Reply With QuoteI don't understand why you have a bolted connection between the beam and the fixed support. Seems to defeat your purpose. In my mind that should be a pivoted connection, not fixed.
Good afternoon,
I'm relatively new to the forum . . and I have a question. Go figure!
I would like an explanation of how to calculate loading on a testing apparatus. It has been way too long since I took those basic engineering courses in college . . . like 40 years ago! I'm a facilities manager, not an engineer.
Let's say I wish to test the (vertical) strength of an assembly made up of a 4" schedule 40 pipe welded to a 12" x 12" x 3/4" base plate which can be fixed to an 8" thick concrete drive using 4 - 3/4" wedge anchors. The top of the pipe has a 6" x 6" x 1/2" plate welded to it and there is a 3” loop made of 3/4" steel rod welded into two holes in the top plate.
Rather than trying to design and fabricate an A-frame or other type of stiff frame to pull vertically, I was thinking perhaps a wide flange beam may work? My idea is to bolt a 16 foot long W12x14 to a fixed pylon. Looking at a simple diagram, the pylon/vertical spacer would be on the left. The assembly being tested would be at the midpoint of the beam, 7 1/2 feet from each end support . On the right side of the beam, a bottle jack of sufficient size with an oil filled 0 - 10,000 lb gauge would be placed under the beam, perhaps also on a spacer if needed. The test assembly could then be attached with 3/8" 7/19 aircraft cable thru the top loop and around the beam above, perhaps with 2 or more loops to decrease any stretch in the cable?
When the jack is pumped, the right side of the beam lifts trying to pull the pipe column assembly up and off of the concrete. As long as the left side pylon is sufficiently sized and bolted to the concrete, I see the beam bending about the midsection as it tries to lift the pipe column thereby testing the assembly. I know that I could put a scale between the beam and the assembly, but I would prefer to do a little calculation beforehand and then proof the calculations with real world testing.
Plausible?! Problems? How would I calculate the upward force at the mid beam position for a given up loading value at the jack? I'll try to attach a diagram . . .
Thanks for any assistance!
allouissious
I don't understand why you have a bolted connection between the beam and the fixed support. Seems to defeat your purpose. In my mind that should be a pivoted connection, not fixed.
The fixed end will skew your numbers big time. Like Boggs says... need to pin that end. Otherwise you'd need to subtract out the forces required to bend the beam from your end results.
If I'm understanding you correctly... you want to see what it takes to yank the piece in the center out of the floor? Or whatever mode of failure happens first.
You're going to want to watch out for the reaction when something lets go when you're doing this. Some things might want to fly when you find the max. this'll take. POP...!
Far as calculating goes... Take the area of the piston inside of your bottle jack and multiply it by the psi at the gage. (Also... you probably want to get a gage with a tell-tale dial in it.... or whatever they call it... a second dial that gets pushed along by the regular dial so that once pressure is lost you still have that tell-tale dial sitting at the maximum psi reached.) So you have piston surface area x's pressure = force of the jack on the beam. Then because of the lever action and it's mechanical advantage you need to calculate for that.
The force you came up with that the jack produced... multiplied by the length (in inches) from the pin centerline to the jack centerline... divided by the distance from the jack centerline to the gadget you are yanking on.
So... say you have bottle jack with a 3" piston in it and you get a reading of 2,000 psi when things come apart...
1.5 (radius of piston) x 1.5 x pie = 7.069 in2
7.069 x 2,000psi = 14,137 lbs. of force
14,137 x 180" = 2,544,660
2,544,660 divided by 90" (or whatever distance is from gadget being yanked to force applied by jack) = 28,274 lbs. of pull force on your gadget.
Next thing you need to do is figure if you're going to have enough beam for this.
In my example above your beam would deflect .81" and see stress of 64,622psi
Not enough beam for this example... You'd bend it.
Myself... and I'm sure some others will disagree with this... I'd estimate what you expect your maximum pull will be and then size your beam for somewhere around 1/8" deflection max. at center with your lengths in your sketch. (The old .01" per foot rule of thumb... 15 foot = .15" max deflection.)
Oh... and welcome to the forum by the way.
Have a good one,
Bob
Umm, I am with my guys on this. Why are you trying to bend the beam or the left support leg?
As I see it, that left end needs to be resting on the left support leg, but contained so it cannot slide off. i.e. pinned as suggested.
Last edited by PinkertonD; 06-15-2012 at 01:33 PM.
THANKS . . for all of the help guys.
As I mentioned, it has been too many years since my undergrad engineering classes! I had wondered about the left side connection also. Pinned or hinged or ?? Thanks for clearing that up.
The beam I had used in the example was just one we had laying around from some previous job and I thought to start there. The calculations were where I was lost. I have a simple beam program on my computer, but I really needed to know the calculation routine to determine the loading so I could “fill in the blanks” to find the actual beam size.
THANKS especially to Bob for your complete explanation of how to calculate the forces. This helps greatly. The tattle tale idea for the gauge is a good one. Our 5000# dynamometer has one, I’ve just not found one on a gauge to fit the hydraulic jack, yet. As for the ‘pop’ issue . . . on somewhat similar testing, I usually use ¼” aircraft cable to restrain any catastrophic failures. Bolts and nuts though, sometimes can become missiles under the ‘wrong’ circumstances! Generally speaking, 5000# is our maximum force applied.
Again, thanks for the help with my problem . . . I’ve enjoyed reading many of the questions . . and answers . . on the site since I found it last year. Y’all keep up the good work!!
How much vertical travel will you require to identify a failure? I am guessing not much, so move the test piece way closer to the left end so the lifting load will be reduced via the mechanical advantage.
Vertical travel on somewhat similar testing has usually been 1" -1 1/2". Our upper load is 5000# max. Not much when you consider 4 - 1/2" wedge anchors set 3" into the concrete.
Thanks for the suggestion!
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