supported incline 1

[fkijoe]

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.

 

[FrenchCAD]

I would say the longest one get an highest load

Cyril Guichard
Mechanical Engineer

 

[CMcF]

If the cables are vertical, then they see an equal load. If they are inclined, then they do not.
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.

 

[brad]

I disagree with FrenchCAD; there is not enough information given to make a judgment. CMcF is correct--it is dependent on the angle of the cables.

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

 

[jbknudsen]

It is almost a loaded question. I suppose if everthing was perfectly balanced, I would tend to believe that the slings will see equal loads. That is if the slings are vertical and the distance from one sling to the other is the same at the top as it is at the bottom (slings to be parallel or vertical). With that said, does it matter if the beam is rotated 10° or 45°. As long as the slings are located in a symmetric position to the beam (for example: both are fastened in 1" from the end of the beam). On another note, I would think that the longer sling would see a slightly higher load because it is heaiver than the shorter sling.

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

 

[dvd]

The shorter cable has the higher load.

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.

 

[quark]

Knudsen is absolutely right (so are CMCF and Bradh).

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,

 

[FrenchCAD]

&quot;I disagree with FrenchCAD; there is not enough information given to make a judgment. CMcF is correct--it is dependent on the angle of the cables.&quot;

My mistake, I misunderstood the problem. I thought the cables were at a 45° angle, not the beam.

Cyril Guichard
Mechanical Engineer

 

[CMcF]

DVD you make a good point, it highlights the normal abuse of beam theory, when people try to apply it to short stubby beams.

 

[Tryan]

Is it hypothetical? If so take the difference of the weight of the cables out of the equation and hanging parallel to each other, and the answer is equal!This was not arrived at by calculations, but by a simple practical experiment. Set up two scales ,two identical pieces of square tube a metre long and two pieces of string. Put the ends of the top tube on each scale, suspending the other piece of tube at 45deg with the string and the scales read the same!!

 

[Speedy]

It's simple trigonometry. The centroid of the beam comes to rest under the piovotal point. The angle of the beam at rest is 45deg, so resolve forces along the cable. The longer cable has the highest load as it is always nearest to the vertical. The ratio of both angles depends on the cable lengths. I assumed the cable to be weightless also.

It's easy when u draw it on CAD.

 

[FrenchCAD]

Isn't it scary to see so many different answers? ;-)

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

 

[brad]

Speedy,
&quot;the longer cable has the highest load as it is always nearest to the vertical&quot; 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 &quot;beam&quot; is not a beam; for a beam as suggested by the poster, the error due to not accounting for centroid is incredibly small.

Brad

 

[pablo02]

Looks like a statics / basic physics review is needed by many -- I must agree most of you, that if the cables are vertical ( & assuming the cable weight is negligible) the load is distributed equally between the cables... it's obvious to us electricals..

 

[brad]

FrenchCAD--
Yes, for most practical purposes.
I guess the French aren't so bad after all .

Best regards,
Brad

 

[FrenchCAD]

I guess the French aren't so bad after all

lol At last, we try not to be ;-)

Cyril Guichard
Mechanical Engineer

 

[MintJulep]

dvd is almost correct.

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.