Impact Force / Maximum Plastic Deformation on Cantilever Beam 4

[Pengy]

Let's say you have a 10' cantilevered beam with an ultimate moment capacity (not design moment @ yield, but capacity @ ultimate stress) of 200 k-ft. Is there an approach to find what the approx. total deflection (elastic+plastic) would be at the end of the cantilever if the point load resulting in the ultimate moment was placed @ some point along the beam (i.e. total deflection with 20k load @ end of cantilever, or total deflection with 40k load @ midpoint of cantilever)?

Ultimately I'm trying to determine how much a freestanding post could deflect in a collision (with the impacting item assuming to have a rigid body with no deformation) to determine the potential impact force.

 

[bones206]

Just because a member deforms plastically doesn't mean it will collapse. Here's a good visual: Link

 

[phamENG]

I think it's important that we remember our analysis techniques are usually conservative simplifications that don't necessarily reflect the way a structural member will behave beyond a certain point. Put another way, our typical analysis tools will help us determine the maximum load effect that can be safely carried, not necessarily the load effect that will cause collapse. (One of the reasons a beam can "fail" by calculation but has been in a building for 80 years with no problems.)

For instance, retired13's statement that "collapse will occur under any further load increase" is acceptable when designing a member to not develop a plastic hinge, but in this case - where we're interested in the real world inelastic behavior and plastic deformations - it doesn't hold up.

Also, as shown in the video bones206 posted, we're not considering a static load. So, even if the section is fully plastified and we assume the stress/strain curve does flatten out at fy (which it doesn't), then plastic deformation will continue until the load is released (in this case, the driver takes his foot off of the accelerator). Since there is no axial load, it would follow that it will never "collapse" as there is no compression exacerbate the moment through P-delta effects. The only collapse would be caused by flexural rupture if the driver doesn't stop and just plows right through.

Making sure something won't fail and understanding how something will fail aren't always the same thing.

 

[r13]

Isn't that conservativism clearly indicated by the codes? For load factor of X, strength reduction factor of Y, the design failure force = Z, then the real word breaking force will be at least equal to Z*X/Y. The contractor in the video shall really address their finding to ASTM, if they can calculate and quantify the finding, and the code authorities to make revolutionary changes to save everybody money.

The assumption of a hinged joint is that the entire cross section is yield, at that very moment, the applied force, and the reactions are at a state of limit equilibrium, with further increase of force, the member collapses (by own weight, I guess). I've no clue what would happen once get into the inelastic range, or if it is useful though.

 

[hardbutmild]

I wouldn't necessarily agree with your last post phamENG. First, at some rotation failure will occur because strains in steel will simply rupture it even without p-delta effects. Second, it really depends on how do you define a failure. It's usually not "when the thing breaks", but rather "when the thing can't fulfil it's function and repair costs are too high to try and repair it".

I think that no one mentioned, and I believe it is self evident, but any holes in your post are undesirable.

 

[phamENG]

hardbutmild - I think we actually do agree with each other when you consider the context of the discussion. The OP is trying to find the deflection at the ultimate moment (I ready that as rupture) to use that to back out a maximum impact force. In other words, these are disposable and failure as defined by this problem is rupture (or "when the thing breaks").

I agree that a floor beam with 6" of deflection over a 20' span is a serviceability failure even if it hasn't yielded. But in this case, a bollard that has deflected 12" over a 10' height but hasn't ruptured sounds to be a-ok, though it will be replaced once they take the forklift driver to the hospital.

 

[hardbutmild]

Sure, I agree. I just figured that it needs to stop a vehicle which could limit the actual deformations to some level lower than failure. But since I have no actual experience on these structures, I'll just spectate this discussion

 

[r13]

If this is an experiment, for which calculation is to be verified by field observation, then I think this is a worthy question to explore. But, forgive me for my overthought of safety, as a structural engineer, don't we check structural stability and safety during construction? Even with the chance to injury a person is null, but the forklift and anything besides it, I suggest the design should be limited to far below the failure mechanism to occur, unless someone can say, "after yield there is still a safety factor of > x.x, and there will be ample time for replacement", with straight face.

 

[bones206]

The OP is just trying to predict deflection, not deciding whether the thing is considered “failed” or not.

 

[r13]

Assume full rigidity can be achieved at the support level, calculate deflection due to M_p. Deflection beyond that point don't serve any purpose other than science study.

 

[bones206]

I wanted to follow up with an example to illustrate the simplified energy method. I would consider this a quick and dirty way to estimate an upper bound deflection. In reality, not all of the kinetic energy is going to convert to strain energy in an idealized beam bending mode. A potentially significant amount could be absorbed as strain energy very local to the point of impact, while some percentage of energy (maybe on the order of 2%-5%) will just damp out.

 

[ozPE]

i'm bit late to the discussion
i agree with bones206, energy equations will do the job

i designed a timber shielding sitting on and spanning between bottom flanges of bridge girders, the design load was 1 or more men falling about 6ft from the top flange to the timber shielding. more man is considered as such, one guy falls and sits there and 1 or 2 more falls depending on the code and engineers judgement. so you have static and impact loading at the same time, assuming they are near each other for worst case. i conservatively assumed the guys are all rigid objects

i had to find the max stress in the timbers..

anyway the area under the force-displacement graph = work done = KE at impact
KE: all parameters are known
in the area formula, only parameter that is not known is the force, so if you calculate the force based on above equation then that is the impact force to use to find stresses.

this does not take into account the steel plastic situation, i never thought about that.