Shear force and bending moment diagram practice problem #2

[@BurskiBags]

you explain this so much better than my professor lmao. Youre a life saver

[@Engineer4Free]

I gochu lol

[@unboxbybil]

Thanks for helping a lot of students like me. I really appreciate your effort and help. Have a great day ahead

[@Engineer4Free]

Thanks Syabil, I'm really glad that I'm able to help students all over the place, and really enjoy hearing that my work is being appreciated. Mission accomplished =).

[@aimanjamil5454]

Most youtube teachers are the BEST. Youre one on them. Thanks. You helped so. Much students each year

[@Engineer4Free]

Thanks Aiman!! =)

[@ii_ragee752]

bro u explained better than my professors ...................i understood this in 11 mins...

[@AWE5OMEANT]

Thanks for the explanation! I couldn't understand it fully when my professor explained it.

[@Engineer4Free]

Happy I can help you out! Check out videos 1-9 here engineer4free.com/structural-analysis and also maybe videos 66-72 here engineer4free.com/statics if you need more insight =)

[@rybread5718]

Absolute badass. Thanks so much! For some reason I have to revisit this video every time I get into a new class with structural analysis.

[@Engineer4Free]

Thx dowggg 💪💪

[@thezeus101]

This has a ton of great basic concepts that you can apply to all problems. Good explanation. Thanks

[@Engineer4Free]

Thanks Andrew!!! You can find all 8 more examples at engineer4free.com/structural-analysis =)

[@dericelvin]

i think we should be paying guys u and study at home instead of wasting tuition fees listening to the professors. u saved my semester...literally

[@Engineer4Free]

I would be okay with that 😏. But either way I'm glad I could help you.

[@ijaaramuu]

How dare you speak like that? I will keep watching! You don't Even know there's nothing that's much you can do alone On courses like strength of Materials and Mechanics! Thank you in Advance for your help dear!

[@alexbeis1861]

Good review video. At 7:29 the slope coming off of 140 should be less than the slope from 0-140 since the rate of change is decreasing not increasing and there was no negative pointload. Some profs might think you dont truly understand what is going on if you draw it like this. Just my 2 cents.

[@Engineer4Free]

Yes you are correct. The way I drew it is not to scale, but if it was, the slope to the right would not be greater than the slope to the left. Sorry if it caused any confusion, but good eye and thanks for pointing it out for others!

[@jessicarapier5330]

This is helping a lot with reinf. concrete. Thank you so much!

[@Engineer4Free]

Great to hear!!

[@nipunalakshan8923]

it's very helpfull to my end exam,thanks a lot

[@Engineer4Free]

Glad to hear it!! 🙂

[@SloppySongs862]

At the beginning when you are summing moments about A why do you multiply the 10kN/m by 6m and 7m? Really enjoy your videos, been stuck on these problems for a while :)

[@Engineer4Free]

Hey George. 10kN/m * 6m gives the magnitude of the resultant force of the uniformly distributed load (udl), it's magnitude = 60kN. This resultant force is in the centre of the udl. The centre of the udl is 7m away from point A. When we take sum of moments about A, the moment caused by the udl about A is euqal to it's resultant force, times the perpendicular distance from A to the line of action of that force, so we get 60kN * 7m = 420kNm. You can write it all in one step as 10kN/m * 6m * 7m to save time. Just make sure that you end up with the correct units of kNm for moments, as expected. Hope that helps! All sfd/bmd videos are here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams

[@SloppySongs862]

Thank you so much! Makes sense now. Have a good Christmas 🎄

[@ashleydefrancesco6162]

This is such an easy way to do it! Love it

[@Engineer4Free]

Awesome, glad to hear it!!

[@beckeras8848]

Shouldn't you have a linear segment of decrease of 20 that corresponds to the rectangle negative area and then a parabolic decrease corresponding to the negative triangle area? at 8:34. I understand that they're one single area but the decrease happened as a result of both a concentrated load + the remainder of the distributed load. Therefore the concentrated load will have a small linear decrease and then the parabolic of the concentrated load will pick up and finish at 90.

