Mechanical Engineering: Distributed Loads on Beams (2 of 17) Find Distributed Load on Beam Ex. 1
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- čas přidán 20. 11. 2015
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In this video I will find the Force on A and B of a distributed load on a beam.
Next video in this series can be seen at:
• Mechanical Engineering...
1 video explains better than 4 weeks of lectures...
realmaco1 yah bro ,they should close the universities ,universities systems are becoming boring ,just wasting 💰
This video is extremely well done. Thanks for passing on some very good information.
What a great video, I understand it perfectly now, thank you.
man you are really good, i have been watching your videos about different subjects. THANKS!
great video big help to understand the concept, it could turn out to be even better if u had explained about SFD & BMD for that particular question. Thank you.
You’re a god sent help man
SO GOOD AT EXPLAINING
clear and straight-forward. Super helpful!! Thank you
Glad it was helpful!
Straight to the point. Thank you
Thank you so much you are great at explaining
this is a great video, but you see, my professor did not really give us simple applications like this one, he started with vertical beams and curved lines of action(i don't know what they're called) and way too many formulas to memorize. Is there any other video on this subject?
thank u so much i didnt attend the lectures but just watched this video i have a final in 2days
All the best on you final! 🙂
Thank you Michel!
That was really great... !
went through 4 previous video lecture, including my main go to in statics channel, and was more lost than ever. Thanks for video, first exam today and this was last topic in ch 4 and easy to follow video.
Glad it helped!
thank u sir ..u r the best....
Very helpful! Thanks :)
thanks, needed this for my final.
Good luck on your final.
this video explains better than 20 pages worth of this stuff
Glad you found our videos and that you find them helpful! 🙂
may god bless you, i dont understand my teacher teach me.And my mother thought im wasting my time on internet.I only i was rich ,sure i will give these gold channel some gift :{
Sir in your playlist of shear on beam these videos weren't there
So I thought there weren't none(distributed force on beam)
If somehow those playlists were linked I would watch those before my xm and it would help me a lot
It would be helpful if there's a way if you have such videos available
Great video. thanks. When you get the total center X=3.33 how do you know from where to measure it? from the left corner or from the right? you choosed left. what is the rule?
I understood the logic, like if we would calculate it by integral the (0,0) would be the left corner... depends on my axis.
It actually doesn't matter, and may depend on the particular situation. Typically we use the left as a reference point.
thank you! really helpful, really easy to follow
Glad it helped!
Simple and basics
Use the formula centroid for trapezoid. Y= 2a+b/a+b multiply by (h/3)
Y= 2(3000)+6000/3000+6000 multiply by (6/3)
Y= 2.66666667
X= 6 - 2.66666667 = 3.33333
Sir, is it a uniformly distributed load
Thank you so much!
Heartfelt thank you. Really really really thank you ....
Glad the videos are helpful
you good at your job!!
Simply awesome man🤘🤘🤘
Thank you teacher.
nice video !! thank you..
Really helpfull thanks alot sir !
Isn't the 2/3 distance of centroid from vertex valid only when measuring along the median?
How can we take that distance along the base?
2/3 from the side with shortest height and 1/3 from longest
cool explanation
Very good clear and informative video
Glad it was helpful!
Thank you sir I'm iraq and I wish learning English language and i love English teachers
You're welcome!
thank you! This helped a lot!
Glad it helped!
thanks
Nicely done!❤
Thank you! 😊
.solve this please sir:
A horizontal beam, simply
supported at its ends, carries a load which varies uniformly from 15 kN/m
at one end to 60 kN/m at the other. Estimate the central deflection if the span is 7 m, the section 450mm deep and the
maximum bending stress 100MN/m2. E = 210GN/m²
thanks sir
why we don't take into consideration the X components of the forces at A and B?
In this example, there are no forces in the x-direction.
how can you know the X of the triangle is 4?
The X value in a triangle is always two thirds of the way along the base length, if you start from the smaller end!
i.e. (2/3) * 6 = 4
For a triangle, the centroid is always gonna be 1/3*(side length) ***on the side with "more x". In this case, it's 4 because while (1/3)*6=2 the triangle has more of on the right side so x=6-2=4.
2 x L / 3
great video
Thank you. Glad you liked it.
Hes amazing
We appreciate the comment.
ty sir
You are welcome. 🙂
Should just have explain moment at the same time. Becuase your actually using the moment of both for areas to find the centroid
if the movement is clockwise, why it is not +27000?
+Gantox Davaajargal
The convention is that a clockwise moment is negative and a counterclockwise moment is positive. You will get the correct answer even if you use the opposite convention, but most everyone follows the standard convention.
Great
What if i find b use moment?
You should be able to use either side a or b.
One thing confusing for me is when I need to decide if it's clockwise or counterclock wise @-@
There are videos in the beginning of the playlist that explain that.
Make any one sign convention use that only throughout the problem.
Awman, Our instructor taught us that if its clockwise rotation, its gonna be positive(+) :(
it doesnt matter you will take the other one -ve
How we know that x1=4 and x2=3 ???
Here you must find the location of the center of mass of the two areas. For the rectangle it is in the middle. For a triangle it is 1/3 the distance from the base to the top.
why is the one-third centroid from the triangle starting from the right side and not from the left side?
+Kevin Ng
The force is greater on the right side.
This is late but so that others see if they're curious too, if's just because there's more mass on the right side and therefore "more x"
Why x1=4?????????
X1= 6 x 2/3 = 4. Because we are calculating the distance to the center of the triangle, which is two thirds in from the left. So we multiply the total distance of the beam (6 meters) by 2/3
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