Cantilever Deflection Under "Sudden" loading... 1

[walterbrennan]

There’s an equation from a text that is giving me something of a fit…

A vessel strikes the tip of a cantilevered finger pier at some oblique angle.

1. The initial condition finds a mass (m) moving along at a constant velocity (v) resulting in a kinetic energy (KE = ½ mv^2); while the cantilever is as yet undeflected, resulting in a stored energy (strain energy, potential energy, spring energy, etc.) equal to zero.

2. Upon contact of the moving load with the cantilever tip, the cantilever deflects until the kinetic energy of the mass is dissipated, leaving the system (briefly) at rest; the deflection having (presumably) converted the kinetic energy into stored energy in the cantilever (i.e. conservation).

Can anyone produce the derivation for the following relationship: Δ = (KE/k)^1/2 ...?

The absorbed energy side of the equation appears to be k x Δ^2, but this just does not look right for the strain energy of a cantilever...

Thanks,

walterbrennan

 

[GregLocock]

Assuming k is the stiffness of the cantilever and delta is its deflection, both at the point of impact, the stored elastic energy is 1/2*k*delta^2, ie a factor of 2 different from what you've quoted






Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[walterbrennan]

That's exactly what I thought, and why I'm asking. The formula I'm asking about comes from a venerable old text on marine engineering (Tobiasson), but I just couldn't make it work against the usual spring stiffness based energy equation.

Strange stuff...

 

[azcats]

Bear with below - I haven't done this type of thing in a long time.

I found stuff online that gets close to the requested derivation, but has that pesky extra 2 in it.

Strain Energy in a cantilever due to bending (U) = (P^2 * L^3)/(6EI) Page 6

k = (3EI)/L^3 Link

Δ = (PL^3)/(3EI)

U = (P^2)/2k [plugging k into the U formula]

Δ = P/k [plugging k into the Δ formula] ---> P = k*Δ

U = (k^2)(Δ^2)/2k = k * (Δ^2)/2 ----> Δ = (2U/k)^.5

I don't suppose by chance that the book you referenced was looking at two posts?

Are there any other energies being neglected? Are we allowed to assume that all the energy is being conserved and this is a perfectly elastic collision? Or is there a typo in the book?

 

[walterbrennan]

Yes, the pesky 2 comes from the fact that the tip-loaded cantilever of uniform cross section stores energy in an equivalent fashion as that of a spring of constant stiffness (i.e. the area under a linear spring force-displacement curve, or the triangle ½ x k x Δ2).

No, unfortunately the author was not considering two posts in the design example... it's a cantilevered finger pier. And he isn't considering other energy losses.

The publication is Marinas and Small Craft Harbors,first edition, by Tobiasson; and the Δ = (KE/k)^1/2 reference is found on page 375 about halfway down the page.

At first I though he might be accounting for the presence of two stringer beams; but a unit load deflection calculation earlier in the example demonstrates that he is checking the finger as a whole; considering the stringer beams as a composite cross-section, for the purposes of resisting the entire transverse load.

If anyone has a second edition of Tobiasson, with the same example problem, I would be interested in knowing if the same Δ = (KE/k)^1/2 relation is indicated there.

Thanks,

PBW

 

[IRstuff]

I vaguely recall during the Challenger investigation that there were a group of people arguing that the mounting struts for the shuttle were under designed by a factor of 2, and that there was something to effect that transient loads were double what a comparable static load might be. Never got a good explanation though.

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[racookpe1978]

Odd. The theoretical problem is a diving board, supported at one end, the other free to move unedr a force perpendicular to the mounting plane.

The real "boat hits pier" problem has the boat impacting horizontally into a long pier regularly supported (at 10-15 foot intervals) by flexible piers mounted into mud and dirt supporting a horizonal "board" that is trying to flex against the strong side of the member.

 

[msquared48]

There is another minor force to consider in the overall system here, in that in addition to the mud to restrain the lateral motion, there is the displacement of the water above the mud by the pile that would effectively act as a damper in the system, absorbing energy and decreasing the lateral deflection.

Mike McCann, PE, SE

 

[glass99]

What about higher frequency vibration? Surely more than one mode of vibration gets engaged. I imagine a real impact is more of a "bang" than a slow winding of a spring. What about conservation of momentum rather than conservation of energy? If the pier is heavy concrete structure, there will be a large energy loss as the two objects move at the same velocity.

One comparison is the ASCE7 loads for car impact into a bollard. The forces are way lower than you would calculate for an elastic cantilever.

ps: I am not a pier expert!

 

[GregLocock]

IRstuff- that's the horrible old 'suddenly applied weight' scenario. It's not wrong, just poorly explained.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[paddingtongreen]

There are so many things that are omitted in the simple statement of the problem, some mentioned above plus, the added mass of the pier once movement starts, the elasticity of the boat, and the local crushing of the pier and/or boat.

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[walterbrennan]

Looks as though I may owe (the late) Mr. Tobiasson something of an apology. He does not, in fact appear to have misapplied the principles of energy conservation, per se… I believe he was operating by rule of thumb.

On further investigation, I noted that Mr. Tobiasson’s own work referenced an older book, Design and Consruction of Ports and Marine Structures (Alonzo DeF. Quinn, 2nd Ed. 1972).

According to Mr. Quinn, the rule of thumb is as follows:

“The energy to be absorbed by the fender system and dock is usually taken to be 1/2E, as the remaining one-half is assumed to be absorbed by the ship and the water, because of the rotation of the center of mass of the ship around the point of contact of the bow with the fender, which is assumed to be at the one-fourth point of the length of the ship.”

Mr. Quinn goes on to elucidate that, for berthing points other than the one-quarter-point, the energy imparted to the fender/dock is necessarily higher than the fifty percent presumed above. For a direct strike in the line of the ship’s travel, for instance, or a side-strike amidships (either of which would orient the reaction vector toward the center of the ship mass), the full energy would be presumed to be imparted to the fender/dock.

Mr. Tobiasson’s example problem appears to invoke the 1/2E rule on the basis that the his model boat, while striking directly in the line of the boat’s travel, is impacting the end of a cantilevered finger pier at an oblique angle. In essence, he appears to presume that some sidesway/rotation of the boat will follow the impact, as the finger deflects laterally; gobbling up some of the energy, along the way. Right, wrong or indifferent, it appears that Mr. Tobiasson was, in fact at least employing a known rationale when he trimmed his design energy in half.

Being in the realm of what I would consider to be a learning exercise (or, at least, an historical study ), I doubt I’d get into too much trouble sharing the particulars from both Tobiasson and Quinn; which I’ve excerpted and attached here in PDF for your consideration.

I've attached excerpts from both references to this post.

Thanks,

PBW