[@niranjanwagh5767]

I always hated this topic in college. But after watching your video all the hate is gone. It was really sad that professors used to make use do manual calculation for SF and BM at every single point instead teaching us that BMD can be drawn using SFD by its area. I seen it first time. And I am watching complete playlist now.

[@chukinning624]

A true lifesaver. Admire.

[@Engineer4Free]

Thanks =) =)

[@kingkabester6479]

life saving video.Thanks

[@Engineer4Free]

Happy to help! Do check out engineer4free.com/structural-analysis for the other examples too 🙂🙂

[@sibusisombele2188]

I really struggle with the calculations.I still don't understand how you got 7m in the calculation for moments about A

[@Engineer4Free]

you need to find the centroid of the uniformly distributed load (udl). It's in the centre of it. The centre of the udl is 7m away from A. This is the distance that you multiply by the resultant force of the udl, to find the moment that the udl causes about A. Hope that helps!

[@sibusisombele2188]

@@Engineer4Free thank you so much

[@Nelly303]

Best explanation!!!🔥🔥🔥🔥

[@Engineer4Free]

Glad it was helpful! More @ engineer4free.com/structural-analysis =)

[@thapeloradebenicholas9938]

A life saver🎉🎉🎉❤thank you sir...

[@dinithisahanika7823]

you are amazing..this was what i need

[@turyamuhakibenjamin6549]

such a clear explanation! thank you so much

[@Pourya.mokhtari]

Thank you so much sir very nice way to draw this diagrams❤️

[@abdirahmanawil7300]

Must be subscribed best

[@Engineer4Free]

🙌

[@zubeirjamaa]

Thank you very much.

[@Engineer4Free]

No problem 🙂

[@abdulghayoor7611]

really helpful. well explained.

[@Engineer4Free]

Great thanks for watching Abdul!

[@khalidhajiali1849]

Is there any video explaining how to use the method you used in the top right with the arrows to determine how where to draw lines for the shear body diagram?

[@Engineer4Free]

Hey Khalid, yes, please see videos 66 - 72 here: engineer4free.com/statics for an explanation. What your asking about is called the positive sign convention for beam bending.

[@vashusharma2846]

So the resultant created by the distributed load which is placed at the x-centroid does not have anything to do with making the shear force diagram (and is that for all cases?)

[@Engineer4Free]

The resultant isn’t a real point load acting on the beam, so don’t consider it in the SFD as one. The resultant can be used though as the centre of the udl for purpose of moment about point A. So you only really consider the resultant in the sum of moments equation, then forget about it after that.

[@vashusharma2846]

Awesome man! Thanks so much for these videos they are saving me more than my damn teacher xD

[@jamestillyard6347]

Could you link me to a vid that explains the initial calculations that you quickly show as i don't understand the distributed load section and where you got the 7m from?

[@Engineer4Free]

Yeah no problem. 7m is the distance from A to the centroid of the distributed load. For videos that introduce centroids and distributed loads, see # 58 - 65 here: engineer4free.com/statics and then after, you can work through all SFD/BMD problems in order here: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams the "long way" takes a slower and more detailed approach and the "fast way" is what you see in this video, intended for use in higher level studies when you just need to slam them out as fast as possible

[@Mark-rb4os]

Other than that, you helped me a lot! thank you! keep it up!

[@Engineer4Free]

Glad to hear that! =)

[@DinerosBackyard]

How did you get 7?

[@osamasarfraz8749]

how do you find the location of maximum and minimum bending moment?

[@Engineer4Free]

The maximum here is x=6. It can be reasoned visually because shear is positive on the left and negative on the right. It means that from x=6 on, to the left the moment must be tending toward zero, or otherwise just constantly falling, making x=6 the local (and absolute) max.

[@Mark-rb4os]

I have a question, where did you get that 7m in the solution when you try to solve for moment at point A?

[@bobawilliam2251]

It's a lenght from A to the middle of the area where "q" strenght is applied

[@Engineer4Free]

Yes, thanks!

[@simengg7824]

Which software and pen are you using in these videos

[@Sarrett.Studios]

Why did you make parabolic an upward curve

[@onurcalisir7213]

For the moment diagram between 140-190 the we are going from FAT stacks to short stack the curve should be the inverse isn't it?

