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
For an Example of a Beam with a Concentrated Moment, Watch: ua-cam.com/video/g1bkB9QFLQ4/v-deo.html For an Example of a Beam with a Mixed Uniform and Point Load, Watch: ua-cam.com/video/D06s-1fYmOw/v-deo.html For more information about the concepts of BMD, SFD and NFD, Watch: ua-cam.com/video/Ncj1tIOq6nM/v-deo.html For an Example of Simple Beam with a uniform Load: ua-cam.com/video/6fvUmS867O0/v-deo.html
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 😭😭😭😭
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)
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
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 ua-cam.com/video/gYw4BCNTHz8/v-deo.html and give me your feedback on the sound quality.
thank you for the thorough explanation. showing the different cases of force direction and distribution was a very helpful refresher.
Glad it was helpful!
how do you work out the moments at point A and B on the inclined beam?
How did you get "YA" Value as 16.5 Kn anyone please help with the reaction
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
thank you so much, professor!
I am happy it was helpful
For an Example of a Beam with a Concentrated Moment, Watch:
ua-cam.com/video/g1bkB9QFLQ4/v-deo.html
For an Example of a Beam with a Mixed Uniform
and Point Load, Watch:
ua-cam.com/video/D06s-1fYmOw/v-deo.html
For more information about the concepts of BMD,
SFD and NFD, Watch:
ua-cam.com/video/Ncj1tIOq6nM/v-deo.html
For an Example of Simple Beam with a uniform
Load:
ua-cam.com/video/6fvUmS867O0/v-deo.html
Thanks !
I am glad the video was useful
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 😭😭😭😭
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)
@@drnafie-structuralengineer4620 great thanks so much dear.
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
Please improve audio
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 ua-cam.com/video/gYw4BCNTHz8/v-deo.html
and give me your feedback on the sound quality.