When you draw a distributed load on a beam like this, you are graphically representing something that's actually happening in real life. It could be something like a force exerted by a rectangular or triangular pile of snow along a beam. If you put your graph's origin at point A, then the "x" axis will be in units of meters and the "y" axis will be in units of kN/m. It's ok to have graphs with axis that aren't in the same units, actually it's quite common. Imagine a car thats travelling at a constant velocity. You can make a graph that is seconds on the x axis and meters per second on the y axis. If the velocity is constant, the m/s will be a flat line. If you looked at an interval of some amount of seconds, the area under the line would be a rectangle, and you could easily find it as A=b*h and it would be in units of meters (seconds * meters/second = meters). It turns out that the area is equal to the distance travelled in meters. Often the area under curves is meaningful and so are the units that it is in. It turns out in the example in this video, where we have a graph with the x axis in meters and the y axis in kN/meter, that the area under the graph will be in units of kN (meters * kN/meter = kN). So if we find the area of the whole graph (between A and B), then that is equal to the total force exerted by the distributed load across that span (and passes through the centroid of the shape), which we can then use in the determination of the reactions. Point of the story is that the "height" of these graphs is not in meters, so it's not the actual depth of snow or whatever is causing the force, it IS the force at any given point, and by calculating areas of shapes with different units on each axis, you often reveal meaningful stuff. Hope that helps, let e know if your still confused!
20KN/m represents the max force magnitude, of which increases at a constant rate (F=(20/6)*x). This creates a right hand triangle with x=6m, y=20KN/m. He says height only to use trigonometric terms to solve for the sum of the distributed forces, which is the area of the triangle.
Thank you very much! My statics class is online this semester, and our lectures are basically the teacher telling us to read her notes. It helps so much to see examples worked through!
Statics already can be tricky, and I'm sure getting taught online doesn't make it any easier, but gladI can help! Incase you haven't already, do check out the full playlist at engineer4free.com/statics and share with any other students in your class that might also find it helpful =)
Amazing and spot-on explanation. I always forget how to find moment with a beam including distributed load, and always relied on symmetry to set reactions equal to each other. This video is much, much appreciated!!
before watching your videos I thought I could never pass my statics course in my whole life but you made me feel confident about this course thank youu so muchhhh
thank you so much man, cleared up a lot of confusion. barely got through statics, so mechanics of materials has been pretty tough. always get stuck on figuring out how to plug in a distributed load into the moment equilibrium equation.
Are you me? I did my Statics course in 2019 and just started Mechanics of Materials last week. So much stuff I had forgotten. This video was so helpful.
Why other problems has to multiply to 1/3 and 2/3 I don’t understand. Do you have a link for a video discussing why use 1/3 or 2/3? I don’t know where did they get that some problems even has 1/2 and I don’t know where did they get that
Hi! Can you please clarify something for me? When you are calculating the moment for let's say B_y in the second problem, you multiplied it by the perpendicular distance being 6m, are you calculating the moment about point A, or are you calculating the moment about the centroid?
Then we'd have a sort of overhanging beam. See videos 1-9 here: engineer4free.com/structural-analysis . A few of those examples involve overhanging beams
How would we know exactly where the point of action? I have a exam on this and based off different triangles, my lecturer will say the point of action is either 1/3 or 2/3 but I am not sure when I would identify the point of action to be either 2/3 or 1/3. Choosing either gives different final answers. I feel as if my question is not clear but if it is, please clarify. Thank you.
Thank you very much for guiding me in the distributed load situation. May you give some real-life situations (example) where the distribution load is in triangle shape?
Hello , i had an exam today the distributed load was a triangle but the load was up instead of down ,in this case is the solution different or it's same as down distributed load
In the triangular distributed load, when summing the forces in the y-direction it is true the Ay + By = 60kN but it is not true that Sum_Fy = Ay + By or Sum_Fy = 60kN. This doesn't end up impacting the final solution but is mathematically incorrect to write.