[@anonfish5509]

Can you do a video expressing the shear force and BM in terms of x?

[@Engineer4Free]

Check out videos 67-72 here: engineer4free.com/statics they cover SFDs & BMDs in terms of x :)

[@ivortenthani6434]

Really helpful

[@johncarter9812]

I don't understand the point at the SFD where from 15 you go straight down to 5. why did we subtract point load 20KN again ???

[@Amjjad]

Great video! But at the end of the shear force diagram, isn't the reaction force of 45 positive? Or are the two sides of the diagram always opposite in signal?

[@ruveyhakaragoz7457]

this video saved my life lol

[@Engineer4Free]

Happy to be of service 🙂

[@BigKofs]

Sorry why is there a need to add another 20 force to get that 5 deflecting downwards

[@Engineer4Free]

There is a 20kN point force located at x=6m (it's blue on the original diagram, and hiding within the distributed load). An externally applied point load will always cause a jump in the SFD, even if it's within a distributed load.

[@thiccmama6989]

Why are the clockwise moment's not negative?

[@Engineer4Free]

If you moved all of the terms on the left hand side of the equation to the right hand side, their signs would be negative, and would satisfy the typical sign convention of taking counter clockwise to be positive. I often skip the step where I write ΣMa=0, and rather just write the expression with all clockwise moments on one side, and all counterclockwise moments on the other side, all with positive signs. It's the same thing.

[@raghadal-ghandour8831]

how is the distributed force = 20KN Should it not be (6*10)= 60KN??

[@Engineer4Free]

Hmmm, look again. The overall moment caused about A by the distributed load is 420kNm. You do have the magnitude of the resultant right, as 60kN. You also need to multiply it my 7m because the resultant is 7m away horizontally from point A.

[@pranalkhyade9626]

In the Ma equation 20Kn(6)+10(6)(7) why have you taken the 7 m ? Someone please reply I have an exam in sometime

[@Engineer4Free]

The centroid of the uniformly distributed load (the rsultant point load) is 7m away from point A, which we are taking the moments about.

[@BigKofs]

Oh so it has to be from the center? Thank you! I was just wondering how the 1m was determined

[@zozz192]

Thank you again

[@Engineer4Free]

You’re welcome again...again!!

[@Cleanmentality]

What program do you use to make your diagrams?

[@Engineer4Free]

Hey, I can't actually remeber which one I used for this exact video, but the full list of my hardware and software is at engineer4free.com/tools you'll be able to determine it based on the publish date of the video

[@TheToddIrvine]

can you provide a link to a video where u explain your mini free body diagrams and internal shear.

[@Engineer4Free]

Yeah check out videos 66-72 here: engineer4free.com/statics =)

[@owler7715]

What if the 20kn point load is somewhere after the center of the uniform dead load? how would the diagrams look like? Im sorry, still puzzled :'(

[@Engineer4Free]

Look at the SFD. The area with the uniformly distributed load (udl) is a sloped line. The slope is the same to the left and the right of the location of the point load, it just jumps by a magnitude equal to the point load. The slope will be the same no matter where the point load is, just the jump will occur wherever it is. If this is seeming too advanced then I strongly recommend taking a few minutes to review the "long way" of doing SFDs and BMDs, which are videos 66-72 here: engineer4free.com/statics and then after watch videos 1-9 here: engineer4free.com/structural-analysis . Work along with the videos if you can and it will help you recognize the method/pattern. The videos in the first link (statics) are the more comprehensive "long way" and the videos in the second link (structural analysis) are the "fast way." The video that we're commenting on is using the "fast way" so that's why it might seem too quick for ya.

[@PREDATOR-cl5be]

hey man , your videos have been a massive help but i have a question, like how did u get 7m distance on the support reactions part?? And do u have any videos to draw shear and bending moment diagrams with loads in triangular shape??? (P.S - Keep up the fantastic job mate)

[@Engineer4Free]

Hey glad to hear it! 7m is the distance between the resultant force of the distributed load, and point A. The resultant force of a uniformly distributed load (udl) is in the centre of it. The udl is 6m wide, so it's resultant force is 3m from either of its sides. the udl starts 4m from point A, so 4m + 3m = 7m. This is the distance from A to the resultant of the udl, which we need in the moment equation. As for triangular loads, no I don't think I did any. I think the only video that I did on triangular loads was this one from my statics series: kzbin.info/www/bejne/lXyQZJuLrtSkhdk but it just talks about finding the reactions.

