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Role of Heavy Block as Table in Drop Shock Test

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aneg

Mechanical
Mar 8, 2019
1
Hello all,

First post here. I started a new job and as I'm reading through how they do shock test I started my mind in a twist, so help me clarify.

They use the method of attaching a unit to a fixture that is bolted to a heavy, tool steel plate. The whole plate is lifted a few inches and dropped, inducing the impulse and shock accelerations through the assembly and into the unit under test. Now where my problem is, is imagining the free body diagram of the UUT's themselves. In that frame of reference, the weight of the block they are attached to has nothing to do with the forces it sees(?). But, in my mind it makes sense that attaching it to a heavy plate would help create heavy shock forces into the unit. So my intuition isn't backed by my FBD analysis.

As I see it mathematically, the UUT has mass, m, the plate has mass, M. Total momentum prior to impact is (m+M)*(2gh)^.5 and therefore the post the block lands on gives an impulse equal to that amount in a half sinusoidal force-time manner, with some peak force. However, the impulse the unit sees is again only dependent on its own momentum prior to impact, m*(2gh)^.5, and therefore peak force is substantially less than what the plate sees.

Again, I know this is a common test setup, so I feel there has to be a reason for the heavy block. Please ask if I need to clarify any of my thoughts, I've been mulling it in my head for a while now and just throwing my thoughts at the screen.

Alex
 
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They use the plate because otherwise you'd have to custom build parts to mount the device being tested to the guides for every single test, which would be expensive.

You'd also have to protect the device from the part of the test fixture that stops the moving assembly- the shock test is designed to load the assembly to a certain g level through its normal mounting, NOT to hit it with a giant hammer and see if it breaks.

The plate provides both functions.
 
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