It's cause the centroid takes into account the length of the moment arm, not just integrating until the force is half the total force. The math is right. Check out this vid if you're still interested, cause this drove me crazy for a few days. ua-cam.com/video/GeQ3y9M10T4/v-deo.html
You need to find the x location of the centroid. Plz watch this intro to centroids: ua-cam.com/video/QK_TuE2lfSc/v-deo.html to get the general idea, and then this video which is an example of finding the x component of the centroid of a non constant distributed load: ua-cam.com/video/QExNiquTAW4/v-deo.html
BUT THE MATH IS DIFFERENT IF THE DISTRIBUTED LOAD HAS ITS MAX FORCE ON THE LEFT SIDE, WHEN YOU MAKE A CUT IT TURNS IT INTO A TRAPEZOID, THERE ARE NO EXAMPLES ON ANY WEBSITE THAT EXPLAIN HOW TO DO THAT
The easiest workaround is to just consider the leftward direction as positive, and start your coordinate axis with a value of zero on the right hand side then, just flipping the problem. The math is all the same if you can wrap your head around leftward being positive. I do that in some later videos in the mechanics of materials course to avoid complicating things that don't need to be complicated
Centroid of a triangle is always 1/3 away from the high side and 2/3 of the way from the short side. Can be found in any table of Centroid for simple shapes. More here: ua-cam.com/video/5oil5x9tVhg/v-deo.html
@@Engineer4Free I mean In The Load distribution acting on the beam for example my beam is carrying a trapezoidal load there's a formula for it to convert unto UDL so that we can solve the FEM as a UDL sorry for my bad english
Oh, you can use symmetry to speed it up, but I wouldn't recommend converting to a udl with an equivalent resultant force because the SFD and BMD would be incorrect. For trapezoid just treat it as a composite shape. See video 65 here: engineer4free.com/statics
I'm not sure I understand you question, can you clarify what you mean? The distributed loads drawn in red are given in the problem, and the goal is to calculate the reaction forces at A and B.
This is a statics problem, so it's implied that the object is in static equilibrium, ie, not accelerating in any direction. The only applied force is vertical down, and the reactions compensate with an equal and opposite force up. There is no applied force with a horizontal component, therefor the pin will not provide any horizontal reaction force.
Technically, you absolutely could calculate Ax though the sum of the forces in the x direction at equilibrium always equals zero. As there is only one force to sum and that sum equals zero, that force must be zero. (SumFx=Ax=0). If you then try to calculate the magnitude of A you'd just be adding zero squared under the root (no effect) so A = Ay. So you'd just be writing a bunch of steps to calculate the magnitude that would have zero effect (no change) in your final answer so you'd be wasting paper, ink or lead, and time.
The top case (rectangular shape, constant distributed load) is often referred to as a "uniformly distributed load" or "UDL" for short. Triangular shaped load like the bottom one can sometimes be called a "non-uniformly distributed load" or "uniformly varying load."
There's a formula you can use to find the centre of mass for any triangle, and it's 1/3(x1+x2+x3) , 1/3(y1+y2+y3) in this example above if you set A as the origin you can see that B and the Point above B all have x values of 6 so Centre of mass from A would be 1/3(0+6+6) which gives you 4m from A. You can watch a video from exam solutions if this isnt clear
Can you please explain how KN/m is a measurement of height?? normally we say something is x inches, or x meteres tall.
When you draw a distributed load on a beam like this, you are graphically representing something that's actually happening in real life. It could be something like a force exerted by a rectangular or triangular pile of snow along a beam. If you put your graph's origin at point A, then the "x" axis will be in units of meters and the "y" axis will be in units of kN/m. It's ok to have graphs with axis that aren't in the same units, actually it's quite common. Imagine a car thats travelling at a constant velocity. You can make a graph that is seconds on the x axis and meters per second on the y axis. If the velocity is constant, the m/s will be a flat line. If you looked at an interval of some amount of seconds, the area under the line would be a rectangle, and you could easily find it as A=b*h and it would be in units of meters (seconds * meters/second = meters). It turns out that the area is equal to the distance travelled in meters. Often the area under curves is meaningful and so are the units that it is in. It turns out in the example in this video, where we have a graph with the x axis in meters and the y axis in kN/meter, that the area under the graph will be in units of kN (meters * kN/meter = kN). So if we find the area of the whole graph (between A and B), then that is equal to the total force exerted by the distributed load across that span (and passes through the centroid of the shape), which we can then use in the determination of the reactions. Point of the story is that the "height" of these graphs is not in meters, so it's not the actual depth of snow or whatever is causing the force, it IS the force at any given point, and by calculating areas of shapes with different units on each axis, you often reveal meaningful stuff. Hope that helps, let e know if your still confused!
20KN/m represents the max force magnitude, of which increases at a constant rate (F=(20/6)*x). This creates a right hand triangle with x=6m, y=20KN/m. He says height only to use trigonometric terms to solve for the sum of the distributed forces, which is the area of the triangle.
The measurement is describing the diagram. Indicating that 20kN is spread out on that beam surface throughout it's 6metres of length
@@Engineer4Free oh my, this was a proper good 11/10 explanation, it all makes sense now
@@shahman1 I gochu 🙌🙌
Thank you very much! My statics class is online this semester, and our lectures are basically the teacher telling us to read her notes. It helps so much to see examples worked through!
Statics already can be tricky, and I'm sure getting taught online doesn't make it any easier, but gladI can help! Incase you haven't already, do check out the full playlist at engineer4free.com/statics and share with any other students in your class that might also find it helpful =)
Amazing and spot-on explanation. I always forget how to find moment with a beam including distributed load, and always relied on symmetry to set reactions equal to each other. This video is much, much appreciated!!
