physics question

[fretburnr]

is this some kind of ANSI testing? if so, there's probably more info to be had still.

[CorvallisBMW]

physics is all metric, but not my engineering classes though. Statics, dynamics and strength of materials all use mixed units, as well as a bunch of the general and introductory engineering classes I've had. hopefully next year in pro school it will be all metric. (yes, I go to the same school as you)

Awesome, what kind of engineering? I'm in IME

[LivingLegend06]

Awesome, what kind of engineering? I'm in IME

I'm a sophmore ME, all ready to apply to pro school next year. Any good advice for me?

[akorcovelos]

It's to verify that the cone will not pull out of the concrete if equipment is dropped on it.

I didn't think this was so complicated, I asked becasue I didn't think I needed to pull my old Strength book. I just need a good number to start with. As far as physics go it can be greatly simplified and still give a valid result.

We certify to ACI, and this is not for certification it's for new product development.

[fretburnr]

okay... 500lb!

it may create an impact of more force than that, but certainly won't create less.

[akorcovelos]

okay... 500lb!

it may create an impact of more force than that, but certainly won't create less.

Yeah, but we want this fixture to be as compact as possible so the question was posed as to the minimum weight that would achieve the 500 lb impact. I guess I'll just start at 500 and work backwards with a impact gage.

[JeffRR]

If your looking to generate a max 500lb load on the cone, then just apply the 500lb as a static load and be done with it. Don't worry about drop height & load as you'd need with a dynamic test.

This is assuming that you've done your homework and feel pretty confident that a 500lb max load is adequate for your structure.

[akorcovelos]

needs to be a drop test, a static test dosn't confirm the cone can resist impact, just load.

[Ral]

It's been a year or so since I took a Structures class, but if I remember right there was a way to calculate the total energy a structure could absorb before failure. If the energy of your block is less than the energy absorbed by the cone, you are golden. the block's energy is mass*g*h ==> 500 lb*2'. The shear/bending stress of the cone must be able to withstand this 1000 ft*lb energy.

but again, it's been a long time since I've done this. (funny, I'm the structures guy for my senior design project)

[LivingLegend06]

needs to be a drop test, a static test dosn't confirm the cone can resist impact, just load.

a static load of 500lbs is more likely to cause failure than any drop test with a maximum force of 500lbs.

just drop 500lbs on it and see what happens.

[akorcovelos]

a static load of 500lbs is more likely to cause failure than any drop test with a maximum force of 500lbs.

just drop 500lbs on it and see what happens.

yeah, I'm at that point now.

[JeffRR]

needs to be a drop test, a static test dosn't confirm the cone can resist impact, just load.

How is a 500lb peak load static or dynamic different other than the time duration at the peak load?

500lb max load static would be a constant load where as a impact test designed to have a 500lb max peak load will hit a peak of 500lb and then decay to a static load provided by the weight of the impactor as it sits on the cone.


Just for the sake of discussion if your 500lb impactor comes to rest in 0.25s, you'll have a 2000lb peak load.

Drop the 500lbs from 2' but be aware that your cone is going to be over design if the 500lb impact load is an accurate max load.

[TwoJ's]

I'm a sophmore ME, all ready to apply to pro school next year. Any good advice for me?

Nice, I'm in my junior year and am a ME major.

[mrsleeve]

Ok after much thinking this is what I came up with using the above parameters


F=ma right? so we figure mass at 500lbs and acceleration at gravity of 32 ft s ^2 (I rounded)

so 500*32 ft s^2= 16000 lbs of force at impact after 1 second of fall.

now the block only falls 2 feet right so it dose not get to a 32ft per second we need figure for the time of fall to figure out the amount of acceleration after 2 feet

2/32 =.0625

so you are only getting 6.25% of the 1 second of acceleration thus you only generate 6.25% of the force

16000lbs*.0625 sec = 1000lbs of force generated at impact after 2 feet fall.

Now if you only need 500lbs at impact you figure 250lbs

250*32=8000lbs

8000*.0625 = 500

I hope all that made sense.

[LivingLegend06]

Ok after much thinking this is what I came up with using the above parameters


F=ma right? so we figure mass at 500lbs and acceleration at gravity of 32 ft s ^2 (I rounded)

so 500*32 ft s^2= 16000 lbs of force at impact after 1 second of fall.

now the block only falls 2 feet right so it dose not get to a 32ft per second we need figure for the time of fall to figure out the amount of acceleration after 2 feet

2/32 =.0625

so you are only getting 6.25% of the 1 second of acceleration thus you only generate 6.25% of the force

16000lbs*.0625 sec = 1000lbs of force generated at impact after 2 feet fall.

Now if you only need 500lbs at impact you figure 250lbs

250*32=8000lbs

8000*.0625 = 500

I hope all that made sense.

you are so wrong on so many points.

did you even read the thread?


on earth the mass of an object can be approximated by taking its weight (lbs) and dividing by the acceleration due to gravity (32.2). by fixing the very first thing you did and using mass instead of weight you can come to the conclusion that a 500lb weight will exert a force of 500lbs on the floor. HOLY FUCKING SHIT BATMAN.