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Force on a Toilet by Obese Individual

Force on a Toilet by Obese Individual

Force on a Toilet by Obese Individual

(OP)
I have a client who has an employee who is in the range of 500 lb.

We wanting to specify a stainless steel toilet, typically used for bariatric patients in a hospital.  The client is not too pleased with the look and wants a vitreous china unit.

We can find floor set vitreous china units rated for 1000 lb and 2000 lb static loads.

Our concern is if the employee were to "fall" and sit down forcefully on the toilet.

I can calculate the momentum he would generate, but not sure how to translate this in to force and add it to his 500 lb weight (force).

Conservation of momentum will not work since the toilet is generally not going to move.

I am thinking of just specifying the 2000 lb unit and be done with it.

Any ideas.

RE: Force on a Toilet by Obese Individual

Force=1/2mV^2  so if you guess around 3ft/sec you get 9*500/2=

2250lbf, good luck with the fatty.

RE: Force on a Toilet by Obese Individual

1/2mV2 is Kinetic energy

However, you can take a similar approach.  Assume that 400 of the pounds (most of weight is above the hips) falls 2ft onto the toilet.  Assume that there's a 3-in stopping distance because of the fat.

400lb*g*2ft/3-in = 3200lbf, that's about 8g's so it's a bit high, but it only lasts for about 44 ms. winky smile

TTFN

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RE: Force on a Toilet by Obese Individual

(OP)
Got some information from the manufacturer.

They drop a 1000 lb bag (for plop factor) from 5 feet and 10 feet.  

That is a lot of momentum that hits that plumbing fixture.

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