Awesome, glad I could help!!
Now i finally understand how it works. Lightbulb went off! Thank you!
Literally a life savor, having a test on friday and couldnt find anything on triangular load thank you 😭😭😭💞💞
Glad I could help, hope it went well!! 🙂
before watching your videos I thought I could never pass my statics course in my whole life but you made me feel confident about this course thank youu so muchhhh
I’m really glad I could help! Good luck!! 🙂
thank you so much man, cleared up a lot of confusion. barely got through statics, so mechanics of materials has been pretty tough. always get stuck on figuring out how to plug in a distributed load into the moment equilibrium equation.
Are you me? I did my Statics course in 2019 and just started Mechanics of Materials last week. So much stuff I had forgotten. This video was so helpful.
This video was everything I needed. Thanks for the upload!
Glad to hear it! Thanks for watching Daniel =)
This video was amazing! Completely helped me solve my problem and other portions of the problem! Thanks!
Glad it helped! There are some more similar examples in videos 58-65 here: engineer4free.com/statics =)
Awesome! I’ll honestly probably have to check those out my statics final is comin up and I have a lot left to learn 😔
Why other problems has to multiply to 1/3 and 2/3 I don’t understand. Do you have a link for a video discussing why use 1/3 or 2/3? I don’t know where did they get that some problems even has 1/2 and I don’t know where did they get that
Why is the centroid of the triangle is at 1/3 ish place of the tall side?
I really appreciate your videos , and I wish you the best of all , please continue on your work ... better than most of the university lecturers 🌸🌸🌸
Thank youu
This is nothing compared to your hard work , wee need more people like you ^-^
Thank you was struggling with triangular loads until I saw this.
Glad you found the video and it helped! More over at engineer4free.com/statics ☺️
Hi! Can you please clarify something for me? When you are calculating the moment for let's say B_y in the second problem, you multiplied it by the perpendicular distance being 6m, are you calculating the moment about point A, or are you calculating the moment about the centroid?
how would you solve it if per say the A and B were not located on each ends. So lets say A is at 2m from the left and b is at 4m from the left.
Then we'd have a sort of overhanging beam. See videos 1-9 here: engineer4free.com/structural-analysis . A few of those examples involve overhanging beams
Thanks Alot, this video helped me really
How would we know exactly where the point of action? I have a exam on this and based off different triangles, my lecturer will say the point of action is either 1/3 or 2/3 but I am not sure when I would identify the point of action to be either 2/3 or 1/3. Choosing either gives different final answers.
I feel as if my question is not clear but if it is, please clarify. Thank you.
It's always 1/3 away from the tall side, and 2/3 away from the small side
@@Engineer4Free thank you very much 💯
From Iraq, thanks a lot for this effort🤗
You’re welcome, thanks for watching!! =)
Thank you very much for guiding me in the distributed load situation.
May you give some real-life situations (example) where the distribution load is in triangle shape?
honestly a lifesaver, thanks for this!!
Glad I can help ya out there, Autonomous Wasabi 🍣
Hello , i had an exam today the distributed load was a triangle but the load was up instead of down ,in this case is the solution different or it's same as down distributed load
In the triangular distributed load, when summing the forces in the y-direction it is true the Ay + By = 60kN but it is not true that Sum_Fy = Ay + By or Sum_Fy = 60kN. This doesn't end up impacting the final solution but is mathematically incorrect to write.
Nice Explanation!
Thanks!
Would it not be (1/sqrt2) of the way over, not 2/3rds. Thus would split the load in half
It's cause the centroid takes into account the length of the moment arm, not just integrating until the force is half the total force. The math is right.
Check out this vid if you're still interested, cause this drove me crazy for a few days.
ua-cam.com/video/GeQ3y9M10T4/v-deo.html
Edit,
I realsie now that the centroid isnt placed to split the mass into two equal amounts, The spread of the mass will influence the centroid ,
glad we so have University of Youtubt
I didn't understand how did you go through the centroid of the triangle thing?
Check out videos 58-60 here engineer4free.com/statics for intro on centroids, including triangles 👌
is the centroid for a triangular always 2/3rds in
1/3 of the base from the 90 degree angle
Yes, always 1/3 from the 90 degree angle or 2/3 from the other angle. Thanks for helping out Ty =)
when the distributed force is not uniform simple shape how can we change it to concentrated force
You need to find the x location of the centroid. Plz watch this intro to centroids: ua-cam.com/video/QK_TuE2lfSc/v-deo.html to get the general idea, and then this video which is an example of finding the x component of the centroid of a non constant distributed load: ua-cam.com/video/QExNiquTAW4/v-deo.html
For triangular calculation center distance couldn't be 4m it will be 2m
BUT THE MATH IS DIFFERENT IF THE DISTRIBUTED LOAD HAS ITS MAX FORCE ON THE LEFT SIDE, WHEN YOU MAKE A CUT IT TURNS IT INTO A TRAPEZOID, THERE ARE NO EXAMPLES ON ANY WEBSITE THAT EXPLAIN HOW TO DO THAT
The easiest workaround is to just consider the leftward direction as positive, and start your coordinate axis with a value of zero on the right hand side then, just flipping the problem. The math is all the same if you can wrap your head around leftward being positive. I do that in some later videos in the mechanics of materials course to avoid complicating things that don't need to be complicated
How were you able to determine the centroid of the traingle?