[@PREDATOR-cl5be]

@@Engineer4Free got it, thanks

[@unknownymous1404]

@@Engineer4Free then where did 3m came from

[@shasbdo3897]

i appreciate this, thank u so much!!!

[@Engineer4Free]

You're welcome!! The full playlist is here: engineer4free.com/structural-analysis =)

[@shasbdo3897]

@@Engineer4Free i will check it out :)

[@utkugorgulu5349]

Sir, can you give an explanation of why you drawed the graph (140-190 in BMD) as concave down rather than concave up

[@Engineer4Free]

Hey thanks for the question. First of all, if the entire BMD appears upside down to you, then it is because we may be going by different conventions. I follow the North American convention of drawing positive values above the axis for a beam. I believe UK and some other countries draw positive values below the axis. Mathematically the other way is less intuitive, because it gets really confusing to take the derivative of a function where positive is in the downward direction, but practically it is quite useful, especially in the analysis of frames where members can be oriented in any direction because the positive values correspond to tension side of the member... The only reason I mention all that is because, if the entire BMD was flipped/mirrored about the axis, then the parabola in that section would be concave up. Following the convention that I do, the parabola should be concave down. There are two ways that I like to interpret why this is. 1) If you numerically integrate the SFD in this region, say we slice it up into 100 vertical pieces, each 0.02m wide. The area of each slice on the SFD translates to the change in magnitude on the BMD across that 0.02m section. The areas on the left side are bigger because they are taller, and the areas of the SFD slices as we move to the right get smaller as the triangle decreases in height, ie, the changes in magnitude on the BMD get smaller. So this is why we are getting greater slopes on the BMD on the left hand side of the region than we are on the right hand side. There is only one way to draw a parabola that satisfies this, and it is as drawn in the video, and that is for the parabola to open downwards, ie concave down. 2) We already know that the BMD is the integral of the SFD, so that means the SFD is the derivative of the BMD. Every value on the SFD is therefore the slope of the BMD at any given value of x. Looking at the first region, the SFD is a constant 35, and the slope of the BMD is a constant 35. The SFD is 35 at x = 4, and 15 at x = 6, so the slope of the BMD at x = 4 is 35, and at x = 6 is 15, ie it is decreasing as we go left to right in this section, which again would mean that it must be concave down. When I looked back over this video to address your comment, I realized that I made a bit of a graphical typo on the slopes at x = 4, 6, 10. I have put the following note in the description of the video: "There is a bit of a graphical typo in this video. The values, shapes, and concavity on the bending moment diagram are correct, but where the BMD equals 140 and 90, I drew the slope as discontinuous. It is in fact continuous at those points. The slope appears to be continuous at the max where BMD = 190, though it is truly discontinuous there. If you took the derivative of the BMD and plotted it (it's the SFD), there is that discontinuity at x = 6m that basically tells us the slope on the LHS is +15 and the slope on the RHS is -5, ie no single slope defines the point on the BMD where x = 6m. If you do the same thing at the other mentioned values (x = 4, x = 10) then you'll see that the slope of the BMD (value on the SFD) is +35 on both sides of x = 4, and -45 on both sides of x = 10, ie both of those points on the BMD have continuous slopes. Yay...calculus! I got a little carried away trying to pronounce the shape of the parabola in the small vertical space that I was working with and as a result, it can be a little misleading if you are really paying attention. There is a really great free tool on bendingmomentdiagram.com/ that you can see a computer generated SFD and BMD for any statically determinate beam. I recommend inputting this beam or any others that you aren't sure of to see the non-shaky-hand drawn and properly scaled version, it's really nice." Anyways, I hope that all helps to clarify why the section in question is concave down, and sorry that there is a bit of slope mix-ups that probably added to the confusion. Cheers, let me know if that clarifies it all!

