Hi, I am re-posting this question.
Hi, I am re-posting this question.
(OP)
Hi, I am re-posting this question.
One of the design requirements for my offshore structure is an impact from a mass with a given velocity. (Mass and Velocity are given.)
I am looking for a realistic way to translate that "impact load" into an "equivalent static load", and geting rid of doing any impact dynamic analysis.
Here is my methodology: m (mass) and V(velocity) are given, so I know the input energy is 0.5MV^2.(eq.1)
On the other hand I know if a force of F applys on my structure, it makes it deflect about "X". So the energy absorbed by the structure at that point is 0.5F.X.(eq.2) . Since it is a linear analysis, I konw F=K.X (eq.3)which K is the stifffness of that degree of freedome on the structure. By replacing (eq.3) into (eq.2) we will have: Absorbed energy=0.5K.X^2 (eq.4)
If we ignore the absorbed energy by the structure due to its damping and inertia, and also the energy lost at the impact, we could say: Input energy=Absorbed energy -> 0.5mv^2=0.5k.x^2
so x=v(m/k)^2. (K is known because we have our static model in place)
F=k.x -> F=m^2.(v/k)
Sorry for this long introduction. But I needed to generate a base for my question.
Do you think this result for F is a realistic value?
I appreciate your feedbacks.
One of the design requirements for my offshore structure is an impact from a mass with a given velocity. (Mass and Velocity are given.)
I am looking for a realistic way to translate that "impact load" into an "equivalent static load", and geting rid of doing any impact dynamic analysis.
Here is my methodology: m (mass) and V(velocity) are given, so I know the input energy is 0.5MV^2.(eq.1)
On the other hand I know if a force of F applys on my structure, it makes it deflect about "X". So the energy absorbed by the structure at that point is 0.5F.X.(eq.2) . Since it is a linear analysis, I konw F=K.X (eq.3)which K is the stifffness of that degree of freedome on the structure. By replacing (eq.3) into (eq.2) we will have: Absorbed energy=0.5K.X^2 (eq.4)
If we ignore the absorbed energy by the structure due to its damping and inertia, and also the energy lost at the impact, we could say: Input energy=Absorbed energy -> 0.5mv^2=0.5k.x^2
so x=v(m/k)^2. (K is known because we have our static model in place)
F=k.x -> F=m^2.(v/k)
Sorry for this long introduction. But I needed to generate a base for my question.
Do you think this result for F is a realistic value?
I appreciate your feedbacks.





RE: Hi, I am re-posting this question.
The gotcha is if the impact time is of the same order as the inverse of the frequency of a resonance in the structure in which case you may get more stress at some frequencies than you were expecting.
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?
RE: Hi, I am re-posting this question.
the point of impact is the initial kinetic enrgey of the impact is ultimately converted into strain energy of the structure, as you have "k" represents the stiffness of the structure, and "x" it's displacement. however you're assuming a constant deceleration, no?
another take is to consider the impact time duration ... KE = F*t, but again a constant deceleration is assumed.
the best way to get an accurate estimate is to run an impact analysis (using Marc for example). when you do this, you'll discover your next problem is the stiffness of the impacting mass, but you might be ablet o show that a high stiffness is conservative.
does the code covering your structure give you any advice ? how has your company done this in the past ? other companies ?? or is this a new requirement ?
how do you deal with wave impacts ?
RE: Hi, I am re-posting this question.
1-If the impact time is less than the Tn , then we can say the structure will get the whole impact, but if imapct time>Tn, then we can say the structure will miss part of the impact energy, right?
2-can I say at the impact moment, the softer object absorbes more energy? Like if I hit a rubber ball into a concrete wall. At the impact moment, at the contact area, the same and equal force will be applied to each object (of coures in opposit directions). Since the rubber ball will deform more, under the same force, its gonna take more energy. (Energy=0.5F.Delta for linear materials, or generally E=the area below Force/deflection curve). what do you think?
thanks again.
RE: Hi, I am re-posting this question.
2 yes
Cheers
Greg Locock
New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?