Distributed weight of inclined beam.

  • #1
Mohmmad Maaitah
88
19
Homework Statement
The beam weighs 280 lb/ft. Determine the internal
normal force, shear force, and moment at point C.
Relevant Equations
Equations of equilibrium
I'm having problem in drawing the distributed load (weight per foot) for the inclined beam as it shows.
should it be rectangular? if so shouldn't the distributed load be vertical down as the resultant weight at the centroid (W)
1702636296255.png
please help me understand how to draw the free body diagram for this one.
1702636361361.png
 
Physics news on Phys.org
  • #2
If I draw it as Triangle it must be wrong (reason I want to is because I could make the Distributed load vertical then).
 
  • #3
The distributed weight is $$\frac{dm}{ds}dsg$$ where ds is differential length along the bar; its direction is vertically downward.
 
  • #4
Chestermiller said:
The distributed weight is $$\frac{dm}{ds}dsg$$ where ds is differential length along the bar; its direction is vertically downward.
I know and I solved the problem, I just want to see a free body diagram including the distributed load acting on the bar like this
(just how to draw the distributed load and I would appreciate a picture)
1702644491006.png
 

Attachments

  • 1702644424783.png
    1702644424783.png
    4.6 KB · Views: 36
  • 1702644452531.png
    1702644452531.png
    6.3 KB · Views: 32
  • #5
Mohmmad Maaitah said:
I know and I solved the problem, I just want to see a free body diagram including the distributed load acting on the bar like this
(just how to draw the distributed load and I would appreciate a picture)
The distributed load should be in the direction of the resultant weight force. Taking a slice of the beam doesn't change its direction, just the magnitude of the resultant.
 
  • #6
erobz said:
The distributed load should be in the direction of the resultant weight force. Taking a slice of the beam doesn't change its direction, just the magnitude of the resultant.
Right, but how do I draw the distributed load going vertically down?
like this:
1702646062261.png
 
  • #7
1702646347846.png


The weight per unit length of the beam is constant ( i.e. a uniformly distributed load), there is no "triangle" shaped distribution here. Those are representative of a changing distributed load over the span.

If you imagine rotating the beam to horizontal the distributed load from its own weight just tends to:
1702647800911.png
 
Last edited:
  • Like
  • Love
Likes Mohmmad Maaitah and Lnewqban
  • #8
Mohmmad Maaitah said:
°°Homework Statement: The beam weighs 280 lb/ft. Determine the internal
normal force, shear force, and moment at point C.
Relevant Equations: Equations of equilibrium

I'm having problem in drawing the distributed load (weight per foot) for the inclined beam as it shows.
should it be rectangular? if so shouldn't the distributed load be vertical down as the resultant weight at the centroid (W)
help me understand how to draw the free body diagram for this one.
Just imagine the extreme case of 90° Inclination.
The beam becomes a column then.
The total weight will still be the same, only that “less equally distributed”.

Please, see:
https://learnaboutstructures.com/Determinate-Frame-Analysis
 
  • Informative
Likes Mohmmad Maaitah
  • #9
I think the common misconception with distributed loads is that the arrows represent forces, when they actually represent the force gradient ## \frac{dF}{dx} ##. This is what I think your mix up is, given that you think the shape of the distribution should be triangular. The force increases linearly as we progress along the beam, but the gradient (the distributed load) is uniform.
 
  • Like
Likes Lnewqban and Mohmmad Maaitah
  • #10
Thank you all for the reply it is a big help!!
 
  • Like
Likes berkeman, Lnewqban and erobz
  • #11
For the case of the imaginary column, think of how normal force, shear force, and moment at point C will change.

Tip: No bending, only compression load that is inversely proportional to the height of the cross-section.
 
  • Informative
Likes Mohmmad Maaitah
  • #12
erobz said:
I think the common misconception with distributed loads is that the arrows represent forces, when they actually represent the force gradient ## \frac{dF}{dx} ##. This is what I think your mix up is, given that you think the shape of the distribution should be triangular. The force increases linearly as we progress along the beam, but the gradient (the distributed load) is uniform.
I find it a bit odd to describe it as a force gradient. I suppose it works if you define F to be the total force over the extent 0 to x.

I would rather describe it as force per unit length.
 
  • Like
Likes erobz and Mohmmad Maaitah

Related to Distributed weight of inclined beam.

What is the distributed weight of an inclined beam?

The distributed weight of an inclined beam refers to the weight per unit length that is uniformly spread along the length of the beam. This distribution takes into account the angle of inclination, which affects the horizontal and vertical components of the load.

How do you calculate the distributed load on an inclined beam?

To calculate the distributed load on an inclined beam, you need to determine the weight per unit length (w) and then resolve it into its horizontal (w_x) and vertical (w_y) components. The vertical component is typically w * cos(θ) and the horizontal component is w * sin(θ), where θ is the angle of inclination.

What factors influence the distributed weight on an inclined beam?

The factors that influence the distributed weight on an inclined beam include the beam's material density, cross-sectional area, length, and the angle of inclination. External factors such as additional loads or supports can also impact the distributed weight.

How does the angle of inclination affect the load distribution on an inclined beam?

The angle of inclination affects the load distribution by altering the horizontal and vertical components of the distributed weight. As the angle increases, the horizontal component of the load increases while the vertical component decreases, impacting the beam's support reactions and internal stresses.

Why is understanding the distributed weight of an inclined beam important in structural engineering?

Understanding the distributed weight of an inclined beam is crucial in structural engineering to ensure the stability and safety of structures. Accurate calculations of load distribution help in designing appropriate supports, selecting suitable materials, and preventing structural failures due to improper load handling.

Similar threads

Replies
41
Views
485
  • Introductory Physics Homework Help
Replies
4
Views
198
  • Introductory Physics Homework Help
Replies
5
Views
520
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
17K
  • Introductory Physics Homework Help
Replies
5
Views
3K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Back
Top