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lllazar wrote:I was wondering - since the base legs take more force, would it be a good idea to make them large than the columns in the upper portion of the tower - say, 3/16 in^2?

Well, ask yourself this: WHY do the 'base legs take more force'?

....The answer is, well, they don't really take MORE force, since obviously the same amount of force is being applied at the top, it's just that they're under more strain (if you do a sketch of the force components, you'll easily see why). In answer to your question then, SIZE really isn't the main matter, because you could have a bigger piece that's less massive and have those legs break just as easily, so: getting down to the question you want to be asking, YES it might very well be a good idea to have stronger base legs than your top ones.

Well yes, but they see they do take more force because if you draw the force vectors in a diagram, they clearly are taking more "strain" as you put it.

And yes, that's pretty much what i meant, should the base be stronger...now i realize that thicker =/= stronger, but essentially i got my answer

Thanks!

dragonfly wrote:

lllazar wrote:I was wondering - since the base legs take more force, would it be a good idea to make them large than the columns in the upper portion of the tower - say, 3/16 in^2?

.... they don't really take MORE force, since obviously the same amount of force is being applied at the top, it's just that they're under more strain (if you do a sketch of the force components, you'll easily see why). In answer to your question then, SIZE really isn't the main matter, because you could have a bigger piece that's less massive and have those legs break just as easily, so: getting down to the question you want to be asking, YES it might very well be a good idea to have stronger base legs than your top ones.

If the upper part of a tower is vertical, the base (legs) are inclined, and the applied load (at the top of the tower) is vertical, then the legs do carry more force than the vertical members. Here is an illustration/explanation: Let's say our (very simplified) tower looks something like this:



Basically, the top part of the tower is a rectangular prism and the legs are inclined (the angle between the legs and the horizontal plane is denoted by "a"). Here, it is not difficult to see that the vertical member (labeled 1) carries a compressive force of P, since a vertical force of P is directly applied to it at the top. But, what about the inclined member (labeled 2)? Is the force in member 2 greater than P? To determine this, let's isolate the joint that connects members 1, 2 and 3 and show the forces acting at that joint. We end up with three forces (see image above). Since the algebraic sum of these force vectors must be zero, we can write:

F1 - F2 sin(a) = 0 and
F3 - F2 cos(a) = 0

where F1 is the axial force in member 1, F2 is the axial force in member 2 and F3 is the axial force in member 3.

From the first equation above, we get F2 = F1/sin(a). Since F1 = P, then F2 = P/sin(a). If a = 30 degrees, then F2 = 2 P. That is, the force in member 2 is two times greater than the force in member 1. Clearly, in this example, regardless of angle a, F2 is always greater than F1.

In engineering, the word "strain" is related to displacement not force. For example, if a tension member has a length of 10 feet and it elongates 1 inch, then its axial strain equals to 1/120. The term "strain" should not be used interchangeably with the term "force."

My intention was not to use the words strain and force interchangeably; quite the opposite. And yes, I'm aware of the physics of the whole thing, so your sensitivity to my word choice may have confused you. When I spoke of force I was literally meaning the amount of weight being placed on the top part of the tower doesn't suddenly magnify to an actual greater mass once it hits the base. In terms of the word 'strain', I meant everything you just tried to explain, but I suppose I was not using 'strain' in the engineering-form you assumed.

dragonfly wrote:My intention was not to use the words strain and force interchangeably; quite the opposite. And yes, I'm aware of the physics of the whole thing, so your sensitivity to my word choice may have confused you. When I spoke of force I was literally meaning the amount of weight being placed on the top part of the tower doesn't suddenly magnify to an actual greater mass once it hits the base.

Still watch your terms. Weight ≠ Mass. Weight is a measurement of force. Mass is a measurement of amount of matter. The mass loaded determines the weight, yes. But when you talk about the base, you're not talking about mass at all. Mass has nothing to do with the forces in the members. The only thing you are concerned about in your base members are the reaction forces at each support and the forces in each member.

And yes, you do get greater forces in certain members than the weight applied, as SLM illustrated above. Diagonal members in the base WILL "take more force" by virtue of design. The force in those members will be greater.

If you do a sketch of the force components as you suggested, you will see that those members do take more force.

In terms of the word 'strain', I meant everything you just tried to explain, but I suppose I was not using 'strain' in the engineering-form you assumed.

Strain is a measurement of deformation per unit area and quite unrelated to what is being discussed here.

Thanks.

Going back to SLM's diagram.....

If you extend the line of member #1 (the vertical upper leg) straight down, and,
call the angle between it and the sloped lower leg (#2) "b", and
call the vertically applied force (down through #1) F1, and the force on #2 F2
Let's say we're talking about a 3-legged tower, with a 15 kg load. Each leg is carrying 1/3 of the load, so F1=5kg

then, F2 = F1 x (1/cos b). You can get to this relationship either thru vector addition, or empirically playing around in the jhu app (mentioned/link to it provided previously)

so, if #2 were vertical, F2 would =1 x F1, or 5kg;

if #2 is at 8 degrees from vertical (b=8 degrees), F2 = F1 x (1/0.9902) = F1 x 1.0099, or 5.05 kg

or if #2 is at at 16 degrees from vertical (b = 16 degrees), F2 = F1 x (1/0.9611) = F1 x 1.0405, or 5.2 kg,

and, if #2 is at an angle of 28 degrees from vertical (b = 28 degrees), F2 = F1 x (1/0.8833) = F1 x 1.1322, or 5.66 kg.....

O_O
My mind cant comperhend stuff like that...

Bubba1960 wrote:O_O
My mind cant comperhend stuff like that...

You are underestimating your mind's ability to comprehend seeming difficult content. Spend more time thinking about it, ask questions, discuss it with your coach or teacher ...

SLM wrote:

Bubba1960 wrote:O_O
My mind cant comperhend stuff like that...

You are underestimating your mind's ability to comprehend seeming difficult content. Spend more time thinking about it, ask questions, discuss it with your coach or teacher ...

also, because this forum is not LaTeX enabled, stuff like this looks a lot messier than it actually is.