Su estilo de explicar es muy sencillo y claro de entender. Me hubiera gustado tener una maestra asi. Explica tan bien que enamora su estilo. Gracias maestra.
Lots of good stuff in this video, but here's the most important point -- the Area Moment of Inertia is proportional to the cube of the height of the beam. Consider standard lumber: a 2x8 beam (oriented with height = 8) vs a 4x4 beam (same length). They have the same amount of wood, but the 4x4 beam will deflect 4 times as much as the 2x8. This is why rafters do not have a square cross sectional area and are often 2x10's or even 2x12's.
Eres increible es la explicacion mas simple y facil de comprender que he visto Sube mas videos para calculo de vigas con perfiles y ejes huecos de maquinas muchas gracias por tu trabajo
this is the best lecture i have heard-- i am having difficultly in under standing theory of elasticity , theory of plasticity , heat transfer by convection & hydyodynamics
If you use your cantilever for a piping systeme support , how do you calculate the dynamic charge of the fluid and then add the dynamic weight to the centalever calculation ?
I didnt understand one thing. how did you calculate the moment provided by the lever while lifting the load at 7.28? The distance L isnt perpendicular to the point of application of applied force. How come is it FxL? SHouldnt you take the perpendicular component of it?
Suppose there is ductile body which is subjected to bending moment . After calculating induced bending stress . How should I check whether it will fail or not.?
2:45. Were each of those two classmates (tall and short) grew 1 inch naturally? Or were they pulled along their lengths and thus both experienced elongation of 1 inch? I can only see strain on them if they were pulled. Btw, it is highly likely that the tall dude came back to his original height once the force was removed whereas the shorty exhibited a plastic deformation and gained 1 inch of much needed height permanently!
6:23 Can someone explain the integral please? Is Integral(y^2) some kind of a formula for area moment of inertia?? And lastly How is it that: Integral (y^2) = (bh^3)/12 ?? Thanks in advance.
+Thor In pure bending around the z axis (axis coming out of the screen) in this case, you have a distribution of stresses on the cross-section that are positive when y > 0 and negative when y < 0 as the y axis is defined in the middle of the beam, so that the beam bends clockwise around the z axis. The resultant moment at the cross-section is the sum of those stresses that depend on y over the cross-section area. The integral alone is the moment of area. The integral of the cross-section area is seen as integrating x from 0 to b and y from -h/2 to h/2: integral(integral(y^2 dy) dx), then the integral about x is independent and sums up to b, the other integral leaves 1/3*(h^3/8 - (-h^3)/8) which is h^3/12.
This is very helpful when im just get in civil engineering, why youtube recomend this now, now i feels like wasting alot of time reading textbook to understand this thing Anyways thanks for the video its very helpful
It was useful for me . I need to know how much load can withstand for simply supported Beam . In that what are measurements i have to take in practical time .
superb...ccan u say why we draw shear force and bending moment?what happens physically if shear force is zero ? and what happens to the beams physically if the bending moment is maximum?? ans it soon pls
Thank you so much, for this video I'm looking for Manual hand calculation for analysis and design for learning complete step by step process please can you share calculation document, please 🙏
Great video and easy to follow, I am confused though, i tried plugging your values into the formula s but couldnt get hte same results as you. possibly because I couldn't read the value exponents from your board (I am not a mechanical engineer,but i need to understand as i am building a canopy over double doors. The values I have as as follows:- F=0.5 N, L=0.25M, b=0.025M h=8x10^-4M, E=69x10^6Pa, I=1.07^-2 M. I am many orders of magnitude out. I get 3.51^-3 mm and 4.67^-3 Mpa respectively. Where have I gone wrong?
Minor niggle. The choice of the failure of the Tacoma Narrows isn't very relevant to the content of this lesson. In that case it was vortex shedding loads causing a dynamic response leading to failure by lateral-torsional buckling. All of these are advanced topics. The video was well presented and much clearer than many others.
Thanks maria... I had underatand beams abit beam better ;) Im a mobil DJ and im building a set up to suspend ligting and audio gear. Most gear would be evently distribuited on 3 beams on a Y shape.. My question is what would be a safe load or max load capacity on this structure... The struccture would be suspended front 3 points.. one at each end of the Y Hope you can help me..
Mam can you tell us how the beam is in condition to carry load or not or if it is able to take load on it how much of its area will be taking how much load on it and for the slab also
Can somebody break down the example calculation steps for both answers 35.4mm and 46.9MPa. Also can't read the value for h; is it 0.8mm denoted in meters?
@enriqueramosvillafan well, yeah, she skipped the explanation of the integral... but there are general rules about it which you can find in tables. it actually just gives you a mm^4 factor which says why the beam breaks at different axis (as if u try to bend the beam normally as shown at around 1:30, or if u try it where u rotate it 90° where u will need much more force to break it (it has more area momentum of inertia..)
