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Finding Bolt Tension on a machines base plate when the machine is eccentrically loaded 2

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Automech2013

Mechanical
Joined
Jan 24, 2013
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US
I'm an engineer and we are installing a piece of equipment we recently purchased. It has some pretty large moment loads and I want to find the tension on the anchor bolts. But don't have a lot of experience in this type of application I'm more of automotive engineer than a structural guy. I found an example problem online that is very similar but don't know how to actually solve for the the 4 bolts tension. Here is what the example is. a Base plate has a moment of 10 kips at 12" off center and the base plate is 12" long the bolts are 2" and 10" from the from the pivot edge of the plate and there are 4 bolts. If you relate that into an equation you have 120In-Kips=(xkips*2inches*2bolts)+(ykips*10"*2bolts). They showed a diagram giving the tension per bolt as 5.77 kips on the bolts 10" away and 1.15 kips on the bolts 2" away so I already had the answers for my "x" and "y". But They said under the diagram a simple calculation can be performed to find these tensions - then didn't show the math. Can anyone please show me how they solved for these tensions?
 
Statics 101
Check the AISC site. It has some base plate manuals that you can download - possibly #1 - I don't remember.

 
Depending on the situation, geometry and loading, you could also get into prying action on the bolts - leveraging that induces additional tension forces on the bolts.

Mike McCann
MMC Engineering
 
Thanks to all. I'm just not sure how they actually solved for the "x" and "y" values. As far as I can remember to solve a 2 variable equation you need to sets of equations. But 5th grade math was a long time ago...LOL.
 
Hi Automech
Can you provide a sketch or give us the link to the example your looking at.
 
You may need to use the quadratic formula, depending what you are looking at. If there is prying involved, you will need to use other math tools too.

Mike McCann
MMC Engineering
 
Thanks Desertfox and here is the link to the example: Also my real life situation is a little different my moment load is downward instead of upward but believe it has the same effect on the baseplate and anchors 2 in tension and 2 in compression.
 
Automech:
The second equation or relationship you need to solve for ‘x’ and ‘y’ is based on their assumption that the base is very stiff. Thus, it rotates about the front edge as a rigid body, so the elongation of the bolts, or the load each takes, will be in a ratio of 2 for the front bolts and 10 for the back bolts, or 1:5. In other words y = 5x, should be plugged into your moment equation to solve for ‘x’. As others have mentioned there can be more to this base plate problem than the simplified approach above, depending on the stiffness/rigidity of the machine base or foot. Dig out your old Engineering Mechanics and Strength of Materials text books, so should have a basic understanding of what you are doing on this kind of problem, before you do it.
 
Thanks guys in the meantime while I was waiting for a response I ended up solving this. Took a minute to think of it these terms but it finally came to me. If you think of the whole thing as a lever with the baseplate edge as the fulcrum at R=0 and you have one force pushing 2" and another at 10". Then I solved for x in terms of y.120=20x+4y and x=10/2y=5y then solve for y.120=20(5y)+4y=104y→y=120/104=15/13≈1.15 and substitute to find "x"x=5(15/13)=75/13≈5.77. But I greatly appreciate all the help. Like I said don't do this every day and it's been a long time since I did this grade school math.
 
This is close enough for your use, but does not consider prying action on the bolt due to the leveraging of the plate against the support. This is dealt with in Blodgett, pages 3.3-8 through 3.3-10 and does involve the solution of a cubic equation.

Mike McCann
MMC Engineering
 
Thanks all..and will definitely check out Blodgett.
 
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