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# Differences between calculation and reality3

## 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

### RE: Differences between calculation and reality

I don't know it is correct or not using energy method to justify the result of cyclic loading/unloading as double the load. You should check the code specified allowable stress for members with repeat/cyclic load condition (provisions on fatigue), which is more close to the reality. Even in the remote likelihood, with the allowable stress is reduced, your beam hasn't reached the breaking point yet, I believe.

### RE: Differences between calculation and reality

(OP)
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

Well that approach really doesn't make any sense. Is there an impact associated with the load when you apply it?

### RE: Differences between calculation and reality

I find that reality and theory generally converge... it's a matter of adjusting the theory.

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

m.piron,

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

if the equation predicts failure, and the part doesn't fail, then the equation is conservative.

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).

### RE: Differences between calculation and reality

(OP)

#### Quote:

Well that approach really doesn't make any sense. Is there an impact associated with the load when you apply it?

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?

#### Quote:

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.

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

#### Quote:

I don't think your method is correct to account for this effect

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

I'm just wondering why

#### Quote:

Can the steel handle 227 MPa ? (Don't know your steel)

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

### RE: Differences between calculation and reality

that doesn't help me (I don't work in your field) … ftu ?

### RE: Differences between calculation and reality

(OP)
Ultimate tensile strength 360MPa
Yield strength 235 MPa

But with my suppose of 2*F, not good

### RE: Differences between calculation and reality

Perhaps your 2xF is what's not being replicated correctly in reality.

### RE: Differences between calculation and reality

Maybe you should outline what this applied load is in reality? It's source, and why it's varying? Then we'll maybe better understand if the 2 factor is appropriate or not.

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

Let me try to verify your assumption of 2*F was incorrect.

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

You've made some fundamental errors in your assumptions. There is no substantial acceleration of the load when you place it on the beam. You place the load on the beam, there is a minute force imbalance in favour of the load, and then immediately after a minute force imbalance in favour of the reaction, almost immediately resolving to equilibrium. Place an accelerometer on your load if you want to verify this (although it's probably not measurable). If you ever have a reaction of 2*F, you are going to rocket the load up, until it leaves the beam, and then it will drop back down on the beam, repeating the cycle. I think it will be obvious that as you load the beam you aren't seeing it vibrate like a guitar string, or a spring.

### RE: Differences between calculation and reality

After refreshing my memory, I find you was err on equating banana to the orange.

- 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

In addition to the likely overestimation of the load, as detailed by others above, realize that the specified yield and ultimate strength are minimums. The actual yield and ultimate are almost certainly higher, probably by at least 10%.

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

### RE: Differences between calculation and reality

there's still something odd about this … does the bar look like it's being stressed to near yield ?

### RE: Differences between calculation and reality

The factor of 2 referred to in the initial post is a conservative "surcharge" to allow for a load that is applied "suddenly" but not "impactively".  It is conservative for several reasons, the main one being that it is (very simply) derived from considering a single-degree-of-freedom dynamic model.  The amount of conservatism when it is applied to a multi-DOF system depends upon the extent to which the distribution of the applied load is "compatible" with the eventual deflected shape.  Please do not ask me to attempt a definition of "compatible", but my gut thinks your loading is highly incompatible with the eventual (static) deflection.

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

Your load isn’t 2F. Your energy logic is all wrong. You can’t equate gravitational potential energy with work. If the situation is static then the load is F.

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

I don't think it is done suddenly, nor dynamic in nature. Below is his response to my original comment.

#### Quote (m.piron)

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

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

#### Quote (Retired)

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".

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

### RE: Differences between calculation and reality

#### Quote (m.piron)

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

Why it resist?

#### Quote (m.piron)

What is wrong in my calculation?
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

2xF is the correct force to use for a impact load, "impacting" from a distance of zero above the object. It is valid for some live loads. No so valid for others.

### RE: Differences between calculation and reality

We don't have "impact" here. Impact is dynamic in nature, it involves velocity and contact distance. In engineering practice, the dynamic load effect is transformed and expressed as a percentage of the moving weight for ease of calculation, usually less than 1 (100% of the moving object).

