Thanks for your reply.
That is right, we have added load distributing large area steel pads to distribute bolt head concen. load to a larger area, and some other changes.
My only intention for posting this was to obtain info on bolt torque so not to exceed rubber compressive load limit.
Mike
Yes, I also imagined torque value will be low, that is why I thought to post my question to make sure I am not making a mistake.
Currently the box is mounted rigidly on the foundation and is cracking at the bolted area due to high vib.
If we use rubber pads, we will have to use lock nuts &...
The pad is sandwiched between a foundation and the rigid box and is "completely" supported (sandwiched) by the foundation and the box (so the bolt load will be distributed almost equally).
Rubber is 40 Durometer. If more than 50 PSi is applied it just compresses and its isolation eff. is...
I seem to be having a fine time calculating bolts tightening torque value for the following situation:
A 1000 lbs rectangular box sits on two rubber pads of 1" thick and 4" wide x 18" long at the ends of box.
There are (4) 1/2" bolts (2 each end) that need to be tightened, but the max allowed...
I went through Roark "formulas for stress & strain" 5th edition and could not find any solution for a beam supported on 14 points.
All of examples & formulae are for simply supported beams.
I have one equation and 14 unkowns.
Any reference to a specific edition or page number?
Hi there!
Would some one give me a hint on how to tackle this issue?
Have a skid of about 40' long that is supported on 14 (7 each side) vibration isolators. The supported unit weighs about 130,000 lbs. Isolators have equal space through out the skid length. CG is at 15' from the left.
Here...
Thanks katmar for replying.
You are correct. This is for exhaust gas at around 650 °F.
Reason I didn't say it was because i was interested in methodoloy, not accuracy at this time.
So, I guess i need to get exh gas density involved in 33 feet to get its pressure loss in PSI?
Am I right...
Hi there!
I am at loss, need helpful hints.
I need to calculate pressure loss thru a 30" ID formed tubing miter cut and welded together to form a 45° & a 90° elbows. 90° has two miter cuts. 30" leg length.
Looking at Crane data book I see that for alpha=45° =15fT, and for alpha=90° K=60fT.
OK...
If I understand your comment correctly, the offset would be in the order of 4,800,000 * 0.6 = 2,880,000 ton-hour (or even half that amount 2,400,000) am I right?
Thanks
Mike
The system is not a turbine, but couple of gas engines driving generators, and using exhaust gas & the jacket water heat to heat up absorbtion chiller's hot water circuit. The customer has this requirement that by running the system for 8000 hours a year & powering 600 ton chiller, we need to...
Hi all,
If I have this Co-Gen unit that can generate enough hot water to power a 600 ton absorbtion chiller and the system runs about 8000 hours a year, then what would equivalent "ton-hours" of electric cooling be?
I am guessing it would be 600 * 8000 = 4,800,000 ton-hour of electric cooling...
Actually,
If you return to
http://www.eng-tips.com/viewthread.cfm?qid=130088&page=1
and see my posting (third posting after yours), you will notice that I replied to my question basically using your method here.
Thanks anyway.
