## Differences between calculation and reality

## Differences between calculation and reality

(OP)

Good morning.

I have the structure in picture. The material of the beam is steel S235.

The real F is 350N, but I continuosly put and remove the load, so I take into account 2*F (found by match the elastic and potential energy, 0.5*k*x^2 = m*g*x).

I found a stress of 453 MPa, so this beam couldn't carry the load.

My question is, I'm using this beam every day, and it's not broken, nor bend.

Why it resist?

What is wrong in my calculation?

Thanks

I have the structure in picture. The material of the beam is steel S235.

The real F is 350N, but I continuosly put and remove the load, so I take into account 2*F (found by match the elastic and potential energy, 0.5*k*x^2 = m*g*x).

I found a stress of 453 MPa, so this beam couldn't carry the load.

My question is, I'm using this beam every day, and it's not broken, nor bend.

Why it resist?

What is wrong in my calculation?

Thanks

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

The load is simply placed over the beam, and when done you can se a little bit of rebound.

That's why I choose to use 2*F

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

Try loading your beam to the stress determined by the plastic section modulus with a phi factor of 1 or 1.05, and, let me know what the results are.

Dik

## RE: Differences between calculation and reality

Cyclic loading in my comment means the action of removing and replacing the load numerical times in a time frame, or the load reverses direction in a regular base, the object is considered static in nature. I don't think your method is correct to account for this effect. If you are a scientist, I wouldn't question your approach, but if you are an engineer, I suggest to find the answer from the steel design code, which is more appropriate.

## RE: Differences between calculation and reality

As others have noted, doubling the load is an impact load, suddenly applied. Can the steel handle 227 MPa ? (Don't know your steel)

227 MPa = 33 ksi … this is an easy stress for any steel I deal with, but possibly your commercially available steels have lower allowables (to design to).

At this stress, aerospace steels wouldn't have any fatigue issue either (particularly with your low Kt).

another day in paradise, or is paradise one day closer ?

## RE: Differences between calculation and reality

The impact is minimum.

If you place a load over a spring, after certain time it stops in the equilibrium position, the travel is x=F/k and the force is F=m*g (I know that "travel" is not the correct term, but I can't translate from the italian "freccia", any help?). But in the beginning, it swing and reach a travel of x=2*F/k. In this condition the load is double the static load.

And the system reach this condition, I can see the beam that swing.

Maybe I must consider it as a static condition, but why?

I'm sorry but I don't understand what did you mean

The system work, and my calculation say that don't work so... my method is certainly not correct.

I'm just wondering why

I wrote in the beginning, is a S235 steel (the old Fe360)

Thank you for your answer

## RE: Differences between calculation and reality

another day in paradise, or is paradise one day closer ?

## RE: Differences between calculation and reality

Yield strength 235 MPa

With statical load, it's ok.

But with my suppose of 2*F, not good

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

You're also working out an elastic stress, but if the stress is greater than the yield stress the section will progressively yield further into the section until the entire section is yielding and you fail the system.

If there are no reductions due to local buckling of the walls of your CHS, then the maximum moment that can be sustained is the plastic moment capacity. Look up the difference between elastic and plactic section modulus.

## RE: Differences between calculation and reality

Place the load, and measure the deflection in the mid-span, then calculate the deflection per beam theory for 1*F, and 2*F. The measured deflection shall be much less than the calculated deflection resulting from 2*F, and quite close to the result from 1*F. Thus, your load is less than 2*F, but remains as 1*F.

If you bother to measure the deflection for 10 years and beyond, at a point of time, you would notice the deflection would be much larger than before, and the calculated value, its the phenomenon of stress relaxation in the beam, in engineering term - fatigue has kicked in.

I'm too far away from my school days, so I wasn't able to pinpoint the mistakes you made in the equation. Some newer graduates with the training on energy method shall be able to provide more insight though.

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

- On the right hand side of your equation, m*g*h is potential energy due to gravity. "h" is the vertical fall distance of an object.

