Slanted / Raker Beams

[BLUEPUPIL]

Imagine a roof’s sloped beam, similar to a rafter. The roof has a 25 degree pitch angle. The live load is uniform and with downward vertical direction, causing bending in the beam. Is it correct (possible), to reduce the loads [DL and LL], multiplying them by 0.90(cosine 25 degrees) in order to calculate max. bending moment and deflection? Based on such approach, we can assume a compression force acting along the longitudinal axis of the beam =LOAD x 0.42(sine 25 degrees).

 

[jayrod12]

Correct but then you have to use the sloped length of rafter to determine max moments instead of the projected length. You'll end up with the same number either way.

Correct on the axial load.

 

[BAretired]

It is not a case of reducing the load; it's a case of resolving the load into its components perpendicular and parallel to the beam axis.

If the end reactions are vertical, the axial force varies along the length of the beam; compression at the low end and tension at the high end.

BA

 

[BLUEPUPIL]

Thank you, jayrod12
The length of horizontal projection of a slanted beam is always shorter than the actual length along the incline. Hence, multiplying the latter with cosine, just equals loads.
Thank you, BAretired
If the upper connection of the slanted beam is pinned, than there is tension at that end. If we suppose that the higher connection is roller, than there is tiny amount of compression at the top, which increases its value down along the inclined beam [uniformly distributed load] and reaches its max. at the low end of the beam.

 

[mmarlow]

I doubt the compression (or tension) in combination with the flexure will make much of a difference.. However, I would assume its in compression, not tension. Buckling due to combined flexure and compression is more concerning in my opinion. So, the top would be a roller, and the bottom would be a pin.

-MMARLOW EIT

 

[BAretired]

If the upper connection of the slanted beam is pinned, than there is tension at that end. If we suppose that the higher connection is roller, than there is tiny amount of compression at the top, which increases its value down along the inclined beam [uniformly distributed load] and reaches its max. at the low end of the beam.

Not so. If the upper connection is a horizontal roller and the lower end is a pin, both reactions are vertical which causes equal and opposite axial forces, compression at the low end and tension at the high end.

If the upper connection is a slanted roller (rolling parallel to the axis of the beam) and the lower end is pinned, there is zero axial force at the high end and compression at the low end.

If the upper connection is a vertical roller and the lower connection is pinned, it's like a ladder leaning against a wall; the reaction at the roller is horizontal while the reaction at the pin has a horizontal component equal and opposite to the wall reaction and a vertical component equal to the total load on the beam.

The question cannot be answered without specifying support conditions. Once specified, it is a matter of elementary statics.

BA