Sizing a beam to withstand falling objects.

  • Thread starter brbeck777
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I need to size a member to withstand large falling rocks, but it seems that if I do not know how far the rock travels after impact, I have insufficient information. I do not have the resources to test falling rocks on an array of members and then measure the distance traveled.

I need a basic equation or some good advice to find the impact force then size a beam accordingly to that force.

Thanks,
Brian
 

Answers and Replies

  • #2
brewnog
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If in doubt,
Make it stout,
Out of stuff you know about.
 
  • #3
pervect
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If the problem were a rope being loaded by a falling weight, I believe you'd work the problem in terms of the amount of energy you wanted the rope to be able to dissipate.

If you try to work the dynamics of the rope jerking to a stop, it is a complicated problem, but if you ask "how much energy can the rope store" it is a simple problem. You look at the effective spring constant of the rope (force/distance), and take the maximum stored energy as .5 k x^2, where x is the deflection of the rope at its yield stress.

You can then add an appropriate (large) safety factor. The goal is that your system be able to handle a certain amount of stored energy, given by the energy of the falling object.

Note that elastic ropes have a huge advantage here over stiffer ropes in this sort of application.

I imagine you could apply the same principles to a beam being loaded by falling objects as you could to a rope. Presumably you'll only have to deal with one falling object every once in a while. If you have objects falling continuously, you might have to consider resonance phenomenon.
 
  • #4
NateTG
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Well, you could always go with the 'tough enough that the rock breaks first' approach.

However it is physically impossible to build a structural element that would withstand the moon hitting it at .5 c, so you've got to have some idea about what's going on, other than 'falling rock'.
 
  • #5
Danger
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I would even suspect, although someone should check me on it, that the shape of the involved rocks would make a difference. A smooth roundish one, for instance, could be equated to something like a bowling ball. I should think that an irregular one of equal mass landing on a jagged edge would impart a far higher 'point impact'. (Sorry, I don't know the proper terms; I mean much more psi.)
 
  • #6
Q_Goest
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What pervect said is basically correct. For any impact, the simple way of determining whether or not something will break is to calculate the member's spring constant and then compare the amount of energy of the 'rock' (kinetic energy of rock) with the energy absorbed by your member when it deflects sufficiently to absorb that energy. Pervect gave the equation for energy absorbed by a spring. To determine the spring rate, simply calculate the amount of deflection you get in some given beam under a given load. The spring rate of the beam will be linear.

Once you know how far the member must deflect in order to absorb the impact, you can back-calculate the stress. The only warning here is that once you exceed the yield stress, different materials do different things. A ductile beam for example will bend and absorb much more energy than a brittle beam. If stress exceeds yield, the analysis gets much more complex.

. . . but it seems that if I do not know how far the rock travels after impact, I have insufficient information.
Note that the additional energy involved AFTER impact is generally neglected.

Regarding what Danger said, the contact area of the impact will have something to do with the local stresses at that spot, but it won't change the analysis of the stress. You can safely neglect the shape of the rock.
 
  • #7
Danger
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Thanks for the clarificatilon, Q.
 
  • #8
Mech_Engineer
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NateTG said:
Well, you could always go with the 'tough enough that the rock breaks first' approach.

However it is physically impossible to build a structural element that would withstand the moon hitting it at .5 c, so you've got to have some idea about what's going on, other than 'falling rock'.

"Give me a lever long enough and a fulcrum on which to place it, and I shall move the world."

-Archimedes


"Give me a beam big enough and a wall on which to mount it, and I shall stop the moon."

-Crazy Civil Engineer


It's theoretically possible, it would just take a really big beam :smile:
 

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