supported incline
supported incline
(OP)
My question may seem simple, but we are having a debate at work that needs to be solved. The problem consists of a rigid beam supported by cables (one at each end)from another horizontal beam. The supported beam is hung in such a manner that it is at a 45deg. angle (one cable is shorter than the other). The question is: Are both cables "seeing" an equal load, or is the cable supporting the lower end of the beam (the longer cable) "seeing" a greater load? I would appreciate any advice given. Thanks.





RE: supported incline
Cyril Guichard
Mechanical Engineer
RE: supported incline
If they are inclined, and you know the angles, you can create a tringle of forces, whith the vertical line being the weight of the suspended beam.
RE: supported incline
As described, this devolves into a simple Statics problem. The cables each exert a vector force along their line.
The horizontal forces in the cables must be equal and opposite, and the vertical forces must combine to equal the weight of the beam.
This is a quick hand calculation.
The quick answer--the cable with the greatest angle from vertical will have the higher load (for the case of only two cables).
Brad
RE: supported incline
If the load was measured at the top of the slings, I think there would be a slightly higher load for the longer sling.
If the load was measured at the bottom of the sling (sling/beam interface) the load should be the same.
My $0.02 worth.
Jesus is the Way, the Life and the Truth.
Jay
RE: supported incline
Since the beam has depth it's centroid doesn't lie on the top chord, it is located at the middle of the web. When you rotate the beam to 45° the centroid is moved to a location closer to the upper end. Make a scale drawing and watch what happens to the centroid: it isn't located at the mid-point between the two cables.
As an extreme case consider a square plate. When hung from the upper corner it's centroid will be located directly below the upper suspension point. The lower cable does not carry any of the load.
RE: supported incline
Case 1
Beam horizontal and strings are perfectly vertical.
S1 = S2 = W/2 where S1 and S2 are tensions in the two cables and W is weight of the beam.
Case 2
Beam inclined to horizontal by angle 'O' and cables are perfectly vertical.
Balancing moments
S1cosO x L = WcosO x L/2 and S2cosO x L = WcosO x L/2
So again S1 = S2 = W/2
(here I am resolving forces perpendicular to the beam)
Case 3
Beam horizontal and cables are at angle O1 and O2 respectively with vertical. Again balancing moments
S1cosO1 x L = W x L/2 and S2cosO2 = W x L/2
if O1<O2 then CosO1>CosO2 and S1<S2 and viceversa.
Ain't that so simple?
Regards,
RE: supported incline
My mistake, I misunderstood the problem. I thought the cables were at a 45° angle, not the beam.
Cyril Guichard
Mechanical Engineer
RE: supported incline
RE: supported incline
RE: supported incline
It's easy when u draw it on CAD.
RE: supported incline
Guess some of us should re-read their notes from school
3 parrallel forces here, so 1 equals the 2 others which are half the first in value. Isn't it correct?
Cyril Guichard
Mechanical Engineer
RE: supported incline
"the longer cable has the highest load as it is always nearest to the vertical" requires an assumption on your part that cannot be made, given the information provided. I think I understand the assumption that you implicitly made, but it was not stated by the poster.
Eight posts later, and CMcF's original post still answered the problem as simply posed, within very reasonable engineering error.
The issue of centroids mentioned above is only relevant from a practical sense if the "beam" is not a beam; for a beam as suggested by the poster, the error due to not accounting for centroid is incredibly small.
Brad
RE: supported incline
RE: supported incline
Yes, for most practical purposes.
I guess the French aren't so bad after all
Best regards,
Brad
RE: supported incline
lol At last, we try not to be
Cyril Guichard
Mechanical Engineer
RE: supported incline
The center of gravity may not lie in the same horizontal plane as the cable attachment points. However dvd appears to assume that the beam is symetrical, which is not stated.
The center of gravity DOES NOT MOVE wrt the ends of the beam however as this is quite impossible if the shape of the beam does not change.
However, if the CG is not in the same horizontal plane of the cable attachments then the force due to gravity, assumed to act through the CG will move wrt the cable attachment points. The direction depends on if the attachment is above or below the CG. As a though experiement to confirm this, imagine a symetrical I beam supported by a SINGLE cable at one end, but not attached at the neutral axis of the beam. The beam will not hang perfectly vertical.
So, the tension in each cable may not be the same in the original question, but insufficient information is provided to fully answer the question.