Differences between calculation and reality 3

[m.piron]

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

 

[r13]

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.

 

[m.piron]

There is no cyclic loading, the beam is static.
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

 

[canwesteng]

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

 

[dik]

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

 

[r13]

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.

 

[rb1957]

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).

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

 

[m.piron]

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?

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

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

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

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



Thank you for your answer

 

[rb1957]

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

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

 

[m.piron]

Ultimate tensile strength 360MPa
Yield strength 235 MPa

With statical load, it's ok.
But with my suppose of 2*F, not good

 

[atrizzy]

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

 

[Agent666]

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.

 

[r13]

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.

 

[canwesteng]

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.

 

[r13]

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?

 

[BridgeSmith]

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

 

[rb1957]

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

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

 

[Denial]

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.

 

[Tomfh]

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.

 

[r13]

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

There is no cyclic loading, the beam is static.
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".