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# RE: loader surcharge at top of retaining wall

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: loader surcharge at top of retaining wall
• From: "Gunnar Hafsteinn Isleifsson" <gunnarhi(--nospam--at)post4.tele.dk>
• Date: Wed, 15 Nov 2000 01:29:37 +0100

```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.

```