Internal Force Diagram - Inclined Beam Example - Normal, Shear and Bending Example

This video shows how to draw bending, shear and moment diagrams for an inclined beam.
This is part of a civil engineering course about structural analysis of determinate structures.
A systematic and simple method for force resolution and handling of uniformly distributed loads on inclined beams is explained.
An Example is given for practice and to further reinforce the ideas presented.
For more information about the concepts and sign convention of Internal force diagram refer to the following lecture:
• Internal Force Diagram...
For an Example of a Beam with a Concentrated
Moment, Watch:
• Internal Force Diagram...
For an Example of a Beam with a Mixed Uniform
and Point Load, Watch:
• Internal Force Diagram...
For an Example of Simple Beam with a uniform
Load:
• Internal Force Diagram...

[@lukethechampion169]

thank you for the thorough explanation. showing the different cases of force direction and distribution was a very helpful refresher.

[@drnafie-structuralengineer4620]

Glad it was helpful!

[@drnafie-structuralengineer4620]

For an Example of a Beam with a Concentrated Moment, Watch:
czcams.com/video/g1bkB9QFLQ4/video.html
For an Example of a Beam with a Mixed Uniform
and Point Load, Watch:
czcams.com/video/D06s-1fYmOw/video.html
For more information about the concepts of BMD,
SFD and NFD, Watch:
czcams.com/video/Ncj1tIOq6nM/video.html
For an Example of Simple Beam with a uniform
Load:
czcams.com/video/6fvUmS867O0/video.html

[@kynan.a.v6488]

how do you work out the moments at point A and B on the inclined beam?

[@cmilemile]

thank you so much, professor!

[@drnafie-structuralengineer4620]

I am happy it was helpful

[@tehsoonguan9343]

Thanks !

[@drnafie-structuralengineer4620]

I am glad the video was useful

[@pavan6794]

How did you get "YA" Value as 16.5 Kn anyone please help with the reaction

[@drnafie-structuralengineer4620]

You can evaluate this reaction in 2 steps.
First take the moment about A, then YA and XA will not be included in the equation, the equation will be as follows:
YB x 9 - 2 x 3 x 7.5 - 3 x 6.5 x 3 + 9 x 2.5 = 0 => this will produce YB= 9
Then take sum of forces in the Y direction =0
YB + YA - 3 x 6.5 - 2 x3 = 0 => YA = 19.5 + 6 -9 = 16.5

[@yirrroy22tyrri82]

Hi dear. Still i dont understand one point in general ( not related to this lesson) i wann know how to know where to place cosine and where to place sine, this confusing me a lot,,, i cannot proceed the rest unless understanding this 😭😭😭😭

[@drnafie-structuralengineer4620]

For the special case of the inclined beam there is a very easy rule.
Suppose that the inclined beam makes an angle theta with the X-direction (horizontal) . and we have forces in the X (horizontal ) and the Y direction (vertical).
Then the normal force will be formed of two components:
The first component is X Cos (theta) ( X is the sum of all the forces in the X directions), the second component is Y Sin (theta) ( Y is the sum of all the Vertical forces) [ recall that theta is between the beam and X, so X takes the Cosine and Y takes the Sine}
The Shear force will also be formed of 2 components but the cosine and sine will be reversed.
we will have X sin( theta) and Y cos (theta)

[@yirrroy22tyrri82]

@@drnafie-structuralengineer4620 great thanks so much dear.

[@haneybebsnedic8885]

Hello, I hope I can help. Based on my understanding, if your angle (theta) is along the X-axis and you're looking for the vertical force, you will use sine (since Sin=opp/hyp); and for horizontal force, you'll use cosine (since it is Cos=adj/hyp).
And if the angle is along the Y-axis, you will use:
Sine for Horizontal force
Cosine for Vertical force

[@abemi869]

Please improve audio

[@drnafie-structuralengineer4620]

I have changed to a new microphone for the newest video. I hope this will improve the sound quality.
You can check the newest video czcams.com/video/gYw4BCNTHz8/video.html
and give me your feedback on the sound quality.