magnitude of BMD will be positive, and where the SFD has a negative area, the change in magnitude of the BMD will be negative. Together, knowing the direction of the change in magnitude, and which side will have a steeper slop, there becomes only one possible way to draw the rough shape of the parabola in each given section. I hope that clears it up for you!

[@beryamin8815]

You’re amazing thank you so much 😊

[@Engineer4Free]

Thankssssss =) =)

[@zozz192]

Thank you very much! 👍

[@Engineer4Free]

You’re welcome!! 🙂