Hi, i would like to work on a project but have a few questions that might be of interest to a few here. If im looking to reduce a tube 75 feet long to one inch I know I will need to reduce it to 75 inches. How will that apply to pounds. For example let say that that pipe can hold 100 pounds , how will I reduce the 100 punds to apply to the for the to apply the reduction for the 75 inch tube?
That was great! So I created something that uses a beam. I ran the numbers on efunda and found that my maximum bending moment stress is 692 MPa. The yield stress for my material (4130 steel alloy) is (gasp!) only 360 MP. So if I understand your video, I should be seeing plastic deformity yet I do not. Is my understanding correct or is the deformity so small that I can't measure it. Thanks again!
well, if I were to take extra classes on structural analysis, I know who I'd consult, the only problem is, I can't contact you as of yet. But, you did well putting things simply, good job!
Any change in the beams cross sectional area, or shortening it's length, will increase it's strength, and decrease its deflection. By adding to the width, height, or shortening the beam, it will decrease deflection.
The practical tips given are good. However, writing the stress formula as sigma=Force/Area while the topic is about "Bending of Beams" can mislead the students. In bending of the Euler-Bernoulli beams, the stress calculation is based on the simple bending formula, sigma=M*c/I, where: M=bending_moment; c=maximum_distance_of_the_material point_from_neutral_plane I=second_moment_of_are_of_cross_section
mam at 2.21 of video you have used yield stress and fracture stress but its yield strength strength and stress are two different things in solid mechanics strength can't be calculated it can only be plot by using ashby plots made by cambridge university or plot in stress strain diagram by breaking test this shows MIT university is just overrated what's with number 1 bullshit another idiot from mit video ruclips.net/video/5uuQ77vAV_U/видео.html he doesn't even know difference between coagulation and flocculation calling Alum as flocculation agent ahahahahhahahahha
Great explanation. I do question the quality of the presentation; I would like to believe that a school as illustrious as MIT would have the budget for a decent camera, lighting and sound. Perhaps a studio? Nonetheless you have done a great job.
Amazing i learned a lot ..btw i think u made a silly mistake for the moment on the rock example...it is force times perpendicular distance from force to calculate moment so the length shown is not right! as the force is shown vertically down
How can I find torque for this application ? www.dropbox.com/s/f18ls68yiil1iso/image001.png?dl=0 Specification : Sheet metal 7 mtrs length and 0.5 to 1.1mm thickness
You are a credit to your profession and your presentation style is both elegant and straightforward.
I'm confused now.. why did I spend 1 year to study this where I can get 100% knowledge in this 10 mins video? Awesome video and thank you so much!
Su estilo de explicar es muy sencillo y claro de entender. Me hubiera gustado tener una maestra asi. Explica tan bien que enamora su estilo.
Gracias maestra.
Lots of good stuff in this video, but here's the most important point -- the Area Moment of Inertia is proportional to the cube of the height of the beam. Consider standard lumber: a 2x8 beam (oriented with height = 8) vs a 4x4 beam (same length). They have the same amount of wood, but the 4x4 beam will deflect 4 times as much as the 2x8. This is why rafters do not have a square cross sectional area and are often 2x10's or even 2x12's.
Same amount of wood matter. One is twice as long but the other is twice as wide.
Wonderful lecture and well explained! Thank you!
wow even me I am undergraduate civil engineering student ,I really appreciate that u post this video clear and easy way .thanks
Eres increible
es la explicacion mas simple y facil de comprender que he visto
Sube mas videos para calculo de vigas con perfiles y ejes huecos de maquinas
muchas gracias por tu trabajo
The way you teach is brilliant!
I am a teacher as well but this is beautiful work and you are gifted
this is the best lecture i have heard-- i am having difficultly in under standing theory of elasticity , theory of plasticity , heat transfer by convection & hydyodynamics
Wish I had such tutorials during my engineering course :)
Great work. Clear and concise.
please upload more videos , ur way of explaining is very much understandable
Does FOS of high loaded structural components shall be calculated based on Yield Stress ?
If you use your cantilever for a piping systeme support , how do you calculate the dynamic charge of the fluid and then add the dynamic weight to the centalever calculation ?
i have few doubts abt monet of inertia. Is there any more videos by Mariya?
Absolutely marvellous explanation! Thank you!
very effective teaching as well as appreciable hard work...
Madam, Thanks for this video, contents of which are very easy to comprehend.
Very nice job and easy to understand. Thank you.
Does this only work for cantilever beams?
Really really good, fantastic and clear explanation of concepts.
am i wrong? the distance for calculation of the moment should be perpendicular, right?
Excellent explanation. It actually made sense!!
I didnt understand one thing. how did you calculate the moment provided by the lever while lifting the load at 7.28? The distance L isnt perpendicular to the point of application of applied force. How come is it FxL? SHouldnt you take the perpendicular component of it?