[@utkugorgulu5349]

Thank you so much for this great answer...

[@gagankhipal7546]

from where the 7m come?

[@BigKofs]

Co ask Can you please explain how it’s 7 and not 4

[@beryamin8815]

You’re champion 🖤🖤😍🌺🙏

[@Engineer4Free]

Thanks Bery =) =)

[@nazdolfdanielsalvacion5188]

How did you get the 7 from the Moment at A

[@Engineer4Free]

7m is the distance from point A to the center of the distributed load. For moments resulting form uniformly distributed loads, we use the resultant force which is located in the center. Because the distributed load is 6m long, the resultant force is 3m from either side. The udl starts 4m from A, so 4 + 3 = 7m, is the d in the M=Fd equation. Got it?

[@mohdhakim2305]

i don't have any idea how to get those eqn😭

[@Engineer4Free]

This video is using "the fast way" and completely skips the explanation on the equations of equilibrium at the beginning. Watch videos 66 - 72 here: engineer4free.com/statics much slower and detailed explanation, including the equations at the beginning for finding reactions 🌴

[@netrixz4170]

thank u for this video!

[@Engineer4Free]

You're welcome!!

[@shewitmisghina6948]

I love it

[@Engineer4Free]

Thanks Shewit =) =)

[@antiquarian1773]

do you have videos where instead of beams you use frames?

[@Engineer4Free]

Hey sorry, I only have videos on statically indeterminate beams, not frames.

[@antiquarian1773]

@@Engineer4Free thanks for the reply

[@simonblue6354]

Hello! I have a question. Is it always the maximum moment is in the point of infelection? Thanks in advance

[@Engineer4Free]

It's not a local max/min on the BMD that indicates an inflection point, it's when the sign is opposite across a point. Check out videos 8 and 9 here: engineer4free.com/structural-analysis for some examples with inflection points

[@elgagabriongoy5392]

This dude does it like my mum in a kitchen 🍳

[@Engineer4Free]

🥪

[@siegeclimbing9818]

is it 1m or 2m

[@Engineer4Free]

Can you be more specific? Not sure what you are referring to.

[@siegeclimbing9818]

the leangth......im sure now its 2m but it looks like 1m

[@Engineer4Free]

Oh, the lengths written in purple from left to right are: 4m, 2m, 4m, 2m which makes the total span of the beam 12m. Sorry that my 2's here look like 1's.

[@siegeclimbing9818]

Engineer4Free haha🤣 not only here almost every time your 2 like 1....But very nice videos they helped me a lot....

[@user-vs5wu3ri8z]

thx boss

[@Engineer4Free]

You're welcome!! More examples here: engineer4free.com/structural-analysis =)

[@tabokakeabonye400]

Yeah that's very gòod

[@Engineer4Free]

Thanks Taboka 🙂

[@merensow200]

sir, how did you find the reactions. This is the problem.

[@Engineer4Free]

I applied the sum of forces in y direction and sum of moments about point A. For the uniformly distributed load, take it's resultant force as a point force that is in the centre of it. You can see the work for it in the top right of canvas in black ink.

[@DEON2017]

You didn't go into depth for the calculations at the beginning of the video so i'm still kind of confused 🤦‍♂ but everything else is cool

[@aymanhassan1248]

Anyone knows which sofware used to draw this?

[@Engineer4Free]

Hey Ayman, you can find a list of all of the hardware and software that I use over at engineer4free.com/tools

[@zeshuli8707]

好厉害 学会了

[@samuelxisun6294]

zeshu li 推狗

[@ansons.8134]

太推了

[@Engineer4Free]

^_^

[@ayman10s12]

So fuckin fast

[@Engineer4Free]

Ayyyy yeah, but it's sort of supposed to be because this is part of a review section of a senior level course on structural analysis ( engineer4free.com/structural-analysis ) that assumes you're familiar with these things. For a slower, and much more detailed set of tutorials on SFD/BMD, check videos # 66-72 here: engineer4free.com/statics

[@chrislonyema2525]

🤤

[@Engineer4Free]

🤤

[@leonblake3386]

very strange bowels

[@babakiani]

thank you very much for the great Content!