Tbh this method was the only method that really helped me understand problems with negative slope. I tried doing the section method calculating the moment and vertical force with respect to x and couldn’t find a solution. With this method it was so straight forward. Thanks for the help.
The solutions to this example problem or the solutions to this book? If you want the solution to this example it is in the video if it is to this book, I’m sorry, I don’t have them.
Thank you for this! My professor told me something really weird about this type of situation with the triangle. Usually the triangles start at zero on the very left, then go up. I think when the professor saw it this way, he got himself confused and solved it a completely different way. Now, I see that he was wrong. I plan on talking to him soon. Again, thank you!
you are gorgeus sir. my old man professor know anything or he is very selfish and greedy. whatever, It helps when information is shared. No one is trying to reinvent gravity to understand quantum physics. But university professors sometimes have the opposite idea. Thank you for your very clear and unambiguous explanation.
Thanks for posting such a helpful information. Q: why did you come up with positive slope in load equation W(x)=1/2x-3? As you mentioned, the line in the problem is going down and slope should be negative. please correct me if I'm wrong. Thanks
Great question! The way it is drawn it is a negative slope but if you draw negative forces below the line then that hypotenuse of the load triangle becomes a positive slope. Doing it that way makes all the equations work out better. I explain that more in this video: th-cam.com/video/bj8jvZ6C8l4/w-d-xo.html Hope this helps!
That would be correct if the load was sloped downwards but I explain in the video why it is actually sloped upwards. I also explain it even better in this video: th-cam.com/video/bj8jvZ6C8l4/w-d-xo.html
Good question! In that step I am summing moments about a point. A moment is a force times by a distance. The 6 is the force and the 7.5 is the distance. Same with the 9 times by 2. Here’s a couple of videos that explain moments and equilibrium equations in more detail: th-cam.com/video/PH13ZBg5Mak/w-d-xo.htmlsi=te31Lzf3g0hoFAhn th-cam.com/video/ho29u38Oc_4/w-d-xo.htmlsi=oe7UuM-wO8mHRJOs Does that help?
Great question. I guess I couldn’t tell you why you didn’t get 3.8 but 0 is also an intercept for that equation because we know the moment is 0 when x=0 so you didn’t necessarily get the wrong answer. Use that moment equation that I came up with in the video and then set it equal to zero and solve for x. Since it is a third order equation you will get three values for the intercepts. One will be 0, one will be 3.8 and another one will be something else that would be outside the range of where we are using the equation (less than 0 or greater than 6). I would recommend using Wolfram Alpha or Symbolab solve for those intercepts for you.
Great question! You need to create an equation for the force on the beam and then you integrate that equation to get the shear equation. Once you have the shear equation you set it equal to zero and solve for the variable (x in this case) to get where along the beam the shear is zero. Does that make sense?
Great question! It comes from setting the shear equation equal to zero and solving for x. I do that in the video at 8:10 th-cam.com/video/imLU3O791PQ/w-d-xo.html Let me know if you have any more questions!
We set the moment function equal to zero and solve for x to find out where the bending moment diagram crosses the axis and when we do that we find that the bending moment is 0 at 3.8 meters along the beam.
Tbh this method was the only method that really helped me understand problems with negative slope. I tried doing the section method calculating the moment and vertical force with respect to x and couldn’t find a solution. With this method it was so straight forward. Thanks for the help.
I’m so glad! Thanks for sharing!
SO CLEAR ! I don't understand why your videos don't show up first, thank you great job
Thanks!
This video made me understand triangular loads perfectly, thank you so much! : )
I’m glad you found it helpful!
The solutions to this example problem or the solutions to this book? If you want the solution to this example it is in the video if it is to this book, I’m sorry, I don’t have them.
the most attractive person in engineering right now. like literally :) ik that im stucked doing statics but this cheers me up
Haha thanks :)
Thank you for this! My professor told me something really weird about this type of situation with the triangle. Usually the triangles start at zero on the very left, then go up. I think when the professor saw it this way, he got himself confused and solved it a completely different way. Now, I see that he was wrong. I plan on talking to him soon.
Again, thank you!