Suppose there is ductile body which is subjected to bending moment . After calculating induced bending stress . How should I check whether it will fail or not.?
How do you calculate in diagonals on a deck beam with 3 posts?
2:45. Were each of those two classmates (tall and short) grew 1 inch naturally? Or were they pulled along their lengths and thus both experienced elongation of 1 inch? I can only see strain on them if they were pulled. Btw, it is highly likely that the tall dude came back to his original height once the force was removed whereas the shorty exhibited a plastic deformation and gained 1 inch of much needed height permanently!
6:23 Can someone explain the integral please? Is Integral(y^2) some kind of a formula for area moment of inertia?? And lastly How is it that: Integral (y^2) = (bh^3)/12
?? Thanks in advance.
+Thor
In pure bending around the z axis (axis coming out of the screen) in this case, you have a distribution of stresses on the cross-section that are positive when y > 0 and negative when y < 0 as the y axis is defined in the middle of the beam, so that the beam bends clockwise around the z axis. The resultant moment at the cross-section is the sum of those stresses that depend on y over the cross-section area. The integral alone is the moment of area.
The integral of the cross-section area is seen as integrating x from 0 to b and y from -h/2 to h/2: integral(integral(y^2 dy) dx), then the integral about x is independent and sums up to b, the other integral leaves 1/3*(h^3/8 - (-h^3)/8) which is h^3/12.
This is very helpful when im just get in civil engineering, why youtube recomend this now, now i feels like wasting alot of time reading textbook to understand this thing
Anyways thanks for the video its very helpful
It was useful for me . I need to know how much load can withstand for simply supported Beam . In that what are measurements i have to take in practical time .
superb...ccan u say why we draw shear force and bending moment?what happens physically if shear force is zero ? and what happens to the beams physically if the bending moment is maximum?? ans it soon pls
Thank you so much, for this video
I'm looking for
Manual hand calculation for analysis and design for learning complete step by step process
please can you share calculation document, please 🙏
Great video and easy to follow, I am confused though, i tried plugging your values into the formula s but couldnt get hte same results as you. possibly because I couldn't read the value exponents from your board (I am not a mechanical engineer,but i need to understand as i am building a canopy over double doors.
The values I have as as follows:-
F=0.5 N, L=0.25M, b=0.025M h=8x10^-4M, E=69x10^6Pa, I=1.07^-2 M.
I am many orders of magnitude out. I get 3.51^-3 mm and 4.67^-3 Mpa respectively.
Where have I gone wrong?
How do I differentiate the width/thickness measurement from the height measurement in a beam?
Thank God. Saved my life.
Minor niggle. The choice of the failure of the Tacoma Narrows isn't very relevant to the content of this lesson. In that case it was vortex shedding loads causing a dynamic response leading to failure by lateral-torsional buckling. All of these are advanced topics.
The video was well presented and much clearer than many others.
Great Stuff. Explained really well. It's a pity you didn't teach me back at College. :)
What is formula of 45 degree marking in beam
thanks mam .........nice presentation and i understand it very smoothly... I want know about mohar's circle
is any presentation available of u r own?
keep posting similar videos ur presentation is exceptional.Great job.
Thanks maria...
I had underatand beams abit beam better ;)
Im a mobil DJ and im building a set up to suspend ligting and audio gear. Most gear would be evently distribuited on 3 beams on a Y shape..
My question is what would be a safe load or max load capacity on this structure...
The struccture would be suspended front 3 points.. one at each end of the Y
Hope you can help me..
Mam can you tell us how the beam is in condition to carry load or not or if it is able to take load on it how much of its area will be taking how much load on it and for the slab also
ruclips.net/video/fRyUf-GY754/видео.html.
plz tell me how to find BSR range(Bending strengh ratio)
Can somebody break down the example calculation steps for both answers 35.4mm and 46.9MPa. Also can't read the value for h; is it 0.8mm denoted in meters?
RSJ gauge installation before bridge approach (?)
@enriqueramosvillafan well, yeah, she skipped the explanation of the integral... but there are general rules about it which you can find in tables.
it actually just gives you a mm^4 factor which says why the beam breaks at different axis (as if u try to bend the beam normally as shown at around 1:30, or if u try it where u rotate it 90° where u will need much more force to break it (it has more area momentum of inertia..)
Thankyou for making this simple, i can now understand it. I am a scaffolder and was really curious to understanding the bendingnof the tubes lol😂
I enjoyed this lecture so much, its unbelievable!! :D
nice presentation.your way of presenting with examples is excellent.nice job
Hi, i would like to work on a project but have a few questions that might be of interest to a few here.
If im looking to reduce a tube 75 feet long to one inch I know I will need to reduce it to 75 inches.
How will that apply to pounds.