- On the left hand side of your equation, k*x²/2 is energy in linear spring, it can be think as work energy - F*x. "x" is the elongation or shortening of the spring.

I think I am missing something here. By the way, what is the "k" value in your equation?

## RE: Differences between calculation and reality

Rod Smith, P.E., The artist formerly known as HotRod10

## RE: Differences between calculation and reality

another day in paradise, or is paradise one day closer ?

## RE: Differences between calculation and reality

We then come to a definition of "suddenly". Again this is highly subjective, but you will get a bit of a feeling for it if you compare the load application time to the natural period of the structure.

## RE: Differences between calculation and reality

Sometimes people use 2F as a kludge to estimate dynamic effects, but this isn’t the case here. Your load is static.

## RE: Differences between calculation and reality

The "little bit of rebound" indicates the beam remains elastic after unloading.

I think the problem is because he has misinterpreted the parameters, k & x for instance, in equating the moment and energy, thus the wrong conclusion - "2*F".

## RE: Differences between calculation and reality

Yes, exactly. He’s concluded that F = 2F, which is clearly wrong.

## RE: Differences between calculation and reality

Because its strength is adequate to carry the loads.

You are using twice the value of the load F.

Are you applying two loads simultaneously or one at a time?

BA

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

During a collision, the impact force can go above 1, when two subjects run into each other with speed, and the deformation (Δ) is very small. The impact force will be infinitely large when Δ approaches zero.

## RE: Differences between calculation and reality

You’re correct, however some codes use 2F as a simplified way of handling dynamic loading - Eg a beam supporting a lift/elevator.

We structural engineers struggle with dynamics, and the codes often resort to giving us static loads to account for dynamic effects, even though it’s not strictly correct.

## RE: Differences between calculation and reality

Thanks for your information. But I would think the use of 2F incorporated safety concerns rather than the true impact effect derived from physics, which happens daily around us, but not well understood, and difficult to evaluate.

## RE: Differences between calculation and reality

My assumption is, in the beginning the system have a certain potential energy, when you release the load the beam deflect, and the potential energy is converted into kinetic and elastic energy. When the beam is at the maximum deflection, all the potential energy will be converted into elastic energy, because speed is zero so there is no kinetik energy.

you're right, I look at the properties of that material, UTS=360 MPa is the minimum, but can reach UTS=510 MPa.

I don't measured it, but looks straight.

Meanwhile, I've done a FEM analysis of that system.

This is the result of the static analysis:

The stress is 224 MPa, the same calculated by hand.

After that, I done a dynamic analysis. In that analysis, I simply placed the load at time 0 (what's happen in reality).

This is the stress over time graph:

As you can see, the maximum stress reached is 350 MPa, 1.6*F. I think that it is less than 2*F due to the damping factor: the potential energy converts into elastic energy and heat due to damping, so the real force is less than calculated.

## RE: Differences between calculation and reality

m*g*h = F*Δ, in which h = object travel distance; Δ = elastic deformation of object(s) at contactRearrange the terms, we get F = m*g*(h/Δ), and simplified to

F = W*(h/Δ)Example: An object weighted 1 lb, dropped 100" to 3 different mediums below, with the measured settlements 0.1", 1", and 100". What is F for each case?

W = 1 lb , h = 100", Δ = [0.01",1",100"]

1) F = 1*(100/0.01) = 10,000 lbs

2) F = 1*(100/1) = 100 lbs

3) F = 1*(100/100) = 1 lb

What is the results explain to us? Feel free to draw your own conclusion.

## RE: Differences between calculation and reality

For a force exerted in motion F = m*a, ie. the pull force on spring.

For a force from fall F = m*g, ie. weight of an object

I hope you can see the difference between the two cases, and apply each thoughtfully.

