Formula or Ratio for Steel Bar Deflection

I found this in about 1 minute, by searching for it. Not sure what you want is in here, but it's a start.

BEAM DEFLECTION FORMULAS

(5) and (9) are the nearest, but neither is quite right.

If the beam is perfectly straight and uniform and the applied force is colinear with it (and the effect of gravity is neglected), it should just squish up into a flat plate.

I agree. If the beam is 200' in diameter :-)

Is this called "shallow drawing"?

If the length is L, the upward deflection will be somewhere between +L/2 and -L/2.

Depends how hard you push, and which way it bends.

Nope, the question is nonsensical anyway. The sag over 100 feet of length will have it lying on the ground in the middle anyway. So deflection will be = to height above ground, or 50 feet if mounted higher than ground level. We'll ignore all the laws that get in the way.

And, if the bar were mounted vertically and you pushed on the other end, it couldn't deflect upwards at all, could it.

I think that nonsensical sounds a lot better than ignorant or stupid, don't you?

Except for the self-weight, this bar is being loaded as a column rather than a beam. Are the ends held so that they remain straight, or will they be allowed to flex off-axis?

"The length, material type, twisting etc. are assumed to not be factors"

There are no constraints, apparently.

About 40% down the page in this Wikipedia article: http://en.wikipedia.org/wiki/Column --there are some schematics of columns. Turning these on their side will depict the OP's question for various types of end connections.

For the first one, both ends fixed, the maximum deflection is about 38-39 ft.
For the second, one end fixed and the other free to rotate, the maximum is ~35 ft.
For the fourth, both ends free to rotate, the maximum deflection is about 41-42 ft.

(Crude scale model trials with 0.060" welding wire.)

Though Euler is commonly "taught so thought of" as vertical - it applies to horizontal. You may note 'g' is not a factor in the math model.

Or as 'hinted' earlier can be applicable to thermal distortion in rail tracks and/or pipelines.

An illustrative dynamic application, is loads from locomotive traction and breaking.

Another is velocity change in long pipelines.

He also applies to chassis crumple rate/behaviour.

Not that I get peeved at dweebs "discovering" the 'math model imagery' they should have got the first time

Nor is there any frigging 'real world' way it will collapse into a 'flat plate'

I don't have any formula,

But a simple question, why it should deflect upward and why not left, right or downward?If it has to deflect in any direction that will be downward due to gravity.Theoretically it will not deflect anywhere if force is exactly applied in its axis and bar is made perfectly straight and uniform. its will experience compression force and finally you will observe its diameter expending.

For practical reasons there will be deflection but that will depend on what angle force is applied,how much is the inherent bend is in bar and so many factors.

Mr. Rakesh,

Regarding Bending, You are right and The bar can bend to any direction, left, right, up or down, or inclind plane, it depends upon the angle of force applied.

But as for as Formula is concerned, the Slenderness Ratio Formula can be used for this purpose as it can be treated as a COLUMN as for as load is concerned and correction for deflection due to self weight (on account of horzontal position) to be acoounted.

Thanks,

DHAYANANDHAN.S

From the OP's statement "...while pushing the bar in towards itself ...", this is a buckling problem. If the bar is support horizontally, the it's conceivable that it will buckle up (although I suspect it would be more likely to buckly down as the weight of the beam would give the beam a downward bend.

Assuming that the bar is thin, it is more a question of how much the sliding end is moved toward the secured end. Whether the ends are allowed to rotate also makes a difference. If gravity is included, it becomes different again.

If this was a thin bar, 100feet long, I would not be surprised to see an offset bump, or even several bumps.

There is no simple, dimensionless, formula for this.

For small bending deflections (less than say 1/50 of the length) the displacement curve is approximated to a simple radius. So, based on this premise, it is simply the same length curve as was the original line. Therefore, look up on the web the length of a segment, compare it to the original length and then create a formula which gives you the deflection.

It can also be done by iteration in autocad or some other software. It is true that a formula can give you the exact answer, but it is often sufficient to get the answer to 3 significant figures.

a neuropsychologist would say this is not a simple question, but, in the most basic analysis: The answer is zero, if the force is centered on the bar's cross section and normal to a plane that is perpendicular to the bar (i.e., also normal) at the point it is secured. There is, in that simplest case, no force being applied that should deflect it. otherwise you will need a formula taking into account the direction of force, the degree to which it is off-centered, and a variety of other matters. In addition to these issues regarding how you are going to apply the force, your method of "securing" it is not a trivial consideration. If the bar is not normal to the plane on which it is secured, the securing method is going to exert some resistance to any deflection, unless it is fixed to the plane at a point, such that it cannot slide, which seems unlikely. I doubt there is a formula that takes into account all these variables simultaneously, and if you eliminate them, as in the "most basic analysis" above, nothing will happen - will it?

Jwatersphd

G'day,

Have a look at

http://www.efunda.com/formulae/solid_mechanics/columns/eccentric.cfm

Also Google Column Load Formulas

RRV

GA

The OP should be looking at the relation between the diametre and the length of the bar, to find what is the minimum diametre for a length that will prevent the deflection due to a small excentricity (which is usually unavoidable). In the construction industry, there are norms that forces the minimum diametre for a column length, no matter what is the loading.

On the other hand, if he is looking at the amount of deflection that will occur, then it will depends on how much force is applied, the material spreecifics, the diametre etc...

For 100 feet of length, 1 inch diameter is a superduper slender "column"; basically spaghetti that could almost be formed into a pretzel via bare hands, perhaps assisted with a couple of posts in the ground.

Absolutely!

Look up these terms;

deflection; yield point.

The answer is 0, if by "etc." you mean gravity and error in locating forces. If the forces remained distributed across the created surface, it would indeed eventually become a flat plate. If you want a theory or formula containing all relative factors, I'm sure someone has one. Included in this theory would be an analysis of materials at the molecular level, which for a bar 100' long would contain a lot of data. I'm guessing you would create some new materials, heat, light and other radiation and sound. People in the die stamping business do a very similar thing.

The elevation of the top of the end product plate would be higher than the original elevation of the top of the bar, so that there may be an answer to your question, but it has nothing to do with deflection. It would have to do with volume of the circular plate. (assuming it is a rod that you started with)

Try this site they have lots of calculators. http://www.calculatoredge.com/

Nice little site :)

Sure, but do they have a calculator that pertains to this situation? If so, could you specify which?

Eventually, almost 50 feet downward, assuming that there is gravity, then rotate it if you want it up. The size of the bar will determine the force and bend radius therefore the almost.

Dear Friend,

The SLENDERNESS RATIO formula will give the deflection either Up-ward or Down-ward. Up-ward or Down-ward depends upon the angle of Force applied. Pl. refer Strength of Materials Text Book for more details.

The deflection depends upon the Force applied, size of the rod, etc. This applies for the Rod in vertical position, and apllied load is truly vertical in the axis vertically, and there will be NO DEFLECTION due to self weight.

I understand from your description, that the steel bar is secured horizontally, and deflection, on account of Self-Weight will take place and is to be also taken in to account and apply the correction.

DHAYANANDHAN.S

GA

The site I found that will be useful is: http://www.ehow.com/how_5113653_calculate-slenderness-ratio.html

It describes it all and gives a step by step methode to determine the dimensions etc.

I hope the OP is still listening and that this is what he is looking for.