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Re: loader surcharge at top of retaining wall
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- Subject: Re: loader surcharge at top of retaining wall
- From: "John Riley" <jpriley485(--nospam--at)peoplepc.com>
- Date: Wed, 15 Nov 2000 00:01:12 -0600
An intuitive way to visualize this is to think of the beam as a spring, loaded at its midspan. When the load is suddenly applied, it deflects a certain distance, then springs back to almost its original position, then down again to almost the initial deflection, and so on until the energy is dissipated. The deflection limits of this "pendulum" movement are zero and the initial deflection. And the "at rest" loaded position will be half way between. This "at rest" loaded position is the static deflection and it is half the dynamic deflection. Assuming elastic behavior, deflection is directly proportional to load, so the dynamic "load" is twice the static load; or another way to say it is that the impact factor is 2. ----- Original Message ----- From: Gunnar Hafsteinn Isleifsson <gunnarhi(--nospam--at)post4.tele.dk> To: <seaint(--nospam--at)seaint.org> Sent: Tuesday, November 14, 2000 6:29 PM Subject: RE: loader surcharge at top of retaining wall Timoshenko's Mechanics of Materials has an example which shows that a statically applied sudden load corresponds to a impact factor equal to 2. Imagine you have a simply supported beam with a mass M held at midspan, so it just touches the beam, without imparting any load. If you suddenly release the load (no dropping height or initial velocity), you get a deflection in the beam that is double that of the deflection from a statically (infinitely slowly) applied load. Thus an impact factor of 2 without any incoming kinetic energy. The above assumes no energy losses, elastic behaviour, etc., etc. Regards, Gunnar Hafsteinn Isleifsson -----Original Message----- From: Sherman, William [mailto:ShermanWC(--nospam--at)cdm.com] Sent: Tuesday, November 14, 2000 19:42 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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- RE: loader surcharge at top of retaining wall
- From: Gunnar Hafsteinn Isleifsson
- RE: loader surcharge at top of retaining wall
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