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

> Materials' specifications are statistical in nature and they are not exact and things don't necessarily fail instantaneously. https://www.azom.com/article.aspx?ArticleID=6022 states a tensile strength of 360 to 510 MPa. Assuming that your transient is not as severe as you analyzed and the beam is at the high end of its strength, your beam might not have any issue.

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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## RE: Differences between calculation and reality

OPmeant the object was placed on a softer medium (in practice) as opposed to a stiff medium (in calculation). So the magnitudes of the first sine wave are different, which is caused by the differences in damping property of the materials (some materials capable of absorbing more energy than the others). Otherwise, I agree withcanwesteng.## RE: Differences between calculation and reality

I'm sorry but you're wrong.

Suppose that I'm holding the weight over the beam,

the weight touch the beam without load it. In that time, there is no load over the beam: the system have only potential energy.Then, I release the weight. Part of the potential energy will be converted into elastic energy, the remaining part will be converted into heat (because of dissipation).

It's not simple to calculate by hand, because that part of energy depend on the velocity.

If part of energy will be converted on heat, the deflection of the beam will be less than the deflection calculated without dissipation.

Less deflection = less maximum stress.

That's why in FEM the maximum stress is 350 MPa, and not 450 MPa as calculated by hand.

I'm pretty sure about that, because I studied a lot car suspension

I don't know how well the simulation match the reality. It's difficult for me to measure it.

The bold part explain why I'm saying there is no impact.

Let me explain with an example: it's not like a car hitting a bumps... it like a car in parking, that you lift (maintaining the contact tire-ground) and then release.

You can't say that this is an impact, but the spring go over the equilibrium point, then rebound, pass throught the equilibrium point, rebound again... (only with old shock adsorber :D).

If you want to calculate the torsional stress of the springs wire... is not the stress at equilibrium, but that reached during the rebound, when the deflection is maximum.

And that without impact.

## RE: Differences between calculation and reality

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

## RE: Differences between calculation and reality

Rod Smith, P.E., The artist formerly known as HotRod10

## RE: Differences between calculation and reality

## RE: Differences between calculation and reality

You certainly are making a lot of work out of a pretty straight forward problem. Is this a Ph.D. thesis? Your first calcs., without the load factor of 2 on the loads, does seem to indicate that calcs. indicate/follow reality, as long as they are reality based. You get a bending stress of about 32ksi, likely well within the yield strength of the pipe material, so it continues to function as intended. Unless you are dropping the loads on the pipe beam from some height, or repeating this loading 100 times a day, you don’t have an impact loading or a likely fatigue issue, so give it a rest. You’ll break that pipe quicker by continuing to beat this dead horse, than by the way it is now being loaded. You could strain gage it or actually measure the deflections at several points to help confirm the basic calc. mentioned above.

## RE: Differences between calculation and reality

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

## RE: Differences between calculation and reality

_{e}^{3}- d_{i}^{3})/6 = 650.7mm^{3}Assumed Fy = 235MPa

Mu = Fy.Z = 235*650.7 = 152,907N-mm

If the impact factor is 2.0, then 2.0F*300 = 700*300 = 210,000N-mm or about 1.37Mu

For impact factor to be 2.0. the loads would need to be simultaneously released onto the beam. That would not be advisable.

It may be that the loads were placed individually and released slowly onto the beam. It is also possible that Fy > 235MPa.

BA

## RE: Differences between calculation and reality

Your arms dampen the weight application. If this is the case, there certainly isn't 2x your weight applied.

If you weigh 140 kg, the distance between your shoulders is more than 400 mm. I'm guessing you spread your hands further apart, this also contributes to a lower bending moment.

## RE: Differences between calculation and reality

Regarding bodily impact forces - When you run you apply more than 2x body weight to your knees (don’t I know about this).

And if you fall from height onto a rigid surface you will experience peak impact forces vastly in excess of 2x your body weight.

Human body can handle brief impact of a few tens of g’s. But once you’re up near 100 g’s it is immediately fatal. This is why people die in car accidents. They simply come to a stop too quickly.