Force-torque analysis

[davidrdguez]

Hi everyone.

Here is a very easy one, but after several years without studying theory of statics mechanics I got lost.

Let's think about a neon sign. It is fixed with a horizontal bar. I want to study this bar. Attached a sketch to show this.

The upper case, one bar; There is the weight of the sign. The bar (black) is fixed in one side and has the force of the weight (P) plus a torque due to the distance to the center of gravity (P*l)

Let's say this bar is not strong enough and need to place more bars (many more). In the lower case of the sketch two.

How would the diagram of loads be for each bar?

The distance between these bars matters

Thank you very much in advance

Greetings!


David

We don't have problems, we have challenges!

 

[rb1957]

the bars would react the moment (P*l) as a couple (upper bar in tension, lower bar in compression). there would still be bending in the bars as they transfer the shear load (P/2) to the fixed end.

if you add a sloping bar then this will carry the shear load (as compression). in this case there would be a horizontal bar and a sloping bar that would intersect at the load P

 

[TheTick]

Because both bars in the lower picture are cantilever mounted, the system is indeterminate. It is not possible to calculate the loads at each joint.

 

[rb1957]

"It is not possible to calculate the loads at each joint."

oh come on ... sure it's redundant but we can solve redundant structures (without resorting to FEA).

sure my solution simplified the reactions the torque P*l would be reacted by the couple (as i proposed) and a small moment in both bars.

inclining the lower bar (and pinning the LH sides) makes the problem determinate.

 

[davidrdguez]

Hi,

First of all, thank you for the answers.

It seems it is more difficult than expected

I can't use a sloping bar. This black "bars" are complicated pieces made out of stainless steel, glass fiber and Teflon.

I set the example with only two bars in order to explain the problem as easy as possible. But actually there are many more shared in 3 (or even 4/5) levels and separated non uniformly (d, d').

Attached another sketch showing this. There is also a front view for better understanding.

I would prefer to do this by hand. The results will be implemented like input for the FEA analysis of the black bars.

Should it be done by any "advance" method?? (non classic mechanics)

David

 

[rb1957]

ok, so the "smily face" shows an end view (along the bars).

are the bars really cantilevered at the LH end ? could we call them pinned ? (makes life much easier)

how accurate an answer are you looking for ?? a simple approach would be to have each bar react an equal amount of the weight. the weight is reacted at the LH end; the offset moment (or "torque")is P*this distance and this would be resolved as a couple between the upper and lower bars and the couple force would be equally distributed amount the available bars (ie /2 at the top).

all this presupposes that the bas are equal in cross-section.

you're only talking about weight, is there no lateral force ?

 

[chicopee]

Your first free body diagram is screwed up. The value P*I should positioned at the fixed end and directed c.c.w.to represent the resisting moment. The load P should have a resisting shear representation at the fixed end and directed upward as a value P.

The second diagram with two beams to support that same load can be analyzed several ways. One is to assume resisting shear and moment to be half of the values of P and PI. Another way is to assume an allowable end deflection(s) to calculate allowable vertical end loads and horizontal tensile and compressive loads on both beams which for the compressive load, column evaluation would be required.

The third arrangement can be evaluated several ways but essentially using the same approach as detailed in item #2.

 

[TheTick]

I know there are other numerical methods, but OP specifically said statics. Statics says "indeterminate".

 

[desertfox]

Hi davidrdguez

Your problem is statically indeterminate however I have asumed all bars are equal in size and properties, by doing this and then assuming that the bottom line of bars becomes the pivot point for the neon sign to pivot on due to its own weight, I can calculate the tension in the remaining four bars spaced at a distance 'd' and 2*'d' above the line of eight blocks, the largest load will be in the bars at 2'd'so if you design all the bars to take this maximum load your design should err on the safe side.

desertfox

 

[davidrdguez]

Here I bring some more pictures of the real design

There are many "bars" here because we were working on wrong assumptions (very big momentum) and many bars were needed. The final design will be something like the "smily face" of my second picture uploaded

In the first one you can see the full assembly:
- we can assume the grey plate is completely fixed
- the pink-blue-green parts are the famous bars
- the yellow plate contai
ns the brown parts. this is the weight to support

The bars are screwed to the grey and yellow plates so I understand they are fixed in both sides. No pinned

rb1957:

I need to be accurate enough to make a FEA analysis of the bars; clamped at the LH side and the forces and momentum on the RH

Agree with distributing the weight in the number of bars. The problem with the momentum will be we have 3 or 4 levels.

There is not lateral force. This is a 2D study

chicopee

Agree with your first paragraph. These are the reactions. But I need to know what happens in the RH side

I’d like to you explain further or address me to some source where I can check this method

TheTick

If there is anther method non static is welcome!

deserfox

The bars will be similar. I like your calculation. I will study it further

Thank you all!!

http://yfrog.com/na02cutjx

 

[rb1957]

"I need to be accurate enough to make a FEA analysis of the bars" ... model the entire structure, rather than an individual bar as i imagine you are doing, then you don't need assumptions about how much load the individual pieces will react.

that said, the problem is quite solvable (with IMHO reasonable accuracy) with simplifying assumptions ... weight load applied to all the bars, pinned ends (at the reaction end, fixed onto the sign).