TRIANGULAR LOAD Shear and Moment Diagrams EXAMPLE PROBLEM

In this video I go through an example problem of drawing shear and moment diagrams of a beam that has a triangular load on it.
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Quick Tips (I explain these in more depth in the video)
- The area under the loading curve = the change in shear (positive area is a positive change and negative area is a negative change)
- The area under the shear function/diagram = the change in moment (positive area is a positive change and negative area is a negative change)
- Positive slope on the loading function = concave up shear function
- Negative slope on the loading function = concave down shear function
- Positive slope on the shear function = concave up moment function
- Negative slope on the shear function = concave down moment function
- Distributed load intensity = the slope of the shear function
- Amount of shear = slope of the moment diagram
- Where the shear diagram crosses the axis is a max or min on the moment diagram
Drawing shear and moment diagrams process:
1) Establish your coordinate system with the positive X direction being along the length of the beam, starting on the left.
2) Solve for the support reactions on the beam using moment equilibrium equations and force equilibrium equations in the Y direction.
3) Draw the free body diagram of the beam with the solved support reactions written as forces as well as the loading scenario.
4) Draw the shear X axis and the moment X axis vertically aligned below the beam to line up the changes of the loading with the shear and moment diagrams. Then label these diagrams with V for shear and M for moment and label the units for each.
5) Plot the shear diagram starting from the left end. As you encounter forces in the loading scenario, draw the shear diagram as outlined in Note A. Label the distance X along the diagram where the shear function crosses the axis.
6) Plot the moment diagram starting from the left end. As you encounter changes in the shear diagram, draw the moment diagram as outlined in Note B. Label magnitude of the moment at the maximum and minimum points along the function as well as what point along the X axis they occur.
Note A:
- A point load (a force at a single point) causes the shear diagram to jump straight up or down; up if the load is pushing up and down if the force is pushing down. This includes reaction forces. Point moments will not directly effect the shear diagram.
- Distributed loads cause the shear diagram to decrease or increase in a way that can be modeled by a function. You can model the distributed load as a function and then integrate that function to get the shear function (see the Quick Tips at the top for more specifics on this).
Note B:
-The moment reactions at the ends of the beams are zero if the beam is supported by pin, roller or free connections. This means that the moment diagram will start and end at zero. However, if there is a fixed support then there will be a moment reaction which will cause the diagram to not start or end at zero, depending on which side the fixed support is on (usually it is on the left end). If there is a support reaction then it will cause a vertical jump in your moment diagram just as a point moment would. Remember that the support reaction is in the opposite direction of the internal moment. If the support reaction is counter clockwise (positive) then the internal forces which you are drawing are negative and thus will have a negative drop.
- A point moment will cause a vertical jump. The direction of the jump being as explained in the previous point.
- The integral of the shear function is the moment function. This only applies to sections of the function that are continuous, in other words, there are no vertical jumps up or down
(see the Quick Tips at the top for more specifics on how to draw the diagram).

[@zEre326]

SO CLEAR ! I don't understand why your videos don't show up first, thank you great job

[@studentengineering]

Thanks!

[@nehazaman7880]

This video made me understand triangular loads perfectly, thank you so much! : )

[@studentengineering]

I’m glad you found it helpful!

[@studentengineering]

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.

[@mrcoffee315]

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.

[@studentengineering]

I’m so glad! Thanks for sharing!

[@rene-rj1jv]

this is the video that made this concept click for me, thanks so much :)

[@studentengineering]

That’s awesome! You’re so welcome

[@lennitajustine1439]

the most attractive person in engineering right now. like literally :) ik that im stucked doing statics but this cheers me up

[@studentengineering]

Haha thanks :)

[@JG-zy9mv]

thank you so so much! this is such a big help to me in my engineering course. -from Philippines

[@studentengineering]

You’re welcome!

[@josuearellanes3974]

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!

[@studentengineering]

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!

[@MiHwAr2149]

thank you man, that was clearly explained

[@studentengineering]

You’re welcome!

[@herseyodtuyeyataygecisyapm7364]

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.

[@studentengineering]

You’re welcome!

[@thumbpolkm3511]

very helpful thank you, trying to study for my midterm rn

[@studentengineering]

You’re welcome! Good luck with your midterm!

[@RepublicaPH]

Me watching your toturial for my structural theory subject.. From 🇵🇭

[@studentengineering]

Awesome! Good luck!

[@martintobar6261]

Thank you. I love you

[@studentengineering]

You’re welcome!

[@thomaschan3492]

Support from Hong Kong🖐🏻☝🏻

[@studentengineering]

Thanks!

[@WersiyCollins]

So clear

[@studentengineering]

I’m glad!

[@odyson2395]

The curve must be outward on the shear diagram right? Since the triangular load is sloping downwards?

[@studentengineering]

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: kzbin.info/www/bejne/mJubm6mQa6hropY

[@michelletavakoli5536]

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

[@studentengineering]

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: kzbin.info/www/bejne/mJubm6mQa6hropY Hope this helps!

[@michelletavakoli5536]

@@studentengineering Thank you so much!

[@mikexbox1]

God bless you

[@studentengineering]

Thanks!

[@henryhenry8713]

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

[@studentengineering]

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: kzbin.info/www/bejne/hnmUZI14nJqAl80si=te31Lzf3g0hoFAhn kzbin.info/www/bejne/nqCVaqhpbbSWlZYsi=oe7UuM-wO8mHRJOs Does that help?

[@raydenebalbuena6475]

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.

[@studentengineering]

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.

[@raydenebalbuena6475]

@@studentengineering Got it. Thank you so much!

[@LuyandaSimphiweBlessing]

HOW DID YOU GET 3.8m ON THE SHEAR DIAGRAM?

[@studentengineering]

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?

[@brendantapadera3464]

Where does 1.76 came from

[@studentengineering]

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 kzbin.info/www/bejne/n56vhmaFbJ5khrM Let me know if you have any more questions!

[@buknoy2705]

where 3.8m came from on BMD

[@studentengineering]

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.

[@denversagario7568]

the (x) transition at 7:00

[@studentengineering]

What about it?

[@engr.aianjay]

Hayssssss my ps is making me crazy. 😆😆

[@studentengineering]

I don’t understand what you’re trying to say

[@Runa36]

It was helpful but next time hide your face because I almost lost concetration . Too cute😍😍😄😍

[@studentengineering]

Haha I’m glad it was helpful

[@AbdalAbuassamen]

w man

[@studentengineering]

I don’t understand

[@arabellealbat2464]

Ure cute

[@studentengineering]

Haha thanks! My wife thinks so too :)