Shear force and bending moment diagram example #5: mixed distributed and point loads

[@RealTejasYagnik]

I feel glad that there is teacher like you in KZbin.

[@Engineer4Free]

Thanks Tejas, means a lot to hear that :)

[@RealTejasYagnik]

Engineer4Free :) ;) ☺

[@Si4koTushxD]

I just want to say, Thank you. You basicly opened my eyes for something really simple. ^_^

[@Engineer4Free]

Glad I could help!! These are simple at heart, just often introduced in a way that seems more complicated than necessary. I've got several videos on SFD and BMD here that you should check out: www.engineer4free.com/blog/ultimate-guide-to-shear-force-and-bending-moment-diagrams

[@ryanhildebrandt8199]

best instruction videos on the web. thumbs up my man

[@Engineer4Free]

Thanks bro!

[@SpongeTMM]

This is a great example and good explanation, thanks!

[@camesjumby]

Very helpful video, but I believe that when you are summing the moments about point A that you should be assuming counter-clockwise as positive, based on how you added up the moments in the equilibrium equation.

[@Engineer4Free]

I skipped the step in ghé sum of moments equation where I write all terms add up to zero. I went to the next step by moving all negative terms to the other side and changing their sign. That's why the term with By is positive on one side, and all other terms are positive, and on the other side of the equation. I did take counter clockwise as positive, as indicated by the little sign top right of the capital M, if you write the full summation equal to zero, and them move terms around, you'll get what I have.

[@pranavsistla7487]

why did you multiply the distributed load by 2m and then 3m in the initial moments equation?

[@alexk9795]

Can you make examples of more complicated structures eg pi shaped structure with beams coming out of the side? It would be really healpful since these are pretty basic exercises!

[@Acladiere]

This is brilliant

[@Engineer4Free]

Thanks Kas! ☺️

[@ruzainiahmedh3521]

thankyou so much .i hv a doubt.if the shear force at the rightmost end is -50N,it implies that shear force at right most end is 50 N vertically upwards .so when considering vertical forces at B.it should add up to a net force of Rb+Sb=50+50=100.then how could B be at rest

[@yashasvisharma4148]

You are better than my teacher 😉.Thanks btw

[@Engineer4Free]

Thanks for the feedback :)

[@yashasvisharma4148]

Sir, Relation between Enthalpy change and Internal energy change please.😫

[@seongminpark7513]

Thank you for the explanation! I have a question about labeling the intervales in between the different applied loads, how would you know if an interval going from linear to parabolic would be a knick (not smooth) or smoothly connected? or also from parabolic to parabolic? (this applies to the moment diagram for this example) Thank you!

[@Engineer4Free]

Good question. You could find the equation in terms of x, and then derive it to find the slope at the point from each side. But that would be a lot of work if it is purely out of interest. Even if the slope is same just on each side of a point though, the overall Shear or Bending Moment equation will be discontinuous anyways, as each interval is governed by different equations. So that may be a good enough observation to note. When I was studying, it usually sufficed to just find the x location and y magnitude at each point of interest and the general shape / form of the interval in-between.

[@CivilThinking]

Thanks, really helpful. Please keep uploading new tutorials. You're an inspiration.

[@Engineer4Free]

Thanks!!! I plan to 🙂🙂

[@baldbipolarbikerboy]

at 10:01 he explains why UDL's are curved in BMD

[@ahmadjaber957]

i have a question, in our course they use slightly more complicated structures like for example 3,4 beams with complicated shapes and forces. Does that affect the way in which i apply these methods?

[@jharden4926]

If i wanted to find the sheet force and bending moment about a point C (lets say it is right in the middle of the distributed force) how would i go about doing so mathematically?

[@MeesBorg]

Im benchwatching your whole static playlist, tomorrow is the exam and this is Verry helpfull!! I saw that you covered dynamics aswell, what are your plans for future subjects? P.s. Do you have a patreon account or something similar where we can support you? This takes a lot of time to make/ edit I reckon, I would be more than happy to support you! :)

[@Engineer4Free]

Hey Maes, thanks for the feedback! I have a few things in the works at the moment. I'd like to finish up some of the courses that I've started but haven't completed yet, like Linear Algebra, Dynamics, and Differential Equations. After that I'll be moving on to some other courses! I don't have Patreon at the moment, just a coffee fund haha: engineer4free.com/coffee I've been looking at Patreon though and think it might be a nice road to go down soon.

[@graysonwomble8820]

@@Engineer4Free you should definitely make a patreon, would love to donate!! your videos are so helpful!

[@santosshresth4419]

thanks for the awesome explainations

[@Engineer4Free]

You're welcome Santos!

[@santosshresth4419]

i would like to tell u to make more of such videos but thanks to god i don't have to study in next semister @@Engineer4Free

[@med7756]

Why can't we make the distributed load 10 time 2 then we move it to the half of the distance then we make division for the distributed load by half and go down 40 then conplete the other half to the end of the roller

[@vladberez]

How do you know which way the parabola goes???

[@poachedeggsunnyside]

I am confused, during the bending moment diagram, why instead of going down from 30kN to 20kN, it going upwards to 40kN?

