TRIANGULAR LOAD Shear and Moment Diagram EXAMPLE PROBLEM // Calculus Explained

[@Gerry363c]

Confusing in finding the sum of the moments..more explanation needed. Thanks by the way

[@studentengineering]

Watch this video and see if it helps: TRIANGULAR LOAD Shear and Moment Diagrams EXAMPLE PROBLEM kzbin.info/www/bejne/n56vhmaFbJ5khrM

[@Gerry363c]

@ thanks

[@Johndavid1451]

Thanks for the video, man! Flipping the triangular load upside down really helped me in understanding the moment and writing the shear equation.

[@studentengineering]

Yea when I finally figured out that it worked like that it made a huge difference in how I understood triangular loads. I’m glad you found it helpful!

[@SignMusic13]

Thanks for the video!

[@studentengineering]

You’re welcome!

[@frozenporcupine]

thanks man. great example used

[@studentengineering]

Thanks!

[@avachar11]

Nice. Thank you

[@studentengineering]

You’re welcome!

[@kenmalacinski8442]

I am confused on why 3 is being plugged in instead of 6. I am getting different boundary conditions but ending up with the same max moment of -2250. I see you are defining the part where the triangular distributed load starts as 0 on the x axis? Why isn't this defined as 3 on the x axis?

[@studentengineering]

I think we just summed moments about a different point. Does that make sense?

[@40Ksigmar]

Why did you use By instead of Ay for the Sum of moments A please?

[@studentengineering]

Great question! When we sum moments about a point we don’t need to include forces that pass through that point. This is because those forces won’t cause rotation about that point. When you sum moments about a point to find reaction forces you want to pick a point that is in the line of action of as many unknown forces as possible so that you don’t have to consider them in that particular equilibrium equation. Besides that reason, it doesn’t really matter which point you choose to sum moments about.

[@youssefelbanna711]

Hello, I love your videos they make this topic so clear! However I have a simple question. When you write the equation w(x) for the triangular distributed load, when is it -300, not 300? I am talking about the intercept or whatever you are going to call it. How do you decide on the sign?

[@studentengineering]

Thanks! I’m glad! When the force is going down I write it as negative and when it is going up I write it as positive.

[@youssefelbanna711]

@@studentengineering Thank you for the response. Usually all distributed loads go downwards, does that mean that I will always have the w(x) intercept as negative?

[@studentengineering]

Yes

[@jamesandrews7426]

Why multiply 450 with 1? I understood the centre of gravity for the right angled is 1/3 its length, but since you took moment at point A, shouldnt it be 450x1/3? Because its 1/3 away from point A?

[@studentengineering]

Great question! You’re right that it should be 1/3 along the triangle’s length. Because the triangle is 3 ft long that comes out to be 1

[@zana.m.rashid6747]

I liked 👍

[@studentengineering]

I’m glad!

[@zana.m.rashid6747]

🔥🔥

[@studentengineering]

Thanks!

[@jacksonhilsabeck9109]

for the moment equation why do you plug in 3 and not 6?

[@studentengineering]

Great question! That would be the case if the moment equation described the bending moment across the whole beam but it doesn’t. The equation describes the moment starting at 3 feet along the beam and going to 6 feet along the beam, moving from left to right. This is where the triangular distributed load occurs and so plugging 3 into the moment equation gives us the moment that occurs 3 feet along the section of the triangular distributed load but 6 feet along the beam. Does that make sense?

[@pepito8685]

😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂

[@studentengineering]

What’s so funny?

[@kevinplug6059]

You ​@@studentengineering made it easy🎉🎉