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

Retired13,

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

Tomfh,

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

(OP)

#### Quote:

After refreshing my memory, I find you was err on equating banana to the orange.

- 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?

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.

#### Quote:

In addition to the likely overestimation of the load, as detailed by others above, realize that the specified yield and ultimate strength are minimums. The actual yield and ultimate are almost certainly higher, probably by at least 10%.

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

#### Quote:

there's still something odd about this … does the bar look like it's being stressed to near yield ?

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

Just for fun, let's drop the term "speed (velocity) in the evaluation of impact, and say the impact is due to the transformation of potential energy to work energy, the equation looks like below,

m*g*h = F*Δ, in which h = object travel distance; Δ = elastic deformation of object(s) at contact
Rearrange 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

#### Quote (m.piron)

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?

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

The dynamic effect in your analysis is so called "impact", which occurs when both time of contact (duration), and elastic deformation (similar to change in length of spring) at contact approach zero. As pointed out before, impact is dynamic in nature, but simplified to a factor that is used in a static manner for ease of calculation. However, your simple acts on placing and moving object from the beam, in general, will not produce significant dynamic effect to reach the state of 2*F. But you are correct on effects of damping, which can be came from the flexibility of the beam, and rigidity of the supports. This is an interesting topic after all.

### RE: Differences between calculation and reality

OP is not correct on the effects of damping - it should hardly have any effect on the first sine wave, only reducing the subsequent oscillations.

### RE: Differences between calculation and reality

Copy from Wikipedia:

#### Quote:

A sine wave or sinusoid is a mathematical curve that describes a smooth periodic oscillation. A sine wave is a continuous wave.

#### Quote:

The term "damped sine wave" describes all such damped waveforms, whatever their initial phase value. The most common form of damping, and that usually assumed, is exponential damping, in which the outer envelope of the successive peaks is an exponential decay curve.

### RE: Differences between calculation and reality

> How well does the dynamic simulation match the real motion? You stated that "I simply placed the load at time 0 (what's happen in reality)," but that cannot be if you also say there's no impact, since the instantaneous loading is an impact. If there is no loading, then the motion is some sort of truncated sinusoid, triangular, or exponential, waveform over some finite time, which would greatly reduce the overshoot. The overshoot is greatest with a step function input.

> 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
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

I think OP meant 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 with canwesteng.

### RE: Differences between calculation and reality

(OP)

#### Quote:

OP is not correct on the effects of damping - it should hardly have any effect on the first sine wave, only reducing the subsequent oscillations.

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

#### Quote:

> How well does the dynamic simulation match the real motion? You stated that "I simply placed the load at time 0 (what's happen in reality)," but that cannot be if you also say there's no impact, since the instantaneous loading is an impact. If there is no loading, then the motion is some sort of truncated sinusoid, triangular, or exponential, waveform over some finite time, which would greatly reduce the overshoot. The overshoot is greatest with a step function input.

> 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.

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

The issue is the "release;" is it instantaneous or not? If not, then your premise is incorrect. Any other input results in the energy spread out over time; if the spread more than 0.1 seconds, then the overshoot is substantially reduced.

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

If the object to be supported is placed, (not dropped) on the pipe you are analyzing (which the previous posts seem to indicate is the case), there is only a small kinetic component to the loading (negligible impact load). In this case, the load on the pipe is only slightly more than the weight of the object, and only for an instant until the pipe stops deflecting.

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

### RE: Differences between calculation and reality

I hope you didn't design car suspensions after all that study then. If you apply a force, F, to a spring, K (or a car suspension), you could have exactly critical damping, and you won't affect the initial displacement at all. Only the rebound effect, or free vibration, will be effected (in this case, this is different for vibrating loads of course, where damping reduces vibration amplitude, but note, that the amplitude of vibrations is always greater than F*k, unless the natural frequency of the spring is smaller than the forcing frequency).

2
M.piron:

### RE: Differences between calculation and reality

Plastic Modulus Z = (de3 - di3)/6 = 650.7mm3
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

Let me guess, you weigh (less than) 140 kg and you work out.
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

Impact factor can be anything. 2x is just a code cookbook factor that’s generally conservative. You likely won’t yield the structure.

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.

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