You’re welcome! It took me a while to figure out that this was an easier way of thinking about it but once I did it made so much more sense!
you are gorgeus sir. my old man professor know anything or he is very selfish and greedy. whatever, It helps when information is shared. No one is trying to reinvent gravity to understand quantum physics. But university professors sometimes have the opposite idea. Thank you for your very clear and unambiguous explanation.
You’re welcome!
this is the video that made this concept click for me, thanks so much :)
That’s awesome! You’re so welcome
Me watching your toturial for my structural theory subject.. From 🇵🇭
Awesome! Good luck!
thank you so so much! this is such a big help to me in my engineering course. -from Philippines
You’re welcome!
very helpful thank you, trying to study for my midterm rn
You’re welcome! Good luck with your midterm!
thank you man, that was clearly explained
You’re welcome!
So clear
I’m glad!
Support from Hong Kong🖐🏻☝🏻
Thanks!
Thanks for posting such a helpful information. Q: why did you come up with positive slope in load equation W(x)=1/2x-3? As you mentioned, the line in the problem is going down and slope should be negative. please correct me if I'm wrong.
Thanks
Great question! The way it is drawn it is a negative slope but if you draw negative forces below the line then that hypotenuse of the load triangle becomes a positive slope. Doing it that way makes all the equations work out better. I explain that more in this video: th-cam.com/video/bj8jvZ6C8l4/w-d-xo.html
Hope this helps!
@@studentengineering Thank you so much!
The curve must be outward on the shear diagram right? Since the triangular load is sloping downwards?
That would be correct if the load was sloped downwards but I explain in the video why it is actually sloped upwards. I also explain it even better in this video:
th-cam.com/video/bj8jvZ6C8l4/w-d-xo.html
I don't understand from first step due to -6.75 and-9.2 from were you have get them from
And :Ay+105-9.6 were you also got them from, please help me
Good question! In that step I am summing moments about a point. A moment is a force times by a distance. The 6 is the force and the 7.5 is the distance. Same with the 9 times by 2. Here’s a couple of videos that explain moments and equilibrium equations in more detail:
th-cam.com/video/PH13ZBg5Mak/w-d-xo.htmlsi=te31Lzf3g0hoFAhn
th-cam.com/video/ho29u38Oc_4/w-d-xo.htmlsi=oe7UuM-wO8mHRJOs
Does that help?
It was helpful but next time hide your face because I almost lost concetration . Too cute😍😍😄😍
Haha I’m glad it was helpful
Thank you. I love you
You’re welcome!
the (x) transition at 7:00
What about it?
I tried setting the moment equal to 0 to get the distance where it crosses the axis but I got 0. How did you get 3.8? I think I made a mistake.
Great question. I guess I couldn’t tell you why you didn’t get 3.8 but 0 is also an intercept for that equation because we know the moment is 0 when x=0 so you didn’t necessarily get the wrong answer. Use that moment equation that I came up with in the video and then set it equal to zero and solve for x. Since it is a third order equation you will get three values for the intercepts. One will be 0, one will be 3.8 and another one will be something else that would be outside the range of where we are using the equation (less than 0 or greater than 6). I would recommend using Wolfram Alpha or Symbolab solve for those intercepts for you.
@@studentengineering Got it. Thank you so much!
God bless you
Thanks!
HOW DID YOU GET 3.8m ON THE SHEAR DIAGRAM?
Great question! You need to create an equation for the force on the beam and then you integrate that equation to get the shear equation. Once you have the shear equation you set it equal to zero and solve for the variable (x in this case) to get where along the beam the shear is zero. Does that make sense?
Where does 1.76 came from
Great question! It comes from setting the shear equation equal to zero and solving for x. I do that in the video at 8:10
th-cam.com/video/imLU3O791PQ/w-d-xo.html
Let me know if you have any more questions!
where 3.8m came from on BMD
We set the moment function equal to zero and solve for x to find out where the bending moment diagram crosses the axis and when we do that we find that the bending moment is 0 at 3.8 meters along the beam.
Hayssssss my ps is making me crazy. 😆😆
I don’t understand what you’re trying to say
w man
I don’t understand
Ure cute
Haha thanks! My wife thinks so too :)