Centroid of a triangle is always 1/3 away from the high side and 2/3 of the way from the short side. Can be found in any table of Centroid for simple shapes. More here: ua-cam.com/video/5oil5x9tVhg/v-deo.html
what is the name of he software you are using for your tutorial?
Hey, a list of all the hardware and software that I use in the tutorials can be found at engineer4free.com/tools
Thank you sir
Hello how to convert the triangular load into uniform load ? is that possible ?
No you don't do that. If you have a triangular load acting on a structure, you just have to deal with it as is.
@@Engineer4Free I mean In The Load distribution acting on the beam for example my beam is carrying a trapezoidal load there's a formula for it to convert unto UDL so that we can solve the FEM as a UDL sorry for my bad english
Oh, you can use symmetry to speed it up, but I wouldn't recommend converting to a udl with an equivalent resultant force because the SFD and BMD would be incorrect. For trapezoid just treat it as a composite shape. See video 65 here: engineer4free.com/statics
@@Engineer4Free I used the WLx/3 for triangular load to UDL if that's the case my whole Load analysis is wrong 😟😟
in the trapezoidal Load my problem is that i dont know the Length of the two triangles
thank you
Thanks!!!! It was very useful
Awesome!! Thanks for watching 🙂🙂
thank you!
You're welcome!! =)
Thank you so much 🌹🌹🇮🇶
You’re welcome!!
Thankyou
Can you please explain how loads on top of A and B were calculated?
I'm not sure I understand you question, can you clarify what you mean? The distributed loads drawn in red are given in the problem, and the goal is to calculate the reaction forces at A and B.
how did we know that there are no any horizontal reaction?
This is a statics problem, so it's implied that the object is in static equilibrium, ie, not accelerating in any direction. The only applied force is vertical down, and the reactions compensate with an equal and opposite force up. There is no applied force with a horizontal component, therefor the pin will not provide any horizontal reaction force.
Technically, you absolutely could calculate Ax though the sum of the forces in the x direction at equilibrium always equals zero. As there is only one force to sum and that sum equals zero, that force must be zero. (SumFx=Ax=0). If you then try to calculate the magnitude of A you'd just be adding zero squared under the root (no effect) so A = Ay. So you'd just be writing a bunch of steps to calculate the magnitude that would have zero effect (no change) in your final answer so you'd be wasting paper, ink or lead, and time.
my one precious semester wasted on this concept..
😭
thanks
ur welcome =)
thank you !!
You're welcome!!
Thank techer.
You're welcome =)
Differenciate between linear varying load and uvl
The top case (rectangular shape, constant distributed load) is often referred to as a "uniformly distributed load" or "UDL" for short. Triangular shaped load like the bottom one can sometimes be called a "non-uniformly distributed load" or "uniformly varying load."
what if the triangle is symmetrical?
There's a formula you can use to find the centre of mass for any triangle, and it's 1/3(x1+x2+x3) , 1/3(y1+y2+y3) in this example above if you set A as the origin you can see that B and the Point above B all have x values of 6 so Centre of mass from A would be 1/3(0+6+6) which gives you 4m from A. You can watch a video from exam solutions if this isnt clear
@@OIOTV-zo9qp thank you!
@@maazahmedpoke No problem Akhi
Thanks for contributing!!!! Really appreciate it :)
@@Engineer4Free Thank you too for your wonderful videos, you're really having a positive impact.
THX
UR WELCOME
👍
Engineer4Free more like CharasmaticGeniusExplainsSimpleConceptsAndSavesLives4Free. Catchyright?
😂🥰 Thanks Jacqui! I think I should change my username now
My engineering professor sends hour-long videos and still fails to make us understand a single concept. I think you should take his place, seriously.
Wish I could! But until that happens, the full playlist is here: engineer4free.com/statics =)
Mechanics of materials Hibbler e.g 1.1 please ed 8
I have some problems in the second figure .. about the distance where the 540 N force is acting .. Please help
You speed quickly
You shouldnot do that
be nice
@@dude443 lmao
CovidImages need to be invested more than half19
Thank you !!
You’re welcome, thanks for watching!! 🙌