For example let say that that pipe can hold 100 pounds , how will I reduce the 100 punds to apply to the for the to apply the reduction for the 75 inch tube?
That was great! So I created something that uses a beam. I ran the numbers on efunda and found that my maximum bending moment stress is 692 MPa. The yield stress for my material (4130 steel alloy) is (gasp!) only 360 MP. So if I understand your video, I should be seeing plastic deformity yet I do not. Is my understanding correct or is the deformity so small that I can't measure it. Thanks again!
hey if you don't mind, I'd like to hear some more about your beam. are there any more details you could share?
can u please upload design calculation for g+1building from safe bearing capacity to beam slab columnZ reinforcement detail and design
well, if I were to take extra classes on structural analysis, I know who I'd consult, the only problem is, I can't contact you as of yet. But, you did well putting things simply, good job!
I prefer to consider strain in m/m - Yes you can cancel the units but it reminds you what it means. Its a change in length per unit of length
does increasing the depth of the beam decrease the deflection?
Any change in the beams cross sectional area, or shortening it's length, will increase it's strength, and decrease its deflection. By adding to the width, height, or shortening the beam, it will decrease deflection.
yes
fantastic, this was the perfect review for me as an ME student thanks!!!
ruclips.net/video/fRyUf-GY754/видео.html..
i am advocate in india-- i have under stood u r lecture very claerly
What if one side of the beam if thicker than the other?
Good explanations that broght me back to college classes times!
know you are the perfect one ........i don't heard this type of lacture you are best....
Very useful for me as a refresher. Thanks!
Wow best lecture, keep going dear.
ruclips.net/video/fRyUf-GY754/видео.html.
Excellent explanation !
The practical tips given are good. However, writing the stress formula as sigma=Force/Area while the topic is about "Bending of Beams" can mislead the students. In bending of the Euler-Bernoulli beams, the stress calculation is based on the simple bending formula,
sigma=M*c/I, where:
M=bending_moment;
c=maximum_distance_of_the_material point_from_neutral_plane
I=second_moment_of_are_of_cross_section
best explanation ever.
Muy buen video Señorita y muy buena pronunciación también
Saludos desde Chile
Regards!
Explanation was awesome.
Useful lecture.
Great presentation. Well done.
Excellent presentation! Great job!
Short and simple informative video.. Thanks
Most hollow beams in bending fail by collapse of the walls and compression face. How can this be analyzed predicted?
You should review the "lever" chapter and in special the necessity to have a point of support, without it you do not have a lever.
Ratio 50%, How to calculate the Beam/Built-up/channel/Angle load bearing Strength as per LRFD/EC3/ASD
+Urs Friend CADD AISI or AA design manual can be a good reference. Consider lateral bucking, local buckling, crippling, shear etc....
Now I'm ready to build my underground bunker, thanks
mam at 2.21 of video you have used yield stress and fracture stress
but its yield strength
strength and stress are two different things in solid mechanics
strength can't be calculated it can only be plot by using ashby plots made by cambridge university or
plot in stress strain diagram by breaking test
this shows
MIT university is just overrated what's with number 1 bullshit
another idiot from mit video
ruclips.net/video/5uuQ77vAV_U/видео.html
he doesn't even know
difference between coagulation and flocculation
calling Alum as flocculation agent
ahahahahhahahahha
very informative and helpful...now i have an idea of what i need to research
Great explanation. I do question the quality of the presentation; I would like to believe that a school as illustrious as MIT would have the budget for a decent camera, lighting and sound. Perhaps a studio? Nonetheless you have done a great job.
Amazing i learned a lot ..btw i think u made a silly mistake for the moment on the rock example...it is force times perpendicular distance from force to calculate moment so the length shown is not right! as the force is shown vertically down
thanks, it's a good lesson for me
really good explanation,thanks wish you all good luck.
if you have another video or Channel plz.
Great lecture! plus demonstration thanks :D
Excellent............good presentation
I need more videos of yours please reply the link
Luv your style easy to follow how to follow you
got lost at the area momentum of inertia...
Mam you got a Fan added to your list tody.
Thku for such a wonderfull explanation
تبارك الله عليك بتوفيق 👌👌🔔👍
helped a lot . thanks for making this
nice presentation!! expecting more on mechanics.
very good teaching technique.
How can I find torque for this application ?
www.dropbox.com/s/f18ls68yiil1iso/image001.png?dl=0
Specification : Sheet metal 7 mtrs length and 0.5 to 1.1mm thickness
excelent ........maria ....clear explnation.
very nice presentation for a layman like myself.
love your video i wish you could make more videos on building design
Excellent video, thanks.
Great presentation, thanks
Excellent video.
Good video, but you might ought to have mentioned safety factor. (1/2 yield or more).
Brilliant! and practical.
Awesome explanation
ruclips.net/video/fRyUf-GY754/видео.html..