Distributed weight of inclined beam.

[Mohmmad Maaitah]

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)

please help me understand how to draw the free body diagram for this one.

 

[Mohmmad Maaitah]

If I draw it as Triangle it must be wrong (reason I want to is because I could make the Distributed load vertical then).

 

[Chestermiller]

The distributed weight is $$\frac{dm}{ds}dsg$$ where ds is differential length along the bar; its direction is vertically downward.

 

[Mohmmad Maaitah]

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)

 

[erobz]

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.

 

[Mohmmad Maaitah]

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:

 

[erobz]

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:

 

[Lnewqban]

°°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

 

[erobz]

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.

 

[Mohmmad Maaitah]

Thank you all for the reply it is a big help!!

 

[Lnewqban]

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.

 

[haruspex]

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.