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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: "Sherman, William" <ShermanWC(--nospam--at)cdm.com>
- Date: Tue, 14 Nov 2000 13:41:35 -0500
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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