[@Engineer4Free]

The change in magnitude of the BMD across one section is equal to the area of the SFD in that same section. If the SFD is positive (above the axis), then the change in magnitude of BMD will be positive. If the area of the SFD is negative (below the axis) then the change in magnitude on the BMD will be negative. The area of the SFD across this region you’re looking at is (10kN)*(1m)=10kNm. It’s positive. So the BMD jumps up 10kNm from 30kNm to 40kNm across the section. It’s linear because there is no distributed load in the region. Just because the SFD “goes down” doesn’t mean that the BMD will too. It’s all about the area being positive or negative. You’ll notice the slope in this region of the BMD is less than the slope in the previous region, but they’re both positive. Greater area in the previous region means great slope, both being positive areas in SFD makes positive slopes on BMD.

[@poachedeggsunnyside]

Engineer4Free thank you sir, i spent whole day trying to understand and i figured it out. Now it is confirmed by you by answering my question! Thank you for your time putting out this tutorial for people to learn :)

[@hayatuberhe3230]

thank u sir but i dont understand why BY on shear force is -50kn cos when we see the reaction the force is acting upward

[@Engineer4Free]

If you draw the free body diagram of point B with a virtual cut just the the left, knowing that By is pointing up, that means that the internal shear needs to point down on the left for the forces in the y direction to sum to zero (8:03). If an internal shear is pointing upwards on the left that is considered positive, so because it is pointing downwards, it is opposite the positive sign convention (0:38) and considered negative. I do recommend watching videos 67 and 68 here: engineer4free.com/statics I go over the positive sign conventions for these types of problems.

[@hayatuberhe3230]

i get it..thank u sir

[@Engineer4Free]

Glad to hear it, thanks for watching!

[@散华-l9m]

really cool

[@rajithasampath8033]

how did you get that UDL=40kNm, as I think it should be 20kNm

[@Engineer4Free]

Hey Rajitha. At 6:00 when I write 40kN, that is the applied point load (the blue one on the original diagram). That has nothing to do with the UDL. At that point in time, we're taking the virtual cut to the right of the point load, so the entire magnitude of the UDL is only 10kNm * 1m = 10kN. At 7:06 I am considering the virtual cut to be just before the end of the beam, so we are now considering the entire UDL, so it's magnitude in that scenario would be 10kN/m * 2m = 20 kN. Hope that make sense. You can check out the other examples in videos 1 - 9 here: engineer4free.com/structural-analysis for some more practice

[@aramahmad778]

thanks, what is the name of this tool used free hand?

[@Engineer4Free]

Yeah I draw everything free hand. You can see the full list of hardware and software that I use here: engineer4free.com/tools 👍👍

[@aramahmad778]

@@Engineer4Free thank u

[@manoharchandanapalli383]

well said sir

[@Engineer4Free]

Thanks Chandanapalli!

[@muhammadzaidibinmohdroni2698]

i dont get at the point B how you get that -50kN at the finishing?

[@OutThBlue5]

Muhammad Zaidi Bin Mohd Roni There is a further load acting from the 40kN arrow to point B. The load is 10kN/m acting along 1m, giving 10x1= 10kN. On the shear force graph its acting negative, therefore -40kN -10kN = -50kN at B. -50kN + (the reaction force) 50kN brings the beam back to equilibrium at 0kN.

[@Engineer4Free]

Yeah thanks for replying! The udl will cause the shear to drop by 10kN/m along its length, and then the point load also makes it jump immediately by an extra 40kN where it acts. The slope of that line on both sides of the jump is the same 👌👌

[@muhammadzaidibinmohdroni2698]

@@OutThBlue5 ok thank you

[@salmanmohd8499]

Thanks

[@Engineer4Free]

You're welcome =)

[@14henrylover]

Thank you sir again really helpful

[@Engineer4Free]

Your welcome!!

[@jackbortz8412]

I thought shear force diagrams and moment diagrams always had to go back to zero?

[@Engineer4Free]

Not necessarily, it entirely depends on the type of structure and type of loading. A simply supported beam with no externally applied moments acting at the ends will have a BMD that starts and finishes at zero. Adding an externally applied moment at either end will change that. Shear force diagram for a simply supported beam will start and finish at values equal in magnitude to the reaction forces at each end. A shear force diagram could be zero at the free end of a cantilever beam subjected to a uniformly distributed load. Basically, it depends. I'd recommend watching videos 1-9 here: engineer4free.com/structural-analysis and working along with them to get practice and see a few different types of structures and loadings.

[@naishinlungpanmei1357]

(Draw the shear force and bending moment diagrams of a simply supported beam of span 6 m and carrying a uniformly distributed load of 50kN / m over the entire length and a concentrated load of 80 kN at 2 m from the left support. Can solve this for me

[@Nemesis-The1]

Hey, aswsome content and very helpfull toturials!! But I have a request from you and it would be great if you could help. so I have this question from an old exam (preparing for exam) that is very hard and I cannot reach the professor who is responsible for it for guidance, but I am glad I found you... I have been trying for 3 days to get it right and hope you can help me solve it. This is the question> imgur.com/a/HwOtedo they ask for SFD and BMD it would be really nice if you could help me with a solution!