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RE: loader surcharge at top of retaining wall
[Subject Prev][Subject Next][Thread Prev][Thread Next]- To: "'seaint(--nospam--at)seaint.org'" <seaint(--nospam--at)seaint.org>
- Subject: RE: loader surcharge at top of retaining wall
- From: "Jones, Mark A (Battle Creek)" <Mark.A.Jones(--nospam--at)jacobs.com>
- Date: Tue, 14 Nov 2000 13:48:19 -0800
Force is not energy. Energy is force through a distance. Truly this problem is not a force problem but a conservation of energy problem: sum(E.initial) = sum(E.final). Usually the energy is approximated fairly easily (KE=m*V). Since this assumes all energy will be transferred to the structure it is conservative, i.e., it doesn't discount because of damping, heat, etc. If you are trying to equate this to a force, then the real bugger is the stopping distance. I recall doing a similar problem a few years ago, what happens when 10,000 ton barge at 5 kts runs into pier. IIRC, I found a solution method in a marine engineering book. Unfortunately, can't remember which one and my previous employer kept the calcs. They were really picky about me not taking engineering data home, something about national security or some such. :) Sure wished I had my calc books now. We did some really interesting and esoteric calculations. I could use some of those equations. Mark Jones Jacobs Engineering -----Original Message----- From: Sherman, William [mailto:ShermanWC(--nospam--at)cdm.com] Sent: Tuesday, November 14, 2000 1:42 PM To: 'seaint(--nospam--at)seaint.org' Subject: RE: loader surcharge at top of retaining wall Christopher Wright wrote: > The 1/4 second estimate sounds like a hipshot that provided stresses > that made someone feel good. Your impact factor of 2 is probably a > little low--it implies a statically applied sudden load without any > incoming kinetic energy or energy loss in the collision. Probably > there is enough energy lost in friction so that you're not too far off, > but you're not all that conservative unless you can live with some damage. Your first statement is a valid statement - but I'm not sure why you say it doesn't account for any kinetic energy. Isn't kinetic energy accounted for in the formula F=Ma? At 10 MPH, the speed equates to 14.7 feet per second. Using my assumed deceleration rate of 0.25 seconds, my equivalent force formula is as follows: F = Mass x acceleration = (weight/32.2 ft/sec^2) x (14.7/0.25 sec) = 1.83 W But I must admit that if the deceleration rate is, for example, 0.10 seconds, the force increases to F = 4.6 W. Thus the force is very dependent upon the rate of deceleration. I'm sure a detailed dynamic equation can be written to determine a more accurate deceleration rate based on the wall stiffness (i.e., a moving object impacting on a spring), although the wall stiffness requires some additional assumptions regarding materials properties (instantaneous modulus of elasticity, cracked section properties, base fixity, etc). All-in-all the best one can do is get an "estimate" of the magnitude of load.
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